text:
\begin{document}
\title[Simplicial complement and  Moment-angle Complexes]
{\bf The homology of simplicial complement and the cohomology of the moment-angle complexes}
\author[X. Wang \& Q. Zheng]{Xiangjun Wang and Qibing Zheng}
\thanks{The authors were supported by NSFC grant No. 10771105 and No. 11071125}
\keywords{Stanley-Reisner face ring, moment-angle complex, $Tor$ algebra, homology of simplicial complement.}
\subjclass[2000]{Primary 13F55, 18G15, Secondary 16E05, 55U10.}
\address{School of Mathematical Sciences and LPMC, Nankai University,
Tianjin, 300071, P.R.China.}
\email{xjwang@nankai.edu.cn}
\address{School of Mathematical Science and LPMC, Nankai University,
Tianjin, 300071, P.R.China}
\email{zhengqb@nankai.edu.cn}
\maketitle

\begin{abstract}
A simplicial complement $\PP=\{\sigma_1, \sigma_2, \cdots , \sigma_s\}$ is a sequence
of subsets of $[m]$ and the simplicial complement $\PP$ corresponds to
an unique simplicial complex $K_\PP$ with vertices in $[m]$. In this paper, we defined
the homology of a simplicial complement $H_{i,\sigma}(\Lambda^{*,*}[\PP], d)$ over a principle ideal domain
$\kk$ and proved that $H_{*,*}(\Lambda[\PP], d)$ is isomorphic to
the $Tor$ of the corresponding face ring $\kk(K_\PP)$ by the Taylor resolution.  As applications, we give methods
to compute the ring structure of $Tor_{*,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$,
$\mbox{link}_{K_\PP}\sigma$, $\mbox{star}_{K_\PP}\sigma$ and the cohomology modules of the generalized moment-angle complexes.
\end{abstract}

\section{Introduction and statement of results}

   The moment-angle complexes have been studied by topologists for many years
(\cf \cite{Porter} \cite{Lopez}). In 1990's
Davis and Januszkiewicz \cite{DJ} introduced quasi-toric manifolds which were being studied
intensively by algebraic geometers. They observed
that every quasi-toric manifold is the quotient of a moment-angle complex by the free action
of a real torus, here the moment-angle complex is denoted by $\mathcal{Z}_K$ corresponding to
an abstract simplicial complex $K$.
The topology of $\ZZ_K$ is complicated and getting more attentions by topologists lately
(\cf \cite{GM} \cite{Hochster} \cite{Baskakov} \cite{Panov} \cite{Franz}).
Recently a lot of work has been done on generalizing the moment-angle complex
$\ZZ_K=\ZZ_K(\underline{D^2}, \underline{S^1})$ to pairs of spaces $(\underline{X}, \underline{A})$
(\cf  \cite{BB}, \cite{BBCG08}, \cite{GTh}, \cite{LP}).
In this paper we study the cohomology of the generalized moment-angle complexes
$\ZZ_K(\underline{X}, \underline{A})$ corresponding to the pairs of spaces
$(\underline{X}, \underline{A})$ with inclusions $A_i\hookrightarrow X_i$ being
homotopic to constant for all $i$.

    Classically the homological algebra aspect of the Stanley-Reisner {\it face ring}
plays an important role in the cohomology of $\ZZ_K$ (\cf \cite{BP}, \cite{Stanley}).
Let $K$ be an abstract simplicial complex with $m$ vertices.
Choose a ground ring $\kk$ with unit (we are mostly interested in a
principal ideal domain).  Let $\kk[\xx]$ be the $\NN^m$ graded polynomial algebra
over $\kk$ on $m$ indeterminates $\xx=\{x_1, x_2, \cdots, x_m\}$. The Stanley-Reisner {\it face ring}
is the quotient ring $\kk(K)=\kk[\xx]/\II_K$, where $\II_K$ is the Stanley-Reisner ideal
generated by the monomials $\xx_\sigma$
corresponding to non-faces $\sigma\not\in K$.
\[
\II_K=\langle \xx_\sigma=x_{i_1}x_{i_2}\cdots x_{i_n}
    |\sigma=\{i_1, i_2, \cdots, i_n\}\not\in K\rangle.
\]
It is well known that the cohomology of $\ZZ_K$  is isomorphic to the $Tor$  of
$\kk(K)$ over $\kk[\xx]$, that is
\begin{align*}
H^{q}(\ZZ_K, \kk)\cong & \bigoplus_{2j-i=q} Tor^{\kk[\xx]}_{i, j}(\kk(K), \kk) & & \mbox{and}\\
 Tor^{\kk[\xx]}_{i, j}(\kk(K), \kk) \cong & Tor^{\kk[\xx]}_{i, j}(\kk, \kk(K)).
\end{align*}

    There are several methods to compute the $Tor$ of $\kk(K)$. For example, one can use the
Koszul resolution $\Lambda[u_1, u_2, \cdots, u_n]\otimes \kk[\xx]$ of $\kk$, then by applying
the functor $\otimes_{\kk[\xx]}\kk(K)$, the homology of $\Lambda[u_1, u_2, \cdots, u_n]\otimes \kk(K)$
is $Tor^{\kk[\xx]}_{i,2j}(\kk, \kk(K))$. On the other hand, one can construct a free resolution
$(R^*, d)$ of $\kk(K)$, then by applying the functor $\otimes_{\kk[\xx]}\kk$, the homology of $R^*\otimes_{\kk[\xx]}\kk$
is $Tor^{\kk[\xx]}_{i,j}(\kk(K), \kk)$. In this paper, we will follow the second method to
give a free resolution of $\kk(K)$ and  compute the homology of the resulting chain complex
$R^*\otimes_{\kk[\xx]}\kk$.

   Let $\PP=\{\sigma_1, \sigma_2, \cdots , \sigma_s\}$  be a sequence of subsets of
$[m]=\{1,2,\cdots,m\}$, which is called a {\it simplicial complement} in this paper. It is easy to see that
$$\II_\PP=\langle \xx_{\sigma_1}, \xx_{\sigma_2}, \cdots, \xx_{\sigma_s}\rangle$$
is an ideal generated by square-free monomials. Thus there is an unique simplicial complex $K_\PP$
such that $\II_{K_\PP}=\II_\PP$.

    Let $\Lambda[\PP]$ be the exterior algebra over $\kk$
generated by $\PP$. Given a generator $\uu=\sigma_{k_1}\sigma_{k_2}\cdots \sigma_{k_q}$ in $\Lambda[\PP]$,
let $S_\uu = \sigma_{k_1}\cup \sigma_{k_2}\cup \cdots \cup \sigma_{k_q}$ be the total subset
of $\uu$. The bi-degree of $\uu$ is defined as
 \[bideg(\uu)=(q, S_\uu),\]
and the monomial
$\xx_{S_\uu}$ determines an unique $\NN^m$ vector $\ba_\uu$ such that $\xx_{S_\uu}=\xx^{\ba_\uu}$.

Let
\[
\partial_i(\uu)
  =\sigma_{k_1}\cdots\widehat{\sigma}_{k_i}\cdots\sigma_{k_q}=
  \sigma_{k_1}\cdots\sigma_{k_{i-1}}\sigma_{k_{i+1}}\cdots\sigma_{k_q}
\]
and for a generator $\uu\otimes\xx^\ba\in \Lambda[\PP]\otimes_\kk\kk[\xx]$ define
\[
\widetilde{d}(\uu\otimes\xx^\ba)=\sum_i(-1)^i\partial_i(\uu)\otimes\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}\xx^\ba,
\]
where $\displaystyle{\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}=\frac{\xx_{S_\uu}}{\xx_{S_{\partial_i(\uu)}}}}$
denote the monomial of $\kk[\xx]$ corresponding to
the subset $S_\uu\setminus S_{\partial_i(\uu)}$.

   It is known that $(\Lambda[\PP]\otimes_\kk\kk[\xx], \w{d})$ is a free
$\kk[\xx]$-resolution of $\kk[\xx]/\II_\PP$  called the Taylor resolution \cite{Mil}.
Then applying the functor $\otimes_{\kk[\xx]}\kk$, we get
$$(\Lambda[\PP]\otimes_\kk\kk[\xx]\otimes_{\kk[\xx]}\kk, \w{d}\otimes 1)=(\Lambda^{*,*}[\PP],d)$$
and

  {\bf Theorem 2.7} {\it Let $\PP=\{\sigma_1, \sigma_2, \cdots, \sigma_s\}$ be a simplicial complement and $K_\PP$
be the corresponding simplecial complex such that $\II_{K_\PP}=\II_\PP$.
Let $(\Lambda^{*,*}[\PP], d)$ be the chain complex induced from the bi-graded exterior algebra on
$\PP$. The differential
$d: \Lambda^{q,*}[\PP]\rightarrow \Lambda^{q-1,*}[\PP]$ is given by
\begin{align*}
d(\uu)= & \sum_i(-1)^i\partial_i(\uu)\cdot\delta_{\partial_i(\uu)}
\end{align*}
where $\delta_{\partial_i(\uu)}=1$ if the total subset $S_\uu=S_{\partial_i(\uu)}$
and $\delta_{\partial_i(\uu)}=0$ if $S_\uu\not=S_{\partial_i(\uu)}$.  Then the homology of
$(\Lambda^{*,*}[\PP],\ d)$ is
\[
H_{*,\sigma}(\Lambda^{*,*}[\PP],\ d)=Tor^{\kk[\xx]}_{*,\sigma}(\kk[\xx]/\II_\PP, \kk)
  =Tor^{\kk[\xx]}_{*,\sigma}(\kk(K_\PP), \kk),
\]
which is called the homology of the simplicial complement $\PP$.}

    Fix a $\sigma\in 2^{[m]}$ and let $\Lambda^{*,\sigma}[\PP]$ be the submodule generated by
$\uu$ with total subset $S_\uu=\sigma$. From the formula for $d(\uu)$, we see that
$(\Lambda^{*,\sigma}[\PP], d)$ is a  sub-complex of
$(\Lambda^{*,*}[\PP], d)$. Then from a theorem of Baskakov, we see that
\[
H_{q,\sigma}(\Lambda^{*,\sigma}[\PP], d)=Tor^{\kk[\xx]}_{q,\sigma}(\kk(K_\PP), \kk)
  =Tor^{\kk[\xx]}_{q,\sigma}(\kk, \kk(K_\PP))=\w{H}^{|\sigma|-q-1}(K_\PP\cap\sigma, \kk),
\]
where $K_\PP\cap\sigma$ is the {\it full sub-complex} of $K_\PP$ consisting
of all simplices of $K_\PP$ which have all of their vertices in $\sigma$, that is
$K_\PP\cap\sigma=\{\tau\cap\sigma|\tau\in K_\PP\}$.

    Notice that both $\kk(K_\PP)$ and $\kk$ are $\kk[\xx]$ algebras, $Tor_{*,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$
has natural algebraic structure and the following isomorphism of algebras holds.
\[
H^*(\ZZ_{K_\PP}, \kk)\cong Tor_{*,*}^{\kk[\xx]}(\kk(K_\PP), \kk).
\]
In section 3, we proved that

{\bf Theorem 3.6} {\it The algebraic structure in $H_{*,*}(\Lambda^{*,*}[\PP], d)\cong Tor_{*,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$
is given by
\[
[c]\times[c']=[c\cdot c']\cdot \delta_{\sigma, \sigma'}
\]
where $[c]\in H_{q,\sigma}(\Lambda^{*,\sigma}[\PP], d)$, $[c']\in H_{j,\sigma'}(\Lambda^{*,\sigma'}[\PP], d)$,
the cycle $c\cdot c'$ is the product of $c$ and $c'$ in the exterior algebra $\Lambda^{*,*}[\PP]$ and
$\delta_{\sigma, \sigma'}=1$ if $\sigma\cap\sigma'=\phi$, $\delta_{\sigma, \sigma'}=0$ if $\sigma\cap\sigma'\not=\phi$.
}

    Fix a subset $\omega\subset [m]$, there is the {\it $\omega$-compression $E_\omega$} defined on the
simplicial complement $\PP$ given by
$$E_\omega\PP=\{\sigma_1\setminus\omega, \sigma_2\setminus\omega, \cdots, \sigma_s\setminus\omega\}.$$
For the subset $\omega$, the {\it link} and {\it star} of the simplicial complex $K_\PP$ corresponding to $\omega$
are the sub-complexes
\begin{align*}
\mbox{star}_{K_\PP}\omega= & \{\tau\in K_\PP|\tau\cup\omega\in K_\PP\}\\
\mbox{link}_{K_\PP}\omega= & \{\tau\in K_\PP|\tau\cup\omega\in K_\PP;\ \tau\cap\omega=\phi\}.
\end{align*}
In section 4 we proved that:

  {\bf Theorem 4.7} {\it Let $\PP$ be a simplicial complement and $K_\PP$ be the simplicial complex corresponding
to $\PP$. Then:}
\begin{enumerate}
\item {\it The $star$,} $\mbox{star}_{K_\PP}\omega=K_{E_\omega\PP}$ {\it is the simplicial
complex corresponding to the simplicial complement $E_\omega\PP$. Thus}
\[
Tor^{\kk[\xx]}_{q,\sigma}(\kk(\mbox{star}_{K_\PP}\omega), \kk)=
    H_{q,\sigma}(\Lambda^{*,\sigma}[E_\omega\PP], d),
\]
\item {\it The $link$,} $\mbox{link}_{K_\PP}\omega=K_{E_\omega\PP}\cap ([m]\setminus\omega)$ {and}
\[
H_{q,[m]\setminus \omega}(\Lambda^{*,[m]\setminus \omega}[E_\omega\PP], d)=
\w{H}^{m-|\omega|-q-1}(\mbox{link}_{K_\PP}\omega, \kk).
\]
\end{enumerate}

