Locally finite basic classical simple Lie superalgebras
1502.04586.tex
\documentclass{amsart}
\usepackage{amssymb}
\usepackage{color}
\usepackage{verbatim} %Allows use of \begin{comment} ... \end{comment}
%\usepackage[notcite,notref]{showkeys}
\newtheorem{Thm}[equation]{Theorem}
\newtheorem{Pro}[equation]{Proposition}
\newtheorem{cor}[equation]{Corollary}
\newtheorem{lem}[equation]{Lemma}
\newtheorem{exa}[equation]{Example}
\newtheorem{con}[equation]{Convention}
\newtheorem{deft}[equation]{Definition}
\newtheorem{rem}[equation]{Remark}
\newtheorem{cons}[equation]{Construction}
\newtheorem{fact}[equation]{Fact}
\newtheorem{note}[equation]{Note}
\newtheorem{tab}[equation]{Table}
\numberwithin{equation}{section}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
%if want to chapter also appear delete the next line
\renewcommand{\thesection}{\arabic{section}}
	\addtolength{\oddsidemargin}{-.875in}
	\addtolength{\evensidemargin}{-.875in}
	\addtolength{\textwidth}{1.75in}

	\addtolength{\topmargin}{-.875in}
	\addtolength{\textheight}{1.75in}



% For each author, add one of the following
\address{%
             %<-------------------
}

%
% The following are for editors only.
%
%\volume{VOLUME}
%\annum{YEAR}
%\issue{ISSUE}
%\editor{NAME OF EDITOR}
%\received{Date article was first received}
%\final{Date final version was received}

% The main body of your article follows
\def\a1s{a_1,\cdots, a_s}
\def\a{\alpha}
\def\ba{\bar\a}
\def\aa{\mathcal A}
\def\ars{\aa^r_\sg}
\def\ac{\alpha^{\vee}}
\def\adot{\dot{\alpha}}
\def\ah{\hat{\a}}
\def\fa{\mathfrak{a}}
\def\fA{\mathfrak{A}}
\def\aff{\hbox{Aff}}
\def\andd{\quad\hbox{and}\quad}
\def\Aut{\hbox{Aut}}
\def\at{\tilde{\a}}
\def\Pia{\Pi(A_{1})}
\def\pia{\pi(\a)}
\def\ad{\hbox{ad}}
\def\pib{\pi(\b)}
\def\fb{\frak{b}}
\def\b{\beta}
\def\bt{\tilde{\b}}
\def\bc{\beta^{\vee}}
\def\bd{\dot{\b}}
\def\bb{\mathcal{B}}
\def\bh{\hat{\b}}
\def\bof{\textbf{b}}
\def\Bl{B_{\ell}}
\def\bl4{B_{\ell\geq4}}
\def\cc{{\mathcal C}}
\def\fc{\mathfrak{C}}
\def\bbbc{{\mathbb C}}
\def\cij{c_{ij}}
\def\d{\delta}
\def\D{\Delta}
\def\dt{\tilde{\delta}}
\def\dj{{\mathcal{D}}_{j}}
\def\di{{\mathcal D}_{i}}
\def\dd{\mathcal D}
\def\fD{\mathfrak D}
\def\delh{\hat{\d}}
\def\dhj{\hat{\d}_{j}}
\def\dhi{\hat{\d}_{i}}
\def\be{{\bf e}}
\def\et{e_\theta}
\def\es{e_\sg}
\def\fe{\mathfrak{E}}
\def\ee{\mathcal{E}}
\def\End{\hbox{End}}
\def\e{\epsilon}
\def\ve{\varepsilon}
\def\vei{\varepsilon_i}
\def\veij{\ve_i+\ve_j}
\def\veji{\ve_j+\ve_i}
\def\veimj{\ve_i-\ve_j}
\def\vejmi{\ve_j-\ve_i}
\def\bbbf{\mathbb{F}}
\def\ff{\mathcal F}
\def\form{(\cdot,\cdot)}
\def\phired{\phi_{ red}}
\def\GL{GL}
\def\gg{{\mathcal G}}
\def\ggzero{\gg_{\bar0}}
\def\ggone{\gg_{\bar1}}
\def\dg{\dot{\gg}}
\def\dgg{\dot{\gg}}
\def\ggi{{\gg_{\bar{i}}}}
\def\ggj{{\gg_{\bar{j}}}}
\def\gd{\gg^\delta}
\def\ggi{\gg_{\ve_i}}
\def\ggmi{\gg_{-\ve_i}}
\def\ggij{\gg_{\ve_i+\ve_j}}
\def\ggimj{\gg_{\ve_i-\ve_j}}
\def\ggmimj{\gg_{-\ve_i-\ve_j}}
\def\ggmij{\gg_{-\ve_i+\ve_j}}
\def\gs{\gg_\sg}
\def\gt{\tilde{\gg}}
\def\gc{\gg_{c}}
\def\gsc{\gc^{\sg}}
\def\gipia{\gg_{\bar{i},\pia}}
\def\gjpib{\gg_{\bar{j},\pib}}
\def\gpia{\gg_{\pia}}
\def\fg{\mathfrak{g}}
\def\fgg{\mathfrak{G}}
\def\dfg{\dot{\fg}}
\def\tfg{\tilde{\mathfrak{g}}}
\def\gspia{\gs_{\pia}}
\def\gspib{\gs_{\pib}}
\def\gspig{\gs_{\pig}}
\def\heart{\hbox{\tiny$\heartsuit$}}
\def\hij{h_{i,j}}
\def\hiuj{h_i^j}
\def\hil{H^{(i)}_{l}}
\def\his{H^{(i)}_{s}}
\def\Ht{\tilde{H}}
\def\Hd{\dot{H}}
\def\Hds{\dot{H}^\star}
\def\hht{\tilde{\mathcal H}}
\def\hh{{\mathcal H}}
\def\dhh{\dot{\hh}}
\def\hhz{\hh^0}
\def\sdhh{\dhh^\star}
\def\hhi{{\hh_{\bar{i}}}}
\def\hhj{{\hh_{\bar{j}}}}
\def\hs{\hh(\sg)}
\def\hhstar{\hh^{\star}}
\def\hsstar{(\hs)^{\star}}
\def\fh{\mathfrak{h}}
\def\tfh{\tilde{\mathfrak{h}}}
\def\fhs{\fh^\sg}
\def\fcj{\mathfrak{J}}
\def\fj{\mathfrak{j}}
\def\ii{\mathcal{I}}
\def\jj{\mathcal{J}}
\def\ind{\hbox{ind}}
\def\kk{\mathcal{K}}
\def\bbbk{\mathbb{K}}
\def\lam{\lambda}
\def\Lam{\Lambda}
\def\LL{\mathcal{L}}
\def\Lt{\tilde{\mathcal L}}
\def\fl{\mathfrak{L}}
\def\LLc{\mathcal{L}_c}
\def\LLcc{\mathcal{L}_{cc}}
\def\LLa{\mathcal{L}_{\a}}
\def\ls{\LL(\sg)}
\def\mod{\hbox{mod}}
\def\bop{\small\hbox{$\bigoplus$}}
\def\ep{\epsilon}
\def\fm{(\cdot,\cdot)}
\def\oro{\omega_\rho}
\def\calq{\mathcal{Q}}
\def\bbbq{\mathbb{Q}}
\def\bq{{\bf q}}
\def\qdl{\dot{Q}_{l}}
\def\qds{\dot{Q}_{s}}
\def\qss{Q_{s}}
\def\qll{Q_{l}}
\def\m{\mathcal{M}}
\def\bbbn{\mathbb{N}}
\def\ral{r_{\alpha}}
\def\bbbr{{\mathbb R}}
\def\calr{\mathcal{R}}
\def\rd{\dot{R}}
\def\rdz{\dot{R}^{\times}}
\def\rt{\tilde{R}}
\def\rtz{\tilde{R}^{\times}}
\def\rds{\dot{R}_{sh}}
\def\rdl{\dot{R}_{lg}}
\def\rde{\dot{R}_{ex}}
\def\rs{R_{sh}}
\def\rl{R_{lg}}
\def\rts{\tilde{R}_{sh}}
\def\rtl{\tilde{R}_{lg}}
\def\rhi{\hat{r}_{i}}
\def\rhj{\hat{r}_{j}}
\def\tr{\hbox{tr}}
\def\tij{t^{(i)}_{-\d_{j}}}
\def\jlql{\jjl \times \qll}
\def\jsqs{\jjs \times k \qss}
\def\1k{\frac{1}{k}}
\def\op{\oplus}
\def\ot{\otimes}
\def\thj{\hat{t}^{(j)}}
\def\thi{\hat{t}^{(i)}}
\def\la{\langle}
\def\ra{\rangle}
\def\apd{\alpha+\dta}
\def\normal{\lhd}
\def\tia{t^{(i)}_{\alpha}}
\def\tii{t^{(i)}}
\def\tia{t^{(i)}_{\alpha}}
\def\jlql{ J_{l}\times\qll}
\def\jsqs{J_{s}\times k\qss}
\def\vp{\varphi}
\def\ts{\theta_{s}}
\def\tl{\theta_{l}}
\def\sub{\subseteq}
\def\sg{\sigma}
\def\pih{\hat{\Pi}}
\def\nfm{\la\cdot |\cdot\ra}
\def\rcross{R^{\times}}
\def\ta{t_\a}
\def\quadd{\quad\quad}
\def\rs{R^{\sg}}
\def\rstimes{(\rs)^{\times}}
\def\rzero{R^{0}}
\def\rszero{(\rs)^{0}}
\def\pf{\noindent{\bf Proof. }}
\def\Phis{\Phi_{sh}}
\def\Phil{\Phi_{lg}}
\def\pig{\pi(\gamma)}
\def\pir{\pi(R)}
\def\pirtimes{\pir^{\times}}
%\def\rttimes{\rt^{\times}}
\def\vpi{\v^{\pi}}
\def\ri{R_{i}}
\def\rs{R(\sg)}
\def\riso{R_{\hbox{iso}}}
\def\rsiso{\rs_{\hbox{iso}}}
\def\rsiiso{(\rsi)_{\hbox{iso}}}
\def\rsi{R^{\sg}_{i}}
\def\rsib{\bar{R}^{\sg}_{i}}
\def\piib{\bar{\Pi}_{i}}
\def\piid{\dot{\Pi}_{i}}
\def\rsid{\dot{R}^{\sg}_{i}}
\def\rniso{R_{\hbox{niso}}}
\def\rre{R_{re}}
\def\rim{R_{ns}}
\def\rsniso{\rs_{\hbox{niso}}}
\def\rsiniso{(\rsi)_{\hbox{niso}}}
\def\rsubt{R_{\hbox{t}}}
\def\sgt{\tilde{\sg}}
\def\stimes{S^\times}
\def\hsi{\hs_{i}}
\def\ti{{\mathcal T}_{i}}
\def\ki{{\mathcal K}_{i}}
\def\gsat{\gs_{\at}}
\def\mi{{\mathcal M}_{i}}
\def\fp{\mathfrak{p}}
\def\gsi{\gs_{i}}
\def\i{{\mathcal I}}
\def\xs{x^{\sg}}
\def\zeq{Z_{\be,\bq}}
\def\kk{\mathcal K}
\def\mna{M_{n}(\aa)}
\def\hd{\dot{\hh}}
\def\fs{\mathfrak{S}}
\def\sfs{\mathfrak{s}}
\def\cals{\mathcal{S}}
\def\slna{sl_n(\aa)}
\def\ft{\mathfrak{t}}
\def\tr{\hbox{tr}}
\def\ks{\kk^{\sg}}
\def\mna{M_{n}(\aa)}
\def\hds{\hd^{\sg}}
\def\xd{x^{\d}}
\def\kd{\kk^{\d}}
\def\ati{\at_i}
\def\gamt{\tilde{\gamma}}
\def\ss{\mathcal{S}}
\def\sszero{\ss_{\bar 0}}
\def\ssone{\ss_{\bar 1}}
\def\sle{(S,L,E)}
\def\slf{(S,L,F)}
\def\ttt{(t_1,t,t_2)}
\def\T{{\mathcal T}}
\def\Lamt{\tilde{\Lam}}
\def\u{{\mathcal U}}
\def\ud{\dot{\mathcal U}}
\def\bv{\bold v}
\def\vrs{v_\sg^r}
\def\vis{v_\sg^i}
\def\vjt{v_\tau^j}
\def\v{{\mathcal V}}
\def\fV{\mathfrak{V}}
\def\vsid{\dot{\v}^{\sg}_{i}}
\def\vsi{\v^{\sg}_{i}}
\def\vsib{\bar{\v}^{\sg}_{i}}
\def\vd{\dot{\mathcal V}}
\def\vs{\v^{\sg}}
\def\vz{{\mathcal V}^{0}}
\def\vdf{\vd_\mathbb{F}}
\def\vdq{\vd_\mathbb{Q}}
\def\vq{\v_\mathbb{Q}}
\def\vdr{\vd_\mathbb{R}}
\def\vzf{\vz_\mathbb{F}}
\def\vzq{\vz_\mathbb{Q}}
\def\vzr{\vz_\mathbb{R}}
\def\vt{\tilde{\mathcal V}}
\def\vsd{\dot{\v}^{\sg}}
\def\w{{\mathcal W}}
\def\wi{\w_{i}}
\def\wt{\tilde{\mathcal W}}
\def\wpi{{\mathcal W}_{\Pi}}
\def\wdot{\dot{\mathcal W}}
\def\wht{\hat{W}}
\def\zn{{\mathbb Z}^{\nu}}
\def\z{{\mathcal Z}}
\def\Zt{\tilde\calz}
\def\calz{{\mathcal Z}}
\def\bbbz{{\mathbb Z}}
\def\ss{\mathcal{S}}
 \def\tph{\tilde{\phi}}
\def\1il{1\leq i\leq\ell}
\def\lg{\hbox{lg}}
\def\sh{\hbox{sh}}
\def\sgn{\hbox{sgn}}
\def\al{\aa_{_{\LL}}}
\def\als{\aa_{_{\LL,\sg}}}
\def\red{\textcolor[rgb]{1.00,0.00,0.00}}
\begin{document}
\centerline{\bf }
\vspace{0.7cm}
\baselineskip=.7cm

\title{Locally finite basic classical simple  Lie superalgebras}
\thanks{2010 Mathematics Subject Classification: Primary  17B67; Secondary 17B65,17B22.}
\keywords{Locally finite Lie superalgebras, Conjugacy classes.}
\thanks{Address: Department of Mathematics, University of Isfahan, Isfahan, Iran,
P.O.Box 81745-163, and School of Mathematics, Institute for Research in
Fundamental
Sciences (IPM), P.O. Box: 19395-5746, Tehran, Iran.\\
Email addresse: ma.yousofzadeh@sci.ui.ac.ir \& ma.yousofzadeh@ipm.ir.\\
This research was in part
supported by a grant from IPM (No. 93170415) and partially carried out in IPM-Isfahan branch.}
\maketitle
\centerline{Malihe Yousofzadeh}
\vspace{2cm}

\textbf{Abstract.} In this work, we study locally finite simple Lie superalgebras containing a Cartan subalgebra and  equipped with an invariant  nondegenerate   even supersymmetric bilinear form. We call these Lie superalgebras  locally finite basic classical simple  Lie superalgebras, classify  them and study the conjugacy classes of their Cartan subalgebras under the group of automorphisms.
\setcounter{section}{-1}         %<-------------------
\section{Introduction}
Following a physical interest in the context of supersymmetries, in 1977, V. Kac \cite{K1} introduced Lie superalgebras (known as $\bbbz_2$-graded Lie algebras in physics). He classified  classical Lie superalgebras which are  finite dimensional  simple Lie superalgebras whose even parts are reductive Lie algebras. These Lie superalgebras  are a generalization of finite dimensional simple Lie algebras over an algebraically closed field of characteristic zero but classical Lie superalgebras are   not necessarily equipped with  nondegenerate invariant bilinear forms while Killing form on a finite dimensional simple Lie algebra over a field of characteristic zero is invariant and nondegenerate. To get a better super version of finite dimensional simple Lie algebras, one can work with those classical Lie superalgebras equipped with even  nondegenerate invariant bilinear forms, called finite dimensional basic classical simple Lie superalgebras (f.d.b.c.s Lie superalgebras for short). A  f.d.b.c.s Lie superalgebra has a weight space decomposition with respect to a Cartan subalgebra $\hh$ of the even part whose root system is  a union of a finite root system and a subset of the dual space $\hh^*$ of $\hh$ consisting of some elements which are  self-orthogonal  with respect to the induced form on $\hh^*.$

The interaction of a finite dimensional simple Lie algebra with its root system is a powerful tool to study the structure of the Lie algebra via its root system. To consider such an approach on Lie superalgebra level, we first need to have an abstract definition of the root system of a Lie superalgebra apart from its connection with the Lie superalgebra. In \cite{you6}, the author introduces the notion of extended affine root supersystems and systematically studies them.
Roughly speaking, a spanning set    $R$ of a  nontrivial vector space over a field $\bbbf$ of characteristic zero, equipped with a  symmetric bilinear form, is called an extended affine root supersystem if the root string property is also satisfied. Such an $R$   is called a locally finite root supersystem if the form is nondegenerate. Locally finite root supersystems which are finite had been studied by V. Serganova in 1996 under the name  generalized root systems \cite{serg}. Almost all generalized root systems appear as the root systems of f.d.b.c.s  Lie superalgebras.  For a locally finite root supersystem $R,$ the self-orthogonal  elements  are  called nonsingular roots and the elements which are not self-orthogonal,  are called real roots. Real roots of $R$ form a locally finite root system (\cite{LN}, \cite{NS}) and   if $R$ is irreducible with nonzero nonsingular part, its nonsingular roots have either one  or two conjugacy classes under the Weyl group action depending on  whether  an arbitrary  nonzero nonsingular root is conjugate to its opposite or not \cite{you2}. This helps  us  to know the structure of a Lie superalgebra $\LL$ whose root system is  an irreducible locally finite root supersystem; more precisely,  as real roots form a locally finite root system, we get that the derived algebra of the even part of $\LL$ is semisimple and considering the conjugacy classes of imaginary roots under the action of the Weyl group, we can show that the odd part of $\LL$ is a completely reducible module for the even part with at most two irreducible constituents; see Theorem \ref{main1}.




Extended affine root supersystems appear as the root systems of the super version of invariant affine reflection algebras \cite{N1} called extended affine Lie superalgebras \cite{you3}. Roughly speaking, a nonzero Lie superalgebra  is called an extended affine Lie superalgebra, if it is equipped with an invariant nondegenerate even supersymmetric bilinear form and that it has a weight space decomposition with respect to a toral subalgebra of its even part whose root vectors satisfy some natural conditions.
Extended affine Lie superalgebras whose corresponding toral subalgebras are  self-centralizing (referred to as Cartan subalgebras) are important in some sense, namely,   finite dimensional basic classical simple Lie superalgebras  and affine Lie superalgebras  are examples of such  Lie superalgebras. We study extended affine Lie superalgebras having a Cartan subalgebra and figure out the properties of  their root spaces; in particular, we find the dimension of root spaces corresponding to so-called  nonisotropic roots.  We then  focus on a special subclass of extended affine Lie superalgebras whose elements are called  locally finite basic classical simple  Lie superalgebras (l.f.b.c.s Lie superalgebras for short); the root system of a l.f.b.c.s. Lie superalgebra is an irreducible  locally finite root supersystem.
 Locally finite basic classical simple Lie superalgebras with zero odd part are exactly locally finite split simple Lie algebras \cite{NS}. We classify l.f.b.c.s Lie superalgebras; due to our classification, a l.f.b.c.s Lie superalgebra with  nonzero odd part is either a finite dimensional basic classical simple Lie superalgebra or   isomorphic to one  and only one of the Lie superalgebras $\mathfrak{osp}(2I,2J)$ ($I,J$   index sets with  $|I\cup J|=\infty,$ $|J|\neq 0$),  $\mathfrak{osp}(2I+1,2J)$ ($I,J$   index sets with  $|I|<\infty,$ $|J|=\infty$) or $\mathfrak{sl}(I_{\bar 0},I_{\bar 1})$  ($I$ an infinite superset with $|I_{\bar 0}|,|I_{\bar 1}|\neq0$); see Subsection \ref{examples}  for the  definitions of these Lie superalgebras.
 We conclude the paper with studding the conjugacy classes of Cartan subalgebras of  l.f.b.c.s. Lie superalgebras under the group of automorphisms.
We show that if $I$  is an infinite set and $J$ is a nonempty set, $\mathfrak{osp}(2I+1,2J)$ is isomorphic to $\mathfrak{osp}(2I,2J)$  while their root systems with respect to their standard  Cartan subalgebras are not isomorphic. This in particular  means that these two standard  Cartan subalgebras are not conjugate. We then  prove that  these are the only representatives for the conjugacy classes of Cartan subalgebras of  $\mathfrak{osp}(2I+1,2J)\simeq\mathfrak{osp}(2I,2J).$ We finally show  that if $\LL$ is a locally finite basic classical simple Lie superalgebra which is not isomorphic to $\mathfrak{osp}(2I+1,2J)\simeq\mathfrak{osp}(2I,2J)$ ($I$ an infinite set  and $J$ a nonempty set), then all Cartan subalgebras of $\LL$ are conjugate under the automorphism group of $\LL.$

\section{Extended Affine Lie Superalgebras And Their Root Systems}\label{extended}
Throughout this work, $\bbbf$ is a field of characteristic zero. Unless otherwise mentioned, all vector spaces are considered over $\bbbf.$
We  denote the dual space of a  vector space $V$ by  $V^*.$  We denote the degree of a homogenous element $v$ of a superspace by $|v|$ and make a convention that if  in an expression, we use $|u|$ for an element $u$ of a superspace, by default we have assumed $u$ is homogeneous.    We denote the group of automorphisms of an abelian  group $A$ or a Lie superalgebra $A$   by $Aut(A)$ and for a subset $S$ of an abelian group  $A,$ by $\la S\ra,$ we mean the subgroup generated by $S.$
 For a set $S,$ by $|S|,$ we mean the cardinal number of $S.$ For a map $f:A\longrightarrow B$ and $C\sub A,$ by $f\mid_{C},$ we mean the restriction of $f$ to $C.$ For two symbols $i,j,$ by $\d_{i,j},$ we mean the Kronecker delta. We  also use  $\biguplus$ to  indicate the disjoint union.
We finally recall that the direct union is, by definition,  the direct limit of a direct system
whose morphisms are inclusion maps.


In this section, we recall the notions of extended affine Lie superalgebras and extended affine root supersystems from \cite{you3} and gather the information we need through the paper regarding them.
In the sequel, by a {\it symmetric form} on an additive abelian group $A,$ we mean a map $\fm: A\times A\longrightarrow \bbbf$ satisfying
\begin{itemize}
\item $(a,b)=(b,a)$ for all $a,b\in A,$
\item $(a+b,c)=(a,c)+(b,c)$ and $(a,b+c)=(a,b)+(a,c)$ for all $a,b,c\in A.$
\end{itemize}
In this case, we set  $A^0:=\{a\in A\mid(a,A)=\{0\}\}$ and call it the {\it radical} of the form $\fm.$ The form is called {\it nondegenerate} if $A^0=\{0\}.$
We note that if the form is nondegenerate, $A$ is torsion free and we can identify $A$ as a subset of $\bbbq\ot_\bbbz A.$ Throughout the paper, if an abelian group  $A$ is  equipped with a nondegenerate symmetric form, we consider $A$ as a subset of $\bbbq\ot_\bbbz A$ without further explanation.
Also if $A$ is a vector space over $\bbbf,$ bilinear forms are used in the usual sense.
\medskip

We call a triple   $(\LL,\hh,\fm)$  a {\it super-toral triple} if
\begin{itemize}
\item {\rm $\LL=\LL_{\bar 0}\op\LL_{\bar 1}$ is  a nonzero Lie superalgebra, $\hh$  is a nonzero subalgebra of $\LL_{\bar 0}$ and  $\fm $ is an invariant  nondegenerate   even supersymmetric bilinear form $\fm$ on $\LL,$}
\item
{\rm $\LL$ has a  weight space decomposition $\LL=\op_{\a\in\hh^*}\LL^\a$ with respect to $\hh$ via the adjoint representation. We note that in this case $\hh$ is abelian; also as $\LL_{\bar 0}$ as well as $\LL_{\bar 1}$ are $\hh$-submodules of $\LL,$ we have $\LL_{\bar 0}=\op_{\a\in \hh^*}(\LL_{\bar 0})^\a$ and $\LL_{\bar 1}=\op_{\a\in\hh^*}(\LL_{\bar 1})^\a$  with $(\LL_{\bar i})^\a:=\LL_{\bar i}\cap \LL^\a,$ $ i=0,1,$}

 \item {\rm the restriction of the form  $\fm$ on $\hh$ is nondegenerate.}
\end{itemize}
 We call  $R:=\{\a\in \hh^*\mid \LL^\a\neq\{0\}\},$ the {\it root system} of $\LL$ (with respect to $\hh$). Each element of $R$ is called  a {\it root.} We refer to elements of $R_{ 0}:=\{\a\in \hh^*\mid (\LL_{\bar 0})^\a\neq\{0\}\}$ (resp. $R_{ 1}:=\{\a\in \hh^*\mid (\LL_{\bar 1})^\a\neq\{0\}\}$) as {\it even roots} (resp. {\it odd roots}). We note that $R=R_{ 0}\cup R_{ 1}.$

Suppose that $(\LL,\hh,\fm)$ is a super-toral triple with corresponding root system $R.$ Since the form is invariant and even, we have   \begin{equation}\label{form-zero}((\LL_{\bar i})^\a,(\LL_{\bar j})^\b)=\{0\},\;\;\; \a,\b\in R,\; i,j\in\{0,1\},\;\; i\neq j,\;\a+\b\neq0.\end{equation} This   in particular implies that for $\a\in R$ and $i\in\{0,1\},$  the form restricted to  $(\LL_{\bar i})^\a+(\LL_{\bar i})^{-\a}$ is nondegenerate.
 Take  $\mathfrak{p}:\hh\longrightarrow \hh^*$ to be  the function  mapping   $h\in\hh$ to $(h,\cdot).$    Since the form is nondegenerate on $\hh,$ the map $\mathfrak{p}$ is one to one (and so onto if $\hh$ is finite dimensional). So for each element $\a$ of the image $\hh^\frak{p}$ of $\hh$ under the map $\mathfrak{p},$  there is a unique $t_\a\in\hh$ representing $\a$ through the form $\fm.$ Now we can transfer the form on $\hh$ to a form on $\hh^\mathfrak{p},$ denoted again by $\fm$  and defined by \begin{equation}\label{form}(\a,\b):=(t_\a,t_\b)\;\;\;(\a,\b\in \hh^\fp).\end{equation}   Although  $R\sub \hh^*=\hh^{\mathfrak{p}}$ if $\hh$ is finite dimensional, in  general, by Lemma 3.1 of \cite{you3},
 $\a\in R$ is  an element  of  $\hh^\mathfrak{p}$ if  there are $i\in\{0,1\},$ $x\in(\LL_{\bar i})^\a$ and $y\in(\LL_{\bar i})^{-\a}$ with  $0\neq[x,y]\in\hh;$ moreover, if $\a\in R\cap \hh^\mathfrak{p},$  $x\in\LL^\a$ and $y\in\LL^{-\a}$ with  $[x,y]\in\hh,$ using the same lemma,      we have \begin{equation}\label{bracket}[x,y]=(x,y)t_\a.\end{equation}

\begin{lem}\label{simple}
Suppose that $\LL=\LL_{\bar 0}\op\LL_{\bar 1}$ is a Lie superalgebra equipped with an invariant nondegenerate even supersymmetric bilinear form $\fm$ such that
\begin{itemize}
\item  $\LL_{\bar 0}$ is a finite dimensional semisimple Lie algebra with the  only  simple ideals $I_1$ and $I_2,$
\item for Cartan subalgebras $\hh_1$ and $\hh_2$ of simple Lie algebras $I_1$ and $I_2$ respectively, $(\LL,\hh:=\hh_1\op\hh_2,\fm)$ is a super-toral triple with  $\LL^0=\hh,$
\item for each $0\neq h\in\hh,$ there is $\d\in R_1$ with $\d(h)\neq 0$ and   $t_\d\not\in \hh_1\cup  \hh_2,$
\end{itemize}
then $\LL$ is a simple Lie superalgebra in the sense that it has no nontrivial ideal (non necessarily $\bbbz_2$-graded).
\end{lem}


\begin{proof} Suppose that $I$ is a nonzero ideal of $\LL.$ Then   by \cite[Pro. 2.1.1]{MP}, $$I=\op_{\a\in R}(I\cap \LL^\a).$$ Through the following steps, we prove that $I=\LL.$

Step 1. $I\cap \hh\neq\{ 0\}:$ Since $I$ is nonzero, there is $\a\in R$  with $\LL^\a\cap I\neq\{0\}.$ If $\a=0,$ there is nothing to prove. Suppose that $\a\neq 0$ and $0\neq x\in \LL^\a\cap I.$ Since the form is nondegenerate on $\LL^\a\op\LL^{-\a},$ there is $y\in\LL^{-\a}$ with $(x,y)\neq 0.$ So (\ref{bracket}) implies that $t_\a\in I\cap \hh$ and so $I\cap \hh\neq\{0\}.$



Step 2. $I=\LL:$ Using Step 1, one finds    $0\neq h\in I\cap \hh.$ We  take $\d\in R_1$ to be such that $\d(h)\neq 0$ and $t_\d\not\in \hh_1\cup\hh_2.$  Since   $\d(h)x=[h,x]$ for all $x\in\LL^{\d},$ we have  $\LL^\d\sub I.$   Now since the form restricted to $\LL^\d\op\LL^{-\d}$ is nondegenerate,  there are $x\in\LL^{\d_1},y\in\LL^{-\d_1}$ such that $(x,y)\neq 0,$  so (\ref{bracket}) implies that $t_\d\in I.$ But $I\cap \LL_{\bar 0}$ is an ideal of the semisimple Lie algebra $I_1\op I_2$ and $t_\d\in (I\cap \hh)\setminus(\hh_1\cup \hh_2),$ so $I\cap \LL_{\bar 0}=I_1\op I_2=\LL_{\bar 0};$ in particular $\hh\sub I.$ Now for each $\a\in R\setminus\{0\},$ there is $k\in \hh$ with $\a(k)\neq0.$ Therefore, for each $x\in\LL^\a,$ $\a(k)x=[k,x]\in I$ and so $\LL^\a\sub I.$ These altogether  complete the proof.\end{proof}

\begin{deft}
{\rm A super-toral triple  $(\LL=\LL_{\bar 0}\op\LL_{\bar 1},\hh,\fm)$ (or $\LL$ if there is no confusion),  with root system $R=R_{ 0}\cup R_{ 1},$ is called an {\it extended affine Lie superalgebra} if}
\begin{itemize}{\rm
\item{(1)}  for each  $\a\in R_i\setminus\{0\}$ ($i\in\{0,1\}$), there are $x_\a\in(\LL_{\bar i})^\a$ and $x_{-\a}\in(\LL_{\bar i})^{-\a}$ such that  $0\neq[x_\a,x_{-\a}]\in\hh,$
      \item{(2)}  for each  $\a\in R$ with $(\a,\a)\neq 0$ and $x\in \LL^\a,$ $ad_x:\LL\longrightarrow\LL,$ mapping $y\in \LL$ to $[x,y],$ is a locally nilpotent linear transformation.}
\end{itemize}
\end{deft}


By \cite[Cor. 3.9]{you3}, the root system of an extended affine Lie superalgebra is an extended affine root supersystem  in the following sense:
\begin{deft}\label{iarr}
{\rm Suppose that $A$ is a nontrivial additive abelian group, $R$ is a subset of $A$ and  $\fm:A\times A\longrightarrow \bbbf$ is  a symmetric form. Set
$$\begin{array}{l}
R^0:=R\cap A^0,\\
\rcross:=R\setminus R^0,\\
\rcross_{re}:=\{\a\in R\mid (\a,\a)\neq0\},\;\;\;\rre:=\rcross_{re}\cup\{0\},\\
\rcross_{ns}:=\{\a\in R\setminus R^0\mid (\a,\a)=0 \},\;\;\; \rim:=\rcross_{ns}\cup\{0\}.
\end{array}$$
We say $(A,\fm,R)$ is an {\it extended affine root supersystem}  if the following hold:
$$\begin{array}{ll}
(S1)& \hbox{$0\in R$ and $\la R\ra= A,$}\\\\
%(S2)&\hbox{the form restricted to $A_{re}:=\hbox{span}_\bbbz\rre$ is nondegenerate,}\\\\
(S2)& \hbox{$R=-R,$}\\\\
(S3)&\hbox{for $\a\in \rre^\times$ and $\b\in R,$ $2(\a,\b)/(\a,\a)\in\bbbz,$}\\\\
(S4)&\parbox{4.5in}{ ({\it root string property}) for $\a\in \rre^\times$ and $\b\in R,$  there are nonnegative  integers  $p,q$  with $2(\b,\a)/(\a,\a)=p-q$ such that \begin{center}$\{\b+k\a\mid k\in\bbbz\}\cap R=\{\b-p\a,\ldots,\b+q\a\};$\end{center}we call  $\{\b-p\a,\ldots,\b+q\a\}$ the {\it $\a$-string through $\b,$}
} \\\\
(S5)&\parbox{4.5in}{for $\a\in \rim$ and $\b\in R$ with $(\a,\b)\neq 0,$
$\{\b-\a,\b+\a\}\cap R\neq \emptyset.$ }
%note that if $(\b,\a)\neq 0,$ then $\b+r\a\in R$ if and only if $\b+r\a\in R^\times.$
%there are positive integers $p,q$ with $p+q\geq1$ such that $\{\b+k\a\mid k\in \bbbz\}\cap R=\{\b+k\a\mid -p\leq k\leq q\}$ and if for some $k\in\bbbz,$ both  $\b+k\a$ and $\b-k\a$ are elements of $R,$ then $(\b+k\a)/2,(\b-k\a)/2\in R.$
\end{array}$$
If there is no confusion, for the sake of simplicity, we say   {\it $R$ is an extended affine root supersystem in $A.$}
Elements of $R^0$ are called {\it isotropic roots,} elements of  $\rre$ are  called  {\it real roots} and  elements of $\rim$ are  called  {\it nonsingular roots}.
A subset $X$ of $R^\times$ is called {\it connected} if each two elements $\a,\b\in X$ are connected in $X$ in the sense that there is a chain $\a_1,\ldots,\a_n\in X$ with $\a_1=\a,$ $\a_n=\b$ and $(\a_i,\a_{i+1})\neq0,$  $i=1,\ldots,n-1.$ An extended affine root supersystem $R$ is called {\it irreducible} if  $\rre\neq\{0\}$ and $\rcross$ is connected (equivalently, $R^\times$ cannot be written as a disjoint union of two nonempty orthogonal subsets).
\begin{comment}
Suppose that $R^\times$ cannot be written as a disjoint union of two nonempty orthogonal subsets. If $R^\times$ is disconnected, then there are $\a,\b$ such that there is no chain connecting $\a,\b$ to each other. Set $A=\{\gamma\mid \hbox{$\gamma$ is connected to $\a$}\}$ and $B=\rcross\setminus A.$ So $A,B$ are nonempty, $(A,B)=\{0\}$ and $\rcross=A\uplus B.$
\end{comment}
An extended affine root supersystem $(A,\fm,R)$ is called a {\it locally finite root supersystem} if the form $\fm$ is nondegenerate and it is called an {\it  affine reflection system} if $\rim=\{0\};$ see \cite{N1}.}
\end{deft}



\begin{lem}\label{abr}
Suppose that $A$ is a nontrivial additive abelian group, $R$ is a subset of $A$ and  $\fm:A\times A\longrightarrow \bbbf$ is  a nondegenerate symmetric form. If  $(A,\fm,R)$ satisfies $(S1)$ and $(S3)-(S5),$ then $(S2)$ is also satisfied. In  particular,  a subset  $S$  of a locally finite root supersystem   $R$ is a locally finite root supersystem in its $\bbbz$-span if \begin{itemize}
\item
the restriction of the form to $\la S\ra$ is nondegenerate,
\item $0\in S,$
\item for $\a\in S\cap\rre^\times, \b\in S$ and $\gamma\in S\cap\rim$ with $(\b,\gamma)\neq 0,$ $r_\a(\b)\in S$ and  $\{\gamma-\b,\gamma+\b\}\cap
    S\neq\emptyset.$
\end{itemize}
\end{lem}
\begin{proof} See \cite[Lem. 1.4 \& Rem. 1.6(ii)]{you6}.\end{proof}

\begin{deft}
{\rm Suppose that $(A,\fm, R)$ is a locally finite root supersystem.
\begin{itemize}
\item
 The subgroup $\w$ of $Aut(A)$ generated by $r_\a$ ($\a\in\rre^\times$) mapping  $a\in A$ to $a-\frac{2(a,\a)}{(\a,\a)}\a,$  is called the {\it Weyl group} of $R.$
\item A subset $S$ of  $R$ is called a  {\it sub-supersystem} if the restriction of the form to $\la S\ra$ is nondegenerate, $0\in S,$ for $\a\in S\cap\rre^\times, \b\in S$ and $\gamma\in S\cap\rim$ with $(\b,\gamma)\neq 0,$ $r_\a(\b)\in S$ and  $\{\gamma-\b,\gamma+\b\}\cap
    S\neq\emptyset.$
\item A sub-supersystem $S$ of $R$ is called  {\it closed} if for $\a,\b\in S$ with $\a+\b\in R,$ we have $\a+\b\in S.$
\item If  $(A,\fm, R)$ is irreducible, $R$ is said to be  of {\it real type} if  $\hbox{span}_\bbbq R_{re}=\bbbq\ot_{\bbbz} A;$ otherwise, we say it is of {\it imaginary type.}
\item  If $\{R_i\mid i\in I\}$ is a class of  sub-supersystems of $R$ which are mutually   orthogonal with respect the form $\fm$ and $R\setminus\{0\}=\uplus_{i\in I}(R_i\setminus\{0\}),$  we say $R$ is {\it the direct sum} of $R_i$'s and write $R=\op_{i\in I}R_i.$
\item The locally finite  root supersystem $(A,\fm,R)$ is called a {\it locally finite root system} if $\rim=\{0\};$ see \cite{LN}.
\item $(A,\fm, R)$ is  said to be {\it isomorphic} to another  locally finite root supersystem $(B,\fm',S)$ if there is a group isomorphism $\varphi:A\longrightarrow B$ and a nonzero scalar $r\in\bbbf$ such that $\varphi(R)=S$ and  $(
a_1,a_2)=r(\varphi(a_1),\varphi(a_2))'$ for all $a_1,a_2\in A.$
\end{itemize}}
\end{deft}
\begin{lem}\label{super-sys}
(a) If $\{(X_i,\fm_i,S_i)\mid i\in I\}$ is a class of   locally finite root supersystems, then for   $X:=\op_{i\in I}X_i$ and $\fm:=\op_{i\in I}\fm_i,$   $(X,\fm,S:=\cup_{i\in I} S_i)$ is a locally finite root supersystem.

(b) Suppose that $( A,\fm, R)$ is a locally finite root supersystem with Weyl group $ \w.$ Then we have the following:

(i) Connectedness is an equivalence relation on $ R\setminus\{0\}.$ Also if $S$ is a connected component of $ R\setminus\{0\},$ then $S\cup\{0\}$ is an irreducible sub-supersystem of $ R.$ Moreover, $ R$ is a direct sum of irreducible sub-supersystems.

(ii) For $ A_{re}:=\la R_{re}\ra$ and $\fm_{re}:=\fm\mid_{_{ A_{re}\times  A_{re}}},$ $( A_{re},\fm_{re}, R_{re})$ is a locally finite root system.

(iii) If $ R$ is irreducible and $ R_{ns}\neq\{0\},$ then $ R_{ns}^\times=\w\d\cup-\w\d$ for each $\d\in  R_{ns}^\times.$
\end{lem}
\begin{proof}  See \cite[\S 3]{you2}.
\end{proof}

\medskip
\begin{lem}\label{lin}
Suppose that $\v$ is a vector space equipped with a symmetric bilinear form and $R$ is a subset of $\v$ such that $(\la R\ra ,\fm_{\la R\ra\times \la R\ra}, R)$ is a locally finite root supersystem. If $\{\a_1,\ldots,\a_n\}\sub \rre$ is $\bbbq$-linearly independent and $\d\in \rim\setminus \hbox{span}_\bbbq \{\a_1,\ldots,\a_n\},$ then $\{\a_1,\ldots,\a_n\}$ and $\{\d,\a_1,\ldots,\a_n\}$ are $\bbbf$-linearly independent.
\end{lem}
\begin{proof}
We just  show that $\{\d,\a_1,\ldots,\a_n\}$ is $\bbbf$-linearly independent; the other statement is similarly proved. Take  $\{1,x_i\mid i\in I\}$  to be  a basis for $\bbbq$-vector space $\bbbf.$
Suppose that $r,r_1,\ldots,r_n\in\bbbf$ and $r\d+\sum_{j=1}^nr_j\a_j=0.$  Suppose that  for $1\leq j\leq n,$ $r_j=s_j+\sum_{i\in I}s_j^ix_i$ with $\{s_j,s_j^i\mid i\in I\}\sub \bbbq.$ We first show  $r=0.$ To the contrary, assume $r\neq 0.$ Without loss of generality, we assume $r=1.$ So $0=\d+\sum_{j=1}^n r_j\a_j=\d+\sum_{j=1}^n(s_j+\sum_{i\in I}s_j^ix_i)\a_j.$ Now for $\a\in \rre,$ we have
$$\frac{2(\d,\a)}{(\a,\a)}+\sum_{j=1}^ns_j\frac{2(\a_j,\a)}{(\a,\a)}+\sum_{j=1}^n\sum_{i\in I}s_j^ix_i\frac{2(\a_j,\a)}{(\a,\a)}=0.$$ This implies that  for $\a\in\rre$ and  $i\in I,$ $\sum_{j=1}^ns_j^i\frac{2(\a_j,\a)}{(\a,\a)}=0$ and so $(\sum_{j=1}^ns_j^i\a_j,\a)=0.$ But it follows from Lemma \ref{super-sys}$(b)(ii)$ that  the form on $\hbox{span}_\bbbq \rre$ is nondegenerate, so $\sum_{j=1}^ns_j^i\a_j=0$ for all $i\in I.$ Now as $\{\a_j\mid 1\leq j\leq n\}$  is $\bbbq$-linearly independent, we have
 $$
s_j^i=0 \;\;\;\;(i\in I,j\in \{1,\ldots,n\}).
$$
Therefore, we get $0=\d+\sum_{j=1}^n r_j\a_j=\d+\sum_{j=1}^n(s_j+\sum_{i\in I}s_j^ix_i)\a_j=\d+\sum_{j=1}^ns_j\a_j.$ Thus we have $\d=-\sum_{j=1}^ns_j\a_j$ which is absurd. This shows that $r=0.$ Now repeating the above argument, one gets that $s_j^i=0$  for all $ i\in I,j\in \{1,\ldots,n\}$  and that $0=\sum_{j=1}^ns_j\a_j.$ Thus we have $s_j=0$ for all $1\leq j\leq n.$ This  implies that $r_j=s_j+\sum_{i\in I}s_j^ix_i=0$ for all $1\leq j\leq n$ and so we are done.
\end{proof}

Using Lemma \ref{super-sys}, to know the classification of  irreducible locally finite root supersystems, we first need to know the classification of locally finite root systems. Suppose that $T$ is a nonempty  index set  with $|T|\geq 2$ and $\u:=\op_{i\in
T}\bbbz\ep_i$ is the free $\bbbz$-module over   the
set $T.$ Define the  form $$\begin{array}{c}\fm:\u\times\u\longrightarrow\bbbf\\
(\ep_i,\ep_j)\mapsto\d_{i,j}, \hbox{ for } i,j\in T
\end{array}$$
and set
\begin{equation}\label{locally-finite}
\begin{array}{l}
\dot A_T:=\{\ep_i-\ep_j\mid i,j\in T\},\\
D_T:=\dot A_T\cup\{\pm(\ep_i+\ep_j)\mid i,j\in T,\;i\neq j\},\\
B_T:=D_T\cup\{\pm\ep_i\mid i\in T\},\\
C_T:=D_T\cup\{\pm2\ep_i\mid i\in T\},\\
BC_T:=B_T\cup C_T.
\end{array}
\end{equation}
These are irreducible locally finite root systems
in their $\bbbz$-span's. Moreover, each irreducible  locally finite root system is either  an irreducible finite root system  or a locally finite root system  isomorphic to one of these locally finite root
systems. We refer to locally finite root systems listed in (\ref{locally-finite}) as  {\it type} $A,D,B,C$
and $BC$  respectively. We note that if $R$ is  an irreducible locally finite
root system as above, then  $(\a,\a)\in\bbbn$ for all
$\a\in R.$ This allows us to  define
$$\begin{array}{l}
R_{sh}:=\{\a\in R^\times\mid (\a,\a)\leq(\b,\b);\;\;\hbox{for all $\b\in R$} \},\\
R_{ex}:=R\cap2 R_{sh}\andd
R_{lg}:= R^\times\setminus( R_{sh}\cup R_{ex}).
\end{array}$$
The elements of $R_{sh}$ (resp. $R_{lg},R_{ex}$) are called {\it
short roots} (resp. {\it long roots, extra-long roots}) of $R$. We point  out that following the usual notation in the literature,  the locally finite root system  of type $A$  is denoted by  $\dot A$ instead of $A,$ as all locally finite root systems listed above are spanning sets for $\bbbf\ot_\bbbz \u$ other than the one of type $A$ which spans a subspace of codimension 1; see \cite{LN} and  \cite[Rem. 1.6(i)]{you6}.


