On the Equivariant Lazard ring and Tom Dieck's equivariant cobordism ring
C. L. Liu
For a torus $G$ of rank $r = 1$, we showed that the canonical ring \homo\ $\lazard_G \to MU_G$, where $\lazard_G$ is the \equi\ Lazard ring and $MU_G$ is Tom Dieck's \equi\ cobordism ring, is surjective. We also showed that the completion map $MU_G \to \widehat{MU}_G \cong MU(BG)$ is injective. Moreover, we showed that the same results hold if we assume a certain algebraic property holds in $\lazard_G$ when $r \geq 2$.
1307.2401.tex
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%Main document 
\begin{document}


%Title
\title{On the Equivariant Lazard ring and Tom Dieck's equivariant cobordism ring}
\author{C. L. Liu}
\begin{abstract}
For a torus $G$ of rank $r = 1$, we showed that the canonical ring \homo\ $\lazard_G \to MU_G$, where $\lazard_G$ is the \equi\ Lazard ring and $MU_G$ is Tom Dieck's \equi\ cobordism ring, is surjective. We also showed that the completion map $MU_G \to \widehat{MU}_G \cong MU(BG)$ is injective. Moreover, we showed that the same results hold if we assume a certain algebraic property holds in $\lazard_G$ when $r \geq 2$.
\end{abstract}

\address{Department of Mathematics, University of Duisburg Essen, Thea-Leymann-Strasse 9, Essen, 45127, Germany} 
\email{chun.liu@uni-due.de}
\date{\today}


\maketitle

\medskip

\medskip









\section{Introduction}

Let $G$ be a compact abelian Lie group. In his paper \cite{complex cobor}, Tom Dieck defined an \equi\ cobordism theory $MU^*_{G}(-)$. While it has many nice properties, for instance, there is a Conner-Floyd natural transformation
$$MU^*_{G}(-) \to K^*_{G}(-),$$
it is rather mysterious from the computational point of view. For example, even when $G$ is a cyclic group of order $n > 2$, the ring structure of $MU_{G}$ is not clear, from the algebraic point of view (When $n=2$, see \cite{complex cobor ring when order two} for an explicit algebraic description of $MU_{G}$). It is worth mentioning that although a description of $MU_G$ is given in Sinha's paper \cite{computation of MU}, it depends on an implicit choice of a basis of $MU_G$ as a free $MU$-module.

In their paper \cite{equi FGL}, Cole, Greenlees and Kriz introduced a notion called \equi\ \fgl\ and showed that there is a representing ring, called \equi\ Lazard ring. By Corollary 14.3 in \cite{equi FGL}, this ring $\lazard_{G}(F)$ (relative to a complete $G$-flag $F$) has a very explicit, algebraic description. Moreover, there is a canonical ring \homo
$$\nu_{G} : \lazard_{G}(F) \to MU_{G}.$$
As an analogue of the non-\equi\ case, Greenlees conjectured that this ring \homo\ is an isomorphism (see section 13 in \cite{equi FGL 2}). Since the general case will follow easily from the case when $G$ is a torus, we will focus on this case. 

There are two parts in this paper. Part 1 (section \ref{sec canonical ring homo}) is devoted to the investigation of the injectivity and surjectivity of the canonical ring \homo\ $\nu_G$. In part 2 (section \ref{sec completion map}), we consider the completion map
$$MU_G \to \widehat{MU}_G \cong MU(BG)$$
\wrt\ the ideal generated by the Euler classes. To be more precise, we are interested in showing its injectivity. For both parts, we have satisfactory results when $\rank{G} = 1$ (see Corollary \ref{cor greenlee conj for rank 1} and Theorem \ref{thm completion map}). When $\rank{G} \geq 2$, 
we manage to establish the same results, under the assumption that a certain algebraic property holds for $\lazard_G(F)$ (see Theorem \ref{thm lazard to MU iso}, \ref{thm completion map}).

\bigskip
\bigskip










\section{Notations and assumptions}
\label{sect notation}

In this paper, $G$ is a split torus of rank $r$. Hence, any $G$-\repn\ over $\C$ can be written as direct sum of 1-dimensional $G$-\repn s. We also fix a complete $G$-universe $\cat{U}$ and a complete $G$-flag $F$ given by 
$$0 = V^0 \subset V^1 \subset V^2 \subset \cdots.$$
We denote the 1-dimensional $G$-characters $V^i/V^{i-1}$ by $\ga_i$. All $G$-characters are 1-dimensional unless stated otherwise. For simplicity, we will denote the \equi\ Lazard ring relative to the flag $F$ by $\lazard_G$.

\bigskip
\bigskip















\section{The canonical ring homomorphism $\nu$} 
\label{sec canonical ring homo}

In this section, we will investigate the injectivity and surjectivity of the canonical ring \homo
$$\nu_G : \lazard_G \to MU_G.$$

We will follow the notation in \cite{equi FGL}. For instance, we will denote the structure constants of $\lazard_G(F)$ by $b^{ij}_s, d(\alpha)^i_s, f^i_{st}$. Let $S_G \subset \lazard_G$ be the multiplicative set generated by the non-trivial Euler classes ($e(\beta) = d(\beta)^1_0$ where $\beta$ is non-trivial) and $K_{G,S}$ be the kernel of the localization map $\lazard_G \to S_G^{-1} \lazard_G$. First, we have the following result regarding the injectivity of the map $\nu_G $.

\begin{prop}
\label{prop lazard to MU inj}
The kernel of $\nu_G : \lazard_G \to MU_G$ is $K_{G,S}$.
\end{prop}

\begin{proof}
By Theorem 1.2 in \cite{computation of MU}, the non-trivial Euler classes in $MU_G$ are regular. Therefore, $\nu_G(K_{G,S})=0$. Consider the following commutative diagram :

\squarediagram{\lazard_G/K_{G,S}}{S_G^{-1} \lazard_G}{MU_G}{S_G^{-1}MU_G}

\noindent The top row is certainly injective and the right column is an iso\morp\ by Proposition 13.2 in \cite{equi FGL 2}. The result then follows.
\end{proof}

For the surjectivity of $\nu_G$, we have a satisfactory result when $r=1$.

\begin{prop}
\label{prop surj of map equi lazard to equi MU}
Suppose $G$ is a split torus of rank 1. Then 
$$\nu_G : \lazard_G \to MU_G$$
is surjective.
\end{prop}

\begin{proof}
Consider the following commutative diagram :

\squarediagramword{\lazard_G}{S_G^{-1} \lazard_G}{MU_G}{S_G^{-1} MU_G}{}{\nu_G}{S_G^{-1}\nu_G}{}

\noindent By Proposition 13.2 in \cite{equi FGL 2}, the map $S_G^{-1}\nu_G$ is an iso\morp. Moreover, the bottom row is injective because Euler classes in $MU_G$ are regular (Theorem 1.2 in \cite{computation of MU}). So, for any element $a \in MU_G$, there exist some non-trivial $G$-characters $\gamma_i$ such that $(\prod_i e(\gamma_i)) a$ is in the image of $\nu_G$. Suppose $\nu_G$ is not surjective. Let $M \defeq MU_G / \lazard_G$. Then there exists an element $0 \neq a \in M$ such that $e(\beta^n) a = 0$, where $\beta \in G^*$ is a generator and $n$ is a positive integer. 

Consider the following commutative diagram :

\begin{center}
$\begin{CD}
@. \lazard_G @>e(\beta)>> \lazard_G @>>> \lazard @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> MU_G @>e(\beta)>> MU_G @>>> MU @>>> 0 
\end{CD}$
\end{center}

\noindent The top row is exact because $\lazard_G / (e(\beta)) \cong \lazard_{\{1\}} \cong \lazard$. By Theorem 1.2 in \cite{computation of MU}, the bottom row is exact. Furthermore, the right column is an isomorphism by Quillen's Theorem. By the Snake Lemma, the map $e(\beta) : M \to M$ is injective. Therefore, the localization map $M \to S_1^{-1} M$ is injective, where $S_m \subset \lazard_G$ denotes the multiplicative set generated by $e(\beta), \ldots, e(\beta^m)$.

