TITLE: Modular arithmetic calculation $366^3\pmod {391}$
QUESTION [0 upvotes]: $366^3\pmod{391}$ Is there an easy way to calculate this without a calculator?

REPLY [3 votes]: Since $391=17\cdot 23$ and $366\equiv 9\pmod{17},366\equiv -2\pmod{23}$, we have:
$$ 366^3 \equiv 15\pmod{17},\qquad 366^3\equiv -8\equiv 15\pmod{23}, $$
hence:
$$ 366^3\equiv \color{red}{15}\pmod{391}.$$


TITLE: intersection and cartesian product
QUESTION [0 upvotes]: Let $  \mathbb{N}\supseteq A, B $.
$|B \times A| = 35,$ and $|A\times B \setminus B\times B| = 14$.
Find $|A \cap B|$.
I figured out that $|B|$ must be 5 and $|A|$ must be 7, but that I got stuck.

REPLY [0 votes]: Following the hint, we have $$|A\times B| = 35 \qquad |(A\setminus B) \times B|=14$$Using that $|A\times B|=|A|\cdot |B|$, and the previous equations, we conclude that $|B|$ has to divide both $14$ and $35$.
Now, note that $|A\setminus B|= |A|-|A\cap B|$.

REPLY [0 votes]: $$|A\times B \setminus B\times B| = 14\\
|A\times B - B\times B| = 14\\
|(A-B)\times B| = 14$$
$$|A\times B - B\times B|=|A\times B -(A\times B)\cap (B\times B)|= \\
|A\times B -(A\cap B)\times B)|=\\
|A\times B |-|(A\cap B)\times B)|=14\\\to 35-|(A\cap B)\times B)|=14\\|(A\cap B)\times B)|=21=3\times7$$note that $A \cap B \subset B \to |A \cap B|\leq |B| \to |A\cap B|=3,|B|=7$


text:
\begin{document}
\title{Powers of $t$-spread principal Borel ideals}
\author {Claudia  Andrei, Viviana Ene, Bahareh Lajmiri}

\address{Claudia Andrei, Faculty of Mathematics and Computer Science, University of Bucharest, Str. Academiei 14,
 010014 Bucharest, Romania} \email{claudiaandrei1992@gmail.com}

\address{Viviana Ene, Faculty of Mathematics and Computer Science, Ovidius University, Bd.\ Mamaia 124,
 900527 Constanta, Romania}  \email{vivian@univ-ovidius.ro}

\address{Bahareh Lajmiri, Department of Mathematics and Computer Science, Amirkabir University of Technology,
424 Hafez Ave, Tehran, Iran} \email{bahareh.lajmiri@aut.ac.ir}

\thanks{This paper was written while the third author visited the Faculty of Mathematics and Computer Science, Ovidius University of Constanta. The research of the third author was partially supported by Amirkabir University of Technology.}

\begin{abstract}
We prove that $t$-spread principal Borel ideals are sequentially Cohen-Macaulay and study their powers. We show that these ideals
possess the strong persistence property and compute their limit depth.
\end{abstract}

\subjclass[2010]{13D02,13H10, 05E40, 13C14}
\keywords{$t$-spread principal Borel ideals, persistence property, limit depth, sequentially Cohen-Macaulay ideals}

\maketitle
\section*{Introduction}

In this paper, we study $t$-spread prinipal Borel ideals. They have been recently introduced in \cite{EHQ}. Let $K$ be a field and $S=K[x_1,\ldots,x_n]$ the polynomial ring in $n$ variables over $K.$ Let $t\geq 0$ be an integer. A monomial $x_{i_1}\cdots x_{i_d}\in S$ with $i_1\leq \cdots \leq i_d$ is called \emph{$t$-spread} if $i_j-i_{j-1}\geq t$ for $2\leq j\leq d.$ For example, every monomial in $S$ is $0$-spread and every squarefree monomial is $1$-spread.

We recall from \cite{EHQ} that a monomial ideal $I\subset S$ with the minimal system of monomial generators $G(I)$
is called \emph{$t$-spread strongly stable} if it satisfies the following condition: for all $u\in G(I)$ and $j\in \supp(u),$ if $i<j$ and $x_i(u/x_j)$ is $t$-spread, then $x_i(u/x_j)\in I.$  A monomial  ideal $I\subset S$ is called \emph{$t$-spread principal Borel} if there exists a monomial $u\in G(I)$ such that $I=B_t(u)$ where $B_t(u)$ denotes
the smallest $t$-spread strongly stable ideal which contains $u.$ For example, for an  integer $d\geq 2,$ if
$u=x_{n-(d-1)t}\cdots x_{n-t}x_n$, then $B_t(u)$ is minimally generated by all the
$t$-spread monomials of degree $d$ in $S.$ In this case, we call $B_t(u)$ the \emph{$t$-spread Veronese ideal} generated in degree $d$ and denote it $I_{n,d,t}.$

Throughout this paper we consider $t$-spread monomial ideals with $t\geq 1. $ In particular, they are squarefree monomial ideals. As it was observed in \cite{EHQ}, if $u=x_{i_1}\cdots x_{i_d}$ is a $t$-spread monomial in $S$ then
a monomial $x_{j_1}\cdots x_{j_d}\in G(B_t(u))$ if and only if the following two conditions hold:
\begin{enumerate}
	\item [(i)] $j_k\leq i_k$ for $1\leq k\leq d,$
	\item [(ii)] $j_k-j_{k-1}\geq t$ for $2\leq k\leq t.$
\end{enumerate}

In \cite[Theorem 1.4]{EHQ} it was shown that every $t$-spread strongly stable ideal has linear quotients with respect to the pure lexicographic
order in $S.$ In addition, in \cite[Corollary 2.5]{EHQ} it was shown that a $t$-spread strongly stable ideal generated in a single degree  is Cohen-Macaulay if and only if it is a $t$-spread Veronese ideal. In particular, it follows that every $t$-spread principal Borel ideal has linear quotients and such an ideal is Cohen-Macaulay if and only if it is $t$-spread Veronese.

Since $B_t(u)$  is a squarefree monomial ideal, we may interpret it as the Stanley-Reisner ideal of a simplicial complex $\Delta.$ In the
proof of Theorem~\ref{ass}, we characterize the facets of $\Delta$. This  gives explicitly  all the generators of the ideal
of the Alexander dual  $\Delta^{\vee}.$ We then derive that the Stanley-Reisner ideal of $\Delta^{\vee}$ has linear
quotients, which yields the sequential Cohen-Macaulay property of
$B_t(u)$ by Alexander duality; see Theorem~\ref{SCM}.

In Section~\ref{two}, we consider the Rees algebra $\MR(B_t(u))$ of a $t$-spread principal Borel ideal $B_t(u).$ In
Proposition~\ref{elexchange}, we show that  $B_t(u)$ satisfies the
$\ell$-exchange property with respect to the sorting order $<_{sort}.$ This allows us to describe in Theorem~\ref{GBRees} the reduced
Gr\"obner basis of the
defining ideal of $\MR(B_t(u))$ with respect to a suitable monomial order. The form of the binomials in this Gr\"obner basis shows that
$B_t(u)$ satisfies an $x$-condition which guarantees  that all the powers of $B_t(u)$ have linear quotients. Moreover, the Rees algebra
$\MR(B_t(u))$ is a normal Cohen-Macaulay domain which implies in turn, by a result of \cite{HQ}, that $B_t(u)$ possesses  the strong
persistence property as considered in \cite{HQ}. Consequently, $B_t(u)$ satisfies the persistence property, which means that
$\Ass(B_t(u))\subset \Ass(B_t(u)^2)\subset \cdots \subset \Ass(B_t(u)^k)\subset \cdots.$ Note that the associated primes of $B_t(u)$ can be read from Theorem~\ref{ass}.

In Section~\ref{three}, we study the limit behavior of the depth for the powers of $t$-spread principal Borel ideals. In Theorem~\ref{limdepth} we show that if the generator $u=x_{i_1}\cdots x_{i_d}$ satisfies the condition $i_1\geq t+1,$ then
$\lim_{k\to\infty}\depth(S/B_t(u)^k)=0.$ In particular, we determine the analytic spread of $B_t(u),$ that is, the Krull dimension of the fiber ring
$\MR(B_t(u))/\mathfrak{m} \MR(B_t(u))$ where $\mathfrak{m}=(x_1,\ldots,x_n).$

\section{$t$-spread principal Borel ideals are sequentially Cohen-Macaulay}

Let $u=x_{i_1}x_{i_2}\cdots x_{i_d}\in S= K[x_1,\ldots, x_n]$ be a $t$-spread monomial. Suppose that $i_d=n$ and $I=B_t(u)$. For a monomial $u\in I$, we denote
$\supp(u)=\{j\in [n]: x_j \mid u\}.$ Let $\Delta$ be the simplicial complex such that $I_{\Delta}=I$. We denote $I^{\vee}$ the Stanley-Reisner ideal of the Alexander dual of $\Delta$.

\begin{Theorem}\label{ass}
  Let $t\geq 1$ be an integer and $I=B_t(u)$ where $u=x_{i_1}x_{i_2}\cdots x_{i_d}$ is  a $t$-spread monomial. We assume that $\bigcup_{v\in G(I)}\supp(v)=[n].$ Then $I^{\vee}$ is generated by the monomials of the following forms
  \begin{equation}\label{form1}
 \prod_{k=1}^{n}x_k/(v_{j_1}\cdots v_{j_{d-1}})
\end{equation}
   with $j_l\leq i_l$ for $1\leq l\leq d-1$ and $j_l-j_{l-1}\geq t$ for $2\leq l\leq d-1$, where $v_{j_k}=x_{j_k}\cdots x_{j_k+(t-1)}\text{ for }1\leq k\leq d-1.$
\begin{equation}\label{form2}
\prod_{k=1}^{i_1}x_k.
\end{equation}
\begin{equation}\label{form3}
\prod_{k=1}^{i_s}x_k/(v_{j_1}\cdots v_{j_{s-1}})
\end{equation}
    with $2\leq s\leq d-1$, $j_l\leq i_l$ for $1\leq l\leq s-1$, $j_l-j_{l-1}\geq t$ for $2\leq l\leq s-1$, where
		$v_{j_k}=x_{j_k}\cdots x_{j_k+(t-1)}\text{ for }1\leq k\leq s-1.$
\end{Theorem}

\begin{proof}
  Let $\Delta$ be the simplicial complex whose Stanley-Reisner ideal is $I$ and let $\mathcal{F}(\Delta)$ be the set of the facets of $\Delta$. We prove that every facet of $\Delta$ is  of one of the following forms:
  \begin{enumerate}\label{facets}
   \item[(i)] $F_1=\{j_1, j_1+1, \ldots, j_1+(t-1), j_2, j_2+1, \ldots, j_2+(t-1),\ldots, j_{d-1}, j_{d-1}+1, \ldots, j_{d-1}+(t-1)\}$, for some $j_1, j_2, \ldots, j_{d-1}$ such that $j_l\leq i_l$ for $1\leq l\leq d-1$ and $j_l-j_{l-1}\geq t$ for $2\leq l\leq d-1$.
   \item[(ii)] $F_2=\{i_1+1, i_1+2,\ldots, n\}$.
   \item[(iii)] $F_3=\{j_1, j_1+1, \ldots, j_1+(t-1),\ldots, j_{s-1}, j_{s-1}+1, \ldots, j_{s-1}+(t-1), j_s, j_s+1, \ldots, n\}$, for some $j_1, j_2, \ldots, j_s$ such that $2\leq s\leq d-1$, $j_l\leq i_l$ for $1\leq l\leq s-1$, $j_s=i_s+1$ and $j_l-j_{l-1}\geq t$ for $2\leq l\leq s$.
  \end{enumerate}

Since $I=B_t(u)$ has the primary decomposition \[I=\bigcap_{F\in\mathcal{F}(\Delta)}P_{[n]\setminus F},\]
  where $P_{[n]\setminus F}$ is the  prime ideal generated by all variables $x_j$ with $j\in[n]\setminus F$, by
	\cite[Corollary 1.5.5]{HHBook}, the statement holds.

  Since $(x_1,\ldots x_{i_1})\in \Min(I)$ by \cite[Theorem 2.4]{EHQ}, we obtain $F_2\in\mathcal{F}(\Delta)$.

  We have $F_1$ and $F_3\in \Delta$, since $x_{F_1}=\prod_{i\in F_1}x_i$ and $x_{F_3}=\prod_{i\in F_3}x_i$ do not belong to $I$. Indeed, if $x_{F_1}\in I$, then there exists $v=x_{k_1}\cdots x_{k_d}\in G(I)$ such that $v\mid x_{F_1}$. Since $v$ is the product of $d$ distinct variables and $F_1$ consists of $d-1$ intervals of the form $[j_r, j_r+(t-1)],$ $1\leq r\leq d-1$, there exist $l\in \{2,\ldots, d\}$ and $1\leq r\leq d-1$ such that $k_{l-1}, k_l\in [j_r,j_r+(t-1)]$. It follows that $k_l-k_{l-1}<t$, which is false.

  If $x_{F_3}\in I$, then there exists $v=x_{k_1}\cdots x_{k_d}\in G(I)$ such that $v\mid x_{F_3}$. Since $k_l-k_{l-1}\geq t$ for $2\leq l\leq s$ and $F_3$ consists of $s-1$ intervals of the form $[j_r, j_r+(t-1)],$ $1\leq r\leq s-1,$ and only one interval of the form $[j_s,n]$, we have $k_s\in [j_s,n]$.  It follows that $j_s=i_s+1\leq k_s$, which is false because $v\in G(I)$ and $k_s\leq i_s$.

