A universality theorem for projectively unique polytopes
and a conjecture of Shephard
We prove that every polytope described by algebraic coordinates is the face of a projectively unique polytope. This provides a universality property for projectively unique polytopes. Using a closely related result of Below, we construct a combinatorial type of -dimensional polytope that is not realizable as a subpolytope of any stacked polytope. This disproves a classical conjecture in polytope theory, first formulated by Shephard in the seventies.
By employing a technique developed by Adiprasito and Ziegler , we prove the following universality theorem for projectively unique polytopes.
For any algebraic polytope , there exists a polytope that is projectively unique and that contains a face projectively equivalent to .
Here, a polytope is algebraic if the coordinates of all of its vertices are real algebraic numbers, and a polytope in is projectively unique if any polytope in combinatorially equivalent to is projectively equivalent to . In other words, is projectively unique if for every polytope combinatorially equivalent to , there exists a projective transformation of that realizes the given combinatorial isomorphism from to .
Theorem I is sharp: We cannot hope that every polytope is the face of a projectively unique polytope. Indeed, it is a consequence of the Tarski-Seidenberg Theorem [3, 10] that every combinatorial type of polytope has an algebraic realization. In particular, every projectively unique polytope, and every single one of its faces, must be projectively equivalent to an algebraic polytope. Hence, a -dimensional polytope with vertices whose set of vertex coordinates consists of algebraically independent transcendental numbers is not a face of any projectively unique polytope.
A consequence of Theorem I is that for every finite field extension over , there exists combinatorial type of polytope that is projectively unique, but not realizable in any vector space over . This extends on a famous result of Perles, who constructed a projectively unique polytope that is not realizable in any rational vector space, cf. [5, Sec. 5.5, Thm. 4].
In the second part of this paper, we consider a conjecture of Shephard, who asked whether every polytope is a subpolytope of some stacked polytope, i.e. whether it can be obtained as the convex hull of some subset of the vertices of some stacked polytope. While he proved this wrong in , he conjectured it to be true in a combinatorial sense.
For every , every combinatorial type of -dimensional polytope can be realized using subpolytopes of -dimensional stacked polytopes.
The conjecture is true for -dimensional polytopes, as seen by Kömhoff in , but remained open for dimensions . On the other hand, Theorem I encourages us to attempt a disproof of Shephard’s conjecture. The idea is to use the universality theorem above to provide a projectively unique polytope that is not a subpolytope of any stacked polytope. Since any admissible projective transformation of a stacked polytope is a stacked polytope, no realization of the polytope provided this way is a subpolytope of any stacked polytope.
Unfortunately, the method of Theorem I is highly ineffective: The counterexample to Shephard’s conjecture it yields is of a very high dimension. We use a refined method, building on the same idea, to present the following result.
There exists a combinatorial type of -dimensional polytope that cannot be realized as a subpolytope of any stacked polytope.
It remains open to decide whether every combinatorial type of -dimensional polytope can be realized as the subpolytope of some stacked polytope.
(U) Universality of projectively unique polytopes
Point configurations, PP configurations and weak projective triples
Definition U.1 (PP configurations, Lawrence equivalence, projective uniqueness).
A point configuration is a finite collection of distinct points in . If is an oriented hyperplane in , then we use resp. to denote the open halfspaces bounded by . If is a polytope in such that then the pair is a polytope–point configuration, or short PP configuration.
A hyperplane is external to if is a face of . Two PP configurations , in are Lawrence equivalent if there is a bijection between the vertex sets of and and the sets and , such that, if is any hyperplane for which the closure of contains , there exists an oriented hyperplane for which the closure of contains and
where denotes the set of vertices of .
A PP configuration in is projectively unique if for any PP configuration in Lawrence equivalent to it, and every bijection that induces the Lawrence equivalence, there is a projective transformation that realizes . A point configuration is projectively unique if the PP configuration is projectively unique, and it is not hard to verify that a polytope is projectively unique if and only if the PP configuration is projectively unique.
