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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.

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Sorry for the confusion. By "quantity" I mean a non-constant function from Hom( V, V\otimesV)^{GL(V)} to a familiar set like the ground ring (I am most interested in the rational numbers, but one could extend to the complex numbers if needed), the natural numbers, {0,1} etc. For this I don't even care about continuity (e.g. rank also is of interest), just "computability" (a vague term, sorry) by hand in small cases and computer in large ones, and having a large enough collection to hopefully distinguish the maps I am interested in. –  Dev Sinha Jun 19 '10 at 9:52
    
OK, it sounds like you are interested in classifying the orbits of $GL(V)$-action on $\operatorname{Hom}(V,V\otimes V),$ as in Tom's answer, together with numerical invariants that distinguish them. If that's indeed the case, since all other answers were dealing with algebraic invariants instead, your best bet would be to post another question clearly stipulating it. –  Victor Protsak Jun 22 '10 at 1:49
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5 Answers

The answer depends on what you mean by "quantities"! As José demonstrated, there are no polynomial absolute invariants, essentially because of homogeneity, however, there are plenty of rational ones, obtained as fractions $F/G,$ where $F$ and $G$ are polynomial relative invariants of the same weight $k:$

$$ F(g\cdot f)=(\det g)^k F(f),\ G(g\cdot f)=(\det g)^k G(f), \quad g\in GL(V).$$

Your dimension count is one of the common false believes. In fact, sanity is restored when working with rational functions, for

$$ \operatorname{tr\ deg} K(X)^G=\operatorname{tr\ deg} K(X)-\dim O_x,$$

where $X$ is an irreducible algebraic variety over an alg. closed field $K$ of char 0 with an action of an algebraic group $G$, $O_x=G\cdot x$ is a generic orbit, $G_x$ is the stabilizer of $x$, $\dim O_x=\dim G-\dim G_x.$ In fact, rational invariants always separate generic orbits.1 What goes wrong with polynomial invariants is that orbits need not be Zariski closed. In your example, due to the presence of dilations, Zariski closure of every orbit contains zero. A $G$-invariant polynomial function $F$ is constant on any orbit $O$ and hence its value at any point $x\in O$ is equal to $F(0),$ so $F$ is constant.

Another way to resolve the issue is to replace the group $G$ with its subgroup $[G,G]$, which is the inter-section of the kernels of all 1-dimensional representations of $G$, thus replace $Gl(V)$ with $SL(V)$ om the example. After the $GL(V)$-equivariant identification $Hom(V,V\otimes V)\simeq V^*\otimes V\otimes V$ and polarization, which replaces a homogeneous degree $d$ polynomial function on $W$ with a multilinear map with $d$ arguments $W^\otimes d\to K$, the question reduces to finding multilinear $SL(V)$-invariants. Classical invariant theory shows that they are all obtained by composing $GL(V)$-invariant contractions $V^* \otimes V\to K,$ $\xi\otimes v\mapsto \xi(v)$ and expansions $K\to V^*\otimes V,$ $1\mapsto \sum e_i^*\otimes e_i$, permutations, symmetrizations, antisymmetrizations, and the $SL(V)$-invariant determinants $V^{\otimes k} \to K,$ $v_1\otimes\ldots\otimes v_k\mapsto \det[v_1|\ldots|v_k],$ $k=\dim V.$ Explicit formulas are a bit messy.


Footnotes

1See Vinberg and Popov's article on invariant theory in Algebraic Geometry 4 volume of the yellow Russian Math Encyclopaedia.

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No nontrivial linear one exists, I'm afraid.

The underlying reason why the trace exists is that there is a $\mathrm{GL}(V)$-equivariant endomorphism of $V$: namely, the identity endomorphism.

By contrast, there cannot be any $\mathrm{GL}(V)$-equivariant linear maps $V \to V \otimes V$.

For definiteness, let us consider $V$ a real vector space.

Consider the action of the $\mathbb{R}^\times$ subgroup of $\mathrm{GL}(V)$ consisting of scalar matrices. Then if $f:V \to V \otimes V$ is linear $$f(\lambda v) = \lambda f(v)$$ for any $v \in V$ and $\lambda \in \mathbb{R}^\times$.

On the other hand, equivariance would say that $$ \lambda \cdot f(v) = \lambda^2 f(v).$$

You can only reconcile both if $f$ is identically zero.

Edit

Unwisely, I had assumed linearity. (The OP did mention other polynomial invariants.) As pointed in comments to this answer, there are of course rational invariants.

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So there's no linear invariant. And for linear maps $V\to V$ the trace is the only linear invariant. But the question was not about linear invariants alone. For linear maps $V\to V\otimes V$ there are in fact no nonconstant polynomial invariants, by the same sort of scaling argument; but shouldn't there be some invariants that are rational functions of the coefficients? –  Tom Goodwillie Jun 19 '10 at 1:07
    
Tom: You've anticipated my answer which I have been conscientiously typing for the past 120 minutes. Due to time constraints and annoying markup bugs, I had to give up on an explicit example. –  Victor Protsak Jun 19 '10 at 1:26
    
Thanks to both -- I've edited my answer to reflect your comments. –  José Figueroa-O'Farrill Jun 19 '10 at 1:38
    
What is meant by the fascinating statement that "the underlying reason why Trace exists is that there is a GL(V)-equivariant endomorphism $V \to V$"? Even if not terribly deep this would be a nice remark to understand. –  T.. Jun 22 '10 at 21:35
    
This argument does not work over $\mathbb{F}_2$, as far as I can tell; does anything interesting happen in that case? –  Daniel Litt Jul 13 '10 at 14:35
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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)

$c^1_{11}c^1_{12}-c^1_{21}c^1_{11}+c^1_{12}c^2_{12}-c^2_{21}c^1_{21}+c^2_{12}c^2_{22} -c^2_{22}c^2_{21}$

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Jose's argument proves that there are no polynomial invariants, i.e. no invariants that are polynomial in the matrix coefficients with respect to a given basis. Victor suggests that there should be rational invariants. Here's an easy one. Let $\alpha : V \to V \otimes V$ be your function. Then there are roughly four (not necessarily symmetric) bivectors $\beta: k \to V\otimes V$ by tracing $\alpha^{\otimes 2} : V^{\otimes 2} \to V^{\otimes 4}$ in various different ways. For generic $\alpha$, this bivector $\beta$ will be nondegenerate, i.e. it will have an "inverse" $\beta^{-1}: V\otimes V \to k$, defined by declaring that one of the traces of $\beta\otimes \gamma$ is the identity map $V \to V$. If $\beta^{-1}$ exists, then it is a (nonsymmetric) inner product on $V$. Now, $\alpha$ determines two natural vectors $\operatorname{tr}_L(\alpha),\operatorname{tr}_R(\alpha)\in V$, by tracing in the two different ways. So some invariants are the squared lengths of $\operatorname{tr}_L(\alpha),\operatorname{tr}_R(\alpha)$ and their two inner products with respect to $\beta^{-1}$.

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I am a bit suspicious. Any rational $GL(V)$-invariant, when written as a quotient of polynomial semi-invariants, must somehow involve determinants of order $\dim V.$ Are they hidden within your "squared length with respect to $\beta^{-1}$" prescription? (A minor terminological point: bivectors are skew-symmetric.) –  Victor Protsak Jun 19 '10 at 7:47
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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.

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