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% ------ MACROS FOR THIS ARTICLE -------
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\begin{document}
\title{Almost regular sequences and Betti numbers}
\author{Annetta Aramova and J\"{u}rgen Herzog}
\date{}
\maketitle
\section*{Introduction}
The concept of algebraic shifting has been introduced by Kalai
\cite{K1} in 1984. The construction assigns to each simplicial
complex $\Delta$ a combinatorially simpler, shifted complex $\Delta^s$
which shares many nice properties with $\Delta$. Kalai showed
for example that
$\Delta$ is Cohen-Macaulay over a field $K$ if and only if
$\Delta^s$ has this property which in turn is the case if and
only if $\Delta^s$ is pure. Algebraic shifting also preserves
the Betti numbers, as noted by Bj\"orner and Kalai \cite{BK1}.
Recently Duval \cite{D1} has shown that a simplicial complex is
sequentially Cohen-Macaulay if and only if the $h$-triangle is
preserved under algebraic shifting. Thus, as these results
indicate, algebraic shifting is
rather stable for many combinatorial properties of simplicial
complexes.
In this paper we attempt to explain algebraic shifting and its
main features in a more algebraic way. The starting
observation, noticed already several times in the literature
(see e.g.\ \cite{G}, \cite{HT}, \cite{D1}), is that
algebraic shifting by its definition
corresponds to the operation of taking
the generic initial ideal of a Stanley Reisner ideal of a
simplicial complex in the exterior algebra.
Using Alexander duality, it is noted in
\cite{HT}
that Kalai's theorem which asserts that Cohen-Macaulayness
is preserved under algebraic shifting, is equivalent to the fact
that the exterior face rings $K\{\Delta\}$ and $K\{\Delta^s\}$
have the same regularity. Recall that the regularity $\reg(M)$ of a
finitely generated graded module $M$ over the exterior algebra $E$ is
defined to be the maximum of the numbers $j$ for which there
exists an integer $i$ with $\Tor_i^R(K,M)_{i+j}\neq 0$. Even
though the projective dimension of $M$
is infinite, unless $M$
is free, this maximum indeed exists, as is remarked in \cite{AHH}.
Kalai's theorem reformulated as a statement about regularity
corresponds to the
well-known theorem of Bayer and Stillman \cite{BS} which says
that an ideal in a polynomial ring
and its generic initial ideal have the same regularity.
Thus it is natural to ask
whether a similar theorem not only holds for monomial,
but more generally for all graded ideals
in the exterior algebra. One of the main results
of this paper states that this is indeed the case.
In fact, it is shown in Theorem \ref{hope} that the
regularity of a submodule
$N\subset F$ of a free module $F$ with
basis and that of its generic initial module $\Gin(N)$ coincide.
The known proof in the polynomial case cannot be imitated
since one of the crucial arguments in that case is that the
regularity can be computed from short
exact sequences whose left term is a module of finite
length. This is no longer true for (finitely generated)
modules over the exterior algebra, since over this algebra
all modules have finite length.
Instead our proof is based on the usage of almost regular
sequences.
A remarkable strengthening of the Bayer-Stillman theorem has
been recently proved by Bayer, Charalambous and Popecsu \cite{BCP}. They
introduced the so-called extremal Betti numbers whose position
is indicated by the lower corners in the Betti diagram displayed
by the computer algebra program MACAULAY.
One of their results, which implies the
Bayer-Stillman theorem, says that the extremal Betti numbers of
an ideal as
well as their positions are preserved when passing to the
generic inital ideal.
In the first two sections we give an alternative
proof of this theorem. Our poof uses almost regular sequences. We
call a sequence of linear forms $l_1,\ldots,l_n$ in
the polynomial ring $S$ almost regular on a graded $S$-module $M$ if for each
$i$, multiplication with $l_i$ on $M/(l_1,\ldots, l_{i-1})M$
has a finite length kernel. For an almost regular sequence the
Koszul homology for each partial sequence has finite length.
This allows to express some of the invariants of the
Koszul homology modules in terms of
this finite length kernels, which in generic coordinates and for
the reverse lexicographic term order are preserved under the passage to the
generic initial module. Though the basic philosophy of this
proof does not so much differ from that of Bayer,
Charalambous and Popescu, it nevertheless gives a hint for a possible
approach of the similar problem for graded modules over the exterior
algebra.
In the Section 3, 4 and 5 we prove the above mentioned Theorem
\ref{hope}. Following our arguments in the polynomial ring, we
have to replace the Koszul complex by the Cartan complex and
have to find a concept similar to almost regular sequences which
makes sense for modules over the exterior algebra. The solution
of the ladder problem is the observation that for a basis
${\bold v}= v_1,\ldots,v_n$ of $E_1$ the Cartan
cohomology $H^\pnt({\bold v};M)$ of a finitely generated graded
$E$-module $M$
has the natural structure of a finitely generated graded module
over a polynomial ring $S$, and that the connecting homomorphisms
attached to the partial sequences $v_1,\ldots, v_i$ of
${\bold v}$ operate on this cohomology in the same way as the
variables of $S$. Thus for generic bases these
connecting homomorpisms are injective for large enough
homological degrees, see \ref{genbas} for a precise statement. This is
the crucial new indregient for the proof of \ref{hope}, while
the rest of the arguments follow the usual trails. We would like
to point out that the finite length kernels which we
consider in the polynomial case are replaced here by what we call
generalized simplicial cohomology of a graded module, and which is just
the cohomology of the complex $\cdots M_{i-1}\to M_i\to
M_{i+1}\to \cdots$ where the chain maps are multiplication with
a general linear form of $E$. The definition is
justified by the fact that in case of a face ring this
cohomology is exactly the ordinary reduced simplicial
cohomology. Some of the basic and well-known properties of this
cohomology are
discussed in Section 3; see also \cite{G} and \cite{AAH}.
Section 4 presents all of the facts on Cartan (co)homology which
is needed in the sequel, and finally in Section 5 we study
generic initial modules.
In Section 6 we give a simplicial interpretation of the Cartan cohomology
of a face ring $K\{\Delta\}$, and show that the multigraded components of this
cohomology can be identified with reduced simplicial cohomology
of certain links of $\Delta$.
Section 7 is devoted to a module theoretic version of the
well-known Alexander duality. Combining the results of Section 6
and 7 we obtain an analogue of the Eagon-Reiner theorem
\cite{ER}. Indeed in \ref{eagonreiner} it is shown that $K\{\Delta\}$
has a linear resolution if and only if the dual simplicial
complex $\Delta^*$ is Cohen-Macaulay over $K$. After these
preparations it is easy to see how the basic facts of algebraic
shifting can be deduced from our theory. This is done in Section
8.
In the final section of this paper we show that the Bayer-
Charalambous-Popescu theorem is also valid for algebraic
shifting. In other words we show (Theorem \ref{final})
that the Stanley-Reisner rings $K[\Delta]$ and $K[\Delta^s]$
have the same extremal Betti numbers. This result may be
considered as a
strong improvement of Kalai's theorem which says that $\Delta$
is Cohen-Macaulay over $K$ if and only if $\Delta^s$ is so. The
difficulty of the proof results from the fact that algebraic
shifting is defined via the exterior algebra while in the
theorem the resolutions over the polynomial ring of $K[\Delta]$
and $K[\Delta^s]$ are compared. In the proof we use a result of
\cite{AAH} in which the resolution of a monomial ideal in the
polynomial ring and the resolution of the corresponding monomial
ideal in the exterior algebra are related. As a second
indregient of our proof we use the surprising fact that the extremal Betti
numbers can be identified as classical module invariants
of Cartan cohomology, namely as dimension and multiplicity.
\section{Almost regular sequences and Koszul homology}
Let $K$ be a field, $V$ an $n$-dimensional vector space and $S$
the symmetric algebra of $V$. The graded maximal ideal of $S$
which is generated by the elements of $S_1$
will be denoted by $\mm$. We further denote by ${\cal M}$ the
category of finitely generated graded modules over $S$.
Let $M\in {\cal M}$;
an element $x\in S_1$ is called {\em almost $M$-regular}, if
the colon module $0:_Mx=\{c\in M\: xc=0\}$ is of finite length.
The set of almost $M$-regular elements is a nonempty open subset of $S_1$.
Indeed, $M/H^0_\mm(M)$ is a module of positive depth,
so that the Zariski open set ${\cal S}\subset S_1$ of regular elements of
$M/H^0_\mm(M)$ in $S_1$ is not empty. For any element $x\in
{\cal S}$ we have that $0:_Mx$ is a finite length module.
Let ${\bold l}=l_1,\ldots,l_m$ be a sequence of linear forms in $S$.
In order to simplify notation we
set $M\langle j\rangle=M/(l_1,\ldots,l_j)M$, and for $i\geq 1$ we let
$H_i(j)$ be the $i$th Koszul homology
$H_i(l_1,\ldots,l_j;M)$ of $M$ with respect to
the sequence $l_1,\ldots,l_j$. We further set $H_i(0)=0$ for
$i>0$ and for $j\geq 1$ we let $H_0(j-1)$ be the
colon ideal $0:_{M_{\langle j-1\rangle}} l_j$.
Observe that, in
our notation, $H_0(j)$ is {\em not} the $0$th Koszul homology.
The sequence ${\bold l}=l_1,\ldots,l_m$ is called an {\em
almost regular $M$-sequence } if for all $j=1,\ldots, m$,
the linear form $l_j$ is almost $M_{\langle j-1\rangle}$-regular.
If all permutations of the sequence ${\bold l}$
are almost $M$-regular, then we call ${\bold l}$ an {\em
unconditioned almost regular $M$-sequence }.
Suppose ${\bold l}=l_1,\ldots,l_m$ is
almost $M$-regular, then all
$H_i(j)$ are modules of finite length and since
$M$ is a graded $S$-module, all $H_i(j)$ are naturally graded.
Now suppose in addition that ${\bold l}$ is a basis of $S_1$.
Then there are graded isomorphisms $H_i(n)_j\iso \Tor_i(K,M)_j$ for
all $i$ and $j$. In particular, the graded $ij$th Betti numbers
$\beta_{ij}$ of $M$ coincide with $\dim_KH_i(n)_j$.