  We claim that $F_i\cup \{j\}\notin \Delta$ for every $j\in [n]\setminus F_i$ and $i\in\{1,3\}$. This will prove that every set of the form (i) or (iii) is a facet of $\Delta.$

  Let $j\in[n]\setminus F_1$. We consider the following cases:
  \begin{enumerate}
    \item[(a)]  $j<j_1.$ Then $x_j x_{j_1+(t-1)}\cdots x_{j_{d-1}+(t-1)}\in I$ because $j<j_1\leq i_1$ and $j_l+(t-1)\leq i_l+(t-1)<i_{l+1}$ for $1\leq l\leq d-1$.
    \item[(b)]  $j\geq j_{d-1}+t.$ Then $x_{j_1} x_{j_2}\cdots x_{j_{d-1}}x_j\in I$.
    \item[(c)] There exists $1\leq l\leq d-2$ such that $j_l+(t-1)<j<j_{l+1}.$ Then \[x_{j_1} x_{j_2}\cdots x_{j_l} x_j x_{j_{l+1}+(t-1)}\cdots  x_{j_{d-1}+(t-1)}\in I,\] since $j_k\leq i_k$ for $1\leq k\leq l$, $j<j_{l+1}\leq i_{l+1}$ and $j_k+(t-1)\leq i_k+(t-1)<i_{k+1}$ for $l+1\leq k\leq d-1$.
  \end{enumerate}
	Therefore, we have proved that for $j\in[n]\setminus F_1$, $F_1\cup\{j\}\not\in \Delta$ which implies that $F_1$ is a facet in $\Delta.$

  Let $j\in[n]\setminus F_3$. We show that $x_j x_{F_3}\in I$, thus $F_3\cup\{j\}\not\in \Delta.$
  \begin{enumerate}
    \item[(a)] If $j<j_1$, then $x_j x_{j_1+(t-1)}\cdots x_{j_{s-1}+(t-1)}x_{j_s+(t-1)}x_{j_s+(2t-1)}\cdots x_{j_s+(d-s)t-1}\in I$ because $j<j_1\leq i_1$, $j_k+(t-1)\leq i_k+(t-1)<i_{k+1}$ for $2\leq k\leq s-1$ and $j_s+(kt-1)=i_s+1+(kt-1)=i_s+kt\leq i_{s+k}$ for $1\leq k\leq d-s$.
    \item[(b)] If there exists $2\leq l\leq s$ such that $j_{l-1}+(t-1)<j<j_l$, then \[x_{j_1} x_{j_2}\cdots x_{j_{l-1}} x_j x_{j_l+(t-1)}\cdots  x_{j_s+(t-1)}\cdots x_{j_s+(d-s)t-1}\in I.\]
        Indeed, $j_k\leq i_k$ for $1\leq k\leq l$. If $l=s$, then $j<j_s=i_s+1$ and $j_s+(kt-1)=i_s+1+(kt-1)=i_s+kt\leq i_{s+k}$ for
				$1\leq k\leq d-s$. In the case that $l<s$, we have $j<j_l\leq i_l$, $j_k+(t-1)\leq i_k+(t-1)<i_{k+1}$ for $l\leq k\leq s-1$ and $j_s+(kt-1)=i_s+1+(kt-1)\leq i_{s+k}$ for $1\leq k\leq d-s$.
  \end{enumerate}

  It remains to show that the sets of the forms (i)--(iii) are the only facets of $\Delta$. This is equivalent to showing that for every face
	$G\in \Delta$, there exists $F\in \mathcal{F}(\Delta)$ of one of the forms (i)--(iii)  which contains $G$.

  Let $G\in \Delta$ and $j_1=\min\{j:j\in G\}$. Inductively, for $l\geq 2$, we set \[j_l=\min\{j:j\in G\text{ and }j\geq j_{l-1}+t\}.\]

  In the case that $j_l\leq i_l$ for all $l$, then the sequence $j_1<j_2<\ldots<j_l<\ldots$ has at most $d-1$ elements. Otherwise, $x_{j_1}\cdots x_{j_d}\in I$, which implies that $G\notin\Delta$, a contradiction. Let $j_1<j_2<\ldots<j_r$ with $r\leq d-1$. Then $G\subset F_1=\{j_1,\ldots, j_1+(t-1),\ldots, j_r,\ldots, j_r+(t-1), i_{r+1},\ldots, i_{r+1}+(t-1),\ldots, i_{d-1},\ldots, i_{d-1}+(t-1)\}\in \mathcal{F}(\Delta)$.

  If there exists $l\leq d$ such that $j_l>i_l$, then we denote by $s$ the smallest index with this property. In the case that $s=1$, $j_1=\min\{j:j\in G\}$ and $G\subset F_2=\{i_1+1, i_1+2,\ldots, n\}\in \mathcal{F}(\Delta)$. If $s>1$, then $G\subset F_3=\{j_1, j_1+1, \ldots, j_1+(t-1),\ldots, j_{s-1}, j_{s-1}+1, \ldots, j_{s-1}+(t-1), i_s+1, i_s+2, \ldots, n\}\in \mathcal{F}(\Delta)$.
\end{proof}

\begin{Theorem}\label{SCM}
Let $t\geq 1$ be an integer, $u=x_{i_1}x_{i_2}\cdots x_{i_d}\in S$  a $t$-spread monomial, and  suppose that $i_d=n.$ Then
the $t$-spread principal Borel ideal  $I=B_t(u)$ is sequentially Cohen-Macaulay.
\end{Theorem}

\begin{proof}
By \cite[Theorem 8.2.20]{HHBook}, it is enough to show that $I^{\vee}$, that is, the Stanley-Reisner ideal of $\Delta^{\vee}$, is componentwise linear. By \cite[Corollary 8.2.21]{HHBook}, it suffices to prove that $I^{\vee}$ has linear quotients.

Let $w_1=\prod_{k=1}^{i_1}x_k$. Let $w_2,w_3,\ldots, w_q$ be the minimal monomial generators of $I^{\vee}$  of the form (\ref{form3}), ordered decreasingly with respect to the  pure lexicographic order. Let $w_{q+1},w_{q+2},\ldots, w_r$ be the remaining minimal monomial generators of $I^{\vee}$, namely those of form (\ref{form1}), ordered decreasingly with respect to the pure lexicographic order.

We show that $I^{\vee}$ has linear quotients with respect to the above order of its minimal generators. This is equivalent to proving the following claim:
\begin{eqnarray}\label{eqn}
 \text{ for every }1<j\leq r \text{ and }g<j, \text{ there exists }w>_{\lex}w_j \text{ and  } \\ \nonumber
1\leq l\leq n \text{  such that }x_l=\frac{w}{\gcd(w, w_j)} \text{ and }x_l\mid \frac{w_g}{\gcd(w_g, w_j)}.
\end{eqnarray}

Since $w_2$ is the largest generator of the form (\ref{form3}) with respect to the pure lexicographic order, we have $x_1x_2\cdots x_{i_1-1}\mid w_2$. Thus, we obtain $x_{i_1}=w_1/\gcd(w_1, w_2)$ and claim (\ref{eqn}) is proved for $g=1<j=2.$

Let $2<j\leq q$, $g=1$ and $w_j=\prod_{k=1}^{i_s}x_k/v_{j_1}v_{j_2}\cdots v_{j_{s-1}}$ with $s\leq d-1$.
If $j_1=i_1$, then we have  $x_{i_1}={w_1}/{\gcd(w_1, w_j)}.$  Let $j_1<i_1$. Then $x_{j_1}\mid w_1/\gcd(w_1, w_j)$. We have to find $w>_{\lex}w_j$ such that $x_{j_1}=w/\gcd(w, w_j)$.

	\begin{enumerate}
    \item[(a)] If  there exists a least integer $1\leq l\leq s-2$ such that $j_{l+1}>j_l+t$, then we take $w=\prod_{k=1}^{i_s}x_k/v_{j_1+1}\cdots v_{j_l+1}v_{j_{l+1}}\cdots v_{j_{s-1}}$.
		\item[(b)] If  $j_{l+1}=j_l+t$ for $1\leq l\leq s-2$, then \[w=\prod_{k=1}^{i_s}x_k/v_{j_1+1}\cdots v_{j_l+1}v_{j_{l+1}+1}\cdots v_{j_{s-1}+1}.\]
  \end{enumerate}

  Let $1<g<j$. Then $w_g>_{\lex}w_j$. Let \[w_g=\frac{\prod_{k=1}^{i_{s'}}x_k}{v_{g_1}v_{g_2}\cdots v_{g_{s'-1}}}\] with $s'\leq d-1$. Thus, we obtain $v_{g_1}v_{g_2}\cdots v_{g_{s'-1}}<_{\lex}v_{j_1}v_{j_2}\cdots v_{j_{s-1}}$. It implies that there exists $h\geq 1$ such that $j_1=g_1$, $j_2=g_2$,\ldots, $j_{h-1}=g_{h-1}$ and $j_h<g_h\leq i_h$. Since $x_{j_h}\mid v_{j_1}v_{j_2}\cdots v_{j_{s-1}}$ and $x_{j_h}\nmid v_{g_1}v_{g_2}\cdots v_{g_{s'-1}}$, we get $x_{j_h}\mid {w_g}/{\gcd(w_g,w_j)}.$ Then we must find a generator $w$ of
	$I^{\vee}$ with $w>_{\lex}w_j$ such that $x_{j_h}=w/\gcd(w,w_j).$
\begin{enumerate}
\item[(a)] If there exists a least integer $h\leq l\leq s-2$ such $j_{l+1}>j_l+t$, then
  \[w=\frac{\prod_{k=1}^{i_s}x_k}{v_{j_1}\cdots v_{j_{h-1}}v_{j_h+1}v_{j_{h+1}+1}\cdots v_{j_l+1}v_{j_{l+1}}\cdots v_{j_{s-1}}}.\]
\item[(b)] If  $j_{l+1}=j_l+t$ for $h\leq l\leq s-2$, then $w={\prod_{k=1}^{i_s}x_k}/{v_{j_1}\cdots v_{j_{h-1}}v_{j_h+1}\cdots v_{j_{s-1}+1}}.$
\end{enumerate}
  With  similar arguments, one proves claim (\ref{eqn}) for $q+1\leq j\leq r$.
\end{proof}

\begin{Example}
Let $I=B_2(x_2x_4x_9)\subset K[x_1,\ldots,x_9].$ Then  \[I^{\vee}=(x_1x_2, x_1x_4, x_3x_4, x_1x_6x_7x_8x_9, x_3x_6x_7x_8x_9, x_5x_6x_7x_8x_9),\] where we ordered the generators as indicated in the proof of the above theorem.
\end{Example}

\section{The Rees algebra of $t$-spread Borel principal ideals}
\label{two}

In this section we consider the
Rees algebra of  $t$-spread Borel principal ideals.

For two monomials $v,w\in S$ of degree $d$, we write $vw=x_{i_1}x_{i_2}\cdots x_{i_{2d}}$ with $i_1\leq i_2 \leq \cdots \leq i_{2d}.$
Then the \emph{sorting} of the pair $(v,w)$ is the pair of monomials $(v',w')$ where $v'=x_{i_1}x_{i_3}\cdots x_{i_{2d-1}}$ and $w'=x_{i_2}x_{i_4}\cdots x_{i_{2d}}.$
The map $\sort:S_d \times S_d \to S_d \times S_d$
with $\sort(v,w)=(v',w')$ is called the \emph{sorting operator}. A subset $B \subset S_{d}$ is called  \emph{sortable} if $\sort(B \times B) \subset B\times B$. An $r$-tuple $(u_1,\ldots,u_r)$
 of monomials  of degree $d$ is sorted if for all $1\leq i<j\leq r,$ the pair $(u_i,u_j)$ is sorted, that is,
$\sort(u_i,u_j)=(u_i,u_j).$ This means that if we write $u_1=x_{i_1}\cdots x_{i_d}, u_2=x_{j_1}\cdots x_{j_d},\ldots, u_r=x_{k_1}\cdots x_{k_d},$ then $(u_1,\ldots,u_r)$ is sorted if and only if $i_1\leq j_1\leq \cdots \leq k_1\leq i_2\leq j_2\leq \cdots \leq k_2\leq \cdots \leq i_d\leq j_d\leq \cdots\leq k_d.$ By \cite[Theorem 6.12]{EHBook}, for every $r$-tuple $(u_1,\ldots,u_r)$
 of monomials of degree $d$, there exists a unique sorted $r$-tuple $(u_1^\prime,\ldots,u_r^\prime)$ such that
$u_1\cdots u_r=u_1^\prime\cdots u_r^\prime.$ When $(u_1^\prime,\ldots,u_r^\prime)=(u_1,\ldots,u_r),$ we say that the product $u_1\cdots u_r$ is sorted.

Let $t\geq 1$ be an integer and $I=B_t(u)$ where $u=x_{i_1}\cdots x_{i_d}$ is a $t$-spread monomial.
It was shown in \cite[Proposition 3.1]{EHQ} that the minimal set of monomial generators of a $t$-spread principal Borel ideal is sortable.
Therefore, if $v,w \in G(I)$, and $\sort(v,w)=(v',w')$, then $v',w' \in G(I)$.

Let $\MR(I)= \bigoplus_{j\geq 0} {I^j}{t^j}$ be the Rees algebra of the ideal $I=B_t(u).$
Since the minimal generators of $B_t(u)$ have the same degree, the fiber
$\MR(I)/\mathfrak{m}\MR(I)$ of the Rees ring $\MR(I)$ is isomorphic to $K[G(I)].$ Let $\psi:T=K[\{t_v: v\in G(I)\}]\to K[G(I)]$ be the homomorphism  given by
$t_{v}\mapsto v$ for $v\in G(I).$
 As it was proved in  \cite[Theorem 3.2]{EHQ}, the set of binomials
\[\MG=\{t_vt_w-t_{v^\prime}t_{w^\prime}: (v,w) \text{ unsorted }, (v^\prime, w^\prime)=\sort(v,w)\}\]
 is a
Gr\"obner basis of the toric ideal $J_u=\ker \psi$ with respect to the sorting order on $T$. For more details about  sorting order we refer to
\cite[Chapter 14]{St95} or \cite[Section 6.2]{EHBook}. The initial monomial of the binomial $t_vt_w-t_{v^\prime}t_{w^\prime}\in \MG$ with respect to the sorting order is $t_vt_w.$

We now  recall the definition of \emph{$\ell$-exchange property} from \cite{HHV05}; see also \cite[Section 6.4]{EHBook} and
\cite[Section 3]{EO}.