Let be a projectively unique PP configuration in . Then there exists a -dimensional polytope on vertices that is projectively unique and that contains as a face.
Here denotes the cardinality of .
Definition U.3 (Framed PP configurations).
Let denote any PP configuration in , and let be any subset of . Let be any PP configuration in Lawrence equivalent to , and let denote the labeled isomorphism inducing the Lawrence equivalence.
The PP configuration is framed by the set if implies . Similarly, a polytope (resp. a point configuration ) is framed by a set if (resp. ) is framed by .
Examples U.4 (Some instances of framed PP configurations).
If is any PP configuration, then frames .
If is any PP configuration framed by a set , then every superset of frames as well.
If is any projectively unique PP configuration, and is a projective basis, then frames .
Any -cube, , is framed by of its vertices, cf. [1, Lem. 3.4].
Definition U.5 (Weak projective triple in ).
A triple of a polytope in , a subset of and a point configuration in is a weak projective triple in if and only if
is a projectively unique point configuration,
frames the polytope , and
some subset of spans a hyperplane , the wedge hyperplane, which does not intersect .
Definition U.6 (Subdirect cone).
Let be a weak projective triple in , seen as a canonical subspace of . Let denote the wedge hyperplane in spanned by points of with . Let denote any point not in , and let denote any hyperplane in such that and separates from . Consider, for every vertex of , the point . Denote by the pyramid
The PP configuration in is a subdirect cone of .
Lemma U.7 ([1, Lem. 5.8.]).
For any weak projective triple the subdirect cone is a projectively unique PP configuration, and the base of the pyramid is projectively equivalent to .
If is a weak projective triple, there exists a projectively unique polytope of dimension on vertices that contains a face projectively equivalent to . ∎
Constructions for projectively unique point configurations and the proof of Theorem I
Let be any algebraic polytope. Our goal for this section is to find a weak projective triple that contains . Applying Corollary U.8 then finishes the proof of Theorem I. The main step towards that goal is to embed into a projectively unique point configuration. In the construction, we will use the following straightforward observation repeatedly.
Let be a projectively unique point configuration, let and let be a point configuration framed by . Then is a projectively unique point configuration.∎
The point configuration is projectively unique for every .
The proof is by induction on . We start proving that is projectively unique. This implies that is projectively unique for any .
is projectively unique: To see that is projectively unique, we start with the folklore observation that the points , together with the origin , form a projectively unique configuration (cf. Figure 1(a)). Furthermore, we claim frames , thereby proving that is projectively unique since is projectively unique.
To see this, notice that the point of is determined as the intersection of the lines and , which are spanned by points of . Similarly, all points that arise as coordinate permutations and/or sign changes from are determined this way. Geometrically, these are the center points of the facets of the cube (cf. Figure 1(b)).
The remaining lattice points of coincide with the midpoints of the edges of said cube. To determine them, let be any edge of and let and be the facets of incident to that edge. Finally, let be the hyperplane spanned by the center point of and the center points of and . The midpoint of is the unique point of intersection of and (cf. Figure 1(c)).
is projectively unique: For , consider the projective basis of consisting of the vertex of , together with the neighboring vertices , , , and the origin (cf. Figure 2(a)). We will see that once the coordinates of the elements in are fixed, then the coordinates of all the remaining lattice points of can be determined uniquely.
Consider the set of points of lying in a common facet of that is incident to ; for example, (cf. Figure 2(b)). Observe that is just an affine embedding of into . As such, is projectively unique, and thus it is determined uniquely if a projective basis for its span is fixed.
Clearly, the points of form an affine basis for the affine span of . Furthermore, the coordinates of the point , are fixed by . Indeed, is the the point of intersection of the line with the hyperplane (cf. Figure 2(c)). To sum up, we have that
the points are determined uniquely from the points of ,
the points are elements of , and
the points form a projective basis for the span of .
