TITLE: On the fundamental group of closed 3-manifolds
QUESTION [5 upvotes]: I know that every finitely presented group can be realized as the fundamental group of a compact, connected, smooth manifold of dimension 4 (or higher). In dimension 2 there are strong restriction on the fundamental group of closed manifolds.
I wanted to know what happen in dimension 3: which are the conditions on a finitely presented group to be realized as the fundamental group of a closed (connected) 3-manifolds? I know there are strong restriction in dimension 3, but is there an "if and only if" characterization of fundamental groups of closed 3-manifolds? In affirmative case, can you suggest me some reference?

REPLY [7 votes]: Igor's suggestion of the recent paper by Aschenbrenner, Friedl, and Wilton is probably the best place to start as it has a good treatment this problem which includes both a summary of recent advances and a litany of open problems.
If you want to work through a classification of geometric and non-hyperbolic manifold groups, Thurston's book "Three-Dimensional Geometry and Topology" (especially sections 4.3, 4.4. and 4.7) is very handy. I have also found the wikipedia article Seifert fiber spaces together with Scott's article: The geometries of 3-manifolds (errata here) to be especially useful in dealing with the groups of Seifert fiber spaces.
Finally, Groves, Manning, and Wilton provide a theoretical algorithm for determining if a group with solvable word problem is a three manifold group in this preprint.


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\vspace*{1in}
\textsf{ \Large Recursive projections of symmetric tensors\\
	and Marcus's proof of the Schur inequality}\\[.2in]

\textsf{S. Gill Williamson}\footnote{\url{http://cseweb.ucsd.edu/~gill}}

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\hspace{1 pt}

\begin{center}
{\Large Abstract}\\[.2in]
\end{center}
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In a 1918 paper  \cite{is:fgh}, Schur proved a remarkable inequality that related group representations, Hermitian forms and  determinants.  He also gave concise necessary and sufficient  conditions for equality.
In  \cite{mm:crs}, Marcus  gave a beautiful short proof of Schur's inequality by applying the Cauchy-Schwarz inequality to symmetric tensors, but he did not discuss the case of equality. In \cite{gw:tch}, Williamson gave an inductive proof of Schur's equality conditions by contracting Marcus's symmetric tensors onto lower dimensional subspaces where they remained symmetric tensors.  Here we unify these results notationally and conceptually, replacing contraction operators with the more geometrically intuitive projection operators.

\section{Introduction}
The following theorem will be the focus of this paper.

\begin{theorem}[\bfseries Schur's theorem for finite groups and hermitian forms]
\label{thm:stf}
Let $G$ be a subgroup of the symmetric group $S_n$ on $\underline{n}=\{1,2, \ldots, n\}$.
Let $H=(h_{ij})$ be an $n\times n$ complex positive definite Hermitian matrix and define
$\,\bG_H$ to be the group generated by all transpositions $(i,j)$ such that $h_{ij}\neq 0$.
Let $\,M$ be a representation of $\,G$ as unitary linear operators on $\,U$,
$\dim(U)=m$, and
let $\,M_H = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n h_{i\sigma(i)}$.
$\,M_H$, called a generalized matrix function, is  positive definite Hermitian, and for $\,u\in U$, $\|u\|=1$,
\begin{equation}
\label{eq:sin}
\det(H)\leq (M_Hu,u)
\end{equation}
\begin{equation}
\label{eq:seq}
\det(H) = (M_Hu,u) \;{\it if\;and\;only\;if}\;\bG_H\subseteq G\,\, {\it and}\,\,
\end{equation}
\[
(M(\sigma)u,u)=\epsilon(\sigma) \;{\it for\;all\;} \sigma\in \bG_H\;\;
{\it where\;\;}  \epsilon(\sigma) {\it \;\;is\; the\; sign\; of\; \sigma.}\\
\]
\end{theorem}
In this section we discuss the basics.
In Section~2 we prove Marcus's generalization of Schur's inequality.
In Section~3 we treat the case of equality.
In Section~4 we discuss the combinatorial lemmas needed for Section~3.
In Section~5 we  give examples of Schur's inequality and discuss the trace version.

\begin{remark}[\bfseries Comments on theorem~\ref{thm:stf}]
\label{rem:comequcon}
\hspace*{1in}\\[4pt]
$\bf(1)$ Throughout the rest of the paper, ${\bf M}_n(\bC)$ denotes the $n\times n$ matrices with entries in $\bC$, the complex numbers.  Likewise, the order (or cardinality) of the finite group $G$ will be denoted by $g$.\\
\hspace*{1in}\\[1pt]
$\bf(2)$ Referring to equality condition~\ref{eq:seq} we have
\[M_H = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n h_{i\sigma(i)}\;\,{\rm implies\; that\;\,}
(M_Hu,u) = \sum_{\sigma\in G} (M(\sigma)u,u) \prod_{i=1}^n h_{i\sigma(i)}.\]
Assume $M(\sigma_1)$ and $M(\sigma_2)$ unitary and $u\in U$ has $\|u\|=1$.
If $|(M(\sigma_1)u,u)|=|(M(\sigma_2)u,u)|=1$ then $u$ is a unit eigenvector of both
$M(\sigma_1)$ and $M(\sigma_2)$.
We have $M(\sigma_j)u=\lambda_ju$, $\lambda_j =e^{ir_j}$, $i=1,2$.
Thus, $(M(\sigma_1)u,u)(M(\sigma_2)u,u) = \lambda_1\lambda_2 =
(M(\sigma_1)M(\sigma_2)u,u) = (M(\sigma_1\sigma_2)u,u).$\\
\hspace*{1in}\\[1pt]
$\bf(3)$ A group $\bG_D$, analogous to $\bG_H$ of theorem~\ref{thm:stf}, can defined for any $D\in {\bf M}_n(\bC).$
Let $K\subseteq \{1,2,\ldots,n\}$ be an orbit of $\bG_D$.
Since $\bG_D$ is generated by transpositions, the restriction $\bG_D|K$ is the symmetric group $S_K$.\\
\end{remark}

\begin{remark}[\bfseries Inner products on tensors]
\label{rem:innproten}
Let ${\underline n}=\{1, \ldots, n\}$.
Denote by ${\underline n}^{\underline n}$  the set of all functions
from ${\underline n}$ to ${\underline n}$.
The notation $\Gamma_n \equiv {\underline n}^{\underline n}$ is also common in this subject.  Note the cardinality, $| \Gamma_n| = | {\underline n}^{\underline n}| =  n^n.$
Let $e_1, \ldots, e_n$ be an orthonormal basis for the unitary space $V$.
Using the inner product in $U$, we have an inner product on
$U\otimes [\otimes^n V]$ which on homogeneous elements is
\[
\left(u_1\otimes x_1\otimes\cdots\otimes x_n, u_2\otimes y_1\otimes\cdots\otimes y_n\right)=
(u_1,u_2)\prod_{i=1}^n (x_i,y_i).
\]
Note that
\begin{equation}
\label{eq:onbasis}
\{e_\alpha\mid e_\alpha=e_{\alpha(1)}\otimes \cdots \otimes e_{\alpha(n)},\,\,\alpha\in{\underline n}^{\underline n}\}
\end{equation}
is an orthonormal basis for $\otimes^n V$.\\
\end{remark}

\begin{definition}[\bfseries Generalized symmetry operators]
\label{def:gsop}
Let $G$ be a subgroup of the symmetric group $S_n$ on $\{1,2, \ldots, n\}$.
Let $M$ be a representation of $G$ as unitary linear operators on a unitary space $U$, $\dim U=m$.
Define an endomorphism of $U \otimes \left[\otimes^nV\right]$ by
$T_G=\sum_{\sigma\in G} M(\sigma)\otimes P(\sigma)$ where $P(\sigma)$ is the permutation operator
defined by
$P(\sigma)(e_\alpha)=e_{\alpha(\sigma^{-1}(1))}\otimes \cdots \otimes e_{\alpha(\sigma^{-1}(n))}$
on the basis and extended (conjugate) linearly to $\otimes^n V$.
On homogeneous tensors
\begin{equation}
\label{eq:psigma}
P(\sigma)(x_1\otimes \cdots \otimes x_n) = x_{\sigma^{-1}(1)}\otimes \cdots \otimes x_{\sigma^{-1}(n)}.
\end{equation}
$T_G$ will be called a {\em generalized symmetry operator} on $U\otimes (\otimes^n V)$
of {\em degree} $m=\dim(U)$ and {\em order} $n=\dim(V)$.
On homogeneous elements
\[
T_G(u\otimes x_1\otimes \cdots \otimes x_n)=
\sum_{\sigma\in G} M(\sigma)u\otimes x_{\sigma^{-1}(1)}\otimes \cdots \otimes x_{\sigma^{-1}(n)}.\\
\]
\end{definition}

\begin{remark}[\bfseries Cauchy-Schwartz on symmetric tensors]
\label{rem:cssymten}
Let $e=e_1\otimes \cdots \otimes e_n$,  $x=x_1\otimes \cdots \otimes x_n$ and $u\in U$, $||u||=1$.
By the Cauchy-Schwarz inequality,

\begin{equation}
\label{eq:csifront}
|\left(T_G(u\otimes x), T_G(u\otimes e)\right)|^2\leq \|(T_G(u\otimes x)\|^2\|(T_G(u\otimes e)\|^2.
\end{equation}

Let $A\in {\bf M}_n(\bC)$, $A=(a_{ij})$, be upper triangular and nonsingular.
Define vectors $x_i$, $i=1,\ldots, n$,  by $x_i=\sum_{j=1}^n a_{ij}e_j$.
Thus, $A=(a_{ij})=((x_i,e_j))$.
We say that $A$ is {\em row--associated} with   the vectors
$x_i$, $i=1,\ldots, n$ and the basis $e_i$, $i=1, \ldots, n$.
Marcus (\cite{mm:crs}) observed that
\[
0< |\left(T_G(u\otimes x), T_G(u\otimes e)\right)|^2= g^2(\det(A))^2
\]
\[
\|(T_G(u\otimes x)\|^2 = g(M_{AA^*}u,u)\;\;{\rm and}\;\;
\|T_G(u\otimes e)\|^2=g.\\
\]
\end{remark}
From definition~\ref{def:gsop} and remarks~\ref{rem:innproten}, \ref{rem:cssymten}, we have the following:

\begin{theorem}[\bfseries Cauchy-Schwartz-Schur inequalities]
\label{thm:inequalities}
Let $u\in U$, $\|u\| = 1$.
Let $H\in{\bf M}_n(\bC)$ be positive definite Hermitian.
Write $H=AA^*$ where $A$ is upper triangular and nonsingular (e.g., using Cholesky decomposition), and
define  $x_i$, $i=1,\ldots, n$, by $A=((x_i,e_j))$ (remark~\ref{rem:cssymten}). The following inequalities are equivalent
\begin{description}
\item[(1) C-S]
$|\left(T_G(u\otimes x), T_G(u\otimes e)\right)|^2\leq \|T_G(u\otimes x)\|^2\|T_G(u\otimes e)\|^2$
\item[(2) Schur H]
$\det(H)\leq (M_{H}u,u)$,
\item[(3)  A form]
$(\det(A))^2 \leq (M_{AA^*}u,u)$.
\end{description}
\begin{proof}
This result follows directly from remark~\ref{rem:cssymten}.\\
\end{proof}
\end{theorem}

Referring to theorem~\ref{thm:inequalities}, we have the following:
\begin{theorem}[\bfseries Cauchy-Schwartz-Schur equalities]
\label{thm:equalities}
Let $u\in U$ and $H=AA^*$ be as in theorem~\ref{thm:inequalities}.
The following equalities are equivalent.
\begin{description}
\item[($0$) C-S Equality condition]
$T_G(u\otimes x) = k T_G(u\otimes e),\;k\neq 0$
\item[($1$) C-S Equality]
$|\left(T_G(u\otimes x), T(u\otimes e)\right)|^2= \|T_G(u\otimes x)\|^2\|T_G(u\otimes e)\|^2$
\item[($2$) Schur H Equality]
$\det(H) = (M_{H}u,u)$.
\item[($3$) A Equality]
$(\det(A))^2 = (M_{AA^*}u,u)$
\item[($4$) Schur H Equality Condition]
$\bG_H\subseteq G, (M(\sigma)u,u)=\epsilon(\sigma), \sigma\in \bG_H$
\item[($5$) A Equality Condition]
$\bG_A\subseteq G, (M(\sigma)u,u)=\epsilon(\sigma), \sigma\in \bG_A$.
\end{description}
\begin{proof}
The equivalence of conditions {\bf($0$)}, {\bf($1$)}, {\bf($2$)} and  {\bf($3$)} follows from
theorem~\ref{thm:inequalities} and the standard condition for equality in  the Cauchy-Schwartz inequality (i.e. the equivalence of {\bf($0$)} and {\bf($1$)}).\\[5pt]

Condition {\bf($5$)} implies {\bf($4$)} (\cite{gw:tch}, Lemma 5.3):\\
{\bf Assume {\bf($5$)} and show {\bf($4$)}}.
Note that $H(i,j)\equiv h_{ij} \neq 0$  implies that
$H(i,j)=\sum_{k=1}^n A(i,k)A^*(k,j) = \sum_{k=1}^n A(i,k)\bar{A}(j,k)\neq 0$.
Thus, there is some $s$ such that $(i,s), (j,s)\in \bG_A$ which, by hypothesis, implies that $(M((i,s))u,u)=-1$ and $(M((j,s))u,u)=-1$.
Thus, $(i,j)=(i,s)(j,s)(i,s)\in \bG_H$ has $(M((i,j))u,u)=-1$,
and item {\bf($4$)} holds  (we use remark~\ref{rem:comequcon} (2)).
Note this also shows that $G_H\subseteq G_A$.\\[5pt]
Condition {\bf ($4$)}  implies {\bf ($2$)} (\cite{gw:tch}, Lemma 5.2): \\
{\bf Assume {\bf($4$)} and show {\bf($2$)}}.
$M_H = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n h_{i\sigma(i)}$, thus
$(M_Hu,u) = \sum_{\sigma\in G} (M(\sigma)u,u) \prod_{i=1}^n h_{i\sigma(i)}$.
Note that $\sigma\in S_n$, $\prod_{i=1}^n h_{i\sigma(i)}\neq 0$ implies
$h_{i\sigma(i)}\neq 0$, $1\leq i \leq n$, which implies, by definition of $\bG_H$,
$(i,\sigma(i))\in\bG_H$, $1\leq i \leq n$.
Recall that $\sigma$ is a product of transpositions of the form $(i,\sigma(i))$.
Each such transposition has $(M((i,\sigma(i)))u,u)=-1$ by our hypothesis.
Thus, $\sigma\in \bG_H\subseteq G$ and $(M(\sigma)u,u)= \epsilon(\sigma)$.
We have shown that
\[
(M_Hu,u) = \sum_{\sigma\in \bG_H} (M(\sigma)u,u) \prod_{i=1}^n h_{i\sigma(i)}=
 \sum_{\sigma\in S_n} \epsilon(\sigma) \prod_{i=1}^n h_{i\sigma(i)}=\det(H).
\]
Finally, item {\bf($0$)} implies {\bf($5$)} is proved in \cite{gw:tch} (Theorem 4.2)
and below (Theorem~\ref{thm:mainresult}).
This completes the proof of theorem~\ref{thm:equalities}.\\
\end{proof}
\end{theorem}

\begin{remark}[\bfseries Basic observations on theorem~\ref{thm:equalities}]
\label{rem:basicstuff}
In showing $(5)$ implies $(4)$ above, we have shown that
$\bG_H\subseteq \bG_A$.
Using the ideas in $(4)$ implies $(2)$, for any matrix $D\in {\bf M}_n(\bC)$,
\[
\sum_{\sigma\in G} M(\sigma)\prod_{i} D(i,\sigma(i))=
\sum_{\sigma\in \bG_D\cap G}M(\sigma)\prod_{i} D(i,\sigma(i)).
\]
From theorem~\ref{thm:equalities}, we will have $(4)$ implies $(5)$.
If we take $G=\bG_H$ in $(4)$ and $M(\sigma)=\epsilon(\sigma)I_m$, we have
$\bG_A\subseteq \bG_H$ from $(5)$.
Reversing this argument using $(5)$ implies $(4)$ we get
$\bG_H\subseteq \bG_A$  (also derived from the argument proving $(5)$ implies
$(4)$ above).  Thus, $\bG_H=\bG_A$. A simple example shows the generating sets $\{(i,j)\mid A(i,j)\neq 0\}$ and $\{(i,j)\mid H(i,j)\neq 0\}$ need not  be the same.
Schur in~\cite{is:fgh} proved $(2)$ of theorem~\ref{thm:inequalities}
and showed the equivalence of $(2)$ and $(4)$ of theorem~\ref{thm:equalities}.
The focus of this paper is to characterize the multilinear algebraic properties of the symmetry operators $T_G$ that result in the equivalence of  $(0)$ and $(5)$
in theorem~\ref{thm:equalities}.  The most interesting structural properties of $T_G$ arise in proving $(0)$ implies $(5)$ (theorem~\ref{thm:mainresult}).
The converse $(5)$ implies $(0)$  follows from theorem~\ref{thm:equalities}.
\end{remark}

\section{Inequalities}

We use the terminology of the previous section.
Let $U$ and $V$ be $m$ and $n$ dimensional unitary spaces with the standard inner products.\\

\begin{remark}[\bfseries Properties of $T_G$]
\label{rem:pot}
To summarize, the $M(\sigma)$ and $P(\sigma)$ are unitary operators: $(M(\sigma))^*=M(\sigma^{-1})$ (by definition of $M$) and
$(P(\sigma))^*=P(\sigma^{-1})$ (simple computation). Thus, $M(\sigma)\otimes P(\sigma)$ is unitary since
 \[
 (M(\sigma)\otimes P(\sigma))^* = (M(\sigma))^*\otimes (P(\sigma))^*=M(\sigma^{-1})\otimes P(\sigma^{-1}).
 \]
 $M$ is defined to be a representation of $G$  as unitary operators on $U$, and $P$ is a representation of $G$ as unitary operators on $\otimes^n V$.
$T_G$ is essentially idempotent, $T_G^2=gT_G$, and Hermitian, $T_G^*=T_G$.\\
\end{remark}

\begin{remark}[\bfseries Inner products as products of associated matrices]
\label{rem:ippm}
Consider the unitary inner product defined by
\begin{equation}
\label{eq:uipxy}
(x_1\otimes \cdots \otimes x_n, y_1\otimes \cdots \otimes y_n)= \prod_{i=1}^n (x_i,y_i).\\
\end{equation}
Let $x_i=\sum_{t=1}^n a_{it}e_t$ and $y_j=\sum_{t=1}^n b_{jt}e_t$ where $e_1, \ldots, e_n$
is the orthonormal basis for $V$.
In the unitary space $V$, we have $(x_i,y_j)=\sum_{t=1}^n a_{it} \bar{b}_{jt}$.
In terms of remark~\ref{rem:cssymten}, $A$ is row--associated with   the vectors
$x_i$, $i=1,\ldots, n$ and the basis $e_i$, $i=1, \ldots, n$, and
$B$ is row--associated with   the vectors $y_i$, $i=1,\ldots, n$ and the basis
$e_i$, $i=1, \ldots, n$.  With the basis $e_i$, $i=1, \ldots, n$ understood, we also refer
to $A$ and $B$ as the matrices associated with
$x= x_1\otimes \cdots \otimes x_n$ and $y_1\otimes \cdots \otimes y_n$ respectively.

