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Let $\mathfrak {so}_{n}$ denote the skew-symmetric complex $n \times n$-matrices and let $M$ denote the symmetric $n \times n$-matrices of trace 0.

As I understand, $M$ is a module over $\mathfrak {so}_n$. What then is its decomposition into irreducibles?

The standard representation of $\mathfrak {so}_n$ has dimension $n$, the adjoint representation dimension $\frac 1 2 n \cdot (n-1)$ and there are two spin representations of small dimension. But I don't see a way how these, together with the trivial representation, should add up to the dimension of $M$.

Edit: This comes from trying to understand the Cartan decomposition $\mathfrak g=\mathfrak k \oplus \mathfrak p$, where $[\mathfrak k,\mathfrak p] \subseteq \mathfrak p$, cf. the wikipedia article on Cartan decomposition. As the associated symmetric should be irreducible, the representation should be irreducible, but my numbers just don't add up.

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    MathOverflow? This seems a legitimate research question.2011-02-18

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I just worked out the weights by hand for the case $n=3$, except I dropped the trace = 0 condition (since it allows for a nicer basis to do computations in).

In this case, we find that the 6d representation is 5-d irreducible rep + Trivial. Of course, the trivial rep is the scalar multiples of the identity, and the complement is the traceless matrices, so the action on $M$ is the unique irreducible 5-d rep.

Without computing higher things by hand, I can't say more than this - but it at least shows you that neither the standard rep nor the adjoint need to show up.

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    @Hans: For $n=3$, I got weights $0, 0, \pm 1, \pm 2$ (there should be 6 of them because the vector space is dim 6). When you say "In dimension 4,...", did you mean when $n=4$? I didn't work out anything for that case.2011-02-19