text:
\begin{document}

\title{Graphs characterized by the second distance eigenvalue}

\author{Rundan Xing, Bo Zhou\footnote{Corresponding author. E-mail: zhoubo@scnu.edu.cn}\\
Department of Mathematics, South China Normal University,\\
Guangzhou 510631, P.R. China}

\date{}
\maketitle

\begin{abstract}
We characterize all connected graphs with second distance eigenvalue less than $-0.5858$. \\ \\
{\bf Keywords:} distance, distance matrix, distance eigenvalues, forbidden subgraph, diameter

\end{abstract}

\section{Introduction}

We consider simple undirected graphs. Let $G$ be a connected graph
with vertex set $V(G)=\{v_1,\dots,v_n\}$. For
$1\le i,j\le n$, the distance between vertices $v_i$ and $v_j$ in $G$,
denoted by $d_G(v_i,v_j)$ or simply $d_{v_iv_j}$, is the length of a
shortest path connecting them in $G$. The distance matrix of $G$ is
the $n\times n$ matrix $D(G)=(d_{v_iv_j})$. Since
$D(G)$ is symmetric, the eigenvalues of $D(G)$ are
all real numbers. The distance eigenvalues of $G$, denoted by
$\lambda_1(G),\dots,\lambda_n(G)$, are the eigenvalues of
$D(G)$, arranged in non-increasing order. For $1\le k\le n$, we call $\lambda_k(G)$ the $k$th distance eigenvalue of $G$.

The study of distance eigenvalues dates back to the classical work
of Graham and Pollack~\cite{GP}, Edelberg et al.~\cite{EGG} and Graham
and Lov\'asz~\cite{GL} in 1970s. Merris~\cite{Me90} studied the
relations between the distance eigenvalues and the Laplacian
eigenvalues of trees. The first distance eigenvalue has received much attention. Ruzieh
and Powers~\cite{SuPo90}
showed that the path $P_n$ is the unique $n$-vertex connected graph
with maximal first distance eigenvalue, while the complete graph
$K_n$ is the unique $n$-vertex connected graph with minimal first
distance eigenvalue. Among others, Stevanovi\'c and Ili\'c~\cite{SI}
showed that the star $S_n$ is the unique $n$-vertex tree with
minimal first distance eigenvalue. The extremal graphs with maximal
or minimal first distance eigenvalues may be found in, e.g.,
\cite{BNP,NP,WZ,XZD,YW}.

The last (least) distance eigenvalue has also received some
attention, see~\cite{LZ,Yu}.

Let $G_1$ and $G_2$ be two vertex-disjoint graphs. $G_1\cup G_2$
denotes the vertex-disjoint union of $G_1$ and $G_2$, and $G_1\vee
G_2$ denotes the graph obtained from $G_1\cup G_2$ by joining each
vertex of $G_1$ and each vertex of $G_2$ using an edge.

In this paper, we characterize all connected graphs with second
distance eigenvalue less than $-0.5858$. We prove the following
result.

\begin{Theorem}\label{conn-graph}
Let $G$ be a connected graph with at least two vertices. Then
$\lambda_2(G)\in(-\infty,-0.5858)$ if and only if $G\cong K_n$ for
some $n\ge 2$, or $G\cong K_1\vee(K_{n_1}\cup K_{n_2})$ for some
$n_1,n_2\ge 1$, or $G\cong K_1\vee(K_{n_1}\cup K_{n_2}\cup K_{n_3})$
for some  $n_1,n_2,n_3\ge 1$, or $G\cong K_1\vee(K_{n_1}\cup
K_{n_2}\cup K_{n_3}\cup K_{n_4})$ for some $n_1,n_2,n_3, n_4$ such
that $1\le n_1\le n_2\le n_3\le n_4$ and one of the following items
holds$:$

(i) $n_1=n_2=1$ and $1\le n_3\le 873$;

(ii) $n_1=n_2=1$, $n_3\ge 874$, and $r(-0.5858)<0$, where
\begin{eqnarray*}
   r(\lambda)
&=&
   \lambda^4-(n_3+n_4)\lambda^3-(3n_3n_4+8n_3+8n_4+5)\lambda^2\\
& &
   -(12n_3n_4+11n_3+11n_4+6)\lambda-(6n_3n_4+4n_3+4n_4+2);
\end{eqnarray*}

(iii) $n_1=1$, $n_2=2$, and $n_3=2$, or $3=n_3\le n_4\le 870$, or $4=n_3\le n_4\le 14$, or $5=n_3\le n_4\le 8$, or $n_3=n_4=6$;

(iv) $n_1=1$, $n_2=3$, and $3=n_3\le n_4\le 7$ or $n_3=n_4=4$;

(v) $n_1=n_2=n_3=2$ and $2\le n_4\le 5$.
\end{Theorem}

\section{Proof of Theorem~\ref{conn-graph}}

For an $n\times n$ symmetric matrix $M$, let
$\mu_1(M),\dots,\mu_n(M)$ be the eigenvalues of $M$, arranged in non-increasing order. Let $A$
be an $n\times n$ symmetric matrix, and $B$ an $m\times m$ principal
submatrix of $A$. The interlacing theorem \cite[pp.~185--186]{HJ}
states that $\mu_{n-m+i}(A)\le\mu_i(B)\le\mu_i(A)$ for $1\le i\le
m$.

Let $G$ be an $n$-vertex connected graph, and $H$ an $m$-vertex
induced subgraph of $G$, where $m\ge 2$. If $H$ is connected and
$d_H(u,v)=d_G(u,v)$ for all $\{u,v\}\subseteq V(H)$, then write
$H\trianglelefteq G$. Obviously, if $H$ is of diameter two, then
$H\trianglelefteq G$. If $H\trianglelefteq G$, then $D(H)$ is a
principal submatrix of $D(G)$, and thus from the interlacing theorem,
$\lambda_2(G)\ge\lambda_2(H)$ and $\lambda_n(G)\le\lambda_m(H)$.

For integer $n\ge 1$, let $nG$ be the vertex-disjoint union of $n$ copies of graph $G$.

Let $I_n$ be the $n\times n$ identity matrix, and $J_{m\times n}$
the $m\times n$ all-one matrix. For convenience, let $J_n=J_{n\times
n}$ and  $\mathbf{1}_n=J_{n\times 1}$.

\begin{Lemma}\label{Kn12-Kn123}
For integers $n_1,n_2,n_3\ge 1$, $\lambda_2(K_1\vee(K_{n_1}\cup K_{n_2}))\in(-1,-0.5858)$ and $\lambda_2(K_1\vee(K_{n_1}\cup K_{n_2}\cup K_{n_3}))\in(-1,-0.5858)$.
\end{Lemma}

\begin{proof}
Let $G_1=K_1\vee(K_{n_1}\cup K_{n_2})$.
We have, with respect to the partition $V(G_1)=V(K_1)\cup V(K_{n_1})\cup V(K_{n_2})$, that
\[
D(G_1)=\left[\begin{array}{ccc}
  0 & \mathbf{1}_{n_1}^{\top} & \mathbf{1}_{n_2}^{\top}\\
   \mathbf{1}_{n_1} & J_{n_1}-I_{n_1} & 2J_{n_1\times n_2}\\
   \mathbf{1}_{n_2} & 2J_{n_2\times n_1} & J_{n_2}-I_{n_2}
   \end{array}\right].
\]
Then the characteristic polynomial of $D(G_1)$ is
\begin{eqnarray*}
   \det(\lambda I_{n_1+n_2+1}-D(G_1))
&=&
   \left|\begin{array}{ccc}
   \lambda & -\mathbf{1}_{n_1}^{\top} & -\mathbf{1}_{n_2}^{\top}\\
   -\mathbf{1}_{n_1} & (\lambda+1)I_{n_1}-J_{n_1} & -2J_{n_1\times n_2}\\
   -\mathbf{1}_{n_2} & -2J_{n_2\times n_1} & (\lambda+1)I_{n_2}-J_{n_2}
   \end{array}\right|\\
&=&
   (\lambda+1)^{n_1+n_2-2}
   \left|\begin{array}{ccc}
   \lambda+n_2+1 & -(\lambda+n_1+1) & 0\\
   -n_2 & \lambda-n_1 & -\lambda\\
   -n_2 & 0 & \lambda+1
   \end{array}\right|\\
&=&
   (\lambda+1)^{n_1+n_2-2}f(\lambda),
\end{eqnarray*}
where
\[
f(\lambda)=\lambda^3-(n_1+n_2-2)\lambda^2-(3n_1n_2+2n_1+2n_2-1)\lambda-2n_1n_2-n_1-n_2.
\]
Obviously, $K_1\vee 2K_1\cong P_3\trianglelefteq G_1$, implying that $\lambda_2(G_1)\ge\lambda_2(P_3)=1-\sqrt{3}>-1$, and thus
$\lambda_2(G_1)$ is the second largest root of the equation
$f(\lambda)=0$. By direct check, we have $f(-1)=n_1n_2>0$ and
$f(-0.5858)=-0.2426n_1n_2-0.1716(n_1+n_2)-0.1005<0$,

and thus $\lambda_2(G_1)\in(-1,-0.5858)$.

Now let $G_2=K_1\vee(K_{n_1}\cup K_{n_2}\cup K_{n_3})$ and $n=n_1+n_2+n_3+1$. The characteristic polynomial of $D(G_2)$ is
\begin{eqnarray*}
& &
   \det(\lambda I_n-D(G_2))\\
&=&
   \left|\begin{array}{cccc}
   \lambda & -\mathbf{1}_{n_1}^{\top} & -\mathbf{1}_{n_2}^{\top} & -\mathbf{1}_{n_3}^{\top}\\
   -\mathbf{1}_{n_1} & (\lambda+1)I_{n_1}-J_{n_1} & -2J_{n_1\times n_2} & -2J_{n_1\times n_3}\\
   -\mathbf{1}_{n_2} & -2J_{n_2\times n_1} & (\lambda+1)I_{n_2}-J_{n_2} & -2J_{n_2\times n_3}\\
   -\mathbf{1}_{n_3} & -2J_{n_3\times n_1} & -2J_{n_3\times n_2} & (\lambda+1)I_{n_3}-J_{n_3}\\
   \end{array}\right|\\
&=&
   (\lambda+1)^{n_1+n_2+n_3-3}
   \left|\begin{array}{cccc}
   \lambda+n_1+1 & \lambda+n_2+1 & 0 & 0\\
   0 & -(\lambda+n_2+1) & \lambda+n_3+1 & 0\\
   -(\lambda-n_1) & -n_2 & -n_3 & \lambda\\
   0 & -n_2 & -n_3 & -(\lambda+1)
   \end{array}\right|\\
&=&
   (\lambda+1)^{n-4}g(\lambda),
\end{eqnarray*}
where
\begin{eqnarray*}
   g(\lambda)
&=&
   \lambda^4-(n_1+n_2+n_3-3)\lambda^3-3(n_1n_2+n_1n_3+n_2n_3+n_1+n_2+n_3-1)\lambda^2\\
& &
   -[5(n_1n_2+n_1n_3+n_2n_3+n_1n_2n_3)+3(n_1+n_2+n_3)-1]\lambda\\
& &
   -3n_1n_2n_3-2(n_1n_2+n_1n_3+n_2n_3)-(n_1+n_2+n_3).
\end{eqnarray*}
Obviously, $K_1\vee 3K_1\cong S_4\trianglelefteq G_2$, implying that $\lambda_2(G_2)\ge\lambda_2(S_4)=2-\sqrt{7}>-1$ and $\lambda_n(G_2)\le\lambda_4(S_4)=-2<-1$, and thus $\lambda_2(G_2)$ and $\lambda_n(G_2)$ are
the second largest and the least roots of the equation $g(\lambda)=0$ respectively. By direct
check, $g(-1)=2n_1n_2n_3>0$ and
\begin{eqnarray*}
   g(-0.5858)
&=&
   -0.0710n_1n_2n_3-0.1005(n_1n_2+n_1n_3+n_2n_3)\\
& &
   -0.0711(n_1+n_2+n_3)-0.0416\\
&<&0.
\end{eqnarray*}
Thus we have $\lambda_2(G_2)\in(-1,-0.5858)$.
\end{proof}

\begin{Lemma}\label{Kn1234}
Let $n_1,n_2,n_3,n_4$ be positive integers with $n_1\le n_2\le n_3\le
n_4$. Then $\lambda_2(K_1\vee(K_{n_1}\cup K_{n_2}\cup K_{n_3}\cup
K_{n_4}))\in(-1,-0.5858)$ if and only if one of the following items holds$:$

(i) $n_1=n_2=1$ and $1\le n_3\le 873$;

(ii) $n_1=n_2=1$, $n_3\ge 874$, and $r(-0.5858)<0$, where
\begin{eqnarray*}
   r(\lambda)
&=&
   \lambda^4-(n_3+n_4)\lambda^3-(3n_3n_4+8n_3+8n_4+5)\lambda^2\\
& &
   -(12n_3n_4+11n_3+11n_4+6)\lambda-(6n_3n_4+4n_3+4n_4+2);
\end{eqnarray*}

(iii) $n_1=1$, $n_2=2$, and $n_3=2$, or $3=n_3\le n_4\le 870$, or $4=n_3\le n_4\le 14$, or $5=n_3\le n_4\le 8$, or $n_3=n_4=6$;

(iv) $n_1=1$, $n_2=3$, and $3=n_3\le n_4\le 7$ or $n_3=n_4=4$;

(v) $n_1=n_2=n_3=2$ and $2\le n_4\le 5$.
\end{Lemma}

\begin{proof}
Let $G=K_1\vee(K_{n_1}\cup K_{n_2}\cup K_{n_3}\cup K_{n_4})$ and $n=n_1+n_2+n_3+n_4+1$.

