4
$\begingroup$

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.

  • 0
    I think the action is irreducible. The symmetric traceless matrices are the adjoint representation.2011-02-18
  • 0
    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 54.2011-02-18
  • 0
    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.2011-02-18
  • 0
    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?2011-02-18
  • 0
    But [g,m]=g*m-m*g is...2011-02-18
  • 0
    Jack: I doubt that the trivial representation is anyhwere in $M$. This would mean that there are nontrivial quadratic forms with trace $0$ that are invariant under conjugation with any matrix from SO. This sounds strange, doesn't it?2011-02-18
  • 0
    I was using matrix multiplication to calculate the eigenvalues, and just stopped once enough matched. Apparently that's not how so(n) acts.2011-02-18
  • 0
    Yeah, it acts through the commutator (because SO acts by conjugation).2011-02-18
  • 0
    MathOverflow? This seems a legitimate research question.2011-02-18

1 Answers 1