Explicit invariant of tensors nonvanishing on the diagonal

The group $SL_n \times SL_n \times SL_n$ acts naturally on the vector space $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ and has a rather large ring of polynomial invariants. The element $$\sum_{i=1}^n e_i \otimes e_i \otimes e_i \in \mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$$ is known to be GIT-semistable with respect to this action. In other words, there is a homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial, of nonzero degree, which is nonvanishing on this element. The proof I know (Theorem 4.7) uses the Hilbert-Mumford criterion and so does not explicitly construct the polynomial.

Can you give an explicit homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial that is nonzero on this element.

One can check from the definition of the hyperdeterminant that the hyperdeterminant vanishes on this element as long as $n>2$. So that won't work.

I don't know how many other natural ways there are for defining an invariant function on $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ for all $n$ simultaneously there are.

My motivation for this problem is to understand semistability of tensor powers of tensors better, and so to understand slice rank of tensor powers of tensors better. I want to give explicit criteria to show that all a tensor's tensor powers are semistable. Tensor powers of the diagonal tensor remain, so the diagonal tensor certainly has this property. Given an explicit invariant, perhaps one could find an explicit neighborhood of the diagonal tensor consisting of tensors that also have this property.

• Is there something called hypertrace? – მამუკა ჯიბლაძე Jan 8 '17 at 18:57
• I am not sure what is the hyperdeterminant, but the following covariant seems to work. For vector spaces $A$, $B$, and $C$ of dimension $n$, there is a natural linear map $A\otimes_k B\otimes_k C \to \text{Hom}_k(A^\vee,\text{Hom}_k(B^\vee,C))$. The $n^{\text{th}}$ exterior algebra determines a homogeneous, degree $n$, polynomial map $A^\vee \to \text{Hom}_k(\wedge^n(B^\vee),\wedge^n(C))$. Altogether, this is a $\textbf{SL}(A)\times\textbf{SL}(B)\times \textbf{SL}(C)$-equivariant map $A\otimes B \otimes C \to \text{Sym}^n_k(A)\otimes \bigwedge^n(B)\otimes \bigwedge^n(C)$ . . . – Jason Starr Jan 8 '17 at 19:06
• . . . For your element of $A\otimes B\otimes C$, with respect to some basis of $A$, the corresponding element of $\text{Sym}^n_k(A)\otimes\bigwedge^n(B)\otimes \bigwedge^n(C)$ seems to be $(a_1 a_2 \cdots a_n)\otimes (b_1\wedge b_2 \wedge \dots \wedge b_n)\otimes(c_1\wedge c_2 \wedge \dots \wedge c_n)$. According to Dolgachev's book, "Lectures on Invariant Theory", Section 11.1, the polynomial $a_1 a_2 \cdots a_n$ is semistable (though definitely it is not properly stable since it has an $n$-dimensional stabilizer), and the determinant "is" an invariant . . . – Jason Starr Jan 8 '17 at 19:12
• @Jason: don't you need to compose this with an invariant of an $n$-ary $n-ic$ that does not vanish on the first construction? – Abdelmalek Abdesselam Jan 8 '17 at 19:12
• . . . According to Mumford's "Geometric Invariant Theory", Section 1.5, p. 48, there is a way to turn the determinant invariant into a $\textbf{SL}(A)$-invariant on $\text{Sym}^n_k(A)$, but I am not sure how effective this is. – Jason Starr Jan 8 '17 at 19:13

