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This is related to Symplectic group over $\mathbb{Z}/p\mathbb{Z}$ is generated by its root subgroups. There I was told that in general, the symplectic group $\text{Sp}_{2n}(R)$ is not generated by its root subgroups and a maximal torus.

My question now is: for which commutative rings $R$ is $\text{Sp}_{2n}(R)$ generated by its root subgroups and a maximal torus?

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Since $\operatorname{Sp}_{2\ell}$ is simply-connected, there is no need for the maximal torus. So the question is about the triviality of $\operatorname{K_1}(\mathsf{C}_\ell, R) = \operatorname{Sp}(2\ell,R)/\operatorname{Ep}(2\ell,R)$, where $\operatorname{Ep}(2\ell,R)$ is the elementrary symplectic group, that is, the subgroup of $\operatorname{Sp}(2\ell,R)$ generated by its root subgroups.

A lot of cases where $\operatorname{K_1}(\mathsf{C}_\ell,R)$ is trivial are not specific fot the symplectic case and hold for all Chevalley groups. Below is a(n incomplete) list. I only consider commutative rings.

  • $R$ is a ring of stable rank 1 (this includes local and semi-local rings, boolean rings, the ring of algebraic integers, the ring of entire functions, the disc algebra). The proof is a combination of surjective stabiliy for $\operatorname{K}_1$ (a paper by M. Stein) and an easy argument for $\operatorname{SL}(2,R)$ (every its element is a product of 4 elementary matrices, this was apparently first noted by H. Bass);
  • $R$ is a Dedekind ring of arithmetic type in a number field which is not totally imaginary (this is Theorem 3.6 from Bass—Milnor—Serre);
  • $R$ is Euclidean (the same as the stable rank 1 case, but with the Euclidean algorithm for $\operatorname{SL}(2,R)$ and no bound on the number of elementary factors);
  • More generally, if $R$ is a ring such that $\operatorname{K}_1({}\cdot{},R)=1$, it is sometimes possible to show that $\operatorname{K}_1({}\cdot{},R[x_1,\ldots,x_n])=1$, see my answer here;
  • There are many papers dealing with the rings of geometric or analytical origin, for example, in this paper by B. Ivarsson, F. Kutzschebauch and E. Løw the authors consider commutative Banach algebras and some rings of continuous functions (they prove that null-homotopic matrices are elementary, but the paper also contains some references for other rings).
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  • $\begingroup$ Many thanks again for your very useful answer. I have been studying the text by Bass-Milnor-Serre and noticed that they defined $\text{Ep}(2l,R)$ to be the subgroup of $\text{Sp}(2l,R)$ generated by all $\begin{pmatrix}I&0\\\sigma&I\end{pmatrix}$ and $\begin{pmatrix}I&\sigma\\0&I\end{pmatrix}$ where $\sigma$ is symmetric. You claim this to be the subgroup of the symplectic group generated by its root subgroups. How can we show this? I am particularly interested in doing this for the root system $C_l$. $\endgroup$
    – user341790
    Commented Aug 21, 2021 at 9:31
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    $\begingroup$ @ErikBarinaga The root subgroups corresponding to the roots involving the long simple root (or its opposite) are of this form, and any other root is a difference of two such, so this can be sorted out by Chevalley commutator formula. Anyway, if this smaller subgroup (of the elementary subgroup) already coincides with the whole symplectic group, then the usual elementary subgroup does as well. $\endgroup$
    – Andrei Smolensky
    Commented Aug 21, 2021 at 13:22
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    $\begingroup$ The method in a paper by M. Stein shows that for any $g\in\operatorname{Sp}(2n,R)$ one can find a product of root elements $f$ such that $fg\in\operatorname{Sp}(2n-2,R)$ (under the natural embedding $\operatorname{Sp}_{2n-2}\to\operatorname{Sp}_{2n})$. This can then be repeated, leading to a product of root elements $h$ such that $hg\in\operatorname{Sp}(2,R)$, while the latter coincides with $\operatorname{SL}(2,R)$, so $hg$ is itself a product of root elements. So $g\in\operatorname{Ep}(2n,R)$ and $\operatorname{K}_1$ is trivial. $\endgroup$
    – Andrei Smolensky
    Commented Aug 25, 2021 at 10:26
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    $\begingroup$ $\operatorname{Sp}_{2n-2}$ embeds in $\operatorname{Sp}_{2n}$ similarly to the embedding of $\operatorname{SL}_{n-1}$ into $\operatorname{SL}_n$ (into the upper left corner). So if an element of $\operatorname{Sp}_{2n}$ has zeroes in the appropriate entries, it is an element of $\operatorname{Sp}_{2n-2}$. And one can make these entries zero by means of multiplying by suitable root elements. $\endgroup$
    – Andrei Smolensky
    Commented Aug 25, 2021 at 12:00
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    $\begingroup$ Better use $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ as generators, these are the root elements (and the root elements of $\operatorname{Sp}_{2n}$ corresponding to the long root have this as a submatrix, and coincide with the identity matrix in all other entries). $\endgroup$
    – Andrei Smolensky
    Commented Sep 22, 2021 at 17:48

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