Abstract
For all integers  and , the Hadamard product  of k elements of  (with K being the complex numbers or real numbers) is the element  which is the coordinate-wise product of  (introduced by Cueto, Morton, and Sturmfels for a model in Algebraic Statistics). This product induces a rational map . When ,  and ,  are irreducible, we prove four theorems for the case , three of them with  as a line. We discuss the existence (non-existence) of a cancellation law for ★-products and use the symmetry group of the Hadamard product. In the second part, we work over . Under mild assumptions, we prove that by knowing , we know . The opposite, i.e., taking and multiplying a set of complex entries that are invariant for the complex conjugation and then seeing what appears in the screen , very often provides real ghosts, i.e., images that do not come from a point of . We discuss a case in which we certify the existence of real ghosts as well as a few cases in which we certify the non-existence of these ghosts, and ask several open questions. We also provide a scenario in which ghosts are not a problem, where the Hadamard data are used to test whether the images cover the full screen.
    Keywords:
                                                                    Hadamard product;                    join;                    secant variety;                    real projective variety;                    Hadamard product of vectors        MSC:
                14N05; 14N07; 14M99
            1. Introduction
Suppose there are  channels, each providing packets of -ples of real (or complex) numbers. Then, for each packet, some device makes the entry-wise multiplication of the k received packets. The screen sees a sequence of packets formed by -ples of numbers (the results of the multiplications made by the device). In [], the authors explained why this occurs in real-life applications and explained some of the key mathematical tools used to handle it. They introduced the name of Hadamard products of element of an -dimensional (real or complex) vector space or of elements of n-dimensional projective spaces. Real-life applications can be found in Algebraic Statistics in the form of statistical models associated with the graphical models called Restricted Boltzmann Machines [,]. Recently, C. Bocci and E. Carlini published a monograph on Hadamard products [], which interested readers may refer to along with the references therein for applications of Restricted Boltzmann Machines in machine learning. Among these references, we use [,,,,,] in the present paper. Of course, Hadamard products of matrices are much older, but these products do not occur in this paper.
This paper has two parts, which are connected by the notion of Hadamard products of elements of a projective space. The title only reflects the second part. The first part, Section 2 and Section 3, consists of “abstract Algebraic Geometry” over an algebraically closed field of characteristic 0; the interested reader will recognize several key features used in the applications over the real number field . In this part, we use the group of symmetries of the Hadamard product and show that it is a strong tool. In the second part, Section 4, we work over  with  as its algebraic closure. Supposing that the “picture” over  is known (for instance, by Algebraic Geometry) and that all of the packets come from k sets (or varieties) that are invariant for the action of the complex conjugation, we want to know whether it is possible to reconstruct the data from the real numbers among the complex numbers on the screen. We provide several counterexamples and explain some mitigation strategies.
Taking a field K and calling  the variables of the vector space  and the homogeneous variables of the associated projective space , the Hadamard product  is the coordinate-wise product, i.e.,
      
      
        
      
      
      
      
     This map strongly depends on the choice of the coordinates. The ★-product induces a rational map  the indeterminacy locus  of which is the set of all  such that  for all i. The set  is Zariski closed in . Taking irreducible algebraic varieties  such that , the rational map  that restricts h to  is a morphism on its Zariski open subset ; hence, over any field K, we see the set  with our screen, which receives and sees only numbers in K. In Algebraic Geometry (even in parts with a strong applied flavor, as in []), when we are interested in, say, , we must first check the algebraic closure  and consider ★ over , where we call  the closure of  ([], Def. 1.3). With this definition,  is an irreducible projective variety defined over K, and as such is defined by finitely many polynomial equations with coefficients in K. Knowing an upper bound for the degrees of these polynomials and knowing a large number of points of , we obtain polynomial equations defining  over  (and, if the points are in , over K). Thus, the set  of K points of the irreducible variety is well defined. It contains the set , although it is often far larger. The “ghosts” in the title arise in the case of  with ; obviously, .
In this paper, we use the following notation (as in [] and most references): for  we set ,  and . A coordinate linear space is the intersection of finitely many coordinate hyperplanes . Because the coordinate system is fixed, our group of symmetries is not the linear group  or its quotient  by the multiples  of the identity matrix, but rather the diagonal group  (which is isomorphic as an algebraic group to the product of  copies of the multiplicative group) or its quotient  by its subgroup . Recall that for all , by definition we have  ([], §1.2). The group  is the set of all transformations  for some .