Let me start with some remarks about the classical symbolic method (without which one cannot understand 19th century invariant theory) and multisymmetric functions. I will use an example first. Take four series of three variables $a=(a_1,a_2,a_3)$, $b=(b_1,b_2,b_3)$, $c=(c_1,c_2,c_3)$ and $d=(d_1,d_2,d_3)$. Now define the polynomial $\mathcal{A}$ in these 12 indeterminates given by $$\mathcal{A}=(bcd)(acd)(abd)(abc)$$ where we used a shorthand notation for determinants $$(bcd)=\left| \begin{array}{ccc} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ d_1 & d_2 & d_3 \\ \end{array} \right|$$ and similarly for the other "bracket factors" $(acd)$, etc. This is an example of multisymmetric function in the "letters" $a,b,c,d$ which is multihomogeneous of multidegree $[3,3,3,3]$. Let me write $M_{[3,3,3,3]}^3$ for the space of such multisymmetric functions. The subscript is the multidegree while the superscript refers to the number of variables in each series. The symmetric group acting is $\mathfrak{S}_4$ which permutes these series. If each series had one variable in it instead of three we would be in the familiar realm of the theory of ordinary symmetric functions. The analogous theory of multisymmetric functions is much more complicated and has been studied classically by Euler, Lagrange, Poisson, Schläfli, Brill, Gordan, Junker and more recently by Angeniol, Dalbec, Briand, Vaccarino, Domokos and others. A nice yet old reference on this is the book on elimination theory by Faa di Bruno I mentioned in this MO answer.

Consider now a generic ternary cubic form $F=F(x_1,x_2,x_3)$. One can write it as $$F(x)=\sum_{i_1,i_2,i_3=1}^3 F_{i_1,i_2,i_3} x_{i_1} x_{i_2} x_{i_3}$$ where $(F_{i_1,i_2,i_3})$ is a symmetric tensor (by which I mean a "matrix" rather than a Bourbaki tensor). This form lives in the symmetric power $S^3(\mathbb{C}^3)$ but I will be (intentionally) sloppy and write $S^3$ instead. The space of degree 4 homogeneous polynomials in the coefficients of $F$ will be denoted by $S^4(S^3)$.

Let me define a linear map $\Phi:S^4(S^3)\rightarrow M_{[3,3,3,3]}^3$ as follows. Given a degree four homogeneous polynomial $C(F)$ first construct its full polarization $$\widetilde{C}(F_1,F_2,F_3,F_4)=\frac{1}{4!} \frac{\partial^4}{\partial t_1\partial t_2\partial t_3\partial t_4} C(t_1F_1+t_2 F_2+t_3 F_3+t_4 F_4)$$ which satisfies $\widetilde{C}(F,F,F,F)=C(F)$. Now set $F_1(x)=a_{x}^3$ where $a_x$ is the classical notation for the linear form $a_1x_1+a_2x_2+a_3x_3$ and similarly specialize to the cubes of linear forms $F_2(x)=b_{x}^3$, $F_3(x)=c_{x}^3$ and $F_4(x)=d_{x}^3$. Finally, define $\Phi(C)=\widetilde{C}(F_1,F_2,F_3,F_4)$.

I will also define another linear map $\Psi: M_{[3,3,3,3]}^3\rightarrow S^4(S^3)$. Let $\mathcal{S}=\mathcal{S}(a,b,c,d)$ be a multisymmetric function in $M_{[3,3,3,3]}^3$. We define the differential operator $$\mathcal{D}=\frac{1}{3!^4} F\left(\frac{\partial}{\partial a_1},\frac{\partial}{\partial a_2},\frac{\partial}{\partial a_3}\right) F\left(\frac{\partial}{\partial b_1},\frac{\partial}{\partial b_2},\frac{\partial}{\partial b_3}\right)$$ $$F\left(\frac{\partial}{\partial c_1},\frac{\partial}{\partial c_2},\frac{\partial}{\partial c_3}\right) F\left(\frac{\partial}{\partial d_1},\frac{\partial}{\partial d_2},\frac{\partial}{\partial d_3}\right)$$ and let $\Psi(\mathcal{S})=\mathcal{D}\ \mathcal{S}$, i.e., the result of applying the differential operator $\mathcal{D}$ to the polynomial $\mathcal{S}$.

