Hadamard Products of Projective Varieties with Errors and Erasures

Abstract

In Algebraic Statistics, M.A. Cueto, J. Morton and B. Sturmfels introduced a statistical model, the Restricted Boltzmann Machine, which introduced the Hadamard product of two or more vectors of an affine or projective space, i.e., the componentwise product of their entries, forcing Algebraic Geometry to enter. The Hadamard product $X\star Y$ of two subvarieties $X,Y\subset {\mathbb{P}}^n$ is defined as the Zariski closure of the Hadamard product of its elements. Recently, D. Antolini and A. Oneto introduced and studied the definition of Hadamard rank, and we prove some results on it. Moreover, we prove some theorems on the dimension and shape of the Hadamard powers of X. The aim is to describe the images of the Hadamard products without taking the Zariski closure. We also discuss several scenarios describing the case in which some of the data, i.e., the variety X, is wrong or it is not possible to recover it.

1. Introduction

Around 2010 in Algebraic Statistics, M.A. Cueto, J. Morton and B. Sturmfels introduced a statistical model which needed an extension of the classical Hadamard product of two matrices, i.e., the Hadamard products of vectors or of elements of a projective space. In Algebraic Statistics, this statistical model is associated to the graphical models called Restricted Boltzmann Machines [1,2,3,4]. C. Bocci and E. Carlini published a monograph on the Hadamard products [5], and the interested reader may get from it and its references the applications of Restricted Boltzmann Machines in machine learning. Among these references, we select [6,7,8,9,10,11] for inclusion in our bibliography. For specific results we use some small part of the newer papers [12,13], but for most of the paper small quoted parts of the book [5] are enough. The preface (pages ix to xiii) of [5] explains the motivations coming from Algebraic Statistics and in particular the Restricted Boltzmann Machines.

Take a field K and call $x_0,\dots ,x_n$ the variables of the vector space $K^{n+1}$ and the homogeneous variables of the associated projective space ${\mathbb{P}}^n\left(K\right)$. The Hadamard product $\star :K^{n+1}\times K^{n+1}\to K^{n+1}$ is the coordinatewise product, i.e.,

$$(a_0,\dots ,a_n)\star (b_0,\dots ,b_n):=(a_0{b}_0,\dots ,a_n{b}_n)$$

for all $(a_0,\dots ,a_n)\in K^{n+1}$ and all $(b_0,\dots ,b_n)\in K^{n+1}$.

Let ${\mathbb{P}}^n\left(K\right)$ be the projective space associated to $K^{n+1}$. As in classical projective geometry, each $p\in {\mathbb{P}}^n\left(K\right)$ has $n+1$ homogeneous coordinates, say $p_0,\dots ,p_n$, not all zero, and we write $p=[p_0:\cdots :p_n]$. Note that $[p_0:\cdots :p_n]=[t{p}_0:\cdots :t{p}_n]$ for all $t\in K\setminus \left\{0\right\}$.

We always take as the field K an algebraically closed field $\mathbb{K}$ and use its Zariski topology (see the second part of Section 5 for other fields, e.g., $\mathbb{R}$). Hence, for instance, ${\mathbb{P}}^n={\mathbb{P}}^n\left(\mathbb{K}\right)$ and $X=X\left(\mathbb{K}\right)$. The ⋆-product induces a rational map $h:{\mathbb{P}}^n\times {\mathbb{P}}^n⤏{\mathbb{P}}^n$ and its indeterminacy locus, ${\mathcal{J}}_n$, is the set of all

$$([a_0:\cdots :a_n],[b_0:\cdots :b_n])\in {\mathbb{P}}^n\times {\mathbb{P}}^n$$

such that $a_i{b}_i=0$ for all i. The set ${\mathcal{J}}_n$ is Zariski closed in ${\mathbb{P}}^n\times {\mathbb{P}}^n$.

Take irreducible algebraic varieties $X,Y\subset {\mathbb{P}}^n$ such that $X\times Y⊈{\mathcal{J}}_n$. The rational map $h_{X,Y}:X\times Y⤏{\mathbb{P}}^n$ induced restricting of h to $X\times Y$ is a morphism on its Zariski open subset $X\times Y\setminus (X\times Y)\cap {\mathcal{J}}_n$. Let $X\star Y$ be the Zariski closure of $h(X\times Y\setminus (X\times Y)\cap {\mathcal{J}}_n)$ in ${\mathbb{P}}^n$. Since $h(X\times Y\setminus (X\times Y)\cap {\mathcal{J}}_n)$ is irreducible, $X\star Y$ is an irreducible projective variety, the Hadamard product, or the ⋆-product of X and Y. Obviously, $X\star Y=Y\star X$. For all integers $k\ge 2$, let $X^{\star k}$ denote the Hadamard product of k copies of X. The variety $X^{\star k}$ is the Zariski closure in ${\mathbb{P}}^n$ of all $c_1\star \cdots \star c_k$ with $c_i\in X$ for all i.

Let $X\subset {\mathbb{P}}^n$ be an integral variety. Take closed subsets $A\subset {\mathbb{P}}^n$ and $B\subset {\mathbb{P}}^n$ in the Zariski topology. We see ${\mathbb{P}}^n$ as our screen. We say that $u\in {\mathbb{P}}^n$ is an erasure if it cannot be seen. We say that $u\in {\mathbb{P}}^n$ is an error if we see it, but it is wrong. In the theory of error-correcting codes, an error costs two erasures ([14], pp. 44–45). We see the elements of X as our data. We apply k times Hadamard products to elements of X or to elements that we think are in X and then see the screen. The elements of A give the errors, while the elements of B give the erasures. There are two different scenarios. Case $\alpha$ is when $A\cup B\subset X$ (a part of our data is not correct or it not possible to access it). In case $\beta$, we drop this assumption, i.e., A and B may come from an enemy. We subdivide cases $\alpha$ and $\beta$ in the following way.

Take an irreducible component W of $A\cup B$. We say that W is in the scenario ${\alpha }_1$ or ${\beta }_1$ if all elements of $W\star X^{\star (k-1)}$ are errors or erasures.

We say that W is the scenario ${\alpha }_2$ or ${\beta }_2$ if all elements of $W^{\star k}$ are errors or erasures. Note that in case $\alpha$, we have $W^{\star k}\subseteq W\star X^{\star (k-1)}$. We may also allow a case intermediate between ${\alpha }_2$ and ${\alpha }_1$ in which we need to avoid a prescribed fraction of the k Hadamard products which comes from W. We discuss these cases in Section 5, where we point out why the new results and definitions of our paper may be used to mitigate these issues.

Computing the dimension of an Hadamard product, e.g., $W^{\star k}$ and $W\star X^{\star (k-1)}$, is a perfect work for an Algebraic Geometer or an expert in Computational Algebra well-versed in [5]. A main contribution for case $\alpha$ is our introduction of two concepts, Hadamard open rank and weak Hadamard rank. It is very important to see the true images before doing the Zariski closure (see Theorem 4 for a case with a linear space).

Even if $A\cup B=\varnothing$, to get all points of the screen, we need the variety $X^{\star k}$ to increase with k until it is ${\mathbb{P}}^n$. This is not always the case, but it is true in many cases. We discuss this assumption in Remark 1. The necessary and sufficient conditions are easy to test. We just point out that in case $\alpha$, to see the full screen we need to compute the first integer k such that $X^{\star k}={\mathbb{P}}^n$. See Section 5 for more details.

Section 2 contains the notation and some remarks.

Section 3 contains three new definitions: Hadamard open rank, weak Hadamard open rank, and weak Hadamard rank. It contains some related new results (Proposition 1 and Theorems 1, 2 and 3) with many examples and observations. The main point of the section is the discussion (with some new results) of the Hadamard rank introduced by D. Antolini and A. Oneto in [13], a very useful definition, far better than the weak Hadamard rank. We feel that this section is useful for many other problems, because it describes what is seen in the Hadamard products before taking the closure in the Zariski topology. At the end of the section, we list three open questions.