For $B=(b_{ij})$, the conjugate transpose $B^*(i,j)=(\bar{b}_{ji})$.
Thus, the inner product $(x_i,y_i)$ becomes
\begin{equation}
\label{eq:ipmtrx}
(x_i,y_j)=\sum_{t=1}^n a_{it} \bar{b}_{jt}=\sum_{t=1}^n A(i,t)B^*(t,j)=AB^*(i,j).
\end{equation}
If $x= x_1\otimes \cdots \otimes x_n=y_1\otimes \cdots \otimes y_n$ then $A=B$ and
$AA^*$ is the Gram matrix of the sequence $x_1, \ldots, x_n$ (i.e. the rows of $A$).\\
\end{remark}

We use the terminology of theorem~\ref{thm:stf} and equation~\ref{eq:ipmtrx}.
\begin{lemma}[\bfseries Marcus's inner product form for  symmetric tensors \cite{mm:crs}]
\label{lem:ipgst}
Let $T_G(u_1\otimes x)$ and $T_G(u_2\otimes y)$ be two generalized symmetric tensors where
$x= x_1\otimes \cdots \otimes x_n$, $y=y_1\otimes \cdots \otimes y_n$, and
$A$ and $B$ are the matrices associated with  $x$ and $y$ respectively
(remark~\ref{rem:ippm}).
Then
\begin{equation}
\label{eq:iptst}
(T_G(u_1\otimes x), T_G(u_2\otimes y)) = g(M_{AB^*}u_1,u_2).
\end{equation}

\begin{proof}
$(T_G(u_1\otimes x), T_G(u_2\otimes y))=(T_G^2(u_1\otimes x), u_2\otimes y)=
g(T_G(u_1\otimes x), u_2\otimes y)$
(for properties of $T_G$, see remark~\ref{rem:pot}).
\[
\left(T_G(u_1\otimes x), u_2\otimes y\right)=
\left(\sum_{\sigma\in G}M(\sigma)u_1\otimes P(\sigma)x, u_2\otimes y\right) =
\]
\[
\sum_{\sigma\in G}(M(\sigma)u_1, u_2)(P(\sigma)x, y) =
\sum_{\sigma\in G}(M(\sigma)u_1, u_2) \prod_{i=1}^n (x_{\sigma^{-1}(i)},y_i)=
\]
\[
\sum_{\sigma\in G}(M(\sigma)u_1, u_2) \prod_{i=1}^n AB^*(\sigma^{-1}(i), i)\,\,({\rm equation}\,\,
\ref{eq:ipmtrx})
\]
This latter equation becomes
\[
\left(\left(\sum_{\sigma\in G}M(\sigma)\prod_{i=1}^n AB^*(\sigma^{-1}(i), i)\right)u_1, u_2\right) =
\left(M_{AB^*}u_1, u_2\right).
\]
\end{proof}
\end{lemma}

\begin{theorem}[\bfseries Marcus's generalization of Schur's inequality \cite{mm:crs}]
\label{thm:mgsi}
Let $A$ and $B$ be $n\times n$ complex matrices.
Let $M_K$ be the generalized matrix function of $K$.

Then, for any $u_1, u_2\in U$,
\begin{equation}
\label{eq:mgsi}
\left| \left(M_{AB^*}u_1,u_2\right)\right|^2 \leq
 \left(M_{AA^*}u_1,u_1\right) \left(M_{BB^*}u_2,u_2\right).
\end{equation}
\begin{proof}
Let $T_G(u_1\otimes x)$ and $T_G(u_2\otimes y)$ be generalized symmetric tensors where
$x= x_1\otimes \cdots \otimes x_n$ and $y=y_1\otimes \cdots \otimes y_n$
are chosen so the the $n\times n$ matrices
$A$ and $B$ are associated with  $x$ and $y$ respectively
(remark~\ref{rem:ippm}).
By the Cauchy-Schwarz inequality, we have
\begin{equation}
\label{eq:csi}
|\left(T_G(u_1\otimes x), T_G(u_2\otimes y)\right)|^2\leq
\end{equation}
\[
\left(T_G(u_1\otimes x), T_G(u_1\otimes  x)\right)\;
\left(T_G(u_2\otimes y), T_G(u_2\otimes y)\right).
\]
From lemma~\ref{lem:ipgst}, equation~\ref{eq:csi} becomes
\begin{equation}
\label{eq:csimf}
g^2|(M_{AB^*}u_1, u_2)|^2 \leq g\,(M_{AA^*}u_1, u_1)\,\,g\,(M_{BB^*}u_2, u_2).\\
\end{equation}
\end{proof}
\end{theorem}

\begin{corollary}[\bfseries Schur's inequality from theorem~\ref{thm:mgsi}]
\label{cor:mpsi}
Let $G$ be a subgroup of the symmetric group $S_n$ of degree $n$.
Let $H=(h_{ij})$ be an $n\times n$ positive definite Hermitian matrix.
Let $M$ be a representation of $G$ as unitary linear operators on $U$,
$\dim(U)=m$, and
let $M_H = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n h_{i\sigma(i)}$.
Then $M_H$ is a positive definite Hermitian transformation on $U$ and, if $u\in U$ has  $||u||=1$, then
\begin{equation}
\label{eq:sinc}
\det(H)\leq (M_Hu,u).
\end{equation}
\begin{proof}
Write
$M_H=\sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n H(i,\sigma(i))$.
\[
(M_H)^* = \sum_{\sigma\in G} (M(\sigma))^* \prod_{i=1}^n \overline{H(i,\sigma(i))}
= \sum_{\sigma\in G} (M(\sigma^{-1})) \prod_{i=1}^n H^*(\sigma(i),i) = M_{H^*}.
\]

Thus, $H^*=H$ implies that $(M_H)^* = M_H$  and hence $M_H$ is Hermitian.
If $H$ is positive definite Hermitian then $\det(H)>0$, and \ref{eq:sinc} will imply
that every eigenvalue of $M_H$ is positive, hence $M_H$ is positive definite Hermitian. To prove~\ref{eq:sinc}, take $H=AA^*$ and let $B=I_n$ in equation~\ref{eq:mgsi}.
 In which case $M_{BB^*}=M_{I_n} = I_m$.
 Assume without loss of generality that $A$ is triangular and take
 $u= u_1=u_2$ to be a unit vector.  In this case, $(M_Au,u) = \det(A)$ and
 equation~\ref{eq:mgsi} becomes $|\det(A)|^2 = \det(H) \leq (M_H u, u)$
 which proves~\ref{eq:sinc}.
\end{proof}
\end{corollary}

\section{Equalities}

The proof of Schur's inequality, corollary~\ref{cor:mpsi}, was obtained from Marcus's tensor form of the Cauchy-Schwartz inequality
\begin{equation}
\label{eq:csi2}
|\left(T_G(u_1\otimes x), T_G(u_2\otimes y)\right)|^2\leq
\end{equation}
\[
\left(T_G(u_1\otimes x), T_G(u_1\otimes  x)\right)\;
\left(T_G(u_2\otimes y), T_G(u_2\otimes y)\right)
\]
by taking $u_1=u_2=u$ and
$y=y_1\otimes  \cdots \otimes y_n = e_1\otimes  \cdots \otimes e_n = e$:
\begin{equation}
\label{eq:csi3}
|\left(T_G(u\otimes x), T_G(u\otimes e)\right)|^2\leq\|T_G(u\otimes x)\|^2\,|T_G(u\otimes e)\|^2.
\end{equation}
Both $T_G(u\otimes x)$ and $T_G(u\otimes e)$ are nonzero, hence equality occurs in equation~\ref{eq:csi3} if and only if
\begin{equation}
\label{eq:csi4}
T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0.\\
\end{equation}\\

 \begin{remark}[\bfseries The case $T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0$, $n=2$ ]
 \label{rem:case2}
 Choose $u\in U,\, \|u\|=1.$
 The nonsingular upper triangular matrix  $A=(a_{ij})=((x_i,e_j))$ is in $\mathbf{M}_2(\bC)$.
 In this case, $x_1=a_{11}e_1 + a_{12}e_2$, $x_2=a_{22}e_2$ and
 $u\otimes x = u\otimes x_1\otimes x_2$.
 From \ref{def:gsop}
 \[
T_G(u\otimes x) =
\sum_{\sigma\in G} M(\sigma)u\otimes  x_{\sigma^{-1}(1)} \otimes x_{\sigma^{-1}(2)}.
 \]
We have assumed that
\begin{equation}
\label{eq:parcastwo1}
T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0.
\end{equation}
If $G=\{\iota, \tau\}$, $\tau = (1,2)$, we compute directly that
 \begin{equation}
 \label{eq:simpcasex}
 T_G(u\otimes x) =
 a_{11}a_{22}\left(u\otimes e_1\otimes e_2 + M(\tau)u\otimes e_2\otimes e_1\right) +
  a_{12}a_{22}Su\otimes e_2\otimes e_2
 \end{equation}
 where
 $
S=I_m+M(\tau).
 $
 For $x=e$ (for which $A=I_2$) we get
 \begin{equation}
 \label{eq:simpcasee}
 T_G(u\otimes e) = u\otimes e_1\otimes e_2 + M(\tau)u\otimes e_2\otimes e_1.
 \end{equation}
 Let $\bG_A$ be as in remark~\ref{rem:comequcon}(3).
 From equations~\ref{eq:simpcasex} and  \ref{eq:simpcasee} we see that
 equation~\ref{eq:parcastwo1} holds (for $G=\{\iota, \tau\}$) if and only if either
\[
(1)\,\bG_A=\{\iota\}\;\;(a_{12}=0)\;\; {\rm or}  \;\;(2)\, \bG_A=\{\iota, \tau\} \;\;(a_{12}\neq 0)\;\;{\rm and}\;\;Su= 0.
\]
SInce $S=I_m+M(\tau)$,  we have $(Su,u)=0$ if and only if $(M(\tau)u, u) = -1$.
 In the trivial case where $G=\{\iota\}$, equation~\ref{eq:parcastwo1} holds if and only if
 $\bG_A=\{\iota\}$ (i.e., $a_{12}=0$):
 \begin{equation}
 \label{eq:simpcasex2}
 T_G(u\otimes x) =
 a_{11}a_{22}u\otimes e_1\otimes e_2 +  a_{12}a_{22}u\otimes e_2\otimes e_2
 \end{equation}
 and
  \begin{equation}
 \label{eq:simpcasee2}
 T_G(u\otimes e) = u\otimes e_1\otimes e_2.
 \end{equation}
 To summarize, in the $n=2$ case, equation~\ref{eq:parcastwo1} holds if and only if
 $\bG_A$ is contained in $G$ and $(M(\sigma)u,u)=\epsilon(\sigma)$ for $\sigma\in \bG_A$.\\
  \end{remark}

\begin{definition}[\bfseries Compatible permutations]
\label{def:comper}
Let $\alpha\in \Gamma_n$.  A permutation $\sigma\in S_n$ will be
$\alpha$-compatible if $\alpha(\sigma(i))\geq i$ for all $i\in \underline{n}$.
Let
\[
S_n^\alpha = \{\sigma\mid \sigma\in S_n, \alpha(\sigma(i))\geq i, i\in\underline{n}\}
\]
denote the set of all $\alpha$-compatible  permutations.\\
\end{definition}

\begin{definition}[\bfseries Restricted $\alpha$-compatible permutations]
\label{def:resalpcomper}
$T_G(u\otimes x)$ has
upper triangular matrix $A=(a_{ij})=((x_i,e_j))$ associated with $x=x_1\otimes \cdots \otimes x_n$
and the basis $\{e_i\mid i=1, \ldots n\}.$
Define
\begin{equation}
\label{eq:resalpcomper}
S_n^\alpha (A)\equiv S_n^\alpha (x,e)=
\{\sigma \mid \sigma\in S_n^\alpha,\; \prod_{i=1}^n a_{i\alpha\sigma(i)}\neq 0\}.
\end{equation}
We call $S_n^\alpha (A)\equiv S_n^\alpha (x,e)$ the $\alpha$-compatible permutations restricted by $A$
or, alternatively,
the $\alpha$-compatible permutations restricted by
$x=x_1\otimes\cdots\otimes x_n$ and the orthonormal basis
$e_i$, $1\leq i\leq  n$.\\
\end{definition}

\begin{remark}[\bfseries Inner products and $\alpha$-compatible permutations]
\label{rem:innprocomper}
Let $u\in U,\, \|u\|=1.$
Let $T_G(u\otimes x)$ be as in definition~\ref{def:resalpcomper}.
Then
\begin{equation}
\left(T_G(u\otimes x), u\otimes e_\alpha\right)=
\sum_{\sigma\in G} \left(M(\sigma)u,u\right)\prod_{i=1}^n \left(x_i, e_{\alpha\sigma(i)}\right).
\end{equation}
Note that $A$ upper triangular implies that
$\prod_{i=1}^n \left(x_i, e_{\alpha\sigma(i)}\right) = 0$
if $\sigma\not\in S_n^\alpha$ (definition~\ref{def:comper}).
In fact, $\prod_{i=1}^n \left(x_i, e_{\alpha\sigma(i)}\right) = 0$ if
$\sigma\not\in S_n^\alpha(x,e)\subseteq S_n^\alpha $
(definition~\ref{def:resalpcomper}) and we have
\begin{equation}
\label{eq:spiforinnpro}
\left(T_G(u\otimes x), u\otimes e_\alpha\right)=
\sum_{\sigma\in G\cap S_n^\alpha (x,e)} \left(M(\sigma)u,u\right)
\prod_{i=1}^n \left(x_i, e_{\alpha\sigma(i)}\right).\\
\end{equation}
\end{remark}

 We now discuss the general case of equality as it relates to collinearity:
\begin{equation}
\label{eq:gencascol}
T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0
\end{equation}
The proofs will follow the ideas and notation developed in remark~\ref{rem:case2}.

\begin{remark}[\bfseries Terminology and notation]
\label{rem:ternot}
If $K\subseteq \underline{n}$ then $S_K={\rm PER}(K)$ denotes all permutations of $K$.
Thus, $S_n=S_{\underline{n}}$.
The set of all functions from $K$ to $K$ is denoted by
$\Gamma_K$.
Thus, $\Gamma_n = \Gamma_{\underline{n}}$.
For $p\in \underline{n}$.
\[
\Gamma_{n,p} = \{\gamma\mid \gamma\in \Gamma_n, \gamma(p)=p,
\gamma(i)\neq p\; {\rm if}\; i\neq p\}.
\]
Let $G_p=\{\sigma\mid \sigma(p)=p\}$ denote the stability subgroup of $G$ at $p$.
Let $K_p = \underline{n}\setminus \{p\} = \{1, \ldots, p-1\}\cup \{p+1, \ldots n\}$.
Let $G'_p = G_p|K_p$ be the restriction of the stability subgroup $G_p$ to $K_p$.
For $\gamma\in \Gamma_{n,p}$ let $\gamma'_p = \gamma | K_p$ and note that
$\Gamma_{K_p}=\{\gamma'_p\mid \gamma \in \Gamma_{n,p}\}$.
Also, $|\Gamma_{n,p}|=|\Gamma_{K_p}|={(n-1)}^{n-1}$, the map $\gamma\mapsto \gamma'_p$
providing the canonical bijection.