For any , let  denote the set of all  such that . If , then the connected group  counts the number of zero-coordinates of P. In particular,  if and only if .
In the following Theorems 1–4, we assume that the base field is algebraically closed and of characteristic 0. Here, we only use the assumption of characteristic 0 in order to freely quote [] in their proofs.
Theorem 1. 
Let  be an integral projective variety not contained in a coordinate hyperplane. Set , , and  with  and . Set  and assume . Then, .
To explain the importance of Theorem 1, we start with 5 positive integers n, x, y, a, and b such that  and . From the existence of the Chow variety of , there is a (not connected in general) quasi-projective family T of pairs  such that X is a closed and irreducible subvariety of dimension x and degree a and that Y is a closed and irreducible subvariety of dimension y and degree b. Hence, T also parameterizes (not one-to-one) all possible varieties . Now, assume  and set . It is known that  (Remark 7); however, a priori it could be very low, even 1. If , then Theorem 1 states that this is not the case under very mild assumptions, as .
Easy examples show that in general we may have  with neither  and  nor  and ; e.g., in the case with , to obtain  it is sufficient that neither X nor Y be a coordinate line. Hence, reconstructing X and Y from  seems to be hopeless, except perhaps if we restrict the class of possible X and Y. We ask the following question.
Question 1: Assume , , and that  for some line L. Is Y a line?
On the ★-decomposition of surfaces, we prove the following three theorems.
Theorem 2. 
Let ,  be an integral surface such that  for some curves X and Y. Assume  for all codimension 3 coordinate linear spaces V. Then:
(a) The rational map  is a morphism.
(b)  for all codimension 2 coordinate linear spaces , i.e., for , each element of  satisfies  for all .
Theorem 3. 
Let ,  be a plane such that X and Y are integral curves  for all codimension 3 coordinate linear spaces V. Then, X and Y are lines and  for all codimension 2 coordinate linear spaces . Moreover, Y is uniquely determined by the pair .
Conversely, take a line ,  such that  for all codimension 2 coordinate linear spaces . Then,  is a plane and  for all codimension 3 coordinate linear subspaces V.
The following result is an example in which we have a “cancellation law” for the ★-product.
Theorem 4. 
Take lines ,  such that  and  for all codimension 2 coordinate linear spaces V. Then, .
We think that cancellation results are not frequent; see Example 3 for a non-uniqueness related to Theorems 3 and 4. This example is a byproduct of the study of symmetries of the ★-product. See the end of Section 3 for the algebraically closed case (Remark 18) and Section 4 (Remark 19) for the case  with the Abelian group .
In the rest of this introduction the main field that we use is the field  of real numbers; however, it is important to also use its algebraic closure . Let  denote the complex conjugation; in addition, we call  the complex conjugation on , , and the subsets of . If  is an algebraic projective variety defined over , then . We perform the construction for  screens. Fix an integer , and consider the map  defined taking the entry-wise product of the k vectors. This induces a rational map  in the following way. Let  denote the set of all  such that  for all . The closed algebraic set  is the indeterminacy locus of h, and for all  we call  point . This ★ product is invariant for the complex conjugations, and the ★ product of k real points (not in ) is an element of . If  are irreducible subvarieties and , by restricting h and then taking the Zariski closure we obtain the ★-products  of k varieties , . As the multiplication map of a field, the ★-products obey the commutative and associative laws. Here, it is important to distinguish the results we see on our screen, i.e., the set of all ,  with K either  or , from its closure in the Zariski topology (which by definition is  if ) or Euclidean topology. The two closures coincide over , but do not coincide not if we work over . For any field K and any , , let  denote the set of all  with  for all i and . We have . If  (or is just algebraically closed), then  is constructable, i.e., a finite union of algebraic varieties (Remark 5, i.e. ([], Ex. II.3.18, Ex. II.3.19). If , then  is semialgebraic and its closure in the Euclidean topology is semialgebraic ([], Prop. 2.2.2 and 2.2.7). Under very natural conditions the real picture is sufficient to describe the complex one (Theorem 5).