  Let $(\underline{X}, \underline{A})=\{(X_i, A_i, x_i)|i\in[m]\}$ denote
a set of triples of $CW$-complexes with base points $x_i\in A_i$ and $K_\PP$ be an abstract simplicial
complex. The generalized moment-angle complex determined by $(\underline{X}, \underline{A})$ and
$K_\PP$ denoted by $\ZZ_{K_\PP}(\underline{X}, \underline{A})$ is defined
to be the $colimit$
$$\ZZ_{K_\PP}(\underline{X}, \underline{A})=
\bigcup_{\sigma\in K_\PP} D(\omega),
$$
where
\begin{align*}
D(\omega)= & Y_1\times Y_2 \times \cdots \times Y_m  &\mbox{and} & &
Y_i=& \left\{\begin{array}{lll}
          X_i & \mbox{if $i\in\omega$}\\
          A_i & \mbox{if $i\not\in\omega$.}
            \end{array}\right.
\end{align*}

    Based on Theorem 4.7 and the decomposition of $\Sigma\ZZ_{K_\PP}(\underline{X}, \underline{A})$ given by
Bahri, Bendersky, Cohen and Gitler (\cf \cite{BB,BBCG08}), we proved in section 5

 {\bf Theorem 5.7} {\it Let $K_\PP$ be an abstract simplicial complex corresponding to a simplicial complement $\PP$
and let $(\underline{X}, \underline{A})=\{(X_i, A_i, x_i)|i\in [m]\}$ denote m choices of
connected, pointed pairs of $CW$-complexes, with the inclusion $A_i\hookrightarrow X_i$ homotopic to constant for
all $i$. If $\w{H}^*(X_i, \kk)$ and $\w{H}^*(A_i, \kk)$ are free $\kk$-modules for all $i$,
then the cohomology of $\ZZ_{K_\PP}(\underline{X}, \underline{A})$ is isomorphic to
\[
H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
\bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau\subset [m]\setminus\omega}
  H_{*,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d) \otimes
  \left(\bigotimes_{i\in\omega}\w{H}^*(X_i, \kk)\right)\otimes
  \left(\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)\right)\right)
\]
as $\kk$-modules, where
\[
H_{*,\tau}(\Lambda^{*,\tau}[E_\omega\PP],d)\cong Tor_{*,\tau}^{\kk[\xx]}(\kk(\mbox{star}_{K_\PP}\omega), \kk).
\]
}

    As applications, we also consider the cohomology of $\ZZ_{K_\PP}(\underline{X}, \underline{A})$
for some special triples of $CW$-complexes $(\underline{X}, \underline{A})$ including
all the $X_i$ are contractible; all the $A_i$ are contractible and
$(\underline{X}, \underline{A})=(\underline{S^2}, \underline{S^1})$.

{\it Acknowledgements } The authors are grateful to Z.~L\"u for his introduction of this topic. The authors
are indebted to V.~M.~Buchstaber for his comments and suggestions, especially for the name of definition
{\it simplicial complement} which is originally called {\it partition.} The authors also thank S. Gitler for his
helpful suggestions.

\section{The homology of simplicial complement and the $Tor$ of face ring}

    Let $[m]=\{1,2,\cdots, m\}$ and $2^{[m]}$ denote the power set of $[m]$.
Choose a ground ring $\kk$ with unit (We usually suppose that $\kk$ is a principle ideal domain).
Let $\kk[\xx]$ be the $\NN^m$ graded polynomial algebra over $\kk$ on $m$ indeterminates
$\xx=\{x_1, x_2, \cdots, x_m\}$. The monomial of $\kk[\xx]$ is expressed as
\[
\xx^\ba=x_1^{a_1}x_2^{a_2}\cdots x_m^{a_m}
\]
for a unique vector $\ba=(a_1, a_2, \cdots, a_m)\in \NN^m$.
For a subset $\sigma=\{i_1, i_2, \cdots, i_n\}$ of $[m]$,
let $\xx_\sigma=x_{i_1}x_{i_2}\cdots x_{i_n}$ be the corresponding monomial in $\kk[\xx]$,
while $\xx_\phi=1$. Then the monomial
$\xx_\sigma$ can be expressed as $\xx^{\ba_\sigma}=\xx_\sigma$ for a unique
vector $\ba_\sigma\in\NN^m$.

  {\bf Definition \nr} {\it A sequence $\PP=\{\sigma_1, \sigma_2, \cdots , \sigma_s\}$ of
subsets of $[m]$ is called a simplicial complement. Given a simplicial complement
$\PP=\{\sigma_1, \sigma_2, \cdots, \sigma_s\}$,
let $\II_\PP$ denote the ideal of $\kk[\xx]$ generated by $\xx_{\sigma_i}$'s
\[
\II_\PP=\langle \xx_{\sigma_1}, \xx_{\sigma_2}, \cdots, \xx_{\sigma_s}\rangle.
\]
Two simplicial complements $\PP$ and $\QQ$  are called equivalent if they generate the same ideal of $\kk[\xx]$,
that is $\II_\PP=\II_\QQ$.}

    Given an abstract simplicial complex $K$, let $\PP_K$ denote the simplicial complement
consists of all the non-faces $\sigma\not\in K$ or equivalently consists of all the
{\it missing faces} in the sense that it is not a simplex of $K$ but all of its proper subsets
are simplices of $K$.  One can easily see from its definition
that $\II_{\PP_K}=\II_K$.

    Furthermore, given a simplicial complement $\PP$, noticed that $\II_\PP$ is
an ideal generated by square-free monomials,
there is a unique simplicial complex $K_\PP$ such that $\II_\PP=\II_{K_\PP}$.

    {\bf Definition \nr} {\it Define $K_\PP$  to be the simplicial complex corresponding to
$\PP$ such that the non-faces $2^{[m]}\setminus K_\PP$ is the
full subset of $2^{[m]}$ consisting of all subsets $\sigma\in 2^{[m]}$ which contain a
subset $\sigma_i$ in $\PP$, that is
\[
2^{[m]}\setminus K_\PP=\{\sigma\in 2^{[m]} \mid\ \mbox{there is a $\sigma_i\in\PP$
 such that $\sigma_i\subset\sigma$}\}.
\]}

     Notice that,
$\xx_\sigma=\xx_{\sigma_i}\cdot\xx_{\sigma\setminus\sigma_i}$ if $\sigma_i\subset\sigma$,
we have $\II_\PP=\II_{K_\PP}$.
The Stanley-Reisner face ring is
\[
\kk(K_\PP)=\kk[\xx]/\II_{K_\PP}=\kk[\xx]/\II_\PP.
\]

    {\bf Proposition \nr} {\it If $\PP$ and $\QQ$ are equivalent simplicial complements,
then
\[
\kk(K_\PP)=\kk[\xx]/\II_\PP=\kk[\xx]/\II_\QQ=\kk(K_\QQ)
\]
and the corresponding simplicial complexes $K_\PP=K_\QQ$. \qb}

    {\bf Remark:}  Notice that if the empty set $\phi$ is an element of the simplicial complement $\PP$, then
$\II_\PP=\langle \xx_\phi=1\rangle=\kk[\xx]$, the simplicial complex $K_\PP=\varnothing$, which is different from the
empty simplicial complex $K=\{\phi\}$.

{\bf Definition \nr} {\it   Given a simplicial complement
$\PP=\{\sigma_1, \sigma_2, \cdots , \sigma_s\}$,
let $\Lambda[\PP]$  be the exterior algebra over $\kk$ on the generators $\sigma_1, \sigma_2, \cdots , \sigma_s$.
Define an $\NN\times 2^{[m]}$-graded $\kk$-module structure on $\Lambda [\PP]$ as follows.
For a generator $\uu=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q}$ of $\Lambda^{*,*}[\PP]$,
let
\begin{align*}
S_\uu = & \sigma_{k_1}\cup \sigma_{k_2}\cup \cdots \cup \sigma_{k_q}
& \mbox{and}& & S_1= & \phi
\end{align*}
denote the union of the subsets $\sigma_{k_j}$'s, which is called the {\it total subset} of $\uu$.
Define the degree of $\uu=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q}$ as
\[
bideg(\uu)=(q, S_\uu)
\]
so that $\Lambda[\PP]$ becomes an $\NN\times 2^{[m]}$ graded $\kk-$module
and $\displaystyle{\Lambda^{*,*}[\PP]=\bigoplus_{q,\sigma}\Lambda^{q,\sigma}[\PP]}$. }

    {\bf Remark:}  $\xx_{S_\uu}$ is the least common multiple of the monomials
$\xx_{\sigma_{k_1}}, \xx_{\sigma_{k_2}}, \cdots, \xx_{\sigma_{k_q}}$ and
$\Lambda^{*,*}[\PP]$ corresponding to the $\xx_{\sigma_1},\ \xx_{\sigma_2},
\ \cdots,\ \xx_{\sigma_s}$ labeled standard $s-1$ simplex $\Delta^{s-1}$ (\cf \cite{Mil} Chapter 4).

    Let $R^{*,*}=\Lambda^{*, *}[\PP]\otimes_\kk \kk[\xx]$ denote the free $\kk[\xx]$-module
generated by $\Lambda^{*,*}[\PP]$. For a generator $\uu=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q}$
of $\Lambda^{q,*}[\PP]$, the monomial in $\kk[\xx]$ corresponding to the total subset
$S_\uu$ is expressed as $\xx_{S_\uu}=\xx^{\ba_{\uu}}$ for an unique $\NN^m$ vector $\ba_\uu\in \NN^m$.
We define the $\NN\times\NN^m$ degree of  $\uu$ as
$$
bideg(\uu)=(q, \ba_{\uu}).
$$
Then $R^{*,*}$ becomes an $\mathbb{N}\times\mathbb{N}^m$ graded module over $\kk$.

   Given a generator $\uu=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q}$
of $\Lambda^{q, *}[\PP]$, denote that
\begin{equation*}
\partial_i(\uu)=\partial_i(\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q})=
\sigma_{k_1}\cdots\wh{\sigma}_{k_i}\cdots\sigma_{k_q}=\sigma_{k_1}\cdots\sigma_{k_{i-1}}
\sigma_{k_{i+1}}\cdots\sigma_{k_q}.
\end{equation*}
Notice from Definition 2.4 that, the total subset $S_\uu=S_{\partial_i(\uu)}\cup (S_{\uu}\setminus S_{\partial_i(\uu)})$,
we define a $\kk[\xx]$-module homomorphism
$$\w{d}: \Lambda^{q,*}[\PP]\otimes_\kk \kk[\xx]\longrightarrow \Lambda^{q-1,*}[\PP]\otimes_\kk \kk[\xx]$$
by setting
\begin{equation}\tag{\nr}
\w{d}(\uu\otimes\xx^\ba)=\sum_{i=1}^q(-1)^i\partial_i(\uu)\otimes
    \xx_{(S_\uu\setminus S_{\partial_i(\uu)})}\xx^\ba,
\end{equation}
where $\displaystyle{\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}=\frac{\xx_{S_\uu}}{\xx_{S_{\partial_i(\uu)}}}}$
is the monomial of $\kk[\xx]$ corresponding to the subset
$S_\uu\setminus S_{\partial_i(\uu)}$.

{\bf Theorem \nr \  (Taylor resolution)} {\it
$(\Lambda^{*,*}[\PP]\otimes_\kk\kk[\xx],\ \ \w{d})$ is a free resolution of $\kk[\xx]/\II_\PP$.
Thus
\[
H_{q,\ba}\left((\Lambda^{*,*}[\PP]\otimes_\kk\kk[\xx])\otimes_{\kk[\xx]}\kk,\ \  \w{d}\otimes 1\right)=
  Tor^{\kk[\xx]}_{q, \ba}(\kk[\xx]/\II_\PP, \kk).
\]
}
\begin{proof}  One may find a proof of this Theorem in Chapter 4 of \cite{Mil}, where it is given by the cellular resolution.
Here we give a proof by induction on the cardinal number $|\PP|=s$.

    Notice that, $\partial_i\partial_j=\partial_{j-1}\partial_i$ for $i<j$, an standard argument
shows that $\w{d}$ is a derivation, that is  $\w{d}\cdot\w{d}=0$.

    For $\Lambda^{0,*}[\PP]\otimes_\kk\kk[\xx]=\kk[\xx]$, we define
the augmentation
\[
\varepsilon: \Lambda^{0,*}[\PP]\otimes_\kk\kk[\xx]\longrightarrow \kk[\xx]/\II_\PP
\]
to be the quotient map. Furthermore from $\w{d}(\sigma_i)=-\xx_{\sigma_i}$
we see that
\[
\xymatrix {\Lambda^{1,*}[\PP]\otimes_\kk\kk[\xx] \ar[r]^-{\w{d}} & \Lambda^{0,*}[\PP]\otimes_\kk\kk[\xx]
  \ar[r]^-{\varepsilon} & \kk[\xx]/\II_\PP\ar[r] & 0}
\]
is exact. Thus we only need to prove that $H_{q,*}(\Lambda^{*,*}[\PP]\otimes_\kk\kk[\xx], \w{d})=0$ for all $q>0$,
and this will be done by induction on the cardinal number $|\PP|=s$.

  If $|\PP|=1$, that is $\PP=\{\sigma_1\}$, it is easy to see that
\[
\xymatrix @C=1.0cm @R=0.4cm
{
0 \ar[r] & \kk\{\sigma_1\}\otimes_\kk\kk[\xx] \ar[r]^-{\w{d}} \ar@{=}[d]
 & \kk[\xx] \ar[r]^-{\varepsilon} \ar@{=}[d] & \kk[\xx]/\II_\PP \\
0 \ar[r] & \Lambda^{1,*}[\{\sigma_1\}]\otimes_\kk\kk[\xx] \ar[r]^-{\w{d}} &
  \Lambda^{0,*}[\{\sigma_1\}]\otimes_\kk\kk[\xx] \ar[r]^-{\varepsilon} &
  \kk[\xx]/\II_\PP
}
\]
is exact and then $H_{1,*}(\Lambda^{*,*}[\{\sigma_1\}]\otimes_\kk\kk[\xx],\ \w{d})=0$.

