Abstract. We give necessary and sufficient conditions for a homogeneous polynomial to be the volume polynomial of compact convex sets, resolving the Heine–Shephard realization problem. For sets in dimension , we prove that the coefficient locus is semialgebraic exactly when , ruling out any finite algebraic characterization for . For all , we nevertheless characterize realizability by coefficient nonnegativity and an explicitly enumerable family of homogeneous quadratic inequalities with rational coefficients. For computable coefficient vectors, these conditions yield finite certificates of nonrealizability.
The nonsemialgebraicity obstruction occurs on an affine slice of ternary cubics whose boundary departs from an algebraic curve by a term of order . The proof uses matrix-valued slope measures to represent planar mixed volume and a logarithmic energy estimate. We also show that finitely many mixed-volume relations can force a nonsemialgebraic realizing set: an irrational-rotation Minkowski series with infinitely many exposed edges, whose area function has the unit circle as a natural boundary. Consequently, some volume polynomials admit no realization by semialgebraic compact convex sets, and hence none by polytopes.
Given nonempty compact convex sets , Minkowski’s theorem makes
a homogeneous polynomial of degree with nonnegative coefficients. The Heine–Shephard realization problem asks which polynomials arise in this way, or equivalently which numerical arrays can occur as all the mixed volumes of a fixed number of convex sets. It originates in work of Heine (1938) and Shephard (1960): Heine characterized three bodies in the plane, and Shephard characterized two bodies in every dimension [5, 9]. These classical characterizations are algebraic.
We resolve the realization problem for every number of sets and every dimension by giving a countable coefficient characterization. We first prove that in dimension at least three the passage from two sets to three precludes any finite algebraic characterization. We then construct an explicitly enumerable family of homogeneous quadratic inequalities with rational coefficients which, together with coefficient nonnegativity, is necessary and sufficient for realizability. The nonsemialgebraicity theorem and the countable characterization are separate results: the former rules out finite algebraic descriptions, while the latter supplies a complete criterion beyond that obstruction.
Let denote the coefficient locus of these volume polynomials, including realizations by
lower-dimensional sets. We keep every coefficient, including pure powers and
repeated-index mixed volumes. A subset of a real vector space is semialgebraic if it is a
finite Boolean combination of polynomial equalities and inequalities, with arbitrary real
coefficients.
Theorem 1.1 (Sharp algebraic threshold).For every and ,
The same nonsemialgebraicity holds when every realizing body is required to be a
full-dimensional polytope. Thus no finite collection of polynomial inequalities, even with
Boolean operations or finitely many auxiliary real variables, describes volume-polynomial
realizability in these dimensions. This answers the semialgebraicity question of Averkov,
Borger and Soprunov [1]. It also places a limit on descriptions by stronger log-concavity
inequalities; the obstruction concerns every finite algebraic description, rather than
a particular proposed family of inequalities. For the relation between volume
polynomials and Lorentzian polynomials, see Brändén–Huh [4], Menges [7] and Huh
[6].
The essential case is a ternary cubic. We specify its six repeated mixed volumes so that
every realization consists of planar bodies with independent plane normals. After
normalization, only the common area of two unit-width planar bodies and their interaction
remain to be optimized. The lower boundary on this affine coefficient slice satisfies
Here and below means that their ratio is bounded above and below by positive
constants. A semialgebraic boundary cannot have the logarithmic gap in (1.1).
The logarithm comes from the geometry of the planar bodies. Their four corner caps
are encoded by positive measures on logarithmic slope space, with area given by the
energy kernel . The operation representing triple mixed volume becomes multiplication of
Fourier matrices. A measure with small energy can retain appreciable total mass only by
spreading over a long interval; the available exponential moments limit that interval to
logarithmic length. Hyperbolic arcs, whose logarithmic slope measures are uniform, attain
the resulting order.
To retain three bodies in higher dimensions, we raise the degree of an existing variable.
The coefficient support forces the enlarged body to have planar fibers, whose Minkowski
average satisfies a sharp area inequality; a cone construction attains the two moments
entering that inequality. This preserves the logarithmic boundary comparison in every
dimension.
The coefficients also constrain the shapes of realizing bodies. In the cubic slice they
force three independent planes and leave a planar optimization problem. Additional
equalities in the planar Minkowski inequality identify projected bodies up to translation;
making three such identifications around a cycle of planes prescribes a linear return map,
which we call projection holonomy.
Theorem 1.2 (Shape forcing).There is a finite system of polynomial conditions
with rational coefficients on , where and , whose projection to is exactly the graph
In every realization, a specified body is affinely equivalent to
It has infinitely many exposed edges and is not semialgebraic. The area function
has rational Taylor coefficients and the unit circle as a natural boundary; in
particular, it is transcendental over .
Consequently some volume polynomials have no realization by semialgebraic convex
sets, and hence none by polytopes of any complexity. The obstruction persists in every
higher dimension after adding one body. A second specialization of the same projection
pattern forces ellipses, so the rigidity is already visible before the contraction series is
introduced.
An exact coefficient criterion remains possible when infinitely many inequalities are
allowed, and it can be given using only rational arithmetic.
