# A different notion of a decomposable symmetric tensor (besides Veronese)

$$\DeclareMathOperator{\complex}{\mathbb{C}}$$ Let $$\bigvee^m(\complex^n)\subseteq (\complex^n)^{\otimes m}$$ denote the space of symmetric tensors, i.e. the set of $$x \in (\complex^n)^{\otimes m}$$ that are invariant under permutations of the $$m$$ factors. The (cone over) the Veronese variety of $$\bigvee^m(\complex^n)$$ is the set of tensors of the form $$x^{\otimes m}$$ for some $$x \in \complex^n$$. This is one notion of the decomposable tensors.

Another notion is the set of tensors in $$\bigvee^m(\complex^n)$$ of the form $$x_1 \vee \dots \vee x_m$$, where $$x_i \in \complex^n$$ for all $$i \in \{1,\dots, m\}$$, and \begin{align} x_1 \vee \dots \vee x_m :=\sum_{\sigma \in S_m} x_{\sigma(1)}\otimes \dots \otimes x_{\sigma(m)}. \end{align}

Does this latter set have a name? Does it form an algebraic variety (like the Veronese does)? If so, what is its dimension, and have its secant varieties been studied?

• Could you clarify what you mean by $(\mathbb{C}^n)^{\otimes m}$ here? (I'm not sure I know what the "m-factors" of a generic element of $(\mathbb{C}^n)^{\otimes m}$ are, but perhaps I'm just naive). Also, I think the $S_n$ in your definition of $x_1\vee \dots \vee x_m$ should be an $S_m$. – DCM Sep 28 at 17:07
• Ah - you fixed the $S_m$ :) – DCM Sep 28 at 17:10
• @DCM Yes, thanks! I'm not sure what the ambiguity is with $(\mathbb{C}^n)^{\otimes m}$. I mean the $m$-fold tensor product of $\mathbb{C}^n$, as a vector space. – doremifasolatido Sep 28 at 17:10
• @AbdelmalekAbdesselam I thought Comon's conjecture was that the tensor rank and Waring rank of a symmetric tensor agree? For tensor rank, the decomposables are $x_1 \otimes \dots \otimes x_m$. For Waring rank, the decomposables are $x^{\otimes m}$. Neither are $x_1 \vee \dots \vee x_m$. – doremifasolatido Sep 28 at 19:15
• $x_1 \vee x_2=x_1 \otimes x_2 + x_2 \otimes x_1$ is often tensor rank 2, but clearly has rank 1 with respect to the notion I am asking about. – doremifasolatido Sep 28 at 21:37

Your $$\vee$$ is essentially multiplication of polynomials. The variety of tensors $$x_1 \vee \dotsb \vee x_m$$ corresponds to polynomials that factor as products of linear factors. Points of the (projective) variety correspond to hyperplane arrangements. Dually, they correspond to cycles of $$m$$ points (in the dual projective space).

This variety has various names including the Chow variety, the variety of completely decomposable forms, and the split variety. For example Hirotachi Abo calls it the "variety of completely decomposable $$d$$-forms" (his $$d$$ is your $$m$$) and denotes it $$\operatorname{Split_d}(\mathbb{P}^n)$$ (also, his $$n$$ is your $$n+1$$). Douglas A. Torrance calls it the Chow variety and uses the same notation, $$\operatorname{Split_d}$$. Yongui Guan calls it the Chow variety and denotes it $$\operatorname{Ch}_d$$. There are plenty, plenty more references which you can find by searching for those terms.

And it is indeed a variety. If you buy the correspondence with zero-dimensional cycles, that's pretty clearly closed–the points might move around and collide, but it's still a zero-cycle. If you want the equations, Guan seems to have some information; that's just what I found online. Better references would be Gelfand-Kapranov-Zelevinsky and Landsberg's book if you have access to them.

Rank with respect to this variety is called Chow rank, split rank, decomposable rank, etc. Catalisano, et al say that for a monomial $$M$$, an $$M$$-decomposition of a form $$F$$ (where $$\deg(F)=\deg(M)$$) is an expression for $$F$$ as a sum of terms given by evaluating $$M$$ at linear forms. Waring rank is $$x_1^d$$-rank, and Chow (or split, or decomposable) rank is $$(x_1 \dotsm x_d)$$-rank.

If I may also respond to one of the comments: Indeed, Comon's conjecture was that the rank and symmetric rank of symmetric tensors would be equal. Symmetric rank is the same as Waring rank. Shitov gave a counterexample to that conjecture. The analogous conjecture for border ranks, that the border rank and border symmetric rank (border Waring rank) of a symmetric tensor are equal, seems to be still open (as far as I know, as of September 2020).