Let $N$ be an Artinian graded module. We set $s(N)=\max\{s\:
N_s\neq 0\}$ if $N\neq 0$ and $s(0)=-\infty$.
Now we introduce the following numbers attached to
$M$ and the basis ${\bold l}=l_1,\ldots,l_n$. We set
\[
r_j=\max\{s(H_i(j))-i\: i\geq 1\}\quad\mbox{and}\quad
s_j=s(H_0(j-1))\quad \mbox{for}\quad j=1,\ldots,n,
\]
and put $r_0=0$.
We observe that $\reg(M)=\max\{r_n,s(M/\mm M)\}$. The main result
of this section is the following:
\begin{Theorem}
\label{similar}
Suppose that the basis ${\bold l}=l_1,\ldots,l_n$ of $S_1$ is an
almost regular $M$-sequence. Then
\begin{enumerate}
\item[(a)] $r_j=\max\{s_1,\ldots,s_j\}$ for $j=1,\ldots,n$. In
particular, $r_1\leq r_2\leq \ldots\leq r_n$.
\item[(b)] Let ${\cal J}=\{j_1,\ldots,j_l\}$, $1\leq
j_1r_{j_{t-1}}$ and $i>j-j_t+1$;
\item[(ii)] $H_{j-j_t+1}(j)_{j-j_t+1+r_{j_t}}\iso H_0(j_t-1)_{r_{j_t}}$;
\item[(iii)] $H_{j-j_t+1}(j)_{j-j_t+1+s}$ is
isomorphic to a submodule of $H_0(j_t-1)_s$ for all $s> r_{j_{t-1}}$;
\item[(iv)] $H_0(j-i)_{r_{j_t}}$ is isomorphic to a factor
module of $H_i(j)_{i+r_{j_t}}$ for all $i$ with \newline
$i> j-j_{t+1}+1$.
\end{enumerate}
\end{enumerate}
\end{Theorem}
Of particular interest is the case $j=n$.
Following \cite{BCP}, a Betti number $\beta_{kk+m}\neq 0$ is called
{\em extremal}, if $\beta_{ii+j}=0$ for all $i> k$ and all
$j> m$. Statement (a) in the following corollary is
implicitly contained in \cite{BCP}.
\begin{Corollary}
\label{beta}
Let the numbers $j_t$ be defined as in {\em \ref{similar}}, and
set $k_t=n-j_t+1$ and $m_t=r_{j_t}$. Then
\begin{enumerate}
\item[(a)] the Betti number $\beta_{ii+j}$ of $M$ is extremal
if and only if
\[
(i,j)\in\{ (k_t,m_t)\: t=1,\ldots,l\}.
\]
Moreover, $\beta_{k_t,k_t+m_t}=\dim_K(0:l_{j_t})_{s_{j_t}}$ for
$t=1,\ldots,l$.
\item[(b)] for all $t=1,\ldots,l$ we have
\begin{enumerate}
\item[(1)]
$\beta_{k_t,k_t+s}\leq \dim_K(0:l_{j_t})_s$ for all $s> m_{t-1}$,
\item[(2)] $\beta_{i,i+m_t}\geq \dim_K(0:l_{n-i+1})_{m_t}$ for all
$i> k_{t+1}$.
\end{enumerate}
\end{enumerate}
\end{Corollary}
\noindent
The following picture displays the Betti diagram of a graded
free resolution in the form of a MACAULAY output. The entry with coordinates
$(i,j)$ is the Betti number $\beta_{ii+j}$. In our picture the
outside corners of the dashed line
give the positions of the extremal Betti numbers. The
preceeding corollary yields some informations on the border Betti
numbers which are those sitting on the dashed line.
$$
\beginpicture
\setcoordinatesystem units <.8mm,.8mm>
\plot 0 0 100 0 /
\plot 10 5 10 -50 /
\put{$$} at 150 0
\put{$\cdot$} at 10 5
\put{$\cdot$} at 10 -50
\put{$j$} at 5 -20
\put{$i$} at 48 4
\put{$\beta_{ii+j}$} at 50 -20
\put{$(k_3,m_3)$} at 42 -45
\put{$(k_2,m_2)$} at 72 -35
\put{$(k_1,m_1)$} at 90 -26
\put{$\cdot$} at 40 -40
\put{$\cdot$} at 70 -30
\put{$\cdot$} at 80 -25
\setdashes <1mm>
\plot 10 -40 40 -40 40 -30 70 -30 70 -25 80 -25 80 0 /
%\vshade 10 -40 0 40 -40 0 40 -30 0 70 -30 0 70 -25 0 80 -25 0
\setdots
\plot 10 -20 44 -20 /
\plot 55 -20 80 -20 /
\setdots
\plot 48 0 48 -17 /
\plot 48 -23 48 -30 /
\endpicture\\
\\
$$
The proof of \ref{similar} uses the following fact:
For each $j$ there is (cf.\ for example \cite{BH}) the following
exact sequence of graded $S$-modules
\begin{eqnarray}
\label{long1}
@>>> H_2(j)@>>> H_1(j-1)(-1)@>>> H_1(j-1)@>>>H_1(j)@>>>
H_0(j-1)(-1)@>>> 0\\
\cdots @>>> H_{i+1}(j)@>>> H_{i}(j-1)(-1)@>>> H_{i}(j-1)@>>>H_{i}(j)@>>>
H_{i-1}(j-1)(-1)@>>> \cdots\nonumber
\end{eqnarray}
\begin{pf}[Proof of {\em \ref{similar}}]
(a) We show by induction of $j$ that $r_j=\max\{r_{j-1},s_j\}$.
For $j=1$ we have $H_i(1)=0$ for $i>1$ and $H_1(1)\iso H_0(0)(-1)$.
Therefore $r_1=s_1$, as asserted.
Now let $j>1$ and assume that $s> \max\{r_{j-1},s_j\}$. By (\ref{long1})
we have the exact sequence
\begin{eqnarray}
\label{long2}
H_i(j-1)_{i+s}@>>> H_i(j)_{i+s}@>>> H_{i-1}(j-1)_{i-1+s}
\end{eqnarray}
for all $i>0$. Since by assumption the left and right homology
groups vanish, it follows that $H_i(j)_{i+s}=0$ for all $i>0$, too.
This shows that $r_j\leq \max\{r_{j-1},s_j\}$.
Suppose that $r_j<\max\{r_{j-1},s_j\}$. In case $r_j0$ be an integer such that $H_i(j-1){i+r_{j-1}}\neq 0$. By (1)
we obtain
the exact sequence
\[
H_{i+1}(j)_{i+1+r_{j-1}}@>>>
H_i(j-1)_{i+r_{j-1}}@>>> H_i(j-1)_{i+1+r_{j-1}}.
\]
This is a contradiction, since
$H_{i+1}(j)_{i+1+r_{j-1}}=0=H_i(j-1)_{i+1+r_{j-1}}$.
Finally in case $r_j>> H_0(j-1)_{s_j}@>>> 0
\]
yields a contradiction, since $H_1(j)_{1+s_j}=0$ while
$H_0(j-1)_{s_j}\neq 0$.
(b) We prove the assertions by induction on $j$. We begin with
statement (i). In case $j=j_t$, we have
$j-1r_{j_{t-1}}$ and $i\geq 1$. Then
sequence (\ref{long2}) implies that $H_i(j)_{i+s}=0$ for $s>r_{j_{t-1}}$
and $i> j-j_t+1=1$, as desired.
In case $j>j_t$, we have $j-1\geq j_t$, and it
follows from the induction hypothesis that $H_i(j-1)_{i+s}=0$
for all $s>r_{j_{t-1}}$ and $i-1>(j-1)-j_t+1$. Hence the exact
sequence (\ref{long2}) yields the assertion.
In order to prove (ii) and (iii) we note that we have
the exact sequence
\[
H_1(j_t-1)_{1+s}@>>> H_1(j_t)_{1+s}@>>>
H_0(j_t-1)_{s}@>>> 0.
\]
By definition, $H_1(j_t-1)_{1+s}=0$ for $s>r_{j_{t-1}}$,
(ii) and (iii) follow for $j=j_t$.
Suppose now that $j>j_t$, and consider the exact sequence
\begin{eqnarray}
\label{long3}
&&H_{j-j_t+1}(j-1)_{j-j_t+1+s} @>>>
H_{j-j_t+1}(j)_{j-j_t+1+s}@>>>
H_{j-j_t}(j-1)_{j-j_t+s}\\
&&@>>> H_{j-j_t}(j-1)_{j-j_t+1+s}\nonumber
\end{eqnarray}
By (i) it follows that $H_{j-j_t+1}(j-1)_{j-j_t+1+s}=0$
for all $s>r_{j_{t-1}}$
This yields $H_{j-j_t+1}(j)_{j-j_t+1+r_{j_t}}\subset
H_{j-j_t}(j-1)_{j-j_t+r_{j_t}}$.
If $s=r_{j_t}$, then also $H_{j-j_t}(j-1)_{j-j_t+1+s}=0$. This
follows if $j-10$. Therefore (\ref{long3})
implies that $H_{i-1}(j-1)_{i-1+r_{j_t}}$ is a factor module of
$H_i(j)_{i+r_{j_t}}$. Induction on $j$ concludes the proof.
\end{pf}
\section{Almost regular sequences and Gin's}
In this section we give an alternative proof of the theorem
of Bayer, Charalambous and S.\ Popescu (see \cite[Theorem
1.6]{BCP}) which, in the special case of an ideal $I$,
asserts that
the extremal Betti numbers of
$I$ and $\Gin(I)$ are the same. We include this proof, since a
modification of it (see Section 5) will
provide a similar statement for modules over the
exterior algebra.
We consider a field extension $L/K$ containing elements
$a_{ij}$, $i,j=1,\ldots,n$, which are algebraically independent
over $K$. Let, as before, $M$ be a finitely
generated graded $S=K[x_1,\ldots,x_n]$-module, and set
$S'=L[x_1,\ldots,x_n]$ and $M'=S'\tensor_SM=L\tensor_KM$. In $S'$ we
consider the `generic' linear forms $l_j=\sum_{i=1}^na_{ij}x_i$,
$j=1,\ldots,n$.
\begin{Proposition}
\label{indeterm}
The sequence $l_1,\ldots,l_n$ is an unconditioned almost regular
$M'$-sequence.
