6
$\begingroup$

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, and let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ and $\phi_{2}:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ be complex Lie algebra morphisms. By composing $\phi_1$ and $\phi_2$ with the adjoint representation of $\mathfrak{g}$, we obtain two representations of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{g}$. My question is then the following.

Question: If these two $\mathfrak{sl}_2(\mathbb{C})$-representations are isomorphic, does it follow that $\phi_1$ and $\phi_2$ are related by an element of the Lie algebra automorphism group $\text{Aut}(\mathfrak{g})$?

To provide some context, suppose that $\mathfrak{g}=\mathfrak{so}_{4n}(\mathbb{C})$. If $\lambda$ is a partition of $4n$ having only even parts with each part appearing an even number of times, then $\lambda$ corresponds to exactly two distinct nilpotent orbits, $\mathcal{O}_1$ and $\mathcal{O}_2$, in $\mathfrak{so}_{4n}(\mathbb{C})$. Now, let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ and $\phi_2:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ be Lie algebra maps satisfying $\phi_1(e)\in\mathcal{O}_1$ and $\phi_2(e)\in\mathcal{O}_2$. As discussed above, each map gives a representation of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{so}_{4n}(\mathbb{C})$. These two representations are actually isomorphic. While $\phi_1$ and $\phi_2$ cannot be related by an inner automorphism of $\mathfrak{so}_{4n}(\mathbb{C})$ (as $\mathcal{O}_1\neq\mathcal{O}_2$), they are nevertheless related by a Lie algebra automorphism.

$\endgroup$
3
$\begingroup$

REVISED VERSION: On further reflection, I think the answer to your question is always "yes" (if $\mathfrak{g}$ is simple, to avoid complications of the type Dave indicates). At first I was confused by the example discussed in the question, but I think the main issue is how the automorphism group interacts with Dynkin diagrams of nilpotent orbits.

In the classical Dynkin-Kostant treatment of nilpotent orbits in a simple Lie algebra $\mathfrak{g}$, the basic strategy is to embed a (nonzero) nilpotent element in a copy of $\mathfrak{sl}_2(\mathbb{C})$. Such an embedding isn't unique, but there is a resulting bijection between conjugacy classes of nilpotents (under the adjoint group of $\mathfrak{g}$) and conjugacy classes of such subalgebras. In turn, the adjoint action of the subalgebra leads to a decorated Dynkin diagram (with vertices labelled $0, 1, 2$) which determines uniquely the given nilpotent orbit. (Some references to textbooks by Carter and Collingwood-McGovern which include Dynkin diagrams are given in my old notes here.)

In your type $D$ example, there are typically pairs of orbits interchanged by an outer automorphism of $\mathfrak{g}$ that comes from a graph automorphism. These orbits have many common properties, and corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ do act equivalently on $\mathfrak{g}$ even though they lead to distinct Dynkin diagrams. The key fact is that these Dynkin diagrams just involve a permutation of labels induced by the outer automorphism. Only in such limited cases can two subalgebras of type $\mathfrak{sl}_2(\mathbb{C})$ act equivalently in the adjoint representation of $\mathfrak{g}$: this is clear from the determination of Dynkin diagrams for each $\mathfrak{g}$.

[Concerning terminology, it's fairly conventional to call two representations equivalent but the associated modules isomorphic. The choice of either representation or module language is usually optional, at least in finite dimensional cases.]

$\endgroup$
  • 1
    $\begingroup$ Jim, the representations afforded by taking $\phi_1$ and $\phi_2$ to be root embeddings won't usually be isomorphic. For example, in type $B_2$, the adjoint representation for a short root decomposes into three isomorphic 3-dimensional irreducibles (plus a 1-dimensional trivial), but the adjoint representation for a long root has a 3-dimensional, two 2-dimensionals, and three 1-dimensionals. (Well, something like that, anyway.) $\endgroup$ – Dave Witte Morris Mar 25 '15 at 22:44
  • $\begingroup$ @Dave: I was just having some second thoughts, so I edited my answer. I may still be oversimplifying. $\endgroup$ – Jim Humphreys Mar 25 '15 at 22:48
  • 1
    $\begingroup$ When $\mathfrak{g}$ is simple, and we consider only root embeddings, I think your analysis is correct, except that we first have to eliminate cases where the representation corresponding to a long root is not isomorphic to the representation corresponding to a short root. Extrapolating from $B_2$, I would think they are usually not isomorphic. $\endgroup$ – Dave Witte Morris Mar 25 '15 at 22:54
  • 1
    $\begingroup$ Jim, when you say the answer is always "yes", do you mean for the simple case? It's easy to cook up counterexamples for the non-simple case, because we just have to take make sure every weight of $\mathfrak{sl}_2$ occurs the same number of times in the two representations. $\endgroup$ – Dave Witte Morris Mar 26 '15 at 20:52
  • $\begingroup$ @Dave: Yes, I'm only considering simple Lie algebras. I've also tried to clarify my answer yet again but will give up edits at this point. $\endgroup$ – Jim Humphreys Mar 26 '15 at 21:14
8
$\begingroup$

No, here is a counterexample.

Let $\mathfrak{g}_1 = \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C})$, and let $\phi_1 \colon \mathfrak{sl}_2(\mathbb{C}) \to \mathfrak{g}_1$ be the diagonal embedding, so $\phi(x) = (x,x,x)$. Then the representation obtained by composing $\phi_1$ with the adjoint representation is the direct sum of three $3$-dimensional irreducible modules.

Let $\mathfrak{g}_2 = \mathfrak{so}_5(\mathbb{C})$, and let $\phi_2$ be the embedding onto $\langle \mathfrak{u}_\alpha, \mathfrak{u}_{-\alpha} \rangle$, where $\alpha$ is a short root. Then the representation obtained by composing $\phi_2$ with the adjoint representation is the direct sum of a trivial representation and precisely three $3$-dimensional irreducible modules.

Now, if we let $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$, then we can think of $\phi_1$ and $\phi_2$ as homomorphisms into $\mathfrak{g}$ and, from the above, we see that their compositions with the adjoint representation are isomorphic. However, the image of $\phi_2$ is contained in the unique ideal of $\mathfrak{g}$ that is isomorphic to $\mathfrak{so}_5(\mathbb{C})$, but the image of $\phi_1$ is not contained in this ideal. So no element of $\mathrm{Aut}(\mathfrak{g})$ can take $\phi_1$ to $\phi_2$.

$\endgroup$
  • $\begingroup$ I am probably misreading something but it sounds like you say that if V denotes the direct sum of three 3-dimensional irreducible $\mathfrak{sl}_2$-modules then V is isomorphic to the direct sum of a trivial representation and V itself. Given that the trivial representation is one-dimensional, how is this possible? $\endgroup$ – Vincent Mar 26 '15 at 20:07
  • 1
    $\begingroup$ What I mean is that the representations of $\phi_1$ and $\phi_2$ on $\mathfrak{g}$ both contain three 3-dimensional irreducibles, and the rest of each representation is a big trivial module. Specifically, each is the direct sum of three 3-dimensional irreducibles and ten 1-dimensional trivial modules. So the two representations are isomorphic. $\endgroup$ – Dave Witte Morris Mar 26 '15 at 20:19
  • $\begingroup$ Thank you! This is exceedingly helpful. In light of the two given answers, I should probably require $\mathfrak{g}$ to be simple. $\endgroup$ – Peter Crooks Mar 27 '15 at 13:57

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.