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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.

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    $\begingroup$ 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. $\endgroup$ – Victor Petrov Dec 11 '19 at 16:01
  • $\begingroup$ almost a duplicate of mathoverflow.net/questions/121715/… $\endgroup$ – Abdelmalek Abdesselam Dec 11 '19 at 18:01
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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.

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  • $\begingroup$ 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! $\endgroup$ – user44191 Dec 11 '19 at 22:03

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