Equivariant Khovanov Homology of Periodic Links
Wojciech Politarczyk
In this paper we study Khovanov homology of periodic links. We construct a variant of the Khovanov homology -- the equivariant Khovanov homology -- which is adapted to the equivariant setting. In the construction of the equivariant Khovanov homology we utilize the action of a finite cyclic group $\z{n$ on the Khovanov complex of a periodic link. Further we prove the invariance with respect to the equivariant Reidemeister moves, show that it is isomorphic to the Khovanov homology modulo $n$-torsion and express values of this homology theory on $p^{n$-periodic trivial links in terms of the group cohomology of $\z{p^{n$, if $p$ is a prime. Additionally we construct a spectral sequence, analogous to the one from \cite{EverittTurner, converging to the equivariant Khovanov homology and we utilize this spectral sequence to compute the equivariant Khovanov homology of torus links $T(n,2)$ with respect to the symmetry of order $2$.
equiv_kh_arxiv.tex
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%opening
\title{Equivariant Khovanov Homology of Periodic Links}
\author{Wojciech Politarczyk}

\begin{document}

\maketitle

\begin{abstract}
In this paper we study Khovanov homology of periodic links. We construct a variant of the Khovanov 
homology -- the equivariant Khovanov homology -- which is adapted to the equivariant setting. In 
the construction of the equivariant Khovanov homology we utilize the action of a finite cyclic 
group $\z{n}$ on the Khovanov complex of a periodic link. Further we prove the invariance with 
respect to the equivariant Reidemeister moves, show that it is isomorphic to the Khovanov homology 
modulo $n$-torsion and express values of this homology theory on $p^{n}$-periodic trivial links in 
terms of the group cohomology of $\z{p^{n}}$, if $p$ is a prime. Additionally we construct a 
spectral sequence, analogous to the one from \cite{EverittTurner}, converging to the equivariant 
Khovanov homology and we utilize this spectral sequence to compute the equivariant Khovanov 
homology 
of torus links $T(n,2)$ with respect to the symmetry of order $2$.
\end{abstract}

\section{Introduction}

One of the main themes in topology is the study of symmetries of certain objects like topological 
spaces or manifolds. In knot theory one is particularly interested in symmetries of knots, that is 
symmetries of the $3$-sphere that preserve the given knot.

In this paper, which is based on the authors PhD Thesis \cite{PolitarczykPhD}, we study knots 
which possess certain kind of symmetry of finite order which is derived from a semi-free action 
of the cyclic group on the $3$-sphere, i.e., for a fixed knot $K$ we are interested in 
diffeomorphisms $f \colon (S^{3},K) \to (S^{3},K)$ of finite order, with the property that $K$ is 
disjoint with the fixed point set of $f$. Due to the resolution of the Smith 
Conjecture in \cite{BassMorgan}, the existence of such symmetry can be rephrased in the following 
way. Let $\rho_{n}$ be the rotation of $\mathbb{R}^{3}$ by the $\frac{2\pi}{n}$ angle about the 
$OZ$ 
axis. We are interested in knots $K \subset \mathbb{R}^{3}$, which are disjoint from the $OZ$ axis 
and invariant under $\rho_{n}$. A knot $K$ is $n$-periodic if it admits such rotational symmetry. 
$n$-periodic links are defined analogously.

The importance of periodic links stems from the fact, that according to \cite{PrzytyckiSokolov}, 
a $3$-manifold $M$ admits an action of the cyclic group of prime order $p$ with the fixed point set 
being an unknot if, and only if, it can be obtained as a surgery on a $p$-periodic link. 
Additionally, $M$ admits a free action of the cyclic group $\z{p}$ if, and only if, it can be 
obtained as a surgery on a link formed by taking a $p$-periodic link $L$ together with the fixed 
point axis $F$. Hence, these two classes of cyclic symmetries of $3$-manifolds are determined by 
symmetries of their Kirby diagrams.

Another possible application of periodic links is, according to Przytycki \cite{Przytycki}, to 
give a unified theory of skein modules for branched and unbranched coverings. Skein module of a 
$3$-manifold $M$ is a certain algebraic objects associated to $M$, which serves as a 
generalization of a certain polynomial link invariant, like the Jones polynomial, for links in $M$. 
For more details on skein modules refer to \cite{Przytycki2}.

There are many techniques at hand to study periodic knots. The first significant results were 
obtained by Trotter in \cite{Trotter}, where the author studies actions of the cyclic group on the 
fundamental group of the complement of the knot, to derive all possible periods of torus links. 
Murasugi studied periodic links with the aid of the Alexander polynomial in \cite{Murasugi2}, 
obtaining very strong criterion for detecting periodicity. In \cite{DavisLivingston} authors  
consider the question whether a Laurent polynomial, which satisfies the congruence of Murasugi, can 
be realized as the Alexander polynomial of a periodic link.

Several authors studied the Jones polynomial of periodic links. The first result in this direction 
was obtained by Murasugi in \cite{Murasugi1}. Besides that, \cites{Przytycki,Traczyk,Yokota} give 
other criteria for detecting periodicity of knots in terms of their Jones polynomial. Several other 
authors \cites{PrzytyckiSikora,Chbili1,Przytycki,Traczyk2,Traczyk3,Yokota2} studied 
$SU_{n}$-quantum polynomials and the HOMFLYPT polynomial of periodic links. A summary of these 
results can be found in \cite{Przytycki3}.

Khovanov in \cite{Khovanov1} made a breakthrough in knot theory, by constructing certain homology 
theory of links, called the Khovanov homology, which categorifies the Jones polynomial, i.e., the 
Jones polynomial can be recovered from the Khovanov homology as an appropriately defined Euler 
characteristic. Hence, it is natural to ask whether this homology theory can be utilized to study 
periodic links. The first such attempt was made in \cite{Chbili2}. However, the author works only 
with $\z{2}$ coefficients due to certain technical problem with signs, which appears along the 
way. Nevertheless, the author obtains an invariant of periodic links and shows, via transfer 
argument, that his invariant is isomorphic to the submodule of fixed points of the induced action 
on the Khovanov homology.

The purpose of this article is to construct and study a generalized version of the invariant from 
\cite{Chbili2}, called the equivariant Khovanov homology of periodic links. If $p$ is an odd prime, 
then the equivariant Khovanov homology from \cite{Chbili2} is isomorphic to 
$\kh_{\z{p}}^{\ast,\ast,1}(D; \z{2})$, see Definition \ref{defn: equiv kh}. This is a consequence 
of the fact, that the group algebra $\GpAAlg{\z{2}}{\z{p}}$ is semi-simple.

In a subsequent paper \cite{Politarczyk} we define equivariant analogues of the Jones polynomial 
and 
study their properties. We prove that they satisfy certain variant of the skein relation and use 
this property to derive certain periodicity criterion, which generalizes the ones from 
\cites{Przytycki,Traczyk}. Moreover, we study the state sum expansion of these polynomials, which 
leads to a proof of the congruence from \cite{Murasugi1}.

The paper is organized as follows. Section \ref{sec: preliminaries} contains a summary of some 
facts 
from representation theory and homological algebra which are essential in further considerations.

In section \ref{sec: periodic links} we show that if $D$ is an $n$-periodic diagram of an 
$n$-periodic link $L$, then the Khovanov complex $\cc(D)$ becomes a cochain complex of graded 
$\cyclicGroupRing{n}$-mo\-dules. This enables us to study the Khovanov homology of periodic links 
with arbitrary coefficients.

In section \ref{subsec: equiv kh z}, with the aid of integral representation theory of cyclic 
groups, we construct the equivariant Khovanov homology -- denoted by 
$\kh_{\z{n}}^{\ast,\ast,\ast}(D)$ -- a triply graded homology theory, where the third grading is 
supported only in positive degrees, which divide $n$. Further, we show that this is indeed an 
invariant of periodic links, utilizing machinery from \cite{BarNatan}.

\begin{myThm}{\ref{thm:invariance_equivariant_homology}}
{\itshape Equivariant Khovanov homology groups are invariants of periodic links, that is, they are 
invariant under equivariant Reidemeister moves.}
\end{myThm}

\noindent For the definition of equivariant Reidemeister moves see \ref{defn: equiv Reidemeister 
moves}. Intuitively, equivariant Reidemeister moves are deformation of periodic link diagrams in 
which we perform simultaneously $n$ Reidemeister moves in such a way that the symmetry of the 
diagram is preserved at all times. Additionally, we cannot cross the fixed point axis while 
performing the equivariant Reidemeister moves.

Next, we investigate the relation between the equivariant Khovanov homology and the 
classical Khovanov homology.

\begin{myThm}{\ref{thm:total_equivariant_homology}}
{\itshape Let $p_{1}, \ldots, p_{s}$ be the collection of all prime divisors of $n$. Define the 
ring $R_{n} = \gpRing{\frac{1}{p_{1}}, \frac{1}{p_{2}}, \ldots, \frac{1}{p_{s}}}.$
Under this assumption there exists a natural map
$$\bigoplus_{d | r} \kh_{\z{n}}^{\ast, \ast, d}(L) \to \kh(L)$$
which, when tensored with $R_{n}$, becomes an isomorphism.}
\end{myThm}

Hence, the equivariant Khovanov homology, after collapsing the third grading, encodes the same 
information as the classical Khovanov homology, modulo torsion of order dividing $n$.

Further, we analyze the structure of this invariant for trivial links. It turns out, that the 
equivariant Khovanov homology of $p^{n}$-periodic trivial link, for some prime $p$, is expressible 
in terms of the group cohomology of subgroups of $\z{p^{n}}$ with coefficients in cyclotomic rings, 
see Proposition \ref{prop: comp equiv kh trivial links}.

Computations of the equivariant Khovanov homology of trivial links suggests that, unlike the 
classical Khovanov homology, $\kh_{\z{n}}^{i,\ast,d}(D)$ can be non-trivial for $i > n_{+}(D)$, 
where $n_{+}(D)$ stands for the number of positive crossings of the diagram $D$. Theorem 
\ref{thm:total_equivariant_homology} implies, that for such values of $i$, the abelian group 
$\kh_{\z{n}}^{i,\ast,d}(D)$ is necessarily torsion. Our next goal is to use an additional algebraic 
structure of the equivariant Khovanov homology to extract some information about the additional 
torsion. This additional algebraic structure manifest itself in the fact that for any $0 \leq s 
\leq 
n$, $\kh_{\z{p^{n}}}^{\ast,\ast,p^{s}}(D)$ is a graded module over the graded ring 
$\Ext{\ast}{\cyclicGroupRing{p^{n}}}{\cyclotomicRing{p^{s}}}{\cyclotomicRing{p^{s}}}$. This ring is 
isomorphic to certain quotient of the polynomial ring $\GpAAlg{\cyclotomicRing{d}}{T_{s}}$. 
Analysis of this structure yields the periodicity result, which can be thought of as an analogue of 
the periodicity of the cohomology groups of the cyclic groups.

\begin{myCor}{\ref{cor: periodicity}}
{\itshape Let $m \in \zzz$ be chosen such that $\kh^{m,\ast}(D) \neq 0$ and $\kh^{i,\ast}(D)$ 
vanishes for $i 
> m$. Let $T_{s}$ denote the cohomology class in the ext ring
$$T_{s} \in \ext{2}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$$
from Proposition 
\ref{prop: properties of ext rings of cylotomic rings}. Then, multiplication by $T_{s}$
$$- \cup T_{s} \colon \equivKh{{p^{n}}}{i}{\ast}{p^{s}}(D) \to 
\equivKh{{p^{n}}}{i+2}{\ast}{p^{s}}(D)$$
is an epimorphism for $i=m$ and isomorphism for $i> m$.}
\end{myCor}

As a consequence we can obtain some information about the additional torsion appearing in the 
equivariant Khovanov homology.

\begin{myCor}{\ref{cor: torsion}}
{\itshape Let $m$ be as in the previous corollary. For $i > m$, $\equivKh{p^{n}}{i}{\ast}{1}(D)$ is 
annihilated by $p^{n}$, and for $1 \leq s \leq n$, $\equivKh{p^{n}}{i}{\ast}{p^{s}}(D)$ is 
annihilated by $p^{n-s+1}$.}
\end{myCor}

In section \ref{subsec: equiv kh q} we study the structure of the rational equivariant Khovanov 
homology. These considerations are sufficient to compute the rational equivariant Khovanov homology 
of torus links $T(n,2)$ and, if $\gcd(n,3)=1$, for torus knots $T(n,3)$, with respect to the 
$\z{d}$-symmetry, provided that $d \mid n$ is odd and greater than $2$. In all of these cases we 
have
\begin{align*}
\kh_{\z{d}}^{\ast,\ast,1}(D; \qqq) &= \kh^{\ast,\ast}(D; \qqq), \\
\kh_{\z{d}}^{\ast,\ast,k}(D; \qqq) &= 0, \quad k>1, \quad k \mid d.
\end{align*}

Theorem \ref{thm:spectral_sequence_equivariant_homology}, which is the main result of Section 
\ref{sec: spectral sequence}, yields a spectral sequence converging to the equivariant Khovanov 
homology of a periodic link. Since the long exact sequence of Khovanov homology, coming from two 
resolutions of a single crossing of $D$, cannot be adapted to the equivariant setting, the spectral 
sequence is supposed to fill in this gap and provide a computational tool. Instead of resolving a 
single crossing, we resolve crossings from a single orbit. We take all possible resolutions of 
these crossings and assemble these data into a spectral sequence. In the non-equivariant setting we 
can recover from this construction the spectral sequence from \cite{EverittTurner}.

The above spectral sequence is used, in Section \ref{sec: computations}, to compute the rational 
$2$-equivariant Khovanov homology of torus links, i.e. the equivariant Khovanov homology with 
respect to the $\z{2}$-symmetry. It turns out, that something analogous happen as in the case of 
symmetries of order $d > 2$. Namely, almost always the only non-trivial part is 
$\kh_{\z{2}}^{\ast,\ast,1}$, with an exception of torus links $T(2n,2)$ for which
$$\kh_{\z{2}}^{i,j,2}(T(2n,2); \qqq) = \left\{ 
\begin{array}{ll}
\qqq, & i = 2n, \quad j = 6n, \\
0, & \mbox{otherwise}. \\
\end{array}
\right.
$$

\paragraph*{Convention} In the remainder part of this article we adopt the convention that all 
links and tangles are oriented unless stated otherwise.

\paragraph*{Acknowledgments} The author is very grateful to Józef Przytycki for bringing 
this problem to his attention, to Maciej Borodzik and Prof. Krzysztof Pawałowski for their 
suggestions and corrections which improved the presentation of the paper.