Section 4 contains several new results on the classical Hadamard products, in particular the dimensions of them. One of the main results (Theorem 4) describes the Hadamard powers of a v-dimensional vector space, and it also gives the true images before taking the closure. The case $v=1$ was known before ([10], Lemma 2.10 and Th. 3.4, [5], Theorem 2.1, [13], Lemma 3.3).

Section 6 adapts to the case of the Hadamard products the classical notions of cones and strange embedded varieties. We only have trivial examples (Examples 4, 5, 6), but we explain why a too-naive extension of the definition of a strange variety would not be very interesting.

Then there are two sections, Section 7 and Section 8, in which we discuss this paper with respect to the older results, quote the main results proved in this paper, and give some suggestions for future works.

2. Notation and Preliminaries

We fix an algebraically closed field $\mathbb{K}$. All algebraic varieties are defined over $\mathbb{K}$. For any integer $n\ge 1$, we fix a system of homogeneous coordinates $x_0,\dots ,x_n$ of ${\mathbb{P}}^n$. Set $H_i:=\{x_i=0\}\subset {\mathbb{P}}^n$, $i=0,\dots ,n$, ${\Delta }_{n-1}:=H_0\cup \dots \cup H_n$ and $U_n:={\mathbb{P}}^n\setminus {\Delta }_{n-1}$. The set $H_0,\dots ,H_n$ are the coordinate hyperplanes. The coordinate linear subspaces are the intersections of finitely many coordinate hyperplanes. Let ${\Delta }_v$ denote the union of the v-dimensional coordinate linear subspaces.

Let $X\subset {\mathbb{P}}^n$ be an integral variety. The variety X is said to be concise if $X\cap U_n\ne \varnothing$, i.e., if X is not contained in a coordinate hyperplane.

Remark 1.

Let $X\subset {\mathbb{P}}^n$ be an integral projective variety. An obvious necessary condition for the existence of a positive integer k such that $X^{\star k}={\mathbb{P}}^n$ is that X is not contained in a coordinate hyperplane, i.e., that X is concise. Assume that X is concise. By [13], Th. 3.3, $X^{\star k}={\mathbb{P}}^n$ for some large integer k if and only if X is contained in a binomial hypersurface. If we know a system of homogeneous equations describing X, then it is easy to check if X is contained in a binomial hypersurface. If $X^{\star k}={\mathbb{P}}^n$ for some large k, then $X^{\star n}={\mathbb{P}}^n$, because for a concise variety, the sequence $dim{X}^{\star k}$ is strictly increasing until $dim{X}^{\star k}=n$, i.e., $X^{\star k}={\mathbb{P}}^n$.

Set $\mathbf{1}:=[1:\cdots :1]$. We have $\mathbf{1}\in U_n$. Let G be the quotient of the diagonal subgroup of $GL(n+1,\mathbb{K})$ by the multiples of the identity. The group G acts on ${\mathbb{P}}^n$ sending each coordinate linear subspace into itself and acting transitively on $U_n$. Thus $U_n$ is the G-orbit of $\mathbf{1}$.

For any set $S\subseteq {\mathbb{P}}^n$, let $\langle S\rangle$ denote its linear span, i.e., the intersection of all hyperplanes of ${\mathbb{P}}^n$ (not only the coordinate hyperplanes) containing S, with the convention $\langle S\rangle ={\mathbb{P}}^n$ if there is no such a hyperplane. Let $\ll S\gg$ denote the coordinate linear span of S, i.e., the minimal coordinate linear subspace of ${\mathbb{P}}^n$ containing S.

Note that we have $X\star Y\subseteq \ll X\gg \cap \ll Y\gg$ for all integral subvarieties X and Y of ${\mathbb{P}}^n$.

We recall the following Terracini Lemma for the Hadamard product ([5], Lemma 1.6) in the form true in arbitrary characteristic ([15], Cor. 1.10).

Lemma 1.

Let $X\subset {\mathbb{P}}^n$ and $Y\subset {\mathbb{P}}^n$ be integral varieties. Fix $p\in X$ and $q\in Y$. Then:

1.

$T_{p\star q}X\star Y\supseteq \langle T_{p}X\star q\cup p\star T_{q}Y\rangle$;

2.

In characteristic 0 if $(p,q)$ is general in $X\times Y$, then $T_{p\star q}X\star Y=\langle T_{p}X\star q\cup p\star T_{q}Y\rangle$.

Proof. 

Mimic the proof of ([15], Cor. 1.10) with ⋆ instead of the addition +. □

As in ([15], Cor. 1.10), the part for non-general points is true even if p is a singular point of X or q is a singular point of Y.

Remark 2.

Let $X\subset {\mathbb{P}}^n$ and $Y\subset {\mathbb{P}}^n$ be integral varieties. As an arbitrary characteristic, the definition of $X\star Y$ as the image of a rational map $X\times Y⤏{\mathbb{P}}^n$ gives $dimX\star Y\le min\{n,dimX+dimY\}$. The integer $min\{n,dimX+dimY\}$ is often called the expected dimension of $X\star Y$: A warning. Set $m:=dimX$ and assume that X is concise. Fix an integer $k\ge 2$. We saw that $dim{X}^{\star k}\le min\{km,n\}$. Often we say that $min\{km,n\}$ is the expected dimension of $X^{\star k}$. However, if there is a positive-dimension subgroup Γ of G such that $g\left(X\right)=X$ for all $g\in \Gamma$, then $dim{X}^{\star k}\le min\{km-{\gamma }^{k-1},n\}$, where $\gamma :=dim\Gamma$. In this case we say, as in [5], that $min\{km-{\gamma }^{k-1},n\}$ is the expected dimension of $X^{\star k}$.

3. The Open and the Weak Hadamard Ranks

Definition 1.

Let $X\subset {\mathbb{P}}^n$ be an integral subvariety. For any $q\in {\mathbb{P}}^n$, the weak Hadamard X-rank $H{r}_X\left(q\right)$ of q is the minimal integer t such that there is $S\in S(X,t)$ such that q is the Hadamard product of the elements of S, with the convention $H{r}_X\left(q\right)=+\infty$ if there is no such t.

For each $q\in {\mathbb{P}}^n$ and any positive integer t, let $\mathcal{S}(X,t,q)$ denote the set of all $S\in S(X,t)$ such that q is the Hadamard product of the elements of S. We have $\mathcal{S}(X,t,q)=\varnothing$ if $t<H{r}_X\left(q\right)$ and hence $\mathcal{S}(X,t,q)=\varnothing$ if $H{r}_X\left(q\right)=+\infty$. Each set $\mathcal{S}(X,t,q)$ is constructible ([16], Ex. II.3.18, Ex. II.3.19). Hence, it makes sense to speak about the dimensions and the irreducible components of $\mathcal{S}(X,t,q)$.

Set $\mathcal{S}(X,0,q)=\mathcal{S}(X,\le 0,q)=\varnothing$. For each positive integer t set

$$\mathcal{S}(X,\le t,q):={\cup }_{x\le t}\mathcal{S}(X,x,q).$$

Each set $\mathcal{S}(X,\le t,q)$ is constructible.

For each $q\in {\mathbb{P}}^n$ such that $H{r}_X\left(q\right)$ is finite, let $\mathcal{S}(X,q)$ denote the set of all finite sets $S\in S(X,H{r}_X\left(q\right))$ such that q is the Hadamard product of the elements of S. We have $\mathcal{S}(X,q)\ne \varnothing$, but note that we are assuming that $H{r}_X\left(q\right)$ is finite.

Remark 3.

Assume $X^{\star k}={\mathbb{P}}^n$ for some positive integer k, and let w be the minimal integer such that $X^{\star w}={\mathbb{P}}^n$. Then $H{r}_X\left(q\right)=w$ for a general $q\in {\mathbb{P}}^n$.

Remark 4.

Set $m:=dimX$ and $w:=\lceil n/m\rceil$. A dimensional count gives $X^{\star (w-1)}\ne {\mathbb{P}}^n$. Assume $X^{\star w}={\mathbb{P}}^n$ and take a general $q\in {\mathbb{P}}^n$. A dimensional count gives $H{r}_X\left(q\right)=w$ and $dim\mathcal{S}(X,q)=mw-n$.