It is not hard to see that $\Phi$ and $\Psi$ are inverses of one another. In the particular case of $\mathcal{A}$ above, the image $\Psi(\mathcal{A})$ is Aronhold's degree four $SL_3$ invariant of plane cubics and $\mathcal{A}$ itself is called the (German) symbolic form of that invariant. The classics usually omitted "$\Psi(\cdots)$" but they knew perfectly well what they were doing. Also note that this formalism, in an essentially equivalent form, was invented earlier by Cayley as I briefly explained in Section 1.11.4 of my article "A Second-Quantized Kolmogorov-Chentsov Theorem". One can easily write $\Psi(\mathcal{A})$ applied to $F$, in "physicsy" tensor contraction notation as $$\epsilon_{i_{1} i_{2} i_{3}}\epsilon_{i_{4} i_{5} i_{6}}\epsilon_{i_{7} i_{8} i_{9}}\epsilon_{i_{10} i_{11} i_{12}} F_{i_{4} i_{7} i_{10}} F_{i_{1} i_{8} i_{11}} F_{i_{2} i_{5} i_{12}} F_{i_{3} i_{6} i_{9}}$$ with summation over all twelve indices in the set $\{1,2,3\}$. I used the standard physics notation for the Levi-Civita tensor $\epsilon_{i_1 i_2 i_3}$ equal to zero if some indices are equal and otherwise to the sign of the permutation $i_1,i_2,i_3$ of $1,2,3$. This notation is consistent with the one I used in my previous answers to these MO questions: Q1, Q2, and Q3. At bottom, this is an intrinsically diagrammatic language and if the reader had trouble following what I said so far, it's probably because I did not draw any pictures or "finite-dimensional Feynman diagrams" (I don't know how to do this on MO). To see these pictures, please have a look at this article about an invariant of cubic threefolds or Section 2 of this one mostly on binary forms. In a nutshell, the above contraction can be encoded in a bipartite graph with 4 $\epsilon$-vertices and 4 $F$-vertices, all of degree three.

Of course the above formalism generalizes with no difficulty (except managing notations) to plethysms of symmetric powers $S^p(S^q(\mathbb{C}^n))$ and similar $\Phi$, $\Psi$ maps giving the isomorphism with the relevant space of multisymmetric functions, i.e., $M_{[q,q,\ldots,q]}^{n}$ (with $q$ repeated $p$ times) or in simpler notation $M_{[q^p]}^{n}$. There is a canonical $GL_n$-equivariant map $$H_{p,q}:S^p(S^q(\mathbb{C}^n))\rightarrow S^q(S^p(\mathbb{C}^n))$$ with "H" standing for Hermite, Hadamard or Howe. I will not define it here but it's the first thing that would come to mind if asked to produce such a map (see this review by Landsberg for a detailed discussion in the wider context of geometric complexity theory). It is not hard to see that a homogeneous multisymmetric function in $M_{[p^q]}^{n}\simeq S^q(S^p(\mathbb{C}^n))$ is in the image of $H_{p,q}$ iff it is a polynomial in the homogeneous elementary multisymmetric functions which belong to $M_{[1^q]}^{n}$. In the context of the Aronhold invariant above, these elementary functions are the coefficients of the polynomial $$P(x_1,x_2,x_3)=a_x b_x c_x d_x$$ and similarly in general.

Finally, I can start discussing Will's excellent question. Let $T_{i_1,i_2,i_3}$ denote the tensor of interest with the action of $(g,h,k)\in (SL_n)^3$ given by $$[(g,h,k)\cdot T]_{i_1,i_2,i_3}=\sum_{j_1,j_2,j_3=1}^{n} g_{i_1 j_1} h_{i_2 j_2} k_{i_3 j_3} T_{j_1,j_2,j_3}\ .$$ If Will's tensor had an even number $r$ of indices with the action of $(SL_n)^r$ the answer would be trivial. Take $n$ copies of $T$, contract all first indices of these copies with one $\epsilon$, then all second indices with another $\epsilon$, etc. For $r=3$ this tentative invariant vanishes identically. One can however keep the first indices free, and this gives a map $T\mapsto \Gamma(T)=F$ producing an $n$-ary $n$-ic $F(x_1,\ldots,x_n)$ corresponding to the symmetric tensor $$F_{i_1\ldots i_n}=\sum_{j_1,\ldots,j_n, k_1\ldots,k_n=1}^{n} \epsilon_{j_1\ldots j_n} \epsilon_{k_1\ldots k_n} T_{i_1,j_1,k_1}\cdots T_{i_n,j_n,k_n}\ .$$ For symmetric $T$'s, such as Will's diagonal tensor $D_{ijk}=\delta_{ij}\delta_{ik}$, this is the Hessian covariant. Jason's idea can thus be rephrased as finding an $SL_n$-invariant $I=I(F)$ of $n$-ary $n$-ics $F$ and proposing $I(\Gamma(T))$ as an answer to Will's question. The main issue is to show $I(\Gamma(D))\neq 0$. Now $\Gamma(D)$ is the form $F(x_1,\ldots,x_n)=n!\ x_1 x_2\cdots x_n$.

