Set Evincing the Ranks with Respect to an Embedded Variety (Symmetric Tensor Rank and Tensor Rank

Abstract

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. We study when a finite set $S\subset X$ evinces the X-rank of the general point of the linear span of S. We give a criterion when X is the order d Veronese embedding $X_{n,d}$ of ${\mathbb{P}}^n$ and $\left|S\right|\le (\genfrac{}{}{0pt}{}{n+\lfloor d/2\rfloor }{n})$. For the tensor rank, we describe the cases with $\left|S\right|\le 3$. For $X_{n,d},$ we raise some questions of the maximum rank for $d\gg 0$ (for a fixed n) and for $n\gg 0$ (for a fixed d).

1. Introduction

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. For any $q\in {\mathbb{P}}^r$, the X-rank $r_X\left(q\right)$ of q is the minimal cardinality of a finite set $S\subset X$ such that $q\in ⟨S⟩$, where $⟨⟩$ denotes the linear span. The definition of X-ranks captures the notion of tensor rank (take as X the Segre embedding of a multiprojective space) of rank decomposition of a homogeneous polynomial (take as X a Veronese embedding of a projective space) of partially symmetric tensor rank (take a complete linear system of a multiprojective space) and small variations of it may be adapted to cover other applications. See [1] for many applications and [2] for many algebraic insights. For the pioneering works on the applied side, see, for instance, [3,4,5,6,7]. The paper [7] proved that X-rank is not continuous and showed why this has practical importance. The dimensions of the secant varieties (i.e., the closure of the set of all $q\in {\mathbb{P}}^r$ with a prescribed rank) has a huge theoretical and practical importance. The Alexander–Hirschowitz theorem computes in all cases the dimensions of the secant varieties of the Veronese embeddings of a projective space ([8,9,10,11,12,13,14]). For the dimensions of secant varieties, see [15,16,17] for tensors and [18,19,20,21,22,23,24,25,26,27] for partially symmetric tensors (i.e., Segre–Veronese embeddings of multiprojective spaces). For the important problem of the uniqueness of the set evincing a rank (in particular for the important case of tensors) after the classical [28], see [29,30,31,32,33,34,35,36,37,38]. See [39,40,41,42,43,44,45,46,47] for other theoretical works.

Let $S\subset X$ be a finite set and $q\in {\mathbb{P}}^r$. We say that S evinces the X-rank of q if $q\in ⟨S⟩$ and $\left|S\right|=r_X\left(q\right)$. We say that S evinces an X-rank if there is $q\in {\mathbb{P}}^r$ such that S evinces the X-rank of q. Obviously, S may evince an X-rank only if it is linearly independent, but this condition is not a sufficient one, except in very trivial cases, like when $r_X\left(q\right)\le 2$ for all $q\in {\mathbb{P}}^r$. Call $r_{X,\max }$ the maximum of all integers $r_X\left(q\right)$. An obvious necessary condition is that $\left|S\right|\le r_{X,\max }$ and this is in very special cases a sufficient condition (see Propositions 1 for the rational normal curve). If S evinces the X-rank of $q\in {\mathbb{P}}^r$, then $q\in ⟨S⟩$ and $q\notin ⟨S^{\prime }⟩$ for any $S^{\prime }⊊S$. For any finite set $S\subset {\mathbb{P}}^r$, set ${⟨S⟩}^{\prime }:=⟨S⟩\backslash \left({\cup }_{S^{\prime }⊊S}⟨S^{\prime }⟩\right)$. Note that ${⟨S⟩}^{\prime }=\varnothing$ if and only if either $S=\varnothing$ or S is linearly dependent (when $\left|S\right|=1$, ${⟨S⟩}^{\prime }=S$ and S evinces itself). In some cases, it is possible to show that some finite $S\subset X$ evinces the X-rank of all points of ${⟨S⟩}^{\prime }$. We say that S evinces generically the X-ranks if there is a non-empty Zariski open subset U of $⟨S⟩$ such that S evinces the X-ranks of all $q\in U$. We say that S totally evinces the X-ranks if S evinces the X-ranks of all $q\in {⟨S⟩}^{\prime }$. We first need an elementary and well-known bound to compare it with our results.

Let $\rho \left(X\right)$ be the maximal integer such that each subset of X with cardinality $\rho \left(X\right)$ is linearly independent. See ([43] Lemma 2.6, Theorem 1.18) and ([42] Proposition 2.5) for some uses of the integer $\rho \left(X\right)$. Obviously, $\rho \left(X\right)\le r+1$ and it is easy to check and well known that equality holds if and only if X is a Veronese embedding of ${\mathbb{P}}^1$ (Remark 1). If $\left|S\right|\le \lfloor \left(\rho \right(X)+1)/2\rfloor$, then S totally evinces the X-ranks (as in [43] Theorem 1.18) while, for each integer $t>\lfloor \left(\rho \right(X)+1)/2\rfloor$ with $t\le r+1$, there is a linearly independent subset of X with cardinality t and not totally evincing the X-ranks ( Lemma 3). Thus, to say something more, we need to make some assumptions on S and these assumptions must be related to the geometry of X or the reasons for the interest of the X-ranks. We do this in Section 3 for the Veronese embeddings and in Section 4 for the tensor rank. For tensors, we only have results for $\left|S\right|\le 3$ (Propositions 3 and 4).