Let $I\subset S$ be a monomial ideal generated in a single degree and $K[\{t_u: u\in G(I)\}]$ the polynomial ring in $|G(I)|$
variables endowed with a monomial order $<.$ Let $P$ be the kernel of the $K$-algebra homomorphism
 \[ K[\{t_u: u\in G(I)\}]\to K[G(I)], t_u\mapsto u, u\in G(I).\]

A monomial $t_{u_1}\cdots t_{u_N}$ is called \emph{standard with respect to $<$} if it does not belong to $\ini_<(P).$

\begin{Definition}\cite{HHV05}\label{defellexchange}
The monomial ideal $I\subset S$ satisfies the $\ell$-exchange property with respect to $<$ if the following condition holds: let
 $t_{u_1}\cdots t_{u_N},t_{v_1}\cdots t_{v_N}$ be two standard monomials  with respect to $<$  of the same degree $N$

satisfying
\begin{itemize}
	\item [(i)] $\deg_{x_i} u_1\cdots u_N=\deg_{x_i}v_1\cdots v_N$ for $1\leq i\leq q-1$ with $q\leq n-1,$
	\item [(ii)] $\deg_{x_q}u_1\cdots u_N<\deg_{x_q}v_1\cdots v_N.$
\end{itemize}
Then there exists  integers $\delta, j$ with $q<j\leq n$ and $j\in \supp(u_\delta)$ such that $x_q u_\delta/x_j\in I.$
\end{Definition}

\begin{Proposition}\label{elexchange}
Let $u=x_{i_1}\cdots x_{i_d}$ be a $t$-spread monomial in $S.$ Then the $t$-spread principal Borel ideal $B_t(u)$ satisfies the
$\ell$-exchange property with respect to the sorting order $<_{sort}.$
\end{Proposition}

\begin{proof}
Let $T=K[\{t_v : v\in G(B_t(u))\}]$ and $t_{u_1}\cdots t_{u_N},t_{v_1}\cdots t_{v_N}\in T$ be two standard monomials  with respect to $<_{\sort}$  of the same degree $N$
such that the two conditions of Definition~\ref{defellexchange} are fulfilled. As the chosen monomials are standard with respect to $<_{\sort}$, it follows that the products
$ u_1\cdots u_N, v_1\cdots v_N$ are sorted. Since $\deg_{x_i} u_1\cdots u_N=\deg_{x_i}v_1\cdots v_N$ for $1\leq i\leq q-1$, we have $\deg_{x_i}(u_\nu)=\deg_{x_i}(v_\nu)$
for all $1\leq \nu\leq N$ and $1\leq i\leq q-1$. The condition $\deg_{x_q}u_1\cdots u_N<\deg_{x_q}v_1\cdots v_N$ implies that there exists
$1\leq \delta\leq N$ such that $\deg_{x_q}(u_\delta)<\deg_{x_q}(v_\delta).$

Let $u_{\delta}=x_{j_1}\cdots x_{j_d}$, $v_{\delta}=x_{\ell_1}\cdots x_{\ell_d}$, and assume that $q=\ell_\mu$ for some $1\leq \mu <d.$ It follows that
$j_1=\ell_1,\ldots,j_{\mu-1}=\ell_{\mu-1}$ and $j_\mu>\ell_\mu=q.$ If $j_\mu\not\in \supp(v_{\delta}),$ then we take $j=j_\mu.$ Then the monomial
$x_qu_{\delta}/x_j$ is $t$-spread, thus it belongs to $B_t(u)$, since $q=\ell_\mu\geq \ell_{\mu-1}+t=j_{\mu-1}+t$ and $j_{\mu+1}\geq j_\mu+t>q+t.$

If $j_\mu\in \supp(v_{\delta}), $ we may choose any $j\in \supp(u_\delta)\setminus \supp(v_{\delta})$ with $j>q.$ We should note that such an integer $j$ exists since
$\deg (u_\delta)=\deg (v_\delta)$ and $\deg_{x_q}(u_\delta)<\deg_{x_q}(v_\delta).$ Then $x_qu_{\delta}/x_j$ is $t$-spread since $q,j_\mu\in\supp(v_{\delta}).$
\end{proof}

The Rees ring $\MR(I)$ where $I=B_t(u)$ has the presentation
\[
\varphi: R=S[\{t_v : v\in G(I)\}]\to \MR(I),
\]
given by
\[
x_i\mapsto x_i, 1\leq i\leq n, t_v\mapsto v t, v\in G(I).
\]
Let $J=\ker \varphi$ be the toric ideal of $\MR(I).$ We consider the sorting order $<_{\sort}$ on the ring $T=K[\{t_v : v\in G(I)\}]$ and the lexicographic order $<_{\lex}$ on the ring $S.$
Let $<$ be the monomial order on $ R=S[\{t_v : v\in G(I)\}]$ defined as follows: if $m_1, m_2$ are monomials in $S$ and $v_1,v_2$ are monomials in $T,$ then
\[
m_1v_1>m_2v_2 \text{ if } m_1>_{\lex} m_2 \text{ or } m_1=m_2 \text{ and } v_1>_{\sort}v_2.
\]

\begin{Theorem}\label{GBRees} Let $I=B_t(u)$ be a $t$-spread principal Borel ideal.
The reduced Gr\"obner basis of the toric ideal $J$ with respect to $<$ consists of the set of binomials $t_vt_w-t_{v^\prime}t_{w^\prime}$ where $(v,w)$ is  unsorted and
 $(v^\prime, w^\prime)=\sort(v,w),$ together with
the binomials of the form $x_i t_v-x_j t_w$ where $i<j,$ $x_iv=x_jw$, and $j$ is the largest integer for which $x_iv/x_j\in G(I).$
\end{Theorem}

\begin{proof}
By Proposition~\ref{elexchange}, we know that the principal Borel ideal $B_t(u)$ satisfies the
$\ell$-exchange property with respect to the sorting order $<_{sort}.$ Thus, the statement follows by applying \cite[Theorem 5.1]{HHV05}.
\end{proof}

\begin{Proposition}\label{linear quotients}
All the powers of $B_t(u)$ have linear quotients. In particular, all the powers of $B_t(u)$ have a linear resolution.
\end{Proposition}

\begin{proof}
By Theorem~\ref{GBRees}, it follows that every binomial in the reduced Gr\"obner basis of the toric ideal $J$  has the degree in variables $x_i$ at most $1$ which means that $B_t(u)$
satisfies the $x$-condition. Then, by \cite[Theorem 10.1.9]{HHBook}, all the powers of $B_t(u)$ have linear quotients.
\end{proof}

Theorem~\ref{GBRees} shows that the Rees algebra $\MR(B_t(u))$ has a quadratic Gr\"obner basis and the initial ideal of the toric ideal $J$ is squarefree. Therefore, we get the following
consequences.

\begin{Corollary}
The Rees algebra $\MR(B_t(u))$ is Koszul.
\end{Corollary}

\begin{Corollary}\label{CMRees}
The Rees algebra $\MR(B_t(u))$ is a normal Cohen-Macaulay domain. In particular, $B_t(u)$ satisfies the strong persistence property.
Therefore, $B_t(u)$ satisfies the persistence property.
\end{Corollary}

\begin{proof}
Since the initial ideal of the toric ideal $J$ of $\MR(B_t(u))$ is squarefree, by a theorem due to Sturmfels \cite{St95}, it follows that $K[B_t(u)]$ is a normal domain. Next, by a theorem of Hochster \cite{Ho72}, it follows that
$K[B_t(u)]$ is Cohen-Macaulay.  The second part follows by using \cite[Corollary 1.6]{HQ}.
\end{proof}

\section{Depth of powers of $t$-spread principal Borel ideals}
\label{three}

In this section, we consider the asymptotic behavior of the depth for $t$-spread principal Borel ideals.

Let $u=x_{i_1}\cdots x_{i_d}$ be the generator of the $t$-spread principal Borel ideal $I=B_t(u)\subset S=K[x_1,\ldots,x_n].$ We would like to compute the limit of
$\depth(S/I^k)$ when $k\to \infty.$
Obviously, we may consider $i_d=n$ since otherwise we may reduce to the study of $\depth S'/I^k$ where $S' =K[x_1,\ldots,x_{i_d}].$

\begin{Theorem}\label{limdepth}
Let $t\geq 1$ be an integer and $I=B_t(u)\subset S$ the $t$-spread principal Borel ideal generated by $u=x_{i_1}\cdots x_{i_d}$ where $t+1\leq i_1<i_2<\cdots <i_{d-1}<i_d=n.$ Then
\[
\depth\frac{S}{I^k}=0, \text{ for }k\geq d.
\]
In particular, the analytic spread of $I$ is $\ell(I)=n.$
\end{Theorem}

\begin{proof}
We first observe that the second claim of the theorem follows since it is known that, for any non-zero graded ideal $I,$ $\lim_{k\to \infty} \depth(S/I^k)=n-\ell(I)$ if the Rees algebra $\MR(I)$
is Cohen-Macaulay; see \cite[Proposition 10.3.2]{HHBook}. But the Rees algebra of $B_t(u)$ is indeed Cohen-Macaulay by Corollary~\ref{CMRees}.

In order to prove the first claim, by Auslander-Buchsbaum theorem, we have to show that
\begin{equation}\label{eq1}
\projdim \frac{S}{I^k}=n \text{ for }k\geq d.
\end{equation}

Since $I$ satisfies the $\ell$-exchange property (Proposition~\ref{elexchange}), by using \cite[Theorem 3.6]{EO}, it follows that $I^k$ has linear quotients with respect to $>_{\lex}$
for all $k\geq 1.$ This implies that we may compute the projective dimension of $S/I^k$ by using \cite[Corollary 8.2.2]{HHBook}. More precisely, if $G(I^k)=\{w_1,\ldots,w_m\}$
with $w_1>_{\lex}\cdots >_{\lex}w_m$ and for $1\leq j\leq m,$ $r_j$ is the number of variables which generate the ideal quotient $(w_1,\ldots,w_{j-1}):w_j,$ then
\[
\projdim \frac{S}{I^k}=\max\{r_1,\ldots,r_m\}+1.
\]

Thus, in order to prove (\ref{eq1}), we only need to find a monomial $w\in G(I^k)$ such that the ideal quotient $(w'\in G(I^k): w'>_{\lex} w):w$ is generated by  $n-1$ variables.
Let $w_0=v_1 v_2\cdots v_{d-1}v_d$ where
\[
v_1=x_1 x_{t+1} \ldots x_{(d-3)t+1} x_{(d-2)t+1} x_n\]
\[
v_2=x_1 x_{t+1} \ldots x_{(d-3)t+1} x_{i_{d-1}} x_n
\]
\[\vdots \]
\[
v_{d-1}=x_1 x_{i_2} \ldots x_{i_{d-2}} x_{i_{d-1}} x_n\]
\[
v_d=u=x_{i_1} x_{i_2} \ldots x_{i_{d-2}} x_{i_{d-1}} x_n.\]

Obviously, all the monomials $v_1,\ldots,v_d$ belong to $G(I).$ Let $k\geq d$ and $w=w_0u^{k-d}\in G(I^k).$ If we show that
$(w'\in G(I^k): w'>_{\lex} w):w\supseteq (x_1,\ldots,x_{n-1})$ then the proof
of equality (\ref{eq1}) is completed.

Let $J=(w'\in G(I^k): w'>_{\lex} w).$ We show that $x_j w\in J$ for all $1\leq j\leq n-1.$ Let $1\leq s\leq d$ such that $i_{s-1}\leq j <i_s,$ where we set $i_0=1.$
We consider the monomial
\[
v'_{d-s+1}=\frac{x_j v_{d-s+1}}{x_{i_s}}=x_1x_{t+1}\cdots x_{(s-2)t+1}x_j x_{i_{s+1}}\cdots x_{i_{d-1}}x_n.
\]
Clearly, $v'_{d-s+1}\in G(I)$ since $j<i_s$ and $v'_{d-s+1}$ is a $t$-spread monomial. Indeed, we have $i_{s+1}-j>i_{s+1}-i_s\geq t$ and
\[j-(s-2)t-1\geq i_{s-1}-(s-2)t-1\geq (i_1+(s-2)t)-(s-2)t-1=i_1-1\geq t.
\]
Let $w'_0$ be monomial obtained from  $w_0$ by replacing the monomial $v_{d-s+1}$ with $v'_{d-s+1}$ and let $w'=w'_0 u^{k-d}.$ Then $w'>_{\lex}w,$ thus $w'\in J$ and since $x_j w=x_{i_s}w',$
we have $x_jw\in J,$ which implies that $x_j\in J:w.$
\end{proof}

The analytic spread $\ell(I)$ of a graded ideal $I\subset S$ is the Krull dimension of the fiber ring $\MR(I)/\mathfrak{m} \MR(I)$ where $\mathfrak{m}=(x_1,\ldots,x_n).$ In our case, $I=B_t(u)$ is generated in the same degree $d.$ Thus the fiber ring is actually $K[G(B_t(u))].$  Therefore, we get the following consequence.

\begin{Corollary}
Let $t\geq 1$ be an integer and $B_t(u)\subset S$ the $t$-spread principal Borel ideal generated by $u=x_{i_1}\cdots x_{i_d}$ where $t+1\leq i_1<i_2<\cdots <i_{d-1}<i_d=n.$
Then $\dim K[G(B_t(u))]=n.$
\end{Corollary}

\begin{Remark}
Note that the order with respect to which we derived that all the powers of $B_t(u)$ have linear quotients in Proposition~\ref{linear quotients} does not coincide with the decreasing lexicographic order that we used in the proof of Theorem~\ref{limdepth}. In fact, the toric ideal of
$K[B_t(u)]$ does not have a quadratic Gr\"obner basis with respect to the lexicographic order as we can see in the following example.