    Inductively suppose that for any simplicial complement $\PP$ with cardinal number $|\PP|=s$, that is
$\PP=\{\sigma_1,\sigma_2,\cdots,\sigma_s\}$ contains $s$ subsets,
the homology $H_{q,*}(\Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx], \w{d})=0$ for $q>0$.
Then for a simplicial complement $\mathbb{Q}$ with $|\mathbb{Q}|=s+1$,
that is
\[
\mathbb{Q}=\{\sigma_1,\sigma_2,\cdots,\sigma_s,\sigma\}=\PP\cup\{\sigma\}.
\]
$\Lambda^{*,*}[\PP]\otimes_\kk\kk[\xx]$ is a sub-complex of $\Lambda^{*,*}[\mathbb{Q}]\otimes_\kk\kk[\xx]$,
then the short exact sequence
\[
\xymatrix {0\ar[r] & \Lambda[\PP]\otimes_\kk\kk[\xx] \ar[r]^-{i} &
  \Lambda[\mathbb{Q}]\otimes_\kk\kk[\xx] \ar[r]
  & (\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx]\ar[r] & 0}
\]
induces a long exact sequence in homologies
\[
\xymatrix @C=0.5cm
{\cdots\ar[r] & H_{q,*}\left(\Lambda[\PP]\otimes_\kk\kk[\xx]\right) \ar[r]^-{i_*} &
  H_{q,*}\left(\Lambda[\mathbb{Q}]\otimes_\kk\kk[\xx]\right) \ar[r]
  & H_{q,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx]\right) \ar[r]^-{\partial} & \cdots}
\]
The quotient complex $(\Lambda^{q+1,*}[\mathbb{Q}]/\Lambda^{q+1,*}[\PP])\otimes_\kk\kk[\xx]$ is expressed as
$\Lambda^{q,*}[\PP]\sigma\otimes_\kk\kk[\xx]$ with generators $\uu\cdot\sigma\otimes\xx^{\ba}$ and  the  induced
differential
\[
\wh{d}: \Lambda^{q,*}[\PP]\sigma\otimes_\kk\kk[\xx]\rightarrow \Lambda^{q-1,*}[\PP]\sigma\otimes_\kk\kk[\xx]
\]
is given by
\[
\wh{d}(\uu\cdot\sigma\otimes\xx^\ba)=\sum_{i=1}^q(-1)^i(\partial_i(\uu))\cdot\sigma\otimes
  \xx_{S_i}\xx^\ba
\]
where $S_i=S_{\uu\cdot\sigma}\setminus S_{(\partial_i(\uu))\cdot\sigma}=
(\sigma_{k_1}\cup\cdots\cup\sigma_{k_q}\cup\sigma)\setminus(\sigma_{k_1}\cup\cdots
 \cup\wh{\sigma}_{k_i}\cup\cdots\cup\sigma_{k_q}\cup\sigma)$.

   By the induction hypothesis, we know that $H_{q,*}(\Lambda[\PP]\otimes_\kk\kk[\xx], \w{d})=0$ for $q>0$. Thus we
only need to prove that $H_{q,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx]\right)=0$ for $q>1$ and
\[
\xymatrix @C=0.6cm
 {0\ar[r] & H_{1,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP]\otimes_\kk\kk[\xx])\right) \ar[r]^-{\partial} &
  H_{0,*}(\Lambda[\PP]\otimes_\kk\kk[\xx])=\kk[\xx]/\II_\PP
 }
\]
is exact or equivalently $\partial$ is a monomorphism.  To do so,
let
\[
\PP'=\{\sigma'_1=\sigma_1\setminus\sigma, \sigma'_2=\sigma_2\setminus\sigma,\cdots,\sigma'_s=\sigma_s\setminus\sigma\}.
\]
Define a homomorphism of $\kk[\xx]$-modules
$f:\Sigma\Lambda[\PP']\otimes_\kk\kk[\xx]\rightarrow (\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx]$
by
\[
f(\sigma'_{k_1}\sigma'_{k_2}\cdots\sigma'_{k_q})=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q}\sigma.
\]
Notice that,
\begin{align*}
  & \left(\sigma_{k_1}\cup\cdots\cup\sigma_{k_q}\cup\sigma\right)\setminus
    \left(\sigma_{k_1}\cup\cdots\cup\wh{\sigma}_{k_i}\cup\cdots\cup\sigma_{k_q}\cup\sigma\right) \\
= & \left((\sigma_{k_1}\setminus\sigma)\cup\cdots\cup(\sigma_{k_q}\setminus\sigma)\right)
     \setminus\left((\sigma_{k_1}\setminus\sigma)\cup\cdots\cup\wh{(\sigma_{k_i}\setminus\sigma)}
     \cup\cdots\cup(\sigma_{k_q}\setminus\sigma)\right)\\
= & (\sigma'_{k_1}\cup\cdots\cup\sigma'_{k_q})\setminus(\sigma'_{k_1}
  \cup\cdots\cup\wh{\sigma}'_{k_i}\cup\cdots\cup\sigma'_{k_q})
\end{align*}
we see that for a monomial $\uu'=\sigma'_{k_1}\sigma'_{k_2}\cdots\sigma'_{k_n}\in \Sigma\Lambda^{*,*}[\PP']$,
$S_{\uu'}\setminus S_{\partial_i(\uu')}=S_{\uu\cdot\sigma}\setminus S_{(\partial_i(\uu))\cdot\sigma}$. Thus
\[
f:\Sigma\Lambda[\PP']\otimes_\kk\kk[\xx]\rightarrow \Lambda[\mathbb{Q}]/\Lambda[\PP]\otimes_\kk\kk[\xx]
\]
is an isomorphism of chain complexes. By induction on $s$ we see that
\[
H_{q+1,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx],\ \wh{d}\right)
  \cong H_{q,*}(\Lambda[\PP']\otimes_\kk\kk[\xx],\ \w{d})=0
\]
for $q>0$ and
\[
H_{1,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx],\ \wh{d}\right)
 \cong H_{0,*}(\Lambda[\PP']\otimes_\kk\kk[\xx],\ \w{d})
 \cong\kk[\xx]/\II_{\PP'}.
\]
$H_{0,*}(\Lambda[\PP']\otimes_\kk\kk[\xx],\ \w{d})\cong\kk[\xx]/\II_{\PP'}$ is
generated by $[\sigma\otimes \xx^\ba]$ with  $\xx^\ba\in\kk[\xx]$
and relations $[\sigma\otimes \xx^\ba]=0$ if $\xx^\ba\in \II_{\PP'}$.
The connecting homomorphism
\[
\xymatrix @C=0.6cm
 {\partial:
  H_{1,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx],\ \wh{d}\right) \ar[r] &
  H_{0,*}\left(\Lambda[\PP]\otimes_\kk\kk[\xx], \w{d}\right)=\kk[\xx]/\II_\PP}
\]
is given by $\partial([\sigma\otimes\xx^\ba])=-[\xx_\sigma\xx^\ba]\in\kk[\xx]/\II_\PP$.

    Fix an $\NN^m$ vector $\mathbf{c}$, we see that
$H_{1,\mathbf{c}}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])\otimes_\kk\kk[\xx],\ \wh{d}\right)$
is generated by $[\sigma\otimes \xx^\ba]$ with an unique $\NN^m$ vector $\ba$ such that
$\xx_\sigma\xx^\ba=\xx^\mathbf{c}$.
If
\[
\partial([\sigma\otimes\xx^\ba])=-[\xx_\sigma\xx^\ba]=0\in
H_{0,*}(\Lambda[\PP]\otimes_\kk\kk[\xx],\ \w{d})=\kk[\xx]/\II_\PP,
\]
then in the $\mathbb{N}^m$ graded $\kk$-module $\kk[\xx]$, the monomial
$\xx_\sigma\xx^\ba\in\II_\PP$. This implies that there is a $\sigma_k\in \PP$ such that
$\xx_\sigma\xx^\ba=\xx_{\sigma_k}\xx^\mathbf{b}$. Thus $\xx_\sigma\xx^\ba$ could be expressed as
$\xx_{\sigma\cup\sigma_k}\xx^\mathbf{c}$.
Noticed that  $\sigma\cup\sigma_k=\sigma\cup\sigma'_k$ where $\sigma'_k=\sigma_k\setminus\sigma$,
one has $\xx_\sigma\xx^\ba=\xx_{\sigma}\xx_{\sigma'_k}\xx^\mathbf{c}$
and
\[
\wh{d}(\sigma'_k\sigma\otimes\xx^\mathbf{c})=-\sigma\otimes\xx_{\sigma'_k}\xx^\mathbf{c}=-\sigma\otimes\xx^\ba.
\]
This implies that $[\sigma\otimes\xx^\ba]=0\in H_{1,*}\left((\Lambda[\mathbb{Q}]/\Lambda[\PP])
\otimes\kk[\xx], \wh{d}\right)$
and the connecting homomorphism $\partial$ is a monomorphism.  The theorem follows.
\end{proof}

    To describe the differential
\[
\xymatrix @R=0.4cm{
  d:  & \Lambda^{q,*}[\PP]\otimes_\kk\kk[\xx]\otimes_{\kk[\xx]}\kk \ar[r]\ar@{=}[d]
      & \Lambda^{q-1,*}[\PP]\otimes_\kk\kk[\xx]\otimes_{\kk[\xx]}\kk  \ar@{=}[d] \\
      & \Lambda^{q,*}[\PP] \ar[r]
      & \Lambda^{q-1,*}[\PP], }
\]
let $\sigma$ be a subset of $[m]$ and let $\Lambda^{*,\sigma}[\PP]$ be the
submodules of $\Lambda[\PP]$ generated by the monomials $\uu$ with $S_\uu=\sigma$.
Then by Definition 2.4, $\Lambda[\PP]$ becomes an  $\NN\times 2^{[m]}$ graded module over $\kk$ and
$$\Lambda^{*,*}[\PP]=\bigoplus_{i,\sigma \in 2^{[m]}}\Lambda^{i,\sigma}[\PP].$$

    {\bf Theorem \nr} {\it Let $\PP=\{\sigma_1, \sigma_2, \cdots, \sigma_s\}$ be a simplicial complement
and $K_\PP$ be the corresponding simplecial complex by Definition 2.2.
Let $(\Lambda^{*,*}[\PP], d)$ be the chain complex induced from the bi-graded exterior algebra on $\PP$,
the differential
$d: \Lambda^{q,*}[\PP]\rightarrow \Lambda^{q-1,*}[\PP]$ is given by
\begin{align*}
d(\uu)= & \sum_i(-1)^i\partial_i(\uu)\cdot\delta_{\partial_i(\uu)}
\end{align*}
where $\delta_{\partial_i(\uu)}=1$ if the total subset $S_\uu=S_{\partial_i(\uu)}$
and $\delta_{\partial_i(\uu)}=0$ if $S_\uu\not=S_{\partial_i(\uu)}$. Then the homology of
$(\Lambda^{*,*}[\PP],\ d)$ is
\[
H_{*,\sigma}(\Lambda^{*,*}[\PP],\ d)=Tor^{\kk[\xx]}_{*,\sigma}(\kk[\xx]/\II_\PP, \kk)
  =Tor^{\kk[\xx]}_{*,\sigma}(\kk(K_\PP), \kk)
\]
which is called the homology of the simplicial complement $\PP$.}
\begin{proof}
    Consider the chain complex $(\Lambda^{q,*}[\PP]\otimes_\kk\kk[\xx]\otimes_{\kk[\xx]}\kk, \w{d}\otimes 1)$.
From (2.5), we see that for a generator $\uu\otimes 1$ of $\Lambda^{q,*}[\PP]\otimes_\kk\kk[\xx]$
\[
\w{d}(\uu\otimes 1)=\sum_{i=1}^q(-1)^i\partial_i(\uu)\otimes\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}
\]
and $\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}=1$ if and only if $S_\uu=S_{\partial_i(\uu)}$.
Notice that, $\kk[\xx]$ acts on $\kk$ by sending all $x_i$'s to $0$, the theorem follows from
\[
d(\uu\otimes_\kk 1\otimes_{\kk[\xx]} 1)=\w{d}(\uu\otimes_\kk 1)\otimes_{\kk[\xx]} 1=
 \sum_{i=1}^q (-1)^i\partial_i(\uu)\otimes_\kk \xx_{(S_\uu\setminus S_{\partial_i(\uu)})}\otimes_{\kk[\xx]}1
\]
and
\[
\partial_i(\uu)\otimes_\kk\xx_{(S_\uu\setminus S_{\partial_i(\uu)})}\otimes_{\kk[\xx]}1=0\in
 \Lambda^{q-1,*}[\PP]\otimes_\kk\kk[\xx]\otimes_{\kk[\xx]}\kk
\]
if $\xx_{S_\uu\setminus S_{\partial_i(\uu)}}\not=1$, that is $S_{\partial_i(\uu)}\not=S_\uu$.
\end{proof}

    Fix a $\sigma\in 2^{[m]}$ and let $\Lambda^{*,\sigma}[\PP]$ be the submodule
generated by the elements $\uu$ with total subset $S_\uu=\sigma$. From the definition
of $\delta_{\partial_i(\uu)}$ we see that $\partial_i(\uu)\cdot\delta_{\partial_i(\uu)}
\in \Lambda^{*,\sigma}[\PP]$, otherwise $\delta_{\partial_i(\uu)}=0$. Then $(\Lambda^{*,\sigma}[\PP],\ d)$
is a sub-complex of $\Lambda^{*,*}[\PP]$ and
\[
Tor_{q,\sigma}^{\kk[\xx]}(\kk[\xx]/\II_\PP,\ \kk)=H_q(\Lambda^{*,\sigma}[\PP], d)
\]

   A precise expression of the Hochster theorem (\cf \cite{BP}) is:

{\bf Theorem:} (Baskakov) {\it There are isomorphisms
$$Tor^{\kk[\xx]}_{q,\sigma}(\kk,\ \kk[\xx]/\II_{K_\PP})\cong \w{H}^{|\sigma|-q-1}(K_\PP\cap\sigma,\ \kk),$$
where $K_\PP\cap\sigma$ is the {\it full sub-complex} of $K_\PP$ consisting
of all simplices of $K_\PP$ which have all of their vertices in $\sigma$, that is
$K_\PP\cap\sigma=\{\tau\cap\sigma|\tau\in K_\PP\}$.}