\section{Preliminaries}
\label{sec: preliminaries}

\subsection{Representation theory}

Before we start, we will briefly recall some notions from integral representation theory that are 
essential in the remainder part of this thesis. The exposition of the material in this section is 
based on \cite{CurtisReiner}.

\begin{defn}
Let $A$ be a semi-simple and finite dimensional $\qqq$-algebra. We say that $\Lambda \subset 
A$ is a \textit{$\zzz$-order} if it is a subring of $A$ which contains the unit and some 
$\mathbb{Q}$-basis of $A$.
\end{defn}

\noindent An example of a $\zzz$-order is the group ring $\gpRing{G}$, of a finite group $G$,
contained in the rational group algebra $\gpAlg{G}$.

\begin{defn}
Let $\Lambda$ be a $\mathbb{Z}$-order in a semi-simple $\qqq$-algebra $A$. We say that $\Lambda$ is 
\textit{maximal} if it is not contained in any other $\mathbb{Z}$-order in $A$.
\end{defn}

\begin{thm}[\cite{CurtisReiner}*{Thm. 26.20}]
Let $A$ be a finite dimensional semi-simple $\mathbb{Q}$-algebra.
\begin{enumerate}
\item Every $\mathbb{Z}$-order $\Lambda \subset A$ is contained in a some maximal order $\Lambda'$. 
\item If $A$ is commutative, then it possesses a unique maximal order.
\item If we are given a Wedderburn decomposition of $A$
$$A = A_1 \oplus \ldots \oplus A_k,$$
then every maximal order $\Lambda' \subset A$ admits a decomposition into a direct sum of ideals
$$\Lambda' = \Lambda'_1 \oplus \ldots \oplus \Lambda'_k,$$
where each $\Lambda'_i \subset A_i$ is a maximal order in $A_i$ for $i = 1, \ldots, k$. In fact, if 
$e_{i}$ is a central idempotent in $A$ which acts as an identity on $A_{i}$ and by $0$ on $A_{j}$ 
for $j \neq i$, then $\Lambda_{i}' = e_{i}\Lambda'$.
\end{enumerate}
\end{thm}

\begin{ex}
The group algebra $\cyclicGroupAlg{n}$ admits the following Wedderburn decomposition
$$\cyclicGroupAlg{n} = \bigoplus_{d | n} \cyclotomicField{d}. $$
The maximal order in $\cyclotomicField{d}$ is the ring of cyclotomic integers $\cyclotomicRing{d}$. 
Therefore, the unique maximal order $\Lambda' \subset \cyclicGroupAlg{n}$ is equal to the following 
direct sum
$$\Lambda' = \bigoplus_{d | n} \cyclotomicRing{d}.$$
\end{ex}

\begin{prop}[\cite{CurtisReiner}*{Prop 27.1}]\label{prop:long_exact_seq_maximal_order}
Let $\Lambda'$ be a maximal order in $\cyclicGroupAlg{n}$. Under this assumption, the following 
chain of inclusions is satisfied.
$$\cyclicGroupRing{n} \subset \Lambda' \subset \frac{1}{n} \cyclicGroupRing{n}.$$
Therefore, there exists an exact sequence
$$0 \to \cyclicGroupRing{n} \to \Lambda' \to M \to 0,$$
where $n \cdot M = 0$.
\end{prop}

\begin{defn}
Let $m \leq n$ be positive integers and suppose that $L$ is a $\cyclicGroupRing{m}$-module and $T$ 
is a $\cyclicGroupRing{n}$-module. Define the \textit{induction of} $L$ to be the 
$\cyclicGroupRing{n}$-module
$$\ind{\cyclicGroupRing{n}}{\cyclicGroupRing{m}} L = L \otimes_{\cyclicGroupRing{m}} 
\cyclicGroupRing{n},$$
where $\cyclicGroupRing{n}$ is considered as a $\cyclicGroupRing{m}$ module via the inclusion
$$\z{m} \hookrightarrow \z{n}.$$
Define the \textit{restriction of} $T$ to be the $\cyclicGroupRing{m}$-module, whose underlying 
abelian group remains the same, however we restrict the action of $\z{n}$ to the action of its 
subgroup isomorphic to $\z{m}$.
\end{defn}


\begin{prop}\label{prop: decomposition of cyclotomic rings}
Let $p$ be a prime and $n$ a positive integer. Choose $0 \leq s, m \leq n$, then
$$\res{\cyclicGroupRing{p^n}}{\cyclicGroupRing{p^{m}}} \cyclotomicRing{p^{n-s}} = \left\{
\begin{array}{ll}
\zzz^{\varphi(p^{n-s})}, & m \leq s, \\
\cyclotomicRing{p^{m-s}}^{p^{n-m}}, & m > s. \\
\end{array}
\right.
$$
\end{prop}
\begin{proof}
Since $\res{\z{p^{n}}}{\z{p^{m}}} \cyclotomicRing{p^{n-s}}$ is the maximal order in 
$\res{\z{p^{n}}}{\z{p^{m}}} \cyclotomicField{p^{n-s}}$, it is sufficient to check this equality for 
the cyclotomic fields. Let $\chi_{p^{n-s},p^{n}}$ denote the character of 
$\cyclotomicField{p^{n-s}}$ treated as an irreducible $\cyclicGroupAlg{p^{n}}$-module.

\begin{lemma}
Consider the cyclic group $\z{p^{n}}$, where $p$ is a prime. Let $0 \leq s <n$, $0 \leq j 
\leq p^{n-s} -1$ and $0 \leq m \leq p^s - 1$. The characters of $\z{p^{n}}$ are given by the 
following formulas. 
\begin{align*}
\chi_{1,p^n}(t^j) &= 1, \\
\chi_{p^{n-s}, p^{n}}(t^{j + m \cdot p^{n-s}}) &= \left\{
\begin{array}{ll}
\phi(p^{n-s}), & j = 0, \\
-p^{n-s-1},    & j \mid p^{n-s-1}, \\
0, & \mbox{otherwise}, \\
\end{array}
\right.\\
\end{align*}
where $0 \leq j \leq p^{n-s} -1$.
\end{lemma}
\begin{proof}
This follows from \cite{CurtisReiner}*{Cor. 7.19} applied to the complex characters of 
$\z{p^{n}}$.
\end{proof}

\noindent The above lemma implies that the following formulas hold.
$$\res{\z{p^{n}}}{\z{p^{m}}} \chi_{p^{n-s}, p^{n}} = \left\{
\begin{array}{ll}
\varphi(p^{n-s}) \chi_{1, p^m}, & m \leq s, \\
p^{n-m} \chi_{p^{m-s},p^{m}},  & m > s. \\
\end{array}
\right.$$

\noindent Hence, the proposition follows.
\end{proof}

\subsection{Ext groups}

The purpose of this section is to sketch the theory of the derived Hom functor. The exposition is 
based on \cite{StacksProjectDerived}.

To fix the notation, assume that $C^{\ast}$, $D^{\ast}$ and $E^{\ast}$ are bounded 
cochain complexes.

\begin{defn}
Let $C^{\ast}$ be a cochain complex. For $n \in \mathbb{Z}$ denote by $C[n]^{\ast}$ a new cochain 
complex obtained from $C^{\ast}$ by applying the following shift
\begin{align*}
C[n]^k &= C^{k-n}, \\
d_{C[n]}^k &= (-1)^{n} d_{C}^{k-n}. \\
\end{align*}
\end{defn}

The category of cochain complexes can be equipped with the structure of the differential graded 
category i.e. morphism sets can be made into cochain complexes themselves.

\begin{defn}
For two cochain complexes, $C^{\ast}$ and $D^{\ast}$, define the Hom complex
$\Hom{\ast}{R}{C^{\ast}}{D^{\ast}}$ to be
$$\Hom{n}{R}{C^{\ast}}{D^{\ast}} = \prod_{p \in \zzz} \hom_{R}(C^{p}, D^{p+n}), \quad n \in \zzz$$
and equip it with the following differential.
$$d^{C^{\ast}, D^{\ast}}(\psi_{p})_{p} = \left( d_{D} \circ \psi_{p} - (-1)^{p} \psi_{p+1} \circ 
d_{C} \right)_{p}$$ 
\end{defn}

\begin{rem}
Notice that since $C^{\ast}$ and $D^{\ast}$ are bounded, the Hom complex  
$\Hom{\ast}{R}{C^{\ast}}{D^{\ast}}$ is also bounded.
\end{rem}

\noindent The next proposition is a mere reformulation of the definition of the chain homotopy.

\begin{prop}
The following equalities hold
$$H^{n}\left( \Hom{\ast}{R}{C^{\ast}}{D^{\ast}} \right) = \left[ C^{\ast}, D^{\ast}[-n] \right],$$
where the outer square brackets denote the set of homotopy classes of chain maps.
\end{prop}

Analogously as in the case of modules, in order to preserve the exactness of the Hom functor, we 
need to replace every cochain complex by its injective or projective resolution.

\begin{defn}
Let $C^{\ast}$ be cochain complex. Let $I^{\ast}$ be bounded below cochain complex of injective 
modules. We say that $I^{\ast}$ is an \textit{injective resolution} of $C^{\ast}$ if there exists a 
quasi-isomorphism
$$C^{\ast} \to I^{\ast}.$$
\end{defn}

Categories of modules have enough injectives, which guarantees that every module possesses an 
injective resolution. It turns out that the same condition is sufficient to construct 
an injective resolution of a cochain complex, provided that the complex is bounded from below, see 
\cite{StacksProjectDerived}*{\StacksTag{013K}{Lemma 18.3}}

\begin{defn}
Let $I^{\ast}$ be an injective resolution of a cochain complex $D^{\ast}$. Define the derived Hom 
complex
$$\operatorname{R}\Hom{\ast}{R}{C^{\ast}}{D^{\ast}} = \Hom{\ast}{R}{C^{\ast}}{I^{\ast}}.$$
Ext groups are defined as the homology of the derived Hom complex.
$$\Ext{n}{R}{C^{\ast}}{D^{\ast}} = H^{n}\left( \Hom{\ast}{R}{C^{\ast}}{I^{\ast}} \right).$$ 
In other words,
$$\Ext{n}{R}{C^{\ast}}{D^{\ast}} = \left[ C^{\ast}, I^{\ast}[-n] \right].$$
\end{defn}

\noindent Properties of classical Ext groups extended to their generalized version.

\begin{prop}\label{prop: properties of ext}
Let $C^{\ast}$ and $D^{\ast}$ be as in the previous definition.
\begin{enumerate}
\item If $C^{'\ast}$ and $D^{'\ast}$ are bounded cochain complexes quasi-isomorphic to $C^{\ast}$ 
and $D^{\ast}$, respectively, then these quasi-isomorphisms induce the following isomorphisms,
$$\Ext{n}{R}{C^{\ast}}{D^{\ast}} \cong \Ext{n}{R}{C^{'\ast}}{D^{'\ast}}, \quad n \in \zzz.$$
\item If
$$0 \to C_{1}^{\ast} \to C_{2}^{\ast} \to C_{3}^{\ast} \to 0$$
is a short exact sequence of bounded cochain complexes, then there exists a long exact 
sequence of Ext groups
$$\ldots \to \Ext{n}{R}{C_{3}^{\ast}}{D^{\ast}} \to \Ext{n}{R}{C_{2}^{\ast}}{D^{\ast}} \to 
\Ext{n}{R}{C_{1}^{\ast}}{D^{\ast}} \to \ldots.$$
\item Analogously, if
$$0 \to D_{1}^{\ast} \to D_{2}^{\ast} \to D_{3}^{\ast} \to 0$$
is a short exact sequence of bounded cochain complexes, then there exists a long exact 
sequence of Ext groups
$$\ldots \to \Ext{n}{R}{C^{\ast}}{D_{1}^{\ast}} \to \Ext{n}{R}{C^{\ast}}{D_{2}^{\ast}} \to 
\Ext{n}{R}{C^{\ast}}{D_{3}^{\ast}} \to \ldots.$$
\item There exist bilinear maps, induced by composition of maps on the cochain level, 
\begin{align*}
\mu \colon \Ext{n}{R}{C^{\ast}}{D^{\ast}} \times \Ext{m}{R}{B^{\ast}}{C^{\ast}} &\to 
\Ext{n+m}{R}{B^{\ast}}{D^{\ast}}. \\
\end{align*}
Hence, $\Ext{\ast}{R}{C^{\ast}}{C^{\ast}}$ and $\Ext{\ast}{R}{D^{\ast}}{D^{\ast}}$ are 
graded rings. Additionally $\Ext{\ast}{R}{C^{\ast}}{D^{\ast}}$ is a graded
$\left(\Ext{\ast}{R}{D^{\ast}}{D^{\ast}},  \Ext{\ast}{R}{C^{\ast}}{C^{\ast}} 
\right)$-bimodule.
\end{enumerate}
\end{prop}

\begin{proof}
See the discussion after the \StacksTag{06XQ}{Definition 27.1} in \cite{StacksProjectDerived}.
\end{proof}


\noindent Classical result of Cartan and Eilenberg provides two spectral sequences which converge 
to the higher derived functors of Hom.

\begin{thm}[\cite{StacksProjectDerived}*{\StacksTag{015J}{Lemma 21.3}}]
\label{thm: Cartan Eilenberg spectral seq}
There are two spectral sequences
$$\{{}_{I}E_{r}^{\ast,\ast}, d_r\}, \{{}_{II}E_{r}^{\ast,\ast}, d_r\}$$
converging to $\Ext{\ast}{R}{C^{\ast}}{D^{\ast}}$ satisfying
\begin{align*}
{}_{I}E_{2}^{p,q} &= \Ext{p}{R}{C^{\ast}}{H^{q}(D)} \\
{}_{II}E_{2}^{p,q} &= H^{q}\left( \Ext{\ast}{R}{C^{\ast}}{D^{p}} \right)\\
\end{align*}
\end{thm}

\noindent Part 4 of Proposition \ref{prop: properties of ext} shows that Ext groups possess a 
multiplicative structure. This additional structure is derived from certain bilinear maps defined 
on the cochain level. The bilinear map is compatible with the filtration of Cartan and Eilenberg, 
thus its existence is manifested in the Cartan-Eilenberg spectral sequence.

\begin{thm}\label{thm: multiplicative structure ext}
For cochain complexes $B^{\ast}$, $C^{\ast}$ and $D^{\ast}$ there are bilinear maps of spectral 
sequences
\begin{align*}
\mu \colon {}_{I}E_{r}^{p,q}(C^{\ast}, D^{\ast}) \times {}_{I}E_{r}^{p',q'}(B^{\ast}, C^{\ast}) 
&\to 
{}_{I}E_{r}^{p+p',q+q'}(B^{\ast}, D^{\ast}) \\
\mu \colon {}_{II}E_{r}^{p,q}(C^{\ast}, D^{\ast}) \times {}_{II}E_{r}^{p',q'}(B^{\ast}, C^{\ast}) 
&\to 
{}_{II}E_{r}^{p+p',q+q'}(B^{\ast}, D^{\ast}) \\
\end{align*}
commuting with differentials i.e.
$$d^{B^{\ast},D^{\ast}}_{r}(\mu(x,y)) = \mu(d^{C^{\ast}, D^{\ast}}_{r}(x),y) + (-1)^{p+q} \mu(x, 
d^{B^{\ast}, C^{\ast}}_{r}(y)),$$
and converging to bilinear maps from Proposition \ref{prop: properties of ext}.
\end{thm}

\noindent When a cochain complex $D^{\ast}$ is equipped with a filtration, the Hom complex 
$\Hom{}{R}{C^{\ast}}{D^{\ast}}$ becomes filtered. This filtration is defined by considering 
homomorphisms whose images are contained in the respective submodule of the filtration of 
$D^{\ast}$. Moreover, the filtration of $D^{\ast}$ induces a filtration on the derived Hom complex. 
This leads to a spectral sequence.

\begin{thm}\label{thm: spectral sequence derived filtration}
Suppose that $D^{\ast}$ is a bounded and filtered cochain complex. Then there exists a spectral 
sequence $\{E_{r}^{\ast,\ast}, d_{r}\}$ converging to $\Ext{n}{R}{C^{\ast}}{D^{\ast}}$ such that
$$E_{1}^{p,q} = \Ext{p+q}{R}{C^{\ast}}{\mathcal{F}_{p}(D^{\ast}) / \mathcal{F}_{p+1}(D^{\ast})}.$$
\end{thm}

The next lemma provides a description of Ext rings of cyclotomic rings. This description will be 
used later to obtain some information about the torsion of the equivariant Khovanov homology.

