Compatibility and realization
of coordinate projection areas

Caius Folkerts

Abstract. Four-coordinate inequalities characterize the coordinate projection areas of convex bodies in dimension four. In every higher dimension, we show that every positive hollow matrix of rank three lies outside the closure of the realization locus, although each of its four-coordinate restrictions is realizable. The obstruction comes from incompatible equality cases on overlapping coordinate subspaces. We also characterize limits produced by bodies of unbounded diameter through weighted triangle inequalities, and realize their strict part by centrally symmetric polytopes with at most vertices. Together these results determine both exact and limiting realizability for every nonnegative hollow matrix of rank at most three.

1. Introduction

Let retain coordinates . For a nonempty compact convex set , write

We identify an array with its symmetric matrix, setting the diagonal entries equal to zero; such a matrix is called hollow. A positive array has strictly positive off-diagonal entries. Convex sets of lower affine dimension are allowed throughout.

In dimension four, Huang, Huh, Michałek, Wang and Wang [2] proved that an array is realizable exactly when

Equivalently, its hollow matrix has at most one positive eigenvalue. Every four-coordinate projection of a higher-dimensional body obeys the same criterion, and Huh [3] asked whether these conditions suffice in dimension five. The difficulty is that individually realizable four-dimensional images must also agree where their coordinate spaces overlap.

Equality in (1) makes this compatibility visible. When the entries are positive and

the quartet carries the partition . We prove that the partitions attached to two such quartets must agree on their common triple.

Theorem 1.1. Let . Every positive hollow symmetric matrix of rank three lies outside , although all of its four-coordinate restrictions belong to .

Compatibility permits at most three boundary quartets on five labels, whereas a positive rank-three matrix places every quartet on the boundary. The theorem excludes approximation as well as exact realization. For example,

is the squared-distance matrix of five consecutive vertices of a regular hexagon of circumradius one. With , and , it has the factorization

The three vectors are independent, so has rank three and signature and lies outside . Small perturbations give a nonempty open set of nonrealizable positive arrays for which every four-coordinate matrix has the strict signature . Thus the failure of the four-coordinate criterion persists away from its equality cases.

Passing from exact realization to its closure requires understanding bodies whose diameters tend to infinity while their projection areas remain bounded. Those bodies become long and thin in a single direction. Their limiting areas are governed by weighted triangle inequalities

Let be the set of nonnegative arrays admitting such a vector .

Theorem 1.2. An array belongs to if and only if it is the limit of the projection-area arrays of convex bodies whose diameters tend to infinity. In particular,

If all weights are positive and all inequalities in (2) are strict, the array itself is realized by centrally symmetric polytopes with at most vertices and arbitrarily large diameter.

For positive weights, division by turns (2) into the ordinary triangle inequalities for . The metric interpretation gives a construction in which distance potentials determine transverse vertices, while a long common axis makes every coordinate projection a diamond of the prescribed area. The same inequalities describe the only possible loss of compactness. At rank three they would make the rescaled entries squared distances of points on a circle with no obtuse triangle, which is impossible on five labels.

Zero entries admit an exact reduction. After deleting zero rows, coordinates with zero mutual area form classes of proportional rows. Collapsing each class preserves both rank and realizability. Combining this reduction with Theorem 1.1 gives a complete criterion at rank at most three. An array is realizable, equivalently lies in the closure, precisely when there are at most four classes.

The projections considered here retain two coordinates. The complementary projections that delete two coordinates have a different realization problem, even in dimension five; their rank-three obstruction is treated in [1].

2. Compatibility of equality cases

A boundary quartet is a positive four-coordinate array for which one of the inequalities (1) is an equality. Its partition is the pair of opposite edges giving the largest product. The first step is to ensure that the equality theorem for mixed volumes can be applied to its realizing body.

Lemma 2.1. Every realization of a boundary quartet in has nonempty interior.

Proof. Positive coordinate dilations make the areas on opposite pairs equal. Explicitly, dilation of coordinate by

changes to , where . Relabeling therefore reduces the data to

Suppose the realizing set has dimension at most three. It is a linear image, up to translation, of a full-dimensional convex body . Denote its four coordinate functionals by . The brightness norm of ,

satisfies , by the Cauchy projection formula [4].

For any relation , taking cross products and applying the triangle inequality gives

Their sums agree, so each inequality is an equality. The largest of the four absolute coefficients is a strict weighted average of two others; propagation shows that all four absolute coefficients are equal. Thus every nonzero relation has all coefficients nonzero, the relation space is one-dimensional, and every triple of the is independent. Changing coordinate signs, which leaves all areas unchanged, we may assume .