Now, consider a similar commutative diagram with $e(\beta^2)$ instead of $e(\beta)$ and localize it \wrt\ $S_1$ :

\begin{center}
$\begin{CD}
@. S_1^{-1} \lazard_G @>e(\beta^2)>> S_1^{-1} \lazard_G @>>> S_1^{-1} \lazard_{\Z / (2)} @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> S_1^{-1} MU_G @>e(\beta^2)>> S_1^{-1} MU_G @>>> S_1^{-1} MU_{\Z / (2)} @>>> 0 
\end{CD}$
\end{center}

\noindent The rows are exact for the same reasons. By Proposition 13.2 in \cite{equi FGL 2}, the right column is an isomorphism. Again, by the Snake Lemma, the map $e(\beta^2) : S_1^{-1} M \to S_1^{-1} M$ is injective. Therefore, $M \to S_1^{-1} M \to S_2^{-1} M$ is injective. Inductively, $M \to S_n^{-1} M$ is injective. But $a \neq 0$ is in its kernel by our assumption. That draws a contradiction and we are done.
\end{proof}

\begin{cor}
\label{cor greenlee conj for rank 1}
Suppose $G$ is a split torus of rank 1 and $H \subset G$ is a closed subgroup.

\noindent \begin{statementslist}
{\rm (1)} & The canonical map $\nu_G : \lazard_G/K_{G,S} \to MU_G$ is an isomorphism. \\
{\rm (2)} & Let $p : \lazard_G \to \lazard_H$ be the canonical map. Then, $\nu_H : \lazard_H \to MU_H$ factors through $\lazard_H / p(K_{G,S})$ and $\nu_H : \lazard_H / p(K_{G,S}) \to MU_H$ is an isomorphism.
\end{statementslist}
\end{cor}

\begin{proof}
Part (1) follows from Proposition \ref{prop lazard to MU inj} and \ref{prop surj of map equi lazard to equi MU}. For part (2), let $\beta \in G^*$ be a generator and $n$ be a generator of the kernel of $\Z \cong G^* \to H^*$. W\withoutlog, $n > 0$. Consider the following commutative diagram :

\begin{center}
$\begin{CD}
@. \lazard_G @>e(\beta^n)>> \lazard_G @>{p}>> \lazard_H @>>> 0 \\
@. @V{\nu_G}VV @V{\nu_G}VV @V{\nu_H}VV @. \\
0 @>>> MU_G @>e(\beta^n)>> MU_G @>>> MU_H @>>> 0 
\end{CD}$
\end{center}

\noindent Since $\nu_G(K_{G,S}) = 0$, we have $\nu_H \circ p(K_{G,S}) = 0$. So $\nu_H$ factors through $\lazard_H / p(K_{G,S})$. Apply $\otimes_{\lazard_G} (\lazard_G / K_{G,S})$ to the top row, we have :

\begin{center}
$\begin{CD}
@. \lazard_G/K_{G,S} @>e(\beta^n)>> \lazard_G/K_{G,S} @>{p}>> \lazard_H/p(K_{G,S}) @>>> 0 \\
@. @V{\nu_G}VV @V{\nu_G}VV @V{\nu_H}VV @. \\
0 @>>> MU_G @>e(\beta^n)>> MU_G @>>> MU_H @>>> 0 
\end{CD}$
\end{center}

\noindent The top row is clearly exact and the bottom row is exact by Theorem 1.2 in \cite{computation of MU}. The result then follows from part (1) and the five Lemma.
\end{proof}

To generalize our results to split torus of higher rank, we need to assume a certain property holds in $\lazard_G$, which is justified by the following generalization of Theorem 1.2 in \cite{computation of MU}. Suppose $H$ is a compact abelian Lie group. We call a set of $H$-characters $\beta_1, \ldots, \beta_s$ linearly independent if their corresponding torus part (as in $\Z^r$, where $r$ is the rank of $H^*$) are $\Z$-linearly independent.

\begin{prop}
\label{prop gen thm by Sinha}
Suppose $H$ is a compact abelian Lie group. Any set of linearly independent $H$-characters $\beta_1, \ldots, \beta_s$ defines a regular sequence $e(\beta_1), \ldots, e(\beta_s)$ in $MU_H$. Moreover, $MU_H / (e(\beta_1), \ldots, e(\beta_s)) \cong MU_{H'}$ where $H' \subset H$ is the closed subgroup corresponding to $H^* / (\beta_1, \ldots, \beta_s)$.
\end{prop}

\begin{proof}
By induction on $s$, see the proof of Theorem 1.2 in \cite{computation of MU} for the case when $s=1$.
\end{proof}

\begin{thm}
\label{thm lazard to MU iso}
Suppose $G$ is a split torus of rank $r \geq 2$. Assume the analogue of Proposition \ref{prop gen thm by Sinha} holds for $\lazard_G$ when $s \leq 2$. Then the canonical map $\nu_G : \lazard_G \to MU_G$ is an isomorphism.
\end{thm}

\begin{proof}
By Proposition \ref{prop lazard to MU inj} and our assumption ($s=1$), it is enough to show the surjectivity. Let $M \defeq MU_G / \lazard_G$. Suppose $\nu_G$ is not surjective. Since $S_G^{-1} M = 0$ (by Proposition 13.2 in \cite{equi FGL 2}), there is an element $0 \neq x \in M$ and a non-trivial $G$-character $\beta$ such that $e(\beta)x=0$ in $M$. By an automorphism of $G$, we may assume $\beta = (n, 0 , \ldots, 0) \in \Z^r \cong G^*$ where $n > 0$. Let $n, x$ be such a pair with minimal $n$.

Let $T \subset \lazard_G$ be the multiplicative set generated by $e(\gamma)$ where $\gamma = (a_1, 0, \ldots, 0)$ and $0 < a_1 < n$. By the minimality of $n$, we have $0 \neq x \in T^{-1}M$ and $e(\beta)x=0$ in $T^{-1}M$. Consider the following commutative diagram of $\lazard_G$-modules :

\medskip

\begin{center}
$\begin{CD}
 @. T^{-1}\lazard_G @>{e(\beta)}>> T^{-1}\lazard_G @>>> T^{-1}\lazard_H @>>> 0 \\
 @. @V{T^{-1}\nu_G}VV @V{T^{-1}\nu_G}VV @V{T^{-1}\nu_H}VV @. \\
0 @>>> T^{-1}MU_G @>{e(\beta)}>> T^{-1}MU_G @>>> T^{-1}MU_H @. 
\end{CD}$
\end{center}

\medskip

\noindent where $H \subset G$ is the closed subgroup corresponding to $G^* / (\beta)$. The rows are exact as before. By the Snake Lemma, the following sequence is exact :
$$\kernel{T^{-1}\nu_H} \to T^{-1}M \stackrel{e(\beta)}{\longto} T^{-1}M.$$
Therefore, it is enough to show $T^{-1}\nu_H$ is injective.

