Let $G$ be a group. An involution is an element $g\in G$ such that $g^2=1$. Let $F$ be a field, $V$ an $F$-vector space and $b:V\times V \rightarrow F$ a nondegenerate alternating bilinear form. The set $\mathrm{Sp}(V,b)=\{ f:V\rightarrow V \mid f \mathrm{\ is\ bijective\ and\ } b(f(x),f(y))=b(x,y) \forall x,y \in V\}$ with the composition of maps is a group called the symplectic group of $(V,b)$. It is well-known that the symplectic group is generated by transvections. Is it also generated by involutions in $\mathrm{Sp}(V,b)$?
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1$\begingroup$ Is $b$ non-degenerate? What about $F=\mathbb{R}$, $V$ 1-dimensional, $b=0$? $\endgroup$– Uri BaderCommented Jul 31, 2016 at 11:52
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2$\begingroup$ Note that the projective symplectic group is simple (unless you're in some very special cases, see below). In this case (clearly having a non-trivial involution) it must be generated by involutions, as these generate a normal subgroup. See groupprops.subwiki.org/wiki/… $\endgroup$– Uri BaderCommented Jul 31, 2016 at 11:56
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$\begingroup$ Alternatively: use the fact that it is generated by $\text{SL}_2$'s. $\endgroup$– Uri BaderCommented Jul 31, 2016 at 12:04
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$\begingroup$ May I ask you to be more specific? $\endgroup$– OliverCommented Jul 31, 2016 at 12:06
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1$\begingroup$ Note that on a two dimensional space the form $(a,c),(b,d)\mapsto ad-bc$ is just the determinant, so $\text{Sp}=\text{SL}$. Thus also in higher dimensions the symplectic group contains many copies of $\text{SL}_2$. In fact, it is generated by these (this, again, could be seen by simplicity, but in fact it holds in a complete generality (I think)). $\endgroup$– Uri BaderCommented Jul 31, 2016 at 12:18
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1 Answer
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Here is an answer based on the many comments by myself and by Nick Gill.
In characteristic 2 the transvections are always involutions, so generation by transvections implies generation by involutions in all dimensions. Below I will assume that the characteristic is not 2.
In dimensions 2 the determinant is a symplectic form, thus $G$ is conjugated to $\text{SL}_2(F)$. Then it is easy to check that the only involutions in $G$ are $1$ and $-1$. In particular, $G$ is not generated by involtions. Below I will assume that the dimension is not 2.
I claim that $G$ is generated by involutions. Note first that $G$ has at least one non-central involution: for example one can view the form as a direct sum of lower dimensional ones, take 1 on one and -1 on the other. Note also that the group generated by all non-central involution is normal and not central. This group must be $G$, as every proper normal subgroup of $G$ is central. Indeed, this is the case for every symplectic group apart of $\text{Sp}(2,\mathbb{F}_2)$, $\text{Sp}(2,\mathbb{F}_3)$ and $\text{Sp}(4,\mathbb{F}_2)$, and by the assumptions above $G$ is not in this list.
