# Symplectic group over integers and finite fields

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 $Sp(2n,\mathbb{Z})$?

• I'm not sure what you mean by your tensor product notation between a group and a field. – YCor Feb 11 '16 at 18:08
• I mean reduction of the coefficients mod p. Fixed. – Nicolas Malebranche Feb 11 '16 at 18:09
• 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. – RP_ Feb 11 '16 at 18:17
• @René: I see. And of course, the fact that it is a group does not matter for the question. – Nicolas Malebranche Feb 11 '16 at 18:20
• Are you asking whether ${\rm Sp}(2n, \mathbb{Z}) \to {\rm Sp}(2n, \mathbb{F}_p)$ is surjective? – 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