We recall the definition of the Hadamard rank introduced in [13] by D. Antolini and A. Oneto, in which we allow the Hadamard product of the same point.

For any positive integer x let $\tilde{S}(X,x)$ denote the set of all finite multisubsets of X of total weight x, i.e., $A\in \tilde{S}(X,x)$ if and only if there is a positive integer $t\le x$, $A\in S(X,t)$, an ordering $\{a_1,\dots ,a_t\}$ of the elements of A and positive integers $m_1,\dots ,m_t$ such that $m_1+\cdots +m_t=x$. We write $\{(a_1,m_1),\dots ,(a_t,m_t)\}$ for the element $A\in \tilde{S}(X,x)$. If $a_1\star \cdots \star a_t$ is well-defined, then the Hadamard product of A is the product of x points of X with each $a_i$ being a ⋆-factor $m_i$ times.

Definition 2.

Let $X\subset {\mathbb{P}}^n$ be an integral subvariety. The Hadamard rank $H{\tilde{r}}_X\left(q\right)$ of q is the minimal integer t such that there is $S\in \tilde{S}(X,t)$, and q is the Hadamard product of the elements of multiset S, with the convention $H{\tilde{r}}_X\left(q\right)=+\infty$ if there is no such t.

Obviously, $H{\tilde{r}}_X\left(q\right)\le H{r}_X\left(q\right)$ for all $q\in {\mathbb{P}}^n$ and $H{\tilde{r}}_X\left(a\right)=H{r}_X\left(a\right)=1$ for all $a\in X$. The following example shows that sometimes strict inequality holds.

Example 1.

Sometimes, $H{\tilde{r}}_X\left(p\right)<H{r}_X\left(p\right)$. Take for instance a general line $L\subset {\mathbb{P}}^n$. The variety $L^{\star 2}$ is a plane, and the multiplication map is a degree 2 finite morphism $h:L\times L\to L^{\star 2}$ ramified on the diagonal ([5], Th. 2.1). Fix a general $p\in L$. Since $p\star p\notin L$, $H{\tilde{r}}_X(p\star p)>1$. We have $H{\tilde{r}}_L(p\star p)=2$. There are no $p_1,p_2\in L$ such that $p_1\ne p_2$ and $p_1\star p_2=p\star p$, because h is a degree 2 finite morphism between smooth varieties, and $p\star p$ is a ramification point of the degree 2 morphism h.

Easy examples given in [13] show that even with a concise variety X, a general point q of ${\mathbb{P}}^n$ may have $H{\tilde{r}}_X\left(q\right)=+\infty$. A major result of [13] is that this is the case if and only if the concise variety X is contained in a binomial hypersurface ([13], Th. 3.38) and their proof works for the weak Hadamard rank.

Remark 5.

Let $X\subset {\mathbb{P}}^n$ be an integral variety. Fix an integer $k\ge 2$ and assume the inequality $dim{X}^{\star k}\ne dim{X}^{\star (k-1)}$. Then $H{r}_X\left(q\right)=H{\tilde{r}}_X\left(q\right)=k$ for a general $q\in X^{\star k}$.

We recall the definition of open rank introduced and studied by J. Jelisiejew in [17].

Definition 3.

For any $q\in {\mathbb{P}}^n$, the open rank or open X-rank $o{r}_X\left(q\right)$ of q is the minimal integer with the following property: for any closed set $B⊊X$, there exists $S\subset X\setminus B$ such that $\#S\le o{r}_X\left(q\right)$ and $q\in \langle S\rangle$, where $\langle \rangle$ denotes the linear span.

We introduce a similar notion in the setup of the Hadamard rank.

Definition 4.

Let $X\subset {\mathbb{P}}^n$ be an integral subvariety. For any $q\in {\mathbb{P}}^n$, the weak open Hadamard rank $H{o}_X\left(q\right)$ is the minimal integer t such that for all closed subset $T⊊X$, there is $S\in S(X\setminus T,t)$ such that q is the Hadamard product of the elements of S, with the convention $H{o}_X\left(q\right)=+\infty$ if there is no such t. If we allow $S\in \tilde{S}(X\setminus T,t)$ we get the definition of the Hadamard open rank $H{\tilde{o}}_X\left(q\right)$.

Remark 6.

We have $H{\tilde{o}}_X\left(q\right)\ge H{r}_X\left(q\right)$ and often strict inequality holds. For instance, if $q\in X$, then $H{r}_X\left(q\right)=1$. Taking $T=\left\{q\right\}$ we see that $H{\tilde{o}}_X\left(q\right)>1$. Obviously, $H{\tilde{o}}_X\left(q\right)\le H{o}_X\left(q\right)$.

Proposition 1.

Let $X\subset {\mathbb{P}}^n$ be an integral and concise curve. Then $H{\tilde{o}}_X\left(q\right)=+\infty$ for all $q\in {\Delta }_{n-1}$.

Proof. 

Since X is a concise integral curve, the set $T:=X\cap {\Delta }_{n-1}$ is a finite set. Assume the existence of a finite set $S\subset X$ whose Hadamard product is q, allowing repetitions of the elements of S. Since we are working over a field and at least one homogeneous coordinate of q is 0, $T\cap S\ne \varnothing$. Hence S cannot be used to check that the Hadamard open rank of q is finite. □

Example 2.

Fix an integer $d\ge 2$ and assume that either $\mathrm{char}\left(\mathbb{K}\right)=0$ or $\mathrm{char}\left(\mathbb{K}\right)$ is a prime not dividing d. Let X be the degree d smooth Fermat plane curve, i.e., set $X:=\{x_0^d+x_1^d+x_2^d=0\}$. It is easy to check that $X^{\star 2}={\mathbb{P}}^2$ (if $\mathrm{char}\left(\mathbb{K}\right)=0$ it is sufficient to use that it is not binomial and apply [6], Th. 5.3). Let q point in $H_i\setminus H_i\cap X$ for some $i\in \{0,1,2\}$. Proposition 1 gives $H{o}_X\left(q\right)=H{\tilde{o}}_X\left(q\right)=+\infty$.

We claim that $H{r}_X\left(q\right)\le 5$ if q is not a coordinate point, and $H{r}_X\left(q\right)\le 6$ if q is a coordinate point. By [13], Prop 3.11, each element of $U_2$ has at most Hadamard rank 4. Moreover, for each $x\in U_2$, it is easy to check that we may get $A\in S(U_2,4)$ whose weak Hadamard product is x.

First assume that $q=[q_0:q_1:q_2]$ is not a coordinate point. With no loss of generality we may assume $q\in H_0\setminus \{[0:1:0],[0:0:1]\}$. Take $p=[0:p_1:p_2]\in X\cap H_0$. Since X is a Fermat curve and either $\mathrm{char}\left(\mathbb{K}\right)=0$ or $\mathrm{char}\left(\mathbb{K}\right)$ is a prime not dividing d, $p_1\ne 0$ and $p_2\ne 0$. Set $o:=[1:q_1/p_1:q_2:p_2]$.

Take $A\in S(X\cap U_2,4)$ such that o is the Hadamard product of A. Obviously, q is the Hadamard product of $A\cup \left\{o\right\}$.

Now assume that q is a coordinate point, say $q=[0:0:1]$. Take $a\in X\cap H_1$ whose Hadamard product is a. Note that $a\ne o$. Obviously, $[0:0:1]$ is the Hadamard product of $A\cup \{o,a\}$.

Theorem 1.

Let $X⊊{\mathbb{P}}^n$ be an integral variety such that $X^{\star k}={\mathbb{P}}^n$ for some positive integer k. Then $H{\tilde{o}}_X\left(q\right)\le 2k$ for all $q\in U_n$.

Proof. 

We adapt the proof of [13], Prop. 3.11.