%
%\begin{lem} (i)\label{direct-sum}
%If $\{(X_i,\fm_i,S_i)\mid i\in I\}$ is a class of   locally finite root supersystems, then for   $X:=\op_{i\in I}X_i$ and $\fm:=\op_{i\in I}\fm_i$   $(X,\fm,S:=\cup_{i\in I} S_i)$ is a locally finite root supersystem. Also each locally finite root supersystem is a direct   sum of irreducible sub-supersystems.
%
%(ii) Suppose that $(A,\fm,R)$ is  an irreducible  locally finite root supersystem, set  $\v:=\bbbf\ot_\bbbz A$ and identify $A$ as a subset of $\v.$  Then  $\v=\hbox{span}_\bbbf \rre$ if and only if $R$ is of real type.
%\end{lem}
%\begin{proof}
%For $(i),$ See \cite[Lem. 3.20]{you2} and for $(ii),$ see \cite[Lem. 1.8]{you6}.
%\end{proof}

%
%
%\begin{lem}\label{imag}Suppose that $(A,\fm,R)$ is  an irreducible  locally finite root supersystem, set  $\v:=\bbbf\ot_\bbbz A$ and identify $A$ as a subset of $\v.$  Then  $\v=\hbox{span}_\bbbf \rre$ if and only if $R$ is of real type.
%\end{lem}
%\begin{proof}
% The form on $A$ induces a bilinear form $\fm_\bbbf:\v\times\v\longrightarrow \bbbf$ mapping $(r\ot a,s\ot b)$ to $rs(a,b).$  By \cite[Lem. 3.21]{you2}, the form $\fm_\bbbf$ is nondegenerate. Also by \cite[Lem,'s 3.21 $\&$ 3.5]{you2}, the form $\fm_\bbbf$ is nondegenerate on $\hbox{span}_\bbbf \rre.$ To complete the proof using \cite[Pro. 3.14]{you2}, we just need to  show if $\d\in\rim^\times \cap \hbox{span}_\bbbf \rre,$ then  $\d$ is an element of the subspace of $\bbbq\ot_\bbbz \la R\ra$ spanned by $\rre$ and for this it is enough to show that $\d\in\hbox{span}_\bbbq \rre\sub \v.$
%\begin{comment}
%if $\d\in \hbox{span}_\bbbq \rre,$ there is $n\in\bbbz\setminus\{0\}$ such that $n\d\in\la \rre\ra.$ So  $n\ot \d=1\ot n\d\in1\ot\la \rre\ra $ and so $1\ot \d\in \hbox{span}_\bbbq(1\ot \rre).$
%\end{comment}
%Fix a basis $\{1,x_j\mid j\in J\}$ for $\bbbq$-vector space $\bbbf$ and let $\d\in \rim^\times\cap \hbox{span}_\bbbf \rre,$  so there are nonzero real roots $\a_1,\ldots,\a_n$ and elements $r_1,\ldots,r_n$ of $\bbbf$ such that $\d=\sum_{i=1}^nr_i\a_i.$ For each $i\in\{1,\ldots,n\},$ $r_i=s_i+\sum_{j\in J}s_i^jx_j$ for some $s_i,s_i^j\in \bbbq$ with at most finitely many nonzero terms. Now for each $\b\in \rre^\times,$ we have
%\begin{eqnarray*}
%\frac{2(\d,\b)}{(\b,\b)}=\sum_{i=1}^nr_i\frac{2(\a_i,\b)}{(\b,\b)}&=&\sum_{i=1}^n(s_i+\sum_{j\in J}s_i^jx_j)\frac{2(\a_i,\b)}{(\b,\b)}\\
%&=&\sum_{i=1}^ns_i\frac{2(\a_i,\b)}{(\b,\b)} +\sum_{i=1}^n\sum_{j\in J}s_i^j\frac{2(\a_i,\b)}{(\b,\b)}x_j\\
%&=&\sum_{i=1}^ns_i\frac{2(\a_i,\b)}{(\b,\b)} +\sum_{j\in J}\sum_{i=1}^ns_i^j\frac{2(\a_i,\b)}{(\b,\b)}x_j.
%\end{eqnarray*}
%Now as $\frac{2(\d,\b)}{(\b,\b)},\frac{2(\a_i,\b)}{(\b,\b)}\in\bbbz,$ we get that
%$\sum_{i=1}^ns_i^j(\a_i,\b)=(\sum_{i=1}^ns_i^j\a_i,\b)_\bbbf=0$ for all $j\in J.$ But the form $\fm_\bbbf$ is nondegenerate on $\hbox{span}_\bbbf \rre,$ therefore, for each $j\in J,$ $\sum_{i=1}^ns_i^j\a_i=0,$ so
%\begin{eqnarray*}\d=\sum_{i=1}^n r_i\a_i=\sum_{i=1}^n(s_i+\sum_{j\in J}s_i^jx_j)\a_i=\sum_{i=1}^ns_i\a_i+\sum_{i=1}^n\sum_{j\in J}s_i^jx_j\a_i=\sum_{i=1}^ns_i\a_i.\end{eqnarray*} This completes the proof.
%\end{proof}
%
%\begin{lem}
%Suppose that $(\LL,\fm,\hh)$ is an extended affine Lie superalgebra with root system $R.$ Set $\v:=\hbox{span}_\bbbf R$ and $A:=\la R\ra$ and denote the induced form on $\v$ and $A$ again by $\fm.$ Then $\v^0=\{0\}$ if and only if $A^0=\{0\}.$
%\end{lem}
%\begin{proof}
%The first implication is immediate, so we assume $A^0=\{0\}$ and show that $\v^0=\{0\}.$ Without loss of generality, we assume $R$ is irreducible?
%Suppose that $R$ is of real type
%\end{proof}

\medskip
In the following two theorems, we give the classification of irreducible locally finite root supersystems.

\begin{Thm}[{\cite[Thm. 4.28]{you2}}]\label{classification I}
Suppose that $T,T'$ are index sets of cardinal numbers greater than $1$ with  $|T|\neq|T'|$ if $T,T'$ are both finite. Fix a symbol $\a^*$ and  pick $t_0\in T$ and $p_0\in T'.$ Consider the free $\bbbz$-module $X:=\bbbz\a^*\op\op_{t\in T}\bbbz\ep_t\op\op_{p\in T'}\bbbz\d_p$  and define the symmetric form $$\fm:X\times X\longrightarrow \bbbf$$ by $$\begin{array}{ll}
(\a^*,\a^*):=0,(\a^*,\ep_{t_0}):=1,(\a^*,\d_{p_0}):=1\\
(\a^*,\ep_t):=0,(\a^*,\d_q):=0&t\in T\setminus\{t_0\},q\in T'\setminus\{p_0\}\\
(\ep_t,\d_p):=0,(\ep_t,\ep_s):=\d_{t,s},(\d_p,\d_q):=-\d_{p,q}&t,s\in T,p,q\in T'.
\end{array}$$
 Take $R$ to be $\rre\cup \rim^\times$ as in the following table:
$${\small
\begin{tabular}{|c|c|c|}
\hline
type &$\rre$&$\rim^\times$
\\\hline
$\dot A(0,T)$& $\{\ep_t-\ep_s\mid t,s\in T\}$&$\pm\w\a^*$\\
\hline
$\dot C(0,T)$& $\{\pm(\ep_t\pm\ep_s)\mid t,s\in T\}$&$\pm\w\a^*$\\
\hline
$\dot A(T,T')$& $\{\ep_t-\ep_s,\d_p-\d_q\mid t,s\in T,p,q\in T'\}$&$\pm\w\a^*$\\
\hline
\end{tabular}}$$
in which  $\w$ is the subgroup of $Aut(X)$ generated by the reflections $r_\a$ $(\a\in \rre\setminus\{0\})$ mapping $\b\in X$ to $\b-\frac{2(\b,\a)}{(\a,\a)}\a,$ then $(A:=\la R\ra,\fm\mid_{A\times A}, R)$  is an irreducible locally finite root supersystem of imaginary type and conversely, each irreducible locally finite root supersystem of imaginary type  is isomorphic to one and only one of these root supersystems.
\end{Thm}

\begin{Thm}[{\cite[Thm. 4.37]{you2}}]
\label{classification II}
 Suppose  $(X_1,\fm_1,S_1),$ $\ldots,(X_n,\fm_n,S_n),$ for some  $ n\in\{2, 3\},$  are irreducible locally finite root systems. Set $X:=X_1\op\cdots\op X_n$ and $\fm:=\fm_1\op\cdots\op\fm_n$ and consider  the locally finite root system  $(X,\fm,S:=S_1\op \cdots\op S_n).$ Take $\w$ to be the Weyl group of $S.$  For $1\leq i\leq n,$ we identify $X_i$ with a subset of $\bbbq\ot_\bbbz X_i$ in the usual manner. If $1\leq i\leq n$ and $S_i$ is a finite root system of rank $\ell\geq 2,$ we take $\{\omega_1^i,\ldots,\omega_\ell^i\}\sub\bbbq\ot_\bbbz X_i$ to be a  set of fundamental weights for $S_i$ and if $S_i$ is one of infinite locally finite root systems $B_T, C_T, D_T$ or $BC_T$  as in (\ref{locally-finite}), by $\omega_1^i,$ we mean $\ep_1,$ where $1$ is a distinguished element of $T.$ Also if $S_i$ is  one of the  finite root systems $\{0,\pm\a\}$ of type $A_1$ or  $\{0,\pm\a,\pm2\a\}$ of type $BC_1,$ we set $\omega_1^i:=\frac{1}{2}\a.$
Consider $\d^*$ and $ R:= R_{re}\cup R_{ns}^\times$ as in the following table:
$${\tiny
\begin{tabular}{|c|l|c|c|c|c|}
\hline
$ n$& $ S_i\;(1\leq i\leq n)$&$ R_{re}$&$\d^*$&$ R_{ns}^\times$&\hbox{type}\\
\hline
$ 2$& $ S_1=A_\ell,\; S_2=A_\ell$ $(\ell\in\bbbz^{\geq1})$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\pm\w\d^*$&$A(\ell,\ell)$\\
\hline
$2$& $ S_1=B_T,\;S_2=BC_{T'}$ $(|T|,|T'|\geq2)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$B(T,T')$\\
\hline
$2$& $ S_1=BC_T,\;S_2=BC_{T'}$ $(|T|,|T'|\geq2)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$BC(T,T')$\\
\hline
$2$& $ S_1=BC_T,\;S_2=BC_{T'}$ $(|T|=1,|T'|=1)$&$S_1\op S_2$&$2\omega_1^1+2\omega_1^2$&$\w\d^*$&$BC(T,T')$\\
\hline
$2$& $ S_1=BC_T,\;S_2=BC_{T'}$ $(|T|=1,|T'|\geq2)$&$S_1\op S_2$&$2\omega_1^1+\omega_1^2$&$\w\d^*$&$BC(T,T')$\\
\hline
$2$& $ S_1=D_T,\;S_2=C_{T'}$ $(|T|\geq3,|T'|\geq2)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$D(T,T')$\\
\hline
$2$& $ S_1=C_T,\;S_2=C_{T'}$ $(|T|,|T'|\geq2)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$C(T,T')$\\
\hline
$2$& $ S_1=A_1,\;S_2=BC_{T}$ $(|T|=1)$&$S_1\op S_2$&$2\omega_1^1+2\omega_1^2$&$\w\d^*$&$B(1,T)$\\
\hline
$2$& $ S_1=A_1,\;S_2=BC_{T}$ $(|T|\geq2)$&$S_1\op S_2$&$2\omega_1^1+\omega_1^2$&$\w\d^*$&$B(1,T)$\\
\hline
$2$& $ S_1=A_1,\;S_2=C_T$ $(|T|\geq 2)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$C(1,T)$\\
\hline
$2$& $ S_1=A_1,\;S_2=B_3$ &$S_1\op S_2$&$\omega_1^1+\omega_3^2$&$\w\d^*$&$AB(1,3)$\\
\hline
$2$& $ S_1=A_1,\;S_2=D_{T}$ $(|T|\geq3)$&$S_1\op S_2$&$\omega_1^1+\omega_1^2$&$\w\d^*$&$D(1,T)$\\
\hline
$2$& $ S_1=BC_1,\;S_2=B_T$ $(|T|\geq2)$&$S_1\op S_2$&$2\omega_1^1+\omega_1^2$&$\w\d^*$&$B(T,1)$\\
\hline
$2$& $ S_1=BC_1,\;S_2=G_2$ &$S_1\op S_2$&$2\omega_1^1+\omega_1^2$&$\w\d^*$&$G(1,2)$\\
\hline
$3$& $ S_1=A_1,\;S_2=A_1,\; S_3=A_1$ &$S_1\op S_2\op S_3$&$\omega_1^1+\omega_1^2+\omega_1^3$&$\w\d^*$&$D(2,1,\lam)$\\
\hline
$3$& $ S_1=A_1,\; S_2= A_1,\; S_3:=C_T$ $(|T|\geq2)$&$S_1\op S_2\op S_3$&$\omega_1^1+\omega_1^2+\omega_1^3$&$\w\d^*$&$D(2,T)$\\
\hline
\end{tabular}
}$$
For $1\leq i\leq n,$ normalize the form $\fm_i$ on $X_i$ such that  $(\d^*,\d^*)=0$ and that for type $D(2,T),$ $(\omega_1^1,\omega_1^1)_1=(\omega_1^2,\omega_1^2)_2.$  Then $(\la  R\ra,\fm\mid_{\la R\ra\times \la R\ra}, R)$ is an irreducible locally finite root supersystem of real type and conversely, if  $( X,\fm, R)$ is an irreducible locally finite root supersystem of real type, it is either an irreducible locally finite root system or  isomorphic to one and only  one of the locally finite root supersystems listed in the above table.
\end{Thm}

%\begin{deft}\label{degree}
%{\rm For an irreducible locally finite root supersystem  $R$ of type $X.$ Set
%$$R_0:=\left\{\begin{array}{ll}
%\{\a\in \rre\mid 2\a\not\in R\}\cup\{0\}&\hbox{if $X\neq BC(T,T')$}\\\\
%\rre\setminus(\rre^2)_{sh}&\hbox{if $X= BC(T,T')$ and $\rre=\rre^1\op\rre^2$ }\end{array}\right.$$
%
%and $$R_1:=R\setminus R_0.$$
%We call elements of $R_0$ (resp. $R_1$) {\it even} (resp. {\it odd}) roots.}
%\end{deft}


\begin{rem}\label{rem3}{\rm
(i) We make  a convention  that from now on, for the types listed in column ``type'' of Theorems \ref{classification I} and \ref{classification II}, we may use  a finite  index set $T$ and its cardinal number in place of each other, e.g.,  if $T$ is  a nonempty finite set of cardinal number $\ell,$ instead of type $B(1,T),$ we may write $B(1,\ell).$
We also mention that as we will see in Example \ref{exa1}, there is  an extended affine Lie superalgebra with nonzero odd part  whose root system is isomorphic to the locally finite  root system of type $BC_T.$ In order to distinguish the root system  of that Lie superalgebra from  the locally finite  root system of type $BC_T,$ we denote it by  $B(0,T);$ more precisely, the underlying vector space of   the locally finite root supersystem of type $B(0,T)$ is  the same as the one of  $BC_T$  and the corresponding  symmetric form is the opposite  of the defined form for type $BC_T.$


%(ii) By \cite[Pro. 2.14]{you3}, if $(\LL,\fm,\hh)$ is  an extended affine Lie superalgebra with $\LL^0\sub\LL_{\bar 0}$ and corresponding root system $R,$ the sets of even and odd roots coincide with $R_0$ and $R_1$ as in Definition \ref{degree}.


(iii) Suppose  $R$ is of type $X=C(T,T')$ or $BC(T,T')$  with $|T'|>1.$ In what follows we collect some information regarding $R$ which we will use in Lemma \ref{dim1-im}. Keep the same notation as in Theorem \ref{classification II} and fix $p,q\in T'$ with $p\neq q$ and $i\in T$ ($i:=0$ if $|T|=1$).  If $|T|=1,$ set $$\ep:=\left\{\begin{array}{ll}
\frac{1}{2}\ep_0&   \hbox{if }X=C(T,T')\\
\ep_0&  \hbox{if }X=BC(T,T').
\end{array}\right.
$$ Then for $\gamma:=\d_p+\d_q,$ $$\theta_1:=\left\{
\begin{array}{ll}
\ep_i+\d_p& \hbox{if }|T|>1\\
\ep+\d_p& \hbox{if }|T|=1
\end{array}
\right. \andd \theta_2:=\left\{
\begin{array}{ll}
\ep_i-\d_p& \hbox{if } |T|>1\\
\ep-\d_p& \hbox{if }|T|=1,
\end{array}
\right.$$ we have $(\theta_1+\theta_2,\gamma)=0,$  $(\theta_1-\theta_2,\theta_1-\theta_2)=2(\theta_1-\theta_2,\gamma),$  $(\theta_1-\theta_2,\gamma)=(\gamma,\gamma),$ $(\gamma,\theta_1),(\gamma,\theta_2)\neq0$ and  the following elements are not roots:
$$\begin{array}{lllll}
2(\theta_1-\theta_2)-\gamma& 2(\theta_1-\theta_2)-3\gamma& \theta_1-\theta_2-3\gamma& 2(\theta_1-\theta_2-\gamma)&
2\theta_1-\theta_2-\gamma\\
\theta_1-2\theta_2-\gamma&2\theta_1-\theta_2-2\gamma&\theta_1-2\theta_2-2\gamma&2\theta_1-\theta_2-3\gamma&
\theta_1-2\theta_2-3\gamma\\
2\theta_1+\theta_2-\gamma&\theta_1+2\theta_2+\gamma&\theta_1-\theta_2+\gamma&\theta_1+\theta_2+\gamma&\theta_1+\theta_2-\gamma\\
 2\theta_2-2\gamma&2\theta_2+2\gamma     &2\theta_1-2\gamma&\theta_1-2\gamma&\theta_2+2\gamma\\
2\theta_1+\theta_2&
2\theta_1-\theta_2&\theta_1+2\theta_2&\theta_1-2\theta_2&2\theta_1-\gamma\\ 2\theta_2+\gamma&
\theta_1+\gamma&\theta_2-\gamma& 2\theta_1&2\theta_2.
\end{array}
$$
}
\end{rem}


\begin{lem}[{\cite[Lem. 2.3]{you6}}]
Suppose that $R$ is an irreducible locally finite root supersystem of type $X$ in an  abelian group $A.$ Then we have the following:

(i) $A$ is a  free abelian group and $R$ contains a   $\bbbz$-basis for $A.$

(ii) If $X\neq A(\ell,\ell),$ $R$ contains a $\bbbz$-basis  $\Pi$  for $A$ satisfying  the partial sum property in the sense that     for each $\a\in \rcross,$ there are $\a_{1},\ldots,\a_{n}\in\Pi$  (not necessarily distinct) and $r_1,\ldots,r_n\in\{\pm1\}$ with $\a=r_1\a_{1}+\cdots+r_n\a_{n}$ and  $r_1\a_{1}+\cdots+r_t\a_{t}\in \rcross,$  for all $1\leq t\leq n.$
 \end{lem}




\begin{deft}
{\rm A subset $\Pi$  of a locally finite root supersystem  $R$ is called an {\it integral base} for $R$ if $\Pi$ is a $\bbbz$-basis for $A.$ An integral base $\Pi$ of $R$ is called a {\it base} for $R$ if it satisfies the partial sum property.}
\end{deft}

\begin{lem}[{\cite[Lem. 2.4($iii$)]{you6}}]\label{base}
If $R$ is an infinite irreducible locally finite root supersystem, then there is a base $\Pi$ for $R$ and a class $\{R_\gamma\mid \gamma\in\Gamma\}$ of finite irreducible  closed sub-supersystems of $R$ of the same type as $R$ such that $R$ is the direct union of $R_\gamma$'s and for each $\gamma\in\Gamma,$ $\Pi\cap R_\gamma$ is a base for $R_\gamma.$
\end{lem}


\begin{lem}[{\cite[Pro. 1.11($iv$)]{you6}}]
\label{type}
Suppose that $(A,\fm , R)$ is an  extended affine root supersystem. Consider the canonical map $\bar{\;}:A\longrightarrow A/A^0$
and denote the induced form on $\bar A:=A/A^0$ by $(\cdot,\cdot\bar{)}.$
Then $(\bar A,(\cdot,\cdot\bar{)}, \bar R)$ is a locally finite root supersystem. Moreover, if $R$ is irreducible, so is $\bar R.$
\end{lem}
\begin{deft}{\rm
Suppose that $(A,\fm , R)$ is an irreducible extended affine root supersystem.  We define the \emph{type} of $R$ to be the type of $\bar R.$}
\end{deft}


\section{Locally Finite Basic Classical  Lie Superalgebras}
Throughout this section, we assume the field  $\bbbf$ is algebraically closed. Let us  summarize some well-known facts regarding finite dimensional  basic classical simple Lie superalgebras (f.d.b.c.s Lie superalgebras for short) which we shall use in the sequel.   For  an irreducible  locally finite root supersystem  $R$ of type $X,$ set
$${}^0R:=\left\{\begin{array}{ll}
\{\a\in \rre\mid 2\a\not\in R\}\cup\{0\}&\hbox{if $X\neq BC(T,T')$}\\\\
\rre\setminus(\rre^2)_{sh}&\hbox{if $X= BC(T,T')$ and $\rre=\rre^1\op\rre^2$ }\end{array}\right.$$
and $${}^1R:=R\setminus {}^0R.$$  If $(\LL,\hh,\fm)$ is an extended affine Lie superalgebra with $\LL^0=\hh$ such that its corresponding root system  $R$ is an irreducible locally finite root supersystem, we have $R_0={}^0R$ and $R_1={}^1R;$ see (\ref{rns}), (\ref{intersect}), Proposition \ref{dim1-re} and Lemma \ref{a11}.
Finite dimensional basic classical simple Lie superalgebras are examples of such  extended affine Lie superalgebras.
A Cartan subalgebra of the even part of a f.d.b.c.s  Lie superalgebra $\LL$ is called a Cartan subalgebra of $\LL.$ It is known that  Cartan subalgebras of $\LL$ are conjugate under the  group of automorphisms  of $\LL;$ see \cite[(3.1.2)]{Muss} and \cite[Cor. 16.4]{Hum}. This together with the classification of f.d.b.c.s Lie superalgebras implies that the corresponding root system of a f.d.b.c.s Lie superalgebra  with respect to an arbitrary  Cartan subalgebra cannot be a finite root supersystem of types  $BC(1,1),$ $C(m,n)$ or $BC(m,n)$ with $n>1.$ Although,  there exists no f.d.b.c.s. Lie superalgebra with root system $T:=C(1,2),$ the root system $R$ of either of $\mathfrak{osp}(3,4)$ or $\mathfrak{osp}(5,2)$ satisfies $R_0=T_0.$
One also  knows that   weight spaces of a f.d.b.c.s Lie superalgebra  corresponding to nonzero roots are one dimensional other than the ones corresponding to nonzero odd roots of a f.d.b.c.s Lie superalgebra of type $A(1,1)$ which are of dimension  2. Moreover, for a f.d.b.c.s Lie superalgebra of type $A(1,1),$   there are nonorthogonal nonsingular  roots $\d_1,\d_2$  such that both   $\d_1+\d_2$ and  $\d_1-\d_2$ are again  roots. This is a phenomena which occurs just for type $A(1,1)$ among all f.d.b.c.s. Lie superalgebras. We finally recall that the root system of   $\mathfrak{osp}(3,2)$ has the same even roots as $A(1,1)$ and that it  contains   nonzero real odd roots while  all nonzero odd roots of $A(1,1)$ are nonsingular.

In this section, we study simple extended affine Lie superalgebras $(\LL,\fm,\hh)$ with $\LL^0=\hh.$  We show that the root system $R$ of  such a Lie superalgebra $\LL$ is an irreducible locally finite root supersystem. We show that $R$ contains a pair $(\d_1,\d_2)$ of nonorthogonal nonsingular roots with $\d_1+\d_2,\d_1-\d_2\in R$ if and only if $R$ is of type $A(1,1);$ in this case, the root spaces corresponding to nonzero nonsingular roots are 2 dimensional while in all other cases the root spaces corresponding to nonzero  roots are 1 dimensional.  We also  prove that the derived subalgebra of the even part of  the Lie  superalgebra $\LL$  is a semisimple Lie algebra and that  $\LL_{\bar 1}$ is a completely reducible $\LL_{\bar 0}$-module with at most two irreducible constituents. We next conclude  that  $\LL$ is a direct union of finite dimensional basic classical simple  Lie    subsuperalgebras.









From now on till the end of this section, we assume $(\LL,\hh,\fm)$ is an extended affine Lie superalgebra with corresponding root system $R$ with $R^\times\neq\emptyset.$
By \cite[Pro. 3.10]{you3}
\begin{equation}\label{rns}
R_0\cap R_{ns}=\{0\}.
\end{equation}
Also for  $\a\in R_i$ ($i=0,1$) with $(\a,\a)\neq 0,$  by \cite[Lem. 3.6]{you3}, there are $e_\a\in (\LL_{\bar i})^\a$ and  $f_\a\in(\LL_{\bar i})^{-\a}$ such that
$(e_\a,f_\a,h_\a:=\frac{2t_\a}{(\a,\a)})$ is an $\mathfrak{sl}_2$-super-triple in the sense that \begin{equation}\label{sl2}[e_\a,f_\a]=h_\a,[h_\a,e_\a]=2e_\a,[h_\a,f_\a]=-2f_\a.\end{equation}  Moreover, the subsuperalgebra $\gg(\a)$ of $\gg$ generated by $\{e_\a,f_\a,h_\a\}$ is either isomorphic to $\mathfrak{sl}_2$ or to $\mathfrak{osp}(1,2)\simeq\mathfrak{spo}(2,1);$ see \cite[\S 2]{you3}.
Now for $\a\in R_0\setminus\{0\}$  with  $(\a,\a)\neq0,$
\begin{equation}\label{auto}
\theta_\a:=\hbox{exp}(\ad_{e_\a}) \hbox{exp}(-\ad_{f_\a}) \hbox{exp}(\ad_{e_\a})
\end{equation}
is an automorphism of $\LL.$ Also one can easily check that $\theta_\a(h)=h-\a(h)h_\a$ for all $h\in \hh.$ Now  let $\b\in R$ and $x\in \LL^\b.$ Suppose that $\theta_\a(x)=\sum_{k\in\bbbz}x_k$ in which $x_k\in  \LL^{\b+k\a}.$ So  for each $h\in \hh,$
 \begin{eqnarray*}
\b(h)\sum_{k\in\bbbz}x_k=\b(h)\theta_\a(x)
&=&\theta_\a[h,x]\\
&=&[\theta_\a(h),\theta_\a(x)]\\
&=&[h-\a(h)h_\a,\sum_{k\in\bbbz}x_k]\\
&=&\sum_{k\in\bbbz}(\b+k\a)(h-\a(h)h_\a)x_k\\
&=&\sum_{k\in\bbbz}(\b(h)-\a(h)\b(h_\a)-k\a(h))x_k.
 \end{eqnarray*} This implies that if $x_k\neq 0$ for some $k\in \bbbz,$ $k=-\b(h_\a)=-\frac{2(\b,\a)}{(\a,\a)},$ i.e., $$\theta_\a(\LL^\b)\sub \LL^{r_\a(\b)},$$ where $r_\a(\b):=\b-\frac{2(\a,\b)}{(\a,\a)}\a.$









\begin{lem}\label{pro-nonzero} Let  $\a,\b\in R,$  then $[\LL^\a,\LL^\b]\neq\{0\}$ if at least one of the following conditions is satisfied:

(i)  $\a\in \rre^\times$ and $\b\in R$ with $\a+\b\in R,$

(ii) $\a,\b\in \rim^\times$ with $\a+\b\in R^\times.$
\end{lem}
\begin{proof}
($i$) Suppose that $\a\in R_i$ ($i\in\{0,1\}$). Fix an $\mathfrak{sl}_2$-super triple $(y_\a,y_{-\a},h_\a),$ with $y_\a\in(\LL_{\bar i})^\a$ and $y_{-\a}\in(\LL_{\bar i})^{-\a},$  corresponding to $\a.$ Take $\gg(\a)$ to be the subsuperalgebra of $\LL$ generated by $\{y_\a,y_{-\a},h_\a\}.$ Consider $\LL$ as a $\gg(\a)$-submodule and set $\v:=\sum_{k\in\bbbz}\LL^{\b+k\a}$ which is  a $\gg(\a)$-submodule of $\LL.$
For each $k\in \bbbz$ and $x\in \LL^{\b+k\a},$ take $\v_k(x)$ to be the submodule of $\v$ generated by $x.$ Then by the proof of Proposition 3.8 of \cite{you3},  $\v_k(x)$ is finite dimensional and so $\v$ is completely reducible as $\gg(\a)$ is a typical f.d.b.c.s Lie superalgebra. Moreover, the action of  $h_\a$ on $\v$ is diagonalizable with the set of eigenvalues $\{\b(h_\a)+2k\mid k\in\bbbz,\LL^{\b+k\a}\neq\{0\}\}$ and  $\v_{\b(h_\a)+2k}=\LL^{\b+k\a}$ for each $k\in \bbbz.$ Now as $\b,\a+\b\in R,$ we have $\b(h_\a)$ and $\b(h_\a)+2$ are eigenvalues for the action of $h_\a$ on $\v.$ So considering \cite[Lem. 2.4]{you3} as well as the module theory of $\frak{sl}_2,$ there is  a finite dimensional irreducible submodule $\u$ of $\v$ such that $\b(h_\a),\b(h_\a)+2$ are eigenvalues for the action of $h_\a$ on $\u,$ therefore using  the $\frak{sl}_2$-module theory together with  \cite[Cor. 2.5]{you3}, we have $[y_\a,\u_{\b(h_\a)}]\neq \{0\}.$ But
$[y_\a,\u_{\b(h_\a)}]\sub [\LL^\a,\LL^\b]$ and so we are done.


($ii$)
Set $\gamma:=\a+\b.$ Using  Lemma \ref{type} and  \cite[Lem. 2.5]{you6}, we have  $\gamma\in\rre^\times $ and  $-\gamma+\a=-\b\in R.$  Now part ($i$)  implies that there are $x\in\LL^{-\gamma}, y\in \LL^{\a}$ with $0\neq[x,y]\in\LL^{-\b}.$ Since the form on $\LL^{\b}\op\LL^{-\b}$ is nondegenerate, there is $z\in \LL^{\b}$ with $0\neq ([x,y],z)=(x,[y,z])$ which implies that $0\neq [y,z]\in[\LL^\a,\LL^{\b}].$
\end{proof}


\begin{Pro}\label{dim1-re} If     $\LL^0=\hh,$ then for $\a\in \rre^\times,$  $\dim(\LL^\a)=1.$  Also  for $\a\in\rre^\times,$ $\a$ is an element of $R_0$ if and only if $2\a\not\in R.$
\end{Pro}
\begin{proof}
We first note that using \cite[Pro. 3.10($iii$)]{you3}, \begin{equation}\label{intersect}R^\times\cap R_0\cap R_1=\emptyset.\end{equation}
Suppose that $\a\in \rre^\times\cap R_{i},$ for some $i\in\{0,1\}.$ One knows that there is an $\frak{sl}_2$-super triple $(e,f,h)$ in which $e\in(\LL_{\bar i})^{\a}=\LL^{\a}$ and $f\in(\LL_{\bar i})^{-\a}=\LL^{-\a};$ see (\ref{intersect}). Take $\gg(\a)$ to be the subsuperalgebra of $\LL$ generated by $\{e,f,h\}$  and set $M:=\LL^{-2\a}\op\LL^{-\a}\op\bbbf t_\a\op\LL^{\a}\op\LL^{2\a}.$ Then by  (\ref{bracket}) and the fact that the only scalar multiples of $\a$ which can be roots are $0,\pm2\a,\pm \a,\pm(1/2)\a,$ we get that  $M$ is a $\gg(\a)$-submodule of $\LL.$ Now using the same argument as in part ($i$) of Lemma \ref{pro-nonzero}, we get that  $M$ is completely reducible. Therefore,  \cite[Lem. 2.4]{you3} together with the $\mathfrak{sl}_2$-module theory implies that the number of irreducible constituents  in the decomposition of $M$ into irreducible submodules is the dimension of the $0$-eigenspace. This means that $M$ is a finite dimensional  irreducible $\gg(\a)$-module and so again  using \cite[Lem. 2.4]{you3} together with the $\mathfrak{sl}_2$-module theory, we get $\dim(\LL^\a)=1.$  Now suppose $\a\in R_0,$ then by (\ref{intersect}), $\LL^{\pm\a}=(\LL_{\bar 0})^{\pm\a}$ and so  $[\LL^\a,\LL^{\a}]=\{0\}=[\LL^{-\a},\LL^{-\a}]$ which in turn implies that $\LL^{-\a}\op\bbbf t_\a\op\LL^{\a}$ is a $\gg(\a)$-submodule of $M.$ But $M$ is irreducible and so $M=\LL^{-\a}\op\bbbf t_\a\op\LL^{\a}.$ This together with \cite[Lem. 3.6]{you3}  completes the proof.
\end{proof}
\begin{lem}\label{restrict}
Let $\LL^0=\hh.$ We have the following statements:

(i) For  $\a,\b\in R$ with $(\a,\b)\neq 0,$ if  $\a-\b\not\in R,$ then for nonzero  elements $x\in\LL^\a,y\in\LL^\b,$ we have $[x,y]\neq0.$

(ii) Suppose   $(\la R\ra,\fm, R)$ is a locally finite root supersystem. Assume  $S$ is a sub-supersystem of $R$ and set $\hh_S:=\hh\cap\sum_{\a\in S^\times}[\LL^\a,\LL^{-\a}].$ Take $\pi:\hh^*\longrightarrow \hh_S^*$ to be the function mapping $\a\in\hh^*$ to $\a\mid_{\hh_S}.$ Then for $\a,\b\in\la  R\ra$ with $\a-\b\in \la S\ra,$ if $\pi(\a)=\pi(\b),$ we have  $\a=\b;$ in particular $\pi$ restricted to $\la S\ra$ is injective.
\end{lem}
\begin{proof}
($i$)  Contemplating  (\ref{intersect}), (\ref{rns}) and (\ref{form-zero}), one finds $i\in\{0,1\}$ with $\LL^\a=(\LL_{\bar i})^\a,$  $\LL^{-\a}=(\LL_{\bar i})^{-\a}$ and  $\LL^{\b}=(\LL_{\bar j})^{\b}.$ Suppose that $x\in\LL^\a,y\in\LL^\b$ are nonzero elements. Using  (\ref{bracket}) together with the fact that the form on $\LL^\a\op\LL^{-\a}$ is nondegenerate, we pick $z\in\LL^{-\a}$ with $[x,z]=t_\a.$ Since $\a-\b\not\in R,$ we have $[z,y]=0.$ So $$0\neq\b(t_\a)y=[t_\a,y]=[[x,z],y]=[x,[z,y]]-(-1)^{|x|}[z,[x,y]]=-(-1)^{|x|}[z,[x,y]].$$ This shows that $[x,y]\neq0$ and so we are done.




($ii$)  Suppose that $\a,\b\in\la R\ra$ and $\a-\b\in \la S\ra.$ If $\a-\b\neq0,$ then since the  form $\fm$ restricted to $\la S\ra$ is nondegenerate, one finds $\gamma\in S$ with $(\a-\b,\gamma)\neq 0.$ This means that $(\a-\b)(t_\gamma)\neq0.$ But by (\ref{sl2}), $t_\gamma\in [\LL^\gamma,\LL^{-\gamma}]\sub \hh_S,$ so $\pi(\a)\neq\pi(\b).$ This completes the proof.
\end{proof}


\begin{lem} \label{final} Suppose that $\LL^0=\hh.$
Assume  $\d_1,\d_2\in \rim$ with $(\d_1,\d_2)\neq 0$ and $\a:=\d_1+\d_2,\b:=\d_1-\d_2\in R.$
Consider (\ref{bracket}) and fix $e\in\LL^\a,$ $f\in\LL^{-\a},$ $x\in\LL^\b$ and $y\in\LL^{-\b}$ with $[e,f]=t_\a$ and $[x,y]=t_\b.$ Then we have the following:

(i) For nonzero elements $a\in\LL^{\d_1}$ and $b\in\LL^{-\d_1},$ we have  $$[f,a],[f,[y,a]],[x,b],[x,[e,b]]\neq0\andd (a,[f,[y,a]])=(b,[x,[e,b]])=0.$$
(ii) For $a,c\in \LL^{\d_1},b,d\in \LL^{-\d_1}$ with
\begin{itemize}
\item $(a,d)=(b,c)=0$\\
\item $(c,d),(a,b)\neq0$\\
\item $(a,[f,[y,c]])=0,$
\end{itemize}
 the  form on each of the  following subspaces
\begin{itemize}
\item $X:=\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]]+\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]]$\\
\item $Y:=\bbbf [f,a]+\bbbf [f,c]+\bbbf [x,b]+\bbbf [x,d]+\bbbf [e,b]+\bbbf [e,d]+\bbbf [y,a]+\bbbf [y,c]$
\end{itemize}
is nondegenerate.
\end{lem}

\begin{proof}
$(i)$
If $\xi:=2\a\in R,$ then $\xi\in \rre^\times$ and we have $$\frac{2(\d_2,\xi)}{(\xi,\xi)}=\frac{4(\d_1,\d_2)}{8(\d_1,\d_2)}=1/2\not\in\bbbz$$ which is a contradiction. So $2\a\not\in R.$ Similarly, $2\b\not\in R.$ This together with Proposition \ref{dim1-re} and (\ref{intersect}) implies that $$\a,\b\in R_0$$ and   $$\begin{array}{ll}\dim\LL_{\bar 0}^\a=\dim\LL^\a=\dim\LL_{\bar 0}^{-\a}=\dim\LL^{-\a}=1,\\\dim\LL_{\bar 0}^\b=\dim\LL^\b=\dim\LL_{\bar 0}^{-\b}=\dim\LL^{-\b}=1;
\end{array}$$ in particular, we have \begin{equation}\label{*}\LL^\a=\bbbf e,\LL^{-\a}=\bbbf f,\LL^\b=\bbbf x,\LL^{-\b}=\bbbf y.\end{equation} We next note that for  $i,j$ with $\{i,j\}=\{1,2\}$ and $r,s\in \{\pm1\}, $  we have $$\frac{2(\d_i,2r\d_i+s\d_j)}{(2r\d_i+s\d_j,2r\d_i+s\d_j)}=\frac{2s(\d_i,\d_j)}{4rs(\d_i,\d_j)}=1/2r\not\in\bbbz.$$ This means that
\begin{equation}\label{not-belong}
2r\d_i+s\d_j\not \in R;\;\;\;\{i,j\}=\{1,2\}\andd r,s\in \{\pm1\} .
 \end{equation}
%Now set $$\kk:=\bbbf t_\a+\bbbf t_\b\op\LL^\a\op\LL^{-\a}\op\LL^\b\op\LL^{-\b}\op\LL^{\d_1}\op\LL^{-\d_1}\op\LL^{\d_2}\op\LL^{-\d_2}.$$ Since $\LL^0=\hh$ and  $\pm2\a,\pm2\b,\pm2\d_1,\pm2\d_2,\pm2\d_1\pm\d_2,\pm2\d_2\pm\d_1\not\in R,$ we get that $\kk$ is a subsuperalgebra of $\LL;$ see \cite[Lem. 2.4]{you3}. We next note that the form restricted to $\kk$ is nondegenerate.
\begin{comment}
One can easily check that the form on $\bbbf t_\a+\bbbf t_\b$ is nondegenerate.

\noindent Claim 1. The form restricted to $T:=\bbbf t_\a\op\bbbf t_\b$ is nondegenerate. Suppose that $r,s\in\bbbf$ and $rt_\a+st_\b$ is an element of the radical of the form restricted to $T.$ Then $r(\a,\a)=(rt_\a+st_\b,t_\a)=0$ as well as $s(\b,\b)=(rt_\a+st_\b,t_\b)=0.$ Therefore we have $r=s=0.$
\end{comment}
Fix $a\in \LL^{\d_1},b\in\LL^{-\d_1},$  then by Lemma \ref{restrict}($i$) and (\ref{not-belong}),  we have $$[f,a]\neq0,\;[f,[y,a]]\neq0,\;[x,b]\neq0,[x,[e,b]]\neq0.$$
Next set $$M:=\bbbf t_\a+\bbbf t_\b\op\LL^\a\op\LL^{-\a}\op\LL^\b\op\LL^{-\b}\op\bbbf a\op\bbbf[f,a]\op\bbbf[y,a]\op\bbbf[f,[y,a]]$$ and to the contrary assume $-([f,a],[y,a])=([a,f],[y,a])=(a,[f,[y,a]])\neq0.$ We recall that  $$\LL^0=\hh\andd \pm2\a,\pm2\b,\pm2\d_1,\pm2\d_2,\pm2\d_1\pm\d_2,\pm2\d_2\pm\d_1\not\in R;$$ see  (\ref{not-belong}),  \cite[Lem. 2.1($ii$)]{you6} and Lemma \ref{type}. Therefore, (\ref{bracket}) together with  Proposition \ref{dim1-re} implies  that $M$ is a subsuperalgebra    of $\LL$ with $$M_{\bar 0}=\bbbf t_\a+\bbbf t_\b\op\LL^\a\op\LL^{-\a}\op\LL^\b\op\LL^{-\b}\andd M_{\bar 1}=\bbbf a\op\bbbf[f,a]\op\bbbf[y,a]\op\bbbf[f,[y,a]].$$
Now as $-([f,a],[y,a])=(a,[f,[y,a]])\neq0,$ the form restricted to $M$ is nondegenerate and so using Lemma \ref{simple}, we get that  $M$ is a finite dimensional basic classical simple Lie superalgebra with $M_{\bar 0}=\mathfrak{sl}_2\op\mathfrak{sl}_2.$ But all odd roots of $M$ are nonsingular, so $M$ is isomorphic to $A(1,1).$ This is a contradiction as  $\dim(M)=10$ while $\dim(A(1,1))=14.$  So $(a,[f,[y,a]])=0.$ Similarly, we have $(b,[x,[e,b]])=0.$






$(ii)$  Using the fact that the form is invariant and supersymmetric together with part ($i$), we have
\begin{equation}\label{equalities}\begin{array}{ll}
  ([e,[x,b]],[f,[y,a]])=-(\d_1,\d_2)^2(b,a),&
([e,[x,d]],[f,[y,a]])=-(\d_1,\d_2)^2(d,a),\\\\
 ([e,[x,b]],[f,[y,c]])=-(\d_1,\d_2)^2(b,c),& ([e,[x,d]],[f,[y,c]])=-(\d_1,\d_2)^2(d,c).
\end{array}\end{equation}
 \begin{comment}
\begin{eqnarray*}
([e,[x,b]],[f,[y,a]])=([[e,[x,b]],f],[y,a])&=& -([f,[e,[x,b]]],[y,a])\\&=&-([[f,e],[x,b]],[y,a])\\
&=&([t_\a,[x,b]],[y,a])\\&=&-(\a,\d_2)([x,b],[y,a])\\
&=&-(\d_1,\d_2)([[x,b],y],a)\\
&=&(\d_1,\d_2)([y,[x,b]],a)\\
&=&(\d_1,\d_2)([[y,x],b],a)\\
&=&-(\d_1,\d_2)([t_\b,b],a)\\
&=&-(\d_1,\d_2)(\b,-\d_1)(b,a)\\
&=&-(\d_1,\d_2)^2(b,a)
\end{eqnarray*}
\end{comment}
and $$-(c,[f,[y,a]])=(a,[f,[y,c ]])=(a,[f,[y,a]])=(c,[f,[y,c]])=0.$$
Now contemplate  the fact that the form  is supersymmetric
 and consider the matrix$$B:=\left(\begin{array}{cc}
A&  \textbf{0}\\
 \textbf{0} &-A^t
 \end{array}\right)$$
in which
{\small $$\begin{array}{rl}A:=&\left(\begin{array}{llll}
(a ,b)&(c ,b)&([e,[x,b]],b)&([e,[x,d]],b)\\
(a ,[f,[y,a ]])&(c ,[f,[y,a ]])&([e,[x,b]],[f,[y,a ]])&([e,[x,d]],[f,[y,a ]])\\
(a ,[f,[y,c ]])&(c ,[f,[y,c ]])&([e,[x,b]],[f,[y,c ]])& ([e,[x,d]],[f,[y,c ]])\\
(a ,d)&(c ,d)&([e,[x,b]],d)& ([e,[x,d]],d)\\
\end{array}\right)\\\\
=&\left(\begin{array}{llll}
(a ,b)&0&0&([e,[x,d]],b)\\
0&0&(\d_1,\d_2)^2(a ,b)&0\\
0&0&0& (\d_1,\d_2)^2(c ,d)\\
0&(c ,d)&-([e,[x,d]],b)& 0\\
\end{array}\right).\end{array}$$}
If $r_1 a+r_2 c+r_3[e,[x,b]]+r_4 [e,[x,d]]+r_5b+r_6 [f,[y,a]]+r_7 [f,[y,c]]+r_8d$  is an element of the radical of the form on $X\times X.$ Then since by (\ref{form-zero}), we have
{\small$$
(\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]],\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]])\sub(\LL^{\d_1},\LL^{\d_1})=\{0\}$$}

and
{\small$$ (\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]],\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])\sub(\LL^{-\d_1},\LL^{-\d_1})=\{0\},$$}
  $(r_1,\ldots,r_8)^t$ ($t$ indicates the transposition) is  a solution for $BX=\textbf{0}.$  But  $$\det(B)=-(\det A)^2=-(\d_1,\d_2)^8(a,b)^4(c,d)^4\neq 0,$$  so $BX=\textbf{0}$ has sole solution $\textbf{0}.$ This in particular implies    that the from restricted to $X\times X$ is nondegenerate.

