Purely algebraic proof for unitarizability of representations of a compact real semisimple Lie algebra

It is well-known that any finite-dimensional complex representations of a compact real semisimple Lie algebra are unitarizable.

We can prove this from the fact that every finite-dimensional representation of a compact group is unitarizable by averaging with a Haar measure.

My questions is: Is there a "purely-algebraic" proof of this statement without using any properties of Lie groups?

Does anyone know a reference or a idea?

I'm keeping in mind that Weyl used these facts to show complete reducibility of representations of complex semisimple Lie algebras and now algebraic proofs of complete reducility are known.

• What do you mean by your first "well-known" statement? The obvious $n$-dimensional complex representation of $\mathfrak{sl}_n(\mathbf{R})$ is not unitarizable, if I understand correctly the meaning of the latter word. – YCor Dec 7 '16 at 6:08
• The "well-known" statement is true for a compact real semisimple Lie algebra and false in other cases. – Victor Protsak Dec 7 '16 at 7:02
• Sorry, I forgot to write "compact". I edited the question. – mizuki Dec 7 '16 at 7:12
• The question looks very hard to answer when you specify "without using any properties of Lie groups". (Note that it's hard to find an algebraic method of proof for complete reducibility in this situation which doesn't rely on passage to Lie algebras, which is a basic aspect of Lie group theory.) – Jim Humphreys Dec 7 '16 at 16:32
• Am I correct to say that you are asking whether the geometric and analytical properties can be "hidden" in the construction of the Killing form? That is, whether it is possible to proceed from the Killing form and the Lie algebra structure to a unitary structure on an arbitrary representation? – user44191 Dec 7 '16 at 21:34

Why not, doc? Take a unitary representation $V$ of $G$. Its tensor power $T^nV$ is unitary as well via the obvious form $$<a\otimes b\otimes \ldots , a^\prime \otimes b^\prime \ldots> = <a,a^\prime><b,b^\prime> \ldots$$ Now the resulting module of the Schur functor $S^\lambda (V)$ is a submodule of $T^n V$, hence, unitary.
You can get all simple modules from a few modules applying Schur functors. All you need to do is to exhibit unitary forms on, excuse my French, Karoubian generators of the tensor category of $G$-modules.