Theorem 1.3 (Countable coefficient characterization).For every , there is an
explicitly enumerable family of homogeneous rational quadratic forms such that
Every nonrealizable vector strictly violates one of these inequalities. For
computable input coefficients, nonrealizability therefore has a finite rational
certificate. Realizable inputs admit a convergent construction by rational polytopes.
The construction rests on a uniform affine normalization and finite rational nets for
convex bodies. Nonrealizability can be detected by a finite search, whereas the realizing
polytopes are obtained by a convergent limiting construction; Section 8 gives the precise
computational statements.
The two-body realization theorem supplies the algebraic side of the threshold; a proof is
included in Appendix A.
2. Coefficients, projections and compactness
Our normalization of mixed volume is
where means that occurs times. In dimension three we write , and in the plane we use
We use symmetry, multilinearity, nonnegativity, Hausdorff continuity and translation
invariance of mixed volume, together with its covariance by the absolute determinant of a
common linear map [8]. Independent translations of the individual bodies leave the volume
polynomial unchanged.
2.1. Reading geometry from coefficients.
The support of a volume polynomial records the dimensions of all partial sums. This
elementary observation lets zero coefficients impose subspace constraints without any
assumptions on the realizing bodies.
Lemma 2.1 (Degree and dimension).Suppose is full dimensional. For every
nonempty , the largest total exponent of the variables in is .
Proof.Put and ; when is the whole index set, take . Identifying the variables in
and setting the others equal to one gives . Its degree is the stated largest exponent
because the coefficients are nonnegative.
Translate into a linear subspace of dimension . For , the projection of to is
full dimensional there, and every nonempty fiber of contains a translate of . Fubini
therefore bounds the volume below by
Boxes adapted to give an upper bound . Hence its degree is . The cases and
follow directly from translation invariance and volume homogeneity. □
For a unit vector , write for the width of in that direction. Widths and projected areas
occur as coefficients through the following extrusion identity.
Lemma 2.2 (Extrusion).If is compact and convex and , then
In particular, if is planar with unit normal , then
and for arbitrary compact convex ,
Proof.Every nonempty fiber of parallel to is an interval, possibly a point, and
adding increases its length by . Integration proves the first assertion. Comparing
coefficients after replacing by a Minkowski sum proves (2.2); slicing perpendicular
to the plane of gives (2.1). □
For planar convex sets of positive area, the Minkowski inequality gives
with equality precisely when and are positively homothetic [8, Theorem 7.2.1]. Thus
equality of both areas and their mixed area identifies the two bodies up to translation.
This form applies to all compact convex sets of positive area; no boundary regularity is
needed.
2.2. A bounded model for realization.
The coefficient locus has a compact model after one normalization. We use the same
model both to pass to limits in the geometric constructions and to obtain finite rational
approximations later.
Lemma 2.3 (Uniform normalization).If , independent translations and a common
linear map of determinant modulus one put the tuple in the form
Proof.Choose a simplex of maximum volume in , and set and . Replacing by any
shows that each coordinate of has absolute value at most one. Since the simplex lies
in a set of volume one, ; consequently
Write with and take . Each contains zero and lies in their sum, which is contained
in . □
Let be the compact metric space of nonempty compact convex subsets of containing
zero, with Hausdorff distance, and let be the coefficient map. Compactness follows from
the uniformly bounded Lipschitz support functions on the unit sphere, and is continuous
by mixed-volume continuity. The image
is therefore compact. The normalization lemma says that every nonzero volume
polynomial, divided by its value at , belongs to this image.
Proposition 2.4.The locus is closed. If a class of tuples is dense for the product
Hausdorff metric, the closure of its coefficient image is .
Proof.Suppose converges to and write , . All coefficients are nonnegative, so implies
, which is realized by points. If , then eventually ; compactness puts in the same
image, and dilating its realizing bodies by realizes . The second assertion now follows
from continuity and density. □
3. An affine slice of ternary cubics
The obstruction will be found in the following family
where and . The pure coefficients vanish, while all six repeated mixed volumes are
positive. This forces each body to be planar, and the asymmetry between the two cyclic
products of repeated coefficients forces the planes into general position. The following
normalization describes every realization of the slice.
Proposition 3.1.The polynomial belongs to if and only if there are planar compact
convex sets , , and satisfying
Proof.In any realization, a zero pure coefficient excludes dimension three, and
a positive repeated coefficient excludes dimensions zero and one. The planes are
pairwise distinct by (2.1). Translate them through the origin. If their normals were
dependent, the three planes would share a line. Projecting perpendicular to this line,
write for a unit vector along the image of the th plane, for the projected width of
its body and for its area. Formula (2.1) would give
and hence
The two sides for (3.1) are and , respectively. Since , the normals are independent.
A common linear map of determinant modulus one can therefore take the planes to
. Denote the resulting widths of the three bodies by . Comparing repeated coefficients
using (2.1) yields
Apply the diagonal map . The widths now have the stated values, and the new
areas are with . The triple mixed volume is divided by , which gives (3.2).
Conversely, a normalized triple as in the proposition has repeated mixed volumes
equal to times the six prescribed values, and its triple mixed volume is . A common
dilation by therefore gives exactly (3.1). □
We shall set , so the coefficient map is affine. This choice restricts the common area to
, since has area inside a rectangle of area . Define
These fibers are nonempty; take and to be unit right triangles and to be the
rectangle of widths . Then and (3.2) gives . Nonnegativity of mixed volume shows
.
