What's the classification of the algebraic subgroups of Sp(4,R)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:18:22Z http://mathoverflow.net/feeds/question/9378 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9378/whats-the-classification-of-the-algebraic-subgroups-of-sp4-r What's the classification of the algebraic subgroups of Sp(4,R)? Matheus 2009-12-19T16:38:43Z 2010-07-06T13:57:10Z <p>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.</p> http://mathoverflow.net/questions/9378/whats-the-classification-of-the-algebraic-subgroups-of-sp4-r/9399#9399 Answer by Greg Kuperberg for What's the classification of the algebraic subgroups of Sp(4,R)? Greg Kuperberg 2009-12-20T01:53:34Z 2009-12-20T01:53:34Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>Finally the solvable part also complexifies and I think that it is contained in a Borel subalgebra at the Lie algebra level.</p> <p>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 Frobeniusâ€“Schur indicator.</p> http://mathoverflow.net/questions/9378/whats-the-classification-of-the-algebraic-subgroups-of-sp4-r/30738#30738 Answer by Michael Giudici for What's the classification of the algebraic subgroups of Sp(4,R)? Michael Giudici 2010-07-06T08:31:13Z 2010-07-06T08:31:13Z <p>The paper `On the subgroup description of classical groups' by Martin Liebeck and Gary Seitz (available at <a href="http://dx.doi.org/10.1007/s002220050270" rel="nofollow">http://dx.doi.org/10.1007/s002220050270</a>) 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.</p> http://mathoverflow.net/questions/9378/whats-the-classification-of-the-algebraic-subgroups-of-sp4-r/30778#30778 Answer by BS for What's the classification of the algebraic subgroups of Sp(4,R)? BS 2010-07-06T13:57:10Z 2010-07-06T13:57:10Z <p>Maximal (connected, I think) subgroups of complex classical groups have been classified by E.B. Dynkin in the early 50's. <a href="http://www.ams.org/mathscinet-getitem?mr=49903" rel="nofollow">Here</a> 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 <a href="http://www.ams.org/mathscinet-getitem?mr=910539" rel="nofollow">here</a>), 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 <a href="http://www.ams.org/mathscinet-getitem?mr=1650328" rel="nofollow">here</a>, 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. </p>