Let $u=x_6 x_8 x_{10}\in K[x_1,\ldots,x_{10}]$ and $B_2(u)$ the $2$-spread principal Borel ideal defined by $u$. Let
$f=t_{u_1}t_{u_2}t_{u_3}-t_{v_1}t_{v_2}t_{v_3}$ with $u_1=x_1x_3x_8$, $u_2=x_1x_7x_9$, $u_3=x_2x_4x_6$ and $v_1=x_1x_3x_9$, $v_2=x_1x_6x_8$, $v_3=x_2x_4x_7$. Then $f$ is a binomial in the toric ideal of $K[B_2(u)]$ whose initial monomial with respect to the lexicographic order is $t_{u_1}t_{u_2}t_{u_3}.$ One can easily see that there is no quadratic monomial in the initial ideal of the toric ideal which divides
$t_{u_1}t_{u_2}t_{u_3}.$ This shows that, with respect to the lexicographic order, the reduced Gr\"obner basis of the toric ideal of
$K[B_2(u)]$ is not quadratic.
\end{Remark}

\begin{Remark}
Condition $i_1\geq t+1$ in the hypothesis of Theorem~\ref{limdepth} is essential. Indeed, let us consider
$I=B_t(u)$ where $u=x_t x_{2t}\cdots x_{dt}$, the $t$-spread Veronese ideal in $n=dt$ variables. We claim that
\[
\depth \frac{S}{I^k}=d-1, \text{ for }k\geq d,
 \] which implies that
\[
\lim_{k\to\infty} \depth \frac{S}{I^k}=d-1.
\]
In particular, this implies that the analytic spread of $I$ is  $\ell(I)=n-d+1=\dim K[B_t(u)].$
Indeed, we show that
\[
\projdim \frac{S}{I^k}=n-d+1, \text{ for }k\geq d.
\] As in the proof of Theorem~\ref{limdepth}, one  considers the monomials \[v_1=x_1x_{t+1}\cdots x_{(d-2)t+1}x_{dt},\ldots,
v_{d-1}=x_1 x_{2t}\cdots x_{dt}, v_d=u=x_t x_{2t}\cdots x_{dt},\] and shows that, for $k\geq d,$
\[(w^\prime\in G(I^k):w^\prime >_{\lex} w):w\supseteq (x_j: j\in [n]\setminus \supp(u)),
\] where $w=v_1\cdots v_d u^{k-d}.$

On the other hand, we show that for every $j\in \supp(u),$ $x_j$ does not appear among the variables which generate the colon ideals
 of $I^k.$ Let us assume that there exists $k\geq 1$ and some monomial $\mu\in I^k,$ $\mu=\mu_1\cdots \mu_k$ with $\mu_1,\ldots, \mu_k\in I$,
such that $x_j\in (\mu^\prime\in I^k: \mu^\prime>_{\lex }\mu)$ for some $j\in \supp(u).$ This implies that there exists
$\mu^\prime=\mu_1^\prime\cdots \mu_k^\prime\in I^k$
and some integer $s>j$ such that $x_j \mu= \mu^\prime x_s.$  Let $j=qt$ for some $1\leq q\leq d. $
Then
\[
\sum_{\ell=(q-1)t+1}^{qt}\deg_{x_{\ell}}(x_j \mu)=\sum_{\ell=(q-1)t+1}^{qt}\deg_{x_{\ell}}(x_j \mu_1\cdots \mu_k)=k+1> k=
\]
\[
=\sum_{\ell=(q-1)t+1}^{qt}\deg_{x_{\ell}}( \mu_1^\prime\cdots \mu_k^\prime x_s)=\sum_{\ell=(q-1)t+1}^{qt}\deg_{x_{\ell}} (\mu^\prime x_s),
\] contradiction. Therefore, the maximal number of variables which generate the colon ideals of $I^k$ for $k\geq d$ is $n-d,$ which implies that $\projdim \frac{S}{I^k}=n-d+1, \text{ for }k\geq d.$

\end{Remark}

config: arxiv


TITLE: Geodesic equations and christoffel symbols
QUESTION [3 upvotes]: I want to learn explicitly proof of the proposition 9.2.3.
Which books or lecture notes I can find? Please give me a suggestion. Thank you:)

REPLY [2 votes]: http://www.math.uh.edu/~minru/4350-11/geodesic.pdf
See pages 3 and 4 of this PDF!


TITLE: Extreme Points of a function
QUESTION [0 upvotes]: I have the equaition $f(x)=x^2(1-x)^4$.
It's derivative is $f'(x)=(1-x)^3(2x-6x^2)$ (as I calculated), and it's second derivative is $f''(x)=(1-x)^2(24x^2-14x+2)$ (again, as I calculated).
When finding where $f'(x)=0$, you get 3 $x$ values: $x_1=0, x_2=1, x_3=\frac{1}{3}$.
When putting them into the second derivative, you get:
$$
f''(0)=2 \\
f''(1) = 0 \\
f''(\frac{1}{3})=0
$$
From this, I thought the only extreme point of $f(x)$ was at $x=0$. However, the points at $x = 1 \ and \ x=\frac{1}{3}$ are also extreme points.
Why is that? How can I find thoes points? Thanks for your help.

REPLY [0 votes]: Every point $t$ where $f'(t)=0$ is an extreme or turning point.
The value of $f''(t)$ simply determines which type:

$f''(t)>0$ gives a trough, or local minimum
$f''(t)=0$ gives a point of inflection
$f''(t)<0$ gives a peak or local maximum


TITLE: Undergraduate-level book on modules, with emphasis on the structure theory of semisimple rings?
QUESTION [2 upvotes]: I am an undergraduate currently taking a representation theory course and we have $3$ chapters on modules, those being

Basics (submodules, quotients, generators/relations, direct sums/products, tensor products)
Finitely-generated modules over PIDs
Semisimple modules (simple/semisimple modules, semisimple rings, Artin-Wedderburn)

For FG Modules over PIDs Jacobson's "Basic Algebra I" seems to be an adequate resource, but his coverage of semisimple rings/modules in II (using category theory and something called Morita contexts) is too high-brow for me. So far, I have found only $2$ other books: Robert Ash's online book on graduate algebra, which is a bit too terse and Weintraub's "Algebra", which on the other hand is a bit overlong and is overly concerned with projective modules and other things.
Is there any book/online resources you could recommend?

REPLY [1 votes]: Bhattacharya, Phani Bhushan, Surender Kumar Jain, and S. R. Nagpaul. Basic abstract algebra. Cambridge University Press, 1994.

covers all these things at an undergraduate level.


TITLE: Geometrical representation of complex numbers.
QUESTION [0 upvotes]: For all real numbers $x$, let the mapping $f(x)=\frac{1}{x-i}$, where $i=\sqrt-1$. If there exist real number $a, b, c$ and $d $ for which $f(a),f(b),f(c)$ and $f(d)$ form a square on the complex plane. Find the area of the square.

My attempt:
Multiplying and dividing $f(x)$ by its conjugate. We get,
$\begin{align}
f(x) =\frac{x+i}{x²+1}\\
\therefore  f(a)=\frac{a+i}{a²+1}
\end{align}$
If we assume center of square as origin then angle between $f(a)$ and $f(b)$ in the complex plane will be $90°$.
So we can write, $\begin{align}
f(a)e^{\frac{iπ}{2}}=f(b)\quad
  (\because|f(a)|=|f(b)|)\\
\therefore  f(b) =if(a)
\end{align}$
Now,$|f(a)-f(b)|=|f(a)-if(a)|$ represents distance between two vertices of square i.e side of square.
$\begin{align}
\therefore Area=|f(a)-if(a)|^2\\
=2|f(a)|^2\\
=\frac{2}{a²+1}
\end{align}$
But the answer given is$\frac{1}{2}$.Please guide me.

REPLY [0 votes]: After getting a clue from other answers, I found another method to solve the problem.
$$f(t)=\left(\frac{t}{t^2+1}\right)+i\left(\frac{1}{t^2+1}\right)$$
Let $h=\frac{t}{1+t^2}$ and $k=\frac{1}{t^2+1}$. After eliminating $t$, we get,
$$h^2+k^2-k=0$$
It is a circle with center $(0,\frac{1}{2})$ and radius $\frac{1}{2}$ units. Now, in a circle infinite squares can be inscribed but all of them have equal side and hence equal area.
In this case $Area=\frac{1}{2}\,sq.unit$


text:
\begin{document}
\textwidth 150mm \textheight 225mm
\title{No signed graph with the nullity $\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-1$ \footnote{This work is supported by the National Natural Science Foundations of China (No. 11901253), the Natural Science Foundation for Colleges and Universities in Jiangsu Province of China (No. 19KJB110009), and the Science Foundation of Jiangsu Normal University (
No. 18XLRX021).}}
\author{{Yong Lu\footnote{Corresponding author.}, Jingwen Wu}\\
{\small  School of Mathematics and Statistics, Jiangsu Normal University,}\\ {\small  Xuzhou, Jiangsu 221116,
People's Republic
of China.}\\
{\small E-mail: luyong@jsnu.edu.com}}

\date{}
\maketitle
\begin{center}
\begin{minipage}{120mm}
\vskip 0.3cm
\begin{center}
{\small {\bf Abstract}}
\end{center}
{\small Let $G^{\sigma}=(G,\sigma)$ be a signed graph and $A(G,\sigma)$ be its adjacency matrix. Denote by $m(G)$ the matching number of $G$. Let $\eta(G,\sigma)$ be the nullity of $(G,\sigma)$. He et al. [Bounds for the matching number and cyclomatic number of a signed graph in terms of rank, Linear Algebra Appl. 572 (2019), 273--291] proved that
$$|V(G)|-2m(G)-c(G)\leq\eta(G,\sigma)\leq |V(G)|-2m(G)+2c(G),$$ where $c(G)$ is the dimension of cycle space of $G$. Signed graphs reaching the lower bound or the upper bound are respectively characterized by the same paper.
In this paper, we will prove that no signed graphs with nullity $|V(G)|-2m(G)+2c(G)-1$. We also prove that there are  infinite signed  graphs with nullity $|V(G)|-2m(G)+2c(G)-s,~(0\leq s\leq3c(G), s\neq1)$ for a given $c(G)$.

\vskip 0.1in \noindent {\bf Key Words}: \ Signed graph,  nullity,  matching number,  the dimension of cycle space. \vskip
0.1in \noindent {\bf AMS Subject Classification (2010)}: \ 05C35; 05C50. }
\end{minipage}
\end{center}

\section{Introduction }

Let $G=(V(G),E(G))$ be a simple graph, where $V(G)$ is the vertex
set of $G$ and $E(G)$ is the edge set of $G$. Let $V(G)=\{v_{1},v_{2},\ldots,v_{n}\}$. Then the \emph{adjacency matrix} $A(G)$ of $G$ is the symmetric $n\times n$ matrix with entries $A(i,j)=1$ (or written as $a_{ij}=1$) if and only if $v_{i}v_{j}\in E(G)$ and zeros elsewhere. The \emph{nullity}(resp. \emph{rank}) of $G$, denoted by $\eta(G)$(resp. $r(G)$), is the multiplicity of 0 (resp. non-zero) eigenvalue of $A(G)$. A \emph{pendant vertex} is defined as a vertex with degree one. The unique neighbour of the pendant vertex is called \emph{quasi-pendant vertex}. If a cycle of $G$ contains only a vertex of degree three, and the remaining vertices are of degree two, then the cycle is called a \emph{pendant cycle}.

Assume that $V_{1}\subseteq V(G)$ and $V_{1}\neq \phi$, Let $G[V_{1}]$ be the \emph{induced subgraph} of $G$, whose vertex set is $V_{1}$ and edge set is the set of those edges of $G$ that have both ends in $V_{1}$. Denoted by $G-V_{1}$ the induced subgraph obtained from $G$ by deleting each vertex in $V_{1}$ and its incident edges. For convenience, if $V_{1}=\{x\}$, we write $G-x$ for $G-\{x\}$. For the induced subgraph $H_{1}$ of $G$, $H_{1}+x$ is defined as the subgraph of $G$ induced by the vertex set $V(H_{1})\cup \{x\}$.

The \emph{matching number} of $G$, denoted by $m(G)$, is the size of a maximum matching of $G$.
Let $M$ be a maximum matching of $G$, if there exists the edge $e\in M$ such that the vertex $v\in e$, $v$ is called \emph{$M$-saturated}, otherwise, $v$ is called \emph{$M$-unsaturated}.  An \emph{$M$-alternating path} of $G$ is defined as a path whose edges are alternately in the edge sets $E\backslash M$ and $M$. An \emph{$M$-augmenting path} is defined as a path whose starting vertex and ending vertex are $M$-unsaturated.

The \emph{distance} between $u$ and $v$, denote by $d(u,v)$, is the length of the shortest path from the vertex $u$ to $v$. The length of the shortest cycle of $G$ is called the \emph{girth} of $G$, denote by $g(G)$.  Let $c(G)$ be \emph{the dimension of cycle space} of $G$, $c(G)=|E(G)|-|V(G)|+\theta(G)$, where $\theta(G)$ is the number of connected components of $G$. If the  cycles (if any) of $G$ are pairwise vertex-disjoint, the acyclic graph $T_{G}$ is obtained from  contracting each cycle of $G$ into a vertex, which is called \emph{cyclic vertex}. Let $W_{G}$ (resp. $U$) be the vertex set consisting of all cyclic vertices (resp. all non-cyclic vertices) in $T_{G}$, $V(T_{G})=W_{G}\cup U$. Furthermore, the graph is obtained from $T_{G}$ by deleting all cyclic vertices, denoted by $[T_{G}]$.

A \emph{signed graph} $G^{\sigma}=(G, \sigma)$ consists of a simple graph $G$ with edge set $E$ and a mapping $\sigma: E\rightarrow\{+, -\}$. $G$ is called the \emph{underlying graph} of $(G, \sigma)$. The \emph{adjacency matrix}  of $(G, \sigma)$, denoted by $A(G, \sigma)=a^{\sigma}_{ij}=\sigma(v_{i}v_{j})a_{ij}$, where $a_{ij}\in A(G)$. We use $r(G, \sigma)$ to denote the \emph{rank} of a signed graph $(G, \sigma)$.

Denote by  $C^{\sigma}_{n}$ a signed cycle of order $n$. The  \emph{sign} $sgn(C^{\sigma}_{n})$ of $C^{\sigma}_{n}$ is defined as $\prod_{e\in E(C^{\sigma}_{n})}\sigma(e)$. If $sgn(C^{\sigma}_{n})=+$ (or $sgn(C^{\sigma}_{n})=-$), then $C^{\sigma}_{n}$ is said to be \emph{positive} (or \emph{negative}). If all the cycles  of $(G,\sigma)$ are positive, then $(G,\sigma)$ is \emph{balanced}, and \emph{unbalanced} otherwise.

\begin{figure}[htbp]
 \centering
  \includegraphics[scale=0.65]{1.eps}
 \caption{$\infty(p,1,q)$, $\infty(p,l,q)$ and $\theta(p,l,q)$.}
\end{figure}

A connected graph $G$ is called \emph{bicyclic} if $|E(G)|=|V(G)|+1$.
The \emph{base} of a bicyclic graph $G$ is the unique connected bicyclic subgraph of $G$ containing no pendant vertices.