    Thus we have the following theorem:

    {\bf Theorem \nr} (Combinatorial Hochster theorem) {\it Let $\PP$ and $K_\PP$ be as above, then
\begin{align*}
H_{q,\sigma}(\Lambda^{*,\sigma}[\PP],\ d)\cong & \w{H}^{|\sigma|-q-1}(K_\PP\cap\sigma,\ \kk)
& \mbox{and} & & H_{q,[m]}(\Lambda^{*,[m]}[\PP],\ d)\cong & \w{H}^{m-q-1}(K_\PP,\ \kk).
\end{align*}
Furthermore the cohomology module of the classical moment-angle complex is
\[
H^{2j-q}(\ZZ_{K_\PP}, \kk)=\bigoplus_{\sigma\subset [m], |\sigma|=j}
 H_{q,\sigma}(\Lambda^{*,*}[\PP],\ d).
\]}

  {\bf Remark:} Here we use the agreement that the cohomology of the empty simplicial complex $\{\phi\}$ is
$\w{H}^{-1}(\{\phi\}, \kk)=\kk$.
\begin{proof} From Theorem 2.7, we see that $H_{q,\sigma}(\Lambda^{*,\sigma}[\PP],\ d)=
Tor_{q,\sigma}^{\kk[\xx]}(\kk(K_\PP), \kk)$.  Then from
\[
Tor_{q,\sigma}^{\kk[\xx]}(\kk(K_\PP), \kk)=Tor_{q,\sigma}^{\kk[\xx]}(\kk,\ \kk(K_\PP))
  = \w{H}^{|\sigma|-q-1}(K_\PP\cap\sigma,\ \kk)
\]
we get the first one. The second one follows from the Hochster Theorem (\cf \cite{BP} for example).
\end{proof}

\section{The algebraic structure on $Tor_{i,\sigma}^{\kk[\xx]}(\kk(K_\PP), \kk)$}

    Let $A$ and $B$ be $\kk[\xx]$ algebras with structure maps
\begin{align*}
\varphi: & A\otimes_{\kk[\xx]}A\rightarrow A, & \varphi': B\otimes_{\kk[\xx]}B\rightarrow B.
\end{align*}
Choose a free resolution $P$ of $A$,
\[
\xymatrix @C=0.5cm{
P: \cdots \ar[r] & P_q \ar[r] & P_{q-1} \ar[r] & \cdots \ar[r] &
  P_1 \ar[r] & P_0 \ar[r] & A,
}
\]
$P\otimes P$ is a free resolution of $A\otimes_{\kk[\xx]}A$.
The structure map $\varphi:A\otimes_{\kk[\xx]}A$ gives a chain map
$\varphi: P\otimes_{\kk[\xx]} P\longrightarrow P$ which is unique up to chain homotopy.
\[
\xymatrix @C=0.4cm {
P\otimes_{\kk[\xx]} P: \ar[d]^-{\varphi} & \cdots \ar[r] & (P\otimes P)_q \ar[r] \ar[d]^-{\varphi_q} &
  (P\otimes P)_{q-1} \ar[r] \ar[d]^-{\varphi_{q-1}} & \cdots \ar[r] &
  (P\otimes P)_1 \ar[r] \ar[d]^-{\varphi_1} & (P\otimes P)_0 \ar[r] \ar[d]^-{\varphi_0} &
  A\otimes_{\kk[\xx]}  A \ar[d]^-{\varphi}\\
P:                                      & \cdots \ar[r] & P_q \ar[r] & P_{q-1} \ar[r] &
  \cdots \ar[r] & P_1 \ar[r] & P_0 \ar[r] & A
}
\]
Applying the functor $\otimes_{\kk[\xx]}(B\otimes_{\kk[\xx]}B)$, we get a natural chain map
\begin{equation}\tag{\nr}
\xymatrix @C=0.5cm{
(P\otimes_{\kk[\xx]} P)\otimes_{\kk[\xx]}(B\otimes_{\kk[\xx]}B) \ar@{=}[r] \ar[d]^-{\varphi\otimes 1} &
 (P\otimes_{\kk[\xx]}B)\otimes_{\kk[\xx]}(P\otimes_{\kk[\xx]}B) \ar[d]^-{\times} \\
P\otimes_{\kk[\xx]}(B\otimes_{\kk[\xx]}B) \ar[r]^-{1\otimes \varphi'} & P\otimes_{\kk[\xx]}B.
}
\end{equation}
On the other hand, there are natural maps
\begin{equation}\tag{\nr}
H_i\left(P\otimes_{\kk[\xx]}B\right)\otimes H_j\left(P\otimes_{\kk[\xx]}B\right)
 \longrightarrow H_{i+j}\left((P\otimes_{\kk[\xx]}B)\otimes_{\kk[\xx]}(P\otimes_{\kk[\xx]}B)\right);
\end{equation}
one maps the cycles $c\in (P\otimes_{\kk[\xx]}B)_i$ and $c'\in(P\otimes_{\kk[\xx]}B)_j$
to
\[
c\otimes c'\in \left((P\otimes_{\kk[\xx]}B)\otimes_{\kk[\xx]}(P\otimes_{\kk[\xx]}B)\right)_{i+j}.
\]
Applying the natural chain map $\times: (P\otimes_{\kk[\xx]}B)\otimes_{\kk[\xx]}(P\otimes_{\kk[\xx]}B)
\rightarrow P\otimes_{\kk[\xx]}B$, we get the algebraic structure
\begin{equation}\tag{\nr}
\xymatrix {
Tor_{i,*}^{\kk[\xx]}(A, B)\otimes Tor_{j,*}^{\kk[\xx]}(A,B) \ar[r]^-{\times} \ar@{=}[d] &
Tor_{i+j,*}^{\kk[\xx]}(A,B) \ar@{=}[d] \\
H_i(P\otimes_{\kk[\xx]}B)\otimes H_j(P\otimes_{\kk[\xx]}B) \ar[r] &
H_{i+j}(P\otimes_{\kk[\xx]}B)
}
\end{equation}

    Notice that both $\kk(K_\PP)$ and $\kk$ are $\kk[\xx]$ algebras,
$Tor_{i,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$ is an algebra in a natural way and the
following isomorphism of algebras holds:
\[
H^*(\ZZ_{K_\PP}, \kk)\cong Tor_{i,*}^{\kk[\xx]}(\kk(K_\PP), \kk).
\]

    To describe the algebraic structure of $Tor_{i,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$,
consider the free $\kk[\xx]$-resolution of $\kk(K_\PP)$ given by Theorem 2.6,
which is denoted by $R^{*,*}$
$$R^{*,*}=\Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx].$$
The tensor product
\[
R^{*,*}\otimes_{\kk[\xx]}R^{*,*}\cong (\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk
  (\kk[\xx]\otimes_{\kk[\xx]}\kk[\xx])\cong
  (\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk\kk[\xx]
\]
gives a free resolution of $\kk(K_\PP)\otimes_{\kk[\xx]}\kk(K_\PP)\cong \kk(K_\PP)$. The algebraic structure
of $\kk(K_\PP)$ is given by
\[
\xymatrix {
\kk(K_\PP)\otimes_{\kk[\xx]}\kk(K_\PP)\cong \kk(K_\PP) \ar[r]^-{1} & \kk(K_\PP).
}
\]
Thus we need to construct a chain map
$\varphi: (\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk \kk[\xx]\rightarrow
  \Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx]$
that makes the following diagram commute
\[
\xymatrix {
(\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk \kk[\xx] \ar[r]^-{\varphi} \ar[d]^-{\varepsilon} &
  \Lambda^{*,*}[\PP] \ar[d]^-{\varepsilon}\\
  \kk(K_\PP)\otimes_{\kk[\xx]} \kk(K_\PP) =\kk(K_\PP) \ar[r]^-{1} & \kk(K_\PP)
}
\]

{\bf Construction \nr} {\it The exterior algebra $\Lambda^{*,*}[\PP]$ has natural product structure.
Given two generators $\uu$ and $\vv$ of $\Lambda^{*,*}[\PP]$ the total subset of
$\uu\cdot\vv$ is $S_{\uu\cdot\vv}=S_{\uu}\cup S_{\vv}$ if $\uu\cdot\vv\not=0$ in
$\Lambda^{*,*}[\PP]$. Thus the monomial $\xx_{S_{\uu\cdot\vv}}$ is a factor of
$\xx_{S_\uu}\cdot\xx_{S_\vv}$ in $\kk[\xx]$.  We define the product
\[
\varphi: (\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk\kk[\xx]
\longrightarrow \Lambda^{*,*}[\PP]\otimes_\kk\kk[\xx]
\]
to be the $\kk[\xx]$-module map induced by
\begin{align}\tag{\nr}
\uu\times\vv = & \uu\cdot\vv \otimes\frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\cdot\vv}}}.
\end{align}
It is apparent that so defined product keeps the second degree ($\NN^m$ degree).
}

{\bf Theorem \nr} {\it The product defined in Construction 3.4 is a chain map. Thus the algebraic structure
in $H_{*,*}(\Lambda^{*,*}[\PP],d)\cong Tor_{*,*}^{\kk[\xx]}(\kk(K_\PP), \kk)$ is given
by
\[
[c]\times [c']=[c\cdot c']\cdot \delta_{\sigma,\sigma'},
\]
where $[c]\in H_{q,\sigma}(\Lambda^{*,\sigma}[\PP], d)$, $[c']\in H_{j,\sigma'}(\Lambda^{*,\sigma'}[\PP], d)$,
the cycle $c\cdot c'$ is product of $c$ and $c'$ in $\Lambda^{*,*}[\PP]$ and $ \delta_{\sigma,\sigma'}=1$
if $\sigma\cap\sigma'=\phi$, $ \delta_{\sigma,\sigma'}=0$ if $\sigma\cap\sigma'\not=\phi$.
}
\begin{proof}
Notice that $\Lambda^{0,*}[\PP]=\kk$, it is apparent that
\[
\xymatrix{
(\Lambda^{0,*}[\PP]\otimes_\kk\Lambda^{0,*}[\PP])\otimes_\kk \kk[\xx]
\ar^-{\varepsilon}[r]\ar^-{\varphi_0}[d] &
\kk(K_\PP)\otimes_{\kk[\xx]} \kk(K_\PP)=\kk(K_\PP) \ar^-{1}[d]\\
\Lambda^{0,*}[\PP]\otimes_\kk \kk[\xx] \ar^-{\varepsilon}[r] & \kk(K_\PP)}
\]
commute.

    To prove that $\varphi$ defined by Construction 3.4 is a chain map, we only need to check that
\[
d(\uu\times\vv)=d(\uu)\times\vv+(-1)^q\uu\times d(\vv)
\]
or equivalently to check that $\Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx]$ is a differential graded algebra,
because the differential
in $(\Lambda^{*,*}[\PP]\otimes_\kk\Lambda^{*,*}[\PP])\otimes_\kk \kk[\xx]$ is given by
$d(\uu\otimes\vv)=d(\uu)\otimes\vv+(-1)^q\uu\otimes d(\vv)$ if $\uu\in\Lambda^{q,*}[\PP]$.

    Let $V$ be the free $\kk$-module generated by $\PP=\{\sigma_1, \sigma_2, \cdots, \sigma_s\}$
and let $T(V)$ be the tensor algebra of $V$, that is
$\displaystyle{T(V)=\bigoplus_{q\geqslant 0} V^{\otimes q}}$. We still use
\begin{align*}
\uu= & \sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_q} & &\mbox{and} &
S_\uu= & \sigma_{k_1}\cup\sigma_{k_2}\cup\cdots\cup\sigma_{k_q}
\end{align*}
to denote the generator of $T(V)$ and the total subset of $\uu$ respectively. The exterior algebra
$$\Lambda^{*,*}[\PP]=T(V)/\langle \uu\vv-(-1)^{pq}\vv\uu, \uu\uu\rangle$$
where $\langle \uu\vv-(-1)^{pq}\vv\uu, \uu\uu\rangle$ is the ideal generated by
$\uu\vv-(-1)^{pq}\vv\uu$ and $\uu\uu$ for any generators $\uu\in T^{q, *}(V)$ and
$\vv\in T^{p,*}(V)$.

    Similarly, the monomial $\xx_{S_\uu}\in \kk[\xx]$ corresponding to the total subset
$S_\uu$ is expressed as $\xx_{S_\uu}=\xx^{\ba_\uu}$ for a unique $\NN^m$ vector $\ba_\uu$.
We also define
\begin{align*}
bideg(\uu) = & (q, \ba_\uu) & & \mbox{and} \\
\partial_i(\uu) = & \sigma_{k_1}\cdots \wh{\sigma}_{k_i}\cdots \sigma_{k_q}.
\end{align*}
Consider the $\NN\times\NN^m$ graded free $\kk[\xx]$-module $T(V)\otimes_\kk \kk[\xx]$,
we define a differential
$$\w{d}: T^{q,*}(V)\otimes_\kk \kk[\xx]\rightarrow
T^{q-1,*}(V)\otimes_\kk \kk[\xx]$$
by setting
$$\w{d}(\uu\otimes\xx^{\ba})=\sum_i(-1)^i\partial_i(\uu)\otimes \frac{\xx_{S_\uu}}{\xx_{S_{\partial_i(\uu)}}}
  \cdot\xx^\ba
$$
as (2.5).