The normalization has reduced the problem to planar geometry, but it has not yet
separated the three bodies. The next section replaces the interaction of and by a single
planar body . Once that replacement is available, optimizing will be an equality case of
one Cauchy–Schwarz inequality.
4. Corner measures and triple mixed volume
4.1. The energy of a corner.
For a planar convex set in its bounding rectangle, the area deficit is the sum of
four corner areas. There is a useful quadratic expression for each of them. Set
so that is a probability measure and
For finite real signed measures put
With the Fourier convention , formula (4.1) identifies with . In particular this energy is
nonnegative and satisfies Cauchy–Schwarz.
Orient the boundary of a convex polygon counterclockwise. A non-axis edge contributes the atom
to a measure , where and . Arrange these four measures in a matrix, with
rows and columns ordered , and write . Axis edges contribute no atom. The
exponential moments of an atom recover its horizontal and vertical lengths, which is
the reason for both the logarithmic coordinate and the square-root weight in
(4.2).
Figure 1: The bounding rectangle separates area into four corner energies. The
hyperbolic arc on the right has constant mass density in logarithmic slope; it will
supply the logarithmic lower bound.
Lemma 4.1 (Corner identities).If has coordinate widths , then
Corner measures add under Minkowski addition. Consequently, for polygons of widths and ,
For arbitrary planar compact convex sets,
Proof.Translate the bounding rectangle to . Its complement consists of four caps,
each bounded by a chain on which the signs of the edge components are fixed. After
reflecting one cap, write the absolute edge components as in slope order. The shoelace
formula gives its area as
The identity
makes this precisely . Summing the caps proves (4.3). The two exponential
moments of (4.2) are and , and each signed horizontal or vertical variation of a
convex boundary equals the corresponding width. Omitting axis edges gives (??).
The edges of a Minkowski sum are obtained by merging edge directions and
adding parallel edge vectors. At a fixed direction the atom location is fixed and its
mass is linear in edge length; thus . Comparing the coefficient of in (4.3) proves
(4.4). Cauchy–Schwarz in the direct sum of the four energy spaces yields (4.5).
Degenerate polygons follow by adding a vanishing axis-parallel rectangle, and polygon
approximation extends the inequality to all compact convex sets. □
4.2. Admissible Fourier matrices.
Call a matrix of finite positive measures admissible if
for each row and column, respectively. Unit-width polygons give admissible matrices,
and their area identity becomes
The moment constraints have a particularly useful operator consequence.
Lemma 4.2.The Fourier matrix of an admissible matrix satisfies for every real
.
Proof.For , Cauchy–Schwarz with the reciprocal weights and gives
□
4.3. A planar body for the interaction of two planes.
We identify bodies in with their coordinates and bodies in with their coordinates.
The next construction turns the triple mixed volume of into a mixed area involving
.
Proposition 4.3.Let and have coordinate widths and positive area. There is
a planar compact convex set of widths such that, for every compact convex ,
Its area satisfies
For polygonal inputs, its corner Fourier matrix is
Proof.Suppose first that are polygons. For each pair of non-axis counterclockwise edges
and with , prescribe an edge of by
Each of its four signed coordinate totals is at most one. For example, the positive
horizontal total is
Add axis edges until each signed total is exactly one. These edges sum to zero, so their
concatenation in angular order is the boundary of a convex polygon, possibly degenerate,
of widths . Indeed, for any fixed edge the successive cross products with the remaining
edges change sign at most once; since their sum is zero, every partial sum lies on the same
side of that edge line.
To identify its mixed-area functional, put . The facet first-variation formula, followed by
planar extrusion, gives
where is the outward unit normal. A facet whose normal has non-axis projection to
must be the sum of two edges; neither whole planar summand can be exposed in such a
direction. The two edges cannot both be vertical, so the common normal has nonzero
-component, which forces . Its area-weighted normal is
Conversely this vector exposes both edges; its restrictions are on and on . The
contribution of this parallelogram facet to (4.12) is therefore exactly the contribution of
(4.11) to .
The difference between the two functionals is supported on the four axis normals.
Translation invariance makes it , and it vanishes on both coordinate unit segments. For
example, with ,
The logarithmic slope of (4.11) is the difference of the input logarithmic slopes, its mass
is their product, and its signs are . Summing over the common row sign gives (4.10). Hence
where the inequality uses Lemma 4.2. Together with (4.7), this proves (4.9).
For arbitrary , approximate them by polygons and correct their widths by diagonal
maps tending to the identity. Translate the resulting ’s into the unit square and take a
Hausdorff convergent subsequence. The limit has widths , and continuity proves (4.8) and
(4.9). □
The remaining body can now be optimized explicitly. Write and .