Let $T' \subset \lazard_G$ be the multiplicative set generated by $e(\gamma)$ where $\gamma = (a_1, \ldots, a_r)$ such that at least one of $a_2, \ldots, a_r$ is non-zero. Consider the following commutative diagram :

\squarediagramword{T^{-1}\lazard_H}{T^{-1}MU_H}{T^{-1}{T'}^{-1}\lazard_H}{T^{-1}{T'}^{-1}MU_H}{T^{-1}\nu_H}{}{}{}

\noindent By our assumption ($s=2$), $\lazard_H \to {T'}^{-1} \lazard_H$ is injective, so is the left column. Notice that $T^{-1}{T'}^{-1}\lazard_H = S_H^{-1} \lazard_H$ and the same for $MU_H$. By Proposition 13.2 in \cite{equi FGL 2}, the bottom row is an isomorphism. Hence, $T^{-1}\nu_H$ is injective and we are done.
\end{proof}

\begin{cor}
\label{cor greenlee conj for higher rank}
Suppose $G$ is a split torus of rank $r \geq 2$. Assume the analogue of Proposition \ref{prop gen thm by Sinha} holds for $\lazard_G$ when $s \leq 2$. Then, for any closed subgroup $H \subset G$, the canonical map $\nu_H : \lazard_H \to MU_H$ is an isomorphism.
\end{cor}

\begin{proof}
Suppose $H \subset G$ is the closed subgroup corresponding to $G^* / (\beta)$ for some non-trivial $G$-character $\beta = (n, 0, \ldots, 0) \in \Z^r \cong G^*$. Consider the following commutative diagram :


\medskip

\begin{center}
$\begin{CD}
\lazard_G @>{e(\beta)}>> \lazard_G @>>> \lazard_H @>>> 0 \\
@V{\nu_G}VV @V{\nu_G}VV @V{\nu_H}VV @. \\
MU_G @>{e(\beta)}>> MU_G @>>> MU_H @>>> 0 
\end{CD}$
\end{center}

\medskip

\noindent The top row is exact as before and the bottom row is exact by Proposition \ref{prop gen thm by Sinha}. By Theorem \ref{thm lazard to MU iso}, the left and middle columns are isomorphisms, then so is the right column. The result for arbitrary $H$ can be shown by repeated applications of the above argument.
\end{proof}

\begin{rmk}
{\rm
By Corollary \ref{cor greenlee conj for rank 1} and \ref{cor greenlee conj for higher rank}, if $G$ is a split torus of rank $r$, $H \subset G$ is a closed subgroup and we assume the analogue of Proposition \ref{prop gen thm by Sinha} holds for $\lazard_G$ when $s \leq 2$ in the case when $r \geq 2$, then the canonical map $\nu_H : \lazard_H \to MU_H$ is surjective. By Theorem 16.1 in \cite{equi FGL}, $\lazard_H$ is generated by the Euler classes $e(\beta)$ and the structure constant $f^1_{ij}$. Therefore, $MU_H$, as a ring, is generated by the Euler classes $e(\beta)$ and $\nu_H(f^1_{ij})$. For a more explicit description of $\nu_H(f^1_{ij})$, see remark 11.4 in \cite{universal equi alg cobor}.
}
\end{rmk}

\bigskip










\section{The completion map}
\label{sec completion map}

Our main goal in this section is to show that the completion map 
$$MU_G \to \widehat{MU}_G \cong MU(BG),$$
with respect to the ideal generated by the Euler classes, is injective. Since we will consider some algebraic cobordism theories, we need to fix a ground field $k$. In this section, all schemes (or $G$-schemes) are over $k$. We will also assume $\char{k} = 0$.

Since there is a ring \homo\ $MU \to MU_G$, one would expect to also have a ring \homo\ $\lazard \to \lazard_G$. But it is not clear how to put a non-\equi\ \fgl\ over $\lazard_G$. Therefore, we will consider the algebraic cobordism rings instead. 

Let $\Omega(-)$ be the algebraic cobordism theory defined in \cite{universal alg cobor}, $\omega(-)$ be the algebraic cobordism theory defined by double point relation, as in \cite{alg cobor by DPR}, $\Omega^G(-)$ be the \equi\ algebraic cobordism theory defined in \cite{universal equi alg cobor} and $\Omega$, $\omega$, $\Omega^G$ be the corresponding cobordism rings over $\spec{k}$. By Theorem 1 in \cite{alg cobor by DPR}, $\omega(-) \cong \Omega(-)$. Also, the canonical ring \homo\ $\lazard \to \Omega$ is an isomorphism, by Theorem 1.2.7 in \cite{universal alg cobor}. 

Since double point relation holds in $\Omega^G(-)$ (Proposition 5.5 in \cite{universal equi alg cobor}), we have a canonical ring \homo 
$$\omega \to \Omega^G$$
(given by putting trivial $G$-action) as desired.

Let $\Omega^G_{\rm Tot}(-)$ be the \equi\ algebraic cobordism theory defined in \cite{homo equi alg cobor}. By Proposition 8.5 in \cite{universal equi alg cobor}, there is a ring \homo\ 
$$\Omega^G \to \Omega^G_{\rm Tot}.$$ 
Moreover, by subsection 3.3.1 in \cite{homo equi alg cobor}, 
$$\Omega^G_{\rm Tot} \cong \Omega[[e_1, \ldots, e_r]],$$
where $\beta_1, \ldots, \beta_r$ is a set of generator of $G^*$ and $e_i$ corresponds to the Euler class $e(\beta_i) = c(\beta_i)[\id_{\pt}]$.

\begin{lemma}
\label{lemma composition preserve Lazard}
The composition
$$\lazard \cong \omega \to \Omega^G \to \Omega^G_{\rm Tot} \cong \omega[[e_1, \ldots, e_r]] \cong \lazard[[e_1, \ldots, e_r]]$$
preserves elements in $\lazard$.
\end{lemma}

\begin{proof}
The result follows from the fact that the composition
$$\omega \to \Omega^G \to \Omega^G_{\rm Tot} \cong \omega[[e_1, \ldots, e_r]]$$
preserves elements in $\omega$.
\end{proof}

Now, let $I \subset \lazard_G$ be the ideal generated by the Euler classes and $\hat{\lazard}_G$, $\hat{\Omega}^G$ be the $I$-adic completions of $\lazard_G$, $\Omega^G$ respectively. Since $\Omega^G_{\rm Tot}$ is $I$-adically complete, we have an induced ring \homo 
$$\hat{\Omega}^G \to \Omega^G_{\rm Tot}.$$

\begin{prop}
\label{prop completion map for equi alg cobor is inj}
Suppose $G$ is a split torus and $\char{k} = 0$.

\noindent \begin{statementslist}
{\rm (1)} & The ring \homo\  $\hat{\Omega}^G \to \Omega^G_{\rm Tot}$ is an isomorphism. \nonumber\\
{\rm (2)} & The completion map $\Omega^G \to \hat{\Omega}^G$ is injective. \nonumber
\end{statementslist}
\end{prop}

\begin{proof}
For part (1), the map $\lazard \cong \omega \to \Omega^G$ induces a map $f : \lazard[e_1, \ldots, e_r] \to \Omega^G$, by sending $e_i$ to $e(\beta_i)$. So, we have
$$\lazard[e_1, \ldots, e_r] \stackrel{f}{\to} \Omega^G \to \hat{\Omega}^G \to \Omega^G_{\rm Tot} \cong \lazard[[e_1, \ldots, e_r]],$$
which is just the completion map \wrt\ the ideal $(e_1, \ldots, e_r)$ (by Lemma \ref{lemma composition preserve Lazard}).

By Theorem 6.12 in \cite{universal equi alg cobor}, the map $\lazard_G \to \Omega^G$ is surjective, so is $\hat{\lazard}_G \to \hat{\Omega}^G$. By Theorem 6.5 in \cite{equi FGL 2}, $\hat{\lazard}_G \cong \lazard[[e_1, \ldots, e_r]]$. Therefore, we have
$$\hat{\lazard}_G \to \hat{\Omega}^G \to \Omega^G_{\rm Tot} \cong \lazard[[e_1, \ldots, e_r]] \cong \hat{\lazard}_G,$$
which is the identity map (can be seen by considering its quotient by $I^n$). Hence, $\hat{\lazard}_G \to \hat{\Omega}^G$ is an isomorphism and so is $\hat{\Omega}^G \to \Omega^G_{\rm Tot}$.

For part (2), consider the composition
$$\lazard[e_1, \ldots, e_r] \stackrel{f}{\to} \Omega^G \to \hat{\Omega}^G \stackrel{g}{\cong} \lazard[[e_1, \ldots, e_r]]$$
($g$ is an isomorphism by part (1)). Since 
$$\lazard[e_1, \ldots, e_r] / (e_1, \ldots, e_r)^n \to \Omega^G / I^n \to \lazard[[e_1, \ldots, e_r]] / (e_1, \ldots, e_r)^n$$
is the identity map, the map $\lazard[e_1, \ldots, e_r] / (e_1, \ldots, e_r)^n \to \Omega^G / I^n$ is injective for all $n \geq 1$. Therefore, $f^{-1}(I^n) = (e_1, \ldots, e_r)^n$. If $a$ is an element in $\cap_{n \geq 1} I^n \subset \Omega^G$, then 
$$f^{-1}(a) \subset f^{-1}(\cap_n I^n) = \cap_n f^{-1}(I^n) = \cap_n (e_1, \ldots, e_r)^n = 0.$$ 
Hence, $a = 0$ and we are done.
\end{proof}







\begin{thm}
\label{thm completion map}
Suppose $G$ is a spit torus of rank $r$ and $k = \C$.