For all positive integers $n,d$ let ${\nu }_{d,n}:{\mathbb{P}}^n\to {\mathbb{P}}^r$, $r=\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-1$, denote the Veronese embedding of ${\mathbb{P}}^n$, i.e., the embedding of ${\mathbb{P}}^n$ induced by the complete linear system $|{\mathcal{O}}_{{\mathbb{P}}^n}\left(d\right)|$. Set $X_{n,d}:={\nu }_{d,n}\left({\mathbb{P}}^n\right)$. At least over an algebraically closed base field of characteristic 0 (i.e., in the set-up of this paper), for any $q\in {\mathbb{P}}^r$, the integer $r_{X_{n,d}}\left(q\right)$ is the minimal number of d-powers of linear forms in $n+1$ variables whose sum is the homogeneous polynomial associated to q.

We prove the following result, whose proof is elementary (see Section 3 for the proof). In its statement, the assumption “$h^1\left({\mathcal{I}}_{\mathcal{A}}\left(\lfloor d/2\rfloor \right)\right)=0$” just means that the vector space of all degree $\lfloor d/2\rfloor$ homogeneous polynomials in $n+1$ variables vanishing on A has dimension $\left(\genfrac{}{}{0pt}{}{n+\lfloor d/2\rfloor }{n}\right)-\left|A\right|$, i.e., A imposes $\left|A\right|$ independent conditions to the homogeneous polynomials of degree $\lfloor d/2\rfloor$ in $n+1$ variables.

Theorem 1.

Fix integers $n\ge 2$, $d>k>2$ and a finite set $A\subset {\mathbb{P}}^n$ such that $h^1\left({\mathcal{I}}_{\mathcal{A}}\left(\lfloor d/2\rfloor \right)\right)=0$. Set $S:={\nu }_{d,n}\left(A\right)$. Then, S totally evinces the ranks for $X_{n,d}$.

A general $A\subset {\mathbb{P}}^n$ satisfies the assumption of Theorem 1 if and only if $\left|A\right|\le \left(\genfrac{}{}{0pt}{}{n+\lfloor d/2\rfloor }{n}\right)$. For much smaller $\left|A\right|$, one can check the condition $h^1\left({\mathcal{I}}_{\mathcal{A}}\left(\lfloor d/2\rfloor \right)\right)=0$ if A satisfies some geometric conditions (e.g., if A is in linearly general position, it is sufficient to assume $\left|A\right|\le n\lfloor d/2\rfloor +1$).

We conclude the paper with some questions related to the maximum of the X-ranks when X is a Veronese embedding of ${\mathbb{P}}^n$.

2. Preliminary Lemmas

Remark 1.

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. Since any $r+2$ points of ${\mathbb{P}}^r$ are linearly dependent, we have $\rho \left(X\right)\le r+1$. If X is a rational normal curve, then $\rho \left(X\right)=r+1$ because any $r+1$ points of X spans ${\mathbb{P}}^r$. Now, we check that, if $\rho \left(X\right)=r+1$, then X is a rational normal curve. This is well known, but usually stated in the set-up of Veronese embeddings or the X-ranks of curves. Set $n:=\dim X$ and $d:=\mathrm{deg}\left(X\right)$. Assume $\rho \left(X\right)=r+1$. Let $H\subset {\mathbb{P}}^r$ be a general hyperplane. If $n>1$, then $X\cap H$ has dimension $n-1>0$ and in particular it has infinitely many points. Any $r+1$ points of $X\cap H$ are linearly dependent. Now, assume $n=1$. Since X is non-degenerate, we have $d\ge n$. By Bertini’s theorem, $X\cap H$ contains d points of X. Since $\rho \left(X\right)=r+1$, $\dim H=r-1$ and $H\cap X\subset H$, we have $d\le r$. Hence, $d=r$, i.e., X is a rational normal curve.

The following example shows, that in many cases, there are are sets evincing X-ranks, but not totally evincing X-ranks or even generically evincing X-ranks.

Example 1.

Let $X\subset {\mathbb{P}}^r$, $r\ge 3$, be a rational normal curve. Take $q\in {\mathbb{P}}^r$ with $r_X\left(q\right)=r$, i.e., take $q\in \tau \left(X\right)\backslash X$, where $\tau \left(X\right)$ is the tangential variety of X ([48]). Take $S\subset X$ evincing the X-rank of q. Thus, $\left|S\right|=r$ and S spans a hyperplane $⟨S⟩$. Since $\dim \tau \left(X\right)=2$ and $\tau \left(X\right)$ spans ${\mathbb{P}}^r$, $⟨S⟩\cap \tau \left(X\right)$ is a proper closed algebraic subset of $⟨S⟩$. Thus, for a general $p\in ⟨S⟩$, we have $r_X\left(p\right)<\left|S\right|$ and hence S does not generically evinces X-ranks.