Lemma 4.4.Suppose has widths and area , while have widths and common area . For
the body above, with , put . Then
If and
equality is attained by
Proof.Apply (4.5) to (4.8). The set in (4.16) has the required widths; adding the
coordinate segments fills the missing widths without changing the non-axis corner
directions. Extrusion gives
which proves both its area condition and equality. □
5. The logarithmic boundary gap
The corner calculus reduces the sharp threshold to one estimate. When and have area ,
the area of their mixed body can fall below , but only by order . We first prove this
upper bound and then construct bodies attaining its order. Throughout this
section denote positive absolute constants, whose values may change between
occurrences; all small-parameter assertions hold below a fixed positive absolute
threshold.
5.1. Energy controls mass logarithmically.
For define the positive measure
A direct integration gives
Thus the energy needed to test total mass on an interval grows linearly with its length.
Exponential tail control turns this length into a logarithm.
Lemma 5.1.Let be a finite real signed measure satisfying . If , then
If , then .
Proof.Writing , the potential in (5.2) and energy Cauchy–Schwarz give
For choose and square; for let . □
5.2. The fourth moment of a Fourier matrix.
For an admissible Fourier matrix , set
For a unit right triangle, the single corner measure is a unit atom, so . The contraction
bound alone gives . The following estimate gains an additional power of the excess ,
because the determinant is itself the Fourier transform of a measure with controlled
exponential moments.
Lemma 5.2.If is admissible and , with sufficiently small, then
Proof.Let be the squared singular values of , and let be those of its nonnegative mass
matrix . Contraction and entrywise domination give
There is nothing to prove when , so assume . We first bound away from one, so that
the fourth-moment defect controls the energy of . The mass estimate will then turn this
energy bound into a bound on that tends to zero with .
To obtain the initial uniform gap when , consider an entry measure of mass and put
Its two exponential moments are at most one, so
One of the two permutation products in is at least . Applying the last inequality to
these two entries and using concavity of the square root yields
Each entry has both exponential moments at most one; the same is true of each
convolution, so . Lemma 5.1 therefore gives
Finally , so . Substitute this in the second inequality of (5.4). □
Corollary 5.3 (Upper bound for the area deficit).If have widths and area , and for
the body of Proposition 4.3, then
for all sufficiently small .
Proof.For polygons, pointwise Hilbert–Schmidt Cauchy–Schwarz followed by scalar
Cauchy–Schwarz gives
Both second moments equal , so the result follows from Lemma 5.2 and (4.13).
For general bodies use the normalized polygon approximations in Proposition 4.3.
Their two area deficits may differ, but both tend to ; applying the fourth-moment
bound to each separately and passing to the limit proves the claim. □
5.3. Hyperbolic arcs attain the logarithmic order.
The test measure suggests using a corner measure spread uniformly over a long
interval. Such a measure has a direct geometric realization. For put
and let
The lower boundary is a convex hyperbolic arc from to , so is convex and has widths .
Its logarithmic slope parameter satisfies
Thus the lower arc has corner measure , while the closing diagonal has corner measure .
These identities can also be obtained by inscribing polygons in the arc and taking their
Riemann-sum limits.
Proposition 5.4 (Lower bound for the area deficit).For every sufficiently small , there
are unit-width polygons of area for which the associated body has area with
Proof.First take identical copies of for . Their Fourier matrices and (4.10) give
Since is even, the two area deficits are
The first formula follows either by integrating the hyperbola or by integrating the
kernel on .
The convolution is supported on and has mass . Testing it against in (5.2) gives
Moreover for , so . Consequently
The positive continuous function tends to zero as , and therefore takes every
sufficiently small positive value.
To obtain polygons with exactly the desired deficit, fix such an and take a nested
sequence of polygons inscribed in its lower arc, retaining the two endpoints. Their areas
increase to , so their deficits tend to from above. Let and set
The diagonal corner measure remains , while the other corner measure is
multiplied by . Thus have deficit exactly , and the deficit of their mixed body
tends to . The measures involved have uniformly bounded slope support, so this
convergence follows directly from the continuous kernel energy. For sufficiently
large , (5.9) retains, for example, the constant in place of . This proves (5.7).
□
5.4. Optimizing the coefficient slice.
We now return to in (3.3). The algebraic comparison curve is
Here is the area deficit at the comparison minimum. The function is the larger
root of , and the identities below show why that quadratic is the appropriate
reference.
Theorem 5.5 (Logarithmic gap).For all sufficiently small ,
The lower bound is witnessed by triples of polygons.
Proof.Normalize any realization of by Proposition 3.1, and write
Thus (4.16) supplies a polygon attaining (5.12), and for its coefficient ,
Since and , this is at least , as required. □
Corollary 5.6.The locus is not semialgebraic.
Proof.If it were, its inverse image under the affine map would be semialgebraic.
The graph of the finite fiber infimum would then be semialgebraic by the
Tarski–Seidenberg theorem; being a lower bound and being approached arbitrarily
closely from above are first-order conditions over the real field [2]. The same would
hold for
A semialgebraic function graph is contained in the zero set of a nonzero
polynomial. Indeed, in a finite polynomial sign description, at least one defining
polynomial must vanish at each graph point; otherwise the signs would be constant
on a neighborhood and the graph would have interior. The product of the nonzero
defining polynomials therefore vanishes on the graph. Write such a relation as
After division by , all terms with tend to zero because . But , so the nonzero
polynomial cannot tend to zero. This is a contradiction. □
6. The threshold in every dimension
Multiplying a cubic volume polynomial by a power of a new variable gives a
higher-dimensional example, but it also adds a body. To keep the sharp threshold at three
bodies we must instead raise the degree of an existing variable. In an arbitrary realization
that body may have varying planar sections, so a direct product construction alone does
not describe the resulting coefficient slice. The appropriate correction is the sharp
second-moment factor for concave functions.