Let $\infty(p,l,q)$ (see Figure 1) be the graph obtained from $C_{p}$, $C_{q}$ and $P_{l}$ by identifying $v$ with $v_{1}$ and $u$ with $v_{l}$, respectively.

 A bicyclic graph $G$ is called an $\infty$-graph, if $G$ contains
$\infty(p,l,q)$ as its base.

Let $P_{p+2}$,~$P_{q+2}$,~$P_{l+2}$ be three paths, where $min\{p,l,q\}\geq0$ and at most one of $p,~l,~q$ is 0. Let $\theta(p,l,q)$ (see Figure 1) be the graph obtained from $P_{p+2}$,~$P_{l+2}$ and $P_{q+2}$ by identifying the three initial vertices and terminal vertices. The bicyclic graph containing
$\theta(p,l,q)$ as its base is called a $\theta$-graph.

A connected graph $G$ is called \emph{tricyclic} if $|E(G)|=|V(G)|+2$.

As shown in Figure 2, there are eight types of bases of tricyclic graph, denoted by $T_{i}$, $i=1,\ldots,8$. The tricyclic graph $G$ can be obtained from $T_{i}$ by attaching trees to its vertices.

\begin{figure}[htbp]
 \centering
  \includegraphics[scale=0.5]{11.eps}
 \caption{$T_{1}$--$T_{8}$}
\end{figure}

Collatz et al. \cite{CS}  had wanted to obtain all graphs of order $n$ with $\eta(G)>0$. Until today, this problem is also unsolved.
In recent years, the investigation on the nullity (or rank) of simple graphs is an significant topic in the theory of graph spectra. In 2014, Wang and Wong
\cite{WLWD} investigated the relationship between the nullity of the simple graph and the matching number, and proved that $|V(G)|-2m(G)-c(G)\leq \eta(G)\leq |V(G)|-2m(G)+2c(G)$. In 2015, Song, Song and Tam \cite {SST} characterized the graphs with the nullity $\eta(G)= |V(G)|-2m(G)+2c(G)$. In 2016, Wang \cite {WL} characterized the graphs with nullity $\eta(G)=|V(G)|-2m(G)-c(G)$. In 2019, Li and Guo \cite{LXGJM} proved that there is no graph with nullity $\eta(G)=|V(G)|-2m(G)+2c(G)-1$, and for fixed $c(G)$, infinitely many connected graphs with nullity $\eta(G)=|V(G)|-2m(G)+2c(G)-s$, where $0\leq s\leq 3c(G)$ and $s\neq1$, are also constructed.

In 2018, Lu, Wang and Zhou \cite{LYWLG} proved that $\eta(G)-2c(G)\leq \eta(G,\sigma) \leq \eta(G)+2c(G)$ for an unbalanced signed graph with $c(G)\geq1$, and all corresponding extremal graphs are characterized.
In 2019, He, Hao and Lai \cite {HSJHRX} proved that for every signed graph, $|V(G)|-2m(G)-c(G)\leq \eta(G,\sigma)\leq |V(G)|-2m(G)+2c(G)$. Moreover, signed graphs reaching the lower bound or the upper bound are respectively characterized in \cite{HSJHRX}. In this paper, we prove that no signed graph with nullity
$\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-1$, and for fixed $c(G)$, infinitely many connected signed graphs with nullity $\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-s$, where $0\leq s\leq 3c(G)$ and $s\neq1$, are also constructed.

Our main results are the following three theorems.

\begin{thm}\label{thm:1.1}

Suppose $(G,\sigma)$ is an unbalanced signed unicyclic graph with an unique cylce $C_{q}^{\sigma}$, then
\begin{enumerate}[(1)]
\item  $\eta(G,\sigma)=|V(G)|-2m(G)-1$, if $q\equiv1(mod~2)$,~$m([T_{G}])=m(T_{G})$~(Lemma 4.3~\cite{HSJHRX});
\item  $\eta(G,\sigma)=|V(G)|-2m(G)+2$, if $q\equiv2(mod~4)$,~$m([T_{G}])=m(T_{G})$~(Lemma 5.3~\cite{HSJHRX});
\item  $\eta(G,\sigma)=|V(G)|-2m(G)$, otherwise.
\end{enumerate}
\end{thm}

\begin{thm}\label{thm:1.2}

For any unbalanced signed graph $(G,\sigma)$, $$\eta(G,\sigma)\neq|V(G)|-2m(G)+2c(G)-1.$$
\end{thm}

\begin{thm}\label{thm:1.3}

Let $(G,\sigma)$ be an unbalanced signed graph. If $c(G)$ is fixed, then there exists infinitely connected unbalanced signed graphs, such that $\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-s$, where $0 \leq s \leq 3c(G), s\neq1$.

\end{thm}

\section{Preliminaries}

\begin{lem} \label{le:2.13}\cite{Bondy}

A matching $M$ of $G$ is a maximum matching if and only if $G$ contains no M-augmenting path.
\end{lem}
\begin{lem} \label{le:2.8}\cite{MXBWD}

Let $G$ be a graph with a pendant vertex $u$, and $v$ is the unique quasi-pendant vertex of $u$. Then $m(G)=1+m(G-v)=1+m(G-u-v)$.

\end{lem}
\begin{lem} \label{le:2.9}\cite{CCJH}

Let $G$ be a simple graph. Then $m(G)-1\leq m(G-v) \leq m(G)$, for any vertex $v \in V(G)$.
\end{lem}

\begin{lem} \label{le:2.10}\cite{MXBWD}

Let $G$ be a graph obtained by joining a vertex of an even cycle $C$ by an edge to a vertex y of a connected graph $H$. Then $m(G)=m(C)+m(H)$.
\end{lem}

\begin{lem} \label{le:2.14}\cite{LXGJM}

For any undirected graph $G$, $\eta(G)\neq|V(G)|-2m(G)+2c(G)-1.$
\end{lem}

\begin{lem} \label{le:2.12}\cite{WDMXB}

Let $G$ be a graph with $x \in V(G)$,

\begin{enumerate}[(1)]
  \item $c(G-x)=c(G)$  if $x$ lies outside any cycle of $G$;
  \item $c(G-x)\leq c(G)-1$  if $x$ lies on a cycle of $G$;
  \item $c(G-x)\leq c(G)-2$  if $x$ is a common vertex of distinct cycles of $G$;
  \item if the cycles of $G$ are pairwise vertex-disjoint, then $c(G)$ precisely equals the number of cycles in $G$.
\end{enumerate}

\end{lem}

\begin{lem} \label{le:2.1}\cite{BJBKDG}

Let $x$ be a vertex of $(G,\sigma)$. Then $\eta(G,\sigma)-1 \leq \eta(G-x,\sigma) \leq \eta(G,\sigma)+1$.
\end{lem}

\begin{lem} \label{le:2.4}\cite{FYZWY}

Let $(G,\sigma)$ be a signed graph containing a pendant vertex $u$, $uv\in E(G)$ and $(H,\sigma)=(G,\sigma)-u-v$. Then $\eta(G,\sigma)=\eta(H,\sigma)$.
\end{lem}

\begin{lem} \label{le:2.2}\cite{HSJHRX}

Let $(G,\sigma)$ be a connected signed graph. Then $|V(G)|-2m(G)-c(G) \leq \eta(G,\sigma) \leq |V(G)|-2m(G)+2c(G)$.
\end{lem}

\begin{lem} \label{le:2.6}\cite{FYZWY}

Let $P_{n}^{\sigma}$ be a signed path. Then,
\begin{equation}
\eta(P_{n}^{\sigma})=\left\{\begin{array}{ll}
0, ~if~n\equiv0(mod~2);\\\\
1, ~if~n\equiv1(mod~2).
\end{array}
\right.\nonumber
\end{equation}
\end{lem}

\begin{lem} \label{le:2.7}\cite{FYZWY}

Let $C^{\sigma}_{p}$ be a signed cycle. Then,

\begin{equation}
\eta(C^{\sigma}_{p})=\left\{\begin{array}{ll}
0, ~if~p\equiv1(mod~2);\\\\
0, ~if~p\equiv0(mod~4)~and~\sigma(C_{p})=-;\\\\
0, ~if~p\equiv2(mod~4)~and~\sigma(C_{p})=+;\\\\
2, ~if~p\equiv0(mod~4)~and~\sigma(C_{p})=+;\\\\
2, ~if~p\equiv2(mod~4)~and~\sigma(C_{p})=-.
\end{array}
\right.\nonumber
\end{equation}
\end{lem}

\begin{lem} \label{le:2.5}\cite{HYPLJS}

Let $(G,\sigma)$ be a signed graph. Then $(G,\sigma)$ is balanced if and only if $(G,\sigma)\thicksim (G,+)$.

\end{lem}

\begin{lem} \label{le:2.11}\cite{LYWLG}

Let $(G,\sigma)$ be a signed graph, if $(G_{1},\sigma)$,$(G_{2},\sigma)$,$\cdots$,$(G_{t},\sigma)$ are all the connected components of $(G,\sigma)$. Then $\eta(G,\sigma)=\sum_{i=1}^{t} \eta(G_{i},\sigma)$.

\end{lem}

\begin{lem} \label{le:2.3}\cite{HSJHRX}

Let $(G,\sigma)$ be a connnected signed graph. Then $\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)$ if and only if all the following conditions hold for $(G,\sigma)$:
\begin{enumerate}[(1)]
\item  the cycles (if any) of $(G,\sigma)$ are pairwise vertex-disjoint.
\item  for each cycle (if any) $C_{q}^{\sigma}$ of $(G,\sigma)$, either $q\equiv0(mod~4)$ and $\sigma(C_{q})=+$ or $q\equiv 2(mod~4)$ and $\sigma(C_{q})=-$.
\item  $m(T_{G})=m([T_{G}])$.
\end{enumerate}
\end{lem}

\section{Proof of Theorem 1.1}
In this section, we  will proved that there is no unbalanced signed unicyclic graph with the nullity $\eta (G,\sigma)=|V(G)|-2m(G)+2c(G)-1$. In order to proof Theorem \ref{thm:1.1}, we need to introduce some definitions and lemmas.

In simple graph, we introduced the concept of elementary graph, either the complete graph with order two $K_{2}$ or the cycle $C_{q},~q\geq3$. If the graph whose each component is elementary graph, the graph is called the basic graph. The set of signed graphs, denoted by $\mathbf{U_{i}}$, is the set of all basic graphs with $i$ vertices in $(G,\sigma)$. For any basic graph $(u,\sigma)\in \mathbf{U_{i}}$, $p(u), ~c(u)$ and $s(u)$ are defined as the number of components, the number of cycles and the number of the negative edges in the cycle of $(u,\sigma)$, respectively.

\begin{defn}\label{de:3.2}
Let $G$ be an unicyclic graph with an unique cycle $C_{q}$. Then,

\begin{enumerate}[(1)]
  \item $E_{1}$: The edge set of $G$ between  $C_{q}$ and $[T_{G}]$.
  \item $F_{1}$: The matching set of $G$ with $m(G)$ edges.
  \item $F_{2}$: The matching set of $[T_{G}]$ with $m([T_{G}])$ edges.
  \item $F_{1}^{'}$: The matching set of $G$ with $m(G)$ edges, each of which has at least one edge in $E_{1}$.
  \item $F_{1}^{''}$: The matching set of $G$ with $m(G)$ edges, and $M\cap E_{1}=\phi$ for all $M\in F_{1}$.
\end{enumerate}
\end{defn}

By Definition \ref{de:3.2}, we have $F_{1}=F_{1}^{'}\cup F_{1}^{''}$. We can obtain the following corollary:

\begin{cor}\label{co:1}
 Let $C_{q}$ be an even cycle. Then,
\begin{enumerate}[(1)]
  \item If $F_{1}^{'}=\phi$, the maximum matching of $G$ is the union of a maximum matching of $C_{q}$ and a maximum matching of $G-C_{q}$. Then  $|F_{1}|=|F_{1}^{''}|=2|F_{2}|$.
  \item If $F_{1}^{'}\neq \phi$, then $|F_{1}|=|F_{1}^{'}|+|F_{1}^{''}|> 2|F_{2}|$.

\end{enumerate}
\end{cor}

\begin{lem} \label{le:3.1}\cite{LYYLH}

Let $(G,\sigma)$ be a signed graph on $n$ vertices and $P_{(G,\sigma)}(\lambda)=\Sigma_{i=0}^{n}a_{i} \cdot \lambda^{n-i}$ be a characteristic polynomial of $A(G,\sigma)$. Then $a_{i}=\sum_{(u,\sigma) \in \mathbf{U_{i}}}(-1)^{p(u)+s(u)}\cdot 2^{c(u)}$, $i \in \{1,2,\ldots,n\}$.

\end{lem}

\begin{lem} \label{le:3.2}\cite{HSJHRX}

Let $(G,\sigma)$ be a connected unbalanced signed unicyclic graph whose unique cycle is $C_{q}^{\sigma}$. Then $\eta(G,\sigma)=|V(G)|-2m(G)+2$ if and only if $q\equiv 2(mod~4)$ and $m(T_{G})=m([T_{G}])$.

\end{lem}
\begin{lem} \label{le:3.3}\cite{HSJHRX}

Let $(G,\sigma)$ be an unicyclic signed graph which contains the unique signed cycle $C_{q}^{\sigma}$. Then $\eta(G,\sigma)=|V(G)|-2m(G)-1$ if and only if $q$ is odd and $m(T_{G})=m([T_{G}])$.

\end{lem}

\begin{lem} \label{le:3.4}\cite{RSZY}

Let $G$ be a graph with at least one cycle. Suppose that all cycles of $G$ are pairwise vertex-disjoint and each cycle is odd, then $m(T_{G})=m([T_{G}])$ if and only if $m(G)=\sum_{C\in \mathcal{L}(G)}m(C)+m([T_{G}])$, where $\mathcal{L}(G)$ denotes the set of all cycles in $G$.

\end{lem}

By Lemma \ref{le:3.4}, we can obtain the following corollary.
\begin{cor}\label{co:2}
Let $G$ be an unicyclic graph with the unique cycle $C_{q}$, $q$ is odd. Then
 $m(T_{G})= m([T_{G}])$ if and only if $m(G)=m([T_{G}])+m(C_{q})$.