A key problem is that our real screen  may see something much larger than  (Examples 8 and 9). We call these real ghosts. We provide some examples in which they arise, and one (Example 9) in which it is certified that they must occur.
We write a subsection (full screen) explaining a different problem. The Hadamard data is a test for determining whether the images cover the full screen. We prove that for this problem ghosts are not an issue. We conclude the section with four mitigating suggestions.
In the Conclusions section, we add a question related to the ones discussed in Section 4. Section 4 explains why the question would certify the nonexistence of real ghosts in that particular setup.
Many thanks are due to the referees for their useful suggestions.
2. The Algebro-Geometric Part, I
In this section and the next one, our base field K is algebraically closed and of characteristic zero, the latter assumption being used only in order to freely quote [].
We write , , , and  instead of , , , and .
For each , let  denote the coordinate point with  and  for all . For each integer , let  denote the union of all x-dimensional coordinate linear subspaces. We have .
Remark 1. 
Take . There is a unique  such that . With the notation from ([], §1.2), we have .
Remark 2. 
Fix  such that  and take  and . For any , if we know  and , then we know  for all . If we also know  and , then we know P.
Remark 3. 
Because  as an algebraic group, the algebraic group  is isomorphic to . Hence, all its connected algebraic subgroups are isomorphic (as algebraic groups) to a product of the multiplicative group.
Remark 4. 
Take a connected one-dimensional algebraic subgroup . Here, G is isomorphic to the multiplicative group. Hence, each orbit of G is either one point or an affine rational curve.
Remark 5. 
Take irreducible varieties  and  such that . This condition is satisfied if . Note that  is a non-empty Zariski open subset of . Because our base field is algebraically closed, the set  is a constructable subset of , i.e., it is a finite union of locally closed subsets of  for the Zariski topology ([], Ex. II.3.18, Ex. II.3.19). Thus, its closure  in  is a projective variety of dimension . Because  is irreducible,  is irreducible.
Remark 6. 
Take integral projective varieties ,  such that , i.e., such that the rational map  is a morphism. Because  is a projective variety,  is closed ([], Th. II.4.9), i.e., in the definition of  we do not need to take the closure.
Remark 7. 
Let  and  be integral and closed subvarieties. Set , , , and . Let  denote the associated embedding and  the Segre embedding. The set  is an integral projective subvariety of dimension  and degree  ([], Cor. 5.1).
Remark 8. 
Take an integral surface , ,  such that  for some irreducible curves X and Y. If  is a morphism, then  divides  (Remark 7).
Lemma 1. 
Let ,  be an integral surface.
- 1
 - If , then W is a rational fibration.
 - 2
 - If , then W is a toric surface.
 
Proof.  
Remark 4 provides the first assertion. Now, assume . Because  is Abelian, we find that W is a rational surface and that  acts on it with an open orbit. Hence, W is a (possibly singular) toric surface with the embedding  being toric.   □
Example 1. 
Fix integral curves  and . Fix any  and . Let  be any element of  such that  and . Set  and . Because  is not defined,  is not a morphism.
★-Powers
In this subsection, we consider the ★-powers  of an integral variety . We always assume . This is not very restrictive, as X is irreducible and we only consider ★-power of X. Hence, if X is contained in the coordinate hyperplanes , , then it is sufficient to consider the variety as a subvariety of the -dimensional projective space . Set . For each  set, ; hence, . We have  for all . For each , set .
Remark 9. 
Take an irreducible variety  such that  and take any  and any variety . Then,  is -equivalent to Y. Hence, we have  for all  and  for all .
The following two results (the first being the particular case  of the second) are proved as in ([], part (1) of Prop. 2.1).
Proposition 1. 
Let  be an integral variety. Assume  and set . We have  for all .
Proof.  
Fix a general . Because  is general, it is a smooth point of . For any  and any , consider the ★-product  of all , . Set . The set  is an m-dimensional vector space containing the point . By the Terracini Lemma for ★ products ([], Lemma 1.6),  is the tangent space of  at . Take A, , and  with , , , and . We first take  and . We have . Taking  and , we obtain . Thus,  is the linear span of three linear spaces containing the point , 1, L of dimension  and 2 of dimension m with a certain order, say,  and . Because , we have .    □
Proposition 2. 