For $n$ even, the smallest degree for an invariant of $n$-ary $n$-ics $F(x_1,\ldots,x_n)$ is $n$ (unique up to scale). Its symbolic form is $$\mathcal{S_n}=(a^{(1)} a^{(2)}\cdots a^{(n)})^n$$ where $a^{(1)}=(a_{1}^{(1)},a_{2}^{(1)}\ldots,a_{n}^{(1)}),\ldots,a^{(n)}=(a_{1}^{(n)},a_{2}^{(n)}\ldots,a_{n}^{(n)})$ are $n$ series of $n$ variables. Let $I_n=\Psi(\mathcal{S_n})$ denote this invariant. One can also write $$I_n(F)=\epsilon_{i_{1,1}\ldots i_{1,n}}\epsilon_{i_{2,1}\ldots i_{2,n}}\cdots\epsilon_{i_{n,1}\ldots i_{n,n}} \ F_{i_{1,1}\ldots i_{n,1}}F_{i_{1,2}\ldots i_{n,2}}\cdots F_{i_{1,n}\ldots i_{n,n}}$$ with summation over indices understood. The evaluation of $I_n(\Gamma(D))$ is a multiple of the number of row-even minus the number of row-odd $n\times n$ Latin squares. Via work of Huang, Rota and Janssen showing that the result is nonzero is equivalent to the (widely open) Alon-Tarsi conjecture.

The following may be new (if not please let me know): For $n$ odd, the smallest degree for an invariant of $F(x_1,\ldots,x_n)$ is $n+1$ (unique up to scale). Its symbolic form is $$\mathcal{S}=(a^{(2)} a^{(3)}\cdots a^{(n+1)})(a^{(1)} a^{(3)}\cdots a^{(n+1)})\cdots(a^{(1)} a^{(2)}\cdots a^{(n)})\ .$$ Again let me define $I_n=\Psi(\mathcal{S_n})$. This is nonzero because the polynomial in the $a^{(i)}$ is not only multihomogeneous of multidegree $[n^{n+1}]$ but also multisymmetric. For instance the transposition $a^{(1)}\leftrightarrow a^{(2)}$ exchanges the first two bracket factors while the other ones pick up a factor $(-1)$. This gives $(-1)^{n-1}=1$ since $n$ is odd. Other elementary transpositions $a^{(i)}\leftrightarrow a^{(i+1)}$ can be treated in the same way. Of course $I_3$ is Aronhold's degree 4 invariant. I don't know if $I_5$ could be of use for the study of quintic threefolds. I boldly (or foolishly?) make the following conjecture which I think is a natural analog of the Alon-Tarsi conjecture in the odd case.

Conjecture: For $n$ odd, $I_n$ does not vanish on $F(x_1,\ldots,x_n)=x_1 x_2\cdots x_n$.

Here is a proof for $n=3$. The Aronhold invariant of degree 4 is the defining equation for the Veronese secant variety $\sigma_{3}(v_3(\mathbb{P}^2))$. Hence, it does not vanish on forms with border rank $>3$. The border rank of the form $x_1 x_2\cdots x_n$ is exactly $2^{n-1}$ as shown by Oeding in this article. The case $n=3$ relevant to the present proof was shown earlier by Landsberg and Teitler. QED.

A final remark regarding the endgame in Jason's argument with a view to producing an explicit invariant of moderate degree: it follows from what I mentioned above that saying that any homogeneous multisymmetric function in $M_{[p^q]}^{n}\simeq S^q(S^p(\mathbb{C}^n))$ can be expressed as a polynomial in the elementary ones is equivalent to the Hermite-Hadamard-Howe map $H_{p,q}$ being surjective and this is known for $p$ large with an explicit bound due to Brion (see the review by Landsberg I mentioned). The problem I was discussing is closer to the $p=q$ case related to the Foulkes-Howe conjecture. The latter is known to be false, but not by much! (see this article).

