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	\title {\brackf  Cross $t$-intersecting families for symplectic polar spaces}
	
	
	\author{Tian Yao\thanks{E-mail: \texttt{yaotian@mail.bnu.edu.cn}}}

	\author{Kaishun Wang\thanks{Corresponding author. E-mail: \texttt{wangks@bnu.edu.cn}}}
	\affil{Laboratory of Mathematics and Complex Systems (Ministry of Education), School of
		Mathematical Sciences, Beijing Normal University, Beijing 100875, China}
	
	
	\date{}
	
	\openup 0.5\jot
	\mathscr{A}ketitle
	
	
	\brackegin{abstract}
		Let $\mathscr{P}$ be a symplectic polar space over a finite field $\mathbb{F}_q$, and $\mathscr{P}_m$ denote the collection of all $k$-dimensional totally isotropic subspace in $\mathscr{P}$.
	Let $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ satisfy $\dim(F_1\cap F_2)\geqslant t$ for any $F_1\in\mathscr{F}_1$ and $F_2\in\mathscr{F}_2$. We say they are cross $t$-intersecting families.  Moreover, we say they are trivial if each member of them contains a fixed $t$-dimensional totally isotropic subspace. In this paper, we show that cross $t$-intersecting families with maximum product of sizes are trivial. We also describe the structure of non-trivial $t$-intersecting families with maximum product of sizes.
		
		

		
		\noindent{\brackf Key words}\ \ cross $t$-intersecting families; symplectic polar spaces.
		
		\
		
		\noindent{\brackf AMS classification:} \   05D05, 05A30, 51A50
		

		
	\end{abstract}
	
	\section{Introduction}
	
		Intersection problems originate from the famous Erd\H{o}s-Ko-Rado Theorem \cite{EKR}. In recent years, intersection problems for mathematical objects which are relative to vector spaces have been caught lots of attention \cite{VHM1,VHM2,AEKR,SPEKR1}.
	
	Let $n$ and $k$ be positive integers with $n\geqslant k$, $V$ an $n$-dimensional vector space over the finite field $\mathbb{F}_q$, where $q$ is a prime power, and ${V\brack k}_q$ denote the family of all $k$-dimensional subspaces of $V$. We usually replace ``$k$-dimensional subspace'' with ``$k$-subspace'' for short. Define the \emph{Gaussian binomial coefficient} by
	$${n\brack k}_q:=\prod_{0\leqslant i<k}\dfrac{q^{n-i}-1}{q^{k-i}-1},$$
	and set ${n\brack0}_q=1$ and ${n\brack c}=0$ if $c$ is a negative integer. Note that the size of ${V\brack k}_q$ is ${n\brack k}_q$. From now on, we will omit the subscript $q$.
	
	Let $t$ be a positive integer. A family $\mathscr{F}\subset{V\brack k}$ is called \emph{$t$-intersecting} if $\dim(F_1\cap F_2)\geqslant t$ for any $F_1,F_2\in\mathscr{F}$. A $t$-intersecting family $\mathscr{F}$ is called \emph{trivial} if there exists a $t$-subspace contained in each element of $\mathscr{F}$. The Erd\H{o}s-Ko Rado Theorem for vector space \cite{VEKR2,VEKR1,VEKR3} shows that a $t$-intersecting subfamily of ${V\brack k}$ with maximum size is trivial when $\dim V>2k$. The structure of non-trivial $t$-intersecting subfamily of ${V\brack k}$ with maximum size was determined via the parameter ``$t$-covering number", see \cite{VHM1,VHM2}. For $\mathscr{F}_1\in{V\brack k_1}$ and $\mathscr{F}_2\subset{V\brack k_2}$, we say they are cross $t$-intersecting if $\dim(F_1\cap F_2)$ holds for any $F_1\in\mathscr{F}_1$ and $F_2\in\mathscr{F}_2$. Recently, Cao et al \cite{CAO} describe the structure of cross $t$-intersecting families with the first and second larges product of sizes.
	
	
	
	
	
	
	
	Let $f$ be a non-degenerate alternating bilinear form defined on a $2\nu$-dimensional vector space $\mathbb{F}_q^{2\nu}$ over $\mathbb{F}_q$.  An $m$-subspace $M$ of $V$ is called \emph{totally isotropic} if $f(x,y)=0$ holds for any $x,y\in M$. We know that $\nu$ is the dimension of the maximal totally isotropic subspaces. Denote the sets of all totally isotropic subspaces and $m$-dimensional totally isotropic subspaces with respect to $f$ by $\mathscr{P}$ and $\mathscr{P}_m$, respectively, where $0\leqslant m\leqslant\nu$. Equipped with the inclusion relation, $\mathscr{P}$ is a \emph{symplectic polar space}, denoted by the same symbol $\mathscr{P}$. The \emph{rank} of $\mathscr{P}$ is the dimension of the maximal totally isotropic subspaces. The symplectic space is one of the six kinds of classical polar spaces with $\nu\geqslant2$ \cite{CPS}.
	
	
	
	A subfamily of $\mathscr{P}_m$ is called \emph{$t$-intersecting} if any two members have a intersection with dimension at least $t$. The maximum sized $t$-intersecting subfamilies of $\mathscr{P}_m$ were widely studied and described. See \cite{SPEKR2,SPEKR1} for $t=1$ and \cite{SPEKRT} for all $t$. Recently, the authors characterized the second largest $t$-intersecting families \cite{YT}. There are also some results for other classical polar spaces, see\cite{012,PLANE,HMQ,MENTION1,SPEKR2,SPEKR1} for more details.
	
	Let $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ satisfy that $\dim(F_1\cap F_2)\geqslant t$ for any $F_1\in\mathscr{F}_1$ and $F_2\in\mathscr{F}_2$. 
	We say they are \emph{cross $t$-intersecting}. 
	Moreover, they are called \emph{trivial} if each member of them contains a fixed $t$-dimensional totally isotropic subspace. 
	In \cite{sp-add},  Ihringer characterized the structure of cross $t$-intersecting subfamilies of $\mathscr{P}_\nu$ with maximum product of sizes.
	
	The first main result of this paper is the following.
	
	\brackegin{thm}\langlebel{trivial}
		Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1\geqslant m_2\geqslant t$ and $2\nu\geqslant2m_1+m_2+1$, and $\mathscr{P}$  a symplectic polar spaces with rank $\nu$. Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are cross $t$-intersecting families with maximum product of sizes. Then there exists a $t$-dimensional totally isotropic subspace contained in each member of $\mathscr{F}_1$ and $\mathscr{F}_2$. 
	\end{thm}
	
	
	Based on Theorem \ref{trivial}, we get a more general theorem, see Theorem \ref{trivial2}.
	
	For a subspace $A$ of $\mathbb{F}_q^{2\nu}$ and a positive integer $a$, write $\mathscr{M}(a,A)=\{F\in\mathscr{P}_a: F\subset A\}$. Let $M\in\mathscr{P}_{m_2+1}$,  $T\in\mathscr{M}(t,M)$ and $S\in\mathscr{P}_{t+1}$. 
	Write
	\brackegin{equation}\langlebel{c1-c4}
		\brackegin{aligned}
			&\mathscr{A}thscr{C}_1(M,T;m_1,t)=\{F\in\mathscr{P}_{m_1}: T\subset F, \dim(F\cap M)\geqslant t+1\},\\
			&\mathscr{A}thscr{C}_2(M,T;m_2)=\{F\in\mathscr{P}_{m_2}: T\subset F\}\cup{M\brack m_2},\\
			&\mathscr{A}thscr{C}_3(S;m_1)=\{F\in\mathscr{P}_{m_1}:S\subset F\},\\
			&\mathscr{A}thscr{C}_4(S;m_2,t)=\{F\in\mathscr{P}_{m_2}:\dim(F\cap S)\geqslant t\}.
		\end{aligned}
	\end{equation}
Observe that $\mathscr{A}thscr{C}_1(M,T;m_1,t)$ and $\mathscr{A}thscr{C}_2(M,T;m_2)$ are cross $t$-intersecting families. So are $\mathscr{A}thscr{C}_3(S;m_1)$ and $\mathscr{A}thscr{C}_4(S;m_2,t)$.	 
Our second main result describe the structure of cross $t$-intersecting families with the second largest product of sizes.	