\end{Proposition}
The proposition will follow easily from:
\begin{Lemma}
\label{almost}
Let $a_1,\ldots, a_n\in L$ be algebraically independent elements over
$K$. Then the linear form $l=\sum_{i=1}^na_ix_i$ is almost
$M'$-regular.
\end{Lemma}
\begin{pf}
The $K$-subalgebra $R=K[a_1,\ldots,a_n]$ of $L$ is isomorphic to
the polynomial ring since the $a_i$ are algebraically
independent over $K$. Set $T=R[x_1,\ldots,x_n]$ and
$C=T\tensor_SM=R\tensor_QM$ with natural grading
$C_i=R\tensor_QM_i$ for all $i$.
We claim that the multiplication map $l\: C_i\to C_{i+1}$ is
injective for all $i\gg 0$. Assume we have already
shown this claim, and let $Q=K(a_1,\ldots,a_n)$ be the
quotient field of $R$. Then $Q\tensor_RC$ is a localization of
$C$, and so the claim implies that $l$ is almost regular for the
module $Q\tensor_RC$, which is isomorphic to $Q\tensor_KM$. By
flatness it follows that $l$ is almost $M'$-regular, too.
In order to prove the claim we let $N=M/H^0_\mm(M)$ and
$D=T\tensor_SN=R\tensor_KN$. Then $\depth(N)>0$ and $N_i=M_i$ for $i\gg 0$.
It follows that $D_i=C_i$ for $i\gg 0$, and it remains to
show that $l$ is $D$-regular. Since the extension
$T/S$ is flat, one has
\[
\Ass_T(D)=\{PT\: P\in \Ass_S(N)\},
\]
see for example \cite[Theorem 23.2]{M}.
Suppose now that $l$ is a zerodivisor on $D$. Then $l\in
(PT)_1$ for some $P\in \Ass_S(N)$,
and so, if the linear forms $l_1,\ldots,l_r\in P_1$ form
a $K$-basis of $P_1$, we get that $l\in Rl_1+Rl_2+\cdots +Rl_r$.
In other words, there exist polynomials $f_i=
f_i(a_1,\ldots,a_n)\in R$ such that
\begin{eqnarray}
\label{compare}
l=\sum_{i=1}^rf_il_i.
\end{eqnarray}
Assigning to the $a_i$ the bidegree $(1,0)$ and to the $x_i$ the
bidegree $(0,1)$, we see that $l=\sum_{i=1}^na_ix_i$ is
bihomogeneous of degree $(1,1)$, and that the $l_i$ are
bihomogeneous of degree $(0,1)$. Therefore we may assume that
the $f_i$ are bihomogeneous of degree $(1,0)$. In other words,
we may assume the $f_i$ are linear forms in $R$.
Let $l_i=\sum_{j=1}^n\alpha_{ji}x_j$ for $i=1,\ldots, r$ with
$\alpha_{ij}\in K$.
Comparing the coefficients of $x_j$ in (\ref{compare}) we then get that
$a_j=\sum_{i=1}^r\alpha_{ji}f_i$ for $j=1,\ldots, n$, and it follows that
$R_1=\sum_{i=1}^rKf_i$. This is a contradiction, since
$r0$.
\end{pf}
\begin{pf}[Proof of {\em \ref{indeterm}}]
We adopt the notation in the proof of \ref{almost}.
The form $l_1$ is almost $M'$-regular, by \ref{almost}. In
order to show that $l_2$ is almost $M'/l_1M'$-regular, we let
$M''=Q\tensor_KM$. Then $M'=L\tensor_QM''$, and hence
$M'/l_1M'=L\tensor_Q(M''/ l_1M'')$. Now we apply \ref{almost} to
the $Q[x_1,\ldots,x_n]$-module $M''/ l_1M''$, the field
extension $L/Q$ and the linear form $l_2$.
Induction on the length of the sequence completes
the proof.
\end{pf}
Let $F$ be a graded free $S$-module with homogeneous basis
${\bold g}=g_1,\ldots, g_m$. The elements of the form $x^ag_i$
are called {\em monomials of $F$}. Here
$x_a=x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ for $a=(a_1,\ldots,a_n)$.
We order the monomials of $F$ in the {\em reverse lexicographic}
order, that is, if $m_1=x^ag_{i}$ and
$m_2=x^bg_j$, then
\[
m_1>m_2\quad\text{if}\quad
\begin{cases}
\deg(x^ag_{i})>\deg(x^bg_j),\\
\text{or}\\
\deg(x^ag_{i})=\deg(x^bg_j),\quad \text{and of one the following
holds:}\\
\text{(i) there exists an $r$ such that}\\
\quad\quad a_rr,\quad \text{or} \\
\text{(ii)}\quad x^a=x^b\quad\text{and}\quad i>> M_{i-1}@>l_v >> M_i@>l_v >> M_{i+1}@>>>\cdots
\]
where $l_v$ denotes left multiplication by $v$.
We denote the $i$th cohomology of this complex by $H^i(M,v)$.
Notice that $H^\pnt(M,v)=\Dirsum_iH^i(M,v)$ is again an object in
${\cal M}_l$. Indeed,
\[
H^\pnt(M,v)=\frac{0:_Mv}{vM},
\]
where $0:_Mv=\{a\in M\: va=0\}$.
It is clear that a short exact
sequence
\[
0@>>> U@>>> M@>>> N@>>> 0,
\]
of finitely generated graded $E$-modules induces the long exact
cohomology sequence
\[
\cdots @>>> H^i(U,v)@>>> H^i(M,v)@>>>
H^i(N,v)@>>>H^{i+1}(U,v)@>>>\cdots
\]
\begin{Definition}
\label{generic}
{\em Let $M\in{\cal M}_l$. An element $v\in E_1$ is called
{\em general for $M$} if $\dim_KH^i(M,v)\leq
\dim_KH^i(M, u)$ for all $i$ and
all $u\in E_1$.}
\end{Definition}
The property of being general for $M$ is an open
condition, that is, there exists a non-empty Zariski open subset
$G\subset E_1$, such that $v\in E_1$ is general if and only if
$v\in G$.
\begin{Definition}
\label{simpl}
{\em Let $v$ be general for $M$. Then $H^\pnt(M,v)$ is called the
{\em simplicial cohomology of $M$.}}
\end{Definition}
Suppose $v$ is general for $M$. Since $H^\pnt(M,v)$,
as a graded vector space,
does not depend on the particular choice of $v$, we simply write
$H^\pnt(M)$.
A particular case is of special interest: we fix a basis
$e_1,\ldots,e_n$ of $V$. The standard basis elements
$e_{k_1}\wedge\cdots\wedge e_{k_s}$ of $E$,
$k_1<\dots l_v >> M$ yields $M^*@>r_v >> N^*$, where $r_v$ denotes
right multiplication.
The graded components of $M^*$ can be easily determined. For a
$K$-vector space $W$ we set $W^\vee=\Hom_K(W,K)$.
\begin{Lemma}
\label{dual}
$(M^*)_i\iso (M_{n-i})^\vee$ for all $i$.
\end{Lemma}
\begin{pf}
Let $\phi\in M^*$ be homogeneous of degree $i$. Then
$\phi(M_j)\subset E_{j+i}$ for all $j$. Note that
$\phi|_{M_{n-i}}$ is a $K$-linear map from $M_{n-i}\to E_n$.
We fix an isomorphism $\eta\: K\to E_n$. Then $\phi|_{M_{n-i}}$ may
be identified with an element in $(M_{n-i})^\vee$. We define
the $K$-linear map $\alpha\: (M^*)_i\to (M_{n-i})^\vee$ by setting
$\alpha(\phi)=\phi|_{M_{n-i}}$.
Suppose that $\alpha(\phi)=0$, and let $a\in M_j$. Then $\phi(a)\in E_{j+i}$.
Suppose that $\phi(a)\neq 0$. Then there exists an element $b\in E_{n-j-i}$
such that $\phi(ba)=b\phi(a)\neq 0$. This is a contradiction, since $ba\in
M_{n-i}$, and since $\phi|_{M_{n-i}}= 0$, by assumption. This
shows that $\alpha$ is injective. In order to see that $\alpha$
is surjective, let $\psi\in (M_{n-i})^\vee$ given, and consider
the graded submodule $N= \Dirsum_{j\geq n-i}M_j$ of $M$. Then
$\psi$ induces a $E$-module homomorphism $\tilde{\phi}$ with
$\tilde{\phi}|_{N_{n-i}}= \eta\circ\psi$ and
$\tilde{\phi}|_{N_{j}}=0$ for $j>n-i$. Since $E$ is injective
$\tilde{\phi}\: N\to E$ can be extended to $\phi\: M\to E$, and
it is clear that $\alpha(\phi)=\psi$.
\end{pf}
Now given $v\in E_1$, then for all $i$ there is a unique $K$-linear map
$\partial_v: (M_{n-i})^\vee\to (M_{n-i-1})^\vee$ making the following diagram
\[
\begin{CD}
(M^*)_i@>r_v >> (M^*)_{i+1}\\
@V \alpha VV @VV \alpha V\\
(M_{n-i})^\vee@>>\partial_v > (M_{n-i-1})^\vee
\end{CD}
\]
commutative. It follows that
\[
(M^\vee,\partial_v)\: \cdots@>>> (M_i)^\vee@>\partial_v >>
(M_{i-1})^\vee@>\partial_v >> (M_{i-2})^\vee@>>>\cdots
\]
is a complex
whose $i$th homology we denote by $H_i(M, v)$. In case $v$ is
general for $M$ we simply write $H_i(M)$ for $H_i(M, v)$, and
call $H_i(M)$ the {\em $i$th simplicial homology of $M$}. The
reader checks easily that for any simplicial complex $\Delta$
one has $H_i(K\{\Delta\})\iso \tilde{H}_{i-1}(\Delta;K)$ for all
$i$.
