# Invariants for $SO_n \backslash \mathfrak{gl}_n / SO_n$

Is there a nice theorem about the algebra of invariants $$\mathbb{C}[\mathfrak{gl}_n]^{SO_n \times SO_n}$$, where the action is by left and right multiplication? I'm hoping for something along the lines of the Chevalley restriction theorem, with a list of generators for the algebra being a nice bonus if possible (the subalgebra should have dimension $$n$$; I'm hoping it's free with $$n$$ generators).

For example, for $$n = 2$$, the algebra of invariants is generated by the determinant and the sum of squares of components.

My slightly-educated guess is that the theorem should be approximately:

$$\mathbb{C}[\mathfrak{gl}_n]^{O_n \times O_n} \simeq \mathbb{C}[\mathfrak{h}]^{W'}$$, where $$W' = W \rtimes C_2^n$$, $$C_2$$ is the cyclic group of order 2, the action of $$W = S_n$$ on $$C_2^n$$ comes from permutation, and the action of $$C_2^n$$ on $$\mathfrak{h}$$ is by negation of a coordinate. Generators for the latter algebra would be symmetric polynomials on the squares of coordinates in $$\mathfrak{h}$$, with degrees $$2, 4, 6, ... 2n$$. Then $$\mathbb{C}[\mathfrak{gl}_n]^{SO_n \times SO_n} = \mathbb{C}[\mathfrak{gl}_n]^{O_n \times O_n}[\text{det}]$$; generators would be the same, except the generator of highest degree (degree $$2n$$) would be replaced with the determinant.

• The main point why Chevalley theorem is true is that any closed (i.e. corresponding to a semisimple element) orbit under the adjoint action meets the maximal torus. In your case you can use Cartan decomposition instead: $G=KAK$, where $K=G^\theta$ and $A$ is a $\theta$-split torus for an involution $\theta$. So I bet that your guess is true. – Victor Petrov Dec 11 '19 at 16:01
• almost a duplicate of mathoverflow.net/questions/121715/… – Abdelmalek Abdesselam Dec 11 '19 at 18:01

Let $$X$$ denote the $$n \times n$$ complex matrices and let $$D$$ be the diagonal $$n \times n$$ complex matrices. We first claim that a generic matrix in $$X$$ factors as $$U\Sigma V$$ with $$U$$ and $$V \in SO(n)$$ and $$\Sigma \in D$$. Proof: Let $$\mu$$ be the multication map $$SO(n) \times D \times SO(n) \to X$$. Then it is not bad to compute that $$D\mu$$ is an isomorphism $$TSO(n) \oplus TD \oplus TSO(n) \to TX$$ at $$(\mathrm{Id}, \mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n),\ \mathrm{Id})$$. So the image contains an analytically open set around such a $$\mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n)$$; the image is also constructible by Chevalley's theorem, so the image contains a Zarsiki open set around such a $$\mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n)$$.
So, an $$SO(n) \times SO(n)$$ invariant function is determined by its restriction to $$D$$. Its restriction to $$D$$ must be invariant for the subgroup of $$SO(n) \times SO(n)$$ which takes $$D$$ to itself, which can be described as the Coxeter group of type $$D_n = S_n \ltimes \{ \pm 1 \}^{n-1}$$: We can permute the $$\sigma$$ in any order, and can multiply and even number of them by $$-1$$.
The invariants for this action are known to be generated by the polynomials $$e_1(\sigma_1^2, \ldots, \sigma_n^2)$$, $$e_2(\sigma_1^2, \ldots, \sigma_n^2)$$, ..., $$e_{n-1}(\sigma_1^2, \ldots, \sigma_n^2)$$, $$\sigma_1 \sigma_2 \cdots \sigma_{n}$$. Since you have already given explicit formulas for invariants of $$n \times n$$ matrices which restrict to these on the diagonal matrices, your formulas are generators as well. $$\square$$
When we see the Coxeter group $$S_n \ltimes \{ \pm 1 \}^{n-1}$$, we should expect its partner the Lie Group $$SO(2n)$$ to be close by. Indeed, consider $$SO(2n)$$ acting by conjugation on $$(2n) \times (2n)$$ skew symmetric matrices. The invariants for that action are known by the Chevalley restriction theorem. A generic skew symmetric matrix can be conjugated to a matrix of the form $$\left[ \begin{smallmatrix} 0 & M \\ -M^T & 0 \end{smallmatrix} \right]$$, and your description is the $$SO(2n)$$ invariants restricted to matrices of this block form.
• I don't think I did give explicit formulae for any non-diagonal matrices, but if I'm not mistaken, it's not hard: let $f_i(A) = a_i(A^\top A)$, where $a_i$ is the $i$th coefficient of the characteristic polynomial. Thanks! – user44191 Dec 11 '19 at 22:03