Fix integral varieties X, Y of  such that  and . For all integers , set . Then, .
Proof.  
Fix a general . Because Q is general, Y is smooth at Q. For any  and any , consider the ★-product  of Q and all , . Set , then repeat the proof of Proposition 1.    □
If X is a binomial hypersurface of type , then  ([], Prop. 6.4). Hence, there are irreducible varieties  such that  for all .
Definition 1. 
Take an irreducible variety  such that . A point  is said to be a strict ★-vertex of X if  for a general . The ★-vertex  of X is the set of all its strict ★-vertices.
Remark 10. 
A point O is a strict ★-vertex of X if and only if  for all . The ★-vertex  of X is a linear subspace, as it is an intersection of linear subspaces . Moreover, if X is not a linear space, then .
The following example taken from [] shows the existence of  with .
Example 2. 
Fix  such that . Let V be the binomial hyperplane . We have  and  for all  ([], Cor. 2.2)). Thus, .
Remark 11. 
Let  be an irreducible variety such that . The Terracini Lemma ([], Lemma 1.6) provides .
Remark 12. 
Let ,  be an irreducible variety such that . Let  denote the ideal generated by all homogeneous polynomials vanishing on V. Because V is irreducible, the ideal  is contained in the inessential ideal  and is saturated. The homogeneous ideal is prime ([], Th. 8.5). Because  and V is irreducible,  for any i. Because  is a prime ideal,  for all i. Hence, the assumption  of ([], Th. 5.3) is satisfied by the variety V.
Proposition 3. 
Take an irreducible . If , , and X is not contained in a binomial hypersurface, then .
Proof.  
As we have assumed that X is not contained in a binomial hypersurface, . Remark 9 provides . Assume  and fix a general . Because  and  is general,  and . Thus,  and  are -equivalent. Because , we obtain ; hence,  for all . Let Y denote the closure in  of the set of all  such that . By Remark 12 and ([], Th. 7.4), i.e., by ([], 5.3), Y is a binomial variety. Thus, X is contained in a binomial hypersurface, contradicting our assumption.    □
Applying Proposition 3 several times, we obtain the following result.
Corollary 1. 
Let  be an integral variety not contained in a binomial hypersurface. Then,  for all .
Remark 13. 
Take an integral variety  and set . Every codimension x linear subspace of  meets X. Hence, each coordinate linear subspace of dimension  meeting X is a non-empty set of dimension at least .
Remark 14. 
Take an integral variety  and set . Because  only has finitely many coordinate linear subspaces, a form of the theorem of Bertini provides the existence of a non-empty Zariski open subset T of  such that  for every .
Proposition 4. 
Take integral varieties ,  and set  and . Assume ; then, there is a non-empty Zariski open subset T of  such that  for all . hence, the restriction of h to  is a surjective morphism . In particular, for all , the variety  is the image of  and we do not need to take the closure in its definition.
Proof.  
We use Remarks 6 and 14 and the fact that the intersection of two non-empty Zariski open subsets of  is a non-empty Zariski open subset of .    □
For all integers , let  denote the set of all integral e-dimensional varieties  such that X is dimensionally transversal to every coordinate linear subspace, i.e., for all  with the convention . For any integral variety , let  be the maximal number of zero-coordinates of some .
Remark 15. 
The case  of the definition of  shows that no  contains a coordinate point.
Lemma 2. 
Fix integers  and let . We have  for a general .
Proof.  
As there are only finitely many coordinate linear subspaces, it is sufficient to use the theorem of Bertini.    □
Proposition 5.  
Fix positive integers e and f such that  and take integral varieties , . Assume . Then, the rational map  induces a morphism  and .
Proof.  
The results follow from the definition of h as the multiplication of all coordinates.    □
Remark 16.  
We may iterate Proposition 5; in particular, if  and , then the variety  is the image of a morphism .
Remark 17. 
Take an integral variety  such that . Because  for every , the rational map  factors through the degree 2 morphism, which is the quotient of  by the order 2 automorphism . Set  and . If , then the upper bound for the integer  coming from Remark 7 is .