Lemma 1.

If $S\subset X$ is a finite set evincing the rank of some $q\in {\mathbb{P}}^r$, then each $S^{\prime }\subset S$, $S^{\prime }\ne \varnothing$, evinces the X-rank of some $q^{\prime }\in {\mathbb{P}}^r$.

Proof. 

We may assume $S^{\prime }\ne S$. Write $S^{″}:=S\backslash S^{\prime }$. Since S evinces the rank of q, S is linearly independent, but $S\cup \left\{q\right\}$ is not linearly independent. Since $S^{\prime }\ne \varnothing$ and $S^{″}\ne \varnothing$, there are unique $q^{\prime }\in ⟨S^{\prime }⟩$ and $q^{″}\in ⟨S^{″}⟩$ such that $q\in ⟨\{q^{\prime },q^{″}\}⟩$. Since S evinces the rank of q, $S^{\prime }$ evinces the rank of $q^{\prime }$. ☐

Lemma 2.

Every non-empty subset of a set evincing generically (resp. totally) X-ranks evinces generically (resp. totally) the X-ranks.

Proof. 

Assume that S evinces generically the X-ranks and call U a non-empty open subset of ${⟨S⟩}^{\prime }$ such that $r_X\left(q\right)=\left|S\right|$ for all $q\in U$; if S evinces totally the X-ranks, take $U:={⟨S⟩}^{\prime }$. Fix $S^{\prime }⊊S$, $S^{\prime }\ne 0$ and set $S^{″}:=S\backslash S^{\prime }$. Let E be the set of all $q\in {⟨S⟩}^{\prime }$ such that $⟨\left\{q\right\}\cup S^{″}⟩\cap U\ne \varnothing$. If $q\in E$, then $r_X\left(q\right)=\left|S^{\prime }\right|$ because $r_X\left(q^{\prime }\right)=\left|S\right|$ for each $q^{\prime }\in ⟨\left\{q\right\}\cup S^{″}⟩\cap U$. Since $S^{\prime }\cap S^{″}=\varnothing$ and $S^{\prime }\cup S^{″}=S$ is linearly independent, E is a non-empty open subset of ${⟨S⟩}^{\prime }$ (a general element of $⟨S⟩$ is contained in the linear span of a general element of $⟨S^{\prime }⟩$ and a general element of $⟨S^{\prime }⟩$). Now, assume $U={⟨S⟩}^{\prime }$. Every element of ${⟨S⟩}^{\prime }$ is in the linear span of an element of ${⟨S^{\prime }⟩}^{\prime }$ and an element of ${⟨S^{″}⟩}^{\prime }$. ☐

Lemma 3.

Take a finite set $S\subset X$, $S\ne \varnothing$.

(a) 

If $\left|S\right|\le \lfloor \left(\rho \right(X)+1)/2\rfloor$, then S totally evinces the X-ranks.

(b) 

For each integer $t>\lfloor \left(\rho \right(X)+1)/2\rfloor$, there is $A\subset X$ such that $\left|A\right|=t$ and A does not totally evince the X-ranks.

Proof. 

Take $q\in {⟨S⟩}^{\prime }$ and assume $r_X\left(q\right)<\left|S\right|$. Take $B\subset X$ evincing the X-rank of q. Since $\left|B\right|<\left|S\right|$, we have $B\ne S$. Since $q\in ⟨S⟩\cap ⟨B⟩$, but no proper subset of either B or S spans q, $S\cup B$ is linearly dependent. Since $\left|B\right|\le \left|S\right|-1$, we have $|B\cup S|\le \rho \left(X\right)$, contradicting the definition of $\rho \left(X\right)$.

Now, we prove part (b). By Lemma 1, it is sufficient to do the case $t=\lfloor \left(\rho \right(X)+1)/2\rfloor +1$. By the definition of the integer $\rho \left(X\right)$, there is a subset $D\subset X$ with $\left|D\right|=\rho \left(X\right)+1$ and D linearly dependent. Write $D=A\bigsqcup E$ with $\left|A\right|=\lfloor \left(\rho \right(X)+1)/2\rfloor +1$ and $\left|E\right|=\rho \left(X\right)+1-\left|A\right|$. Note that $\left|A\right|>\left|E\right|$. Since $\left|A\right|\le \rho \left(X\right)$ (remember that $\rho \left(X\right)\ge 2$), both A and E are linearly independent. Since $A\cup E$ is linearly dependent, there is $q\in ⟨A⟩\cap ⟨E⟩$. Since $\left|D\right|=\rho \left(X\right)+1$, every proper subset of D is linearly independent. Hence, $⟨A^{\prime }⟩\cap ⟨E⟩=\varnothing$ for all $A^{\prime }⊊A$. Thus, $q\in {⟨A⟩}^{\prime }$. Since $\left|E\right|<\left|A\right|$, A does not evince the X-rank of q. ☐

Remark 2.