\brackegin{thm}\langlebel{non-trivial}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1\geqslant m_2\geqslant t+1$ and $2\nu\geqslant2m_1+m_2+t+3$, and $\mathscr{P}$  a symplectic polar spaces with rank $\nu$.	Suppose that $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are non-trivial cross $t$-intersecting families with maximum product of sizes. 
	\brackegin{itemize}
		\item[\rm{(1)}] If $m_2>2t$ or $(m_1,m_2,t)=(2,2,1),(3,2,1)$, then there exist $M\in\mathscr{P}_{m_2+1}$ and $T\in\mathscr{M}(t,M)$ such that
		\brackegin{itemize}
			\item[\rm{(\romannumeral1)}] $\mathscr{F}_1=\mathscr{A}thscr{C}_1(M,T;m_1,t)$, $\mathscr{F}_2=\mathscr{A}thscr{C}_2(M,T;m_2)$; or
			\item[\rm{(\romannumeral2)}] $m_1=m_2$ and $\mathscr{F}_1=\mathscr{A}thscr{C}_2(M,T;m_1)$, $\mathscr{F}_2=\mathscr{A}thscr{C}_1(M,T;m_2,t)$.
		\end{itemize}
		\item[\rm{(2)}] If $m_2\leqslant2t$ and $(m_1,m_2,t)\neq(2,2,1),(3,2,1)$, then there exists $S\in\mathscr{P}_{t+1}$ such that
		\brackegin{itemize}
			\item[\rm{(\romannumeral1)}] $\mathscr{F}_1=\mathscr{A}thscr{C}_3(S;m_1)$, $\mathscr{F}_2=\mathscr{A}thscr{C}_4(S;m_2,t)$; or
			\item[\rm{(\romannumeral2)}] $m_1=m_2$ and $\mathscr{F}_1=\mathscr{A}thscr{C}_4(S;m_1,t)$, $\mathscr{F}_2=\mathscr{A}thscr{C}_3(S;m_2)$.
		\end{itemize}
	\end{itemize}
\end{thm}

\section{Some Lemmas}

In this section, we give some useful lemmas in preparation for the proof of Theorems \ref{trivial} and \ref{non-trivial}.
\brackegin{lem}\langlebel{gaosifangsuo}
	Let $m$ and $i$ be positive integers with $i<m$. Then the following hold.
	\brackegin{itemize}
		\item[\rm{(1)}] $q^{m-i}<\frac{q^m-1}{q^i-1}<q^{m-i+1}$ and $q^{i-m-1}<\frac{q^i-1}{q^m-1}<q^{i-m}$;
		\item[\rm{(2)}] $q^{i(m-i)}<{m\brack i}<q^{i(m-i+1)}$.
	\end{itemize}
\end{lem}

For $1\leqslant m\leqslant \nu-1$, let $Q$ be an $m$-subspace of $\mathbb{F}_q^{2\nu}$ and $\alpha_1,\dots,\alpha_m$ any basis of $Q$. Note that the rank of the matrix $(f(\alpha_i,\alpha_j))_{m\times m}$ is even and independent of the choice of the basis. We say $Q$ is of  \emph{type $(m,s)$} if the rank of the matrix $(f(\alpha_i,\alpha_j))_{m\times m}$ is $2s$. Note that $Q$ is of type $(m,0)$ if and only if $Q$ is totally isotropic.
	
For positive integers $m_1$ and $ m$ with $m_1\leqslant m\leqslant\nu$, let $N'(m_1;m;2\nu)$ be the number of members of $\mathscr{P}_{m}$ containing a fixed member of $\mathscr{P}_{m_1}$. By \cite[Theorem 1]{JS1}, the size of $\mathscr{P}_m$ is ${\nu\brack m}\prod_{i=0}^{m-1}(q^{\nu-i}+1)$, from which we derive that
\brackegin{equation*}\langlebel{JS}
	N'(m_1;m;2\nu)=\prod\limits_{i=1}^{m-m_1}\dfrac{q^{2(\nu-m+i)}-1}{q^{i}-1}.
\end{equation*}
By \cite[Theorem 9 in Chapter 2]{JS2}, the number of members of $\mathscr{P}_m$ contained in a fixed $(m+1,1)$-type subspace is $q+1$.	
	
Let $\mathscr{F}\subset\mathscr{P}_m$. For $T\in\mathscr{P}$, if $\dim(T\cap F)\geqslant t$ holds for each $F\in\mathscr{F}$, we say $T$ is a \emph{$t$-cover} of $\mathscr{F}$. Let $\tau_t(\mathscr{F})$ denote the minimum dimension of $\mathscr{F}$'s $t$-covers. From \cite[Lemma 2.4]{CAO}, we derive the following Lemma.	
	
\brackegin{lem}\langlebel{sp-fangsuo}
	Let $\nu$, $m$, $s$ and $t$ be positive integers with  $\nu>m\geqslant s\geqslant t$. Suppose $\mathscr{F}\subset\mathscr{P}_k$, $X$ is a $t$-cover of $\mathscr{F}$  with dimension $x$ and $S\in\mathscr{P}_s$. If $\dim(X\cap S)=y<t$, then there exists $R\in\mathscr{P}_{s+t-y}$ such taht $S\subset R$ and
	$$|\mathscr{F}_S|\leqslant{x-t+1\brack1}^{t-y}|\mathscr{F}_R|.$$
\end{lem}	

\brackegin{lem}\langlebel{sp-yibanshangjie}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $\nu>m_i\geqslant t$, $i\in\{1,2\}$,  and $2\nu\geqslant2m_1+m_2-t$. Suppose $\mathscr{F}\subset\mathscr{M}p$ and $\mathscr{G}\subset\mathscr{M}mp$ are cross $t$-intersecting. Then
	$$|\mathscr{F}|\leqslant{\tau_t(\mathscr{F})\brack t}{m_2-t+1\brack1}^{\tau_t(\mathscr{G})-t}N'(\tau_t(\mathscr{G});m_1;2\nu).$$
\end{lem}
	\brackegin{proof}
	Let $S$ be a $t$-cover of $\mathscr{F}$ with dimension $\tau_t(\mathscr{F})$. From
	\brackegin{equation}\langlebel{sp-yibanshangjie-f1}
		\mathscr{F}=\brackigcup\limits_{W\in{S\brack t}}\mathscr{F}_W,
	\end{equation} 
	we get
	$$|\mathscr{F}|\leqslant{\tau_t(\mathscr{F})\brack t}N'(t;m_1;2\nu),$$
	which implies that the desired result holds for $\tau_t(\mathscr{G})=t$. In the following, we assume that $\tau_t(\mathscr{G})>t$.

	Let $W_1\in{S\brack t}$ with $\mathscr{F}_{W_1}\neq\emptyset$. We first give an upper bound for $|\mathscr{F}_{W_1}|$.
Since $\tau_t(\mathscr{G})>t$, there exists $G\in\mathscr{G}$ such that $\dim(G\cap W_1)<t$. Notice that $G$ is a $t$-cover of $\mathscr{F}$. By Lemma \ref{sp-fangsuo}, there exists a $(2t-\dim(W_1\cap G))$-dimensional totally isotropic subspace  $W_2$ such that
$$|\mathscr{F}_{W_1}|\leqslant{m_2-t+1\brack1}^{\dim W_2-\dim W_1}|\mathscr{F}_{W_2}|.$$
By $|\mathscr{F}_{W_1}|>0$, we have $|\mathscr{F}_{W_2}|>0$, which implies that $\dim W_2\leqslant m_1$.
If $\dim W_2<\tau_t(\mathscr{G})$, there exists $G'\in\mathscr{G}$ with $\dim(W_2\cap G')<t$. 
Using Lemma \ref{sp-fangsuo} repeatedly, we get a series of totally isotropic subspaces $W_1,W_2,\dots,W_u$ with $\dim W_{u-1}<\tau_t(\mathscr{G})\leqslant\dim W_u\leqslant m_1$ and 
$$|\mathscr{F}_{W_i}|\leqslant{m_2-t+1\brack1}^{\dim W_{i+1}-\dim W_i}|\mathscr{F}_{W_{i+1}}|$$
for each $i\in\{1,\dots,u-1\}$.
Hence
$$|\mathscr{F}_{W_1}|\leqslant{m_2-t+1\brack1}^{\dim W_u-t}|\mathscr{F}_{W_u}|\leqslant{m_2-t+1\brack1}^{\dim W_u-t}N'(\dim W_u;m_1;2\nu).$$
From Lemma \ref{gaosifangsuo} and $2\nu\geqslant2m_1+m_2-t$, for each $a\in\{0,\dots,m_1-1\}$, we obtain
$$\dfrac{N'(a;m_1;2\nu)}{N'(a+1;m_1;2\nu)}=\dfrac{q^{2(\nu-a)}-1}{q^{m_1-a}-1}\geqslant q^{2\nu-2m_1+1}\geqslant q^{m_2-t+1}\geqslant{m_2-t+1\brack1}.$$
Note that $\dim W_u\geqslant\tau_t(\mathscr{G})$. We have
\brackegin{equation*}
	\brackegin{aligned}
		|\mathscr{F}_{W_1}|\leqslant{m_2-t+1\brack1}^{\dim W_u-t}N'(\dim W_u;m_1;2\nu)\leqslant{m_2-t+1\brack1}^{\tau_t(\mathscr{G})-t}N'(\tau_t(\mathscr{G});m_1;2\nu).
	\end{aligned}
\end{equation*}
Together with \eqref{sp-yibanshangjie-f1}, we get
$$|\mathscr{F}|\leqslant\sum_{W\in{S\brack t}}|\mathscr{F}_W|\leqslant{\tau_t(\mathscr{F})\brack t}{m_2-t+1\brack1}^{\tau_t(\mathscr{G})-t}N'(\tau_t(\mathscr{G});m_1;2\nu),$$
as desired.
\end{proof}