If the tensor $x=x_1\otimes\ldots\otimes x_n$ has associated matrix $A=(a_{ij})=((x_i, e_j))$, then
define the tensor $x^p$ and vectors $x^p_i$, $i\in K_p$,
to have associated  matrix $A(p\mid p)$ in the same sense.
Here we use the standard notation which defines $A(p\mid p)$ to be the submatrix of $A$ gotten by deleting  row $p$ and column  $p$ from $A$.
Thus,
\[
x^p=x^p_1\otimes \cdots\otimes x^p_{p-1}\otimes x^p_{p+1} \otimes\cdots \otimes x^p_n.
\]
Let $T_G=\sum_{\sigma\in G} M(\sigma)\otimes P(\sigma)$ be a generalized
symmetry operator on $U\otimes [\otimes^n V]$ where $\dim(V)=n$ and $\dim(U)=m$.
Note that $T_G$ has {\em degree} $m$ and {\em order} $n$
(definition~\ref{def:gsop}).
$V$ has orthonormal basis $\{e_i\mid 1\leq i\leq n\}$ and $\otimes^n V$ has
orthonormal basis $\{e_\alpha \mid \alpha \in \Gamma_n\}$ where
$e_\alpha = e_{\alpha(1)} \otimes \cdots \otimes e_{\alpha(n)}$.
Let $T_{G'_p} = \sum_{\sigma\in G'_p} M(\sigma)\otimes P(\sigma)$ act on $U\otimes [\otimes^{n-1} V'_p]$.
$V'_p$ has orthonormal basis $\{e_i\mid i\in K_p\}$ and
$\otimes^{n-1} V'_p$  has orthonormal basis
$\{e_{\gamma'_p} \mid {\gamma'_p} \in \Gamma_{K_p}\}$ where
$e_{\gamma'_p} = \otimes_{i\in K_p}e_{\gamma'_p(i)}$.
$T_{G'_p}$ has degree $m$ and order $n-1$.\\
\end{remark}

\begin{remark}[\bfseries Maximal row spike functions]
\label{rem:jump}
In the following discussion we will use the  notions of
{\em spike functions} (definition~\ref{def:spifun}),
 {\em maximal row spike  functions},  $\alpha_{rc}$ (definition~\ref{def:maxrowspifun}),
 and the characterization of permutations compatible with maximal row spike functions
 (lemma~\ref{lem:rescomperspifun}).
The statements of these two definitions and the lemma are all that will be needed to understand the proofs that follow.
 In fact, lemma~\ref{lem:rescomperspifun} can easily be proved directly from these two definitions.  This material has been included in a separate section in order to state and prove lemma~\ref{lem:spifuncomper} which is the natural combinatorial setting for this discussion.
\end{remark}
\begin{lemma}[\bfseries If columns $A^{(c)}$, $c>1$, have $r<c$ with $a_{rc}\neq 0$]
\label{lem:allcolspifnc}
Choose $u\in U$, $\|u\|=1$. Suppose, for some $k\neq 0$, $T_G(u\otimes x) = k T_G(u\otimes e)$, and, for all $1<c\leq n$,
column $A^{(c)}$ has some $r<c$ with $a_{rc}\neq 0$. Then
$\bG_A=G=S_n$ and $(M(\sigma)u,u)=\epsilon(\sigma)$ for all $\sigma$.

\begin{proof}
The condition on the columns $A^{(c)}$ implies that every column ($c>1$) has a {\em maximum row spike function}
 $\alpha_{rc}$ (definition~\ref{def:maxrowspifun}).
Consider the equation
\begin{equation}
\label{eq:spiforide}
\left(T_G(u\otimes x), u\otimes e_{\alpha_{rc}}\right) =
k \left(T_G(u\otimes e), u\otimes e_{\alpha_{rc}}\right).
\end{equation}
By equation~\ref{eq:spiforinnpro},
\begin{equation}
\label{eq:spiforinnpro2}
 \left(T_G(u\otimes x), u\otimes e_{\alpha_{rc}}\right)=
\sum_{\sigma\in G\cap S_n^{\alpha_{rc}} (x,e)} \left(M(\sigma)u,u\right)\prod_{i=1}^n \left(x_i, e_{\alpha_{rc}\sigma(i)}\right).
\end{equation}
By lemma~\ref{lem:rescomperspifun}, $S_n^{\alpha_{rc}} (x,e) = \{\iota, \tau\}$ where
$\iota$ is the identity permutation and $\tau=(r,c)$ is a transposition.
Thus, $\left(T_G(u\otimes x), u\otimes e_{\alpha_{rc}}\right)=$
\begin{equation}
\label{eq:spiforinnpro3}
(I_mu,u)\left(\prod_{i\neq r} (x_i,e_i)\right)(x_r,e_c)\;\;{\rm if}\;\;\tau\not\in G\;\;{\rm and}
\end{equation}
\begin{equation}
\label{eq:spiforinnpro4}
\left[(I_mu,u)+(M(\tau)u,u)\right]\left(\prod_{i\neq r} (x_i,e_i)\right)(x_r,e_c)\;\;{\rm if}\;\;\tau\in G.
\end{equation}
Note that by the definition of the maximal row spike function,  $\alpha_{rc}$, we have
$(x_r,e_c)=a_{rc}\neq 0$.
However, $\left(T_G(u\otimes e), u\otimes e_{\alpha_{rc}}\right)=0$.
Thus, equation~\ref{eq:spiforinnpro4}, not equation~\ref{eq:spiforinnpro3}, must hold, and
$\left[(u,u)+(M(\tau)u,u)\right] = 0$ or $(M(\tau)u,u)=\epsilon(\sigma)$.
By hypothesis, for every column, $A^{(c)}$, $c>1$, there is a maximum row spike  function
 $\alpha_{rc}$.
Thus, we have shown that for every $c$, $1<c\leq n$, there is an $r<c$ such that
$\tau=(r,c)\in \bG_A$ satisfies $\left[(u,u)+(M(\tau)u,u)\right] = 0$ or $(M(\tau)u,u)=\epsilon(\sigma)$.
This set of transpositions, $\tau$, generates $S_n$ (a trivial induction).
We have proved that $(M(\tau)u,u)=\epsilon(\sigma)$ for all $\sigma\in \bG_A=S_n=G$
(note remark~\ref{rem:comequcon} here).\\
\end{proof}
\end{lemma}

\begin{lemma}[\bfseries General case where column $A^{(p)}$ has only $a_{pp}\neq0$]
\label{lem:procasgen}
Choose $u\in U$, $\|u\|=1$. Assume for $k\neq 0$, $T_G(u\otimes x) = k T_G(u\otimes e)$.
Using the terminology of remark~\ref{rem:ternot}, assume that $x$ is associated with
 the upper triangular and nonsingular matrix $A=(a_{ij})$.
Assume for some $p$, $1<p\leq n$, the only nonzero entry in column $A^{(p)}$ is the diagonal entry $a_{pp}$.
Then, for any $\gamma\in \Gamma_{n,p}$,
\begin{equation}
 \label{eq:procasgen}
 \left(T_G(u\otimes x), u\otimes e_\gamma\right)=
 a_{pp}\left(T_{G'_p}(u\otimes x^p), u\otimes e_{\gamma'_p}\right).
 \end{equation}

 \begin{proof}
 Consider $u\otimes e_\gamma$, $\gamma\in \Gamma_{n,p}$.
Then,
\begin{equation}
 \label{eq:procasn1}
 \left(T_G(u\otimes x), u\otimes e_\gamma\right)=
 \sum_{\sigma\in G}(M(\sigma)u,u)\prod_{i=1}^n(x_{\sigma^{-1}(i)}, e_{\gamma(i)})
 \end{equation}
 where
 \[
 \prod_{i=1}^n(x_{\sigma^{-1}(i)}, e_{\gamma(i)}) =
 (x_{\sigma^{-1}(1)}, e_{\gamma(1)})\cdots (x_{\sigma^{-1}(p)}, e_{\gamma(p)})\cdots (x_{\sigma^{-1}(n)}, e_{\gamma(n)}).
 \]
 Note that $(x_{\sigma^{-1}(p)}, e_{\gamma(p)})=(x_{\sigma^{-1}(p)}, e_p)\neq 0$
 requires $\sigma^{-1}(p)=p$.
 Thus, $(x_{\sigma^{-1}(p)}, e_p)=(x_p, e_p)=a_{pp}\neq 0.$
 Hence the sum in equation~\ref{eq:procasn1} can be taken over $G_p$,
 the  stabilizer subgroup of $G$ at $p$.
 We have
 \begin{equation}
  \label{eq:procasn3}
 \left(T_G(u\otimes x), u\otimes e_\gamma\right)=
 a_{pp}\sum_{\sigma\in G_p}(M(\sigma)u,u)\prod_{i\neq p} (x_{\sigma^{-1}(i)}, e_{\gamma(i)}).
 \end{equation}

As in remark~\ref{rem:ternot}, let $G'_p$ be $G_p$ restricted to $K_p$, and
 let $x^p=x^p_1\otimes \ldots x^p_{p-1}\otimes x^p_{p+1} \otimes x^p_n$
be the tensor that has matrix $A(p\mid p)$ with respect to the orthonormal basis
$\{e_i\mid i\in K_p\}$.
Let  $\gamma'_p$ denote $\gamma$ restricted to $K_p$.

 The unitary representation $M$ of $G$ restricts in the obvious way to $G_p$ and ${G'}_p$.
 Let $V'_p = \langle e_i\mid i\in K_p\rangle$.
 The generalized symmetry operator
 $T_{G'_p} = \sum_{\sigma\in G'_p}M(\sigma)\otimes P(\sigma)$
 acting on  $U\otimes [\otimes V'_p]$ allows us to reformulate equation~\ref{eq:procasn3}
 as follows:
 \begin{equation}
 \label{eq:procasgen2}
 \left(T_G(u\otimes x), u\otimes e_\gamma\right)=
 a_{pp}\left(T_{G'_p}(u\otimes x^p), u\otimes e_{\gamma'_p}\right).\\
\end{equation}
\end{proof}
\end{lemma}

\begin{remark}[\bfseries Inductive step using equation~\ref{eq:procasgen}]
\label{rem:sumprocasgen}
From  the terminology of remark~\ref{rem:ternot} we note that
$T_{G'_p} = \sum_{\sigma\in G'_p} M(\sigma)\otimes P(\sigma)$ acts on $U\otimes [\otimes^{n-1} V'_p]$.
$V'_p$ has orthonormal basis $\{e_i\mid i\in K_p\}$ and
$\otimes^{n-1} V'_p$  has orthonormal basis
$\{e_{\gamma'_p} \mid {\gamma'_p} \in \Gamma_{K_p}\}$ where
$e_{\gamma'_p} = \otimes_{i\in K_p}e_{\gamma'_p(i)}$.
Thus, $T_{G'_p}$ has degree $m$ and order $n-1$ and will be used in the inductive step (on $n$) in the next theorem.
\end{remark}

\begin{theorem}
\label{thm:mainresult}
Let $u\in U$, $\|u\|=1$.
Let $A=(a_{ij})=((x_i,e_j))$
be the upper triangular nonsingular matrix associated with
$x=x_1\otimes \cdots \otimes x_n$  and basis $e_i$, $1\leq i\leq n$.
If $\,T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0$, then  $\bG_A$ is a
subgroup of $G$ and $(M(\sigma)u,u)=\epsilon(\sigma)$ for all $\sigma\in \bG_A$.
\begin{proof}
If for all $c$, $1<c\leq n$, column $A^{(c)}$ has some $r<c$ with $a_{rc}\neq 0$
then the result follows from lemma~\ref{lem:allcolspifnc}.
Otherwise, for some $p$, $1<p\leq n$, the only nonzero entry in column $A^{(p)}$ is the diagonal entry $a_{pp}$.
We apply lemma~\ref{lem:procasgen}, in particular
 \begin{equation}
 \label{eq:mainresult1}
 \left(T_G(u\otimes x), u\otimes e_\gamma\right)=
 a_{pp}\left(T_{G'_p}(u\otimes x^p), u\otimes e_{\gamma'_p}\right)
\end{equation}
and for $x=e$:
\begin{equation}
 \label{eq:mainresult2}
 \left(T_G(u\otimes e), u\otimes e_\gamma\right)=
 \left(T_{G'_p}(u\otimes e^p), u\otimes e_{\gamma'_p}\right).\\
\end{equation}
The proof is by induction on the order $n$ of $T_G$ (i.e., $n=\dim(V)$).
The base case, $n=2$, is established in  remark~\ref{rem:case2}.
Assume the theorem is true for generalized symmetry operators of degree $m$ and order $n-1$.
Note that $T_G(u\otimes x) = k T_G(u\otimes e)\,, \;k\neq 0$, implies that
$(T_G(u\otimes x), u\otimes e_\gamma)=k(T_G(u\otimes e), u\otimes e_\gamma)$
for all $\gamma\in \Gamma_{n}$.  Thus, by equations~\ref{eq:mainresult1} and \ref{eq:mainresult2}
\begin{equation}
\label{eq:mainresult3}
a_{pp}(T_{G'_p}(u\otimes x^p), u\otimes e^p_{\gamma'_p})=
k(T_{G'_p}(u\otimes e^p), u\otimes e^p_{\gamma'_p})
\end{equation}
for all $\gamma'_p\in\Gamma_{K_p}$.
But, $\{u\otimes e_{\gamma'_p} \mid {\gamma'_p} \in \Gamma_{K_p}\}$
spans $U\otimes [\otimes^{n-1}V_p]$ (\ref{rem:ternot}).
Thus,
\begin{equation}
\label{eq:mainresult4}
(T_{G'_p}(u\otimes x^p))= k_p(T_{G'_p}(u\otimes e^p)),\;\;k_p=k/a_{pp}.
\end{equation}
$T_{G'_p}$ has degree $m$ and order $n-1$.
Likewise,
\begin{equation}
\label{eq:mainresult5}
T_{G'_1}(u\otimes x^1) = k_1 T_{G'_1}(u\otimes e^1),\;\;k_1=k/a_{11}.
\end{equation}
$T_{G'_1}$ has degree $m$ and order $n-1$.
The tensor $x^1$ is associated with the
upper triangular nonsingular submatrix $B=A(1|1)$.
By \ref{eq:mainresult5} and the induction hypothesis, $\bG_B$ is a subgroup of $G'_1$, the stabilizer subgroup of $G$ at $1$ restricted to $K_1$, and $(M(\sigma)u,u)=\epsilon(\sigma)$ for $\sigma\in G_B$.
Let $C=A(p|p)$. $C$ is also upper triangular and nonsingular.
By \ref{eq:mainresult4} and the induction hypothesis, $\bG_C$ is a subgroup of $G'_p$, the stabilizer subgroup of $G$ at $p$ restricted to $K_p$, and $(M(\sigma)u,u)=\epsilon(\sigma)$ for $\sigma\in \bG_C$.
Observe that  the union of the set of generating transpositions
$\{(i,j)\mid i<j,\, B(i,j)\neq 0\}$,
together with the set $\{(i,j)\mid i<j,\, C(i,j)\neq 0\}$, equals the set
$\{(i,j)\mid i<j,\, A(i,j)\neq 0\}$.  Here we use the fact that $a_{1p}= 0$.
Thus, $\bG_A$ is a subgroup of $G$ and $(M(\sigma)u,u)=\epsilon(\sigma)$ for all $\sigma\in G_A$.
We use remark~\ref{rem:comequcon}(2) here.
\end{proof}
\end{theorem}

\section{Remarks about compatible permutations}

We combine here some interesting combinatorial lemmas about compatible permutations.
We have used only  lemma~\ref{lem:rescomperspifun} which can be proved independently as an exercise.

\begin{definition}[\bfseries Spike functions]
\label{def:spifun}
Let $r,c\in{\underline{n},\;r<c}$.
The $r,c$~{\em spike function}, $\alpha_{rc}$, is defined by
\[
\alpha_{rc}(r) =c\;{\rm and}\; \alpha_{rc}(i) =i,\; i\neq r.
\]\\
\end{definition}

