Rank-three obstructions to
coordinate projection volumes

Proof. Every three-label principal determinant is , so that principal matrix has one positive and two negative eigenvalues. Interlacing and rank three give the asserted signature. Write . The zero diagonal gives , while positive off-diagonal entries force all to have the same nonzero sign. Taking them positive and writing gives the representation. Positivity excludes coincident circle points. □

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. □

Proof. Two weights 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. Thus any nonzero nonnegative solution gives a positive one.

For positive weights, the matrix has rank three and satisfies the ordinary triangle inequalities. Lemma 2.2 would therefore give five points on a circle with no obtuse triangle, a contradiction. □

Proof. Let converge. Adding sufficiently small Euclidean balls allows us to assume each has interior without changing the limit. Choose a maximal-volume inscribed simplex and write

where . The upper inclusion follows because replacing any simplex vertex by a point of cannot increase the simplex volume; every barycentric coordinate has absolute value at most one. Pass to a subsequence with , where has interior.

Let and denote its columns by . Covariance of mixed volume gives

For arbitrary independent directions, extrusion along segments gives

The fixed inball in therefore bounds every -column wedge of . These wedges are the columns of , so the product of the largest singular values of is uniformly bounded.

We now suppress the index and write the singular values as . Choose a fixed such that for every . For , set

At one has . At one has . By continuity, choose with . In a left singular basis of , let have eigenvalues . Then

and at most eigenvalues of are smaller than one.

Return to the sequence and replace by and its marked matrix by . Equation (3) is unchanged, by covariance and . The new bodies and directions are bounded. Take limits and . Since at least eigenvalues of each are one, . Thus has dimension at least , and continuity of mixed volumes proves the required representation.

Conversely, start with any and marked matrix in (3). Choose invertible , put , and define

Then . The projection tuples of are therefore the mixed-volume tuples of , which converge to the prescribed tuple. This proves the converse and the closure formula. □

Proof. Approximate uniformly by smooth support functions . For a sufficiently large constant , the function is a support function, since its spherical Hessian plus its value times the identity is positive semidefinite. Thus for a convex body , and positivity gives . Weak continuity proves the support inclusion. The support-function criterion used here is recalled in [5, Lemma 4.5].

To identify the measures, write the common mixed volume as . Alexandrov–Fenchel applied to gives

Letting and interchanging yields for every . Differences of support functions contain all functions on the sphere, so the measures agree.

The equality theorem of Shenfeld and van Handel [5, Theorems 2.2 and 4.4] gives on , for some and . Its hypotheses require interior only for the repeated body ; the comparison bodies may be lower dimensional. Integrate against , using the support inclusion and its zero first moment. Since , the resulting identity forces . □

Proof. Assume such a representation. Use the cyclic ordering from Lemma 2.1, and let and describe the polygonal cone (1). Define

Then

The are centered positive measures supported on . Since , each is concentrated on .

For every nonzero , its positive and negative parts both contain at least two nonzero coordinates. Otherwise the equation at the sole positive or negative coordinate would contradict positivity of the off-diagonal entries. Since , the two zonotopes below have equal, strictly positive -self-volumes and mixed volume,

Lemma 4.1 applied to each row of gives a linear map such that

The first identity means that the vector of measure densities lies in the polygonal cone. More explicitly, take and write . Then and almost everywhere. Triangulating a compact section of (1) gives nonnegative measurable coefficients of its extreme rays. Integrating those coefficients against produces finite positive measures such that

For , the coefficient is positive, so domination by confines to . Thus only the two endpoint functions may be nonzero on the measure assigned to an edge.

Fix and let be the other cycle neighbor of . In (7) for , only the two edges incident to can contribute to an integral involving . The edge has coefficient , leaving only . Centeredness and (5) give

The same holds for .

Let and . All signed endpoint moments vanish, and the other coordinate functions vanish on , so . Moreover, any relation with one zero coefficient has all coefficients zero. Indeed, on the relation reduces to ; if one endpoint coefficient is zero, the positive absolute moment at the other endpoint forces its coefficient to vanish too. Propagating around the polygon proves the claim. A vector space of relations with this property has dimension at most one. There are marked vectors in , hence

Integrating the second identity in (6) against gives

where denotes column of . The right-hand sides span a space of dimension at least two. To see this, let and take a row functional annihilating . The row satisfies

Positive coefficients determine every coordinate from a single one, so the solution space has dimension at most one; for odd it is zero, since a nonzero solution would alternate signs around an odd cycle. The map is injective. Therefore , and for odd . Either bound gives when , whereas (8) confines to , of dimension at most one. This contradiction proves the proposition. □

Proof. If , apply the determinant-one shear fixing and sending to . It sends to and the component of in to . Extrusion along the normal segment gives

which is the claimed identity. Interchange if necessary; if both normal components vanish, all the bodies lie in and both sides are zero.

The functional is a seminorm by the Cauchy projection formula [4]. Its triangle inequality, applied to

gives the weighted inequalities. If all vanished, every entry would vanish, so the weights are nonzero for a nonzero array. □

Proof of Theorem 1.1. Suppose . Theorem 3.1 with gives a representation (4) with . Proposition 4.2 excludes dimension . In dimension , translate into a hyperplane and apply Lemma 5.1. Choose five labels including one with a positive weight. Their principal matrix has rank exactly three, because every three-label determinant is positive, and its inherited weights contradict Corollary 2.3. This proves exclusion from the closure.

For strict perturbations, write as in Lemma 2.1, where . Set

The inner quadratic form is strictly negative on for every , since there it equals . It has a positive value on , so has signature . Its entries remain positive, and as ; because lies outside the closed realization locus, so do all sufficiently small perturbations. The positive strict-signature locus and the complement of are open, giving the asserted open sets of counterexamples. □