Is sl(3,R) a subalgebra of sp(4,R)?

[Kurret]

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

Does this mean that sl(n,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,R) has dimension 10 and sl(3,R) has dimension 8. Is sl(3,R) a subalgebra of sp(4,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.

 

[lpetrich]

First, note that those are noncompact Lie algebras. All of the noncompact simple ones can be analytically continued to compact ones. For these ones:
SL(n,R) -> SU(n)
Sp(2n,R) -> Sp(2n)
giving
SL(3,R) -> SU(3)
Sp(4,R) -> Sp(4)
Wikipedia has a List of simple Lie groups with these results and more.

So subalgebra results and the like will carry over. Let's now see if SU(3) can be a subalgebra of Sp(4).

R. Slansky's book Group Theory for Unified Model Building has a table of maximal subalgebras of all the compact simple Lie algebras with rank 8 or less. It's Table 14, book page 86. That includes all of them that one is likely to run into, and also all 5 exceptional algebras. All other subalgebras are subalgebras of the maximal ones. Here are those for Sp(4), with how they were found:
SU(2) * U(1) -- root demotion
SU(2) * SU(2) -- Dynkin-diagram extension and root removal
SU(2) -- height subalgebra, as I call it

SU(3) is nowhere in sight. So SU(3) cannot be a subalgebra of Sp(4), and thus, SL(3,R) cannot be a subalgebra of Sp(4,R).