The following lemma characterizes the spike function compatible permutations
(\cite{gw:tch}, Lemma 3.3, p. 339).\\

\begin{lemma}[\bfseries Spike function compatible permutations]
\label{lem:spifuncomper}
Let $\alpha_{rc}$ be a spike function (\ref{def:spifun}).
The set of all $\alpha_{rc}$-compatable permutations, $S_n^{\alpha_{rc}}$, consists of the identity permutation, $\iota$, together with all $\sigma$ which have only one cycle of length greater than one.
Moreover, that cycle can be written in the form  $(r_1,\ldots , r_p)$ where
$r=r_1 < \cdots < r_p\leq c$.
Thus, $|S_n^{\alpha_{rc}}|=2^{c-r}$.

\begin{proof}
Let $(r_1,\ldots , r_p)$ be a nontrivial cycle of $\sigma$.
Assume without loss of generality that $r_1 = \min\{r_1,\ldots , r_p\}$.
Suppose $r_1\in \{1, \ldots, r-1, r+1, \ldots n\}$ (i.e., $r_1\neq r$).
Then $\alpha_{rc}(\sigma(r_p)) = \alpha_{rc}(r_1) = r_1 < r_p$ which
contradicts the assumption $\sigma\in S_n^{\alpha_{rc}}$.
Thus, $r_1=r$ which implies there is only one nontrivial cycle of $\sigma$.
If $r_p>c$, then
$\alpha_{rc}(\sigma(r_p)) = \alpha_{rc}(r) = c < r_p$, again contradicting
$\sigma\in S_n^{\alpha_{rc}}$.
Thus, $r_p\leq c$ and we need only show the strictly increasing property:
$r=r_1 < \cdots < r_p$.  By minimality of $r_1$, we have $r_1< r_2$.
Suppose we have $r_1< r_2 < \cdots < r_{t-1} > r_t$ for some $2< t \leq p$.
Then $\alpha_{rc}(\sigma(r_{t-1})) = \alpha_{rc}(r_t) = r_t < r_{t-1}$
which contradicts the assumption $\sigma\in S_n^{\alpha_{rc}}$.
Thus the unique nontrivial cycle of $\sigma$ is of the form
$(r_1,\ldots , r_p)$ which we may assume satisfies $r=r_1 < \cdots < r_p\leq c$.
These cycles can be constructed by choosing a nonempty subset of
$\{r+1, \ldots, c\}$.  This can be done in $2^{c-r} - 1$ ways.  Counting the
identity permutation, this gives $|S_n^{\alpha_{rc}}|=2^{c-r}$.\\
\end{proof}
\end{lemma}

\begin{definition}[\bfseries Maximal row spike functions]
\label{def:maxrowspifun}
We use the terminology of remark~\ref{rem:innprocomper}.
Let $A=(a_{ij}) = ((x_i, e_j))$ be the upper triangular nonsingular matrix associated with  $x_1\otimes\dots\otimes x_n$ and basis $e_1, \ldots, e_n$.
Suppose column $A^{(c)}$ has at least two nonzero entries.
Let $r=\max\{i\mid i<c, a_{ic} \neq 0\}$.
We call the spike  function $\alpha_{rc}$ the {\em maximal row} spike function for column $c$.  If  $\{i\mid i<c, a_{ic} \neq 0\}$ is empty, we say column $c$ has no maximal row spike function.
\end{definition}

We next characterize the sets $S_n^\alpha(x,e)$ for maximal row spike functions
$\alpha = \alpha_{rc}$.\\

\begin{lemma}[\bfseries Permutations compatible with maximal row spike functions]
\label{lem:rescomperspifun}
We use definition~\ref{def:maxrowspifun} and the terminology of
remark~\ref{rem:innprocomper}.
Let $A=(a_{ij})=((x_i,e_j))$ be the upper triangular matrix associated with $x=x_1\otimes \cdots \otimes x_n$
and the orthonormal basis $\{e_i\mid i=1, \ldots n\}.$
Let $\alpha_{rc}$ be a maximal row spike function for column $A^{(c)}$.
Then
\begin{equation}
S_n^{\alpha_{rc}}(x,e) = \{\iota, \tau\}
\end{equation}
where $\iota$ is the identity and $\tau=(r,c)$.

\begin{proof}
Note that $S_n^\alpha(x,e)\subseteq S_n^{\alpha_{rc}}$.
By lemma~\ref{lem:spifuncomper}, $\sigma\in S_n^{\alpha_{rc}}$ implies
$\sigma=(r_1,\ldots , r_p)$ where  $r=r_1 < \cdots < r_p\leq c$.
If $r_p<c$ then
\[
A(r_p, \alpha_{rc}(\sigma(r_p)) = A(r_p, \alpha_{rc}(r)) = A(r_p, c) = 0
\]
 by maximality of $r$.
Thus, by equation~\ref{eq:resalpcomper}, $\sigma\not\in S_n^{\alpha_{rc}}(x,e).$
Thus, $r_p=c$.
If $p>2$, we have $1< r_{p-1} < c=r_p$.
\[
A(r_{p-1}, \alpha_{rc}(\sigma(r_{p-1})) = A(r_{p-1}, \alpha_{rc}(c)) = A(r_{p-1}, c) = 0
\]
by the maximality of $r$.
Again by equation~\ref{eq:resalpcomper}, $\sigma\not\in S_n^{\alpha_{rc}}(x,e).$
Thus, $p=2$.
We have shown that  $S_n^{\alpha_{rc}}(x,e) = \{\iota, \tau\}$ where
$\iota$ is the identity and $\tau=(r,c)$.\\
\end{proof}

\end{lemma}
\section{Character form and examples of Schur's Theorem}
The following definition and terminology will be useful:
\begin{definition} [\bfseries Schur generalized matrix function]
\label{def:sgmf}
Let $G$ be a subgroup of the symmetric group $S_n$ of degree $n$.
Let $K=(k_{ij})$ (alternatively, $k_{ij}\equiv K(i,j)$) be an $n\times n$ matrix with complex entries.
Let $M$ be a representation of $G$ as unitary linear operators on a unitary space $U$,
$\dim U=m$.
Then the {\em Schur generalized matrix function}, $M_K$, is defined as follows:
\begin{equation}
\label{eq:stmf}
M_K = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n K(\sigma^{-1}(i), i) =
\sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n K(i,\sigma(i)).
\end{equation}
\end{definition}
\hspace*{1in}\\

\begin{remark}[\bfseries Character or trace form of Schur's theorem]
\label{rem:cfst}
We use the notation of theorem~\ref{thm:stf}.
$M_H$ is positive definite Hermitian.
Let $\mathcal{E}_{M_H}$ denote the multiset of eigenvalues of $M_H$ (i.e., eigenvalues with multiplicities).
We have $\lambda >0$ for all $\lambda\in\mathcal{E}_{M_H}$ and
\begin{equation}
\label{eq:trcform}
\sum_{\lambda\in\mathcal{E}_{M_H}} \lambda = {\rm Tr}(M_H) =
\sum_{\sigma\in G}{\rm Tr}(M(\sigma))\prod_{i=1}^n h_{i\sigma(i)}.
\end{equation}
Let $u_{\rm min}$ be a unit eigenvector corresponding to $\lambda_{\rm min}$,
the minimum eigenvalue in $\mathcal{E}_{M_H}$.
From theorem~\ref{thm:stf} we have
$\det(H)\leq (M_Hu_{\rm min},u_{\rm min})=\lambda_{\rm min}$.
From equation~\ref{eq:trcform}, $m\det(H)\leq m\lambda_{\rm min}\leq {\rm Tr}(M_H)$.
Thus, $m\det(H) = {\rm Tr}(M_H)$ if and only if
$\mathcal{E}_{M_H}=\{\det(H), \ldots, \det(H)\}$ (i.e., $\det(H)$ has multiplicity $m$).
In this case, $M_H=\det(H)I_m$, $I_m$ the identity.
We have shown
\begin{equation}
\label{eq:trcform2}
m\det(H)\leq {\rm Tr}(M_H) \; {\rm and}\;
\end{equation}
\[
m\det(H) = {\rm Tr}(M_H) \iff M_H=\det(H)I_m.
\]
As an example, take $G=S_3$, the symmetric group on $\{1,2,3\}$, and take
\[
M(\sigma)=
\left(
\begin{array}{cc}
\epsilon(\sigma)&0\\
0&1
\end{array}
\right).
\]
Let $H$ be a $3\times 3$ positive definite Hermitian matrix such that the
permanent, ${\rm per}(H) > \det(H)$.  Then
\[
M_H=
\left(
\begin{array}{cc}

\det(H)&0\\
0&{\rm per}(H)
\end{array}
\right).
\]
In this case,
\[
(M_Hu_{\rm min},u_{\rm min})=\lambda_{\rm min} = \det(H).
\]
Note that the equality condition of \ref{eq:trcform2} does not hold:
$2 \det(H) < {\rm Tr}(M_H) = \det(H) + {\rm per}(H)$.
To compare this with the equality condition~\ref{eq:seq}, note that  $\bG_H\subseteq S_3$ holds trivially.
Let $u=(u_1,u_2)\in U$ have $\|u\|=1$.
Condition~\ref{eq:seq} states that $(M_Hu,u) = \det(H)$ if and only if
$(M(\sigma)u,u)=\epsilon(\sigma)$ for all $\sigma\in \bG_H$.
In this case these conditions hold if and only if $u=(e^{ir},0)$.
Note that ${\rm per}(H) > \det(H)$ implies $H$ has at least one off diagonal element.

\end{remark}

\begin{remark}[\bfseries Examples of Schur's theorem]
\label{rem:exscth}
We consider an example of $M_H = \sum_{\sigma\in G} M(\sigma) \prod_{i=1}^n h_{i\sigma(i)}$.
Take $G=S_3$, the symmetric group on $\{1,2,3\}$.
Take the unitary representation to be the following:
\[
M(e)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\;\;
M((123))=\left(\begin{array}{cc}\frac{-1}{2}&\frac{+\sqrt{3}}{2}\\\frac{-\sqrt{3}}{2}&\frac{-1}{2}\end{array}\right)\;\;
M((132))=\left(\begin{array}{cc}\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\\frac{+\sqrt{3}}{2}&\frac{-1}{2}\end{array}\right)\;\;
\]
\[
M(23)=\left(\begin{array}{cc}\frac{-1}{2}&\frac{+\sqrt{3}}{2}\\\frac{+\sqrt{3}}{2}&\frac{1}{2}\end{array}\right)\;\;
M((13))=\left(\begin{array}{cc}\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\\frac{-\sqrt{3}}{2}&\frac{1}{2}\end{array}\right)\;\;
M((12))=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\;\;
\]

The Schur generalized matrix function, $M_H$ is
\[
M_H=
\left(\begin{array}{cc}1&0\\0&1\end{array}\right)
h_{11}h_{22}h_{33}\;\;+\;\;
\left(\begin{array}{cc}\frac{-1}{2}&\frac{+\sqrt{3}}{2}\\\frac{-\sqrt{3}}{2}&\frac{-1}{2}\end{array}\right)
h_{12}h_{23}h_{31}\;\;+\;\;
\left(\begin{array}{cc}\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\\frac{+\sqrt{3}}{2}&\frac{-1}{2}\end{array}\right)
h_{13}h_{21}h_{32}\;\;+\;\;
\]
\[
\left(\begin{array}{cc}\frac{-1}{2}&\frac{+\sqrt{3}}{2}\\\frac{+\sqrt{3}}{2}&\frac{1}{2}\end{array}\right)
h_{11}h_{23}h_{32}\;\;+\;\;
\left(\begin{array}{cc}\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\\frac{-\sqrt{3}}{2}&\frac{1}{2}\end{array}\right)
h_{13}h_{22}h_{31}\;\;+\;\;
\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)
h_{12}h_{21}h_{33}.
\]
Let
\[
H=
\left(
\begin{array}{ccc}
1&0&0\\
0&3&i\\
0&-i&1
\end{array}
\right)
\;\;\;\;{\rm which\;\; gives\;\;}\;\;
M_H =
\begin{pmatrix}

\frac{5}{2}&\frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2}&\frac{7}{2}
\end{pmatrix}

\]
The eigenvalues $\mathcal{E}_{M_H}=\{2, 4\}$, $\det(H)=2<{\rm Trace}(M_H)=6$.
Also,
\[
(M_Hu_{\rm min}, u_{\rm min}) = \lambda_{\rm min} = 2 = \det(H)
\]
By theorem~\ref{thm:stf}, equation~\ref{eq:seq},
$\det(H) = (M_Hu_{\rm min},u_{\rm min})$ implies that
\[
\bG_H\subseteq G\,\, {\rm and}\,\,(M(\sigma)u_{\rm min},u_{\rm min})=
\epsilon(\sigma),\, \sigma\in \bG_H.
\]
By definition (in theorem~\ref{thm:stf}), $\bG_H$ is the group generated by
$\{e, (2,3)\}$.
We have, trivially, $(M(e)u_{\rm min},u_{\rm min})=1$.
To compute $(M(2,3)u_{\rm min},u_{\rm min})$ we evaluate the minimum eigenvector
$u_{\rm min}$ for $M_H$ and get $u_{\rm min} = (\frac{-\sqrt{3}}{2}, \frac{1}{2})$
and compute $M((23))u_{\rm min}$ where
\[
M((23))=\left(\begin{array}{cc}\frac{-1}{2}&\frac{+\sqrt{3}}{2}\\\frac{+\sqrt{3}}{2}&\frac{1}{2}\end{array}\right)
\]
getting $-u_{\rm min}$ so $(M((23))u_{\rm min}, u_{\rm min})=-1=\epsilon((23))$ as required by the equality condition~\ref{eq:seq}.\\
\end{remark}

{\bf Acknowledgment:}  The author wishes to thank Dr. Tony Trojanowski for many helpful observations and specific suggestions that improved this paper.\\

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\setlength{\baselineskip}{12pt}

\title{Sequentially Cohen-Macaulay Rees algebras}
\author{Naoki Taniguchi}
\address{Department of Mathematics, School of Science and Technology, Meiji University, 1-1-1 Higashi-mita, Tama-ku, Kawasaki 214-8571, Japan}
\email{taniguti@math.meiji.ac.jp}
\urladdr{http://www.isc.meiji.ac.jp/~taniguci/}

\author{Tran Thi Phuong}
\address{Department of Mathematics, School of Science and Technology, Meiji University, 1-1-1 Higashi-mita, Tama-ku, Kawasaki 214-8571, Japan}
\email{sugarphuong@gmail.com}

\author{Nguyen Thi Dung}
\address{Thai Nguyen University of Agriculture and Forestry, Thai Nguyen, Vietnam}
\email{xsdung050764@gmail.com}

\author{Tran Nguyen An}
\address{Thai Nguyen University of Pedagogical, Thai Nguyen, Vietnam}
\email{antrannguyen@gmail.com}

\thanks{2010 {\em Mathematics Subject Classification.}13A30, 13D45, 13E05, 13H10}
\thanks{{\em Key words and phrases:} Dimension filtration, Sequentially Cohen-Macaulay module, Rees module}

\thanks{The first author was partially supported by Grant-in-Aid for JSPS Fellows 26-126 and by JSPS Research Fellow. The second author was partially supported by JSPS KAKENHI 26400054. The third and the fourth author were partially supported by a Grant of Vietnam Institute for Advanced Study in Mathematics (VIASM) and Vietnam National Foundation for Science and Technology Development (NAFOSTED). }

\begin{abstract}
This paper studies the question of when the Rees algebras associated to arbitrary filtration of ideals are sequentially Cohen-Macaulay. Although this problem has been already investigated by \cite{CGT}, their situation is quite a bit of restricted, so we are eager to try the generalization of their results.
\end{abstract}

\maketitle
\tableofcontents
\section{Introduction}

The notion of sequentially Cohen-Macaulay property was originally introduced by R. P. Stanley (\cite{St}) for Stanley-Reisner algebras and then it has been furiously explored by many researchers, say D. T. Cuong, N. T. Cuong, S. Goto, P. Schenzel and others (see \cite{CC, CGT, GHS, Sch}), from the view point of not only combinatorics, but also commutative algebra. The purpose of this paper is to investigate the question of when the Rees algebras are sequentially Cohen-Macaulay, which has a previous research by \cite{CGT}. In \cite{CGT} they gave a characterization of the sequentially Cohen-Macaulay Rees algebras of $\m$-primary ideals (\cite[Theorem 5.2, Theorem 5.3]{CGT}). However their situation is not entirely satisfactory, so we are eager to analyze the case where the ideal is not necessarily $\m$-primary. More generally we want to deal with the sequentially Cohen-Macaulayness of the Rees modules since the sequentially Cohen-Macaulay property is defined for any finite modules over a Noetherian ring. Thus the main problem of this paper is when the Rees modules associated to arbitrary filtration of modules are sequentially Cohen-Macaulay.