    Similar to Construction 3.4, $T(V)\otimes_\kk \kk[\xx]$ has algebraic structure given by
\[
\uu\times\vv=\uu\cdot\vv \otimes \frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\cdot\vv}}}.
\]
From
\begin{align*}
  & \w{d}(\uu\times\vv) = \w{d}(\uu\vv)\otimes \frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\vv}}}\\
= & \sum_i(-1)^i\partial_i(\uu)\vv\otimes\frac{\xx_{S_{\uu\vv}}}{\xx_{S_{\partial_i(\uu)\vv}}}\cdot
    \frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\vv}}}
+(-1)^q\sum_i(-1)^i\uu\partial_i(\vv)\otimes  \frac{\xx_{S_{\uu\vv}}}{\xx_{S_{\uu\partial_i(\vv)}}}\cdot
    \frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\vv}}}
\end{align*}
and
\begin{align*}
  & \w{d}(\uu)\times\vv + (-1)^q\uu\times\w{d}(\vv)\\
= & \sum_i(-1)^i\partial_i(\uu)\times\vv\otimes \frac{\xx_{S_\uu}}{\xx_{S_{\partial_i(\uu)}}}
+(-1)^q\sum_i(-1)^i\uu\times\partial_i(\vv)\otimes \frac{\xx_{S_\vv}}{\xx_{S_{\partial_i(\vv)}}}\\
= & \sum_i(-1)^i\partial_i(\uu)\vv\otimes
    \frac{\xx_{S_{\partial_i(\uu)}}\cdot\xx_{S_\vv}}{\xx_{S_{\partial_i(\uu)\vv}}}\cdot
    \frac{\xx_{S_\uu}}{\xx_{S_{\partial_i(\uu)}}}
+(-1)^q\sum_i(-1)^i\uu\partial_i(\vv)\otimes
    \frac{\xx_{S_\uu}\cdot\xx_{S_{\partial_i(\vv)}}}{\xx_{S_{\uu\partial_i(\vv)}}}\cdot
    \frac{\xx_{S_\vv}}{\xx_{S_{\partial_i(\vv)}}}
\end{align*}
we see that $\w{d}(\uu\times\vv)=\w{d}(\uu)\times\vv+(-1)^q\uu\times\w{d}(\vv)$ and
 $T(V)\otimes_\kk \kk[\xx]$ is a differential graded algebra.  It is easy to check
that the ideal $\langle \uu\vv-(-1)^{pq}\vv\uu, \uu\uu\rangle$ is invariant under the differential
$\w{d}$. Thus $\{T(V)\otimes_\kk \kk[\xx], \w{d}\}$ induces the differential graded algebraic structure
on $\Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx]$ as desired.

    To prove the second part of the Theorem, applying the functor $\otimes_{\kk[\xx]}\kk$,
 we see that in $(\Lambda^{*,*}[\PP]\otimes_\kk \kk[\xx])\otimes_{\kk[\xx]}\kk$
\[
(\uu\times\vv\otimes_\kk 1)\otimes_{\kk[\xx]} 1=(\uu\cdot\vv\otimes_\kk 1)\otimes_{\kk[\xx]}
 \frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\vv}}}=0
\]
if the monomial $\displaystyle{\frac{\xx_{S_\uu}\cdot\xx_{S_\vv}}{\xx_{S_{\uu\vv}}}\not=1}$
or equivalently $S_\uu\cap S_\vv\not=\phi$.
\end{proof}

   {\bf Example:} Compute the $Tor^{\kk[\xx]}_{*,*}(\kk(K), \kk)$ and  the cohomology of $\ZZ_K$
corresponding to $K$ (See Figure 1)

  There is the simplicial complement $\PP=\left\{\sigma_1=\{1,5\},\ \sigma_2=\{2,4\},
\ \sigma_3=\{1,2,3\},\ \sigma_4=\{3,4,5\}\right\}$ corresponding to the simplicial complex  $K=K_\PP$.

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\end{picture}
\end{center}\vspace{5mm}

From Theorem 2.7, we compute the differentials in $\Lambda^{*,*}[\PP]$;
\begin{align*}
d(\sigma_1\sigma_2\sigma_3\sigma_4)= & -\sigma_2\sigma_3\sigma_4 +\sigma_1\sigma_3\sigma_4
                                       -\sigma_1\sigma_2\sigma_4 +\sigma_1\sigma_2\sigma_3;\\
d(\sigma_1\sigma_3\sigma_4) = & -\sigma_3\sigma_4.
\end{align*}
And these are all the non-trivial differentials. Thus the $Tor^{\kk[\xx]}_{*,*}(\kk(K), \kk)$
is a free $\kk$-module generated by
\[
\left\{\begin{array}{ll}
         \sigma_1\sigma_2\sigma_4\\
         \sigma_1\sigma_2\sigma_3
         \end{array}
        \begin{array}{ll}
         \sigma_1\sigma_2\\
         \sigma_1\sigma_3\\
         \sigma_1\sigma_4\\
         \sigma_2\sigma_3\\
         \sigma_2\sigma_4
        \end{array}
        \begin{array}{ll}
         \sigma_1\\
         \sigma_2\\
         \sigma_3\\
         \sigma_4
        \end{array}
        \begin{array}{ll}
         1
        \end{array}
\right\}.
\]

    Consider the algebraic structure of $Tor_{*,*}^{\kk[\xx]}(\kk(K), \kk)$, we see from Theorem 3.6
that
\[
\sigma_1\times\sigma_2=\sigma_1\sigma_2
\]
is the only non-trivial product.  By the Hochester Theorem, we see that $H^*(\ZZ_K, \kk)$ is a free $\kk$-module with Poincar\`{e}
series $1+2x^3+2x^5+5x^6+2x^7$.

\section{The simplicial complements and the simplicial complexes}

   In \cite{CL} X. Cao and Z.~L\"{u} introduced a $Z/2$-algebra
$$2^{[m]^*}=\{f|f:2^{[m]}\rightarrow Z/2=\{0,1\}\}$$
consists of all $Z/2$ valued functions on the power set $2^{[m]}$ with addition $(f+g)(\sigma)=f(\sigma)+g(\sigma)$
and multiplication $(f\cdot g)(\sigma)=f(\sigma)\cdot g(\sigma)$.

    Let $f: 2^{[m]}\rightarrow Z/2$ be a function in $2^{[m]*}$.  We define
$$supp(f)=\{\sigma\in 2^{[m]}\mid f(\sigma)=1\}=f^{-1}(1)\subset 2^{[m]}$$
which is called the {\it support} of $f$.

    On the other hand, given a simplicial complex $K$,
there is the {\it characteristic function} $f_K\in 2^{[m]^*}$ of $K$ defined by
\begin{align}\tag{\nr}
f_K(\sigma)= & \left\{\begin{array}{ll}
                         1 & \mbox{if $\sigma\in K$}\\
                         0 & \mbox{otherwise}
                  \end{array}\right. ,
\end{align}
with $supp(f_K)=K$.

    Given a simplex $\sigma\in 2^{[m]}$, there are functions $\delta_\sigma$ and $\mu_\sigma\in 2^{[m]*}$
defined as
\begin{align}\tag{\nr}
\delta_\sigma(\tau)= & \left\{\begin{array}{ll}
                       1 & \mbox{if $\tau=\sigma$}\\
                       0 & \mbox{otherwise.}
                             \end{array}\right.
& \mu_{\sigma}(\tau)= & \left\{\begin{array}{ll}
                           1 & \mbox{if $\sigma\subset\tau$}\\
                           0 & \mbox{otherwise}
                          \end{array}\right.
\end{align}
From (4.2), one can easily show that any $f\in 2^{[m]*}$ can be expressed as
\[
f=\sum_{\sigma\in 2^{[m]}}f(\sigma)\delta_\sigma=\sum_{\sigma\in supp(f)}
  \delta_\sigma.
\]
Thus $\{\delta_\sigma\mid \sigma\in 2^{[m]}\}$ forms a basis of $2^{[m]^*}$.

{\bf Theorem \nr} {\it Let $\PP=\{\sigma_1,\sigma_2,\cdots,\sigma_s\}$ be a simplicial complement,
$K_\PP$ be the simplicial complex corresponding to $\PP$ by Definition 2.2 and $f_{K_\PP}$ be
the  characteristic function corresponding to
$K_\PP$ by (4.1). Then
\[
f_{K_\PP}=\prod_{i=1}^{s}(1+\mu_{\sigma_i})=\sum_{\uu\in\Lambda[\PP]}\mu_{S_\uu}
=\sum_{\sigma\in 2^{[m]}}\left(\sum_i dim_\kk\Lambda^{i,\sigma}[\PP]\right)\mu_\sigma,
\]
where the first sum runs over all the $\kk$-module generators $\uu$ of $\Lambda^{*,*}[\PP]$.}
\begin{proof}
    Notice that, $1(\tau)=1$ for any $\tau\in 2^{[m]}$ and $(1+\mu_{\sigma})(\tau)=0$
if and only if $\sigma\subset\tau$, we see that for any $\tau\in 2^{[m]}$
\[
f_{K_\PP}(\tau)=(1+\mu_{\sigma_1})(\tau)\cdot(1+\mu_{\sigma_2})(\tau)\cdot\cdots\cdot
       (1+\mu_{\sigma_s})(\tau)=0
\]
if and only if there exists an $i$ such that $(1+\mu_{\sigma_i})(\tau)=0$.
This is equivalent to that there is a $\sigma_i\in\PP$ such that $\sigma_i\subset\tau$.
By Definition 2.2, $\tau$ is a non-face of $K_\PP$. Thus
$\displaystyle{f_{K_\PP}=\prod_{i=1}^{s}(1+\mu_{\sigma_i})}$.

    For the second part of the theorem, notice that
for any subset $\omega\in 2^{[m]}$,
$(\mu_{\sigma}\cdot\mu_{\tau})(\omega)=\mu_{\sigma}(\omega)\cdot\mu_{\tau}(\omega)=1$
if and only if $\sigma\subset\omega$ and $\tau\subset\omega$ $\Longleftrightarrow$
$\sigma\cup\tau\subset\omega$. One has
\[
\mu_{\sigma}\cdot\mu_{\tau}=\mu_{\sigma\cup\tau}.
\]
Thus
\begin{align*}
\prod_{i=1}^{s}(1+\mu_{\sigma_i})= &
    1+\sum_{\{i_1,i_2\cdots,i_q\}\subset [m]}\mu_{\sigma_{i_1}}\mu_{\sigma_{i_2}}\cdots\mu_{\sigma_{i_q}}
    = \sum_{\uu\in\Lambda[\PP]}\mu_{S_\uu}
\end{align*}
by Definition 2.4.

    Fix a $\sigma\in 2^{[m]}$, $\Lambda^{*,\sigma}[\PP]$ is the $\kk-$submodule generated
by $\uu$ with total subset $S_\uu=\sigma$,
$$\sum_{S_\uu=\sigma}\mu_{S_\uu}=
 \left(\sum_i \mbox{dim}_\kk\Lambda^{i,\sigma}[\PP]\right)\mu_\sigma.$$
The third part of the Theorem follows.
\end{proof}

  {\bf Definition \nr} {\it Fix a subset $\omega\in [m]$, there is the map $\varepsilon_\omega: 2^{[m]}\longrightarrow 2^{[m]}$ defined
by $\varepsilon_\omega(\tau)=\tau\cup\omega$. The $\omega$-compression
$E_\omega: 2^{[m]^*}\longrightarrow 2^{[m]^*}$ is defined by
\[
E_\omega(f)=f\circ\varepsilon_\omega.
\]
For the simplicial complement $\PP=\{\sigma_1, \sigma_2, \cdots, \sigma_s\}$,  we
define the  $\omega$-compression of $\PP$ by
\[
E_\omega\PP=\{\sigma_1\setminus \omega, \sigma_2\setminus \omega, \cdots, \sigma_s\setminus \omega\}.
\]}

    From $(f+g)\circ\varepsilon_\omega=f\circ\varepsilon_\omega+g\circ\varepsilon_\omega$ and
$(f\cdot g)\circ\varepsilon_\omega=(f\circ\varepsilon_\omega)\cdot(g\circ\varepsilon_\omega)$,
we see that $E_\omega: 2^{[m]^*}\longrightarrow 2^{[m]^*}$ is a homomorphism of $Z/2$-algebra.

    {\bf Lemma \nr} {\it Let $\PP$ be a simplicial complement and $\displaystyle{f_{K_\PP}=\prod_{i=1}^s(1+\mu_{\sigma_i})}$
be the characteristic function of the simplicial complex $K_\PP$ by Theorem 4.3. Then
\[
E_\omega(f_{K_\PP})=\prod_{i=1}^s(1+\mu_{\sigma_i\setminus\omega})=f_{K_{E_\omega\PP}},
\]
i.e. $E_\omega(f_{K_\PP})$ is the characteristic function of the simplicial complex $K_{E_\omega\PP}$
corresponding to the simplicial complement $E_\omega\PP$.
}
\begin{proof}
    The $\omega$-compression $E_\omega: 2^{[m]^*} \longrightarrow 2^{[m]^*}$ is a homomorphism
of $Z/2$-algebra. Thus
\[
E_\omega(f_{K_\PP})=E_\omega\left(\prod_{i=1}^s(1+\mu_{\sigma_i})\right)=
    \prod_{i=1}^s\left(E_\omega(1)+E_\omega(\mu_{\sigma_i})\right).
\]

    For the function $\mu_\sigma$, take a $\tau\in 2^{[m]}$, we have
\[
E_\omega(\mu_\sigma)(\tau)=1\Longleftrightarrow\mu_{\sigma}(\varepsilon_\omega(\tau))=1
 \Longleftrightarrow\sigma\subset\tau\cup \omega\Longleftrightarrow(\sigma\setminus \omega)\subset\tau.
\]
This implies that
\begin{align}\tag{\nr}
E_\omega(\mu_{\sigma})= & \mu_{\sigma\setminus \omega}.
\end{align}
Then from
$E_\omega(1)=1$, we see that
$\displaystyle{E_\omega(f_{K_\PP})=\prod_{i=1}^s\left(1+E_\omega(\mu_{\sigma_i})\right)=
 \prod_{i=1}^s\left(1+\mu_{\sigma_i\setminus \omega}\right)=f_{K_{E_\omega\PP}}}.$
\end{proof}

    For an arbitrary simplex $\omega\in K$, define its {\it link} and {\it star} as the
sub-complexes
\begin{align*}
  & \mbox{star}_K\omega = \{\tau\in K|\omega\cup\tau\in K\};
  & \mbox{link}_K\omega = \{\tau\in K|\omega\cup\tau\in K, \omega\cap\tau=\phi\}.
\end{align*}