Edit: I learned from C. Ikenmeyer that the conjecture about the nonvanishing of $I_n$ for $n$ odd is not new and is stated in Remark 3.26 in https://arxiv.org/abs/1511.02927 However, I noticed recently that this is equivalent to the Alon-Tarsi conjecture for $n$ even. Indeed, the above invariants $I_n$ satisfy the following identity: $$I_n(F(x_1,\ldots,x_{n-1})x_n)=(-1)^{\frac{n}{2}}\frac{n!}{n^n}\ I_{n-1}(F)$$ for all $n$ even and all forms $F$ which only depend on the first $n-1$ variables.

• There is no "problematic endgame" in my answer. There exists an integer $d_0$ such that $D^d$ is a restriction of a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant form for every $d\geq d_0$. Gotzmann's bound gives an upper bound for $d_0$. I believe that $d_0$ might equal $n!$, but Gotzmann appears to give a bound closer to $n! n^{n(n-1)}$. In terms of writing down an explicit invariant form, I doubt it makes much difference whether $d_0$ equals $n!$ or $(n!)^{(n!)^{(n!)}}$: neither of these will be useful for computations. – Jason Starr Jan 10 '17 at 2:27
• Thank you for this really detailed answer! It looks like it reduces my problem to some combinatorial question. Unfortunately it's a bit disappointing for me, because my hope was to take this explicit invariant and calculate it on other forms. If proving the nonvanishing on the diagonal is already a tricky combinatorial problem, I don't expect understanding a more general form will get any easier. But I will still try to compute it on the forms I care about. – Will Sawin Jan 10 '17 at 8:26
• I didn't know the case of 4-tensors was so much simpler. I will try to do computations with the invariant you mention there and see if I get anywhere. – Will Sawin Jan 10 '17 at 8:27
• @Jason: Sorry, just removed the problematic word. I didn't mean to say the argument was incorrect but simply that there were remaining problems to be solved as far as arriving at a fully explicit invariant of reasonable degree, which is what I think Will needs. – Abdelmalek Abdesselam Jan 10 '17 at 12:08
• @Will: Don't be discouraged! Jason's approach plus my two cents is only one possible strategy factoring through $\Gamma$. It is the first thing to try because the invariant theory of "quantics" like $F$ is much more developed than that of tensors like $T$. Another possibility is to avoid using $\Gamma$ and build a graph of $T$'s and $\epsilon$'s by hand. The next available degree to try to do that is $2n$ (since the tentative invariant of degree $n$ vanishes for $r$ odd). It may be possible that a lucky arrangement of $T$'s and $\epsilon$'s... – Abdelmalek Abdesselam Jan 10 '17 at 12:12

Edit. There was a mistake in the original formulation below of Theorem 2.4, p. 134 of Gelfand-Kapranov-Zelevinsky (now corrected). That theorem only applies after passing to the standard open affines. The upshot is that we know that there exists some integer $d_0>0$ such that the determinant $D_{A,k}^{2d}$ gives a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant polynomial that is nonzero on Will Sawin's tensor for all $d\geq d_0$. However, it is not immediately clear what exponent $d_0$ is necessary. An upper bound does follow from the Castelnuovo-Mumford regularity of the symmetric power $(\mathbb{P}(A))^n//\mathfrak{S}_n$ as a subvariety of $\mathbb{P}\text{Sym}^n_k(A)$, and there are bounds on this by Gotzmann. However, the Gotzmann bound is certainly too large for any effective computations.