Let $R$ be a commutative Noetherian ring, $M \neq (0)$ a finitely generated $R$-module with $d = \dim_R M < \infty$.
Then we consider a filtration
$$
\calD : D_0 :=(0) \subsetneq D_1 \subsetneq  D_2 \subsetneq  \ldots \subsetneq D_{\ell} = M
$$
of $R$-submodules of $M$, which we call {\it the dimension filtration of $M$}, if $D_{i-1}$ is the largest $R$-submodule of $D_i$ with $\dim_R D_{i-1}< \dim_R D_i$ for $1\le i \le \ell$, here $\dim_R(0) = -\infty$ for convention. We note here that our notion of dimension filtration is based on \cite{GHS} and slightly different from that of the original one given by P. Schenzel (\cite{Sch}), however let us adopt the above definition throughout this paper. Then we say that $M$ is  {\it a sequentially Cohen-Macaulay $R$-module}, if the quotient module $C_i = D_i/D_{i-1}$ of $D_i$ is a Cohen-Macaulay $R$-module for every $1 \le i \le \ell$. In particular, a Noetherian ring  $R$ is called {\it a sequentially Cohen-Macaulay ring}, if $\dim R < \infty$ and $R$ is a sequentially Cohen-Macaulay module over itself.

Let us now state our results, explaining how this paper is organized. In Section 2 we sum up the notions of the sequentially Cohen-Macaulay properties and filtrations of ideals and modules.
In Section 3 we shall give the proofs of the main results of this paper, which are stated as follows.

Suppose that $R$ is a local ring with maximal ideal $\m$. Let $\calF = \{F_n\}_{n\in \mathbb{Z}}$ be a filtration of ideals of $R$ such that $F_1\neq R$, $\mathcal M = \{M_n\}_{n\in \mathbb{Z}}$ a $\calF$-filtration of $R$-submodules of $M$. Then we put
$$
\mathcal R =\sum_{n\geq0} F_nt^n  \subseteq R[t], \ \ \mathcal R^{\prime}  =\sum_{n\in\mathbb Z} F_nt^n  \subseteq R[t,t^{-1}], \ \
\mathcal G =\mathcal R^{\prime}/t^{-1}\mathcal R^{\prime}
$$
and call them {\it the Rees algebra, the extended Rees algebra and the associated graded ring of $\calF$}, respectively. Similarly we set
$$
\mathcal R ( \calM) =\sum_{n\geq0} t^n \otimes M_n \subseteq  R[t]\otimes_R M, \ \
\mathcal R^{\prime} (\calM) = \sum_{n\in\mathbb Z}  t^n \otimes M_n \subseteq R[t,t^{-1}]\otimes_R M
$$
and
$$
\mathcal G(\calM) =\mathcal R^{\prime}(\calM)/t^{-1}\mathcal R^{\prime}(\calM)
$$
which we call {\it the Rees module, the extended Rees module and the associated graded module of $\calM$}, respectively. Here $t$ stands for an indeterminate over $R$.
We also assume that $\calR $ is a Noetherian ring and $\calR(\calM)$ is a finitely generated $\calR$-module. Set
$$
\calD_i = \{M_n\cap D_i\}_{n\in \mathbb Z}, \ \
\calC_i = \{[(M_n\cap D_i)+D_{i-1}]/D_{i-1}\}_{n\in \mathbb Z}.
$$
for every $1 \le i \le \ell$. Then $\mathcal D_i$ (resp. $\mathcal C_i$) is a $\calF$-filtration of $R$-submodules of $D_i$ (resp. $C_i$).
With this notation the main results of this paper are the following, which are the natural generalization of the results \cite[Theorem 5.2, Theorem 5.3]{CGT}.

\begin{thm}\label{1.2} The following conditions are equivalent.
\begin{enumerate}[{\rm (1)}]
\item $\mathcal R^{\prime} (\calM)$ is a sequentially Cohen-Macaulay $\mathcal R^{\prime} $-module.
\item $\mathcal G(\calM)$ is a sequentially Cohen-Macaulay $\calG$-module and $\{\mathcal G (\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $\calG(\calM)$.
\end{enumerate}
When this is the case, $M$ is a sequentially Cohen-Macaulay $R$-module.
\end{thm}

Let $\M$ be a unique graded maximal ideal of $\calR$. We set
$$
\rma(N) = \max \{n\in \Bbb Z \mid [\H^t_\M(N)]_n \neq (0) \}
$$
for a finitely generated graded $\calR$-module $N$ of dimension $t$, and call it {\it the a-invariant of $N$} (see \cite[DEFINITION (3.1.4)]{GW}). Here $\{[\H^t_\M(N)]_n\}_{n \in \Bbb Z}$ stands for the homogeneous components of the $t$-th graded local cohomology module $\H^t_\M(N)$ of $N$ with respect to $\M$.

\begin{thm}\label{1.3} Suppose that $M$ is a sequentially Cohen-Macaulay $R$-module and $F_1 \nsubseteq \p$ for every $\p \in \Ass_R M$. Then the following conditions are equivalent.
\begin{enumerate}[{\rm (1)}]
\item $\calR (\calM)$ is a sequentially Cohen-Macaulay $\calR$-module.
\item $\calG(\calM)$ is a sequentially Cohen-Macaulay $\calG$-module, $\{\calG (\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $\calG(\calM)$  and  $\rma(\calG (\calC_i))<0$ for every $1 \le i \le \ell$.
\end{enumerate}
When this is the case, $\calR'(\calM)$ is a sequentially Cohen-Macaulay $\calR^{\prime}$-module.
\end{thm}

In Section 4 we focus our attention on the case of graded rings. In the last section we will explore the application of Theorem \ref{6.7} to the Stanley-Reisner algebras of shellable complexes (Theorem \ref{7.4}).

\section{Preliminaries}
In this section we summarize some basic results on sequentially Cohen-Macaulay properties and filtration of ideals and modules, which we will use throughout this paper.

Let $R$ be a Noetherian ring, $M \ne (0)$ a finitely generated $R$-module of dimension $d$. We put
$$
\Assh_R M = \{\p \in \Supp_R M \mid \dim R/\p = d\}.
$$
For each $n \in \Bbb Z$, there exists the largest $R$-submodule $M_n$ of $M$ with $\dim_R M_n \le n$.
Let
\begin{eqnarray*}
\calS (M) &=& \{\dim_R N \mid N \ \text{is~an~}R\text{-submodule~of~}M, N \ne (0)\} \\
                 &=& \{\dim R/\p \mid \p \in \Ass_RM \}.
\end{eqnarray*}
We set $\ell = \sharp \calS (M)$ and write $\calS (M) = \{d_1< d_2 < \cdots < d_\ell = d\}$.
Let $D_i = M_{d_i}$ for each $1 \le i \le \ell$.
We then have a filtration
$$
\calD : D_0 :=(0) \subsetneq D_1 \subsetneq  D_2 \subsetneq  \ldots \subsetneq D_{\ell} = M
$$
of $R$-submodules of $M$, which we call {\it the dimension filtration of $M$}.
We put $C_i = D_i/D_{i-1}$ for every $1 \le i \le \ell$.

\begin{defn}[\cite{Sch, St}]\label{3.1}
We say that $M$ is {\it a sequentially Cohen-Macaulay $R$-module}, if $C_i$ is Cohen-Macaulay for every $1 \le i \le \ell$. The ring $R$ is called {\it a sequentially Cohen-Macaulay ring}, if $\dim R < \infty$ and $R$ is a sequentially Cohen-Macaulay module over itself.
\end{defn}

\noindent The typical examples of sequentially Cohen-Macaulay ring is the Stanley-Reisner algebra $k[\Delta]$ of a shellable complex $\Delta$ over a field $k$. Also every one-dimensional Noetherian local ring is sequentially Cohen-Macaulay. Moreover, if $M$ is a Cohen-Macaulay module over a Noetherian local ring, then $M$ is sequentially Cohen-Macaulay, and the converse holds if $M$ is unmixed.

\if0
 We now remark the characterizations of the dimension filtration.
Let
$$
(0) = \bigcap_{\p \in \Ass_R M}M(\p)
$$
be a primary decomposition of $(0)$ in $M$, where $M(\p)$ stands for the $R$-submodule of $M$ with $\Ass_R M/M(\p) = \{ \p \}$ for all $\p \in \Ass_R M$.

We begin with the following well-known facts, which play an important role in our paper.

\begin{prop}$($\cite[Proposition 2.2, Corollary 2.3]{Sch}$)$\label{3.2}
The following assertions hold true.
\begin{enumerate}[$(1)$]
\item $D_i = \bigcap_{\dim R/\p \ge d_{i+1}}M(\p)$ for $0 \le i \le \ell-1$.
\item $\Ass_R C_i = \{ \p \in \Ass_R M \mid \dim R/\p = d_i\}$, $\Ass_R D_i = \{ \p\in \Ass_R M \mid \dim R/\p \le d_i\}$ for all $1 \le i \le \ell$ and $\Ass_R M/D_i = \{ \p\in \Ass_R M \mid \dim R/\p \ge d_{i+1}\}$ for all $1 \le i \le \ell-1$.
\end{enumerate}
\end{prop}

\begin{thm}$($\cite[Theorem 2.3]{GHS}$)$\label{3.3}
Let $\mathcal M = \{M_i \}_{0 \leq i \leq t}~ (t>0)$ be a family of $R$-submodules of $M$ such that
\begin{enumerate}[{\rm (1)}]
\item $M_0 = (0) \subsetneq  M_1 \subsetneq M_2 \subsetneq \ldots \subsetneq M_t = M $ and
\item $\dim _RM_{i-1} < \dim_RM_{i}$ for all $1 \leq i \leq t$.
\end{enumerate}
Assume that $\Ass_R M_i/M_{i-1} = \Assh_R M_i/M_{i-1}$ for all $1\leq i \leq t$. Then $t=\ell$ and $M_i = D_i$ for every $0\leq i \leq \ell$.
\end{thm}

We then have the following.

\begin{cor}$($\cite[Proposition 4.3]{Sch}$)$\label{3.4}
Suppose that $R$ is a local ring. Then $M$ is a sequentially Cohen-Macaulay $R$-module if and only if $M$ admits a Cohen-Macaulay filtration, that is, a family $\mathcal M = \{M_i\}_{1\leq i\leq t}~ (t>0)$ of $R$-submodules of $M$ with
$$
M_0 = (0) \subsetneq  M_1 \subsetneq M_2 \subsetneq \ldots \subsetneq M_t = M
$$
such that
\begin{enumerate}[{\rm (1)}]
\item $\dim _R M_{i-1} < \dim_R M_{i}$ for all $1 \leq i \leq t$.
\item $M_i/M_{i-1}$ is a Cohen-Macaulay $R$-module for all $1\leq i \leq t$.
\end{enumerate}

\end{cor}

\fi

Firstly let us note the non-zerodivisor characterization of sequentially Cohen-Macaulay modules.

\begin{prop}\label{3.5}
Let $(R, \m)$ be a Noetherian local ring, $M \neq (0)$ a finitely generated $R$-module. Let $x \in \m$ be a non-zerodivisor on $M$. Then the following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $M$ is a sequentially Cohen-Macaulay $R$-module.
\item $M/xM$ is a sequentially Cohen-Macaulay $R/(x)$-module and $\{D_i/xD_i\}_{0 \le i \le \ell}$ is the dimension filtration of $M/xM$.
\end{enumerate}
\end{prop}

\begin{proof}
Notice that $x \in \m$ is a non-zerodivisor on $C_i$ and $D_i$ for all $1\le i\le \ell$ (See \cite[Corollary 2.3]{Sch}). Therefore we get a filtration
$$
D_0/xD_0 = (0) \subsetneq D_1/xD_1 \subsetneq \cdots \subsetneq D_{\ell}/xD_{\ell} = M/xM
$$
of $R/(x)$-submodules of $M/xM$. Then the assertion is a direct consequence of \cite[Theorem 2.3]{GHS}.
\end{proof}

\noindent The implication $(2) \Rightarrow (1)$ is not true without the condition that $\{D_i/xD_i\}_{0 \le i \le \ell}$ is the dimension filtration of $M/xM$. For instance, let $R$ be a $2$-dimensional Noetherian local domain of depth $1$. Then $R/(x)$ is sequentially Cohen-Macaulay for every $0 \neq x \in R$, but $R$ is not. Besides this, let $I$ be an $\m$-primary ideal in a regular local ring $(R, \m)$ of dimension $2$. Then $I$ is not a sequentially Cohen-Macaulay $R$-module, even though $I/xI$ is, where $0 \neq x \in \m$.
These examples show that \cite[Theorem 4.7]{Sch} is not true in general.

From now on, we shall quickly review some preliminaries on filtrations of ideals and modules. Let $R$ be a commutative ring, $\calF = \{F_n\}_{n \in \Bbb Z}$ a filtration of ideals of $R$, that is, $F_n$ is an ideal of $R$, $F_n \supseteq F_{n+1}$, $F_m F_n \subseteq F_{m+n}$ for all $m, n \in \Bbb Z$ and $F_0 = R$. Then we put
$$
\calR=\calR(\calF) = \sum_{n \ge 0}F_nt^n \subseteq R[t], \ \ \calR'=\calR'(\calF) = \sum_{n \in \Bbb Z}F_nt^n \subseteq R[t, t^{-1}]
$$
and call them {\it the Rees algebra, the extended Rees algebra of $R$ with respect to $\calF$}, respectively. Here $t$ stands for an indeterminate over $R$.

Let $M$ be an $R$-module, $\calM=\{M_n\}_{n \in \Bbb Z}$ an $\calF$-filtration of $R$-submodules of $M$, that is, $M_n$ is an $R$-submodule of $M$, $M_n \supseteq M_{n+1}$, $F_m M_n \subseteq M_{m+n}$ for all $m, n \in \Bbb Z$ and $M_0 = M$. We set
$$
\calR(\calM)= \sum_{n \ge 0} t^n \otimes M_n \subseteq R[t] \otimes_R M, \ \ \calR'(\calM) = \sum_{n \in \Bbb Z} t^n \otimes M_n \subseteq R[t, t^{-1}] \otimes_R M
$$
which we call  {\it the Rees module, the extended Rees module of $M$ with respect to $\calM$}, respectively, where
$$
t^n \otimes M_n = \{ t^n \otimes x \mid x \in M_n\} \subseteq R[t, t^{-1}] \otimes_R M
$$
for all $n \in \Bbb Z$. Then $\calR(\calM)$ (resp. $\calR'(\calM)$) is a graded module over $\calR$ (resp. $\calR'$).

If $F_1 \neq R$, then we define {\it the associated graded ring $\calG$ of $R$ with respect to $\calF$ and the associated graded module $\calG(\calM)$ of $M$ with respect to $\calM$} as follows.
$$
\calG = \calG(\calF) = \calR'/{u\calR'}, \ \  \calG(\calM) = \calR'(\calM)/{u\calR'(\calM)},
$$
where $u = t^{-1}$. Then $\calG(\calM)$ is a graded module over $\calG$ and the composite map
$$
\psi : \calR(\calM) \overset{i}{\longrightarrow} \calR'(\calM) \overset{\varepsilon}{\longrightarrow} \calG(\calM)
$$
is surjective and $\Ker {\psi} = u\calR'(\calM) \cap \calR(\calM) = u[\calR(\calM)]_{+}$, where $[\calR(\calM)]_{+} = \sum_{n>0}t^n \otimes M_n$.

For the rest of this section, we assume that $F_1\neq R $, $\calR = \calR(\calF)$ is Noetherian and $\calR(\calM)$ is finitely generated.
Then we have the following. The proof of Proposition \ref{2.3} is based on the results \cite[Proposition 5.1]{CGT}. Since it plays an important role in this paper, let us give a brief proof for the sake of completeness.

\begin{prop}\label{2.3}
The following assertions hold true.
\begin{enumerate}[$(1)$]
\item Let $P \in \Ass_{\calR}\calR(\calM)$. Then $\p \in \Ass_R M$, $P = \p R[t]\cap \calR$ and
\begin{eqnarray*}
\dim \calR/P = \left\{
 \begin{array}{l}
  \dim R/{\p} + 1 \ \ \ \text{if} \ \dim R/{\p} <\infty,  F_1 \nsubseteq \p, \\
  \dim R/{\p}    \ \ \ \ \ \   \ \ \text{otherwise},
 \end{array}
\right.
\end{eqnarray*}
where  $\p = P \cap R$.
\item $\p R[t] \cap \calR \in \Ass_{\calR} \calR(\calM)$ for every $\p \in \Ass_R M$.
\item Suppose that $M \neq (0)$, $d = \dim_R M < \infty$ and there exists $\p \in \Assh_R M$ such that $F_1 \nsubseteq \p$. Then $\dim_{\calR}\calR(\calM) = d + 1$.
\end{enumerate}
\end{prop}

\begin{proof}
$(1)$~~ Let $P \in \Ass_{\calR}\calR(\calM)$. Then $P \in \Ass_{\calR}R[t]\otimes_R M$, so that $P = Q \cap \calR$ for some
$$
Q \in \Ass_{R[t]}R[t]\otimes_R M = \bigcup_{\p \in \Ass_R M}\Ass_{R[t]}R[t]/{\p R[t]}.
$$
Thus there exists $\p \in \Ass_R M$ such that $\p = Q \cap R$ and $Q = \p R[t]$. Therefore $P = \p R[t] \cap \calR$, $\p = P \cap R$. Put $\overline{R} = R/{\p}$. Then $\overline{\calF}= \{F_n \overline{R}\}_{n \in \Bbb Z}$ is a filtration of ideals of $\overline{R}$ and $\calR/P \cong \calR(\overline{\calF})$ as graded $R$-algebras. Hence the assertion holds by \cite[Part II, Lemma (2.2)]{GN}.

