Question

Hi! I would like to know if there is an explicit classification of the algebraic (i.e., Zariski closed) subgroups of the symplectic group Sp(4,R) and/or more generally Sp(2n,R) somewhere in the literature.

Answer 1

On the one hand, I could not find a published answer with a cursory search. On the other hand, as Ben says, you could work out the answer "by hand". Instead of writing down a sheer list, which might be complicated (and I haven't done the work), I'll write down the main ingredients.

A Zariski-closed subgroup $H$ of any connected semisimple Lie group $G$ has three pieces: (1) finite, (2) connected semisimple, and (3) connected solvable. The Zariski topology forces $H$ to have only finitely many components; if $H_0$ is the connected subgroup, then $H/H_0$ is the finite piece. Then the Lie algebra of $H_0$ has a Levi decomposition, so that you get the other two pieces. The way to analyze the question is to chase down the possibilities for all three pieces.

I think that the finite part always lifts to a slightly larger finite subgroup of $H$. This is not true for groups in general, but I think that it is true in context. Then this finite group is contained in a maximal compact group of $G$. Happily, the compact core of $\text{Sp}(4,\mathbb{R})$ is $\text{SU}(2)$, and the finite subgroups are classified by simply laced Dynkin diagrams.

A semisimple, connected subgroup of $G$ corresponds to a semisimple Lie subalgebra, and that complexifies. The Lie algebra $\text{sp}(4,\mathbb{C})$ does not have very many inequivalent semisimple subalgebras. From looking a rank, they are isomorphic to $\text{sl}(2,\mathbb{C})$ or $\text{sl}(2,\mathbb{C}) \oplus \text{sl}(2,\mathbb{C})$. I am confusing myself a little with the possible positions of the former, although I know there are only a few. The latter embeds in only one way. Then you would work backwards to get the real forms of these complex subalgebras; again there wouldn't be very many.

Finally the solvable part also complexifies and I think that it is contained in a Borel subalgebra at the Lie algebra level.

As for the more general question, for $\text{Sp}(2n,\mathbb{R})$, there is a tidy converse answer that also shows you that you can't expect a tidy answer for all fixed $n$. Namely, if $G$ is any algebraic group, you can classify its anti-self-dual (or symplectically self-dual) representations. Every algebraic group will have some, because every algebraic group has representations in $\text{GL}(n,\mathbb{R})$. A more interesting case is when $G$ has an irreducible symplectically self-dual representation. For that purpose, you check that the irreducible representation is real, and then check the Frobenius–Schur indicator.

Answer 2

The paper `On the subgroup description of classical groups' by Martin Liebeck and Gary Seitz (available at http://dx.doi.org/10.1007/s002220050270) gives a structure description of the closed subgroups of classical groups over an algebraically closed field. The subgroups are either the stabilisers of a subspace, subspace decomposition or tensor product decomposition, or a classical group, or modulo scalars is the normaliser of an elementary abelian $r$-group, or modulo scalars is almost simple. It generalises the result by Aschbacher in the finite field case.

Answer 3

Maximal (connected, I think) subgroups of complex classical groups have been classified by E.B. Dynkin in the early 50's. Here is a link to the MR of the russian paper, translated in Amer. Math. Soc. Transl., Series 2, Vol. 6 (1957), 245-378 (which isn't in MR). Then T.M. Selim (found in the citations of Dynkin's paper) undertook the real case (see here), but the summary in MR has strange notations and what is exactly proved is not clear to me. The paper by Liebeck and Seitz cited in Michael's answer has MR here, but the link to the article in MR (given in Michael's answer) seems broken. By the way, it is in Inventiones 134 (1998), no. 2, 427--453.