Consequently, is a projectively unique point configuration, since is projectively unique. We can repeat this argumentation for all point configurations
is projectively unique. Moreover, since the last vertex of a cube of dimension is determined by the remaining ones by (cf. [1, Lem 3.4], compare also Example U.4(iv)), the configuration is projectively unique as well. By symmetry,
is also projectively unique. Clearly, and intersect along a projective basis: for instance, the set lies in both and and forms a projective basis as desired. Thus, the point configuration is projectively unique. ∎
Embedding vertex sets of algebraic polytopes
We start with a point configuration that shows that it is enough to fix each coordinate of a point to frame it.
For each point in the positive orthant of , there is a point configuration that contains and is framed by the points
We denote by the translation of by the all-ones vector. Moreover, let denote the diagonal matrix with diagonal entries . Notice that is projectively unique (by Proposition U.10), contains and the set . Since every projective transformation fixing is the identity, the subset frames . ∎
Finally, we only need to frame each coordinate of the point. The idea is to realize the defining polynomial of any real algebraic number in a functional arrangement (cf. [8, Def. 9.6]), which conversely determines the real algebraic number.
For a function , a functional arrangement for is a -parameter family of point configurations in such that the following conditions hold:
For all in , the functional arrangement contains the output point , the input points and the set .
For all , the set frames the point configuration .
For the last condition, let denote any point configuration Lawrence equivalent the functional arrangement , where is the bijection of points that induces the Lawrence equivalence.
For all , if is the identity on and , we have .
Hence, a functional arrangement essentially computes a function and its inverse by means of its point-line incidences alone. An just as like functions, they can be combined:
Let and , , denote functional arrangements for functions and , respectively. Then
is a functional arrangement for the function from to .∎
Every integer coefficient polynomial is realized by a functional arrangement .
The proof of this fact is based on the classical von Staudt constructions (, compare also [12, Ch. 5], [11, Sec. 11.7] or [8, Sec. 5]), which are a standard tool to encode algebraic operations in point-line incidences.
To construct the desired functional arrangements, notice that every integer coefficient polynomial in variable can be written using , and , combined by addition and multiplication. Hence, thanks to Lemma U.13, it suffices to provide:
A functional arrangement for the function computing the addition of two real numbers.
A functional arrangement for the function computing the product of two real numbers.
Both functional configurations are shown below. We invite the reader to check that they indeed form functional arrangements for addition and multiplication.
By switching output and input points of these functional arrangements, we also obtain functional arrangements and for and . ∎
Let us construct a functional arrangement for . Using Lemma U.13, this arrangement can be written as combination of the functional arrangements for addition, subtraction and multiplication:
Figure 5 shows the evaluations of this functional arrangement at and .
The point configuration , as given above, is framed by . Hence, it enables us to compute from . Similar point configurations for any algebraic number are given in the following corollary.
For each real algebraic number , there is a point configuration framed by such that .
If is algebraic of degree (i.e. is rational), let be such that and set . We then define .
If is instead of degree (i.e. is irrational, but algebraic), and is an integer coefficient polynomial with root , then let and denote rational numbers with the property that is the only root of in the interval . Then the desired point configuration is given by
which contains by construction, and it is framed by . Indeed, let be a configuration Lawrence equivalent to , where the equivalence is induced by the bijection . If is the identity on , then by Definition U.12(iii) and since , we obtain , and . Here must be a root of , which lies in the interval by Lawrence equivalence. This root is unique, so . To sum up, determines , which together with frames by Definition U.12(ii). ∎
Let be any point in , , with algebraic coordinates. Then there is a projectively unique point configuration containing and .
Conclusion of proof
Proof of Theorem I.
Let denote a algebraic polytope in . We assume that , since if we can realize as a face of some -dimensional pyramid. By dilation and translation, we may assume that lies in the interior of the cube . Consider now the point configuration
where is the point configuration provided by Corollary U.17. Set and . With this, we have that is a weak projective triple. Indeed,
obviously frames (cf. Example U.4(i)), and
since , we have , and hence any of the facet hyperplanes of can be chosen as wedge hyperplane for the triple.