It follows from our definition of simplicial homology for $M$,
that
\[
H_i(M)\iso H^{n-i}(M^*,r_v)\quad\mbox{for all}\quad i,
\]
where $v$ is general for $M$.
We let ${\cal M}$ be the category of graded left and right
$E$-modules, satisfying $ax=(-1)^{\deg a\deg x}xa$ for all
homogeneous elements $a\in E$ and $x\in M$. For example, any
graded ideal $I\subset E$ belongs to ${\cal M}$.
Observe that for $M\in {\cal M}$
the complexes $(M^*,l_v)$ and $(M^*,r_v)$ are
isomorphic. The complex isomorphism $\sigma\: (M^*,l_v)\to
(M^*,r_v)$ is given by $\sigma_i=(-1)^i\id_{(M^*)_i}$ for all $i$.
Thus we have the following result:
\begin{Proposition}
\label{duality}
For all $M\in {\cal M}$ and all $i$ one has
\[
H_i(M)\iso H^{n-i}(M^*).
\]
\end{Proposition}
\section{Generic bases and Cartan (co)homology}
In this section we study Cartan homology. This homology corresponds to
Koszul homology for symmetric algebras. We recall a few facts
from \cite{AHH}.
Let ${\bold v}=v_1,\cdots,v_m$ be a sequence of elements of degree
$1$ in $E$. The Cartan complex $C_\lpnt({\bold v};E)$ of the sequence ${\bold v}$ with values
in $E$ is defined as
the complex whose $i$-chains $C_i({\bold v};E)$
are the elements of degree $i$ of the free divided power
algebra $C_\lpnt({\bold v};E)= E\langle x_1,\ldots,x_m\rangle$.
Recall that $C_\lpnt({\bold v};E)$ is the
polynomial ring over $E$ in the set of variables
\[
x_i^{(j)},\quad i=1,\ldots,m,\quad j=1,2,\ldots
\]
modulo the relations
\[
x_i^{(j)}x_i^{(k)}=\frac{(j+k)!}{j!k!}x_i^{(j+k)}.
\]
We set $x_i^{(0)}=1$ and $x_i^{(1)}=x_i$ for $i=1,\ldots, m$.
The algebra $C_\lpnt({\bold v};E)$ is a free $E$-module with basis $x^{(a)}=
x_1^{(a_1)}x_2^{(a_2)}\ldots x_m^{(a_m)}$, $a\in\NN^m$. We say
that $x^{(a)}$ has degree $i$ if $|a|=i$ where $|a|=a_1+\ldots +a_m$.
Thus $C_i({\bold v};E)=\Dirsum_{|a|=i}Ex^{(a)}$.
The $E$-linear differential $\partial$ on $C_\lpnt({\bold v};E)$
is defined a follows: for
$x^{(a)}=x_1^{(a_1)}\cdots x_m^{(a_m)}$ we set
\[
\partial(x^{(a)})=\sum_{a_i>0}v_ix_1^{(a_1)}\cdots x_i^{(a_i-1)}\cdots
x_m^{(a_m)}.
\]
It is easily verified that $\partial\circ \partial =0$, so that
$(C_\lpnt({\bold v};E),\partial)$ is indeed
a complex. Moreover,
$\partial$ is an $E$-derivation, that is, $\partial$ is
$E$-linear and
\[
\partial(g_1g_2)=g_1\partial(g_2)+\partial(g_1)g_2
\]
for any two homogeneous elements $g_1$ and $g_2$ in $C_\lpnt({\bold v};E)$.
These rules imply that the cycles $Z_\lpnt({\bold v};E)$ of
$C_\lpnt({\bold v};E)$ form
a divided power algebra, and that the boundaries $B_\lpnt({\bold
v};E)$ form an ideal
in $Z_\lpnt({\bold v};E)$, so that the homology $H_\lpnt({\bold v};E)$ of
$C_\lpnt({\bold v};E)$ inherits a natural
structure of a divided power algebra. Let $M$ be left $E$-module;
then $C_\lpnt({\bold v};M)=C_\lpnt({\bold v};E)\tensor_EM$ is called the
{\em Cartan complex
of $M$ with respect to the sequence ${\bold v}$}. The homology of
$C_\lpnt({\bold v};M)$ will be denoted by $H_\lpnt({\bold v};M)$. Note, that
$H_\lpnt({\bold v};M)$ has a natural left $H_\lpnt({\bold v};E)$-module
structure.
For each $j=1,\ldots,m-1$ there exists an exact sequence of
complexes
\[
0@>>> C_\lpnt(v_1,\ldots,v_j;M)@>\iota >>
C_\lpnt(v_1,\ldots,v_{j+1};M)@>\tau >>
C_\lpnt(v_1,\ldots,v_{j+1};M)(-1)@>>> 0,
\]
where $\iota$ is a natural inclusion map, and where $\tau$ is
given by
\[
\tau(g_0+g_1x_{j+1}+\cdots + g_kx_{j+1}^{(k)})=
g_1+g_2x_{j+1}+\cdots + g_kx_{j+1}^{(k-1)},
\qquad g_i\in C_{k-i}(v_1,\ldots,v_j;M).
\]
From this exact sequence one obtains immediately the following long exact
sequences for the Cartan homology.
\begin{Proposition}
\label{long}
Let $M\in{\cal M}_l$; then for all $j=1,\ldots,m-1$ there
exists a long exact sequence of graded left $E$-modules
\begin{eqnarray*}
{}\cdots@>>>&& H_i(v_1,\ldots,v_j;M)@>\alpha_i >> H_i(v_1,\ldots,v_{j+1};M)
@>\beta_i >>H_{i-1}(v_1,\ldots,v_{j+1};M)(-1)\\
@>\delta_{i-1} >>&&
H_{i-1}(v_1,\ldots,v_j;M)@>>>H_{i-1}(v_1,\ldots,v_{j+1};M)@>>> \cdots.
\end{eqnarray*}
Here $\alpha_i$ is induced by the inclusion map $\iota$, $\beta_i$
by $\tau$, and $\delta_{i-1}$ is the connecting homomorphism,
which acts as follows: if
$z=g_0+g_1x_{j+1}+\cdots +g_{i-1}x_{j+1}^{(i-1)}$ is a cycle in
$C_{i-1}(l_1,\ldots,l_{j+1};M)$, then
$\delta_{i-1}([z])=[g_0v_{j+1}]$.
\end{Proposition}
For any finitely generated left $E$-module $M$, the
Cartan cohomology with respect to the sequence ${\bold
v}=v_1,\ldots,v_m$ is defined to be the homology of the cocomplex
$C^\pnt({\bold v};M)=\Hom_E(C_\lpnt({\bold v};E), M)$. Explicitly, we
have
\[
C^\pnt({\bold v};M): 0@>\partial^0 >> C^0(M) @>\partial^1
>> C^1(M) @>>>\ldots,
\]
where the cochains $C^\pnt({\bold v};M)$ and cochain maps
$\partial^\pnt$ can be described as follows: the elements of
$C^i({\bold v};M)$ may be identified with all homogeneous polynomials
$\sum_am_ay^a$ of degree $i$ in the variables $y_1,\ldots,y_m$ with
coefficients $m_a\in M$, and where as usual for $a\in \NN^n$,
$y^a$ denotes the monomial $y_1^{a_1}y_2^{a_2}\dots y_n^{a_n}$.
The element $m_ay^a\in C^\pnt({\bold v};M)$ is defined by the
mapping property
\[
m_ay^a(x^{(b)})=
\begin{cases} m_a&\mbox{for}\quad b=a,\\
0&\mbox{for}\quad b\neq a.
\end{cases}
\]
After this identification the cochain maps are simply
multiplication by the element $y_{\bold v}=\sum_{i=1}^nv_iy_i$.
In other words, we have
\[
\partial^i\: C^i({\bold v};M)@>>> C^{i+1}({\bold v};M), \quad
f\mapsto y_{\bold v}f.
\]
In particular we see that $C^\pnt({\bold v};E)$ may be identified with the
polynomial ring $E[y_1,\ldots,y_m]$, and that $C^\pnt({\bold v};M)$ is a
finitely generated $C^\pnt({\bold v};E)$-module. It is obvious that
cocycles and coboundaries of $C^\pnt({\bold v};M)$ are
$E[y_1,\ldots,y_m]$-submodules of $C^\pnt({\bold v};M)$.
As $E[y_1,\ldots,y_m]$ is noetherian, it follows that the Cartan cohomology
$H^\pnt({\bold v};M)$ of $M$ is a finitely generated (graded)
$E[y_1,\ldots,y_m]$-module.
Cartan homology and cohomology are related as follows:
\begin{Proposition}
\label{related}
Let $M\in{\cal M}$. Then
\[
H_i({\bold v};M)^*\iso H^i({\bold v};M^*)\quad \mbox{for all
}\quad i.
\]
\end{Proposition}
\begin{pf}
Since $E$ is injective, the functor
$(-)^*$ commutes with homology and we obtain
\begin{eqnarray*}
H_i({\bold v};M)^*&\iso& H^i(\Hom_E(C_i({\bold v};M), E))\iso\\
H^i(\Hom_E(C_i({\bold v};E), M^*)&\iso&H^i({\bold v};M^*).
\end{eqnarray*}
\end{pf}
\begin{Proposition}
\label{dlong}
Let $M\in{\cal M}_l$. Then for all $j=1,\ldots,m-1$ there
exists a long exact sequence of graded left $E$-modules
\begin{eqnarray*}
{}\cdots@>>> H^{i-1}(v_1,\ldots,v_{j+1};M)@>>> H^{i-1}(v_1,\ldots,v_j;M)
@>>>H^{i-1}(v_1,\ldots,v_{j+1};M)(-1)\\
@>y_{j+1}>> H^i(v_1,\ldots,v_{j+1};M)@>>>H^i(v_1,\ldots,v_j;M)@>>> \cdots
\end{eqnarray*}
\end{Proposition}
\begin{pf} It is immediate that such a sequence exists. We only
show that the map
\[
H^{i-1}(v_1,\ldots,v_{j+1};M)(-1)\to
H^i(v_1,\ldots,v_{j+1};M)
\]
is indeed multiplication by
$y_{j+1}$. We show this on the level of cochains. In order to
simplify notation we set
$C_i=C_i(v_1,\ldots,v_{j+1};E)$ for all $i$, and let
\[
\gamma\:\Hom_E(C_{i-1},M)\to \Hom_E(C_{i},M)
\]
be the map
induced by $\tau\: C_i\to C_{i-1}$, where
\[
\tau(x^{(b)})=
\begin{cases} x_1^{(b_1)}\cdots
x_{j+1}^{(b_{j+1}-1)}&\mbox{if}\quad b_{j+1}>0,\\
0&\mbox{otherwise}.
