4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ "completely factorizable" = "completely reducible" = "direct sum of irreducibles"? $\endgroup$ – darij grinberg Aug 6 '19 at 11:34
  • $\begingroup$ Yes, completely reducible $\endgroup$ – liorz Aug 6 '19 at 11:36
  • 2
    $\begingroup$ 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.) $\endgroup$ – YCor Aug 6 '19 at 11:41
1
$\begingroup$

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!!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why does the non-semisimplicity of $A_K$ imply that $\pi_K$ is not completely reducible? $\endgroup$ – darij grinberg Aug 6 '19 at 16:01
  • $\begingroup$ Why The fact that lie algebra Ak is not semisimple promise that πK is not reducible? $\endgroup$ – liorz Aug 6 '19 at 16:19
  • $\begingroup$ @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.) $\endgroup$ – YCor Aug 6 '19 at 19:03
  • $\begingroup$ Doc, do I have to explain everything? It follows from Nakayama Lemma, I reckon... $\endgroup$ – Bugs Bunny Aug 6 '19 at 19:05
1
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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($\mathfrak{g}$)". $\endgroup$ – YCor Aug 17 '19 at 8:21
  • $\begingroup$ I edited the last sentence. I hope it is clearer now. $\endgroup$ – Thorsten Heidersdorf Aug 17 '19 at 11:23
  • 1
    $\begingroup$ 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.) $\endgroup$ – YCor Aug 17 '19 at 11:28
  • $\begingroup$ There is a very easy explanation: I was careless. I edited the answer one more time. $\endgroup$ – Thorsten Heidersdorf Aug 17 '19 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.