\section{Proof of Theorem \ref{trivial}}

To prove Theorem \ref{trivial}, we need the following two lemmas.
\brackegin{lem}\langlebel{sp-dijian}
	Let $\nu$, $b$, $c$ and $t$ be positive integers with $2\nu\geqslant 2b+c+1$ and $\nu>b\geqslant t+1$, $c\geqslant t$. For $x\in\{t,\dots,b\}$, let
	$$g_{b,c}(x)={x\brack t}{c-t+1\brack1}^{x-t}N'(x;b;2\nu).$$
	Then $g_{b,c}(x)$ is decreasing with respect to $x$.
\end{lem}	
\brackegin{proof}
	By Lemma \ref{gaosifangsuo} and $2\nu\geqslant 2b+c+1$, for each $x\in\{t,\dots,b-1\}$, we have
	$$\dfrac{g_{b,c}(x+1)}{g_{b,c}(x)}=\dfrac{(q^{x+1}-1)(q^{c-t+1}-1)(q^{b-x}-1)}{(q^{x-t+1}-1)(q-1)(q^{2(\nu-x)}-1)}<q^{2b+c+1-2\nu}\leqslant1.$$
	Then $g_{b,c}(x+1)<g_{b,c}(x)$, as desired.
\end{proof}

\brackegin{lem}\langlebel{sp-cover}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $\nu>m_i\geqslant t$, $i\in\{1,2\}$. Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are cross $t$-intersecting. For each $i\in\{1,2\}$, let $\mathscr{S}_i$ denote the set of all $t$-covers of $\mathscr{F}_i$ with dimension $\tau_t(\mathscr{F}_i)$. Then $\mathscr{S}_1$ and $\mathscr{S}_2$ are cross $t$-intersecting.
\end{lem}
\brackegin{proof}
	Let $S_1\in\mathscr{S}_1$ and $S_2\in\mathscr{S}_2$. It is sufficient to show that $\dim(S_1\cap S_2)\geqslant t$.	
	
Since $\mathscr{F}_1$ and $\mathscr{F}_2$ are cross $t$-intersecting, we have $\dim S_2=\tau_t(\mathscr{F}_2)\leqslant m_1<\nu$. 	
Then there exist two $(\tau_t(\mathscr{F}_2)+1)$-dimensional totally isotropic subspaces $Y_1$ and $Y_2$ containing $S_2$  such that $S_2=Y_1\cap Y_2$ and there exists no maximal totally isotropic subspace contains both of them. 	
Therefore, there exists $k\in\{1,2\}$ such that $Y_k\cap S_1=S_2\cap S_1$. Similarly, it is routine to check that there exist $F_1\in\mathscr{M}p$ and $F_2\in\mathscr{M}mp$ such that $S_1\subset F_2$, $S_2\subset F_1$ and $F_1\cap F_2=S_1\cap S_2$.	
By the maximality of $\mathscr{F}_1$ and $\mathscr{F}_2$, since $\mathscr{F}_1\cup\{F_1\}$ and $\mathscr{F}_2$ are still cross $t$-intersecting, we have $F_1\in\mathscr{F}_1$. Similarly, we have $F_2\in\mathscr{F}_2$.
Thus $\dim(S_1\cap S_2)=\dim(F_1\cap F_2)\geqslant t$, as desired.
\end{proof}	

\brackegin{proof}[\brackf Proof of Theorem \ref{trivial}]
	
	Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are cross $t$-intersecting families.
	Assume that $\tau_t(\mathscr{F}_1)=\tau_t(\mathscr{F}_2)=t$. Let $T_1$ and $T_2$ be $t$-covers of $\mathscr{F}_1$ and $\mathscr{F}_2$ with dimension $t$, respectively. By Lemma \ref{sp-cover} , we have $T_1=T_2:=T$.
	Then 
	$$|\mathscr{F}_1|\leqslant N'(t;m_1;2\nu),\quad|\mathscr{F}_2|\leqslant N'(t;m_2;2\nu),$$
	and two equalities hold at the same time if and only if $\mathscr{F}_i=\{F\in\mathscr{P}_{m_i}:  T\subset F\}$ for each $i\in\{1,2\}$.

To finish our proof, it is sufficient to show 
\brackegin{equation}\langlebel{sp-f1f2<nn}
	|\mathscr{F}_1||\mathscr{F}_2|<N'(t;m_1;2\nu)N'(t;m_2;2\nu)
\end{equation}
if $(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))\neq(t,t)$. By Lemma \ref{sp-yibanshangjie} and $2\nu\geqslant2m_1+m_2+1$, $m_1\geqslant m_2$, we have
\brackegin{equation*}
	\brackegin{aligned}
		|\mathscr{F}_1||\mathscr{F}_2|\leqslant&\leqslantft({\tau_t(\mathscr{F}_1)\brack t}{m_1-t+1\brack1}^{\tau_t(\mathscr{F}_1)-t}N'(\tau_t(\mathscr{F}_1);m_2;2\nu)\right)\\
		&\cdot\leqslantft({\tau_t(\mathscr{F}_2)\brack t}{m_2-t+1\brack1}^{\tau_t(\mathscr{F}_2)-t}N'(\tau_t(\mathscr{F}_2);m_1;2\nu)\right).
	\end{aligned}
\end{equation*}
Note that $t\leqslant\tau_t(\mathscr{F}_1)\leqslant m_2$ and $t\leqslant\tau_t(\mathscr{F}_2)\leqslant m_1$.
Together with $(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))\neq(t,t)$ and Lemma \ref{sp-dijian}, \eqref{sp-f1f2<nn} follows.	
\end{proof}

Based on Theorem \ref{trivial},  we obtain a more general theorem.

\brackegin{thm}\langlebel{trivial2}
	Let $d$, $\nu$, $t$, $m_1$,\dots, $m_d$ be positive integers with $d\geqslant2$, $m_1\geqslant m_2\geqslant\cdots\geqslant m_d\geqslant t$ and $2\nu\geqslant 2m_1+m_2+1$. If $\mathscr{F}_1\subset\mathscr{M}p$,\dots, $\mathscr{F}_d\subset\mathscr{P}_{m_d}$ satisfy that $\dim(F_1\cap\cdots\cap F_d)\geqslant t$ 
	for any $F_i\in\mathscr{F}_i$, $i=1,\dots,d$. If $\prod_{i=1}^d|\mathscr{F}_i|$ reaches to the maximum value, then there exists  $T\in\mathscr{P}_t$ such that $\mathscr{F}_i=\{F\in\mathscr{P}_{m_i}: T\subset F\}$ for each $i\in\{1,\dots,d\}$.
\end{thm}
\brackegin{proof}
	For distinct $i,j\in\{1,\dots,d\}$, $\mathscr{F}_i$ and $\mathscr{F}_j$ are cross $t$-intersecting families. Then by
	Theorem \ref{trivial}, we have
	$$|\mathscr{F}_i||\mathscr{F}_j|\leqslant N'(t;m_i;2\nu)N'(t;m_j;2\nu).$$
	Therefore
\brackegin{equation}\langlebel{sp-cross-prod}
	\brackegin{aligned}
		\leqslantft(\prod_{s=1}^d|\mathscr{F}_s|\right)^{d-1}&=\prod_{1\leqslant i<j\leqslant d}|\mathscr{F}_i||\mathscr{F}_j|\\
		&\leqslant\prod_{1\leqslant i<j\leqslant d}N'(t;m_i;2\nu)N'(t;m_j;2\nu)\\
		&=\leqslantft(\prod_{s=1}^dN'(t;m_s;2\nu)\right)^{d-1},
	\end{aligned}
\end{equation}
and equality holds if and only if $|\mathscr{F}_i||\mathscr{F}_j|=N'(t;m_i;2\nu)N'(t;m_j;2\nu)$ for any distinct $i,j\in\{1,\dots,d\}$.

