The tensor product of some (finite dimensional real) vector spaces is acted on by the direct product of their general linear groups. I would like to know if there are explicit invariants in the case of 3 vector spaces. For one vector space there are two orbits: 0 vector, and nonzero vector. For two vector spaces, $T\in U\otimes V \cong Hom(U^*,V)$ there are finitely many orbits characterized by $rank(T)$. For 3 vector spaces the dimension of $U\otimes V\otimes W$ is $uvw$ and the dimension of $GL(U)\times GL(V) \times GL(W)$ is $u^2+v^2+w^2$ so that usually the space of orbits has positive dimension. Any references would be most welcome. I am particularly interested in the case U,V have dimension 4 and W has dimension 8.

2$\begingroup$ Are you asking for information about the ring of invariants (or covariants) (in the sense of geometric invariant theory) or for a list of normal forms with parameters? In the former case, I think that the GIT quotient (essentially, the space of orbits of the semistable vectors) is, in principle, understood. While I'm not sure that your particular interest ($u=v=w=4$) is worked out in the literature, the case $u=v=w=3$ is known (and the list of normal forms with parameters is known). The case $u=v=w=4$ (with $18$ moduli) is as hard as homogeneous quartics in $4$ variables, so good luck. $\endgroup$ – Robert Bryant Apr 17 '14 at 11:20

$\begingroup$ I see that the OP has just upped the ante: Now, the case of particular interest is $(u,v,w) = (4,4,8)$ instead of $(4,4,4)$. Of course, this case will have even more invariants, though they may be harder to find. The expected dimension of the moduli space in this new case is $33$. Still, in principle, the GIT machinery will say something, though whether it will be of any use is another matter. $\endgroup$ – Robert Bryant Apr 17 '14 at 15:29
In the world of exterior differential systems, an element of a triple tensor product is called a tableau. The known invariants of tableaux are complicated; see the book Exterior Differential Systems by Bryant, Chern, Gardner, Goldschmidt and Griffiths. There is no classification of tableaux.

2$\begingroup$ To provide a little more detail, any first order PDE system with constant coefficients can be represented as a tableau, and any tableau can be treated as such a PDE system. In particular, the characteristic variety of the PDE system is an invariant. I think I recall that that every real algebraic projective variety occurs as the characteristic variety of a tableau, but I am not sure. $\endgroup$ – Ben McKay Apr 16 '14 at 21:07
For what it's worth, in the case when $U,V$, and $W$ all have dimension $2$ (i.e., a case that is much simpler than the $4$dimensional one you're interested in), it is known that there are exactly six orbits. In particular, every vector is in the orbit of exactly one of these six vectors (where $\{\mathbf{e}_1,\mathbf{e}_2\}$ is some fixed basis of $U,V,W$):
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1$
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2$
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2$
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1$
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2$
 $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2 + \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1$
Furthermore, a generic vector in $U \otimes V \otimes W$ belongs to the orbit of the vector 5 above: the other orbits all have measure $0$.

1$\begingroup$ It should be noted that your list of normal forms is complete when the ground field is $\mathbb{C}$, but not when the ground field is $\mathbb{R}$. For example, in the latter case, there are two open orbits. Also, although this is a trivial remark, you forgot to list the $0$orbit. $\endgroup$ – Robert Bryant Apr 17 '14 at 11:24

1$\begingroup$ Whoops, thanks Robert! I have a tendency to translate problems in my head into $\mathbb{C}$, without always remembering to translate them back. $\endgroup$ – Nathaniel Johnston Apr 17 '14 at 12:01