Original post.
I am just writing up my comments above as an answer. Let $k$ be a commutative, unital ring. Let $A$ be a free $k$-module of rank $n$. Denote by $\mathbb{P}(A)$ the Proj of the graded $k$-algebra $$S^\bullet_k(A^\vee) = \bigoplus_{d\geq 0} \text{Sym}^d_k(A^\vee).$$
For every integer $m\geq 0$, denote by $(S^\bullet_k(A^\vee))^{\otimes m}$ the $m$-fold tensor product $k$-algebra, $$(S^\bullet_k(A^\vee))^{\otimes m} = S^\bullet_k(A^\vee)\otimes_k \cdots \otimes_k S^\bullet_k(A^\vee).$$ This is naturally a $\mathbb{Z}^m$-graded algebra, and it has a natural action of the symmetric group $\mathfrak{S}_m$. Denote by $(S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}$ the $\mathbb{Z}$-graded, $\mathfrak{S}_m$-fixed subalgebra, $$S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots, 1 \rangle} = \bigoplus_{d\geq 0} \text{Sym}^d_k(A^\vee)\otimes_k \cdots \otimes_k \text{Sym}^d_k(A^\vee).$$ Now assume that $m!$ is invertible in $k$, so that $\text{Sym}^m_k(A)^\vee$ is canonically isomorphic to $\text{Sym}^m_k(A^\vee)$ as $k$-vector spaces. Then the transpose of the $m$-multilinear $k$-module homomorphism, $$A^m \to \text{Sym}^m_k(A), \ \ (a_1,\dots,a_m)\mapsto a_1\cdots a_m,$$ induces a $k$-algebra homomorphism, $$S_k^\bullet(\text{Sym}^m_k(A^\vee)) \to (S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}.$$ This, in turn, defines a $k$-morphism, $$\gamma': (\mathbb{P}(A))^m\to \mathbb{P}(\text{Sym}^m_k(A)),\ \ ([a_1],\dots,[a_m]) \mapsto [a_1\cdots a_m]$$ where $(\mathbb{P}(A))^m$ is shorthand for the $m$-fold self-fiber product $\mathbb{P}(A)\times_{\text{Spec}\ k}\cdots \times_{\text{Spec}\ k}\mathbb{P}(A)$. The morphism $\gamma'$ is invariant for the natural action of the symmetric group $\mathfrak{S}_m$ on $(\mathbb{P}(A))^m$. Thus, $\gamma'$ factors through the quotient of this group action. Denote by $\gamma$ the induced $k$-morphism, $$\gamma: (\mathbb{P}(A))^m//\mathfrak{S}_m \to \mathbb{P}(\text{Sym}^m_k(A)).$$

It is a classical fact that $\gamma$ is a closed immersion. This is explicitly stated and proved in Proposition 2.1 and Theorem 2.2, Chapter 4, pp. 132-133 of Gelfand-Kapranov-Zelevinsky. In fact, Gelfand-Kapranov-Zelevinsky prove much more: for every nonzero $x_i\in A^\vee$, denoting by $\gamma_m(x_i)\in \text{Sym}^m_k(A^\vee)$ the associated $m^{\text{th}}$ power, then the induced $k$-algebra homomorphism of affine opens, $$S_k^\bullet(\text{Sym}^m_k(A^\vee))[1/\gamma_m(x_i)] \to [(S_k^\bullet(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}]^{\mathfrak{S}_m}[1/(x_{1,i}\cdots x_{m,i})],$$ is a surjective $k$-algebra homomorphism. This is essentially Theorem 2.4 on p. 134. From this it follows that for every $\mathfrak{S}_m$-invariant, homogeneous element $D$ of degree $r>0$ in $(S_k^\bullet(A^\vee))^{\otimes m}_{(r,\dots,r)}$, there exists an integer $e_0$ such that for every $e\geq e_0$, $D$ times the image of $\text{Sym}^e_k(\text{Sym}^m_k(A^\vee))$ is in the image of $\text{Sym}^{e+r}_k(\text{Sym}^m_k(A^\vee))$. Thus there exists an integer $d>0$ such that $D^d$ is in the image of $\text{Sym}^{dr}_k(\text{Sym}^m_k(A^\vee))$.

In particular, for $m=n$ (assuming that $n!$ is invertible in $k$), the determinant map, $$\text{det}_{A,k}: A^n \to \bigwedge^n_k(A),$$ defines an element $D_{A,k} \in (S_k^\bullet(A^\vee))^{\otimes n}_{(1,\dots,1)}$. This element is not $\mathfrak{S}_m$-invariant. However, the square, $D_{A,k}^2$ is $\mathfrak{S}_m$-invariant. Thus, $D_{A,k}^{2d}$ is in the image $I_{2d}$ of $\text{Sym}_k^{2d}(\text{Sym}_k^n(A^\vee))$ for some integer $d>0$.