Next suppose that  $$P:=\underbrace{r_1 [f,a]+r_2 [f,c]+r_3[x,b]+r_4 [x,d]}_{m:=}+\underbrace{r_5[e,b]+r_6 [e,d]+r_7 [y,a]+r_8[y,c]}_{n:=}$$ is an element of the radical of the form restricted to $Y\times Y.$
Using (\ref{form-zero}), we have
\begin{equation}\label{equal4}\begin{array}{ll}
(r_1 [f,a]+r_2 [f,c]+r_3[x,b]+r_4 [x,d],\bbbf [e,b]+\bbbf [e,d]+\bbbf [y,a]+\bbbf [y,c])=\{0\},\\
 (r_5[e,b]+r_6 [e,d]+r_7 [y,a]+r_8[y,c],\bbbf [f,a]+\bbbf [f,c]+\bbbf [x,b]+\bbbf [x,d])=\{0\}.\end{array}\end{equation}
But by  the Jacobi superidentity, we have
 \begin{equation}\label{equal5}
\begin{array}{ll}
\;[f,[e,[x,b]]]=(\d_1,\d_2)[x,b]&[f,[e,[x,d]]]=(\d_1,\d_2)[x,d]\\
\;[e,[f,[y,a]]]=(\d_1,\d_2)[y,a]&[e,[f,[y,c]]]=(\d_1,\d_2)[y,c]\\
\;[f,[e,b]]=(\d_1,\d_2)b&[f,[e,d]]=(\d_1,\d_2)d\\
\;[e,[f,a]]=(\d_1,\d_2)a&[e,[f,a]]=(\d_1,\d_2)c.
\end{array}
 \end{equation}
\begin{comment}
\begin{eqnarray*}
\;[f,[e,[x,b]]]=[[f,a],[x,b]]=-[t_\a,[x,b]]=(\d_1,\d_2)[x,b].\\
\;[f,[e,[x,d]]]=[[f,a],[x,d]]=-[t_\a,[x,d]]=(\d_1,\d_2)[x,d].  \\
\;[f,[e,b]]=[[f,e],b]=-[t_\a,b]=(\d_1,\d_2)b\\
\;[f,[e,d]]=[[f,e],d]=-[t_\a,d]=(\d_1,\d_2)d\\
\;[e,f,[y,a]]=[[e,f],[y,a]]=[t_\a,[y,a]]=(\d_1,\d_2)[y,a]\\
\;[e,f,[y,c]]=[[e,f],[y,c]]=[t_\a,[y,c]]=(\d_1,\d_2)[y,c]\\
\;[e,f,a]=[[e,f],a]=[t_\a,a]=(\d_1,\d_2)a\\
\;[e,f,a]=[[e,f],c]=[t_\a,c]=(\d_1,\d_2)c.\end{eqnarray*}
\end{comment}
Now as the form is invariant, (\ref{equal4}) and (\ref{equal5}) implies that
{\small$$\begin{array}{ll}
(r_1 a+r_2 c+(\d_1,\d_2)^{-1}(r_3[e,[x,b]]+r_4 [e,[x,d]]),\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\},\\
 (r_5 b+r_6 d+(\d_1,\d_2)^{-1}(r_7 [f,[y,a]]+r_8[f,[y,c]]),\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]])=\{0\}.\end{array}$$}
 This  together with (\ref{form-zero}) implies that  $$\begin{array}{l}m':=r_1 a+r_2 c+r_3(\d_1,\d_2)^{-1}[e,[x,b]]+r_4(\d_1,\d_2)^{-1} [e,[x,d]]\andd\\ n':=r_5 b+r_6 d+r_7 (\d_1,\d_2)^{-1}[f,[y,a]]+r_8(\d_1,\d_2)^{-1}[f,[y,c]]\end{array}$$ are elements of the radical of the form on $X\times X.$
Therefore, $m'=n'=0$ and so $m=[f,m']=0 $ and $n=[e,n']=0.$ Thus $P=0.$ This completes the proof.
%Therefore, for $r_1,r_2,r_3,r_4,r_5,r_6,r_7,r_8\in\bbbf,$ we have
%\begin{equation}\label{non}\begin{array}{rl}
%-r_3(\d_1,\d_2)^2(b,a)=&(r_1 a+r_2 c+r_3[e,[x,b]]+r_4 [e,[x,d]],[f,[y,a]])\\
%-r_4(\d_1,\d_2)^2(d,c)=&(r_1 a+r_2 c+r_4 [e,[x,d]],[f,[y,c]])\\
%r_1(a,b)=&(r_1 a+r_2 c,b)\\
%r_2(c,d)=&(r_2 c,d)
%\end{array}\end{equation}and
%\begin{equation}\label{non'}\begin{array}{rl}
%r_5(b,a)=&(r_5b+r_6 d+r_7 [f,[y,a]]+r_8 [f,[y,c]],a)\\
%r_6(d,c)=&(r_6 d+r_7 [f,[y,a]]+r_8 [f,[y,c]],c)\\
%-r_7(\d_1,\d_2)^2(b,a)=&(r_7 [f,[y,a]]+r_8 [f,[y,c]],[e,[x,b]])\\
%-r_8(\d_1,\d_2)^2(c,d)=&(r_8 [f,[y,c]],[e,[x,d]]).
%\end{array}\end{equation}
%We next note that for $\a_1,\a_2\in R,$ $(\LL^{\a_1},\LL^{\a_2})=\{0\}$ if $\a_1+\a_2\neq0.$
%Now if $$r_1 a+r_2 c+r_3[e,[x,b]]+r_4 [e,[x,d]]+r_5b+r_6 d+r_7 [f,[y,a]]+r_8 [f,[y,c]]$$ is an element of the radical of the form restricted to $$\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]]+\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]],$$ we have  $$(r_1 a+r_2 c+r_3[e,[x,b]]+r_4 [e,[x,d]],\bbbf b+\bbbf d+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\}$$ and
%  $$(r_5b+r_6 d+r_7 [f,[y,a]]+r_8 [f,[y,c]],\bbbf a+\bbbf c+\bbbf [e,[x,b]]+\bbbf [e,[x,d]])=\{0\}.$$ So (\ref{non}) and (\ref{non'}) imply that $r_1=r_2=r_3=r_4=r_5=r_6=r_7=r_8=0.$ The last assertion follows from the same argument.
\end{proof}




\begin{Pro}\label{dim1} Suppose that  $\LL^0=\hh.$ Then for  $\d\in R_{ns}^\times,$ $\dim(\LL^\d)=1$ if there is $\eta\in \rim$ with $\d-\eta\not\in R$ and $(\d,\eta)\neq0;$ otherwise,  $\dim(\LL^\d)=2.$ In particular, if $R$ is an irreducible locally finite root supersystem of imaginary type, $\dim(\LL^\d)=1$ for  $\d\in\rim^\times.$
\end{Pro}
\begin{proof}
Suppose that $\d\in\rim^\times.$ If there is $\eta\in\rim$ with  $(\d,\eta)\neq0$  and $\d-\eta\not\in R,$ then  $\d+\eta\in \rre^\times$ and so we have $\dim(\LL^{\d+\eta})=1$ by Proposition  \ref{dim1-re}. Use (\ref{bracket}) to fix $x\in\LL^\eta$ and $y\in\LL^{-\eta}$ with $[x,y]=t_\eta$ and consider the map $\varphi:\LL^\d\longrightarrow \LL^{\d+\eta}$ mapping $a\in\LL^{\d}$ to $[x,a].$ Then by Lemma \ref{restrict}($i$), the map $\varphi$ is an injective linear map
\begin{comment}
; indeed if $ a\in\LL^\d$ and $[x,a]=0,$  using  the fact that  $[y,a]\in\LL^{\d-\eta}=\{0\},$ we have  $$(\d,\eta)a=[t_\eta,a]=[[x,y],a]=[x,[y,a]]-(-1)^{|x||y|}[y,[x,a]]=0$$
which implies that $a=0.$ So $\varphi$ is an injective linear map
 \end{comment}
from $\LL^\d$ into a one dimensional vector space and so it is onto, in particular, $\dim(\LL^\d)=\dim(\LL^{\d+\eta})=1.$


Now suppose for each nonsingular root $\eta\in \rim,$ either $(\d,\eta)=0$ or $\d-\eta\in R.$ Then using \cite[Lem. 1.3($ii$)]{you6}, there is $\eta\in \rim$ with $\d+\eta,\d-\eta\in R_{re}^\times.$ Set $$\d_1:=\d,\d_2:=\eta\in\rim^\times\andd\a:=\d_1+\d_2,\b:=\d_1-\d_2\in \rre^\times.$$
 %
% Suppose that $I$ is an ideal of $M,$ then $I\cap M=((\bbbf t_\a+\bbbf t_\b)\cap I)\op(\LL^\a\cap I)\op(\LL^{-\a}\cap I)\op(\LL^\b\cap I)\op(\LL^{-\b}\cap I)\op(\bbbf a\cap I)\op(\bbbf[f,a]\cap I)\op(\bbbf[y,a]\cap I)\op(\bbbf[f,[y,a]]\cap I).$ If either $a\in I$ or $[f,[y,a]]\in I,$ then $[a,[f,[y,a]]]=(a,[f,[y,a]])t_{\frac{\a+\b}{2}}\in I$ and so $t_{\frac{\a+\b}{2}}\in I.$ This implies that $e,f,x,y\in I$ and so $t_\a,t_\b\in I.$ Now it follows that $M=I.$ Similarly if either $[y,a]\in I$ or $[f,a]\in I,$ using the same argument as above together with the fact that $([a,y],[f,a])=(a,[y,[f,a]])\neq 0$ we get that $M=I.$ Also if $I\cap M_{\bar 0}\neq \{0\},$ one gets that $M=I.$ So $M$ is a simple Lie superalgebra. So each invariant bilinear form on $M$ is either identically zero or it is nondegenerate. So the form $\fm$ restricted to $M$ is a nondegenerate even supersymmetric bilinear form. This means that $M$ is a basic classical Lie superalgebra with root system $\{0,\pm\a,\pm\b,\frac{\pm\a\pm\b}{2}\}$ which is absurd by the classification of finite dimensional basic classical simple  Lie superalgebras.
As in the proof of Lemma  \ref{final}, we have
\begin{equation}
\label{r-re}
\a,\b\in R_0.
\end{equation}Since the form is nondegenerate on $\LL^{\d_1}\op\LL^{-\d_1},$ we  pick  $a\in\LL^{\d_1}$ and $b\in \LL^{-\d_1}$ with $(a,b)\neq0.$ Next considering (\ref{bracket}), we fix $e\in\LL^\a,$ $f\in\LL^{-\a},$ $x\in\LL^\b$ and $y\in\LL^{-\b}$ with $[e,f]=t_\a$ and $[x,y]=t_\b$ and set $$\begin{array}{llll}
a_1:=a\in\LL^{\d_1}& a_2:=[f,a]\in\LL^{-\d_2},&a_3:=[y,a]\in\LL^{\d_2},&a_4:=[f,[y,a]]\in\LL^{-\d_1},\\
b_1:=b\in\LL^{-\d_1}& b_2:=[e,b]\in\LL^{\d_2},&b_3:=[x,b]\in\LL^{-\d_2},&b_4:=[e,[x,b]]\in\LL^{\d_1}.
\end{array}$$
Then using \cite[Lem. 2.1($ii$)]{you6}, (\ref{r-re}),  (\ref{not-belong}), (\ref{rns}) and (\ref{bracket}), one can see that $M:=M_{\bar 0}\op M_{\bar 1}$ with
$$\begin{array}{l}
M_{\bar 0}:=\bbbf t_\a+\bbbf t_\b+\bbbf e+\bbbf f+\bbbf x+\bbbf y\andd
M_{\bar1}:=\sum_{i=1}^4(\bbbf a_i+\bbbf b_i)
\end{array}$$
is a  finite dimensional  subsuperalgebra of $\LL.$ It follows from Lemma \ref{final}($i$) and (\ref{equalities}) that  the form restricted to $M$ is nondegenerate and so by Lemma \ref{simple}, $M$ is a finite dimensional basic classical  simple Lie superalgebra with  $M_{\bar 0}\simeq\mathfrak{sl}_2\op\mathfrak{sl}_2.$ So   $M$ is a finite dimensional basic classical simple Lie superalgebra isomorphic to $A(1,1)$ as all odd roots are nonsingular. Therefore, its root spaces corresponding to nonzero nonsingular roots are 2-dimensional. This in particular implies that $a_1$ and $b_4$ are linearly independent and so  $dim (\LL^{\d_1})\geq 2.$  Now to the contrary assume $c\in \LL^{\d_1}$ is  linearly independent from $\{a_1,b_4\}.$ Since  the form is supersymmetric and invariant, we have $(a,[f,[y,c]])=-(c,[f,[y,a]]).$
%\begin{eqnarray*}
%(a,[f,[y,c]])=([a,f],[y,c])=-([f,a],[y,c])
%&=&-([[f,a],y],c)\\&=&([y,[f,a]],c)\\
%&=&([f,[y,a]],c)\\
%&=&-(c,[f,[y,a]]).
%\end{eqnarray*}
Now contemplating (\ref{equalities}) and Lemma \ref{final}($i$), we  set {\small $$\begin{array}{rl}A:=&\left(\begin{array}{lll}
(a,b)&(c,b)&([e,[x,b]],b)\\
(a,[f,[y,a]])&(c,[f,[y,a]])&([e,[x,b]],[f,[y,a]])\\
(a,[f,[y,c]])&(c,[f,[y,c]])&([e,[x,b]],[f,[y,c]])\\
\end{array}\right)\\\\
=&\left(\begin{array}{lll}
(a,b)&(c,b)&0\\
0&(c,[f,[y,a]])&([e,[x,b]],[f,[y,a]])\\
(a,[f,[y,c]])&0&([e,[x,b]],[f,[y,c]])\\
\end{array}\right)\\\\
=&\left(\begin{array}{lll}
(a,b)&(c,b)&0\\
0&-(a,[f,[y,c]])&(\d_1,\d_2)^2(a,b)\\
(a,[f,[y,c]])&0&(\d_1,\d_2)^2(c,b)\\
\end{array}\right).\end{array}$$} Since $det(A)=0,$ the system $AX=\textbf{0}$ has a nonzero solution $(r,s,k)^t$ ($t$ indicates  the transposition).
Setting  $c':=ra+sc+k[e,[x,b]],$ we have $c'\neq 0$ as  $\{a,c,[e,[x,b]]\}$ is linearly independent. Also  since $A(r,s,k)^t=0,$ $$(c',b)=(c',[f,[y,a]])=(c',[f,[y,c]])=0.$$ Suppose that  $s=0,$ %$(r,k)^t$ is the nonzero solution for $BX=0$ where {\small\begin{eqnarray*}B:=\left(\begin{array}{ll}
%(a,b)&([e,[x,b]],b)\\
%(a,[f,[y,a]])&([e,[x,b]],[f,[y,a]])\\
%\end{array}\right)&=&\left(\begin{array}{lll}
%(a,b)&0\\
%0&([e,[x,b]],[f,[y,a]])\\
%\end{array}\right)\\
%&=&\left(\begin{array}{lll}
%(a,b)&0\\
%0&(\d_1,\d_2)^2(a,b)\\
%\end{array}\right),\end{eqnarray*}} which is absurd as $B$ is invertible.
then $A(r,s,k)^t=0$ implies that $r(a,b)=k(\d_1,\d_2)^2(a,b)=0.$ So we have $r=s=k=0$ which is absurd.
 Thus $s\neq 0.$ This implies that $\{a,c',[e,[x,b]]\}$ is linearly independent. Next considering the fact that  the form is nondegenerate  on $\LL^{\d_1}\op\LL^{-\d_1},$ one finds  $d\in \LL^{-\d_1}$ with $(c',d)\neq 0.$ If $(a,d)=0,$ take $a':=a,$ otherwise set $a':=-\frac{(c',d)}{(a,d)}a+c',$ then $(a',d)=0$ and $-(a',[f,[y,c']])=(c',[f,[y,a']])=0.$
Now using Lemma \ref{final}($ii$), it is not hard to see that  the form is nondegenerate on $K\times K$ where   $K:=K_{\bar 0}\op K_{\bar 1}$ with  $$K_{\bar 0}:=\bbbf t_\a+\bbbf t_\b+\LL^\a+\LL^{-\a}+\LL^{\b}+\LL^{-\b}$$and
$$\begin{array}{rl}K_{\bar 1}:=&\bbbf a'+\bbbf c'+\bbbf [e,[x,b]]+\bbbf [e,[x,d]]+\bbbf b+\bbbf d+\bbbf [f,[y,a']]+\bbbf [f,[y,c']]\\
+&\bbbf [f,a']+\bbbf [f,c']+\bbbf [x,b]+\bbbf [x,d]+\bbbf [e,b]+\bbbf [e,d]+\bbbf [y,a']+\bbbf [y,c'].\end{array}$$
So by Lemma \ref{simple}, $K$
 is a finite dimensional basic classical simple Lie superalgebra with $K_{\bar 0}\simeq\mathfrak{sl}_2\op\frak{sl}_2.$ Since all odd roots are nonsingular, we get that $K$ is of type $A(1,1)$ which is absurd due to its dimension.
This completes the proof of the first assertion.  Now suppose $R$ is an irreducible locally finite root supersystem of imaginary type and $\d\in \rim^\times,$ then by \cite[Lem. 1.3($ii$)]{you6}, there is $\eta\in \rim$ with $(\d,\eta)\neq0;$ on the other hand, we know from the classification theorem for imaginary types that there are no two nonorthogonal  nonsingular roots whose both summation and subtraction are again roots. So there is $s\in\{\pm1\}$ with $\d+s\eta\in R$ and $\d-s\eta\not\in R.$ Now the result follows from the first assertion.
\begin{comment}
Set $$\begin{array}{l}N_{\bar 0}:=\bbbf t_\a+\bbbf t_\b+\bbbf e+\bbbf f+\bbbf x+\bbbf y,\\
N_{\bar 1}:=\sum_{i=1}^4(\bbbf a_i+\bbbf b_i)+\bbbf c+\bbbf [f,c]+\bbbf [y,c]+\bbbf [f,[y,c]]
\end{array}$$ and take $N:=N_{\bar 0}\op N_{\bar 1}.$ Then  $N$  is a subsuperalgebra of $\LL$ having  a weight space decomposition $N=\op_{\gamma\in\{0,\pm\a,\pm\b,\pm\d_1,\pm\d_2\}} N^{\gamma}$ with respect to $\bbbf t_\a\op\bbbf t_\b$ with $N^\gamma=N\cap \LL^\gamma$ for each $\gamma\in \{0,\pm\a,\pm\b,\pm\d_1,\pm\d_2\}.$ If the form on $N$ is nondegenerate, then  as before, one gets that $N$  is a finite dimensional basic classical simple Lie superalgebra isomorphic to $A(1,1)$ which is a contradiction as the dimension of  $N^{\d_1}=\LL^{\d_1}\cap N$ is $3.$ So the form is degenerate on $N.$ Since the form is even and nondegenerate on $N_{\bar 0},$ (\ref{form-zero}) implies that the form is degenerate either  on $N^{\d_1}\op N^{-\d_1}$ or on $N^{-\d_2}\op N^{-\d_2}.$  If the form is degenerate on $N^{\d_1}\op N^{-\d_1},$ then
there is an element  $ra+sc+k[e,[x,b]]+r'b+s'[f,[y,a]]+k'[f,[y,c]]$  of the radical of the form on $N^{\d_1}+N^{-\d_1}.$ Therefore by (\ref{form-zero}), one of the following two cases can happen:
\begin{itemize}
\item[(1)] $ra+sc+k[e,[x,b]]\neq 0$ and \\
$(ra+sc+k[e,[x,b]],\bbbf b+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\},$\\
\item[(2)] $r'b+s'[f,[y,a]]+k'[f,[y,c]]\neq 0$ and\\
 $(r'b+s'[f,[y,a]]+k'[f,[y,c]],\bbbf a+\bbbf c+\bbbf [e,[x,b]]) =\{0\}.$
\end{itemize}
 If (2) happens, $ r'b+s'[f,[y,a]]+k'[f,[y,c]]\in \LL^{-\d_1}$ is nonzero. Since $(\d_1,\a)\neq0,-\d_1-\a\not\in R$ and $(\b,-\d_1+\a)\neq0,-\b+(-\d_1+\a)\not\in R,$ Lemma \ref{restrict} implies that {\small\begin{equation}\label{final2}r'[x,[e,b]]+s'[x,[e,[f,[y,a]]]]+k'[x,[e,[f,[y,c]]]]=[x,[e,r'b+s'[f,[y,a]]+k'[f,[y,c]]]]\neq0.\end{equation}}
We next note that  as $[x,e]\in\LL^{2\d_1}=\{0\},$ the Jacobi superidentity implies that \begin{equation}\label{equal3}\begin{array}{l}
\;[e,[x,[f,[y,a]]]]=[x,[e,[f,[y,a]]]]\in(\bbbf\setminus\{0\})a\andd\\
\; [e,[x,[f,[y,c]]]]=[x,[e,[f,[y,c]]]]\in(\bbbf\setminus\{0\})c.\end{array}\end{equation} So we have
\begin{eqnarray*}&&(r'b+s'[f,[y,a]]+k'[f,[y,c]],\bbbf [x,[e,[f,[y,a]]]]+\bbbf [x,[e,[f,[y,c]]]]+\bbbf [x,[e,b]])\\
&=&(r'b+s'[f,[y,a]]+k'[f,[y,c]],\bbbf a+\bbbf c+\bbbf [e,[x,b]]) \\
&=&\{0\}.
 \end{eqnarray*}This together with the fact that  the form is invariant implies that  $$(r'[e,[x,b]]+s'[e,[x,[f,[y,a]]]]+k'[e,[x,[f,[y,c]]]],\bbbf b+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\}.$$  So  by (\ref{equal3}) and (\ref{final}), without loss of generality,  we always may assume $(1)$ occurs, i.e., $$ra+sc+k[e,[x,b]]\neq 0\andd(ra+sc+k[e,[x,b]],\bbbf b+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\}.$$
Therefore, contemplating (\ref{equalities}) and Lemma \ref{final}($i$), for {\small $$\begin{array}{rl}A:=&\left(\begin{array}{lll}
(a,b)&(c,b)&([e,[x,b]],b)\\
(a,[f,[y,a]])&(c,[f,[y,a]])&([e,[x,b]],[f,[y,a]])\\
(a,[f,[y,c]])&(c,[f,[y,c]])&([e,[x,b]],[f,[y,c]])\\
\end{array}\right)\\\\
=&\left(\begin{array}{lll}
(a,b)&(c,b)&0\\
0&(c,[f,[y,a]])&([e,[x,b]],[f,[y,a]])\\
(a,[f,[y,c]])&0&([e,[x,b]],[f,[y,c]])\\
\end{array}\right),\end{array}$$} the system $AX=0$ has a nonzero solution $(r,s,k)^t$ ($t$ indicates the transposition) and so the determinate of $A$ is zero. But using (\ref{equalities}), since the form is invariant and supersymmetric, we have $$\det(A)=-2(\d_1,\d_2)^2(a,b)(b,c)(a,[f,[y,c]])=2(\d_1,\d_2)^2(a,b)(b,c)(c,[f,[y,a]]).$$ So $(b,c)(a,[f,[y,c]])=-(b,c)(c,[f,[y,a]])=0$ as we desired.
Next suppose the form is degenerate on $N^{\d_2}\op N^{-\d_2}.$ Using the same argument as above, we find $r,s,k\in\bbbf$ such that  {\small $$r[f,a]+s[f,c]+k[x,b]\neq 0\andd(r[f,a]+s[f,c]+k[x,b],\bbbf [e,b]+\bbbf [y,a]+\bbbf [y,c])=\{0\}.$$} But $$[x,b]=k'[f,e,[x,b]]\andd b=s' [f,[e,b]]\;\;\;\;\hbox{ for some } k',s'\in \bbbf\setminus\{0\}.$$ So the invariancy of the form implies that $$ra+sc+kk'[e,[x,b]]\neq 0\andd(ra+sc+kk'[e,[x,b]],\bbbf b+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\}.$$ Therefore, we return to the previous case  and so we are done.



Now we proceed with the proof in the following  three  cases:

\medskip

\noindent {\bf Case 1.} $(b,c)\neq 0$ and $(c,[f,[y,a]])=(a,[f,[y,c]])=0:$   We have using Lemma \ref{final}($i$) that $0=(ra+sc+k[e,[x,b]],[f,[y,a]])=-k(\d_1,\d_2)^2(a,b),$ so we have  $k=0.$ Therefore $a':=ra+sc$ is an element of the radical of the form on $N.$ Since the form is nondegenerate on $\LL^{\d_1}\op\LL^{-\d_1},$ one finds $d\in \LL^{-\d_1}$ with $(a',d)\neq0.$ We note that $r,s\neq0$ and consider the matrix $$\left(\begin{array}{cc}
 r(a,d)&s(c,d)\\
 r(a,b)&s(c,b)
 \end{array}\right)=\left(\begin{array}{cc}
 r(a,d)&s(c,d)\\
 r(a,b)&-r(a,b)
 \end{array}\right)$$
  which is nonsingular as $(ra+sc,d)=(a',d)\neq0.$ Therefore there is $p,q\in\bbbf$ such that $$\left(\begin{array}{cc}
 r(a,d)&s(c,d)\\
 r(a,b)&s(c,b)
 \end{array}\right)\left(\begin{array}{c}
p\\
q
 \end{array}\right)=\left(\begin{array}{c}
 0\\
1
 \end{array}\right).$$ Now setting $r':=rp,$  $s':=sq$ and $c':=r'a+s'c,$ we have $(c',b)\neq0$ and $(c',d)=0.$ Using Lemma \ref{final}($ii$), we get that the form on each of the following subspaces
\begin{itemize}
\item $\bbbf a'+\bbbf c'+\bbbf [e,[x,b]]+\bbbf [e,[x,d]]+\bbbf b+\bbbf d+\bbbf [f,[y,a']]+\bbbf [f,[y,c']]$\\
\item $\bbbf [f,a']+\bbbf [f,c']+\bbbf [y,a']+\bbbf [y,c']+\bbbf [e,b]+\bbbf [e,d]+\bbbf [x,b]+\bbbf [x,d]$
\end{itemize}
is nondegenerate.
This in turn implies that for $$K_{\bar 0}:=\bbbf t_\a+\bbbf t_\b+\LL^\a+\LL^{-\a}+\LL^{-\a}+\LL^{-\a}$$and
$$\begin{array}{rl}K_{\bar 1}:=&\bbbf a'+\bbbf c'+\bbbf [e,[x,b]]+\bbbf [e,[x,d]]+\bbbf b+\bbbf d+\bbbf [f,[y,a']]+\bbbf [f,[y,c']]\\
+&\bbbf [f,a']+\bbbf [f,c']+\bbbf [y,a']+\bbbf [y,c']+\bbbf [e,b]+\bbbf [e,d]+\bbbf [x,b]+\bbbf [x,d],\end{array}$$
$K:=K_{\bar 0}\op K_{\bar 1}$ is a finite dimensional basic classical simple Lie superalgebra  of type $A(1,1)$ which is absurd due to its dimension.



\medskip

\noindent {\bf Case 2.} $(b,c)= 0$ and $(c,[f,[y,a]])\neq0:$ We know that  there is $r,s,k\in\bbbf$ with  $$ra+sc+k[e,[x,b]]\neq 0\andd(ra+sc+k[e,[x,b]],\bbbf b+\bbbf [f,[y,a]]+\bbbf [f,[y,c]])=\{0\}.$$ Since $0=(ra+sc+k[e,[x,b]], b)=r(a,b),$ we get that $r=0.$ Exchanging the role of $a$ and $[e,[x,b]]$ as well as the role of $b$ and $[f,[y,a]]$ in the previous argument,  we get a contradiction in this case as well.


\medskip

\noindent {\bf Case 3.}
$(c,b)=(c,[f,[y,a]])= 0:$ In this case   $c$ is an element of the radical of the form on $N.$ Set $c':=c$ and pick $d\in\LL^{-\d_1}$ with $(c,d)=(c',d)\neq0.$ If $(a,d)=0,$ take $a':=a,$ otherwise set $a':=-\frac{(c,d)}{(a,d)}a+c,$ then for $K$ defined as in Case 1, $K$ is a finite dimensional basic classical simple Lie superalgebra,  a contradiction as before. This completes the proof.
\end{comment}
\end{proof}

\begin{lem}\label{dim1-im} Suppose that    $\LL^0=\hh.$ If    $R $ is an  irreducible extended affine root supersystem of type $X\neq A(1,1),BC(1,1),$ then for $\d_1,\d_2\in \rim$ with $(\d_1,\d_2)\neq0,$ there is a  unique $r\in\{\pm 1\}$ with $\d_1+r\d_2\in R.$
\end{lem}
\begin{proof}
Suppose that $\d_1,\d_2\in \rim$ with $(\d_1,\d_2)\neq0.$ Since $R$ is an extended affine root supersystem, there is at least one $r\in\{\pm1\}$ with $\d_1+r\d_2\in R.$ Now to the contrary, we assume $\a:=\d_1+\d_2,\b:=\d_1-\d_2\in R$ and get a contradiction.  Since $X\neq A(1,1),BC(1,1)$ and  there are two non-orthogonal  nonsingular roots whose summation and subtraction are again roots, using Lemma \ref{type} together with Theorems \ref{classification I} and \ref{classification II}, we get that  $R$ is of  one of the types $C(1,T)$ or $C(T,T'),BC(T,T')$ with $|T'|>1.$ Considering Lemma \ref{type} and  Theorem \ref{classification II} together with Remark \ref{rem3}(iii), there is $\gamma\in \rre$ such that $(\a,\gamma)=0,$ $(\gamma,\d_1),(\gamma,\d_2)\neq0,$ $(\b,\b)=2(\b,\gamma),$ $\frac{2(\b,\gamma)}{(\gamma,\gamma)}=2$ and that   the following elements  are not elements of $R:$
$$\begin{array}{lllll}
2(\d_1-\d_2)-\gamma& 2(\d_1-\d_2)-3\gamma& \d_1-\d_2-3\gamma& 2(\d_1-\d_2-\gamma)&
2\d_1-\d_2-\gamma\\
\d_1-2\d_2-\gamma&2\d_1-\d_2-2\gamma&\d_1-2\d_2-2\gamma&2\d_1-\d_2-3\gamma&
\d_1-2\d_2-3\gamma\\
2\d_1+\d_2-\gamma&\d_1+2\d_2+\gamma&\d_1-\d_2+\gamma&\d_1+\d_2+\gamma&\d_1+\d_2-\gamma\\
 2\d_2-2\gamma&2\d_2+2\gamma     &2\d_1-2\gamma&\d_1-2\gamma&\d_2+2\gamma\\
2\d_1+\d_2&
2\d_1-\d_2&\d_1+2\d_2&\d_1-2\d_2&2\d_1-\gamma\\ 2\d_2+\gamma&
\d_1+\gamma&\d_2-\gamma& 2\d_1&2\d_2.
\end{array}
$$
\begin{comment}

By Lemma \ref{radical}, there is   a subspace $\dot\v$ of $\v$ with  $\v=\dot\v\op\v^0$ such that $\dot R:=\{\dot\a\in\dot\v\mid \exists \sg\in\v^0; \dot\a+\sg\in R\}$ is a locally finite root supersystem in its $\bbbz$-span of the same type as $R$ and there is a class $\{S_{\dot\a}\mid \dot\a\in \dot R\}$ of subsets of $\v^0$ such that $R=\cup_{\dot\a\in \dot R}(\dot\a+S_{\dot\a})$ and
$$\left\{\begin{array}{ll}
0\in S_{\dot\a}& \hbox{if $\dot R$ is of real type and $\dot\a\in (\dot R_{re})_{red}:=\dot R_{re}\setminus\{2\a\mid\a\in \dot R_{re}\}\cup\{0\}$}\\
0\in S_{\dot\a}& \hbox{if $\dot R$ is of imaginary type and $\dot\a\in \dot R.$}
\end{array}\right.$$
There are $\dot\d_1,\dot\d_2\in\dot R_{ns}^\times,$ $\sg\in S_{\dot\d_1}$ and $\tau\in S_{\dot\d_2}$ such that $\d_1=\dot\d_1+\sg$ and $\d_2=\dot\d_2+\tau.$ Set $$\dot\a:=\dot\d_1+\dot\d_2,\dot\b:=\dot\d_1-\dot\d_2,\a:=\dot\a+(\sg+\tau),\b:=\dot\b+(\sg-\tau)$$
Considering Remark \ref{rem3}, we know that there is $\gamma\in \dot R\cap \rre$ such that $(\gamma,\d_1),(\gamma,\d_2)\neq0,$ $\frac{2(\b,\gamma)}{(\gamma,\gamma)}=2$ and  the following elements  are not elements of $R:$
$$\begin{array}{lllll}
2(\d_1-\d_2)-\gamma& 2(\d_1-\d_2)-3\gamma& \d_1-\d_2-3\gamma& 2(\d_1-\d_2-\gamma)&
2\d_1-\d_2-\gamma\\
\d_1-2\d_2-\gamma&2\d_1-\d_2-2\gamma&\d_1-2\d_2-2\gamma&2\d_1-\d_2-3\gamma&
\d_1-2\d_2-3\gamma\\
2\d_1+\d_2-\gamma&\d_1+2\d_2+\gamma&\d_1-\d_2+\gamma&\d_1+\d_2+\gamma&\d_1+\d_2-\gamma\\
 2\d_2-2\gamma&2\d_2+2\gamma     &2\d_1-2\gamma&\d_1-2\gamma&\d_2+2\gamma\\
2\d_1+\d_2&
2\d_1-\d_2&\d_1+2\d_2&\d_1-2\d_2&2\d_1-\gamma\\ 2\d_2+\gamma&
\d_1+\gamma&\d_2-\gamma& 2\d_1&2\d_2.
\end{array}
$$
\end{comment}
%
%$$\begin{array}{lll}
%\dot\b-2\dot\gamma\in(\dot R_{re})_{sdiv}& \dot\gamma-\dot\b\in(\dot R_{re})_{sdiv}&\dot\gamma-\dot\b\not\in\dot R\\
%\dot\gamma+\dot\d_1\not\in\dot R&\dot\gamma-\dot\d_1\in\dot R&\dot\d_1\pm\d_2\in \dot R\\
%\dot\gamma-\dot\d_2\not\in\dot R&\dot\gamma+\dot\d_2\in\dot R&\dot\a\pm\dot\b\not\in\dot R\\
%\dot\a+\dot\d_1\not\in\dot R&\dot\a+\dot\d_2\not\in\dot R&\dot\a\pm\dot\gamma\not\in\dot R\\
%\dot\b-\dot\gamma\in\dot R& \dot\b+\dot\gamma\not\in \dot R& \dot\b +\dot\d_2\in \dot R\\
%\dot\b-\dot\d_1\in\dot R& \dot\b+\dot\d_1\not\in \dot R& \dot\b -\dot\d_2\not\in \dot R\\
%\end{array}$$
Since $(\gamma,\d_1),(\gamma,\d_2)\neq0$ and  $\gamma+\d_1\not\in  R,$ $\gamma-\d_2\not\in R,$ we have $\eta:=\d_1-\gamma,\zeta:=-\gamma-\d_2\in R.$ We next note that $\frac{2(\b,\gamma)}{(\gamma,\gamma)}=2$ and that $\b+\gamma\not\in R,$ so we have using the root string property that $\b-\gamma,\b-2\gamma\in R_{re}^\times.$ In particular, we have $$\eta,\zeta\in \rim^\times\andd \eta+\zeta=\b-2\gamma,\eta-\zeta=\a\in R.$$
Fix $e\in\LL^\a,$ $f\in\LL^{-\a},$ $x\in\LL^\b,$ $y\in\LL^{-\b},$   $m\in \LL^{\gamma}$ and $n\in \LL^{-\gamma}$ with $[e,f]=t_\a,$  $[x,y]=t_\b$ and  $[m,n]=t_\gamma.$
 Since the form on  $\LL^{\d_1}\op\LL^{-\d_1}$ is nondegenerate, one can find $a\in\LL^{\d_1}$ and  $b\in \LL^{-\d_1}$ with $(a,b)\neq 0.$ Take $a_i,b_i,$ $1\leq i\leq 8,$ to be as in the following table:


\begin{center}
\begin{tabular}{|l|l||l|l|}
\hline
$a_1$&$a\in\LL^{\d_1}$& $b_1$& $b\in\LL^{-\d_1}$\\
\hline
$a_2$&$[f,a]\in\LL^{-\d_2}$& $b_2$& $[e,b]\in\LL^{\d_2}$\\
\hline
$a_3$&$[y,a]\in\LL^{\d_2}$& $b_3$& $[x,b]\in\LL^{-\d_2}$\\
\hline
$a_4$&$[f,[y,a]]\in\LL^{-\d_1}$& $b_4$& $[e,[x,b]]\in\LL^{\d_1}$\\
\hline
$a_5$&$[n,a]\in\LL^{\eta}$& $b_5$& $[m,b]\in\LL^{-\eta}$\\
\hline
$a_6$&$[n,[f,a]]\in\LL^{\zeta}$& $b_6$& $[m,[e,b]]\in\LL^{-\zeta}$\\
\hline
$a_7$&$[m,[y,a]]\in\LL^{-\zeta}$& $b_7$& $[n,[x,b]]\in\LL^{\zeta}$\\
\hline
$a_8$&$[m,[f,[y,a]]]\in\LL^{-\eta}$& $b_8$& $[n,[e,[x,b]]]\in\LL^{\eta}$\\
\hline
\end{tabular}
\end{center}
and note that for $1\leq i\leq 8,$ $(a_i,b_i)\neq 0$  as the form is invariant as well as supersymmetric and $(a,b)\neq0.$ This in turn implies that $[a_i,b_i]\neq 0$ by (\ref{bracket}).




 Next we note that if $t_\b=kt_\gamma,$ then $2(\b,\gamma)=(\b,\b)=k(\b,\gamma)=k(\gamma,\gamma)=(\b,\gamma)$ which is a contradiction. Also if $t_\a\in \hbox{span}_\bbbf \{t_\b,t_\gamma\},$ one concludes $(t_\a,t_\a)=0$ which is absurd, so $\{t_\a,t_\b,t_\gamma\}$ is a linearly independent subset of $\hh.$  Now set $$K:=\bbbf t_\a+\bbbf t_\b+\bbbf t_\gamma+\bbbf t_{\b-\gamma}+\bbbf t_{\b-2\gamma}=\bbbf t_\a\op\bbbf t_\b\op\bbbf t_\gamma$$ and   $$M=M_{\bar 0}\op M_{\bar 1}$$ with
$$\begin{array}{ll}
M_{\bar 0}:=K\bigoplus\op_{\theta\in\{\pm\a,\pm\b,\pm\gamma,\pm(\b-\gamma),\pm(\b-2\gamma)\}}\LL^\theta,&
M_{\bar 1}:=\sum_{i=1}^8(\bbbf a_i+\bbbf b_i).
 \end{array}$$
We claim that $M$ is a subsuperalgebra of $\LL.$
To prove our claim, we need to know the multiplication table of the elements of a basis of $M.$ Since $[\LL_{\bar 1},\LL_{\bar 1}]\sub \LL_{\bar 0},$ we first show that for $z,z'\in \{a_i,b_i\mid 1\leq i\leq 8\},$ $[z,z']$ is an element of $M_0.$ In the following tables, we denote the weight $[z,z']$ for all $z,z'\in\{a_i,b_i\mid 1\leq i\leq 8\};$ in fact the   $(r,s)$-th entry of each table denotes the summation of the weights of $r$-th entry of the first column and $s$-th entry of the first row; the red parts denote the elements which are not roots.
\medskip


{\tiny\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
&$a_1$&$a_2$&$a_3$&$a_4$&$a_5$&$a_6$&$a_7$&$a_8$\\
\hline
$a_1$&${\red {2\d_1}}$&$\b$&$\a$&$\d_1-\d_1$&${\red {2\d_1-\gamma}}$&$\d_1-\d_2-\gamma$&${\red {\d_1+\d_2+\gamma}}$&$\gamma$\\
\hline
$a_2$&$\b$&${\red {-2\d_2}}$&$\d_2-\d_2$&$-\a$&$\d_1-\d_2-\gamma$&${\red {-2\d_2-\gamma}}$&$\gamma$&${\red  {-\d_1-\d_2+\gamma}}$\\
\hline
$a_3$&$\a$&$\d_2-\d_2$&${\red {2\d_2}}$&$-\b$&${\red {\d_1+\d_2-\gamma}}$&$-\gamma$&${\red {2\d_2+\gamma}}$&$\d_2-\d_1+\gamma$\\
\hline
$a_4$&$\d_1-\d_1$&$-\a$&$-\b$&${\red {-2\d_1}}$&$-\gamma$&${\red {-\d_1-\d_2-\gamma}}$&$-\d_1+\d_2+\gamma$&${\red  {-2\d_1+\gamma}}$\\
\hline
$a_5$&${\red {2\d_1-\gamma}}$&$\d_1-\d_2-\gamma$&${\red {\d_1+\d_2-\gamma}}$&$-\gamma$&${\red {2\d_1-2\gamma}}$&$\d_1-\d_2-2\gamma$&$\a$&$\eta-\eta$\\
\hline
$a_6$&$\d_1-\d_2-\gamma$&${\red {-2\d_2-\gamma}}$&$-\gamma$&${\red {-\d_1-\d_2-\gamma}}$&$\d_1-\d_2-2\gamma$&${\red {-2\d_2-2\gamma}}$&$ \zeta-\zeta$&$-\a$\\
\hline
$a_7$&${\red {\d_1+\d_2+\gamma}}$&$\gamma$&${\red {2\d_2+\gamma}}$&$-\d_1+\d_2+\gamma$&$\a$&$\zeta-\zeta$&${\red{ 2\d_2+2\gamma}}$&$-\d_1+\d_2+2\gamma$\\
\hline
$a_8$&$\gamma$&${\red {-\d_1-\d_2+\gamma}}$&$\d_2-\d_1+\gamma$&${\red {-2\d_1+\gamma}}$&$\eta-\eta$&$-\a$&$-\d_1+\d_2+2\gamma$&${\red {2\gamma-2\d_1}}$\\
\hline
$b_1$&$\d_1-\d_1$&$-\a$&$-\b$&${\red {-2\d_1}}$&$-\gamma$&${\red {-\d_1-\d_2-\gamma}}$&$-\d_1+\d_2+\gamma$&${\red {-2\d_1+\gamma}}$\\
\hline
$b_2$&$\a$&$\d_2-\d_2$&${\red {2\d_2}}$&$-\b$&${\red {\d_1+\d_2-\gamma}}$&$-\gamma$&${\red {2\d_2+\gamma}}$&$\d_2-\d_1+\gamma$\\
\hline
$b_3$&$\b$&${\red {-2\d_2}}$&$\d_2-\d_2$&$-\a$&$\d_1-\d_2-\gamma$&${\red  {-2\d_2-\gamma}}$&$\gamma$&${\red {-\d_1-\d_2+\gamma}}$\\
\hline
$b_4$&${\red {2\d_1}}$&$\b$&$\a$&$\d_1-\d_1$&${\red {2\d_1-\gamma}}$&$\d_1-\d_2-\gamma$&${\red {\d_1+\d_2+\gamma}}$&$\gamma$\\
\hline
$b_5$&$\gamma$&${\red {-\d_1-\d_2+\gamma}}$&$\d_2-\d_1+\gamma$&${\red {-2\d_1+\gamma}}$&$\eta-\eta$&$-\a$&$-\d_1+\d_2+2\gamma$&${\red {2\gamma-2\d_1}}$\\
\hline
$b_6$&${\red {\d_1+\d_2+\gamma}}$&$\gamma$&${\red {2\d_2+\gamma}}$&$-\d_1+\d_2+\gamma$&$\a$&$\zeta-\zeta$&${\red{ 2\d_2+2\gamma}}$&$-\d_1+\d_2+2\gamma$\\
\hline
$b_7$&$\d_1-\d_2-\gamma$&${\red {-2\d_2-\gamma}}$&$-\gamma$&${\red {-\d_1-\d_2-\gamma}}$&$\d_1-\d_2-2\gamma$&${\red {-2\d_2-2\gamma}}$&$\zeta-\zeta$&$-\a$\\
\hline
$b_8$&${\red {2\d_1-\gamma}}$&$\d_1-\d_2-\gamma$&${\red {\d_1+\d_2-\gamma}}$&$-\gamma$&${\red {2\d_1-2\gamma}}$&$\d_1-\d_2-2\gamma$&$\a$&$\eta-\eta$\\
\hline
\end{tabular}
}