Put , and . These vectors are independent, and the six areas give

An integral of absolute linear forms satisfies Hlawka’s inequality

which follows by integrating the scalar absolute-value inequality. Apply it to , , . The displayed values cancel, leaving , contrary to independence and the fact that is a norm. □

We use mixed area measures with the normalization

They are positive and multilinear, and their first moment is zero. These facts, and continuity under Hausdorff convergence, hold without interior assumptions [5, Section 4.1].

Lemma 2.2. Fix and put . If , then

For the coordinate segment , with deleting coordinate , one also has

where embeds the unit sphere of in .

Proof. Let the common mixed volume be . Alexandrov–Fenchel applied to gives

Divide by and let to obtain . Interchanging proves equality. The defining integral then identifies the measures, since differences of support functions contain all functions on the sphere [5, Lemma 4.5].

For (3), extrusion along gives

Apply the defining integral in dimension and use on . □

A nonzero normal direction is 1-extreme for if it cannot be written as the sum of three linearly independent normals at a common boundary point. The full-dimensional equality theorem of Shenfeld and van Handel [5, Theorem 2.2] says that, when has interior and , equality in

holds exactly when is linear on all 1-extreme normals of , for some . The comparison bodies need not have interior. If is not a point, then , since otherwise would agree with a point on every regular supporting normal of , and those normals positively span .

Proposition 2.3. For the positive coordinate projection areas of one convex set, the partitions of two boundary quartets sharing three labels agree on those labels.

Proof. Let the shared triple be , and write the quartets as and . Their projections have interior by Lemma 2.1; their common projection onto is therefore three-dimensional.

For either quartet , form the complementary matrix

It is the matrix of . Its kernel has a vector whose signs are the two blocks of the boundary partition. To see the signs, use the positive diagonal normalization in Lemma 2.1. The normalized matrix has kernel vector when its largest pair is , and undoing the normalization preserves signs.

The rectangles and have equal positive mixed self-volume and mixed volume. Lemma 2.2 consequently gives

Orient so the omitted coordinate, or , has positive sign. The two negative coordinates then form the two-element block of the restricted partition on . Each is supported on , while the measure of the omitted segment is by (3). Positivity of the measures therefore confines to the union of the two hyperplanes indexed by that block.

If the restricted partitions differ, their two-element blocks have a single common label . Intersecting the resulting support restrictions inside gives

Write . The Cauchy projection formula and centeredness now imply

Hence has two-dimensional exposed faces in both directions . Projection commutes with the exposed face in any direction belonging to the projection space, so the corresponding faces of also have dimension at least two. Both normals are 1-extreme, since three normals at a common point whose sum is would all annihilate the two-dimensional direction space of that face, and could not be independent in .

Apply the equality theorem to the rectangles associated with . The label belongs to its negative block, so has zero width in direction whereas has positive width. Adding the support-function equalities at and cancels the translation and equates these widths up to the positive factor , a contradiction. □

Corollary 2.4. Among five coordinate labels with positive projection areas, at most three quartets are boundary quartets. If at least two are boundary quartets, their partitions extend uniquely to a partition of the five labels into blocks of sizes two and three, and every boundary quartet omits a label from the larger block. The bound three is attained.

Proof. Suppose two boundary quartets omit . Their common triple has a partition , so their partitions are and . Both extend to . A boundary quartet omitting would have to induce both and by compatibility, which is impossible. The same holds for omission of , leaving at most the other three quartets.

For equality, take the convex hull of the square on the first block and the cube on the second. Its within-block areas are and , respectively, and every cross-block projection is a diamond of area . Exactly the three quartets omitting a coordinate of the second block are boundary quartets. □

Corollary 2.5. A positive hollow symmetric matrix of rank three and order at least five is not realizable as a coordinate projection-area matrix.

Proof. Every three-label principal determinant is , so that principal matrix has signature . Interlacing and rank three give the full signature and signature on every quartet. The four-dimensional criterion then places every quartet on its boundary, contradicting Corollary 2.4. □

3. Long convex bodies and weighted metrics

Bounded projection areas permit a body to become arbitrarily long only if its transverse dimensions shrink. The next lemma records the resulting limit, without imposing interior assumptions on the bodies.

Lemma 3.1. Suppose and . There are a unit vector and a seminorm on such that

Consequently , with weights .

Proof. Translate the midpoint of a diameter chord to the origin and write its endpoints as , with . For every , the triangle with these endpoints and projects to a triangle of area

The identity bounds the transverse component of by , for a fixed . The diameter condition also gives .

Write , where , and let be the set of points so obtained. These bodies lie in a fixed compact cylinder, with and , and contain the segment . After passage to a subsequence,

in Euclidean and Hausdorff distance. For , put and define

Each fiber is nonempty, and widths are even and subadditive, so is a seminorm.

