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](https://arxiv.org/pdf/1605.06702v3.pdf)) 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.



Explicit invariant of tensors nonvanishing on the diagonal
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](https://mathoverflow.net/questions/117415/old-books-still-used/117417#117417).


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"](https://arxiv.org/abs/1604.05259).
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](https://mathoverflow.net/questions/121715/invariant-polynomials-for-a-product-of-algebraic-groups),
[Q2](https://mathoverflow.net/questions/257709/ring-of-invariants-of-operatornamesl-6-acting-on-lambda3-mathbb-c6),
and [Q3](https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality).
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](https://arxiv.org/abs/1402.2669) about an invariant of cubic threefolds or Section 2
of [this one](https://arxiv.org/abs/0904.1734) 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](https://arxiv.org/abs/1305.7387) 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](https://arxiv.org/abs/1608.02530). The case $n=3$ relevant to the present proof was shown earlier by
[Landsberg and Teitler](https://arxiv.org/abs/0901.0487). 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](https://arxiv.org/abs/1509.03944)).


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.