3. The Algebro-Geometric Part, II: Proofs of the Theorems
Proof of Theorem 1:  
Because ,  and . Fix . For each ,  is -equivalent to X. Hence, there is  such that  is an isomorphism. Because  and , there are only finitely many , , such that . Hence, the minimal algebraic subgroup G of  containing all  has positive dimension. By assumption, there is  such that . Note that . Because , we find that  has a codimension 1 component. By ([], Th.2.2.5), this codimension 1 component has degree at most . Hence, .    □
Fix an integer s such that  and s integers . Set , , and . Let  denote the morphism defined by the formula
      
      
        
      
      
      
      
    
      with  if  and  if . Each fiber of  is an affine space of dimension s and its closure in  is an s-dimensional linear projective space.
Proof of Theorem 2:  
Because , the set  is finite for each irreducible component  of . Thus, we may assume that  is finite. Fix . Because X and Y are projective curves,  and  for each coordinate hyperplane .
Claim 1:.
Proof of Claim 1: Assume  (for instance) for some codimension 3 coordinate linear subspace V and take . Fix any . The point  is well-defined; hence, , which is a contradiction.
Claim 1 means that each  has  for at most 2 indices i. Because each  has  homogeneous coordinates,  is defined for all , i.e.,  is a morphism.
Assume that part (b) fails and take  such that there are  with . Without loss of generality, we may assume . Fix . Because Y is a projective curve, . Fix . Then, from step (a) we have . The point  is contained in the codimension 3 linear subspace , which is a contradiction.    □
Proof of Theorem 3:  
Set , . By Theorem 2 , is a morphism and each element of  has at most one zero as a coordinate. Moreover, W has exactly  points ,  such that  with two zero-coordinates, i.e., the points . Fix  and take  and . The set  is a line containing the n points , . Because  is a morphism and Y is irreducible, the closed set  is either a point or the line . Write  with  and  for all , and let  be the automorphism induced by the multiplication by 1 on the i-th coordinate and the multiplication by  on the coordinate of .
Assume  for some . Writing  with , in this case Y is the line , as P has a unique 0 among its coordinates. Because L is the line spanned by  and , we have a contradiction.
Now, assume . Because P has a unique zero among its coordinates, Y is contained in the plane , i.e., the plane spanned by  and . Because  is a plane, it contains at most three coordinate points. Fix j such that  and take . We find that Y is contained in the plane  spanned by  and . Because , Y is the line .
In the same way, we can see that X is a line. Because X is a line and , the points  are distinct and uniquely determined by X. Because Y is uniquely determined by the lines , Y is uniquely determined by W and X.   □
Proof of Theorem 4:  
Set . The variety W is a plane ([], Th. 2.1). Because  for each codimension 2 coordinate linear subspace V, the coordinate of each element of W has at most two zeroes. We first apply Theorem 3 with  and then with  and .    □
Example 3. 
Take a general line ,  and set . Because , W is a plane ([], Th. 2.1) and . Let  be the element induced by multiplication by  for the variable  and 1 for the other variables. Set . For general L, we have . Obviously, . If , we can see that in Theorem 3 W does not determine the set , although  determines Y. The same is true if we take an order 2 element of  instead of α.
Question 2: Take a general line ,  and take any line  such that . Is there an order 2 element  such that ?
Remark 18. 
Take  and set , with  as an integral variety; here, W does not change if we permute . Take , , such that . We have . If we restrict this to powers of the same variety X, the same holds if  for all . If X is not fixed by any non-trivial element of , then we need to take  for all , i.e., we need to take a k-root of the identity map. Over an algebraically closed field K with characteristic 0, the group  has  roots of unity.
Question 3: Fix 5 positive integers n, x, y, a, and b such that . Let  be the family of all irreducible  such that , , and  ([], Cor. 5.1).
(a) Provide upper or lower bounds for the dimension of the set of all  such that  for some integral varieties X and Y with , ,  and .
(b) Fix . What is the dimension  of all pairs  such that , , ,  and ? What is the maximum of all integers ? Is  for some ?
(d) The same questions may be raised for the ★-products of  varieties.
4. Over the Real Numbers
We take  and . On  and , we have the Euclidean topology and the Zariski topology, both of which are useful. The Euclidean topology has a basis formed by the open balls, which are semialgebraic open sets. Recall that in the Euclidean topology the closure and open part of a semialgebraic set are semialgebraic ([], Prop. 2.2.2) and that images of semialgebraic sets by a regular mapping are semialgebraic ([], Prop. 2.2.7).