Thus, by Corollary U.8, there exists a projectively unique polytope that contains a face projectively equivalent to . ∎
(S) Subpolytopes of stacked polytopes
In this section, we disprove Shephard’s conjecture. While this alone could be done using Theorem I (with arguments slightly differing from those below), we here use a refined argumentation to construct combinatorial types of -dimensional polytopes that are not realizable as subpolytopes of -dimensional stacked polytopes (Theorem II). Instead of Theorem I, we will use the following result of Below.
Let be any algebraic -dimensional polytope. Then there is a polytope of dimension that contains a face that is projectively equivalent to in every realization of .
The remainder of this section is concerned with the proof of Theorem II. For the convenience of the reader, we retrace, in a higher generality and with improved quantitative bounds, Shephard’s ideas that lead him to discover -dimensional polytopes that are not subpolytopes of stacked polytopes . We recall some notions.
A polytope is -stacked if it is the connected sum (cf. [11, Sec. 3.2]) of polytopes with at most facets each. With this, a -stacked -dimensional polytope is simply a stacked polytope.
Let denote the Hausdorff distance between compact convex subsets of , cf. [13, Sec. 1.8]. Let denote the metric ball in with center and radius . Shephard’s main observations are:
Lemma S.3 (cf. [14, (i) & (ii)]).
Let be any polytope with . Then, for any polytope with , we have .
Equivalently, if is any polytope with , , then for any subpolytope of .
Sketch of Proof.
By assumption, we have . Hence, every face of can be enclosed in some ball of radius and so every point in is at distance at most from some vertex of . Let now be any vertex of not in , and let denote the point of intersection of the line segment with . Then , and since contains no vertex of in the interior, we have
If , this can be used to estimate the euclidean norm of as , and hence if , then . This gives , or . ∎
Lemma S.4 (cf. [14, (iii) & (iv)]).
For any -stacked -dimensional polytope , , we have
Sketch of Proof.
Assume . As observed in the proof Lemma S.3, the edges of have length at most .
Now, the polytope can be written as the connected sum of polytopes , each of which has at most facets. Let be any one of the that contains the origin. We claim that has an edge of length at least . Indeed, since contains the origin, is has two vertices that enclose an angle at least with respect to the origin. Furthermore, , so these two vertices are at least at distance from each other. Since the graph of each polytope is connected, there must be a path of edges in from to , and so one of these edges must be of length or more. Finally, Sperner’s Theorem shows that a polytope with facets has at most edges, so that , which gives the desired bound. To combine the two observations, notice that since , all edges of are edges of , so
Combining the two lemmas above, we recover Shephard’s main result.
Corollary S.5 (cf. ).
For any subpolytope of a -stacked -dimensional polytope, , we have
In particular, every -polytope that approximates closely is not the subpolytope of any stacked polytope. We now only need to add a simple observation to Shephard’s ideas:
For is a subpolytope of a -stacked polytope , then any face of is a subpolytope of a -stacked polytope as well.
It suffices to prove this in the case where is a facet of . Let denote the hyperplane spanned by . Recall that is obtained as the connected sum of polytopes , and so is the connected sum of the polytopes . Now every single one of the polytopes has most facets, and every facet of is obtained as the intersection of a facet of with , so is -stacked. Observing that is a subpolytope of finishes the proof. ∎
Conclusion of proof
Proof of Theorem Ii.
Let be any -dimensional polytope with . By Corollary S.5, is not a subpolytope of any -stacked polytope, and the same holds for any polytope projectively equivalent to .
Theorem II now provides a polytope of dimension that contains a face that is projectively equivalent to in every realization of . Assume now that some polytope combinatorially equivalent to is a subpolytope of some stacked polytope. By Proposition S.6, any face of is a subpolytope of some -stacked polytope. But the face of corresponding to is projectively equivalent to , and hence not obtained by deleting vertices of a -stacked polytope. A contradiction. ∎
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