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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?

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up vote 5 down vote accepted

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 $A_0.$

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

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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, 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.

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