The characteristic polynomial of $D(G)$ is
\begin{eqnarray*}
& &
   \det(\lambda I_n-D(G))\\
&=&
   \left|\begin{array}{ccccc}
   \lambda & -\mathbf{1}_{n_1}^{\top} & -\mathbf{1}_{n_2}^{\top} & -\mathbf{1}_{n_3}^{\top} & -\mathbf{1}_{n_4}^{\top}\\
   -\mathbf{1}_{n_1} & (\lambda+1)I_{n_1}-J_{n_1} & -2J_{n_1\times n_2} & -2J_{n_1\times n_3} & -2J_{n_1\times n_4}\\
   -\mathbf{1}_{n_2} & -2J_{n_2\times n_1} & (\lambda+1)I_{n_2}-J_{n_2} & -2J_{n_2\times n_3} & -2J_{n_2\times n_4}\\
   -\mathbf{1}_{n_3} & -2J_{n_3\times n_1} & -2J_{n_3\times n_2} & (\lambda+1)I_{n_3}-J_{n_3} & -2J_{n_3\times n_4}\\
   -\mathbf{1}_{n_4} & -2J_{n_4\times n_1} & -2J_{n_4\times n_2} & -2J_{n_4\times n_3} & (\lambda+1)I_{n_4}-J_{n_4}\\
   \end{array}\right|\\
&=&
   (\lambda+1)^{n_1+n_2+n_3+n_4-4}\cdot\\
& &
   \left|\begin{array}{ccccc}
   -(\lambda+n_1+1) & \lambda+n_2+1 & 0 & 0 & 0\\
   0 & -(\lambda+n_2+1) & \lambda+n_3+1 & 0 & 0\\
   0 & 0 & -(\lambda+n_3+1) & \lambda+n_4+1 & 0\\
   \lambda-n_1 & -n_2 & -n_3 & -n_4 & \lambda\\
   0 & -n_2 & -n_3 & -n_4 & -(\lambda+1)
   \end{array}\right|\\
&=&
   (\lambda+1)^{n-5}h_{n_1,n_2,n_3,n_4}(\lambda),
\end{eqnarray*}
where
\begin{eqnarray*}
& &
   h_{n_1,n_2,n_3,n_4}(\lambda)\\
&=&
   \lambda^5-(n_1+n_2+n_3+n_4-4)\lambda^4-[3(n_1n_2+n_1n_3+n_1n_4+n_2n_3+n_2n_4+n_3n_4)\\
& &
   +4(n_1+n_2+n_3+n_4)-6]\lambda^3-[5(n_1n_2n_3+n_1n_2n_4+n_1n_3n_4+n_2n_3n_4)\\
& &
   +8(n_1n_2+n_1n_3+n_1n_4+n_2n_3+n_2n_4+n_3n_4)+6(n_1+n_2+n_3+n_4)-4]\lambda^2\\
& &
   -[7n_1n_2n_3n_4+8(n_1n_2n_3+n_1n_2n_4+n_1n_3n_4+n_2n_3n_4)\\
& &
  +7(n_1n_2+n_1n_3+n_1n_4+n_2n_3+n_2n_4+n_3n_4)+4(n_1+n_2+n_3+n_4)-1]\lambda\\
& &
   -4n_1n_2n_3n_4-3(n_1n_2n_3+n_1n_2n_4+n_1n_3n_4+n_2n_3n_4)\\
& &
   -2(n_1n_2+n_1n_3+n_1n_4+n_2n_3+n_2n_4+n_3n_4)-(n_1+n_2+n_3+n_4).
\end{eqnarray*}
Obviously, $K_1\vee 4K_1\cong S_5\trianglelefteq G$, implying that $\lambda_2(G)\ge\lambda_2(S_5)=3-\sqrt{13}>-1$, and thus $\lambda_2(G)$ is the
second largest root of the equation $h_{n_1,n_2,n_3,n_4}(\lambda)=0$.

If $n_1\ge 3$, then $K_1\vee 4K_3\trianglelefteq G$, and thus $\lambda_2(G)\ge\lambda_2(K_1\vee
4K_3)=-0.5830>-0.5858$.
If $n_2\ge 4$, then $K_1\vee (K_1\cup 3K_4)\trianglelefteq G$, and thus $\lambda_2(G)\ge\lambda_2(K_1\vee (K_1\cup
3K_4))=-0.5855>-0.5858$.
To obtain the result, we need only to consider the cases when $n_1\le 2$ and $n_2\le 3$.

\noindent {\bf Case 1.} $n_1=1$.

\noindent {\bf Case 1.1.} $n_2=1$. If $n_4=1$, then $G=K_1\vee 4K_1\cong S_5$ with $\lambda_2(G)=3-\sqrt{13}\in(-1,-0.5858)$. Suppose that $n_4\ge 2$. We have $h_{n_1,n_2,n_3,n_4}(\lambda)=h_{1,1,n_3,n_4}(\lambda)=(\lambda+2)r_{1,1,n_3,n_4}(\lambda)$, where
\begin{eqnarray*}
   r_{1,1,n_3,n_4}(\lambda)
&=&
   \lambda^4-(n_3+n_4)\lambda^3-(3n_3n_4+8n_3+8n_4+5)\lambda^2\\
& &
   -(12n_3n_4+11n_3+11n_4+6)\lambda-(6n_3n_4+4n_3+4n_4+2).
\end{eqnarray*}
Then $\det(\lambda
I_n-D(G))=(\lambda+1)^{n-5}(\lambda+2)r_{1,1,n_3,n_4}(\lambda)$. Recall that $\lambda_2(G)>-1$ is the second largest root of $h_{1,1,n_3,n_4}(\lambda)=0$, and thus $\lambda_2(G)$ is also the second largest root of $r_{1,1,n_3,n_4}(\lambda)=0$. Obviously, $K_1\vee(3K_1\cup K_2)\trianglelefteq G$, and then $\lambda_n(G)\le\lambda_6(K_1\vee(3K_1\cup K_2))=-2.6288<-2$, implying that $\lambda_n(G)$ is the least root of $r_{1,1,n_3,n_4}(\lambda)=0$. Note that $\lambda_1(G)>0$ is the largest root of $r_{1,1,n_3,n_4}(\lambda)=0$. Since $r_{1,1,n_3,n_4}(-1)=3n_3n_4>0$, we have $\lambda_2(G)\in (-1,-0.5858)$ if and only if
$r_{1,1,n_3,n_4}(-0.5858)<0$. If $1\le n_3\le 873$, then
\begin{eqnarray*}
   r_{1,1,n_3,n_4}(-0.5858)
&=&
   (1.1508\times 10^{-4}n_3-0.1005)n_4-0.1005n_3-0.0832\\
&<&
   -0.1005n_3-0.0832\\
&<&
   0.
\end{eqnarray*}
Thus $\lambda_2(G)\in(-1,-0.5858)$ if and only if items~(i)
or~(ii) holds.

\noindent {\bf Case 1.2.} $n_2=2$. Then $n_4\ge n_3\ge 2$.

Suppose first that $n_3=2$. If $n_4=2$, then $G=K_1\vee(K_1\cup 3K_2)$ with
$\lambda_2(G)=-0.5925\in(-1,-0.5858)$. Assume that $n_4\ge 3$.
We have
$h_{n_1,n_2,n_3,n_4}(\lambda)=h_{1,2,2,n_4}(\lambda)=(\lambda+3)s_{1,2,2,n_4}(\lambda)$,
where
\[
s_{1,2,2,n_4}(\lambda)=\lambda^4-(n_4+4)\lambda^3-(16n_4+26)\lambda^2-(38n_4+32)\lambda-(17n_4+11).
\]
Then $\det(\lambda
I_n-D(G))=(\lambda+1)^{n-5}(\lambda+3)s_{1,2,2,n_4}(\lambda)$.
Obviously, $K_1\vee(K_1\cup 2K_2\cup K_3)\trianglelefteq G$,
and then $\lambda_n(G)\le\lambda_9(K_1\vee(K_1\cup
2K_2\cup K_3))=-3.6122<-3$, which, together with the fact that
$\lambda_2(G)>-1$, implies that $\lambda_2(G)$ and $\lambda_n(G)$
are the second largest and the least roots of
$s_{1,2,2,n_4}(\lambda)=0$ respectively. By direct check,
$s_{1,2,2,n_4}(-1)=6n_4>0$ and
$s_{1,2,2,n_4}(-0.5858)=-0.0292n_4-0.2547<0$, implying that
$\lambda_2(G)\in (-1,-0.5858)$.

Suppose that $n_3=3$. We have
\begin{eqnarray*}
   h_{n_1,n_2,n_3,n_4}(\lambda)
&=&
   h_{1,2,3,n_4}(\lambda)\\
&=&
   \lambda^5-(n_4+2)\lambda^4-(22n_4+51)\lambda^3-(109n_4+150)\lambda^2\\
& &
   -(176n_4+148)\lambda-70n_4-46.
\end{eqnarray*}
Recall that $\lambda_2(G)$ is the second largest root of
$h_{1,2,3,n_4}(\lambda)=0$. Since $n_4\ge 3$, we have by direct
check that $h_{1,2,3,n_4}(-3.9)=32.1839n_4-89.9612>0$,
$h_{1,2,3,n_4}(-3)=-10n_4+20<0$, $h_{1,2,3,n_4}(-1)=18n_4>0$, and
$h_{1,2,3,n_4}(0)=-70n_4-46<0$. Thus $\lambda_2(G)\in (-1,0)$, and
$\lambda_2(G)\in (-1,-0.5858)$ if and only if
$h_{1,2,3,n_4}(-0.5858)=9.5128\times 10^{-4}n_4-0.8281<0$, i.e.,
$3\le n_4\le 870$.

Suppose that $n_3=4$. If $n_4\ge 15$, then $K_1\vee (K_1\cup K_2\cup
K_4\cup K_{15})\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_1\vee (K_1\cup K_2\cup K_4\cup
K_{15}))=-0.58577>-0.5858$.
Hence, together with Tabel~\ref{tab1}, we have  $\lambda_2(G)\in(-1,-0.5858)$  if and only if
$4\le n_4\le 14$.

\begin{table}[h]
\small \caption{The second distance eigenvalue of $K_1\vee (K_1\cup K_2\cup K_4\cup K_{n_4})$ for $4\le n_4\le 14$.}\label{tab1}
\bigskip
\renewcommand{\arraystretch}{1.6}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\noalign{\hrule height 0.5pt}

$n_4$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$\\
\noalign{\hrule height 0.4pt}
\hline $\lambda_2$ & $-0.5877$ & $-0.5872$ & $-0.5869$ & $-0.5866$ & $-0.5864$ & $-0.5863$\\
\noalign{\hrule height 0.4pt}
$n_4$ & $10$ & $11$ & $12$ & $13$ & $14$ &\\
\noalign{\hrule height 0.4pt}
\hline $\lambda_2$ & $-0.5861$ & $-0.58604$ & $-0.58595$ & $-0.58588$ & $-0.58582$ &\\
\noalign{\hrule height 0.4pt} \hline
\end{tabular}
\end{table}

Suppose that $n_3=5$. If $n_4\ge 9$, then $K_1\vee (K_1\cup K_2\cup
K_5\cup K_9)\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_1\vee (K_1\cup K_2\cup K_5\cup
K_9))=-0.58576>-0.5858$.
Hence, together with Tabel~\ref{tab2}, we have $\lambda_2(G)\in(-1,-0.5858)$  if and only if
$5\le n_4\le 8$.

\begin{table}[h]
\small \caption{The second distance eigenvalue of $K_1\vee (K_1\cup K_2\cup K_5\cup K_{n_4})$ for $5\le n_4\le 8$.}\label{tab2}
\bigskip
\renewcommand{\arraystretch}{1.6}
\centering
\begin{tabular}{|c|c|c|c|c|}
\noalign{\hrule height 0.5pt}

$n_4$ & $5$ & $6$ & $7$ & $8$\\
\noalign{\hrule height 0.4pt}
\hline $\lambda_2$ & $-0.5867$ & $-0.5864$ & $-0.5861$ & $-0.5859$\\
\noalign{\hrule height 0.4pt} \hline
\end{tabular}
\end{table}

Suppose finally that $n_3\ge 6$. If $n_4\ge 7$, then $K_1\vee
(K_1\cup K_2\cup K_6\cup K_7)\trianglelefteq G$, and thus $\lambda_2(G)\ge\lambda_2(K_1\vee (K_1\cup
K_2\cup K_6\cup K_7))=-0.58576>-0.5858$.  By direct check, $\lambda_2(K_1\vee (K_1\cup K_2\cup
2K_6))=-0.5860\in(-1,-0.5858)$. Thus $\lambda_2(G)\in(-1,-0.5858)$  if and only if
$n_3=n_4=6$.