Fix with . Rotate the coordinate plane to align its first axis with , then apply the area-preserving map . The resulting image of is the image of under a linear map converging to

Continuity of planar area under Hausdorff convergence and integration over the first coordinate give

If , use the same rotation when and any planar axis when . The transformed projection lies in a rectangle with side lengths at most and , where ; its area tends to zero. Thus the formula holds for every pair.

Finally,

Applying gives (2) with , and these weights are not all zero. □

The converse is an exact construction. For a compact convex set , denote its width functional by

Planar extrusion gives, for every ,

Indeed, adding a segment to a planar convex set increases each nonempty fiber parallel to that segment by its length; integrating the fibers gives the width term. This remains true for degenerate planar sets.

Proposition 3.2. If and is any witness in (2), there is a centrally symmetric polytope containing the origin such that

Thus is realizable for every , by bodies of diameter at least .

Proof. Let , , and choose . The weighted inequalities imply

For , the numbers form a pseudometric. Consider its anchored Lipschitz polytope

It is compact and centrally symmetric. For any fixed , the distance potential belongs to and satisfies . Hence the range of on is exactly .

In , take

For , the functional becomes , of width . For , , its width is , and for two labels in both sides vanish. Equation (4) proves the assertion about realizations and diameters. □

Theorem 3.3. The set is closed and consists exactly of the limits of projection-area arrays along sequences with diverging diameter. Moreover,

The approximating bodies for every point of may be chosen to be full-dimensional centrally symmetric polytopes.

Proof. Normalize witnesses for (2) by . Compactness of the simplex shows that the inequalities persist under limits, proving closedness. Lemma 3.1 and Proposition 3.2 establish the characterization by diverging diameter. Adding a sufficiently small cube to each approximating polytope gives full-dimensional centrally symmetric approximants with the same limit.

For an arbitrary convergent sequence of realizable arrays, either the realizing bodies have a subsequence of bounded diameter or they have a subsequence whose diameters tend to infinity. In the former case, translation and Blaschke selection give a compact convex limit realizing the array; in the latter, the limit belongs to . The converse inclusion follows from Proposition 3.2. □

4. Exact realizations from strict metrics

A limiting prism becomes an exact realization when its transverse vertices can be assigned independently to the coordinate planes. Strict triangle inequalities provide precisely the separation needed for this construction.

Theorem 4.1. Suppose and satisfy

Then is realized by centrally symmetric polytopes with at most vertices and arbitrarily large diameter.

Proof. Put and . For each pair , define

The distance potential attains the difference on its designated pair. On every other pair , it satisfies the strict inequality

Indeed, twice the difference is . Each term has absolute value at most , and at least one of is outside , making that bound strict.

We seek a polytope with long diagonal whose projection in plane is the diamond with transverse vertices . Set and . Then

Every other projected transverse vector has a unique expression , and taking determinants gives

Since there are only finitely many pairs, for all sufficiently large one has simultaneously. This is exactly the condition that lie in the desired diamond.

Consequently

has that diamond as every coordinate-plane projection, of area . There are at most vertices, and . □

The strict realizations are dense in . To include zero weights, first take the polytope from Proposition 3.2, replace by , and use the arrays

They satisfy the weighted inequalities with positive weights and converge to . For positive weights the associated pseudometric becomes a strict metric by adding to every off-diagonal distance. The resulting exact realizations approach the original array. Thus the metric construction describes the whole escape set through the closure of an explicit family of polytopes.

5. Rank three and the failure of approximation

The compatibility obstruction excludes exact rank-three realizations. To exclude approximation, it remains to show that rank-three arrays cannot satisfy the weighted triangle inequalities.

Lemma 5.1. A positive hollow symmetric matrix of rank three satisfying the triangle inequalities is the squared-distance matrix of at most four points on a circle. If there are four points, they form a rectangle.

Proof. We first use an elementary property of four-point metrics. If is such a metric, then

To see this, relabel so that is the largest of the three sums on opposite pairs. For a subset , let be the matrix equal to one across the cut and zero elsewhere. Then

where

The coefficients are nonnegative, and when .

Now let have rank three and satisfy the hypotheses. Its principal matrix on the first three labels is invertible, since . Set for . The rank condition gives

For each , the principal matrix on has kernel vector . Its coordinate sum must be zero. Otherwise the all-ones vector could be shifted by a multiple of this kernel vector to have coordinate sum zero without changing its strictly positive quadratic value, contrary to the four-point inequality. Thus every column of has coordinate sum one.

Choose a nondegenerate planar triangle with squared side lengths , and place its circumcenter at the origin. Such a triangle exists because the triangle inequalities for the positive numbers imply strict triangle inequalities for their square roots. Write its vertices as , with , and put . The sum-one condition gives

The zero diagonal therefore puts every on the same circle, and the displayed identity becomes . Positive off-diagonal entries make the points distinct.