$(2)$~~Let $\p \in \Ass_R M$. We write $\p = (0):_R x$ for some $x \in M$. Then $(0):_{\calR}\xi = \p R[t] \cap \calR $ where
$\xi = 1 \otimes x \in [\calR(\calM)]_0$.

$(3)$~~Follows from the assertions $(1)$, $(2)$.
\end{proof}

\begin{cor}
Suppose that $R$ is a local ring and $M \neq (0)$. Then
\begin{eqnarray*}
\dim_{\calR}\calR(\calM) = \left\{
 \begin{array}{l}
  \dim_R M + 1 \ \ \ \text{if} \ \ \text{there exists }\p \in \Assh_R M \text{ such that }  F_1 \nsubseteq \p, \\
  \dim_R M    \ \ \ \ \ \   \ \ \text{otherwise}.
 \end{array}
\right.
\end{eqnarray*}
\end{cor}

Similarly we are able to determine the structure of associated prime ideals of the extended Rees modules.

\begin{prop}\label{2.4}
The following assertions hold true.
\begin{enumerate}[$(1)$]
\item Let $P \in \Ass_{\calR'}\calR'(\calM)$. Then $\p \in \Ass_R M$, $P = \p R[t, t^{-1}]\cap \calR'$ and $\dim \calR/P = \dim R/{\p} + 1$, where $\p = P \cap R$.
\item $\p R[t, t^{-1}] \cap \calR' \in \Ass_{\calR'} \calR'(\calM)$ for every $\p \in \Ass_R M$.
\item Suppose that $M \neq (0)$. Then $\dim_{\calR'} \calR'(\calM) = \dim_R M + 1$.
\end{enumerate}
\end{prop}

Apply Proposition \ref{2.4}, we get the following.

\begin{cor}\label{2.5}
Suppose $R$ is a local ring and $M \neq (0)$.
Then $\dim_{\calG}\calG(\calM) = \dim_R M$.
\end{cor}

\if0
\begin{proof}
Since $R$ is local, $\calR'$ is an $H$-local ring. Let $\mathfrak{N}$ be the unique graded maximal ideal of $\calR'$. Then we get $\calR'(\calM)_{\mathfrak{N}} \neq (0)$ and $u \in \mathfrak{N}$.
Thanks to the Nakayama's lemma, $\calG(\calM)_{\mathfrak{N}} \neq (0)$, whence $\calG(\calM)\neq (0)$. Hence we have $\dim_{\calG}\calG(\calM) = \dim_R M$ by Proposition \ref{2.4}.
\end{proof}
\fi

\section{Proof of Theorem \ref{1.2} ~ and Theorem \ref{1.3}}

This section aims to prove Theorem \ref{1.2} and Theorem \ref{1.3}. In what follows, let $(R, \m)$ be a Noetherian local ring, $M \neq (0)$ a finitely generated $R$-module of dimension $d$. Let $\calF = \{F_n\}_{n\in \mathbb{Z}}$ be a filtration of ideals of $R$ with $F_1\neq R$, $\mathcal M = \{M_n\}_{n\in \mathbb{Z}}$ a $\calF$-filtration of $R$-submodules of $M$. We put $\a = \calR(\calF)_{+} = \sum_{n>0}F_nt^n$.

Throughout this section we assume that $\calR=\calR(\calF) $ is a Noetherian ring and $\calR(\calM)$ is finitely generated. Let $1 \le i \le \ell$. We set
$$
\calD_i = \{M_n\cap D_i\}_{n\in \mathbb Z}, \ \
\calC_i = \{[(M_n\cap D_i)+D_{i-1}]/D_{i-1}\}_{n\in \mathbb Z}.
$$
Then $\mathcal D_i$ (resp. $\mathcal C_i$) is a $\calF$-filtration of $R$-submodules of $D_i$ (resp. $C_i$). Look at the following exact sequence
$$
0 \to [\calD_{i-1}]_n \to [\calD_i]_n \to [\calC_i]_n \to 0
$$
of $R$-modules for all $n \in \Bbb Z$. We then have the exact sequences
$$
0 \to \calR(\calD_{i-1}) \to \calR(\calD_i) \to \calR(\calC_i) \to 0
$$
$$
0 \to \calR'(\calD_{i-1}) \to \calR'(\calD_i) \to \calR'(\calC_i) \to 0 \ \ \text{and}
$$
$$
0 \to \calG(\calD_{i-1}) \to \calG(\calD_i) \to \calG(\calC_i) \to 0
$$
of graded modules. Since $\calR(\calD_i)$ is a finitely generated $\calR$-module, so is $\calR(\calC_i)$.

\begin{lem}$($cf. \cite[Proposition 5.1]{CGT}$)$\label{5.9}
$\{\calR'(\calD_i)\}_{0\le i \le \ell}$ is the dimension filtration of $\calR'(\calM)$. If  $F_1 \nsubseteq \p$ for every $\p \in \Ass_R M$, then $\{\calR(\calD_i)\}_{0\le i \le \ell}$ is the dimension filtration of $\calR(\calM)$.
\end{lem}

\begin{proof}
Let $1 \le i \le \ell$.  Then $\dim_{\calR'}\calR'(\calD_i) = d_i + 1$, since $D_i \neq (0)$. Let $P \in \Ass_{\calR'}\calR'(\calC_i)$. Thanks to Proposition \ref{2.4}, we then have $\dim \calR'/P = d_i + 1 = \dim_{\calR'}\calR'(\calC_i)$. By using \cite[Theorem 2.3]{GHS}, $\{\calR'(\calD_i)\}_{0\le i \le \ell}$ is the dimension filtration of $\calR'(\calM)$.
Similarly we obtain the last assertion.
\end{proof}

We now ready to prove Theorem \ref{1.2}.

\begin{proof}[Proof of Theorem \ref{1.2}]
The equivalence of conditions $(1)$ and $(2)$ is similar to the proof of Proposition \ref{3.5}. Let us make sure of the last assertion. Look at the following exact sequences
$$
0 \to \calR'(\calC_i) \overset{\varphi}{\to} R[t, t^{-1}]\otimes_R C_i \to X=\Coker{\varphi}  \to 0
$$
of graded $\calR'$-modules for $1 \le i \le \ell$. Since $\calR'(\calC_i)$ is a Cohen-Macaulay $\calR'$-module and $X_u= (0)$, we have $R[t, t^{-1}]\otimes_R C_i$ is Cohen-Macaulay. Therefore $M$ is a sequentially Cohen-Macaulay $R$-module, because $C_i$ is Cohen-Macaulay.
\end{proof}

From now on, we focus our attention on the proof of Theorem \ref{1.3}. To do this, we need some auxiliaries.

\begin{lem}\label{5.1}
Let $P \in \Spec \calR$ such that $P \nsupseteq \a$. If $\calG(\calM)_P \neq (0)$ $($resp. $\calR(\calM)_P \neq (0)$ and $P \supseteq u\a )$, then $\calR(\calM)_P \neq (0)$ $($resp. $\calG(\calM)_P \neq (0) )$.
When this is the case, the following assertions hold true.
\begin{enumerate}[$(1)$]
\item $\calR(\calM)_P$ is a Cohen-Macaulay $\calR_P$-module if and only if $\calG(\calM)_P$ is a Cohen-Macaulay $\calG_P$-module.
\item $\dim_{\calR_P}\calR(\calM)_P = \dim_{\calR_P}\calG(\calR)_P + 1$.
\end{enumerate}
\end{lem}

\begin{proof}
Let $P \in \Spec \calR$ such that $P \nsupseteq \a$, but $P \supseteq u\a$. We choose a homogeneous element $\xi = at^n \in \a \setminus P$ where $n > 0$, $a \in F_n$.
Then we get $x=u\xi  = at^{n-1} \in P$, since $P \supseteq u\a$.
\begin{claim}\label{claim}
If $Q \in \Ass_{\calR}\calR(\calM)$ such that $Q \subseteq P$, then $x \notin Q$. Therefore $x$ is a non-zerodivisor on $\calR(\calM)_P$.
\end{claim}

\begin{proof}[Proof of Claim \ref{claim}]
We assume that there exists $Q \in \Ass_{\calR}\calR(\calM)$ such that $Q \subseteq P$, but $x \in Q$. Write $Q = (0):_{\calR} \eta$ where $\eta = t^{\ell}\otimes m \ (\ell \in \Bbb Z,~m \in M_{\ell})$. Then we have $\xi = at^n \in (0):_{\calR} \eta = Q \subseteq P$, which implies a contradiction.
\end{proof}

Since $P \nsupseteq \a$, we get $\calR_P = \calR'_P$ and $\calR(\calM)_P = \calR'(\calM)_P$. Therefore
$$
(u\a) \calR_P = (u \a)\calR'_P = u \calR'_P = x\calR'_P \ \ \text{and} \ \ (u\a)\calR(\calM) \subseteq u[\calR(\calM)]_{+}.
$$
Hence $[u\calR(\calM)_{+}]_P = x \calR'(\calM)_P = x\calR(\calM)_P$, so that
$$
\calR(\calM)_P/{x\calR(\calM)_P} \cong \calG(\calM)_P
$$
as $\calR_P$-modules.
On the other hand, let $P \in \Spec \calR$ such that $\calG(\calM)_P \neq (0)$. Then $P \supseteq u \a$, since $u \a = u\calR' \cap \calR = \Ker (\calR \overset{i}{\to} \calR' \overset{\varepsilon}{\to} \calG)$. Therefore the assertions immediately come from the above isomorphism.
\end{proof}

Here we need the following fact, which was originally given by G. Faltings.

\begin{fact}[\cite{F}]\label{5.3} Let $I$ be an ideal of $R$ and $t \in \Bbb Z$. Consider the following two conditions.
\begin{enumerate}[$(1)$]
\item There exists an integer $\ell > 0$ such that $I^{\ell}{\cdot}\H^i_{\m}(M) = (0)$ for each $i \neq t$.
\item $M_\p$ is a Cohen Macaulay $R_\p$-module and $t = \dim_{R_\p}M_\p + \dim R/\p$ for every $\p\in\Supp_RM$ but $\p \nsupseteq I$.
\end{enumerate}
Then the implication $(1) \Rightarrow (2)$ holds true. The converse holds, if $R$ is a homomorphic image of a Gorenstein local ring.
\end{fact}

Let $\M$ be a unique graded maximal ideal of $\calR$.
Although a part of the proof of Proposition \ref{5.4} is due to the result \cite{TI}, we note the brief proof for the sake of completeness.

\begin{prop}\label{5.4}
Suppose that $\H^i_\M(\calG(\calM))$ is a finitely graded $\calR$-module for all $i \neq d$. Then $\H^i_\M(\calR(\calM))$ is a finitely graded $\calR$-module for all $i \neq d+1$.
\end{prop}

\begin{proof}
Passing to the completion and taking the local duality theorem, it is enough to show that there exists an integer $\ell > 0$ such that $\a^{\ell}{\cdot}\H^i_\M(\calR(M)) = (0)$ for every $i \neq d+1$. To see this, let $P \in \Supp_\calR \calR(M)$ such that $P \nsupseteq \a$ and $P \subseteq \M$. Put $L = u \a = u\calR' \cap \calR$.

\begin{claim}\label{claim2}
$\sqrt{P^* + L} \nsupseteq \a$.
\end{claim}

\begin{proof}[Proof of Claim \ref{claim2}]
Suppose that $P^* + L \supseteq \a^{\ell}$ for some $\ell > 0$. Since $\calR/{\a^{\ell}}$ is finitely graded, we can choose an integer $s>0$ such that $[\calR/{\a^{\ell}}]_n = (0)$ for all $n \ge s$. Then
$$
\calR_n = F_nt^n \subseteq [P^*]_n + F_{n+1}t^n
$$
for all $n \ge s$. On the other hand, for each $n \ge 0$, we set
$$
I_n=\{a \in R \mid at^n \in P^*\}.
$$
Then $I_n$ is an ideal of $R$ and $I_n \subseteq F_n$ and $I_n \supseteq I_{n+1}$ for all $n \ge 0$.
Hence $F_n \subseteq I_n + F_k$ for all $n\ge s$, $k \in \Bbb Z$.
Since $\calR$ is Noetherian, we get $\calR^{(d)} = R[F_dt^d]$ for some $d>0$, so that $F_{d\ell} = (F_d)^{\ell}$ for all $\ell > 0$.
We then have
$$
F_n \subseteq \bigcap_{\ell >0} [I_n + (F_d)^{\ell}] = I_n
$$
for all $n \ge s$, whence $\calR_n \subseteq P^*$. Thus
$$
\a^s \subseteq \sum_{n \ge s}\calR_n \subseteq P^* \subseteq P.
$$
which is impossible, because $\a \nsubseteq P$.
\end{proof}

 Therefore we can take $Q \in \Min_\calR \calR/[P^* + L]$ such that $\a \nsubseteq Q \subseteq \M$. Then $\calR(\calM)_Q \neq (0)$, because $\calR(\calM)_{P^*} \neq (0)$ and $P^* \subseteq Q$. Thanks to Lemma \ref{5.1}, $\calG(\calM)_Q \neq (0)$. Then  $\calG(\calM)_Q$ is Cohen-Macaulay and $\dim_{\calR_Q}\calG(\calM)_Q + \dim \calR_{\M}/{Q\calR_{\M}} = d$ by using Fact \ref{5.3}. Hence
 $ \calR(\calM)_Q$ is Cohen-Macaulay and $\dim_{\calR_Q}\calR(\calM)_Q + \dim \calR_{\M}/{Q\calR_{\M}} = d +1$ by Lemma \ref{5.1}.

Since $P^* \subseteq Q$, $\calR(M)_{P^*}$ is Cohen-Macaulay, so is $\calR(M)_P$. We also have
\begin{align*}
d+1 &=  \dim_{\calR_Q}\calR(M)_Q + \dim \calR_{\M}/{Q\calR_{\M}}\\
    &= (\dim_{\calR_{P^*}}\calR(M)_{P^*} + \dim \calR_{Q}/{P^{*}\calR_{Q}} ) + (\dim \calR_{\M}/{P^{*}\calR_{\M}} - \dim \calR_{Q}/{P^{*}\calR_{Q}})\\
    &=  \dim_{\calR_{P^*}}\calR(M)_{P^*} + \dim \calR_{\M}/{P^{*}\calR_{\M}} \\
    &= \dim_{\calR_{P}}\calR(M)_{P} + \dim \calR_{\M}/{P\calR_{\M}}.
\end{align*}

Thanks to Fact \ref{5.3} again, there exists $\ell > 0$ such that
$$
\a^{\ell}{\cdot}\H^i_\M(\calR(\calM)) = (0) \ \ \text{for each} \ \ i \neq d+1
$$
which shows $\H^i_\M(\calR(\calM))$ is finitely graded.
\end{proof}

We set
$$
\rma(N) = \max \{n\in \Bbb Z \mid [\H^t_\M(N)]_n \neq (0) \}
$$
for a finitely generated graded $\calR$-module $N$ of dimension $t$, and call it {\it the a-invariant of $N$} (see \cite[DEFINITION (3.1.4)]{GW}). With this notation we have the following.

\begin{lem}\label{5.6}
The following assertions hold true.
\begin{enumerate}[$(1)$]
\item $[\H^{d+1}_{\M}(\calR(\calM))]_n = (0)$ for all $n\ge 0$.
\item If $[\H^{d+1}_{\M}(\calR(\calM))]_{-1} = (0)$, then $\H^{d+1}_{\M}(\calR(\calM))= (0)$.
\end{enumerate}
Consequently $\rma(\calR(\calM)) = -1$, if $\dim_{\calR}\calR(\calM) = d+1$.
\end{lem}

\begin{proof}
We look at the following exact sequences
$$
0 \to L \to \mathcal R(\calM) \to M \to 0
$$
$$
0 \to L(1) \to \mathcal R(\calM) \to \mathcal G(\calM) \to 0
$$
of graded $\calR$-modules, where $L = \calR(\calM)_+$. By applying the local cohomology functors to the above sequences, we get
$$
\H^d_\m(M) \to \H^{d+1}_\M(L) \to \H^{d+1}_\M(\calR(\calM)) \to 0
$$
and
$$
\H^d_\M(\calG(\calM)) \to \H^{d+1}_\M(L)(1) \to \H^{d+1}_\M(\calR(\calM)) \to 0.
$$
Thus
\begin{align*}
[\H^{d+1}_\M(L)]_n &\cong [\H^{d+1}_\M(\calR(\calM))]_n  \ \ \text{for} \ \ n \neq 0, \ \text{and} \\
[\H^{d+1}_\M(L)]_{n+1} &\to [\H^{d+1}_\M(\calR(\calM))]_n \to 0  \ \ \text{for} \ \ n \in \Bbb Z.
\end{align*}
Therefore $[\H^{d+1}_\M(\calR(\calM))]_n =(0)$ for $n \ge 0$, because $\H^{d+1}_\M(\calR(\calM))$ is Artinian.
Moreover we have
$$
[H^{d+1}_\M(\calR(\calM))]_{-1} \to [\H^{d+1}_\M(\calR(\calM))]_n \to 0
$$
for $n < 0$, so we get the assertion $(2)$.
\end{proof}

We finally arrive at the following Theorem \ref{5.7} which is a module version of the results \cite[Part II, Theorem (1.1)]{GN}, \cite[Theorem 1.1]{V} (see also \cite[Theorem 1.1]{TI}, \cite[Theorem (1.1)]{GS}).

\begin{thm}\label{5.7} The following conditions are equivalent.
\begin{enumerate}[{\rm (1)}]
\item $\calR(\calM)$ is a Cohen-Macaulay $\calR$-module and $\dim_{\calR}\calR(\calM) = d+1$.
\item $\H^i_\M(\calG(\calM)) = [\H^i_\M(\calG(\calM))]_{-1}$ for every $i < d$ and $\rma(\calG(\calM)) < 0$.
\end{enumerate}
When this is the case, $[\H^i_\M(\calG(\calM))]_{-1}\cong \H^i_{\m}(M)$ as $R$-modules for all $i < d$.
\end{thm}

\begin{proof}
Consider the following exact sequences
$$
(*)~~~~   \cdots \to \H^i_{\m}(L) \to \H^i_\M(\calR(\calM)) \to \H^i_\m(M) \to \H^{i+1}_\M(L) \to \H^{i+1}_\M(\calR(\calM)) \to \cdots
$$
$$
(**)~~~~\cdots \to \H^i_{\m}(L)(1) \to \H^i_\M(\calR(\calM)) \to \H^i_\M(\calG(\calM)) \to \H^{i+1}_\M(L)(1) \to \H^{i+1}_\M(\calR(\calM)) \to \cdots
$$
for each $i < d$.

