For $H=\left( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix} \right)$ and a commutative ring $F$, the symplectic group $Sp(2n,F)$ is the set of all matrices $M\in F^{2n\times 2n}$ such that $MHM^T = H$. (Here, $M^T$ means the transposed matrix.)

My question is: for $\mathbb{F}_p$ a finite field of prime order, is $Sp(2n,\mathbb{F}_p) = Sp(2n,\mathbb{Z})\otimes\mathbb F_p$, the reduction mod $p$ of $Sp(2n,\mathbb{Z})$?

  • 2
    $\begingroup$ I'm not sure what you mean by your tensor product notation between a group and a field. $\endgroup$ – YCor Feb 11 '16 at 18:08
  • 1
    $\begingroup$ I mean reduction of the coefficients mod p. Fixed. $\endgroup$ – Nicolas Malebranche Feb 11 '16 at 18:09
  • 2
    $\begingroup$ The problem was not so much the notation (which is fine), but the fact that you define $\operatorname{Sp}(2n,R)$ as a group, rather than an $R$-scheme. I'm sure this is what you meant, but it isn't what you're writing. $\endgroup$ – RP_ Feb 11 '16 at 18:17
  • $\begingroup$ @René: I see. And of course, the fact that it is a group does not matter for the question. $\endgroup$ – Nicolas Malebranche Feb 11 '16 at 18:20
  • 3
    $\begingroup$ Are you asking whether ${\rm Sp}(2n, \mathbb{Z}) \to {\rm Sp}(2n, \mathbb{F}_p)$ is surjective? $\endgroup$ – Piotr Achinger Feb 11 '16 at 18:31

Yes: $\mathrm{Sp}(2n, \mathbb F_p)$ is generated by its root subgroups. Each root subgroup is cyclic, and generated by an element that lifts in an obvious way to $\mathrm{Sp}(2n, \mathbb Z)$.

More concretely, if I haven't messed up the calculations in your form (I'm used to a different one), then, writing $E_{i j}$ for the matrix that has a 1 in its $(i, j)$th entry and 0 elsewhere, we have that $\mathrm{Sp}(2n, \mathbb F_p)$ is generated by:

  • $\{1 + t(E_{i j} - E_{(i + n)(j + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i - e_j$) for $1 \le i < j \le n$,
  • $\{1 + t(E_{i(j + n)} + E_{j(i + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i + e_j$) for $1 \le i \ne j \le n$,
  • $\{1 + t E_{i(i + n)} : t \in \mathbb F_p\}$ (corresponding to the root $2e_i$) for $1 \le i \le n$, and
  • the transposes of the various groups above (corresponding to the negative roots).

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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