\begin{prop}\label{prop: properties of ext rings of cylotomic rings}
Let $p$ be a prime and let $0 \leq s \leq n$ be integers. Denote by $\Phi_{p^{s}}(t)$ the 
$p^{s}$-th cyclotomic polynomial. Under these assumptions, there exists an isomorphism
$$\ext{\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}} \cong 
\gpRing{\xi_{p^{s}}}[T_{s}] / (\Phi_{p^{s},p^{n}}(\xi_{p^{s}}) T_{s}),$$
where
$$\Phi_{p^{s},p^{n}}(t) = \frac{t^{p^{n}} - 1}{\Phi_{p^{s}}(t)}$$
and $T_{s} \in \ext{2}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$ is a class 
represented by the following Yoneda extension
$$0 \to \gpRing{\xi_{p^{s}}} \to \gpRing{\z{p^{n}}} \stackrel{\Phi_{p^{s}}(t)}{\to} 
\gpRing{\z{p^{n}}} \to \gpRing{\xi_{p^{s}}} \to 0.$$
Additionally, for any $\gpRing{\z{p^{n}}}$-module $N$, multiplication by $T_{s}$
$$- \cup T_{s} \colon \ext{i}{\gpRing{\z{p^{n}}}}{\cyclotomicRing{p^{s}}}{N} \to 
\ext{i+2}{\gpRing{\z{p^{n}}}}{\cyclotomicRing{p^{s}}}{N}$$
is an isomorphism for $i > 0$ and epimorphism for $i=0$.
In particular
$$\ext{2i}{\cyclicGroupRing{p^{n}}}{\cyclotomicRing{p^{s}}}{\cyclotomicRing{p^{s}}} = \left \{
\begin{array}{ll}
\z{p^{m}}, & i>0, s=0, \\
\z{p^{m-s+1}}, & i>0, s > 0, \\
\cyclotomicRing{p^{s}}, & i = 0, \\
0, & \mbox{otherwise}. \\
\end{array}
\right.$$
\end{prop}
\begin{proof}
The first part follows from \cite{Wall}*{Lemma 1.1}. To prove the second part, notice that
\begin{align*}
\Phi_{p^{s},p^{m}}(\xi_{p^s}) &= \lim_{z \to \xi_{p^s}} \frac{z^{p^m}-1}{\Phi_{p^s}(z)} = 
\lim_{z \to \xi_{p^s}} (z^{p^{s-1}}-1) \frac{z^{p^m}-1}{z^{p^{s}}-1} = \\
&= p^{m-s} (\xi_{p}-1).
\end{align*}
by de L'Hospital rule. Since the algebraic norm of $\xi_{p}-1$ is equal to $p$, it follows readily 
that
$$\cyclotomicRing{p^{s}} / (\Phi_{p^{s},p^{n}}(\xi_{p^{s}}))  \cong \z{p^{n-s+1}}.$$
\end{proof}

\section{Periodic links}
\label{sec: periodic links}

Let us start with the recollection of the definition of a periodic link.

\begin{defn}
Let $n$ be a positive integer, and let $L$ be a link in $S^{3}$. We say that $L$ is $n$-periodic, 
if there exists an action of the cyclic group of order $n$ on $S^3$ satisfying the following 
conditions. 
\begin{enumerate}
\item The fixed point set, denoted by $F$, is the unknot.
\item $L$ is disjoint from $F$.
\item $L$ is a $\z{n}$-invariant subset of $S^3$.
\end{enumerate}
\end{defn}

\begin{figure}
\centering
\import{./}{splot1.pdf_tex}
\caption{Borromean rings are $3$-periodic. The fixed point axis $F$ is marked with a dot.}
\label{fig: bor rings 3-periodic}
\end{figure}

\begin{ex}
Borromean rings provide an example of a $3$-periodic link. The symmetry is visualized on Figure 
\ref{fig: bor rings 3-periodic}. The dot marks the fixed point axis.
\end{ex}

\begin{ex}\label{ex: periodic torus links}
Torus links constitute an infinite family of periodic links. In fact, according to \cite{Murasugi2},
the torus link $T(m,n)$ is $d$-periodic if, and only if, $d$ divides either $m$ or $n$.
\end{ex}

Periodic diagrams of periodic links can be described in terms of planar algebras. For the 
definition of planar algebras see \cite{BarNatan}. Take an $n$-periodic planar diagram $D_{n}$ 
with $n$ input disks, like the one on Figure \ref{fig: planar diagram periodic}. Choose a tangle $T$ 
which possesses enough endpoints, and glue $n$ copies of $T$ into the input disks of $D_{n}$. In 
this way, we obtain a periodic link whose quotient is represented by an appropriate closure of $T$. 
See Figure \ref{fig: planar diagram periodic link} for an example.

\begin{figure}
\centering
\import{./}{planar_algebra_periodic.pdf_tex}
\caption{$4$-periodic planar diagram.}
\label{fig: planar diagram periodic}
\end{figure}

\begin{figure}
\centering
\import{./}{planar_algebra_periodic_link.pdf_tex}
\caption{Torus knot $T(3,4)$ as a $4$-periodic knot obtained from the planar diagram from Figure 
\ref{fig: planar diagram periodic}}
\label{fig: planar diagram periodic link}
\end{figure}

Using this description of periodic links, it is possible to exhibit a cobordism which induces 
an action of $\z{n}$ on the Khovanov bracket $\khbracket{D}$, as defined in \cite{BarNatan}, where 
$D$ is a periodic link diagram. First, notice that we can assume that $D$ represents a link in 
$D^{2} \times I$ and the symmetry comes from a rotation of the $D^{2}$ factor, where $D^{2}$ denotes 
the $2$-disk. In order to construct the cobordism, notice that the diffeomorphism, denote it by 
$f$, generating the $\z{n}$-symmetry of $D^{2} \times I$, is isotopic to the identity. Indeed, this 
isotopy can be chosen in such a way that it changes the angle of rotation linearly from $0$ to 
$\frac{2\pi}{n}$. Denote this isotopy by $H$. The cobordism in question is the trace of $H$.
$$\Sigma_{H} = \{(H(x, t), t) \in D^{2} \times I \times I \colon x \in L, \quad t \in 
I \}.$$
Cobordism $\Sigma_{H}$ is diffeomorphic to the cylinder $S^{1} \times I$, however it is not 
isotopic, rel boundary, to the cylinder, which is equal to $\Sigma_{H_{0}}$, where $H_{0}$ 
denotes the constant isotopy from the identity to the identity. However, $\Sigma_{H}$ is 
invertible in $\Kobh{2\ell}$, because the composition $\Sigma_{\overline{H}} \circ \Sigma_{H}$, 
where $\overline{H}(\cdot, t) = H(\cdot, 1-t)$, is isotopic to $\Sigma_{H_{0}}$, rel boundary.

\begin{figure}
\centering
\includegraphics{tqft_image.pdf}
\caption{Periodic Kauffman state with $3$ components and symmetry of order $2$. Middle cylinder 
contains the fixed point axis $F$.}
\label{fig: periodic Kauffman state}
\end{figure}

Before proceeding further, one remark is in order. During the construction of the Khovanov's 
bracket $\khbracket{D}$ it was necessary to multiply each summand of the differential with $\pm 1$. 
This particular choice of signs forces us to do the same with maps between complexes.

Recall from \cite{BarNatan} that we need to number all crossings of $D$ and based on that impose 
an ordering of the set of crossings. Consider two vector spaces -- $W$ spanned by the crossings of 
$D$ and its exterior algebra $V = \Lambda^{\ast} W$. Each Kauffman state of $D$ can be 
labeled with a unique vector from the distinguished basis of $V$, consisting of vectors of the form
$$c_{i_{1}} \wedge c_{i_{2}} \wedge \ldots \wedge c_{i_{k}}, \quad i_{1} < i_{2} < \ldots < i_{k}$$
where $c_{i_{j}} \in \cross{D}$ is a crossing of $D$. Choose a tangle $T$ and let $W_{T}$ be the 
vector space associated to $T$ and $D = D_{n}(T, \ldots, T)$ with $W_{D}$ defined analogously. Under 
this assumptions 
$$W_{D} \cong W_{T}^{n}, \quad \mbox{ and } \quad \Lambda^{\ast} W_{D} \cong \left(\Lambda^{\ast} 
W_{T}\right)^{\otimes n}.$$
Symmetry of $D$ induces an action of $\z{n}$ on $\Lambda^{\ast} W_{D}$, which permutes factors in 
the above decomposition. Cobordism $\Sigma_{H}$ discussed above induces a map
$$\Sigma_{H} \colon \khbracket{D} \to \khbracket{D}$$
which permutes Kauffman states of $D$. This permutation is compatible with the induced action 
on $\Lambda^{\ast} W_{D}$. Geometrically, the map $\Sigma_{H}|_{\khbracket{D}^{r}}$ is 
induced by a ``permutation'' cobordisms similar to the one from Figure \ref{fig: periodic Kauffman 
state}. However, additional sign is needed to assure that this map commutes with the differential. 
Let us define
\begin{align}\label{eqn: permutation map}
\psi \colon \left(\Lambda^{\ast} W_{T}\right)^{\otimes n} &\to \left(\Lambda^{\ast} 
W_{T}\right)^{\otimes n} \\
\psi \colon x_{1} \otimes x_{2} \otimes \ldots \otimes x_{n} &\mapsto (-1)^{\alpha} x_{2} 
\otimes \ldots \otimes 
x_{n} \otimes x_{1}, \quad x_{i} \in W_{T}, \nonumber
\end{align}
where
$$\alpha = (n-1) n_{-}(T) + \deg x_{1} (\deg x_{2} + \deg x_{3} + \ldots + \deg x_{n}).$$
Above, $n_{-}(T)$ stands for the number of negative crossings of $T$. Automorphism $\psi$ maps any 
vector from the distinguished basis of $\Lambda^{\ast} W_{D}$ to $\pm 1$ multiplicity of some other 
vector from the basis.
$$\psi(v) = \sign(\psi,v) w$$
We can utilize these signs to change the definition of $\Sigma_{H}$ as follows.
\begin{align}
\Sigma_{H}|_{D_{v}} &\colon D_{v} \to D_{w}, \nonumber \\
\Sigma_{H}|_{D_{v}} &= \sign(\psi,v) \Sigma_{v,w}, \label{eqn: Kauffman states}
\end{align}
where $\Sigma_{v,w}$ denotes the appropriate permutation cobordism. This discussion leads to the 
following proposition.

\begin{prop}\label{prop: periodic cochain complex}
If $D$ is a periodic link diagram, then $\cc(D)$ is a complex of graded $\gpRing{\z{n}}$-modules.
\end{prop}

\begin{rem}
This sign convention was implicitly described in \cite{BarNatan}.
\end{rem}

\begin{proof}[Proof of Prop. \ref{prop: periodic cochain complex}]
The only thing left to prove, is the commutativity of $\Sigma_{H}$ and the differential of 
$\cc(D)$. Geometric properties of the Khovanov bracket imply that the components of both maps 
commute up to sign. Hence, the only thing left to check is that all these maps really commute.

Let $x_{1}, \ldots, x_{n} \in V$ be homogeneous vectors. Consider the following linear maps
\begin{align*}
d_{D} \colon \left(\Lambda^{\ast} W_{T}\right)^{\otimes n} &\to \left(\Lambda^{\ast} 
W_{T}\right)^{\otimes n}, \\
d_{D} \colon x_{1} \otimes \ldots \otimes x_{n} &\mapsto \sum_{i=1}^{n}
(-1)^{\alpha_{i}}x_{1} \otimes \ldots \otimes d_{T}(x_{i}) \otimes 
\ldots \otimes x_{n}, \\
\sigma_{i} \colon \left(\Lambda^{\ast} W_{T}\right)^{\otimes n} &\to \left(\Lambda^{\ast} 
W_{T}\right)^{\otimes n}, \\
\sigma_{i} \colon x_{1} \otimes \ldots \otimes x_{n} &\mapsto 
(-1)^{\deg x_{i} \cdot \deg x_{i+1}}
x_{1} \otimes \ldots \otimes x_{i+1} \otimes x_{i} \otimes \ldots \otimes x_{n},\\
\widetilde{\sigma_{i}} &= (-1)^{n_{-}(T)} \sigma_{i}. 
\end{align*}
where $1 \leq i \leq n-1$, $\alpha_{i} = (-1)^{\deg(x_{n}) + \ldots + \deg(x_{i+1})}$ and
$$d_{T}(w) = \sum_{v \in \cross{T}} w \wedge v.$$
Notice that the map $\psi$ from (\ref{eqn: permutation map}) is expressible as the composition of 
the maps $\widetilde{\sigma}_{i}$ in the following way.
$$\psi = \widetilde{\sigma}_{n-1} \circ \widetilde{\sigma}_{n-2} \circ \ldots \circ 
\widetilde{\sigma}_{1}.$$
The map $d_{D}$, on the other hand, corresponds to the differential, i.e. let
$$d_{r-n_{-}(D)} \colon \cc^{r - n_{-}, \ast}(D) \to \cc^{r+1-n_{-}(D)}(D)$$
be the differential in the Khovanov complex. It can be expressed in the following form
$$d_{r-n_{-}(D)} = \sum_{(v,w)} \sign(v,w) \Sigma_{(v,w)},$$
where the summation extends over pairs $(v,w)$ consisting of $v \in \cross{D}$ and $w$ a vector 
from the distinguished basis of $\Lambda W_{D}$, such that $v \wedge w \neq 0$. Also, 
$\Sigma_{(v,w)}$ denotes the appropriate permutation cobordism.
It is not hard to check, that the coefficient $\sign(v,w)$ is equal to the coefficient of $v \wedge 
w$ in $d_{D}(v)$.

Therefore, it is sufficient to check that for $1 \leq i \leq n-1$ the following equality holds
$$\sigma_{i} \circ d_{D} = d_{D} \circ \sigma_{i}.$$
This can be verified by an elementary calculation.

\end{proof}

\noindent Let us now analyze the structure of the cochain complex $\cc(D)$.

\begin{defn}\label{defn: Kauffman states of periodic diagram}
\begin{enumerate}
\item Let $\mathcal{S}_{r}(D)$ denote the set of Kauffman states of $D$ which were obtained by 
resolving exactly $r$ crossings with the $1$-smoothing.
\item For $d \mid n$, let $\mathcal{S}^{d}(D)$ denote the set of Kauffman states which inherit 
a symmetry of order $d$ from the symmetry of $D$, that is Kauffman states of the form
$$D_{n}(\underbrace{T_1, \ldots, T_{\frac{n}{d}}, T_{1}, \ldots, T_{\frac{n}{d}}, 
\ldots, T_{1}, \ldots, T_{\frac{n}{d}}}_{\text{$d$ copies}}),$$
where $T_{1}, \ldots, T_{\frac{n}{d}}$ are distinct resolutions of $T$.
\item For a Kauffman state $s$, write $\Iso_{D}(s) = \z{d}$ if, and only if $s \in 
\mathcal{S}^{d}(D)$.
\item Define $\mathcal{S}_{r}^{d}(D) = \mathcal{S}^{d}(D) \cap \mathcal{S}_{r}(D)$.
\item Notice that there is an induced action of $\z{\frac{n}{d}}$ on $\mathcal{S}^{d}(D)$, indeed 
since all Kauffman states in $\mathcal{S}^{d}(D)$ are fixed by $\z{d}$. Define 
$\overline{\mathcal{S}}_{r}^{d}(D)$ to be the quotient of $\mathcal{S}_{r}^{d}(D)$ by 
the action of $\z{\frac{n}{d}}$.
\end{enumerate}
\end{defn}

\begin{rem}
If $\mathcal{S}^{d}_{r}(D)$ is non-empty, then $d \mid \gcd(n,r)$.
\end{rem}

\begin{defn}
  Let $\zzz_{-}$ be the following $\cyclicGroupRing{n}$-module.
$$\zzz_{-} = \left\{
\begin{array}{ll}
\cyclotomicRing{2}, & 2 \mid n, \\
\zzz, & 2 \nmid n.
\end{array}
\right.$$
In other words, if $n$ is even, the generator of the cyclic group $\z{n}$ acts on $\zzz_{-}$ 
by multiplication by $-1$, otherwise it is the trivial module.
\end{defn}


\begin{lemma}\label{lemma: action on Khovanov complex}
Let $\mathcal{T}$ be any TQFT functor whose target is the category of $R$-modules, for $R$ 
a commutative ring with unit. If $s_{1}, \ldots, s_{\frac{n}{d}} \in \mathcal{S}_{r}^{d}(D)$, for 
$d \mid \gcd(n,r)$ and $d \geq 1$, are Kauffman states constituting a single orbit, then
$$\bigoplus_{i=1}^{\frac{n}{d}} \mathcal{T}(\khbracket{s_{i}}) \cong \ind{\z{n}}{\z{d}} 
\left( \mathcal{T}(\khbracket{s_{1}}) \otimes_{\zzz} \zzz_{-}^{\otimes s(n,r,d)} \right).$$
as $\GpAAlg{R}{\z{n}}$-modules, where
$$s(n,r,d) = \frac{(n-1)n_{-}(D) + r(d-1)}{d}$$
\end{lemma}
\begin{proof}
We will prove that if $s \in \mathcal{S}_{r}^{d}(D)$, then $\Sigma_{H}^{\frac{n}{d}}(s) = 
(-1)^{s(n,r,d)}s$. The orbit of $s$ consists of $\frac{n}{d}$ Kauffman states which 
are permuted by the action of $\z{n}$. Hence, the lemma will follow from Proposition
\cite{CurtisReiner}*{Prop. 10.5} once we determine the induced action of $\z{d}$ on 
$T(\khbracket{s_{1}})$. Since $T(\khbracket{s_{1}})$ possesses a natural action of $\z{d}$, the 
induced action will differ from this one by a certain sign. Appearance of this sign is a 
consequence of our sign convention.