	Note that the product of sizes of families $\{F\in\mathscr{P}_{m_i}: S\subset F\}$, $i=1,\dots,d$, where $S\in\mathscr{P}_t$, reaches to the upper bound of \eqref{sp-cross-prod}. 
Therefore, by assumption and Theorem \ref{trivial}, for distinct $i,j\in\{1,\dots,d\}$, there exists $T_{i,j}\in\mathscr{P}_t$ such that 
$$\mathscr{F}_i=\{F\in\mathscr{P}_{m_i}:  T_{i,j}\subset F\},\quad\mathscr{F}_j=\{F\in\mathscr{P}_{m_j}:  T_{i,j}\subset F\}.$$
If there exists $j'\in\{1,2,\dots,d\}$ such that $T_{i,j'}\neq T_{i,j}$, we have
$$\mathscr{F}_i\subset\{F\in\mathscr{P}_i:  T_{i,j}+T_{i,j'}\subset F\}.$$
	Together with $2\nu\geqslant2m_1+m_2+1$, $m_1\geqslant m_i$ and $\dim(T_{i,j}+T_{i,j'})\geqslant t+1$, we get
$$N'(t+1;m_i;2\nu)<N'(t;m_i;2\nu)=|\mathscr{F}_i|\leqslant N'(t+1;m_i;2\nu),$$
a contradiction.  Therefore, there exists $T\in\mathscr{P}_t$ such that $T_{i,j}=T$ for any distinct $i,j\in\{1,\dots,d\}$. Then the desired result follows.
\end{proof}

\section{Proof of Theorem \ref{non-trivial}}	

Let $\nu>m_i\geqslant t+1$, $i\in\{1,2\}$. 
Suppose $M\in\mathscr{P}_{m_2+1}$, $T\in\mathscr{M}(t,M)$ and  $S\in\mathscr{P}_{t+1}$. Let 	$\mathscr{A}thscr{C}_1(M,T;m_1,t)$, 
$\mathscr{A}thscr{C}_2(M,T;m_2)$, 
$\mathscr{A}thscr{C}_3(S;m_1)$ and
$\mathscr{A}thscr{C}_4(S;m_2,t)$ are families defined in \eqref{c1-c4}. By \cite[Theorem 3.11]{JS}, $|\mathscr{A}thscr{C}_1(M,T;m_1,t)|\cdot|\mathscr{A}thscr{C}_2(M,T;m_2)|$ and $|\mathscr{A}thscr{C}_3(S;m_1)|\cdot|\mathscr{A}thscr{C}_4(S;m_2,t)|$ are independent on the choice of $M$, $T$ and $S$. 	

Write
\brackegin{equation*}
	\brackegin{aligned}
		c_1(\nu,m_1,m_2,t)&=|\mathscr{A}thscr{C}_1(M,T;m_1,t)|\cdot|\mathscr{A}thscr{C}_2(M,T;m_2)|,\\
		c_2(\nu,m_1,m_2,t)&=|\mathscr{A}thscr{C}_3(S;m_1)|\cdot|\mathscr{A}thscr{C}_4(S;m_2,t)|.
	\end{aligned}
\end{equation*}	
It is routine to check that
\brackegin{equation}\langlebel{cp3cp4-size}
	\dfrac{c_2(\nu,m_1,m_2,t)}{N'(t+1;m_1;2\nu)}={t+1\brack1}N'(t;m_2;2\nu)-q{t\brack1}N'(t+1;m_2;2\nu).
\end{equation}
	In the following, we show some inequalities for $c_1(\nu,m_1,m_2,t)$ and $c_2(\nu,m_1,m_2,t)$.

	For $S\in\mathscr{P}_s$, $T\in{S\brack t}$ and $j\in\{t,t+1,\dots,s\}$, write
\brackegin{equation*}\langlebel{xla}
	\brackegin{aligned}
		\mathscr{L}_j(S,T;m)&=\{(I,F)\in\mathscr{P}_j\times\mathscr{P}_m:  T\subset I\subset S, I\subset F\},\\
		\mathscr{A}_j(S,T;m)&=\{F\in\mathscr{P}_m:  T\subset F, \dim(F\cap S)=j\}
	\end{aligned}
\end{equation*}
and
$$s_0(\nu,m,s,t)={s-t\brack1}N'(t+1;m;2\nu)-q{s-t\brack2}N'(t+2;m;2\nu).$$

	\brackegin{lem}\langlebel{c1>c0}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $\nu>m_i\geqslant t+1$, $i\in\{1,2\}$. 
	Then 
	$$c_1(\nu,m_1,m_2,t)>s_0(\nu,m_1,m_2+1,t)N'(t;m_2;2\nu).$$ 
	Moreover, if $2\nu\geqslant m_1+2m_2-t+4$,then
	$$c_1(\nu,m_1,m_2,t)>\leqslantft({m_2-t+1\brack1}-q^{-2}\right)N'(t+1;m_1;2\nu)N'(t;m_2;2\nu).$$
\end{lem}
	\brackegin{proof}
	Let $M\in\mathscr{P}_{m_2+1}$ and $T\in{M\brack t}$. 
	For each $j\in\{t+1,\dots,m_2+1\}$, by double counting $|\mathscr{L}_j(M,T;m_1)|$, we have
	\brackegin{equation}\langlebel{sp-suanliangci-formu}
		|\mathscr{L}_j(M,T;m_1)|={m_2-t+1\brack j-t}N'(j;m_1;2\nu)=\sum_{i=j}^{m_2+1}{i-t\brack j-t}|\mathscr{A}_i(M,T;m_1)|.
	\end{equation}
Then
\brackegin{equation*}
	\brackegin{aligned}
		s_0(\nu,m_1,m_2+1,t)=&\ |\mathscr{L}_{t+1}(M,T;m_1)|-q|\mathscr{L}_{t+2}(M,T;m_1)|\\
		=&\sum_{i=t+1}^{m_2+1}{i-t\brack1}|\mathscr{A}_i(M,T;m_1)|-q\sum_{i=t+2}^{m_2+1}{i-t\brack2}|\mathscr{A}_i(M,T;m_1)|\\
		=&\sum_{i=t+1}^{t+2}|\mathscr{A}_i(M,T;m_1)|+\sum_{i=3}^{m_2+1}\leqslantft({i-t\brack1}-q{i-t\brack2}\right)|\mathscr{A}_i(M,T;m_1)|.
	\end{aligned}
\end{equation*}
	Note that ${i-t\brack1}<q{i-t\brack2}$ for $i\geqslant t+3$. Then
\brackegin{equation*}
	\brackegin{aligned}
		s_0(\nu,m_1,m_2+1,t)&\leqslant|\mathscr{A}_{t+1}(M,T;m_1)|+|\mathscr{A}_{t+2}(M,T;m_1)|\\
		&=|\{F\in\mathscr{M}p:  T\subset F, \dim(F\cap M)\in\{t+1,t+2\}\}|\\
		&\leqslant|\{F\in\mathscr{M}p:  T\subset F, \dim(F\cap M)\geqslant t+1\}|\\
		&\leqslant|\mathscr{A}thscr{C}_1(M,T;m_1,t)|,
	\end{aligned}
\end{equation*}
from which we get  $c_1(\nu,m_1,m_2,t)>s_0(\nu,m_1,m_2+1,t)N'(t;m_2;2\nu)$. 

	Together with Lemma \ref{gaosifangsuo} and $2\nu\geqslant m_1+2m_2-t+4$, we have

\brackegin{align*}
	\dfrac{c_1(\nu,m_1,m_2,t)}{N'(t;m_2;2\nu)}&>s_0(\nu,m_1,m_2+1,t)\\
	&=\leqslantft({m_2-t+1\brack1}-q{m_2-t+1\brack2}\dfrac{q^{m_1-t-1}-1}{q^{2(\nu-t-1)}-1}\right)N'(t+1;m_1;2\nu)\\
	&\geqslant\leqslantft({m_2-t+1\brack1}-q^{m_1+2m_2-t+2-2\nu}\right)N'(t+1;m_1;2\nu)\\
	&\geqslant\leqslantft({m_2-t+1\brack1}-q^{-2}\right)N'(t+1;m_1;2\nu),
\end{align*}

as desired.
\end{proof}

\brackegin{lem}\langlebel{cp1-cp2}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1\geqslant m_2\geqslant t+1$ and $2\nu\geqslant2m_1+m_2+t+2$.
	\brackegin{itemize}
		\item[\rm{(1)}] If $m_2>2t$ or $(m_1,m_2,t)=(2,2,1),(3,2,1)$, then $c_1(\nu,m_1,m_2,t)>c_2(\nu,m_1,m_2,t)$.
		\item[\rm{(2)}] If $m_2\leqslant 2t$ and $(m_1,m_2,t)\neq(2,2,1),(3,2,1)$, then $c_1(\nu,m_1,m_2,t)<c_2(\nu,m_1,m_2,t)$.
	\end{itemize}
\end{lem}
\brackegin{proof}
	Write $c_3(\nu,m_1,m_2,t)=c_2(\nu,m_1,m_2,t)-c_1(\nu,m_1,m_2,t)$.
	
	(1) Suppose $m_2>2t$. By $2\nu\geqslant2m_1+m_2+t+2\geqslant m_1+2m_2-t+4$, \eqref{cp3cp4-size} and Lemma \ref{c1>c0}, we have
\brackegin{equation*}
	\brackegin{aligned}
		\dfrac{c_1(\nu,m_1,m_2,t)}{N'(t;m_2;2\nu)}>\ q{m_2-t\brack1}N'(t+1;m_1;2\nu)>{t+1\brack1}N'(t+1;m_1;2\nu)>\dfrac{c_2(\nu,m_1,m_2,t)}{N'(t;m_2;2\nu)}.
	\end{aligned}
\end{equation*}
which implies that $c_1(\nu,m_1,m_2,t)>c_2(\nu,m_1,m_2,t)$.	
	
	Assume  that $(m_1,m_2,t)=(2,2,1)$.
By \eqref{cp3cp4-size}, it is routine to check that
$$c_3(\nu,2,2,1)=\leqslantft({2\brack1}N'(1;2;2\nu)-q\right)-{2\brack1}\leqslantft(N'(1;2;2\nu)+q^2\right)<0.$$
Then the desired result follows.	