Since $X\ne {\mathbb{P}}^n$, $k>1$. Since $X^{\star k}={\mathbb{P}}^n$, X is concise. Up to an element of G we may assume $\mathbf{1}\in X$. Fix $q\in U_n$ and a closed $T⊊X$. Since $\mathbf{1}\in X$, $X^{\star k}={\mathbb{P}}^n$ and X is the Zariski closure of $X\cap U_n$, there is a non-empty Zariski open subset $U\subseteq U_n$ such that for each $p\in U$ there is $S_p\in S(\mathit{X}\setminus \mathit{T},\mathit{k})$ with p the Hadamard product of the elements of $S_p$. The rational map $[x_0:\cdots :x_n]⤏[{\displaystyle \frac{1}{x_0}}:\cdots :{\displaystyle \frac{1}{x_n}}]$ induces a morphism $u:U_n\to U_n$ such that $u ○ u$ is the identity map. Since $q\in U_n$, the ⋆-product with q induces an automorphism of $U_n$. Set $V:=U\cap (q\star U)\cap u(q\star U)$ and take $y\in V$. Take $A\in S(X\setminus T,k)$ and $B_y\in S(X\setminus T,k)$ such that $q\star y$ is the Hadamard product of the elements of A, and $y^{-1}$ is the Hadamard product of the elements of q. We have $q=(q\star y)\star y^{-1}$. If $A\cap B_y=\varnothing$, then we may take $A\cup B_y$. Now assume $A\cap B_y\ne \varnothing$. In this case, we get $A\cup B_y\in \tilde{S}(X\setminus T,2k)$, taking each element of $A\cap B_y$ with multiplicity 2. □

Remark 7.

Set $m:=dimX$. Take a point $q\in {\mathbb{P}}^n$ and a positive integer t such that we have $dim\mathcal{S}(X,\le t,q)<m$. Since ${\cup }_{S\in \mathcal{S}(X,\le t,q)}S$ is contained in a proper closed subset of X, we have $H{\tilde{o}}_X\left(q\right)>t$.

Remark 8.

Set $m:=dimX$ and $w:=\lceil n/m\rceil$. Assume $X^{\star w}={\mathbb{P}}^n$ and take a general $q\in {\mathbb{P}}^n$. Remark 4 gives $dim\mathcal{S}(X,\le w,q)=n-wm<m$. Thus, Remark 7 gives $H{o}_X\left(q\right)>w$. Theorem 1 gives $H{\tilde{o}}_X\left(q\right)\le 2w$.

Theorem 2.

Let $X\subset {\mathbb{P}}^n$, $n\ge 2$, be an integral curve such that $X^{\star n}={\mathbb{P}}^n$. Then $H{o}_X\left(p\right)=n+1$ for a general $p\in {\mathbb{P}}^n$.

Proof. 

Since $X^{\star n}={\mathbb{P}}^n$, X is concise and $X^{\star (n-1)}$ is a hypersurface of ${\mathbb{P}}^n$. A dimensional count gives $dim\mathcal{S}(X,n+1,p)=1$ for a general $p\in {\mathbb{P}}^n$, that $\mathcal{S}(X,p)$ is finite for a general $p\in {\mathbb{P}}^n$ and that $\mathcal{S}(X,q)$ is finite for a general $q\in X^{\star (n-1)}$. Remark 7 gives $H{\tilde{o}}_X\left(p\right)>n$. Thus to conclude the proof, it is sufficient to prove that $H{o}_X\left(p\right)\le n+1$. Since we have $dim\mathcal{S}(X,n+1,p)=1$, it is sufficient to prove that for every irreducible one-dimensional set $\Gamma \subseteq \mathcal{S}(X,n+1,p)$, we have ${\cap }_{S\in \Gamma }S=\varnothing$. Assume ${\cap }_{S\in \Gamma }S\ne \varnothing$ and take $a\in {\cap }_{S\in \Gamma }S$. For each $S\in \Gamma$, set $S_a:=S\setminus \left\{a\right\}$. For each $S_a$, $S\in \Gamma$ is an element of $S(X,n)$. Since p is general, $p\in U_n$ and hence $a\in X\cap U_n$. Set $q_1:=a^{-1}q$ and $x:=r_X\left(q_1\right)$. First assume $H{r}_X\left(q_1\right)=n$. By assumption, $\mathcal{S}(X,q_1,n)$ contains a one-dimensional family isomorphic to $\Gamma$. Since $dimX=1$ and $dim{\mathbb{P}}^n=n$, varying the general point $p\in {\mathbb{P}}^n$, we get at least a locally closed hypersurface $F\subset {\mathbb{P}}^n$ such that each $q_1\in F$ has Hadamard rank x and $dim\mathcal{S}(X,n,q_1)\ge 1$. Since $\mathcal{S}(X,p)$ is finite for a general $p\in {\mathbb{P}}^n$, we get $x=n-1$. Since $dimF+dim\mathcal{S}(X,n,q_1)\ge n$, we get $dim{X}^{\star n}\le n-1$, a contradiction. □

Remark 9.

Take a variety X and finite sets $A\subset X$ and $B\subset X$ as in case α and q such that $x:=H{o}_X\left(q\right)<+\infty$. Taking T containing $A\cup B$ in the definition of weak Hadamard open rank, we get the existence of $S\in S(X\setminus (A\cup B),y)$ for some $y\le x$ such that q is the Hadamard product of the elements of S.

Question 1: Let $X\subset {\mathbb{P}}^n$ be an integral variety such that $X^{\star x}={\mathbb{P}}^n$ for some x and let w be the minimal integer such that $X^{\star w}={\mathbb{P}}^n$. Set $m:=dimX$. Compute the minimum and the maximum of $H{r}_X\left(p\right)$ and $H{\tilde{r}}_X\left(p\right)$ for $p\in U_n$ (either for a fixed interesting X or for all varieties X with given w and m).

Question 2: Take $X\subset {\mathbb{P}}^n$ and $q\in {\mathbb{P}}^n$ such that $H{\tilde{o}}_X\left(q\right)<+\infty$. Is $H{o}_X\left(q\right)<+\infty$?

Question 3: Take $X\subset {\mathbb{P}}^n$ and assume $H{o}_X\left(q\right)<+\infty$ for all $q\in {\mathbb{P}}^n$. How large can $H{o}_X\left(q\right)-H{\tilde{o}}_X\left(q\right)$ be for some $q\in {\mathbb{P}}^n$?

4. Classical Hadamard Products

For all $p\in {\mathbb{P}}^n$, the Hadamard border rank of p with respect to X is the minimal positive integer x such that $p\in X^{\star x}$ with the convention that it is $+\infty$ if $p\notin X^{\star x}$ for all positive integers x.

Lemma 2.

Let $D\subset {\mathbb{P}}^n$, $n\ge 2$, be an integral curve such that $D\cap {\Delta }_{n-2}=\varnothing$. Then we have $D^{\star n}={\mathbb{P}}^n$.

Proof. 

Since $D\cap {\Delta }_{n-2}=\varnothing$, for all $x\in \{2,\dots ,n\}$ the restriction to $D^x$ of the multiplication map is a morphism. Since $D^x$ is projective, its image is $D^{\star x}$, without taking the closure. Since $X\cap U_n\ne \varnothing$, up to a diagonal change of coordinates, we may assume that $\mathbf{1}\in X$. Assume that the lemma is false and take the minimal integer $x<n$ such that $dim{D}^{\star x}=dim{D}^{\star (x+1)}$. Since $\mathbf{1}\in X$, $D^{\star x}=D^{\star (x+1)}$. The assumption $D\cap {\Delta }_{n-2}=\varnothing$ is equivalent to say that the homogeneous coordinates of each point of D have at most 1 zero. Since $h_{D^{x+1}}$ is a morphism, $D^{\star (x+1)}$ contains points with $x+1$ entries 0 as their homogeneous coordinates. Since each element of $h_{D^x}$ has at most x zeroes as its coordinates, we get a contradiction. □

The following result is a partial extension of ([12], Th. 1.4) to the positive characteristic case. The two proofs are different, because a key quotation used in [12] is not characteristic free. Indeed, in [12] and in [5], characteristic 0 is always assumed.

Theorem 3.