Take $X\subset {\mathbb{P}}^r$ such that $r_X\left(q\right)\le 2$ for all $q\in {\mathbb{P}}^r$ (e.g., by [49], we may take most space curves). Any set $S\subset X$ with $\left|S\right|=2$ evinces its X-ranks if and only if X contains no line.

3. The Veronese Embeddings of Projective Spaces

Let ${\nu }_{d,n}:{\mathbb{P}}^n\to {\mathbb{P}}^r$, $r:=-1+\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)$, denote the Veronese embedding of ${\mathbb{P}}^n$. Set $X_{n,d}:={\nu }_{d,n}\left({\mathbb{P}}^n\right)$.

Proposition 1.

Let $X\subset {\mathbb{P}}^d$, $d\ge 2$, be the rational normal curve.

(a) 

A non-empty finite set $S\subset X$ evinces some rank of ${\mathbb{P}}^d$ if and only if $\left|S\right|\le d$.

(b) 

A non-empty finite set $A\subset X$ totally evinces the X-ranks if and only if $\left|A\right|\le \lfloor (d+2)/2\rfloor$.

Proof. 

By a theorem of Sylvester’s ([48]), every $q\in {\mathbb{P}}^d$ has X-rank at most d. Thus, the condition $\left|S\right|\le d$ is a necessary condition for evincing some rank. By Lemma 1 to prove part (a), it is sufficient to prove it when $\left|S\right|=d$. Take any connected zero-dimensional scheme $Z\subset X$ with $\mathrm{deg}\left(Z\right)=2$ and $S\cap Z=\varnothing$. Thus, $\mathrm{deg}(Z\cup S)=d+2$. Since $X\cong {\mathbb{P}}^1$, $\mathrm{deg}\left({\mathcal{O}}_{\mathcal{X}}\left(1\right)\right)=d$ and X is projectively normal, we have $h^1\left({\mathcal{I}}_{\mathcal{S}\cup \mathcal{Z}}\left(1\right)\right)=1$ and $h^1\left({\mathcal{I}}_{\mathcal{W}}\left(1\right)\right)=0$ for each $W^{\prime }⊊S\cup Z$. This is equivalent to say that the line $⟨Z⟩$ meets $⟨S⟩$ at a unique point, q and $q\ne Z_{\mathrm{red}}$. By Sylvester’s theorem, $r_X\left(q\right)=d$ ([48]). Since $q\in ⟨S⟩$ and $\left|S\right|=d$, S evinces the X-rank of q.

If $A\ne \varnothing$ and $\left|A\right|\le \lfloor (d+2)/2\rfloor$, then A totally evinces the X-ranks by part (a) of Lemma 3 and the fact that $\rho \left(X\right)=d+1$. Now, assume $d\ge \left|A\right|>\lfloor (d+2)/2\rfloor$. Fix a set $E\subset X\backslash A$ with $\left|E\right|=d+2-\left|A\right|$. Adapt the proof of part (b) of Lemma 3. ☐

Proposition 2.

Fix a set $S\subset X_{n,d}$, $n\ge 2$, with $\left|S\right|=d+1$. The following conditions are equivalent:

1. 

there is a line $L\subset {\mathbb{P}}^n$ such that $|S\cap L|>\lfloor (d+2)/2\rfloor$;

2. 

S evinces no $X_{n,d}$-rank;

3. 

there is $q\in {⟨S⟩}^{\prime }$ such that S does not evince the $X_{n,d}$-rank of q.

Proof. 

Obviously, (2) implies (3). If $X^{\prime }\subset X$ is a subvariety and $q\in ⟨X^{\prime }⟩$, we have $r_{X^{\prime }}\left(q\right)\ge r_X\left(q\right)$. Thus, Sylvester’s theorem ([48]) and Lemma 2 show that (1) implies (2).

Now, assume the existence of $q\in {⟨S⟩}^{\prime }$ such that S does not evince the X-rank of q, i.e., $r_X\left(q\right)\le d$. Take $A\subset {\mathbb{P}}^n$ such that $\nu \left(A\right)=S$ and take $B\subset {\mathbb{P}}^n$ such that ${\nu }_d\left(B\right)$ evinces the X-rank of q. Since $q\in {⟨S⟩}^{\prime }$, (Ref. [50] Lemma 1) gives $h^1({\mathbb{P}}^n,{\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right))>0$. Since $|A\cup B|\le 2d+1$, (Ref. [51] Lemma 34) gives the existence of a line $L\subset {\mathbb{P}}^n$ such that $|L\cap (A\cup B\left)\right|\ge d+2$. Let $H\subset {\mathbb{P}}^n$ be a general hyperplane containing L. Since H is general and $A\cup B$ is a finite set, we have $H\cap (A\cup B)=L\cap (A\cup B)$. Since $|L\cap (A\cup B\left)\right|\ge d+2$, we have $|A\cup B\backslash H\cap (A\cup B\left)\right|\le d-1$ and hence $h^1({\mathbb{P}}^n,{\mathcal{I}}_{A\cup B\backslash H\cap (A\cup B})(d-1))=0$. By ([52] Lemma 5.2), we have $A\backslash A\cap H=B\backslash B\cap H$. ☐

See [53,54] for some results on the geometry of sets $S\subset X_{n,d}$ with controlled Hilbert function and that may be useful to extend Proposition 2.