When $m_1\geqslant3$, it is routine to check that
\brackegin{equation*}
	\brackegin{aligned}
		c_1(\nu,m_1,2,1)&=\leqslantft({2\brack1}N'(2;m_1;2\nu)-qN'(3;m_1;2\nu)\right)(N'(1;2;2\nu)+q^2),\\
		c_2(\nu,m_1,2,1)&=N'(2;m_1;2\nu)\leqslantft({2\brack1}N'(1;2;2\nu)-q\right).
	\end{aligned}
\end{equation*}
Then
\brackegin{equation}\langlebel{sp-m21}
	\dfrac{c_3(\nu,m_1,2,1)}{qN'(2;m_1;2\nu)}=\leqslantft(\dfrac{q^{2\nu-2}-1}{q-1}+q^2\right)\dfrac{q^{m_1-2}-1}{q^{2\nu-4}-1}-q^2-q-1.
\end{equation}
Suppose $(m_1,m_2,t)=(3,2,1)$. By $2\nu\geqslant2m_1+m_2=8$,  \eqref{sp-m21} and Lemma \ref{gaosifangsuo}, we have
\brackegin{equation*}
	\brackegin{aligned}
		\dfrac{c_3(\nu,3,2,1)}{qN'(2;3;2\nu)}&<\dfrac{q^{2\nu-2}-1}{q^{2\nu-4}-1}-q^2-q=\dfrac{q^2-1}{q^{2\nu-4}-1}-q<q^{-2}-q<0.
	\end{aligned}
\end{equation*}
Then the desired result follows.

	(2) Suppose $m_2=2t$ and $t=1$. By assumption, we have $m_1\geqslant4$. From Lemma \ref{gaosifangsuo}, $2\nu\geqslant2m_1+m_2+t+2$ and \eqref{sp-m21}, we obtain
\brackegin{equation*}
	\brackegin{aligned}
		\dfrac{c_3(\nu,m_1,2,1)}{qN'(2;m_1;2\nu)}&>q^2\cdot q^{m_1-3}+q^2\cdot q^{m_1+1-2\nu}-q^2-q-1\\
		&=q^{m_1-1}+q^{m_1+3-2\nu}-q^2-q-1\\
		&>q^3-q^2-q-1\\
		&>0.
	\end{aligned}
\end{equation*}
Then $c_1(\nu,m_1,2,1)<c_2(\nu,m_1,2,1)$.

Suppose $m_2=2t$ and $t\geqslant2$. We have $m_2\geqslant t+2$. Let $M\in\mathscr{P}_{m_2+1}$ and $T\in{M\brack t}$.
From \eqref{sp-suanliangci-formu}, we obtain
\brackegin{equation*}
	\brackegin{aligned}
		{t+1\brack1}N'(t+1;m_1;2\nu)&=\sum_{j=t+1}^{2t+1}{j-t\brack1}|\mathscr{A}_j(M,T;m_1)|\\
		&=|\mathscr{A}thscr{C}_1(M,T;m_1,t)|+\sum_{j=t+2}^{2t+1}q{j-t-1\brack1}|\mathscr{A}_j(M,T;m_1)|.
	\end{aligned}
\end{equation*}
Set
$$\alpha=\sum_{j=t+2}^{2t+1}q{j-t-1\brack1}|\mathscr{A}_j(M,T;m_1)|.$$
We have $\alpha\geqslant q|\mathscr{A}_{t+2}(M,T;m_1)|$ and
$$c_1(\nu,m_1,2t,t)=\leqslantft({t+1\brack1}N'(t+1;m_1;2\nu)-\alpha\right)\leqslantft(N'(t;2t;2\nu)+q^{t+1}{t\brack1}\right).$$	
Together with \eqref{cp3cp4-size}, we get
\brackegin{equation}\langlebel{sp-ca1}
	\dfrac{c_3(\nu,m_1,2t,t)}{N'(t+1;m_1;2\nu)}>\dfrac{\alpha N'(t;2t;2\nu)}{N'(t+1;m_1;2\nu)}-q{t\brack1}N'(t+1;2t;2\nu)-q^{t+1}{t\brack1}{t+1\brack1}.
\end{equation}	
By Lemma \ref{gaosifangsuo}, we have
\brackegin{equation}\langlebel{sp-ca2}
	N'(t+1;m_1;2\nu)\leqslant\prod_{i=1}^{m_1-t-1}q^{2(\nu-m_1)+i+1}=q^{2(m_1-t-1)(\nu-m_1)+\frac{(m_1-t-1)(m_1-t+2)}{2}}.
\end{equation}
Assume that $m_1=2t$. By Lemma \ref{gaosifangsuo} and \cite[Theorem 2.10]{QY}, we have
\brackegin{align*}
	\alpha\geqslant&\ q|\mathscr{A}_{t+2}(M,T;m_1)|\\

	\geqslant&\ q{t+1\brack2}\cdot\leqslantft(q^{\frac{(t-2)(t-1)}{2}+(t-2)(\nu-2t-1)+(t-2)(\nu-2t)}{t-1\brack1}\right)\\

	\geqslant&\ q^{2(t-2)(\nu-2t)+\frac{t^2-t+2}{2}}{t\brack1}.
\end{align*}
Together with \eqref{sp-ca1}, \eqref{sp-ca2}  and $2\nu\geqslant2m_1+m_2=6t$, we obtain
\brackegin{align*}
	\dfrac{c_3(\nu,2t,2t,t)}{{t\brack1}N'(t+1;2t;2\nu)}&>\dfrac{q^{2(t-2)(\nu-2t)+\frac{t^2-t+2}{2}}(q^{2(\nu-t)}-1)}{q^t-1}-q^{2(t-1)(\nu-2t)+\frac{t^2+t}{2}}-q^{2t+2}\\
	&\geqslant\ q^{2(t-1)(\nu-2t)+\frac{t^2+t+2}{2}}-q^{2(t-1)(\nu-2t)+\frac{t^2+t}{2}+1}\\
	&=\ 0.
\end{align*}
Now assume that $m_1\geqslant 2t+1$. By Lemma \ref{gaosifangsuo} and \cite[Theorem 2.10]{QY}, we have
\brackegin{align*}
	\alpha\geqslant&\ q|\mathscr{A}_{t+2}(M,T;m_1)|\\

	\geqslant&\ q{t+1\brack2}\cdot q^{\frac{(t-1)t}{2}+\frac{(m_1-2t-1)(m_1-2)}{2}+2(m_1-t-2)(\nu-m_1)}\\

	\geqslant&\ q^{2(m_1-t-2)(\nu-m_1)+\frac{t^2+t}{2}+\frac{(m_1-2t-1)(m_1-2)}{2}}{t\brack1}.
\end{align*}
Together with \eqref{sp-ca1}, \eqref{sp-ca2}, $2\nu\geqslant2m_1+m_2+t\geqslant7t+2$ and
$$N'(t;2t;2\nu)\geqslant\prod_{i=1}^tq^{2(\nu-2t)+i}=q^{2t(\nu-2t)+\frac{t(t+1)}{2}}\geqslant q^{4\nu-8t+3},$$
we get
\brackegin{align*}
	\dfrac{c_3(\nu,m_1,2t,t)}{{t\brack1}N'(t+1;m_1;2\nu)}&>\leqslantft(\dfrac{\alpha}{{t\brack1}N'(t+1;m_1;2\nu)}-\dfrac{q(q^t-1)}{q^{2(\nu-t)}-1}-\dfrac{q^{2t+2}}{q^{4\nu-8t+3}}\right)N'(t;2t;2\nu)\\

	&\geqslant\leqslantft(1-q^{-1}-q^{7t-3-2\nu}\right)q^{3t+2-2\nu}N'(t;2t;2\nu)\\
	&>\ 0.
\end{align*}
Then $c_1(\nu,m_1,2t,t)<c_2(\nu,m_1,2t,t)$.