\medskip

{\tiny\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
&$b_1$&$b_2$&$b_3$&$b_4$&$b_5$&$b_6$&$b_7$&$b_8$\\
\hline
$b_1$&${\red {-2\d_1}}$&$-\b$&$-\a$&$\d_1-\d_1$&${\red {-2\d_1+\gamma}}$&$-\d_1+\d_2+\gamma$&${\red {-\d_1-\d_2-\gamma}}$&$-\gamma$\\
\hline
$b_2$&$-\b$&${\red {2\d_2}}$&$\d_2-\d_2$&$\a$&$\d_2-\d_1+\gamma$&${\red {2\d_2+\gamma}}$&$-\gamma$&${\red {\d_1+\d_2-\gamma}}$\\
\hline
$b_3$&$-\a$&$\d_2-\d_2$&${\red {-2\d_2}}$&$\b$&${\red {-\d_1-\d_2+\gamma}}$&$\gamma$&${\red {-2\d_2-\gamma}}$&$\d_1-\d_2-\gamma$\\
\hline
$b_4$&$\d_1-\d_1$&$\a$&$\b$&${\red {2\d_1}}$&$\gamma$&${\red {\d_1+\d_2+\gamma}}$&$\d_1-\d_2-\gamma$&${\red {2\d_1-\gamma}}$\\
\hline
$b_5$&${\red {-2\d_1+\gamma}}$&$\d_2-\d_1+\gamma$&${\red {-\d_1-\d_2+\gamma}}$&$\gamma$&${\red {2\gamma-2\d_1}}$&$-\d_1+\d_2+2\gamma$&$-\a$&$\eta-\eta$\\
\hline
$b_6$&$-\d_1+\d_2+\gamma$&${\red {2\d_2}}{\red {+\gamma}}$&$\gamma$&${\red {\d_1+\d_2+\gamma}}$&$-\d_1+\d_2+2\gamma$&
${\red {2\d_2+2}} $ ${\red {\gamma}}$&$\zeta-\zeta$&$\a$\\
\hline
$b_7$&${\red {-\d_1-\d_2-\gamma}}$&$-\gamma$&${\red {-2\d_2-\gamma}}$&$\d_1-\d_2-\gamma$&$-\a$&$\zeta-\zeta$&${\red {-2\d_2-2\gamma}}$&$\d_1-\d_2-2\gamma$\\
\hline
$b_8$&$-\gamma$&${\red {\d_1+\d_2-\gamma}}$&$\d_1-\d_2-\gamma$&${\red {2\d_1-\gamma}}$&$\eta-\eta$&$\a$&$\d_1-\d_2-2\gamma$&${\red {2\d_1-2\gamma}}$\\
\hline
\end{tabular}
}

\medskip
Using these two tables together with (\ref{bracket}), we get that $[M_{\bar 1},M_{\bar 1}]\sub M_{\bar 0}.$  Now we want to show that $[M_{\bar 0},M_{\bar 1}]\sub M_{\bar 1}.$ Considering Lemmas \ref{final}($i$), \ref{restrict}($i$) and Proposition \ref{dim1-re}, we have
$$\begin{array}{l}
 \LL^{\b-\gamma}=\LL^{\d_1-\d_2-\gamma}=\bbbf [a_2,a_5]=\bbbf[b_3,b_8]=\bbbf [a_1,a_6]=\bbbf[b_4,b_7],\\\\
 \LL^{-\b+\gamma}=\LL^{-\d_1+\d_2+\gamma}=\bbbf [a_3,a_8]= \bbbf[b_2,b_5]=\bbbf [b_1,b_6]=\bbbf[a_4,a_7].
 \end{array}$$
Also  contemplating Lemma \ref{pro-nonzero} as well as Proposition \ref{dim1-re} and
 setting  $$u:=[a_2,a_5],\;\; v:=[a_3,a_8],\;\;z:=[m,v],\;\;w:=[n,u],$$ we get  that
$$
 \LL^{2\gamma-\b}=\LL^{2\gamma-\d_1+\d_2}=\bbbf z\andd
 \LL^{\b-2\gamma}=\LL^{-2\gamma+\d_1-\d_2}=\bbbf w,
$$ as $(\d_2,\eta),(\zeta,\d_1)\neq0$ and $\eta+\d_2,\d_1-\zeta\not\in R.$
%Note that $v,u,m,n\in\LL_{\bar 0}$ and so $\b-\gamma,\b-2\gamma\in\rre.$
We consider the following tables; as before, the $(r,s)$-th entry of  the following tables denotes the summation of the weights of $r$-th entry of the first column and $s$-th entry of the first row and the red parts denote the elements which are not roots.
\medskip
\begin{center}
{\tiny\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
&$e$&$f$&$x$&$y$&$m$&$n$\\
\hline
$a_1$&${\red {2\d_1+\d_2}}$&$-\d_2$&${\red {2\d_1-\d_2}}$&$\d_2$&${\red {\d_1+\gamma}}$&$\d_1-\gamma$ \\
\hline
$a_2$& $\d_1$& ${\red {-\d_1-2\d_2}}$&${\red {\d_1-2\d_2}}$& $-\d_1$&${\red {\gamma-\d_2}}$&$-\d_2-\gamma$\\
\hline
$a_3$&${\red {\d_1+2\d_2}}$& $-\d_1$& $\d_1$&${\red {2\d_2-\d_1}}$&$\gamma+\d_2$&${\red {-\gamma+\d_2}}$\\
\hline
$a_4$&$\d_2$&${\red {-2\d_1-\d_2}}$&$-\d_2$&${\red {\d_2-2\d_1}}$&$\gamma-\d_1$&${\red {-\d_1-\gamma}}$\\
\hline
$a_5$&${\red {2\d_1+\d_2-\gamma}}$&$-\gamma-\d_2$&${\red {2\d_1-\d_2-\gamma}}$&${\red {\d_2-\gamma}}$&$\d_1$&${\red {\d_1-2\gamma}}$\\
\hline
$a_6$&$\d_1-\gamma$& ${\red {-\d_1-2\d_2-\gamma}}$& ${\red {\d_1-2\d_2-\gamma}}$& ${\red {-\d_1-\gamma}}$& $-\d_2$& ${\red {-\d_2-2\gamma}}$\\
\hline
$a_7$& ${\red {\d_1+2\d_2+\gamma}}$&$\gamma-\d_1$& ${\red {\d_1+\gamma}}$& ${\red {\gamma-\d_1+2\d_2}}$&  ${\red {2\gamma+\d_2}}$& $\d_2$\\
\hline
$a_8$&$\gamma+\d_2$&${\red {-2\d_1-\d_2+\gamma}}$&${\red {-\d_2+\gamma}}$&${\red {-2\d_1+\d_2+\gamma}}$&${\red {-\d_1+2\gamma}}$&$-\d_1$\\
\hline
$b_1$&$\d_2$&${\red {-2\d_1-\d_2}}$&$-\d_2$&${\red{\d_2-2\d_1}}$&$\gamma-\d_1$&${\red {-\d_1-\gamma}}$\\
\hline
$b_2$&${\red {\d_1+2\d_2}}$& $-\d_1$& $\d_1$&${\red {2\d_2-\d_1}}$&$\gamma+\d_2$&${\red {-\gamma+\d_2}}$\\
\hline
$b_3$& $\d_1$& ${\red {-\d_1-2\d_2}}$&${\red{\d_1-2\d_2}}$& $-\d_1$&${\red{\gamma-\d_2}}$&$-\d_2-\gamma$\\
\hline
$b_4$&${\red {2\d_1+\d_2}}$&$-\d_2$&${\red {2\d_1-\d_2}}$&$\d_2$&${\red {\d_1+\gamma}}$&$\d_1-\gamma$\\
\hline
$b_5$&$\gamma+\d_2$&${\red {-2\d_1-\d_2+\gamma}}$&${\red {-\d_2+\gamma}}$&${\red {-2\d_1+\d_2+\gamma}}$&${\red {-\d_1+2\gamma}}$&$-\d_1$\\
\hline
$b_6$& ${\red {\d_1+2\d_2+\gamma}}$&$\gamma-\d_1$& ${\red {\d_1+\gamma}}$& ${\red {\gamma-\d_1+2\d_2}}$&  ${\red {2\gamma+\d_2}}$& $\d_2$\\
\hline
$b_7$&$-\gamma+\d_1$& ${\red {-\d_1-2\d_2-\gamma}}$& ${\red {-\gamma+\d_1-2\d_2}}$& ${\red {-\d_1-\gamma}}$ & $-\d_2$&  ${\red {-2\gamma-\d_2}}$\\
\hline
$b_8$&${\red {2\d_1+\d_2-\gamma}}$&$-\gamma-\d_2$&${\red {2\d_1-\d_2-\gamma}}$&${\red {\d_2-\gamma}}$&$\d_1$&${\red {\d_1-2\gamma}}$\\
\hline
\end{tabular}
}
\end{center}
\medskip

\begin{center}
{\tiny\begin{tabular}{|l|l|l|l|l|}
\hline
&$u$&$v$&$z$&$w$\\
\hline
$a_1$& ${\red {2\d_1-\d_2-\gamma}}$&$\d_2+\gamma$&${\red {\d_2+2\gamma}}$&${\red {2\d_1-\d_2-2\gamma}}$\\
\hline
$a_2$&${\red {\d_1-2\d_2-\gamma}}$&$-\d_1+\gamma$&${\red {-\d_1+2\gamma}}$&${\red {\d_1-2\d_2-2\gamma}}$\\
\hline
$a_3$&$\d_1-\gamma$& ${\red {\gamma-\d_1+2\d_2}}$&${\red {2\gamma-\d_1+2\d_2}}$&${\red {\d_1-2\gamma}}$\\
\hline
$a_4$& $-\d_2-\gamma$&${\red {\gamma-2\d_1+\d_2}}$&${\red {2\gamma-2\d_1+\d_2}}$&${\red {-\d_2-2\gamma}}$\\
\hline
$a_5$&${\red {2\d_1-\d_2-2\gamma}}$&$\d_2$&$\d_2+\gamma$& ${\red {2\d_1-\d_2-3\gamma}}$\\
\hline
$a_6$& ${\red {\d_1-2\d_2-2\gamma}}$& $-\d_1$& $\gamma-\d_1$& ${\red {\d_1-2\d_2-3\gamma}}$\\
\hline
$a_7$&  $\d_1$&${\red {-\d_1+2\d_2+2\gamma}}$&${\red {-\d_1+2\d_2+3\gamma}}$& $-\gamma+\d_1$\\
\hline
$a_8$&$-\d_2$&${\red {-2\d_1+\d_2+2\gamma}}$& ${\red {-2\d_1+\d_2+3\gamma}}$&$-\d_2-\gamma$\\
\hline
$b_1$& $-\d_2-\gamma$&${\red {\gamma-2\d_1+\d_2}}$&${\red {2\gamma-2\d_1+\d_2}}$&${\red {-\d_2-2\gamma}}$\\
\hline
$b_2$&$\d_1-\gamma$& ${\red {\gamma-\d_1+2\d_2}}$& ${\red {-\d_1+2\d_2+2\gamma}}$&${\red {\d_1-2\gamma}}$\\
\hline
$b_3$&${\red {\d_1-2\d_2-\gamma}}$&$-\d_1+\gamma$&${\red {-\d_1+2\gamma}}$&${\red {\d_1-2\d_2-2\gamma}}$\\
\hline
$b_4$& ${\red {2\d_1-\d_2-\gamma}}$&$\d_2+\gamma$&${\red {\d_2+2\gamma}}$&${\red {2\d_1-\d_2-2\gamma}}$\\
\hline
$b_5$&$-\d_2$&${\red {-2\d_1+\d_2+2\gamma}}$& ${\red {-2\d_1+\d_2+3\gamma}}$&$-\d_2-\gamma$\\
\hline
$b_6$&   $\d_1$&${\red {-\d_1+2\d_2+2\gamma}}$&${\red {-\d_1+2\d_2+3\gamma}}$& $-\gamma+\d_1$\\
\hline
$b_7$&${\red {\d_1-2\d_2-2\gamma}}$& $-\d_1$& $\gamma-\d_1$& ${\red {\d_1-2\d_2-3\gamma}}$\\
\hline
$b_8$&${\red {2\d_1-\d_2-2\gamma}}$&$\d_2$&$\d_2+\gamma$& ${\red {2\d_1-\d_2-3\gamma}}$\\
\hline
\end{tabular}
}\end{center}
\medskip

Now one can easily check the following multiplication table:

\medskip
\begin{center}
{\tiny\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
&$e$&$f$&$x$&$y$&$m$&$n$\\
\hline
$a_1$&$0$&$-a_2$&$0$&$-a_3$&$0$&$-a_5$ \\
\hline
$a_2$& $-(\a,\d_1)a_1$&$0$&$0$&$-a_4$&$0$&$-a_6$\\
\hline
$a_3$&$0$&$-a_4$&$-(\b,\d_1)a_1$&$0$&$-a_7$&$0$\\
\hline
$a_4$&$-(\d_2,\a)a_3$&$0$&$-(\b,\d_1)a_2$&$0$&$-a_8$&$0$\\
\hline
$a_5$&$0$&$-a_6$&$0$&$0$&$-(\d_1,\gamma)a_1$&$0$\\
\hline
$a_6$&$-(\a,\d_1 )a_5$&$0$&$0$&$0$&$(\gamma,\d_2)a_2$&$0$\\
\hline
$a_7$&$0$&$-a_8$&$0$&$0$&$0$&$(\d_2,\gamma)a_3$\\
\hline
$a_8$&$-(\d_2,\a)a_7$&$0$&$0$&$0$&$0$&$-(\d_1,\gamma)a_4$\\
\hline
$b_1$&$-b_2$&$0$&$-b_3$&$0$&$-b_5$&$0$\\
\hline
$b_2$&$0$&$-(\a,\d_1)b_1$&$-b_4$&$0$&$-b_6$&$0$\\
\hline
$b_3$& $-b_4$&$0$&$0$&$-(\d_1,\b)b_1$&$0$&$-b_7$\\
\hline
$b_4$&$0$&$-(\d_2,\a)b_3$&$0$&$-(\b,\d_2)b_2$&$0$&$-b_8$\\
\hline
$b_5$&$-b_6$&$0$&$0$&$0$&$0$&$-(\gamma,\d_1)b_1$\\
\hline
$b_6$&$0$&$-(\a,\d_1)b_5$&$0$&$0$&$0$&$(\d_2,\gamma)b_2$\\
\hline
$b_7$&$-b_8$&$0$&$0$&$0$&$(\d_2,\gamma)b_3$&$0$\\
\hline
$b_8$&$0$&$-(\a,\d_2)b_7$&$0$&$0$&$-(\gamma,\d_1)b_4$&$0$\\
\hline
\end{tabular}
}\end{center}
\medskip




%\begin{comment}
Take $r_1,\ldots,r_7,s_1,\ldots,s_6, k_1,\ldots, k_4\in\bbbf\setminus\{0\}$ to be such that
$$\begin{array}{llllll}
\;[a_8,a_1]=r_1m,&[a_3,a_1]=r_2e,&[a_3,a_2]=r_3t_{\d_2},&v=s_1[a_4,a_7],&
[a_7,a_2]=s_2 m,&[a_4,a_2]=s_3f,\\
\;[a_5,a_4]=s_5n,&[a_8,a_5]=s_7t_\eta,&[a_7,a_6]=s_6t_\zeta,&[b_2,b_3]=k_2t_{\d_2},&[b_8,b_5]=k_3t_{\eta},&[b_4,b_1]=k_1t_{\d_1},\\
\;[b_7,b_6]=k_4t_\zeta,&v=r_7[b_2,b_5]&v=r_6[b_1,b_6],&u=r_4[a_1,a_6],&
u=r_5[b_4,b_7],& u=s_4[b_3,b_8].
\end{array}$$




\begin{comment}


{\tiny\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
&$e$&$f$&$x$&$y$&$m$&$n$&$u$&$v$&$z$&$w$\\
\hline
$a_1$&$2\d_1+\d_2$&$-\d_2$&$2\d_1-\d_2$&$\d_2$&$\d_1+\gamma$&$\d_1-\gamma$& $2\d_1-\d_2-\gamma$&$\d_2+\gamma$&$\d_2+2\gamma$&$2\d_1-\d_2-2\gamma$\\
\hline
$a_2$& $\d_1$& $-\d_1-2\d_2$&$\d_1-2\d_2$& $-\d_1$&$\gamma-\d_2$&$-\d_2-\gamma$&$\d_1-2\d_2-\gamma$&$-\d_1+\gamma$&$-\d_1+2\gamma$&$\d_1-2\d_2-2\gamma$\\
\hline
$a_3$&$\d_1+2\d_2$& $-\d_1$& $\d_1$&$2\d_2-\d_1$&$\gamma+\d_2$&$-\gamma+\d_2$&$\d_1-\gamma$& $\gamma-\d_1+2\d_2$&$-\d_1+2\d_2+2\gamma$& $+\d_1-2\gamma$\\
\hline
$a_4$&$\d_2$&$-2\d_1-\d_2$&$-\d_2$&$\d_2-2\d_1$&$\gamma-\d_1$&$-\d_1-\gamma$& $-\d_2-\gamma$&$\gamma-2\d_1+\d_2$&$2\gamma-2\d_1+\d_2$&$-\d_2-2\gamma$\\
\hline
$a_5$&$2\d_1+\d_2-\gamma$&$-\gamma-\d_2$&$2\d_1-\d_2-\gamma$&$\d_2-\gamma$&$\d_1$&$\d_1-2\gamma$&$2\d_1-\d_2-2\gamma$&$\d_2$&$\d_2+\gamma$& $2\d_1-\d_2-3\gamma$\\
\hline
$a_6$&$\d_1-\gamma$& $-\d_1-2\d_2-\gamma$& $\d_1-2\d_2-\gamma$& $-\d_1-\gamma$& $-\d_2$& $\d_2-2\gamma$& $\d_1-2\d_2-2\gamma$& $-\d_1$& $\gamma-\d_1$& $\d_1-2\d_2-3\gamma$\\
\hline
$a_7$& $\d_1+2\d_2+\gamma$&$\gamma-\d_1$& $\d_1+\gamma$& $\gamma-\d_1+2\d_2$&  $2\gamma-\d_2$& $\d_2$&  $\d_1$&$-\d_1+2\d_2+2\gamma$&$-\d_1+2\d_2+3\gamma$& $-\gamma+\d_1$\\
\hline
$a_8$&$\gamma+\d_2$&$-2\d_1-\d_2+\gamma$&$-\d_2+\gamma$&$-2\d_1+\d_2+\gamma$&$-\d_1+2\gamma$&$-\d_1$&$-\d_2$&$-2\d_1+\d_2+2\gamma$& $-2\d_1+\d_2+3\gamma$&$-\d_2-\gamma$\\
\hline
$b_1$&$\d_2$&$-2\d_1-\d_2$&$-\d_2$&$\d_2-2\d_1$&$\gamma-\d_1$&$-\d_1-\gamma$& $-\d_2-\gamma$&$\gamma-2\d_1+\d_2$&$2\gamma-2\d_1+\d_2$&$-\d_2-2\gamma$\\
\hline
$b_2$&$\d_1+2\d_2$& $-\d_1$& $\d_1$&$2\d_2-\d_1$&$\gamma+\d_2$&$-\gamma+\d_2$&$\d_1-\gamma$& $\gamma-\d_1+2\d_2$&  $-\d_1+2\gamma$&$2\gamma-\d_1+2\d_2$\\
\hline
$b_3$& $\d_1$& $-\d_1-2\d_2$&$\d_1-2\d_2$& $-\d_1$&$\gamma-\d_2$&$-\d_2-\gamma$&$\d_1-2\d_2-\gamma$&$-\d_1+\gamma$&$-\d_1+2\gamma$&$\d_1-2\d_2-2\gamma$\\
\hline
$b_4$&$2\d_1+\d_2$&$-\d_2$&$2\d_1-\d_2$&$\d_2$&$\d_1+\gamma$&$\d_1-\gamma$& $2\d_1-\d_2-\gamma$&$\d_2+\gamma$&$\d_2+2\gamma$&$2\d_1-\d_2-2\gamma$\\
\hline
$b_5$&$\gamma+\d_2$&$-2\d_1-\d_2+\gamma$&$-\d_2+\gamma$&$-2\d_1+\d_2+\gamma$&$-\d_1+2\gamma$&$-\d_1$&$-\d_2$&$-2\d_1+\d_2+2\gamma$& $-2\d_1+\d_2+3\gamma$&$-\d_2-\gamma$\\
\hline
$b_6$& $\d_1+2\d_2+\gamma$&$\gamma-\d_1$& $\d_1+\gamma$& $\gamma-\d_1+2\d_2$&  $2\gamma-\d_2$& $\d_2$&  $\d_1$&$-\d_1+2\d_2+2\gamma$&$-\d_1+2\d_2+3\gamma$& $-\gamma+\d_1$\\
\hline
$b_7$&$\d_1+2\d_2$& $-\d_1$& $\d_1$&$2\d_2-\d_1$&$\gamma+\d_2$&$-\gamma+\d_2$&$\d_1-\gamma$& $\gamma-\d_1+2\d_2$&  $-\d_1+2\gamma$&$2\gamma-\d_1+2\d_2$\\
\hline
$b_8$&$\d_2$&$-2\d_1-\d_2$&$-\d_2$&$\d_2-2\d_1$&$\gamma-\d_1$&$-\d_1-\gamma$& $-\d_2-\gamma$&$\gamma-2\d_1+\d_2$&$2\gamma-2\d_1+\d_2$&$-\d_2-2\gamma$\\
\hline
\end{tabular}
}

\end{comment}

Then  we have


\begin{eqnarray*}
[a_1,v]=-[v,a_1]=-[[a_3,a_8],a_1]
&=&-[a_3,[a_8,a_1]]-[a_8,[a_3,a_1]]\\
&=&-[a_3,r_1m]-[a_8,r_2e]\\
&=&-r_1[a_3,m]-r_2[a_8,e]\\
&=&r_1a_7+r_2(\a,\d_2)a_7,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_2,v]=-[v,a_2]=-s_1[[a_4,a_7],a_2]
&=&-s_1[a_4,[a_7,a_2]]-s_1[a_7,[a_4,a_2]]\\
&=&-s_1[a_4,s_2m]-s_1[a_7,s_3f]\\
&=&s_1(s_2+s_3)a_8,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_3,u]=-[u,a_3]=-[[a_2,a_5],a_3]&=&-[a_2,[a_5,a_3]]-[a_5,[a_2,a_3]]\\&=&[[a_2,a_3],a_5]\\
&=&r_3(\d_2,\eta)a_5,
\end{eqnarray*}



\begin{eqnarray*}
[a_4,u]=-[u,a_4]=-[[a_2,a_5],a_4]&=&-[a_2,[a_5,a_4]]-[a_5,[a_2,a_4]]\\
&=&-s_5[a_2,n]-s_3[a_5,f]\\
&=&(s_5+s_3)a_6
\end{eqnarray*}



\begin{eqnarray*}
\;[a_5,v]=-[v,a_5]=-[[a_3,a_8],a_5]&=&-[a_3,[a_8,a_5]]-[a_8,[a_3,a_5]]\\
&=&s_7(\eta,\d_2)a_3
\end{eqnarray*}



\begin{eqnarray*}
\;[a_6,v]=-[v,a_6]=-s_1[[a_4,a_7],a_6]&=&-s_1[a_4,[a_7,a_6]]-s_1[a_7,[a_4,a_6]]\\
&=&-s_6s_1(\zeta,\d_1)a_4
\end{eqnarray*}



\begin{eqnarray*}
\;[a_5,z]=-[z,a_5]=-[[m,v],a_5]&=&-[m,[v,a_5]]+[v,[m,a_5]]\\
&=&s_7(\eta,\d_2)[m,a_3]+(\d_1,\gamma)[v,a_1]\\
&=&s_7(\eta,\d_2)a_7-r_1(\d_1,\gamma)a_7
-r_2(\a,\d_2)(\d_1,\gamma)a_7,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_6,z]=-[z,a_6]=-[[m,v],a_6]&=&-[m,[v,a_6]]+[v,[m,a_6]]\\
&=&-s_6s_1(\zeta,\d_1)[m,a_4]-(\gamma,\d_2)[v,a_2]\\
&=&(-s_6s_1(\zeta,\d_1)+s_1(s_2+s_3)(\d_2,\gamma))a_8,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_7,u]=- [u,a_7]=-r_4[[a_1,a_6],a_7]&=&-r_4[a_1,[a_6,a_7]]-r_4[a_6,[a_1,a_7]]\\
&=&s_6r_4[t_\zeta,a_1]=s_6r_4(\zeta,\d_1)a_1,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_7,w]=-[w,a_7]&=&-[[n,u],a_7]\\
&=&-[n,[u,a_7]]+[u,[n,a_7]]\\
&=&(r_4s_6(\zeta,\d_1)-r_3(\d_2,\gamma)(\d_2,\eta))a_5,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_8,u]=-[u,a_8]=-[[a_2,a_5],a_8]&=&-[a_2,[a_5,a_8]]-[a_5,[a_2,a_8 ]]\\
&=&s_7[t_\eta,a_2]=-s_7(\eta,\d_2)a_2,
\end{eqnarray*}



\begin{eqnarray*}
\;[a_8,w]=-[w,a_8]&=&-[[n,u],a_8]\\
&=&-[n,[u,a_8]]+[u,[n,a_8]]\\
&=&s_7(\eta,\d_2)[n,a_2]+[u,(\d_1,\gamma)a_4]\\
&=&(s_7(\eta,\d_2)-(\d_1,\gamma)(s_5+s_3))a_6,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_1,u]=-[u,b_1]=-r_5[[b_4,b_7],b_1]&=&-r_5[b_4,[b_7,b_1]]-r_5[b_7,[b_4,b_1]]\\
&=&r_5k_1(\zeta,\d_1)b_7,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_2,u]=-[u,b_2]=-s_4[[b_3,b_8],b_2]&=&- s_4[b_3,[b_8,b_2]]-s_4[b_8,[b_3,b_2]]\\
&=&s_4(\eta,\d_2)k_2b_8,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_3,v]=-[v,b_3]=-r_7[[b_2,b_5],b_3]&=&-r_7[b_2,[b_5,b_3]]-r_7[b_5,[b_2,b_3]]\\
&=&-r_7k_2(\eta,\d_2)b_5,
\end{eqnarray*}

\begin{eqnarray*}
[b_4,v]=-[v,b_4]=-r_6[[b_1,b_6],b_4]&=&-r_6[b_1,[b_6,b_4]]-r_6[b_6,[b_1,b_4]]\\
&=&-r_6k_1(\zeta,\d_1)b_6,\end{eqnarray*}



\begin{eqnarray*}
\;[b_5,u]=-[u,b_5]=-s_4[[b_3,b_8],b_5]&=&-s_4[b_3,[b_8,b_5]]-s_4[b_8,[b_3,b_5]]\\
&=&k_3s_4[t_\eta,b_3]\\
&=&-s_4k_3(\d_2,\eta)b_3,\end{eqnarray*}



\begin{eqnarray*}
\;[b_5,w]=-[w,b_5]=-[[n,u],b_5]&=&-[n,[u,b_5]]+[u,[n,b_5]]\\&=&-s_4k_3(\d_2,\eta)[n,b_3]+[u,(\gamma,\d_1)b_1]\\
&=&(- s_4k_3(\d_2,\eta)-(\gamma,\d_1)k_1r_5(\zeta,\d_1))b_7,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_6,u]=-[u,b_6]=-r_5[[b_4,b_7],b_6]&=&-r_5[b_4,[b_7,b_6]]-r_5[b_7,[b_4,b_6]]\\
&=&r_5k_4[t_\zeta,b_4]\\
&=&r_5k_4(\d_1,\zeta)b_4,\end{eqnarray*}



\begin{eqnarray*}
\;[b_6,w]=-[w,b_6]=-[[n,u],b_6]&=&-[n,[u,b_6]]+[u,[n,b_6]]\\
&=&r_5k_4(\d_1,\zeta)[n,b_4]-(\d_2,\gamma)[u,b_2]\\
&=&(r_5k_4(\d_1,\zeta)+(\d_2,\gamma)k_2s_4(\eta,\d_2))b_8,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_7,v]=-[v,b_7]=-r_6[[b_1,b_6],b_7]&=&-r_6[b_1,[b_6,b_7]]-r_6[b_6,[b_1,b_7]]\\
&=&r_6k_4[t_\zeta,b_1]=-r_6k_4(\d_1,\zeta)b_1,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_7,z]=-[z ,b_7]=-[[m,v],b_7]&=&-[m,[v,b_7]]+[v,[m,b_7]]\\
&=&-r_6k_4(\d_1,\zeta)[m,b_1]-(\d_2,\gamma)[v,b_3]\\
&=&(-r_6k_4(\d_1,\zeta)-(\d_2,\gamma)r_7k_2(\eta,\d_2))b_5,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_8,v]=-[v,b_8]=-r_7[[b_2,b_5],b_8]&=&-r_7[b_2,[b_5,b_8]]-r_7[b_5,[b_2,b_8]]\\
&=&k_3r_7[t_\eta,b_2]=k_3r_7(\d_2,\eta)b_2,
\end{eqnarray*}



\begin{eqnarray*}
\;[b_8,z]=-[[m,v],b_8]&=&-[m,[v,b_8]]+[v,[m,b_8]]\\
&=&k_3r_7(\d_2,\eta)[m,b_2]+(\gamma,\d_1)[v,b_4]\\
&=&(k_3r_7(\d_2,\eta)+r_6k_1(\zeta,\d_1)(\gamma,\d_1 ))b_6.
\end{eqnarray*}

These altogether imply that $[M_{\bar 0},M_{\bar 1}]\sub M_{\bar 1}.$  Therefore, to complete the proof of our claim, we just need to show $[M_{\bar 0}, M_{\bar 0}]\sub M_{\bar 0}$ but it is immediate using (\ref{bracket}) together with the following table; as before, the $(r,s)$-th entry of  the following table denotes the summation of the weights of $r$-th entry of the first column and $s$-th entry of the first row and the red parts denote the elements which are not roots:

\medskip
\begin{center}
{\tiny\begin{tabular}{|l|l|l|l|l|l|}
\hline
&$e$&$x$&$m$&$u$&$w$\\
\hline
$e$&${\red{2\a}}$&${\red{2\d_1}}$&${\red{\d_1+\d_2+\gamma}}$&${\red{2\d_1-\gamma}}$&${\red{2\d_1-2\gamma}}$\\
\hline
$x$&${\red{2\d_1}}$&${\red{2\b}}$&$\d_1-\d_2+\gamma$&${\red{2(\d_1-\d_2)-\gamma}}$&${\red{2(\d_1-\d_2-\gamma)}}$\\
\hline
$m$&${\red{\d_1+\d_2+\gamma}}$&$\d_1-\d_2+\gamma$&${\red{2\gamma}}$&$\b$&$\b-\gamma$\\
\hline
$u$&${\red{2\d_1-\gamma}}$&${\red{2(\d_1-\d_2)-\gamma}}$&$\b$&${\red{2(\b-\gamma)}}$&${\red{2(\d_1-\d_2)-3\gamma}}$\\
\hline
$w$&${\red{2\d_1-2\gamma}}$&${\red{2(\d_1-\d_2-\gamma)}}$&$\b-\gamma$&${\red{2(\d_1-\d_2)-3\gamma}}$& ${\red{2(\b-2\gamma)}}$\\
\hline
$f$&$\a-\a$&${\red{-2\d_2}}$&${\red{-\d_1-\d_2+\gamma}}$&${\red{-2\d_2-\gamma}}$&${\red{-2\d_2-2\gamma}}$\\
\hline
$y$&${\red{2\d_2}}$&$\b-\b$&$-\b+\gamma$&$-\gamma$&${\red{-2\gamma}}$\\
\hline
$n$&${\red{\d_1+\d_2-\gamma}}$&$\b-\gamma$&$\gamma-\gamma$&$\b-2\gamma$&${\red{\d_1-\d_2-3\gamma}}$\\
\hline
$v$&${\red{2\d_2+\gamma}}$&$\gamma$&$2\gamma-\b$&$\b-\gamma-\b+\gamma$&$-\gamma$\\
\hline
$z$&${\red{2\d_2+2\gamma}}$&${\red{2\gamma}}$&${\red{-\d_1+\d_2+3\gamma}}$&$\gamma$&$\b-2\gamma-\b+2\gamma$\\
\hline
\end{tabular}
}\end{center}
%\end{comment}

\medskip

 Now setting  $\ep:=\a/2,$ $\theta_1:=\b/2$ and $\theta_2:=-\gamma+\b/2,$ $M$ has a weight space decomposition with respect to $K$ with the set of weights
\begin{eqnarray*}
S&:=&\{0,\pm\a,\pm\b,\pm\gamma,\pm(\b-\gamma),\pm(\b-2\gamma),\pm\d_1,\pm \d_2,\pm\eta,\pm\zeta\}\\
&=&\{0,\pm 2\ep,\pm2\theta_1,\pm2\theta_2,\pm(\theta_1\pm\theta_2),\pm(\ep\pm\theta_1), \pm(\ep\pm\theta_2)\}
\end{eqnarray*}
that is  a locally finite root supersystem of type $C(1,2).$ It follows from  Lemma \ref{final}($i$) that for $\theta\in\{\pm\d_1,\pm \d_2,\pm\eta,\pm\zeta\},$   the form on $M^\theta\op M^{-\theta}$ is nondegenerate. Also the corresponding matrix of the from restricted to $K$ with respect to the ordered basis $\{t_\a,t_\b,t_\gamma\}$ is $$\left(\begin{array}{ccc}
(\a,\a)&(\a,\b)&(\a,\gamma)\\
(\b,\a)& (\b,\b)& (\b, \gamma)\\
(\gamma,\a)& (\gamma,\b)& (\gamma,\gamma)
\end{array}\right)=\left(\begin{array}{ccc}
(\a,\a)&0&0\\
0& 2(\b,\gamma)& (\b, \gamma)\\
0& (\gamma,\b)& (\b,\gamma)
\end{array}\right)$$ whose determinant is $(\a,\a)(\b,\gamma)^2\neq0.$ This implies that the form is nondegenerate on $K$ and so the form is nondegenerate on $M.$ We note that $\hbox{span}_\bbbf\{t_\theta\mid \theta\in \{\pm \d_1,\pm\d_2,\pm\zeta,\pm\eta\}\}=\hbox{span}_\bbbf\{t_\theta\mid \theta\in \{\pm \a,\pm\b,\pm\gamma\}=K$  and the form is nondegenerate on $K,$ so for each $0\neq h\in K,$ there is $\theta \in  \{\pm \d_1,\pm\d_2,\pm\zeta,\pm\eta\}$ such that $\theta(h)=(h,t_\theta)\neq0.$ Now as $M_{\bar 0}$ is a finite dimensional semisimple Lie algebra of type $A_1\op C_2,$ using Lemma \ref{simple},  $M$ is a finite dimensional basic classical simple Lie superalgebra with  $\LL_{\bar 0}$ isomorphic to $A_1\op C_2.$ Therefore, using the classification of f.d.b.c.s Lie superalgebras, $M$ is either isomorphic to $\mathfrak{osp}(3,4)$ or $\mathfrak{osp}(5,2).$ But this makes a contradiction as there are nonzero real odd roots for each of $\mathfrak{osp}(3,4)$ or $\mathfrak{osp}(5,2)$ while odd roots of $M$ are all nonsingular.  This completes the proof.
 \begin{comment}
 We can see that $M_{\bar 0}$ is a semisimple Lie algebra with root system $A_1\op C_2;$ in fact setting $\ep:=\a/2,$ $\theta_1:=\b/2$ and $\theta_2:=-\gamma+\b/2,$ the root system of $M_{\bar 0}$ with respect to $K:=\bbbf t_\a+\bbbf t_\b+\bbbf t_\gamma+\bbbf t_{\b-\gamma}+\bbbf t_{\b-2\gamma}=\bbbf t_\a+\bbbf t_\b+\bbbf t_\gamma$ is  $\{0,\pm 2\ep,\pm2\theta_1,\pm2\theta_2,\pm(\theta_1\pm\theta_2)\}.$ Also if $I$ is a nonzero graded ideal of $M,$ $I\cap M_{\bar 0}$ is an ideal of $M_{\bar 0}.$ If $I\cap M_{\bar 0}\neq \{0\},$ either  $\bbbf t_\a\cap I\neq\{0\}$ or $(\bbbf t_\b+\bbbf t_\gamma+\bbbf t_{\b-\gamma}+\bbbf t_{\b-2\gamma})\cap I\neq\{0\}.$
In the former case, since $(\d_1,\a),(\d_2,\a),(\eta,\a),(\zeta,\a)\neq0,$ $M_{\bar 1}\sub I.$ This together with \cite[Lem. 2.4]{you3} and   the fact that $(a_i,b_i)\neq0,$  implies that  $t_{\d_1},t_{\d_2},t_{\eta},t_\zeta\in I,$ so
$\bbbf t_\b+\bbbf t_\gamma\sub I$ and so $M=I.$ In the latter case $\op_{\theta\in\{\pm\b,\pm(\b-\gamma),\pm(\b-2\gamma)\}}\LL^\theta\sub I$ and so $\bbbf t_\b+\bbbf t_\gamma\sub I.$  Since $(\d_1,\gamma),(\d_2,\gamma)\neq0,$ $(\LL^{\d_1}\cap M)\op(\LL^{\d_2}\cap M)\sub I.$ Therefore, $t_{\d_1},t_{\d_2},t_\eta,t_{\zeta}\in I$ which in turn implies that $t_\a\in I$ and $M_{\bar 1}\sub I$ and so $M=I.$ Now suppose $M_{\bar 1}\cap I\neq\{0\},$ since $M$ has a weight space decomposition with respect to $\hh\cap M,$ we have $M^\theta \cap I\neq \{0\}$ for some $\theta\in\{\pm\d_1,\pm \d_2,\pm\eta,\pm\zeta\}.$ But it follows from  (\ref{*}) that the form on $M^\theta\op M^{-\theta}$ is nondegenerate, and so if $0\neq x\in M^\theta \cap I,$ there is $y\in M^{-\theta }$ such that $[x,y]=(x,y)t_\theta\neq 0,$ so $t_\theta\in I.$ So $I\cap M_{\bar 0}\neq \{0\}.$ Therefore, $M=I.$ So we get that $M$ is a finite dimensional basic classical simple Lie superalgebra with  Cartan subalgebra $K=\hh\cap M$
and corresponding root system $S=\{0,\pm\a,\pm\b,\pm\gamma,\pm(\b-\gamma),\pm(\b-2\gamma),\pm\d_1,\pm\d_2,\pm\eta,\pm\zeta\}=\{0,\pm 2\ep,\pm2\theta_1,\pm2\theta_2,\pm(\theta_1\pm\theta_2),\pm(\ep\pm\theta_1), \pm(\ep\pm\theta_2)\}$ which is a root supersystem of type $C(1,2)$ which is a contradiction due to the classification of finite dimensional basic classical simple Lie superalgebras. This completes the proof.
\end{comment}
%\begin{eqnarray*}
%([[a,y],f],[x,[e,b]])=([a,y],[f,[x,[e,b]]])=(a,[y,[f,[x,[e,b]]]])\neq 0.
%\end{eqnarray*}So as above we get that $$\bbbf t_\a+\bbbf t_\b\op\LL^\a\op\LL^{-\a}\op\LL^\b\op\LL^{-\b}\op\hbox{span}_\bbbf\{a,[f,a],[y,a],[f,[y,a]], b,[e,b],[x,b],[x,[e,b]]\}$$ is a finite dimensional  basic classical simple  Lie superalgebra with root system $\{0,\pm\a,\pm\b,\frac{\pm\a\pm\b}{2}\}$ which is again absurd.
%Step 2. $(a,[f,[y,a]])=0:$ Since the form on  $\LL^{\d_1}\op\LL^{-\d_1}$ is nondegenerate, one can find $b\in \LL^{-\d_1}$ with $(a,b)\neq 0.$ If $(b, [x,[e,b]])\neq 0,$ we get a contradiction as in the previous step. So suppose that      $(b, [x,[e,b]])= 0,$ then $[y,[f,[x,[e,b]]]]$ is a nonzero scalar multiple of $b$ and
%\begin{eqnarray*}
%([[a,y],f],[x,[e,b]])=([a,y],[f,[x,[e,b]]])=(a,[y,[f,[x,[e,b]]]])\neq 0.
%\end{eqnarray*}So as above we get that $$\bbbf t_\a+\bbbf t_\b\op\LL^\a\op\LL^{-\a}\op\LL^\b\op\LL^{-\b}\op\hbox{span}_\bbbf\{a,[f,a],[y,a],[f,[y,a]], b,[e,b],[x,b],[x,[e,b]]\}$$ is a finite dimensional  basic classical simple  Lie superalgebra with root system $\{0,\pm\a,\pm\b,\frac{\pm\a\pm\b}{2}\}$ which is again absurd.
\end{proof}


