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This is a question about the answer in this other post: Symplectic group over integers and finite fields.

In general, for any ring $R$, the symplectic group $\text{Sp}(2n,R)$ is generated by its root subgroups and a maximal torus $T$.

Why is $\text{Sp}(2n,\mathbb{Z}/p\mathbb{Z})$ generated only by its root subgroups, where $p$ is a prime number?

A reference to a book that discusses this would already make me very happy, but I haven't been able to find one...

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    $\begingroup$ Because $\operatorname{Sp}_{2n}$ is simply-connected and $\mathbb{Z}/p\mathbb{Z}$ is a field. The standard reference is "Lectures on Chevalley groups" by R. Steinberg. $\endgroup$ Aug 18, 2021 at 12:38
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    $\begingroup$ Also, the symplectic group over a commutative ring is usually not generated by its elementary subgroup and a maximal torus. $\endgroup$ Aug 18, 2021 at 12:41
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    $\begingroup$ The elementary subgroup is by definition the subgroup generated by the root subgroups. $\operatorname{Sp}(2n,\mathbb{Z})$ is generated by its root subgroups as well. The simplest way to see this is perhaps by means of the stable rank condition, see "Stability theorems for $K_1$, $K_2$ and related functors modeled on Chevalley groups" by M. Stein. It essentially shows an explicit way to decompose any symplectic matrix into a product of elementary root unipotents by reducing it to a matrix in $\operatorname{SL}_2$ and finishing it off with by Euclidean algorithm. $\endgroup$ Aug 18, 2021 at 14:08
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    $\begingroup$ Also, I now think that Steinberg notes is not the best source for this particular question (over fields), because he defines Chevalley groups as genetated by the root subgroups. R. Carter does the same in "Simple groups of Lie type", but also provides the identification with the classical groups. $\endgroup$ Aug 18, 2021 at 14:10
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    $\begingroup$ Another good source for symplectic (and other classical) groups over rings is "The Classical groups and $K$-theory" by Hahn an O'Meara. $\endgroup$ Aug 18, 2021 at 14:13

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I'll spell out Andrei Smolensky's argument a bit (and remove this from the unanswered list).

(1) For any field $k$, the group $\operatorname{SL}_2(k)$ is generated by its elementary subgroups. In particular, you can see that the torus is generated by elementary subgroups because $$ \begin{bmatrix} 1&0 \\ a-a^2&1 \\ \end{bmatrix} \begin{bmatrix} 1&-1/a \\ 0&1 \\ \end{bmatrix} \begin{bmatrix} 1&0\\ -1+a&1 \\ \end{bmatrix} \begin{bmatrix} 1&1 \\ 0&1 \\ \end{bmatrix}= \begin{bmatrix} 1/a&0 \\ 0&a \\ \end{bmatrix}.$$

(2) For each pair of roots $\pm \alpha$ of a reductive group $G$, we get a map from $\operatorname{SL}_2$ to $G$. In particular, the torus of $\operatorname{SL}_2$ maps to the one parameter subgroup in the torus of $G$ corresponding to $\pm \alpha^{\vee}$. So, if the co-roots span the co-weight lattice (which is equivalent to $G$ being simply connected), then the tori of these various $\operatorname{SL}_2$'s will generate the torus of $G$.

Concretely, the co-roots of the symplectic group are $\pm e_i \pm e_j$ and $\pm e_k$, inside the co-weight lattice $\mathbb{Z}^n = \bigoplus \mathbb Z e_i$, and it is clear that these span.

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  • $\begingroup$ The title of the question is a bit different from its body, but, just to be clear, this is information only over a field, not over a ring, right? $\endgroup$
    – LSpice
    Aug 18, 2021 at 19:16
  • $\begingroup$ Also, I think I might prefer to call what you've written down the cocharacter lattice of $\operatorname{Sp}_{2n}$ (or the coroot lattice of type $\mathsf C_n$), since it is only of index $2$ in what I would call the coweight lattice, by which I mean the integral dual to the root lattice (which also contains $\tfrac1 2\sum e_i$). $\endgroup$
    – LSpice
    Aug 18, 2021 at 19:22

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