Because the multiplication in  and  is continuous for the Euclidean topology, if the channels have small errors or if the multiplications have small errors, then the observant reader can easily see that the picture in the screen is “near” to the true one. If we know the degrees and dimensions of the expected picture, it is easy to see the best approximation of the picture in the screen.
Recall that  is the complex conjugation and that if  is defined over , then . Because  and  are compact topological spaces for the Euclidean topology,  is compact. Moreover,  is a real algebraic variety of dimension at most . However, in many cases  (e.g., take the real conic ). If X is irreducible over  but not over , then the real algebraic set  has dimension . Hence, in the next statements, e.g., Theorem 5, the assumptions imply that the real algebraic varieties are irreducible over .
Remark 19. 
Take , , defined over , and set . The variety W and its real structure are not changed if we permute the ★-factors. Moreover, W and its real structure are not changed if we take ,  such that  and use  instead of . Now, assume  for all i. This was previously considered over  in Remark 18; the main difference over  is that if k is odd (resp. even), then the set of all  (such that ) has cardinality 1 (resp. ).
Assume that  is defined over  and that its embedding is defined over , i.e., assume that  is scheme-theoretically cut out by finitely homogeneous polynomials with real coefficients. Because the embedding of X is defined over , we have . The set  is a compact real algebraic variety of dimension at most . The set  may be empty, e.g., the real smooth conic . For any X, let  denote the set of its smooth points. If  contains a smooth point p of , i.e., if , then the real algebraic set  in a neighborhood of p for the Euclidean topology is a smooth differentiable manifold of dimension . For any irreducible quasi-projective variety W defined over , we have  if and only if  is Zariski dense in . For any set  (resp. ), let  and  denote the closure of S for the Zariski and Euclidean topologies, respectively. If  is constructable, i.e., a finite union of sets which are locally closed for the Zariski topology, then .
Remark 20. 
For any semialgebraic set , the Euclidean closure  and Zariski closure  have the same dimension and S contains a non-empty open ball (for the Euclidean topology) of  ([], Prop. 2.8.4 or Cor. 2.8.9).
If  for all , then we may use the ★-product form of the Terracini Lemma ([], Lemma 1.6), ([], Lemma 2.12), ([], Lemma 4.13). If  for all , then we have the following result.
Theorem 5. 
Fix an integer  and integral varieties ,  such that  for all i. Set . Then,  is uniquely determined by one of the following semi-algebraic sets:
- (a)
 - (b)
 - (c)
 - (d)
 - the Zariski closure of A in .
 
Moreover, any set of homogeneous polynomials with real coefficients describing  have  as their common solutions in .
Obviously,  if and only if .
Example 4. 
If , then it is sufficient to have  for all  to obtain ; hence, . If , then we always have .
Example 5. 
Take . Let  be a line such that . We have , and the reducible conic  is defined over . We obtain the same answer taking  with R,  as lines defined over  and intersecting , except that  is a single point while  contains a non-empty open subset of .
Example 6. 
Take  and L as a line such that  and L meets no coordinate line. It is easy to check that  is a smooth quadric. Obviously, , i.e., Q is defined over . Because , σ exchanges the two rulings of Q and .
Proposition 6. 
Fix an integer  and integral projective varieties , . Assume  for all i. Then, .
Proof.  
For any smooth point p of some variety , let  denote its Zariski tangent space. The set  is a complex projective space containing p and . Now, assume that W is defined over  and that . The set  is the complex vector space associated with the  real vector space . By the Terracini Lemma ([], Lemma 1.6), the dimension over  (the irreducible projective variety ) may be computed in the following way. There is a non-empty Zariski open subset U of  with the property that for all  the integer  is the same, where  is the ★-product of  and all , . By the Terracini Lemma ([], Lemma 1.6), this integer is . Because ,  is Zariski dense in , in the Zariski open set U we may find  with  for all i. Note that  is the complexification of the real projective space spanned over  by all real projective spaces , where  is the ★-product of  and all , .    □
Lemma 3. 
Take  and geometrically integral projective varieties , , defined over  such that  for all i. Then, the semialgebraic set  has real dimension , while  is the Zariski closure in  of .
Proof.  