\noindent
\begin{statementslist}
{\rm (1)} & The canonical map $\Omega^G \to MU_G$ is injective. \nonumber\\
{\rm (2)} & The kernel of the canonical, surjective map $\lazard \to \Omega^G$ is $K_{G,S}$. \nonumber
\end{statementslist}

\medskip

If $r \geq 2$ and we assume the analogue of Proposition \ref{prop gen thm by Sinha} holds for $\lazard_G$ when $s \leq 2$, then

\noindent
\begin{statementslist}
{\rm (3)} & $\lazard_G \iso \Omega^G \iso MU_G$ \nonumber\\
{\rm (4)} & The completion maps $\lazard_G \to \hat{\lazard}_G$ and $MU_G \to \widehat{MU}_G$ are both injective. \nonumber\\
{\rm (5)} & $MU_G$ is an integral domain. \nonumber
\end{statementslist}

\medskip

If $r = 1$, then

\noindent
\begin{statementslist}
{\rm (3')} & $\Omega^G \iso MU_G$ \nonumber\\
{\rm (4')} & The completion map $MU_G \to \widehat{MU}_G$ is injective. \nonumber\\
{\rm (5')} & $MU_G$ is an integral domain. \nonumber
\end{statementslist}
\end{thm}

\begin{proof}
For part (1), consider the following commutative diagram :
\squarediagram{\Omega^G}{\hat{\Omega}^G}{MU_G}{\widehat{MU}_G}
The right column is the composition $\hat{\Omega}^G \to \Omega^G_{\rm Tot} \cong \hat{\lazard}_G \to \widehat{MU}_G$, which is an isomorphism (by part (1) of Proposition \ref{prop completion map for equi alg cobor is inj} and Proposition 13.3 in \cite{equi FGL 2}). The result then follows from part (2) of Proposition \ref{prop completion map for equi alg cobor is inj}. 

For part (2), notice that we have
$$\lazard_G \stackrel{a}{\to} \Omega^G \stackrel{b}{\to} MU_G$$
such that $a$ is surjective (Theorem 6.12 in \cite{universal equi alg cobor}) and $b$ is injective (by part (1)). The result then follows from Proposition \ref{prop lazard to MU inj}. 

Part (3) follows from Corollary \ref{cor greenlee conj for higher rank} and part (2). Part (4) follows from part (2) of Proposition \ref{prop completion map for equi alg cobor is inj} and part (3). Part (5) follows from part (4) and the fact that 
$$\widehat{MU}_G \cong \lazard[[e_1, \ldots, e_r]]$$ 
is an integral domain. Part (3'), (4'), (5') follow from similar arguments as those for part (3), (4), (5) respectively.
\end{proof}


\begin{rmk}
{\rm 
Many results are only true for split torus. Even for cyclic group of order $n$, $\lazard_G$ can behave quite badly (so is $MU_G$). For example, if $\beta \in G^*$ is a generator, $0 \neq e(\beta) \in \lazard_G$ is not regular (because $0 = e(\beta^n) = e(\beta)(n + a e(\beta))$ for some $a$). Moreover, if $n = 6$, one can show that $S_G^{-1} \hat{\lazard}_G$ is the zero ring. In particular, that means $\phi$ is not injective when $n = 6$, because $S_G^{-1} \lazard_G$ is never the zero ring by Corollary 6.4 in \cite{equi FGL 2}. 
}
\end{rmk}









\begin{comment}

Old approach that didn't work

\begin{prop}
\label{prop localization wrt I}
The completion map $\phi$ factors through $(1+I)^{-1} \lazard_G$ and the induced map 
$$(1 + I)^{-1} \lazard_G \to \hat{\lazard}_G$$ 
is injective.
\end{prop}

\begin{proof}
According to section 6.B in \cite{equi FGL 2}, 
$$\hat{\lazard}_G \cong \lazard[[e(\beta_1), \ldots, e(\beta_r)]]$$
where $\beta_i$ are generators of $G^*$. So elements in $1+I$ are invertible in $\hat{\lazard}_G$. Therefore, $\phi$ factors through $(1+I)^{-1} \lazard_G$.

Let $R \defeq (1+I)^{-1} \lazard_G$ and
$$f : R\{\{F\}\} \to R[[\overline{y}]]$$ 
be the $R$-algebra \homo\ which sends $y(V^1) = y(\alpha_1) = y(\epsilon)$ to $\overline{y}$. It is well-defined because 
$$y(V^n) = y(\alpha_1) \cdots y(\alpha_n)$$
and $F$ is a complete flag. Let $c_{nj} \in R$ be the elements such that
$$f(y(V^n)) = \sum_{j \geq 0} c_{nj} \overline{y}^j.$$




\medskip

\noindent Claim 1 : $c_{nj} \equiv \delta^n_j$ mod $I$.

Since $c_{0j} = \delta^0_j$, the claim is true for $n=0$. For $n \geq 0$, we have
\begin{eqnarray}
f(y(V^{n+1})) &=& f(y(V^n)) \cdot f(y(\alpha_{n+1})) \nonumber\\
&=& (\sum_{j \geq 0} c_{nj} \overline{y}^j) \cdot \sum_{k \geq 0} d(\alpha_{n+1})^1_k\, f(y(V^k)) \nonumber\\
&=& (\sum_{j \geq 0} c_{nj} \overline{y}^j) \cdot \sum_{k \geq 0} d(\alpha_{n+1})^1_k \sum_{l \geq 0} c_{kl} \overline{y}^l \nonumber
\end{eqnarray}
\noindent Therefore,
$$c_{n+1 m} = \sum_{\stackrel{j,k,l \geq 0}{j + l = m}} c_{nj}\, d(\alpha_{n+1})^1_k\, c_{kl}$$
(It is a finite sum because, for all $l \geq 0$, $c_{kl} = 0$ for sufficiently large $k$). The result then follows from the induction assumption and the fact that $d(\beta)^1_k \equiv \delta^1_k$ mod $I$ (Corollary 16.6 in \cite{equi FGL}). \claimend

\medskip





{\color{red} need some fix here. Infinite dimensional}

By claim 1, $f$ is a $R$-algebra isomorphism and it induces a $R$-algebra isomorphism 
$$R\{\{F\}\} \hat{\otimes}_R R\{\{F\}\} \to R[[u,v]].$$ 
Let $F[u,v] \in R[[u,v]]$ be the element corresponding to $\Delta y(V^1) \in R\{\{F\}\} \hat{\otimes}_R R\{\{F\}\}$. By the fact that $R\{\{F\}\}$ over $R$ defines a $(G,F)$-\fgl, we have $F(u,0)=u$, $F(u,v) = F(v,u)$ and $F(u,F(v,w)) = F(F(u,v),w)$, namely, $F$ defines a non-\equi\ \fgl\ over $R$. Therefore, there is a canonical ring \homo\ $\lazard \to (1+I)^{-1} \lazard_G$. We can then define a ring \homo
$$g : (1 + I)^{-1} \lazard[e_1, \ldots, e_r] \to (1+I)^{-1} \lazard_G$$
which sends $e_i$ to $e(\beta_i)$. Hence, we have
$$(1 + I)^{-1} \lazard[e_1, \ldots, e_r] \to (1+I)^{-1} \lazard_G \to \hat{\lazard}_G.$$