Proof of Theorem 1:

Set $k:=\lfloor d/2\rfloor$. Note that $h^1\left({\mathcal{I}}_{\mathcal{A}}\left(x\right)\right)=0$ for all $x\ge k$ and in particular $h^1\left({\mathcal{I}}_{\mathcal{A}}(d-k)\right)=0$. Fix $q\in {⟨{\nu }_{d,n}\left(A\right)⟩}^{\prime }$ and assume $r_{X_{n,d}}\left(q\right)<\left|A\right|$. Fix $B\subset {\mathbb{P}}^n$ such that ${\nu }_{d,n}\left(B\right)$ evinces the $X_{n,d}$-rank of q. Since $h^1\left({\mathcal{I}}_{\mathcal{A}}\left(k\right)\right)=0$ and $\left|A\right|>\left|B\right|$, we have $h^0\left({\mathcal{I}}_{\mathcal{B}}\left(k\right)\right)>h^0\left({\mathcal{I}}_{\mathcal{A}}\left(k\right)\right)$. Thus, there is $M\in |{\mathcal{O}}_{{\mathbb{P}}^n}\left(k\right)|$ containing B, but with $A⊈M$, i.e., $A\backslash A\cap M\ne \varnothing$, while $B\backslash B\cap M=\varnothing$. Since $h^1\left({\mathcal{I}}_{\mathcal{A}}(d-k)\right)=0$, we have $h^1\left({\mathcal{I}}_{\mathcal{A}\backslash \mathcal{A}\cap \mathcal{M}}(d-k)\right)=0$. Since $h^1\left({\mathcal{I}}_{\mathcal{A}}\left(d\right)\right)=0$, ${\nu }_{d,n}\left(A\right)$ is linearly independent. Since ${\nu }_{d,n}\left(B\right)$ evinces a rank, it is linearly independent. Grassmann’s formula gives $\dim ⟨{\nu }_{d,k}\left(A\right)⟩\cap ⟨{\nu }_{d,b}\left(B\right)⟩=|A\cap B|+h^1\left({\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right)\right)-1$. We have $A\cup B=\left(\right(A\cup B)\cap M)\cup (A\backslash A\cap M)$. Since $A\backslash A\cap B$ is a finite set, we have $h^2\left({\mathcal{I}}_{\mathcal{A}\backslash \mathcal{A}\cap \mathcal{B}}(d-k)\right)=h^2\left({\mathcal{O}}_{{\mathbb{P}}^n}(d-k)\right)=0$. Since $h^1\left({\mathcal{I}}_{\mathcal{A}\backslash \mathcal{A}\cap \mathcal{M}}(d-k)\right)=0$, the residual exact sequence (also known as the Castelnuovo’s sequence)

$$0\to {\mathcal{I}}_{\mathcal{A}\backslash \mathcal{A}\cap \mathcal{B}}(d-k)\to {\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right)\to {\mathcal{I}}_{\mathcal{M}\cap (\mathcal{A}\cup \mathcal{B}),\mathcal{M}}\left(d\right)\to 0$$

gives $h^1\left({\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right)\right)=h^1(\mathcal{M},{\mathcal{I}}_{\mathcal{M}\cap (\mathcal{A}\cup \mathcal{B})}\left(d\right))$. Since M is projectively normal, $h^1(M,{\mathcal{I}}_{\mathcal{M}\cap (\mathcal{A}\cup \mathcal{B})}\left(d\right))=h^1\left({\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right)\right)$. Thus, the Grassmann’s formula gives $\dim ⟨{\nu }_{d,n}(A\cap M)⟩\cap ⟨{\nu }_{d,n}(B\cap M)⟩=|A\cap B\cap M|+h^1\left({\mathcal{I}}_{\mathcal{A}\cup \mathcal{B}}\left(d\right)\right)-1$. Since $B\subset M$, we get $⟨{\nu }_{d,n}(A\cap M)⟩\cap ⟨{\nu }_{d,n}(B\cap M)⟩=⟨{\nu }_{d,k}\left(A\right)⟩\cap ⟨{\nu }_{d,b}\left(B\right)⟩$. Since $A\cap M⊋A$, we get $q\notin {⟨{\nu }_{d,n}\left(A\right)⟩}^{\prime }$, a contradiction. ☐

4. Tensors, i.e., the Segre Varieties

Fix an integer $k\ge 2$ and positive integers $n_1,\dots ,n_k$. Set $Y:={\prod }_{i=1}^k{\mathbb{P}}^{n_i}$ (the Segre variety) and $N:=-1+{\prod }_{i=1}^k(n_i+1)$. Let $\nu :Y\to {\mathbb{P}}^N$ denote the Segre embedding. Let ${\pi }_i:Y\to {\mathbb{P}}^{n_i}$ denote the projection on the i-th factor. For any $i\in \{1,\dots ,k\}$, set $Y\left[i\right]:={\prod }_{h\ne i}{\mathbb{P}}^{n_h}$ and call ${\eta }_i:Y\to Y\left[i\right]$ the projection which forgets the i-th component. Let $\nu \left[i\right]:Y\left[i\right]\to {\mathbb{P}}^{N_i}$, $N_i:=-1+{\prod }_{h\ne i}(n_h+1)$ denote the Segre embedding of $Y\left[i\right]$. A key difficulty is that $\rho \left(\nu \right(Y\left)\right)=2$ because $\nu \left(Y\right)$ contains lines.

