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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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I think the action is irreducible. The symmetric traceless matrices are the adjoint representation. – Jack Schmidt Feb 18 '11 at 21:02
What confuses me, though, is that $M$ has dimension $\frac 1 2 n (n+1)-1$, right? For example, if $n=10$, we get 45 versus 54. – Hans Feb 18 '11 at 21:33
You're right. And the difference in dimension is n-1, which is a little irritating. Could the rest really be a bunch of trivial reps? I must have something wrong. – Jack Schmidt Feb 18 '11 at 21:41
Ok, dumb question: g=[0,1,0;-1,0,0;0,0,0] is in so(n), and m=[1,0,0;0,0,0;0,0,-1] is in M, but g*m=[0,0,0;-1,0,0;0,0,0] is not in M, right? – Jack Schmidt Feb 18 '11 at 21:53
But [g,m]=gm-mg is... – Hans Feb 18 '11 at 22:24

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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To calculate the weights for all symmetric matrices is a good idea. However, in the case n=3 I find the weights to be $\pm 1$ (each of dimension 2) and 0 (of dimension 2). In dimension 4, I find them to be 2a, 2b,a+b,a-b and their negatives and 0 (of multiplicity 3). This suggests a decomposition 2a + 2b + (a+b) + trivial. – Hans Feb 19 '11 at 7:29
@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. – Jason DeVito Feb 19 '11 at 17:31

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