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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.)

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  • $\begingroup$ What precisely is the (trivial) answer in the $GL$ case? $\endgroup$ – Mikhail Borovoi Apr 19 '14 at 19:44
  • $\begingroup$ For quivers without oriented cycles any invariant function is constant, so by trivial answer I mean basic field. $\endgroup$ – Sasha Pavlov Apr 21 '14 at 10:50
  • $\begingroup$ I think you can find the answer to your question in the book "The Classical Groups: Their Invariants and Representations" by Hermann Weyl (I don't have this book on my table). $\endgroup$ – Mikhail Borovoi Apr 22 '14 at 12:32
  • $\begingroup$ The determinants of your matrices $M_1$, ... $M_m$ are polynomial invariants in the $SL$ case, and a natural guess would be that the algebra of polynomial invariants is generated by these determinants. Is this correct? $\endgroup$ – Mikhail Borovoi Apr 22 '14 at 13:49
  • $\begingroup$ We can do even more general constriction: lets take coefficients of polynomial $det(z_1 M_1+\ldots+z_m M_m)$, they are all invariant and determinants correspond to coefficients of $z_i^n$. But simple examples for small $m$ and $n$ shows that there are other invariants, not covered by this construction. $\endgroup$ – Sasha Pavlov Apr 26 '14 at 0:49
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I think the result in general is unknown. I have tried the case m = n = 3 (arXiv:0906.5525v2, sorry in french). Bruno Blind

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  • $\begingroup$ The present question should be a bit more approachable than what you studied in your article. Here the actions on the left and the right are with two different groups. $\endgroup$ – Abdelmalek Abdesselam Apr 29 '14 at 14:08
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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.

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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".

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