\begin{lem}\label{a11}
Suppose that $\LL^0=\hh$ and   the corresponding root system $R$ of $\LL$ is irreducible with  no isotropic roots.  If there are nonsingular roots $\d_1,\d_2$ with $(\d_1,\d_2)\neq0$ and $\d_1+\d_2,\d_1-\d_2\in R,$ then  $R$ is  an irreducible  locally finite root supersystem, in its $\bbbz$-span, of type $A(1,1).$
\end{lem}
\begin{proof}
By Lemma \ref{dim1-im}, $R$ is either of type $A(1,1)$ or $BC(1,1).$  We show that $R$ cannot be of type $BC(1,1).$  To the contrary, suppose that $R$ is of type $BC(1,1),$ then by \cite[Lem. 1.13]{you6}, $R$ is an irreducible  locally finite root supersystem.  We assume that   $R=\{0,\pm\ep_0,\pm\d_0,\pm2\ep_0,\pm2\d_0,\pm\ep_0\pm\d_0\}.$  Then $$\LL=\hh\op\sum_{\a\in\{\pm\ep_0,\pm\d_0,\pm2\ep_0,\pm2\d_0,\pm\ep_0\pm\d_0\}}\LL^\a.$$ We note that by (\ref{rns}) and Proposition  \ref{dim1-re}, for $\a\in\{\pm\ep_0\pm\d_0,\pm\ep_0,\pm\d_0\},$ $\LL^\a\sub \LL_{\bar 1}.$  Set $\d_1:=r\ep_0+s\d_0$ and $\d_2:=r\ep_0-s\d_0$ for some $r,s\in\{\pm1\}$ and note that  $[\LL^{\ep_0}+\LL^{-\ep_0}+\LL^{\d_0}+\LL^{-\d_0},\LL^{\d_1}+\LL^{-\d_1}+\LL^{\d_2}+\LL^{\d_2} ]\sub \LL_{\bar 0}\cap (\LL^{\ep_0}+\LL^{-\ep_0}+\LL^{\d_0}+\LL^{-\d_0})$ and  $[\LL^{\pm\ep_0},\LL^{\pm\d_0}]\sub \LL_{\bar 0}\cap\LL^{\pm\ep_0\pm\d_0}.$  Therefore
\begin{equation}\label{com}
\begin{array}{l}[\LL^{\pm\ep_0},\LL^{\pm\d_0}]=[\LL^{\ep_0}+\LL^{-\ep_0}+\LL^{\d_0}+\LL^{-\d_0},\LL^{\d_1}+\LL^{-\d_1}+\LL^{\d_2}+\LL^{\d_2} ]=\{0\}.
\end{array}
\end{equation}
Take $\a,\b,e,f,x,y,a_1,\ldots,a_4,b_1,\ldots,b_4$ to be  as in the proof of Proposition \ref{dim1}; contemplating Proposition \ref{dim1-re} and (\ref{com}), as in Proposition \ref{dim1}, one can check that $$M= M_{\bar 0}\op M_{\bar 1}\hbox{ with } \left\{\begin{array}{l}M_{\bar0}:=\bbbf t_\a\op\bbbf t_\b\op\displaystyle{\sum_{\gamma\in\{\pm2\ep_0,\pm 2\d_0\}}\LL^\gamma}
\\
M_{\bar1}:=\LL^{\ep_0}\op\LL^{-\ep_0}\op\LL^{\d_0}\op\LL^{-\d_0}+\sum_{i=1}^4(\bbbf a_i+\bbbf b_i)
\end{array}
\right.$$ is a finite dimensional  basic classical simple Lie superalgebra. But the even part of $M$ is  isomorphic to $\mathfrak{sl}_2\op\mathfrak{sl}_2$ with Cartan subalgebra $\bbbf t_\a+\bbbf t_\b$ and the root system of $M$ with respect to $\bbbf t_\a+\bbbf t_\b$ is $BC(1,1).$ This   is  a contradiction using  the classification  of finite dimensional basic classical simple Lie superalgebras. So  to complete the proof, we assume $R$ is of type $A(1,1)$ and  show that $R$ is a locally finite root supersystem.  Take $\v:=\hbox{span}_\bbbf R$ and denote the induced form on $\v$ again by $\fm.$
Using the same argument as in \cite[Lem. 3.10]{you2}, one can see that $\bar R_{re}$ is locally finite in its $\bbbf$-span in the sense that it intersects each finite dimensional subspace of $\hbox{span}_\bbbf \bar R_{re}$ in a finite set. So using Lemmas 3.10, 3.12 and 3.21  of \cite{you2}, we get that $\bar R$ is an irreducible locally finite root supersystem in its $\bbbz$-span. Also using \cite[Lem. 3.5]{you2}; we get that $\bar R_{re}$ is a locally finite root system and the restriction of the form  $(\cdot,\cdot\bar)$ to $\bar\v_{re}:=\hbox{span}_\bbbf \bar R_{re}$ is nondegenerate. Therefore we have
\begin{equation}\label{non*}
\parbox{4.3in}{the restriction of the form $(\cdot,\cdot\bar)$ to $\bar\v_{\bbbq}:=\hbox{span}_\bbbq \bar R_{re}$ is nondegenerate.}
 \end{equation} Since  $\bar R_{re}$ is a locally finite root system, by \cite[Lem. 5.1]{LN2},  it contains a $\bbbz$-linearly independent subset $T$ such that \begin{equation}\label{ref*}\w_TT=(\bar R_{re})^\times_{red}=\bar R_{re}\setminus\{2\bar \a\mid\a\in R_{re}\},\end{equation} in which by $\w_T,$ we mean  the subgroup of the Weyl group of $\bar R_{re}$ generated by $r_{\bar\a}$ for all $\bar\a\in T.$  On the other hand, we know there is a subset $\Pi$ of $R$ such that $\bar \Pi$ is  a $\bbbz$-basis for $\hbox{span}_\bbbz\bar R;$ see \cite[Lem. 2.3]{you6}. This allows us to define the linear isomorphism  $$\varphi:\hbox{span}_\bbbq \bar R\longrightarrow \bbbq\ot_\bbbz \hbox{span}_\bbbz\bar R$$ mapping $\bar \a$ to $1\ot\bar \a$ for all $\a\in\Pi.$
Now suppose that  $\bar R$ is of real type, then
\begin{eqnarray*}
\varphi(\hbox{span}_\bbbq \bar R_{re})
=
\hbox{span}_\bbbq (1\ot \bar R_{re})=\bbbq\ot\hbox{span}_\bbbz\bar R=\varphi(\hbox{span}_\bbbq \bar R)
\end{eqnarray*}
which in turn implies that $\hbox{span}_\bbbq \bar R=\hbox{span}_\bbbq \bar R_{re}.$ Therefore,  $\hbox{span}_\bbbq \bar R=\hbox{span}_\bbbq T$ and so $\hbox{span}_\bbbf \bar R=\hbox{span}_\bbbf T .$ But $T$ is $\bbbz$-linearly independent and so it is $\bbbq$-linearly independent. We now prove that $T$ is $\bbbf$-linearly independent.  Suppose that $\{\bar \a_1,\ldots,\bar \a_n\}\sub T$ and $\{r_1,\ldots,r_n\}\sub\bbbf$ with  $\sum_{i=1}^nr_i\bar \a_i=0.$ Take  $\{a_j\mid j\in J\}$ to be  a basis for $\bbbq$-vector space $\bbbf.$ For each $1\leq i\leq n ,$ suppose $\{r^j_i\mid j\in J\}\sub\bbbq$ is such that $r_i=\sum_{j\in J}r^j_ia_j.$ Then for each $\bar\a\in T,$ we have
\begin{eqnarray*}
0=\sum_{i=1}^nr_i\frac{2(\bar \a_i,\bar\a)^{\bar{}}}{(\bar\a,\bar\a)^{\bar{}}}=\sum_{i=1}^n\sum_{j\in J}r^j_ia_j\frac{2(\bar \a_i,\bar\a)^{\bar{}}}{(\bar\a,\bar\a)^{\bar{}}}=\sum_{j\in J}\sum_{i=1}^nr^j_i\frac{2(\bar \a_i,\bar\a)^{\bar{}}}{(\bar\a,\bar\a)^{\bar{}}}a_j.
\end{eqnarray*}
Since $\frac{2(\bar \a_i,\bar\a)^{\bar{}}}{(\bar\a,\bar\a)^{\bar{}}}\in\bbbz,$ we get that for each $j\in J$ and $\bar\a\in T,$ $$(\sum_{i=1}^nr^j_i\bar \a_i,\bar\a)^{\bar{}}=\sum_{i=1}^nr^j_i(\bar \a_i,\bar\a)^{\bar{}}=0.$$ So by (\ref{non*}), $\sum_{i=1}^nr^j_i\bar \a_i=0$  for all $j\in J.$ But $T$ is $\bbbq$-linearly independent  and so $r_i^j=0$ for all $1\leq i\leq n$ and $ j\in J.$
This  means that  \begin{equation}\label{lin*}
\parbox{2in}{$T$ is  $\bbbf$-linearly independent.}
 \end{equation}
Next suppose that $\bar R$ is of imaginary type and fix $\a^*\in\rim^\times.$ Using a modified version of the above argument together with \cite[Lem 3.14]{you2} (see also \cite[Lem. 3.21]{you2}), we get that \begin{equation}\label{lin2*}
\parbox{2.5in}{$T\cup\{\bar{\a^*}\}$ is  $\bbbf$-linearly independent.}
\end{equation}
For each element $\a\in T,$ we fix a preimage $\dot\a\in R$ of $\a$ under $\bar{\;}$ and set
$$K:=\left\{\begin{array}{ll}
\{\dot\a\mid \a\in T\}&\hbox{if $\bar R$ is of real type,}\\
\{\dot\a\mid \a\in T\}\cup\{\a^*\}&\hbox{if $\bar R$ is of imaginary  type.}
\end{array}\right.$$
We have  using \cite[Pro. 3.14]{you2} together with (\ref{ref*}) that $\bar \v=\hbox{span}_\bbbf \bar K.$ Therefore setting  $\dot \v:=\hbox{span}_\bbbf K$ and using (\ref{lin*}) and (\ref{lin2*}), we get that  $\v=\dot\v\op \v^0.$ We set $\dot R:=\{\dot\a\in\dot \v\mid \exists\sg\in\v^0, \dot \a+\sg\in R\},$ then  $\dot R$ is a locally finite root supersystem in its $\bbbz$-span isomorphic to $\bar R.$ Also since $K\sub R\cap\dot R,$ $-K\sub R\cap\dot R.$ So the subgroup $\w_K$ of the Weyl group of $R$  generated by the reflections based on  real roots of $K,$ we have
$$\w_K(\pm K)\sub R\cap \dot R\andd \pm\w_KK=\left\{\begin{array}{ll}(\dot R_{re})_{red}^\times&\hbox{if $\bar R$ is of real type,}\\
\dot R^\times& \hbox{if $\bar R$ is of imaginary  type.}\end{array}\right.$$%\begin{eqnarray*}
%\hbox{span}_\bbbq(1\ot \dot R)=\hbox{span}_\bbbq (1\ot S)=\bbbq\ot \hbox{span}_\bbbz S=\bbbq\ot \hbox{span}_\bbbz \dot R_{re}= \hbox{span}_\bbbq 1\ot \dot R_{re}.
%\end{eqnarray*}
We finally  set $S_{\dot \a}:=\{\sg\in \v^0\mid \dot\a+\sg\in R\}$ for $\dot\a\in \dot R.$ Then $R=\cup_{\dot\a\in\dot R}(\dot\a+S_{\dot\a}).$ Now using  Case 1 of the proof of  Lemma 1.13 of \cite{you6}, we have
\begin{equation}\label{s-0}
S_{\dot\a}\sub R^0=\{0\};\;\; \;\;\dot\a\in \dot R_{re}.
 \end{equation}  Suppose that $\d_1,\d_2\in\rim$ with $(\d_1,\d_2)\neq0$ and $\d_1+\d_2,\d_1-\d_2\in R,$ then there are $\dot\d_1,\dot\d_2\in\dot R_{ns}, \sg\in S_{\dot\d_1}, \tau\in S_{\dot\d_2}$ such that $\d_1=\dot\d_1+\sg$ and $\d_2=\dot\d_2+\tau.$ Since $\d_1\pm\d_2\in R,$ we get using (\ref{s-0}) that $\sg+\tau,\sg-\tau=0$ and so $\sg=\tau=0.$ Therefore, taking  $\dot R=\{0,\pm2\ep_0,\pm2\d_0,\pm\ep_0\pm\d_0\},$  we have $\ep_0\pm\d_0\in R.$ Now if $r\in\{\pm1\}$ and $\d\in S_{\ep_0+r\d_0},$  since $(\ep_0-r\d_0,\ep_0+r\d_0+\d)\neq 0,$  either $2r\d_0+\d\in R$ or $2\ep_0+\d\in R$ which together with  (\ref{s-0}) implies that  $\d=0.$ This shows $R\sub \dot R$ and so $\v=\dot\v.$  Therefore $R=\dot R$ is a locally finite root supersystem in its $\bbbz$-span of type $A(1,1).$
  \end{proof}


\begin{lem}\label{irr}
Suppose that $\LL^0=\hh$ and that $\LL$ is simple.  Then $R$ is an irreducible locally finite root supersystem.
\end{lem}

\begin{proof}
Set $\v:=\hbox{span}_\bbbf R$ and denote the induced form on $\v$ again by $\fm.$  As in Lemma \ref{a11}, there is a subspace $\dot\v$ of $\v,$ a subset $\dot R$ of $\dot\v$ and a class $\{S_{\dot\a}\mid \dot \a\in \dot R\}$ of subsets of $\v^0,$ the radical of the form $\fm,$ such that $\v=\dot\v\op\v^0$ and $\dot R$ is a locally finite root supersystem in $\dot\v$ isomorphic to the image  $\bar R$ of $R$ in $\v/\v^0$ under the canonical projection map, with $R=\cup_{\dot\a\in \dot R} (\dot\a+S_{\dot\a}).$ Take $K$ to be the subsuperalgebra of $\LL$ generated by $\cup_{\a\in \rcross}\LL^\a.$ One can check that
\begin{equation}\label{equal}
K=\sum_{\a\in  R^\times}\LL^{\a}+\sum_{\dot\a\in\dot R^\times}\sum_{\sg\in S_{\dot\a}}\sum_{\tau\in S_{-\dot\a}}[\LL^{\dot\a+\sg},\LL^{-\dot\a+\tau}].
\end{equation}Since $K$ is a nonzero ideal of $\LL,$ we have $K=\LL.$ Now to the contrary, assume that $R$ is not irreducible, then there are nonempty subsets  $A_1,A_2$ of $\rcross$  such that $\rcross  =A_1\uplus A_2$ and that $(A_1,A_2)=\{0\}.$ Now if $\a,\b\in A_i$ ($i=1,2$) are such that $\a+\b\in R,$ then for $j\in\{1,2\}$ with $j\neq i,$ we have  $(\a+\b, A_j)=\{0\},$ so either $\a+\b\in A_i$
\begin{comment}
if $\a+\b\in A_j,$ then $(\a+\b, R)=\{0\}$ which implies that $\a+\b\in R^{0},$ a contradiction.
\end{comment}
or $\a+\b\in R^0.$ Therefore, either $\a+\b\in A_i$ or $\a=\dot\a+\sg$ and $\b=-\dot\a+\tau$ for some $\dot\a\in\dot R^\times$ and $\sg\in S_{\dot\a},\tau\in S_{-\dot\a}.$ Also if   $\{i,j\}=\{1,2\},$ $\a\in A_i$ and $\b\in A_j,$ then   $(\a+\b,A_i)=(\a, A_i)\neq\{0\}$ and $(\a+\b,A_j)=(\b,A_j)\neq\{0\}.$
\begin{comment}
$(\a+\b,A_i)=(\a,A_i)\neq\{0\}.$ If  $(\a,A_i)=\{0\},$ we have $(\a,R^\times)=\{0\},$ so $\a\in R^0,$ a contradiction.
\end{comment}
This implies that $\a+\b$  neither belongs to $R^\times$ nor belongs to $R^0.$ So $\a+\b\not\in R.$ Therefore, for $\dot A_i:=\{\dot\a\in \dot R\mid \dot\a+S_{\dot\a}\sub A_i\}$ ($i=1,2$), $\sum_{\a\in  A_i}\LL^{\a}+\sum_{\dot\a\in\dot A_i}\sum_{\sg\in S_{\dot\a}}\sum_{\tau\in S_{-\dot\a}}[\LL^{\dot\a+\sg},\LL^{-\dot\a+\tau}]$ is a nontrivial ideal of $\LL=K.$
\begin{comment}
We note that if $\{i,j\}=\{1,2\}$ and  $\dot\a+\sg\in A_i,$ then $(\dot \a,A_j)=\{0\}$ and so $(\dot\a+\tau, A_j)=\{0\}$ for each $\tau\in S_{\dot\a}.$  This implies that $\dot\a+S_{\dot\a}\sub A_i.$ Similarly $-\dot\a+S_{-\dot\a}\sub A_i.$
\end{comment}
 This makes a contradiction, so $R$ is an irreducible extended affine root supersystem. We next note that if $\d\in R^0\setminus\{0\},$ then  $t_\d$ is a nonzero element of the center of $\LL=K$ which is a contradiction. So $R^0=\{0\}$ and
\begin{equation}\label{equal2}
\LL=\sum_{\a\in \rcross}\LL^\a\op\sum_{\a\in\rcross}[\LL^\a,\LL^{-\a}].
\end{equation}
Now if $R$ is not of type $ A(\ell,\ell),$ \cite[Lem. 1.13]{you6} implies that $R$ is a locally finite root supersystem in its $\bbbz$-span. Also if $R$ is of type $A(\ell,\ell),$ Lemma \ref{dim1} and Proposition \ref{dim1-re}  imply that $\LL$ is finite dimensional, in particular, $\hh=\sum_{\a\in\rcross}[\LL^\a,\LL^{-\a}]=\sum_{\a\in\rcross}\bbbf t_\a$ is finite dimensional. Therefore, $\hh^*=\hbox{span}_\bbbf R$ and the induced form on $\hh^*$ is nondegenerate. But  $\v=\hbox{span}_\bbbf R=\hh^*$, and so the form on $\v$ is nondegenerate.  This in  turn implies that  the form restricted to $\la R\ra$ is nondegenerate. This means that $R$ is an irreducible locally finite root supersystem.
\end{proof}


\begin{lem}\label{semisimple}
Suppose that $\LL^0=\hh.$ If the root system  $R$ of $\LL$ is a locally finite root supersystem in its $\bbbz$-span, then the ideal  $K:=\sum_{\a\in R^\times }\LL^\a+\sum_{\a\in R^\times}[\LL^\a,\LL^{-\a}]$ of $\LL$ is a direct sum of  simple ideals.\end{lem}
\begin{proof}
Set $A:=\la R\ra,$  $\v:=\hbox{span}_\bbbf R$ and denote the induced form on $\v$ again by $\fm.$ Since $R$ is a locally finite root supersystem in $A,$ the form restricted to $A$ is nondegenerate and so using the same argument as in \cite[Lem. 3.21]{you2}, the form on $\v$ is also nondegenerate.
Suppose that $R=\op_{i\in I} S_i$ is the decomposition of $R$ into irreducible sub-supersystems and set $\LL(i):=\op_{\a\in S_i^\times}\LL^{\a}\op\sum_{\a\in S_i^\times}[\LL^\a,\LL^{-\a}].$ Then $K=\sum_{i\in I}\LL(i).$
%Set $$\hh(i):=\sum_{\a\in S_i^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in S_i}\bbbf t_{\a}\;\;\;\; (i\in I).$$ We claim that $\sum_{i\in I} \hh(i)$ is a direct sum; in fact if  $i\in I$ and $h=\sum_{\a\in S_i}r_\a t_\a=\sum_{\b\not\in S_i}s_\b t_\b,$ then for each $\gamma\in S_i$ and $\eta\in R\setminus S_i,$ we have
%$$\begin{array}{l}
%(\sum_{\a\in S_i}r_\a\a,\gamma)=(h,t_\gamma)=\sum_{\b\not\in S_i}s_\b(t_\b,t_\gamma)=\sum_{\b\not\in S_i}s_\b(\b,\gamma)=0\\
%(\sum_{\a\in S_i}r_\a\a,\eta)=(h,t_\eta)=\sum_{\a\in S_i}r_\a(t_\a,t_\eta)=\sum_{\a\in S_i}r_\a(\a,\eta)=0.
%\end{array}$$ This implies that  $\sum_{\a\in S_i}r_\a\a\in \v^0=\{0\}$ and  so $h=0.$ Therefore $K=\op_{i\in I}\LL(i).$
We prove that each $\LL(i)$ $(i\in I)$ is a simple  ideal of $\LL.$
We note that each $S_i$ is  a closed sub-supersystem of $R,$
\begin{comment}
for $i,j\in I$ with $i\neq j,$ $\a\in S_i^\times$ and $\b\in R_j^\times,$  there are $\gamma\in S_i$ and $\eta\in R_j$
with $(\a,\gamma)\neq0$ and $(\b,\eta)\neq0,$ so $(\a+\b,\gamma)\neq 0$ and $(\a+\b,\eta)\neq0.$ This implies that $\a+\b\not\in R.$  Also if $\zeta,\xi\in S_i$ with $\zeta+\xi\in R\setminus\{0\}.$ Since for each $k\in I\setminus\{i\},$ $(\zeta+\xi, R_k)=\{0\},$ we get that $\zeta+\xi\in S_i.$
\end{comment}
so  $\LL(i)$ is an ideal of $\LL.$
Suppose that $i\in I$ and note that
$\LL(i)$ has a weight space decomposition $\LL(i)=\op_{\a\in S_i}\LL(i)^\a$  with respect to $\hh(i):=\sum_{\a\in S_i^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in S_i}\bbbf t_{\a}$ with $\LL(i)^\a=\LL^\a$ for $\a\in S_i^\times$ and $\LL(i)^0=\hh(i);$ see Lemma \ref{restrict}($ii$).
Now suppose that $J$ is a nonzero  ideal of $\LL(i),$ then $J=\op_{\a\in S_i}(J\cap \LL(i)^\a).$ We claim that there is $\a\in S_i^\times$ with $J\cap \LL(i)^\a\neq\{0\}.$ To the contrary, assume $J=J\cap \LL(i)^0$ and suppose  $h:=\sum_{\a\in S_i}r_\a t_\a\in \LL(i)^0$ is a nonzero element of $J,$ then since  $\sum_{\a\in S_i}r_\a\a$ is a nonzero element of  $\la S_i\ra,$ one finds $\b\in S_i$ with $(\sum_{\a\in S_i}r_\a\a,\b)\neq 0,$ therefore, $[h,\LL^\b]=(\sum_{\a\in S_i}r_\a\a,\b)\LL^\b\neq \{0\}$ and so $\LL^\b\sub J,$  a contradiction.
Now fix  $\a\in S_i^\times$ with  $J\cap \LL^\a=J\cap \LL(i)^\a\neq\{0\},$  then $t_\a\in [\LL^{-\a},\LL^\a\cap J]\sub J.$ So for each $\b\in S_i$ with $(\a,\b)\neq 0,$ we have $[t_\a,\LL^\b]=(\b,\a)\LL^\b\neq \{0\}$ which in turn implies that $\LL^\b\sub J.$ Now for an arbitrary $\b\in S_i^\times,$ as  $S_i$ is irreducible, there are a chain $\a_0:=\a,\a_1,\ldots,\a_n:=\b$ of elements of $S_i^\times$ such that $(\a_i,\a_{i-1})\neq 0$ for all $1\leq i\leq n,$ so $\LL^{\a_i}\sub J$ for all $1\leq i\leq n,$ in particular  $\LL^\b\sub J.$ So $\LL(i)\sub J.$ This completes the proof.
\end{proof}

%\begin{lem}
%If $R$ is irreducible, then $K$ is locally finite basic classical ?simple Lie superalgebra
%\end{lem}
%We now show that each $K$ is locally finite. We know that $R$ is the direct union of its finite irreducible  subsystems, say $R=\cup_{\lam\in \Lam} R^\lam.$ It is enough to show that each $\sum_{\a\in (R^\lam)^\times}\LL^\a+\sum_{\a\in (R^\lam)^\times}[\LL^\a,\LL^{-\a}]$ is a finite dimensional simple Lie subsuperalgebra. Now $\LL^\lam$



\begin{lem}\label{direct-union}
Suppose that $\LL$ is simple and  $\LL^0=\hh.$ If $\LL$ is finite dimensional, it is a basic classical simple Lie superalgebra and if $\LL$ is infinite dimensional,  then  it is a direct union of finite dimensional basic classical simple Lie superalgebras; in  particular $\LL$ is locally finite. Moreover, up to a scalar multiple, there is a unique  invariant nondegenerate even supersymmetric  bilinear form on $\LL.$
\end{lem}
\begin{proof}
 We first note that $R$ is an irreducible locally finite root supersystem by Lemma \ref{irr}. We now assume $L$ is infinite dimensional. Contemplating Lemma 2.4($iii$) of  \cite{you6}, we get that  $R$ is a direct union of its irreducible closed finite sub-supersystems of the same type as $R,$  say  $R=\cup_{\gamma\in \Gamma} R(\gamma),$ where $\Gamma$ is  a nonempty index set. For each $\gamma\in\Gamma,$ set $\LL(\gamma):=\sum_{\a\in  R(\gamma)^\times}\LL^\a+\sum_{\a\in R(\gamma)^\times}[\LL^\a,\LL^{-\a}].$ Considering   Lemma \ref{dim1} and Proposition  \ref{dim1-re} and using  the same argument as in Lemma \ref{semisimple}, one can see that
\begin{comment}
We prove that $\LL(\gamma)$ is a simple Lie subsuperalgebra of $\LL.$  Since $R(\gamma)$ is a closed sub-supersystem, $\LL(\gamma)$ is a subsuperalgebra. Suppose that $I$ is a nonzero ideal of $\LL(\gamma),$ then there is $\a\in R(\gamma)$ such that $\LL^\a\cap I\neq\{0\}.$  If $\a=0,$ there is $\{r_\b\mid \b\in R(\gamma)\}\sub \bbbf$ such that $0\neq \sum_{\b\in R(\gamma)}r_\b t_\b\in I.$ Since the form on $\la R(\gamma)\ra$ is nondegenerate, one finds $\eta\in R(\gamma)$ with $(\sum_{\b\in {}^\eta R}r_\b \b, \eta)\neq0.$ Now $0\neq (\sum_{\b\in  R(\gamma)}r_\b \b, \eta)\LL^\eta=[\sum_{\b\in  R(\gamma)}r_\b t_\b,\LL^\eta]\sub I$ and so $\LL^\eta\sub I.$ So if $I$ is a nonzero ideal of $\LL,$ there is a nonzero $\eta\in \LL(\gamma)$ with $\LL^\eta\sub I.$ This in particular implies that $t_\eta\in I.$ Now for $\zeta \in R(\gamma)$ with $(\zeta, \eta)\neq 0,$ we have $(\zeta,\eta)\LL^\zeta=[t_\eta,\LL^\zeta]\sub I$ and so $\LL^\zeta\sub I.$ Now as $R(\gamma)$ is irreducible,  it follows that $\LL^\zeta\sub I$ for all $\zeta\in R(\gamma)$ and so $I=\LL(\gamma).$
 \end{comment}
  $\LL(\gamma)$ is a finite dimensional simple subsuperalgebra. We next  show that $\LL(\gamma)$ is a finite dimensional  basic classical simple Lie superalgebra. Since the form restricted to $\LL(\gamma)$ is nonzero and even,  we just need to show $\LL(\gamma)_{\bar0}$ is a  reductive Lie algebra  or equivalently $\LL(\gamma)_{\bar 1}$ is a completely reducible $\LL(\gamma)_{\bar0}$-module.
We carry out this throughout the following two cases:
\smallskip

\underline{$R$ is of real type:}
In this case, $ \hbox{span}_\bbbf\{t_\a\mid\a\in R(\gamma)\}=\hbox{span}_\bbbf\{t_\a\mid \a\in R(\gamma)_{re}\}.$ This in turn implies that  $$\sum_{\a\in R(\gamma)^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in  R(\gamma)_{re}^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in  R(\gamma)_{\bar0}^\times}[\LL^\a,\LL^{-\a}].$$ Suppose that $R(\gamma)_{\bar 0}= \op_{i=1}^n \Phi_i,$ where $n$ is a positive integer, is the decomposition of the finite root system $R(\gamma)_{\bar 0}$ into irreducible subsystems.  Using the same argument as in Lemma \ref{semisimple}, we get that  for each $1\leq i\leq n,$ $\sum_{\a\in \Phi_i^\times}\LL^\a+\sum_{\a\in \Phi_i^\times}[\LL^\a,\LL^{-\a}]$ is a finite dimensional simple ideal of $\LL(\gamma)_{\bar 0}$ and so $\LL(\gamma)_{\bar 0}$ is a semisimple Lie algebra.

\smallskip

\underline{$R$ is of imaginary type:} Take $\w$ to be the Weyl group of $R$ and fix $\d^*\in \rim^\times.$ Without loss of generality, we assume each $R(\gamma)$ contains  $\d^*.$
 We recall from Lemma  \ref{super-sys}($iii$)  that $\rim^\times=\pm\w\d^*$ and note  that  $-\d^*\not\in \w\d^*$  as otherwise $-\d^*\in \d^*+\la \rre\ra$ which contradicts the fact that $R$ is of imaginary type. For $\gamma\in \Gamma,$ take $\w_\gamma$ to be the Weyl group of $R_\gamma$ and set $S_1:=\w_\gamma\d^*$ as well as  $S_2:=-\w_\gamma\d^*.$  We note that  by Proposition \ref{dim1-re} and (\ref{rns}), $ R_{\bar 1}= \rim^\times$ and   claim that for $i=1,2,$ $\sum_{\a\in S_i}(\LL_{\bar 1})^\a$ is an irreducible $\LL(\gamma)_{\bar 0}$-module. By \cite[Lem. 4.6]{you2},  $\sum_{\a\in S_i}(\LL_{\bar 1})^\a$ is an $\LL(\gamma)_{\bar 0}$-submodule of  $\LL_{\bar 1}.$
 Suppose that $U$ is a nonzero $\LL(\gamma)_{\bar 0}$-submodule of $\sum_{\a\in S_i}(\LL_{\bar 1})^\a.$  Then  $U= \sum_{\a\in S_i}((\LL_{\bar 1})^\a\cap U).$ Since $U\neq\{0\},$ there is $\gamma\in S_i$ such that $U\cap(\LL_{\bar 1})^\gamma\neq\{0\}.$ So $(\LL_{\bar 1})^\gamma\sub U$ as   by Proposition \ref{dim1}, we have $\dim(\LL_{\bar 1})^\gamma=1.$ Suppose that $\b\in S_i,$ then there are $\gamma_1,\ldots,\gamma_n\in (R_\gamma)_{\bar 0}\setminus\{0\}$ such that $\b=r_{\gamma_1}\cdots r_{\gamma_n}(\gamma).$ Now considering (\ref{auto}), we have $\theta_{\gamma_1}\cdots\theta_{\gamma_n}(\LL_{\bar 1})^\gamma\sub(\LL_{\bar 1})^\b\cap U.$ This together with the fact that $\dim((\LL_{\bar 1})^\b)=1,$ implies that $(\LL_{\bar 1})^\b\sub U.$  So $U=\sum_{\b\in S_i}(\LL_{\bar 1})^\b.$  Therefore $(\LL_\gamma)_{\bar 1}=S_1\op S_2$ is completely reducible $(\LL_\gamma)_{\bar0}-$module.

Finally, suppose that $\LL$ is finite dimensional, using the same argument as above, if $R$ is of real type, $\LL_{\bar 0}$ is a finite dimensional semisimple Lie algebra and if $R$ is of imaginary type, $\LL_{\bar 1}$ is a  completely reducible $\LL_{\bar 0}$-module, so  $\LL$ is a finite dimensional  basic classical simple Lie superalgebra.
The last assertion follows easily form the facts that $\LL$ is a direct union of finite dimensional basic classical simple Lie superalgebras and that nonzero invariant  forms  on such Lie superalgebras are proportional.
\end{proof}

\begin{deft}
{\rm A nonzero  Lie superalgebra $\LL=\LL_{\bar 0}\op\LL_{\bar 1}$   over an algebraically closed field is called a {\it locally finite basic classical simple Lie superalgebra} if
\begin{itemize}
\item $\LL$ is   locally finite and  simple,
\item $\LL_{\bar 0}$ has a nontrivial subalgebra $\hh$ with respect to which $\LL$ has a weight space decomposition $\LL=\sum_{\a\in \hh^*}\LL^\a$ via the adjoint representation with corresponding root system $R$ such that $\LL^0=\hh$ and $\rcross\neq \emptyset,$
    \item  $\LL$ is equipped with an invariant nondegenerate even supersymmetric  bilinear form.
\end{itemize}
}
\end{deft}

\begin{Thm}\label{main1}
Suppose that $\LL$ is a nontrivial Lie superalgebra, then $\LL$ is a locally finite basic classical simple  Lie superalgebra if and only if
it contains a subalgebra $\hh$ (refereed to as a {\it Cartan subalgebra}) and that it  is equipped with a  bilinear form $\fm$ such that $(\LL,\hh,\fm)$ is a   simple extended affine Lie superalgebra  with corresponding root system $R$ such that $R^\times\neq\emptyset$ and $\LL^0=\hh.$
 In this case,  we have

(i) the root system $R$ of $\LL$ is an irreducible locally finite root supersystem,

(ii) $\LL$ is a direct union of finite dimensional  basic classical simple  Lie  superalgebras,

(iii) $[\LL_{\bar 0},\LL_{\bar 0}]$ is a semisimple Lie algebra,

(iv) if $\LL_{\bar 1}\neq\{0\},$ it  is a completely reducible $\LL_{\bar 0}$-module with at most two irreducible constituents.
\end{Thm}
\begin{proof}  Suppose that $\LL$ is  a locally finite simple Lie superalgebra equipped with an invariant nondegenerate even supersymmetric bilinear form $\fm$ and that $\LL_{\bar 0}$ has a nontrivial subalgebra $\hh$ with respect to which $\LL$ has a weight space decomposition $\LL=\sum_{\a\in \hh^*}\LL^\a$ via the adjoint representation with corresponding root system $R$ such that $\LL^0=\hh$ and $\rcross\neq \emptyset.$   Use the same notation as in the text and suppose   $\a\in R_i\setminus\{0\}$ for some $i\in\{0,1\}.$ Since the form is nondegenerate and even, we get that the form on $(\LL_{\bar i})^\a\op(\LL_{\bar i})^{-\a}$ is nondegenerate and so there are $x\in (\LL_{\bar i})^\a$ and $y\in (\LL_{\bar i})^{-\a}$ such that $(x,y)\neq 0.$ Now  $[x,y]\in\hh$ and for each $h\in\hh,$ $$(h,[x,y])=([h,x],y)=\a(h)(x,y).$$ This implies that $[x,y]\neq0.$  Also as $\LL$ is locally finite, for each $x\in \LL^\a,y\in \LL^\b$ $(\a\in R\setminus\{0\},\b\in R),$  the subsuperalgebra of $\LL$ generated by $x,y$ is finite dimensional, so there is a positive integer $n$ such that $(ad_x)^n(y)\in\LL^{n\a+\b}$ equals to zero. This means that  $ad_x$ is locally nilpotent. Altogether, we concluded that  $(\LL,\fm,\hh)$ is a simple  extended affine Lie superalgebra.
This  together with Lemma \ref{direct-union} completes the proof of the first assertion.

Now suppose $\LL$ is a locally finite basic classical simple Lie superalgebra, $(i)$ and $(ii)$ follow from Lemmas \ref{irr} and \ref{direct-union}. For $(iii),$ we note that if $R_0=\op_{i=1}^n \Phi_i$ is the decomposition of $R_0$ into irreducible subsystems, then we have $[\LL_{\bar 0},\LL_{\bar 0}]=\op_{i=1}^n\LL(i)$ in which $\LL(i):=\sum_{\a\in\Phi_i\setminus\{0\}}(\LL_{\bar 0})^\a+\sum_{\a\in\Phi_i\setminus\{0\}}[(\LL_{\bar 0})^\a,(\LL_{\bar 0})^{-\a}]$ is  a simple ideal. Finally for ($iv$), we first suppose $R$ is of imaginary type, then by Proposition \ref{dim1-re} and  (\ref{rns}), $R_1=\rim\setminus\{0\}$ and by Proposition \ref{dim1}, $\dim \LL^\d=1$ for $\d\in\rim^\times.$ Now one can use the same argument as in  the proof of  Lemma \ref{direct-union} to get that $\LL_{\bar 1}$ is completely reducible $\LL_{\bar 0}$-module with two irreducible constituent. Next suppose $R$ is of type $A(\ell,\ell),$ then by Propositions \ref{dim1} and \ref{dim1-re}, $\LL$ is finite dimensional and so by Lemma \ref{direct-union} and the theory of finite dimensional basic classical simple Lie superalgebras, we are done. Finally, suppose $R$ is of  a real type  other than type $A(\ell,\ell).$ Then by Propositions \ref{dim1-re}, \ref{dim1} and  Lemma \ref{a11},  \begin{equation}\label{dim}\dim(\LL^\a)=1 \;\;\;\forall\a\in R^\times.\end{equation}  Also using the classification theorem for real types, we have the following:
\begin{equation}\label{submod}
\parbox{4in}{If $\{\b\in R_1\mid 2\b\in R\}\neq\emptyset,$ then  for   $\a\in \rim^\times,$  there is $\gamma\in R_0^\times$  such that $\gamma+\a\in R_1$ and $2(\gamma+\a)\in R;$ also for $\b\in R_1$ with $2\b\in R,$ there is $\eta\in R_0^\times$  such that $\eta+\b\in \rim^\times.$}
\end{equation}
We next set  $S_1:=\sum_{\a\in \rim^\times} \LL^\a$ and $S_2:=\sum_{\a\in  R_{re}^\times; 2\a\in R}\LL^\a$ and suppose that   $U$ is a submodule of $\LL_{\bar 0}$-module $\LL_{\bar 1}=S_1\op S_2,$ then $U=\sum_{\a\in R_1}(\LL^\a\cap U).$
If $U\neq\{0\},$  then $U\cap \LL^\a\neq\{0\}$ for some $\a\in \rim^\times$ or for some $\a\in R^\times$ with $2\a\in R.$ Now
(\ref{submod})  together with  Lemma  \ref{pro-nonzero} and (\ref{dim}) imply that  $U\cap S_1\neq\{0\}$ and if $S_2\neq\{0\},$ $U\cap S_2\neq\{0\}.$
Now  considering  (\ref{dim}) and using  the same argument as in the  proof of Lemma \ref{direct-union} together with the fact that  all roots $\b\in R_1$ with $2\b\in R$ and also  all nonzero nonsingular roots  are conjugate under the Weyl group action, we get that $S_1+S_2\sub U.$ So $U= \LL_{\bar 1}.$ This completes the proof.
\end{proof}


\begin{lem}\label{real-semisimple}
Suppose that $(\LL,\fm,\hh)$ is a locally finite basic classical simple Lie superalgebra with corresponding root system $R.$ Then $\LL_{\bar 0}$ is a semisimple Lie algebra  if and only if $R$ is of real type.
\end{lem}
\begin{proof}
There is nothing to prove if $\LL$ is a Lie algebra. So suppose that $\LL_{\bar 1}\neq\{0\}.$ If $R$ is of type $B(0,T ),$ take $\d:=0$ and otherwise, take $\d$ to be a fix nonzero nonsingular root.  We know that $\rim=\{0\}\cup\pm \w\d$ in which $\w$ is the Weyl group of $R.$ Also $\LL=\sum_{\a\in R^\times }\LL^\a+\sum_{\a\in R^\times}[\LL^\a,\LL^{-\a}]$ as  $\LL$  is simple. This implies that   $\hh= \sum_{\a\in R^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in R}\bbbf t_\a=\sum_{\a\in \rre}\bbbf t_\a+\bbbf t_\d.$ So $$\LL_{\bar 0}=\sum_{\a\in R_0^\times}(\LL_{\bar 0})^\a\op \hh=\sum_{\a\in R_0^\times}\LL^\a\op \hh=\sum_{\a\in R_0^\times}\LL^\a\op \sum_{\a\in R_{re}^\times}[\LL^\a,\LL^{-\a}]+\bbbf t_\d$$ and  by (\ref{rns}) and Theorem \ref{main1}, $$K:=[\LL_{\bar0},\LL_{\bar0}]=\sum_{\a\in R_0^\times}\LL^\a\op \sum_{\a\in R_0^\times}[\LL^\a,\LL^{-\a}]=\sum_{\a\in R_0^\times}\LL^\a\op \sum_{\a\in \rre^\times}[\LL^\a,\LL^{-\a}]$$ is a semisimple ideal of $\LL_{\bar 0}.$  Also contemplating \cite[Lem. 3.5]{you2}, the induced form on $\hbox{span}_\bbbf R$ restricted to $\hbox{span}_\bbbf \rre$ is nondegenerate. Now it follows that
$t_\d\in \hbox{span}_\bbbf \{t_\a\mid\a\in\rre\}$ if and only if $t_\d\in \hbox{span}_\bbbq \{t_\a\mid\a\in\rre\};$ see \cite[Lem. 1.8]{you6}. So we have
\begin{eqnarray*}
\hh=\sum_{\a\in \rre}\bbbf t_\a\Leftrightarrow t_\d\in \hbox{span}_\bbbf \{t_\a\mid\a\in\rre\}&\Leftrightarrow &t_\d\in \hbox{span}_\bbbq \{t_\a\mid\a\in\rre\}\\
&\Leftrightarrow &\d\in \hbox{span}_\bbbq \{\a\mid\a\in\rre\}\\
&\Leftrightarrow & \hbox{ $\exists n\in\bbbz\setminus\{0\}$ with $ n\d\in \la \rre\ra$}\\
&\Leftrightarrow& \d\in \bbbq \ot_\bbbz \la \rre\ra\\
&\Leftrightarrow& \hbox{$R$ is of real type.}
\end{eqnarray*}
 Now if $R$ is of real type, then $\hh=\sum_{\a\in \rre}\bbbf t_\a= \sum_{\a\in R_{re}^\times}[\LL^\a,\LL^{-\a}]$ and so  $\LL_{\bar 0}=K$ which is  a semisimple Lie algebra.  Conversely, suppose $\LL_{\bar 0}$ is a semisimple Lie algebra and to the contrary assume $R$ is of imaginary type. Then $\LL_{\bar 0}=K\op\bbbf t_\d.$ Since $\LL_{\bar 0}$ is a semisimple Lie algebra and  $K$ is an ideal of $\LL_{\bar 0}$ of codimension 1, there is a $1$-dimensional ideal $I$ of $\LL_{\bar 0}$ such that $\LL_{\bar 0}=K\op I.$ Fix $0\neq x\in I.$ We have $I=\bbbf x.$ If $I\sub \hh,$ for each $\gamma\in R_0^\times,$ we have $[I,(\LL_{\bar 0})^\gamma]\sub I\cap (\LL_{\bar 0})^\gamma \sub \hh\cap(\LL_{\bar 0})^\gamma=\{0\};$ also we have $[I,\hh]\sub[\hh,\hh]=\{0\}.$ This means that $I$ is a central ideal of $\LL_{\bar 0}$ which is absurd. So  $x \not\in \hh.$  Therefore $x=\sum_{\a\in R_0^\times}x_\a+\sum_{\a\in\rre} r_\a t_\a+r t_\d$ in which $r_\a\in \bbbf$ $(\a\in\rre),$  $r\in\bbbf\setminus\{0\}$ and for each $\a\in R_0^\times,$  $x_\a\in(\LL_{\bar 0})^\a$   with $x_\gamma\neq 0$ for some $\gamma\in R_0\setminus\{0\}.$   Since  $I=\bbbf x$ is an ideal, we have  {\small \begin{eqnarray}
 [x_\gamma,(\LL_{\bar 0})^{-\gamma}]+\sum_{\a\in R_0^\times\setminus\{\gamma\}}[x_\a,(\LL_{\bar 0})^{-\gamma}]+\sum_{\a\in\rre} [r_\a t_\a,(\LL_{\bar 0})^{-\gamma}]+r [t_\d,(\LL_{\bar 0})^{-\gamma}]\hspace{-2mm}&=&\hspace{-2mm}[x,(\LL_{\bar 0})^{-\gamma}]\nonumber\\
 &\sub& \bbbf x\label{final3};\end{eqnarray}}  but $x_\gamma\neq0$ and the form restricted to $(\LL_{\bar 0})^\gamma\op(\LL_{\bar 0})^{-\gamma}$ is nondegenerate, so  we have $[x_\gamma,(\LL_{\bar 0})^{-\gamma}]=\bbbf t_\gamma$ by (\ref{bracket}). Now (\ref{final3}) implies  that  $t_\gamma\in \bbbf(\sum_{\a\in\rre} r_\a t_\a+r t_\d)\cap (\sum_{\a\in\rre}\bbbf t_\a)$ which is  a contradiction as $R$ is of imaginary type. This completes  the proof.
%\bigskip
%
%\textcolor[rgb]{1.00,0.00,0.50}{Also if $R$ is of imaginary type and $\LL$ is finite dimensional, since $\hh=\bbbf t_{\d}\op \sum_{\a\neq0}[\LL^{\a},\LL^{-\a}],$ we get that the center of $\LL_{\bar 0}$ is one dimensional.}
\end{proof}

\subsection{Examples}\label{examples}
\begin{comment}
Suppose that $\v=\v_0\op\v_1$ is a superspace and fix bases $\{v_i\mid i\in I\}$ and  $\{v_j\mid j\in J\}$ for $\v_0$ and $\v_1$ respectively, in which  $I$ and $J$ are two disjoint nonempty  index sets. Consider the Lie superalgebra  $\mathfrak{gl}(\v)$ and for $j,k\in I\uplus J,$
define
\begin{equation}\label{elementary2}e_{j,k}:\v\longrightarrow\v;\;\;
v_i\mapsto \d_{k,i}v_j\;\;\; (i\in I\uplus J),\end{equation} then
$\mathfrak{gl}(I\uplus J):=\hbox{span}_\bbbf\{e_{j,k}\mid j,k\in I\uplus J\}$ is a
Lie subsuperalgebra of $\mathfrak{gl}(\v).$
For $A=\displaystyle{\sum_{i,j\in I\uplus J}a_{i,j}e_{i,j}}\in\mathfrak{gl}(I\uplus J),$ define the {\it supertrace} of $A$ to be $str(A):=\sum_{i\in I}a_{i,i}-\sum_{j\in J}a_{j,j}.$
\begin{exa}{\rm
Suppose that $I,J$ are two disjoint index sets with cardinal numbers greater than $1$ such that $|I|\neq|J|$ if $I,J$ are both finite. Consider the Lie superalgebra  $\gg:=\mathfrak{gl}(I\uplus J).$
Set $$\gg:=\mathfrak{sl}(I,J):=\{A\in\gg\mid str(A)=0\},$$ then $\gg$ is a Lie subsuperalgebra of $\gg.$}

\end{exa}

\begin{exa}{\rm
For two disjoint  index sets  $I,J$ with $J\neq\emptyset,$ suppose that $\{0,i,\bar i,\mid i\in I\cup J\}$ is a superset
with $|0|=|i|=|\bar i|=0$ for $i\in I$ and $|j|=|\bar j|=1$ for $j\in J.$  Take $\mathfrak{u}$ to be a vector superspace with a basis
$\{v_i\mid i\in I\cup\bar I\cup J\cup\bar J\cup\{0\}\}$  and $$|v_i|:=|i|;\;\;i\in I\cup\bar I\cup J\cup\bar J\cup\{0\}$$ in which by $\bar I$ (resp. $\bar J$), we mean $\{\bar i\mid i\in I\}$ (resp. $\{\bar j\mid j\in J\}$). Take $\fm$ to be the skew supersymmetric bilinear form $\fm$ on $\mathfrak{u}$ defined by   \begin{equation}\label{form1}(v_0,v_0):=1,(v_i,v_j):=0, (v_{\bar i},v_{\bar j}):=0,(v_i, v_{\bar j})=(-1)^{|i||j|}(v_{\bar j},v_i):=\d_{i,j}\end{equation} for $i,j\in I\cup J.$
Now for $j,k\in\{0\}\cup I\cup \bar I\cup J\cup \bar J,$
define
\begin{equation}\label{elementary2}e_{j,k}:\mathfrak{u}\longrightarrow\mathfrak{u};\;\;
v_i\mapsto \d_{k,i}v_j,\;\;\; (i\in \{0\}\cup I\cup \bar I\cup J\cup \bar J).\end{equation}  Then
\begin{equation}\label{elemnatry2}\mathfrak{ls}(\mathfrak{u}):=\hbox{span}_\bbbf\{e_{j,k}\mid j,k\in \{0\}\cup I\cup \bar I\cup J\cup \bar J\}\end{equation} is a
Lie subsuperalgebra of $\End_\bbbf(\mathfrak{u}).$
If $I$ is nonempty, we set $$\mathfrak{v}:=\hbox{span}_\bbbf\{v_i\mid i\in I\cup\bar I\cup J\cup\bar J\}$$ and  make a convention that we denote $\fm\mid_{\mathfrak{v}\times\mathfrak{v}}$ again by $\fm$ and for  $j,k\in I\cup \bar I\cup J\cup \bar J,$ $e_{j,k}\mid_{\mathfrak{v}}$ is also denoted by $e_{j,k}.$ Therefore $$\mathfrak{ls}(\mathfrak{v}):=\hbox{span}_\bbbf\{e_{j,k}\mid j,k\in I\cup \bar I\cup J\cup \bar J\}$$ is a Lie subsuperalgebra of $\End_\bbbf(\mathfrak{v}).$
For $\gamma=\mathfrak{u}$ or $\gamma=\mathfrak{v},$ and for $i=0,1,$ take  $$\mathfrak{osp}(\gamma)_{\bar i} := \{s \in \mathfrak{ls}(\gamma)_{\bar i}\mid (su , v) = -(-1)^{|s||u|}(u,sv),\; \forall u, v \in \mathfrak{\gamma}\}$$ and  set
 $$\mathfrak{osp}(\gamma):=\mathfrak{osp}(\gamma)_{\bar 0}\op\mathfrak{osp}(\gamma)_{\bar 1}.$$