Take $\mathbb{K}$ of an arbitrary characteristic. Let $X⊊{\mathbb{P}}^n$ be an integral and of a concise variety. Let $Y\subset {\mathbb{P}}^n$ be an integral curve such that $Y\cap {\Delta }_{n-1}=\varnothing$. Then $dimX\star Y=dimX+1$.

Proof. 

Up to a diagonal change of coordinates, we may assume $\mathbf{1}\in Y$. By assumption, $X\ne {\mathbb{P}}^n$. Since $dimX\times Y=dimX+1$, it is sufficient to prove that $dimX\star Y>dimX$. Assume $dimX\star Y=dimX$. Since $\mathbf{1}\in Y$, we get $X\star Y=X$. Hence $X\star Y^{\star x}=X$ for all $x>1$. Taking $x=n$ and applying Lemma 2, we get a contradiction because X is concise. □

Lemma 3.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Let $X\subset {\mathbb{P}}^n$ be an integral m-dimensional variety such that $X\cap {\Delta }_{n-m-1}=\varnothing$. Then $T_{p}X\cap {\Delta }_{n-m-1}=\varnothing$ for a general $p\in X_{\mathrm{reg}}$.

Proof. 

Take an irreducible component $U:=H_{i_1}\cap \cdots \cap H_{i_{m+1}}$ of ${\Delta }_{n-m-1}$.

Let ${\ell }_U:{\mathbb{P}}^n\setminus U\to {\mathbb{P}}^m$ denote the linear projection from U. Since $U\cap X=\varnothing$, the induced rational map $f:{\ell }_{|X}:X\to {\mathbb{P}}^m$ is a morphism. Let V be an $(n-m)$-dimensional linear subspace of ${\mathbb{P}}^n$ containing U. Since $U\cap X=\varnothing$, $V\cap X$ has dimension 0. Thus f has finite fibers. Since X is projective, f is a finite morphism. Since $\mathrm{char}\left(\mathbb{K}\right)=0$, the differential of f is injective at a general $p\in X_{\mathrm{reg}}$, i.e., $T_{p}X\cap U=\varnothing$. Since ${\Delta }_{n-m-1}$ has finitely many irreducible components, $T_{p}X\cap {\Delta }_{n-m-1}=\varnothing$ for a general $p\in X_{\mathrm{reg}}$. □

Proposition 2.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Fix integers $m\ge v>0$. Let $X\subset {\mathbb{P}}^n$ be an integral m-dimensional variety such that $X\cap {\Delta }_{n-m-1}=\varnothing$ and $dim\langle X\rangle \ge 2m$. Take a general element $S\in S(X,v+1)$ and set $V:=\langle S\rangle$. Then $V\cap {\Delta }_{n-v-1}=\varnothing$.

Proof. 

Since $dim\langle X\rangle \ge v$ and S is general in X, $dim\langle S\rangle =v$.

(a) First we do the case $v=m$. Fix a general $P\in X_{\mathrm{reg}}$. Let $(2P,X)$ denote the closed subscheme of X with ${\left({\mathcal{I}}_{P,X}\right)}^2$ as its ideal sheaf. We have $deg(2P,X)=m+1$ and $T_{P}X=\langle (2P,X)\rangle$. Lemma 3 gives $T_{p}X\cap {\Delta }_{n-m-1}=\varnothing$. Since $(2P,X)$ is a flat limit of a family of elements of $S(X_{\mathrm{reg}},m+1)$ and ${\Delta }_{n-m-1}$ has finitely many irreducible components, we get this case.

(b) Assume $v<m$. Let Y be the intersection of X with a general codimension $m-v$ linear subspace. Since ${\Delta }_{n-v-1}$ has finitely many irreducible components, the theorem of Bertini gives $Y\cap {\Delta }_{n-v-1}=\varnothing$ ([18], I.6.3). Apply step (a) to Y. □

Remark 10.

Let $f:{\mathbb{P}}^m\to W$ be a surjective morphism with W being not a point. Then $dimW=m$ and there is no integral curve $T\subset {\mathbb{P}}^m$ such that $f\left(T\right)$ is a point.

The case $v=1$ of the next theorem is [10], Lemma 2.10 and Th. 3.4. See also [5], Theorem 2.1, and [13], Lemma 3.3.

Theorem 4.

Fix integers $v\ge 1$ and $n\ge v+1$. Set $w:=\lfloor n/v\rfloor$. Let $V\subset {\mathbb{P}}^n$ be a linear space such that $dimV=v$ and $V\cap {\Delta }_{n-v-1}=\varnothing$. Then:

(i)

For all positive integers $x\le v$, the restriction to $V^x$ of the multiplication map is a morphism.

(ii)

For each $p\in {\mathbb{P}}^n$ of Hadamard border rank at most w with respect to V the Hadamard border rank and the a Hadamard rank coincide.

(iii)

$dim{V}^{\star x}=mx$ for all $x\le w$.

(iv)

$V^{\star (w+1)}={\mathbb{P}}^n$.

Proof. 

Since $V\cap {\Delta }_{n-v-1}=\varnothing$, up to a diagonal transformation we may assume that $\mathbf{1}\in V$. For all integers $0\le e\le v$, let $\mathcal{V}\left(e\right)$ denote the set of all e-dimensional linear subspaces A of V such that $\mathbf{1}\in A$. Thus $\mathcal{V}\left(0\right)=\left\{\mathbf{1}\right\}$ and $\mathcal{V}\left(v\right)=\left\{V\right\}$. Each $\mathcal{V}\left(e\right)$ is an irreducible variety.

Observation 1: Since $V\cap {\Delta }_{n-v-1}=\varnothing$ and ${\Delta }_x$ is a union of finitely many linear subspaces of dimension x, for each $x\in \{1,\dots ,v-1\}$ we have $V\cap {\Delta }_{n-x-1}\ne \varnothing$ and $dimV\cap {\Delta }_{n-x-1}=v-1-x$.

Observation 2: Fix $e\in \{1,\dots ,v-1\}$. By Observation 1 we have $A\cap {\Delta }_{n-e-1}=\varnothing$ for a general $A\in \mathcal{V}\left(e\right)$.

Now we prove part (iii). Assume that part (iii) fails and let x be the minimal integer for which it fails. Thus $x\ge 2$ and $dim{V}^{\star (x-1)}=(x-1)v$. Set $c:=dim{V}^{\star x}-(x-1)v$. By assumption we have $0\le c\le v-1$.

Claim 1: For all $e=0,\dots ,c$, we have $dim(V^{\star (x-1)}\cap A_e)=(x-1)v+e$ for a general $A_e\in \mathcal{V}\left(e\right)$.

Proof of Claim 1: We use induction on the integer e. If $e=0$, then $A_0=\mathbf{1}$ and hence Claim 1 is true for $e=0$. Assume $e>0$ and that Claim 1 is true for the integer $e-1$. Fix a general $A\in \mathcal{V}(e-1)$. Thus $dim{V}^{\star (x-1)}\star A=(x-1)v+e-1$. Let $\mathcal{V}(e,A)$ denote the set of all $B\in \mathcal{V}\left(e\right)$ containing A. Fix a general $B\in \mathcal{V}(e,A)$. To prove Claim 1 by contradiction we may assume $dim{V}^{\star (x-1)}\star B=(x-1)v+e-1$. Since $\mathbf{1}\in A$, we get $V^{\star (x-1)}\star B=V^{\star (x-1)}\star A$. By Observation 2 every element of V (resp. A, resp. B) has at most v (resp. $e-1$, resp. e) zeroes as its homogeneous coordinates). For each $E,F,G\subset \{0,\dots ,n\}$ such that $\#E=v$, $\#F=e-1$ and $\#G=E$ there are $q\in V$ (resp. $q_1\in A$, resp. $q_2\in B$) with zeros at the coordinate in E (resp. F, resp. G). Recall that $f:=h_{|V^x}$ is a morphism. Thus the maximum number of zero coordinate of some $q_2\in V^{\star (x-1)}\star A$ (resp. $q_3\in V^{\star (x-1)}\star B$) is $(x-1)v+e-1$ (resp. $(x-1)v+e$), contradicting the equality $V^{\star (x-1)}\star B=V^{\star (x-1)}\star A$.