\end{cases}
\]
Our assertion is that
$\gamma$ is multiplication by $y_{j+1}$.
For all $x^{(b)}\in C_i$ we have
$\gamma(my^a)(x^{(b)})=my^a(\tau(x^{(b)}))$.
This implies that
\[
\gamma(my^a)(x^{(b)})=\begin{cases} m&\mbox{if}\quad
(b_1,\ldots,b_{j+1})=(a_1,\ldots,a_{j+1}+1),\\
0&\mbox{otherwise}.
\end{cases}
\]
Hence we see that $\gamma(my^a)=my^ay_{j+1}$, as desired.
\end{pf}
\begin{Lemma}
\label{trans}
Let $v_1,\ldots,v_n$ and $w_1,\ldots,w_n$ be two bases of $E_1$,
and suppose that $w_j=\sum_{i=1}^na_{ij}v_i$ for $j=1,\ldots,n$.
Then there is an isomorphism of graded $E$-modules
$$
\phi\: H^\pnt(v_1,\ldots,v_{n};M)\to H^\pnt(w_1,\ldots, w_{n};M)
$$
and a linear $E$-automorphism $\alpha: E[y_1,\ldots,y_{n}]\to
E[y_1,\ldots,y_{n}]$ with $\alpha(y_j)=\sum_{i=1}^na_{ji}y_i$
for $j=1,\ldots,n$, such that
\[
\phi(fc)=\alpha(f)\phi(c)
\]
for all $f\in E[y_1,\ldots,y_n]$ and all $c\in
H^\pnt(v_1,\ldots,v_{n};M)$
\end{Lemma}
\begin{pf}
The linear $E$-automorphism $\alpha$ satisfies
$\alpha(y_{\bold v})=y_{\bold w}$, and so $\alpha$ induces a
complex isomorphism
\[
C^\pnt(v_1,\ldots,v_n;M)\To C^\pnt(w_1,\ldots,w_n;M), \quad
g(y_1,\ldots,y_m)\mapsto g(\alpha(y_1),\ldots,\alpha(y_n)),
\]
which induces the graded isomorphism
$\phi\: H^\pnt(v_1,\ldots,v_n;M)\to
H^\pnt(w_1,\ldots,w_n;M)$ with the desired properties.
\end{pf}
Now let $L/K$ be a field extension containing the algebraically
independent elements $a_{ij}$ over $K$, $i,j=1,\ldots,n$. We set
$E'=L\tensor_KE$ and for any $M\in {\cal M}$ we set $M'=L\tensor_KM$.
\begin{Proposition}
\label{genbas}
Let ${\bold v}=v_1,\ldots,v_n$ be a basis of $E_1$ and consider the
`generic basis' ${\bold w}=w_1,\ldots,w_n$ of $E'_1$ where
$w_j=\sum_{i=1}^na_{ij}v_i$ for $j=1,\ldots,n$. Then the
sequence $y_1,\ldots,y_n$ is an unconditioned almost regular for the
$L[y_1,\ldots,y_n]$-module $H^\pnt(w_1,\ldots,w_n;M')$.
\end{Proposition}
\begin{pf}
Let $\phi\: H^\pnt(v_1,\ldots,v_n;M')\to
H^\pnt(w_1,\ldots,w_n;M')$ and $\alpha\: E'[y_1,\ldots,y_n]\to
E'[y_1,\ldots,y_n]$ be
defined as in \ref{trans} for the base change
$w_j=\sum_{i=1}^na_{ij}v_i$ ($j=1,\ldots,n$), and let $l_j\in
E[y_1,\ldots,y_n]$ be the unique linear form with
$\alpha(l_j)=y_j$ for $j=1,\ldots,n$. Then it follows from
\ref{trans} that $y_1,\ldots,y_n$ is an unconditioned almost
regular sequence for $H^\pnt(w_1,\ldots,w_n;M')$ if and only if
$l_1,\ldots,l_n$ has this property for
$H^\pnt(v_1,\ldots,v_n;M')$. Let $(b_{ij})$ the inverse matrix
of the matrix $(a_{ij})$. Then $l_j=\sum_{i=1}^nb_{ji}y_i$ for
$j=1,\ldots,n$. Since the entries of the matrix $(a_{ij})$ are
algebraically independent over $K$, the same is true for the
entries of the inverse matrix $(b_{ij})$. Therefore, by
\ref{indeterm}, $l_1,\ldots, l_n$ is an unconditioned
almost regular
$H^\pnt(v_1,\ldots,v_n;M')$-sequence, and the proposition follows.
\end{pf}
Proposition \ref{genbas} has a remarkable consequence.
\begin{Corollary}
\label{dualgen}
Let ${\bold w}=w_1,\ldots,w_n$ be a
generic basis of $E'_1$ as in {\em (\ref{genbas})}. Then
the natural maps
\[
H_i(w_1,\ldots,w_{j+1};M')
@>\beta_i >>H_{i-1}(w_1,\ldots,w_{j+1};M')(-1)
\]
are surjective for all $j=0,\ldots,n-1$ and all $i\gg 0$.
\end{Corollary}
\begin{pf}
The assertion will follow from the claim that the multiplication
map
\[
y_{j+1}\: H^{i-1}(w_1,\ldots,w_{j+1};(M')^*)(-1)@>>>
H^{i}(w_1,\ldots,w_{j+1};(M')^*)
\]
is injective for all all $j=0,\ldots,n-1$ and all $i\gg 0$.
Indeed,
the exact functor $(-)^*$ turns $H_i(w_1,\ldots,w_{j+1};M')$ into
$H^i(w_1,\ldots,w_{j+1};(M')^*)$ (by \ref{related}) and $\beta_i$
into multiplication by $y_{j+1}$.
In order to prove the claim first notice that $(M')^*=(M^*)'$.
Thus \ref{genbas} applied to $M^*$ yields that
$y_n,\ldots,y_1$ is an almost regular $H^\pnt({\bold
w};(M')^*)$-sequence. Therefore the multipication map
\begin{eqnarray}
\label{mult}
H^{i-1}({\bold w};(M')^*)/
(y_{n-j+1},\ldots,y_n)H^{i-2}({\bold w};(M')^*)\\
@> y_{n-j} >> H^{i}({\bold w};(M')^*)/
(y_{n-j+1},\ldots,y_n)H^{i-1}({\bold w};(M')^*) \nonumber
\end{eqnarray}
is injective for all $j=0,\ldots, n-1$ and all $i\gg 0$.
In particular it follows from \ref{dlong} that
\[
0@>>> H^{i-1}({\bold w};(M')^*)@>y_{n} >>
H^i({\bold w};(M')^*)@>>>
H^i(w_1,\ldots,w_{n-1};(M')^*)@>>>0
\]
is exact for $i\gg 0$. Thus we have that the $i$th
component of $H^\pnt(w_1,\ldots,w_{n-1};(M')^*)$ and of
$H^\pnt({\bold w};(M')^*)/(y_n)H^\pnt({\bold w};(M')^*)$
coincide for $i\gg 0$.
Similarly it follows from \ref{dlong},
(\ref{mult}) and induction on $j$ that
the $i$th
component of $H^\pnt(w_1,\ldots,w_{n-j};(M')^*)$ and of
$H^\pnt({\bold w};(M')^*)/(y_{n-j+1},\ldots, y_n)
H^\pnt({\bold w};(M')^*)$
coincide for $i\gg 0$. This together with (\ref{mult})
completes the proof of the claim.
\end{pf}
\begin{Definition}
\label{genericbasis}
{\em We call a basis $v_1,\ldots,v_n$ {\em generic for $M$} if
the natural maps
\[
H_i(v_1,\ldots,v_{j+1};M)
@>\beta_i >>H_{i-1}(v_1,\ldots,v_{j+1};M)(-1)
\]
are surjective for all $j=0,\ldots,n-1$ and all $i\gg 0$.}
\end{Definition}
Corollary \ref{dualgen} shows that after suitable extension of
the base field there exists always a generic basis.
We now fix $M\in {\cal M}$ and a sequence ${\bold
v}=v_1,\ldots,v_n$ in $E_1$.
Similarly as in Section 1 we set $M{\langle j-1\rangle}=M/(v_1,\ldots,
v_{j-1})M$ and put
$H_i(j)=H_i(v_1,\ldots,v_j;M)$ for $i>0$ and
$H_0(j)=H(M{\langle j-1\rangle},v_j)$ for $j=1,\ldots,n$.
Furthermore we set $H_i(0)=0$ for all $i$.
Notice that $H_0(j)$ is {\em not} the
$0$th Cartan homology of $M$ with respect to $v_1,\ldots, v_j$,
but is the cohomology of $M{\langle j-1\rangle}$ with respect to $v_j$ as
defined in Section 2. From \ref{long} we obtain immediately the
following long exact sequence of graded $E$-modules
\begin{eqnarray}
\label{long4}
H_2(j)@>>> H_1(j)(-1)@>>> H_1(j-1)@>>> H_1(j)@>>>
H_0(j)(-1)@>>> 0\\
\cdots @>>> H_i(j-1)@>>> H_i(j)@>>> H_{i-1}(j)(-1)@>>>
H_{i-1}(j-1)@>>>\cdots \nonumber
\end{eqnarray}
Finally we set
\[
r_j=\max\{s(H_i(j))-i\: i\geq 1\}\quad \mbox{and}\quad
s_j=s(H_0(j))\quad \mbox{for}\quad j=1,\ldots,n,
\]
and we put $r_0=0$.
