Consider tri-linear forms, $\{A_{ijk}\}$ where $i=1,..,n_1$, $j=1,..,n_2$, $k=1,..n_3$, over a field of zero characteristic, up to the equivalence $A\to (U_1,U_2,U_3)(A)$, by three matrices. What is known about the classification? (Complete) invariants? Unlike the case of bi-linear forms, in general the classification has moduli.

I've found only an old paper of Thrall, dealing with partial cases. Is there some general (and more modern) exposition/introduction/lecture notes?

My particular question: for which triples $(n_1,n_2,n_3)$ is the classification discrete? (The obvious necessary condition, $n_1n_2n_3\le n^2_1+n^2_1+n^2_3$, is certainly not sufficient.)

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    $\begingroup$ In the case $n_1=n_2=n_3$, isn't it the same as classification of non-associative algebras? That was worked on a lot by A.Albert, etc. $\endgroup$ Feb 10, 2013 at 14:34
  • $\begingroup$ MathSciNet shows 2 references on the paper you cited, have a look! $\endgroup$ Feb 10, 2013 at 14:36
  • $\begingroup$ Note that, because there is a $2$-parameter group of scalings in your equivalence group that acts trivially, you actually have that $$ n_1^2+n_2^2+n_3^2-2\ge n_1n_2n_3 $$ is the obvious necessary condition. (Thus, for example, $n_1=n_2=n_3=3$ must have invariants, though that wouldn't be obvious without subtracting the $2$.) $\endgroup$ Feb 10, 2013 at 14:51
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    $\begingroup$ @Dmitry: Also, could you be a bit more explicit about what you mean by 'discrete'? For example, in the case $n_1=n_2=n_3=2$, there are only a finite number of orbits when the ground field is $\mathbb{R}$ or $\mathbb{C}$, but there are infinitely many distinct orbits when the ground field is $\mathbb{Q}$. Would you count the latter case as 'discrete'? $\endgroup$ Feb 10, 2013 at 17:06
  • $\begingroup$ @Robert: being ignorant I did not think that the field matters much. :( Could you give more details? A reference? $\endgroup$ Feb 10, 2013 at 20:09

2 Answers 2


I'll phrase my answer for an algebraically closed field.

As soon as all of the $n_i$ are at least 3 you get moduli. So the interesting cases are $(2,2,n)$ and $(2,3,n)$ which I claim have finitely many orbits.

Given a tensor of type $(2,2,n)$, you can always write it as $\sum e_i \otimes f_j \otimes h_{i,j}$ where $i=1,2$, $j=1,2$. Then the $h_{i,j}$ span a 4-dimensional subspace of the $n$-dimensional factor, and just using the $GL_n$ action, you can always do a transformation and assume this lies in a fixed $4$-dimensional subspace, and you're still free to use the full $GL_4$ in the stabilizer to further analyze orbit equivalence. Hence you have reduced to showing it for $(2,2,4)$. Similar remarks apply to reducing $(2,3,n)$ to $(2,3,6)$.

For $(2,2,4)$, you can replace $GL_2 \times GL_2$ acting on the $2 \times 2$ factor with $SO_4$ acting on its vector representation (you have essentially not lost any freedom by doing so). So the space you're dealing with is $4 \times 4$ matrices $V \to W$. Since one of these $4$ dimensional spaces (call it $W$) has an orthogonal form, it makes sense to take the dual to get $W \to V^\*$ (since we can identify $W = W^\*$ while respecting the group action). The two invariants that classify orbits are the rank of this matrix and the rank of the composition $V \to W \to V^*$ (left to reader). (In fact, this is true more generally whenever you consider a space of matrices between some vector space and some orthogonal (or symplectic) vector space under the action of the general linear group and the corresponding classical group.)

For $(2,3,6)$, it is more involved. There is a basic invariant, the determinant (by flattening this tensor to a $6 \times 6$ matrix). I claim that any two tensors that have nonzero determinant are in the same orbit: this is just the statement that any two invertible matrices can be transformed into one another via row operations. So now we're left with those matrices with determinant 0, which I also claim has finitely many orbits.

By a similar trick as above, we can reduce to showing that tensors of format $(2,3,5)$ have finitely many orbits. I don't think I have a direct way of seeing this, so I'll just appeal to a result of Vinberg about Z-gradings on simple Lie algebras: given a nonnegative integer valued function on the simple roots of a simple Lie algebra $\mathfrak{g}$, you can coarsen the grading given by roots by just evaluating this function on the root.

The degree 0 piece is then a reductive Lie algebra, and the result is that the simply-connected form of this Lie algebra acts on each degree $i$ piece with finitely many orbits. The case of $(2,3,5)$ comes from setting $\mathfrak{g}$ to be of type $\mathrm{E}_8$ and taking the function which just picks out the coefficient of the trivalent node of the corresponding Dynkin diagram. The degree 0 piece is $\mathfrak{sl}_2 \times \mathfrak{sl}_3 \times \mathfrak{sl}_5 \times \mathbf{C}$ and the degree 1 piece is exactly the space of $2 \times 3 \times 5$ tensors.