The assertion on the dimensions is true by Proposition 13. For any variety  defined over , the semialgebraic set  has real dimension at most . Because  is irreducible and contains , we have  as the Zariski closure in  of .   □
Proof of Theorem 5:  
The Zariski closure inside  is provided by the polynomials with real coefficients, providing the Zariski closure in (c).
The semialgebraic set in (b) is the Euclidean closure in  of the semialgebraic set in (a). The semialgebraic set in (c) is the Zariski closure inside  of both the set in (b) and the set in (d). Thus, it is sufficient to prove that  is the Zariski closure in  of the semialgebraic set . This is true by Lemma 3.    □
Thus, we can see that with mild assumptions on , , the ★-product of the complexifications is uniquely determined by the real ★-product of their real parts. The following examples shows that the converse does not hold, even when we avoid taking the closure.
Example 7. 
Take . Let  be the real conic  with  and let  be a general real line. Obviously,  contains the complement of finitely many points of , while .
Example 8. 
Take irreducible  defined over  and take any . Set . Because ★ is commutative, ; hence, . In many cases (though not all cases, e.g., this not true if X is a line), there are no  such that .
When we have an even number of ★-powers of the same variety, the following construction provides many ghosts.
Example 9. 
Fix an integral variety  defined over  (such that  and the rational map  is birational) onto its image up to the exchange map (Remark 17); i.e., setting  and , we have  and . Fix any  such that . The set of all such P contains a non-empty Euclidean open subset of . Note that  for any  such that . Hence, the set  contains a Euclidean -dimensional manifold U containing no point of . We may also assume , restricting the manifold U if necessary. It is reasonable to say that U is a part of a real ghost of the complex ★-product. It only arises here because we processed the complex data before restriction to the real screen .
4.1. Full Screens
Now, we consider a different problem. The task is checking whether  is Zariski dense in  using the real screen .
For instance, we do not need the shape of , as we can run a test to be sure that our k screens cover all the screen. Having  means that k channels are sufficient to provide all of the data we will need. Now, we prove that we can perform the test using the real screen .
Lemma 4. 
Let ,  be a non-empty Zariski open subset such that . Then,  is a Zariski dense open subset of .
Proof.  
Let  denote the variables in  (hence, ). The set  is Zariski closed in ; hence, it is the zero-set of finitely many polynomials . For any , let  denote the polynomial obtained from f, taking the complex conjugation of all coefficients of f. Set  and . Note that  and  are uniquely determined by the polynomials  and  and that  and  have real coefficients. Because ,  is the zero locus of the polynomials  and , , and consequently of the polynomials  and , we have  with real coefficients. The set  is Zariski open in  because its complement is the zero-locus of  real polynomials.
Hence, the lemma is true for the integer m if and only if . In the inductive proof on m, it can be directly seen that  is Zariski dense in .
Both parts of the lemma are true for , as  is a finite set. Thus, we may assume  and use induction on the integer m. Let  be the Grassmannian of all one-dimensional linear subspaces of . The variety  is defined over , , and . Let  be the set of all  such that . Because U is a non-empty Zariski open subset of , V is a non-empty Zariski open subset of . Since , . Per the inductive assumption,  contains a Zariski open subset of . For each , the set  is finite; hence,  is contained in a Zariski subset of  of dimension at most . The set  is a proper Zariski closed subset of . Thus, the lemma is true for the integer m.    □
Proposition 7. 
Take ,  defined over  and such that . Then,  is Zariski dense in .
Proof.  
Apply Lemma 4. Note that if f in its proof is homogeneous, then , , and  are homogeneous.    □
4.2. Non-Existence of Real Ghosts
In ([], §5.1), there is a section called “Large dimensions” in which (under suitable generality assumptions) the variety  and the Segre embedding of  are projectively equivalent. The assumptions are very strong; each  is embedded in a projective subspace , and these k linear spaces  are linearly independent, i.e., they span a linear space of dimension . Because  with strict inequality if  is not a linear space, the ambient space  is huge. If we take each  defined over  with the embedding  defined over , then we will have no real ghost, as in this case  and .
In ([], §5.2) for  there is a proof in which, by using a specific linear projection (hence, a smaller n, though not by much), we still find that .
The first question below (Question 4) is over an arbitrary field.
Question 4: Find a better lower bound for n. Extend the case  to the case  using linear projections.