6.1. Averaging planar sections.For put
If is a convex body, write for uniform averaging on .
Lemma 6.1.Every bounded nonnegative concave function on satisfies
If are the fibers of a compact convex set over , and is their Minkowski average, defined
by , then
Widths and mixed area with a fixed body commute with this average.
Proof.Inequality (6.2) is the case of Berwald’s inequality [3]; the following level-set
proof also identifies its sharp constant. We may assume . Let and . Brunn–Minkowski
makes concave on , with right limit one at zero. Dilating towards a point where
approaches gives , so .
Compare with on , extended by zero. They have the same integral. Concavity
implies that is nonincreasing where ; therefore has at most one sign change, from
positive to negative. If is a crossing point, then
Layer-cake integration now gives
which is (6.2). A plateau at the maximum only adds a downward jump to and
leaves the comparison unchanged.
Convexity of the total set gives . Thus is concave, and planar Brunn–Minkowski
applied to Minkowski averages yields . Combine this with (6.2). The averages can be
formed as limits of finite Minkowski averages; fiber support functions are bounded
and concave in the interior of , and have uniform Lipschitz bounds in their direction
variable. This also justifies the asserted linearity of widths and mixed area. □
6.2. The lifted coefficient slice.
Fix , put , and define
Proposition 6.2.Every realization of (6.4) gives planar bodies of widths and a
planar body of widths such that, for some ,
In particular,
Conversely, every normalized cubic triple in Proposition 3.1lifts to a realization of
(6.4) with the same . Polygonal triples lift to polytopes.
Proof.Let realize (6.4). Lemma 2.1 reads its monomial support as
Translate the affine hulls through zero and put , . Since the total sum is full
dimensional, is a plane. The positive coefficients of and ensure that neither of the
other two planes is contained in .
Choose a complement to in , so . Let be the projection of and write its fibers as ,
with . Set , . In product coordinates of determinant modulus one, Fubini gives
The factor is the Jacobian in the quotient coordinate.
The three planes have independent normals within . Otherwise they would share a line,
and the cyclic product calculation in Proposition 3.1 would still apply to the averaged
coefficients. Indeed, the relevant directional widths of every fiber are a single width
transverse to the common line, multiplied by fixed angular factors; averaging therefore
preserves their ratio and the cyclic-product identity. Replacing the area and width of by
their respective averages would make the two cyclic products equal, whereas their
prescribed ratio is .
We may therefore take these planes to and use the remaining coordinates for . Write
and
Comparing the six repeated coefficients in (6.6) with (6.4) gives
Hence , and . Apply the diagonal map on , leaving fixed, and put . In the new
coordinates,
Both have widths . Lemma 6.1 puts , while Proposition 4.3 and linearity of mixed
area give
The corner inequality (4.5) proves (6.5) and the other forward assertions.
For the converse, start with a normalized cubic triple . On the standard simplex , set
Its first two normalized moments are
Define the cone
and embed at quotient coordinate zero. Formula (6.6) and the two moments give
A common dilation by realizes the required polynomial. The cone is a polytope
whenever is a polygon. □
The proof shows the role of the correction factor; averaging arbitrary sections gives the
universal inequality, and the cone in (6.8) attains both moments that enter it. Thus the lift
preserves the boundary comparison even though an arbitrary realization need not have
constant sections.
Proof of Theorem 1.1. The case is Corollary 5.6. For , restrict (6.4) to and define
The lifted rectangle-and-triangles example shows that every fiber is nonempty.
The forward part of Proposition 6.2 supplies exactly (5.12), with and ; therefore
the upper estimate in Theorem 5.5 holds for . Its polygonal lower witnesses lift by
the converse part of the proposition, with unchanged . Consequently
The same semialgebraic-graph argument proves that is not semialgebraic.
Restriction to three variables maps onto for every ; one direction discards bodies,
and the other adjoins points. A semialgebraic source would have a semialgebraic
image. The one-body locus is , and the two-body locus is semialgebraic by Shephard’s
theorem, proved in Appendix A.
Finally, full-dimensional polytope tuples are Hausdorff dense in all compact
convex tuples; approximate by polytopes and add a vanishing cube. Proposition 2.4
identifies the closure of their coefficient image with . Since the closure of
a semialgebraic set is semialgebraic, their coefficient locus is nonsemialgebraic
whenever . □
Remark 6.3. The same boundary gap excludes definability in any polynomially
bounded expansion of the ordered real field, even with real parameters. If the locus
were definable, so would be for large , with . Hence
would be a finite definable function satisfying , contrary to polynomial
boundedness. No monotonicity of is required. The assertion also holds for any
Hausdorff-dense realization class, because topological closure is definable.