  {\bf Theorem \nr} {\it Let $\PP$ be a simplicial complement and $K_\PP$ be the simplicial complex corresponding
to $\PP$. Then:}
\begin{enumerate}
\item {\it The $star$,} $\mbox{star}_{K_\PP}\omega=K_{E_\omega\PP}$ {\it is the simplicial
complex corresponding to the simplicial complement $E_\omega\PP$. Thus}
\[
H_{q,\sigma}(\Lambda^{*,\sigma}[E_\omega\PP], d)=
  Tor^{\kk[\xx]}_{q,\sigma}(\kk(\mbox{star}_{K_\PP}\omega), \kk),
\]
\item {\it The $link$,} $\mbox{link}_{K_\PP}\omega=K_{E_\omega\PP}\cap ([m]\setminus\omega)$ {and}
\[
H_{q,[m]\setminus \omega}(\Lambda^{*,[m]\setminus \omega}[E_\omega\PP], d)=
 \w{H}^{m-|\omega|-q-1}(\mbox{link}_{K_\PP}\omega, \kk).
\]
\end{enumerate}

\begin{proof}
    Let $f_{K_\PP}$ and $f_{K_{E_\omega\PP}}$ be the characteristic functions of $K_\PP$ and
$K_{E_\omega\PP}$ respectively. Then a simplex $\tau\in \mbox{star}_{K_\PP}\omega$ if and only if
\begin{align*}
\tau\cup\omega\in K_\PP & \Longleftrightarrow f_{K_\PP}(\tau\cup\omega)=1
\Longleftrightarrow f_{K_\PP}\circ\varepsilon_\omega(\tau)=1
\Longleftrightarrow E_\omega(f_{K_\PP})(\tau)=f_{K_{E_\omega\PP}}(\tau)=1
\end{align*}
Thus the characteristic function of $\mbox{star}_{K_\PP}\omega$ is $f_{K_{E_\omega\PP}}$ and
\begin{align*}
\mbox{star}_{K_\PP}\omega= & \{\tau\in K_\PP|\omega\cup\tau\in K_\PP\}= K_{E_\omega\PP},  \\
\mbox{link}_{K_\PP}\omega= & \{\tau\in K_\PP|\omega\cup\tau\in K_\PP, \tau\cap\omega=\phi\} =
  K_{E_\omega\PP}\cap([m]\setminus\omega).
\end{align*}
From Theorem 2.7 we see that
\begin{align*}
Tor^{\kk[\xx]}_{q,\sigma}(\kk(\mbox{star}_{K_\PP}\omega), \kk)= & Tor^{\kk[\xx]}_{q,\sigma}(\kk(K_{E_\omega\PP}), \kk)
    =H_{q,\sigma}(\Lambda^{*,\sigma}[E_\omega\PP], d)
\end{align*}
and from Theorem 2.8 we get
\begin{align*}
\w{H}^{m-|\omega|-q-1}(\mbox{link}_{K_\PP}\omega,\ \kk) = &
\w{H}^{m-|\omega|-q-1}(K_{E_\omega\PP}\cap([m]\setminus\omega),\ \kk) =
H_{q,[m]\setminus \omega}(\Lambda^{*,[m]\setminus \omega}[E_\omega\PP], d).
\end{align*}
The theorem follows.
\end{proof}

\section{The cohomology of the generalized moment-angle complex}

    In this section we consider the cohomology module of the generalized moment-angle complex.
Recall from \cite{BB} \cite{BBCG08},

  {\bf Definition \nr} {\it Let $(\underline{X}, \underline{A})=\{(X_i, A_i, x_i)|i\in[m]\}$ denote
a set of triples of $CW$-complexes with base points $x_i\in A_i$ and $K_\PP$ be an abstract simplicial
complex. The generalized moment-angle complex determined by $(\underline{X}, \underline{A})$ and
$K_\PP$ denoted by $\ZZ_{K_\PP}(\underline{X}, \underline{A})$ is defined using the functor
\[
D: K_\PP\longrightarrow CW_*
\]
as follows: For every $\omega\in K_\PP$, let
\begin{align*}
D(\omega)= & Y_1\times Y_2 \times \cdots \times Y_m &\mbox{where} & &
Y_i=& \left\{\begin{array}{lll}
          X_i & \mbox{if $i\in\omega$}\\
          A_i & \mbox{if $i\not\in\omega$.}
            \end{array}\right.
\end{align*}
The generalized moment-angle complex is
\[
\ZZ_{K_\PP}(\underline{X}, \underline{A})=\bigcup_{\omega\in K_\PP} D(\omega).
\]}

    Let $Y_1\wedge Y_2 \wedge \cdots \wedge Y_m $ be the smash product given by
the quotient space
\[
Y_1\times Y_2\times\cdots\times Y_m/S(Y_1\times Y_2\times\cdots\times Y_m)
\]
where $S(Y_1\times Y_2\times\cdots\times Y_m)$ is the subspace of the product with at least
one coordinate given by the base-point $x_i\in Y_i$. The {\it generalized smash moment-angle
complex} is defined to be the image of $\ZZ_{K_\PP}(\underline{X}, \underline{A})$
in $X_1\wedge X_2\wedge\cdots\wedge X_m$, that is
\[
\wh{\ZZ}_{K_\PP}(\underline{X}, \underline{A})=\bigcup_{\omega\in K_\PP} \wh{D}(\omega)
\]
where
\begin{align*}
\wh{D}(\omega)= & Y_1\wedge Y_2 \wedge \cdots \wedge Y_m  &\mbox{subject to} & &
Y_i=& \left\{\begin{array}{lll}
          X_i & \mbox{if $i\in\omega$}\\
          A_i & \mbox{if $i\not\in\omega$.}
            \end{array}\right.
\end{align*}

    Given a non-empty subset $I=\{i_1, i_2, \cdots, i_k\}\subset [m]$ and a family of pairs
$(\underline{X}, \underline{A})$, define
\[
(\underline{X_I}, \underline{A_I})=\{(X_{i_j}, A_{i_j})|i_j\in I\}
\]
which is the subfamily of $(\underline{X}, \underline{A})$ determined by $I$.  It is known from
\cite{BB,BBCG08} that:
\begin{align}\tag{\nr}
H: \Sigma(\ZZ_{K_\PP}(\underline{X}, \underline{A}))\longrightarrow &
  \Sigma\left(\bigvee_{I\subset [m]}\wh{\ZZ}_{K_\PP\cap I}(\underline{X_I}, \underline{A_I})\right)
\end{align}
is a natural pointed homotopy equivalence.  To describe it more precisely, let
$\omega\subset I$ and
\begin{align*}
\wh{D}_I(\omega)= & Y_{i_1}\wedge Y_{i_2} \wedge \cdots\wedge Y_{i_k}
 & \mbox{with} & & Y_{i_j}=\left\{\begin{array}{ll}
                           X_{i_j} & \mbox{if $i_j\in \omega$}\\
                           A_{i_j} & \mbox{if $i_j\not\in\omega$.}
                   \end{array}\right.
\end{align*}
Then
\[
\wh{\ZZ}_{K_\PP\cap I}(\underline{X_I}, \underline{A_I})=\bigcup_{\omega\in K_\PP\cap I}\wh{D}_I(\omega).
\]

    Associated to a simplicial complex $K_\PP$, there is a partial ordered set ({\it poset}) $\overline{K_\PP}$
with point $\sigma$ in $\overline{K_\PP}$ corresponding to a simplex $\sigma\in K_\PP$ and order given by
reverse inclusion of simplices. Thus $\sigma_1\leqslant \sigma_2$ in $\overline{K_\PP}$ if and only if
$\sigma_2\subseteq\sigma_1$ in $K_\PP$.  Given an $\omega\in\overline{K_\PP}$ there are further posets given
by
\begin{align*}
\overline{K_\PP}_{<\omega} = & \{\tau\in\overline{K_\PP}|\tau<\omega\}=\{\tau\in\overline{K_\PP}|\omega\varsubsetneq\tau\}
 & \mbox{and} \\
\overline{K_\PP}_{\leqslant\omega} = & \{\tau\in\overline{K_\PP}|\tau\leqslant\omega\}
 =\{\tau\in\overline{K_\PP}|\omega\subset\tau\}.
\end{align*}

    Given a poset $P$, the {\it order complex} $\Delta(P)$ is the simplicial complex with vertices given by set of
points of $P$ and $k$-simplices given by ordered $(k+1)$-tuples $(p_0, p_1, \cdots, p_k)$ with
$p_0<p_1<\cdots<p_k$.  It follows that $\Delta(\overline{K_\PP})_{<\phi}=K'_{\PP}$ is the barycentric subdivision
of $K_\PP$.

    Given a simplicial complex $K_\PP$, we use $|K_\PP|$ to denote its geometric realization.
The symbol $|K_\PP|\ast Y$ denotes the join of $K_\PP$ and $Y$. If $Y$ is a pointed $CW$-complex, then
$|K_\PP|\ast Y$ has the homotopy type of $\Sigma |K_\PP|\wedge Y$.

  {\bf Theorem (Bahri, Bendersky, Cohen and Gitler \cite{BB,BBCG08} 2.12)} {\it Let $K_\PP$ be an abstract simplicial
complex and $\overline{K_\PP}$
its associated poset. Let $(\underline{X}, \underline{A})=\{(X_i, A_i, x_i)|i\in [m]\}$ denote m choices of
connected, pointed pairs of $CW$-complexes, with the inclusion $A_i\hookrightarrow X_i$ homotopic to constant for
all $i$. Then there is a homotopy equivalence
\[
\wh{\ZZ}_{K_\PP}(\underline{X}, \underline{A})\longrightarrow \bigvee_{\omega\in K_\PP}
|\Delta(\overline{K_\PP}_{<\omega})|\ast\wh{D}(\omega).
\]
}

    From (5.2), we see that if $A_i\hookrightarrow X_i$ are null-homotopic for all $i$, then
\begin{align*}
\Sigma(\ZZ_{K_\PP}(\underline{X}, \underline{A}) \simeq &
  \Sigma\left(\bigvee_{I\subset [m]} \wh{\ZZ}_{K_\PP \cap I}(\underline{X_I}, \underline{A_I})\right)\\
   \simeq &  \Sigma\left(\bigvee_{I\subset [m]}\left(\bigvee_{\omega\in K_\PP\cap I}
  |\Delta(\overline{K_\PP\cap I}_{<\omega})|\ast\wh{D}_I(\omega)\right)\right).\notag
\end{align*}
Fix an $\omega\in K_\PP$, it is easy to see that $\omega\in K_\PP\cap I$ if and only if
$\omega\subset I$. Thus $\Sigma(\ZZ_{K_\PP}(\underline{X}, \underline{A})$ has the homotopy type of
\begin{align}\notag
\left(\bigvee_{I\subset [m]}\left(\bigvee_{\omega\in K_\PP\cap I}
  |\Delta(\overline{K_\PP\cap I}_{<\omega})|\ast\wh{D}_I(\omega)\right)\right) = &
\left(\bigvee_{\omega\in K_\PP}\left(\bigvee_{\omega \subset I\subset [m]}
  |\Delta(\overline{K_\PP\cap I}_{<\omega})|\ast\wh{D}_I(\omega)\right)\right)\\
 \simeq & \left(\bigvee_{\omega\in K_\PP}\left(\bigvee_{\omega \subset I\subset [m]}
  \Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|\wedge\wh{D}_I(\omega)\right)\right)\tag{\nr}
\end{align}

  {\bf Remark \nr} If $\omega$ is a maximum face of $K_\PP\cap I$
in the sense that $\omega\in K_\PP\cap I$ but it is not a proper subset of any other simplices,
then $\overline{K_\PP\cap I}_{<\omega}=\{\tau\in \overline{K_\PP\cap I}|\omega\varsubsetneq\tau\}=\varnothing$.
The simplicial complex $\Delta(\overline{K_\PP\cap I}_{<\omega})=\{\phi\}$ is the empty simplicial complex.
Here we use the agreement
\begin{align*}
|\{\phi\}|\ast \wh{D}_I(\omega)= & \wh{D}_I(\omega) & \mbox{and} & &
\Sigma|\{\phi\}|\wedge \wh{D}_I(\omega) = & \wh{D}_I(\omega).
\end{align*}
Combine with the agreement in Theorem 2.8, $\w{H}^{-1}(|\{\phi\}|, \kk)=\kk$, we have
\begin{align*}
\w{H}^0(\Sigma|\{\phi\}|, \kk)= & \kk & \mbox{and} & & \w{H}^*(\Sigma|\{\phi\}|\wedge\wh{D}_I(\omega), \kk)= &
\w{H}^*(\wh{D}_I(\omega), \kk)
\end{align*}

   Consider the reduced cohomology of $\ZZ_{K_\PP}(\underline{X}, \underline{A})$, we see from (5.3) that:
\begin{align}\tag{\nr}
\w{H}^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)= &
  \bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\omega\subset I\subset[m]}
  \w{H}^*\left(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|\wedge\wh{D}_I(\omega), \kk\right)\right).
\end{align}

   Furthermore suppose that there is no
$Tor$ problem in the K\"{u}nneth formulae for the cohomology of
$\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|\wedge\wh{D}_I(\omega)$
(For example, take $\kk$ to be a field or suppose that $\w{H}^*(X_i, \kk)$ and $\w{H}^*(A_i, \kk)$ are free
$\kk$-modules for all $i$).
Then from
\begin{align*}
\wh{D}_I(\omega)= & Y_{i_1}\wedge Y_{i_2}\wedge\cdots\wedge Y_{i_k}\simeq
 \left(\bigwedge_{i\in\omega}X_i\right)\bigwedge\left(\bigwedge_{j\in I\setminus\omega} A_j\right),\\
H^*(\wh{D}_I(\omega), \kk) = & \left(\bigotimes_{i\in\omega} \w{H}^*(X_i, \kk)\right) \otimes\left(\bigotimes_{j\in I\setminus\omega}
  \w{H}^*(A_j, \kk)\right)
\end{align*}
which is a free $\kk$-module, we have
\begin{align}\tag{\nr}
 & \w{H}^*\left(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|\wedge\wh{D}_I(\omega), \kk\right)
=  \w{H}^*\left(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|, \kk\right)\otimes\w{H}^*\left(\wh{D}_I(\omega), \kk\right)\\
= & \w{H}^*(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|,\kk)\otimes
  \left(\bigotimes_{i\in\omega} \w{H}^*(X_i, \kk)\right) \otimes\left(\bigotimes_{j\in I\setminus\omega}
  \w{H}^*(A_j, \kk)\right).\notag
\end{align}