 One knows that $\mathfrak{ls}(\gamma),$ $\gamma=\mathfrak{u},\mathfrak{v},$ is a Lie subsuperalgebra of $\End_\bbbf(\gamma).$
Set\begin{equation}\label{h}\fh:=\hbox{span}_\bbbf\{h_t,d_k\mid t\in I,\;k\in J\}\end{equation} in which for  $t\in I$ and $ k\in J,$ $$h_t:=e_{t,t}-e_{\bar t,\bar t}\andd d_k:=e_{k,k}-e_{\bar k,\bar k}$$ and for $i\in I$ and $j\in J,$ define
$$\begin{array}{lll}
\begin{array}{c}
\ep_i:\fh\longrightarrow \bbbf\\
h_t\mapsto \d_{i,t},\;\;\; d_k\mapsto 0,
\end{array}&&
\begin{array}{c}
\d_j:\fh\longrightarrow \bbbf\\
h_t\mapsto 0,\;\;\; d_k\mapsto \d_{j,k},
\end{array}
\end{array}$$ in which $t\in I$ and $k\in J.$
Then   for $\gamma=\mathfrak{u},\mathfrak{v},$ $\mathfrak{osp}(\gamma)$ has a weight space decomposition with respect to $\fh.$
Taking $R(\gamma)$  to be the set of weights of $\mathfrak{osp}(\gamma)$  with respect to $\fh,$  we have
\begin{equation}\label{root}\begin{array}{l}R(\mathfrak{u})=\{\pm\ep_r,\pm(\ep_r\pm\ep_s),\pm\d_p,\pm(\d_p\pm\d_q),\pm(\ep_r\pm\d_p)\mid  r, s\in I,\;p,q\in J,\;r\neq s\},\\
\\R(\mathfrak{v})=\{\pm(\ep_r\pm\ep_s),\pm(\d_p\pm\d_q),\pm(\ep_r\pm\d_p)\mid  r, s\in I,\;p,q\in J,\;r\neq s\}\end{array}\end{equation} in which $\pm(\ep_r\pm\ep_s)$'s
 are disappeared if $|I|=1$ and $\pm\ep_r$'s, $\pm(\ep_r\pm\ep_s)$'s as well as $\pm(\ep_r\pm\d_p)$'s are disappeared if $|I|=0.$
 Moreover, for $r, s\in I,\;p,q\in J,\;r\neq s$ and $p\neq q,$ we have
$$\begin{array}{ll}
\mathfrak{osp}(\gamma)^{\ep_r+\ep_s}=\hbox{span}_\bbbf  (e_{r,\bar s}- e_{s,\bar r}),&
\mathfrak{osp}(\gamma)^{-\ep_r-\ep_s}=\hbox{span}_\bbbf  (e_{\bar r,s}- e_{\bar s,r}),\\\\
\mathfrak{osp}(\gamma)^{\ep_r-\ep_s}=\hbox{span}_\bbbf  (e_{r,s}- e_{\bar s,\bar r)},&
\mathfrak{osp}(\gamma)^{\d_p+\d_q}=\hbox{span}_\bbbf (e_{p,\bar q}+ e_{p,\bar q}),\\\\
\mathfrak{osp}(\gamma)^{-\d_p-\d_q}=\hbox{span}_\bbbf (e_{\bar p,q}+ e_{\bar q,p}),&
\mathfrak{osp}(\gamma)^{\d_p-\d_q}=\hbox{span}_\bbbf (e_{p, q}- e_{\bar q,\bar p}),\\\\
\mathfrak{osp}(\gamma)^{\ep_r+\d_p}=\hbox{span}_\bbbf (e_{r,\bar p}- e_{p,\bar r}),&
\mathfrak{osp}(\gamma)^{-\ep_r-\d_p}=\hbox{span}_\bbbf (e_{\bar r,p}+ e_{\bar p,r}),\\\\
\mathfrak{osp}(\gamma)^{\ep_r-\d_p}=\hbox{span}_\bbbf (e_{r,p}+ e_{\bar p,\bar r}),&
\mathfrak{osp}(\gamma)^{-\ep_r+\d_p}=\hbox{span}_\bbbf (e_{\bar r,\bar p}- e_{p,r}),\\\\
\mathfrak{osp}(\gamma)^{2\d_p}=\hbox{span}_\bbbf e_{p,\bar p},&
\mathfrak{osp}(\gamma)^{-2\d_p}=\hbox{span}_\bbbf e_{\bar p,p},\\\\
\mathfrak{osp}(\mathfrak{u})^{\ep_r}=\hbox{span}_\bbbf  (e_{r,0}- e_{0,\bar r}),&
\mathfrak{osp}(\mathfrak{u})^{-\ep_r}=\hbox{span}_\bbbf  (e_{\bar r,0}- e_{0,r}),\\\\
\mathfrak{osp}(\mathfrak{u})^{\d_p}=\hbox{span}_\bbbf (e_{0,\bar p}- e_{p,0}),&
\mathfrak{osp}(\mathfrak{u})^{-\d_p}=\hbox{span}_\bbbf (e_{0,p}+ e_{\bar p,0}).
\end{array}$$
Define $$\fm:\mathfrak{osp}(\gamma)\times\mathfrak{osp}(\gamma)\longrightarrow \bbbf;\;\;\; (x,y)\mapsto str(xy)\;\;\;\;\;(x,y\in\mathfrak{osp}(\gamma)).$$
Then $\mathfrak{osp}(\gamma)$ is a locally finite basic classical Lie superalgebra whose root system is of type $X$ as in the following table:


\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$X$&$(|I|,|J|)$&$\gamma$&$X$&$(|I|,|J|)$&$\gamma$\\
\hline
$B(0,J)$& $(0,\geq1)$&$\mathfrak{u}$&$\dot C(0,J)$ & $(1,\geq2)$& $\mathfrak{v}$\\
\hline
$B(1,J)$&$(1,\geq 1)$&$\mathfrak{u}$&$D(2,1,\a)$ & $(2,1)$& $\mathfrak{v}$\\
\hline
$B(I,1)$ & $(\geq2,1)$&$\mathfrak{u}$&$D(2,J)$ & $(2,\geq2)$& $\mathfrak{v}$\\
\hline
$B(I,J)$&$(\geq 2,\geq2)$&$\mathfrak{u}$&$D(1,I)$ & $(\geq3,1)$&$\mathfrak{v}$\\
\hline
$\dot A(0,J)$&$(1,1)$&$\mathfrak{v}$&
$D(I,J)$ & $(\geq3,\geq2)$&$\mathfrak{v}$\\
\hline
\end{tabular}
\end{center}}
\end{exa}
\begin{exa}
{\rm Suppose that $I$ is a superset and take $$\gg:=\{X\in M_I(\bbbf)\mid str(X)=0\}.$$


}
\end{exa}
\begin{exa}
{\rm
Suppose that $\v=\v_{\bar 0}\op\v_{\bar 1}$ is a superspace. Set $I$ to be the ideal the subsuperalgebra $\gg:=\{T\in End(\v)\mid str(T)=0\}$ of the Lie superalgebra $End(\v)$ generated by the identity endomorphism if $\dim(\v_{\bar 0})=\dim(\v_{\bar 1})<\infty$ and take it to be the zero ideal otherwise, then $L/I$ is a locally finite basic classical Lie superalgebra of type $X$ as in the following table:
\begin{center}
\begin{tabular}{|c|l|}
\hline
$X$ & $(\dim(\v_{\bar 0}),\dim(\v_{\bar 1}))$\\
\hline
$\dot A(0,J)$& $(0,|J|)$\\
\hline
$ A(\ell,\ell)$ &$(\ell,\ell)\;\; (\ell<\infty)$\\
\hline
$\dot A(I,J)$ & $(|I|,|J|)\;\;(|I|\neq|J|\hbox{ if $|I|,|J|<\infty$})$\\
\hline
\end{tabular}
\end{center}
}
\end{exa}
\end{comment}

For a unital associative superalgebra $\aa$ and nonempty index supersets $I,J,$ by an $I\times J$-matrix
with entries in $\aa,$ we mean a map $A:I\times J\longrightarrow \aa.$ For $i\in I,j\in J,$ we set
$a_{ij}:=A(i,j)$ and call it the  {\it $(i,j)$-th entry} of $A.$ By a convention, we denote the matrix
$A$ by $(a_{ij}).$  We also denote the set of all $I\times J$-matrices  with entries in $\aa$ by
$\aa^{I\times J}.$  If $I=J,$ we denote $\aa^{I\times J}$ by $\aa^I.$  For $A=(a_{ij})\in\aa^{I\times J},$ the matrix
$B=(b_{ij})\in\aa^{J\times I}$ with  $$b_{ij}:=\left\{ \begin{array}{ll} a_{ji}& |i|=|j|\\
a_{ji}& |i|=1,|j|=0\\
-a_{ji}& |i|=0,|j|=1
\end{array}
\right.$$  is called the
{\it supertransposition} of $A$ and denoted by $A^{st}.$
 If $A=(a_{ij})\in\aa^{I\times J}$ and
$B=(b_{ij})\in \aa^{J\times K}$ are such that for all $i\in I$ and $k\in K,$ at most for finitely many $j\in J,$ $a_{ij}b_{jk}$'s are nonzero, we define the product $AB$ of $A$ and $B$ to be the
$I\times K$-matrix $C=(c_{ik})$ with $c_{ik}:=\sum_{j\in J}a_{ij}b_{jk}$ for all $i\in I,k\in K.$
We note that if $A,B,C$ are three matrices such that $AB,$ $(AB)C,$ $BC$ and $A(BC)$ are defined,
then $A(BC)=(AB)C.$ We make a convention that if $I$ is a disjoint union of subsets
$I_1,\ldots, I_t$ of $I,$ then for an $I\times I$-matrix $A,$ we write
$$A=\left [\begin{array}{ccc} A_{1,1}&\cdots& A_{1,t}\\
A_{2,1}&\cdots&A_{2,t}\\
\vdots&\vdots&\vdots\\
A_{t,1}&\cdots&A_{t,t}\\
\end{array}\right ]$$
in which for $1\leq r,s\leq t,$ $A_{r,s}$ is an $I_r\times I_s$-matrix whose $(i,j)$-th entry
coincides with $(i,j)$-th entry of $A$ for all $i\in I_r,j\in I_s.$ In this case, we say that $A\in \aa^{I_1\uplus\cdots\uplus I_t}$ and note that the defined matrix product  obeys the product of block matrices. If $\{a_i\mid i\in I\}\sub \aa,$ by
$\hbox{diag}(a_i),$ we mean an $I\times I$-matrix whose $(i,i)$-th entry is $a_i$ for all $i\in I$ and other
entries are zero. If $\aa$ is unital, we set $1_I:=\hbox{diag}(1_\aa).$ A matrix $A\in\aa^I$ is called {\it invertible} if there is a matrix $B\in\aa^I$ such that $AB$ as well as  $BA$ are defined and $AB=BA=1_I;$ such a $B$ is unique and denoted by $A^{-1}.$
 For $i\in I,j\in J$ and $a\in \aa,$ we define $E_{ij}(a)$ to be a matrix in $\aa^{I\times J}$ whose $(i,j)$-th entry is
$a$ and other entries are zero and if $\aa$ is unital, we set $$e_{i,j}:=E_{i,j}(1).$$  Take $M_{I\times J}(\aa)$ to be the subspace of $\aa^{I\times J}$ spanned by
$\{E_{ij}(a)\mid i\in I,j\in J,a\in A\}.$ $M_{I\times J}(\aa)$  is a superspace with $M_{I\times J}(\aa)_{\bar i}:=\hbox{span}_\bbbf\{E_{r,s}(a)\mid |r|+|s|+|a|=\bar i\},$ for $i=0,1.$ Also with respect to the  multiplication of matrices, the vector superspace
$M_{I\times I}(\aa)$ is an associative $\bbbf$-superalgebra and so is a Lie superalgebra under the Lie bracket
$[A,B]:=AB-(-1)^{|A||B|}BA$ for all $A,B\in M_{I\times I}(\aa).$ We denote  this Lie superalgebra  by
$\mathfrak{pl}_I(\mathcal{A}).$ We note that for $X,Y\in \mathfrak{pl}_I(\mathcal{A}),$ $(XY)^{st}=(-1)^{|X||Y|}Y^{st}X^{st}.$
 For an element $X\in \mathfrak{pl}_I(\mathcal{A}),$ we set $str(X):=\sum_{i\in I}(-1)^{|i|}x_{i,i}$ and call it the {\it supertrace} of $X.$

\begin{lem}\label{iso1}
(i) Suppose that $Q$ is a homogeneous element of $\bbbf^I,$ then $\gg_Q:=\{X\in \mathfrak{pl}_{I}(\bbbf)\mid X^{st}Q=-(-1)^{|X||Q|}QX\}$ is a Lie subsuperalgebra of $\mathfrak{pl}_{I}(\bbbf).$

(ii) If $Q_1,Q_2$ are homogeneous elements of $\bbbf^I$  and $T$ is an invertible homogeneous element of $\bbbf^I$ of degree zero such that   $Q_2=T^{st}Q_1 T,$ then $\gg_{Q_1}$ is isomorphic to $\gg_{Q_2}$ via the isomorphism mapping $X$ to $T^{-1}XT.$

(iii) Suppose that $I$ and $J$ are two supersets and $\eta:I\longrightarrow J$ is a bijection preserving the degree. For a matrix $A=(A_{ij})$ of $\bbbf^I,$ define $A^\eta\in\bbbf^J$ to be $(A^\eta_{ij})$ with $A^\eta_{ij}=A_{\eta^{-1}(i)\eta^{-1}(j)}.$ If $Q$ is a homogeneous element of $\bbbf^I$ and $Q':=Q^\eta,$ then the Lie superalgebra $\gg_Q:=\{X\in \mathfrak{pl}_I(\bbbf) \mid X^{st}Q=-(-1)^{|X||Q|}QX\}$ is isomorphic to the Lie superalgebra  $\gg_{Q'}:=\{X\in \mathfrak{pl}_J(\bbbf)\mid X^{st}Q'=-(-1)^{|X||Q'|}Q'X\}.$
\end{lem}

\begin{proof}
$(i),(ii)$ It is easy to verify.

$(iii)$ Suppose that  matrices $A,B\in\bbbf^I$ are  such that $AB$ is defined, then for $i,j\in I,$  we have
\begin{eqnarray*}
(A^\eta B^\eta)_{\eta(i)\eta(j)}=\sum_{t\in I}A^\eta_{\eta(i)\eta(t)}B^\eta_{\eta(t)\eta(j)}=\sum_{t\in I}A_{it} B_{tj}=(AB)_{ij}=(AB)^\eta_{\eta(i)\eta(j)}.
 \end{eqnarray*}
This in particular  implies that if $A,B,C,D\in\bbbf^I$ are such that $AB$ and $CD$ are defined and $AB=CD,$ then $A^\eta B^\eta=C^\eta D^\eta.$
 Moreover, as $\eta$ preserves the degree, we have $(A^{st})^\eta=(A^\eta)^{st}.$ Now it is easy to see that the function $\theta:\gg_Q\longrightarrow \gg_{Q'}$ mapping $X$ to $X^\eta$ is an isomorphism.
\end{proof}


\begin{exa}\label{exa1}
{\rm
For two disjoint  index sets  $I,J$ with $J\neq\emptyset,$ suppose that $\{0,i,\bar i,\mid i\in I\cup J\}$ is a superset
with $|0|=|i|=|\bar i|=0$ for $i\in I$ and $|j|=|\bar j|=1$ for $j\in J.$  We set $\dot I:=I\cup\bar I,$ $\dot I_0:=\{0\}\cup I\cup\bar I$ and $\dot J:=J\cup \bar J$ in which $$\bar I:=\{\bar i\mid i\in I\}\andd \bar J:=\{\bar j\mid j\in J\}.$$ For $\mathcal{I}=\dot I\cup\dot J$ or $\mathcal{I}=\dot I_0\cup\dot J,$ we set $$Q_\mathcal{I}:=\left(\begin{array}{cc}M_1&0\\
0&M_2\end{array}\right)$$ in which $$ M_2:=\sum_{j\in J}(e_{j,\bar j}-e_{\bar j, j})\andd M_1:=\left\{\begin{array}{ll}
-2e_{0,0}+\sum_{i\in I}(e_{i,\bar i}+e_{\bar i,i})& \hbox{if }  \hbox{\small $\mathcal{I}=\dot I_0\cup\dot J$}\\
\sum_{i\in I}(e_{i,\bar i}+e_{\bar i, i})& \hbox{if } \hbox{\small $I\neq\emptyset, \mathcal{I}=\dot I\cup\dot J.$}
\end{array}\right.$$
Now by Lemma \ref{iso1}, $$\gg_\i:=\gg_{Q_\i}=\{X\in \mathfrak{pl}_\i(\bbbf)\mid X^{st}Q_\i=-Q_\i X\}$$ is a Lie subsuperalgebra of $\mathfrak{pl}_\i(\bbbf)$ which  we refer to as  $\mathfrak{osp}(2I,2J)$ or $\mathfrak{osp}(2I+1,2J)$ if  $\i=\dot I\cup\dot J$ or  $\i=\dot I_0\cup\dot J$ respectively.
Set\begin{equation}\label{h}\fh:=\hbox{span}_\bbbf\{h_t,d_k\mid t\in I,\;k\in J\}\end{equation} in which for  $t\in I$ and $ k\in J,$ $$h_t:=e_{t,t}-e_{\bar t,\bar t}\andd d_k:=e_{k,k}-e_{\bar k,\bar k}$$ and for $i\in I$ and $j\in J,$ define
$$\begin{array}{lll}
\begin{array}{c}
\ep_i:\fh\longrightarrow \bbbf\\
h_t\mapsto \d_{i,t},\;\;\; d_k\mapsto 0,
\end{array}&&
\begin{array}{c}
\d_j:\fh\longrightarrow \bbbf\\
h_t\mapsto 0,\;\;\; d_k\mapsto \d_{j,k},
\end{array}
\end{array}$$  $(t\in I,k\in J).$ One sees that   $\gg_\i$ has a weight space decomposition with respect to $\fh.$
Taking $R(\i)$  to be the corresponding  set of weights,  we have
\begin{equation}\label{root}\begin{array}{l}R(\dot I_0\cup\dot J)=\{\pm\ep_r,\pm(\ep_r\pm\ep_s),\pm\d_p,\pm(\d_p\pm\d_q),\pm(\ep_r\pm\d_p)\mid  r, s\in I,\;p,q\in J,\;r\neq s\},\\
\\R(\dot I\cup\dot J)=\{\pm(\ep_r\pm\ep_s),\pm(\d_p\pm\d_q),\pm(\ep_r\pm\d_p)\mid  r, s\in I,\;p,q\in J,\;r\neq s\}\end{array}\end{equation} in which $\pm(\ep_r\pm\ep_s)$'s
 are disappeared if $|I|=1$ and $\pm\ep_r$'s, $\pm(\ep_r\pm\ep_s)$'s as well as $\pm(\ep_r\pm\d_p)$'s are disappeared if $|I|=0.$
 Moreover, for $r, s\in I,\;p,q\in J$  with $r\neq s$ and $p\neq q,$ we have
$$\begin{array}{ll}
(\gg_\i)^{\ep_r+\ep_s}=\hbox{span}_\bbbf  (e_{r,\bar s}- e_{s,\bar r}),&
(\gg_\i)^{-\ep_r-\ep_s}=\hbox{span}_\bbbf  (e_{\bar r,s}- e_{\bar s,r}),\\\\
(\gg_\i)^{\ep_r-\ep_s}=\hbox{span}_\bbbf  (e_{r,s}- e_{\bar s,\bar r)},&
(\gg_\i)^{\d_p+\d_q}=\hbox{span}_\bbbf (e_{p,\bar q}+ e_{p,\bar q}),\\\\
(\gg_\i)^{-\d_p-\d_q}=\hbox{span}_\bbbf (e_{\bar p,q}+ e_{\bar q,p}),&
(\gg_\i)^{\d_p-\d_q}=\hbox{span}_\bbbf (e_{p, q}- e_{\bar q,\bar p}),\\\\
(\gg_\i)^{\ep_r+\d_p}=\hbox{span}_\bbbf (e_{r,\bar p}+ e_{p,\bar r}),&
(\gg_\i)^{-\ep_r-\d_p}=\hbox{span}_\bbbf (e_{\bar r,p}- e_{\bar p,r}),\\\\
(\gg_\i)^{\ep_r-\d_p}=\hbox{span}_\bbbf (e_{r,p}- e_{\bar p,\bar r}),&
(\gg_\i)^{-\ep_r+\d_p}=\hbox{span}_\bbbf (e_{\bar r,\bar p}+ e_{p,r}),\\\\
(\gg_\i)^{2\d_p}=\hbox{span}_\bbbf e_{p,\bar p},&
(\gg_\i)^{-2\d_p}=\hbox{span}_\bbbf e_{\bar p,p},\\\\
\mathfrak{osp}(2I+1,2J)^{\ep_r}=\hbox{span}_\bbbf  ( e_{0,\bar r}+2e_{r,0}),&
\mathfrak{osp}(2I+1,2J)^{-\ep_r}=\hbox{span}_\bbbf  (e_{0,r}+2e_{\bar r,0}),\\\\
\mathfrak{osp}(2I+1,2J)^{\d_p}=\hbox{span}_\bbbf ( e_{0,\bar p}-2e_{ p,0}),&
\mathfrak{osp}(2I+1,2J)^{-\d_p}=\hbox{span}_\bbbf (e_{0,p}+2 e_{\bar p,0}).
\end{array}$$
Define $$\fm:\gg_{\i}\times\gg_\i\longrightarrow \bbbf;\;\;\; (x,y)\mapsto str(xy)\;\;\;\;\;(x,y\in\gg_\i).$$
Then $(\gg_\i,\fh,\fm)$ is a locally finite basic classical simple Lie superalgebra whose root system is an irreducible locally finite root supersystem  of type $X$ as in the following table:


\begin{center}
\begin{tabular}{|c|c|c||c|c|c|}
\hline
$X$&$(|I|,|J|)$&$\i$&$X$&$(|I|,|J|)$&$\i$\\
\hline
$B(0,J)$& $(0,\geq1)$&$\dot I_0\cup\dot J$&$\dot C(0,J)$ & $(1,\geq2)$& $\dot I\cup\dot J$\\
\hline
$B(1,J)$&$(1,\geq 1)$&$\dot I_0\cup\dot J$&$D(2,1,\a)$ & $(2,1)$& $\dot I\cup\dot J$\\
\hline
$B(I,1)$ & $(\geq2,1)$&$\dot I_0\cup\dot J$&$D(2,J)$ & $(2,\geq2)$& $\dot I\cup\dot J$\\
\hline
$B(I,J)$&$(\geq 2,\geq2)$&$\dot I_0\cup\dot J$&$D(1,I)$ & $(\geq3,1)$&$\dot I\cup\dot J$\\
\hline
$\dot A(0,2)$&$(1,1)$&$\dot I\cup\dot J$&
$D(I,J)$ & $(\geq3,\geq2)$&$\dot I\cup\dot J$\\
\hline
\end{tabular}
\end{center}
We refer to $\fh$ as the {\it standard Cartan subalgebra} of $\gg_\i.$ We note that $(\gg_\i)_{\bar 0}$ is centerless unless  $\i=\dot I\cup \dot J$ with  $| I|=1;$ see Lemma \ref{real-semisimple}. In this case, suppose $I=\{1\},$ then  for a  fixed index  $j\in J,$ $t_{\ep_1+\d_j}-(1/2)t_{2\d_j}$ is a nonzero central element of the even part of $\gg_\i.$
\hfill{$\diamondsuit$}}
\end{exa}

\begin{lem}
\label{cor2}
Suppose that $I,J$ are two nonempty  index sets with $|I|=\infty,$ then $\mathfrak{osp}(2I,2J)\simeq\mathfrak{osp}(2I+1,2J).$
\end{lem}

\begin{proof}
 Consider the following matrices of $\bbbf^{\{0\}\uplus I\uplus \bar I\uplus J\uplus\bar J}:$
{\scriptsize$$S:=\left (\begin{array}{ccccc}
1&0&0&0&0\\
0&I&I&0&0\\
0&I&-I&0&0\\
0&0&0&I&0\\
0&0&0&0&I
\end{array}\right), Q_e:=\left (\begin{array}{ccccc}
-2&0&0&0&0\\
0&2I&0&0&0\\
0&0&-2I&0&0\\
0&0&0&0&I\\
0&0&0&-I&0
\end{array}\right),Q:=\left (\begin{array}{ccccc}
-2&0&0&0&0\\
0&0&I&0&0\\
0&I&0&0&0\\
0&0&0&0&I\\
0&0&0&-I&0
\end{array}\right),$$}
then we  have $S^{st} Q S=Q_e.$ Also for matrices  {\scriptsize$$S':=\left (\begin{array}{cccc}
I&I&0&0\\
I&-I&0&0\\
0&0&I&0\\
0&0&0&I
\end{array}\right), Q_o:=\left (\begin{array}{cccc}
2I&0&0&0\\
0&-2I&0&0\\
0&0&0&I\\
0&0&-I&0
\end{array}\right),Q':=\left (\begin{array}{cccc}
0&I&0&0\\
I&0&0&0\\
0&0&0&I\\
0&0&-I&0
\end{array}\right)$$} of $\bbbf^{I\uplus \bar I\uplus J\uplus\bar J},$
we have   ${S'}^{st} Q' S'=Q_o.$ Now by Lemma \ref{iso1}, $\gg_{Q}\simeq\gg_{Q_e},$  $\gg_{Q'}\simeq\gg_{Q_o},$ and   $\gg_{Q_e}\simeq\gg_{Q_o}.$ This completes the proof.
\end{proof}




\begin{exa}\label{exa2}
{\rm
Suppose that $J$ is a superset  with $J_{\bar 0},J_{\bar 1}\neq\emptyset.$ Set $\gg:=\mathfrak{sl}(J_{\bar 0},J_{\bar 1})=\{X\in \mathfrak{pl}_J(\bbbf)\mid str(X)=0\}$ and  $\hh:=\hbox{span}_\bbbf\{e_{i,i}-e_{r,r},e_{j,j}-e_{s,s},e_{i,i}+e_{j,j}\mid i,r\in J_{\bar 0},j,s\in J_{\bar 1} \}.$
For $t\in J_{\bar 0}, k\in J_{\bar 1},$ define $$\begin{array}{ll}\ep_t:\hh\longrightarrow \bbbf,\\
\hbox{\small$e_{i,i}-e_{r,r}\mapsto \d_{i,t}-\d_{r,t},\;e_{j,j}-e_{s,s}\mapsto 0,e_{i,i}+e_{j,j}\mapsto \d_{i,t},$}\\
\\\d_k:\hh\longrightarrow \bbbf,\\
\hbox{\small$e_{i,i}-e_{r,r}\mapsto 0,\;e_{j,j}-e_{s,s}\mapsto -\d_{j,k}+\d_{k,s},e_{i,i}+e_{j,j}\mapsto -\d_{k,j},$}\end{array}
$$ ($i,r \in J_{\bar 0}, j,s\in J_{\bar 1}$). Also define $$\fm:\gg\times\gg\longrightarrow \bbbf;\;\; (X,Y)\mapsto str(XY).$$ If $|J|<\infty$ and $|J_{\bar 0}|=|J_{\bar 1}|,$ take  $K:=\bbbf \sum_{j\in J}e_{jj}$  and note that it is a subset of the radical of the form $\fm.$ So it induces a bilinear form on $\gg/K$ denoted again by $\fm.$  Set $$\mathfrak{sl}_s(J_{\bar0},J_{\bar1}):=\left\{\begin{array}{ll}
\gg/K&\hbox{if $|J|<\infty$ and $|J_{\bar 0}|=|J_{\bar 1}|$}\\
\gg& \hbox{otherwise}.
\end{array}\right.$$
Then $(\LL:=\mathfrak{sl}_s(J_{\bar0},J_{\bar1}),\fm,\hh/K)$ is a locally finite basic classical simple Lie superalgebra with root system $$\{\ep_i-\ep_j,\d_p-\d_q,\pm(\ep_i+\d_p)\mid i,j\in J_{\bar 0}, p,q\in J_{\bar 1}\}$$  which is an irreducible  locally finite root supersystem of type $X$ as in the following table:


\begin{center}
\begin{tabular}{|c|c|}
\hline
$X$&$(|J_{\bar 0}|,|J_{\bar 1}|)$\\
\hline
$\dot A(0,J_{\bar 1})$& $(1,\geq2)$\\
\hline
$\dot A(0,J_{\bar 0})$& $(\geq2,1)$\\
\hline
$\dot A(J_{\bar 0}, J_{\bar 1})$&
$(\geq2,\geq2)$\\
&
$|J_{\bar0}|\neq |J_{\bar 1}| \hbox{ if $J_{\bar0},J_{\bar 1}$ are both finite }$\\
\hline
$A(\ell,\ell)$ & $(\ell,\ell)$\;\;\; $(\ell\in\bbbz^{\geq1})$\\
\hline
\end{tabular}
\end{center}
Also if $(|J_{\bar 0}|,|J_{\bar 1}|)\neq (1,1),$ for $ i,j\in J_{\bar 0}$ and $ p,q\in J_{\bar 1}$ with $i\neq j$ and $p\neq q,$ we have
$$
\begin{array}{ll}
\LL^{\ep_i-\ep_j}=\bbbf e_{i,j},& \LL^{\d_p-\d_q}=\bbbf e_{q,p},\\
\LL^{\ep_i+\d_p}=\bbbf e_{i,p},& \LL^{-\ep_i-\d_p}=\bbbf e_{p,i}.
\end{array}
$$
We refer to $\hh/K$ as the {\it standard Cartan subalgebra} of $\LL=\mathfrak{sl}_s(J_{\bar0},J_{\bar1}).$ We now  need to discuss the center of $\LL_{\bar0}$ for our future purpose. We recall from finite dimensional theory of Lie superalgebras that  if $|J_{\bar 0}|, |J_{\bar 1}|<\infty,$ $\LL_{\bar 0}$ has nontrivial  center if and only if $|J_{\bar 0}|\neq |J_{\bar 1}|$ and that in this case, it has a one dimensional center.  Now suppose  $|J_{\bar 0}\cup J_{\bar 1}|=\infty,$ say $|J_{\bar 0}|=\infty.$ Fix $i_0\in J_{\bar 0}$ and $j_0\in J_{\bar 1}.$ Then $\{e_{i,i}-e_{i_0,i_0},e_{j,j}-e_{j_0,j_0},e_{i_0,i_0}+e_{j_0,j_0}\mid i\in J_{\bar 0}\setminus\{i_0\},j\in J_{\bar 1}\setminus\{j_0\}\}$ is  a basis for $\hh.$ Suppose $i_1,\ldots,i_\ell$ are distinct elements of $J_{\bar0}\setminus\{i_0\}$ and $j_1,\ldots,j_n$ are distinct elements of
$J_{\bar1}\setminus\{j_0\}.$ If $z=\sum_{t=1}^\ell r_t(e_{i_t,i_t}-e_{i_0,i_0})+\sum_{t=1}^n s_t(e_{j_t,j_t}-e_{j_0,j_0})+k(e_{i_0,i_0}+e_{j_0,j_0})$  (where $\sum_{t=1}^n s_t(e_{j_t,j_t}-e_{j_0,j_0})$ is disappeared if $|J_{\bar 1}|=1$)
 is an element of the center of $\LL_{\bar 0},$ then for each $i\in J_{\bar 0}\setminus\{i_0\},$  $[z,(\LL_{\bar 0})^{\ep_{i}-\ep_{i_0}}]=\{0\}.$ Now if $i=i_s$ for some  $s\in \{1,\ldots,\ell\},$ we get   $r_s+(\sum_{t=1}^\ell r_t)-k=0$ and if $i\not\in \{i_1,\ldots,i_\ell\},$ we get $\sum_{t=1}^\ell r_t-k=0.$ Therefore we have $r_s=0$ for all $s\in\{1,\ldots,\ell\}$ and so $k=0.$ This shows that $\LL$ is centerless  if $|J_{\bar 1}|=1.$ If  $|J_{\bar 1}|>1,$ $[z,\LL^{\ep_{i_0}+\d_{j_0}}]=\{0\}.$ This implies  that $\sum_{t=1}^n s_t=0 .$ We also have  $[z,\LL^{\d_{j}-\d_{j_0}}]=\{0\}$ for all $j\in J_{\bar 1}\setminus\{j_0\}.$ Now it follows that $s_t=0$ for all $t\in \{1,\ldots,n\}.$ This means that $z=0.$
\hfill{$\diamondsuit$}
}
\end{exa}

\begin{lem}\label{class1}
For  index sets  $I,J$ with $|J|\neq 0$ and a superset $T$ with $T_{\bar 0},T_{\bar 1}\neq\emptyset,$  set $$\mathfrak{a}_{I,J}:=\mathfrak{osp}(2I,2J) (\hbox{if $I\neq\emptyset$}), \mathfrak{b}_{I,J}:=\mathfrak{osp}(2I+1,2J),\mathfrak{c}_T:=\mathfrak{sl}_s(T_{\bar0},T_{\bar1}).$$
 Suppose that $\gg$ and $\LL$ are two Lie superalgebras of the class $$\{\mathfrak{a}_{I,J},\mathfrak{b}_{I,J},\mathfrak{c}_T\mid I,J,T\}.$$ Then $\gg$ and $\LL$ are isomorphic if and only if (up to changing the role of $\gg$ and $\LL$) one of the following holds:
 \begin{itemize}
\item there are index sets $I,J,I',J'$ with $|I|=|I'|\neq0,$ $|J|=|J'|\neq0,$ $\gg=\mathfrak{a}_{I,J}$ and $\LL=\mathfrak{a}_{I',J'},$
\item there are index sets $I,J,I',J'$ with   $|I|=|I'|,$ $|J|=|J'|\neq0,$ $\gg=\mathfrak{b}_{I,J}$ and $\LL=\mathfrak{b}_{I',J'},$
\item there are supersets $I,J$ with $|I_{\bar 0}|=|J_{\bar 0}|\neq0,$ $|I_{\bar 1}|=|J_{\bar 1}|\neq0,$ or $|I_{\bar 0}|=|J_{\bar 1}|\neq0,$ $|I_{\bar 1}|=|J_{\bar 0}|\neq0$ such that $\gg=\mathfrak{c}_{I}$ and $\LL=\mathfrak{c}_{J},$
\item there are   index sets $I,J$ with $|I|=|J|=1$ and a superset $T$ with $|T_{\bar0}|=1, |T_{\bar1}|=2$  or $|T_{\bar0}|=2, |T_{\bar1}|=1$ such that $\gg=\mathfrak{a}_{I,J}$ and $\LL=\mathfrak{c}_T,$
\item there are index sets $I,J$ with $J\neq\emptyset,|I|=\infty,$ $\gg=\mathfrak{a}_{I,J}$ and $\LL=\mathfrak{b}_{I,J}.$
 \end{itemize}
Moreover, in each of  the first three cases, the mentioned isomorphism can be chosen such that the standard Cartan subalgebra of $\gg$ is mapped  to the standard Cartan subalgebra of $\LL.$
\end{lem}
\begin{proof}
We first note that for two Lie algebras  $\mathfrak{k}_1$ and $\mathfrak{k}_2$  such that  $[\mathfrak{k}_1,\mathfrak{k}_1]$ and $[\mathfrak{k}_2,\mathfrak{k}_2]$ are semisimple with the complete  sets  of simple ideals $\{\mathfrak{k}_1^1,\ldots,\mathfrak{k}_1^n\}$ and $\{\mathfrak{k}_2^1,\ldots,\mathfrak{k}_2^m\}$ respectively,  if $\mathfrak{k}_1$ and $\mathfrak{k}_2$ are isomorphic, we have
\begin{equation}\label{fact}
\begin{array}{ll}
\bullet& \hbox{\small$[\mathfrak{k}_1,\mathfrak{k}_1]$ and $[\mathfrak{k}_2,\mathfrak{k}_2]$ are isomorphic,}\\
\bullet&\hbox{\small$\mathfrak{k}_1$ is centerless if and only if $\mathfrak{k}_2$ is centerless,}\\
\bullet& \hbox{\small$m=n$ and under a permutation of indices $\mathfrak{k}_1^i$ is isomorphic to $\mathfrak{k}_2^i$ for $i\in\{1,\ldots, n\}.$}
\end{array}
\end{equation}Now take $\aa$ to be one of the Lie superalgebras $\mathfrak{a}_{I,J},\mathfrak{b}_{I,J},\mathfrak{c}_T.$ We have already seen that if  $\aa$  is  infinite dimensional, then the  even part of $\aa$ is centerless if and only if $\aa\neq\mathfrak{a}_{I,J}$ for some infinite index set $J$ and an index set $I$ with $|I|=1.$
Next  suppose that $\gg$ and $\LL$ are as in the statement and assume they are  isomorphic, then  we have $\gg_{\bar 0}\simeq \LL_{\bar 0}.$ We also know that  $[\gg_{\bar 0},\gg_{\bar 0}]$ as well as $[\LL_{\bar 0},\LL_{\bar 0}]$ are semisimple Lie algebras by Theorem \ref{main1}. Using these together with (\ref{fact}),  Lemmas \ref{iso1} and \ref{cor2}, classification of basic classical simple Lie superalgebras  and \cite[Pro.'s VI4,VI6]{NS}, we are done.
\begin{comment}
Note that $B(T,1)$ and $F(4)$  have isomorphic even parts but they are not isomorphic as their dimensions are different. $\mathfrak{a}_{I,J}$ with $|I|=1,|J|=1$ is isomorphic to $\mathfrak{sl}(T_{\bar0},T_{\bar 1})$ with $|T_{\bar0}|=1,|T_{\bar 1}|=2$ as both are f.d.b.c.s Lie superalgebra with isomorphic even parts and the same dimension (due to the classification of f.d.b.c.s Lie superalgebras).
\end{comment}
\end{proof}


\section{Classification Theorem}
In this section, we classify locally finite basic classical simple Lie superalgebras and study the conjugacy classes of their Cartan subalgebras under the group of automorphisms. To this end, we need to know Chevalley bases for f.d.b.c.s. Lie superalgebras.
Chevalley bases for f.d.b.c.s Lie superalgebras  were introduced in 2001 by I. Kenji and  K. Yoshiyuki  \cite{KY} using  the fact that a  f.d.b.c.s Lie superalgebra is a contragredient  Lie superalgebra and its Cartan matrix is symmetrizable. We define Chevalley bases in a  somehow different manner from the one they have defined.
Throughout this section we assume the field  $\bbbf$ is  algebraically closed.

\begin{lem}\label{lin2}
Suppose that $(\gg_1,\fm_1,\hh_1),(\gg_2,\fm_2,\hh_2)$ are two locally finite  basic classical simple Lie superalgebras with  corresponding root systems $R_1,R_2$ respectively. For $i=1,2,$ denote the induced form on $\hbox{span}_\bbbf R_i\sub \hh_i^*$ again by $\fm_i.$ If  $R_1$ and $R_2$ are isomorphic, say via $f:\la R_1\ra\longrightarrow \la R_2\ra$ with $( f(\a), f(\a'))_2=k(\a,\a')_1$ for all $\a,\a'\in R$ and some $k\in \bbbf\setminus\{0\},$ then there is a linear isomorphism $\tilde f:\hbox{span}_\bbbf R_1\longrightarrow \hbox{span}_\bbbf R_2$ whose restriction to $\la R_1\ra$ coincides with  $f$ and for $\a,\a'\in \hbox{span}_\bbbf R_1,$ $(\tilde f(\a),\tilde f(\a'))_2=k(\a,\a')_1.$
\end{lem}
\begin{proof}
We know that $R_1$ is of real type if and only if for each nonsingular root $\d,$ there exists a nonzero integer $n$ with $n\d\in \la (R_1)_{re}\ra$ or equivalently $\hbox{span}_\bbbq (R_1)_{re}=\hbox{span}_\bbbq R_1.$ Now  fix a basis $\{\a_i\mid i\in I\}\sub (R_1)_{re}$ for $\hbox{span}_\bbbq (R_1)_{re}$ as well as a nonzero nonsingular root  $\d$ of $R_1$ if $R_1$ is of imaginary type. Set $$B:=\left\{\begin{array}{ll}
\{\a_i\mid i\in I\} & \hbox{if $R_1$ is of real type}\\
\{\d,\a_i\mid i\in I\} & \hbox{if $R_1 $ is of imaginary type.}
\end{array}\right.$$ Then by Lemma \ref{lin}, $B$ is $\bbbf$-linearly independent and so by Lemma \ref{super-sys}$(b)(iii)$, it is a basis for both $\hbox{span}_\bbbf R_1$ and $\hbox{span}_\bbbq R_1.$ Similarly, $f(B)$ is a basis for $\hbox{span}_\bbbf R_2.$ We define the linear transformation  $\tilde f$ mapping $\a\in B$ to $f(\a).$ It is immediate that $(\tilde f(\a),\tilde f(\a'))=k(\a,\a')$ for $\a,\a'\in \hbox{span}_\bbbf R_1.$ Now if $\a\in R_1\sub\hbox{span}_\bbbq B,$  $\a=\sum_{j=1}^n\frac{r_j}{s_j}\b_j$ where $r_j,s_j\in\bbbz$ and $\b_1,\ldots,\b_n\in B,$ so for $s=s_1\cdots s_n$ and $r'_j=r_js/s_j$ ($1\leq j\leq n$), we have $s\a=\sum_{j=1}^nr'_j\b_j.$ Therefore, we have $sf(\a)=\sum_{j=1}^nr'_jf(\b_j).$ Thus, we have $f(\a)=\tilde{f}(\a).$
\end{proof}
\subsection{Chevalley bases  for basic classical simple Lie superalgebras}
Suppose that  $\gg$ is a finite dimensional basic classical simple Lie superalgebra of type $X\neq A(1,1)$  with a Cartan subalgebra $\hh$ and  corresponding root system $R=R_0\cup R_1$ such that $\gg_{\bar 1}\neq\{0\}.$
In what follows for   $\a\in R^\times$ with $(\gg_{\bar i})^\a\neq \{0\}$ ($i\in\{0,1\}$),  we set  $|\a|:=\bar i.$ Now we want to define a total ordering on  $\v:=\hbox{span}_\bbbq R.$ We fix a basis $\{v_1,\ldots,v_m\}$ for $\v.$ For $u=r_1v_1+\cdots+r_mv_m\in\v,$ we say $0\prec u$ if $u\neq0$ and that the first nonzero $r_i,$ $1\leq i\leq m,$ is positive; next for $u,v\in\v,$ we say $u\prec v$ if $0\prec v-u.$ We set $R^+:=R\cap \{v\in\v\mid 0\prec v\}$ as well as $R^-:=-R^+.$ Elements of $R^+$ are called {\it positive} and elements of $R^-$ are called {\it negative}. As usual,  for $u,v\in\v,$ we say $u\preceq v$ if either $u=v$ or $u\prec v.$ Fix  an invariant  nondegenerate   even supersymmetric  bilinear  form $\fm$ on $\gg.$ We denote the induced  nondegenerate symmetric bilinear   form on $\hh^*$ again by $\fm.$   We recall that  for $\a\in\hh^*,$ $t_\a$ indicates  the unique element of $\hh$ representing $\a$ through the form $\fm.$ For  $\a\in \hh^*,$ set
 $$\sg_\a:=\left\{\begin{array}{ll}
  -1 & \a\in R_1\cap R^-\\
  1& \hbox{otherwise.}
\end{array}\right.$$
Next fix $r\in \bbbf\setminus\{0\}$  and for each $\a\in R^\times,$  set
$$ h_\a:=rt_\a.$$
One can see that
\begin{equation}
\label{identity}
\sg_{-\a}=(-1)^{|\a|}\sg_\a\andd h_\a=-h_{-\a}\;\;\;(\a\in R^\times).
\end{equation}
For $\a\in R^+,$ fix $Y_{\a}\in \gg^{\a}$ and $Y_{-\a}\in\gg^{-\a}$ such that $[Y_\a,Y_{-\a}]=h_\a.$ We note that for $\a\in R^\times,$ $[Y_\a,Y_{-\a}]=\sg_\a h_\a.$
\begin{deft}
{\rm
A set $\{X_\a,h_{i}\mid \a\in \rcross,i=1,\ldots,\ell\}$ is called a {\it Chevalley basis} for $\gg$ if
\begin{itemize}
\item there are a nonzero scalar $r$ and a   subset $\{\b_1,\ldots,\b_\ell\}$ of $R^\times$ such that $\{h_1:=h_{\b_1},\ldots,h_\ell:=h_{\b_\ell}\}$ is a basis for $\hh$ where for $\a\in R^\times,$ by $h_\a,$ we mean $rt_\a,$
\item for each $\a\in R^\times,$ $X_{\a}\in \gg^{\a},$
\item for each $\a\in R^\times,$ $[X_{\a},X_{-\a}]=\sg_\a h_\a.$
\end{itemize}
}\end{deft}


Suppose that $\{X_\a,h_{i}\mid \a\in \rcross,i=1,\ldots,\ell\}$ is a {\it Chevalley basis} for $\gg.$ We know from Lemma \ref{pro-nonzero} that  if $\a,\b\in \rcross$ such that $\a+\b\in \rcross,$ then $[\gg^\a,\gg^\b]\neq \{0\}.$ This together with the fact that $\dim(\gg^{\a+\b})=1$ implies that there is a nonzero scalar $N_{\a,\b}$ with $[X_\a,X_\b]=N_{\a,\b}X_{\a+\b};$ we also interpret $N_{\a,\b}$ as zero for $\a,\b\in R^\times$ with  $\a+\b\not\in R.$ We refer to $\{N_{\a,\b}\mid \a,\b\in R^\times\}$ as a set of {\it structure constants} for $\gg$ with respect to $\{X_\a,h_{i}\mid \a\in \rcross,i=1,\ldots,\ell\}$. Using a modified argument as in \cite[Pro. 7.1]{ca}, one can see the following proposition but for the convenience of readers, we give its proof in details; see also \cite[\S 3]{KY}.
\begin{Pro}\label{constant}
Keep the same notation as above;  we have the following:

(i) If $\a,\b\in \rcross,$ then  $N_{\a,\b}=-(-1)^{|\a||\b|}N_{\b,\a}.$


(ii)  If $\a,\b\in \rcross$ with $\a+\b\in \rcross,$ then for $s_{\a,\b}:=\sg_\a\sg_{\a+\b},$ we have   $$N_{\a,\b}=s_{\a,\b}N_{\b,-\a-\b}=\sg_\a\sg_{\a+\b}N_{\b,-\a-\b}.$$


(iii) Suppose that $\a,\b\in R^\times$ with  $\a+\b\in R^\times,$ then  $$N_{\a,\b}N_{-\a,-\b}=r_{\a,\b}:=\sg_{\b}\sg_{\b+\a}\sg_\a(-1)^{|\b||\a|}\sum_{i=0}^p(-1)^{i|\a|}(\b-i\a)(h_\a);$$ where $p=0$ if $\a,\b\in \rim$ and  otherwise, $p$ is the largest nonnegative integer such that  $\b-p\a\in R.$


(iv) If $\a,\b,\gamma,\d\in R^\times$ with $\a+\b+\gamma+\d=0$ such that each pair is not the opposite  of the one  another, then $$(-1)^{|\a||\gamma|}\sg_{\a+\b} N_{\a,\b}N_{\gamma,\d}+(-1)^{|\a||\b|}\sg_{\b+\gamma} N_{\b,\gamma}N_{\a,\d}+(-1)^{|\b||\gamma|}\sg_{\a+\gamma} N_{\gamma,\a}N_{\b,\d}=0.$$
\end{Pro}
\begin{proof}
$(i)$ We have $$N_{\a,\b}X_{\a+\b}=[X_\a,X_\b]=-(-1)^{|X_\a||X_\b|}[X_\b,X_\a]=-(-1)^{|\a||\b|}N_{\b,\a}X_{\a+\b}.$$ This completes the proof.

