Tensor product modulo general linear group Suppose $V_1$ and $V_2$ are finite-dimensional vector spaces. Then $GL(V_1)\times GL(V_2)$ acts on $V_1\otimes V_2$, and the orbits under this action are precisely the sets of tensors of fixed rank. A consequence is that the number of orbits is $1+\min\{\dim(V_1),\dim(V_2)\}$.
I am interested in what happens when we extend from two vector spaces to three. What can be said about the set of $GL(V_1)\times GL(V_2)\times GL(V_3)$ orbits in $V_1\otimes V_2\otimes V_3$? Are there infinitely many orbits?
 A: Nice question! This isn't an answer but it's slightly too long to be a comment. I already don't know the answer when the vector spaces all have dimension $2$ which is a little shocking to me. After playing around with the analogue of row reduction ("plane reduction") for a $2 \times 2 \times 2$ tensor I've managed to get things down to a one-parameter family that I don't know how to reduce further. Really this should be written as a $2 \times 2 \times 2$ cube but until I learn how to typeset that, here's one way of writing it: name the vector spaces $X, Y, Z$ and give them bases $\{ x_1, x_2 \}, \{ y_1, y_2, \}, \{ z_1, z_2 \}$. Then I don't know whether the one-parameter family of tensors
$$x_1 y_1 z_1 + x_2 y_1 z_2 + x_1 y_2 z_2 + t x_2 y_2 z_2$$
(with $\otimes$ written as concatenation to save notation) consists of elements in distinct orbits or not. Anyone know the answer one way or another? Kronecker normal form for matrix pencils may be relevant but I don't understand it yet.
A: If the dimension are high enough, at least if all dimensions are greater or equal then $3$, there are cases with continuous families of orbits.
To see this, we transform the problem in a statement about quiver representations. A quiver $Q=(Q_0,Q_1)$ is a finite directed graph with vertices $Q_0$ and arrows $Q_1$. A representation $(V,f)$  of $Q$ assigns to each vertex $i \in Q_0$ a vector space $V_i$ and to each arrow $\alpha: i \rightarrow j$ a linear map $f_\alpha: V_i \rightarrow V_j$. Two representations are equivalent if the linear maps on the arrows are related by simultaneous conjugation on the vertices. A representation is called indecomposable, if it is  not equivalent to the sum of two nontrivial representations.
Quivers can be divided into three
classes, depending on whether there are finitely many (finite type) or infinitely many (depending on one parameter (tame) or arbitrary many parameters (wild)) orbits of indecomposable representations.
For all $d \in \mathbb{N}$ there is the Kronecker quiver $K_d$:
$$
1 \stackrel{d}{\longrightarrow} 2
$$
It consists of two vertices joined by $d$ arrows.
The quiver $K_1$ is of finite type. Fixing vector spaces $V_1$ and $V_2$ on the two vertices, a representation is an element in $\mathrm{Hom}(V_1,V_2)$ with two representations being equivalent iff they are conjugated. This produces the finitely many orbits in this case.
Given a third vector space $W$ of dimension $d$, fixing a basis of $W$ gives an isomorphism  $W \otimes V_1^* \otimes V_2 \simeq \mathrm{Hom}(V_1,V_2)^d$,
which is the representation space of the quiver $K_d$. A representation is given by a $d$-tuple $(f_1, \ldots, f_d)$ of linear maps from $V_1$ to $V_2$.
This is a tame quiver for $d=2$ and wild for $d>2$. The group $\mathrm{Gl_d}$ acts on the representation space, by acting on the tuples of linear maps. The action preserves the set of indecomposable representations. For dimensions $(n_1,n_2)$ of $(V_1,V_2)$, by Kac theorem on indecomposable representations, this set depends on $1-<n_1,n_2>=1+d n_1 n_2 - n_1^2 -n_2^2$ parameters.
Here $<.,.>$ is the Euler form of the quiver.
If we pick $d$=3 and $n_1=n_2=n$, the number of parameters is $n^2+1$.
Taking the quotient by the action of $\mathrm{Gl}_3$ heuristically reduces the dimension by $3^2=9$, so for all $n$ with $n^2+1> 9$ we expect a continuous family of orbits.
However, to make the argument precise seems complicated, mainly because the space of indecomposables is not a nice geometrically known object. Here is a sketch of how I think it can be done. In
https://arxiv.org/abs/0802.2147 , Lemma 2.3 a parametrization of an open
subset of the indecomposables of $K_3$ is described, it is a quotient $M$ of an affine space by identifying finitely many points. For each point $f \in M$, there is an open subset $U_f$ of $\mathrm{Gl}_3$, such that the action
of $g \in U_f$ on $f$ is again conjugated by an element in $M$.
The action of $U_f$ on $M$ is given by matrix inversion and multiplication, thus it  is polynomial in the entries. In total, the $\mathrm{Gl}_3$ action on $M$ gives a foliation with polynomial leaves, thus the number of leaves is uncountable.
The case $d=2$ as discussed by Qiaochu Yuan, is special since $K_2$ is of tame representation type. I dont know the statement in this case.
