Let $m,n\in\mathbb{N}$, $l$ be a prime number, let $J$ be the standard symplectic matrix


$$J=\left[
\begin{array}[cc]
\\0 & I\_n \\
-I\_n & 0\\
\end{array}\right]$$


Let $$\mathrm{Sp}(2n,\mathbb{Z}/l^m)=\{A\in\mathrm{Gl}\_{2n}(\mathbb{Z}/l^m)|A^TJA=J\}.$$


Let $N\in \mathrm{Mat}\_{2n}(\mathbb{Z}/l^m)$ be a skew symmetric matrix, such that for any $A\in\mathrm{Sp}(2n,\mathbb{Z}/l^m)$ $$A^TNA=N$$


Then is it true $N$ is a scalar multiple of $J$?


If not, will the conclusion be true for $m$ large enough or replace $\mathbb{Z}/l^m$ by $\mathbb{Z}$?



Invariant two form under symplectic group
Yes. Let $(q\_k,p\_k)$ be a symplectic basis to $\mathbb{R}^{2n}$ so that $J(q\_k)=p\_k$ and $J(p\_k)=-q\_k$ for all $k$. Consider matrices $A\_k$ and $B\_{k,l}$ such that


* $A\_k (q\_k)=p\_k \quad$and$\quad A\_k (p\_k) = -q\_k$
* $A\_k (q\_l)=q\_l \quad$and$\quad A\_k (p\_l) = p\_l$ for $l \neq k$
* $B\_{k,l}(q\_k)=q\_l, \quad B\_{k,l}(q\_l)=q\_k, \quad B\_{k,l}(p\_k)=p\_l,\quad B\_{k,l}(p\_l)=p\_k\quad$ and $B\_{k,l}$ is the identity away from the subspace spanned by the vectors $q\_k,p\_k,q\_l,p\_l$.


The matrices $A\_k$ and $B\_{k,l}$ are symplectic, as can easily be checked. If $E\_k$ is the subspace spanned by the vectors $q\_k,p\_k$, we have $E\_k=\text{Ker}(A\_k^2+I)$. Also, note that $A\_k^{T}=A\_k^{-1}$ and $B\_{k,l}^T=B\_{k,l}$.


Now, since these matrices are symplectic, we can subsitute them into the formula for $N$ to get $A\_k^{-1} N A\_k = N$ and $B\_{k,l} N B\_{k,l}= N$. Taking the first equation and applying it to the vector $q\_k$, we have $A\_k^{-1}NA\_k(q\_k)=N(q\_k)$, so $N(p\_k)=A\_kN(q\_k)$. Similarly, $N(q\_k)=-A\_kN(p\_k)$. In particular, $A\_k^2 N(q\_k)=-N(q\_k)$ and $A\_k^2N(p\_k)=-N(p\_k)$. Hence, $N(q\_k), N(p\_k) \in E\_k$, and so there are numbers $a\_k,b\_k,c\_k,d\_k$ such that
$$ N(q\_k)=a\_k q\_k+b\_k p\_k \\ N(p\_k)=c\_k q\_k+d\_k p\_k .$$
Applying $A\_k$ to either of these equations and using $N(q\_k)=-AN(p\_k)$ or $N(p\_k)=AN(q\_k)$, respectively, we have that $a\_k=d\_k$ and $c\_k=-b\_k$. Applying $B\_{k,l}$ to these formulas for each $k,l$ shows that $a\_k=a\_l=:a$ and $b\_k=b\_l =: b$. So putting this all together, we get $N=aI+bJ$. Since $N$ is skew symmetric, it follows that $a=0$, and hence the result.


