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 [
1], 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 [
1,
2]. Recently, C. Bocci and E. Carlini published a monograph on Hadamard products [
3], 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 [
4,
5,
6,
7,
8,
9] 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 [
3]), when we are interested in, say,
, we must first check the algebraic closure
and consider ★ over
, where we call
the closure of
([
3], 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 [
3] 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
([
3], §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 [
3] 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. ([
10], Ex. II.3.18, Ex. II.3.19). If
, then
is semialgebraic and its closure in the Euclidean topology is semialgebraic ([
11], 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 [
3].
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 ([3], §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 ([10], 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 ([10], 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 ([3], 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 ([
12], 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 ([
3], 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
([
3], 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 [
3] shows the existence of
with
.
Example 2. Fix such that . Let V be the binomial hyperplane . We have and for all ([3], Cor. 2.2)). Thus, . Remark 11. Let be an irreducible variety such that . The Terracini Lemma ([3], 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 ([13], Th. 8.5). Because and V is irreducible, for any i. Because is a prime ideal, for all i. Hence, the assumption of ([4], 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 ([
3], Th. 7.4), i.e., by ([
4], 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 ([
14], 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 ([
3], 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 ([3], 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
([
3], 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 ([
11], Prop. 2.2.2) and that images of semialgebraic sets by a regular mapping are semialgebraic ([
11], 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 ([11], Prop. 2.8.4 or Cor. 2.8.9). If
for all
, then we may use the ★-product form of the Terracini Lemma ([
3], Lemma 1.6), ([
8], Lemma 2.12), ([
9], 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 ([
3], 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 ([
3], 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 ([
3], §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 ([
3], §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.