Take a general $U\in \mathcal{V}\left(c\right)$. Taking the integer c in Claim 1, we get the equality $dim{V}^{\star (x-1)}\star U=dim{V}^{\star x}$. Since $\mathbf{1}\in U$, we get $V^{\star (x-1)}\star U=V^{\star x}$. Recall that $f:=h_{|V^x}$ is a morphism. Observation 2 gives that the maximal number of zero-entries of an element of $V^{\star (x-1)}\star U$ (resp. $V^{\star x}$) is $(x-1)v+c$ (resp. $xv$), a contradiction.

Now we prove part (iv). If $wm=n$, then part (iii) gives part (iv). Now assume $wn<n$ and set $c:=n-xv$. Let W be a general element of $\mathcal{V}\left(c\right)$. Observation 2 shows that each element of W has at most c zeros as coordinates. Thus the restriction to $V^x\times W$ of the multiplication map is a morphism. With this observation, the proof of part (iii) gives part (iv), just using $V^{\star x}$ instead of $V^{\star (x-1)}$. □

5. Erasures, Forbidden Parts of the Screen and Other Fields

In the first part of this section, we discuss the part of the screen killed by bad data, either erasures or fake data. On the final screen, erasures are data which we do not see (a part of the screen which is obviously down) or a part which is flagged as unreliable, while fake data are things we see, but we do not know if they are correct.

In the second part of the section, we consider other fields or drops the assumption that X is projective.

Both parts of this section are just suggestions for the interested reader. On these topics, we have neither theorems nor applications.

5.1. Erasures

Remark 11.

Take a variety X and finite unions of proper subvarieties $A\subset X$ and $B\subset X$ as in case α and $q\in {\mathbb{P}}^n$ such that $x:=H{o}_X\left(q\right)<+\infty$. Taking T containing $A\cup B$ in the definition of the weak Hadamard open rank, we get the existence of $S\in S(X\setminus (A\cup B),y)$ for some $y\le x$ such that q is the Hadamard product of the elements of S. Thus all existence results for open Hadamark rank and weak Hadamard open rank give solutions S such that $S\cap (A\cup B)=\varnothing$. Non-existence results cannot be applied, but often the proofs we gave help. For instance, in the proof of Proposition 1, we see for which A, B and $q\in {\Delta }_{n-1}$ we have S giving $H{\tilde{o}}_X\left(q\right)<+\infty$ and with $S\cap (A\cup B)=\varnothing$.

Now we describe what part of the full screen is covered by faulty or fake data after $k\ge 2$ steps. We fix an irreducible component W of $A\cup B$. Set $m:=dimX$ and $a:=dimW$. We always assume $a<m$, because in the case $a\ge m$ we see no way to get safe information and in case $\alpha$ we have $W=X$, so no safe data at all.

We first assume $X^{\star k}={\mathbb{P}}^n$.

In cases ${\alpha }_1$ and ${\beta }_1$ we need to avoid all points of $W\star X^{\star (k-1)}$. We may assume that W is concise, because if $W\subset {\Delta }_{n-1}$, then W is contained in a coordinate hyperplane, say $W\subseteq H_i$, and hence $W\star X^{\star (k-1)}\subseteq H_i$. Now assume that W is concise and take $p\in W\cap U_n$. Since $p\in U_n$, $dimp\star X^{\star (k-1)}=dim{X}^{\star (k-1)}$. Thus we need to take as k the first integer such that $X^{\star k}={\mathbb{P}}^n$. This is a perfect job for an Algebraic Geometer expert of [5] (we did it as a small case for a linear space V in Theorem 4, where we also proved and stated that the images do not need closures; they are images of the multiplication of subsets of X).

In cases ${\alpha }_2$ and ${\beta }_2$, we need to avoid all points of $W^{\star k}$. Here it is sufficient to assume $ka<n$.

Now we interpolate between ${\alpha }_1$ and ${\alpha }_2$. For a real number $0<ϵ<1$, we take $b:=\lfloor ϵk\rfloor$ and need to avoid $W^{\star b}\star X^{\star (k-b)}$. Of course, after fixing the integer $k\ge 2$, we need $ϵ\le 1/k$ to avoid filling in ${\mathbb{P}}^n$ with the data partially coming from W.

If we fix $k\ge 2$ with $X^{\star k}\ne {\mathbb{P}}^n$, we get the same result, except in case ${\alpha }_1$, where we need to assume $dim{X}^{\star k}>dim{X}^{\star (k-1)}$. For ${\alpha }_2$, everything works, i.e., a Zariski open subset $X^{\star k}$ is correct, if $X^{\star k}$ has the expected dimension $km$. Again, this is easy for certain varieties X and a good problem for a specific X arising in the future.

Now we call K the base field, because we do not assume that it is algebraically closed and never take the closure. We work in $K^{n+1}$ because we look at the single coordinates, not just as homogeneous coordinates. We call ${\pi }_i:K^{n+1}\to K$, $i=0,\dots ,n$, the $n+1$ projections. Hence for $a=(a_0,\dots ,a_n)\in K^{n+1}$ we see each $a_i:={\pi }_i\left(a\right)$ and we may ask if $a_i$ is correct or not. We have a set $\tilde{X}\subset K^{n+1}$ and only use the elements of $\tilde{X}$ for the multiplication map. For instance, we start with an integral variety $X\subset {\mathbb{P}}^n\left(K\right)$ and take at $\tilde{X}$ the affine cone of X. If $\tilde{X}$ comes from k, it is feasible for an Algebraic Geometer and/or an expert of [5] to compute the dimension of the set obtained from $\tilde{X}$ making k multiplications.

We fix a real number $ϵ$ such that $0<ϵ<1$ and, after we see $a\in K^{n+1}$, we accept a if and only a has at most $\lfloor (n+1)ϵ\rfloor$ not certified coordinates. This is the standard situation for error-correcting codes of block length $n+1$. This case scales well with n. We set ${\tilde{X}}_n:=\tilde{X}$. We see $K^{n+1}$ as the coordinate hyperplane $H_{n+1}=\{x_{n+1}=0\}$ of $K^{n+2}$. We take ${\tilde{X}}_{n+1}\subset K^{n+2}$ such that ${\tilde{X}}_{n+1}\cap H_{n+1}={\tilde{X}}_n$. Note that for any $a=(a_0,\dots ,a_{n+1})\in K^{n+2}$, we may check the first $n+1$ coordinates of a using ${\tilde{X}}_n$. And so on. We get a family $K^{x+1}$, $x\ge n$ of linear spaces and ${\tilde{X}}_x\subset K^{x+1}$ such that ${\tilde{X}}_x={\tilde{X}}_{x+1}\cap H_{x+1}$ for all $x\ge n$. We may take the same $ϵ$ for all $x\ge n$ or use a family ${\left\{{ϵ}_x\right\}}_{x\ge n}$ of small positive real numbers. This is the standard setup for convolutional codes ([14], Ch. 14).

5.2. Different “Varieties”

Now we describe some modifications of our default setup (projective varieties over an algebraically closed field).

First assume $\mathbb{K}=\mathbb{C}$ but that X is not a complex projective variety. Suppose you want to work with complex analytic varieties. By GAGA or the older theorem of Chow, you need to allow non-closed analytic subvarieties of ${\mathbb{P}}^n\left(\mathbb{C}\right)$. Their image by holomorphic maps may be very bad, and their closures in the classical Euclidean topology may even be worse. One should look at images without taking the closures. We did it for projective subvarieties, but for non-closed subsets of ${\mathbb{P}}^n\left(\mathbb{C}\right)$, conditions like $X\cap {\Delta }_{n-m-1}$ are not useful.

One could modify the definition of open rank, Hadamard open rank and weak Hadamard open rank allowing sets T which are not algebraic, for instance, finite unions of locally closed differential submanifolds of dimension $2dimX-1$, but in this case the difficulties of taking closures are even worse. If X is closed, conditions like $X\cap {\Delta }_{n-m-1}$ are meaningful and they may be used.

