Let $G=GL_n$ be the general linear group (let's say over an algebraically closed field of char $=0$). Let's denote as $T$ the torus of diagonal matrices: is there an explicit description of the invariant functions $$\mathbb{C}[G \times G]^T $$ where $T$ acts by simultaneous conjugation?
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1$\begingroup$ This ring contains the ring of invariant functions under the simultaneous action of the full conjugation by $G$, and this ring is generated by the traces of words in the two "letters" and their inverses (I believe this is a result of Procesi). Perhaps your ring is similarly generated by the individual diagonal entries of these words (in addition to the trace, which is the sum of the diagonal entries). $\endgroup$– Jason StarrMay 17, 2022 at 13:42
1 Answer
Call the two matrices ${}_1A$ and ${}_2A$.
Your ring can be expressed by taking the $T$-invariants of the free ring in $2n^2$ variables $\mathbb C[{}_kA_{ij}]_{1\leq i,j\leq n, 1\leq k \leq 2}$ and then inverting $\det {}_1 A$ and $\det {}_2 A$.
The $T$-invariants of the free ring are the ring of functions on an affine toric variety. It has an explicit combinatorial structure. A basis for it is given by the $T$-invariant monomials.
A generating set is given by the monomials that can't be decomposed into two monomials. These have the following form: For $k \leq n$ a natural number, $i_1,\dots ,i_k$ distinct indices from $1$ to $n$, and $\epsilon_1,\dots, \epsilon_k$ values from $1$ to $2$, the element $$ \prod_{j=1}^k {}_{\epsilon_j}A_{i_j i_{j+1}}$$ is such a monomial, and it's not too hard to see how they have this form.
To see the ring has this form, simply note that any function on $G \times G$ is a polynomial in the entries divided by some power of the two determinants, and since the determinants are $T$-invariant, any invariant polynomial must be an invariant polynomial in the entries divided by some power of the two determinants.