Suppose $m_2<2t$. We have $t\geqslant2$ and
\brackegin{equation*}
	\brackegin{aligned}
		c_1(\nu,m_1,m_2,t)&\leqslant{m_2-t+1\brack1}N'(t+1;m_1;2\nu)\leqslantft(N'(t;m_2;2\nu)+q^{m_2-t+1}{t\brack1}\right)\\
		&\leqslant{t\brack1}N'(t+1;m_1;2\nu)\leqslantft(N'(t;m_2;2\nu)+q^{t}{t\brack1}\right).
	\end{aligned}
\end{equation*}
By Lemma \ref{gaosifangsuo}, we have
\brackegin{equation}\langlebel{sp-jishu-xiajie}
	N'(t;m_2;2\nu)\geqslant\dfrac{q^{2(\nu-m_2+1)}-1}{q-1}\geqslant q^{2\nu-2m_2+1}.
\end{equation}
Together with $2\nu\geqslant2m_1+m_2+t+1\geqslant2m_2+2t$, $t\geqslant2$, \eqref{cp3cp4-size} and Lemma\ref{gaosifangsuo}, we get

\brackegin{align*}
	&\dfrac{c_3(\nu,m_1,m_2,t)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	\geqslant&\leqslantft({t+1\brack1}-q{t\brack1}\cdot\dfrac{q^{m_2-t}-1}{q^{2(\nu-t)}-1}\right)-{t\brack1}\leqslantft(1+\dfrac{q^{t}{t\brack1}}{N'(t;m_2;2\nu)}\right)\\

	>&\ q^t\leqslantft(1-q^{m_2+t+1-2\nu}-q^{2m_2+2t-1-2\nu}\right)\\
	\geqslant&\ q^t\leqslantft(1-q^{-2m_1}-q^{-1}\right)\\
	>&\ 0,
\end{align*}

Then $c_1(\nu,m_1,m_2,t)<c_2(\nu,m_1,m_2,t)$.
\end{proof}

\brackegin{lem}\langlebel{cp-cp'}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1>m_2\geqslant t+1$ and $2\nu\geqslant2m_1+m_2+2$. The following hold.
	\brackegin{itemize}
		\item[\rm{(1)}] $c_1(\nu,m_1,m_2,t)>c_1(\nu,m_2,m_1,t)$.
		\item[\rm{(2)}] $c_2(\nu,m_1,m_2,t)>c_2(\nu,m_2,m_1,t)$.
	\end{itemize}
\end{lem}
\brackegin{proof}
	(1) By Lemmas \ref{gaosifangsuo}, \ref{c1>c0} and $2\nu\geqslant2m_1+m_2+2\geqslant m_1+2m_2-t+4$, we have

	\brackegin{align*}
		\dfrac{c_1(\nu,m_1,m_2,t)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\geqslant{m_2-t+1\brack1}-q^{-2}=q{m_2-t\brack1}+1-q^{-2}.
	\end{align*}

Since $m_1>m_2\geqslant t+1$, by \eqref{sp-jishu-xiajie}, we obtion $N'(t+1;m_1;2\nu)\geqslant q^{2\nu-2m_1+1}$.
Together with $2\nu\geqslant2m_1+m_2+2$, we get

\brackegin{align*}
	&\dfrac{c_1(\nu,m_2,m_1,t)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	\leqslant&\leqslantft(N'(t;m_1;2\nu)+q^{m_1-t+1}{t\brack1}\right){m_1-t+1\brack1}\dfrac{N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	\leqslant&{m_1-t+1\brack1}\leqslantft(\dfrac{N'(t;m_1;2\nu)N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}+\dfrac{q^{m_1+1}N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\right)\\
	\leqslant&{m_1-t+1\brack1}\dfrac{N'(t;m_1;2\nu)N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}+q^{4m_1+m_2+1-4\nu}\\
	\leqslant&{m_1-t+1\brack1}\dfrac{N'(t;m_1;2\nu)N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}+q^{-m_2-3}.
\end{align*}

Then by $2\nu\geqslant2m_1+m_2+2$ and $m_1>m_2\geqslant t+1$, we have

\brackegin{align*}
	&\ \dfrac{c_1(\nu,m_1,m_2,t)-c_1(\nu,m_2,m_1,t)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	>&\ q{m_2-t\brack1}+(1-q^{-2}-q^{-m_2-3})-{m_1-t+1\brack1}\dfrac{N'(t;m_1;2\nu)N'(t+1;m_2;2\nu)}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\

	=&\ 1-q^{-2}-q^{-m_2-3}-\dfrac{q^{m_2-t}-1}{q^{m_1-t}-1}\\

	>&\ 0.
\end{align*}

Then the desired result follows.

(2) From \eqref{cp3cp4-size} and $m_1>m_2$, we obtain
\brackegin{equation*}
	\brackegin{aligned}
		\dfrac{c_2(\nu,m_1,m_2,t)-c_2(\nu,m_2,m_1,t)}{{t+1\brack1}N'(t+1;m_1;2\nu)N'(t+1;m_2;2\nu)}&=\dfrac{N'(t;m_2;2\nu)}{N'(t+1;m_2;2\nu)}-\dfrac{N'(t;m_1;2\nu)}{N'(t+1;m_1;2\nu)}\\
		&=(q^{2(\nu-t)}-1)\leqslantft(\dfrac{1}{q^{m_2-t}-1}-\dfrac{1}{q^{m_1-t}-1}\right)\\
		&>0,
	\end{aligned}
\end{equation*}
as desired.
\end{proof}

To present the proof of Theorem \ref{non-trivial} briefly, we prove the following two lemmas. Write
$$c_0(\nu,m_1,m_2,t)=q{m_2-t\brack1}N'(t+1;m_1;2\nu)N'(t;m_2;2\nu).$$
\brackegin{lem}\langlebel{sp-cross-upper-other}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1,m_2\geqslant t+1$ and $2\nu\geqslant m_1+m_2+1+\mathscr{A}x\{m_1+t+2,m_2\}$. Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are non-trivial cross $t$-intersecting families with $\tau_t(\mathscr{F}_2)\geqslant\tau_t(\mathscr{F}_1)$ and $(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))\neq(t,t+1)$. Then $|\mathscr{F}_1||\mathscr{F}_2|<c_1(\nu,m_1,m_2,t)$.
\end{lem}
\brackegin{proof}
	Suppose $\tau_t(\mathscr{F}_1)=t$. By assumption, we have $m_1\geqslant\tau_t(\mathscr{F}_2)\geqslant t+2$. By Lemmas \ref{gaosifangsuo}, \ref{sp-yibanshangjie}, \ref{sp-dijian} and $2\nu\geqslant 2m_1+m_2+t+3\geqslant m_1+m_2+2t+5$, we have
	\brackegin{equation*}
		\brackegin{aligned}
			\dfrac{|\mathscr{F}_1||\mathscr{F}_2|}{c_0(\nu,m_1,m_2,t)}&\leqslant\dfrac{\leqslantft({\tau_t(\mathscr{F}_2)\brack t}{m_2-t+1\brack1}^{\tau_t(\mathscr{F}_2)-t}N'(\tau_t(\mathscr{F}_2);m_1;2\nu)\right)N'(t;m_2;2\nu)}{q{m_2-t\brack1}N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
			&\leqslant\dfrac{{t+2\brack2}{m_2-t+1\brack1}^2N'(t+2;m_1;2\nu)}{q{m_2-t\brack1}N'(t+1;m_1;2\nu)}\\
			&=\dfrac{(q^{m_1-t-1}-1)(q^{m_2-t+1}-1)}{q(q^{2(\nu-t-1)}-1)(q^{m_2-t}-1)}{t+2\brack2}{m_2-t+1\brack1}\\
			&<q^{m_1+m_2+2t+5-2\nu}\\
			&\leqslant1.
		\end{aligned}
	\end{equation*}
	Then the desired result follows from Lemma \ref{c1>c0}.

Suppose $\tau_t(\mathscr{F}_1)\geqslant t+1$. We have $\tau_t(\mathscr{F}_2)\geqslant t+1$ and $m_1\geqslant t+1$. By Lemma \ref{gaosifangsuo},  \ref{sp-yibanshangjie}, \ref{sp-dijian} and $2\nu\geqslant\mathscr{A}x\{m_1+m_2+2t+4,2m_1+m_2+1,m_1+2m_2+1\}$, we obtain

\brackegin{align*}
	\dfrac{|\mathscr{F}_1||\mathscr{F}_2|}{c_0(\nu,m_1,m_2,t)}&\leqslant\dfrac{{t+1\brack1}^2{m_1-t+1\brack1}{m_2-t+1\brack1}N'(t+1;m_1;2\nu)N'(t+1;m_2;2\nu)}{q{m_2-t\brack1}N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	&=\dfrac{(q^{m_2-t+1}-1)}{q(q^{2(\nu-t)}-1)}{t+1\brack1}^2{m_1-t+1\brack1}\\
	&<q^{m_1+m_2+2t+3-2\nu}\\
	&<1.
\end{align*}

Then the desired result follows from Lemma \ref{c1>c0}.
\end{proof}	

	\brackegin{lem}\langlebel{sp-cross-upper-tt+1}
	Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $\nu>m_i\geqslant t+1$, $i\in\{1,2\}$, and $2\nu\geqslant m_1+2m_2+3$. Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$ and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are maximal non-trivial  cross $t$-intersecting families with $(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))=(t,t+1)$. Then one of the following holds.
	\brackegin{itemize}
		\item[\rm{(1)}] $\mathscr{F}_1=\mathscr{A}thscr{C}_1(M,T;m_1,t)$ and $\mathscr{F}_2=\mathscr{A}thscr{C}_2(M,T;m_2)$ for some $M\in\mathscr{P}_{m_2+1}$ and $T\in{M\brack t}$.
		\item[\rm{(2)}] $\mathscr{F}_1=\mathscr{A}thscr{C}_3(S;m_1) $and $\mathscr{F}_2=\mathscr{A}thscr{C}_4(S;m_2,t)$ for some $S\in\mathscr{P}_{t+1}$.
		\item[\rm{(2)}h] $|\mathscr{F}_1||\mathscr{F}_2|<c_1(\nu,m_1,m_2,t)$.
	\end{itemize}
\end{lem}
	\brackegin{proof}
	Let $T$ be a $t$-cover of $\mathscr{F}_1$ with dimension $t$ and $\mathscr{S}$ denote the set of all $t$-covers of $\mathscr{F}_2$ with dimension $t+1$. By Lemma \ref{sp-cover}, each element of $\mathscr{S}$ contains $T$. Let $M$ be a subspace of $\mathbb{F}_q^{2\nu}$ generated by $\brackigcup_{S\in\mathscr{S}}S$. For each $F\in\mathscr{F}_2\setminus(\mathscr{F}_2)_T$ and $S\in\mathscr{S}$, we have 
	$\dim(F\cap T)=t-1$, $\dim(F\cap S)=t$ and
	$$m_2+1=\dim(T+F)\leqslant\dim(S+F)=m_2+1.$$
	Then $T+F=S+F$, which implies that $T+F=M+F$. Thus $\dim(F\cap M)=\dim M-1$ and $t+1\leqslant\dim M\leqslant m_2+1$.		
	Note that the type of $M$ is $(\dim M,0)$ or $(\dim M,1)$.