 {\bf Theorem \nr} {\it Let $K_\PP$ be an abstract simplicial complex corresponding to a simplicial complement $\PP$
and let $(\underline{X}, \underline{A})=\{(X_i, A_i, x_i)|i\in [m]\}$ denote m choices of
connected, pointed pairs of $CW$-complexes, with the inclusion $A_i\hookrightarrow X_i$ homotopic to constant for
all $i$. If $\w{H}^*(X_i, \kk)$ and $\w{H}^*(A_i, \kk)$ are free $\kk$-modules for all $i$,
then the cohomology  of $\ZZ_{K_\PP}(\underline{X}, \underline{A})$ is isomorphic to
\[
H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
\bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau}
  H_{*,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d) \otimes
  \left(\bigotimes_{i\in\omega}\w{H}^*(X_i, \kk)\right)\otimes
  \left(\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)\right)\right)
\]
as $\kk$-modules,  where
\[
H_{*,\tau}(\Lambda^{*,\tau}[E_\omega\PP],d)\cong Tor_{*,\tau}^{\kk[\xx]}(\kk(\mbox{star}_{K_\PP}\omega), \kk)
\]
subject to $\tau\subset [m]\setminus \omega$.
}

\begin{proof}
    Given an $\omega\subset I$, from its definition we see that the posets
\begin{align*}
\overline{K_\PP\cap I}_{\leqslant \omega}= & \{\tau\in K_\PP\cap I|\omega\subset \tau\}
  = \{\tau\in K_\PP|\omega\subset\tau\subset I\} \hspace{10mm}\mbox{and}\\
\overline{\mbox{link}_{K_\PP\cap I}\omega} = & \{\tau'\in K_\PP\cap I|\omega\cup\tau'\in K_\PP\cap I, \tau'\cap\omega=\phi\}\\
  = & \{\tau'\in K_\PP|\omega\cup\tau'\in K_\PP, \tau'\cap\omega=\phi,\ \mbox{and}\ \tau'\subset I\}
     =\overline{(\mbox{link}_{K_\PP}\omega)\cap I}_{\leqslant \phi}.
\end{align*}
There is a one to one correspondence between the posets $\Psi: \overline{K_\PP\cap I}_{<\omega}\longrightarrow
\overline{(\mbox{link}_{K_\PP}\omega)\cap I}_{<\phi}$ given by $\Psi(\tau)=\tau\setminus\omega$.
Thus the order complex $\Delta(\overline{K_\PP\cap I}_{<\omega})=((\mbox{link}_{K_\PP}\omega)\cap I)'$
is the barycentric subdivision of
$(\mbox{link}_{K_\PP}\omega)\cap I$ and then
\begin{align}\tag{\nr}
\w{H}^*(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|, \kk)=
\w{H}^{*-1}(|\Delta(\overline{K_\PP\cap I}_{<\omega})|, \kk)=
  \w{H}^{*-1}((\mbox{link}_{K_\PP}\omega)\cap I, \kk).
\end{align}

    Recall from Theorem 4.7 that $\mbox{link}_{K_\PP}\omega=K_{E_\omega\PP}\cap([m]\setminus\omega)$, we have
\[(\mbox{link}_{K_\PP}\omega)\cap I =
K_{E_\omega\PP}\cap([m]\setminus\omega)\cap I=K_{E_\omega\PP}\cap (I\setminus\omega).
\]
Then from Theorem 2.8 and (5.8), we have
\begin{align*}
H_{q,I\setminus\omega}(\Lambda^{*,I\setminus\omega}[E_\omega\PP], d)
 =  & \w{H}^{|I\setminus\omega|-q-1}((\mbox{link}_{K_\PP}\omega)\cap I, \kk)
 =  \w{H}^{|I\setminus\omega|-q}(\Sigma|\Delta(\overline{K_\PP\cap I}_{<\omega})|, \kk)
\end{align*}

    From the definition of $E_\omega\PP=\{\sigma_1\setminus\omega, \sigma_2\setminus\omega, \cdots,
\sigma_s\setminus\omega\}$, we see that total subset $S_\uu$ of any generator $\uu\in\Lambda^{*,*}[E_\omega\PP]$
is contained in $[m]\setminus\omega$. Thus the homology of the simplicial complement
$H_{q,\tau}(\Lambda^{*,\tau}[E_\omega\PP],d)$ is concentrated in $\tau\subset[m]\setminus\omega$.
Denote the non-empty set $I$ by $\tau\cup\omega$ for any $\omega\subset I\subset[m]$,
we have
\[
H_{q,\tau}(\lambda^{*,\tau}[E_\omega\PP], d)=\w{H}^{|\tau|-q}(\Sigma|\Delta
 (\overline{K_\PP\cap (\tau\cup\omega)}_{<\omega})|, \kk).
\]

  Apply this formula to (5.6) and (5.5) we get
\begin{align*}
 & \w{H}^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
  \bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau\subset[m]\setminus\omega}
  \w{H}^*\left(\Sigma|\Delta(\overline{K_\PP\cap (\omega\cup\tau)}_{<\omega})|\wedge\wh{D}_{\omega\cup\tau}(\omega), \kk\right)\right) \\
 = &  \bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau\subset[m]\setminus\omega}
 H_{q, \tau}(\Lambda^{*, \tau}[E_\omega\PP], d)\otimes
  \left(\bigotimes_{i\in\omega} \w{H}^*(X_i, \kk)\right) \otimes\left(\bigotimes_{j\in \tau}
  \w{H}^*(A_j, \kk)\right)\right)
\end{align*}
where $I=\omega\cup\tau\not=\phi$.

    The theorem follows from $H^0(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=\kk$ and the agreement by taking
$I=\omega=\tau=\phi$ to be the empty set
\[
H_{0,\phi}(\Lambda^{*,\phi}[E_\phi\PP], d)
  \otimes\left(\bigotimes_{i\in\phi}\w{H}^*(X_i, \kk)\right)
  \otimes\left(\bigotimes_{j\in\phi}\w{H}^*(A_j, \kk)\right)=\kk.
\]
\end{proof}

   {\bf Remark:} If the abstract simplicial complex $K_\PP$ is given by an abstract simplicial complement $\PP$,
we might not know if a subset $\omega$ is a simplex of $K_\PP$ or not. This is not a problem, because
while $\omega\not\in K_\PP$, the empty set $\phi$ is an element of $E_\omega\PP$ and
$H_{*,*}(\Lambda^{*,*}[E_\omega\PP], d)\equiv 0$. Thus Theorem 5.7 could be written as
\begin{align}\notag
  & H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)\\
= & \bigoplus_{\omega\in[m]}\left(\bigoplus_{\tau\in[m]}
  H_{*,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d) \otimes
  \left(\bigotimes_{i\in\omega}\w{H}^*(X_i, \kk)\right)\otimes
  \left(\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)\right)\right)\tag{\nr}
\end{align}

  {\bf Corollary \nr} {\it If all the $A_i$ are contractible and $\w{H}^*(X_i, \kk)$ are free $\kk$-modules.
Then
\[
H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
 \bigoplus_{\omega\in K_\PP}\left(\bigotimes_{i\in \omega} \w{H}^*(X_i, \kk)\right)
\]
}

\begin{proof}
If all the $A_i$ are contractible then $\displaystyle{\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)=0}$ for any non-empty set $\tau$.
The corollary follows from $H_{0,\phi}(\Lambda^{*,\phi}[E_\omega\PP],d)=\kk$ if $\omega\in K_\PP$.
\end{proof}

 {\bf Corollary \nr} {\it If all the $X_i$ are contractible and $\w{H}^*(A_i, \kk)$ are free
$\kk$-modules. Then
\begin{align*}
H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)= &
 \bigoplus_{\tau}\left(H_{*,\tau}(\Lambda^{*,\tau}[\PP], d)
 \otimes\left(\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)\right)\right).
\end{align*}
Furthermore take $X_i=D^2$ and $A_i=S^1$ for all $i$,
\begin{align*}
H^{r}(\ZZ_{K_\PP}, \kk)=H^{2|\tau|-q}(\ZZ_{K_\PP}(\underline{D^2}, \underline{S^1}), \kk)= &
 \bigoplus_{2|\tau|-q=r}H_{q,\tau}(\Lambda^{*,\tau}[\PP], d),
\end{align*}
where $H_{q,\tau}(\Lambda^{*,\tau}[\PP], d)=Tor_{q,\tau}^{\kk[\xx]}(\kk(K_\PP), \kk)$.
}

\begin{proof}
    If all the $X_i$ are contractible, then $\displaystyle{\bigotimes_{i\in\omega}\w{H}^*(X_i, \kk)=0}$ for any
non-empty set $\omega$. The corollary follows from
\[
H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
 \bigoplus_{\tau}H_{*,\tau}(\Lambda^{*,\tau}[E_\phi\PP], d)
 \otimes\left(\bigotimes_{i\in\phi}\w{H}^*(X_i,\kk)\right)\otimes
  \left(\bigotimes_{j\in\tau}\w{H}^*(A_j, \kk)\right)
\]
and the agreement $\displaystyle{\bigotimes_{i\in\phi}\w{H}^*(X_i,\kk)=\kk}$.
\end{proof}

  {\bf Proposition \nr} {\it Let $X_i=S^2$ and $A_i=S^1$ all $i$. Then
\[
H^{r}(\ZZ_{K_\PP}(\underline{S^2}, \underline{S^1}), \kk) =
  \bigoplus_{\begin{array}{c}\omega\in K_\PP\\ 2|\omega|+2|\tau|-q=r\end{array}}
  \left(\bigoplus_{\tau}H_{q,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d)\right).
\]
}

\begin{proof}
  Noticed that $\wh{D}_{\omega\cup\tau}(\omega)$ is the $2|\omega|+|\tau|$ sphere if $\omega\cap\tau=\phi$,
$$\wh{D}_{\omega\cup\tau}(\omega)
 =\left(\bigwedge_{i\in\omega}S^2\right)\wedge\left(\bigwedge_{j\in\tau}S^1\right)
 =S^{2|\omega|+|\tau|},
$$
we see from (5.5) that
$$H^*(\ZZ_{K_\PP}(\underline{X}, \underline{A}), \kk)=
  \bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau\subset[m]\setminus\omega}
  \w{H}^*\left(\Sigma^{2|\omega|+|\tau|+1}|\Delta(\overline{K_\PP\cap (\omega\cup\tau)}_{<\omega}|, \kk\right)\right)
$$
including $\omega=\tau=\phi$. The result follows from
$$\w{H}^{|\tau|-q}\left(\Sigma|\Delta(\overline{K_\PP\cap (\omega\cup\tau}_{<\omega})|, \kk\right)
 =H_{q,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d).$$
\end{proof}

  We finish this paper by giving an example.

{\bf Example \nr} Let $m=6$ and the simplicial complement $\PP=\{\sigma_1=\{1,2\}, \sigma_2=\{3,4\}, \sigma_3=\{5,6\}\}$.
The corresponding simplicial complex is a triangulation of the sphere $S^2$ (see Finger 2)

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    The cohomology ring of $\ZZ_K=\ZZ_{K_\PP}(\underline{D^2}, \underline{S^1})$ could be easily got from
the homology of the simplicial complement  $\PP=\{\sigma_1=\{1,2\}, \sigma_2=\{3,4\},$ $\sigma_3=\{5,6\}\}$,
\[
H_{*,*}(\Lambda^{*,\tau}[\PP], d)\cong\Lambda^{*,*}[\{\sigma_1, \sigma_2, \sigma_3\}]
\]
as an exterior algebra over $\kk$. Thus the cohomology ring $H^*(\ZZ_K, \kk)$ is the exterior algebra
on $\PP$ with non-trivial products.
The Poincar\`{e} series of $H^*(\ZZ_K, \kk)$ is  $1+3x^3+3x^6+x^9$ and the
total Betti number of $\ZZ_K$ is $8$.

