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Let $\sigma$ a finite-order automorphism of a finite-dimensional complex simple Lie algebra $g$. Denote the order of sigma by $k$ and fix a $k^{th}$ root of unity $\omega$. It is well know that $k=1,2 \text{ or } 3$ and that $$g=\oplus_{j\in \mathbb Z_k} g_j$$ where $g_j=\{ x \in g \mid \sigma(x)=\omega^jx\}.$ Moreover, $g_0$ is a simple Lie algebra.

QUESTION: Let $\lambda$ a weight of $g$ and $V(\lambda)$ the irreducible representation of weight $\lambda$. Denote by $V(\lambda)_{g_0}$ the $g_0$-module obtained from $V(\lambda)$ by restricting the action of $g$ to $g_0$. Is $V(\lambda)_{g_0}$ reducible as a $g_0$-module for all $\lambda$?


Note: The results mentioned can be found in the Kac book.

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Isn't that restriction on the order $k$ true for $\sigma$ in the group of outer automorphisms? –  Mariano Suárez-Alvarez Mar 30 '11 at 5:52
@Mariano I am sorry, I didn't understand your question. Can you redo it? Thanks, –  Angelo Mar 30 '11 at 13:59
As Mariano points out, the work of Kac starts with an outer (graph) automorphism of order at most 3 and the fixed point subalgebra. The "inner" automorphisms given by the adjoint group action are more varied and don't involve these types of gradings. Aside from that, to ask for reducibility of all irreducible modules is asking for too much even when $g_0$ is a proper subalgebra. At some point you have to look at the details of each type $A, D, E_6$ (with rank at least 2). –  Jim Humphreys Mar 31 '11 at 12:39

1 Answer 1

Yes if $k=1$. No if $k>1$; the adjoint representation is a counterexample.

For the edited question: isn't the vector representation of SL(n) (and its dual) a counterexample for $k=2$?

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@Bruce: Ok, I agree that it may be the answer. But are you saying the answer is negative for all $g$ of type $A_n$, $D_n$ and $E_6$? (all types that admit a non trivial automorphism) Please, clarify me why! I guess that a bit of calculation is necessary, is not it? For instance, if $g=sl_3$ what we have ? –  Angelo Mar 30 '11 at 13:57
I think I misunderstood the question. I have taken the liberty of making an edit which I hope makes it clearer. –  Bruce Westbury Mar 31 '11 at 7:59
@Angelo: I don't understand what "universal" means here. –  Jim Humphreys Mar 31 '11 at 12:42
@Angelo: You started with "the" irreducible module (unique up to isomorphism), so "universal" is meaningless here no matter whether modules are finite or infinite dimensional. Universal highest weight modules (= Verma modules) in the BGG category are infinite dimensional and rarely irreducible. Also, "categorical" here depends on which category of modules you are working in. –  Jim Humphreys Apr 1 '11 at 13:18
@Jim: I wrote something wrong in my last comment, sorry you and Bruce!. I was talking only about irreducible highest weight modules and I used the term UNIVERSAL. The term universal is valid and important in another context that was on my mind, about representations of loop algebras. On the other hand, I know these modules are infinite dimensional and rarely irreducible. But I disagree that the question is out of context. –  Angelo Apr 3 '11 at 4:06

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