TITLE: What are the invariants of $U\otimes V\otimes W$ under action of $GL(U)\times GL(V) \times GL(W)$
QUESTION [10 upvotes]: 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 non-zero 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.

REPLY [4 votes]: 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 $0$.