Firstly we assume that $\calR(\calM)$ is a Cohen-Macaulay $\calR$-module of dimension $d+1$. Then
$$
\H^i_\m(M) \cong \H^{i+1}_\M(L) \ \ \ \text{and} \ \ \ \H^i_\M(\calG(\calM)) \cong \H^{i+1}_\M(L)(1)
$$
for $i < d$. Therefore we get $\H^i_\M(\calG(\calM)) = [\H^i_\M(\calG(\calM))]_{-1}$ and $[\H^i_\M(\calG(\calM))]_{-1}\cong \H^i_{\m}(M)$ as $R$-modules. Since $\calR(\calM)$ is Cohen-Macaulay, we have
$$
0 \to \H^d_\m(M) \to \H^{d+1}_\M(L) \to \H^{d+1}_\M(\calR(\calM)) \to 0
$$
$$
0 \to \H^d_\M(\calG(\calM)) \to \H^{d+1}_\M(L)(1).
$$
Therefore $\rma(\calG(\calM)) < 0$ by using Lemma \ref{5.6}.

Conversely, let $i<d$. Thanks to the above sequences $(*)$, $(**)$ and our hypothesis, we get
\begin{align*}
[\H^{i+1}_\M(L)]_{n+1} &\cong [\H^{i+1}_\M(\calR(\calM))]_n \\
[\H^{i+1}_\M(L)]_{n+1} &\cong [\H^{i+1}_\M(\calR(\calM))]_{n+1}
\end{align*}
for each $n \ge 0$. Hence $[\H^i_\M(\calR(\calM))]_n = (0)$ for $n \ge 0$, since $\H^{i+1}_\M(\calR(\calM))$ is Artinian.
Moreover, we then have
$$
0 \to [\H^{i+1}_\M(\calR(\calM))]_n \to [\H^{i+1}_\M(\calR(\calM))]_{n-1}.
$$
for $n < 0$ by above sequences $(*)$ and $(**)$. Thanks to Proposition \ref{5.4}, $\H^{i+1}_\M(\calR(\calM))$ is a finitely graded $\calR$-module for $i < d$.  Whence $[\H^{i+1}_\M(\calR(\calM))]_n = (0)$, which shows $\H^{i+1}_\M(\calR(\calM)) = (0)$ for all $i <d $. Hence $\calR(\calM)$ is a Cohen-Macaulay $\calR$-module of dimension $d+1$.
\end{proof}

\begin{cor}\label{5.8} \label{key} Suppose that $M$ is a Cohen-Macaulay $R$-module. Then the following conditions are equivalent.
\begin{enumerate}[{\rm (1)}]
\item $\calR(\calM)$ is a Cohen-Macaulay $\calR$-module and $\dim_{\calR}\calR(\calM) = d+1$.
\item $\calG(\calM)$ is a Cohen-Macaulay $\calG$-module and $\rma (\calG(\calM))< 0$.
\end{enumerate}
\end{cor}

We now reach the goal of this section.

\begin{proof}[Proof of Theorem \ref{1.3}]
Thanks to Lemma \ref{5.9}, $\calR(\calM)$ is a sequentially Cohen-Macaulay $\calR$-module if and only if $\calR(\calC_i)$ is Cohen-Macaulay for every $1 \le i \le \ell$. The latter condition is equivalent to saying that $\calG(\calC_i)$ is a Cohen-Macaulay $\calG$-module and $\rma(\calG(\calC_i)) < 0$ for all  $1\le i \le \ell$ by Corollary \ref{5.8}. Hence we get the equivalence between $(1)$ and $(2)$.
\end{proof}

We close this section by stating the ring version of Theorem \ref{1.2} and Theorem \ref{1.3}. Let $(R, \m)$ be a Noetherian local ring, $\calF=\{F_n\}_{n \in \Bbb Z}$ a filtration of ideals of $R$ such that $F_1 \neq R$. We assume that $\calR = \calR(\calF)$ is a Noetherian ring. Let $\{D_i\}_{0 \le i \le \ell}$ be the dimension filtration of $R$. Then $\calD_i = \{F_n \cap D_i\}_{n \in \Bbb Z}$ (resp. $\calC_i = \{[F_n \cap D_i + D_{i-1}]/D_{i-1}\}_{n \in \Bbb Z}$) is a $\calF$-filtration of $D_i$ (resp. $C_i$) for all $1 \le i \le \ell$.

\begin{thm}
The following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $\calR'$ is a sequentially Cohen-Macaulay ring.
\item $\calG$ is a sequentially Cohen-Macaulay ring and $\{\calG(\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $\calG$.
\end{enumerate}
When this is the case, $R$ is a sequentially Cohen-Macaulay ring.
\end{thm}

\begin{thm}
Suppose that $R$ is a sequentially Cohen-Macaulay ring and $F_1 \nsubseteq \p$ for every $\p \in \Ass R$. Then the following conditions are equivalent.
\begin{enumerate}[{\rm (1)}]
\item $\calR$ is a sequentially Cohen-Macaulay ring.
\item $\calG$ is a sequentially Cohen-Macaulay ring, $\{\calG (\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $\calG$ and  $\rma(\calG (\calC_i))<0$ for all $1 \le i \le \ell$.
\end{enumerate}
When this is the case, $\calR'$ is a sequentially Cohen-Macaulay ring.
\end{thm}

\section{Sequentially Cohen-Macaulay property in $E^{\natural}$}

In this section let $R=\sum_{n \ge 0}R_n$ be a $\Bbb Z$-graded ring. We put $F_n= \sum_{k \ge n}R_k$ for all $n \in \Bbb Z$. Then $F_n$ is a graded ideal of $R$, $\calF = \{F_n\}_{n\in \Bbb Z}$ is a filtration of ideals of $R$ and $F_1 := R_+ \neq R$. Let $E$ be a graded $R$-module with $E_n = (0)$ for all $n<0$. Put $E_{(n)} = \sum_{k \ge n}E_k$ for all $n \in \Bbb Z$. Then $E_{(n)}$ is a graded $R$-submodule of $E$, $\calE=\{E_{(n)}\}_{n \in \Bbb Z}$ is an $\calF$-filtration of $R$-submodules of $E$ and $E_{(0)} = E$.
Then we have $R = \calG(\calF)$ and $E = \calG(\calE)$. Set $R^{\natural} := \calR(\calF)$ and $E^{\natural} := \calR(\calE)$.

Suppose that $R$ is a Noetherian ring and $E \neq (0)$ is a finitely generated graded $R$- module with $d = \dim_R E < \infty$. Notice that $R^{\natural}$ is Noetherian and $E^{\natural}$, $\calR'(\calE)$ are finitely generated.

We note the following.

\begin{lem}\label{6.2}
The following assertions hold true.
\begin{enumerate}[$(1)$]
\item $\dim_{\calR'}\calR'(\calE) = d + 1$.
\item Suppose that there exists $\p \in \Assh_R E$ such that $F_1 \nsubseteq \p$. Then $\dim_{R^{\natural}}E^{\natural} = d + 1$.
\end{enumerate}
\end{lem}

\begin{proof}
See Proposition \ref{2.3}, Proposition \ref{2.4}.
\end{proof}

Let $D_0 \subsetneq D_1 \subsetneq \ldots \subsetneq D_{\ell} = E$ be the dimension filtration of $E$. We set $C_i = D_i/{D_{i-1}}$, $d_i = \dim_R D_i$ for every $1\le i\le \ell$. Then $D_i$ is a graded $R$-submodule of $E$ for all $0 \le i \le \ell$.
 Let $1 \le i \le \ell$. Then from the exact sequence
$$
0 \to [D_{i-1}]_{(n)} \to [D_i]_{(n)} \to [C_i]_{(n)} \to 0
$$
of graded $R$-modules for all $n \in \Bbb Z$, we get the exact sequences
$$
0 \to \calR(\calD_{i-1}) \to \calR(\calD_i) \to \calR(\calC_i) \to 0
$$
$$
0 \to \calR'(\calD_{i-1}) \to \calR'(\calD_i) \to \calR'(\calC_i) \to 0 \ \ \text{and}
$$
$$
0 \to \calG(\calD_{i-1}) \to \calG(\calD_i) \to \calG(\calC_i) \to 0
$$
of graded modules, where $\calD_i = \{[D_i]_{(n)} \}_{n\in \mathbb Z}, \ \
\calC_i = \{[C_i]_{(n)}\}_{n\in \mathbb Z}$. By the same technique as in the proof of Lemma \ref{5.9}, we obtain the dimension filtration of $\calR'(E)$ and $E^{\natural}$ as follows.

\begin{lem}\label{6.4}
$\{\calR'(\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $\calR'(\calE)$. If $F_1 \nsubseteq \p$ for every $\p \in \Ass_R E$, then $\{\calR(\calD_i)\}_{0 \le i \le \ell}$ is the dimension filtration of $E^{\natural}$.
\end{lem}

Hence we get the following, which characterize the sequentially Cohen-Macaulayness of $\calR'(\calE)$.

\begin{prop}\label{6.5}
The following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $\calR'(\calE)$ is a sequentially Cohen-Macaulay $\calR'$-module.
\item $E$ is a sequentially Cohen-Macaulay $R$-module.
\end{enumerate}
\end{prop}

\begin{proof}
$(1) \Rightarrow (2)$~~Follows from the fact that $C_i = \calG(\calC_i)$ for each $1\le i\le \ell$.

$(2) \Rightarrow (1)$~~We get $\calG(\calC_i)$ is Cohen-Macaulay for all $1\le i \le \ell$. Let $Q \in \Supp_{\calR'}\calR'(\calC_i)$. We may assume $u \notin Q$. Then $\calR'(\calC_i)_u = R[t, t^{-1}]\otimes_R C_i$ is Cohen-Macaulay since $C_i$ is Cohen-Macaulay. Hence $\calR'(\calC_i)_Q$ is a Cohen-Macaulay $\calR'_Q$-module.
\end{proof}

Now we study the question of when $E^{\natural}$ is sequentially Cohen-Macaulay. The key is the following.

\begin{lem}\label{6.6}
Suppose $R_0$ is a local ring, $E$ is a Cohen-Macaulay $R$-module and $F_1 \nsubseteq \p$ for some $\p \in \Assh_R E$. Then the following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $E^{\natural}$ is a Cohen-Macaulay $R^{\natural}$-module.
\item $\rma(E) < 0$.
\end{enumerate}
\end{lem}

\begin{proof}
Let $P = \m R + R_+$, where $\m$ denotes the maximal ideal of $R_0$. Then $P \supseteq F_1$. Since $R_+(E_{(n)}/{E_{(n+1)}}) = (0)$, $R_+(F_n/{F_{n+1}}) = (0)$ for all $n \in \Bbb Z$, we have
$$
E = \calG(\calE) \cong \calG(\calE_P), \ \ R = \calG(\calF) \cong \calG(\calF_P).
$$
Suppose that $E^{\natural}$ is a Cohen-Macaulay $R^{\natural}$-module. Then $\calR(\calE_P)$ is Cohen-Macaulay and $\dim_{\calR(R_P)}\calR(\calE_P)= d + 1$, whence $\calG(\calE_P)$ is Cohen-Macaulay and $\rma(\calG(\calE_P))<0$. Therefore we get $\rma(E) < 0$.

On the other hand, suppose that $\rma(E) < 0$. Then $\calR(\calE_P)$ is a Cohen-Macaulay $\calR(R_P)$-module of dimension $ d+1$. Thus $\calR(\calE)_P$ is Cohen-Macaulay. Now we regard $\calR$ as a ${\Bbb Z}^2$-graded ring with the ${\Bbb Z}^2$-grading as follows:
$$
\calR_{(i,j)} =
\left\{
\begin{array}{cc}
R_i t^j &    i \ge j \ge 0\\
(0) & otherwise.
\end{array}
\right.
$$
Moreover we set
$$
\calR(\calE)_{(i,j)} =
\left\{
\begin{array}{cc}
t^j \otimes E_i &    i \ge j \ge 0\\
(0) & otherwise,
\end{array}
\right.
$$
where $t^j \otimes  E_i =\{t^j \otimes x \mid x \in E_i \}$. Then $\calR(\calE)$ is a ${\Bbb Z}^2$-graded $\calR$-module with the above grading $\calR(\calE)_{(i,j)}$. Notice that $\calR_{(0,0)} = R_0$ is a local ring, so that $\calR$ is $H$-local. Let $L$ be the $H$-maximal ideal of the ${\Bbb Z}^2$-graded ring $\calR$. Then we get $P \subseteq L$, whence $L \cap R = P$. Therefore $\calR(\calE)_L$ is a Cohen-Macaulay $\calR_L$-module, so that $E^{\natural}$ is Cohen-Macaulay.
\end{proof}

Our answer is the following.

\begin{thm}\label{6.7}
Suppose that $R_0$ is a local ring, $E$ is a sequentially Cohen-Macaulay $R$-module and $F_1 \nsubseteq \p$ for every $\p \in \Ass_R E$. Then the following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $E^{\natural}$ is a sequentially Cohen-Macaulay $R^{\natural}$-module.
\item $\rma(C_i) < 0$ for all $1 \le i \le \ell$.
\end{enumerate}
\end{thm}

\section{Application --Stanley-Reisner algebras--}

In this section, let $V=\{ 1, 2, \ldots, n\}~(n>0)$ be a vertex set, $\Delta$ a simplicial complex on $V$ such that $\Delta \neq \emptyset$. We denote $\calF(\Delta)$ a set of facets of $\Delta$ and  $m=\sharp \calF(\Delta) ~(>0)$ its cardinality. Let $S=k[X_1, X_2, \ldots, X_n]$ a polynomial ring over a field $k$, $R =k[\Delta]= S/{I_{\Delta}}$ the {\it Stanley-Reisner ring of $\Delta$} of dimension $d$, where $I_{\Delta}=(X_{i_1}X_{i_2}\cdots X_{i_r}\mid \{ i_1<i_2<\cdots <i_r \}\notin \Delta)$ the {\it Stanley-Reisner ideal of $R$}.

We consider the Stanley-Reisner ring $R=\sum_{n \ge 0}R_n$ as a $\Bbb Z$-graded ring and put
$$
I_n= \sum_{k \ge n}R_k = \m^n \ \ \text{for all} \ \ n \in \Bbb Z
$$
where $\m = R_+ = \sum_{n>0}R_n$ is a graded maximal ideal of $R$. Then $\calI = \{I_n\}_{n\in \Bbb Z}$ is a $\m$-adic filtration of $R$ and $I_1 := R_+ \neq R$.

If $\Delta$ is shellable, then $R$ is a sequentially Cohen-Macaulay ring, so by Proposition \ref{6.5} we get the following.

\begin{prop}
If $\Delta$ is shellable, then $\calR'(\m)$ is a sequentially Cohen-Macaulay ring.
\end{prop}

Notice that $\p \nsupseteq I_1$ for every $\p \in \Ass R$ if and only if $F \neq \emptyset$ for all $F \in \calF(\Delta)$, which is equivalent to saying that $\Delta \neq \{\emptyset\}$.

