$SL(n) \times SL(n)$-invariants of $m$-tuples of matrices

Question

I work over field of complex numbers. Let $G=SL(n) \times SL(n)$, and $(A,B) \in G$ acts on $m$-tuples of matrices $M_{n \times n}(\mathbb{C})^{\oplus m}$ as follows $$ (A,B) \cdot (M_1, \ldots, M_m) \mapsto (A^{-1}M_1 B, \ldots, A^{-1}M_m B) $$ Where can I find a description of the ring of invariants? For me it is very important that I consider $SL$ case (opposed to $GL$ case where this is just representations of a Kronecker quiver with $n$ arrows and the answer is trivial.)

Answer 1

I think the result in general is unknown. I have tried the case m = n = 3 (arXiv:0906.5525v2, sorry in french). Bruno Blind

Answer 2

The answer in the generality of quiver representations is given by

1. H. Derksen, J. Weyman, Semi-invariants of quivers and saturation for Littlewood-Richardson coecients, J. Amer. Math. Soc. 13 (2000), no. 3, 467--479.

2. A. Schoeld, M. Van den Bergh, Semi-invariants of quivers for arbitrary dimension vectors, Indag. Math. (N.S.) 12 (2001), no. 1, 125--138.

3. M Domokos, A Zubkov, Semiinvariants of quivers as determinants.

In your special case, I can tell you the generators. Consider the expansion of $$\det(\sum_{k=1}^m \Lambda_k \otimes M_k),$$ where $\Lambda_k$ is a $d\times d$-matrix of indeterminants, and $\otimes$ is the Kronecker product. Then all coefficients in the expansion are invariants. They generates all invariants if $d$ is big enough. In fact, for fixed $d$, the coefficients linearly span the space of degree $dn$ invariants.

Answer 3

My answer to MO question Invariant polynomials for a product of algebraic groups might help a bit. Here there should be invariants of degree $d$ if and only if $d$ is divisible by $n$. For $d=kn$, a linearly generating set can be given in terms of bipartite graphs with colored edges. The vertex set $V$ with $2k$ elements is partitioned into $I$, $J$, each with $k$ elements. All vertices have degree $n$. Finally the edges are colored with $m$ colors. The corresponding invariant is obtained as in my MO answer above by assigning tensors $\epsilon_{a_1,\ldots,a_n}$ to each vertex of $I$, assigning tensors $\eta_{b_1,\ldots,b_n}$ to the vertices of $J$ and contracting indices as indicated by the graph. However one needs to put in the middle of an $a$-$b$ contraction a matrix element $(M_c)_{a,b}$ where $c$ is the color of that edge. The invariants obtained from the multilinear expansion of ${\rm det}(z_1 M_1+\cdots+z_m M_m)$ correspond to the graphs with $k=1$, i.e., just two vertices with a multiple edge repeated $n$ times.

The above is just a consequence of the Cayley-Clebsch Theorem, a.k.a, the first fundamental theorem of invariant theory for $SL_n$. Now one should be able to reduce this infinite generating set using the Grassmann-Plücker relation and the straightening algorithm, but I don't know how complicated this gets in the present situation. A possible entry point in this business: Chapters 8 and 9 from the book "Young Tableaux" by Fulton. You can also search with the keywords "standard monomial theory".