Let $A \defeq (1+I)^{-1} \lazard[e_1, \ldots, e_r]$ and $J_n \defeq g^{-1}(I^n)$. Notice that 
$$R/I^n \cong \hat{\lazard}_G / I^n  \cong \lazard[[e_1, \ldots, e_r]] / I^n \cong \lazard[e_1, \ldots, e_r] / I^n \cong A / I^n.$$ 
Therefore, $J_n = I^n$ and the completion of $A$ \wrt\ $J_n$ is the same as the completion of $R$ \wrt\ $I^n$, which is $\hat{\lazard}_G$. Since $\lazard[e_1, \ldots, e_r] \to \hat{\lazard}_G$ is injective, $A \to \hat{\lazard}_G$ is injective, i.e, $\cap_{i \geq 1} J_i = 0$. If $a \in \cap_{i \geq 1} I^i \subset R$, then $g^{-1}(a) \subset \cap_{i \geq 1} J_i = 0$. Therefore, $a = g (g^{-1}(a)) = 0$. Hence, $R \supset \cap_{i \geq 1} I^i = 0$ and we are done.
\end{proof}

\begin{thm}
\label{thm kernel of the completion map}
Suppose $G$ is a torus. Then
$$\kernel{\phi} = \cap_{i \geq 1} I^i = K_{G,S} = K_{G,I}.$$
\end{thm}

\begin{proof}
By Proposition \ref{prop localization wrt I}, $\kernel{\phi} = K_{G,I}$. Since Euler classes are regular in $\hat{\lazard}_G \cong \lazard[[e(\beta_1), \ldots, e(\beta_r)]]$ (see section 6.B in \cite{equi FGL 2}), we have $K_{G,S} \subset \kernel{\phi}$. So it remains to show $K_{G,I} \subset K_{G,S}$.

Suppose $a \in K_{G,I}$. Then $(1+b)a = 0$ in $\lazard_G$ for some $b \in I$. Then, $(1+b)a = 0$ in $S_G^{-1} \lazard_G$, which is an integral domain by Corollary 6.4 in \cite{equi FGL 2}. If $1+b \neq 0$ in $S_G^{-1} \lazard_G$, then $a = 0$ and we are done. Suppose $1+b = 0$ in $S_G^{-1} \lazard_G$. Then $(1+I)^{-1} S_G^{-1} \lazard_G$ is the zero ring. By Proposition \ref{prop localization wrt I}, $(1+I)^{-1} S_G^{-1} \lazard_G$ is a subring of $S_G^{-1} \hat{\lazard}_G$. So, $S_G^{-1} \hat{\lazard}_G$ is also the zero ring. It then contradicts with the fact that $\hat{\lazard}_G$ is an integral domain and non-trivial Euler classes are non-zero elements.
\end{proof}

\begin{rmk}
\label{rmk odd behavior for non torus}
{\rm 
Many results are only true for torus. Even for cyclic group of order $n$, $\lazard_G$ can behave quite badly. For example, if $\beta \in G^*$ is a generator, $0 \neq e(\beta) \in \lazard_G$ is not regular (because $0 = e(\beta^n) = e(\beta)(n + a e(\beta))$ for some $a$). Moreover, if $n = 6$, one can show that $S_G^{-1} \hat{\lazard}_G$ is the zero ring. In particular, that means $\phi$ is not injective when $n = 6$ ($S_G^{-1} \lazard_G$ is never the zero ring by Corollary 6.4 in \cite{equi FGL 2}). Hence, the completion map $\phi$ is not faithfully flat when $G$ is the cyclic group of order 6.
}
\end{rmk}

\begin{cor}
\label{cor about equi lazard and its map to equi MU}
Suppose $G$ is a torus.

\noindent \begin{statementslist}
{\rm (1)} & The completion map $\phi$ factors through $\lazard_G/K_{G,S}$ and $\phi : \lazard_G/K_{G,S} \to \hat{\lazard}_G$ is injective. \\
{\rm (2)} & The ideal $\kernel{\phi} = K_{G,S} = K_{G,I}$ in $\lazard_G$ is a prime ideal. 
\end{statementslist}
\end{cor}

\begin{proof}
Part (1) follows directly from Theorem \ref{thm kernel of the completion map}. Part (2) follows from part (1) and the fact that $\hat{\lazard}_G$ is an integral domain. 
\end{proof}

Although the completion map $\phi$ is not faithfully flat in general (see Remark \ref{rmk odd behavior for non torus}), it is always flat.

\begin{prop}
\label{prop completion map flat}
For any compact abelian Lie group $G$, the completion map $\phi : \lazard_G \to \hat{\lazard}_G$ is flat.
\end{prop}

\begin{proof}
First of all, it is a non-trivial result because $\lazard_G$ is not Noetherian. Let $A \defeq \Z[d(\alpha)^i_s, b^{i,j}_s, f^i_{s,t}]$ ($i,j,s,t \geq 0$, $\alpha \in G^*$) and $J$ be the kernel of the surjective map $A \to \lazard_G$, i.e., $J$ is the ideal generated by the defining relations. Let $B$ be the completion of $A$ \wrt\ $I$, the ideal generated by $e(\alpha) = d(\alpha)^1_0$. Then the completion map $A \to B$ is obviously flat. Therefore, the map
$$\lazard_G \cong A/J \to B/JB$$
is flat.

Clearly, $(B/JB)\{\{F\}\}$ over $B/JB$ defines a $(G,F)$-\fgl\ and $B/JB$ is complete \wrt\ $IB$. By the universality of $\hat{\lazard}_G$, there exists a ring \homo\ $\hat{\lazard}_G \to B/JB$ which preserves all the structure constants. On the other hand, for all $n \geq 1$, $B / (JB + I^n B)$ is naturally isomorphic to $A/ (J + I^n) \cong \lazard_G / I^n$. So, it induces a ring \homo\ $B/JB \to \hat{\lazard}_G$, which also preserves all the structure constants. Hence, $B/JB \cong \hat{\lazard}_G$ and we are done.
\end{proof}

We then have the following consequences regarding Tom Dieck's equivariant cobordism ring.

\begin{cor}
Suppose $G$ is a torus of rank $r$ and $H \subset G$ is a closed subgroup. If $r \geq 2$, we further assume the analogue of Proposition \ref{prop gen thm by Sinha} holds for $\lazard_G$ for $s \leq 2$. Then we have the following results :

\noindent \begin{statementslist}
{\rm (1)} & The completion map $MU_H \to \widehat{MU}_H$ is flat. \\
{\rm (2)} & The completion map $MU_G \to \widehat{MU}_G$ is flat and injective. \\
{\rm (3)} & $MU_G$ is an integral domain.
\end{statementslist}
\end{cor}

\begin{proof}
\noindent (1) \tab Consider the following commutative diagram :

\squarediagramword{\lazard_G}{\hat{\lazard}_G}{\lazard_H}{\hat{\lazard}_H}{\phi}{p}{p'}{\phi}

\noindent where $p,p'$ are the canonical maps. By Theorem \ref{thm kernel of the completion map}, $\phi(K_{G,S}) = 0$. Therefore, $\phi\circ p(K_{G,S}) = 0$. Since $\phi : \lazard_H \to \hat{\lazard}_H$ is flat (Proposition \ref{prop completion map flat}), so is its induced map $\lazard_H / p(K_{G,S}) \to \hat{\lazard}_H$. By Corollary \ref{cor greenlee conj for rank 1}, $\lazard_H / p(K_{G,S}) \to MU_H$ is an isomorphism.  The result then follows from the fact that $\widehat{MU}_G \cong \hat{\lazard}_G$ (Proposition 13.3 in \cite{equi FGL 2}).

\medskip

\noindent (2) \tab The flatness follows from part (1) and the injectivity follows from Corollary \ref{cor greenlee conj for rank 1}, \ref{cor greenlee conj for higher rank}, \ref{cor about equi lazard and its map to equi MU} and the fact that $\widehat{MU}_G \cong \hat{\lazard}_G$.