Lemma 4.

Let $S\subset Y$ be any finite set such that there is $i\in \{1,\dots k\}$ with ${\eta }_{i|S}$ not injective. Then, $\nu \left(S\right)$ evinces no rank.

Proof. 

By Lemma 1, we reduce to the case $\left|S\right|=2$, say $S=\{a,b\}$ with $a=(a_1,\dots ,a_k)$, $b=(b_1,\dots ,b_k)$ with $a_i=b_i$ if and only if $i>1$. Since all lines of Y are contained in one of the factors of Y and all lines of $\nu \left(Y\right)$ are images of lines of Y, we get $S\subset \nu \left(Y\right)$. Thus, each element of $⟨\nu \left(S\right)⟩$ is contained in $\nu \left(Y\right)$ and hence it has rank 1. Since $\left|S\right|>1$, $\nu \left(S\right)$ evinces no rank. ☐

Lemma 5.

Let $S\subset Y$ such that there are $S^{\prime }\subseteq S$ and $i\in \{1,\dots ,k\}$ with $|S^{\prime }|=3$, ${\nu }_i\left({\eta }_i\left(S^{\prime }\right)\right)$ linearly dependent and ${\pi }_i\left(S^{\prime }\right)\subset {\mathbb{P}}^{n_i}$ linearly dependent. Then, $\nu \left(S\right)$ evinces no rank.

Proof. 

Let $Q\subset {\mathbb{P}}^3$ be a smooth quadric surface. Q is projectively equivalent to the Segre embedding of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and each point of ${\mathbb{P}}^3$ has at most Q-rank 2 by [47] (Proposition 5.1). By Lemma 1, we may assume $S^{\prime }=S$. By Lemma 4, we may assume that ${\eta }_{i|S}$ is injective. Thus, $|{\eta }_{i|S}|=3$. Since ${\nu }_i\left({\eta }_i\left(S\right)\right)$ is not linearly independent and it has cardinality 3, it is contained in a line of ${\nu }_i\left(Y\left[i\right]\right)$. Thus, ${\eta }_i\left(S\right)$ is contained in a line of one of the factors of $Y\left[i\right]$. By assumption, ${\pi }_i\left(S\right)$ is contained in a line of ${\mathbb{P}}^{n_i}$. Thus, S is contained in a subscheme of Y isomorphic to ${\mathbb{P}}^1\times {\mathbb{P}}^1$. Since each point of ${\mathbb{P}}^3$ has Q-rank $\le 2$ and $\left|S\right|=3$, $\nu \left(S\right)$ evinces no rank. ☐

Remark 3.

Fix a finite set $A\subset Y$ such that $S:=\nu \left(A\right)$ is linearly independent. S evinces no tensor rank if there is a multiprojective subspace $Y^{\prime }\subset Y$ such that $A\subset Y^{\prime }$ and $\left|S\right|$ is larger than the maximum tensor rank of $\nu \left(Y^{\prime }\right)$.

Note that Lemmas 4 and 5 may be restated as a way to check for very low $\left|S\right|$ if there is some $Y^{\prime }$ as in Lemma 3 exists.

Proposition 3.

Take $S\subset \nu \left(Y\right)$ with $\left|S\right|=2$. Let $Y^{\prime }$ be the minimal multiprojective subspace of Y containing S. The following conditions are equivalent:

1. 

S evinces no rank;

2. 

S does not generically evince ranks;

3. 

S does not totally evince ranks;

4. 

$Y^{\prime }\cong {\mathbb{P}}^1$.

Proof. 

Since any two distinct points of ${\mathbb{P}}^N$ are linearly independent (i.e., $⟨S⟩$ is a line) and $\nu \left(Y\right)$ is the set of all points with $\nu \left(Y\right)$-rank 1, S evinces no rank if and only if $⟨S⟩\subset \nu \left(Y\right)$. Use the fact that the lines of $\nu \left(Y\right)$ are contained in one of the factors of $\nu \left(Y\right)$. Since $\nu \left(Y\right)$ is cut out by quadrics, if $⟨S⟩⊈\nu \left(Y\right)$, then $|⟨S⟩\cap \nu (Y\left)\right|\le 2$. Since $S\subset ⟨S⟩\cap \nu \left(Y\right),$ we see that all points of $⟨S⟩\backslash S$ have rank 2 ☐

Proposition 4.