The {\em regularity of $M$} is defined by $\reg(M)=
\sup\{j-i\: \Tor_i^E(K,M)_j\}$. In \cite{AHH} it is noted that
$\reg (M)$ is a finite integer. If ${\bold v}$ is a basis of
$V$, then in terms of the above notation we
have $\reg(M)=\max\{r_n, s(M/\mm M)\}$ where $\mm$ is the graded
maximal ideal of $E$.
\begin{Theorem}
\label{exterior}
Let $M\in{\cal M}$. Then
\begin{enumerate}
\item[(a)] $r_j\leq \max\{s_1,\ldots,s_j\}$ for $j=1,\ldots,n$.
\item[(b)] If ${\bold v}$ is a generic basis for $M$, then equality
holds in {\em (a)} for all $j=1,\ldots,n$. Moreover, one has
\[
r_j=s(H_i(j))-i\quad\mbox{for all} \quad i\gg 0.
\]
\end{enumerate}
\end{Theorem}
\begin{pf} (a) We show by induction on $j$ that $r_j\leq \max\{
r_{j-1},s_j\}$. For $j=1$ we have $H_i(1)_{i+s_1}\iso
H_0(1)_{s_1}$ for all $i\geq 0$. Therefore $r_1=s_1$, as
desired.
Now let $j>1$, and suppose that $r_j>\max\{r_{j-1},s_j\}$. Then
$r_j>r_{j-1}$. Consider the exact sequence
\begin{eqnarray}
\label{long5}
\cdots @>>> H_i(j-1)_{i+s}@>>> H_i(j)_{i+s}@>>> H_{i-1}(j)_{i-1+s}@>>>
H_{i-1}(j-1)_{i+s}@>>>\cdots,
\end{eqnarray}
which is deduced from (\ref{long4}). Let $i\geq 1$ be an integer with
$H_i(j)_{i+r_j}\neq 0$. Setting $s=r_j$ and using our assumption
$r_j>r_{j-1}$, we conclude from (\ref{long5}) that
$H_i(j)_{i+r_j}\iso H_{i-1}(j)_{i-1+r_j}$ for all $i\geq 2$. In
particular we have $H_1(j)_{1+r_j}\iso H_i(j)_{i+r_j}\neq 0$.
From (\ref{long4}) we also get the exact sequence
\[
H_1(j-1)_{1+r_j}@>>> H_1(j)_{1+r_j}@>>> H_0(j)_{r_j}@>>> 0.
\]
By our assumption, $H_1(j-1)_{1+r_j}=0$, so that $H_0(j)_{r_j}\iso
H_1(j)_{1+r_j}\neq 0$. This implies that $s_j\geq r_j$, which is
a contradiction since $r_j>\max\{r_{j-1},s_j\}\geq s_j$.
(b) We proceed by induction on $j$. For $j=1$ we have already
seen in part (a) of the proof that the statement is correct. So
now let $j>1$, and
suppose $r_j<\max\{r_{j-1},s_j\}$. Since $H_1(j)_{1+s_j}\to
H_o(j)_{s_j}\to 0$ is exact, it follows that $r_j\geq s_j$.
Therefore our hypothesis implies that $r_j>> H_i(j-1)_{i+s}@>>> H_i(j)_{i+s}@>>> H_{i-1}(j)_{i-1+s}@>>>
0
\end{eqnarray}
is exact for all $i\gg 0$. This implies that
$H_i(j)_{i+r_{j-1}}\neq 0$, and so $r_j\geq r_{j-1}$, a
contradiction.
It remains to be shown that $H_i(j)_{1+r_j}\neq 0$ for $i\gg 0$.
If $r_j=r_{j-1}$, then this follows from (\ref{long6}). So now
suppose that $r_j> r_{j-1}$, then $s_j=r_j>r_{j-1}$, and the
same reasoning as in the proof of (a) shows that
$H_i(j)_{i+s_j}\iso H_0(j)_{s_j}$ for all $i\geq 0$. This
concludes the proof of the theorem.
\end{pf}
\section{Generic initial modules and regularity}
In this section we prove an analogue to the Bayer-Stillman
theorem \cite{BS} on the regularity of generic initial modules.
Let $F$ be a graded free $E$-module with homogeneous basis
${\bold g}=g_1,\ldots, g_m$. Given a basis
${\bold e}=\{e_1,\ldots,e_n\}$ of $V$, one calls the elements
of the form $e_{i_1}\wedge\ldots\wedge e_{i_{k}}g_{i}$ with $1\leq
i_1\beta_i >>H_{i-1}(e_n,\ldots,e_{n-j+1};W)(-1)
\]
are surjective for all $j=0,\ldots,n-1$ and all $i\gg 0$, where
either $W=M$ or $W=\bar{M}$.
Set $s_j=s(H^\pnt(M\langle j\rangle, e_{n-j}))$ and $\bar{s}_j=
s(H^\pnt(\bar{M}\langle j\rangle, e_{n-j}))$. By \ref{bar} we have
$s_j=\bar{s}_j$ for all $j$, and by (b) and \ref{exterior} we
have $r_n=\max\{s_1,\ldots,s_n\}$ and $\bar{r}_n
=\max\{\bar{s}_1,\ldots.\bar{s}_n\}$. The desired conclusion follows.
\end{pf}
\section{Simplicial complexes and Cartan (co)homology}
In this section we describe Cartan cohomology of a module $M$ as
generalized simplicial cohomology of a certain attached module
$M'$, and interpret Cartan cohomology of a face ring
$K\{\Delta\}$ in terms of simplicial cohomology of links.
Let $M\in {\cal M}_l$, and let $S=K[y_1,\ldots,y_n]$ be the
polynomial ring. Set
$V'=V\tensor_KS$, $E'=E\tensor_KS$ and $M'=M\tensor_KS$. Then
$E'$ may be viewed as the exterior algebra of the free
$S$-module $V'$ of rank $n$, and $M'$ is a graded left
$E'$-module. We assign to each variable $y_i$ the degree $-1$.
Then $E'$ is a bigraded $K$-algebra and $M'$ a bigraded
$E'$-module:
\[
M'_{(i,j)}=M_i\tensor_KS_j\quad\text{for all}\quad i,j\in\ZZ
\]
We fix a basis ${\bold v}=v_1,\ldots,v_n$ of $V$.
Then ${\bold v}$ is also a basis of the free $S$-module
$V'$, and we set $y_{\bold v}=\sum_{i=1}^nv_iy_i$. Notice that
$y_{\bold v}$ is of bidegree $(1,-1)$ and hence of total degree
$0$.
With these conventions and
notation the Cartan cochains $C^j({\bold v};M)$ of $M$ with respect to
${\bold v}$ may be identified with
$M'_{(*,-j)}=\Dirsum_iM'_{(i,-j)}$, and the chain maps of
$C^\pnt({\bold v};M)$ identify with multiplication by $y_{\bold
v}$ on $M'$. As $y_{\bold v}$ is bihomogeneous,
the generalized simplicial cohomology $H^\pnt(M',y_{\bold v})$
is naturally bigraded, that is, the $ik$th component
$H^i(M',y_{\bold
v})_k$ of $H^\pnt(M',y_{\bold v})$ is the $i$th cohomology of the complex
\[
({M'}_k, y_{\bold v})\:\ldots @>>> {M'}_{(i-1,k+1)}@>
l_{y_{v}}>> {M'}_{(i,k)}@> l_{y_{v}}>>
{M'}_{(i+1,k-1)}@>>>\ldots.
\]
Our discussions so far yield the following result:
\begin{Proposition}
\label{inter}
Let $M\in {\cal M}_l$. Then
\[
H^i({\bold v};M)_j\iso H^j(M', y_{\bold v})_{j-i}\quad
\text{for all}\quad i,j\in\ZZ.
\]
\end{Proposition}
We now specialize to the situation that $M=K\{\Delta\}$ is the
face ring of a simplicial complex. So $K\{\Delta\}=E/J_{\Delta}$
where $J_{\Delta}$ is the ideal generated by all monomials
$e_F$ in the basis ${\bold e}=e_1,\ldots,e_n$, where $F\not \in
\Delta$, see Section 2. As usual one assigns to a monomial $e_F$
the vector $\deg e_F\in \ZZ^n$ with $i$th entry $0$ if $i\not\in
F$, and $1$ otherwise. We call $\deg e_F$ the multidegree of
$e_F$. Since $J_{\Delta}$ is generated by monomials it is a
multigraded module, and hence so is $K\{\Delta\}$. Assigning to
$y_{i}$ the multidegree $(0,\ldots,-1,0,\ldots,0)$ with $-1$ as
the $i$th entry, the element $y_{\bold e}$ is homogeneous of
multidegree $(0,\ldots,0)$, and hence the complex $C^\pnt({\bold
e};K\{\Delta\})$ is a multigraded complex. For each $a\in \ZZ^n$
we denote by $H^\pnt({\bold e};K\{\Delta\})_a$ the $a$th graded
component of the Cartan cohomology.
We decompose $a$ into $a=a_+-a_-$ where $a_+\in \ZZ^n$ is the
vector whose $i$th component is $a_i$ if $a_i>0$, and is $0$ if
$a_i\leq 0$, while $a_-=a-a_+$. We also set $G_a=\{i\: a_i>0\}$.
Finally for any $b\in\ZZ^n$ we set $|b|=\sum_ib_i$.
Let $\Delta$ be a simplicial complex and $F\in \Delta$ a face.
Recall that the simplicial subcomplex
\[
\lk_{\Delta}F=\{G\in \Delta\: G\union F\in \Delta,\quad G\sect
F=\emptyset \}
\]
of $\Delta$ is called the {\em link of $F$}.
With the
notation introduced we have:
\begin{Proposition}
\label{bass}
Let $\Delta$ be a simplicial complex, and $a\in\ZZ^n$. Then
$H^\pnt({\bold e};K\{\Delta\})_a\newline =0$ if $a_j>1$ for some $j$, and
\[
H^i({\bold e};K\{\Delta\})_a=\tilde{H}^{i-|a_-|-1}(\lk_{\Delta} G_a;K),
\]
if $a_{j}\leq 1$ for all $j$.
