How to Prove Semisimplicity in Classical Linear Lie-Algebras?

[CompuChip]

[SOLVED] Exercises on Lie-algebras

I'm doing some homework on Lie-algebra's, and there are several things I am missing. Let me list them below in one thread.

1. The statement is "If L is a classical linear Lie-algebra, then L is semisimple." I have reduced this to proving Rad L $\subseteq$ Z(L) -- then I have that Rad L = Z(L), so L is reductive and I can use an earlier result to conclude that Rad L = 0 so L is semisimple. But I don't know how to show this. Since what I need to prove does not hold in general, I should plug in some (common) property of classical linear Lie-algebras, but I can't really think of anything useful here... any ideas?
   
2. Finally, I am required to prove that If any Lie-algebra L satisfies Rad L = Z(L), then all finite-dimensional representations of L in which Z(L) is represented by semisimple endomorphisms are completely reducible (Humphreys' book on Introduction to Lie-algebras and Representation theory, exercise 6.5d). So I took such a representation and found that I can simultaneously diagonalize all elements in the center, but I don't really see how this will give me any information on the reducibility of the entire representation.

I know that I haven't really posted a lot of work, but I did try to reduce the problem to the points I'm really stuck at. I obviously don't expect a fully worked answer, but I'd really appreciate some hints or general steps to solve these problems. Thanks you very much!

 

[CompuChip]

I solved and removed one of the problems. The other two remain, so any responses are welcome!

 

[matt grime]

What is your definition of 'classical'?

 

[CompuChip]

They are $$\mathfrak{sl}(\ell, \mathsf{F}}, \mathfrak{sp}(2\ell, \mathsf{F}}$$ and $$\mathfrak{o}(\ell, \mathsf{F}}$$ for a field $\mathsf{F}$, that is:

• $$\mathsf{A}_\ell$$, the special linear algebra of traceless endomorphisms of an (l+1)-dimensional vector space
• $$\mathsf{B}_\ell$$, the orthogonal algebra of all endomorphisms x of a (2l + 1)-dimensional vector space which satisfy f(x(v), w) = -f(v, x(w)) for the bilinear form f whose matrix is $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_\ell \\ 0 & I_\ell & 0 \end{pmatrix}$$.
• $$\mathsf{C}_\ell$$, the symplectic algebra of all endomorphisms x of a (2l)-dimensional vector space which satisfy f(x(v), w) = -f(v, x(w)) for the bilinear form f whose matrix is $$\begin{pmatrix} 0 & I_\ell \\ -I_\ell & 0 \end{pmatrix}$$.
• $$\mathsf{D}_{\ell \ge 2}$$, the orthogonal algebra which is constructed similarly as $$\mathsf{B}_\ell$$ but with $$s = \begin{pmatrix} 0 & I_\ell \\ I_\ell & 0 \end{pmatrix}$$.

Actually, I don't see any common property of these algebra's, except that we had to prove once that they all satisfy L = [L, L] so I thought I'd use that. But that doesn't really help (e.g. I tried proving that [Rad L, L] = 0 using the Jacobi identity but that just got me 0 = 0. I also tried dividing out something, like Rad L or Z(L) or maybe even Z(L) from Rad(L), but that didn't give anything useful either).

 

[CompuChip]

OK, so I have solved #1 as well, basically by looking at the proof in a set of course notes online, and I must admit it is non-trivial (I would never have thought of it myself, and I wasn't even close.)

As for the remaining problem, I have reduced it to showing that the
$$V_\lambda = \{ v \in V | \phi(z) v = \lambda(z) v \}$$
is completely reducible, where vector space V is the module, phi is the representation as in the question, and lambda runs over the linear forms on Z(L).

Woot, 400th post. That's fast.[/size]

[edit]OK, this was as good as trivial. Of course, the eigenspaces V_\lambda are all one-dimensional, so the proof is as good as finished.
Thanks for responding anyway matt.
Still in time for the deadline tomorrow
[/edit]