Take $S\subset \nu \left(Y\right)$ with $\left|S\right|=3$ and $\nu \left(S\right)$ linearly independent. Write $S=\nu \left(A\right)$ with $A\subset Y^{\prime }$. Let $Y^{\prime }$ be the minimal multiprojective subspace of Y containing A. Write $Y^{\prime }={\mathbb{P}}^{m_1}\times \dots {\mathbb{P}}^{m_s}$ with $s\ge 1$ and $m_1\ge \dots \ge m_s>0$. We have $m_1\le 2$.

If ${\eta }_{i|A}$ is injective for all i and either $m_2=2$ or $s\ge 4$ or $m_1=2$ and $s=3$, then S totally evinces its ranks. In all other cases for a general $E\in Y^{\prime }$ with $\left|E\right|=3$, $\nu \left(E\right)$ does not generically evince its ranks.

Proof. 

If ${\eta }_{i|A}$ is not injective for some i, then S evinces no rank by Lemma 4. Thus, we may assume that each ${\eta }_{i|A}$ is injective for all i. Each factor of $Y^{\prime }$ is the linear span of ${\pi }_i\left(A\right)$ in ${\mathbb{P}}^{n_i}$. Hence, $m_1\le 2$. Omitting all factors which are points, we get the form of $Y^{\prime }$ we use. If $Y^{\prime }={\mathbb{P}}^1$ (resp. ${\mathbb{P}}^2$, resp. ${\mathbb{P}}^1\times {\mathbb{P}}^1$), then each point of $⟨S⟩$ has rank 1 (resp. 1, resp. $\le 2$). Thus, in these cases, S evinces no rank. If either $Y^{\prime }={\mathbb{P}}^2\times {\mathbb{P}}^1$ or $Y^{\prime }={\left({\mathbb{P}}^1\right)}^3$, then ${\sigma }_2({\mathbb{P}}^2\times {\mathbb{P}}^1)={\mathbb{P}}^5$ and ${\sigma }_2({({\mathbb{P}}^1)}^3)={\mathbb{P}}^7$ ([23,26]). Thus, the last assertion of the proposition is completed.

(a) Assume $s\ge 2$ and $m_2=2$. Taking a projection onto the first two factors, we reduce to the case $s=2$ (this reduction step is used only to simplify the notation). Take a $H\in |{\mathcal{O}}_{{\mathcal{Y}}^{\prime }}(1,0)|$ containing B (this is possible because $h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(1\right)\right)=3>\left|\mathcal{B}\right|$). Since $Y^{\prime }$ is the minimal multiprojective subspace of Y containing A, we have $A\backslash A\cap H\ne \varnothing$. Since $B\backslash B\cap H=\varnothing$, (Ref. [52] Lemma 5.1) gives $h^1\left({\mathcal{I}}_{A\backslash A\cap H}(0,1)\right)>0$. Thus, either there is $A^{\prime }\subset A$ with $|A^{\prime }|=2$ and ${\eta }_{1|A^{\prime }}$ not injective (we excluded this possibility) or $|A\backslash A\cap H|=3$ (i.e., $A\cap H=\varnothing$) and ${\eta }_1\left(A\right)\subset {\mathbb{P}}^2$ is contained in a line R. Set $M:={\mathbb{P}}^2\times R$. We get $A\subset M$ and hence A is a contained in a proper multiprojective subspace, contradicting the definition of $Y^{\prime }$.

(b) Assume $s\ge 3$ and $m_1=2$. By part (a), we may assume $m_2=1$. Taking a projection, we reduce to the case $s=3$, i.e., $Y^{\prime }={\mathbb{P}}^2\times {\mathbb{P}}^1\times {\mathbb{P}}^1$. Take H as in step (a). As in step (a), we get $A\cap H=\varnothing$ and ${\eta }_1\left(A\right)$ contained in a line R of the Segre embedding of ${\mathbb{P}}^1\times {\mathbb{P}}^1$, contradicting the definition of $Y^{\prime }$.

(c) Assume $s\ge 4$. By step (b), we may assume $m_1=1$. Taking a projection onto the first four factors of $Y^{\prime }$, we reduce to the case $Y^{\prime }={\left({\mathbb{P}}^1\right)}^4$. Fix any $H\in |{\mathcal{O}}_{Y^{\prime }}(1,1,0,0)|$ containing B. Assume for the moment $A⊈H$. By ([52] Lemma 5.1), we have $h^1\left({\mathcal{I}}_{A\backslash A\cap H}(0,0,1,1)\right)>0$, i.e., either there are $a=(a_1,a_2,a_3,a_4)\in A$, $b=(b_1,b_2,b_3,b_4)\in A$ with $a\ne b$ and $(a_3,a_4)=(b_3,b_4)$ of $A\cap H=\varnothing$ and the projection of A onto the last 2 factors of $Y^{\prime }$ is contained in a line. The last possibility is excluded by the minimality of $Y^{\prime }$. Thus, $a,b\in A$ exists. Set $A:=\{a,b,c\}$ and write $c=(c_1,c_2,c_3,c_4)$. Permuting the factors of $Y^{\prime }$, we see that, for each $E\subset \{1,2,3,4\}$, there is $A_E\subset A$ with $|A_E|=2$ and ${\pi }_E\left(A_E\right)$ is a singleton, where ${\pi }_E:Y^{\prime }\to {\mathbb{P}}^1\times {\mathbb{P}}^1$ denote the projection onto the factors of $Y^{\prime }$ corresponding to E. Since the cardinality of the set $\mathcal{S}$ of all subset of $\{1,2,3,4\}$ with cardinality 2 is larger than the cardinality of the set of all subsets of A with cardinality 3, there are $E,F\in \mathcal{S}$ such that $E\ne F$ and $A_E=A_F$. If $E\cap F\ne \varnothing$, say $E\cap F=\left\{i\right\}$, then ${\eta }_{i|A}$ is not injective, contradicting our assumption. If $E\cap F=\varnothing$, we have $E\cup F=\{1,2,3,4\}$. Since $A_E=A_F$, we get $|A_E|=1$, a contradiction. ☐