$(ii)$ Set $\gamma:=-\a-\b$  and consider  the Jacobi superidentity for $X_\a,X_\b,X_\gamma,$ then  we have
{\small
\begin{eqnarray*}
0&=&(-1)^{|\a||\gamma|}[[X_\a,X_\b],X_\gamma]+(-1)^{|\gamma||\b|}[[X_\gamma,X_\a],X_\b]+(-1)^{|\b||\a|}[[X_\b,X_\gamma],X_\a] \\
&=&(-1)^{|\a||\gamma|}[N_{\a,\b}X_{\a+\b},X_\gamma]+(-1)^{|\gamma||\b|}[N_{\gamma,\a}X_{\gamma+\a},X_\b]+(-1)^{|\b||\a|}[N_{\b,\gamma}X_{\b+\gamma},X_\a] \\
&=&(-1)^{|\a||\gamma|}N_{\a,\b}\sg_{-\gamma}h_{-\gamma}+(-1)^{|\gamma||\b|}N_{\gamma,\a}\sg_{-\b}h_{-\b}+(-1)^{|\b||\a|}N_{\b,\gamma}\sg_{-\a}h_{-\a}\\
&=&-(-1)^{|\a||\gamma|}N_{\a,\b}(-1)^{|\gamma|}\sg_{\gamma}h_{\gamma}-(-1)^{|\gamma||\b|}N_{\gamma,\a}(-1)^{|\b|}\sg_{\b}h_{\b}-(-1)^{|\b||\a|}N_{\b,\gamma}(-1)^{|\a|}\sg_{\a}h_{\a}\\
&=&(-1)^{|\a||\gamma|}N_{\a,\b}(-1)^{|\gamma|}\sg_{\gamma}(h_{\a}+h_\b)-(-1)^{|\gamma||\b|}N_{\gamma,\a}(-1)^{|\b|}\sg_{\b}h_{\b}-(-1)^{|\b||\a|}N_{\b,\gamma}(-1)^{|\a|}\sg_{\a}h_{\a}.
\end{eqnarray*}}
So as $|\a|+|\b|=|\gamma|,$ we have
{\small \begin{equation}
\label{y}
(-1)^{|\b||\gamma|}N_{\a,\b}\sg_{\gamma}(h_{\a}+h_\b)=(-1)^{|\a||\b|}N_{\gamma,\a}\sg_{\b}h_{\b}+(-1)^{|\gamma||\a|}N_{\b,\gamma}\sg_{\a}h_{\a}.
\end{equation}}
Now if $\a,\b$ are linearly independent,  $\{h_\a,h_\b\}$ is linearly independent and  so we get
\begin{eqnarray*}
(-1)^{|\b||\gamma|}N_{\a,\b}\sg_{\gamma}=(-1)^{|\a||\b|}N_{\gamma,\a}\sg_{\b}=(-1)^{|\gamma||\a|}N_{\b,\gamma}\sg_{\a}.
\end{eqnarray*}
Therefore, as $|\a|+|\b|=|\gamma|,$ we have $$N_{\a,\b}\sg_{\gamma}=(-1)^{|\gamma|}\sg_{\a}N_{\b,\gamma}.$$ If $\a,\b$ are linearly dependent, then  using \cite[Lem. 2.1($ii$)]{you6} and \cite[Pro. 3.8($i$)]{you3}, one of the following cases can happen: $\a=\pm\b,$ $\a=\pm 2\b$ or $\b=\pm 2\a$ but since $\a+\b\in R\setminus\{0\},$ we just have $\a=\b,$ $\b=-2\a$ or $\a=-2\b.$
In the first case, since $2\a=\a+\b\in R\setminus\{0\},$ we have  $|\a|=|\b|=1$ and $|\gamma|=0.$  So  (\ref{y}) together with part $(i)$ implies that
$$N_{\a,\b}\sg_{\gamma}2h_\b=-N_{\gamma,\a}\sg_{\b}h_{\b}+N_{\b,\gamma}\sg_{\a}h_{\b}=2N_{\b,\gamma}\sg_{\b}h_{\b}$$
so we get $$N_{\a,\b}\sg_{\gamma}=(-1)^{|\gamma|}\sg_{\a}N_{\b,\gamma}.$$
In  the second case, $|\a|=|\gamma|=1$ and $|\b|=0$ and so the result is  immediate using part $(i).$
In the last case,  $|\b|=|\gamma|=1$ and $|\a|=0.$ Therefore  (\ref{y}) together with part $(i)$ implies that
$$-2N_{\gamma,\a}\sg_{\gamma}=-N_{\gamma,\a}\sg_{\gamma}-N_{\gamma,\a}\sg_{\b}=N_{\a,\b}\sg_{\gamma}-N_{\gamma,\a}\sg_{\b}=-2N_{\b,\gamma}\sg_{\a}.$$
This completes the proof.

$(iii)$ We first assume $\a,\b\in \rre.$ Let $0\leq i\leq p$ and set $\b':=\b-i\a.$ We make a convention that
$$\parbox{4.2in}{
for $i=p,$ we set $N_{\a-\b',-\a}=N_{-\a,\a-\b'}=N_{\a,-\a+\b'}=N_{\a-\b',-\a}:=0$
}
$$ and consider the Jacobi superidentity for $X_\a,X_{-\a},X_{\b'}.$ Then we have
 \begin{eqnarray*}
0&=&[[X_\a,X_{-\a}],X_{\b'}]-[X_\a,[X_{-\a},X_{\b'}]]+(-1)^{|\a|}[X_{-\a},[X_\a,X_{\b'}]]\\
&=&[\sg_\a h_\a,X_{\b'}]-N_{-\a,{\b'}}N_{\a,-\a+{\b'}}X_{\b'}+(-1)^{|\a|}N_{-\a,\a+{\b'}}N_{\a,{\b'}}X_{\b'}.
\end{eqnarray*}
This implies that $$\sg_\a{\b'}(h_\a)=N_{-\a,{\b'}}N_{\a,-\a+{\b'}}-(-1)^{|\a|}N_{-\a,\a+{\b'}}N_{\a,{\b'}}.$$
Using part ($ii$) respectively for $(-{\b'},-\a)$ and $(\a-{\b'},-\a),$  we get $$\sg_{{\b'}+\a} N_{-{\b'},-\a}=(-1)^{|{\b'}+\a|}\sg_{-{\b'}} N_{-\a,\a+{\b'}}\andd \sg_{\b'} N_{\a-{\b'},-\a}=(-1)^{|{\b'}|}\sg_{\a-{\b'}} N_{-\a,{\b'}}.$$

Thus we get  \begin{eqnarray*}N_{\a,{\b'}} N_{-\a,\a+{\b'}}&=&(-1)^{|{\b'}+\a|}\sg_{-{\b'}}\sg_{{\b'}+\a}N_{\a,{\b'}} N_{-{\b'},-\a}\\
&=&-(-1)^{|\a|+|{\b'}|}(-1)^{|{\b'}|}(-1)^{|\a||{\b'}|}\sg_{{\b'}}\sg_{{\b'}+\a}N_{\a,{\b'}} N_{-\a,-{\b'}}\\
&=&-(-1)^{|\a|}(-1)^{|\a||{\b'}|}\sg_{{\b'}}\sg_{{\b'}+\a}N_{\a,{\b'}} N_{-\a,-{\b'}}
\end{eqnarray*}
and
\begin{eqnarray*}
N_{-\a,{\b'}} N_{\a,-\a+{\b'}}&=& (-1)^{|{\b'}|}\sg_{\a-{\b'}} \sg_{\b'} N_{\a-{\b'},-\a}N_{\a,-\a+{\b'}}\\
&=& -(-1)^{|{\b'}||\a|}\sg_{\b'}\sg_{{\b'}-\a} N_{\a,-\a+{\b'}}N_{-\a,\a-{\b'}}.
\end{eqnarray*}
So we have
\begin{eqnarray*}
(-1)^{|\a||{\b'}|}\sg_\a{\b'}(h_\a)&=&(-1)^{|\a||{\b'}|}N_{-\a,{\b'}}N_{\a,-\a+{\b'}}-(-1)^{|\a||{\b'}|}(-1)^{|\a|}N_{-\a,\a+{\b'}}N_{\a,{\b'}}\\
&=&\sg_{{\b'}}\sg_{{\b'}+\a} N_{\a,{\b'}}N_{-\a,-{\b'}}-\sg_{\b'}\sg_{{\b'}-\a} N_{\a,-\a+{\b'}}N_{-\a,\a-{\b'}}.
\end{eqnarray*}
%
%\begin{eqnarray*}
%(-1)^{|\a||\b-i\a|}\sg_\a(\b-i\a)(h_\a)&=&\sg_{\b-i\a}\sg_{(\b-i\a)+\a} N_{\a,(\b-i\a)}N_{-\a,-(\b-i\a)}\\
%&-&\sg_{\b-i\a}\sg_{(\b-i\a)-\a} N_{\a,-\a+(\b-i\a)}N_{-\a,\a-(\b-i\a)}\\
%(-1)^{|\a||\b-(i+1)\a|}\sg_\a(\b-(i+1)\a)(h_\a)&=&\sg_{\b-(i+1)\a}\sg_{(\b-(i+1)\a)+\a} N_{\a,(\b-(i+1)\a)}N_{-\a,-(\b-(i+1)\a)}\\
%&-&\sg_{\b-(i+1)\a}\sg_{(\b-(i+1)\a)-\a} N_{\a,\b-i\a-2\a}N_{-\a,-\b+i\a+2\a}\\
%&=&\sg_{\b-(i+1)\a}\sg_{\b-i\a} N_{\a,(\b-(i+1)\a)}N_{-\a,-(\b-(i+1)\a)}\\
%&-&\sg_{\b-(i+1)\a}\sg_{\b-i\a-2\a} N_{\a,\b-i\a-2\a}N_{-\a,-\b+i\a+2\a}.
%\end{eqnarray*}
%

Therefore, we have   \begin{eqnarray*}
\sg_\a\sum_{i=0}^p(-1)^{|\b-i\a||\a|}(\b-i\a)(h_\a)&=&\sum_{i=0}^p\sg_{\b-i\a}\sg_{\b-(i-1)\a} N_{\a,\b-i\a}N_{-\a,-(\b-i\a)}\\
&-&\sum_{i=0}^p\sg_{\b-i\a}\sg_{\b-(i+1)\a} N_{\a,\b-(i+1)\a}N_{-\a,-\b+(i+1)\a}\\
&=&\sg_{\b}\sg_{\b+\a} N_{\a,\b}N_{-\a,-\b}\\
&+&\sum_{i=1}^p\sg_{\b-i\a}\sg_{\b-(i-1)\a} N_{\a,\b-i\a}N_{-\a,-(\b-i\a)}\\
&-&\sum_{i=0}^{p-1}\sg_{\b-i\a}\sg_{\b-(i+1)\a} N_{\a,\b-(i+1)\a}N_{-\a,-\b+(i+1)\a}\\
&=&\sg_{\b}\sg_{\b+\a} N_{\a,\b}N_{-\a,-\b}\\
&+&\sum_{i=0}^{p-1}\sg_{\b-i\a}\sg_{\b-(i+1)\a} N_{\a,\b-(i+1)\a}N_{-\a,-\b+(i+1)\a}\\
&-&\sum_{i=0}^{p-1}\sg_{\b-i\a}\sg_{\b-(i+1)\a} N_{\a,\b-(i+1)\a}N_{-\a,-\b+(i+1)\a}\\
&=&\sg_{\b}\sg_{\b+\a} N_{\a,\b}N_{-\a,-\b}.
\end{eqnarray*}
This means that  \begin{eqnarray*}
 N_{\a,\b}N_{-\a,-\b}&=&\sg_{\b}\sg_{\b+\a}\sg_\a(-1)^{|\b||\a|}\sum_{i=0}^p(-1)^{i|\a|}(\b-i\a)(h_\a)
\end{eqnarray*}
as we desired. We next  assume $\a,\b\in\rim,$ then since $\a+\b\in R,$ we have $\a-\b\not\in R;$ see Proposition \ref{dim1-im}. Then the  Jacobi superidentity  for $X_\a,X_{-\a},X_\b$ together with parts $(i)$ , $(ii)$ turns into
\begin{eqnarray*}
0&=&[\sg_\a h_\a,X_\b]+(-1)^{|\a|}N_{-\a,\a+\b}N_{\a,\b}X_\b\\
&=&\sg_\a \b(h_\a)X_\b+(-1)^{|\a|}N_{-\a,\a+\b}N_{\a,\b}X_\b\\
&=&\sg_\a \b(h_\a)X_\b+(-1)^{|\a|}(-1)^{|\b+\a|}\sg_{\b+\a}\sg_{-\b} N_{-\b,-\a}N_{\a,\b}X_\b\\
&=&\sg_\a \b(h_\a)X_\b+(-1)^{|\a|}(-1)^{|\b|}(-1)^{|\b+\a|}\sg_{\b+\a}\sg_{\b} N_{-\b,-\a}N_{\a,\b}X_\b\\
&=&\sg_\a \b(h_\a)X_\b-(-1)^{|\a|}(-1)^{|\b|}(-1)^{|\b+\a|}(-1)^{|\a||\b|}\sg_{\b+\a}\sg_{\b} N_{-\a,-\b}N_{\a,\b}X_\b.
\end{eqnarray*}
This implies that $$N_{-\a,-\b}N_{\a,\b}=(-1)^{|\a||\b|}\sg_\a\sg_{\b+\a}\sg_{\b} \b(h_\a)$$ and so we are done.



$(iv)$  In what follows if $\eta_1,\eta_2\in\hh^*$ are such that either $\eta_1$ or $\eta_2$ is not a nonzero root, we define $N_{\eta_1,\eta_2}$ to be zero. Suppose that $\a,\b,\gamma,\eta$ are as in the statement.  Considering the Jacobi superidentity for $X_\a,X_\b$ and $X_\gamma,$ we have
%To have the following, we need to have the assumption that   each pair is not the opposite of one another.
\begin{eqnarray*}
N_{\a,\b}N_{\a+\b,\gamma}X_{\a+\b+\gamma}&=&[[X_\a,X_\b],X_{\gamma}]\\
&=&[X_\a,[X_\b,X_\gamma]]-(-1)^{|\a||\b|}[X_\b,[X_\a,X_\gamma]]\\
&=&N_{\b,\gamma}N_{\a,\b+\gamma}X_{\a+\b+\gamma}-(-1)^{|\a||\b|}N_{\a,\gamma}N_{\b,\a+\gamma}X_{\a+\b+\gamma}.
\end{eqnarray*}

Since $(\a+\b)+\gamma+\d=0,$ part ($ii$) implies that
$$N_{\a+\b,\gamma}=(-1)^{|\d|}\sg_{\d}\sg_{\a+\b}N_{\gamma,\d}.$$ Note that if $\a+\b$ is not a root, the above equality is nothing but the natural identity $0=0.$
Similarly, we have $$N_{\b+\gamma,\a}=(-1)^\d\sg_\d\sg_{\b+\gamma}N_{\a,\d}\andd N_{\a+\gamma,\b}=(-1)^\d\sg_\d\sg_{\a+\gamma}N_{\b,\d}.$$
Therefore, we have
\begin{eqnarray*}
(-1)^{|\d|}\sg_{\d}\sg_{\a+\b}N_{\a,\b}N_{\gamma,\d}&=&N_{\a,\b}N_{\a+\b,\gamma}\\
&=&N_{\b,\gamma}N_{\a,\b+\gamma}-(-1)^{|\a||\b|}N_{\a,\gamma}N_{\b,\a+\gamma}\\
&=&-(-1)^{|\d|}(-1)^{|\a||\b|+|\a||\gamma|}\sg_\d\sg_{\b+\gamma}N_{\b,\gamma}N_{\a,\d}\\
&+&(-1)^{|\a||\b|}(-1)^{|\d|}(-1)^{|\a||\b|+|\gamma||\b|}\sg_\d\sg_{\a+\gamma}N_{\a,\gamma}N_{\b,\d}
\end{eqnarray*}
and so we get $$\sg_{\a+\b}N_{\a,\b}N_{\gamma,\d}
=-(-1)^{|\a||\b|+|\a||\gamma|}\sg_{\b+\gamma}N_{\b,\gamma}N_{\a,\d}+(-1)^{|\gamma||\b|}\sg_{\a+\gamma}N_{\a,\gamma}N_{\b,\d}.
$$ Thus we have $$(-1)^{|\a||\gamma|}\sg_{\a+\b}N_{\a,\b}N_{\gamma,\d}
+(-1)^{|\a||\b|}\sg_{\b+\gamma}N_{\b,\gamma}N_{\a,\d}+(-1)^{|\gamma||\b|}\sg_{\a+\gamma}N_{\gamma,\a}N_{\b,\d}=0.$$ This completes the proof.
\end{proof}

%\begin{proof}
%One can use a modified argument as in \cite[Pro. 7.1]{Ca} to get the result. However, the modification is not straightforward, to make the paper shorter, we omit the proof and refer the reader to the version of the present paper posted in ArXiv \cite{}
%\end{proof}

\medskip
We  know  that there are  roots $\a,\gamma$ such that $\a\neq\pm\gamma$ and  $(\a,\gamma)\neq0.$ So either $\a+\gamma\in \rcross$ or $\a-\gamma\in \rcross.$ Replacing $\gamma$ with $-\gamma$ if necessary, we assume $\eta:=-(\a+\gamma)\in\rcross.$ Since $\a+\gamma+\eta=0,$   either two of $\a,\gamma,\eta$ are positive or two of $-\a,-\gamma,-\eta$ are positive.
Selecting this pair of positive roots in an appropriate order, we get a  pair $(\eta_1,\eta_2)$ among the 12 pairs
$$\begin{array}{ll}(\a,\gamma),(\a,\eta),(\gamma,\eta),(\gamma,\a),(\eta,\a),(\eta,\gamma),\\
(-\a,-\gamma),(-\a,-\eta),(-\gamma,-\eta),(-\gamma,-\a),(-\eta,-\a),(-\eta,-\gamma)
\end{array}$$
such that $0\prec \eta_1\preceq \eta_2;$  following \cite{ca}, we call such a pair a {\it special pair}. More precisely, a pair $(\a,\b)$ of elements of $R^\times$ is called a {\it special pair} if $0\prec\a\preceq\b$ and $\a+\b\in R.$  A special pair $(\a,\b)$  is called an {\it extraspecial pair} if for each  special pair $(\d,\gamma)$ with $\a+\b=\d+\gamma,$ we get  $\a\preceq\d.$
\begin{comment}
*******
Take $\aa$ to be the set of all extraspecial pairs $(\a,\b)$ of $R^\times$  and fix a subset $ \bb:=\{N_{\a,\b}\mid (\a,\b)\in\aa\}$ of arbitrary nonzero scalars. In what follows we show that for appropriate choices of root vectors, the corresponding set of structure  constants  is uniquely determined in  terms of
 the elements of $\bb.$

******
\end{comment}
\begin{lem}\label{arbitrary}
Suppose that $\aa$ is the set of all extraspecial pairs  $(\a,\b)$ of $R^\times$  and $\{N_{\a,\b}\mid (\a,\b)\in \aa\}$ is an arbitrary set of nonzero scalars. Then there is $\{e_\a\in\gg^\a\setminus\{0\}\mid \a \in R^+\}$ such that $[e_\a,e_\b]=N_{\a,\b}e_{\a+\b}$ for all $(\a,\b)\in \aa.$
\end{lem}
\begin{proof}
Suppose that $R^+=\{\a_1,\ldots,\a_n\}$ with $\a_1\prec\ldots\prec\a_n$ and take  $t$ to be   the smallest index such that $\a_t$ is the summation of the components of an extraspecial pair. We choose  arbitrary elements  $e_{\a_i}\in \gg^{\a_i},$ for $1\leq i\leq t-1.$ We know that there is a unique extraspecial pair $(\a,\b)$ with $\a_t=\a+\b,$ so there is a unique pair $(i,j)$ with $i\leq j<t$ such that $\a_t=\a_i+\a_j$ and define $e_{\a_t}=N_{\a_i,\a_j}^{-1}[e_{\a_i},e_{\a_j}].$ Now using an induction process, we can complete the proof; indeed, suppose that $t<r\leq n$  and that $\{e_{\a_s}\mid 1\leq s\leq r-1\}$  with the desired property has been chosen. If $\a_r$ is not the summation of the components of an extraspecial pair, we choose $e_{\a_r}$ arbitrary, but otherwise we pick the unique pair $(i',j')$ with $i'\leq j'<r-1$ such that $\a_r=\a_{i'}+\a_{j'}.$ Now we define  $e_{\a_r}=N_{\a_{i'},\a_{j'}}^{-1}[e_{\a_{i'}},e_{\a_{j'}}].$ This completes the proof.
\end{proof}

\begin{Thm}\label{com-cheval}
Suppose that $\gg$ and $\LL$ are two finite dimensional  basic classical simple Lie superalgebras with Cartan subalgebras $\hh$ and $T$ and corresponding root systems $R=R_0\cup R_1$ and $S=S_0\cup S_1$ respectively which are not of type $A(1,1).$  Suppose that $\fm$ (resp. $\fm'$) is an invariant  nondegenerate   even supersymmetric bilinear form on $\gg$ (resp. $\LL$) and denote the induced forms on $\hh^*$ and $T^*$ again by $\fm$ and $\fm'$ respectively. Suppose that $(\la R\ra ,\fm,R)$ and $(\la S\ra,\fm',S)$ are isomorphic finite root supersystems, say via $f:\la R\ra\longrightarrow \la S\ra.$ Then we have the following:

(i) There are Chevalley bases $\{h_i, e_\a\mid\a\in \rcross,1\leq i\leq \ell\}$ and $\{t_i,x_{\b}\mid \b\in S^\times, 1\leq i\leq \ell\}$ for $\gg$ and $\LL$ with corresponding sets of structure constants $\{N_{\a,\b}\mid \a,\b\in \rcross\}$ and $\{M_{\gamma,\eta}\mid \gamma,\eta\in S^\times\}$ respectively  such that $N_{\a,\b}=M_{f(\a),f(\b)}$ for all $\a,\b\in \rcross.$

(ii) $\{N_{\a,\b}\mid \a,\b\in \rcross\}$ is completely  determined in terms of $N_{\a,\b}$'s for extraspecial pairs $(\a,\b).$

(iii) There is an isomorphism from $\gg$ to $\LL$ mapping $\hh$ to $\T$ and  $e_\a$ to $x_{f(\a)}$ for all $\a\in R\setminus\{0\}.$
\end{Thm}
\begin{proof}
$(i),(ii)$ Suppose that  $k\in\bbbf\setminus\{0\}$ is such that $(f(\a),f(\b))'=k(\a,\b)$ for $\a,\b\in R.$ Fix $r,s\in\bbbf\setminus\{0\}$ such that $r=sk.$ This implies that $r(\a,\b)=sk(\a,\b)=s(f(\a),f(\b))'$ for all $\a,\b\in R.$ Use Lemma \ref{lin2} to extend the map $f$ to a linear isomorphism, denoted again by $f,$ from $\hh^*=\hbox{span}_\bbbf R$ to $T^*=\hbox{span}_\bbbf S$ with
\begin{equation}
\label{compatible}
r(\a,\b)=sk(\a,\b)=s(f(\a),f(\b))'\;\;\;\;\; (\a,\b\in \hh^*).
\end{equation}
For $\a\in \hh^*,$ take $t_\a$ to be the unique element of $\hh$ representing $\a$ through $\fm$ and for $\b\in T^*,$ take $t'_\b$ to be the unique element of $T$ representing $\b$ through $\fm'.$ Next set
\begin{equation}
\label{h alpha}
h_\a:=rt_\a\andd h'_\b:=st'_\b\;\;\;\;(\a\in R,\;\; \b\in S).
\end{equation}
Fix a total  ordering $``\preceq"$ on $\hbox{span}_\bbbq R$ as at the beginning of  this subsection and transfer it through $f$ to a total  ordering, denoted again by  $``\preceq",$ on $\hbox{span}_\bbbq S.$ For $\a\in \hh^*$ and $\b\in T^*,$ set
\begin{equation}
\label{sigma}
\sg_\a:=\left\{\begin{array}{ll}
-1& \hbox{if $\a\in R^-\cap R_1$}\\
1& \hbox{otherwise}
\end{array}\right.\andd \sg'_\b:=\left\{\begin{array}{ll}
-1& \hbox{if $\b\in S^-\cap S_1$}\\
1& \hbox{otherwise.}
\end{array}\right.
\end{equation}
Suppose that $\aa$ is the set of all extraspecial pairs of $R,$ then
{\small$$\{(f(\a),f(\b))\mid (\a,\b)\in \aa\}=\{(\eta,\gamma)\mid (\eta,\gamma)\hbox{ is an extraspecial pair of $S$}  \}.$$}  Fix  a subset $\{N_{\a,\b}\mid (\a,\b)\in\aa\}$ of nonzero scalars and set $M_{f(\a),f(\b)}:=N_{\a,\b},$ for all $(\a,\b)\in\aa.$ Using Lemma \ref{arbitrary}, one can find $\{e_\a\in\gg^\a\setminus\{0\}\mid\a\in R^+\}$ and $\{x_{\b}\in\LL^\b\setminus\{0\}\mid \b\in S^+\}$ such that $$[e_\a,e_\b]= N_{\a,\b}e_{\a+\b}\andd [x_{f(\a)},x_{f(\b)}]=M_{f(\a),f(\b)}x_{f(\a)+f(\b)};\;\;\;(\a,\b)\in\aa.$$
Now for each $\a\in R^+$ and $\gamma\in S^+,$ choose $e_{-\a}\in\gg^{-\a}$ and $x_{-\gamma}\in \LL^{-\gamma}$ such that $$[e_\a,e_{-\a}]= h_\a\andd [x_\gamma,x_{-\gamma}]= h'_\gamma\;\;\;\;\; (\a\in R^+,\gamma\in S^+)$$ and note that we have
$$
[e_\a,e_{-\a}]=\sg_\a h_\a\andd [x_\gamma,x_{-\gamma}]=\sg'_\gamma h'_\gamma\;\;\;\;\; (\a\in R^\times,\gamma\in S^\times).
$$

Now for each pair $(\a,\b)$ of $\rcross$ with $\a+\b\in \rcross$ and $(\a,\b)\not \in \aa,$ take $N_{\a,\b}$ to be the   unique  nonzero element of $\bbbf$ with $[e_\a,e_\b]=N_{\a,\b}e_{\a+\b};$ also for each pair $(\gamma,\eta)$ of $S^\times$ with $\gamma+\eta\in S^\times$ such that $(\gamma,\eta)$ is not an extraspecial pair, take  $M_{\gamma,\eta}$ to be  the unique nonzero element of $\bbbf$ with $[x_\gamma,x_\eta]=M_{\gamma,\eta}e_{\gamma+\eta}.$
Fix $\{\b_1,\ldots,\b_\ell\}$ such that $\{h_i:=h_{\b_i}\mid 1\leq i\leq \ell\}$ is a basis for $\hh$ and set $t_i:=h'_{f(\b_i)}.$ Then $\{h_i, e_\a\mid\a\in \rcross,1\leq i\leq \ell\}$ and $\{t_i,x_{\b}\mid \b\in S^\times, 1\leq i\leq \ell\}$ are Chevalley bases  for $\gg$ and $\LL$ respectively.
We   complete the proof through the following two steps.
As usual we also set $N_{\a,\b}:=0$ and $M_{\gamma,\eta}:=0$ if $\a,\b\in\rcross$  and $\gamma,\eta\in S^\times$ with $\a+\b\not\in \rcross$ and $\gamma+\eta\not\in S^\times.$ By a conventional notation in these cases, we set $N_{\zeta,\a+\b}:=0$ and $M_{\xi,\eta+\gamma}:=0$ for $\zeta\in R^\times$ and $\xi\in S^\times.$



\underline{ Step 1.} For a special pair $(\a,\b),$ we have  $N_{\a,\b}=M_{f(\a),f(\b)}$  and  $N_{\a,\b}$ is  determined in terms of $N_{\a',\b'}$'s for extraspecial pairs $(\a',\b'):$ Suppose that $(\a,\b)$ is a special pair which is not an extraspecial pair. So there is a unique extraspecial pair $(\gamma,\d)$ with $\a+\b=\gamma+\d .$ We  have $\a+\b+(-\gamma)+(-\d)=0.$ Therefore by Proposition \ref{constant}, we have {\small $$(-1)^{|\a||-\gamma|}\sg_{\a+\b} N_{\a,\b}N_{-\gamma,-\d}+(-1)^{|\a||\b|}\sg_{\b-\gamma} N_{\b,-\gamma}N_{\a,-\d}+(-1)^{|\b||-\gamma|}\sg_{\a-\gamma} N_{-\gamma,\a}N_{\b,-\d}=0.$$ } So,
{\small \begin{eqnarray}
 N_{\a,\b}
 %&=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} N_{-\gamma,-\d}^{-1} N_{\b,-\gamma}N_{\a,-\d}\\
% &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}N_{-\gamma,-\d}^{-1} N_{-\gamma,\a}N_{\b,-\d}\\
% &=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} N_{-\gamma,\gamma-\b}(-(-1)^{|\a||\d|})s_{-\d,\a}N_{\a,\d-\a}\\
% &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} (-(-1)^{|\a||\gamma|})s_{\a,-\gamma}N_{-\gamma,\gamma-\a}(-(-1)^{|\b||\d|})s_{-\d,\b}N_{\b,\d-\b}\\
% &=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} r^{-1}_{-\gamma,\gamma-\b}N_{\gamma,\b-\gamma}(-(-1)^{|\a||\d|})s_{-\d,\a}N_{\a,\d-\a}\\
% &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} (-(-1)^{|\a||\gamma|})s_{\a,-\gamma}r^{-1}_{-\gamma,\gamma-\a}N_{\gamma,\a-\gamma}(-(-1)^{|\b||\d|})s_{-\d,\b}N_{\b,\d-\b}\\
% &=&-(-1)^{|\a||\b|+|\a||\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} N_{-\gamma,-\d}^{-1} N_{\b,-\gamma}N_{\a,-\d}\\
% &-&(-1)^{|\b||\gamma|+|\a||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}N_{-\gamma,-\d}^{-1} N_{-\gamma,\a}N_{\b,-\d}\\
% &=&(-1)^{|\a||\d|+|\a||\b|+|\a||\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} N_{-\gamma,\gamma-\b}s_{-\d,\a}N_{\a,\d-\a}\\
% &-&(-1)^{|\b||\d|+|\b||\gamma|+|\a||\gamma|+|\a||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} s_{\a,-\gamma}N_{-\gamma,\gamma-\a}s_{-\d,\b}N_{\b,\d-\b}\\
% &=&(-1)^{|\a||\d|+|\a||\b|+|\a||\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} r^{-1}_{-\gamma,\gamma-\b}N_{\gamma,\b-\gamma}s_{-\d,\a}N_{\a,\d-\a}\\
% &-&(-1)^{{|\b||\d|}+|\a||\gamma|+|\b||\gamma|+|\a||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} s_{\a,-\gamma}r^{-1}_{-\gamma,\gamma-\a}N_{\gamma,\a-\gamma}s_{-\d,\b}N_{\b,\d-\b}\\
%
 &=&-(-1)^{|\a||\b|+|\a||\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} N_{-\gamma,-\d}^{-1} N_{\b,-\gamma}N_{\a,-\d}\nonumber\\
 &-&(-1)^{|\b||\gamma|+|\a||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}N_{-\gamma,-\d}^{-1} N_{-\gamma,\a}N_{\b,-\d}\nonumber\\
 &=&(-1)^{|\a|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} N_{-\gamma,\gamma-\b}s_{-\d,\a}N_{\a,\d-\a}\nonumber\\
 &-&(-1)^{|\b||\d|+|\b||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} s_{\a,-\gamma}N_{-\gamma,\gamma-\a}s_{-\d,\b}N_{\b,\d-\b}.\nonumber
 \end{eqnarray}}
Since $N_{\a,\b}\neq0,$ either $N_{-\gamma,\gamma-\b}N_{\a,\d-\a}\neq 0$ or $N_{-\gamma,\gamma-\a}N_{\b,\d-\b}\neq0.$ So
{\small \begin{equation}\label{yyy}
N_{\a,\b}=\left\{\begin{array}{ll}(-1)^{|\a|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} r_{-\gamma,\gamma-\b}N^{-1}_{\gamma,\b-\gamma}s_{-\d,\a}N_{\a,\d-\a}&N_{-\gamma,\gamma-\a}N_{\b,\d-\b}=0\\\\
 -(-1)^{{|\b||\d|}+|\b||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} s_{\a,-\gamma}r_{-\gamma,\gamma-\a}N^{-1}_{\gamma,\a-\gamma}s_{-\d,\b}N_{\b,\d-\b}&N_{-\gamma,\gamma-\b}N_{\a,\d-\a}= 0\\\\
 (-1)^{|\a|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma} r_{-\gamma,\gamma-\b}N^{-1}_{\gamma,\b-\gamma}s_{-\d,\a}N_{\a,\d-\a}\\
 -(-1)^{{|\b||\d|}+|\b||\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} s_{\a,-\gamma}r_{-\gamma,\gamma-\a}N^{-1}_{\gamma,\a-\gamma}s_{-\d,\b}N_{\b,\d-\b}& \hbox{otherwise}.
 \end{array}\right.
\end{equation}}
But $\gamma\prec \a\preceq\b\prec\d,$ so if each of pairs $(\gamma,\b-\gamma), (\a,\d-\a), (\gamma,\a-\gamma) $ and $(\b,\d-\b)$  is a pair of roots, it is a pair of positive roots whose sum of its  components is less than $\a+\b=\gamma+\d$ with respect to the ordering $``\preceq"$. Now we use induction on $\a+\b$ with respect to the ordering $\preceq$  to complete the proof. If $(\a,\b)$ is a special  pair such that $\a+\b$ is as small as possible with respect to $``\preceq",$ it follows from the above  argument  that $(\a,\b)$ is an extraspecial pair and so there is nothing to prove. Next suppose $(\a,\b)$ is a special pair and that we have the result for  $N_{\zeta_1,\zeta_2},$ where   $(\zeta_1,\zeta_2)$ is a special pair satisfying  $\zeta_+\zeta_2\prec\a+\b.$ If $(\a,\b)$ is extraspecial, there is nothing to prove, otherwise we are done using (\ref{yyy}) together with  (\ref{compatible}) and   the induction hypothesis.



\underline{Step 2.}   For $\a,\b\in R^\times$ with $\a+\b\in R^\times,$  we have  $N_{\a,\b}=M_{f(\a),f(\b)}$  and  $N_{\a,\b}$ is completely determined in terms of $N_{\a',\b'}$'s for extraspecial pairs $(\a',\b'):$ Suppose that $\a,\b\in R^\times$ and  $\a+\b\in R^\times,$ then  for $\gamma:=-(\a+\b),$ we have $\a+\b+\gamma=0,$ so as we have seen before, there is a  special pair $(\eta_1,\eta_2)$ among the 12 pairs
$$\begin{array}{ll}(\a,\b),(\a,\gamma),(\b,\gamma),(\b,\a),(\gamma,\a),(\gamma,\b),\\
(-\a,-\b),(-\a,-\gamma),(-\b,-\gamma),(-\b,-\a),(-\gamma,-\a),(-\gamma,-\b).
\end{array}$$
Now using Proposition \ref{constant}, for each pair $(\gamma_1,\gamma_2)$ of these 12 pairs,  $N_{\gamma_1,\gamma_2}=M_{f(\gamma_1),f(\gamma_2)}$ are uniquely determined in terms of $N_{\eta_1,\eta_2}=M_{f(\eta_1),f(\eta_2)}.$ Now we get the result using  Step 1 together with (\ref{compatible}).


$(iii)$ Use the same notation as above. Define $\theta:\gg\longrightarrow \LL$ mapping $h_i=h_{\b_i}$ to $t_i=h'_{f(\b_i)}$ and $e_\a$ to $x_{f(\a)}$ for all $\a\in\rcross$ and $1\leq i\leq \ell.$ We claim that $\theta$ is a Lie superalgebra isomorphism. We first note that by Proposition \ref{dim1-re}, (\ref{intersect}) and (\ref{rns}), $f(R_0)=S_0$ and $f(R_1)=S_1.$  Therefore, we have $\theta(\gg_{\bar i})\sub \LL_{\bar i}$ for $i=0,1.$ Now we need to show $\theta[x,y]=[\theta(x), \theta(y)]$ for all $x,y\in \gg.$ If $x=h_{\b_i}$ and $y=e_\a,$ for some $1\leq i\leq \ell$ and $\a\in \rcross,$ by (\ref{compatible}), we have
\begin{eqnarray*}\theta[h_{\b_i},e_\a]=\theta(\a(h_{\b_i})e_\a)=\a(h_{\b_i})\theta(e_\a)=f(\a)(h'_{f(\b_i)})x_{f(\a)}
&=&[h'_{f(\b_i)},x_{f(\a)}]\\
&=&[\theta(h_{\b_i}),\theta(e_\a)].\end{eqnarray*} Next suppose  $\a,\b\in \rcross.$ If $\a+\b\not\in R,$ then $f(\a)+f(\b)\not\in S$ and so   $[e_\a,e_\b]=0$ and $[\theta(e_\a),\theta(e_\b)]=[x_{f(\a)},x_{f(\b)}]=0,$ also if $\a+\b\in\rcross,$ then by part ($i$),
\begin{eqnarray*}\theta[e_\a,e_\b]=\theta(N_{\a,\b}e_{\a+\b})=N_{\a,\b}\theta(e_{\a+\b})=M_{f(\a),f(\b)}x_{f(\a+\b)}
&=&[x_{f(\a)},x_{f(\b)}]\\
&=&[\theta(e_\a),\theta(e_\b)].\end{eqnarray*}
Finally, for $\a\in\rcross,$ if $h_\a=\sum_{i=1}^\ell r_ih_{\b_i}$ for some $r_i\in \bbbf$ $(1\leq i\leq \ell),$ we get $\a=\sum_{i=1}^\ell r_i\b_i$ and so $f(\a)=\sum_{i=1}^\ell r_if(\b_i)$ which in turn implies that \begin{equation}\label{final5}h'_{f(\a)}=\sum_{i=1}^\ell r_ih'_{f(\b_i)}=\sum_{i=1}^\ell r_i\theta(h_{\b_i})=\theta(h_\a).\end{equation} Therefore, we have
\begin{eqnarray*}\theta[e_\a,e_{-\a}]=\theta(\sg_\a h_\a)=\sg_\a h'_{f(\a)}=\sg_{f(\a)} h'_{f(\a)}=[x_{f(\a)},x_{-f(\a)}]=[\theta(e_\a),\theta(e_{-\a})].
\end{eqnarray*}
This completes the proof.
\end{proof}
\begin{comment}

\begin{Thm}
Suppose that $\aa$ is the set of all extraspecial pairs  $(\a,\b)$ of $R\setminus\{0\}$  with $\a+\b\in R$ and $\{N_{\a,\b}\mid (\a,\b)\in \aa\}$ is an arbitrary subset of nonzero scalars. Suppose that $\{e_\a\mid \a \in R^+\}$ has been chosen as in Lemma \ref{arbitrary} and for $\a\in R^-,$ choose $e_{\a}$ such that $[e_\a,e_{-\a}]=\sg_\a h_\a$ for all $\a\in R\setminus\{0\}.$ For each pair $(\gamma,\eta)$ of elements of $R\setminus\{0\},$ take $N_{\gamma,\eta}$ to be zero if $\gamma+\eta\not\in R$ and if $\gamma+\eta\in R\setminus\{0\},$ take it to be such that $[e_\gamma,e_\eta]=N_{\gamma,\eta}e_{\gamma+\eta}.$ Then $\{N_{\gamma,\eta}\mid \gamma,\eta\in R\setminus\{0\},\gamma+\eta\in R\setminus\{0\}\}$ is  uniquely determined in terms of root   just depends on $\{N_{\a,\b}\mid (\a,\b)\in \aa\},$ $\{|\a|,\sg_\a\mid \a\in R^\times\},$ $\{r_{\a,\b}\mid \a,\b\in \rim^\times\hbox{ or } \a\in \rre^\times,\b\in R\}$ and $\{s_{\a,\b}\mid .....\}.$
\end{Thm}
For $(\a,\b)\in\aa$ with $\a+\b\in R,$ fix $X_{\pm\a}\in\gg^{\pm\a},X_{\pm\b}\in\gg^{\pm\b}$ such that $[X_\a,X_{-\a}]=\sg_\a h_\a$ and $[X_\b,X_{-\b}]=\sg_\b h_\b.$  Also set $X_{\a+\b}:=N_{\a,\b}^{-1}[X_\a,X_\b].$ Since $(\a,\b)$ is the only extraspecial pair whose the sum of its components is $\a+\b, $ $X_{\a+\b}$ is well-defined. Now choose $X_{-(\a+\b)}\in\gg^{-(\a+\b)}$ such that $[X_{\a+\b},X_{-\a-\b}]=\sg_{\a+\b}h_{\a+\b}.$ Also choose $X_\gamma$ for $\gamma\in R^\times\setminus(\pm\aa\cup\{\pm(\a+\b)\mid(\a,\b)\in \aa\})$
such that $[X_\gamma,X_{-\gamma}]=\sg_\gamma h_\gamma.$  Let $\a,\b\in R^\times$ be such that $\a+\b\in R^\times,$ then for $\gamma:=-(\a+\b),$ we have $\a+\b+\gamma=0,$ so either two of $\a,\b,\gamma$ are positive or two of $-\a,-\b,-\gamma$ are positive.
Selecting this pair of positive roots in an appropriate order, we get a special pair $(\eta_1,\eta_2)$ among the 12 pairs
$$\begin{array}{ll}(\a,\b),(\a,\gamma),(\b,\gamma),(\b,\a),(\gamma,\a),(\gamma,\b),\\
(-\a,-\b),(-\a,-\gamma),(-\b,-\gamma),(-\b,-\a),(-\gamma,-\a),(-\gamma,-\b).
\end{array}$$
Now using Proposition ..., for each pair $(\gamma_1,\gamma_2)$ of these 12 pairs,  $N_{\gamma_1,\gamma_2}$  is uniquely determined in terms of $N_{\eta_1,\eta_2}.$

*******************

In particular $\{N_{\gamma,\d}\mid \hbox{$(\gamma,\d)$ is not extraspecial}\} $  just depends on $\{N_{\a,\b}\mid (\a,\b)\in \aa\},$  $\{|\a|\mid \a\in R\setminus\{0\}\},$ $\{\sg_\a\mid \a\in R\setminus\{0\}\},$ $\{\b(h_\a)\mid \a,\b\in R\setminus\{0\}\}.$

****************


Next suppose that $(\a,\b)$ is a special pair with $\a+\b\in R.$ We use induction on $\a+\b$ to prove that $N_{\a,\b}$  can be uniquely determined in terms of ....