\medskip

\noindent (3) \tab It follows from part (2) and the fact that $\widehat{MU}_G \cong \hat{\lazard}_G$ is an integral domain.
\end{proof}




\end{comment}












\begin{comment}
Old proof that didn't work.




\begin{lemma}
Suppose $c_1, \ldots, c_n$ are some elements in $A \defeq (\Z - 0)^{-1} \lazard$ and for $m > 1$, let
$F^{(m)}(x) \in \lazard[[x]]$ be the ``$m \cdot x$'' \wrt\ the \fgl. Let 
$$F^{(m)}(x) = mx + b_2 x^2 + b_3 x^3 + \cdots.$$
Then $A$ is an integral domain and $b_k$ is algebraically independent over $\Q(c_1, \ldots, c_n, b_1, \ldots, b_{k-1})$ (subfield of $Q(A)$) for all sufficiently large $k$.
\end{lemma}

\begin{proof}
First of all, if $B \defeq \Q[y_2, y_3, \ldots]$ ($y_i$ are free commutative variables) and $g(x) \defeq x + y_2 x^2 + y_3 x^3 + \cdots \in B[[x]]$, then $F_B(x,x') \defeq g(g^{-1}(x) + g^{-1}(x'))$ defines a \fgl\ over $B$ and the classifying map $\psi : A \to B$ is an isomorphism. ({\color{red} find the reference of this fact}) So, $A$ is an integral domain and $\psi$ sends $F^{(m)}(x)$ to $g(mg^{-1}(x))$.

Let $g^{-1}(x) \defeq x + d_2 x^2 + d_3 x^3 + \cdots$. One can show that for $k \geq 2$, 
$$y_k + d_k \in \Q(y_2 , \ldots, y_{k-1}) = \Q(d_2 , \ldots, d_{k-1}).$$
So,
$$g(mg^{-1}(x)) = m(x + d_2 x^2 + \cdots) + y_2 m^2(x + d_2 x^2 + \cdots)^2 + y_3m^3(x + d_2 x^2 + \cdots)^3 + \cdots.$$
For all $k \geq 2$,
$$\psi(b_k) = m d_k + y_2 m^2 (d_{k-1} + d_2 d_{k-2} + \cdots + d_{k-1}) + \cdots + y_k m^k.$$
Therefore, $\psi(b_k) - m d_k - m^k y_k \in \Q(y_2 , \ldots, y_{k-1})$. We then have $\psi(b_k) - (m^k-m) y_k \in \Q(y_2 , \ldots, y_{k-1})$. Hence, $\Q(y_2 , \ldots, y_{k-1}) = \Q(\psi(b_2) , \ldots, \psi(b_{k-1}))$ and we are done.
\end{proof}



First of all, by Corollary 6.4 and Theorem 6.5 in \cite{equi FGL 2}, $S^{-1} \lazard_G$ and $\hat{\lazard}_G$ are both integral domains and $e(\beta) \neq 0$ if $\beta \neq \epsilon$. Therefore,
$$\nil{\lazard_G} \subset K_S \subset \cap_{i \geq 1} I^i.$$
Since $\nil{\lazard_G} \subset \cap_{i \geq 1} I^i$, the completions of $\lazard_G$ and $\reduced{(\lazard_G)}$ \wrt\ $I$ agree. Hence, it is enough to show that the completion map
$$\reduced{\phi} : \reduced{(\lazard_G)} \to \hat{\lazard}_G$$
is injective. By $\reduced{\phi}$, we know that $0 \neq e(\beta) \in \reduced{(\lazard_G)}$ if $\beta \neq \epsilon$.

Since $1+I \subset \hat{\lazard}_G$ are invertible, $\phi$ factors through $(1+I)^{-1} \lazard_G$.

\begin{lemma}
\label{lemma localization ID}
$$(1 + I)^{-1} \lazard_G \to \hat{\lazard}_G$$ 
is injective. In particular, $(1 + I)^{-1} \lazard_G$ is an integral domain.
\end{lemma}

\begin{proof}
Let $R \defeq (1+I)^{-1} \lazard_G(F)$ and
$$f_n : A_n \defeq \prod_{0 \leq i \leq n} R \overline{y}^i \to \prod_{0 \leq i \leq n} R y(V^i) \defeq B_n$$
($A_n$ is the truncated polynomial ring over $R$ and $B_n$ is the truncated version of $R\{\{F\}\}$) be the $R$-algebra \homo\ sending $\overline{y}$ to $y(V^1)$. Let $c_{i,j} \in R$ be the elements satisfying the equality $f_n(\overline{y}^i) = \sum_j c_{i,j} y(V^j)$. Then, $c_{0,j} = \delta^0_j$ and $c_{1,j} = \delta^1_j$.

Since $f_n(\overline{y}^{i+1}) = y(V^1)f_n(\overline{y}^i)$, we have
$$c_{i+1,j} = \sum_k c_{i,k} b^{1,k}_j.$$
Since the canonical map $\lazard_G(F) \to \lazard$ sends $b^{i,j}_k$ to $\delta^{i+j}_k$, $b^{1,k}_j \equiv \delta^{k+1}_j$ mod $I$. Therefore, inductively, one can show $c_{i,j} \equiv \delta^i_j$ mod $I$. So, the determinant of the matrix representing $f_n$ is congruent to 1 mod $I$ and hence, invertible, i.e., $f_n$ is an isomorphism. The maps $f_n$ commute with the inverse system and induces a $R$-algebra isomorphism
$$f : R[[\overline{y}]] \to R\{\{F\}\}.$$ 
We also have a $R$-algebra isomorphism $R[[u,v]] \to R\{\{F\}\} \hat{\otimes}_R R\{\{F\}\}$. 

Let $F[u,v] \in R[[u,v]]$ be the element corresponding to $\Delta y(V^1) \in R\{\{F\}\} \hat{\otimes}_R R\{\{F\}\}$. By the above isomorphisms and the fact that $R\{\{F\}\}$ over $R$ defines a $(G,F)$-\fgl, $F(u,0)=u$, $F(u,v) = F(v,u)$ and $F(u,F(v,w)) = F(F(u,v),w)$, namely, $F$ defines a non-\equi\ \fgl\ over $R$. Therefore, there is a canonical ring \homo\ $\lazard \to (1+I)^{-1} \lazard_G(F)$.

Let $\beta_1, \ldots, \beta_r \in G^*$ be a set of generators. Then we have a ring \homo
$$g : (1 + I)^{-1} \lazard[e(\beta_1), \ldots, e(\beta_r)] \to (1+I)^{-1} \lazard_G(F)$$
($e(\beta_i)$ are free variables on the left). In the completion $\hat{\lazard}_G$, $1+I$ is already invertible. So, we have
$$(1 + I)^{-1} \lazard[e(\beta_1), \ldots, e(\beta_r)] \to (1+I)^{-1} \lazard_G \to \lazard[[e(\beta_1), \ldots, e(\beta_r)]].$$
To clarify, denote $I \subset \lazard[e(\beta_1), \ldots, e(\beta_r)]$ by $J$. Since 
$$g : \lazard \cong (1 + J)^{-1} \lazard[e(\beta_1), \ldots, e(\beta_r)] / J \to (1+I)^{-1} \lazard_G / I \cong \lazard,$$
$g^{-1}(I) = J$. If $a \in \cap I^i$, then $g^{-1}(a) \subset \cap J^i = 0$ (non-empty). Therefore, $a = g \circ g^{-1}(a) = 0$. Hence, $\cap I^i = 0$, i.e., $(1+I)^{-1} \lazard_G \to \hat{\lazard}_G$ is injective. The second part of the statement follows from the fact that $\hat{\lazard}_G$ is an integral domain.
\end{proof}

{\color{red} It may be possible to fix the argument so that the completion map is injective for any $G$.}

\begin{lemma}
\label{lemma ann square}
Suppose $A$ is a commutative ring, $a \in A$ and $b \in \ann{a}$. The following diagram commutes and all maps are surjective.

\begin{center}
$\begin{CD}
A @>>> A / \ann{a} @>>> A / \ann{a^2} \\
@| @AAA @AAA \\
A @>>> A / (b) @>>> (A / (b)) / \ann{a}
\end{CD}$
\end{center}

\noindent (Zero rings are allowed here).
\end{lemma}

\begin{proof}
The only non-trivial part is the well-definedness of the map $(A / (b)) / \ann{a} \to A / \ann{a^2}$. Suppose $c \in \ann{a} \subset A / (b)$. Then, $ca \in (b) \subset A$. Hence, $ca^2 = 0$.
\end{proof}

\begin{lemma}
\label{lemma Euler class regular}
For all $\beta \neq \epsilon$, the element $e(\beta) \in \reduced{(\lazard_G)}$ is regular.
\end{lemma}

\begin{proof}
Suppose $e(\beta) a = 0$ in $\reduced{(\lazard_G)}$. By change of basis, we may assume $\beta_1, \ldots, \beta_r$ is a basis of $G^*$ and $\beta = \beta_1^m$ for some positive integer $m$. 