7. Projection holonomy and forced shapes
The sharp threshold was obtained by optimizing a coefficient slice. We now use
equalities among coefficients to determine a body itself. The point is that (2.2)
turns a mixed-volume equality into equality of two planar projections, while the
equality case of Minkowski’s inequality removes all freedom except translation.
Following such identifications around three planes produces a return map on the first
one.
7.1. A three-plane return map.
Put and fix the segment directions
Parallel projection along these directions, successively between , gives
Their planar determinants have modulus one, and their return map is
Thus, if the first two projection comparisons identify with and with , a third
comparison of with imposes
for a translation . The rest of the construction makes these directions and comparisons
recoverable from the coefficients alone.
Use nine body labels
and write for the mixed-volume entry with the indicated labels. These entries are fixed
rational multiples of the ordinary cubic coefficients. First impose
The following table specifies nine further coefficient values
Lemma 7.1 (Calibration).Every realization of (7.4)–(7.5) can be normalized by
independent translations and a common linear map of determinant modulus one so
that
Proof.The positive basis coefficient makes the total sum full dimensional. By
Lemma 2.1, the six labels in (7.4) are segments or points, and the nonzero coefficients
in the table exclude points. Their triple coefficients are absolute determinants, so the
basis coefficient normalizes as stated.
The table gives the absolute coordinates of the three remaining segment vectors.
Coordinate reflections and reversals of segment vectors can simultaneously arrange
and ; fix the second coordinate, then orient the third and first coordinates in turn.
Reverse the last vector to make its third coordinate positive; it has the form with .
Its determinant with is . The possible absolute values are , so the last coefficient in
(7.5) selects uniquely. Reversing a segment vector merely translates that segment.
□
Next impose the three zero-width conditions
and the two positive comparison equalities
For parameters and , the final comparison is
All the conditions are polynomial with rational coefficients; entries not listed are
free.
Proposition 7.2 (Exact functional equation).The coefficient conditions (7.4)–(7.8)
have a realization if and only if there is a positive-area planar compact convex set satisfying
In every realization, after the normalization of Lemma 7.1and further independent
translations, the last three bodies are in their respective coordinate planes. Moreover
Proof.Calibrate the six segments. Extrusion turns (7.6) into zero widths, so
translations put , , . The positive self-entries in (7.7) ensure that all three have
positive relative area. By (2.2) and the equality case of (2.3), the first two comparisons
identify the corresponding projected bodies up to translation. Centering each body
at its area centroid removes these translations simultaneously, since an invertible
linear map between two planes carries area centroids to area centroids. Hence and
.
The projected bodies are therefore translates. Lifting that relation to along gives . The
matrix preserves the positive definite quadratic form
and its eigenvalues are nonreal. Thus is invertible for . After centering and
translating by the unique solution of the resulting linear translation equation, we
obtain (7.9). Translate by the corresponding images to retain their relations with
.
Conversely, place a solution in , its two indicated images in , and the six calibrated
segments in . The first two projection comparisons are identities, and the third follows
from (7.9), since is a translate of . The determinant and zero-width conditions hold by
construction, so the actual volume polynomial of this tuple satisfies every condition.
Extrusion in direction proves (7.10). □
7.2. Irrational rotation and contraction.
The eigenvalues of are , where . This angle is an irrational multiple of . Indeed, if were
a root of unity, then would be an algebraic integer; the recurrence
gives monic integral polynomials for with . A suitable would have the nonintegral
rational root , a contradiction.
For , equation (7.9) says . By (7.11), a linear change of coordinates makes an irrational
rotation. Its powers are dense in the rotation group, so continuity of the support function
makes a disk in those coordinates. Every positive-area solution is therefore an ellipse. This
gives a coefficient pattern whose every realization has three elliptic bodies, and in
particular no all-polytope realization.
For and , the same equation instead determines a unique body, with an area that
records every iterate of the return map.
Theorem 7.3 (The forced Minkowski series).For , the unique compact convex solution
of is
It has positive area and infinitely many exposed edges in distinct directions. Its area is
The function has rational Taylor coefficients and a holomorphic extension to the unit
disk with the unit circle as a natural boundary. It is transcendental over , and is not a
semialgebraic set.
Proof.In the norm defined by , is an isometry, so the tails of (7.12) are uniformly
bounded by . The sum is consequently a compact convex set; its first two summands
form a nondegenerate parallelogram. Separating the first term gives the functional
equation. Conversely, iterating that equation for any compact solution gives
The last term converges to zero in Hausdorff distance, proving uniqueness.
For planar vectors , the area of their zonotope is
This follows by expanding its quadratic area polynomial; the pure terms vanish
and each mixed coefficient is a parallelogram area. Cayley–Hamilton gives
Apply the zonotope formula to and pass to the Hausdorff limit. Summing first
over gives (7.13).
The lines are pairwise distinct, because a repeated line would make a nonzero
integral multiple of a multiple of . A normal perpendicular to exposes the entire
th segment of the sum and a unique endpoint of every other segment. The endpoint
series converges, so the exposed face is a translate of , a genuine edge. A semialgebraic
planar set can have boundary segments on only finitely many lines; its boundary lies
in the union of finitely many polynomial zero sets, and any polynomial vanishing on
a nontrivial line segment has that line as a factor. Thus is not semialgebraic.