   Consider the cohomology of $\ZZ_{K_\PP}(\underline{S^2}, \underline{S^1})$. We start from
computing the homology of the simplicial complement $E_\omega\PP$ with $\omega\in K$:
\begin{enumerate}
\item Take $\omega=\phi$,  the homology
of the simplicial complement $E_\phi\PP$ is
$\Lambda^{*,*}[\{\sigma_1, \sigma_2, \sigma_3\}].$
Thus the submodule
\[
\bigoplus_{\tau}H_{*,\tau}(\Lambda^{*,\tau}[E_\phi\PP], d)=\Lambda^{*,*}[\{\sigma_1,\sigma_2,\sigma_3\}]
\]
is a free $\kk$-module with Poincar\`{e} series $1+3x^3+3x^6+x^9$.
\item Take $\omega=\{1\}$ (similarly, take any of its 6 $0$-simplex $\omega=\{i\}$, $i\in[6]$),
the homology
of the simplicial complement $E_{\{i\}}\PP$ is $\Lambda^{*,*}[E_{\{i\}}\PP]$.
Thus the submodules
\[
\displaystyle{\bigoplus_{\tau}H_{*,\tau}(\Lambda^{*,\tau}[E_{\{i\}}\PP], d)}
\]
is the free $\kk$-module with Poincar\`{e} series $x^2(1+x+2x^3+2x^4+x^6+x^7)$.
\item Take $\omega=\{1,3\}$ and similarly any of its 12 $1$-simplex $\omega=\{i_1, i_2\}$, the homology
of the corresponding simplicial complement is $\Lambda^{*,*}[E_{\{i_1,i_2\}}\PP]$. The submodule
\[
\displaystyle{\bigoplus_{\tau}H_{*,\tau}(\Lambda^{*,\tau}[E_{\{i_1, i_2\}}\PP], d)}
\]
is the free $\kk$-module with Poincar\`{e} series $x^4(1+2x+x^2+x^3+2x^4+x^5)$.
\item Take $\omega=\{1,3,5\}$ and similarly any of its 8 $2$-simplex $\omega=\{i_1, i_2,i_3\}$, the homology
of the corresponding simplicial complement is $\Lambda^{*,*}[E_{\{i_1,i_2,i_3\}}\PP]$. The submodule
\[
\displaystyle{\bigoplus_{\tau}H_{*,\tau}(\Lambda^{*,\tau}[E_{\{i_1, i_2\,i_3\}}\PP], d)}
\]
is the free $\kk$-module with Poincar\`{e} series $x^6(1+3x+3x^2+x^3)$.
\end{enumerate}

   Thus
\[
H^*(\ZZ_{K_\PP}(\underline{S^2}, \underline{S^1}), \kk)=
 \bigoplus_{\omega\in K_\PP}\left(\bigoplus_{\tau}H_{q,\tau}(\Lambda^{*,\tau}[E_\omega\PP], d)\right)
\]
is a free $\kk$-module with Poincar\`{e} series
\begin{align*}
  & (1+3x^3+3x^6+x^9)\\
+ & 6x^2(1+x+2x^3+2x^4+x^6+x^7) \\
+ & 12x^4(1+2x+x^2+x^3+2x^4+x^5)\\
+ & 8x^6(1+3x+3x^2+x^3)\\
= & 1+6x^2+9x^3+12x^4+36x^5+35x^6+36x^7+54x^8+27x^9
\end{align*}
The total Betti number of $\ZZ_{K_\PP}(\underline{S^2}, \underline{S^1})$ is $216$.


TITLE: Labeling the eigenstates of the many electron Hamiltonian
QUESTION [2 upvotes]: I'm taking a course on atomic and molecular physics and there is a chapter about finding suitable quantum numbers that label the eigenstates of a given Hamiltonian. The lecture notes say that it basically comes down to finding a complete set of operators that commute with the Hamiltonian and with each other. The eigenstates can then be labeled using the eigenvalues of these operators. This makes sense to me. However it brings up a few questions:

For the hydrogen atom (and ignoring spin-orbit coupling etc), why are the eigenstates labeled by $n$, $l$ and $m_l$ instead of $n$, $l$, $m_l$ and $m_s$
The course notes contain the following sentence: "The eigenstates of the many-electron system can thus be labeled by $L$ and $S$". With $L$ the total azimuthal quantum number and $S$ the total spin quantum number. But $\hat{L}_z$ and $\hat{S}_z$ also commute with the Hamiltonian in addition to $\hat{L}^2$ and $\hat{S}^2$. So why aren't $M_L$ and $M_S$ necessary as well in order to uniquely label the eigenstates? (Again, this part of the lecture notes ignores everything related to spin-orbit coupling)

REPLY [1 votes]: I think they are skipping over a few things. Usually spin in hydrogen atom orbitals is semi-ignored and you just say that there are 2 electrons in each state, and each state is labeled $n$, $l$, and $m_l$.
The next question is a little more complicated. Indeed $H$, $L^2$, $S^2$, $L_z$, and $S_z$ all commute. However a state that has a definite value for these quantities does not necessarily have a well defined total angular momentum defined $J = L + S$. This has the operator,
\begin{align*}
J^2 &= (L + S)^2, \\
J^2 &= L^2 + S^2 + 2L\cdot S.
\end{align*}
Note that $J^2$ does not commute with $L_z$ and $S_z$. So instead we choose the set of quantum numbers, $J^2$, $L^2$, $S^2$, and $J_z$. It is usually prefered to consider a state with well defined angular momentum than one with well defined $L_z$ and $S_z$.


TITLE: $\cos x\cdot \sin x$ integral problem
QUESTION [0 upvotes]: When integrating $\cos x\cdot \sin x$ it is found that this function could be the derivative of $3$ different possible functions ...
My question is, would that be considered a mathematical uncertainty?

REPLY [1 votes]: While the answers look different, they are all equivalent anti-derivatives as each differs by a constant amount from the others.
Recall the identity $\sin^{2} x + \cos^{2} x = 1$. We can re-write this as $\sin^{2} x = 1 – \cos^{2} x.$.

So the answer of $\frac{\sin2 x}{2}$ + C can be written as $\frac{1 – \cos2 x}{2}$ + C = $\frac{1}{2} – \frac{\cos2 x}{2}$ + C. The constant $\frac{1}{2}$ can be grouped with the arbitrary constant C to give :$\frac{\cos 2x}{2}$ + C. In other words, we’ve determined these two answers are equivalent.

So the third answer of -$\frac{\cos(2x)}{4}$ + C can be re-written as: $\frac{2 \cos2 x – 1}{4}$ + C = -$\frac{\cos2 x}{4}$ + $\frac{1}{4}$ + C. The constant $\frac{1}{4}$ can be grouped with the arbitrary constant C so the answer is -$\frac{\cos2 x}{2}$ + C, and so this answer is also equivalent to the other answers.

In fact if you draw the graphs of all the three anti-derivatives, we get,
 .
As you can see, the graphs are all vertical translations of one another–each function differs from another by a constant amount.
Anti-derivatives that differ by a constant are equivalent to each other, and so the solutions are really $3$ ways of writing the same anti-derivative. Hope it helps.

REPLY [0 votes]: Would you consider the formulae
$$\cos^2x+\sin^2x=1\quad\text{or}\quad \sum_{k=1}^n k^3=\biggl(\sum_{k=1}^n k\biggr)^2$$
as reflecting a ‘mathematical uncertainty’?

REPLY [0 votes]: There's a mathematical ambiguity, but no uncertainty. Let me try to explain:
We speak informally of "the integral of $f$", typically meaning the indefinite integral. The Fundamental Theorem tells us that generally this indefinite integral is a function $F$ with $F' = f$. But in saying this, I've already led you down the garden path, for "the indefinite integral" is meaningless -- it's an indefinite integral. For any function $F$ that has the property that $F' = f$, and any constant $C$, there's another function
$$
G(x) = F(x) + C
$$
that also has the property that $G' = f$, i.e., it's another indefinite integral of $F$.
So why do we talk about "the indefinite integral"? Because it's convenient, and everyone who uses it on a regular basis knows that they're being informal "up to constants". But that doesn't help the beginner one bit!


TITLE: Using Open Covers to prove a set is compact
QUESTION [2 upvotes]: If $(x_n)$ is a sequence converging to a limit $L$ , let $Y = \lbrace x_n:n \epsilon \mathbb{N}\rbrace \cup\lbrace L \rbrace $. Using open covers, prove that Y is compact.
Any help would be appreciated, have no idea where to start.

REPLY [1 votes]: Let $\{O_i: i \in I\}$ be any open cover of $Y$.
This means there is some $i_0 \in I$ such that $L \in O_{i_0}$, as $L \in Y$ must be covered.
By definition of convergence and as $O_{i_0}$ is an open neighbourhood of the limit $L$, there is some $N \in \Bbb N$ such that
$$\forall n \ge N: x_n \in O_{i_0}$$
So that one open set already contains almost all points of $Y$, and for each $n < N$ we can find $i(n) \in I$ such that $x_n \in O_{i(n)}$ (all points must be covered!) and hence we have covered all points of $Y$ by the finite subcover
$$\{O_{i_0}\} \cup \{O_{i(n)}: n < N\}$$
and as the starting cover was arbitrary, $Y$ is compact.


TITLE: Möbius transformations are differentiable
QUESTION [1 upvotes]: Let's consider the Möbius transformation
$$
f:\mathbb{C}\cup \{\infty\}\rightarrow\mathbb{C}\cup\{\infty\},z\mapsto \frac{az+b}{
-\overline{b}z+\overline{a}},
$$
with $a,b\in\mathbb{C}$ such that $|a|^{2}+|b|^{2}=1$.
I want to check that, considering $\mathbb{C}\cup \{\infty\}$ as a differentiable manifold, $f$ is a differentiable map. My first question is:
If I prove that $f$ is a holomorphic map considering $\mathbb{C}\cup \{\infty\}$ as a complex manifold, would this imply that $f$ is differentiable?
The next question is:
How can I prove that $f$ is holomorphic?
My problem is that I haven't found an atlas that allows me to prove $f$ is holomorphic.

REPLY [2 votes]: $\newcommand{\Cpx}{\mathbf{C}}$As you surely know, $z$ is a holomorphic local coordinate on $U_{\infty}$, the complement of $\{\infty\}$, and $w = 1/z$ is a holomorphic local coordinate on $U_{0}$, the complement of $\{0\}$. Your atlas for the sphere will be the maximal holomorphic atlas compatible with these two charts.
Pre- and post-composition of a Möbius transformation $T$ with either chart map is still a Möbius transformation. To prove a general Möbius transformation $T$ is holomorphic (hence smooth) as a mapping of the Riemann sphere, it therefore suffices to restrict $T$ to the complement of its pole in $\Cpx$, where the calculation is elementary.
In more detail, separately restrict $f$ to the following four sets, which cover the sphere:

The intersection $U_{\infty} \cap f^{-1}(U_{\infty})$ (a.k.a., the complement of the pole of $f$ in $\Cpx$);
The intersection $U_{0} \cap f^{-1}(U_{\infty})$ (a.k.a., the complement of the pole of $f$ in $U_{0}$);
The intersection $U_{\infty} \cap f^{-1}(U_{0})$ (a.k.a., the complement of the zero of $f$ in $\Cpx$);
The intersection $U_{0} \cap f^{-1}(U_{0})$ (a.k.a., the complement of the zero of $f$ in $U_{0}$);

In each case, pre- and post-compose $f$ with $z$ or $w$ to get a mapping from $\Cpx$ to itself.


TITLE: Coefficients of Taylor Polynomial
QUESTION [0 upvotes]: Given a function $f(x) \in C^3 $ around $0$ for which $f(x)=ax+bx^2+R_2(x)$ is its Maclaurin sequence.
Define:
$g(x)= \frac{f(x)}{x} $ when $x\neq 0 $ and $g(0)=1$ .
Given that $a=1$. Is $g'(0)=b$?
I tried substituting the expression into $g$ . Such that, for every $x\neq 0 $ , we have:
$g(x)=1+bx+\frac{R_2(x)}{x} $ .
How can I prove that $b=g'(0)$ ?
Thanks everyone

REPLY [0 votes]: $$g(x)=1+bx+xe_2(x)\;\;,\;\;\text{with}\;\;e_2(x)\xrightarrow[x\to 0]{}0\;$$
following Peano's Remainder + Theorem , and so
$$g'(0)=\lim_{x\to 0}\frac{g(x)-g(0)}x=\lim_{x\to 0}\frac{1+bx+xe_2(x)-1}x=$$
$$=\lim_{x\to 0}\left(b+e_2(x)\right)=b$$


TITLE: Real non trivial zeros of Dirichlet L-functions
QUESTION [4 upvotes]: When dealing with the prime number theorem in arithmetic progressions, one cannot exclude the possible presence of a real zero close to $1$ for at most one real character mod $q$. On the other hand, it is also known that the Riemann $\zeta$ function does not vanish on (0, 1).
Are there any result showing that some Dirichlet L-functions (attached to a non-principal character) do not vanish far from $1$ ? I think non-vanishing at $1/2$ is still open in general, but maybe it is known in some cases.

REPLY [3 votes]: Perhaps the most comprehensive result in this direction is due to Conrey and Soundararajan (https://arxiv.org/abs/math/0111013 or https://link.springer.com/article/10.1007/s00222-002-0227-x).  They prove (among other things) that the density of odd positive squarefree integers $d\leq x$ such that $L(s,\chi_{8d})>0$ for all $s\in[0,1]$ is at least $\frac{4}{5\pi^2}$.  For work on nonvanishing at $s=1/2$, Soundararajan (https://arxiv.org/abs/math/9902163 or https://www.jstor.org/stable/2661390?seq=1#page_scan_tab_contents) obtains more robust results.


TITLE: Equivalence relation and equivalence classes specific question
QUESTION [0 upvotes]: I have a particular question in regard to my preparation for an upcoming exam. Unfortunatelly I have no one to ask or check/compare my answers, so here it is:
On the set $U = \{1, 2, 3, 4, 5, 6\}$ define the relation
$$R = \{(i, i) \;|\; i ∈ U\} ∪ \{(1, 2),(2, 1),(2, 3),(3, 2),(1, 3),(3, 1),(4, 6),(6, 4)\}$$
Show that R is an equivalence relation. Establish what are the equivalence
classes of R, in particular, how many equivalence classes R has, and how many
elements each of them has.
So $R = \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(1, 2),(2, 1),(2, 3),(3, 2),(1, 3),(3, 1),(4, 6),(6, 4)\}$
An equivalence relation is
(1) reflexive - which R is
(2) symmetric
(3) transitive.
Here is where I got confused, because I think that R is not an equivalence relation, because for it to be symmetric (definiton of symmetric is $\forall x,y\in U \;:\; (x,y)\in R \Longleftrightarrow (y,x)\in R$) but there is no pair including 5, and 5 is in U, hence its not symmetric. Or should I only work with the tuples given and not consider every element in U? I hope I explained my concerns correctly.
Also, if it is the latter, and R is indeed an equivalence relation, then are these the corect equivalence classes?
$[1]=[2]=[3]= \{1,2,3\}$ and $[4] = [6] = \{4,6\}$ .
Thank you in advance.

REPLY [0 votes]: For symmetry it is enough to ask by each tuple $(a,b)$ in $R$ the question: "is $(b,a)$ also in $R$?"
If the answer is "yes" for all tuples in $R$ then $R$ is symmetric.
You forgot equivalence class $[5]=\{5\}$.
Further it is okay.