We note that as $R$ is a finite set, for each special pair $(\a,\b),$ there is a unique extraspecial pair $(\eta,\zeta)$ with $\a+\b=\eta+\zeta.$ Now suppose that with respect to the total ordering $"\prec",$  $\a+\b$ is as small as possible among the special pairs $(\a,\b).$ Then $(\a,\b)$ is an extraspecial pair as otherwise there is an extraspecial pair $(\eta,\zeta)$ with $\a+\b=\eta+\zeta.$ Now $\eta+\b\prec\a+\b$ that is a contradiction. So we have nothing to prove. Next suppose $(\a,\b)$ is a special pair and that $N_{\zeta_1,\zeta_2}$ is uniquely determined in terms of $N_{}$... for $(\zeta_1,\zeta_2)$ with $\zeta_+\zeta_2\prec\a+\b.$ If $(\a,\b)$ is extraspecial, there is nothing to prove, otherwise there is a unique extraspecial pair $(\gamma,\d)$ with $\a+\b=\gamma+\d .$  So we have $\a+\b+(-\gamma)+(-\d)=0.$ Therefore, we have $$(-1)^{|\a||-\gamma|}\sg_{\a+\b} N_{\a,\b}N_{-\gamma,-\d}+(-1)^{|\a||\b|}\sg_{\b-\gamma} N_{\b,-\gamma}N_{\a,-\d}+(-1)^{|\b||-\gamma|}\sg_{\a-\gamma} N_{-\gamma,\a}N_{\b,-\d}=0.$$ Therefore,
{\tiny \begin{eqnarray*}
 N_{\a,\b}&=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} N_{-\gamma,-\d}^{-1} N_{\b,-\gamma}N_{\a,-\d}\\
 &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}N_{-\gamma,-\d}^{-1} N_{-\gamma,\a}N_{\b,-\d}\\
 &=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma,\gamma-\b} N_{-\gamma,\gamma-\b}(-(-1)^{|\a||\d|})s_{-\d,\a,\d-\a}N_{\a,\d-\a}\\
 &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} (-(-1)^{|\a||\gamma|})s_{\a,-\gamma,\gamma-\a}N_{-\gamma,\gamma-\a}(-(-1)^{|\b||\d|})s_{-\d,\b,\d-\b}N_{\b,\d-\b}\\
 &=&-(-1)^{|\a||\b|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\b-\gamma} r^{-1}_{\gamma,\d}N_{\gamma,\d}s_{\b,-\gamma,\gamma-\b} r^{-1}_{-\gamma,\gamma-\b}N_{\gamma,\b-\gamma}(-(-1)^{|\a||\d|})s_{-\d,\a,\d-\a}N_{\a,\d-\a}\\
 &-&(-1)^{|\b||-\gamma|}(-1)^{|\a||-\gamma|}\sg_{\a+\b}\sg_{\a-\gamma}r^{-1}_{\gamma,\d}N_{\gamma,\d} (-(-1)^{|\a||\gamma|})s_{\a,-\gamma,\gamma-\a}r^{-1}_{-\gamma,\gamma-\a}N_{\gamma,\a-\gamma}(-(-1)^{|\b||\d|})s_{-\d,\b,\d-\b}N_{\b,\d-\b}\\
 \end{eqnarray*}}

But $\gamma\prec \a\prec\b\prec\d,$ so  $(\gamma,\b-\gamma), (\a,\d-\a), (\gamma,\a-\gamma) $ and $(\b,\d-\b)$ or in the opposite order  are special  pairs such that the sum of its components is less than $\a+\b=\gamma+\d$ with respect to the ordering "$\prec$". So the proof is completed using the induction hypothesis.


\begin{Thm}
Suppose that $\gg$ and $\LL$ are two locally finite basic classical simple Lie superalgebras with Cartan subalgebras $\hh$ and $T$ and corresponding root systems $R$ and $S$ respectively.  Suppose that $\fm$ (resp. $\fm'$) is an invariant  nondegenerate   even supersymmetric bilinear form on $\gg$ (resp. $\LL$) and denote the induced forms on $\hh^*$ and $T^*$ again by $\fm$ and $\fm'$ respectively. Suppose that $(\hh^*,\fm,R)$ and $(T^*,\fm',S)$ are isomorphic finite root supersystems, then there is an isomorphism from $\gg$ onto $\LL$ preserving the root spaces.
\end{Thm}
\begin{proof}
Suppose that $f:\hh^*\longrightarrow T^*$ is the isomorphism from $R$ to $S$ and $k\in\bbbf\setminus\{0\}$ is such that $(f(\a),f(\b))'=k(\a,\b)$ for $\a,\b\in R.$ Fix $r,s\in\bbbf\setminus\{0\}$ such that $r=sk.$ We note that this implies that
\begin{equation}
\label{compatible}
r(\a,\b)=sk(\a,\b)=s(f(\a),f(\b))';\;\;\;\; (\a\in R,\;\;\b\in S).
\end{equation}
For $\a\in \hh^*,$ take $t_\a$ to be the unique element of $\hh$ representing $\a$ through $\fm$ and for $\b\in T^*,$ take $t_\b$ to be the unique element of $T$ representing $\b$ through $\fm'.$ Next set
\begin{equation}
\label{h alpha}
h_\a:=rt_\a\andd h'_\b:=st'_\b\;\;\;\;(\a\in R,\;\; \b\in S).
\end{equation}
Fix a total  ordering $``\preceq"$ on $\hbox{span}_\bbbq R$ which is  compatible with $\bbbq$ and transfer it to a total  ordering, denoted again by  $``\preceq",$ on $\hbox{span}_\bbbq S.$ For $\a\in R^\times$ and $\b\in S^\times,$ set
\begin{equation}
\label{sigma}
\sg_\a:=\left\{\begin{array}{ll}
-1& \hbox{if $\a\in R^-\cap R_1$}\\
1& \hbox{otherwise}
\end{array}\right.\andd \sg'_\b:=\left\{\begin{array}{ll}
-1& \hbox{if $\b\in R^-\cap R_1$}\\
1& \hbox{otherwise.}
\end{array}\right.
\end{equation}
Suppose that $\aa$ is the set of extraspecial pairs of $R\setminus\{0\},$ then $\{(f(\a),f(\b))\mid (\a,\b)\in \aa\}$ is the set of extraspecial pairs of $S.$ Fix  a subset $\{N_{\a,\b}\mid (\a,\b)\in\aa\}$ of nonzero scalars. Using Lemma \ref{}, one can find $\{e_\a\mid\a\in R^\times\}$ and $\{x_{\b}\mid \b\in S\setminus\{0\}\}$ with
\begin{equation}
\begin{array}{ll}
\; [e_\a,e_{-\a}]=\sg_\a h_\a,\; [x_\gamma,x_{-\gamma}]=\sg'_\gamma h'_\gamma& \a\in R^+,\gamma\in S^+,\\
\; [e_\a,e_\b]=N_{\a,\b}e_{\a+\b},\;[x_{f(\a)},x_{f(\b)}]=N_{\a,\b}x_{f(\a)+f(\b)}& (\a,\b)\in\aa.
\end{array}
\end{equation}
 Using ...., we know that each $N_{\gamma,\d}$  if $(\gamma,\d)$ is not extraspecial  just depends
 on $\{N_{\a,\b}\mid (\a,\b)\in \aa\},$  $\{|\a|\mid \a\in R\setminus\{0\}\},$ $\{\sg_\a\mid \a\in R\setminus\{0\}\},$ $\{\b(h_\a)\mid \a,\b\in R\setminus\{0\}\}.$  But we know that for $\a,\b\in R\setminus\{0\},$ $|\a|=|f(\a)|,$ $\sg_\a=\sg_{f(\a)}$ and $\b(h_\a)=f(\b)(h_{f(\a)}).$ This in particular implies that $N_{\gamma,\d}=N_{f(\gamma),f(\d)}$ for all $\gamma,\d\in R^\times$  with $\gamma+\d\in R\setminus\{0\}.$

Now for each $\a\in R^+,$ fix $e_{-\a}\in \gg^{-\a}$ and $x_{-f(\a)}\in \LL^{-f(\a)}$ such that $$[e_\a,e_{-\a}]=\sg_\a h_\a\andd [x_{f(\a)},x_{-f(\a)}]=\sg'_{f(\a)} h'_{f(\a)}.$$

Since $f$ is an isomorphism from $R$ to $S,$ we get that $f(R_0)=S_0;$ so there is an isomorphism $\theta:\gg_{\bar 0}\longrightarrow \LL_{\bar 0}$ with $\theta(\gg^\a)=\LL^{f(\a)}$ for all $\a\in R_0.$


\textcolor[rgb]{1.00,0.00,0.50}{
Note: Since $\a(h_\b)=f(\a)(h'_{f(\b)})$ and $N_{\gamma,\eta}=N_{f(\gamma),f(\eta)}$ for all $\gamma,\eta\in R\setminus\{0\},$ if $\theta_\a(e_\b)=k e_{r_\a(\b)},$ then
$\theta'_{f(\a)}(x_{f(\b)})=k x_{r_{f(\a)}(f(\b))}$}

We fix $\d^*\in \rim\setminus\{0\}$ and if $R_1\cap \rre^\times \neq\{0\},$ we fix $\a^*\in R_1\cap \rre^\times.$

Fix a basis $\{\a_1,\ldots,\a_\ell\}$ of $R_0,$ then there is an isomorphism $\theta$ of $[\gg_{\bar 0},\gg_{\bar 0}]$ such that $\theta(e_\a)=x_{f(\a)}$ for each $\a\in\{\a_1,\ldots,\a_\ell\}.$
If $R$ is of imaginary type we know that the center of $\gg$ and $\LL$ are one dimensional, we also know that $\gg^0=\gg^0_{\bar 0}=\sum_{\a\in R_0\setminus\{0\}}[\gg^\a,\gg^{-\a}]\op \bbbf t_{\d^*}$
 and that $\LL^0=\LL^0_{\bar 0}=\sum_{\a\in R_0\setminus\{0\}}[\LL^\a,\LL^{-\a}]\op \bbbf t'_{f(\d^*)}.$ Recall (\ref{compatible}) and fix $z\in Z(\gg_{\bar 0})$ and $z'\in Z(\LL_{\bar 0})$ such that there are $h\in ...$ and $h'\in ..$
with $$z=h+rt_{\d^*}\andd z'=h'+st'_{f(\d^*)}.$$
Extend $\theta$ on $[\gg_{\bar 0},\gg_{\bar 0}]$ to $\gg_{\bar 0}$ by $\theta(z)=z'.$ Then $$h'+st'_{f(\d^*)}=z'=\theta(z)=\theta(h)+\theta(rt_{\d^*})$$ and so $$\theta(rt_{\d^*})=st'_{f(\d^*)}.$$
 Now for each $\b\in R_1\cap \rre^\times,$ (if it is nonempty) there is $(c,\b_1,\ldots,\b_m)$   in which $c\in\{1,-1\},$ $\b_1,\ldots,\b_m\in\{\a_1,\ldots,\a_\ell\}$ with $\b=r_{\b_1}\cdots r_{\b_m}(c\b^*).$  We fix such a tuple and denote it by $\chi_\b.$
We know that there is $M_\b\in\bbbf\setminus\{0\}$ such that $\theta_{\b_1}\cdots \theta_{\b_m}(e_{c\b^*})=M_\b e_{\b}.$ We know
set
$$\tilde \theta(e_\b):=M_\b^{-1}\theta_{f(\a_1)}\cdots \theta_{f(\a_m)}(x_{cf(\b^*)}).$$

\end{proof}
%\end{comment}

Suppose that $\gg$ is a finite dimensional basic classical simple Lie superalgebra with a Cartan subalgebra $\hh$ and corresponding root system $R.$ For a group homomorphism $\phi:\la R\ra\longrightarrow \bbbf\setminus\{0\},$ the linear transformation    $\tilde\phi:\gg\longrightarrow \gg$ mapping $x\in\gg^\a$ ($\a\in R$) to $\phi(\a)x$ is a superalgebra automorphism referred to as a {\it diagonal automorphism}.


\begin{lem}\label{cor}
Keep the same notations and assumptions as in Theorem \ref{com-cheval} and its proof. Suppose that $\Pi$ is an integral base for $R$ and fix nonzero elements  $f_\a\in\gg^\a$ and $y_{\a}\in \LL^{f(\a)}$ for all $\a\in \Pi.$ Then there is  an isomorphism  from $\gg$ to $\LL$ mapping $f_\a$ to $y_\a$ and $h_\a$ to $h'_{f(\a)}$ for all $\a\in\Pi.$  Moreover, if $\Pi$ is a base, then such an  isomorphism  is unique.
\end{lem}
\begin{proof}
Consider the Chevalley bases $\{e_\a,h_i\mid \a\in R^\times,1\leq i\leq \ell \}$ and $\{x_\b,t_i\mid \b\in S^\times,1\leq i\leq \ell\}$  as well as the isomorphism $\theta:\gg\longrightarrow \LL$ as in Theorem \ref{com-cheval}. Since $f_\a\in \gg^{\a}=\bbbf e_\a,$ there is $k_\a\in\bbbf\setminus\{0\}$ such that $f_\a=k_\a e_\a.$  Similarly, there is a nonzero scalar $k'_\a$ such that $y_\a=k'_\a x_{f(\a)}.$ Define $\phi:\la R\ra\longrightarrow\bbbf\setminus\{0\}$ mapping $\a\in\Pi$ to $k^{-1}_\a$ and $\phi':\la S\ra\longrightarrow\bbbf\setminus\{0\}$ mapping $f(\a)\in f(\Pi)$ to $k'_\a.$ Now  using (\ref{final5}), one can see that isomorphism  $\psi:=\tilde \phi'\circ\theta\circ \tilde \phi$ has  the desired properties. Next suppose  $\Pi$ is a base for $R$ and $\psi$ and $\psi'$ are two isomorphisms from $\gg$ to $\LL$ mapping $f_\a$ to $y_\a$ and $h_\a$ to $h'_{f(\a)}$ for all $\a\in \Pi.$ Then $\varphi:=\psi^{-1}\circ\psi'$ is an automorphism of $\gg$ mapping $f_\a$ to $f_\a$ and $h_\a$ to $h_\a,$ for $\a\in\Pi.$ Since  for $\a\in\Pi,$ $\varphi(f_\a)=f_\a,$ we have $\varphi(e_\a)=e_\a.$ On the other hand, as $\varphi$ is identity on $\hh,$  $\varphi$ preserves the root spaces. This together with the fact that   $[e_\a,\varphi(e_{-\a})]=[\varphi(e_\a),\varphi(e_{-\a})]= \varphi(h_\a)=  h_\a$ for all $\a\in \Pi,$ implies that $\varphi(e_{-\a})=e_{-\a}.$ Now suppose that $\a\in R^\times,$ since $\Pi$ is a base, there are $r_1,\ldots,r_n\in\{\pm1\}$ and $\a_{i_1},\ldots,\a_{i_n}\in \Pi$ such that $\a=r_1\a_{i_1}+\cdots+r_n\a_{i_n}$ and that  $r_1\a_{i_1}+\cdots+r_t\a_{i_t}\in\rcross$ for all $1\leq t\leq n.$ This together with Lemma \ref{pro-nonzero} and the fact that root spaces corresponding to nonzero roots are one dimensional, implies that $\gg^\a=\bbbf[e_{r_n\a_n},\ldots,[e_{r_3\a_3},[e_{r_2\a_2},e_{r_1\a_1}]]\cdots].$ But $\varphi(e_\a)=e_\a$ and $\varphi(e_{-\a})=e_{-\a}$ for all $\a\in \Pi,$ so the restriction of $\varphi $ to $\gg^\a$ is identity. This completes the proof.
\end{proof}


\subsection{Extension Theorem}
\begin{Pro}\label{extention}
Suppose that $(\gg,\fm,\hh)$ and $(\LL,\fm',T)$ are two infinite dimensional  locally finite basic classical simple Lie superalgebras with corresponding root systems $R$ and $S$ respectively. Denote the induced forms on $\v:=\hbox{span}_\bbbf R$ and $\u:=\hbox{span}_\bbbf S$ again by $\fm$ and $\fm'$ respectively. Suppose that $(\la R\ra,\fm,R)$ and $(\la S\ra,\fm',S)$ are isomorphic locally finite root supersystems, then there is an isomorphism from $\gg$ to $\LL$ mapping $\hh$ onto $T.$
\end{Pro}
\begin{proof}
Locally finite basic classical simple Lie superalgebras with zero odd part are exactly locally finite split simple Lie algebras in the sense of \cite{NS}. So contemplating \cite[Thm. IV.4]{NS}, we may assume $\LL_{\bar 1}\neq \{0\}.$ Suppose that $f:\la R\ra\longrightarrow \la S\ra$ is the isomorphism from $R$ to $S$ with  $(f(\a),f(\b))'=k(\a,\b)$ for $\a,\b\in R$ and some  nonzero scalar $k.$
Using Lemma \ref{lin2}, we extend $f$ to a linear isomorphism, denoted again by $f,$ from $\hbox{span}_\bbbf R$ to  $\hbox{span}_\bbbf S$ with $(f(\a),f(\b))'=k(\a,\b)$ for $\a,\b\in \hbox{span}_\bbbf R.$
Fix $r,s\in\bbbf\setminus\{0\}$ such that $r=sk.$ Therefore, we have
$$
r(\a,\b)=sk(\a,\b)=s(f(\a),f(\b))'\;\;\;\;\; (\a,\b\in \hbox{span}_\bbbf R).
$$
For $\a\in \hbox{span}_\bbbf R,$ take $t_\a$ to be the unique element of $\hh$ representing $\a$ through $\fm$ and for $\b\in \hbox{span}_\bbbf S,$ take $t'_\b$ to be the unique element of $T$ representing $\b$ through $\fm'$ and  set
$$
h_\a:=rt_\a\andd h'_\b:=st'_\b\;\;\;\;(\a\in R,\;\; \b\in S).
$$
By  Lemma \ref{base}, there is a base $\Pi$ for $R$ and a class $\{R_\gamma\mid \gamma\in\Gamma\}$ of finite irreducible  closed sub-supersystems of $R$ of the same type as $R$ such that $R$ is the direct union of $R_\gamma$'s and for each $\gamma\in\Gamma,$ $\Pi\cap R_\gamma$ is a base for $R_\gamma.$
Now  $\Pi':=f(\Pi)$ is  a base for $S$ and  $S=\cup_{\gamma\in \Gamma} S_\gamma$ in which $S_\gamma:=f(R_\gamma)$ is a finite irreducible  closed sub-supersystem of $S$ and $\Pi'_\gamma:=\Pi'\cap S_\gamma$ is  a base for $S_\gamma.$  For each $\gamma\in\Gamma,$ set $$\gg(\gamma):=\sum_{\a\in R_\gamma^\times}\gg^\a\op\sum_{\a\in R_\gamma^\times}[\gg^\a,\gg^{-\a}]\andd \hh(\gamma):=\sum_{\a\in R_\gamma^\times}[\gg^\a,\gg^{-\a}]$$ and
$$\LL(\gamma):=\sum_{\a\in S_\gamma^\times}\LL^\a\op\sum_{\a\in S_\gamma^\times}[\LL^\a,\LL^{-\a}]\andd T(\gamma):=\sum_{\a\in S_\gamma^\times}[\LL^\a,\LL^{-\a}].$$
Then as in the proof of  Lemma \ref{direct-union}, $\gg(\gamma)$ is  a finite dimensional  basic classical simple Lie superalgebra with Cartan subalgebra $\hh(\gamma)$ and corresponding root system $R_\gamma$ and  $\LL(\gamma)$ is  a finite dimensional  basic classical simple Lie superalgebra with Cartan subalgebra $T(\gamma)$ and corresponding root system $S_\gamma.$ Now fix $\{f_\a\in\gg^\a\mid\a\in \Pi\}$ and $\{y_\a\in\LL^{f(\a)}\mid \a\in \Pi\},$ By Lemma \ref{cor}, for each $\gamma\in\Gamma,$ there is a unique isomorphism $\theta_\gamma$ from $\gg(\gamma)$ to $\LL(\gamma)$ mapping $f_\a$ to $y_\a$ and $h_\a$ to $h'_{f(\a)}$ for $\a\in \Pi_\gamma.$   Now for $\gamma_1,\gamma_2\in\Gamma$ with $R_{\gamma_1}\sub R_{\gamma_2},$ $\theta_{\gamma_1},\theta_{\gamma_2}\mid_{\gg(\gamma_1)}$ are isomorphisms from $\gg(\gamma_1)$ to $\LL(\gamma_1)$ mapping $f_\a$ to $y_\a$ and $h_\a$ to $h'_{f(\a)}$ for all $\a\in \Pi_\gamma,$ therefore, we have $\theta_{\gamma_1}=\theta_{\gamma_2}\mid_{\gg(\gamma_1)}$ by Lemma \ref{cor}.  This allows us to define the isomorphism $\theta:\gg\longrightarrow \LL$ by $\theta(x)=\theta_\gamma(x)$ if $x\in\gg(\gamma).$ The isomorphism  $\theta$ maps $\hh$ onto $T.$ This completes the proof.
\end{proof}


\begin{cor}
\label{conjugate}
Suppose that $\LL$ is a locally finite basic classical simple Lie superalgebra. Assume  $\hh$ and  $T$ are two Cartan subalgebras of $\LL$ with corresponding  root systems $R$ and $S$ respectively. Then $\hh$ and $T$ are conjugate under $Aut(\LL)$ if and only if  $R$ and $S$ are isomorphic locally finite root supersystems.
\end{cor}
\begin{proof}
Assume $\phi:\LL\longrightarrow \LL$ is a Lie superalgebra automorphism such that $T=\phi(\hh).$  Define the bilinear form $\fm':\LL\times\LL\longrightarrow \bbbf$ with  $(x,y)':=(\phi^{-1}(x),\phi^{-1}(y))$ for all $x,y\in \LL.$  The linear isomorphism $\phi\mid_{\hh}:\hh\longrightarrow T$ induces a linear isomorphism $\dot\phi:\hh^*\longrightarrow T^*$ mapping $\a\in\hh^*$ to $\a\circ(\phi|_T)^{-1}.$ For each $h\in\hh,$ $\a\in R$ and $x\in\LL^\a, $ we have  $[\phi(h),\phi(x)]=\phi[h,x]=\a(h)\phi(x)=\dot\phi(\a)(\phi(h))\phi(x).$ Now it follows that   $(\LL,\fm',T)$ is an extended affine Lie superalgebra, $\phi(\LL^\a)=\LL^{\dot\phi(\a)}$ and that $S=\dot\phi(R).$  For $\b\in \hbox{span}_\bbbf S,$ take $t'_\b$ to be the unique element of $T$ representing $\b$ through $\fm'$ and for $\a\in \hbox{span}_\bbbf R,$ take $t_\a$ to be the unique element of $\hh$ representing $\a$ through $\fm.$ One can easily check that $\phi(t_\a)=t'_{\dot\phi(\a)}$  and  $(t_\a,t_\b)=(t'_{\dot\phi(\a)},t'_{\dot\phi(\b)})'$ for all $\a,\b\in R.$ These all together imply that $\dot\phi\mid_{\la R\ra}$ defines an isomorphism from $R$ to $S.$ The reverse part follows from Proposition \ref{extention}, \cite[Cor. IV.5]{NS} and finite dimensional theory of Lie superalgebras.
\begin{comment}
For type $A(1,1),$ if $\LL$ is locally finite basic classical simple Lie superalgebra whose corresponding root system is of type $A(1,1),$ we know from our results that $\dim(\LL)=10.$ We also know that $\LL$ is f.d.b.c.s and so by the classification of these algebras, $\LL$ is isomorphic to one of $A(m,n),.... .$
And so $\LL_0$ is isomorphic to the even part of the corresponding algebra. But we know from the root system of the even part of $\LL,$ that $\LL_0\simeq sl_2\op sl_2.$ So either $\LL\simeq A(1,1)$ or $\LL\simeq B(1,1).$ The latter is absurd due to its dimension.

Each two Cartan subalgebra are conjugate in finite dimensional case.
\end{comment}
\end{proof}



Using \cite[Thm. IV.6]{NS}, one knows the classification of locally finite split simple Lie algebras, i.e., locally finite basic classical simple Lie superalgebras with zero odd part. In what follows using  Theorem \ref{main1}, Examples \ref{exa1}, \ref{exa2},  Proposition \ref{extention} and  Lemma \ref{class1},  we  give the classification of locally finite basic classical simple Lie superalgebras with nonzero odd part:

\begin{Thm} Each locally finite basic classical simple Lie superalgebra with nonzero odd part is either a finite dimensional basic classical simple Lie superalgebra or   isomorphic to one  and only one of the Lie superalgebras $\mathfrak{osp}(2I,2J)$ ($I,J$   index sets with  $|I\cup J|=\infty,$ $|J|\neq 0$),  $\mathfrak{osp}(2I+1,2J)$ ($I,J$   index sets with  $|I|<\infty,$ $|J|=\infty$) or $\mathfrak{sl}(I_{\bar 0},I_{\bar 1})$  ($I$ an infinite superset with $I_{\bar 0},I_{\bar 1}\neq\emptyset$).
\end{Thm}



\begin{Pro}Suppose that $\LL$ is an infinite dimensional  locally finite basic classical simple Lie superalgebra with nonzero odd part, then  if for  an infinite index set  $I$ and a nonempty index set $J,$ $\LL\simeq \mathfrak{osp}(2I+1,2J)\simeq\mathfrak{osp}(2I,2J),$  there are two conjugacy classes for Cartan subalgebras of $\LL$ under  $Aut(\LL);$ otherwise  all Cartan subalgebras of $\LL$ are conjugate under  $Aut(\LL),$ i.e., there is just one conjugacy class  for Cartan subalgebras of $\LL$ under  $Aut(\LL).$
\end{Pro}

\begin{proof}We first assume $I$ is an infinite index set, $J$ a nonempty index set and $\LL\simeq \mathfrak{osp}(2I+1,2J)\simeq\mathfrak{osp}(2I,2J).$ We know form Example \ref{exa1} that there are Cartan subalgebras $\hh_1$ and $\hh_2$ for $\LL$ with corresponding root systems $R_1$ of type $B(I,J)$ and $R_2$ of type $D(I,J)$  respectively; in particular thanks to Corollary \ref{conjugate}, there are at least two conjugacy classes for Cartan subalgebras of $\LL$ under $Aut(\LL).$ We next note that there is    a decomposition  $\LL_{\bar 0}=\gg^1\op\gg^2$ for $\LL_{\bar 0}$ into simple ideals in which  $\gg^1$ is isomorphic to $\mathfrak{o}(2I+1,\bbbf)\simeq \mathfrak{o}(2I,\bbbf)$ and  $\gg^2$  is isomorphic to $\mathfrak{sp}(J,\bbbf);$ see \cite{NS} for the notations.  By \cite[Cor. VI.8]{NS} and finite dimensional theory of Lie algebras, there are two  conjugacy classes for Cartan subalgebras of $\gg^1$ under $Aut(\gg^1)$ and there is just one conjugacy class for Cartan subalgebras of $\gg^2$ under $Aut(\gg^2).$
Therefore, up to $Aut(\gg^1)$-conjugacy,  $\hh_1\cap \gg^1,\hh_2\cap \gg^1$ are the only non-conjugate  Cartan subalgebras of $\gg^1;$ also  $\hh_1\cap \gg^2,\hh_2\cap \gg^2$  are $Aut(\gg^2)$-conjugate Cartan subalgebras of $\gg^2$  and in fact  up to $Aut(\gg^2)$-conjugacy,  $\hh_1\cap \gg^2$ is  the only  Cartan subalgebra of $\gg^2.$  Now suppose that $T$ is a Cartan subalgebra of $\LL$ with corresponding root system $S=S_0\cup S_1.$ We want  to show that $T$ is either conjugate to $\hh_1$ or to  $\hh_2.$ Since $T\cap \gg^1$ is a Cartan subalgebra of $\gg^1$ and $T\cap \gg^2$ is a Cartan subalgebra of $\gg^2,$ there are $i\in\{1,2\}$ and  $\phi_1\in Aut(\gg^1),\phi_2\in Aut(\gg^2)$ such that $\phi_1(T\cap \gg^1)=\hh_i\cap\gg^1$ and $\phi_2(T\cap \gg^2)=\hh_i\cap\gg^2.$ So $\phi_1\op\phi_2$ is an automorphism of $\LL_{\bar0}$ mapping $T=(T\cap \gg^1)\op(T\cap \gg^2)$ to $\hh_i=(\hh_i\cap \gg^1)\op(\hh_i\cap \gg^2).$
 This implies that $(R_i)_0$ is isomorphic to $S_0.$ So using the  classification tables of locally finite root supersystems, Proposition \ref{dim1-re}, Lemma \ref{a11} and the fact that $|R_i|,|S|=\infty,$ $R_i$ is isomorphic to $S.$
\begin{comment}
Note that $B(T,1)$ and $AB(1,3)$ have isomorphic even parts but we do not consider them as $AB(1,3)$ is finite.  Also $B(1,1)$ and $A(1,1)$ have isomorphic even parts but we do not consider them as they are finite. Also $B(1,T)$ and $C(1,T)$ for $|T|\geq2$ which again we do not consider as  $C(1,T)$ cannot be the root system of a basic classical Lie superalgebra and also rank subsystem of type $B$ in the even part of these two are not infinite.
\end{comment}
Therefore there is an automorphism of $\LL$ mapping $T$ to $\hh_i$ by proposition \ref{extention}.  This implies that there are exactly two conjugacy classes for Cartan subalgebras of $\LL$ under $Aut(\LL).$

Next suppose that  $\LL $ is one of the Lie superalgebras  $ \mathfrak{osp}(2I,2J),\mathfrak{osp}(2I+1,2J)$ where $I,J$ are   index sets with  $0\neq|I|<\infty,$ $|J|\neq 0$ or   $\mathfrak{sl}(I_{\bar 0},I_{\bar 1})$ where   $I$ is an infinite superset with $|I|=\infty$ and  $I_{\bar 0},I_{\bar 1}\neq\emptyset.$
Take $\fh$ to be the standard Cartan subalgebra of $\LL$ introduced in  Examples \ref{exa1} and \ref{exa2} and consider its corresponding root system $R.$ Next suppose that $T$ is another Cartan subalgebra of $\LL$ and take  $S$ to be the corresponding root system of $\LL$ with respect to $T.$ Then $S$ is an irreducible locally finite root supersystem with $|S|=\infty.$ From the classification tables of locally finite root supersystems and Lemmas \ref{dim1-im} and \ref{a11}, $S$ is isomorphic to the root system of one of  the Lie superalgebras $\mathfrak{a}_{I',J'},\mathfrak{b}_{I',J'},\mathfrak{c}_{T'}$ introduced in Lemma \ref{class1}. Call this Lie superalgebra $\gg$ and take $\hh$ to be its standard Cartan subalgebra, so by Proposition \ref{extention}, there is an isomorphism  $\phi:\LL\longrightarrow  \gg$ mapping $T$ to $\hh.$  Now since $\LL\simeq\gg,$ using Lemma \ref{class1}, there is an isomorphism $\psi$ form $\LL$ to $\gg$ mapping $\fh$ to $\hh.$ Therefore, $\psi^{-1}\circ \phi$ is an automorphism of $\LL$ mapping $T$ to $\fh.$ This completes the proof.\end{proof}
\begin{thebibliography}{99}
%\bibitem{AABGP} B. Allison, S. Azam, S. Berman, Y. Gao and  A. Pianzola, {\it Extended affine Lie algebras and their root systems}, Mem. Amer. Math.
%Soc. {\bf 603} (1997) 1--122.
%
%\bibitem{AKY} S. Azam, V. Khalili and  M. Yousofzadeh, {\it Extended affine root systems of type BC}, J. Lie Theory {\bf 15(1)} (2005) 145--181.
%\bibitem{AYY} S. Azam, H. Yamane and M. Yousofzadeh, {\it Reflectable bases for affine reflection systems}, J.  Algebra {\bf 371} (2012) 63--93.
%\bibitem{ABG1} B.N. Allison, G. Benkart and Y. Gao,
%\newblock {\em Central extensions of Lie algebras graded by finite root systems},
%\newblock  Math. Ann.  {\bf 316 } (2000),  no. 3, 499--527.
%
%
%\bibitem{ABG2}
%B.N. Allison, G. Benkart and Y. Gao,
%\newblock {\em Lie algebras graded by the root systems $BC_r,$ $r\geq2$},
%\newblock  Mem. Amer. Math. Soc.  {\bf 158 } (2002),  no. 751, x+158.
%


%\bibitem{BZ}  G. Benkart and E. Zelmanov, \newblock {\em Lie algebras graded by finite root systems and intersection matrix
%algebras},
%\newblock Invent. Math.  {\bf 126}  (1996),  no. 1, 1--45.
%
%
%
%
%\bibitem{BS} G. Benkart and O. Smirnov, \newblock {\em Lie algebras graded by the root system $BC_1$},
%\newblock J. Lie theory, {\bf 13} (2003), 91--132.
%
%

%
%\bibitem{BE1} G. Benkart and A. Elduque, {\it Lie superalgebras graded by
%the root systems
%$C(n), D(m,n), D(2,1;\a), F(4),$ and $ G(3),$} Canad. Math. Bull. Vol. 45 (4), (2002), 509--524.
%\bibitem
%%[BE2]
%{BE2} G. Benkart and A. Elduque,  {\it Lie superalgebras graded by the root system $A(m, n),$} J. Lie Theory {\bf 13} (2003), 387--400.
%
%\bibitem
%%[BE3]
%{BE3} G. Benkart and A. Elduque,  {\it Lie superalgebras graded by the root system $B(m,n),$ Selecta Math. (N.S.) {\bf 9} (2003), no. 3, 313--360.}
%
%







%\bibitem{BM} S. Berman and R. Moody, \newblock {\em Lie algebras graded by finite root systems and the intersection matrix algebras of
% Slodowy},
%\newblock Invent. Math.  {\bf 108}  (1992),  no. 2, 323--347.



\bibitem{ca} R. Carter,  Lie algebras of finite and affine type,
Cambridge Studies in Advanced Mathematics, 96. Cambridge University Press, Cambridge, 2005.

%\bibitem{GN} E. Garcia and E. Neher, {\it Jordan superpairs covered by grids and their Tits-Kantor-
%Koecher superalgebras}, comm. in alg., Vol. 32, No. 6, pp. 2149--2175, 2004.
%\bibitem{HY}
%I. Heckenberger and  H. Yamane, {\it A generalization of Coxeter groups, root systems, and Matsumoto's theorem, Mathematishe Zeitschrift,} 259 (2008), 255--276.
%\bibitem{Hof} G. Hofmann, {\it Weyl groups with Coxeter presentation and presentation by conjugation}, J. Lie Theory 17 (2007) 337--355.
%\bibitem{Hoy} K. Hoyt, Kac-Moody superalgebras of finite growth, Ph.D. thesis, UC Berkeley, Berkeley, 2007.
\bibitem{Hum} J.E. Humphreys, {Introduction to Lie algebras and
representation theory}, Spring Verlag, New York, 1972.
\bibitem{KY} K. Iohara and  Y. Koga, {\it Central extensions of Lie superalgebras}, Comment. Math. Helv. 86 (2011), no. 4, 985-�986.
%\bibitem{K} V. Kac, Infinite dimensional Lie algebras, third edition, Cambridge  University Press, 1990.
\bibitem{K1} V. Kac, {\it Lie superalgebras}, Adv. Math. {\bf 26} (1977), 8--96.
%\bibitem{K2} V. Kac, {\it A Sketch of Lie Superalgebra Theory}, Commun. math. Phys. 53  (1977), 31--64.

\bibitem{LN}  O. Loos and   E. Neher, {\it Locally finite root systems}, Mem. Amer. Math. Soc. {\bf 171} (2004), no. 811, x+214.
\bibitem{LN2}  O. Loos,  E. Neher, {\it  Reflection systems and partial root systems},
Forum Math. 23 (2011), no. 2, 349--411.

\bibitem{MP} R.V. Moody and  A. Pianzola, Lie algebras with triangular decompositions. Canadian Mathematical Society Series of Monographs and Advanced Texts. A Wiley-Interscience Publication. John Wiley \& Sons, Inc., New York, 1995.


\bibitem{MY} J. Morita and Y. Yoshii, {\it Locally extended affine Lie algebras}, J. Algebra {\bf 301(1)} (2006), 59--81.


\bibitem{Muss} I.M.   Musson, Lie superalgebras and enveloping algebras, Graduate Studies in Mathematics, 131. American Mathematical Society, Providence, RI, 2012.
\bibitem{NS} K.H. Neeb and N. Stumme,
\newblock {\em The classification of locally finite split simple Lie algebras},
\newblock J. Reine angew. Math. {\bf 533} (2001), 25--53.
\bibitem{N1} E. Neher, {\it Extended affine Lie algebras and other generalizations of affine Lie algebras- a survey}, Developments and trends in infinite-dimensional Lie theory, 53--126, Prog. Math., 228, Birkhauser Boston, Inc., Boston, MA, 2011.
%\bibitem{N2} E. Neher, \newblock {\em Lie algebras graded by $3$-graded root systems and
%Jordan pairs covered by grids},
%\newblock Amer. J.  Math. {\bf 118} (1996), 439--491.



%\bibitem{P} I. Penkov, {\it Classically Semisimple Locally Finite Lie
%Superalgebras}, Forum Math. {\bf 16} (2004), no. 3, 431--446.
%%
%\bibitem{Se} G.B. Seligman, Rational methods in Lie algebras,  M. Dekker Lect. Notes in
%pure and appl. math. {\bf 17}, New York, 1976.


\bibitem{serg} V. Serganova, {\it On generalizations of root systems}, comm. in algebra, {\bf 24(13)}  (1996), 4281--4299.
%\bibitem{van-de} J.W. Van de Lour, {\it A classification of Contragrediant Lie superalgebras of finite growth}, Com. in Algebra, {\bf 17(8)} (1989), 1815--1841.
%\bibitem{yos2}  Y. Yoshii, \newblock {\em Root systems extended by an abelian
%group, their Lie algebras and the classification of Lie algebras
%of type $B$},
%\newblock J. Lie theory {\bf 14 } (2004), 371--394.

\bibitem{van-de}  J.W. Van de Lour,  {\it A classification of Contragrediant Lie superalgebras of finite growth}, {\it Comm. in Algebra}, {\bf 17(8)} (1989), 1815--1841.

%\bibitem{you1} M. Yousofzadeh, {\it Structure of root graded Lie algebras}, J. Lie Theory, {\bf{22}} (2012), 397--435.

%\bibitem{you4} M. Yousofzadeh, {\it Central extension of root graded Lie algebras},
\bibitem{you2}  M. Yousofzadeh, {\it Locally finite root supersystems,}  arXiv:1309.0074.

\bibitem{you3}  M. Yousofzadeh, {\it Extended affine Lie superalgebras,} arXiv:1309.3766.

%\bibitem{you5}  M. Yousofzadeh, {\it Root Graded Lie superalgebras,}  preprint.
\bibitem{you6} M. Yousofzadeh, {\it Extended affine root supersystems,} arXiv:1502.03607.

\end{thebibliography}


\end{document}

1502.04586.extracted_bib
{AABGP B. Allison, S. Azam, S. Berman, Y. Gao and A. Pianzola, {Extended affine Lie algebras and their root systems, Mem. Amer. Math. %Soc. {603 (1997) 1--122. % %
1502.04586.extracted_bib
{AKY S. Azam, V. Khalili and M. Yousofzadeh, {Extended affine root systems of type BC, J. Lie Theory {15(1) (2005) 145--181. %
1502.04586.extracted_bib
{AYY S. Azam, H. Yamane and M. Yousofzadeh, {Reflectable bases for affine reflection systems, J. Algebra {371 (2012) 63--93. %
1502.04586.extracted_bib
{ABG1 B.N. Allison, G. Benkart and Y. Gao, %\newblock {\em Central extensions of Lie algebras graded by finite root systems, %\newblock Math. Ann. {316 (2000), no. 3, 499--527. % % %
1502.04586.extracted_bib
{ABG2 %B.N. Allison, G. Benkart and Y. Gao, %\newblock {\em Lie algebras graded by the root systems $BC_r,$ $r\geq2$, %\newblock Mem. Amer. Math. Soc. {158 (2002), no. 751, x+158. % %
1502.04586.extracted_bib
{BZ G. Benkart and E. Zelmanov, \newblock {\em Lie algebras graded by finite root systems and intersection matrix %algebras, %\newblock Invent. Math. {126 (1996), no. 1, 1--45. % % % % %
1502.04586.extracted_bib
{BS G. Benkart and O. Smirnov, \newblock {\em Lie algebras graded by the root system $BC_1$, %\newblock J. Lie theory, {13 (2003), 91--132. % % % %
1502.04586.extracted_bib
{BE1 G. Benkart and A. Elduque, {Lie superalgebras graded by %the root systems %$C(n), D(m,n), D(2,1;\a), F(4),$ and $ G(3),$ Canad. Math. Bull. Vol. 45 (4), (2002), 509--524. %
1502.04586.extracted_bib
%%[BE2] %{BE2 G. Benkart and A. Elduque, {Lie superalgebras graded by the root system $A(m, n),$ J. Lie Theory {13 (2003), 387--400. % %
1502.04586.extracted_bib
%%[BE3] %{BE3 G. Benkart and A. Elduque, {Lie superalgebras graded by the root system $B(m,n),$ Selecta Math. (N.S.) {9 (2003), no. 3, 313--360. % % %
1502.04586.extracted_bib
{BM S. Berman and R. Moody, \newblock {\em Lie algebras graded by finite root systems and the intersection matrix algebras of % Slodowy, %\newblock Invent. Math. {108 (1992), no. 2, 323--347.
1502.04586.extracted_bib
{ca R. Carter, Lie algebras of finite and affine type, Cambridge Studies in Advanced Mathematics, 96. Cambridge University Press, Cambridge, 2005. %
1502.04586.extracted_bib
{GN E. Garcia and E. Neher, {Jordan superpairs covered by grids and their Tits-Kantor- %Koecher superalgebras, comm. in alg., Vol. 32, No. 6, pp. 2149--2175, 2004. %
1502.04586.extracted_bib
{HY %I. Heckenberger and H. Yamane, {A generalization of Coxeter groups, root systems, and Matsumoto's theorem, Mathematishe Zeitschrift, 259 (2008), 255--276. %
1502.04586.extracted_bib
{Hof G. Hofmann, {Weyl groups with Coxeter presentation and presentation by conjugation, J. Lie Theory 17 (2007) 337--355. %
1502.04586.extracted_bib
{Hoy K. Hoyt, Kac-Moody superalgebras of finite growth, Ph.D. thesis, UC Berkeley, Berkeley, 2007.
1502.04586.extracted_bib
{Hum J.E. Humphreys, {Introduction to Lie algebras and representation theory, Spring Verlag, New York, 1972.
1502.04586.extracted_bib
{KY K. Iohara and Y. Koga, {Central extensions of Lie superalgebras, Comment. Math. Helv. 86 (2011), no. 4, 985-�986. %
1502.04586.extracted_bib
{K V. Kac, Infinite dimensional Lie algebras, third edition, Cambridge University Press, 1990.
1502.04586.extracted_bib
{K1 V. Kac, {Lie superalgebras, Adv. Math. {26 (1977), 8--96. %
1502.04586.extracted_bib
{K2 V. Kac, {A Sketch of Lie Superalgebra Theory, Commun. math. Phys. 53 (1977), 31--64.
1502.04586.extracted_bib
{LN O. Loos and E. Neher, {Locally finite root systems, Mem. Amer. Math. Soc. {171 (2004), no. 811, x+214.
1502.04586.extracted_bib
{LN2 O. Loos, E. Neher, {Reflection systems and partial root systems, Forum Math. 23 (2011), no. 2, 349--411.
1502.04586.extracted_bib
{MP R.V. Moody and A. Pianzola, Lie algebras with triangular decompositions. Canadian Mathematical Society Series of Monographs and Advanced Texts. A Wiley-Interscience Publication. John Wiley \& Sons, Inc., New York, 1995.
1502.04586.extracted_bib
{MY J. Morita and Y. Yoshii, {Locally extended affine Lie algebras, J. Algebra {301(1) (2006), 59--81.
1502.04586.extracted_bib
{Muss I.M. Musson, Lie superalgebras and enveloping algebras, Graduate Studies in Mathematics, 131. American Mathematical Society, Providence, RI, 2012.
1502.04586.extracted_bib
{NS K.H. Neeb and N. Stumme, \newblock {\em The classification of locally finite split simple Lie algebras, \newblock J. Reine angew. Math. {533 (2001), 25--53.
1502.04586.extracted_bib
{N1 E. Neher, {Extended affine Lie algebras and other generalizations of affine Lie algebras- a survey, Developments and trends in infinite-dimensional Lie theory, 53--126, Prog. Math., 228, Birkhauser Boston, Inc., Boston, MA, 2011. %
1502.04586.extracted_bib
{N2 E. Neher, \newblock {\em Lie algebras graded by $3$-graded root systems and %Jordan pairs covered by grids, %\newblock Amer. J. Math. {118 (1996), 439--491. %
1502.04586.extracted_bib
{P I. Penkov, {Classically Semisimple Locally Finite Lie %Superalgebras, Forum Math. {16 (2004), no. 3, 431--446. %% %
1502.04586.extracted_bib
{Se G.B. Seligman, Rational methods in Lie algebras, M. Dekker Lect. Notes in %pure and appl. math. {17, New York, 1976.
1502.04586.extracted_bib
{serg V. Serganova, {On generalizations of root systems, comm. in algebra, {24(13) (1996), 4281--4299. %
1502.04586.extracted_bib
{van-de J.W. Van de Lour, {A classification of Contragrediant Lie superalgebras of finite growth, Com. in Algebra, {17(8) (1989), 1815--1841. %
1502.04586.extracted_bib
{yos2 Y. Yoshii, \newblock {\em Root systems extended by an abelian %group, their Lie algebras and the classification of Lie algebras %of type $B$, %\newblock J. Lie theory {14 (2004), 371--394.
1502.04586.extracted_bib
{van-de J.W. Van de Lour, {A classification of Contragrediant Lie superalgebras of finite growth, {Comm. in Algebra, {17(8) (1989), 1815--1841. %
1502.04586.extracted_bib
{you1 M. Yousofzadeh, {Structure of root graded Lie algebras, J. Lie Theory, {22 (2012), 397--435. %
1502.04586.extracted_bib
{you4 M. Yousofzadeh, {Central extension of root graded Lie algebras,
1502.04586.extracted_bib
{you2 M. Yousofzadeh, {Locally finite root supersystems, arXiv:1309.0074.
1502.04586.extracted_bib
{you3 M. Yousofzadeh, {Extended affine Lie superalgebras, arXiv:1309.3766. %
1502.04586.extracted_bib
{you5 M. Yousofzadeh, {Root Graded Lie superalgebras, preprint.
1502.04586.extracted_bib
{you6 M. Yousofzadeh, {Extended affine root supersystems, arXiv:1502.03607.