TITLE: Invariants of exterior powers
QUESTION [9 upvotes]: Let $\mathfrak{g}$ be the Lie algebra of $GL(n,\mathbb{R})$. Let $\theta(X) = - X^T$ be the Cartan involution on $\mathfrak{g}$; it induces decomposition as $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ of $+1$-eigenspace $\mathfrak{k}$ and $-1$ eigenspace $\mathfrak{p}$. Then $\mathfrak{k}$ is the Lie algebra of $K = O(n)$. I want to compute the dimensions of $K$-invariants $\textrm{Hom}_K(\wedge^q \mathfrak{p}, \mathbb{C})$, which, I suppose, is equal to $(W/ \mathfrak{k} W)^*$ where $W = \wedge^q \mathfrak{p}$, where $\mathfrak{p}$ is viewed as $\mathfrak{k}$-module by adjoint-representation. Could you someone point a way further?

REPLY [9 votes]: To offer a slightly more geometric viewpoint on the same, the space $\bigoplus_q \mathop{\mathrm{Hom}}_K(\Lambda^q(\mathfrak{p}),\mathbb{C})$, which is the direct sum of all spaces you are considering, is the cohomology of the Lie algebra $\mathfrak{gl}_n$ relative to its subalgebra $\mathfrak{o}_n$, which, by a standard argument, is the same as $H^*(U(n)/O(n))$. It is well known (it really is; if you need some reference, see, for example, page 28 of lecture notes of Haefliger - http://www1.mat.uniroma1.it/people/piazza/HaeDC.pdf) to be the exterior algebra generated by elements $\xi_{4k-3}$, for all $k$ such that $2k-1\le n$. In fact, $\xi_{d}$ is the anti-symmetrisation of the trace of product of $d$ matrices. This confirms the very nice answer given by Abdelmalek Abdesselam.