One could take closed subsets of ${\mathbb{P}}^n\left(\mathbb{C}\right)$ which are only algebraic with respect to the variables of ${\mathbb{P}}^n\left(\mathbb{C}\right)$ (call it $z_0,\dots ,z_n$ instead of $x_0,\dots ,x_n$ as elsewhere in this paper) and their complex conjugate ${\overline{z}}_0,\dots ,{\overline{z}}_n$. In this case the images and closures are controlled by Tarski–Seidenberg theorem [19]), but other issues may arise [20].

Now assume that $\mathbb{K}={\overline{\mathbb{F}}}_p$ is the algebraic closure of a finite field. In this case, the variety X is defined over a finite field ${\mathbb{F}}_q$, with q as p-power, i.e., X is defined by homogeneous equations with coefficients contained in ${\mathbb{F}}_q$. For a given $u\in {\mathbb{P}}^n\left({\mathbb{F}}_{q^e}\right)$, it would be very interesting to get a low positive integer f such that u is the Hadamard product of finitely many elements of $X\left({\mathbb{F}}_{q^{ef}}\right)$.

There is the discrete version of the Hadamard product. Here we call K the base field [4] and in this case, it is often used to work over a finite field ([4], §2.2).

6. Cones and Strange Hadamard Varieties

Let $X\subset {\mathbb{P}}^n$ be an integral variety. Set $m:=dimX$. For any $p\in X_{\mathrm{reg}}$ let $T_{p}X\subseteq {\mathbb{P}}^n$ denote the Zariski tangent space of X. Since p is a smooth point of X, the set $T_{p}X\subseteq {\mathbb{P}}^n$ is an m-dimensional linear space. Let $V\subset {\mathbb{P}}^n$ be a linear subspace, $V\ne \varnothing$. A variety $X\subset {\mathbb{P}}^n$ is said to be strange and V is said to be the strange vertex of X is ${\cap }_{p\in X_{\mathrm{reg}}}{T}_{p}X=V$ [21,22,23]. To be strange with the strange vertex containing V, it is sufficient that V is contained in $T_{p}X$ for all p in a non-empty Zariski open subset of $X_{\mathrm{reg}}$. Cones are strange, and their strange vertex contains their vertex (as cones). Take a linear space $V\subset {\mathbb{P}}^n$, $V\ne \varnothing$, and set $a:=dimV$. Let ${\ell }_V:{\mathbb{P}}^n\setminus V\to {\mathbb{P}}^{n-a-1}$ denote the linear projection from V. Assume $X⊈V$ and set $\mu :={\ell }_{V|X\setminus V\cap X}$. The differential of $\mu$ at a general point of $X_{\mathrm{reg}}\setminus X_{\mathrm{reg}}\cap V$ has kernel of dimension $a+1$ if and only if X is strange and V is contained in the strange vertex of X. Thus, the characteristic 0 part of the Terracini Lemma ([15], Cor. 1.10) gives the well-known fact that in characteristic zero, X is strange if and only if X is a cone and that its vertex as a cone is its strange vertex. In all positive characteristics, there are strange curves which are not lines, i.e., strange curves which are not cones [21,22,23,24]. Similar examples exist for higher-dimensional varieties [21,22,23]. Strange varieties X which are not cones have many interesting properties. You may call it pathologies, if you prefer to say that they are the counterexamples for many positive characteristic extensions of geometric theorems which are true in characteristic zero [24].

From the point of view of the classical secant varieties, the cones have bad behavior in any characteristic. Indeed, take an integral and non-degenerate m-dimensional variety $X\subset {\mathbb{P}}^n$. A dimensional count gives that the secant variety $\mathrm{Sec}\left(X\right)$ of X has dimension at most $min\{2m+1,n\}$. Usually the integer $min\{2m+1,n\}$ is called the expected dimension of $\mathrm{Sec}\left(X\right)$. However, if X is a cone with a b-dimensional vertex, we have $dim\mathrm{Sec}\left(X\right)\le 2m-b$. Hence cones are secant defective. Note that $\mathrm{Sec}\left(V\right)=V$ for all linear spaces V. In characteristic 0, the inequality $dim\mathrm{Sec}\left(X\right)\le 2m-b$ also follows from the characteristic 0 part of the Terracini lemma ([15], Cor. 1.11). Strange curves which are not lines have all their secant varieties of the expected dimension ([15], Prop. 1.3, [25], Cor. 1.5 and Cor. 1.10). Often cones are not bad with respect to the Hadamard product as shown by the following example.

Example 3.

A line is a cone. Take a line $L\subset {\mathbb{P}}^n$ such that $L\cap {\Delta }_{n-2}=\varnothing$. It was proved in ([5], Th. 2.1) (and it works in arbitrary characteristic) that $dim{L}^{\star s}=s$ for all positive integers $s\le n$.

Let $W\subseteq {\mathbb{P}}^n$ be a coordinate linear space, i.e., the intersection of some coordinate hyperplanes, say $W=H_{i_1}\cap \cdots \cap H_{i_s}$ with $0\le i_1<\cdots <i_s\le n$. We allow the case $W={\mathbb{P}}^n$, i.e., the case $s=0$. Let $W_0$ denote the set of all $p=[p_0:\cdots :p_n]\in W$ such that $p_i\ne 0$ if $i\notin \{i_1,\dots ,i_s\}$. Note that ${{\mathbb{P}}^n}_0=U_n$. Let $G_W$ be the set of all diagonal $g\in G$ such that $g\left(W\right)=W$. Note that $W_0$ is an open orbit of W for the action of $G_W$ on W induced by the action of G on ${\mathbb{P}}^n$ and that W is the Zariski closure of $W_0$ in ${\mathbb{P}}^n$.

Let $X\subset {\mathbb{P}}^n$ be an integral projective variety. Recall that $\ll X\gg$ is the minimal coordinate linear space containing X with the convention $\ll X\gg ={\mathbb{P}}^n$ if X is not contained in a coordinate hyperplane. We just define the set $\ll X{\gg }_0$. Note that $X_{\mathrm{reg}}\cap \ll X{\gg }_0$ is open and Zariski dense in $\ll X\gg$. Thus $\ll X\gg$ is the minimal coordinate linear space containing the quasi-projective variety $X_{\mathrm{reg}}\cap \ll X{\gg }_0$. Note that if $p\in X_{\mathrm{reg}}$, then $T_{p}X\subseteq \ll X\gg$.

The Terracini lemma for the Hadamard product (Lemma 1) suggests the following extension of the notion of strangeness adapted to the Hadamard product.

Definition 5.

Let $X\subset {\mathbb{P}}^n$ be an integral projective variety. X is said to be Hadamard strange with a strange Hadamard vertex of dimension at least b if for each $p\in X_{\mathrm{reg}}\cap \ll X{\gg }_0$ there is a b-dimensional linear subspace $V_p$ of $T_{p}X$ such that $V_p\star q=V_q\star p$ for all

$$(p,q)\in X_{\mathrm{reg}}\cap \ll X{\gg }_0\times X_{\mathrm{reg}}\cap \ll X{\gg }_0.$$

In this case we say that the family ${\left\{V_p\right\}}_{p\in X_{\mathrm{reg}}\cap \ll X{\gg }_0}$ is contained in the strange Hadamard vertex of X.

Note that the strange Hadamard vertex (if nonempty) is not a subset of ${\mathbb{P}}^n$ but a family of linear subspaces of ${\mathbb{P}}^n$. In Definition 5, we have $b\le dimX$. We only have trivial examples in which we are allowed a fixed linear space, e.g., the following ones.

Example 4.

Take a line $L\subset {\mathbb{P}}^n$ containing a coordinate point, o, and such that $L\cap U_n\ne \varnothing$. Obviously, $T_{p}L=L$ for all $p\in L$. Note that $L\cap {\Delta }_{n-1}=\left\{o\right\}$ and that $q\star L=p\star L=L\star L$ for all $p,q\in L\setminus \left\{o\right\}$. Thus $T_{p}L\star q=p\star T_{q}L$ for all $p,q\in L\cap U_n$. The constant family $\left\{L\right\}$ is the unique one-dimensional strange vertex of L.