The goal of this section is the following. Here $| F_i |$ denotes the cardinality of $F_i$.

\begin{thm}\label{7.4}
Suppose that $\Delta$ is shellable with shelling order $F_1, F_2, \ldots, F_m \in \calF(\Delta)$ such that $\dim F_1 \ge \dim F_2 \ge \cdots \ge \dim F_m$
 and $\Delta \neq \{\emptyset\}$. Then the following conditions are equivalent.
\begin{enumerate}[$(1)$]
\item $\calR(\m)$ is a sequentially Cohen-Macaulay ring.
\item $m=1$ or if $m \ge 2$, then $| F_i |> \sharp \calF(\Delta_1 \cap \Delta_2)$ for every $2 \le i\le m$, where $\Delta_1 =\left< F_1, F_2, \ldots, F_{i-1}\right>$, $\Delta_2 = \left< F_i\right>$.
\end{enumerate}
\end{thm}

\begin{proof}
Thanks to Theorem \ref{6.7}, $\calR$ is sequentially Cohen-Macaulay if and only if $\rma(C_i) < 0$ for all $1 \le i \le \ell$, where $\{D_i\}_{0 \le i \le \ell}$ is the dimension filtration of $R$, $C_i = D_i/D_{i-1}$ and $d_i = \dim_R D_i$ for all $1 \le i\le \ell$.
If $m=1$, then $R = k[\Delta] \cong k[X_i \mid i \in F_1]$, which is a polynomial ring, so that $\ell = 1$ and $\rma(R) = - | F_1 | < 0$. Hence $\calR$ is a Cohen-Macaulay ring by Lemma \ref{6.6}.

Suppose that $m > 1$ and the assertion holds for $m-1$.
We put $\Delta_1=\left< F_1, F_2, \ldots, F_{m-1}\right>$ and $\Delta_2=\left< F_m\right>$. If $\ell = 1$, then $\Delta$ is pure. Look at the following exact sequence
$$
0 \to S/{I_{\Delta}} \to S/{I_{\Delta_1}}\oplus S/{I_{\Delta_2}} \to S/{I_{\Delta_1} + I_{\Delta_2}} \to 0
$$
of graded $R$-modules. We then have
$$
S/{I_{\Delta_1} + I_{\Delta_2}} \cong k[\Delta_2]/(\overline{\xi})
$$
for some monomials $\xi \in I_{\Delta_1}\setminus I_{\Delta_2}$ in $X_1, X_2, \ldots, X_n$ with $0< \deg \xi = \sharp \calF(\Delta_1 \cap \Delta_2)$. Therefore $\rma (S/{I_{\Delta_1} + I_{\Delta_2}}) = \sharp \calF(\Delta_1 \cap \Delta_2) - | F_m |$. We put $\m= R_+$. Then we have the exact sequence of local cohomology modules as follows
$$
0 \to \H^{d-1}_\m(S/{I_{\Delta_1} + I_{\Delta_2}} ) \to \H^d_\m (S/{I_{\Delta}}) \to \H^d_\m (S/{I_{\Delta_1}}) \oplus \H^d_\m (S/{I_{\Delta_2}}) \to 0.
$$
Thus $\rma(R) = \max\{\sharp \calF(\Delta_1 \cap \Delta_2) - | F_m |, \rma(k[\Delta_1]), \rma(k[\Delta_2])\}$. Hence $\calR$ is sequentially Cohen-Macaulay if and only if $\sharp \calF(\Delta_1 \cap \Delta_2) < | F_m |$ and $\rma(k[\Delta_1])<0$. By using the induction arguments, we get the equivalence between $(1)$ and $(2)$.

Suppose now that $\ell>1$. Consider the following exact sequence
$$
0 \to I_{\Delta_1}/I_{\Delta} \to S/I_{\Delta} \to S/I_{\Delta_1} \to 0
$$
of graded $R$-modules. Then we have
$$
I_{\Delta_1}/I_{\Delta} \cong {I_{\Delta_1}+I_{\Delta_2}}/I_{\Delta_2} = I_{\Delta_1 \cap \Delta_2}/I_{\Delta_2} = (\overline{\xi})
$$
where $\xi \in I_{\Delta_1}\setminus I_{\Delta_2}$ is a homogeneous element with $0< \deg \xi = \sharp \calF(\Delta_1 \cap \Delta_2)=:t$. Therefore $I_{\Delta_1}/I_{\Delta} \cong S/I_{\Delta_2}(-t)$, so that
$$
0 \to S/I_{\Delta_2}(-t) \overset{\sigma}{\longrightarrow} S/I_{\Delta} \overset{\varepsilon}{\longrightarrow} S/I_{\Delta_1} \to 0.
$$
We put $L=\Im \sigma$. Then $L \neq (0)$, $\dim_R L = d_1$ and $\rma(L) = t- | F_m |$. We notice here that $L \subseteq D_1$. Now we set ${D_i}' = \varepsilon(D_i)$ for every $1 \le i \le \ell$. Then ${D_1}' \subsetneq {D_2}' \subsetneq \ldots \subsetneq {D_{\ell}}' = k[\Delta_1]$ and ${C_i}' := {D_i}'/{{D_{i-1}}'} \cong C_i$ for all $2 \le i \le \ell$. Hence $\rma(C_i) = \rma({C_i}')$ for $2 \le i \le \ell$.

\vspace{1em}
{\bf \underline{Case 1}} \ \  $L \subsetneq D_1$\ \  (i.e., ${D_1}' \neq (0)$)
\vspace{1em}

In this case ${D_0}':= (0) \subsetneq {D_1}' \subsetneq {D_2}' \subsetneq \ldots \subsetneq {D_{\ell}}' = k[\Delta_1]$ is the dimension filtration of $k[\Delta_1]$.
Look at the following exact sequence
$$
0 \to L \to D_1 \to D_1' \to 0
$$
of $R$-modules. Then $\rma(D_1) = \max\{\rma(L), \rma({D_1}')\}$.

\vspace{1em}
{\bf \underline{Case 2}} \ \ $L = D_1$  \ \  (i.e., ${D_1}' = (0)$)
\vspace{1em}

Similarly $(0) = {D_1}' \subsetneq {D_2}' \subsetneq \ldots \subsetneq {D_{\ell}}' = k[\Delta_1]$ is the dimension filtration of $k[\Delta_1]$.

Summing up, in any case $\calR$ is a sequentially Cohen-Macaulay ring if and only if $\rma(L) < 0$ and the assertion $(1)$ holds for the ring $k[\Delta_1]$. Hence we get the equivalence of conditions $(1)$ and $(2)$ by using the induction hypothesis.
\end{proof}

\begin{rem}\label{7.2}
If $\Delta$ is shellable, then we can take a shelling order $F_1, F_2, \ldots, F_m \in \calF(\Delta)$ such that $\dim F_1 \ge \dim F_2 \ge \cdots \ge \dim F_m$.
\end{rem}

Apply Theorem \ref{7.4}, we get the following.

\begin{cor}
Under the same notation in Theorem \ref{7.4}. Suppose that $| F_m | \ge 2$. If $\left< F_1, F_2, \ldots, F_{i-1}\right>\cap \left< F_i\right>$ is a simplex for every $2 \le i\le m$, then $\calR(\m)$ is a sequentially Cohen-Macaulay ring.
\end{cor}

Let us give some examples.

\begin{ex}
Let $\Delta = \left< F_1, F_2, F_3\right>$, where $F_1 = \{1,2,3\}$, $F_2=\{2,3, 4\}$ and  $F_3=\{4. 5\}$. Then $\Delta$ is shellable with shelling order $F_1, F_2, F_3 \in \calF(\Delta)$. Then
$$
\left< F_1\right> \cap \left< F_2\right>, \ \ \left< F_1, F_2\right> \cap \left< F_3\right>
$$ are simplexes, so that $\calR(\m)$ is a sequentially Cohen-Macaulay ring.
$$
\input{pic02.tex}
$$
\vspace{0.01cm}
\end{ex}

\begin{ex}
Let $\Delta = \left< F_1, F_2, F_3, F_4\right>$, where $F_1 = \{1,2,5\}$, $F_2=\{2,3\}$, $F_3=\{3,4\}$ and $F_4=\{4,5\}$. Notice that $\Delta$ is a shellable simplicial complex with shelling order $F_1, F_2, F_3, F_4 \in \calF(\Delta)$. We put $\Delta_1 = \left< F_1, F_2, F_3\right>$ and  $\Delta_2 = \left< F_4\right>$.  Then
$$
\sharp \calF(\Delta_1 \cap \Delta_2) = 2 = | F_4 |,
$$
whence $\calR(\m)$ is not sequentially Cohen-Macaulay.
$$
\input{pic01.tex}
$$
\end{ex}

\if0
\vspace{0.5cm}

\section{Appendix --Idealization of sequentially Cohen-Macaulay modules--}

Let $(R, \m)$ be a Noetherian local ring. We begin with the following.

\begin{lem}\label{lemma1}
Let $M$ and $N$ be finitely generated $R$-modules. Then $[M \oplus N]_n = M_n \oplus N_n$ for each $n \in \Bbb Z$.
\end{lem}

\begin{proof}
We have $[M \oplus N]_n \supseteq M_n \oplus N_n$, since $\dim_R(M_n\oplus N_n) = \max\{\dim_RM_n, \dim_RN_n\} \le n$. Let $p: M\oplus N \to M, (x,y) \mapsto x$ be the first projection. Then $p(L_n) \subseteq M_n$, since $\dim_Rp(L_n) \le \dim_RL_n \le n$. We similarly have $q(L_n) \subseteq N_n$, where $q: M\oplus N \to N, (x,y) \mapsto y$ denotes the second projection. Hence $[M\oplus N]_n \subseteq M_n \oplus N_n$ as claimed.
\end{proof}

The following two results are the key in this section.

\begin{prop}\label{prop1} Let $M$ and $N$ $(M, N \ne (0))$ be  finitely generated $R$-modules. Then $M\oplus N$ is a sequentially Cohen-Macaulay $R$-module if and only if both $M$ and $N$ are sequentially Cohen-Macaulay $R$-modules.
\end{prop}

\begin{proof} We set $L = M \oplus N$ and $\ell = \sharp \calS (L)$. Then $\calS (L) = \calS (M) \cup \calS (N)$, as $\Ass_RL = \Ass_RM \cup \Ass_RN$. Hence if $\ell = 1$, then $\calS (L) = \calS (M) = \calS (N)$ and $\dim_RL = \dim_RM = \dim_RN$. Therefore when $\ell = 1$, $L$ is a sequentially Cohen-Macaulay $R$-module if and only if $L$ is a Cohen-Macaulay $R$-module, and the second condition is equivalent to saying that the $R$-modules $M$ and $N$ are Cohen-Macaulay, that is $M$ and $N$ are sequentially Cohen-Macaulay $R$-modules. Suppose that $\ell > 1$ and that our assertion holds true for $\ell - 1$. Let
$$
D_0=(0) \subsetneq D_1 \subsetneq  D_2 \subsetneq  \cdots \subsetneq D_{\ell} = L
$$
be the dimension filtration of $L = M \oplus N$, where $\calS (L) = \{d_1< d_2 < \cdots < d_\ell\}$. Then $\{D_i/D_1\}_{1 \le i \le \ell}$ is the dimension filtration of $L/D_1$ and hence $L$ is a sequentially Cohen-Macaulay $R$-module if and only if $D_1$ is a Cohen-Macaulay $R$-module and $L/D_1$ is a sequentially Cohen-Macaulay $R$-module. Because
$$
D_1 = \left\{
\begin{array}{lc}
M_{d_1} \oplus (0) & (d_1 \in \calS (M) \setminus \calS (N)),\\
\vspace{0.5mm}\\
M_{d_1} \oplus N_{d_1} & (d_1 \in \calS (M) \cap \calS (N)),\\
\vspace{0.5mm}\\
(0) \oplus N_{d_1} & (d_1 \in \calS (N) \setminus \calS (M))
\end{array}
\right.
$$ by Lemma \ref{lemma1},  the hypothesis on $\ell$ readily shows  the second condition is equivalent to saying that the $R$-modules $M$ and $N$ are sequentially Cohen-Macaulay.
\end{proof}

Throughout this section let $A$ be a Noetherian local ring and assume that $A$ is a module-finite $R$-algebra.

\begin{thm} \label{thm2}
Let $M\neq (0)$ be a finitely generated $A$-module. Then the following assertions hold true.
\begin{enumerate}
\item[$(1)$] Let $n \in \Bbb Z$ and let $M_n$ denote the largest $R$-submodule of $M$ with $\dim_RM_n \le n$. Then $M_n$ is the largest $A$-submodule of $M$ with $\dim_AM_n \le n$.
\item[$(2)$] The dimension filtration of $M$ as an $A$-module coincides with that of $M$ as an $R$-module.
\item[$(3)$] $M$ is a sequentially Cohen-Macaulay $A$-module if and only if $M$ is a sequentially Cohen-Macaulay $R$-module.
\end{enumerate}
\end{thm}

\begin{proof}
Let $n \in \Bbb Z$ and $X$ denote the largest $A$-submodule of $M$ with $\dim_AX \le n$. Then $X \subseteq M_n$, since $\dim_RX =\dim_AX \le n$. Let $Y = AM_n$. Then $\dim_AY \le n$. In fact, let $\p \in \Ass_RY$. Then since $[M_n]_\p \subseteq Y_\p = A_\p{\cdot} [M_n]_\p \subseteq M_\p$, we see $[M_n]_\p \ne (0)$, so that $\p \in \Supp_RM_n$. Hence $\dim R/\p \le \dim_RM_n \le n$. Thus $\dim_AY = \dim_RY \le n$, whence $M_n \subseteq Y \subseteq X$, which shows $X = M_n$. Therefore assertions (1) and  (2) follows. Since $\dim_AL = \dim_RL$ and $\depth_AL = \depth_RL$ for every finitely generated $A$-module $L$, we get assertion (3).
\end{proof}

We summarize consequences.

\begin{cor}\label{thm3} $A$ is a sequentially Cohen-Macaulay local ring if and only if $A$ is a sequentially Cohen-Macaulay $R$-module.
\end{cor}

\begin{cor}\label{cor1}
Let $M$ be a finitely generated $A$-module and assume that $R$ is a direct summand of $M$ as an $R$-module. If $M$ is a sequentially Cohen-Macaulay $A$-module, then $R$ is a sequentially Cohen-Macaulay local ring.
\end{cor}

\begin{proof}\label{cor2}
We write $M = R \oplus N$ where $N$ is an $R$-submodule of $M$. Since $M$ is a sequentially Cohen-Macaulay $A$-module,  by Theorem \ref{thm2} it is a sequentially Cohen-Macaulay $R$-module as well, so that  by Proposition \ref{prop1}, $R$ is a sequentially Cohen-Macaulay local ring.
\end{proof}

\begin{cor}\label{cor3}
Suppose that $R$ is a direct summand of $A$ as an $R$-module. If $A$ is a sequentially Cohen-Macaulay local ring, then $R$ is a sequentially Cohen-Macaulay local ring.
\end{cor}

We consider the invariant subring $R = A^G$.

\begin{cor}\label{cor4}
Let $A$ be a Noetherian local ring, $G$ a finite subgroup of $\operatorname{Aut}A$. Assume that the order of $G$ is invertible in $A$.  If $A$ is a sequentially Cohen-Macaulay local ring, then the invariant subring $R=A^G$ of $A$ is a sequentially Cohen-Macaulay local ring.
\end{cor}

\begin{proof}
Since the order of $G$ is invertible in $A$, $A$ is a module-finite extension of $R= A^G$ such that $R$ is a direct summand of $A$ (see \cite{BR} and reduce to the case where $A$ is a reduced ring). Hence the assertion follows from Corollary \ref{cor3}.
\end{proof}

\begin{rem} In the setting of Corollary \ref{cor4} let $\{D_i\}_{0 \le i \le \ell}$ be the dimension filtration of $A$. Then each $D_i$ is a $G$-stable ideal of $A$  (compare with Theorem \ref{thm2} (1)) and the dimension filtration of $R$ is given by a refinement of $\{D_i^G\}_{0 \le i \le \ell}$.
\end{rem}

The goal of this section is the following.

\begin{cor}\label{cor5}
Let $R$ be a Noetherian local ring, $M\neq (0)$ a finitely generated $R$-module.  We put $A = R\ltimes M$ the idealization of $M$ over $R$. Then the following conditions are equivalent.
\begin{enumerate}
\item[$(1)$] $A = R \ltimes M$ is a sequentially Cohen-Macaulay local ring.
\item[$(2)$] $A=R\ltimes M$ is a sequentially Cohen-Macaulay $R$-module.
\item[$(3)$] $R$ is a sequentially Cohen-Macaulay local ring and $M$ is a sequentially Cohen-Macaulay $R$-module.
\end{enumerate}
\end{cor}

\fi

\vspace{0.5cm}

\begin{ac}
The authors would like to thank Professor Shiro Goto for his valuable advice and comments.
\end{ac}