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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 $S^2V$?

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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

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