Since $\tau_t(\mathscr{F}_2)=t+1$, there exists $F_{2,1}\in\mathscr{F}_2$ such that $T\not\subset F_{2,1}$. Write
$$H:=T\vee F_{2,1}.$$
We have
$$S=T+(S\cap F_{2,1})\subset H,\quad\dim H=m_2+1.$$
For each $F\in\mathscr{F}_1$, since $T\subset F$, $T\not\subset F_{2,1}$ and $\dim(F\cap F_{2,1})\geqslant t$, we have
$$
	\dim(F\cap H)\geqslant\dim T+\dim(F\cap F_{2,1})-\dim(T\cap F_{2,1})\geqslant t+1.
$$
Therefore
\brackegin{equation}\langlebel{sp-f1-str}
	\mathscr{F}_1\subset\{F\in\mathscr{P}_{m_1}: T\subset F, \dim(F\cap H)\geqslant t+1\}.
\end{equation}

\mathscr{E}dskip
\noindent{\brackf Case 1. $\dim M=t+1$.}
\mathscr{E}dskip

Let $S$ be the unique member of $\mathscr{S}$. Since $T\subset S$, for $F\in\mathscr{F}_1$, either $S\subset F$ or $S\cap F=T$ holds.

Suppose $S$ is contained in each member of $\mathscr{F}_1$. We have
$$\mathscr{F}_1\subset\mathscr{A}thscr{C}_3(S;m_1),\quad\mathscr{F}_2\subset\mathscr{A}thscr{C}_4(S;m_2,t).$$
Together with the maximality of $\mathscr{F}_1$ and $\mathscr{F}_2$, (2) holds.

Now suppose there exists $F_{1,1}\in\mathscr{F}_1$ with $S\cap F_{1,1}=T$. 
Let $I\in\mathscr{M}(t+1,H)$ with  $T\subset I\not\subset S$. 
	Since $I$ is not a $t$-cover of $\mathscr{F}_2$, there exists $F_{2,2}\in\mathscr{F}_2$ such that $\dim(I\cap F_{2,2})<t$. 
	 Note that $\dim(T\cap F_{2,2})\geqslant t-1$.  
	We have $\dim(I\cap F_{2,2})=t-1$. 
	Since $F_{2,2}$ is a $t$-cover of $\mathscr{F}_1$, by Lemma \ref{sp-fangsuo}, we obtain
$$|(\mathscr{F}_1)_I|\leqslant{m_2-t+1\brack1}N'(t+2;m_1;2\nu).$$
Note that $S\subset H$.
Then by \eqref{sp-f1-str}, we have
\brackegin{align*}
	|(\mathscr{F}_1)_T\setminus(\mathscr{F}_1)_S|&\leqslant\sum_{I\in\mathscr{M}(t+1,H),\ T\subset I\not\subset S}|(\mathscr{F}_1)_I|\\
	&\leqslant\leqslantft({m_2-t+1\brack1}-1\right){m_2-t+1\brack1}N'(t+2;m_1;2\nu)\\
	&=q{m_2-t\brack1}{m_2-t+1\brack1}N'(t+2;m_1;2\nu).
\end{align*}
Together with Lemma \ref{gaosifangsuo} and $2\nu\geqslant m_1+2m_2-t+4\geqslant m_1+t+1$, we get
\brackegin{equation}\langlebel{sp-cross-m=t+1-f1}
	\brackegin{aligned}
		|\mathscr{F}_1|\leqslant&\ |(\mathscr{F}_1)_S|+|(\mathscr{F}_1)_T\setminus(\mathscr{F}_1)_S|\\
		\leqslant&\ N'(t+1;m_1;2\nu)+q{m_2-t\brack1}{m_2-t+1\brack1}N'(t+2;m_1;2\nu)\\
		=&\leqslantft(1+q{m_2-t\brack1}{m_2-t+1\brack1}\dfrac{q^{m_1-t-1}-1}{q^{2(\nu-t-1)}-1}\right)N'(t+1;m_1;2\nu)\\
		<&\ (1+q^{m_1+2m_2-t+3-2\nu})N'(t+1;m_1;2\nu)\\
		\leqslant&\ (1+q^{-1})N'(t+1;m_1;2\nu).
	\end{aligned}
\end{equation}
	Let $T'\in{S\brack t}\setminus\{T\}$. Note that $\dim(F_{1,1}\cap T')=t-1$. Since $F_{1,1}$ is a $t$-cover of $\mathscr{F}_2$, by Lemma \ref{sp-fangsuo}, we have
$$
	|(\mathscr{F}_2)_{T'}\setminus(\mathscr{F}_2)_S|\leqslant\leqslantft({m_1-t+1\brack1}-1\right)N'(t+1;m_2;2\nu)=q{m_1-t\brack1}N'(t+1;m_2;2\nu).
$$
	Then by Lemma \ref{gaosifangsuo} and $2\nu\geqslant m_1+m_2+t+4\geqslant m_2+t$, we obtain
\brackegin{align*}
	|\mathscr{F}_2|&=|(\mathscr{F}_2)_{T}|+\sum_{T'\in{S\brack t}\setminus\{T\}}|(\mathscr{F}_2)_{T'}\setminus(\mathscr{F}_2)_S|\\
	&\leqslant\ N'(t;m_2;2\nu)+q^2{t\brack1}{m_1-t\brack1}N'(t+1;m_2;2\nu)\\
	&=\leqslantft(1+q^2{t\brack1}{m_1-t\brack1}\dfrac{q^{m_2-t}-1}{q^{2(\nu-t)}-1}\right)N'(t;m_2;2\nu)\\
	&\leqslant\ (1+q^{m_1+m_2+t+2-2\nu})N'(t;m_2;2\nu)\\
	&\leqslant\ (1+q^{-2})N'(t;m_2;2\nu).
\end{align*}
Together with \eqref{sp-cross-m=t+1-f1} and Lemma \ref{gaosifangsuo}, we get
$$\dfrac{|\mathscr{F}_1||\mathscr{F}_2|}{c_0(\nu,m_1,m_2,t)}\leqslant\dfrac{(1+q^{-1})(1+q^{-2})}{q{m_2-t\brack1}}\leqslant q^{t+1-m_2}\leqslant1.$$
Then (3) follows from Lemma \ref{c1>c0}.

	\mathscr{E}dskip
\noindent{\brackf Case 2. $t+2\leqslant\dim M\leqslant m_2$.}
\mathscr{E}dskip