To study the area, put . The absolutely convergent Fourier series
gives, for ,
The points are distinct and dense on the unit circle. Since , absolute summability of
allows dominated convergence in (7.14), giving
Thus is unbounded near a dense set of boundary points and cannot be continued
across any boundary arc. The rational factor in (7.13) does not change this
conclusion.
For completeness, this also rules out a polynomial relation with arbitrary real
coefficients without invoking general facts about algebraic functions. If , substitute (7.13),
clear the power of , and divide by the highest power of . Along with , all lower terms tend
to zero. The leading coefficient polynomial must therefore vanish at every , which is
impossible unless it is zero.
Finally, , where , and . These values are rational, so (7.13) has rational Taylor
coefficients. □
Proof of Theorem 1.2. Take , , the finite coefficient conditions above, and .
Proposition 7.2 and Theorem 7.3 show both that every realizing tuple has the stated
body and area, and that every occurs. This proves the exact projection statement
and all its asserted properties. □
7.3. Persistence under direct-sum suspension.
For shape forcing, one additional body can carry all the new dimensions. A monomial
in its variable forces every realization to split into complementary subspaces, preserving
the three-dimensional obstruction.
Proposition 7.4.For every cubic polynomial in variables and every ,
The equivalence also holds when both sides require realizations by polytopes, or by
semialgebraic compact convex sets.
Proof.Embed a realizing tuple for in and a unit-volume -cube in a complementary
subspace. Volumes multiply, giving the forward implication.
Conversely, suppose is realized by . The zero polynomial is immediate, so assume
and put . The two-body polynomial is
Lemma 2.1 gives and ; their spans are complementary because the total sum is
full dimensional. Each lies in the span of after translation, and product volume
in these complementary coordinates gives for some . A common dilation absorbs
. Restriction to the three-dimensional subspace, linear maps and dilations preserve
both polytopality and semialgebraicity. □
It follows that arrays with no all-semialgebraic realization occur for in dimension three
and for in every higher dimension. These are shape obstructions for particular arrays,
distinct from the sharp three-body obstruction to a finite algebraic description of the
entire coefficient locus.
8. A countable arithmetic characterization
The preceding obstructions rule out finite algebraic descriptions and exact replacement by
polytopes. Approximation behaves differently; the bounded model in Section 2 has explicit
finite rational nets. Their error bounds give both an exact infinite coefficient criterion and
finite certificates outside the locus.
Fix and put
For a coefficient vector , write . All constructions below depend only on ; the input
vector enters only when the resulting inequalities are evaluated.
8.1. Finite rational nets.
For an integer , let
For each tuple of subsets of this grid containing zero, compute the coefficient vector of
the rational polytopes . Let
This is a finite rational set that can be generated by rational arithmetic. We give
an explicit formula and error estimate, so the definition involves no realization
test.
For a finite rational point set , its convex-hull volume is computed as follows. Return
zero if its affine rank is less than . Otherwise enumerate the supporting hyperplanes
determined by affinely independent -subsets of , retaining those for which all points lie on
one side, and record their complete active point sets. These are precisely the facets.
Recursively dissect each facet by coning its facets to the average of its input points,
and cone the resulting boundary simplices to the average of . At each stage,
affine coordinates and supporting equations are obtained by rational Gaussian
elimination. The average lies in the relative interior, so these cones cover the
hull with disjoint relative interiors. Its volume is therefore the rational number
where the sum is over the resulting simplices. Duplicate facet point sets are discarded.
This finite procedure includes coplanar and lower-dimensional inputs.
For a homogeneous polynomial , finite differences give
Indeed, the difference operator of multi-order kills every other monomial of total
degree and takes to . Since convex hull commutes with Minkowski addition, the required
grid coefficients are explicitly
Lemma 8.1 (Quantitative nets).For the compact image in (2.4),
Consequently, for every ,
Proof.Round every point of to a nearest grid point, with a fixed rule at ties, and
let be the convex hull of the resulting finite set. Then and .
We use the elementary estimate
when have sup-norm Hausdorff distance at most . To prove it, add the coordinate
segments of one at a time. Extrusion bounds the successive volume increments
by ; their sum is the right side of (8.5). The two Hausdorff inclusions give the
assertion.
For , the sums and lie in and are within in Hausdorff distance. The sum of the
absolute weights in (8.2) is at most . Thus every coefficient changes by at most
This proves the net assertion, including its independence from . The distance
comparison follows by choosing a closest point in the compact set and then a net point.
□
Theorem 8.2 (Finite tests at every scale).A vector belongs to if and only if either , or
and, for every ,
Each test is a finite Boolean combination of rational linear inequalities. A nonrealizable
vector with fails every sufficiently fine test.
Proof.If is nonzero and realizable, Lemma 2.3 puts in , and Lemma 8.1 gives every
test. Conversely, the tests put at distance zero from the compact set . A realizing
tuple for , dilated by , realizes . If , its positive distance from that set exceeds for all
sufficiently large , so the corresponding tests fail. Finally, the minimum over a finite
set is the finite disjunction, over , of the inequalities for all . □
8.2. Homogeneous quadratic inequalities.
The finite tests can be expressed as a single countable family of quadratic
inequalities by excluding rational balls outside the compact image. Admit a triple
when the finite rational test
holds, and associate to it the homogeneous quadratic form
The rational constants in these forms are computed once from (8.3); clearing a positive
common denominator makes each form integral if desired.