Example 5.

Take a linear space $V\subset {\mathbb{P}}^n$ containing a coordinate point, o, and such that $V\cap U_n\ne \varnothing$. Obviously, $T_{p}V=V$ for all $p\in V$. Since $p\star o=q\star o$ for all $p,q\in U_n$, the constant family $\left\{o\right\}$ is a 0-dimensional strange Hadamard vertex of V.

Example 6.

Take a strange variety $X\subset {\mathbb{P}}^n$ such that $X\cap U_n\ne \varnothing$, and a coordinate point o is contained in the strange vertex of X. Since $o\in T_{p}X$ for all $p\in X_{\mathrm{reg}}$ and $o\star u=o$ for all u such that $o\star u$ is defined, the constant family $\left\{o\right\}$ is contained in the strange vertex of X. This example shows that in every positive characteristic, there are Hadamard strange varieties which are not cones.

Remark 12.

Let $X\subset {\mathbb{P}}^n$ be an integral variety, which is not concise. X is Hadamard strange as a subset of ${\mathbb{P}}^n$ if and only if it is Hadamard strange as a subvariety of $\ll X\gg$, and the strange Hadamard vertices are the same for ${\mathbb{P}}^n$ and $\ll X\gg$. Thus often a result for concise subvarieties of a projective space gives, with no effort, a result for the non-concise ones.

The following example gives our motivation for defining the strange vertex not as a fixed linear subspace of ${\mathbb{P}}^n$ but as a family of linear subspaces.

Example 7.

Take a line $L\subset {\mathbb{P}}^n$ such that $L\cap {\Delta }_{n-2}=\varnothing$. Note that L is smooth and concise and that $T_{p}L=L$ for all $p\in L$.

Assume that L is strange and that one of its strange vertices is a b-dimensional linear subspace $E\subset {\mathbb{P}}^n$. We have $b\le 1$. Assume first that $b=1$. Hence $E=L$. We get $p\star L=q\star L$ for a general $(p,q)\in L\times L$. We fix p and get $L^{\star 2}=p\star L$, a contradiction. Now assume $b=0$. Hence E is a point. We get $p\star E=q\star E$ for a general $(p,q)\in L\times L$. Note that $E\in L$. Fix $p\in L\cap U_n$. Varying $q\in L\cap U_n$, we get $p\star E=E\star L$. Since $p\in U_n$, we get $E\in {\Delta }_1$. Since $E\in L$, we get a contradiction.

Remark 13.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Let $X\subset {\mathbb{P}}^n$ be an integral strange variety with a b-dimensional strange vertex. The Terracini Lemma for the Hadamard product (Lemma 1) gives $dim{X}^{\star 2}\le 2dimX-b-1$.

Proposition 3.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Let $X⊊{\mathbb{P}}^n$ be a strange and concise Hadamard variety. Set $m:=dimX$ and assume $X\cap {\Delta }_{n-m-1}=\varnothing$. There is no linear space $V\subset {\mathbb{P}}^n$, $V\ne \varnothing$ such that the constant family $\left\{V\right\}$ is contained in the Hadamard strange vertex of X.

Proof. 

Assume that V exists. Fix $(p,q)\in X_{\mathrm{reg}}\cap U_n\times X_{\mathrm{reg}}\cap U_n$. We get $p\star V=q\star V$. We fix $p\in X_{\mathrm{reg}}\cap U_n$ and vary q. Since $X_{\mathrm{reg}}\cap U_n$ is Zariski dense in X, we get $p\star V=X\star V$. Lemma 3 gives $T_{q}X\cap {\Delta }_{n-m-1}=\varnothing$ and hence $p\star V\cap {\Delta }_{n-m-1}=\varnothing$. Hence V is concise. Taking $u\in V\cap U_n$ we get $p\star V\supseteq X\star u$ and hence $dimV=m$. Since $V\subseteq T_{p}X$, we get $X=T_{p}X=V$. Since we proved the case $m=1$ in Example 7, we may assume $m\ge 2$. Since $p\star V=X\star V$, we get $dim{X}^{\star 2}=X$. Since we are in characteristic 0 and X is concise and irreducible, [13], Th. 7.2, implies that X is contained in a binomial integral hypersurface, W. Let $x^{\alpha }-c{x}^{\beta }$ be an equation of W with $c\ne 0$, $\alpha =(a_0,\dots ,a_n)\in {\mathbb{N}}^{n+1}$ and $\beta =(b_0,\dots ,b_n)\in {\mathbb{N}}^{n+1}$ such that $a_0+\cdots +a_n=b_0+\cdots +b_n$. Since W is integral, $a_i\ne 0$ implies $b_i=0$. Hence there are $i,j\in \{0,\dots ,n\}$ such that $H_i\cap H_j\subset W$. Since $X\cap {\Delta }_{n-m-1}=\varnothing$, we exclude the case $m=n-1$, i.e., the case $X=W$. Hence $n\ge 3$. Using $X\cap H_i$ in $H_i$, we conclude by induction on n. □

7. Discussion

Around 2010 in Algebraic Statistics, M.A. Cueto, J. Morton and B. Sturmfels introduced a statistical model involving the extension of the classical Hadamard product of two matrices to the coordinatewise product of a vector with $n+1$ coordinates or of two points of a projective space ${\mathbb{P}}^n$ with a fixed system of coordinates. In Algebraic Statistics, this statistical model is associated to the graphical models called Restricted Boltzmann Machines [1,2,3,4]. For any projective varieties $X,Y\subset {\mathbb{P}}^n$, the Hadamard product $X\star Y$ of X and Y is the Zariski closure of the point of ${\mathbb{P}}^n$, which are the Hadamard product of $p\in X$ and $q\in Y$. This definition is extended to the product of $k\ge 2$ subvarieties. For a review with proofs of the results up to 2024, see the book [5]. More work was done recently [12,13,20].

In the main part of this paper, we prove some new results on this classical setup and discuss an important new definition introduced by D. Antolini and A. Oneto in [13], the Hadamard rank of $q\in {\mathbb{P}}^n$ with respect to X. Some of our results solve (in particular cases) a very important problem: describe the part of ${\mathbb{P}}^n$ obtained taking by Hadamard products of k copies of X before taking the Zariski closure.

Our data is the variety X. In the Introduction and in Section 5, we consider several scenarios describing how to handle the case in which some part of X is either wrong (and hence it gives errors when used for the Hadamard products) or not retrievable.

8. Conclusions

Let $X\subset {\mathbb{P}}^n$ be an integral projective variety. A key notion recently introduced by D. Antolini and A. Oneto is the definition of Hadamard X-rank $H{\tilde{r}}_X\left(q\right)$ of all points $q\in {\mathbb{P}}^n$ [13]. We introduce some new related concepts (open Hadamard rank $H{\tilde{o}}_X\left(q\right)$, weak open Hadamard rank $H{o}_X\left(q\right)$ and weak Hadamard rank $H{r}_X\left(q\right)$) and prove the results on them. Theorem 1 gives an upper bound for the open Hadamard rank $H{\tilde{o}}_X\left(q\right)$ of the concise variety $X\subset {\mathbb{P}}^n$ for all $q\in U_n$. The restriction $q\in U_n$ is important because we prove that $H{o}_X\left(q\right)=+\infty$ if $dimX=1$ and $q\in {\Delta }_{n-1}$ (Proposition 1). We prove that $H{o}_X\left(q\right)=n+1$ for a general q if X is a curve such that $X^{\star n}={\mathbb{P}}^n$. For and on the classical definitions for the Hadamard product, we prove a new theorem for linear subspaces $V\subset {\mathbb{P}}^n$, extending the known case $dimV=1$ (Theorem 4). We also extend a known result to the positive characteristic case (Theorem 3). It seems important to do these extensions to be able to work with finite fields. At the end of Section 3, we give three open questions.

Section 5 contains suggestions for new work for the interested reader.