From this case, we have $\lambda_2(G)\in(-1,-0.5858)$ if and only if item~(iii) holds.

\noindent {\bf Case 1.3.} $n_2=3$.

Suppose first that $n_3=3$. If
$n_4\ge 8$, then $K_1\vee (K_1\cup 2K_3\cup K_8)\trianglelefteq G$,
and thus $\lambda_2(G)\ge\lambda_2(K_1\vee
(K_1\cup 2K_3\cup K_8))=-0.58576>-0.5858$.
Hence, together with Table~\ref{tab3}, we have
$\lambda_2(G)\in(-1,-0.5858)$ if and only if $3\le n_4\le 7$.

\begin{table}[h]
\small \caption{The second distance eigenvalue of $K_1\vee (K_1\cup 2K_3\cup K_{n_4})$ for $3\le n_4\le 7$.}\label{tab3}
\bigskip
\renewcommand{\arraystretch}{1.6}
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\noalign{\hrule height 0.5pt}

$n_4$ & $3$ & $4$ & $5$ & $6$ & $7$\\
\noalign{\hrule height 0.4pt}
\hline $\lambda_2$ & $-0.5878$ & $-0.5870$ & $-0.5865$ & $-0.5862$ & $-0.5859$\\
\noalign{\hrule height 0.4pt} \hline
\end{tabular}
\end{table}

Now suppose
that $n_3\ge 4$. If $n_4\ge 5$, then $K_1\vee (K_1\cup K_3\cup
K_4\cup K_5)\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_1\vee (K_1\cup K_3\cup K_4\cup
K_5))=-0.58575>-0.5858$.
By direct check, $\lambda_2(K_1\vee (K_1\cup K_3\cup
2K_4))=-0.5862\in(-1,-0.5858)$. Thus $\lambda_2(G)\in(-1,-0.5858)$ if and only if $n_3=n_4=4$.

From this case, we have $\lambda_2(G)\in(-1,-0.5858)$ if and only if item~(iv) holds.

\noindent {\bf Case 2.} $n_1=2$. If
$n_3\ge 3$, then $K_1\vee (2K_2\cup 2K_3)\trianglelefteq G$, and
thus $\lambda_2(G)\ge\lambda_2(K_1\vee(2K_2\cup 2K_3))=\sqrt{2}-2>-0.5858$.

If $n_4\ge 6$, then $K_1\vee (3K_2\cup
K_6)\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_1\vee (3K_2\cup K_6))=-0.5856>-0.5858$.

Hence, together with Tabel~\ref{tab4}, we have $\lambda_2(G)\in(-1,-0.5858)$ if and only if item~(v) holds.
\end{proof}

\begin{table}[h]
\small \caption{The second distance eigenvalue of $K_1\vee (3K_2\cup K_{n_4})$ for $2\le n_4\le 5$.}\label{tab4}
\bigskip
\renewcommand{\arraystretch}{1.6}
\centering
\begin{tabular}{|c|c|c|c|c|}
\noalign{\hrule height 0.5pt}

$n_4$ & $2$ & $3$ & $4$ & $5$\\
\noalign{\hrule height 0.4pt}
\hline $\lambda_2$ & $-0.5887$ & $-0.5872$ & $-0.5864$ & $-0.5859$\\
\noalign{\hrule height 0.4pt} \hline
\end{tabular}
\end{table}

Let $C_n$ be the cycle on $n$ vertices.

Let $G$ be a graph. Let $|G|=|V(G)|$. For $e\in E(G)$, let $G-e$ be the graph obtained from $G$ by deleting the edge $e$.
Let $\overline{G}$ be the complement of $G$.

\begin{Lemma}\label{com}\cite[p.~10]{Me2001}
Let $G$ be a graph with at least two vertices. If both $G$ and $\overline{G}$ are connected, then $G$ contains an induced subgraph isomorphic to $P_4$.
\end{Lemma}

Now we are ready to prove our main result.

\begin{Proof}
If $G$ is not  complete, then

$P_3\trianglelefteq G$, and thus $\lambda_2(G)\ge
\lambda_2(P_3)=1-\sqrt{3}>-1$. For $n\ge 2$,
$D(K_n)=J_n-I_n$, and thus $\lambda_2(K_n)=-1$. Thus
$\lambda_2(G)\in(-\infty, -1]$ if and only if $G\cong K_n$ for some
$n\ge 2$.

Suppose that $G$ is a connected graph with
$\lambda_2(G)\in(-1,-0.5858)$.

Suppose that $\overline{G}$ is connected. By Lemma~\ref{com}, $G$
contains an induced subgraph isomorphic to $P_4$. Then
$P_4\trianglelefteq G$ or $C_5\trianglelefteq G$, and thus $\lambda_2(G)\ge
\min\{\lambda_2(P_4),\lambda_2(C_5)\}=\lambda_2(P_4)=-0.58579>-0.5858$,
a contradiction. It follows that $\overline{G}$ is disconnected. Let
$H_1,\dots,H_k$ be the (connected) components of $\overline{G}$,
where $k\ge 2$. Let $G_i=\overline{H_i}$, where $1\le i\le k$.
Obviously, $G_i$ is an induced subgraph of $G$ for $1\le i\le k$,
and $G\cong\vee_{i=1}^k G_i$.

\noindent {\bf Claim 1.} If  $|G_i|\ge 2$, then $G_i$ is disconnected in $G$, where $1\le i\le k$.

Suppose that $|G_i|\ge 2$ and $G_i$ is a connected subgraph of $G$ for some $i$
with $1\le i\le k$. Note that $H_i=\overline{G_i}$ is connected in
$\overline{G}$. By Lemma~\ref{com}, $G_i$ contains an induced
subgraph isomorphic to $P_4$. Since $k\ge 2$, we have
$K_1\vee P_4\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_1\vee P_4)=-0.3820>-0.5858$, a
contradiction. This proves Claim $1$.

If $|G_i|\ge 2$ for each $i$ with $1\le i\le k$, then by Claim $1$,
$G_i$ is disconnected in $G$, which, by noting that $k\ge
2$, implies that $2K_1\vee 2K_1\cong C_4\trianglelefteq G$, and thus $\lambda_2(G)\ge\lambda_2(C_4)=0$, a
contradiction. Thus, $|G_i|=1$ for some $i$ with $1\le i\le k$, say
$i=1$.

Since $\lambda_2\left(\vee_{i=1}^{n}K_1\right)=\lambda_2(K_n)=-1\not\in(-1,-0.5858)$,
we have $|G_j|\ge 2$ for some $j$ with $2\le j\le k$, say $j=2$. By
Claim~$1$, $G_2$ is disconnected in $G$. If $k\ge 3$, then
$K_4-e\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(K_4-e)=\frac{3-\sqrt{17}}{2}>-0.5858$, a contradiction. It
follows that $k=2$, i.e., $G\cong G_1\vee G_2=K_1\vee G_2$, where $G_2$
is a disconnected subgraph of $G$.

If $G_2$ has at least five components, then $K_1\vee 5K_1\cong S_6\trianglelefteq G$, and thus
$\lambda_2(G)\ge\lambda_2(S_6)=4-\sqrt{21}>-0.5858$, a contradiction.
If $G_2$ has a component which is not complete, then there exists a vertex $u$ in this component which has two
nonadjacent neighbors $v$ and $w$, and thus the subgraph of $G$
induced by $V(G_1)\cup\{u,v,w\}$ is isomorphic to $K_4-e$, implying
that $K_4-e\trianglelefteq G$, also a contradiction. Thus
$G\cong K_1\vee(K_{n_1}\cup\dots\cup K_{n_r})$, where $2\le r\le 4$ and
$n_i\ge 1$ for each $i$ with $1\le i\le r$. Now the result follows from Lemmas~\ref{Kn12-Kn123}
and~\ref{Kn1234}.
\end{Proof}

\bigskip

\noindent {\bf Acknowledgement.} This work was supported by  the
Specialized Research Fund for the Doctoral Program of Higher
Education of China (No.~20124407110002) and the National Natural
Science Foundation of China (No.~11071089).


text:
\begin{document}

\title{\Large Computing j-multiplicity}

\author{$\mbox{Koji Nishida}^\ast$
\and
$\mbox{Bernd Ulrich}^\dagger$}
\date{
\begin{flushleft}
\small
\hspace{10ex} ${}^\ast$\,Department of Mathematics and Informatics, Chiba University, \\
  \hspace*{15ex} 1-33 Yayoi-Cho, Inage-Ku, Chiba-Shi, 263-8522, Japan; \\
  \hspace*{15ex} e-mail: nishida@math.s.chiba-u.ac.jp \\
\hspace{10ex} ${}^\dagger$\,Department of Mathematics, Purdue University, \\
  \hspace*{15ex} West Lafayette, IN 47907-1395, USA; \\
  \hspace*{15ex} e-mail: ulrich@math.purdue.edu
\end{flushleft}
}

\maketitle

\begin{abstract}
The j-multiplicity is an invariant that can be defined for
any ideal in a Noetherian local ring $(R, \gm)$.
It is equal to the Hilbert-Samuel multiplicity
if the ideal is $\gm$-primary.
In this paper we explore the computability of the j-multiplicity,
giving another proof for the length formula and
the additive formula.
\end{abstract}

\section{Introduction}
Let $(R, \gm)$ be a Noetherian local
ring of of dimension $d > 0$ and let $I$ be a proper
ideal of $R$.
We set
\[
\jm{I} = \lim_{n \rightarrow \infty}\,
\frac{(d - 1)!}{n^{d-1}}\hspace{1ex}
\length{R}{\lc{0}{\gm}{I^n/I^{n+1}}}
\]
and call it the {\it j-multiplicity} of $I$,
where $\lc{0}{\gm}{\hspace{1ex}\cdot\hspace{1ex}}$
denotes the $0$-th local cohomology functor.
It is obvious that $\jm{I}$ coincides with
the usual multiplicity $\mult{I}{R}$ if $I$
is an $\gm$-primary ideal.
Moreover, the j-multiplicity enjoys a lot of
properties similar to those of the usual multiplicity
(see \cite{am}, \cite{fm1}, \cite{fm2}, \cite{fov}).

In this paper we show some computation of the j-multiplicity
of certain ideals using the length formula which was first
proved by Achilles and Manaresi in \cite[(3.8)]{am}.
We will give another proof for the formula
taking a different approach.
The length formula can be stated as follows:
If $A/\gm$ is an infinite field,
then choosing sufficiently general elements
$a_1, \dots\,, a_{d-1}, a_d$ of $I$, we have
\[
\jm{I} =
\length{R}{R\, /\, {((a_1, \dots\,, a_{d-1}) :_R I^n)+a_dR}}
\]
for $n \gg 0$.
Let $\ga = (a_1, \dots\,, a_{d-1})$ and
$\gb = \ga :_R I^n$ for $n \gg 0$.
Then the equality above means
$\jm{I} = \mult{a_dR}{R / \gb} =
\mult{I}{R / \gb}$
since $R / \gb$ is the zero ring or
a Cohen-Macaulay ring of dimension one.
If $R$ itself is Cohen-Macaulay and
$I$ satisfies a certain Artin-Nagata property
(cf. \cite{u}),
then $I \cap \gb = \ga$, and so we have an exact sequence
$0 \rightarrow I / \ga \rightarrow
R / \gb \rightarrow R / I + \gb \rightarrow 0$,
which implies
$\mult{a_dR}{R / \gb} = \mult{a_dR}{I / \ga} =
\length{R}{I / \ga + a_dI}$\,.
Thus we get
\[
\jm{I} = \length{R}{I / \ga + a_dI}\,,
\]
which is the length formula of Flenner and Manaresi
proved in \cite{fm2}.

Section 2 of this paper is devoted to recalling
some known facts about j-multiplicity in the graded case.
Our proofs are slightly simpler than the original ones.
In Section 3, we prove the length formula
and give some corollaries.
Applying our argument, we will give yet another proof
for the additivity of the j-multiplicity in
\ref{3.8}.
Moreover, we consider the associativity formula
of j-multiplicity in \ref{3.10}.
In the last section, we investigate some concrete
examples in order to illustrate how to use
the formulas given in section 3.

