TITLE: Explicit invariants (under change of basis) of maps $V \to V \otimes V$.
QUESTION [7 upvotes]: It is standard to construct numbers associated to a linear transformation $f: V \to V$  of a finite-dimensional vector space which are invariant under change of basis.  The coefficients of the characteristic polynomial are such, and it is quite simple to see for example that the trace is invariant by equating it with the map which takes $\sum v_i \otimes w_i \in V \otimes V^* \cong Hom(V, V)$ and then evaluates $\sum w_i(v_i)$.
My question is: can one explicitly associate quantities to maps $V \to V \otimes V$ which are invariant under change of basis?  A simple dimension count indicates there should be plenty of invariants, since the orbits will have dimension roughly $d^3 - d^2$, where $d$ is the dimension of $V$, but I have yet to construct a single one.

REPLY [5 votes]: This is essentially a rephrazing of Victor Protsak's reply (which I read too late).
In characteristic zero the answer is in principle given by classical invariant theory. Let $(e^i)$, $i=1,...,n$ be a basis for V. Then
coordinates on the space $Hom(V,V \otimes V)$ are given by tensors $c^i_{jk}$. Such a
tensor sends the vector $\lambda_i e^i$ to $c^i_{jk}\lambda_i e^j \otimes e^k$ (using the summation condition for repeated indices).
In addition let $\epsilon^{i_1,...,i_n}$ be the tensor which is $1$ if the $i_1,...,i_n$
form an even permutation of $1,...,n$, which is $-1$ for an odd permutation and which is zero otherwise. Similarly for $\epsilon_{i_1,...,i_n}$. 
SL(V) invariants are now given by expressions in $c,\epsilon$ which have no free indices
(it is possible to do this via graphs). One has to be careful since many expressions
are zero for symmetry reasons.
There are clearly no linear invariants. If $dim(V)=2$ then I believe $\epsilon^{kl}c^i_{kj}c^j_{il}$ is an honest quadratic invariant. Another one
is $\epsilon^{kl}c^i_{ik}c^j_{lj}$. I.e. the determinant of the left partial
trace with the right partial trace.
EDIT:
The two dimensional case can also be viewed in a less abstract way. If $dim V=2$ then
$V\cong V^\ast$ as $SL(V)$ representation. Furthermore by Clebsch Gordan we have $V^* \otimes V\otimes V\cong
S^3V\oplus V\oplus V$. This is the problem of finding the generating concomitants for binary cubic forms. The answer can be found in Grace and Young. In this setting I
see only one quadratic invariant, namely the determinant between the two copies of V.
This means that the two quadratic invariants identified above are linearly
dependent which does not appear obvious. But they are indeed. If you write them
out explicitly you get (up to sign)
$c^1_{11}c^1_{12}-c^1_{21}c^1_{11}+c^1_{12}c^2_{12}-c^2_{21}c^1_{21}+c^2_{12}c^2_{22}
-c^2_{22}c^2_{21}$

REPLY [2 votes]: I've been thinking about the case when $V$ is $2$-dimensional. Here, I claim, is a parametrization of the $GL(V)$-orbits of 'generic' linear maps $F:V\to V\otimes V$:
To numbers $p, a, b, c$ we associate a map which acts on a basis $v, w$ by:
$F(v)= -p(v\otimes v)+ w\otimes w$
$F(w)= (v\otimes w-w\otimes v)+ a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$.
To put a generic $F$ in this form, write it as $F_++F_-$, symmetric plus antisymmetric. $F_-$ is given by $x\mapsto v\otimes x-x\otimes v$ for a unique $v\in V$. Assume $v\ne 0$. 
Choose $w$ such that $v,w$ is a basis, but remember we are free to replace $w$ by $sv+tw$ for any scalars $s$ and $t\ne 0$. Think of the symmetric tensor $F_+(v)$ as a homogeneous quadratic polynomial in indeterminates $v,w$. Assume it has a $w^2$ term. Replacing $w$ by suitable $tw$ we can make that term $w^2$. Now replacing $w$ by suitable $sv+w$ ("completing the square") we can eliminate the $vw$ term. So $F_+(v)= -pv^2+w^2=-p(v\otimes v)+ w\otimes w$ for some $p$. And $F_+(w)= av^2+2bvw+cw^2=a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$ for some $a,b,c$. This gives the formulas above for $F$.
I used up all the choices I had except that $w$ can still be changed to $-w$, which would change $p,a,b,c$ to $p,-a,b,-c$. So the "parametrization" is actually two to one.