Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n

Question

Let $M_n$ be the set of $n$-by-$n$ matrices with complex entries, viewed as a variety over $k=\mathbb{C}$. Equip $M_n$ with the conjugation action of $\mathrm{GL}(n)=\mathrm{GL}(n,\mathbb{C})$. Consider $A:=\mathrm{Mor}^{\mathrm{GL}(n)}(M_n \oplus M_n, M_n)$, the set of $\mathrm{GL}(n)$-equivariant maps (of algebraic varieties) $M_n \oplus M_n \to M_n$, where $GL(n)$ acts diagonally on $M_n \oplus M_n$. Then the (standard) algebra structure of $M_n$ (with multiplication given by matrix multiplication) induces an algebra structure on $A$.

The following maps belong to $A$:

1. $M_n \oplus M_n \to M_n \colon (A,B) \mapsto A$;
2. $M_n \oplus M_n \to M_n \colon (A,B) \mapsto B$;
3. $M_n \oplus M_n \to M_n \colon (A,B) \mapsto f(A,B)I_n$, where $I_n$ is the identity matrix and $f$ is an element of the ring of invariants $k[M_n \oplus M_n]^{\mathrm{GL}(n)}$.

Do they generate $A$ as an algebra?

Answer 1

The answer is affirmative not only in the case of 2 matrices, but also in the case of any number of matrices; in fact, an analogous statement is true for quiver representations (in characteristic 0).

The original question can be restated as follows.

Let $P$ be the space of polynomial functions of 2 $n\times n$ matrices, with the adjoint action of $GL_n$ and the ring of invariants $I.$ Consider the space $\text{Hom}_{GL_n}(M_n,P)$ as an $I$-module. Is it true that it is generated by the products of matrices?

For the case of any number of generic matrices $A_1,\ldots,A_k,$ Procesi proved that over a field $k$ of characteristic 0, $I$ is spanned by the traces of the products of matrices. Formally, consider words in the free monoid with $k$ generators, substitute the generic matrices, and take a trace.

Procesi, C. The invariant theory of n×n matrices. Advances in Math. 19 (1976), no. 3, 306–381

The statement follows by adjoining an extra generic matrix $A_0$ and converting an $M_n$-space into a $GL_n$-invariant forming a product with $A_0$ and taking the trace, then undoing the trace of the term in the trace polynomial from Procesi's theorem containing $A_0.$

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Here is a vast generalization due to Le Bruyn and Procesi. Given a finite quiver $Q$ and a dimension vector $\alpha,$ consider the corresponding representation space $R(Q,\alpha)$ with the action of the algebraic group $GL(\alpha)$ and the space $P$ of polynomial functions on $R.$ (If the quiver consists of a single vertex with $k$ loops and $\alpha=n$ then the representation space is given by $k$ generic $n\times n$ matrices with the simultaneous conjugation action by $GL_n.$) Then, over a field of characteristic 0, the algebra $I$ of polynomial invariants is spanned by the traces of matrix products over oriented cycles in $Q$ and for any pair of vertices $(i,j)$ of $Q,$ the space $\text{Hom}_{GL(\alpha)}(\text{Hom}_k(V_i,V_j),P)$ is generated as an $I$-module by the products over oriented paths connecting $i$ with $j.$

Lieven Le Bruyn, Claudio Procesi, Semisimple representations of quivers. Trans. Amer. Math. Soc. 317 (1990), no. 2, 585–598

Answer 2

If you consider pairs of unitaries instead, and the group, that you get out of a similar construction, the answer is negative. That is an analogous question but in a slightly different context. The new question is only interesting if one studies it for all dimensions at once.

To be more precise, in http://arxiv.org/abs/1003.4093, it was shown that there are exotic families of continuous maps $$\phi_n \colon U(n) \times U(n) \to U(n),$$ such that:

1) $\phi_{n+m}(U \oplus V,U' \oplus V') = \phi_n(U,V) \oplus \phi_n(U',V')$,

2) $\phi_{nm}(U \otimes V,U' \otimes V') = \phi_n(U,V) \otimes \phi_n(U',V')$, and

3) $\phi_n(AUA^{-1},AVA^{-1}) = A\phi_n(U,V)A^{-1}$.

Here, exotic means that $\phi_n(U,V)$ is not given by evaluating the pair of unitaries at a word $w \in {\mathbb F}_2$, where ${\mathbb F}_2$ denotes the free group on two generators.