\section{Graded case}
Let $G = \oplus_{n \geq 0}\,G_n$ be a positively
graded Noetherian ring such that
$(G_0, \gn)$ is local and $G = G_0[\,G_1\,]$.
We assume that $G_0 / \gn$ is infinite.
Let $L = \oplus_{n \geq 0}\,L_n$
be a finitely generated graded $G$-module.
We set $W = \lc{0}{\gn G}{L}$ and take an integer $k \gg 0$.
Then, as $W \cap \gn^kL = 0$,
$W$ can be imbedded into $L / \gn^k L$,
and so $\kdim{G}{W} \leq \ell(G, L)$,
where $\ell(G, L)$ denotes the Krull dimension of
$L / \gn L$ as a $G$-module.
Furthermore, as $W$ is a finitely generated graded
$G / \gn^k G$-module and as $G / \gn^k G$
is a standard algebra over an Artinian local ring,
there exists a polynomial $P(n)$ in $n$
such that $\length{G_0}{W_n} = P(n)$ for $n \gg 0$.
Now we take a positive integer $d$ such that
$\kdim{G}{L} \leq d$.
Then the degree of $P(n)$ is at most $d-1$
and the coefficient of $n^{d-1}$ is
$\alpha / (d-1)!$ for some integer $\alpha \geq 0$.
We denote this integer $\alpha$ by $\jmult{d}{G}{L}$.
Hence we have
\[
\jmult{d}{G}{L} =
\lim_{n \rightarrow \infty}
\frac{(d-1)!}{n^{d-1}}\hspace{1ex} \length{G_0}{W_n} \geq 0\,.
\]
It is obvious that
$\jmult{d}{G}{L} \neq 0$ if and only if
$\kdim{G}{W} = d$.
Furthermore, we have the following.

\begin{lem}\label{2.1}
{\rm (\cite[6.1.6]{fov})}\hspace{1ex}
$\jmult{d}{G}{L} \neq 0$ if and only if
$\ell(G, L) = d$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
Suppose $\ell(G, L) = d$.
Then there exists $P \in {\rm Supp}_{\,G}\,L / \gn L$ such that
$\dim\, G / P = d$.
Because $P \in {\rm Supp}_{\,G}\,L$ and
$\kdim{G}{L} \leq d$, we have $P \in \ass{G}{L}$.
Hence $P = \ann{G}{x}$ for some $x \in L$.
We notice that $\gn \cdot x = 0$ as $\gn G \subseteq P$.
This means $x \in W$,
and so $P \in \ass{G}{W}$.
Thus we get $\kdim{G}{W} = d$.
The converse is obvious.

\vspace{1em}
Let $f \in G$.
We say that $f$ is $G_+$-filter regular for $L$
if $f \not\in P$ for any $P \in \ass{G}{L}$ with
$G_+ \not\subseteq P$.
In the case where $f \in G_1$,
it is well known that $f$ is
$G_+$-filter regular for $L$ if and only if
the multiplication map
$L_{n-1} \homom{f} L_n$
is injective for $n \gg 0$.

\begin{lem}\label{2.2}
Let $f \in G_1$ be $G_+$-filter regular for $L$.
Then we have the following assertion.

\begin{itemize}
\item[{\rm (i)}]
$f$ is $G_+$-filter regular for $L/W$.

\item[{\rm (ii)}]
$fL_{n-1} \cap W_n = fW_{n-1}$ for $n \gg 0$.

\item[{\rm (iii)}]
If $\kdim{G_0}{L_n} < \kdim{G}{L}$ for all $n \geq 0$,
then $f$ is an ssop for $L$.
\end{itemize}
\end{lem}

{\it Proof.}\hspace{0.5ex}
Because $W = 0 :_{L} \gn^k G$ for some $k > 0$,
we have $\ass{G}{L / W} \subseteq \ass{G}{L}$,
which implies (i).
Then we get (ii) since the multiplication map
$L_{n-1} / W_{n-1} \homom{f}
L_n / W_n$ is injective for $n \gg 0$.
In order to prove (iii),
take $r \gg 0$ so that $L$ can be
generated by the elements in $L_0, L_1, \dots\,, L_r$
and put $L' = \oplus_{n > r}\,L_n$.
Then
\[
G_0 \cap (L' :_G L) \subseteq
\cap_{n = 0}^r\,\ann{G_0}{L_n} \subseteq \ann{G}{L}\,,
\]
and so $G_0 \cap (L' :_G L) \subseteq Q$
for any $Q \in {\rm Supp}_{\,G}\,L$.
Hence, if $Q \in {\rm Supp}_{\,G}\,L$ contais $G_+$,
we have $L' :_G L \subseteq Q$,
and so $Q \in {\rm Supp}_{\,G}\,L / L'$.
This implies that if
$\assh{G}{L} \cap {\mathrm V}(G_+) \neq \phi$,
then $\dim_{\,G}\,L / L' = \dim_{\,G}\,L$,
which implies $\dim_{\,G_0}\,L_n =
\dim_{\,G}\,L$ for some $0 \leq n \leq r$.
Thus we get (iii).

\begin{lem}\label{2.3}
Let $d \geq 2$ and $\kdim{G_0}{L_n} < \kdim{G}{L}$
for any $n \geq 0$.
Then, choosing $f \in G_1$ generally, we have
$\kdim{G}{L / fL} \leq d - 1$ and
$\jmult{d}{G}{L} = \jmult{d-1}{G}{L / fL}$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
Let us choose $f \in G_1$ so that $f$ is $G_+$-filter
regular for $L$.
Then $\kdim{G}{L / fL} \leq d - 1$ by (iii) of \ref{2.2}.
Furthermore, as $\gn G$ contains a non-zero-divisor
of $L / W$, we have $\ell(G, L / W) \leq d - 1$,
and hence we can choose $f \in G_1$ so that
$\ell(G, L / fL + W) \leq d - 2$.
This means $\jmult{d-1}{G}{L / fL + W} = 0$ by \ref{2.1}.
Now we take $n \gg 0$.
Then $fL_{n-1} \cap W_n = fW_{n-1}$ by (ii) of \ref{2.2},
and so we have an exact sequence
\[
0 \ra W_n / fW_{n-1} \ra L_n / fL_{n-1}
\ra L_n / {fL_{n-1}+W_n} \ra 0
\]
Because $W_n / fW_{n-1}$ is a $G_0$-module of finite length,
we get the exact sequence
\[
0 \ra W_n / fW_{n-1} \ra \lc{0}{\gn}{L_n / fL_{n-1}}
\ra \lc{0}{\gn}{L_n / fL_{n-1}+W_n} \ra 0\,.
\]
Therefore $\jmult{d-1}{G}{L / fL} =
\jmult{d-1}{G}{W / fW}$.
On the other hand, as
\[
0 \ra W_{n-1} \homom{f} W_n \ra W_n / fW_{n-1} \ra 0
\]
is exact for $n \gg 0$, we have

\begin{eqnarray*}
\jmult{d-1}{G}{W / fW} & = &
 \lim_{n \rightarrow \infty}
   \frac{(d-2)!}{n^{d-2}}\hspace{1ex}
   \length{G_0}{W_n/fW_{n-1}}  \\
 & = &
 \lim_{n \rightarrow \infty}
   \frac{(d-2)!}{n^{d-2}}\hspace{1ex}
   (\length{G_0}{W_n} - \length{G_0}{W_{n-1}}) \\
 & = &
 \jmult{d}{G}{L}\,,
\end{eqnarray*}

and the proof is complete.

\section{Local case}

Let $I$ be a proper ideal of a Noetherian local ring
$(R, \gm)$ such that $\kdim{}{R} > 0$ and
$R / \gm$ is an infinite field.
Let $M$ be a finitely generated $R$-module and
$d$ be a positive integer such that $\kdim{R}{M} \leq d$.
Then ${\rm gr}_{I}\,(M) = \oplus_{n \geq 0}\,I^nM / I^{n+1}M$
is a finitely generated graded ${\rm gr}_{I}\,(R)$-module
whose Krull dimension is at most $d$.
We set
\[
\jmult{d}{I}{M} := \jmult{d}{{\rm gr}_{I}\,(R)}{{\rm gr}_{I}\,(M)}
\]
and call it the {\it j-multiplicity} of $I$
with respect to $M$. Hence we have
\[
\jmult{d}{I}{M} = \lim_{n \rightarrow \infty}\,
\frac{(d - 1)!}{n^{d-1}}\hspace{1ex}
\length{A}{\lc{0}{\gm}{I^nM/I^{n+1}M}}\,,
\]
and so, if $\kdim{R}{M} = d$ and $M / IM$
has finite length, $\jmult{d}{I}{M}$ coincides with
the usual multiplicity $\mult{I}{M}$.
We denote
$\jmult{\dim R}{I}{R}$ by $\jm{I}$.
Let $\ell(I, M) := \ell({\rm gr}_{I}\,(R), {\rm gr}_{I}\,(M))$
and $\ell(I) = \ell(I, R)$.

\begin{lem}\label{3.1}
$\jmult{d}{I}{M} \neq 0$ if and only if
$\ell(I, M) = d$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
This follows immediately from \ref{2.1}.

\begin{lem}\label{3.2}
Let $d \geq 2$ and $\kdim{R}{M/IM} < \kdim{R}{M}$.
Then, for a general element $a \in I$,
we have $\kdim{R}{M/aM} \leq d - 1$
and $\jmult{d}{I}{M} = \jmult{d-1}{I}{M/aM}$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
We apply \ref{2.3} setting $G = {\rm gr}_{I}\,(R)$ and
$L = {\rm gr}_{I}\,(M)$.
For any $n \geq 0$, $L_n$ is a homomorphic image
of the direct sum of finite number of copies of $M/IM$,
and hence we have
\[
\kdim{G_0}{L_n} \leq \kdim{R}{M/IM} < \kdim{R}{M} = \kdim{G}{L}\,.
\]
Now we choose a general element $a \in I$.
Let $f$ be the initial form of $a$ in $G$.
Then $\kdim{G}{L/fL} \leq d - 1$ and
$\jmult{d}{I}{M} = \jmult{d-1}{G}{L/fL}$ by \ref{2.3}.
Because ${\rm gr}_{I}\,(M/aM)$ is a homomorphic image of
$L/fL$, we have $\kdim{R}{M/aM} \leq d - 1$.
Moreover, we may assume that $a$ is a superficial element of $I$
with respect to $M$,
and so $[L/fL]_n \cong [{\rm gr}_{I}\,(M/aM)]_n$ for $n \gg 0$.
Then we have $\jmult{d-1}{G}{L/fL} = \jmult{d-1}{I}{M/aM}$
by the definition of j-multiplicity and the proof is complete.

\vspace{1em}
For an $R$-submodule $K$ of $M$, we set $K :_M I^\infty =
\bigcup_{n \geq 0} (K :_M I^n)$.

\begin{lem}\label{3.3}
Let $\ic{M} = M / {0 :_M I^\infty}$.
Then $\jmult{d}{I}{M} = \jmult{d}{I}{\ic{M}}$.
Furthermore we have $\kdim{R}{\ic{M}/I\ic{M}} < \kdim{R}{\ic{M}}$
if $\ic{M} \neq 0$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
Let $n \gg 0$.
Then $I^nM \cap (0 :_M I^\infty) = 0$,
which implies $I^n\ic{M}/I^{n+1}\ic{M} \cong I^nM/I^{n+1}M$,
and hence the first assertion follows.
On the other hand, as $I$ contains a non-zero-divisor of $\ic{M}$,
we get the second assertion.

\begin{lem}\label{3.4}
Let $a$ be a general element of $I$.
We put $\ic{M} = M/{(0 :_M I^\infty)}$.
Then we have $\kdim{R}{\ic{M}/a\ic{M}} \leq d - 1$ and
$\jmult{d}{I}{M} = \jmult{d-1}{I}{\ic{M}/a\ic{M}}$.
\end{lem}

{\it Proof.}\hspace{0.5ex}
If $\ic{M} \neq 0$,
the assertion immediately follows from \ref{3.2} and \ref{3.3}.
If $\ic{M} = 0$, then $\jmult{d}{I}{M} = 0$
since $I^nM = 0$ for $n \gg 0$,
and so the assertion is still true.

\begin{lem}\label{3.5}
Let $1 \leq i < d$.
Then, choosing general elements
$a_1, \dots\,, a_i$ of $I$,
we have $\kdim{R}{M/((a_1, \dots\,, a_i)M :_M I^\infty)}
\leq d - i$ and
\[
\jmult{d}{I}{M} =
\jmult{d-i}{I}{M/{(a_1, \dots\,, a_i)M :_M I^\infty}}\,.
\]
\end{lem}

{\it Proof.}\hspace{0.5ex}
We use induction on $i$.