Theorem 8.3.For every ,
The family is effectively enumerable by rational arithmetic, and every nonrealizable
vector strictly violates a displayed inequality.
Proof.The zero vector satisfies all the inequalities. For a nonzero realizable vector,
and . Choose with . For every admitted triple, (8.7) gives , and multiplication by
proves the quadratic inequality. Coefficient nonnegativity is standard.
Conversely, suppose the coefficients are nonnegative and , so . If , let . Choose
rational with
and then with . Every is at distance greater than from , so this triple is admitted.
But , and (8.8) is strictly negative, a contradiction. Hence and is realizable.
To enumerate the family, enumerate rational and integers , generate the finite list
, and retain the triples passing (8.7). Every step is finite rational arithmetic. Every
nonrealizable vector therefore has either a negative coefficient or a strictly negative
value of an admitted quadratic form. □
For computable coefficients, certified rational approximations and interval arithmetic
eventually certify any fixed strict violation. Dovetailing the enumeration of inequalities
with increasing precision therefore rejects every nonrealizable input after a finite
computation and never rejects a realizable one. This procedure works directly on
, including when it is not known whether ; no equality test or division by is
required.
8.3. Constructing realizing bodies.
The tests also organize positive witnesses. Let be a normalized input, , supplied with
rational approximations satisfying
At level , a state is a tuple of grid polytopes with , whose coefficient vector satisfies
Join consecutive states when
A node is a path of states starting at level one. Every level is finite, and all its tests are
rational; for rational polytopes, sup-norm Hausdorff distance at most is equivalent to the
two inclusions into the other polytope plus . Each inclusion reduces to finitely many
rational convex-hull membership tests, solved by enumerating affinely independent subsets
and checking their barycentric coordinates.
Proposition 8.4.The state tree has nodes at every height if and only if . Every
infinite path converges to an exact realizing tuple, with Hausdorff error at level at
most .
If the approximations are computable, selecting the terminal state of the
lexicographically least path at each height gives a rational algorithm that eventually
rejects a nonrealizable input and otherwise produces an entire sequence converging to
a realizing tuple. No computable rate for this latter sequence is asserted.
Proof.For , round the same realizing tuple at every level. Its Hausdorff errors are
at most and its coefficient errors at most , so these rounded tuples form an infinite
path satisfying (8.10) and (8.11).
Conversely, a finitely branching tree with nodes at every height has an infinite
path; a node with arbitrarily long extensions has a child with that property, and this
observation can be iterated from the root. Along any infinite path, the successive
Hausdorff bounds have tail sum . The path therefore converges to compact convex
bodies in , with the stated error. Its coefficients tend to by (8.10), so continuity
makes the limit an exact realization.
Order grid subsets, states and paths lexicographically. If a height has no node,
the input is rejected; otherwise output the terminal state of its least path. The
preceding equivalence proves the rejection assertion. For a realizable input, fix a
depth . Among the finitely many depth- nodes, those extending to height form a
decreasing nonempty family as , so eventually this family stabilizes. The depth- prefix
of the chosen path therefore stabilizes as well. All sufficiently late outputs are within
of that prefix. Since this tends to zero with , the whole output sequence is Hausdorff
Cauchy and converges to a realizing tuple. The argument does not bound the height
at which a prefix stabilizes, which explains the absence of an effective convergence
rate. □
Theorem 8.3 and Proposition 8.4 prove Theorem 1.3.
Appendix A. The two-body realization theorem
We include a simplex proof of the two-body theorem of Shephard [9], with zero coefficients
allowed. Besides completing the threshold statement, the proof constructs the realizing
bodies directly from the ordered ratios of consecutive mixed volumes.
Proposition A.1.A homogeneous polynomial
belongs to if and only if
Every such polynomial is realized by two simplices, possibly lower dimensional.
Proof.For necessity, add times a ball to each realizing body. All resulting
mixed volumes are positive, and Alexandrov–Fenchel makes their successive ratios
nonincreasing. They therefore satisfy (A.1); let .
For sufficiency, the inequalities with give adjacent log-concavity. The remaining
inequalities exclude internal zero blocks; if lies between positive entries, then fails.
Thus the positive support, if nonempty, is an interval , and the ratios decrease on
that interval.
The basic construction uses
We claim that
For positive , consider a section at last coordinate . Allocating an additional height
to the first simplex replaces by , which contains it. Thus the largest section
uses the maximal possible first height . Integrating sections gives the recurrence
Induction proves (A.2); zero values of follow by continuity.
When all are positive, choose and scale both simplices by . For support , put and for
. The required pair is
To check the boundary case without separate zero-pattern computations, extend the
sequence by for and for . For small , its ratios decrease from through the prescribed
to . The positive construction uses common scale . Apply to both simplices
the diagonal map with factors on the first coordinates and on the others; its
determinant is one. The simplices converge to (A.3), so continuity gives the desired
polynomial. Empty vertex lists are omitted, and points realize the zero polynomial.
□
In degree three, the criterion is simply
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