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As I understand it, the symplectic Lie group $Sp(2n,\mathbb{R})$ of $2n \times 2n$ symplectic matrices is generated by the matrices in http://en.wikipedia.org/wiki/Symplectic_group#Infinitesimal_generators .

Does this mean that $sl(n,\mathbb{R})$ is a subalgebra of the corresponding lie algebra, since in that formula we can truncate by removing the matrices $B$ and $C$ and enforce that $A$ is traceless?

Also, $sp(4,\mathbb{R})$ has dimension $10$ and $sl(3,\mathbb{R})$ has dimension $8$. Is $sl(3,\mathbb{R})$ a subalgebra of $sp(4,\mathbb{R})$ or not?

As a more practical question, where should I look if I want to look up these kind of standard results on standard Lie groups? Googling did not take me very far.

jon
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  • $SL(n,\mathbb{R})$ is not a Lie algebra, did u mean $\mathfrak{sl}(n,\mathbb{R})$? I think you're mixing up Lie algebras and Lie groups. – Qidi Mar 10 '15 at 16:38
  • yes, I have edited my question. – jon Mar 10 '15 at 17:04

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You have it backwards, in fact $\mathrm{Sp}(2n, \mathbb R) \subseteq \mathrm{SL}(2n, \mathbb R)$ (see here) so $\mathfrak{sp}(2n, \mathbb R) \subseteq \mathfrak{sl}(2n, \mathbb R)$.

You can see that each of those infinitesimal generators has trace $0$ (you don't need $A$ to be traceless for this to be true), but not every trace $0$ matrix must be written in that form.

Community
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Jim
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  • No I really meant that for example sl(2,R) is a subalgebra of sp(4,R) (what you state is obviously true). If you look at the generators in the wiki article, if we truncate only to the A matrices, these form a representation of gl(2,R) since it is a block diagonal matrix and the A is any 2x2 matrix, right? – jon Mar 11 '15 at 16:19
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    Ahhh, I understand what you mean now. Yes, you're correct, $\mathfrak{gl}(n, \mathbb R)$ (and hence $\mathfrak{sl}(n, \mathbb R)$) is indeed isomorphic to a subalgebra of $\mathfrak{sp}(2n, \mathbb R)$. – Jim Mar 11 '15 at 19:14
  • For your other question, $\mathfrak{sp}(4, \mathbb R)$ and $\mathfrak{sl}(3, \mathbb R)$ are types $B_2$ and $A_2$ respectively. Since they both have maximal tori of dimension $2$ if $\mathfrak{sl}(3, \mathbb R) \subseteq \mathfrak{sp}(4, \mathbb R)$ then the root system $A_2$ would be contained in the root system $B_2$. But it's not, so $\mathfrak{sl}(3, \mathbb R) \subsetneq \mathfrak{sp}(4, \mathbb R)$. – Jim Mar 11 '15 at 19:16
  • I don't know of a reference where you could look up this sort of stuff. Maybe Humphrey's Lie algebras book? Maybe Fulton & Harris's representation theory book? – Jim Mar 11 '15 at 19:17
  • Thank you, but I dont exactly follow your argument. How exactly do you see that they both have tori of dimension 2? and if I understand it correctly, if 1) they both have the same maximal torus and 2) one is contained in the other, then the root system should be contained in the other. What theorem do you cite here? – jon Mar 11 '15 at 23:32
  • I know both maximal tori are dimension $2$ because I happen to know what the maximal tori are in both cases: the diagonal matrices. – Jim Mar 12 '15 at 00:35
  • As for (1) and (2) I don't know of a particular named theorem, but all maximal tori are conjugate so if there's an embedding $\mathfrak{sl}(3, \mathbb R) \subseteq \mathfrak{sp}(4, \mathbb R)$ then you can assume that the tori that you're computing the root systems from is the same torus for either. Then it's clear that a weight space for $\mathfrak{sl}(3, \mathbb R)$ is also a weight space in $\mathfrak{sp}(4, \mathbb R)$. – Jim Mar 12 '15 at 00:38