Let us consider the case where $i = 1$.
We set $\ic{M} = M/(0 :_M I^\infty)$ and
$M' = \ic{M}/a_1\ic{M}$.
Then by \ref{3.3} and \ref{3.4} we have $\kdim{R}{M'} \leq d - 1$
and
$\jmult{d}{I}{M} = \jmult{d-1}{I}{M'}
= \jmult{d-1}{I}{M'/(0 :_{M'} I^\infty)}\,$.
Because the kernel of the composition of the canonical surjections
$M \rightarrow \ic{M} \rightarrow M' \rightarrow
M'/(0 :_{M'} I^\infty)$ is $a_1M :_M I^\infty$,
we get $\kdim{R}{M/(a_1M :_M I^\infty)} \leq d - 1$
and
$\jmult{d}{I}{M} = \jmult{d-1}{I}{M/(a_1M :_M I^\infty)}\,$.

Now we suppose $1 < i < d$ and set
$N = M/((a_1, \dots\,, a_{i-1})M :_M I^\infty)$.
Then $\kdim{R}{N} \leq d - i +1$ and
$\jmult{d}{I}{M} = \jmult{d-i+1}{I}{N}$ by the inductive hypothesis.
Furthermore, by the assertion in the case where $i = 1$,
we have $\kdim{R}{N/(a_iN :_N I^\infty)} \leq d - i$
and $\jmult{d-i+1}{I}{N} =
\jmult{d-i}{I}{N/(a_iN :_N I^\infty)}$.
on the other hand, the kernel of the composition
of the canonical surjections
$M \rightarrow N \rightarrow N/(a_iN :_N I^\infty)$
is $(a_1, \dots\,, a_i)M :_M I^\infty$.
Therefore we get the required assertion.

\begin{thm}\label{3.6}
Choosing general elements
$a_1, \dots\,, a_{d-1}, a_d$ of $I$, we have
\begin{eqnarray*}
\jmult{d}{I}{M} & = &
 \mult{I}{M\,/\,((a_1, \dots\,, a_{d-1})M :_M I^\infty)} \\
      & = &
 \length{R}{M\,/\,{((a_1, \dots\,, a_{d-1})M :_M I^\infty) + a_dM}}\,.
\end{eqnarray*}
\end{thm}

{\it Proof.}\hspace{0.5ex}
We put $N = M/{((a_1, \dots\,, a_{d-1})M :_M I^\infty)}$.
Then $\kdim{R}{N} \leq 1$ and $\jmult{d}{I}{M} =
\jmult{1}{I}{N}$ by \ref{3.5}.
Therefore it is enough to show that
$\jmult{1}{I}{N} = \length{R}{N/{a_dN}}$.
This is obvious if $N = 0$.
So, let us consider the case where $N \neq 0$.
In this case, we may assume that $a_d$ is an $N$-regular
element and $a_dR$ is a reduction of $I$ with respect to $N$.
Moreover, $N/IN$ has finite length.
Therefore we have $\jmult{1}{I}{N} =
\mult{I}{N} = \mult{a_dR}{N} = \length{R}{N/{a_d}N}$
and the proof is complete.

\begin{rem}\label{3.11}
Suppose that $I$ is generated by $d$ elements
and $b \in I \setminus \gm I$.
Then we can choose general elements
$a_1, \dots\,, a_{d-1}$ of $I$
so that they generate $I$ together with $b$.
In this case we have
\[
\jmult{d}{I}{M} =
\length{R}{M\,/\,((a_1, \dots\,, a_{d-1})M :_M I^\infty}) + bM\,.
\]
\end{rem}

Let $S$ be a Noeterian ring and $s$ an integer.
We say that an ideal $J$ of $S$
satisfies $G_s$
if $\mu_{S_P}(JS_P) \leq \height{S}{P}$
for any $P \in {\rm V}(J)$ with $\height{S}{P} < s$.
An ideal $K$ of $S$ is called an
$s$-residual intersection of $J$
if $K = \ga :_S J$ for some $s$-generated ideal $\ga \subseteq J$
with $\height{S}{(\ga :_S J)} \geq s$.
If $K$ is an $s$-residual intersection of $J$ with
$\height{S}{(J + K)} \geq s + 1$,
then $K$ is called a geometric $s$-residual intersection of $J$.

\begin{cor}\label{3a}
Let $S = \oplus_{n \geq 0} S_n$ be a
$d$-dimensional standard algebra such that $S_0 = k$
is an infinite field.
We put $\gn = S_+$ and $R = S_\gn$.
Let $0 < r \in \zz$ and $J$ be an ideal of $S$
generated by homogeneous elements of degree $r$.
We put $I = JR$.
Suppose that $\ell(I) = d$ and
$J$ satisfies $G_d$.
Then, if $f_1, \dots\,, f_{d-1} \in J$ are
general homogeneous elements of degree $r$, we have
\[
\jm{I} = r \cdot
\mult{}{S\,/\,((f_1, \dots\,, f_{d-1}) :_S J)}\,,
\]
where $\mult{}{\ast}$ denotes the multiplicity
with respect to $\gn$.
\end{cor}

{\it Proof.}\hspace{0.5ex}
We put $\ga = (f_1, \dots\,, f_{d-1})$ and
$T = S\,/\,(\ga :_S J^\infty)$.
As $\ell(I) = d$, we have $T \neq 0$,
and so $T$ is a Cohen-Macaulay $S$-module
with $\kdim{S}{T} = 1$.
We notice that $\mult{}{T} = \length{k}{T_i}$ for $i \gg 0$.
On the other hand, by \ref{3.6} it follows that
$T\,/\,JT$ has finite length and
$\jm{I} = \mult{J}{T}$.
Now we take a homogeneous element $f_d \in J$
of degree $r$ so that $f_dS$ is a reduction of $J$
with respect to $T$.
Then $\jm{I} = \length{S}{T\,/\,f_dT}$.
From the exact sequence
$0 \ra T(-r) \stackrel{f_d}{\ra}
T \ra T\,/\,f_dT \ra 0$
of graded $S$-modules, we get
$\length{k}{[T\,/\,f_dT]_i} =
\length{k}{T_i} - \length{k}{T_{i-r}}$
for any $i \in \zz$.
Therefore, taking $n \gg 0$, we have
\[
\length{S}{T\,/\,f_dT} = \sum_{i = n-r+1}^n T_i =
r \cdot \mult{}{T}\,.
\]
Thus we get $\jm{I} = r \cdot \mult{}{T}$,
and hence it is enough to show that
$\mult{}{T} = \mult{}{T'}$,
where $T' = S\,/\,\ga :_S J$.
Because $J$ satisfies $G_d$,
$J + (\ga :_S J)$ is $\gn$-primary.
Hence $(\ga :_S J^\infty)\,/\,(\ga :_S J)$
has finite length as it is annihilated by
$J^\nu + (\ga :_S J)$ for $\nu \gg 0$.
Therefore we get $\mult{}{T} = \mult{}{T'}$
from the exact sequence
$0 \ra (\ga :_S J^\infty)\,/\,(\ga :_S J) \ra T' \ra T \ra 0$,
and the proof is complete.

\vspace{1em}

We say that an ideal $J$ in a Noetherian ring $S$ is
({\it weakly}) {\it $s$-residually $S_2$}
if for every $i$ with $\height{S}{J} \leq i \leq s$
and every (geometric) $i$-residual intersection $K$ of $J$,
$R\,/\,K$ is $S_2$.

\begin{cor}\label{3b}
Let $R$ be Cohen-Macaulay and $d = \kdim{}{R}$.
Suppose that $I$ satisfies $G_d$ and $I$ is
weakly $(d-2)$-residually $S_2$.
Then, choosing general elements
$a_1, \dots\,, a_{d-1}, a_d$ of $I$, we have
\[
\jm{I} = \length{R}{R\,/\,((a_1, \dots\,, a_{d-1}) :_R I) + a_dR}\,.
\]
\end{cor}

{\it Proof.}\hspace{0.5ex}
We put $\ga = (a_1, \dots\,, a_{d-1})$ and $N = R\,/\,(\ga :_R I)$.
As $I$ satisfies $G_d$, $\ga :_R I$ is a geometric
$(d - 1)$-residual intersection of $I$ by \cite[1.4]{u},
and so $N$ is the zero module or
a Cohen-Macaulay $R$-module with $\kdim{R}{N} = 1$
(cf. \cite[3.4]{ceu}).
In particular, $\ga R_\gp = IR_\gp$
for any $\gp \in {\rm V}(I) \setminus \{ \gm \}$
and $\gm \not\in \ass{R}{N}$.
Hence $\ga :_R I = \ga :_R I^\infty$, and so
$\jm{I} = \mult{I}{N} =
\length{R}{N\,/\,a_dN}$ by \ref{3.6},
which means the required equality.

\vspace{1em}

Next we give another proof for the additivity of j-multiplicity
which was first proved by Flenner, O'Carroll and Vogel in
\cite[6.1.7]{fov}.
For that purpose, we prepare the following result.

\begin{lem}\label{3.7}
Let $d = 1$.
We put $\pp = \{ \gp \in {\rm Spec}\,R \mid
\mbox{$\kdim{}{R/\gp} = 1$ and $I \not\subseteq \gp$}\}$.
Then
\[
\jmult{1}{I}{M} = \sum_{\gp \in \pp}\,
\length{R_\gp}{M_\gp}\cdot \mult{I}{R/\gp}\,.
\]
\end{lem}

{\it Proof.}\hspace{0.5ex}
As $\jmult{1}{I}{M} = \jmult{1}{I}{M/(0 :_M I^\infty)}$
by \ref{3.3} and as $M_\gp = (M/(0 :_M I^\infty))_\gp$
for any $\gp \in \pp$,
we may assume ${0 :_M I^\infty} = 0$ and $M \neq 0$.
Then $I$ contains an $M$-regular element.
Hence $\kdim{R}{M} = 1$ and $M/{IM}$ has finite length.
Therefore we have
\[
\jmult{1}{I}{M} = \mult{I}{M} = \sum_{\gp \in \assh{R}{M}}\,
\length{R_\gp}{M_\gp}\cdot \mult{I}{R/\gp}\,.
\]
As $I$ contains an $M$-regular element,
$I \not\subseteq \gp$ for any $\gp \in \assh{R}{M}$,
and so $\assh{R}{M} = \pp \cap {\rm Supp}_{\,R}\,M$.
Thus we get the required equality.

\vspace{1em}

The proof of the next result was suggested by S. Goto.

\begin{thm}\label{3.8} \hspace{0.5ex}
If $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$
is an exact sequence of $R$-modules, then
$\jmult{d}{I}{M} = \jmult{d}{I}{L} + \jmult{d}{I}{N}\,$.
\end{thm}

{\it Proof.}\hspace{0.5ex}
We use induction on $d$.
If $d = 1$, then $\length{R_\gp}{M_\gp} =
\length{R_\gp}{L_\gp} + \length{R_\gp}{N_\gp}$
for any $\gp \in {\rm Spec}\,R$ with $\kdim{}{R/\gp} = 1$,
and hence we get $\jmult{1}{I}{M} = \jmult{1}{I}{L}
+ \jmult{1}{I}{N}$ by \ref{3.7}.
So, let us consider the case where $d \geq 2$.
We set $\ic{L} = L/(0 :_L I^\infty)$,
$\ic{M} = M/(0 :_M I^\infty)$ and $\ic{N} = N/(0 :_N I^\infty)$.
Let $a$ be a sufficiently general element of $I$.
By \ref{3.4} it is enough to show
\begin{equation}\label{eq:1}
\jmult{d-1}{I}{\ic{M}/{a\ic{M}}} =
\jmult{d-1}{I}{\ic{L}/{a\ic{L}}} +
\jmult{d-1}{I}{\ic{N}/{a\ic{N}}}\,.
\end{equation}
Now we consider the exact sequences
\begin{eqnarray*}
 & & 0 \ra E \ra \ic{M} \ra \ic{N} \ra 0 \hspace{2ex}\mbox{and}  \\
 & & 0 \ra \ic{L} \ra E \ra F \ra 0\,,
\end{eqnarray*}
where $E = (L :_M I^\infty)/(0 :_M I^\infty)$ and
$F = (L :_M I^\infty)/((0 :_M I^\infty) + L)$.
As $a$ can be chosen so that it is a non-zero-divisor on
$\ic{M}$ and $\ic{N}$, we get the exact seuences
\begin{eqnarray*}
 & & 0 \ra E/{aE} \ra \ic{M}/{a\ic{M}} \ra \ic{N}/{a\ic{N}}
    \ra 0  \hspace{2ex}\mbox{and}     \\
 & & 0 \ra {0 :_F a} \ra \ic{L}/{a\ic{L}} \ra E/{aE}
    \ra F/{aF} \ra 0 \,.
\end{eqnarray*}
Then, the inductive hypothesis implies
\begin{eqnarray*}
\jmult{d-1}{I}{\ic{M}/{a\ic{M}}} & = &
  \jmult{d-1}{I}{E/{aE}} + \jmult{d-1}{I}{\ic{N}/{a\ic{N}}}  \\
\jmult{d-1}{I}{E/{aE}} & = &
  \jmult{d-1}{I}{X} + \jmult{d-1}{I}{F/{aF}} \hspace{2ex}
  \mbox{and}   \\
\jmult{d-1}{I}{\ic{L}/{a\ic{L}}} & = &
  \jmult{d-1}{I}{0 :_F a} + \jmult{d-1}{I}{X}\,,
\end{eqnarray*}
where $X$ denotes the kernel of $E/{aE} \rightarrow F/{aF}$.
Because $I^nF = 0$ for $n \gg 0$,
we have $\jmult{d-1}{I}{F/{aF}} = \jmult{d-1}{I}{0 :_F a} = 0$,
and so $\jmult{d-1}{I}{E/{aE}} = \jmult{d-1}{I}{X} =
\jmult{d-1}{I}{\ic{L}/{a\ic{L}}}$.
Therefore we get the required equality (\ref{eq:1})
and the proof is complete.