\end{cor}

Next, for the unicyclic graph with an even cycle, similar to Lemma \ref{le:3.4}, we give the following lemma.

\begin{lem} \label{le:3.5}

Let $G$ be an unicyclic graph with a cycle $C_{q}$, $q$ is even. Then $m(T_{G})=m([T_{G}])$ if and only if $m(G)=m(C_{q})+m([T_{G}])$ and $M\cap E_{1}=\phi$ for all $M\in F_{1}$.

\end{lem}

\noindent\textbf{Proof.} Suppose that $G-C_{q}$ has $p$ connected components, says $S_{1},S_{2},\ldots,S_{p}$. Since $[T_{G}]=G-C_{q}$, then $S_{1},S_{2},\ldots,S_{p}$ are also all connected components of $[T_{G}]$. Assume that the vertex $v_{c}$ of $T_{G}$ is correspond to the cycle $C_{q}$ of $G$, which implies that $W_{G}=\{v_{c}\}$ in $T_{G}$.

In $T_{G}$, let $\widetilde{S_{k}}$ ($k=1,\ldots,p$) be the subgraph induced by $V(S_{k})$ and all vertices adjacent to it. Similarly, in $G$, let $\widehat{S_{k}}$ be the subgraph induced by $V(S_{k})$ and all vertices adjacent to it. Since $G$ is an unicyclic graph, then there exists at most an edge between $V(S_{k})$ and $V(C_{q})$,
thus for all $k=1,\ldots,p$, the subgraph $\widetilde{S_{k}}$ and $\widehat{S_{k}}$ are isomorphic.

\textbf{Necessity:}
Note that $m(T_{G})=m([T_{G}])$, and  $m([T_{G}])=\sum_{k=1}^{p}m(S_{k})$. Then each maximum matching of $[T_{G}]$ is the maximum matching of $T_{G}$, denoted by $M_{T}$ the maximum matching of $[T_{G}]$, thus $|M_{T}|=\sum_{k=1}^{p}m(S_{k})$. We can prove $m(S_{k})=m(\widetilde{S_{k}})$ for $k=1,\ldots,p$. Otherwise, there exists $m(S_{k_{0}})<m(\widetilde{S_{k_{0}}})$, where $k_{0}\in \{1,\ldots,p\}$. For $M_{T}$, replace the maximum matching of $S_{k_{0}}$ to the maximum matching of $\widetilde{S_{k_{0}}}$, we obtained a matching of $T_{G}$, denoted by $M_{T}^{'}$. Then£¬
$|M_{T}^{'}|= m(\widetilde{S_{k_{0}}})+\sum_{k\neq k_{0}}m(\widetilde{S_{k}}) > m(S_{k_{0}})+\sum_{k\neq k_{0}}m(S_{k})= |M_{T}|$.

This contradicts $|M_{T}|=m(T_{G})\geq |M_{T}^{'}|$. Hence $m(S_{k})=m(\widetilde{S_{k}})$, for all $k=1,\ldots,p$. Since the subgraph $\widetilde{S_{k}}$ and $\widehat{S_{k}}$ are isomorphic, then $m(\widehat{S_{k}})=m(S_{k})=m(\widetilde{S_{k}})$, for all $k=1,\ldots,p$. Let $M_{1}$ be a maximum matching of $G$. Then
$m(G)=|M_{1}|=|M_{1}\cap E(C_{q})|+\sum_{k=1}^{p}|M_{1}\cap E(\widehat{S_{k}})| \leq m(C_{q})+\sum_{k=1}^{p}m(\widehat{S_{k}})=m(C_{q})+\sum_{k=1}^{p}m(S_{k})$. By $m(T_{G})=m([T_{G}])$, we have $m(G)\geq m(C_{q}) + \sum_{k=1}^{p}m(S_{k})$.

Thus $$m(G)= m(C_{q}) + \sum_{k=1}^{p}m(S_{k}),$$

that is, $$m(G)= m(C_{q}) + m([T_{G}]).$$
Next, we need to prove $M\cap E_{1}=\phi$ for any maximum matching $M$ of $G$, $M\in F_{1}$. Suppose on the contrary, there exists $M_{2} \in F_{1}$ such that $ M_{2} \cap E_{1} \neq \phi$. Let $uv \in E_{1} \cap M_{2}$, where $v\in V(C_{q}), u\in V(S_{k_{1}})$, and $k_{1} \in \{1,\ldots,p\}$. Since $C_{q}$ has a perfect matching, which shows that $m(C_{q}-v)=m(C_{q})-1<m(C_{q})$. In $\widehat{S_{k_{1}}}$, $u$ is a quasi-pendant vertex, by Lemma \ref{le:2.8}, $m(S_{k_{1}}-u)=m(\widehat{S_{k_{1}}}-u-v)=m(\widehat{S_{k_{1}}})-1=m(S_{k_{1}})-1$.
Then,
$|M_{2}|=m(C_{q}-v)+(1+m(\widehat{S_{k_{1}}}-u-v))+\sum_{k\neq k_{1}}|M_{2} \cap \widehat{S_{k}}|<m(C_{q})+\sum_{k=1}^{p}m(\widehat{S_{k}})=m(C_{q})+\sum_{k=1}^{p}m(S_{k})=m(G)$. This contradicts $m(G)=|M_{2}|$, thus
$M\cap E_{1}=\phi$ for all $M\in F_{1}$.

\textbf{Sufficiency:}
Note that $m(G)= m(C_{q}) + \sum_{k=1}^{p}m(S_{k})$. We assume that $m(T_{G})\neq m([T_{G}])$, that is, $m(T_{G})=m([T_{G}])+1$ ($G$ is an unicyclic graph). Which implies that each maximum matching of $T_{G}$ must cover the vertex $v_{c}$ of $W_{G}$. There exists a maximum matching $M_{T}^{''}$ of $T_{G}$ such that the edge $wv_{c}\in M_{T}^{''}$, and the vertex $w$ is adjacent to $v_{0}$ the vertex of $C_{q}^{\sigma}$ in $G$.
Since $q$ is even, then $m(C_{q}-v_{0})=m(C_{q})-1$. Assume that $M_{3}$ is a matching of $G$, which is the union of $M_{T}^{''}$ and a maximum matching of $C_{q}-v_{0}$. Then $|M_{3}|=m(T_{G})+m(C_{q}-v_{0})=m([T_{G}])+1+m(C_{q})-1=m(G)$, which implies that $M_{3}$ is a maximum matching of $G$ and $M_{3}\cap E_{1} \neq \phi$, a contradiction. Thus $m(T_{G})= m([T_{G}])$.\quad $\square$

Next, we give Lemmas \ref {le:3.6} and \ref {le:3.7} by applying the characteristic polynomial of the signed graph.
\begin{lem} \label{le:3.6}

Let $(G,\sigma)$ be an unbalanced signed unicyclic graph, which contains the unique cycle $C_{q}^{\sigma}$. If $q$ is odd and $m(T_{G})\neq m([T_{G}])$, then $\eta(G,\sigma)=|V(G)|-2m(G)$.
\end{lem}

\noindent\textbf{Proof.}
Let $|V(G)|=n,~m(G)=m~ and~ m(G-C_{q})=l$. By Lemma \ref{le:3.1}, we have the characteristic polynomial of $(G,\sigma)$, $P_{(G,\sigma)}(\lambda)=\sum_{i=0}^{n}a_{i} \cdot \lambda^{n-i}$,~$a_{i}=\sum_{(u,\sigma) \in \mathbf{U_{i}}}(-1)^{p(u)+s(u)}\cdot 2^{c(u)}$, for any $i \in \{1,2,\ldots,n\}$.
Since $q$ is odd, then $m(C_{q})=\frac{q-1}{2}$, and $C_{q}^{\sigma}$ has not a perfect matching. Note that $m(T_{G})\neq m([T_{G}])$, by Corollary \ref{co:2}, we have $m(G)>m([T_{G}])+m(C_{q})=l+\frac{q-1}{2}$, that is, $2m+1>q+2l$. If $i> 2m$, then $(G,\sigma)$ contains no basic subgraphs with $i$ vertices and $a_{i}=0$. Thus  $P_{(G,\sigma)}(\lambda)=\lambda^{n}+a_{1}\lambda^{n-1}+\cdots +a_{2m}\lambda^{n-2m}=\lambda^{n-2m}(\lambda^{2m}+a_{1}\lambda^{2m-1}+\cdots +a_{2m})$. Therefore
$\eta(G,\sigma)\geq |V(G)|-2m(G)$.

Next, we need to prove $a_{2m}\neq 0$. Note that $a_{2m}=\sum_{(u,\sigma) \in \mathbf{U_{2m}}}(-1)^{p(u)+s(u)}\cdot 2^{c(u)}$, since $C_{q}^{\sigma}$ is an odd cycle, the basic subgraphs of $(G,\sigma)$ with $2m$ vertices contain no cycle. Then $s(u)=c(u)=0,~p(u)=m$, we have $a_{2m}=\sum_{(u,\sigma) \in \mathbf{U_{2m}}}(-1)^{m}\neq 0$. Thus $\eta(G,\sigma)=|V(G)|-2m(G)$.\quad $\square$
\begin{lem} \label{le:3.7}

Let $(G,\sigma)$ be an unbalanced signed unicyclic graph with the unique cycle $C_{q}^{\sigma}$, and $q$ is even. If  $m(T_{G})= m([T_{G}])$ and $q\equiv 0(mod~4)$; or $m(T_{G})\neq m([T_{G}])$, then $\eta(G,\sigma)=|V(G)|-2m(G)$.
\end{lem}

\noindent\textbf{Proof.}
Let $F_{3}=\{u|u=C_{q}\cup M, M \in F_{2}\}$, $m(G)=m$ and $m(G-C_{q})=l$. $C_{q}^{\sigma}$ has odd number of negative edges. Note that $G$ is a bipartite graph. By Lemma \ref{le:3.1}, then the characteristic polynomial of $(G,\sigma)$ can be expressed by $P_{(G,\sigma)}(\lambda)=\sum_{i=0}^{\lfloor \frac{n}{2} \rfloor}b_{i} \cdot \lambda^{n-2i}$, $b_{i}=\sum_{(u,\sigma) \in \mathbf{U_{2i}}}(-1)^{p(u)+s(u)}\cdot 2^{c(u)}$, for any $i \in \{1,2,\ldots,\lfloor \frac{n}{2} \rfloor\}$.
By the similar proof as in Lemma \ref{le:3.6}, if $i>m$, then $(G,\sigma)$ contains no basic subgraphs with $2i$ vertices and $b_{i}=0$. Hence $$P_{(G,\sigma)}(\lambda)=\lambda^{n}+b_{1}\lambda^{n-2}+\cdots +b_{m}\lambda^{n-2m}=\lambda^{n-2m}(\lambda^{2m}+b_{1}\lambda^{2m-2}+\cdots +b_{m}).$$ Therefore $\eta(G,\sigma)\geq |V(G)|-2m(G)$.

Next, in order to get the result $\eta(G,\sigma)=|V(G)|-2m(G)$, we need to prove $b_{m}\neq 0$.

\textbf{(1).} $m(T_{G})= m([T_{G}])$, and $q\equiv 0(mod~4)$.

By Lemma \ref {le:3.5}, we have $m(G)=m([T_{G}])+m(C_{q})$. Note that $m=\frac{q}{2}+l$. Since $2m=q+2l$ and the definition of $\mathbf{U_{2m}}$, by the similar proof as in Lemma \ref{le:3.6}, we have $\mathbf{U_{2m}}=F_{1}^{\sigma}\cup F_{3}^{\sigma}$. If $(u,\sigma)\in F_{1}^{\sigma}$, then $p(u)=m,~s(u)=c(u)=0$.
If $(u,\sigma)\in F_{3}^{\sigma}$, then $p(u)=l+1,~c(u)=1$. Let $s(u)=2a+1,~a\in \mathbf{Z^{+}}$ (since $C_{q}^{\sigma}$ has an odd number of negative edges), then

$b_{m}=\sum_{(u,\sigma)\in F_{1}^{\sigma}}(-1)^{m}+\sum_{(u,\sigma)\in F_{3}^{\sigma}}(-1)^{l+1+2a+1}\cdot 2^{1}$ = $(-1)^{m}\cdot|F_{1}^{\sigma}|+(-1)^{l}\cdot2\cdot|F_{3}^{\sigma}|=(-1)^{l}((-1)^{m-l}|F_{1}^{\sigma}|+2\cdot|F_{3}^{\sigma}|)$. Since $m-l=\frac{q}{2}$ and $q\equiv 0(mod~4)$, we have $m-l\equiv 0(mod~2)$. Thus $b_{m}=(-1)^{l}(|F_{1}^{\sigma}|+2\cdot|F_{3}^{\sigma}|)\neq 0$.

\textbf{(2).} $m(T_{G})\neq m([T_{G}])$.

Equivalently, $m(T_{G})=m([T_{G}])+1$, then each maximum matching of $T_{G}$ must cover the unique cyclic vertex. By Lemma \ref{le:3.5}, we mainly discuss the following two cases.

\textbf{Case 1.} $m(G)\neq m(C_{q}^{\sigma})+m([T_{G}])$,
that is, $m(G)> m(C_{q}^{\sigma})+m([T_{G}])$,~$m>\frac{q}{2}+l,~2m>q+2l$. Then the basic subgraphs with $2m$ vertices have no cycle. Which shows that $F_{3}^{\sigma}=\phi$ and $\mathbf{U_{2m}}=F_{1}^{\sigma}$. Then $p(u)=m$ and $s(u)=c(u)=0$ for any $(u,\sigma) \in \mathbf{U_{2m}}$. Thus
$b_{m}=\sum_{(u,\sigma)\in F_{1}^{\sigma}}(-1)^{m}=(-1)^{m}|F_{1}^{\sigma}|\neq 0$.