First of all, by Lemma \ref{lemma localization ID}, $(1+I)^{-1} \lazard_G$ is an integral domain. Therefore, $(1+I)^{-1} \lazard_G = (1+I)^{-1} \reduced{(\lazard_G)}$. Also, for all non-trivial $\gamma$, $e(\gamma) \neq 0$ in $(1+I)^{-1} \reduced{(\lazard_G)}$ because 
$$(1+I)^{-1} \reduced{(\lazard_G)} / (f^1_{s,t}) = (1+I)^{-1} \Z[e(\beta_1), \ldots, e(\beta_r)]$$
(by Proposition A.1 in \cite{equi FGL}) and $e(\gamma) \neq 0$ here. Since $e(\beta_1^m)a = 0$ in $(1+I)^{-1} \reduced{(\lazard_G)}$, $a = 0$ in $(1+I)^{-1} \reduced{(\lazard_G)}$. Hence, $(1 + p e(\beta_1))a = 0$ in $\reduced{(\lazard_G)}$ for some $p$.

If $m = 1$, $(1 + p e(\beta_1))a = 0$ in $\reduced{(\lazard_G)}$ implies $a = 0$ and we are done.

Now, suppose $m > 1$. Assume the statement is true for $1 \leq i < m$. Let $S_{m-1} \subset \reduced{(\lazard_G)}$ be the multiplicative set generated by $e(\beta_1), \ldots, e(\beta_1^{m-1})$ and $B_{m-1} \defeq S_{m-1}^{-1} \reduced{(\lazard_G)}$. It is then enough to show that  $a = 0$ in $B_{m-1}$. 

Apply Lemma \ref{lemma ann square}\ with $A = B_{m-1}$, $a = a$ and $b = e(\beta_1^m)$, we have a surjective ring \homo
$$(B_{m-1} / e(\beta_1^m)) / \ann{a} \to B_{m-1} / \ann{a^2}.$$
Notice that $\lazard_H \cong \lazard_G / (e(\beta_1^m))$ where $H \subset G$ is the subgroup corresponding to $G^* \to G^* / (\beta_1^m)$. We have $B_{m-1} / e(\beta_1^m) \cong S_{m-1}^{-1} \lazard_H / \nil{\lazard_G}.$ 

{\color{red} Suppose the rank of $G$ is 1.} Since $S_{m-1}^{-1} \lazard_H$ is an integral domain (Corollary 6.4 in \cite{equi FGL 2}), 
$$B_{m-1} / e(\beta_1^m) \cong S_{m-1}^{-1} \lazard_H$$ 
is an integral domain.

If $a = 0$ in $B_{m-1} / e(\beta_1^m)$, then $(B_{m-1} / e(\beta_1^m)) / \ann{a}$ is the zero ring and so is $B_{m-1} / \ann{a^2}$. That means $a^2 = 0$ in $B_{m-1}$ and we are done because $B_{m-1}$ is reduced. Since $(1 + p e(\beta_1))a = 0$ in $B_{m-1} / e(\beta_1^m)$, which is an integral domain, it is enough to show the following claim.

\vtab

\noindent Claim 1 : For all $p \in \lazard_G$, $1 + p e(\beta_1) \neq 0$ in $S_{m-1}^{-1} \lazard_H$.

For simplicity, denote $e(\beta_1)$ by $e$. Consider
$$C \defeq S_{m-1}^{-1} \lazard_H^m [v^{-1}] = S_{m-1}^{-1} R(H)[v,v^{-1}].$$
($m$ means multiplicative). Here, $e(\beta_1^i) = v^{-1}(1-\beta_1^i)$. It is enough to show $1 + p e \neq 0$ in $C$. Notice that the grading on $\lazard_G$ induces a grading on $\lazard_H^m$ with $\deg e(\gamma) = 1$ and $\deg v = \deg f^1_{1,1}= -1$. 

Since $\beta_1 = 1 - ve$ and $\beta_1^m = 1$ , we have
$$C \cong S_{m-1}^{-1} \Z [e, v,v^{-1}] / ((1 - v e)^m - 1) \cong S_{m-1}^{-1} \Z [e, x,x^{-1}] / ((1 - x)^m - 1)$$
where $x \defeq ev$ (degree 0). Notice that once $e$ is inverted, the localization \wrt\ $S_{m-1}$ is the same as inverting $\beta_1^i - 1 = (1-x)^i - 1$ for $1 \leq i < m$. Let $y \defeq 1-x$. Then, 
$$C \cong T^{-1} \Z [e, e^{-1}, y] / (y^m - 1)$$ 
where $T$ is the multiplicative set generated by $y^i - 1$ for $1 \leq i < m$.

Suppose $1 + p e = 0$ in $C$. Then, $p$ is homogeneous with degree -1. Since $p$ comes from $\lazard_G$, 
$$p = v f'(ev) = ((1-y) / e) f(y) \in C$$
for some polynomials $f', f$ with integer coefficients. Therefore, $1 + (1-y) f(y) = 0$ in $C$. Let $D \defeq T^{-1} \Z [y] / (y^m - 1)$. Then $C \cong D[e,e^{-1}]$. So, $1 + (1-y) f(y) = 0$ in $D$. Since there is a surjective map $D \to T^{-1} \Z [\mu_m]$ where $\mu_m$ is the primitive $m$-th \rou\ and $\Z[\mu_m] \subset \C$ is an integral domain, it is enough to show $1 + (1-\mu_m) f(\mu_m) \neq 0$ in $D = \Z[\mu_m]$, i.e., $(1- \mu_m)^{-1} \notin D$. 

{\color{red} Later}


\vtab





\end{proof}

\begin{prop}
\label{prop completion map flat}
For any compact abelian Lie group $G$, $\phi : \lazard_G \to \hat{\lazard}_G$ is flat.
\end{prop}

\begin{proof}
Some works need to be done because $\lazard_G$ is non-Noetherian. Let $A \defeq \Z[d(\alpha)^i_s, b^{i,j}_s, f^i_{s,t}]$ ($i,j,s,t \geq 0$, $\alpha \in G^*$) and $J$ be the kernel of the surjective map $A \to \lazard_G$, i.e., $J$ is the ideal generated by the defining relations. Let $B$ be the completion of $A$ \wrt\ $I$ (ideal generated by $e(\alpha) = d(\alpha)^1_0$). Then the completion map $A \to B$ is flat. Therefore, 
$$\lazard_G \cong A/J \to B/JB$$
is flat.

Clearly, $(B/JB)\{\{F\}\}$ over $B/JB$ defines a $(G,F)$-\fgl\ and $B/JB$ is complete \wrt\ $IB$. By the universality of $\hat{\lazard}_G$, there exists a ring \homo\ $\hat{\lazard}_G \to B/JB$ which preserves all the structure constants. On the other hand, for all $n \geq 1$, $B / (JB + I^n B)$ is naturally isomorphic to $A/ (J + I^n) \cong \lazard_G / I^n$. So, it induces a ring \homo\ $B/JB \to \hat{\lazard}_G$, which also preserves all the structure constants. Hence, $B/JB \cong \hat{\lazard}_G$ and we are done.
\end{proof}

\begin{thm}
\label{thm kernel of the completion map}
$$\nil{\lazard_G} = K_S = \cap_{i \geq 1} I^i.$$
\end{thm}

\begin{proof}
As mentioned before, it is enough to show that $\reduced{\phi} : \reduced{(\lazard_G)} \to \hat{\lazard}_G$ is injective. By Lemma \ref{lemma Euler class regular}, the map 
$$\reduced{(\lazard_G)} \to S^{-1} \reduced{(\lazard_G)} \cong S^{-1} \lazard_G$$
is injective (The isomorphism follows from the fact that $\nil{\lazard_G} \subset K_S$). Since $S^{-1} \lazard_G$ is an integral domain, so is $\reduced{(\lazard_G)}$. 