\begin{thm}\label{3.9} \hspace{0.5ex}
${\displaystyle
\jmult{d}{I}{M} = \sum_{\gp \in \assh{R}{M}}
\length{R_\gp}{M_\gp}\cdot \jmult{d}{I}{R/\gp}\,.}$
\end{thm}

{\it Proof.}\hspace{0.5ex}
Similarly as the proof of the additive formula for
the usual multiplicity, by \ref{3.8} we can prove
the required equality using induction on
$\sigma(M) =
\sum_{\gp \in \assh{R}{M}}\,\length{R_\gp}{M_\gp}$.

\begin{thm}\label{3.10}
Let $\ell(I) = \kdim{}{R} = d$ and $1 \leq i < d$.
Choosing general elements
$a_1, \dots\,, a_i$ of $I$,
we set $\ga = (a_1, \dots\,, a_i)$ and
$\pp = \{\,\gp \in \Min{R}{R / \ga} \mid
\mbox{$\height{R}{\gp} = i$ and $\kdim{}{R / \gp} = d - i$}\,\}$.
Then we have the following asertions.
\begin{itemize}
\item[{\rm (i)}]
$\pp \neq \phi$
\item[{\rm (ii)}]
$R_\gp$ is Cohen-Macaulay for any $\gp \in \pp$.
\item[{\rm (iii)}]
${\displaystyle
\jm{I} = \sum_{\gp \in \pp}\,
\jm{\ga R_\gp}\cdot\jm{I + \gp\,/\,\gp}\,.
}$
\end{itemize}
\end{thm}

{\it Proof.}\hspace{0.5ex}
Let $S = R / ((a_1, \dots\,, a_i) :_R I^\infty)$.
Then by \ref{3.5} and \ref{3.9} we have
$\kdim{R}{S} \leq d - i$ and
\[
\jm{I} = \sum_{\gp \in \assh{R}{S}}\,
\length{R_\gp}{S_\gp}\cdot\jmult{d-i}{I}{R / \gp}\,.
\]
Because $\jm{I} \neq 0$ by \ref{3.1},
there exists $\gp \in \assh{R}{S}$ such that
$\kdim{}{R / \gp} = d - i$.
Then $\height{R}{\gp} \leq i$,
and so $I \not\subseteq \gp$ by \cite[7.2]{aht}.
Therefore $S_\gp = R_\gp / \ga R_\gp$.
Here we notice that we can choose
$a_1, \dots\,, a_i$ so that
$a_j$ is regular on $R / ((a_1, \dots\,, a_{j-1}) :_R I^\infty)$
for any $1 \leq j \leq i$.
Then $a_1, \dots\,, a_i$ is a regular sequence on $R_\gp$,
and hence $R_\gp$ is a Cohen-Macaulay ring with
$\kdim{}{R_\gp} = i$.
Thus we have seen (i) together with (ii).
Furthermore we get (iii) since
$\pp = \{\,\gp \in \assh{R}{S} \mid
\kdim{}{R / \gp} = d - i\,\}$ and
$\jm{\ga R_\gp} = \mult{\ga R_\gp}{R_\gp} =
\length{R_\gp}{S_\gp}$ for any $\gp \in \pp$.

\section{Examples}
In this section we give some examples in order to
illustrate how to use \ref{3.6}.

\begin{ex}\label{4.1}
Let $S$ be a $3$-dimensional Cohen-Macaulay local ring
and $x, y, z$ be a system of parameters for $S$.
We set $R = S/{(x^2-yz)S}$ and $I = (x, y)R$.
Then we have
\[
\jm{I} = \length{S}{S/{(x, y, z)S}}\,.
\]
\end{ex}

{\it Proof.}\hspace{0.5ex}
If $\gp \in \ass{R}{R/I}$, then $\height{R}{\gp} = 1$
as $R/I = S/{(x, y)S}$ is a $1$-dimensional Cohen-Macaulay
ring. So the image of $z$ in $R$ becomes unit in $R_\gp$,
which means $IR_\gp = xR_\gp$.
Hence $I$ is generically a complete intersection.
On the other hand, if $f, g \in S$ satisfy
$xf - yg = (x^2 - yz)h$ for some $h \in S$,
we get $x(f - xh) = y(g - zh)$.
Hence there exists $\ell \in S$ such that
$f - xh = y\ell$ and $g - zh = x\ell$,
since $x, y$ form an $S$-regular sequence.
This means that we have an exact sequence
\begin{equation}
\label{4.2}
\begin{array}{ccccccc}
R^2 & \ra & R^2 & \ra & I & \ra & 0\,. \\
 & \left(\begin{array}{cc}
   y & x \\
   x & z
   \end{array}\right) & & \left(\begin{array}{cc}
                          x & -y
                          \end{array}\right) & & &
\end{array}
\end{equation}
Now we choose a general element $\xi$ of $I$
so that $I = \xi R +yR$.
Then $\xi R :_R y = \xi R :_R I \subseteq \xi R :_R I^\infty$.
Let $\gq \in \ass{R}{R/(\xi R :_R y)}
\subseteq \ass{R}{R / {\xi R}}$.
Then $\height{R}{\gq} = 1$.
If $I \subseteq \gq$,
we have $IR_\gq = \xi R_\gq$
as $I$ is generically a complete intersection,
and so $(\xi R :_R I^\infty)R_\gq = R_\gq =
(\xi R :_R y)R_\gq$.
If $I \not\subseteq \gq$,
we have $(\xi R :_R I^\infty)R_\gq = \xi R_\gq =
(\xi R :_R y)R_\gq$
since $y \not\in \gq$.
Thus we get $\xi R :_R I^\infty = \xi R :_R y$,
and hence
\[
\jm{I} = \length{R}{R/{(\xi R :_R y) + yR}}
\]
by \ref{3.6} and \ref{3.7}.
Here we notice that $\xi$ can be written as the
image of $x + uy$ for some unit $u$ of $S$.
It is enough to show
$\xi R :_R y = (x + uy, z + ux)R$.
One can easily check that the right hand side is
contained in the left hand side.
Let us take any $\rho \in \xi R :_R y$
and write $y\rho = (x + uy)\eta$ for some $\eta \in R$.
Then $x\eta - y(\rho - u\eta) = 0$,
and so from the exact sequence (\ref{4.2}) we get
\[
\left(\begin{array}{c}
\eta \\
\rho - u\eta
\end{array}\right)
=
\left(\begin{array}{cc}
y & x \\
x & z
\end{array}\right)\,
\left(\begin{array}{c}
\sigma \\
\tau
\end{array}\right)
\]
for some $\sigma, \tau \in R$.
This means
$\rho = \sigma (x + uy) + \tau (z + ux)$.
Thus we see $\xi R :_R y = (x + uy, z + ux)R$
and the proof is complete.

\begin{ex}\label{4.3}
Let $(R, \gm)$ be a $3$-dimensional Cohen-Macaulay local ring
and $I$ an ideal of height two which is generated by
the maximal minors of the matrix
\[
\varphi = \left(\begin{array}{lll}
x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22}& x_{23}
\end{array}\right)
\]
with entries in $R$.
We set $S = R/I$ and asume that each one of
$x_{11}, x_{12}$ and $x_{13}$ is a system of parameters for $S$.

\begin{itemize}
\item[{\rm (i)}]
For some units $u_1, u_2$ and $u_3$ of $R$, we have
\[
\jm{I} = \length{S}{
S/{(x_{11}u_1+x_{12}u_2+x_{13}u_3, x_{21}u_1+x_{22}u_2+x_{23}u_3)S}}\,.
\]

\item[{\rm (ii)}]
Suppose that $R$ is a Gorenstein ring.
Then $(x_{1j}, x_{2j})S$ is a canonical ideal of $S$
for $j = 1, 2, 3$.
On the other hand, any canonical ideal of $S$ is
written in the form
$(x_{11}a_1+x_{12}a_2+x_{13}a_3, x_{21}a_1+x_{22}a_2+x_{23}a_3)S$
for some elements $a_1, a_2$ and $a_3$ of $R$.

\item[{\rm (iii)}]
Suppose that $S$ is an integral domain
whose normalization is a DVR.
Then we have
\[
\jm{I} = \min\{\,
\length{S}{S/{(x_{1j}, x_{2j})S}}\, \}_{j = 1, 2, 3}\,.
\]
Moreover, if $R$ is a Gorenstein ring, we get
\[
\jm{I} = \min\{\,
\length{S}{S/\omega} \mid
\mbox{$\omega$ is a canonical ideal of $S$}\, \}\,.
\]
\end{itemize}
\end{ex}

{\it Proof.}\hspace{1ex}
Let $f_j$ be the determinant of the $2 \times 2$ matrix
derived from $\varphi$ deleting the $j$-th column.
It is well known that

\begin{equation}\label{4.4}
\begin{array}{ccccccccc}
0 & \ra & R^2 & \ra & R^3 & \ra & I \ra & 0 \\
  &  &  & \trmatrix{\,\varphi} &  &
  (f_1\hspace{1ex}{-f_2}\hspace{1ex}f_3) & & &
\end{array}
\end{equation}
is an exact sequence.
In particular, we have that $R/I$ is a Cohen-Macaulay ring and

\begin{equation}\label{4.5}
x_{i1}f_1 - x_{i2}f_2 + x_{i3}f_3 = 0
\hspace{2ex}\mbox{for $i =1, 2$}\,.
\end{equation}

If $\gp \in \assh{R}{R/I}$,
then each one of $x_{11}, x_{12}$ and $x_{13}$
is a unit in $R_\gp$, and so the equality (\ref{4.5})
for $i = 1$ implies that
$IR_\gp$ is generated by any two of $f_1, f_2$ and $f_3$.
Hence $I$ is generically a complete intersection.
On the other hand, for any $j = 1, 2, 3$,
we have that $x_{1j}, x_{2j}, f_j$ is
a system of parameters for $R$ as
$(x_{1j}, x_{2j}, f_j) \supseteq (x_{1j}) + I$,
and so $x_{1j}, x_{2j}$ is an $R$-regular sequence.
Moreover, we see that any two of
$f_1, f_2$ and $f_3$ form an $R$-regular sequence.
In fact, for example,
if there exists a height one prime ideal $\gq$
of $R$ containing $(f_1, f_2)$,
then $f_3 \not\in \gq$ as $\height{R}{I} = 2$,
and so from (\ref{4.5}) we get
$(x_{13}, x_{23}) \subseteq \gq$,
which is impossible as $x_{13}, x_{23}$
is an $R$-regular sequence.

(i)\hspace{1ex}
Let $g_1$ and $g_2$ be sufficiently general elements of $I$.
Then
\[
\jm{I} = \length{S}{S/{((g_1, g_2) :_R I^\infty})S}
\]
by \ref{3.6}.
Let $\gp \in \ass{R}{R/((g_1, g_2) :_R I)} \subseteq
\ass{R}{R/(g_1, g_2)}$.
Then $\height{R}{\gp} = 2$ since $g_1, g_2$
is an $R$-regular sequence.
Therefore, if $I \subseteq \gp$,
we have $IR_\gp = (g_1, g_2)R_\gp$, and so
$((g_1, g_2) :_R I^\infty)R_\gp =
((g_1, g_2) :_R I)R_\gp = R_\gp$.
Even if $I \not\subseteq \gp$, we have
$((g_1, g_2) :_R I^\infty)R_\gp =
((g_1, g_2) :_R I)R_\gp = (g_1, g_2)R_\gp$.
Thus we get $(g_1, g_2) :_R I^\infty
= (g_1, g_2) :_R I$.
We write
\[
g_1 = v_1f_1 + v_2f_2 + v_3f_3 \hspace{2ex}
\mbox{and} \hspace{2ex}
g_2 = w_1f_1 + w_2f_2 + w_3f_3\,,
\]
where $v_j$ and $w_j$ are units of $R$.
We may assume that all maximal minors of the matrix
\[
\psi = \left(\begin{array}{lll}
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3
\end{array}\right)
\]
are units of $R$.
We set
\[
h_1 = v_2g_2 - w_2g_1 \hspace{2ex}
\mbox{and} \hspace{2ex}
h_2 = v_1g_2 - w_1g_1\,.
\]
Then $(g_1, g_2) = (h_1, h_2)$.
Moreover, we have
\[
h_1 = -u_3f_1 + u_1f_3 \hspace{2ex}
\mbox{and} \hspace{2ex}
h_2 = u_3f_2 + u_2f_3\,,
\]
where $u_j$ denotes the determinant of the matrix
derived from $\psi$ deleting the $j$-th column.
Hence we get $(h_1, h_2, f_3) = I$, and so
$(g_1, g_2) :_R I = (h_1, h_2) :_R f_3$.