Finally, if we assume that $k$ is a field of characteristic $0$, then $\textbf{SL}(A)$ is linearly reductive. Thus, the induced $k$-linear map, $$\text{Sym}^{2d}_k(\text{Sym}^m_k(A^\vee))^{\textbf{SL}(A)} \to I_{2d}^{\textbf{SL}(A)},$$ is surjective. In particular, the $\textbf{SL}(A)$-invariant element $D_{A,k}^{2d}$ is in the image of $\left( \text{Sym}_k^{2d}(\text{Sym}_k^n(A^\vee)) \right)^{\textbf{SL}(A)}$. So there exists some $\textbf{SL}(A)$-invariant element $\Delta_{A,k}$ mappsing to $D_{A,k}^{2d}$.

Now put this together with the tensor operation from the comments. For $k$-vector spaces $A$, $B$, and $C$ of dimension $n$, for every $k$-linear map, $$t:A^\vee \to \text{Hom}_k(B,C),$$ there is an associated element $$\widetilde{t} \in \text{Sym}^n(A) \otimes_k \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)),$$ giving the polynomial map, $$A^\vee\to \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)), \ \ a \mapsto \text{det}(t(a)).$$ The associated map $$\text{Hom}_k(A^\vee,\text{Hom}_k(B,C))\to \text{Sym}^n(A)\otimes_k \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)), \ \ t\mapsto \widetilde{t},$$ is a degree $n$, homogeneous $k$-polynomial map that is equivariant for the natural actions of $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$. Via the duality of $\text{Sym}^n(A)$ and $\text{Sym}^n(A^\vee)$, the polynomial $\Delta_{A,k}$ determines a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant homogeneous polynomial of degree $2dn$ on $\text{Hom}_k(A^\vee,\text{Hom}_k(B,C))$.

For the specific tensor $t$ that you wrote down, with respect to appropriate bases for $A$, $B$, and $C$, for every $(a_1,\dots,a_n)\in A$, $t(a_1,\dots,a_n)$ is the linear transformation whose matrix representative is the diagonal $n\times n$ matrix with entries $(a_1,\dots,a_n)$. Thus, $\widetilde{t}$ equals $(a_1\cdots a_n)\otimes (b_1\wedge \dots \wedge b_n)\otimes (c_1\wedge \dots c_n)$. In particular, since $a_1\cdots a_n$ is in the image of $\gamma$, $\Delta_{A,k}$ restricts on $\widetilde{t}$ as a power of the determinant. Since $a_1\cdots a_n$ is a product of $n$ linearly independent linear polynomials, the determinant is nonzero. Thus, $\Delta_{A,k}$ is a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant, homogeneous polynomial of degree $2n$ that is nonzero on your tensor.

• Thanks! But I think this argument proves too much. You claim that the image of $\left( \text{Sym}_k^2(\text{Sym}_k^n(A^\vee)) \right)^{\textbf{SL}(A)}$ contains a certain element, in particular, it is nonzero. This certainly implies that $\text{Sym}_k^n(A^\vee)$ is self dual as a representation of $\textbf{SL}(A)$. But this is false unless $\dim A=2$, as we can see from highest weight theory - the representation and its dual have different highest weights. – Will Sawin Jan 9 '17 at 8:14
• I may have misread Theorem 2.4, p. 134 of GKZ. Maybe the exponent $d$ is necessary after all. – Jason Starr Jan 9 '17 at 8:21
• I did, indeed, misread Theorem 2.4. So now I return to my original intuition: I suspect that the $n!$ power of the determinant is in the image of the pullback map $\gamma^*$ of homogeneous coordinate rings. I also suspect that someone previously must have written an explicit power of the determinant as a polynomial of the elementary symmetric polynomials. I will try to track down a reference. – Jason Starr Jan 9 '17 at 9:17
• come on people this is precisely the question I was asking in my comments above about why $d=n!$ should work. – Abdelmalek Abdesselam Jan 9 '17 at 13:07
• I wrote $d=n!$ initially because of Noether's bound from invariant theory of finite groups. Now I think it is the correct bound based on trying to extend the determinant polynomial from $n$ to $n+1$. – Jason Starr Jan 9 '17 at 13:17