The Kauffman state $s_{1}$ corresponds to a vector of the form
$$w = \underbrace{v \otimes v \otimes \ldots \otimes v}_{d},$$
where $v = v_{1} \otimes v_{2} \otimes \ldots \otimes v_{\frac{n}{d}}$ and $v_{1}, 
\ldots, v_{\frac{n}{d}} \in \Lambda^{\ast}W_{T}$ belong to the distinguished basis. 
Consequently, according to the adopted sign convention,
\begin{align*}
&\psi^{\frac{n}{d}}(w) = (-1)^{k(d-1) + \frac{n_{-}(T)n(n-1)}{d}} w = (-1)^{\frac{r(d-1) + 
n_{-}(D)(n-1)}{d}} w,
\end{align*}
where $k = \deg v_{1} + \deg v_{2} + \ldots + \deg v_{\frac{n}{d}}.$
\end{proof}

\begin{cor}\label{cor: action on Khovanov complex}
If $\mathcal{T}$ is as in the previous lemma and $0 \leq r \leq n_{+} + n_{-}$, then
\begin{align*}
&\mathcal{T}(\khbracket{D}^{r - n_{-}}) = \\
&\bigoplus_{d \mid \gcd(n,r)} \bigoplus_{s 
\in \overline{\mathcal{S}}_{r}^{d}} \ind{\z{n}}{\z{d}} \left( \mathcal{T}(\khbracket{s}) 
\otimes_{\zzz} \zzz_{-}^{\otimes s(n,r,d)} \right)\{ r + n_{+}(D) - n_{-}(D)\}.
\end{align*}
\end{cor}

\begin{rem}
In the above formula there is a small ambiguity, since we identified a Kauffman state belonging to 
an orbit with this orbit. This notational shortcut does not cause any problems since all Kauffman 
states belonging to the same orbit yield isomorphic summands in $\khbracket{D}^{r - n_{-}}$. We 
will use this convention in the remainder part of this thesis.
\end{rem}

\begin{proof}[Proof of Cor. \ref{cor: action on Khovanov complex}]
Since for distinct $d_1, d_2 \mid \gcd(n,r)$ the sets $\mathcal{S}_{r}^{d_1}(D)$, 
$\mathcal{S}_{r}^{d_2}(D)$ are disjoint, the corollary follows readily from Lemma \ref{lemma: 
action on Khovanov complex}.
\end{proof}

Let now $D = D_{n}(T, \ldots, T)$ be an $n$-periodic link diagram obtained from a tangle $T$. 
Suppose that we choose another tangle $T'$, which differs from $T$ by a single application of one 
of the Reidemeister moves. Form another link diagram $D' = D_{n}(T', \ldots, T')$.

\begin{defn}\label{defn: equiv Reidemeister moves}
If $T$ and $T'$ are tangles differing by a single Reidemeister move and $D = D_{n}(T, \ldots, T)$, 
$D' = D_{n}(T', \ldots, T')$ are the corresponding $n$-periodic links diagrams, then we say that 
$D'$ was obtained from $D$ by an application of the equivariant Reidemeister move.
\end{defn}

The above definition yields the question about the relationship between $\khbracket{D}$ and 
$\khbracket{D'}$ in $\Kobh{0}$. The following theorem from \cite{BarNatan} answers this question.

\begin{thm}\label{thm: reidemeister homotopy equivalence}
Any Reidemeister move induces a map,
$$\mathcal{R} \colon \khbracket{D} \to \khbracket{D'},$$
which is an isomorphism in $\Kobh{0}$. In particular, this map induces a chain homotopy 
equivalence after application of any TQFT functor.
\end{thm}

\noindent However, in the equivariant setting we obtain a considerably weaker invariance result.

\begin{thm}\label{thm: reidemeister quasi-isomorphism}
If $D$ and $D'$ are as above and $\mathcal{T}$ is a TQFT functor whose target is the category of 
$R$-modules, then the map
$$\mathcal{R} \colon \khbracket{D} \to \khbracket{D'},$$
induced by an equivariant Reidemeister move, yields a quasi-isomorphism
$$\mathcal{T}(\mathcal{R}) \colon \mathcal{T}(\khbracket{D}) \to \mathcal{T}(\khbracket{D'})$$
in the category of cochain complexes of $\GpAAlg{R}{\z{n}}$-modules.
\end{thm}
\begin{proof}
It is sufficient to prove, that $\mathcal{T}(\mathcal{R})$ is a morphism in the category of 
$\GpAAlg{R}{\z{n}}$-modules, because Theorem \ref{thm: reidemeister homotopy equivalence} implies 
that it automatically induces an isomorphism on homology.

To check this condition, refer to the proof of \cite{BarNatan}*{Thm. 2}. The bracket 
$\khbracket{D}$ is constructed along the lines of the formal tensor product of copies of 
the complex $\khbracket{T}$. Each collection of morphisms
$$f_{i} \colon \khbracket{T} \to \khbracket{T'},$$
for $i=1, \ldots, n$, yield a morphism
$$f_{1} \otimes \ldots \otimes f_{n} \colon \khbracket{D} \to \khbracket{D'}.$$
Taking into account the symmetry of $D$ and $D'$, we obtain the following commutative diagram
\begin{center}
\begin{tikzpicture}
\matrix(m)[matrix of math nodes, row sep = 1cm, column sep=2.5cm]
{
  \khbracket{D} & \khbracket{D'} \\
  \khbracket{D} & \khbracket{D'} \\
};
\path[->, font=\scriptsize]
(m-1-1) edge node[above] {$f_{1} \otimes \ldots \otimes f_{n}$} (m-1-2)
(m-1-1) edge node[left] {$\Sigma_{D}$} (m-2-1)
(m-1-2) edge node[right] {$\Sigma_{D'}$} (m-2-2)
(m-2-1) edge node[above] {$f_{2} \otimes \ldots \otimes f_{n} \otimes f_{1}$} (m-2-2);
\end{tikzpicture}
\end{center}
where $\Sigma_{D}$ and $\Sigma_{D'}$ denote the automorphisms of complexes induced by the action of 
$\z{n}$. Since $R$ is of the form $\mathcal{R} = \mathcal{R}' \otimes \ldots \otimes \mathcal{R}'$, 
where
$$\mathcal{R}' \colon \khbracket{T} \to \khbracket{T'}$$
is induced by a single Reidemeister move, it follows that $\mathcal{T}(\mathcal{R})$ is 
a morphism in the category of $\GpAAlg{R}{\z{n}}$-modules.
\end{proof}

\section{Equivariant Khovanov homology}
\label{sec: equiv kh}

\subsection{Integral equivariant Khovanov homology}
\label{subsec: equiv kh z}

Let $L$ be an $n$-periodic link. It was shown earlier, that under this assumption, the Khovanov 
complex $\cc(D)$, where $D$ is an $n$-periodic diagram of $L$, admits the structure of a cochain 
complex of graded $\gpRing{\z{n}}$-modules. We could try to obtain an invariant of $L$ by defining 
the equivariant Khovanov homology to be
$$H^{\ast,\ast}(\Hom{}{\cyclicGroupRing{n}}{M}{\cc(D)}),$$
for some $\cyclicGroupRing{n}$-module $M$. Unfortunately, Theorem \ref{thm: reidemeister 
quasi-isomorphism} indicates that this approach might not work. This is indeed the case, which is 
illustrated by the following example.

\begin{figure}
\centering
\import{./}{unknot_positive_curls.pdf_tex}
\caption{$2$-periodic diagram of the unknot and its Khovanov bracket. Gray arrows indicate the 
$\z{2}$-action. Black dot stands for the fixed point axis.}
\label{fig: 2-periodic diagram unknot}
\end{figure}

\begin{ex}\label{ex: non homotopy equivalent complexes}
Consider the $2$-periodic diagram $D$ from Figure \ref{fig: 2-periodic diagram unknot}. 
We will show that due to the lack of exactness of the Hom functor, the ordinary cohomology with 
coefficients depends on the chosen diagram. Since
$$\Hom{}{\gpRing{\z{2}}}{\zzz_{-}}{M} = \{x \in M \colon t \cdot x = -x\},$$
we obtain
\begin{align*}
&\Hom{}{\gpRing{\z{2}}}{\zzz_{-}}{\cc^{1,\ast}(D)} = \\
&= \left \langle \Vector{\One \otimes \One}{-\One 
\otimes \One}, \Vector{\One \otimes X}{-X \otimes \One}, \Vector{X \otimes \One}{- \One \otimes X}, 
\Vector{X \otimes X}{- X \otimes X} \right \rangle \\
&\Hom{}{\gpRing{\z{2}}}{\zzz_{-}}{\cc^{2,\ast}(D)} = \left \langle \One, X \right \rangle. \\
\end{align*}
Inspection of the differential $d \colon \cc^{1,\ast}(D) \to \cc^{2,\ast}(D)$ yields
\begin{align*}
d \colon \Vector{\One \otimes \One}{- \One \otimes \One} &\mapsto -2 \cdot \One, \\
d \colon \Vector{\One \otimes X}{- X \otimes \One} &\mapsto -2 \cdot X,
\end{align*}
therefore
$$H^{2,\ast}\left( \Hom{}{\gpRing{\z{2}}}{\zzz_{-}}{\cc(D)} \right) = \z{2}\{5\} \oplus 
\z{2}\{3\}.$$
On the other hand,
$$H^{2, \ast}\left( \Hom{}{\gpRing{\z{2}}}{\zzz_{-}}{\cc(U)} \right) = 0,$$
where $U$ denotes the crossingless diagram of the unknot.
\end{ex}

\noindent The above example shows the necessity of considering the derived functors
$\Hom{\ast}{\cyclicGroupRing{n}}{M}{-}$ and their homology. This is due to the fact, that the 
Khovanov complex of a periodic link is built from permutation modules, which are in general neither 
projective nor injective. This causes the discrepancy visible in the previous example.

\begin{defn}\label{defn: equiv kh}
Define the equivariant Khovanov homology of an $n$-periodic diagram $D$ to be the following 
triply-graded module, for which the third grading is supported only for $d \mid n$.
$$\kh_{\z{n}}^{\ast,\ast,d}(L) = \Ext{\ast,\ast}{\cyclicGroupRing{n}}{\gpRing{\xi_d}}{\cc(D)}.$$
It is worth to notice, that since $\cc(D)$ is a complex of graded modules, Ext groups become also 
naturally graded, provided that we regard $\gpRing{\xi_{d}}$ as a graded module concentrated in 
degree~$0$.
\end{defn}

To check the equivariant Khovanov homology is independent of the choice of periodic diagram, it is 
sufficient to check its invariance under the equivariant Reidemeister moves, see Definition 
\ref{defn: equiv Reidemeister moves}.

\begin{thm}\label{thm:invariance_equivariant_homology}
Equivariant Khovanov homology is an invariant of a periodic link, that it is invariant under the
equivariant Reidemeister moves.
\end{thm}
\begin{proof}
Theorem \ref{thm: reidemeister quasi-isomorphism} implies that a single application of an 
equivariant Reidemeister move to a periodic link diagram, yields a quasi-isomorphism of the 
corresponding Khovanov complexes. Proposition \ref{prop: properties of ext} implies, that this 
quasi-isomorphism induces an isomorphism of the equivariant Khovanov homology.
\end{proof}

\noindent One of the first questions, regarding the properties of the equivariant Khovanov 
homology, we can ask, is the question about its relation to the classical Khovanov homology. The 
answer is given in the following theorem.

\begin{thm}\label{thm:total_equivariant_homology}
Let $p_{1}, \ldots, p_{s}$ be the collection of all prime divisors of $n$. Define the ring $R_{n} = 
\gpRing{\frac{1}{p_{1}}, \frac{1}{p_{2}}, \ldots, \frac{1}{p_{s}}}.$
There exists a natural map
$$\bigoplus_{d | r} \kh_{\z{n}}^{\ast, \ast, d}(L) \to \kh(L),$$
which, when tensored with $R_{n}$, becomes an isomorphism.
\end{thm}
\begin{proof}
From Theorem \ref{thm: Cartan Eilenberg spectral seq} it follows that
$$\Ext{\ast,\ast}{\gpRing{\z{n}}}{\gpRing{\z{n}}}{\cc(D)} \cong \kh^{\ast,\ast}(D).$$
Indeed, entries in the $E_{2}$ page of the \mbox{Cartan-Eileberg} spectral sequence 
${}_{I}E_{\ast}^{\ast,\ast}$ are equal to
\begin{align*}
&\ext{p,\ast}{\gpRing{\z{n}}}{\gpRing{\z{n}}}{\kh^{q,\ast}(D)} = \\
&= \left \{
\begin{array}{ll}
\Hom{}{\gpRing{\z{n}}}{\gpRing{\z{n}}}{\kh^{q,\ast}(D)}, & p = 0, \\
0, & p > 0.\\
\end{array}
\right.
\end{align*}
Hence, the spectral sequence collapses at $E_{2}$. Since
$$\Hom{}{\gpRing{\z{n}}}{\gpRing{\z{n}}}{\kh^{q,\ast}(D)) \cong \kh^{q, \ast}(D)},$$
we obtain the desired conclusion.

The short exact sequence from Proposition 
\ref{prop:long_exact_seq_maximal_order} implies that
$$\Lambda' \otimes_{\zzz} R_{n} = \gpRing{\z{n}} \otimes_{\zzz} R_{n} = \GpAAlg{R_{n}}{\z{n}},$$
where $\Lambda'$ denotes the maximal order in $\gpAlg{\z{n}}$. Therefore, 
\begin{align*}
&\Ext{r,\ast}{\gpRing{\z{n}}}{\Lambda'}{\cc(D)} \otimes_{\zzz} R_{n} \cong \\
&\cong \Ext{r,\ast}{\GpAAlg{R_{n}}{\z{n}}}{\Lambda' \otimes_{\zzz} R_{n}}{\cc(D) \otimes_{\zzz} 
R_{n}} \cong \\
&\cong \Ext{r,\ast}{\GpAAlg{R_{n}}{\z{n}}}{\GpAAlg{R_{n}}{\z{n}}}{\cc(D) \otimes_{\zzz} R_{n}} 
\cong \kh^{r, \ast}(D) \otimes_{\zzz} R_{n},
\end{align*}
because $R_{n}$ is flat over $\zzz$. The last step of the proof consist of noticing, that since 
$\Lambda' = \bigoplus_{d \mid n} \gpRing{\xi_{d}}$, hence
$$\Ext{\ast,\ast}{\gpRing{\z{n}}}{\Lambda'}{\cc(D)} = \bigoplus_{d | r} \kh_G^{\ast, \ast, d}(L).$$
\end{proof}

Until the end of this section we will restrict our attention to the case of $p^{n}$-periodic links, 
where $p$ is an odd prime. This restriction is needed to perform the computations of the Khovanov 
homology of trivial links.

Components of periodic trivial links can be divided into two categories. The first category 
consists of components, which are invariant under the $\z{n}$ action, i.e. they link non-trivially 
with the fixed point axis $F$ and the generator of the cyclic groups rotates them by the angle 
$\frac{2\pi}{p^{n}}$. The second category consists of components, which are freely 
permuted by $\z{n}$. Let $T_{kp^{n}+f}$ denote a crossingless $p^{n}$-periodic diagram of the 
trivial link consisting of $f$ fixed circles, i.e. components from the first category, and $k$ 
orbits of free components.