Let us take any $\rho \in (h_1, h_2) :_R f_3$.
Then there exist $\xi, \eta \in R$ such that
\begin{eqnarray*}
\rho f_3 & = & \xi h_1 + \eta h_2 \\
         & = & \xi (-u_3 f_1 + u_1f_3) + \eta (u_3f_2 + u_2f_3)\,,
\end{eqnarray*}
which induces
\[
\xi u_3 f_1 + \eta u_3(-f_2) + (\rho - \xi u_1 - \eta u_2) f_3 = 0\,.
\]
Therefore, using the exact sequence (\ref{4.4}),
we get
\[
\left(\begin{array}{c}
\xi u_3 \\
\eta u_3 \\
\rho - \xi u_1 - \eta u_2
\end{array}\right)
=
\sigma
\left(\begin{array}{c}
x_{11} \\
x_{12} \\
x_{13}
\end{array}\right)
+ \tau
\left(\begin{array}{c}
x_{21} \\
x_{22} \\
x_{23}
\end{array}\right)
\]
for some $\sigma, \tau \in R$.
This implies $u_3\rho = \sigma y_1 + \tau y_2$,
where $y_i = x_{i1}u_1 + x_{i2}u_2 + x_{i3}u_3$
for $i = 1, 2$.
Thus we see $(h_1, h_2) :_R f_3 \subseteq (y_1,\, y_2)$.
On the other hand, the converse inclusion holds since
$y_3f_3 = x_{i1}h_1 + x_{i2}h_2$
for $i = 1, 2$,
and so the proof of (i) is complete.

(ii)\hspace{1ex}
Let us prove that $(x_{11}, x_{21})S$
is a canonical ideal of $S$.
By the same argument one can prove that
$(x_{12}, x_{22})S$ and $(x_{13}, x_{23})S$
are also canonical ideals of $S$.

Let $\gp \in \ass{R}{R/(f_2, f_3)}$.
Then $\height{R}{\gp} = 2$ as
$f_2, f_3$ is an $R$-regular sequence.
Hence we have $IR_\gp = (f_2, f_3)R_\gp$
when $I \subseteq \gp$.
On the other hand, if $I \not\subseteq \gp$,
$f_1$ is a unit of $R_\gp$, and so
$(x_{11}, x_{21})R_\gp = (f_2, f_3)R_\gp$ by (\ref{4.5}).
Thus we get $(x_{11}, x_{21}) \cap I = (f_2, f_3)$.
Then
\begin{eqnarray*}
(x_{11}, x_{21})S & \cong &
       (x_{11}, x_{21})\, /\, {(x_{11}, x_{21}) \cap I} \\
 & = & (x_{11}, x_{21})\, /\, (f_2, f_3)  \\
 & = & ((f_2, f_3) :_R f_1)\, /\, (f_2, f_3) \\
 & \cong & {\rm Hom}_{\,R}(S, R / (f_2, f_3))\,.
\end{eqnarray*}
Because $S$ is a homomorphic image of the Gorenstein local
ring $R/(f_2, f_3)$, it follows that
${\rm Hom}_{\,R}(S, R / (f_2, f_3))$ is the canonical module of $S$,
and so $(x_{11}, x_{21})S$ is a canonical ideal of $S$.

Now let $\omega$ be any canonical ideal of $S$.
Let $Q$ be the total quotient ring of $S$.
Then there exists $\alpha \in Q$ such that
$\omega = \alpha \cdot (x_{11}, x_{21})S$.
Because $S :_Q (x_{11}, x_{21})S =
{\ic{x_{11}}}^{\,-1}(x_{11}S :_S x_{21})$,
where $\ic{\,\ast\,}$ denotes the reduction mod $I$,
we write $\alpha = {\ic{x_{11}}}^{\,-1}\cdot\ic{y}$
with $y \in R$ such that
$\ic{y} \in x_{11}S :_S x_{21}$.
Then $x_{21}y \in (x_{11}) + I =
(x_{11}, x_{12}x_{21}, x_{13}x_{21}, f_1)$,
and so there exist $z, z' \in R$ such that
$x_{21}(y - x_{12}z - x_{13}z') \in (x_{11}, f_1)$.
This means $y - x_{12}z - x_{13}z' \in (x_{11}, f_1)$
since $x_{11}, x_{21}, f_1$ is an $R$-regular sequence.
Hence $y \in (x_{11}, x_{12}, x_{13}, f_1) =
(x_{11}, x_{12}, x_{13})$. Therefore
$y = x_{11}a_1+ x_{12}a_2 + x_{13}a_3$ for some
$a_1, a_2, a_3 \in R$. Then we have
\begin{eqnarray*}
\omega & = &
       {\ic{x_{11}}}^{\,-1}\cdot\sum_{j=1}^3
         \ic{x_{1j}}\,\ic{a_j} \cdot (x_{11},\, x_{21})S  \\
 & = & (\,\sum_{j=1}^3 \ic{x_{1j}}\,\ic{a_j}\,, \,
        \sum_{j=1}^3
        {\ic{x_{11}}}^{\,-1}\ic{x_{1j}}\,\ic{x_{21}}\,\ic{a_j}\,)\,.
\end{eqnarray*}
Because ${\ic{x_{11}}}^{\,-1}\ic{x_{1j}}\,\ic{x_{21}} =
\ic{x_{2j}}$ for $j = 1, 2, 3$, we get
\[
\omega = (\,\sum_{j=1}^3 x_{1j}a_j\,, \,\sum_{j=1}^3 x_{2j}a_j\,)S
\]
and the proof of (ii) is complete.

In order to prove (iii), we need the following lemma.

\begin{lem}\label{4.6}
Let $(S, \gn)$ be a $1$-dimensional Cohen-Macaulay local ring
and $T$ be an extension ring of $S$ such that
$T / S$ has finite length as an $S$-module.
Let $\ga$ be an $\gn$-primary ideal of $S$ and $v$
be a non-zero-divisor of $T$
such that $\ga v \subseteq S$.
Then $\ga v$ is an $\gn$-primary ideal of $S$ and
\[
\length{S}{S / {\ga v}} \geq \length{S}{S / \ga}\,,
\]
where the equality holds if $v$ is a unit of $T$.
\end{lem}

{\it Proof.}\hspace{1ex}
Let $\gc = S :_S T$,
which is an $\gn$-primary ideal of $S$.
We take an $S$-regular element $u \in \ga$
and set $\gb = u\gc$,
which is an $\gn$-primary ideal of $S$ contained in $\ga$.
Let us consider the exact sequence
\[
0 \ra {\ga v} / {\gb v} \ra S / {\gb v} \ra S / {\ga v} \ra 0\,.
\]
As $v$ is a non-zero-divisor of $T$,
we have ${\ga v} / {\gb v} \cong \ga / \gb$, and so
\[
\length{S}{S / {\ga v}} =
\length{S}{S / {\gb v}} - \length{S}{\ga / \gb}\,.
\]
Now we notice $\gb v = (u\gc)v = u(\gc v)
\subseteq u\gc = \gb$.
Moreover, if $v$ is a unit of $T$,
$\gb = (u\gc)v^{-1}v = u(\gc v^{-1})v \subseteq
u\gc v = \gb v$.
Thus we get the required assertion.

\vspace{1em}
\noindent
{\it Proof of (iii) of \ref{4.3}.}\hspace{1ex}
Let $T$ be the normalization of $S$.
Let $tT$ be the maximal ideal of $T$.
We write $\ic{x_{ij}} = t^{k_{ij}}\alpha_{ij}$,
where $0 \leq k_{ij} \in \zz$ and
$\alpha_{ij}$ is a unit of $T$.
Replacing the columns of $\varphi$,
we may assume $k_{11} \leq k_{12}$ and $k_{11} \leq k_{13}$.

Because $\ic{x_{11}}\,\ic{x_{22}} - \ic{x_{12}}\,\ic{x_{21}} = 0$
and
$\ic{x_{11}}\,\ic{x_{23}} - \ic{x_{13}}\,\ic{x_{21}} = 0$,
we have
\begin{equation}\label{4.7}
t^{k_{11}+k_{22}}\alpha_{11}\alpha_{22} =
t^{k_{12}+k_{21}}\alpha_{12}\alpha_{21}
\hspace{2ex}\mbox{and}\hspace{2ex}
t^{k_{11}+k_{23}}\alpha_{11}\alpha_{23} =
t^{k_{13}+k_{21}}\alpha_{13}\alpha_{21}\,.
\end{equation}
Now we put
$\beta_j = t^{k_{1j} - k_{11}} {\alpha_{11}}^{-1} \alpha_{1j}
\in T$ for $j = 2, 3$.
Then by (\ref{4.7}) we have
$\ic{x_{ij}} = \ic{x_{i1}}\cdot \beta_j$
for any $i = 1, 2$ and $j = 2, 3$,
and so $(x_{1j}, x_{2j})S =
\beta_j \cdot (x_{11}, x_{21})S$ for $j = 2, 3$.
Therefore by \ref{4.6} we get
$\length{S}{S / (x_{11}, x_{21})S} \leq
\length{S}{S / (x_{1j}, x_{2j})S}$ for $j = 2, 3$.

Next we prove
$\jm{I} = \length{S}{S / (x_{11}, x_{21})S}$.
Let $u_1, u_2$ and $u_3$ be units of $S$ stated in (i).
Then $\jm{I} = \length{S}{S / (y_1, y_2)S}$,
where $y_i = x_{i1}u_1 + x_{i2}u_2 + x_{i3}u_3$.
Here we put $\gamma = \ic{u_1} +
t^{k_{12} - k_{11}}{\alpha_{11}}^{-1}\alpha_{12}\ic{u_2} +
t^{k_{13} - k_{11}}{\alpha_{11}}^{-1}\alpha_{13}\ic{u_3}$.
Then $\ic{y_1} = \ic{x_{11}}\gamma$
and $\ic{y_2} = \ic{x_{12}}\gamma$
by (\ref{4.7}).
Let us notice that the units $v_j$ and $w_j$
stated in the proof of (i) can be chosen so that
$\gamma$ is a unit in $T$.
Hence we get $\length{S}{S / (y_1, y_2)S} =
\length{S}{S / (x_{11}, x_{21})S}$ by \ref{4.6}.

The last assertion of (iii) can be proved by
the similar argument using (ii),
and so the proof of \ref{4.3} is complete.

\begin{ex}\label{3c}
Let $R = K[[X, Y, Z]]$ and $T = K[[t]]$
be formal power series rings over an infinite field $K$.
Let $\varphi : R \ra T$ be the homomorphism of $K$-algebras
defined by $\varphi(X) = t^k$, $\varphi(Y) = t^\ell$,
$\varphi(Z) = t^m$, where $k, \ell, m$ are positive integers
such that ${\rm GCD}\{ k, \ell, m \} = 1$.
We put $\gp = {\rm Ker}(\varphi)$.
Then $\gp$ is generated by the maximal minors of the matrix
of the form
\[
\left(\begin{array}{lll}
X^\alpha & Y^{\beta'} & Z^{\gamma'} \\
Y^\beta  & Z^\gamma   & X^{\alpha'}
\end{array}\right)\,,
\]
where $\alpha, \beta, \gamma, \alpha', \beta', \gamma'$
are positive integers {\rm (cf. \cite{h})}.
By replacing the variables X, Y and Z suitably,
we may assume
$k\alpha = \min\{ k\alpha, \ell\beta, m\gamma,
k\alpha', \ell\beta', m\gamma' \}$.
Then we have
\[
\jm{\gp} = \alpha\beta(\gamma + \gamma')\,.
\]
\end{ex}

{\it Proof.}\hspace{0.5ex}
We put $S = R / \gp = K[[t^k, t^\ell, t^m]]$.
By (iii) of \ref{4.3},
$\jm{\gp}$ is the minimum of the lengths of
$S / (t^{k\alpha}, t^{\ell\beta})S$,
$S / (t^{\ell\beta'}, t^{m\gamma})S$ and
$S / (t^{m\gamma'}, t^{k\alpha'})S$.
Here we notice that $\ell(\beta + \beta') = k\alpha + m\gamma$
as $Y^{\beta + \beta'} - X^\alpha Z^\gamma \in \gp$.
This means $m\gamma = \ell\beta' - k\alpha + \ell\beta$,
and so $(t^{\ell\beta'}, t^{m\gamma})S =
t^\delta \cdot (t^{k\alpha}, t^{\ell\beta})S$,
where $\delta = \ell\beta' - k\alpha \geq 0$.
Hence we have $\length{S}{S / {(t^{k\alpha}, t^{\ell\beta})S}}
\leq \length{S}{S / {(t^{\ell\beta'}, t^{m\gamma})S}}$
by \ref{4.6}.
Similary we get
$\length{S}{S / {(t^{k\alpha}, t^{\ell\beta})S}} \leq
\length{S}{S / {(t^{m\gamma'}, t^{k\alpha'})S}}$.
Therefore $\jm{\gp} = \length{S}{S / {(t^{k\alpha}, t^{\ell\beta})S}}$.
Because $S / {(t^{k\alpha}, t^{\ell\beta})S} \cong
R / {\gp + (X^\alpha, Y^\beta)R} =
R / {(X^\alpha, Y^\beta, Z^{\gamma + \gamma'})R}$,
we get the required equality.