Let $\lazard_G \supset P \defeq \cap_{i \geq 1} I^i = \phi^{-1}(0)$ (prime ideal because $\hat{\lazard}_G$ is an integral domain). Consider the commutative diagrams :

\begin{center}
$\begin{CD}
\lazard_G @>{\pi}>> \reduced{(\lazard_G)} \\
@| @V{\reduced{\phi}}VV \\
\lazard_G @>{\phi}>> \hat{\lazard}_G
\end{CD}$
\tab $\stackrel{{\rm Localization}}{\longto}$ \tab 
$\begin{CD}
(\lazard_G)_P @>{\pi_P}>> (\reduced{(\lazard_G)})_P \\
@| @V{(\reduced{\phi})_P}VV \\
(\lazard_G)_P @>{\phi_P}>> Q(\hat{\lazard}_G)
\end{CD}$
\end{center}

\noindent ($Q(\hat{\lazard}_G)$ is the quotient field). Denote all the localization maps by $i_P$. Since $\reduced{(\lazard_G)}$ is an integral domain, $i_P : \reduced{(\lazard_G)} \to (\reduced{(\lazard_G)})_P$ is injective. By Proposition \ref{prop completion map flat}, $\phi$ is flat. So $\phi_P$ is faithfully flat. Therefore, it's injective. $Q(\hat{\lazard}_G)$ is a field, so $(\lazard_G)_P$ is an integral domain. $\pi_P$ is an iso\morp\ because $$(\reduced{(\lazard_G)})_P = (\lazard_G)_P / i_P(\nil{\lazard_G}) = (\lazard_G)_P.$$
Therefore, $(\reduced{\phi})_P$ is injective. Since $i_P : \reduced{(\lazard_G)} \to (\reduced{(\lazard_G)})_P$ and $(\reduced{\phi})_P$ are both injective, so is $\reduced{\phi}$.
\end{proof}

\begin{cor}
\label{cor about equi lazard and its map to equi MU}
A list of results.

\noindent \begin{statementslist}
{\rm (1)} & $\reduced{\phi} : \reduced{(\lazard_G)} \to \hat{\lazard}_G$ is injective. \\
{\rm (2)} & $\reduced{(\lazard_G)}$ is an integral domain. \\
{\rm (3)} & The canonical ring \homo\ $\lazard_G \to MU_G$ factors through $\reduced{(\lazard_G)}$. \\
{\rm (4)} & The canonical ring \homo\ $\reduced{(\lazard_G)} \to MU_G$ is injective. \\
\end{statementslist}
\end{cor}

\begin{proof}
(1) follows directly from Theorem \ref{thm kernel of the completion map}. (2) follows from (1) and the fact that $\hat{\lazard}_G$ is an integral domain. For (3), by Theorem 1.2 in \cite{computation of MU}, the Euler classes are regular elements. Therefore, $0 = K_S \subset MU_G$. 

(4) : Consider the following commutative diagram :

\squarediagram{\reduced{(\lazard_G)}}{\hat{\lazard}_G}{MU_G}{\hat{MU}_G}

\noindent The right column is an iso\morp\ by Proposition 13.3 in \cite{equi FGL 2}. By (1), the left column is injective. 
\end{proof}

\begin{prop}
\label{prop surj of map equi lazard to equi MU}
Suppose $G$ is a split torus of rank 1. Then, the canonical ring \homo
$$\nu : \lazard_G \to MU_G$$
is surjective.
\end{prop}

\begin{proof}
There is a commutative diagram :

\squarediagram{\lazard_G}{S^{-1} \lazard_G}{MU_G}{S^{-1} MU_G}

\noindent The right column is an iso\morp\ by Proposition 13.2 in \cite{equi FGL 2}\ and the bottom row is injective because Euler classes in $MU_G$ are regular. So, for any element $a \in MU_G$, there exist some non-trivial $\gamma_i$ such that $(\prod_i e(\gamma_i)) a$ is in the image of the left column. Suppose the left column is not surjective. Let $M \defeq MU_G / \lazard_G$. Then there exists an element $0 \neq a \in M$ such that $e(\beta^n) a = 0$, where $\beta \in G^*$ is a generator and $n$ is a positive integer. 

Consider the following commutative diagram :

\begin{center}
$\begin{CD}
@. \lazard_G @>e(\beta)>> \lazard_G @>>> \lazard @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> MU_G @>e(\beta)>> MU_G @>>> MU @>>> 0 
\end{CD}$
\end{center}

\noindent The top row is trivially exact. The bottom row is exact by Theorem 1.2 in \cite{computation of MU}. The right column is an isomorphism. By the Snake Lemma, the map $e(\beta) : M \to M$ is injective. Therefore, the localization map $M \to S_1^{-1} M$ ($S_m \subset \lazard_G$ is the multiplicative set generated by $e(\beta), \ldots, e(\beta^m)$) is injective.

Now, consider a similar commutative diagram with $e(\beta^2)$ instead of $e(\beta)$ and localize it by $S_1$ :

\begin{center}
$\begin{CD}
@. S_1^{-1} \lazard_G @>e(\beta^2)>> S_1^{-1} \lazard_G @>>> S_1^{-1} \lazard_{\Z / (2)} @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> S_1^{-1} MU_G @>e(\beta^2)>> S_1^{-1} MU_G @>>> S_1^{-1} MU_{\Z / (2)} @>>> 0 
\end{CD}$
\end{center}

\noindent The rows are exact for the same reasons. By Theorem 13.2 in \cite{equi FGL 2}, the right column is an isomorphism. Again, by Snake Lemma, the map $e(\beta^2) : S_1^{-1} M \to S_1^{-1} M$ is injective. Therefore, $M \to S_1^{-1} M \to S_2^{-1} M$ is injective. Inductively, $M \to S_n^{-1} M$ is injective. But $a \neq 0$ is in its kernel by our assumption. That draws a contradiction and we are done.
\end{proof}

\begin{cor}
Suppose $G$ is a split torus of rank 1 and $H \subset G$ is a subgroup.

\noindent \begin{statementslist}
{\rm (1)} & $\nu : \reduced{(\lazard_G)} \to MU_G$ is an isomorphism. \\
{\rm (2)} & $\nu : \reduced{(\lazard_H)} \to MU_H$ is an isomorphism.
\end{statementslist}
\end{cor}

\begin{proof}
(1) follows from Corollary \ref{cor about equi lazard and its map to equi MU} and Proposition \ref{prop surj of map equi lazard to equi MU}. For (2), let $\beta \in G^*$ be a generator and $n$ be a generator of the kernel of $\Z \cong G^* \to H^*$. W\withoutlog, $n > 0$. Consider the commutative diagram :

\begin{center}
$\begin{CD}
@. \reduced{(\lazard_G)} @>e(\beta^n)>> \reduced{(\lazard_G)} @>{\pi}>> \reduced{(\lazard_H)} @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> MU_G @>e(\beta^n)>> MU_G @>>> MU_H @>>> 0 
\end{CD}$
\end{center}

\noindent The bottom row is exact by Theorem 1.2 in \cite{computation of MU}. The top row may not be exact, but it is a complex and $\pi$ is surjective. By (1), the left and the middle columns are isomorphism. By some diagram chasing, one can show the right column is injective. It's also surjective because $\reduced{(\lazard_G)} \to MU_G \to MU_H$ is surjective.
\end{proof}

\begin{rmk}
{\rm
Many results can be drawn from these, but let's try to generalize it to split torus of arbitrary rank, or try to show $\lazard_G$ is reduced first.
}
\end{rmk}

\end{comment}



\bigskip

\bigskip













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