In order to express the equivariant Khovanov homology of trivial links, let us define the following 
family of polynomials.

\begin{defn}\label{defn:diff polynomials trivial links}
Define a sequence of Laurent polynomials i.e. elements of the ring $\gpRing{q,q^{-1}}$.
\begin{align*}
\mathcal{P}_{0}(q) &= q+q^{-1}\\
\mathcal{P}_{n}(q) &= \frac{1}{p^n}\sum_{\stackrel{1 \leq k \leq 
p^n - 1}{\gcd(k,p^n)=1}}\binom{p^n}{k} q^{2k - p^n} + \\
&+ \frac{1}{p^n}\sum_{1 \leq s < n} \sum_{\stackrel{1 \leq k \leq p^n-1}{\gcd(k,p^n) = p^s}} 
\left(\binom{p^n}{k} - \binom{p^{n-s}}{k'}\right) q^{2k - p^n},
\end{align*}
where $k' = k/p^{s}$ and $n \geq 1$.
\end{defn}

\begin{defn}
If $M^{\ast}$ is a graded and free $\zzz$-module, then define its quantum dimension as in 
\cite{Turner2}.
$$\qdim M^{\ast} = \sum_{i} q^{i} \dim M^{i} \otimes \qqq .$$
\end{defn}

\begin{prop}\label{prop: equiv kh of trivial links}
The Khovanov complex $\cc(T_{kp^n})$ can be decomposed into a direct sum of permutation modules in 
the following way
$$\cc^{0,\ast}(T_{kp^n}) = \bigoplus_{s = 0}^{n} \ind{\z{p^{n}}}{\z{p^{n-s}}} M_{s}^{k},$$
where $M_{s}^{k}$ is a free abelian group and a trivial \mbox{$\z{p^{n-s}}$-module} satisfying
$$\qdim M_{s}^{k} = \sum_{\ell = 1}^{k} p^{s \cdot( \ell - 1)} 
\mathcal{P}_{s}(q^{p^{n-s}})^{\ell}  \sum_{\stackrel{0 \leq i_0, \ldots, i_{s-1} \leq k}{i_0 + 
\ldots + i_{s-1} = k - \ell}} \prod_{j=0}^{s-1} \left( p^j \mathcal{P}_{j}(q^{p^{n-j}}) 
\right)^{i_j}.$$
\end{prop}

\begin{proof}
Let
$$\mathcal{A} = \gpRing{X}/(X^{2}),$$
where
\begin{align*}
\deg \One &= 1, \\
\deg X &= -1. \\
\end{align*}
Since the induced action on
$$\cc^{\ast,\ast}(T_{kp^n}) = \cc^{0,\ast}(T_{kp^n}) = \underbrace{\mathcal{A}^{\otimes k} \otimes 
\mathcal{A}^{\otimes k} \otimes \ldots \otimes \mathcal{A}^{\otimes k}}_{p^n}$$
permutes the factors in the tensor product above, thus it is sufficient to consider 
the restriction of this action to the basis of $\mathcal{A}^{\otimes kp^n}$ consisting of tensor 
products of vectors $\One$ and $X$.

Let us start with the case $k = 1$. Denote by $\ell$ the following map
\begin{align*}
\ell \colon \mathcal{A}^{\otimes p^n} &\to \mathcal{A}^{\otimes p^n}, \\
\ell \colon x_{1} \otimes x_{2} \otimes \ldots \otimes x_{p^n} &\mapsto x_{2} \otimes \ldots 
\otimes x_{p^n} \otimes x_{1}.
\end{align*}
In order to obtain the desired decomposition of $\cc^{0,\ast}(T_{p^n})$ it is sufficient to 
decompose the basis of $\mathcal{A}^{\otimes p^n}$ into disjoint union of orbits. Observe that if 
$v = x_{1} \otimes \ldots \otimes x_{p^n}$ satisfies $\ell^{p^{s}}(v) = v$, for some $s \leq n$, 
then $v$ is completely determined by its first $p^{s}$ factors $x_{1}, \ldots, x_{p^{s}}$.

\begin{lemma}
The set of basis vectors satisfying the following conditions, for fixed $s \leq n$,
\begin{enumerate}
  \item $\Iso(v) = \z{p^{n-s}}$, 
  \item $\deg v = p^{n-s} (2k - p^{s})$, for $1 \leq k \leq p^{s} - 1$
\end{enumerate}
has cardinality 
$$= \left\{
\begin{array}{ll}
\binom{p^s}{k}, & \gcd(k,p^{s}) = 1, \\
\binom{p^{s}}{k} - \binom{p^{s-u}}{k'}, & \gcd(k,p^{s}) = p^{u},
\end{array}
\right.$$
where $k' = k/p^{u}$. In particular, $\qdim M_{s}^{1} = \mathcal{P}_{s}(q^{p^{n-s}})$.
\end{lemma}
\begin{proof}
Notice first, that if a vector $v$ is fixed by $\ell^{p^{s}}$, then necessarily \mbox{$p^{n-s} 
\mid \deg v$}. If $k$ is not divisible by $p$, then $v$ automatically satisfies the first 
condition and cardinality of the set of such vectors is equal to $\binom{p^{s}}{k}$.

On the other hand, when $\gcd(k,p^{s}) = p^{u}$, there are vectors $v$ such that $\deg v = p^{n-s} 
(2k - p^n)$ and $\Iso(v)$ contains properly $\z{p^{n-s}}$. There are exactly 
$\binom{p^{s-u}}{k/p^{s}}$ such vectors. Therefore, the overall cardinality in this case is equal 
to $\binom{p^{s}}{k} - \binom{p^{s-u}}{k'}$.
\end{proof}

To perform the inductive step, notice that if $v \in \mathcal{A}^{(k-1)p^n}$ and $w \in 
\mathcal{A}^{p^{n}}$ satisfy $\Iso(v) = \z{p^{n-s}}$ and $\Iso(w) = \z{p^{n-s'}}$, then $\Iso(v 
\otimes w) = \z{p^{\min(n-s,n-s')}}$. When $\Iso(v) = \z{p^{n-s}}$, then the orbit of $v$ can be 
identified with $\z{p^{s}}$, with the action coming from the following quotient map
$$\z{p^{n}} \to \z{p^{s}}.$$
The product of two orbits $\z{p^{s}}$ and $\z{p^{s'}}$, with the diagonal action, consists of 
several orbits. This decomposition is given below.
\begin{equation}\label{eqn: equality of z n sets}
\z{p^{s}} \times \z{p^{s'}} = \bigsqcup_{i=1}^{p^{s'}} \z{p^{s}},
\end{equation}
if $s \geq s'$. Thus,
\begin{align*}
M_{s}^{k} &= \bigoplus_{0 \leq s' < s} \left( \left(M_{s}^{k-1} \otimes_{\zzz} M_{s'}^{1} 
\right)^{p^{s'}} \oplus 
\left(M_{s'}^{k-1} \otimes_{\zzz} M_{s}^{1}\right)^{p^{s'}} \right) \\
&\oplus \left(M_{s}^{k-1} \otimes_{\zzz} M_{s}^{1} \right)^{p^{s}}. \\
\end{align*}
Consequently,
\begin{align*}
\qdim M_{s}^{k} &= \sum_{0 \leq s' < s} p^{s'} ( \qdim M_{s}^{k-1} \cdot \qdim M_{s'}^{1} \\
&+ \qdim M_{s'}^{k-1} \cdot \qdim M_{s}^{1} ) + \qdim M_{s}^{k-1} \cdot \qdim M_{s}^{1}.
\end{align*}
Expanding this inductive formula yields the desired result.
\end{proof}

\begin{cor}\label{cor: equiv complex of trivial link}
The Khovanov complex $\cc(T_{kp^{n}+f})$ decomposes in the following way.
$$\cc^{0,\ast}(T_{kp^n +f}) = \bigoplus_{s=0}^{n} \ind{\z{p^{n}}}{\z{p^{n-s}}} M_{s}^{k,f},$$
where $M_{s}^{k,f}$ is a trivial \mbox{$\z{p^{n-s}}$-module} such that
$$\qdim M_{s}^{k,f} = (q+q^{-1})^{f} \qdim M_{s}^{k}.$$
\end{cor}
\begin{proof}
Corollary follows from Proposition \ref{prop: equiv kh of trivial links}. Indeed, because
$$\cc^{0,\ast}(T_{kp^{n}+f}) = \cc^{0,\ast}(T_{kp^{n}}) \otimes_{\zzz} \cc^{0,\ast}(T_{f}).$$
Arguing as in the proof of Proposition \ref{prop: equiv kh of trivial links} we can consider orbits 
of vectors which are tensor products of $\One$ and $X$. Every such orbits is a tensor product of an 
orbit from $\cc^{0,\ast}(T_{kp^{n}})$ and a trivial orbit from $\cc^{0,\ast}(T_{f})$. This 
additional trivial orbit from $\cc^{0,\ast}(T_{f})$ preserves the isotropy group of the orbit from 
$\cc^{0,\ast}(T_{kp^{n}})$, therefore
$$M^{k,f}_{s} = M^{k}_{s} \otimes_{\zzz} \mathcal{A}^{\otimes f},$$
which concludes the proof.
\end{proof}

Analysis of the Khovanov complex of the trivial link yields the following result giving explicit 
formulas for the equivariant Khovanov homology of trivial links.

\begin{prop}\label{prop: comp equiv kh trivial links}
The equivariant Khovanov homology of the trivial link $T_{kp^{n}+f}$ is given below.
$$\kh_{\z{p^n}}^{0,\ast,p^{n-u}}(T_{kp^n + f}) = \bigoplus_{s=0}^{n} W_{s,u}^{k,f},$$
where
$$ W_{s,u}^{k,f} = \left\{
\begin{array}{ll}
\bigoplus_{i} H^{\ast}(\z{p^{n-s}}, \zzz^{d_{i}})^{\varphi(p^{n-u})} \{i\}, & n - s \leq u, \\
\bigoplus_{i} H^{\ast}(\z{p^{n-s}}, \cyclotomicRing{p^{n-s-u}}^{d_{i}})^{p^{s}} \{i\}, & n - s 
> u. \\
\end{array}
\right.$$
and
$$\qdim M^{k,f}_{s} = \sum_{i} d_{i} q^{i},$$
for some non-negative integers $d_{i}$.
\end{prop}
\begin{proof}
From corollary \ref{cor: equiv complex of trivial link} one obtains the following equality
\begin{align*}
&\ext{\ast,\ast}{\z{p^{n}}}{\gpRing{\xi_{p^{n-u}}}}{\cc^{0,\ast}(T_{kp^n + f})} = \\
&= \bigoplus_{s=0}^{n} 
\ext{\ast,\ast}{\z{p^{n}}}{\gpRing{\xi_{p^{n-u}}}}{\ind{\z{p^n}}{\z{p^{n-s}}} 
M_{s}^{k,f}}.
\end{align*}
From Shapiro's Lemma and Proposition \ref{prop: decomposition of cyclotomic rings}
\begin{align*}
&\ext{\ast,\ast}{\z{p^{n}}}{\gpRing{\xi_{p^{n-u}}}}{\ind{\z{p^n}}{\z{p^{n-s}}} 
M_{s}^{k,f}} \cong \\
&\cong \ext{\ast,\ast}{\z{p^{s}}}{\res{\z{p^n}}{\z{p^{n-s}}}\gpRing{\xi_{p^{n-u}}}}{M_{s}^{k,f}} 
\cong \\
&\cong \left\{
\begin{array}{ll}
\ext{\ast,\ast}{\z{p^{s}}}{\zzz^{\varphi(p^{n-u})}}{M_{s}^{k,f}}, & n - s \leq u \\
\ext{\ast,\ast}{\z{p^{s}}}{\gpRing{\xi_{p^{n-s-u}}}^{p^{s}}}{M_{s}^{k,f}}, & n - s > u. \\
\end{array}
\right. \\
\end{align*}
The rest follows from \cite{Brown}*{Prop. 2.2}.
\end{proof}

\begin{cor}
When $k = 0$, then the following equality holds.
$$\kh_{\z{p^{n}}}^{\ast,\ast,p^{s}}(T_{f}) = \bigoplus_{i = 0}^{f} H^{\ast}\left(\z{p^{n}}, 
\cyclotomicRing{p^{s}}\right)^{\binom{f}{i}}\{2i - f\}.$$
\end{cor}

\subsection{Torsion in the equivariant Khovanov homology}

Theorem \ref{thm:total_equivariant_homology} together with Proposition \ref{prop: equiv kh of 
trivial links} imply that the equivariant Khovanov homology contains an abundance of torsion not 
present in the classical Khovanov homology. However, we can use the additional algebraic structure 
of the equivariant Khovanov homology to obtain some information about this extra torsion. To be 
more precise, Proposition \ref{prop: properties of ext} implies that the equivariant Khovanov 
homology $\kh_{\z{p^{n}}}^{\ast,\ast,p^{s}}(D)$ is a module over the Ext ring 
$\ext{\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$. 
This is the key to understand the additional torsion.

\begin{prop}
For any $r > 0$, the ring
$\ext{\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$ acts on 
${}_{II}E_{r}$, where $\{{}_{II}E_{r}, d_{r}\}$ is the \mbox{Cartan-Eilenberg} spectral sequence 
converging to $\kh_{\z{p^{n}}}^{\ast,\ast,p^{s}}(D)$ from Theorem \ref{thm: Cartan Eilenberg 
spectral seq}. To be more precise, the ring
$$\ext{\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$$
acts on the columns of the spectral sequence.
$${}_{II}E_{r}^{i,j} \otimes_{\zzz} 
\ext{k}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}} \to {}_{II}E_{r}^{i,j + 
k}.$$
Moreover, on $E_{1}$ page this action agrees with the natural action of the Ext ring on the module
$\ext{\ast,\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\cc^{i,\ast}(D)}$.
\end{prop}
\begin{proof}
It is a direct consequence of Theorem \ref{thm: multiplicative structure ext}, because the 
appropriate version of the Cartan-Eilenberg spectral sequence for 
$$\ext{\ast}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$$
collapses at the $E_{1}$ page.
\end{proof}

\begin{cor}\label{cor: periodicity}
Let $m \in \zzz$ be chosen such that $\kh^{m,\ast}(D) \neq 0$ and $\kh^{i,\ast}(D)$ vanishes for $i 
> m$. Let $T_{s}$ denote the cohomology class in the ext ring
$$T_{s} \in \ext{2}{\gpRing{\z{p^{n}}}}{\gpRing{\xi_{p^{s}}}}{\gpRing{\xi_{p^{s}}}}$$
from Proposition 
\ref{prop: properties of ext rings of cylotomic rings}. Then, multiplication by $T_{s}$
$$- \cup T_{s} \colon \equivKh{{p^{n}}}{i}{\ast}{p^{s}}(D) \to 
\equivKh{{p^{n}}}{i+2}{\ast}{p^{s}}(D)$$
is an epimorphism for $i=m$ and isomorphism for $i> m$.
\end{cor}

\begin{rem}
This is an analogue of \cite[Lemma 1.1]{Wall}.
\end{rem}

\begin{proof}
The action of the class $T_{s}$ can be described in terms of the dimension shifting, see 
\cite{MacLane}. Since the class $T_{s}$ is represented by the following Yoneda extension
$$0 \to \cyclotomicRing{p^{s}} \to \cyclicGroupRing{p^{n}} \to \cyclicGroupRing{p^{n}} \to 
\cyclotomicRing{p^{s}},$$
we can split this exact sequence into two short exact sequences
\begin{align}
&0 \to \cyclotomicRing{p^{s}} \to \cyclicGroupRing{p^{n}} \to M \to 0, \label{align: seq1}\\
&0 \to M \to \cyclicGroupRing{p^{n}} \to \cyclotomicRing{p^{s}} \to 0, \label{align: seq2}
\end{align}
where $M$ is a submodule of $\cyclicGroupRing{p^{n}}$. Let 
\begin{align*}
&\delta_{1} \colon \kh_{\z{p^{n}}}^{i,\ast,p^{s}} \to 
\Ext{i+1,\ast}{\cyclicGroupRing{p^{n}}}{M}{\cc(D)}\\
&\delta_{2} \colon \Ext{i,\ast}{\cyclicGroupRing{p^{n}}}{M}{\cc(D)} \to 
\kh_{\z{p^{n}}}^{i+1,\ast,p^{s}}(D),
\end{align*}
be the connecting homomorphisms from the long exact cohomology sequences derived from the short 
exact sequences (\ref{align: seq1}) and (\ref{align: seq2}), respectively. The map
$$-\cup T_{s} \colon \kh_{\z{p^{n}}}^{i,\ast,p^{s}}(D) \to \kh_{\z{p^{n}}}^{i+2,\ast,p^{s}}(D)$$
is equal to the composition $\delta_{2} \circ \delta_{1}$.

