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.

  • $\begingroup$ I think the action is irreducible. The symmetric traceless matrices are the adjoint representation. $\endgroup$ – Jack Schmidt Feb 18 '11 at 21:02
  • $\begingroup$ 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. $\endgroup$ – Hans Feb 18 '11 at 21:33
  • $\begingroup$ 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. $\endgroup$ – Jack Schmidt Feb 18 '11 at 21:41
  • $\begingroup$ 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? $\endgroup$ – Jack Schmidt Feb 18 '11 at 21:53
  • $\begingroup$ But [g,m]=gm-mg is... $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Hans Feb 19 '11 at 7:29
  • $\begingroup$ @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. $\endgroup$ – Jason DeVito Feb 19 '11 at 17:31

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.