\textbf{Case 2.} $m(G)=m(C_{q}^{\sigma})+m([T_{G}])$ and $M\cap E_{1}\neq \phi,~\exists~M\in F_{1}$ (equivalently, $F_{1}^{'}\neq \phi$).
 Note that $m=l+\frac{q}{2}$, there exists some basic subgraphs with $2m$ vertices, such that contain $C_{q}^{\sigma}$ as a subgraph. Then
$\mathbf{U_{2m}}=F_{1}^{\sigma}\cup F_{3}^{\sigma}$. If $(u,\sigma)\in F_{1}^{\sigma}$, then $p(u)=m, s(u)=c(u)=0$.
If $(u,\sigma)\in F_{3}^{\sigma}$, we have $p(u)=l+1,~c(u)=1$. Let $s(u)=2a+1,~a\in \mathbf{Z^{+}}$ (since $C_{q}^{\sigma}$ has an odd number of negative edges). Then

$b_{m}=\sum_{(u,\sigma)\in F_{1}^{\sigma}}(-1)^{m}+\sum_{(u,\sigma)\in F_{3}^{\sigma}}(-1)^{l+1+2a+1}\cdot 2^{1}$ = $(-1)^{m}\cdot|F_{1}^{\sigma}|+(-1)^{l}\cdot2\cdot|F_{3}^{\sigma}|=(-1)^{l}((-1)^{m-l}|F_{1}^{\sigma}|+2\cdot|F_{3}^{\sigma}|)$. Since $q$ is even, we mainly discuss the following two subcases.

\textbf{Subcase 2.1.} If $q\equiv 0(mod~4)$, then $m-l=\frac{q}{2}\equiv 0(mod~2)$. Thus $$b_{m}=(-1)^{l}\cdot(|F_{1}^{\sigma}|+2\cdot|F_{3}^{\sigma}|)\neq 0$$

\textbf{Subcase 2.2.} If $q\equiv 2(mod~4)$, then $m-l=\frac{q}{2}\equiv 1(mod~2)$. Thus

$b_{m}=(-1)^{l+1}(|F_{1}^{\sigma}|-2\cdot|F_{3}^{\sigma}|)=(-1)^{l+1}(|F_{1}^{\sigma}|-2\cdot|F_{2}^{\sigma}|)$, $|F_{3}|=|F_{2}|$. Note that $F_{1}^{'} \neq \phi$, by Corollary \ref{co:1}, we have $|F_{1}|>2|F_{2}|$. Hence $b_{m}\neq 0$,

Based on the above conclusions, we have $b_{m}\neq 0$. Thus $\eta(G,\sigma)=|V(G)|-2m(G)$.\quad $\square$

By Lemmas \ref{le:3.2}, \ref{le:3.3}, \ref{le:3.6} and \ref{le:3.7}, we obtained Theorem \ref{thm:1.1}.

\section{Proof of Theorem 1.2}

In this section, we mainly investigate the unbalanced signed graphs. In order to proof Theorem \ref{thm:1.2}, we need to introduce some definitions and lemmas.

Let $G$ be the graph with some pendant vertices, and $G$ contains at least a cycle. For any pendant vertex $u$, we will give definitions of two types of pendant vertex $u$:
\begin{defn}\label{de:2}

\begin{enumerate}[(1)]
  \item If the quasi-pendant vertex of $u$ does not lie on the cycle, then $u$ is of Type \uppercase\expandafter{\romannumeral1}.
  \item If the quasi-pendant vertex of $u$ lies on the cycle, then $u$ is of Type \uppercase\expandafter{\romannumeral2}.
\end{enumerate}
\end{defn}

\begin{lem} \label{le:4.1}

Let $(G,\sigma)$ be an unbalanced signed graph with a pendant vertex $u$, and $v$ is adjacent to $u$. If $u$ is of Type \uppercase\expandafter{\romannumeral1}, then
$\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-s$ if and only if $\eta (G-u-v,\sigma)=|V(G-u-v)|-2m(G-u-v)+2c(G-u-v)-s$. $(0\leq s\leq 3c(G))$.
\end{lem}

\textbf{Proof.} By Lemmas \ref {le:2.8} and \ref {le:2.12},
$$|V(G)|=|V(G-u-v)|+2,$$
$$m(G)=m(G-u-v)+1,$$
$$c(G)=c(G-u-v).$$

By Lemma \ref {le:2.4},

$\eta (G-u-v,\sigma)=\eta (G,\sigma)=|V(G)|-2m(G)+2c(G)-s=|V(G-u-v)|+2-2(m(G-u-v)+1)+2c(G-u-v)-s=|V(G-u-v)|-2m(G-u-v)+2c(G-u-v)-s$.\quad $\square$

\begin{lem} \label{le:4.2}

Let $(G,\sigma)$ be an unbalanced signed graph with a pendant vertex $u$, and $v$ is adjacent to $u$. If $u$ is of Type \uppercase\expandafter{\romannumeral2}, then
$\eta(G,\sigma)\leq |V(G)|-2m(G)+2c(G)-2$.
\end{lem}

\textbf{Proof.} We prove the result by contradiction. From Lemma \ref{le:2.2}, suppose on the contrary, there exists some unbalanced signed graphs $(H,\sigma)$ with the nullity $\eta(H,\sigma) =|V(H)|-2m(H)+2c(H)-s$, $s=0,1$. Since $v$ lies on the cycle of $(H,\sigma)$, by Lemmas \ref {le:2.8} and \ref {le:2.12}, we have
$$|V(H)|=|V(H-u-v)|+2,$$
$$m(H)=m(H-u-v)+1,$$
$$c(H)\geq c(H-u-v)+1.$$
By Lemma \ref {le:2.4}, we have $\eta (H-u-v,\sigma)=\eta (H,\sigma)=|V(H)|-2m(H)+2c(H)-s\geq |V(H-u-v)|+2-2(m(H-u-v)+1)+2c(H-u-v)+2-s\geq |V(H-u-v)|-2m(H-u-v)+2c(H-u-v)+1$.
Which contradicts Lemma \ref{le:2.2}. Therefore, for any unbalanced signed graph $(G,\sigma)$ with a pendant vertex $u$  of Type \uppercase\expandafter{\romannumeral2}, then
$\eta(G,\sigma)\leq |V(G)|-2m(G)+2c(G)-2$.\quad $\square$

\begin{lem} \label{le:4.3}

Let $(G,\sigma)$ be an unbalanced signed graph without pendant vertices. If $\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)$,  $c(G)\geq2$, then there exists a vertex $x$ on the cycle in $(G,\sigma)$ such that $\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x)$.
\end{lem}

\textbf{Proof.}
\textbf{(1).} $g(G)=3$,~$c(G)\geq 2$.

Denote by $C_{q}^{\sigma}$ the cycle of $(G,\sigma)$ and $q=3$. Since $c(G)\geq2$, there exists a vertex $x$ on another cycle in $(G,\sigma)$ such that $C_{q}^{\sigma}$ is a subgraph of $(G-x,\sigma)$. Which shows that $(G-x,\sigma)$ does not satisfy in Lemma \ref{le:2.3}(2), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{(2).} $g(G)\geq4$,~$c(G)\geq 2$.

Since $\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)$, $(G,\sigma)$ does not satisfy at least one of the three conditions in Lemma \ref{le:2.3}.

\textbf{Case 1.} If $(G,\sigma)$ does not satisfy Lemma \ref{le:2.3}(1).

 Note that $(G,\sigma)$ contains at least two vertex-joint cycles, called $C_{k}^{\sigma}$ and $C_{s}^{\sigma}$ $(k,s\geq4)$. Let $G[C_{k}^{\sigma},C_{s}^{\sigma}]$ be the graph induced by $V(C_{k}^{\sigma})$ and $V(C_{s}^{\sigma})$.

 \textbf{Subcase 1.1.} $c(G)=2$.

 Since $(G,\sigma)$ contains no pendant vertices, then $(G,\sigma)$ is the union of an $\infty$-graph (or a $\theta$-graph) and some isolated vertices (if any), which implies that $G[C_{k}^{\sigma},C_{s}^{\sigma}]$ is either $\infty^{\sigma}(p,1,q)$ or $\theta^{\sigma}(p,l,q)$.

\textbf{Subcase 1.1.1.} $G[C_{k}^{\sigma},C_{s}^{\sigma}]=\theta^{\sigma}(1,1,1)$.

Since $(G,\sigma)$ is unbalanced, $(G,\sigma)$ contains at least an unbalanced cycle, says $C_{q}^{\sigma}$, and $q=4$. Let $x$ be a vertex of another cycle in $(G,\sigma)$, $x\notin V(C_{q}^{\sigma})$. Then $C_{q}^{\sigma}$ is a subgraph of $(G-x,\sigma)$. Which shows that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(2), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Subcase 1.1.2.} $G[C_{k}^{\sigma},C_{s}^{\sigma}] \neq\theta^{\sigma}(1,1,1)$.

As shown in Figure 1, there exists a vertex $x$ on the cycle in $(G,\sigma)$, such that $(G-x,\sigma)$ contains a pendant vertex with Type \uppercase\expandafter{\romannumeral2}. By Lemma \ref{le:4.2}, we can get that $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Subcase 1.2.} $c(G)\geq 3$.

\textbf{Subcase 1.2.1.} In $(G,\sigma)$, there exists a vertex $x$ on the cycle such that $x\notin G[C_{k}^{\sigma},C_{s}^{\sigma}]$.

 Which implies that $G[C_{k}^{\sigma},C_{s}^{\sigma}]$ is a subgraph of $(G-x,\sigma)$. Then $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(1), hence $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$
For example, as shown in Figure 2, the signed graphs with $T_{i}(i=1,\ldots,4)$ as a underlying graph, which contains a vertex $x$ on the cycle and $x\notin G[C_{k}^{\sigma},C_{s}^{\sigma}]$.

\textbf{Subcase 1.2.2.} In $(G,\sigma)$, let $x$ be a vertex on any cycle, then $x\in G[C_{k}^{\sigma},C_{s}^{\sigma}]$.

Which implies that any cycle of $(G,\sigma)$ is the subgraph of $G[C_{k}^{\sigma},C_{s}^{\sigma}]$. Since $c(G)\geq3$, $(G,\sigma)$ contains one of eight types of bases of tricyclic graphs as a underlying subgraph. As shown in Figure 2, the graph $T_{j}$ can be regard as two vertex-joint cycles, where $j=5,\ldots,8$. Which implies that the tricyclic graphs $T_{j}$ is a underlying subgraph of $G[C_{k}^{\sigma},C_{s}^{\sigma}]$. As shown in Figure 2, there exists a
vertex $x$ of $T_{j}$ such that $T_{j}-x$ also contains two vertex-joint cycles. Hence there exists a vertex $x$ on the cycle of $(G,\sigma)$ such that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(1), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Case 2.} $(G,\sigma)$ satisfies Lemma \ref{le:2.3}(1) but does not satisfy Lemma \ref{le:2.3}(2).

\textbf{Subcase 2.1.} There exists at least a balanced cycle in $(G,\sigma)$.

\textbf{Subcase 2.1.1.} There exists a balanced cycle in $(G,\sigma)$, says $C_{p}^{\sigma}$ and $p \neq0(mod~4)$. Let $x$ be a vertex on another cycle and $x\notin V(C_{p}^{\sigma})$ (since all cycles are pairwise vertex-disjoint), then $C_{p}^{\sigma}$ is the subgraph of $(G-x,\sigma)$. Which shows that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(2), then$$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Subcase 2.1.2.} The length of all balanced graph is the multiple of $4$. Then $(G,\sigma)$ contains at least an unbalanced cycle in $(G,\sigma)$, says $C_{q}^{\sigma}$ and $q \neq2(mod~4)$. Let $x$ be a vertex of the balanced cycle and $x\notin V(C_{q}^{\sigma})$ (Since all cycles are pairwise vertex-disjoint), then $C_{q}^{\sigma}$ is the subgraph of $(G-x,\sigma)$. Which shows that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(2), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Case 2.2.} Cycles of $(G,\sigma)$ are unbalanced.

$(G,\sigma)$ contains at least an unbalanced cycle, says $C_{q}^{\sigma}$ and $q \neq2(mod~4)$. Let $x$ be a vertex of another unbalanced cycle in $(G,\sigma)$ and $x\notin V(C_{q}^{\sigma})$ (all cycles are pairwise vertex-disjoint). Since $c(G)\geq2$, then $C_{q}^{\sigma}$ is the subgraph of $(G-x,\sigma)$. Which shows that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(2), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Case 3.} $(G,\sigma)$ satisfies (1) and (2) of Lemma \ref{le:2.3}, but dose not satisfy (3) of Lemma \ref{le:2.3}.

Note that $m(T_{G})\neq m([T_{G}])$, that is, $m(T_{G})\geq m([T_{G}])+1$.

If $E(T_{G})=\phi$, then $(G,\sigma)$ is the union of some vertex-disjoint signed cycles and isolated vertices. Hence $m(T_{G})=m([T_{G}])=0$, a contradiction. Therefore, we only consider $E(T_{G})\neq\phi$. In $T_{G}$, each maximum matching must cover at least a pendant vertex. Otherwise, there exists an $M$-augmenting path in $G$, which contradicts Lemma \ref{le:2.13}. Let $u$ be a pendant vertex of $T_{G}$. Since $(G,\sigma)$ contains no pendant vertices, we have $u \in W_{G}$. Suppose that $C_{q}^{\sigma}$ the pendant cycle of $(G,\sigma)$ is correspond to the vertex $u$ of $T_{G}$. Let $v$ be the unique vertex with degree three in $C_{q}^{\sigma}$, and let $x$ be a vertex on the cycle $C_{q}^{\sigma}$. Then $T_{G-x}$ is the graph obtained from $T_{G}$ and $C_{q}^{\sigma}-x$ by identifying $u$ and $v$ as a vertex, we mainly discuss the following two subcases.

\textbf{Subcase 3.1.} Each maximum matching of $T_{G}$  cover all pendant vertices.

Let $x$ be a vertex on the cycle $C_{q}^{\sigma}$, and $x$ is adjacent to $v$. Noted that $C_{q}^{\sigma}$ is an even cycle, then $C_{q}^{\sigma}-v-x$ is an odd path and has a perfect matching. By the definition of $T_{G}$, which shows that the maximum matching of $T_{G-x}$ is the union of the maximum matching of $T_{G}$ and the maximum matching of $C_{q}^{\sigma}-v-x$. Then $$m(T_{G-x})=m(T_{G})+m(C_{q}^{\sigma}-v-x).$$ Hence each maximum matching of $T_{G-x}$ must cover some vertices in $W_{G-x}$,
we have $m(T_{G-x})\neq m([T_{G-x}])$. Which shows that $(G-x,\sigma)$ does not satisfy Lemma \ref{le:2.3}(3), then $$\eta(G-x,\sigma)\neq |V(G-x)|-2m(G-x)+2c(G-x).$$

\textbf{Subcase 3.2.} There exists some maximum matchings of $T_{G}$, denoted by $M_{i}(T_{G})$ $(i=1,\ldots,r)$, such that the pendant edge $wv\notin M_{i}(T_{G})$, where $r\geq 1$ and $v$ is a pendant vertex of $T_{G}$.

