# Are finite-dimensional real representations of semisimple real Lie algebras completely reducible?

Suppose $$\mathfrak{g}$$ is a real form of a semisimple Lie algebra $$\mathfrak{g}_\mathbb{C} = \mathfrak{g} \otimes_\mathbb{R} \mathbb{C}$$. Then we have the following:

• There is an equivalence of monoidal categories between the category of finite-dimensional complex representations of $$\mathfrak{g}$$ and the category of finite-dimensional complex representations of $$\mathfrak{g}_\mathbb{C}$$.
• One can classify the irreducible real representations of $$\mathfrak{g}$$. These are either restrictions to $$\mathfrak{g}$$ of irreducible complex representations of $$\mathfrak{g}_\mathbb{C}$$ (which remain irreducible over $$\mathfrak{g}$$) or real forms of irreducible complex representations of $$\mathfrak{g}_\mathbb{C}$$ (in this case there is a real structure on the underlying space of the representation that commutes with the action of $$\mathfrak{g}$$). See, for example, Theorem 1 on page 65 of Onishchik, Lectures on Real Semisimple Lie Algebras.

What I'd like to know is if the category of finite-dimensional real representations of $$\mathfrak{g}$$ is semisimple. In other words, is every finite-dimensional real representation of $$\mathfrak{g}$$ completely reducible? Despite spending quite some time searching through the literature, I can't seem to find the answer to this question.

• Yes, this is more generally true for arbitrary fields of char zero. And is easy to derive from the alg. closed case. And actually follows from algebra: a rep is semisimple iff the subalgebra it generates is a semisimple algebra. And being a semisimple finite-dim algebra (in char zero) is invariant by field extensions (if $L$ is an extension of $K$, and $A$ is a finite-dim $K$-algebra, $A$ is a semisimple $K$-algebra iff $A\otimes_K L$ is a semisimple $L$-algebra.
– YCor
Sep 27, 2022 at 19:12
• Yes. In fact I think you don't even need characteristic zero; a semisimple $K$-algebra might become non-semisimple upon tensoring with $L$, but the reverse cannot happen. A short exact sequence of $A$-modules must split if it yields a split exact sequence of $(A\otimes_KL)$-modules. The point is that a system of linear equations over $K$ must have a solution over $K$ if it has a solution over $L$. Sep 27, 2022 at 20:56

## 1 Answer

As pointed out by YCor in the comments, the answer is Yes. A reference has been pointed out to me: Chapter III, Section 7, Theorem 8 in Jacobson's book Lie algebras.