Since
$$\Ext{i,\ast}{\cyclicGroupRing{p^{n}}}{\cyclicGroupRing{p^{n}}}{\cc(D)} \cong \kh^{i,\ast}(D),$$
see the proof of \ref{thm:total_equivariant_homology}, we deduce that $\delta_{1}$ and $\delta_{2}$ 
are epimorphisms for $i = m$ and isomorphisms for $i>m$, because $\kh^{i,\ast}(D)=0$ for $i 
> m$. Hence the corollary follows.
\end{proof}

\begin{cor}\label{cor: torsion}
Let $m$ be as in the previous corollary. For $i > m$, $\equivKh{p^{n}}{i}{\ast}{1}(D)$ is 
annihilated by $p^{n}$, and for $1 \leq s \leq n$, $\equivKh{p^{n}}{i}{\ast}{p^{s}}(D)$ is 
annihilated by $p^{n-s+1}$.
\end{cor}
\begin{proof}
This follows from previous corollary and the fact that $p^{n} T_{0} = 0$ and $p^{n-s+1}T_{s} = 0$, 
for $1 \leq s \leq n$. 
\end{proof}

\subsection{Rational equivariant Khovanov homology}
\label{subsec: equiv kh q}

Since the rational group algebra $\cyclicGroupAlg{n}$ is semi-simple and artinian, the algebraic 
structure of the equivariant Khovanov homology with rational coefficients simplifies considerably.

\begin{prop}\label{prop: structure rational equiv kh}
If $D$ is an $n$-periodic link diagram, then
$$\cc^{\ast,\ast}(D; \qqq) \cong \bigoplus_{d \mid n} \cc_{\z{n}}^{\ast,\ast,d}(D; \qqq),$$
where
$$\cc_{\z{n}}^{\ast,\ast,d}(D; \qqq) = 
\Hom{}{\cyclicGroupAlg{n}}{\cyclotomicField{d}}{\cc^{\ast,\ast}(D; \qqq)}.$$
Moreover,
$$\cc_{\z{n}}^{\ast,\ast,d}(D; \qqq) = \cc^{\ast,\ast}(D; \qqq) \cdot e_d,$$
and
$$\kh_{\z{n}}^{\ast,\ast,d}(D; \qqq) = \kh^{\ast,\ast}(D; \qqq) \cdot e_{d} = H^{\ast,\ast}(\cc(D; 
\qqq) \cdot e_{d}),$$
where $e_d$ is a central idempotent from $\cyclicGroupAlg{n}$, which annihilates 
$\cyclotomicField{d'}$, for $d' \mid n$ and $d \neq d'$, and acts as an identity on 
$\cyclotomicField{d}$.
\end{prop}
\begin{proof}
The proposition is a consequence of the Wedderburn decomposition and Schur's Lemma. For more 
details refer to \cite{CurtisReiner}*{Chap. 1}.
\end{proof}

\begin{ex}
The rational Khovanov homology of the trivial link $T_{kp^{n}+f}$, for some prime $p$, which 
possesses $k$ free orbits of components and $f$ fixed circles, is given by the following formula
$$\kh_{\z{p^{n}}}^{0,\ast,p^{n-u}}(T_{kp^{n}+f}; \qqq) = \bigoplus_{s=n-u}^{n} 
(M_{s}^{k,f})^{\varphi(p^{n-u})} \otimes \qqq,$$
where $M_{s}^{k,f}$ was defined in Proposition \ref{prop: equiv kh of trivial links}.
\end{ex}

\noindent The above proposition has the following corollary.

\begin{cor}\label{cor: vanishing of equiv kh}
Suppose that $D$ is an $n$-periodic diagram of a link. Choose $d \mid n$. If for any $i,j$ we have 
$\dim_{\qqq} \kh^{i,j}(D; \qqq) < \varphi(d)$, where $\varphi$ denotes the Euler's totient 
function, 
then $\kh_{\z{n}}^{\ast,\ast,d}(D; \qqq) = 0$.
\end{cor}
\begin{proof}
Indeed, since $\kh_{\z{n}}^{\ast,\ast,d}(D; \qqq)$ is a $\cyclotomicField{d}$ vector space, it 
follows readily that $\dim_{\qqq} \kh_{\z{n}}^{\ast,\ast,d}(D; \qqq)$ is divisible by $\dim_{\qqq} 
\cyclotomicField{d} = \varphi(d)$.
\end{proof}

\noindent The above corollary can be used to compute the equivariant Khovanov homology in some 
cases.

\begin{cor}
Let $T(n,2)$ be the torus link. Let $d > 2$ be a divisor of $n$. According to Example
\ref{ex: periodic torus links}, $T(n,2)$ is $d$-periodic. Let $d' > 2$ and $d' \mid d$.
$$\kh^{\ast,\ast,d'}_{\z{d}}(T(n,2); \qqq) = 0.$$
\end{cor}
\begin{proof}
Indeed, because according to \cite{Khovanov1}*{Prop. 35} for all $i,j$ we have
$$\dim_{\qqq} \kh^{i,j}(T(n,2); \qqq) \leq 1$$
and $\varphi(d') > 1$ if $d' > 2$.
\end{proof}

\begin{cor}
Let $\gcd(3,n) = 1$. The $3$-equivariant Khovanov homology 
$\kh^{\ast,\ast,3}_{\z{3}}(T(n,3);\qqq)$ of the torus knot $T(n,3)$ vanishes.

\noindent If $d > 2$ divides $n$, $d' > 2$ and $d' \mid d$, then 
$\kh^{\ast,\ast,d'}_{\z{d}}(T(n,3); \qqq) = 0$.
\end{cor}
\begin{proof}
  Indeed, because \cite{Turner2}*{Thm. 3.1} implies that for all $i,j$ we have
  $$\dim_{\qqq}   \kh^{i,j}(T(n,3); \qqq) \le 1,$$
  provided that $\gcd(3,n)=1$.
\end{proof}

\section{The spectral sequence}
\label{sec: spectral sequence}

Computation of the classical Khovanov homology is usually done with the aid of the long exact 
sequence of Khovanov homology, applied to a chosen link crossing, which is then resolved in two 
different ways. However, resolution of a single crossing kills the symmetry of a periodic link 
diagram in question. Instead, we need to resolve a whole orbit of crossings to obtain another 
periodic diagram. But this is not sufficient to recover the equivariant Khovanov homology. We have 
to take into account all possible resolutions of a chosen orbit of crossings. These data are 
organized into a spectral sequence, the construction of which is the main goal of this chapter. 
During the construction, we obtain a filtration which yields the spectral sequence from 
\cite{EverittTurner}. The spectral sequence that we obtain is applied to the computation of 
the $2$-equivariant Khovanov homology of torus links $T(n,2)$.

Start with a link $L$ and its $n$-periodic diagram $D$. Choose a subset of crossings $X 
\subset \cross{D}$.

\begin{defn}
Let $\alpha \colon \cross(D) \to \{0,1,x\}$ be a map.
\begin{enumerate}
\item If $i \in \{0,1,x\}$ define $|\alpha|_{i} = \#\alpha^{-1}(i)$. 
\item Define the support of $\alpha$ to be $\supp \alpha = \alpha^{-1}(\{0,1\})$.
\item Define also the following family of maps
\begin{align*}
\mathcal{B}_k(X) &= \{\alpha \colon \cross(D) \to \{0,1,x\} \quad | \quad \supp \alpha = X, \quad 
|\alpha|_{1} 
= k \}.
\end{align*}
\item  Denote by $D_{\alpha}$ the diagram obtained from $D$ by resolving crossings from 
$\alpha^{-1}(0)$ by $0$-smoothing and from $\alpha^{-1}(1)$ by $1$-smoothing. 
\end{enumerate}
\end{defn}

First, we will work with $\ccc(D)$, which is defined in the following ways
$$\ccc(D) = \cc(D)[n_{-}(D)]\{2n_{-}(D)-n_{+}(D)\},$$
where $n_{+}(D)$ and $n_{-}(D)$ denote the number of positive and negative crossings of $D$, 
respectively. We will construct a filtration of this complex. The filtration on $\cc(D)$ will be 
obtained from this one by an appropriate shift in degree.

Fix a crossing $c \in \cross(D)$ and consider three maps
$$\alpha_{0},\alpha_{1},\alpha_{x} \colon \cross(D) \to \{0,1,x\},$$
which attain different value at $c$, i.e.  $\alpha_{1}(c) = 1$, $\alpha_{x}(c) = x$ and 
$\alpha_{0}(c) = 0$, and are identical otherwise. These data yield the  following short exact 
sequence of complexes.
\begin{equation} \label{eqn:short_exact_seq_res_crossing}
0 \to \ccc(D_{\alpha_{1}})[1]\{1\} \to \ccc(D_{\alpha_{x}}) \to \ccc(D_{\alpha_{0}}) \to 0,
\end{equation}
as in \cite{Turner2}. However, there exists a chain map
$$\delta_{c} \colon \ccc(D_{\alpha_{0}}) \to \ccc(D_{\alpha_{1}})\{1\},$$
such that $\ccc(D_{\alpha_{x}}) = \cone(\delta_{c})$, where $\cone(\delta_{c})$ denotes the 
algebraic mapping cone of $\delta_{c}$, and (\ref{eqn:short_exact_seq_res_crossing}) is the 
corresponding short exact sequence of complexes. The map $\delta_{c}$ is obtained as follows. We 
identify $\ccc(D_{\alpha_{0}})$ and $\ccc(D_{\alpha_{1}})$ with submodules of 
$\ccc(D_{\alpha_{x}})$ ``generated'' by Kauffman states with $c$ resolved by $0$- or $1$-smoothing, 
respectively. As a graded module $\ccc(D_{\alpha_{x}})$ splits in the following way
$$\ccc(D_{\alpha_{x}}) = \ccc(D_{\alpha_{0}}) \oplus \ccc(D_{\alpha_{1}})\{1\}[1].$$
with $\ccc(D_{\alpha_{1}})$ being a subcomplex. If $\pi_{1}$ denotes the projection 
of $\ccc(D_{\alpha_{x}})$ onto $\ccc(D_{\alpha_{1}})$ and $i_{0}$ denotes the inclusion of 
$\ccc(D_{\alpha_{0}})$, then
$$\delta_{c} = \pi_{1} \circ d \circ i_{0}.$$

When we consider two crossings $c$ and $c'$, we obtain the following bicomplex
\begin{center}
\begin{tikzpicture}
\matrix(m)[matrix of math nodes, column sep=1cm]
{
                      & \ccc(D_{\alpha_{10}})\{1\} & \\
\ccc(D_{\alpha_{00}}) & \oplus                     & \ccc(D_{\alpha_{11}})\{2\} \\
                      & \ccc(D_{\alpha_{01}})\{1\} & \\
};
\path[->]
(m-2-1) edge node {} (m-1-2)
(m-1-2) edge node {} (m-2-3)
(m-2-1) edge node {} (m-3-2)
(m-3-2) edge node {} (m-2-3);
\end{tikzpicture}
\end{center}
where $\alpha_{00}$, $\alpha_{10}$, $\alpha_{01}$, $\alpha_{11}$ differ only at $c$ or $c'$ and
\begin{align*}
\alpha_{00}(c) &= \alpha(c') = 0, \\
\alpha_{10}(c) &= 1, \quad \alpha_{10}(c') = 0, \\
\alpha_{01}(c) &= 0, \quad \alpha_{01}(c') = 1, \\
\alpha_{11}(c) &= \alpha_{11}(c') = 1.
\end{align*}
The horizontal maps are defined analogously as in the previous case. The total complex of 
the above bicomplex is equal to the shifted Khovanov complex of $D_{\alpha_{xx}}$, where 
$\alpha_{xx}$ agrees with $\alpha_{00}$, $\alpha_{10}$, $\alpha_{01}$ and $\alpha_{11}$ outside $c$ 
and $c'$ and
$$\alpha_{xx}(c) = \alpha_{xx}(c') = x.$$
Continuing this procedure we obtain the following bicomplex
$$N^{i,j,k} = \left\{
\begin{array}{ll}
\bigoplus_{\alpha \in \mathcal{B}_{i}(X)} \limits
\ccc^{j,k}(D_{\alpha})\{i\}, & 0 \leq i \leq \# X, \\
0, & \mbox{otherwise}. \\
\end{array}
\right.$$
The total complex of $N$, denoted by $\Tot(N)^{\ast,\ast}$ and defined by
$$\Tot(N)^{i,j} = \bigoplus_{k+l = i} N^{k,l,j},$$
is equal to the shifted Khovanov complex $\ccc(D)$.

\begin{defn}
Let $M^{\ast,\ast,\ast}$ be a bicomplex of graded $\zzz$-modules such that
$$M^{i,j,k} = \left\{
\begin{array}{ll}
\bigoplus_{\alpha \in \mathcal{B}_{i}(X)} \limits
\cc^{j,k}(D_{\alpha})[c(D_{\alpha})]\{i + 3 c(D_{\alpha}) + \# X\}, & 0 \leq i \leq 
\# X, \\
0, & \mbox{otherwise}, \\
\end{array}
\right.$$
where $c(D_{\alpha}) = n_{-}(D_{\alpha}) - n_{-}(D)$. Vertical differentials
$$M^{i,j, \ast} \to M^{i,j+1,\ast}$$
are sums of $\pm 1$ multiplicities of differentials in the respective Khovanov complexes. 
Horizontal differentials
$$M^{i,j, \ast} \to M^{i+1,j,\ast},$$
on the other hand, are induced from the appropriate horizontal maps in the  bicomplex 
$N^{\ast,\ast,\ast}$.
\end{defn}

\begin{prop}
The total complex of $M^{\ast,\ast,\ast}$ is equal to the Khovanov complex $\cc(D)$.
\end{prop}
\begin{proof}
Since the total complex of $N^{\ast,\ast,\ast}$ is equal to $\ccc(D)$ and
$$\cc(D) = \ccc(D)[-n_{-}(D)]\{n_{+}(D) - 2 n_{-}(D)\},$$
we only need to check, that the application of the appropriate shift to $N$ results in $M$.
\begin{align*}
&N^{i,j,k}[-n_{-}(D)]\{n_{+}(D)-2n_{-}(D)\} = \\
&= \bigoplus_{\alpha \in \mathcal{B}_{i}(X)} \ccc^{j,k}(D_{\alpha})[-n_{-}(D)]\{i + n_{+}(D) - 
2n_{-}(D)\} = \\
&= \bigoplus_{\alpha \in \mathcal{B}_{i}(X)} \cc^{j,k}(D_{\alpha})[c(D_{\alpha})]\{i + 
2c(D_{\alpha}) + n_{+}(D) - n_{+}(D_{\alpha})\} = \\
&= \bigoplus_{\alpha \in \mathcal{B}_{i}(X)} \cc^{j,k}(D_{\alpha})[c(D_{\alpha})]\{i + 
3c(D_{\alpha}) \# X\} = M^{i,j,k},
\end{align*}
because
\begin{align*}
&n_{+}(D) - n_{+}(D_{\beta}) = \\
&= \# \cross(D) - n_{-}(D) - (\# \cross(D) - \#X - n_{-}(D_{\beta})) = \\
&= c(D_{\beta}) + \#X.
\end{align*}
\end{proof}

\begin{defn}\label{defn: filtration}
Let
$$\mathcal{F}_{i}(X) = \Tot(\bigoplus_{j \geq i} M^{j,\ast,\ast}),$$
for $0 \leq i \leq \# X$. The family $\{\mathcal{F}_{i}(D)\}_{i}$ is a filtration of the Khovanov 
complex. This filtration is the column filtration of the bicomplex $M^{\ast,\ast,\ast}$, see 
\cite{McCleary}*{Thm. 2.15}.
\end{defn}