\end{Proposition}
\begin{pf}
Let $a_F\in \ZZ^n$ be the vector with entry $1$ if $i\in F$, and $0$
otherwise. Then the elements $a_Fy^b$ with $F\in\Delta$, $a_F+b=a$ and
$-|b|=i$ form a $K$-basis ${\cal B}$ of $C^i({\bold
e};K\{\Delta\})_a$. Since
all entries of $b$ are $\leq 0$, it follows that $a_F+b$ has entries $\leq
1$. Hence if some $a_i>1$, then
the corresponding basis is empty. This proves
the first part of the statement.
Now suppose that $a_i\leq 1$ for all $i$. Then, by the assignment
$a_Fy^b\mapsto F$, the basis ${\cal B}$ is in bijection with the set ${\cal
S}=\{F\in [n]\: G_a\subset F,\quad |F|=|a|+i\}$. Finally
the assignment $F\mapsto F\setminus G_a$ establishes a bijection between the
set ${\cal S}$ and the set ${\cal T}=\{F\in \lk_{\Delta}G_a\:
|F|=-|a_-|+i\}$. Identifying ${\cal B}$ with ${\cal T}$, the chain map of
$C^\pnt({\bold e};K\{\Delta\})$ identifies with the chain map of the augmented
oriented chain complex of $\lk_{\Delta}G_a$, shifted by $-|a_-|$. This
proves the assertions of the proposition.
\end{pf}
\begin{Corollary}
\label{cohenmac}
Let $K$ be a field, and let
$\Delta$ be a $(d-1)$-dimensional simplicial complex.
The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $\Delta$ is Cohen-Macaulay over $K$;
\item[(b)] $H^i({\bold e};K\{\Delta\})_j=0$
for all $i\geq 0$ and all $j\neq d-i$;
\item[(c)] there exists an integer $t$ such that $H^i({\bold
e};K\{\Delta\})_j=0$ for all $i\geq 0$ and all $j\neq t-i$.
\end{enumerate}
\end{Corollary}
\begin{pf}
(a)\implies (b): We have $H^i({\bold e};K\{\Delta\})_j=0$ for
all $i\geq 0$ and all $j\neq d-i$ if an only if
$H^i({\bold e};K\{\Delta\})_a=0$ for
all $i\geq 0$ and all $a\in \ZZ^n$ with $|a|\neq d-i$. It
suffices to consider $a\in\ZZ^n$ with $a_j\leq 1$ for all $j$
and $G_a\in\Delta$, since otherwise $H^i({\bold
e};K\{\Delta\})_a=0$ anyway. For such $a\in\ZZ^n$ we have
$|a_+|=|G_a|$, and hence
\begin{eqnarray}
\label{equiv}
&& |a|\neq d-i\quad\iff\quad |a_+|-|a_-|\neq d-i\quad \iff
\quad i-|a_-|-1\neq d-|G_a|-1.
\end{eqnarray}
A version of Reisner criterion (see \cite[Page 235, (8)]{BH}
says that a $(d-1)$-dimensional simplicial complex is
Cohen-Macaulay over $K$ if and only if
$\tilde{H}_{i-|F|-1}(\lk_{\Delta};K)=0$ for all $F\in \Delta$
and all $i\neq d$. Thus the equivalent conditions (\ref{equiv})
together with \ref{bass} yield the assertion.
(b)\implies (c) is trivial.
(c)\implies (a): For $i=0$ the condition implies that
$H^0({\bold e}; K\{\Delta\})_j\newline =0$ for $j\neq t$. Since
$H^0({\bold e}; K\{\Delta\})$ can be identified with the socle
of $K\{\Delta\}$, and since a $K$-basis of the socle of
$K\{\Delta\}$ is given by the residue classes of the monomials
$e_F$, where $F$ is a facet of $\Delta$, it follows that
$\Delta$ is pure and that $t=d$. Now our hypothesis and the
equivalent conditions (\ref{equiv}) imply that $\Delta$ is
Cohen-Macaulay over $K$.
\end{pf}
\section{Alexander duality}
Let $M\in {\cal M}_l$, and let $(F, \partial)$ a be a minimal
graded free resolution of $M$ as a left $E$-module.
Notice that if $M\in {\cal M}$,
then $(F,\partial)$ is a minimal graded free
resolution of $M$ as a right $E$-module as well, if we define
the right $E$-module structure of the free modules $F_i$ in the
obvious way, by setting $xa=(-1)^{\deg x\deg a}ax$ for all
homogeneous elements $a\in E$ and $x\in F_i$.
We denote by $\Omega^i(M)$ the $i$th syzygy module of $M$. In
other words, $\Omega^i(M)=\Im \partial_i$.
We set $D(M)=\Omega^1(M)^*$.
It is easily seen that $M\mapsto D(M)$ is a contravariant functor in
the stable module category of $E$. In the sequel
we shall use the
following fact:
\begin{Lemma}
\label{free}
For all $M \in {\cal M}$ there exists a free $E$-module $G$ such
that $M\iso D(D(M))\dirsum G$.
\end{Lemma}
\begin{pf}
Let $0\to \Omega^1(M)\to F\to M\to 0$ be a minimal free
presentation of $M$. Then $0\to M^*\to F^*\to D(M)\to 0$ is a
not necessarily free presentation of $D(M)$. Therefore, there
exists a free $E$-module $H$ such $M^*=\Omega^1(D(M))\dirsum H$.
Applying the functor $(\; )^*$ the assertion follows.
\end{pf}
Simplicial (co)homology of $M$ and $D(M)$ are related as follows
\begin{Proposition}
\label{cohom}
Let $M\in {\cal M}$. Then $H_i(M)\iso H^{n-i-1}(D(M))$ and
$H_i(D(M))\iso H^{n-i-1}(M)$ for all $i$.
\end{Proposition}
\begin{pf}
Since $H_i(F)=0$ for any free $E$-module and all
$i$, it follows from
the exact sequence $0\to \Omega^1(M)\to F\to M\to 0$ that
$H_i(M)\iso H_{i-1}(\Omega^1(M))$ for all $i$. the first assertion
follows now from \ref{duality}, while the second follows from
the first and from \ref{free}, observing that the simplicial
homology and cohomology of a free module is trivial.
\end{pf}
We will now see that Alexander duality is a special case of the
preceeding proposition. Let $\Delta$ be a simplicial complex on
the vertex set $[n]$. The {\em dual simplicial complex
$\Delta^{*}$ of $\Delta$} is defined as follows:
\[
\Delta^{*}:= \{ F \subset [n] ; [n]\setminus F \not\in \Delta \}.
\]
We observe the following simple fact
\begin{Lemma}
\label{simple}
$D(K\{\Delta\})\iso K\{\Delta^*\}$.
\end{Lemma}
\begin{pf}
The exact sequence
\[
0\to J_\Delta\to E\to K\{\Delta\}\to 0
\]
yields the exact sequence
\[
0\to K\{\Delta\}^*\to E\to D(K\{\Delta\})\to 0.
\]
Since $K\{\Delta\}^*=0:J_\Delta=J_{\Delta^*}$, the assertion
follows.
\end{pf}
As a consequence of \ref{cohom} and \ref{simple} we now obtain
\begin{Corollary}[Alexander Duality]
\label{alexander}
Let $\Delta$ be a simplicial complex on the vertex set $[n]$.
Then $\tilde{H}_i(\Delta;K)\iso \tilde{H}^{n-i-3}(\Delta^*;K)$
for all $i$.
\end{Corollary}
The next result relates the Betti numbers of $M\in {\cal M}$ to
the Bass numbers of $D(M)$. Recall that the numbers
\[
\mu_{ij}=\dim_K\Ext^i_E(K,M)_j
\]
are called the {\em graded Bass numbers of $M$}. It is clear that
for any basis ${\bold v}=v_1,\ldots,v_n$ of $E_1$ the $K$-vector
spaces $H^i({\bold v};M)_j$ and $\Ext^i(K,M)_j$ are isomorphic
for all $i$ and $j$ so that the Bass numbers can be computed with Cartan
cohomology.
\begin{Proposition}
\label{bettibass}
Let $M\in{\cal M}$. Then for all $i>0$ and all $j$ we have
\[
\beta_{ij}(M)=\mu_{i-1,n-j}(D(M)).
\]
\end{Proposition}
\begin{pf}
Let ${\bold v}$ be a basis of $E_1$. Then $H_i({\bold v};M)\iso
H_{i-1}({\bold v};\Omega^1(M))$ for all $i>0$, so that
$H_i({\bold v};M)\iso H^{i-1}({\bold v};D(M))^*$, by
\ref{cohom}. Hence the assertion follows from \ref{dual}
\end{pf}
A module $M\in {\cal M}$ is said to have a {\em linear
resolution} if there exists an integer $c$ such that
$\beta_{ij}(M)=0$ for all $i\geq 0$ and $j\neq i+c$.
We say that $K\{\Delta\}$ has a linear resolution if
the defining ideal $J_{\Delta}$ has a linear resolution.
The following corollary is the analogue of the Eagon-Reiner
Theorem \cite{ER}.
\begin{Corollary}
\label{eagonreiner}
Let $\Delta$ be a simplicial complex. Then the following
conditions are equivalent:
\begin{enumerate}
\item[(a)] $K\{\Delta\}$ has a linear resolution;
\item[(b)] the dual simplicial complex $\Delta^*$ is
Cohen-Macaulay over $K$.
\end{enumerate}
\end{Corollary}
\begin{pf}
Let $\Delta^*$ be $(d-1)$-dimensional. Then by \ref{cohenmac}
$\Delta^*$ is Cohen-Macaulay over $K$ if and only if
$\mu_{ij}(K\{\Delta^*\})=0$ for all $i\geq 0$ and all $j\neq
d-i$. Since by \ref{simple} we have that
$D(K\{\Delta\})=K\{\Delta^*\}$, we conclude from \ref{bettibass}
that $\beta_{ij}(K\{\Delta\})=0$ for all $i>0$ and $j\neq
(n-d)+i-1$.
\end{pf}
\section{Algebraic shifting}
In this section we note the obvious fact that algebraic shifting,
introduced by Kalai \cite{K1}, is Gr\"obner basis theory applied
to monomial ideals, and we show that the main properties of
algebraic shifting are direct
consequences of our theory developed in the previous
sections.