Question 5: Address Question 4 over  using general linear projections and prove that real ghosts do not appear after a general real projection.
Question 4 (or Question 5) over  is raised again as Question 7 in the Conclusions section. For any linear space , let  denote the linear projection. Suppose that for a fixed  and integers ,  we have a positive solution for a certain integer n, i.e., we find a complex linear space  such that , , and  (taking ) and where the induced morphism  is an isomorphism. Let  be the Grassmannian of all -dimensional complex linear subspaces of . The set of all linear spaces  satisfying all of the conditions satisfied by V is a Zariski open subset U of . Because , , and  has lower dimension, a random  should be in U. We can certify that we have  defined over ; indeed, the set  is Zariski dense in . Hence, we have  defined over . Moreover,  is Zariski open and Zariski dense in , meaning that a random element of  should work. We have seen that if  and  are arbitrary integral varieties defined over , then the varieties  have no real ghosts.
Question 6: Fix  defined over . Assume  and take general . Then, prove that the variety  has no real ghost.
Example 9 shows that if  and , then in Question 6 it is essential to take a general . Indeed,  has real ghosts for a general .
4.3. Suggestion for Mitigations and Killing the Ghosts
Below, we present four suggestions.
(a) If the complex data are mixed, as in Example 8, then our best suggestion is to use a small channel to send (and save) a small part of . This should be sufficient to guess the ghost part.
(b) Taking the setup in Example 9, to identify some real ghost, send random points ; because these points P are random, they are in  and hopefully . In this case, at minimum  is outside the interior of , and is probably in the interior of a ghost.
(c) Supposing that we know the exact the variety  and preprocessing is available, it is possible to describe the set  (or at least its Euclidean closure) using k real channels. In this way, it is possible to determine whether or not new data coming from k complex channels belong to ghosts.
(d) Supposing that we know the dimensions and degrees (very low) of the varieties , only very few data from k distinct (real or complex) channels (without mixing the data) are needed to reconstruct , in which case we are back to (c). For instance, in order to reconstruct a line (assuming that we are looking for a line), it is sufficient to know two of its points; moreover, if the points have only a small error from the “true” point, the reconstructed line will be near the true line.
5. Conclusions
The Hadamard product of  elements of an n-dimensional (real or complex) projective space with a fixed coordinate system is the coordinate-wise product of these elements. There are k channels, each of them providing packets of -ples of real (or complex) numbers. Then, some device makes the entry-wise multiplication of the k packets received by the k channels and shows it on another device, called the screen. The screen sees a sequence of -ples of numbers. Now, assume that the data on the i-th channel are restricted to a set . From  embedded varieties , we obtain an irreducible variety ; its definition requires it to take a closure in the Zariski topology. In Section 2 and Section 3, we prove several theorems for complex projective varieties. In the fourth section, we study the case of real algebraic varieties. Under mild assumptions, we prove that by taking real data from k channels and then multiplying them, we can reconstruct the data over both  and over . We show that this “reconstruction” works even if small errors are introduced by the channels or multiplications. If we instead start from k real varieties , take k channels obtaining data for the complex points of , and multiply to obtain elements of , then the part appearing on  contains the products of the real data. However, there are sometimes large parts, which we call real ghosts, that do not come from the real points of the varieties. We provide examples in which real ghosts are certified to occur and prove cases in which we certify that real ghosts do not occur. The latter case is related to Question 7 below.
Fix integers  and , , take , and let  be a linear subspace of  such that . For any linear space , let  denote the linear projection. Assume  for all i (this is always satisfied if V is general and  for all i); then, we obtain k linear spaces , .
Question 7: Find a “large” integer s such that, for a general s-dimensional linear space V, the rational map  is an isomorphism.
In the fourth section (Over the Real Numbers, subsection Full Screens), we consider a different problem. We test whether the complex images cover the full complex screen  using only the real screen . For this query, it does not matter whether there are real ghosts.
Funding
This research received no external funding.
Data Availability Statement
No new data were created or analyzed in this study.
Acknowledgments
The author is a member of Gruppo Nazionale per le Strutture Algebriche e Geometriche e loro Applicazioni of Istituto di Alta Matematica, Rome, Italy.
Conflicts of Interest
The author declares no conflicts of interest.
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