Let $y$ be a vertex on the cycle $C_{p}^{\sigma}$ and $d(v_{0},y)=2$, where $C_{p}^{\sigma}$ the cycle of $(G,\sigma)$ is correspond to $v$ the vertex of $T_{G}$, and $v_{0}$ is the unique vertex with degree three in $C_{p}^{\sigma}$. By the definition of $T_{G-y}$, which shows that the maximum matching of $T_{G-y}$ is the union of $M_{i}(T_{G})$ $(i\in \{1,\ldots,r\})$ and the maximum matching of $C_{p}^{\sigma}-y$. Then $$m(T_{G-y})=m(T_{G})+m(C_{p}^{\sigma}-y).$$ For any $i\in \{1,\ldots,r\}$, then $M_{i}(T_{G})$ must cover some vertices in $W_{G-y}$ by Lemma \ref{le:2.13}. Which implies that each maximum matching of $T_{G-y}$ must cover some vertices in $W_{G-y}$, we have $m(T_{G-y})\neq m([T_{G-y}])$. Which shows that $(G-y,\sigma)$ does not satisfy Lemma \ref{le:2.3}(3), then $\eta(G-y,\sigma)\neq |V(G-y)|-2m(G-y)+2c(G-y)$. \quad $\square$

\begin{lem} \label{le:4.4}

Let $(G,\sigma)$ be an unbalanced signed graph without pendant vertices. Then $$\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)-1.$$
\end{lem}

\textbf{Proof.} We apply induction on $c(G)$ to prove the result.

If $c(G)=1$, then $G=C_{k}\cup (n-k)K_{1}$~$(3\leq k\leq n)$. By Theorem \ref{thm:1.1}, we have $$\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)-1.$$

If $c(G)\geq 2$, assume that the conclusion is true for $c(G)\leq k$. Next, we need to prove the conclusion is true for $c(G)= k+1$. Suppose on the contrary, there exists some unbalanced signed graphs $(H,\sigma)$ with $c(H)=k+1$ such that $\eta(H,\sigma)=|V(H)|-2m(H)+2c(H)-1$.

Let $x$ be any vertex on the cycle of $(H,\sigma)$. For the signed graph $(H-x,\sigma)$, by Lemmas \ref{le:2.9} and \ref{le:2.12}, we have
$$|V(H)|=|V(H-x)|+1,$$
$$m(H)\leq m(H-x)+1,$$
$$c(H)\geq c(H-x)+1.$$
By Lemma \ref{le:2.1}, we have $\eta(H-x,\sigma)+1\geq \eta(H,\sigma)=|V(H)|-2m(H)+2c(H)-1 \geq|V(H-x)|+1-2(m(H-x)+1)+2(c(H-x)+1)-1=|V(H-x)|-2m(H-x)+2c(H-x)$.
By Lemma \ref{le:2.2}, then $$\eta(H-x,\sigma)=|V(H-x)|-2m(H-x)+2c(H-x)-s,~s=0,1$$.

If $(H-x,\sigma)$ is balanced, by Lemmas \ref{le:2.14} and \ref {le:2.5}, we have
$\eta(H-x,\sigma) \neq|V(H-x)|-2m(H-x)+2c(H-x)-1$.
We mainly consider that $(H-x,\sigma)$ is unbalanced in the following cases.

\textbf{Case 1.} $(H-x,\sigma)$ contains no pendant vertices.

In this case, $(H-x,\sigma)$ is also unbalanced and $c(H-x)\leq c(H)-1=k$. Applying the induction hypothesis to $(H-x,\sigma)$,
then $\eta(H-x,\sigma)\neq|V(H-x)|-2m(H-x)+2c(H-x)-1$.

\textbf{Case 2.} $(H-x,\sigma)$ contains some pendant vertices.

\textbf{Subcase 2.1.} $(H-x,\sigma)$ contains at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}.

By Lemma \ref{le:4.2}, we have $\eta(H-x,\sigma) \leq|V(H-x)|-2m(H-x)+2c(H-x)-2$.

\textbf{Subcase 2.2.} All the pendant vertices of $(H-x,\sigma)$ are of Type \uppercase\expandafter{\romannumeral1}.

Suppose that $(H-x,\sigma)$ contains $p$ pendant vertices. For the pendant vertices of Type \uppercase\expandafter{\romannumeral1}, by using Lemma \ref{le:2.4} repeatedly, after $p$ steps,
we obtain a subgraph $(H_{1},\sigma)$ of $(H,\sigma)$. If $(H_{1},\sigma)$ contains no pendant vertices or at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}, then $(H_{1},\sigma)$ is as we required and we are done. Otherwise, in $(H_{1},\sigma)$, for the pendant vertex of Type \uppercase\expandafter{\romannumeral1}, we continue to use Lemma \ref{le:2.4} repeatedly, we obtain a subgraph $(H_{2},\sigma)$ of $(H_{1},\sigma)$. If $(H_{2},\sigma)$ contains no pendant vertices or at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}, then $(H_{2},\sigma)$ is as we required and we are done. Otherwise, repeating the above steps until we obtain an unbalanced sign graph $(H_{0},\sigma)$ that meets the requirements.

\textbf{(a).} $(H_{0},\sigma)$ contains at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}.

By Lemma \ref{le:4.2}, we have $$\eta(H_{0},\sigma)\leq |V(H_{0})|-2m(H_{0})+2c(H_{0})-2.$$  Next, by Lemma \ref{le:4.1},

$$\eta(H_{0},\sigma)\leq |V(H_{0})|-2m(H_{0})+2c(H_{0})-2$$
if and only if $$\eta(H-x,\sigma)\leq |V(H-x)|-2m(H-x)+2c(H-x)-2.$$

\textbf{(b).} $(H_{0},\sigma)$ contains no pendant vertices.

Since $c(H_{0})=c(H-x)\leq c(H)-1=k$ and $H_{0}^{\sigma}$ is unbalanced, applying the induction hypothesis to $H_{0}^{\sigma}$, then $\eta(H_{0},\sigma)\neq |V(H_{0})|-2m(H_{0})+2c(H_{0})-1$.

By Lemma \ref{le:4.1},
$$\eta(H_{0},\sigma)\neq |V(H_{0})|-2m(H_{0})+2c(H_{0})-1$$ if and only if $$\eta(H-x,\sigma)\neq |V(H-x)|-2m(H-x)+2c(H-x)-1.$$

Based on the above results. Let $x$ be any vertex on the cycle of $(H,\sigma)$. For the nullity $\eta(H-x,\sigma)$ of signed graph $(H-x,\sigma)$, either $\eta(H-x,\sigma)\neq |V(H-x)|-2m(H-x)+2c(H-x)-1$ or  $\eta(H-x,\sigma)\leq |V(H-x)|-2m(H-x)+2c(H-x)-2$.

If $\eta(H-x,\sigma)\neq |V(H-x)|-2m(H-x)+2c(H-x)-1$, then $\eta(H-x,\sigma)= |V(H-x)|-2m(H-x)+2c(H-x)$.
On the other hand, since $\eta(H,\sigma)\neq |V(H)|-2m(H)+2c(H)$, by Lemma \ref{le:4.3}, there exists a vertex $y$ on the cycle of $(H,\sigma)$ such that $\eta(H-x,\sigma)\neq |V(H-x)|-2m(H-x)+2c(H-x)$, a contradiction.

If $\eta(H-x,\sigma)\leq |V(H-x)|-2m(H-x)+2c(H-x)-2$, which contradicts equation $\eta(H-x,\sigma)=|V(H-x)|-2m(H-x)+2c(H-x)-s,~s=0,1$.

Therefore, for any unbalanced signed graph $(G,\sigma)$ without pendant vertices, $\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)-1$.\quad $\square$

\textbf{The Proof of Theorem \ref{thm:1.2}:} Noted that the unbalanced signed graph $(G, \sigma)$ contains at least an unbalanced cycle, and $c(G)\geq1$.

\textbf{Case 1.} $(G,\sigma)$ has no pendant vertices, we can obtained the result by Lemma \ref{le:4.4}.

\textbf{Case 2.} $(G,\sigma)$ has some pendant vertices.

By the definitions of two types of pendant vertices, we mainly discuss the following two subcases.

\textbf{Subcase 2.1.} All pendant vertices are of Type \uppercase\expandafter{\romannumeral1}.

Suppose that $(G,\sigma)$ contains $p$ pendant vertices. For the pendant vertex of Type \uppercase\expandafter{\romannumeral1}, by using Lemma \ref{le:2.4} repeatedly, after $p$ steps, we obtain a subgraph $(G_{1},\sigma)$ of $(G,\sigma)$. If $(G_{1},\sigma)$ contains no pendant vertices or at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}, then $(G_{1},\sigma)$ is as we required and we are done. Otherwise, in $(G_{1},\sigma)$, for the pendant vertex of Type \uppercase\expandafter{\romannumeral1}, we continue to use Lemma \ref{le:2.4} repeatedly, we obtain a subgraph $(G_{2},\sigma)$ of $(G_{1},\sigma)$. If $(G_{2},\sigma)$ contains no pendant vertices or at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}, then $(G_{2},\sigma)$ is as we required and we are done. Otherwise, repeating the above steps until we obtain an unbalanced sign graph $(G_{0},\sigma)$ that meets the requirements.

\textbf{(a).} $(G_{0},\sigma)$ contains no pendant vertices.

By Lemma \ref{le:4.4}, we have $\eta(G_{0},\sigma)\neq |V(G_{0})|-2m(G_{0})+2c(G_{0})-1$.

\textbf{(b).} $(G_{0},\sigma)$ contains at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}.

By Lemma \ref{le:4.2}, we have $$\eta(G_{0},\sigma)\leq |V(G_{0})|-2m(G_{0})+2c(G_{0})-2.$$ Hence $$\eta(G_{0},\sigma)\neq |V(G_{0})|-2m(G_{0})+2c(G_{0})-1.$$ Next, by Lemma \ref{le:4.1}, we have

$\eta(G_{0},\sigma)\neq |V(G_{0})|-2m(G_{0})+2c(G_{0})-1$ if and only if $\eta(G,\sigma)\neq |V(G)|-2m(G)+2c(G)-1$.

\textbf{Subcase 2.2.} There exists at least a pendant vertex of Type \uppercase\expandafter{\romannumeral2}.

By Lemma \ref{le:4.2}, we have $\eta(G,\sigma)\neq|V(G)|-2m(G)+2c(G)-1$.\quad $\square$

Combining with Lemmas \ref{le:2.14}, \ref{le:2.5} and Theorem \ref{thm:1.2}, we can obtained the following corollary.

\begin{cor}\label{co:3}
For any signed graph $(G,\sigma)$, then $$\eta(G,\sigma)\neq|V(G)|-2m(G)+2c(G)-1.$$
\end{cor}

\section{Proof of Theorem 1.3}

In order to  proof Theorem \ref{thm:1.3}, we need to introduce some definitions.

Let $K_{1,c+1}^{\sigma}$ be a signed star, $x$ be the center vertex of $K_{1,c+1}^{\sigma}$ and $y_{0},\ldots, y_{c}$ be pendant vertices of $K_{1,c+1}^{\sigma}$,~respectively.
$O_{1}^{\sigma},\ldots,O_{l_{1}}^{\sigma}$ be a balanced cycle of size 3, $O_{l_{1}+1}^{\sigma},\ldots,O_{l_{1}^{'}}^{\sigma}$ be an unbalanced cycle of size 6. Let $O_{l_{1}^{'}+1}^{\sigma},\ldots,O_{c}^{\sigma}$ be the signed graph with the pendant vertex of Type \uppercase\expandafter{\romannumeral2}, which obtained from $C_{4}^{+}$ a balanced cycle of size 4 by attaching $K_{2}^{\sigma}$.
~Next, we construct a new signed graph $G^{\sigma}$ by identifying $y_{i}$ with a vertex of $O_{i}^{\sigma}$, $y_{j}$ with a pendant vertex of $O_{j}^{\sigma}$, where $i=1,\ldots,l_{1}^{'},~j=l_{1}^{'}+1,\ldots,c$.

\textbf{The Proof of Theorem \ref{thm:1.3}:} According to the definition of $G^{\sigma}$, let $l_{2}=l_{1}^{'}-l_{1},~l_{3}=c-l_{1}^{'}$. In this case, there exists $l_{1}$ balanced cycles $C_{3}^{+}$, $l_{2}$ unbalanced cycles $C_{6}^{-}$ and $l_{3}$ balanced graph $F^{\sigma}$, which is obtained from $C_{4}^{+}$ a balanced cycles of size 4 by attaching $K_{2}^{\sigma}$. Which implies that $y_{0}$ is the unique pendant vertex of $G^{\sigma}$.
Then
$$|V(G^{\sigma})|=3l_{1}+6l_{2}+5l_{3}+2,$$
$$m(G^{\sigma})=l_{1}+3l_{2}+2l_{3}+1,$$
$$c(G^{\sigma})=l_{1}+l_{2}+l_{3}.$$

By Lemmas \ref{le:2.4} and \ref{le:2.11}, we obtain that $\eta(G^{\sigma})=\eta(G^{\sigma}-x-y_{0}))=2l_{2}+l_{3}=|V(G^{\sigma})|
-2m(G^{\sigma})+2c(G^{\sigma})-(3l_{1}+2l_{3})$. Since $l_{i}\geq 0$, $3l_{1}+2l_{3}$ can take over any integer between 0 and $3c(G)$ expect for 1.  \quad $\square$

Combining with Lemma \ref{le:2.5} and Theorem \ref{thm:1.3}, we can obtained the following corollary.
\begin{thm}\label{th:5.1}
For any signed graph $(G,\sigma)$, if $c(G)$ is fixed, then there exists infinitely connected signed graphs, such that $\eta(G,\sigma)=|V(G)|-2m(G)+2c(G)-s$, where $0\leq s\leq 3c(G)$, $s\neq1$.

\end{thm}