Let $\Delta$ be a simplicial
complex on the vertex set $[n]=\{1,\ldots,n\}$, and
$K\{\Delta\}=E/J_{\Delta}$ the face ring of $\Delta$. Let $L/K$
be a field extension containing the algebraically independent
elements $a_{ij}$ ($i,j=1,\ldots,n$) over $K$, and let
$w_1,\ldots,w_n$ with $w_j=\sum_{i=1}^na_{ij}e_i$ for
$j=1,\ldots,n$ be a generic basis
of $E_1'$ where $E'=L\tensor_KE$. For $S=\{i_1<\ldots j$.
\end{enumerate}
\end{Corollary}
Let $M\in {\cal M}$.
After a suitable field extension we may assume that $M$ admits a
generic basis ${\bold v}= v_1,\ldots,v_n$ (cf.\ Section 4).
For $k=1,\ldots,n$ we set as
in the previous sections $H_i(k)=H_i(v_1,\ldots,v_k;M)$ for all
$i> 0$, and $H_0(k)=H(M/(v_1,\ldots,v_{k-1})M, v_k)$. Moreover
we let $H_i(0)=0$ for $i>0$.
We fix an integer $j$. By \ref{genbas} there exists an integer $i_0$
such that for all $i\geq i=i_0$ and all $k=1,\ldots,n$ the
sequences
\begin{eqnarray}
\label{short}
0@>>> H_{i+1}(k-1)_{(i+1)+j}@>>> H_{i+1}(k)_{(i+1)+j}@>>>
H_i(k)_{i+j}@>>> 0
\end{eqnarray}
are exact.
Set $h_i^k=\dim_KH_i(k)_{i+j}$, and $c_k=h_{i_0}^k$ for
$k=1,\ldots,n$. The exact sequences (\ref{short}) yield the
equations
\begin{eqnarray}
\label{equations}
h_{i+1}^k=h_{i+1}^{k-1}+h_i^k
\end{eqnarray}
for all $i\geq i_0$, and $k=1,\ldots,n$. It follows from
(\ref{equations}) that
\[
h_{i_0+i}^n={i+n-2\choose n-1}c_1+{i\choose n-3}c_2+\cdots
+{i\choose 1}c_{n-1}+c_n\quad\text{for all} \quad i\geq 0.
\]
Since $\beta^E_{ii+j}(M)=h^n_i$ for all $i$, we see that
\[
P_j(t)=t^{i_0+1}\sum_{i=1}^n\frac{c_i}{(1-t)^{n-i+1}} +Q(t),
\]
where $Q(t)$ is a polynomial. Thus we obtain:
\begin{Proposition}
\label{whatis}
Let $d_j$ and $e_j$ be defined as in {\em \ref{rational}}.
Then
\[
d_j(M)=n+1-\min\{i\:
c_i\neq 0\}\quad\text{and}\quad e_j(M)=c_{n-d_j+1}.
\]
\end{Proposition}
In order to relate the invariants $d_j$ and $e_j$ to the
generalized simplicial homology modules $H_0(k)$ we need the following
\begin{Lemma}
\label{better}
Let $1\leq l\leq n$ and $j$ be integers. The following
conditions are equivalent:
\begin{enumerate}
\item[(a)]
\begin{enumerate}
\item[(1)] $H_0(k)_j=0$ for $kj$ and all $k\leq
l+j-j'$.
\end{enumerate}
\item[(b)] For all $i\geq 0$ we have
\begin{enumerate}
\item[(1)] $H_i(k)_{i+j}=0$ for $kj$ and all $k\leq
l+j-j'$.
\end{enumerate}
\item[(c)] Condition {\em (b)} is satisfied for some $i$.
\end{enumerate}
Moreover, if the equivalent conditions hold, then
$H_i(l)_{i+j}\iso H_0(l)_j$ for all $i\geq 0$.
\end{Lemma}
\begin{pf}
In our proof we will have to refer to the exact sequence (\ref{long5})
from Section 4, namely to
\begin{eqnarray}
\label{exact}
H_i(k-1)_{i+j'}@>>> H_i(k)_{i+j'}@>>> H_{i-1}(k)_{(i-1)+j'}@>>>
H_{i-1}(k-1)_{(i-1)+
(j'+1)}
\end{eqnarray}
(a)\implies (b): We prove (b) by induction on $i$. For $i=0$,
there is nothing to show. So now let $i>0$ and assume that (1)
and (2) hold for $i-1$. By (\ref{exact}) we
have the exact sequence
\[
H_i(l)_{i+j}@>>> H_{i-1}(l)_{(i-1)+j}@>>>
H_{i-1}(l-1)_{(i-1)+(j+1)}.
\]
Since $l-1\leq l+j-(j+1)$, we have
$H_{i-1}(l-1)_{(i-1)+(j+1)}=0$ by induction hypothesis. Also by
induction hypothesis, $H_{i-1}(l)_{(i-1)+j}\neq 0$; therefore,
$H_i(l)_{i+j}\neq 0$.
Now let $k>> H_{i}(k)_{i+j}@>>>
H_{i-1}(k)_{(i-1)+j}.
\]
By induction hypothesis we have
$H_{i-1}(k)_{(i-1)+j}=0$. Now by induction on $k$
we may assume that $H_i(k-1)_{i+j}=0$. Therefore,
$H_{i}(k)_{i+j}=0$, and this shows (1).
In order to prove (b)(2), we let $j'>j$ and $k\leq l+(j-j')$,
and consider the exact sequence
\[
H_i(k-1)_{i+j'}@>>> H_{i}(k)_{i+j'}@>>>
H_{i-1}(k-1)_{(i-1)+j'},
\]
from which the assertion follows by induction on $i$ and $k$.
(c)\implies (a). We show that if the conditions (1) and (2) hold
for $i>0$, then they also hold for $i-1$. Therefore backwards
induction yields the desired conclusion.
We begin with the proof of (2) for $i-1$ by induction on $k$.
For $k=0$, there is nothing to show. Now let $j'>j$, and $0>> H_{i-1}(k)_{(i-1)+j'}@>>>
H_{i-1}(k-1)_{(i-1)+
(j'+1)}.
\]
Since $k-1\leq l+j-(j'+1)$ it follows by our induction
hypothesis that $H_{i-1}(k-1)_{(i-1)+(j'+1)} =0$. On the other
hand, by assumption we have $H_i(k)_{i+j'}=0$, and hence
$H_{i-1}(k)_{(i-1)+j'}=0$.
In order to prove (1) for $i-1$ we consider the exact sequence
\[
H_i(l-1)_{i+j}@>>> H_i(l)_{i+j}@>>> H_{i-1}(l)_{(i-1)+j}@>>>
H_{i-1}(l-1)_{(i-1)+(j+1)}.
\]
Since $l-1\leq l+j-(j+1)$, we know from (2) (which we have
already shown for $i-1$) that $H_{i-1}(l-1)_{(i-1)+(j+1)}=0$. By
our assumption we have $H_i(l-1)_{i+j}=0$, and hence
\[
H_{i-1}(l)_{(i-1)+j}\iso H_i(l)_{i+j}\neq 0.
\]
That $H_{i-1}(k)_{(i-1)+j}=0$ for $kj$. By \ref{whatis}
this condition is equivalent to
\[
\min\{k\: H_{i_0}(k)_{i_0+j'}\neq 0\}>l+(j-j'),
\]
where $l=\min\{k\: H_{i_0}(k)_{i_0+j}\neq 0\}$. This in turn is
equivalent to
\[
H_{i_0}(k)_{i_0+j'}=0\quad\text{for}\quad k\leq l+(j-j'),
\]
which means that $(l,j)$ is a distinguished pair.
From \ref{interpret} and \ref{whatis} it follows that $l=n+1-i-j$.
Finally, \ref{interpret}, \ref{whatis} and \ref{better} imply that
\[
\beta_{ii+j}(K[\Delta])=e_j(K\{\Delta\})=c_l=\dim_KH_0(l)_j.
\]
\end{pf}
We can now prove the main result of this section.
\begin{Theorem}
\label{final}
Let $\Delta$ be a simplicial complex. Then
\begin{enumerate}
\item[(a)] the $ij$th Betti number of $K[\Delta]$ is extremal if
and only if the $ij$th Betti number of $K[\Delta^s]$ is
extremal;
\item[(b)] the corresponding extremal Betti numbers of
$K[\Delta]$ and of
$K[\Delta^s]$ coincide.
\end{enumerate}
\end{Theorem}
\begin{pf} After a suitable field extension and a generic change
of bases we may assume, as in the proof of \ref{hope}, that
for the basis $e_1,\ldots,e_n$ we have that $\ini(J_{\Delta})=
\Gin(J_{\Delta})=J_{\Delta^s}$, and that $e_n,\ldots, e_1$ is
a generic basis for $K\{\Delta\}$ as well as for $K\{\Delta^s\}$.
We let $H_0(k)=H(K\{\Delta\}/(e_n,\ldots,e_{n-k+1},e_{n-k})K\{\Delta\},
e_{n-k})$. The corresponding homology modules for $K\{\Delta^s\}$ will
be denoted by $H^s_0(k)$. It follows from \ref{bar} that for
all $k=1,\ldots,n$ the
homology modules $H_0(k)$ and $H_0^s(k)$ have the same Hilbert
function. Since the Hilbert functions of these modules determine
uniquely the distinguished pairs $(l,j)$, all
assertions of the theorem follow from \ref{nonvanish}.
\end{pf}
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\ \\
\noindent
Annetta Aramova\\
Permanent address:\\
Institute of Mathematics, Bulgarian Academy of Sciences,\\
Sofia 1113, Bulgaria\\
E-mail:algebra@@bgearn.acad.bg\\
Present address:\\
FB 6 Mathematik und Informatik, Universit\"at -- GHS -- Essen\\
Postfach 103764, 45117 Essen, Germany\\
E-mail:a.aramova@@uni-essen.de
\ \\ \\
\noindent
J\"urgen Herzog\\
FB6 Mathematik und Informatik\\
Universit\"at GH-Essen\\
45117 Essen\\
Germany\\
juergen.herzog@@uni-essen.de
\end{document}