\noindent The following theorem was first proved in \cite{EverittTurner}.

\begin{thm}\label{thm:construction}
Let $D$ be a link diagram and let $X \subset \cross(D)$. The pair $(D, X)$ determines a spectral 
sequence
$$\{E^{\ast,\ast,\ast}_r, d_r\}$$
of graded modules converging to $\kh^{\ast,\ast}(D)$ such that
$$E_1^{i,j,\ast} = \bigoplus_{\beta \in \mathcal{B}_i(X)} \kh^{j,\ast}(D_{\beta})[c(D_{\beta})]\{i 
+ 3c(D_{\beta}) + \#X\},$$
where $c(D_{\beta}) = n_{-}(D_{\beta}) - n_{-}(D)$.
\end{thm}
\begin{proof}
This is the spectral sequence associated to the column filtration of the bicomplex $M$ as in 
\cite{McCleary}*{Thm. 2.15}.
\end{proof}

Suppose now, that $D$ is an $n$-periodic link diagram. If $X \subset \cross{D}$ is invariant, under 
the action of $\z{n}$, then for any $0 \leq k \leq \#X$ there exists an induced action on 
$\mathcal{B}_k(X)$. Hence, each member $\mathcal{F}_{k}(X)$, for $0 \leq k \leq \#X$, of the 
filtration is invariant under the action of $\z{n}$. This discussion leads to the following 
conclusion.

\begin{prop}\label{prop:invariant_filtration}
If $X \subset \cross{D}$ is an invariant subset, then every member of the filtration, from 
Definition \ref{defn: filtration} is a $\cyclicGroupRing{n}$-subcomplex of $\cc(D)$.
\end{prop}

From now on, we will assume that $X$ consists of a single orbit of crossings. We will perform 
analysis of the quotients $\mathcal{F}_{i}(X)/\mathcal{F}_{i+1}(X)$ to determine their structure 
as $\cyclicGroupRing{n}$-modules.

\begin{defn}
Let $0 \leq i \leq n$ and $d \mid \gcd(n,i)$. Analogously as in Definition \ref{defn: Kauffman 
states of periodic diagram}, denote by $\mathcal{B}_{i}^{d}(X)$ the subset of $\mathcal{B}_{i}(X)$ 
consisting of maps satisfying $\Iso(\alpha) = \z{d}$. Also denote by 
$\overline{\mathcal{B}}_{i}^{d}(X)$ the quotient of $\mathcal{B}_{i}^{d}(X)$ by $\z{n}$.
\end{defn}

\begin{lemma}
If $\alpha \in \mathcal{B}_{i}^{d}(X)$, then $D_{\alpha}$ is $\z{d}$-periodic.
\end{lemma}
\begin{proof}
The Lemma follows readily, because such diagrams have similar structure as the Kauffman states 
belonging to $\mathcal{S}_{i}^{d}(D)$, as in Definition \ref{defn: Kauffman states of periodic 
diagram}.
\end{proof}

\begin{prop}\label{prop:quotients_of_filtration_as_ZG_modules}
Suppose, that $n$ is odd, then for $0 \leq i \leq n$
\begin{align*}
&\mathcal{F}_{i}(X) / \mathcal{F}_{i+1}(X) = \\
&= \bigoplus_{d \mid \gcd(n,i)} \bigoplus_{\alpha \in 
\overline{\mathcal{B}}_{i}^{d}(X)} \ind{\z{n}}{\z{d}} \left( 
\cc(D_{\alpha})[t(\alpha)]\{q(\alpha)\} 
\right),
\end{align*}
where
\begin{align*}
t(\alpha) &= c(D_{\alpha}) + i, \\
q(\alpha) &= i + 3c(D_{\alpha}) + n.
\end{align*}
\end{prop}
\begin{proof}
To proof of the Lemma uses an adaptation of the argument from the proof of Lemma \ref{lemma: 
action on Khovanov complex} and Corollary \ref{cor: action on Khovanov complex}.
\end{proof}

\begin{prop}\label{prop:quotient 2}
Let $n = 2$. Under this assumption we have
\begin{align*}
\mathcal{F}_{0}(X) / \mathcal{F}_{1}(X) &= \cc(D_{\alpha_{00}})[t(\alpha_{00})]\{q(\alpha_{00})\} 
\otimes_{\zzz} \zzz_{-}^{\otimes \frac{c(D_{\alpha_{00}})}{2}}, \\
\mathcal{F}_{1}(X) / \mathcal{F}_{2}(X) &= \cc(D_{\alpha_{10}})[t(\alpha_{10})]\{q(\alpha_{10})\} 
\otimes_{\zzz} \cyclicGroupRing{2}, \\
\mathcal{F}_{2}(X) &= \cc(D_{\alpha_{11}})[t(\alpha_{11})]\{q(\alpha_{11})\} \otimes_{\zzz} 
\zzz_{-}^{\otimes \frac{c(D_{\alpha_{11}})}{2} + 1}.
\end{align*}
\end{prop}
\begin{proof}
The middle equality follows easily, because complexes corresponding to $D_{\alpha_{10}}$ and 
$D_{\alpha_{01}}$ are freely permuted by $\z{2}$.

To check the first and third inequality, recall from Equation \ref{eqn: permutation map} from 
Section \ref{sec: periodic links} the sign convention. We have two distinguished crossings 
$c_{1}, c_{2} \in \supp \alpha_{00}$. Let us denote by $T$, as in Section \ref{sec: periodic 
links}, the tangle from which $D$ was constructed. Let $x_{1}, x_{2} \in \Lambda^{\ast}W_{T}$ be 
such that $x_{1} \otimes x_{2}$ corresponds to certain Kauffman state of $D_{\alpha_{00}}$. 
According to our convention, the permutation map acts on $\cc(D_{\alpha_{00}})$ as follows
$$x_{1} \otimes x_{2} \mapsto (-1)^{\frac{n_{-}(D_{\alpha_{00}})}{2} + \deg x_{1} \deg x_{2}} x_{2} 
\otimes x_{1},$$
whereas on $\cc(D)$ it acts as follows.
$$x_{1} \otimes x_{2} \mapsto (-1)^{\frac{n_{-}(D)}{2} + \deg x_{1} \deg x_{2}} x_{2} \otimes 
x_{1}.$$
Comparison of the two coefficients yields, that we need to twist the action of $\z{2}$ on 
$\cc(D_{\alpha_{00}})$ by $\zzz_{-}^{\otimes \frac{c(D_{\alpha_{00}})}{2}}$.

Analogous argument applied to $D_{11}$ yields the third equality.
\end{proof}

Let $p$ be an odd prime and $n > 0$ an integer. We state the next theorem only for $2$-periodic and 
$p^{n}$-periodic links, since these cases will be of importance in the remainder part of this paper 
and in the subsequent paper \cite{Politarczyk}. The statement in other cases can be analogously 
derived, however we omit it due to its technical complication, which dims the whole idea of the 
spectral sequence.

\begin{thm}\label{thm:spectral_sequence_equivariant_homology}
Let $L$ be a $p^{n}$-periodic link, where $p$ is an odd prime, and let $X \subset \cross{D}$ 
consists of a single orbit. Under this assumption, for any $0 \leq s \leq n$ there exists a 
spectral sequence $\{{}_{p^{n-s}}E_{r}^{\ast,\ast}, d_r\}$
of graded modules converging to $\kh_{\z{p^{n}}}^{\ast,\ast,p^{n-s}}(D)$ with
\begin{align*}
{}_{p^{n-s}}E^{0,j}_{1} &= \kh_{\z{p^n}}^{j,\ast,p^{n-s}}(D_{\alpha_0})[c(D_{\alpha_0})]\{ 
q(\alpha_{0}) \}, \\
{}_{p^{n-s}}E^{p^{n}, j}_{1} &= \kh_{\z{p^n}}^{j, \ast, p^{n-s}} (D_{\alpha_{1}}) 
[c(D_{\alpha_1})] \{ q(\alpha_{1}) \},\\
{}_{p^{n-s}}E^{i,j}_{1} &= \bigoplus_{0 \leq v \leq u_{i}} \bigoplus_{\alpha \in 
\overline{\mathcal{B}}_{i}^{p^{v}}(X)} \kh_{\z{p^{v}}}^{j, \ast, 
k(v,s)}(D_{\alpha})[c(D_{\alpha})]\{q(\alpha)\}^{\ell(v,s)}
\end{align*}
for $0 < i < p^{n}$. Above we used the following notation $i = p^{u_{i}} g$, where $\gcd(p,g) = 1$ 
and $\alpha_{0}$, $\alpha_{1}$ are the unique elements of $\mathcal{B}_0(X)$ and 
$\mathcal{B}_{p^{n}}(X)$, respectively, and
\begin{align*}
q(\alpha) &= i + 3c(D_{\alpha}) + p^{n}, \\
k(s,v) &= \left\{
\begin{array}{ll}
1, & v \leq s, \\
p^{v-s}, & v > s, \\
\end{array}
\right. \\
\ell(s,v) &= \left\{
\begin{array}{ll}
\varphi(p^{n-s}), & v \leq s, \\
p^{n-v}, & v > s, \\
\end{array}
\right.
\end{align*}

The $E_{1}$ pages of the respective spectral sequences for $2$-periodic links are given below.
\begin{align*}
{}_{1}E_{1}^{0,j} &= \kh_{\z{2}}^{j,\ast,s(D_{00})}(D_{00})\{3 c(D_{00}) + 2\}, \\
{}_{1}E_{1}^{1,j} &= \kh^{j,\ast}(D_{01})\{3 c(D_{01}) + 3\}, \\
{}_{1}E_{1}^{2,j} &= \kh_{\z{2}}^{j,\ast,3 - s(c(D_{11}))}(D_{11})\{3 c(D_{11}) + 4\}, \\
{}_{2}E_{1}^{0,j} &= \kh_{\z{2}}^{j,\ast,3-s(D_{00})}(D_{00})\{3 c(D_{00}) + 2\}, \\
{}_{2}E_{1}^{1,j} &= \kh^{j,\ast}(D_{01})\{3 c(D_{01}) + 3\}, \\
{}_{2}E_{1}^{2,j} &= \kh_{\z{2}}^{j,\ast,s(c(D_{11}))}(D_{11})\{3 c(D_{11}) + 4\},
\end{align*}
where $s \in \{1,2\}$ and
$$s(D_{\alpha}) \equiv \frac{c(D_{\alpha})}{2} \pmod {2}$$
\end{thm}
\begin{proof}
In the odd case apply Theorem \ref{thm: spectral sequence derived filtration} to the filtration 
$\mathcal{F}_{\ast}(X)$. Use Proposition \ref{prop:quotients_of_filtration_as_ZG_modules} and  
Eckamann-Shapiro Lemma to compute the entries in the $E_{1}$ page as in the proof of Proposition 
\ref{prop: equiv kh of trivial links}.

In the even case apply Theorem \ref{thm: spectral sequence derived filtration} and Proposition 
\ref{prop:quotient 2}.
\end{proof}

\section{Sample computations}
\label{sec: computations}

The purpose of this section is to compute the rational $2$-equivariant Khovanov homology of torus 
links $T(n,2)$. Before we start, however, let us define the equivariant Khovanov and Jones 
polynomials. Although we defer the study of their properties to \cite{Politarczyk}, we define them 
here to simplify the statements of the results presented in this section.

\begin{defn}\label{defn:equivariant_jones_polynomial}
Let $L$ be an $n$-periodic link. For $d \mid n$, define the $d$-th equivariant Khovanov polynomial 
of $L$ as follows
$$\khp_{n,d}(L)(t,q) = \sum_{i,j}t^{i} q^{j} \dim_{\gpAlg{\xi_d}} \kh^{i,j,d}(L; \qqq)$$
and the $d$-th equivariant Jones polynomial of $L$ as
$$\jones_{n,d}(L)(q) = \khp_{n,d}(L)(-1,q).$$
Note that above we take the dimension of the respective vector space over the cyclotomic field 
$\cyclotomicField{d}$.
\end{defn}

Let us start with $T(2,2)$, which serves as a basis for further calculations.

\begin{figure}
  \centering
  \import{./}{hopf_ckh.pdf_tex}
  \caption{Anticommutative cube for $T(2,2)$}
  \label{fig:img_hopf_cube}
\end{figure}

\begin{ex}\label{ex: computation for hopf link}
Consider the Hopf link as depicted on Figure \ref{fig:T(n,2) link} for $n=2$. Its Khovanov bracket 
is depicted on Figure \ref{fig:img_hopf_cube}. Figures \ref{fig: computations kh1 of Hopf link} and 
\ref{fig: computation of kh2 Hopf link} depict $\cc_{\z{2}}^{\ast,\ast,1}(T(2,2))$ and 
$\cc_{\z{2}}^{\ast,\ast,2}(T(2,2))$, respectively. Equivariant Khovanov and Jones polynomials of 
$T(2,2)$ are given below.
\begin{align*}
  \khp_{2,1}(T(2,2))(t,q) &= 1 + q^2 + t^2 q^4 \\
  \khp_{2,2}(T(2,2))(t,q) &= t^2 q^6 \\
  \jones_{2,1}(T(2,2))(q) &= 1 + q^2 + q^4 \\
  \jones_{2,2}(T(2,2))(q) &= q^6 \\
\end{align*}
\end{ex}

\begin{figure}
\centering
\begin{tikzpicture}
  \draw[xstep=3cm, ystep=2cm, gray, thin] (-0.5, -0.5) grid (9.5, 6.5);
  \draw[thin, gray] (0 , 0) -- (-0.5, -0.5);

  \node at (-0.25, -0.5) {$i$};
  \node at (-0.5, -0.25) {$j$};

  \foreach \x in {0, 1, 2}
    \node at (3 * \x + 1.5, -0.5) {$\x$};

  \foreach \y in {0, 2, 4}
    \node at (-0.5, \y + 1) {$\y$};

  \node (n00) at (1.5, 1) {$X \otimes X$};
  \node (n02_1) at (1.5, 2.5) {$1 \otimes X - X \otimes 1$};
  \node (n02) at (1.5, 3.5) {$1 \otimes X + X \otimes 1$};
  \node (n04) at (1.5, 5) {$1 \otimes 1$};

  \node (n12) at (4.5, 3) {$(X,X)$};
  \node (n14) at (4.5, 5) {$(1,1)$};

  \node (n24) at (7.5, 5) {$1 \otimes X - X \otimes 1$};

  \path[->] (n02) edge (n12)
            (n04) edge (n14);

\end{tikzpicture}
\caption{Computation of $\kh_{\z{2}}^{\ast,\ast,1}(T(2,2); \qqq)$.}
\label{fig: computations kh1 of Hopf link}
\end{figure}

\begin{figure}
\centering
\begin{tikzpicture}
  \draw[xstep=3cm, ystep=2cm, gray, thin] (-0.5, -0.5) grid (9.5, 6.5);
  \draw[thin, gray] (0 , 0) -- (-0.5, -0.5);

  \node at (-0.25, -0.5) {$i$};
  \node at (-0.5, -0.25) {$j$};

  \foreach \x in {0, 1, 2}
    \node at (3 * \x + 1.5, -0.5) {$\x$};

  \foreach \y in {2, 4, 6}
    \node at (-0.5, \y - 1) {$\y$};

  \node (n12) at (4.5, 1) {$\frac{1}{2} (-X,X)$};
  \node (n14) at (4.5, 3) {$\frac{1}{2} (-1,1)$};

  \node (n22) at (7.5, 1) {$X \otimes X$};
  \node (n24) at (7.5, 3) {$1 \otimes X + X \otimes 1$};
  \node (n26) at (7.5, 5) {$1 \otimes 1$};

  \path[->] (n12) edge (n22)
            (n14) edge (n24);

\end{tikzpicture}
\caption{Computation of $\kh_{\z{2}}^{\ast,\ast,2}(T(2,2); \qqq)$.}
\label{fig: computation of kh2 Hopf link}
\end{figure}

