# Complete reducibility and field extension

Let $$\pi$$ be a representation of a Lie algebra $$L$$ in a finite-dimensional linear space $$V$$ over the field $$F$$. Let $$K$$ be a field extension of $$F$$. Let $$\pi_K=\pi\otimes K$$ be the corresponding representation of $$L_K$$ in $$V_K$$. Why if $$\pi_K$$ is completely reducible then $$\pi$$ must be completely reducible too?

• "completely factorizable" = "completely reducible" = "direct sum of irreducibles"? – darij grinberg Aug 6 '19 at 11:34
• Yes, completely reducible – liorz Aug 6 '19 at 11:36
• It boils down to a question on (associative unital) subalgebras $A$ of $M_n(F)$: is it true that if $A_K$ is completely reducible (in the sense that it makes $K^n$ a completely reducible $A_K$-module), then so is $A$? (The converse is true for separable extensions and not in general.) – YCor Aug 6 '19 at 11:41

## 2 Answers

Amazingly, I cannot see an elementary solution. I believe there should be one.

Otherwise, one can expand the comment of YCor with some standard Ring Theory. Let $$A$$ be the image of $$U(L)$$ in $$End_F(V)$$. Then $$A_K = A \otimes_F K$$ is the image of $$U(L_K)$$.

Now suppose $$\pi$$ is not completely reducible. This means that $$A$$ is not a semisimple algebra. This means that its Jacobson radical $$J$$ is non-zero. Since $$A$$ is finite-dimensional, the Jacobson is nilpotent: $$J^n=0$$ for some $$n>1$$. Then $$J_K$$ is a nilpotent ideal of $$A_K$$ so that $$A_K$$ is not a semisimple algebra. Hence, $$\pi_K$$ is not completely reducible.

PS Let us dig into the last implication. Suppose $$\pi_K$$ is completely reducible but the Jacobson $$J(A_K)\neq 0$$. As $$V_K$$ is a faithful $$A_K$$-module, $$J(A_K)V_K\neq 0$$. By complete reducibility there exists a $$A_K$$-direct complement $$V_K = DirectComplmnt \oplus J(A_K)V_K$$. By Nakayama Lemma, $$V_K = DirectComplmnt$$. Contradiction!!

• Why does the non-semisimplicity of $A_K$ imply that $\pi_K$ is not completely reducible? – darij grinberg Aug 6 '19 at 16:01
• Why The fact that lie algebra Ak is not semisimple promise that πK is not reducible? – liorz Aug 6 '19 at 16:19
• @darijgrinberg this is standard. Here it can be seen because a nilpotent ideal acts trivially on any irreducible representation, and hence on any completely reducible representation. (If necessary, see Bourbaki, Algebra Chap VIII, §10.2.) – YCor Aug 6 '19 at 19:03
• Doc, do I have to explain everything? It follows from Nakayama Lemma, I reckon... – Bugs Bunny Aug 6 '19 at 19:05

It is worth pointing out that there are general statements that semisimplicity can be checked over an algebraic closure. The following statement is used:

Let $$k$$ be a field and $$A$$ a $$k$$-linear abelian category such that $$End(X)$$ is finite dimensional for every object $$X$$ in $$A$$. Then $$A$$ is semisimple if and only if $$End(X)$$ is a semisimple $$k$$-algebra for all $$X$$ in $$A$$.

(a proof can be e.g. found in Proposition 5.14 of Milne's lecture notes on Lie algebras: https://www.jmilne.org/math/CourseNotes/LAG.pdf)

Furthermore Lemma 5.11 in Milne's notes says that

Let $$A$$ be a $$k$$-algebra. If $$K \otimes_k A$$ is semisimple for some field $$K$$ containing $$k$$, then $$A$$ is semisimple; conversely, if $$A$$ is semisimple, then $$K \otimes_k A$$ is semisimple for all fields $$K$$ separable over $$k$$.

Edited after YCor's comment: If one applies these statements to the special case of the category $$A = Rep(\mathfrak{g})$$ over a field $$k$$, one gets that if $$Rep(\mathfrak{g}_K)$$ is semisimple for some extension field $$K$$ of $$k$$, then $$Rep(\mathfrak{g})$$ is semisimple. This answers the initial question. If $$V$$ is semisimple in $$Rep(\mathfrak{g})$$, then $$V \otimes K$$ might not be semisimple, but this holds if either $$K$$ is separabel over $$k$$ or if $$dim(V) < p$$ where $$char(k) = p$$ (see the comment in Milne's notes - this is originally due to Serre).

• I don't know what is meant by your last sentence "This allows to pass wlog to the algebraic closure for categories such as Rep($\mathfrak{g}$)". – YCor Aug 17 '19 at 8:21
• I edited the last sentence. I hope it is clearer now. – Thorsten Heidersdorf Aug 17 '19 at 11:23
• What I don't see is why Rep($\mathfrak{g}$) semisimple over $k$ implies Rep($\mathfrak{g}$) semisimple over $K$ when $K$ is a nonseparable extension of $k$. (Although, in positive characteristic, it might be true that both hold iff the only finite-dimensional quotient of $\mathfrak{g}$ is $\{0\}$, but this seems to be a quite orthogonal argument.) – YCor Aug 17 '19 at 11:28
• There is a very easy explanation: I was careless. I edited the answer one more time. – Thorsten Heidersdorf Aug 17 '19 at 16:19