The metric inequalities now say that every triangle of these points is nonobtuse. For five points in cyclic order, some two consecutive arc gaps sum to less than , so their three endpoints form an obtuse triangle. Hence there are at most four points. If there are four, the opposite angles of their inscribed quadrilateral sum to and are all at most , so the quadrilateral is a rectangle. □

Corollary 5.2. A positive hollow symmetric matrix of rank three on five labels admits no nonzero nonnegative weights satisfying (2).

Proof. Suppose such weights exist. Two of them cannot vanish, since an inequality with those weights on the right and a positive weight on the left would fail. If exactly one weight vanishes, replace it by a sufficiently small positive number. Inequalities with that weight on the right are only relaxed, while those with it on the left still hold because their right sides are strictly positive. We may therefore assume all weights are positive.

Division by shows that satisfies the ordinary triangle inequalities. Positive diagonal congruence preserves rank, so Lemma 5.1 would realize by five points on a circle with no obtuse triangle, a contradiction. □

Proof of Theorem 1.1. For five labels, Corollary 2.5 excludes and Corollary 5.2 excludes . Theorem 3.3 therefore excludes their union . For , any five-label principal matrix still has rank three, since its three-label principal determinants are positive. Projection onto those five coordinates takes into , giving the same obstruction. Every four-coordinate restriction is realizable by (1). □

Corollary 5.3. For every , there is a nonempty open set of positive arrays disjoint from on which every four-coordinate principal matrix has signature .

Proof. Choose any positive hollow rank-three matrix of order . Its signature gives a factorization . The zero diagonal gives , and positive off-diagonal entries force all to have the same nonzero sign. Taking , and putting , , , we obtain

Set . The inner form is strictly negative on , where it equals , and has a positive value on . Thus has signature , and the same argument on any four labels gives signature . Theorem 1.1 excludes from the closed realization locus, so all sufficiently small positive perturbations remain outside it. Positivity and these strict signatures persist on an open neighborhood. □

6. Zero areas and a complete rank-three criterion

The local four-coordinate inequalities determine how zero entries can occur. They correspond to coordinate replication, so no additional geometric obstruction is concealed in the zero pattern.

Proposition 6.1. Let be a nonnegative hollow symmetric matrix whose four-label principal matrices have at most one positive eigenvalue. After deleting zero rows and columns, the relation

is an equivalence relation, and its classes are precisely the positive proportionality classes of rows. Choose one representative from each class, and let be the resulting principal matrix. Then is positive, , and is realizable if and only if is realizable. The same equivalence holds for membership in the closures of the respective realization loci.

Proof. If , the determinant on labels is

The spectral hypothesis makes this determinant nonpositive, so it vanishes. If rows are nonzero, this identity first gives a common positive column and then their positive proportionality. Conversely, proportional rows have , by examining column . This proves the equivalence relation.

There are positive factors , equal to one at representatives, such that . Define by on nonzero rows and on removed rows. Then , while is a principal matrix, proving equality of ranks. A body realizing projects to one realizing ; conversely, the image under of a body realizing has the prescribed areas. These fixed linear maps also preserve limits of area arrays, proving the closure assertion. □

Theorem 6.2. Let be a nonnegative hollow symmetric matrix of rank at most three. Its area array is realizable if and only if it lies in the closure of the realization locus, and these conditions hold exactly when its quotient in Proposition 6.1 has at most four classes. The all-zero matrix is realized by a point.

Proof. The matrix has at most one positive eigenvalue. For rank at most two this follows from zero trace. If rank is three and two eigenvalues were , the third would be , giving

whereas entrywise nonnegativity gives . Thus Proposition 6.1 applies. A quotient on at most four labels is realizable by the four-dimensional theorem, padding with zero coordinates if necessary. A quotient on at least five labels is positive and has rank three, so Theorem 1.1 excludes even limiting realization. □

References

[1]
Caius Folkerts, Rank-three obstructions to coordinate projection volumes, manuscript, 2026.
[2]
Daoji Huang, June Huh, Mateusz Michałek, Botong Wang and Shouda Wang, Realizations of homology classes and projection areas, arXiv 2505.08881, 2025.
[3]
June Huh, Volume polynomials, arXiv 2601.13249, 2026.
[4]
Rolf Schneider, Convex Bodies: The Brunn–Minkowski Theory, second expanded edition, Encyclopedia of Mathematics and its Applications 151, Cambridge University Press, 2014.
[5]
Yair Shenfeld and Ramon van Handel, The extremals of Minkowski’s quadratic inequality, Duke Mathematical Journal 171 (2022), no. 4, 957–1027.