# Invariants and orbits of $n$-tensors

My question may be absolutely elementary and is probably answered in 19th century. A reference or a short clear argument would be highly appreciated.

Let $V_1, \ldots V_n$ be finite dimensional vector spaces over the same field (may assume complex numbers). What are $GL(V_1)\times \ldots \times GL(V_n)$-orbits on $V_1 \otimes \ldots \otimes V_n$?

The only invariant of an orbit I can see is "a multirank" $(k_1, \ldots k_n)$ where $k_i$ is the dimension of support of an element in $V_i$. The multirank satisfies inequalities $k_i \leq \prod_{j\neq i} k_j$. Would it be too naive to suggest that orbits are in 1-1 correspondence with legal multiranks?

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$n\leq 2$ has elementary answers and I am mostly interested in $n=3$ and $n=4$... –  Bugs Bunny Jun 15 '11 at 21:23
One place to look is in some recent work of J. M. Landsburg, Z. Teitler, et al. As for the last question, I'm afraid it is indeed too naive, since already for 2-dimensional vector spaces, with n=5, the dimension of the tensor product is larger than the dimension of the group (so there will be infinitely many orbits). –  Dave Anderson Jun 16 '11 at 1:01
I can't resist mentioning that when $n=0$ there are lots of orbits. (A trivial group is acting on a $1$-dimension vector space.) –  Tom Goodwillie Jun 16 '11 at 4:08
More seriously, when $n=3$ and all $V_i$ are $2$-dimensional there is a homogeneous degree $4$ polynomial function $V_1\otimes V_2\otimes V_3\to \mathbb C$ that scales by square of determinant when any one of the $GL(V_i)$ acts. The set where it does not vanish is an orbit with multirank $(2,2,2)$ but there is also another orbit corresponding to multirank $(2,2,2)$. –  Tom Goodwillie Jun 16 '11 at 4:28

Here is a start, suppose that $V_i$ is $\mathbb C^{k_i}$ (and restricting to $k_1,k_2,\dots,k_n, n\geq 2$). The tuples $(k_1,k_2,\dots,k_n)$ for which the action of $GL_{k_1}\times\cdots\times GL_{k_n}$ on $\mathbb{C}^{k_1}\otimes \cdots\otimes \mathbb{C}^{k_n}$ has only finitely many orbits are $(k,l),(2,2,k),(2,3,k)$, for positive integers $k,l$. This was proven in

V. G. Kac, "Some remarks on nilpotent orbits", J. Algebra 64 (1980), 190–213.

These orbits are classified in "Orbits and their closures in the spaces $\mathbb{C}^{k_1}\otimes \cdots\otimes \mathbb{C}^{k_r}$" by P.G. Parfenov (MR). This paper doesn't seem to be online, but I believe you can find a summary in section 5 here.

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I don't think that's right, Bugs. Say that $V_1$ and $V_2$ are $d$-dimensional and $V_3=\mathbb C^2$. Then an element of $V_1\otimes V_2$ is a linear map $V_1^\star\to V_2$, generically an isomorphism; an element of $V_1\otimes V_2\otimes V_3$ is an ordered pair $(A,B)$ of these; the unordered $d$-tuple of eigenvalues of $B\circ A^{-1}$ is an invariant of the $GL(V_1)\times GL(V_2)$-action; and an element of $GL(V_3)$ will just perform some fractional linear transformation on all of these numbers, so that if $d\ge 4$ then there is a complex invariant here.

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I might mention that the problem of classifying ordered pairs of linear transformations over $\mathbb{C}$ up to the mentioned equivalence was solved by Weierstrass and Kronecker way back when. This seems to be a forgotten thread of basic linear algebra, but can still be found in Chapter XII of Gantmacher's 'Theory of Matrices' (a book I would greatly recommend to everyone). –  Keerthi Madapusi Pera Jun 16 '11 at 3:30
@Keerthi: that's not quite forgotten! It is called nowadays the representation theory of quivers---the Russians, characteristically romantic in their choice of words, call this problems of linear algebra. The result of Weierstrass and Kronecker is nowdayds presented as the classification of indecomposable modules for the Kronecker quiver, the simplest non-trivial case of the extraordinarily elaborate representation theory of hereditary finite dimensional algebras. –  Mariano Suárez-Alvarez Jun 16 '11 at 3:54
Mariano--Thanks for correcting my ignorant perception of its status! I remember looking for references about it a long time ago, and not having any luck until I stumbled upon Gantmacher's text. Clearly, I didn't have the right formulation. –  Keerthi Madapusi Pera Jun 16 '11 at 4:09