MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathfrak{g}$ be a simple complex Lie algebra. Let $\mathfrak{g}\subset\mathfrak{so}(V)$ be an orthogonal irreducible representation. It can be shown that the number of $\mathfrak{g}$-submodules of $V\otimes V$ isomorphic to $\mathfrak{g}$ equals to the number of non-zero labels on the Dynkin diagram defining the representation $\mathfrak{g}\subset\mathfrak{so}(V)$. Since $\wedge^2V\simeq\mathfrak{so}(V)$, one exemplar $\mathfrak{g}$ is included in $\wedge^2V$.

Using the Lie package, I checked many representations of $\mathfrak{g}=\mathfrak{so}(n,C)$, and for all of them all $\mathfrak{g}$ are included in $\wedge^2V$. But for $\mathfrak{g}=\mathfrak{sl}(n,C)$ this is not the case.

Can one say what number of $\mathfrak{g}$-modules isomorphic to $\mathfrak{g}$ are in $\wedge^2V$, and what number of them are in $S^2V$?

share|cite|improve this question
When you say "number of" you mean multiplicity, right? (This is not equivalent to the literal number, as in the cardinality of the set of submodules isomorphic to a given module.) – Qiaochu Yuan Oct 3 '12 at 18:24
Yes, multiplicity. Thank you! – Anton Galaev Oct 3 '12 at 18:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.