Let us also state the following proposition from \cite{Khovanov1}, which describes the Khovanov 
homology of torus links $T(n,2)$.

\begin{prop}\label{prop:kh_T_2_2n} %*{Prop. 35}
The Khovanov polynomial of $\kh(T(n,2))$ is equal to
\begin{align*}
\khp(T(2k,2)) &= q^{2k-2} + q^{2k} + t^2 q^{2k+2} (1 + tq^4) \sum_{j = 0}^{k-2} t^{2j} q^{4j} \\
&+ t^{2k} q^{6k - 2} + t^{2k} q^{6k}, \\
\khp(T(2k+1,2)) &= q^{2k-1} + q^{2k+1} + t^{2}q^{2k+3}(1+tq^{4}) \sum_{j = 0}^{k-1} t^{2j} q^{4j}, 
\end{align*}
for $k > 1$.
\end{prop}

\begin{thm}\label{thm: computation equiv Kh T_n_2}
Khovanov polynomials of the $2$-equivariant Khovanov homology of torus links $T(n,2)$ are given 
below.
\begin{align*}
\khp_{2,1}(T(2n+1, 2)) &= \khp(T(2n+1,2)) \\
\khp_{2,2}(T(2n+1, 2)) &= 0\\
\khp_{2,1}(T(2n, 2)) &= \khp(T(2n,2)) - t^{2n}q^{6n} \\
\khp_{2,2}(T(2n, 2)) &= t^{2n} q^{6n} \\
\end{align*}
\end{thm}

\begin{figure}
  \centering
  \import{./}{T_n_2.pdf_tex}
 \caption{The $2$-periodic diagram of $T(n,2)$. The chosen orbit of crossings is marked with red 
circles.}
  \label{fig:T(n,2) link}
\end{figure}

\begin{figure}
  \centering
  \def\svgwidth{\columnwidth}
  \import{./}{T_n_2_spectral_seq.pdf_tex}
  \caption{Bicomplex associated to the $2$-periodic diagram of $T(n,2)$ from figure \ref{fig:T(n,2) 
link}.}
  \label{fig:T(n,2) bicomplex}
\end{figure}

\begin{figure}
\centering
\import{./}{T_n_2_D_01.pdf_tex}
\caption{Diagram $D'$ isotopic to the diagram of the $D_{01}$.}
\label{fig: T_n_2 D' diag}
\end{figure}

\begin{figure}
\centering
\import{./}{T_n_2_D_01_res.pdf_tex}
\caption{$0$- and $1$- smoothings of the diagram $D'$, respectively.}
\label{fig: T_n_2 D' diag res}
\end{figure}

Consider the $2$-periodic diagram $D$ of $T(n,2)$ from Figure \ref{fig:T(n,2) link} with the chosen 
orbit marked with red circles. Orient the diagram so that all crossings are positive. The 
associated bicomplex in $\Kob{0}$ is depicted on Figure \ref{fig:T(n,2) bicomplex}. 

\begin{lemma}\label{lemma: equiv kh comp 1}
The zeroth column of the $E_{1}$ page of the spectral sequence from Theorem 
\ref{thm:spectral_sequence_equivariant_homology} applied to the $2$-periodic diagram $D$, 
has the following form.
\begin{align*}
{}_{1}E_{1}^{0,j,k} &= \kh_{\z{2}}^{j,k-1,1}(T(n-1,2)) \oplus \kh_{\z{2}}^{j,k-3,1}(T(n-1,2)), \\
{}_{2}E_{1}^{0,j,k} &= \kh_{\z{2}}^{j,k-1,2}(T(n-1,2)) \oplus \kh_{\z{2}}^{j,k-3,2}(T(n-1,2)). \\
\end{align*}
\end{lemma}
\begin{proof}
From figure \ref{fig:T(n,2) bicomplex} it is not hard to see, that the diagram $D_{00}$ represents 
the split sum $T(n-1,2) \sqcup U$, where $U$ denotes the unknot. Additionally, $D_{00}$ inherits 
orientation from $D$, therefore $c(D_{00})=0$, because $D$ was oriented so that all crossings 
are positive. This concludes the proof.
\end{proof}

\begin{lemma}\label{lemma: equiv kh comp 2}
The second column of the $E_{1}$ page of the spectral sequence from theorem 
\ref{thm:spectral_sequence_equivariant_homology} applied to the $2$-periodic diagram $D$, has teh 
following form.
\begin{align*}
{}_{1}E_{1}^{2,j,k} &= \kh_{\z{2}}^{j,k-4,2}(T(n-2,2)) \\
{}_{2}E_{1}^{2,j,k} &= \kh_{\z{2}}^{j,k-4,1}(T(n-2,2)) \\
\end{align*}
\end{lemma}
\begin{proof}
From figure \ref{fig:T(n,2) bicomplex} it follows that $D_{11} = T(n-2,2)$. It is not hard to check 
that we can orient $D_{11}$ in such a way that all crossings are positive, therefore $c(D_{11})=0.$ 
This finishes the proof.
\end{proof}

\begin{lemma}\label{lemma: equiv kh comp 3}
The first column of the $E_{1}$ page of the spectral sequence from theorem 
\ref{thm:spectral_sequence_equivariant_homology} applied to 
the $2$-periodic diagram $D$ has the following form.
$${}_{1}E_{1}^{1,j,k} = {}_{2}E_{1}^{1,j,k} = \left\{
\begin{array}{ll}
\kh^{j,k-4}(T(2k-2,2)), & j <  2k-2, \\
\qqq\{6k-4\}, & j = 2k-2, \\
\qqq\{6k\}, & j = 2k-1, \\
0, & j > 2k-1,\\
\end{array}
\right.
$$
if $n = 2k$ and
$${}_{1}E_{1}^{1,j,k} = {}_{2}E_{1}^{1,j,k} = \left\{
\begin{array}{ll}
\kh^{j,k-4}(T(2k-1,2)), & j < 2k, \\
\qqq\{6k+1\} \oplus \qqq\{6k+3\}, & j = 2k, \\
0, & j > 2k, \\
\end{array}
\right.
$$
if $n = 2k+1$.
\end{lemma}
\begin{proof}
Let us denote by $D_{01}$ one of the diagrams in the middle column of Figure \ref{fig:T(n,2) 
bicomplex}. First, let us compute $c(D_{01})$. It is not hard to see, that $D_{01}$ can be oriented 
in such a way that all crossings are positive. Therefore $c(D_{01})=0$.

Let us denote by $D'$ the diagram from Figure \ref{fig: T_n_2 D' diag}. Orient it, so that all 
crossings are positive. Quick inspection shows that $D_{01}$ and $D'$ are isotopic. In order to 
prove the lemma, let us compute $\kh(D')$. To do this we will use the long exact sequence of 
Khovanov homology with respect to the crossing marked on Figure \ref{fig: T_n_2 D' diag}. Let 
$D_{0}'$ and $D_{1}'$ denote the $0$- and $1$-smoothing of $D'$, respectively, as on Figure 
\ref{fig: T_n_2 D' diag res}. The $0$-smoothing is the torus link $T(n-2,2)$ with $c(D_{0}') = 0$.  
On the other hand, $D_{1}'$ is a diagram of the unknot with $c(D_{1}') = n-2$.

Consider first the case $n = 2k$, for $k > 1$. In the respective long exact sequence almost all 
terms corresponding to $\kh(D_{1}')$ vanish. There are only two non-vanishing terms. Further 
inspection of the long exact sequence yields that there can be only one possibly non-vanishing 
morphism in this sequence.
\begin{align*}
&\qqq = \kh^{2k-2,6k-6}(T(2k-2,2)) \stackrel{\delta}{\to} \kh^{0,-1}(U) = \qqq \\
\end{align*}
Suppose that $\delta = 0$ and notice that if $n = 2k$, then $D'$ represents a knot. It is not 
hard to see, that the Khovanov homology of this knot is concentrated only on two diagonals $j-2i = 
2k-3, 2k-1$, regardless of whether $\delta$ vanishes or not. Further, if $\delta$ vanishes, then
\begin{equation}
\khp(D') = q \khp(T(2k-2,2)) + t^{2k-1}q^{6k-7} + t^{2k-1}q^{6k-5}.
\label{eqn: khp D'}
\end{equation}
On the other hand, \cite{Lee}*{Thm. 4.4} and \cite{Rasmussen}*{Prop. 3.3} imply that
\begin{equation}
\khp(D') \equiv q^{s(D')} (q + q^{-1}) \pmod{(1 + tq^{4})},
\label{eqn: khp hthin knot}
\end{equation}
for some integer $s(D')$. However, from (\ref{eqn: khp D'}) it follows that
\begin{align*}
&\khp(D') \equiv q^{2k-3} + q^{2k-1} + \\
&+ q^{-8k+8}(q^{6k-7} + q^{6k-5}) (1-q^{-4}) \pmod{(1 + tq^{4})},
\end{align*}
which contradicts (\ref{eqn: khp hthin knot}). Thus, $\delta$ must be non-trivial.

If $n = 2k+1$, there is also only one case to consider. Namely
$$\qqq = \kh^{2k-1,6k-3}(T(2k-1,2)) \to \kh^{0,-1}(U) = \qqq.$$
Notice that $D'$ represents a $2$-component link, whose Khovanov homology is concentrated on two 
diagonals. Therefore, analogously as in the previous case, \cite{Lee}*{Thm. 4.4} and 
\cite{Rasmussen}*{Prop. 3.3} imply that
\begin{equation}
\khp(D') \equiv q^{s}(q + q^{-1}) + t^{\ell} q^{s'}(q+q^{-1}) \pmod{(1+tq^{4})},
\end{equation}
where $\ell$ denotes the linking number of the components of $D'$. Argument analogous as in the 
even case yields that now $\delta$ must vanish.
\end{proof}

\begin{proof}[Proof of Thm. \ref{thm: computation equiv Kh T_n_2}]
The proof is inductive. The first case was done in Example \ref{ex: computation for hopf link}.

Consider first $T(2n+1,2)$. From lemmas \ref{lemma: equiv kh comp 1}, \ref{lemma: equiv 
kh comp 2} and \ref{lemma: equiv kh comp 3} we can derive the $E_{1}$ page spectral sequence 
${}_{2}E_{\ast}^{\ast,\ast,\ast}$, which is depicted on Figures \ref{fig: E_1 page equiv kh 
T_2n+1_2}. In order to finish the computation we need to apply Proposition \ref{prop: structure 
rational equiv kh} and Proposition \ref{prop:kh_T_2_2n}. Since
\begin{align*}
\kh^{2n+1,\ast}(T(2n+1,2)) &= \qqq\{6n+3\}, \\
\kh^{2n,\ast}(T(2n+1,2)) &= \qqq\{6n-1\}, \\
\end{align*}
if follows easily from Proposition \ref{prop:kh_T_2_2n} that the differential
$$d_{1}^{0,2n} \colon {}_{2}E_{1}^{0,2n,\ast} \to {}_{2}E_{1}^{1,2n,\ast}$$
is an isomorphism. Analogous comparisons of grading  of ${}_{2}E_{1}^{1,k}$ and 
$\kh^{k+1,\ast}(T(2n+1,2))$ yield that ${}_{2}E_{2}^{\ast,\ast,\ast}$ is zero. Thus, 
$$\kh_{\z{2}}^{\ast,\ast,2}(T(2n+1,2)) = 0,$$
and consequently 
$$\kh_{\z{2}}^{\ast,\ast,1}(T(2n+1,2)) = \kh^{\ast,\ast}(T(2n+1,2)).$$

Consider now $T(2n,2)$. The $E_{1}$ page of the spectral sequence is presented on 
Figure \ref{fig: E_1 page equiv kh T_2n+1_2}. Comparison of gradings of 
${}_{2}E_{1}^{\ast,\ast,\ast}$ and gradings of $\kh^{\ast,\ast}(T(2n-2,2))$ yields that the only 
non-zero entry of ${}_{2}E_{2}^{\ast,\ast,\ast}$ is 
$${}_{2}E_{2}^{1,2n-1,6n} = \qqq.$$
Therefore,
$$\kh_{\z{2}}^{\ast,\ast,2}(T(2n,2)) = \qqq[2n]\{6n\}.$$

\begin{figure}
\centering
\begin{tikzpicture}
\draw[xstep=4cm, ystep=6cm, gray!30, very thick] (-1.5, -1) grid (8.7, 6.7);
\draw[gray!30, very thick] (-1.5,4) -- (8.7,4);

\draw[->, black, very thick] (-1.5, -1.5) -- (-1.5, 7);
\draw[->, black, very thick] (-2, -1) -- (9, -1);

\node at (0, -1.25) {$0$};
\node at (4, -1.25) {$1$};
\node at (8, -1.25) {$2$};

\node at (-1.75, 0) {$0$};
\node at (-1.75, 2) {$\vdots$};
\node at (-2.25, 4) {$2n-1$};
\node at (-2.25, 6) {$2n$};

\node[fill=white, inner sep=5pt, align=center] at (0, 6) {$\qqq\{6n+3\}$ \\ $\oplus$ \\ 
$\qqq\{6n+1\}$};
\node[fill=white, inner sep=5pt, align=center] at (4, 6) {$\qqq\{6n+3\}$ \\ $\oplus$ \\
$\qqq\{6n+1\}$};
\node[draw, black, very thin, fill=white, inner sep=5pt, minimum height=5cm] at (4,2) 
{$\kh(T(2n-1,2))\{4\}$};
\node[draw, black, very thin, fill=white, inner sep=5pt, minimum height=5cm] at (8,2) 
{$\kh(T(2n-1,2))\{4\}$};
\end{tikzpicture}
\caption{${}_{2}E_{1}^{\ast,\ast,\ast}$ of $T(2n+1,2)$.}
\label{fig: E_1 page equiv kh T_2n+1_2}
\end{figure}

\begin{figure}
\centering
\begin{tikzpicture}
\draw[xstep=4cm, ystep=6cm, gray!30, very thick] (-1.5, -1) grid (8.7, 6.7);
\draw[gray!30, very thick] (-1.5,5) -- (8.7,5);
\draw[gray!30, very thick] (-1.5,4) -- (8.7,4);

\draw[->, black, very thick] (-1.5, -1.5) -- (-1.5, 7);
\draw[->, black, very thick] (-2, -1) -- (9, -1);

\node at (0, -1.25) {$0$};
\node at (4, -1.25) {$1$};
\node at (8, -1.25) {$2$};

\node at (-1.75, 0) {$0$};
\node at (-1.75, 2) {$\vdots$};
\node at (-2.25, 4) {$2n-3$};
\node at (-2.25, 5) {$2n-2$};
\node at (-2.25, 6) {$2n-1$};

\node[fill=white, inner sep=5pt] at (4, 6) {$\qqq\{6n\}$};
\node[fill=white, inner sep=5pt] at (4, 5) {$\qqq\{6n-4\}$};
\node[fill=white, inner sep=5pt] at (8, 5) {$\qqq\{6n-4\}$};
\node[draw, black, very thin, fill=white, inner sep=5pt, minimum height=5cm] at (4,2) 
{$\kh(T(2n-2,2))\{4\}$};
\node[draw, black, very thin, fill=white, inner sep=5pt, minimum height=5cm] at (8,2) 
{$\kh(T(2n-2,2))\{4\}$};
\end{tikzpicture}
\caption{${}_{2}E_{1}^{\ast,\ast,\ast}$ of $T(2n,2)$.}
\label{fig: E_1 page equiv kh T_2n_2}
\end{figure}
\end{proof}

\bibliography{biblio}

\noindent  \textsc{Department of Mathematics and Computer Science, Adam Mickiewicz 
University in Poznań,}

\noindent \textsc{e-mail:} \textit{politarw@amu.edu.pl}

\end{document}

equiv_kh_arxiv.bbl
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\bib{PrzytyckiSokolov}{article}{
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}

\bib{PrzytyckiSikora}{incollection}{
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\bib{Traczyk3}{article}{
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\bib{Traczyk2}{article}{
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\bib{Yokota2}{article}{
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\end{biblist}
\end{bibdiv}
