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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 $\mathbb{Z}$?

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

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  • $\begingroup$ 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

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