Assume that there exists $F_{2,3}\in\mathscr{F}_2\setminus(\mathscr{F}_2)_T$ with $T+F_{2,3}\neq H$. Observe that $\dim(T\vee F_{2,3})=m_2+1$ and $\dim(F\cap(T\vee F_{2,3}))\geqslant t+1$ for each $F\in\mathscr{F}_1$. 
Set
$$W=(T+F_{2,3})\cap H,\quad w=\dim W,$$
$$\mathscr{M}u=\leqslantft\{(U_1,U_2)\in{T+F_{2,3}\brack t+1}\times{H\brack t+1}:  T\subset U_1\not\subset W, T\subset U_2\not\subset W\right\}.$$
We have
\brackegin{equation}\langlebel{sp-uu1u2}
	\mathscr{F}_1\subset\leqslantft(\brackigcup_{U\in{W\brack t+1},\ T\subset U}(\mathscr{F}_1)_U\right)\cup\leqslantft(\brackigcup_{(U_1,U_2)\in\mathscr{M}u}(\mathscr{F}_1)_{U_1+U_2}\right).
\end{equation}
For each $(U_1,U_2)\in\mathscr{M}u$, since $U_1\subset T+F_{2,3}$ and $U_1\not\subset W$, we have $U_1\not\subset H$ and $U_1\neq U_2$. Together with $T\subset U_1\cap U_2$, we get $\dim(U_1+U_2)=t+2$. 
Thus
$$\leqslantft|\brackigcup_{(U_1,U_2)\in\mathscr{M}u}(\mathscr{F}_1)_{U_1+U_2}\right|\leqslant\leqslantft({m_2-t+1\brack1}-{w-t\brack1}\right)^2N'(t+2;m_1;2\nu).$$
Then by $\leqslantft|\leqslantft\{U\in{W\brack t+1}:  T\subset U\right\}\right|={w-t\brack1}$ and \eqref{sp-uu1u2}, we obtain
\brackegin{align*}
	|\mathscr{F}_1|&\leqslant{w-t\brack1}N'(t+1;m_1;2\nu)+\leqslantft({w-t\brack1}-{m_2-t+1\brack1}\right)^2N'(t+2;m_1;2\nu)\\
	&=x^2N'(t+2;m_1;2\nu)+xN'(t+1;m_1;2\nu)+{m_2-t+1\brack1}N'(t+1;m_1;2\nu),
\end{align*}
where $x={w-t\brack1}-{m_2-t+1\brack1}$.
By $2\nu\geqslant m_1+m_2+3$ we have
$${w-t\brack1}-{m_2-t+1\brack1}\geqslant-q^{m_2-t+1}\geqslant-q^{2\nu-m_1-t-2}\geqslant-\dfrac{N'(t+1;m_1;2\nu)}{2N'(t+2;m_1;2\nu)}.$$
Together with $w\leqslant m_2$, $N'(t+2;m_1;2\nu)\geqslant0$ and the property of quadratic function, we obtain
$$|\mathscr{F}_1|\leqslant{m_2-t\brack1}N'(t+1;m_1;2\nu)+q^{2m_2-2t}N'(t+2;m_1;2\nu).$$
Then by Lemma \ref{gaosifangsuo} and $2\nu\geqslant m_1+2m_2-t+1$ we get
\brackegin{equation}\langlebel{sp-cross-m>t+1-f1}
	\brackegin{aligned}
		|\mathscr{F}_1|
		&\leqslant\leqslantft({m_2-t\brack1}+\dfrac{q^{2m_2-2t}(q^{m_1-t-1}-1)}{q^{2(\nu-t-1)}-1}\right)N'(t+1;m_1;2\nu)\\
		&<\leqslantft({m_2-t\brack1}+q^{m_1+2m_2-t+1-2\nu}\right)N'(t+1;m_1;2\nu)\\
		&\leqslant\leqslantft({m_2-t\brack1}+1\right)N'(t+1;m_1;2\nu).
	\end{aligned}
\end{equation}

Set $k:=\dim M$. We have
$$\mathscr{F}_2\subset\{F\in\mathscr{P}_{m_2}:  T\subset F\}\cup\{F\in\mathscr{P}_{m_2}:  T\not\subset F, \dim(F\cap M)=k-1\},$$
which implies that
\brackegin{equation*}
	|\mathscr{F}_2|\leqslant N'(t;m_2;2\nu)+q^{k-t}{t\brack1}N'(k-1;m_2;2\nu).
\end{equation*}
By Lemma \ref{gaosifangsuo} and $2\nu\geqslant m_1+2m_2>m_2+k$, we have
$$\dfrac{q^{k-t}N'(k-1;m_2;2\nu)}{q^{k-t+1}N'(k;m_2;2\nu)}=\dfrac{q^{2(\nu-k+1)}-1}{q(q^{m_2-k+1}-1)}\geqslant q^{2\nu-m_2-k}\geqslant1,$$
which implies that
\brackegin{equation*}
	|\mathscr{F}_2|\leqslant N'(t;m_2;2\nu)+q^{2}{t\brack1}N'(t+1;m_2;2\nu).
\end{equation*}
Together with \eqref{sp-cross-m>t+1-f1}, $m_2\geqslant k\geqslant t+2$, $2\nu\geqslant m_1+2m_2+3$ and Lemma \ref{gaosifangsuo}, we obtain

\brackegin{align*}
	&\dfrac{c_0(\nu,m_1,m_2,t)-|\mathscr{F}_1||\mathscr{F}_2|}{N'(t+1;m_1;2\nu)N'(t;m_2;2\nu)}\\
	\geqslant&{m_2-t\brack1}-1-q^2{t\brack1}\leqslantft({m_2-t\brack1}+1\right)\dfrac{q^{m_2-t}-1}{q^{2(\nu-t)}-1}\\
	=&{m_2-t\brack1}\leqslantft(1-\dfrac{q-1}{q^{m_2-t}-1}-\dfrac{q^2(q^t-1)(q^{m_2-t}-1)}{(q-1)(q^{2(\nu-t)}-1)}-\dfrac{q^2(q^t-1)}{q^{2(\nu-t)}-1}\right)\\
	>&{m_2-t\brack1}(1-q^{-1}-q^{m_2+2t+2-2\nu}-q^{3t+2-2\nu})\\
	\geqslant&{m_2-t\brack1}(1-q^{-1}-q^{-m_1-m_2+2t-1}-q^{-m_1-2m_2+3t-1})\\
	>&{m_2-t\brack1}(1-q^{-1}-q^{-3}-q^{-4})\\
	>&\ 0.
\end{align*}

Then (3) followis from Lemma \ref{c1>c0}.

Now suppose that $T+G=H$ holds for each $G\in\mathscr{F}_2\setminus(\mathscr{F}_2)_T$. 
We have
$$\mathscr{F}_2\subset\{F\in\mathscr{P}_{m_2}: T\subset F\}\cup{H\brack m_2}.$$
Note that $\dim H=m_2+1$ and \eqref{sp-f1-str} holds. We will investigate cross $t$-intersecting families under this case in Case 3.

\mathscr{E}dskip
\noindent{\brackf Case 3. $\dim M=m_2+1$.}
\mathscr{E}dskip

In this case, we have $M=H$. We have
$$\mathscr{F}_1\subset\{F\in\mathscr{P}_{m_1}:  T\subset F, \dim(F\cap M)\geqslant t+1\},\quad\mathscr{F}_2\subset\{F\in\mathscr{P}_{m_2}:  T\subset F\}\cup{M\brack m_2}.$$
If $M$ is totally isotropic, by the maximality of $\mathscr{F}_1$ and $\mathscr{F}_2$, we have
$$\mathscr{F}_1=\mathscr{A}thscr{C}_1(M,T;m_1,t),\quad\mathscr{F}_2=\mathscr{A}thscr{C}_2(M,T;m_2),$$
i.e., (1) holds. 

Now suppose that $M$ is not totally isotropic. Note that the type of $M$ is $(m_2+1,1)$. Since $M=T+F_{2,1}$, it is routine to check that
$$|\{S\in\mathscr{M}(t+1,M): T\subset S\}|={m_2-t\brack1}.$$
Together with \eqref{sp-f1-str}, we have
$$|\mathscr{F}_1|\leqslant{m_2-t\brack1}N'(t+1;m_1;2\nu).$$
Then by $2\nu\geqslant m_1+2m_2-t+2$, it is routine to check that
$$|\mathscr{F}_1|<q{m_2-t\brack1}N'(t+1;m_1;2\nu)\leqslant s_0(\nu,m_1,m_2+1,t).$$
Since $M$ is of type $(m_2+1,1)$, we have
$$|\mathscr{F}_2|\leqslant N'(t;m_2;2\nu)+(q+1)<N'(t;m_2;2\nu)+q^{m_2-t+1}{t\brack1}.$$
Then from Lemma \ref{c1>c0}, we obtain
$$|\mathscr{F}_1||\mathscr{F}_2|<s_0(\nu,m_1,m_2+1,t)\leqslantft(N'(t;m_2;2\nu)+q^{m_2-t+1}{t\brack1}\right)\leqslant c_1(\nu,m_1,m_2,t),$$
i.e., (3) holds.
	\end{proof}

	
	
	\brackegin{proof}[\brackf Proof of Theorem \ref{non-trivial}]
		Let $\nu$, $m_1$, $m_2$ and $t$ be positive integers with $m_1\geqslant m_2\geqslant t+1$ and $2\nu\geqslant2m_1+m_2+t+3$. Suppose $\mathscr{F}_1\subset\mathscr{P}_{m_1}$  and $\mathscr{F}_2\subset\mathscr{P}_{m_2}$ are non-trivial cross $t$-intersecting families with maximum product of sizes. Note that
		\brackegin{equation*}\langlebel{sp-fif2-c1c2}
			|\mathscr{F}_1||\mathscr{F}_2|\geqslant\mathscr{A}x\{c_1(\nu,m_1,m_2,t),c_1(\nu,m_2,m_1,t),c_2(\nu,m_1,m_2,t),c_2(\nu,m_2,m_1,t)\}.
		\end{equation*}
	
		By Lemma \ref{sp-cross-upper-other}, we have$(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))=(t,t+1)$ or $(\tau_t(\mathscr{F}_1),\tau_t(\mathscr{F}_2))=(t+1,t)$. Together with Lemmas \ref{cp1-cp2}, \ref{cp-cp'} and \ref{sp-cross-upper-tt+1}, we finish our proof.
	\end{proof}
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
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\end{document}