Remark 4.

Take a finite $S\subset \nu \left(Y\right)$ and fix $q\in {⟨\nu \left(S\right)⟩}^{\prime }$. Let $A\subset Y$ be the subset with $\nu \left(A\right)=S$. It is easier to prove that S evinces the rank of q if we know that the minimal multiprojective subspace of Y containing A is the minimal multiprojective subspace $Y^{″}$ of Y with $q\in ⟨\nu \left(Y^{″}\right)⟩$. Note that this is always true if $Y^{″}=Y$, i.e., if the tensor q is concise.

5. Questions on the Case of Veronese Varieties

Let $r_{\max }(n,d)$ denote the maximum of all $X_{n,d}$-ranks (in [55,56] it is denoted with $r_{\max }(n+1,d)$). The integer $r_{\max }(n,d)$ depends on two variables, n and d. In this section, we ask some question on the asymptotic behavior of $r_{\max }(n,d)$ when we fix one variable, while the other one goes to $+\infty$.

Let $r_{\mathrm{gen}}(n,d)$ denote the $X_{n,d}$-rank of a general $q\in {\mathbb{P}}^r$. These integers do not depend on the choice of the algebraically closed base field $\mathbb{K}$ with characteristic 0. The diagonalization of quadratic forms gives $r_{\max }(n,2)=r_{\mathrm{gen}}(n,2)=n+1$. The integers $r_{\mathrm{gen}}(n,d)$, $d>2$, are known by an important theorem of Alexander and Hirschowitz ([8,9,10,11,12,13]); with four exceptional cases, we have $r_{\mathrm{gen}}(n,d)=d\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)/(n+1)\rceil$. An important theorem of Blekherman and Teitler gives $r_{\max }(n,d)\le 2r_{\mathrm{gen}}(n,d)$ (and even $r_{\max }(n,d)\le 2r_{\mathrm{gen}}(n,d)-1$ with a few obvious exceptions) ([57,58]). In particular, for a fixed n, we have

$$\frac{1}{(n+1)!}\le \underset{d\to +\infty }{\lim \inf }{r}_{\max }(n,d)/d^n\le \underset{d\to +\infty }{\lim \sup }{r}_{\max }(n,d)/d^n\le \frac{2}{(n+1)!}.$$

It is reasonable to ask if ${\lim \inf }_{d\to +\infty }{r}_{\max }(n,d)/d^n$ exists and its value. Of course, it is tempting also to ask a more precise information about $r_{\max }(n,d)$ for $d\gg 0$. In the case $n=2$, De Paris proved in [55,56] that $r_{\max }(2,d)\ge \lfloor (d^2+2d+5)/4\rfloor$ ([56] Theorem 3), which equality holds if d is even ([56] (Proposition 2.4)) and suggested that equality holds for all d. Since $r_{\max }(2,d+1)\ge r_{\max }(2,d)$ even for odd d, the integer $r_{\max }(2,d)$ grows like $d^2/4$. Thus, there is an interesting interval between the general upper bound of [57] (which, in this case, has order $d^2/3$) and $r_{\max }(2,d)$. There are very interesting upper bounds for the dimensions of the set of all points with rank bigger than the generic one ([59]).

What are

$$\underset{n\to +\infty }{\lim \sup }\frac{(n+1)!r_{\max }(n,d)}{d^n}\mathrm{and}\underset{n\to +\infty }{\lim \inf }\frac{(n+1)!r_{\max }(n,d)}{d^n}?$$

For all $d\ge 3$, study $r_{\max }(n,d)-r_{\max }(n,d-1)$ and compare for $d\gg 0$$r_{\max }(n,d)-r_{\max }(n,d-1)$ with $r_{\max }(n-1,d)$ and $r_{\mathrm{gen}}(n-1,d)$. Of course, this is almost exactly known when $n=2$ by Sylvester’s theorem ([48]) and De Paris ([55,56]), but $r_{\max }(2,d)-r_{\max }(2,d-1)$ for $d\gg 0$ is both ∼$r_{\mathrm{gen}}(1,d)$ and ∼$r_{\max }(1,d)/2$ and so we do not have any suggestion for the case $n>2$.