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

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1. Mar 10, 2015

### 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.

2. Mar 15, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Mar 17, 2015

### 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).