# Complete reducibility of a field extension of an lie algebra representation

Let $\mathfrak{g}$ be a lie algebra over a field $k$ with characterstic $0$ and $k\subset k'$ a finite field extension. Suppose $\mathfrak{g}\otimes k'$ has the property, that all finite dimensional representations are completely reducible.

Does $\mathfrak{g}$ then also has this property?

I am especially interested in the case $k=\mathbb{R}$ and $k'=\mathbb{C}$.

Edit:

In http://www.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf Proposition 5.3 on page 98 seems to give a proof for the $\mathbb{R}\subset\mathbb{C}$-case, but I don't really see why the argument works: If $$\pi\colon\mathfrak{g}\rightarrow\mathfrak{gl}(V)$$ is a real representation of a real Lie-Algebra, he seems to argue, that if the induced complex representation $$\pi_\mathbb{C}\colon\mathfrak{g}\rightarrow\mathfrak{gl}(V\otimes\mathbb{C})$$ is completely reducible, then also $\pi$ is. But how does that work? How does a decomposition of invariant subspaces of $V\otimes\mathbb{C}$ give me a decomposition of $V$? The subspaces of $V\otimes\mathbb{C}$ could lay "crooked", don't they?

• I agree that the argument in the PDF you linked is unclear, and I think you would do everyone a favor if you notify the author. Feb 4 '14 at 21:23

## 1 Answer

EDIT: I have got your question wrong -- sorry! I have proved the converse of your statement. (It is now Theorem 2 below.) Here is your claim:

Theorem 1. Let $k$ be a field of characteristic $0$. Let $\mathfrak g$ be a $k$-Lie algebra. Let $k'/k$ be a finite extension of fields. Assume that every finite-dimensional $k'\otimes_k \mathfrak g$-module (where $k'\otimes_k \mathfrak g$ is considered as a $k'$-Lie algebra) is completely reducible. Then, every finite-dimensional $\mathfrak g$-module is completely reducible.

Let me first show a converse of Theorem 1:

Theorem 2. Let $k$ be a field of characteristic $0$. Let $\mathfrak g$ be a $k$-Lie algebra. Assume that every finite-dimensional $\mathfrak g$-module is completely reducible. Let $k'/k$ be a finite extension of fields. Then, every finite-dimensional $k'\otimes_k \mathfrak g$-module (where $k'\otimes_k \mathfrak g$ is considered as a $k'$-Lie algebra) is completely reducible.

Proof of Theorem 2.

1st step: Exercise 7.3.5 in Weibel's An Introduction to Homological Algebra (applied at degree $1$) that $\mathrm{Ext}^1_{U\mathfrak g}\left(M, N\right) \cong H^1\left(\mathfrak g, \mathrm{Hom}_k\left(M, N\right)\right)$ for any two finite-dimensional $\mathfrak g$-modules $M$ and $N$ (where $U\mathfrak g$ denotes the universal enveloping algebra of $\mathfrak g$, and where $H^1$ denotes the first Lie algebra cohomology). Hence, all finite-dimensional $\mathfrak g$-modules are completely reducible if and only if every finite-dimensional $\mathfrak g$-module $P$ satisfies $H^1\left(\mathfrak g, P\right) = 0$.

2nd step: But every finite-dimensional $\mathfrak g$-module $P$ satisfies $H^1\left(\mathfrak g, P\right) \cong \operatorname*{Der}\left(\mathfrak g, P\right) / \operatorname{Der}_{\operatorname*{Inn}}\left(\mathfrak g, P\right)$ (by Theorem 7.4.7 in loc. cit.). Hence, $H^1\left(\mathfrak g, P\right) = 0$ is equivalent to $\operatorname*{Der}\left(\mathfrak g, P\right) = \operatorname{Der}_{\operatorname*{Inn}}\left(\mathfrak g, P\right)$.

But the statement that $\operatorname*{Der}\left(\mathfrak g, P\right) = \operatorname{Der}_{\operatorname*{Inn}}\left(\mathfrak g, P\right)$ for a fixed $\mathfrak g$-module $P$ is invariant under base extension (more precisely, under replacing $k$, $\mathfrak g$ and $P$ by $k'$, $k'\otimes_k \mathfrak g$ and $k'\otimes_k P$) (because this statement can be formulated as the image of one linear map being contained in the kernel of another linear map, and the matrices representing these two linear maps can be taken to be the same over $k$ and over $k'$). Hence, if every finite-dimensional $\mathfrak g$-module $P$ satisfies $H^1\left(\mathfrak g, P\right) = 0$, then every finite-dimensional $\mathfrak g$-module $P$ satisfies $H^1_{k'}\left(k'\otimes_k \mathfrak g, k'\otimes_k P\right) = 0$ (where the subscript $k'$ in $H^1_{k'}$ means that Lie algebra cohomology -- as well as Lie algebras and modules -- are taken over the ground ring $k'$).

3rd step: But we are not done yet. We need to prove (according to the 1st step) that every $k'\otimes_k \mathfrak g$-module $Q$ satisfies $H^1_{k'}\left(k'\otimes_k \mathfrak g, Q\right) = 0$. If every finite-dimensional $k'\otimes_k \mathfrak g$-module had the form $k'\otimes_k P$ for some finite-dimensional $\mathfrak g$-module $P$, then this would follow from the 2nd step. But this is not the case. However, every finite-dimensional $k'\otimes_k \mathfrak g$-module $Q$ is isomorphic to a direct addend of the finite-dimensional $k'\otimes_k \mathfrak g$-module $k'\otimes_k P$ for some $\mathfrak g$-module $P$ (namely, for $P = Q$ (considered as a $\mathfrak g$-module by restriction) -- while this looks like a foul trick, it is a valid proof!). This completes the proof, because $H^1_{k'}$ is additive in the second argument.

Proof of Theorem 1 in the general case. Follow the proof of Theorem 2 "backwards", ignoring the 3rd step.

• Puh, I hoped to get a more lie-theoretic proof of the $\mathbb{R}\subset\mathbb{C}$-case and less homological algebra. I made an edit to my question. I would be thankful if you would take a look.
– Tina
Feb 4 '14 at 20:50
• Sorry, I just realized that I proved the converse of your question. I've fixed this in the above post now. Now it doesn't address your issue with homological algebra -- I don't know how to make it less complicated, or else I would have done it... Feb 4 '14 at 21:22