\vspace{1em}

\begin{ex}\label{3d}
Let $S, J$ and $I$ be as in \ref{3a}.
Moreover we assume that ${\rm Proj}(S)$ is Cohen-Macaulay
and $JS_P$ is weakly $(d - 3)$-residually $S_2$ for any
$P \in {\rm V}(J) \setminus \{\,\gn\,\}$.
Then, if $\gb \subseteq J$ is an ideal generated by
$d - 1$ homogeneous elements of degree $r$ such that
$\height{S}{(\gb :_S J)} \geq d - 1$,
\[
\jm{I} = r \cdot \mult{}{S / (\gb :_S J)}\,.
\]
\end{ex}

{\it Proof.}\hspace{0.5ex}
By \ref{3a} we have $\jm{I} = r \cdot \mult{}{S / {\ga :_S J}}$,
where $\ga$ is an ideal generated by $d - 1$ {\it general} homogeneous
elements of $J$ of degree $r$.
On the other hand, by \cite[2.1 and 3.1]{u} we have
$\mult{}{S / (\ga :_S J)} = \mult{}{S / (\gb :_S J)}$.
Therefore we get the required equality.

\begin{ex}\label{3e}
Let $S = \oplus_{n \geq 0}\,S_n$ be a $d$-dimensional
Gorenstein standard algebra such that $S_0$ is an
infinite field.
Let $I$ be a homogeneous ideal of
generically a complete intersection such that
$S / I$ is a $2$-dimensional Cohen-Macaulay ring.
Let $J$ be a reduction of $I$ generated by homogeneous
elements $g_1, \dots\,, g_{d-1}, g_d$ of the same degree.
We set $\gn = S_{+}$ and $\gb = (g_1, \dots\,, g_{d-1})$.
Then, if $(\gb :_S J) + J$ is $\gn$-primary,
we have
\[
\jm{IS_\gn} = \length{S}{S / {(\gb :_S J) + J}}\,.
\]
\end{ex}

{\it Proof.}\hspace{0.5ex}
First we show that $J$ satisfies $G_d$ and
$J$ is $(d - 2)$-residually $S_2$.
We take any $P \in {\rm V}(J) = {\rm V}(I)$
with $\height{S}{P} \leq d - 1$.
Then $\gb :_S J \not\subseteq P$, and so
$\gb S_P = JS_P$.
In particular, $\mu_{S_P}(JS_P) \leq d - 1$.
If $\height{S}{P} = d - 2$,
$P \in \Min{S}{S / I}$,
and so $IS_P$ is a complete intersection,
which means $IS_P = JS_P$ as $JS_P$ is a
reduction of $IS_P$.
Hence $\mu_{S_P}(JS_P) = d - 2$ if $\height{S}{P} = d - 2$.
Hence $J$ satisfies $G_d$.
In order to prove that $J$ is $(d - 2)$-residually $S_2$,
it is enough to show that if $\gc \subseteq J$ is
$(d - 2)$-generated ideal with $\height{S}{(\gc :_S J)} \geq d - 2$,
then $S / (\gc :_S J)$ is Cohen-Macaulay.
Because $\height{S}{(\gc :_S J)} \geq d - 2$ and
$\height{S}{J} = d - 2$, we have $\height{S}{\gc} = d - 2$,
and so $\gc$ is generated by a regular sequence.
We take any $Q \in \ass{S}{S / (\gc :_S I)}$.
Then $\height{S}{Q} = d - 2$ as $S / (\gc :_S I)$ is a
$2$-dimensional Cohen-Macaulay ring (cf. \cite[1.3]{ps}).
If $I \subseteq Q$, then $Q \in \Min{S}{S / I}$,
and so $IS_Q = JS_Q$,
which means $(\gc :_S I)S_Q = (\gc :_S J)S_Q$.
On the other hand, if $I \not\subseteq Q$, we have
$J \not\subseteq Q$, and so
$(\gc :_S I)S_Q = \gc S_Q = (\gc :_S J)S_Q$.
Consequently, $\gc :_S I = \gc :_S J$,
and hence $S / (\gc :_S J)$ is Cohen-Macaulay.

Therefore, by \ref{3b} and \ref{3.11}
we have
\[
\jm{JS_\gn} =
\length{S}{S / ((f_1, \dots\,, f_{d-1}) :_S J) + J}
\]
for {\it general} homogeneous elements
$f_1, \dots\,, f_{d-1} \in J$ of degree $r$.
We put $\ga = (f_1, \dots\,, f_{d-1})$.
Then, as $\ga :_S J$ is a geometric $(d - 1)$-residual
intersection of $J$, we have
$(\ga :_S J) \cap J = \ga$ by \cite[3.4]{ceu}.
Thus we get
\begin{equation}\label{4.10}
0 \ra S / \ga \ra S / (\ga :_S J) \oplus S / J
\ra S / {(\ga :_S J) + J} \ra 0
\end{equation}
of graded $S$-modules.
Similary, as $\gb :_S J$ is also a geometric
$(d - 1)$-residual intersection of $J$,
we have the exact sequence
\begin{equation}\label{4.11}
0 \ra S / \gb \ra S / (\gb :_S J) \oplus S / J
\ra S / {(\gb :_S J) + J} \ra 0
\end{equation}
of graded $S$-modules.
Here we notice that by \cite[2.2]{ceu}
the Hilbert function of $S / \ga$ (resp. $S / (\ga :_S J)$)
coincides with that of $S / \gb$ (resp. $S / (\gb :_S J)$).
Therefore by (\ref{4.10}) and (\ref{4.11}) we see that
the Hilbert functions of $S / {(\ga :_S J) + J}$ and
$S / {(\gb :_S J) + J}$ are same,
and hence the lengths of those algebras are same.
Therefore the proof is complete as
$\jm{IS_\gn} = \jm{JS_\gn}$ by \cite[2.10]{fm2}

\begin{ex}\label{3f}
Let $k$ be an infinite field.

\noindent
\begin{itemize}
\item[{\rm (1)}]
Let $S = k[X_1, X_2, X_3, X_4, X_5]$ be a polynomial ring and
$I$ be the ideal generated by the maximal minors of the matrix
\[
\left(\begin{array}{llll}
X_1 & X_2 & X_3 & X_4 \\
X_2 & X_3 & X_4 & X_5
\end{array}\right)\,.
\]
We set $\gn = S_+$. Then $\jm{IS_\gn} = 4$.

\item[{\rm (2)}]
Let $S = k[X_1, X_2, X_3, X_4, X_5, X_6, X_7]$
be a polynomial ring and $I$ be the ideal generated
by the maximal minors of the matrix
\[
\left(\begin{array}{lllll}
X_1 & X_2 & X_3 & X_5 & X_6 \\
X_2 & X_3 & X_4 & X_6 & X_7
\end{array}\right)\,.
\]
We set $\gn = S_+$.
Then $\jm{IS_\gn} = 10$.

\end{itemize}
\end{ex}

{\it Proof.}\hspace{1ex}
(1)\hspace{0.5ex}
It is well known that $I$ is a perfect ideal
of height $3$ which is generically a complete intersection.
Let $\Delta_{ij}$ be the determinant of the matrix
\[
\left(\begin{array}{ll}
X_i & X_j \\
X_{i+1} & X_{j+1}
\end{array}\right)
\]
for $1 \leq i < j \leq 4$.
We set $\ga = (\Delta_{12} - \Delta_{34},
\Delta_{13}, \Delta_{14}, \Delta_{24})$ and
$J = \ga + \Delta_{23}R$.
Then $J$ is a reduction of $I$.
Moreover, we have
$\ga :_R \Delta_{23} = (X_2, X_4, X_5 - X_1, {X_1}^2, X_1X_5)$,
and so
$(\ga : \Delta_{23}) + \Delta_{23}R =
(X_2, X_4, X_5 - X_1, {X_1}^2, X_1X_5, {X_3}^2)$,
which is an $\gn$-primary ideal.
Hence $\jm{IS_\gn} =
\length{S}{S / (\ga :_S \Delta_{23}) + \Delta_{23}S} = 4$
by \ref{3e}.

(2) This assertion can be proved by \ref{3d}.

config: arxiv


TITLE: Is a Mersenne-Prime always of the form $3n + 1$?
QUESTION [2 upvotes]: Examples are:
$M_3 = 7 = 3\times 2 + 1$
$M_5 = 31 = 3\times 10 + 1$
$M_7 = 127 = 3\times 42 + 1$
$M_{13} = 8191 = 3\times2730 + 1$

REPLY [7 votes]: If $p$ is an odd prime, and more generally if it is an odd number, then $2^p-1\equiv (-1)^p-1\equiv -2\equiv 1\pmod{3}$.
The Mersenne prime $3$ is not of the form $3n+1$.

REPLY [5 votes]: Every number of the form $2^{2n+1}-1$ is of such form, because
$$2^{2n+1}-2\equiv2\cdot(4^n-1)\equiv2\cdot(1^n-1)\equiv0\ \ (\text{mod}\,3)$$
And because every prime except $2$ is odd, it's true for every Mersenne prime except the first.

REPLY [1 votes]: $2^{2n+1} - 1 \equiv 1 \pmod 3$ is true for all odd numbers $2n + 1$.
$2^{2n+1} \equiv 2 \pmod 3$
$2^{2n} \equiv 1 \pmod 3$
$4^n \equiv 1^n \equiv 1 \pmod 3$
Alternatively, you could show the cyclic pattern of the powers of two $\pmod 3$: $1, 2, 1, 2, 1, 2, \dots$, so it's clear that $2^k \equiv 1$ for $k$ even and $2$ for $k$ odd.


TITLE: Confusion about word "vector" in "basis vectors" of General Relativity
QUESTION [0 upvotes]: If a vector is defined as a rank-1 tensor, it should be invariant under change of basis, while its components transform. So why a basis vector $\boldsymbol{e}_\alpha$ is called "vector" even though transforms $\tilde{\boldsymbol{e}}^\alpha=\partial_\beta\tilde{x}^\alpha\boldsymbol{e}^\beta$ as the component of a vector?

REPLY [2 votes]: I will present the same answer as Slereah, but by couching it another way perhaps I can add some further clarity.
Considered as a set of geometrical objects, the basis vectors are just that: vectors. Let's call them ${\bf e}_{(a)}$ where the index $a$ runs from 0 to 3.
And suppose ${\bf v}$ is some other vector.
When you change from one chart to another (changing your coordinates) none of the vectors change at all. ${\bf e}_{(a)}$ are still what they were, and $\bf v$ is still what it was. Of course they don't change, because by changing coordinates we are not doing anything to any of the vectors; we are just 'looking' at them from a different 'direction'. (Note: to get clarity here you should avoid the practice of saying that $v^a$ is 'a vector'; it is not: it is the set of components of a vector relative to some given basis.)
But what does change then? The essential point is that the set of vectors ${\bf e}_{(a)}$ always forms a basis, but it is not necessarily the coordinate basis. If these vectors form the coordinate basis for one choice of coordinates, then they do not for most other choices of coordinates. So what to do then? The normal practice is to write
$$
{\bf v} = \sum_a v^a {\bf e}_{(a)} = \sum_a v'^a {\bf e}'_{(a)}
$$
where the vectors ${\bf e}'_{(a)}$ are not the same as the vectors ${\bf e}_{(a)}$. Rather, they are the members of the coordinate basis for the new coordinates. So in moving from ${\bf e}_{(a)}$ to ${\bf e}'_{(a)}$ one is NOT transforming a vector. Rather one is replacing one set of vectors by another set of vectors. The operation of obtaining $v'^a$ from $v^a$ is ordinarily called a 'transformation' and that is a reasonable use of language. The relationship is the standard one,
$$
v'^a = \frac{\partial x'^a}{\partial x^\mu} v^\mu.
$$