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 $\mathbb{Z}$?

  • 1
    $\begingroup$ is $l$ a prime? $\endgroup$ – Venkataramana Jan 22 '18 at 4:59

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.

  • $\begingroup$ Thanks! Also may I know if there is a reference for such results? $\endgroup$ – Qixiao Jan 23 '18 at 13:28
  • $\begingroup$ Sadly I do not know of a reference. $\endgroup$ – David Hughes Jan 23 '18 at 15:50

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.