Last are the examples of $(2,2,2)$, $(2,2,3)$, $(2,3,3)$, and $(2,3,4)$. These also follow from Vinberg's result (pick the coefficient of the trivalent node in the Dynkin diagram of type $\mathrm{D}_4$, $\mathrm{D}_5$, $\mathrm{E}_6$, and $\mathrm{E}_7$, respectively), or maybe can be deduced from the larger examples.

This old blog post I wrote about Vinberg's result might be helpful: http://concretenonsense.wordpress.com/2010/03/01/nilpotent-orbits-in-graded-lie-algebras/

One last remark: The Vinberg result that I cited works over the complex numbers, but the finiteness of the orbits should be valid in any characteristic (in principle, the number of orbits can vary though).

Edit: I forgot about $(2,4,4)$. Such a tensor can be interpreted as a $4 \times 4$ matrix of linear forms in two variables $x,y$, and hence their determinant is a quartic binary form (and you can get all such in this way by putting the irreducible factors as the diagonal entries). These have moduli (for instance, generic quartic binary forms produce elliptic curves by taking the associated double cover of the projective line). Any other format not discussed either has all $n_i \ge 3$ or at least two $n_i \ge 4$ and hence has moduli since it contains a case that we already discussed.

  • $\begingroup$ Thanks! Still more questions: 1. Where is this written? (Instead of writing down the reasoning in my paper I'd prefer just to cite some text) 2. Suppose, for a given group acting on a space, there is just one open dense orbit. Does it imply that all the orbits are discrete (no moduli)? I cannot think of any counterexample, being ignorant. Or, maybe there are some additional (not too restrictive) conditions ? $\endgroup$ Feb 10, 2013 at 20:08
  • $\begingroup$ Possible reference is Kac's paper "Some remarks on nilpotent orbits" which cites other papers: ams.org/mathscinet-getitem?mr=575790 (but note that $2 \times 2 \times n$ is treated as $4$-dimensional orthogonal space tensored with $n$-dimensional space as I wrote it). If a representation has a dense open orbit, it does NOT imply finitely many orbits. These are called prehomogeneous vector spaces. The irreducible ones were studied by Sato-Kimura: ams.org/mathscinet-getitem?mr=430336 Examples of infinitely many orbits is on p.150. $\endgroup$
    – Steven Sam
    Feb 11, 2013 at 3:03
  • $\begingroup$ For an explicit example from the paper: consider the action of $SL_6 \times GL_{19}$ on $\bigwedge^3 \mathbf{C}^6 \otimes \mathbf{C}^{19}$. $\endgroup$
    – Steven Sam
    Feb 11, 2013 at 3:04

This should really be a comment on the answer of Steven Sam, but it doesn't fit in the box. The systematic way to reduce to the cases described by Sam is via castling transformations.

It is shown in a paper of P.G. Parfenov that the group $G = GL(k_{1}, \mathbb{C})\times \dots \times GL(k_{r}, \mathbb{C})$ has finitely many orbits in $V = \mathbb{C}^{k_{1}}\otimes \dots \otimes \mathbb{C}^{k_{r}}$ if and only if $(k_{1}, \dots, k_{r})$ is one of $(n)$, $(m, n)$, $(2, 2, n)$, or $(2, 3, n)$, where $n \geq 3$. This is published in:


(This also follows from Theorem $2$ of V.G. Kac's paper "Some remarks on nilpotent orbits")

That $G$ have finitely many orbits on $V$ means that $(G, V)$ is prehomogeneous ($G$ has a Zariski dense orbit), but the converse is not true. One can ask also when $(G, V)$ is prehomogeneous. A necessary condition was stated above by R. Bryant in a comment to the original question (for $r > 3$ the condition becomes $\sum_{i}k_{i}^{2} - \prod_{i}k_{i} \geq r- 1$). In general this can be resolved by using the Sato-Kimura classification of reduced irreducible prehomogeneous vector spaces. Here reduced means reduced with respect to castling equivalence, which is a notion of equivalence of prehomogeneous vector spaces based on the duality of Grassmannians (see Sato-Kimura and Kimura's book on prehomogeneous vector spaces). One can show that every $(G, V)$ that is prehomogeneous is castling equivalent to such a space with finitely many orbits, that is in the list due to Parfenov mentioned above (I have not been able to find this statement as such published anywhere, but it is known). For general formats $(k_{1}, \dots, k_{r})$ I don't know how to state this cleanly simply as a condition on the $k_{i}$'s, but in practice it is straightforward to check (the necessary details can be gleaned from section 3 of Sato-Kimura). For example, if $r = 3$ then the dimensions have the form $(a, b, c)$. If $ab > c$ then this format is castling equivalent to $(a, b, c - ab)$, and via this remark, one can reduce to a castling equivalent format in which (after reordering) $c \leq 2$.

In some cases the fundamental relative invariant is a hyperdeterminant. This occurs, for example, for the formats $(2, 2, 2)$ (the classical Cayley hyperdeterminant) and $(2, 3, 3)$. See chapter 14 of the book of Gelfand, Kapranov, and Zelevinsky.


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