# Equipartitions with Wedges and Cones

A famous result about mass partitions is the so called Ham-Sandwich theorem. It states that any d mass distributions in R^d can be simultaneously bisected by a single hyperplane. In this work, we study two related questions. The first one is, whether we can bisect more than d masses, if we allow for bisections with more general objects such as cones, wedges or double wedges. We answer this question in the affirmative by showing that with all of these objects, we can simultaneously bisect d+1 masses. For double wedges, we prove a stronger statement, namely that d families of d+1 masses each can each by simultaneously bisected by some double wedge such that all double wedges have one hyperplane in common. The second question is, how many masses we can simultaneously equipartition with a k-fan, that is, k half-hyperplanes in R^d, emanating from a common (d-2)-dimensional apex. This question was already studied in the plane, our contribution is to extend the planar results to higher dimensions. All of our results are proved using topological methods. We use some well-established techniques, but also some newer methods. In particular, we introduce a Borsuk-Ulam theorem for flag manifolds, which we believe to be of independent interest.

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## 1 Introduction

Equipartitions of point sets and mass distributions are essential problems in combinatorial geometry. The general goal is the following: given a number of masses in some space, we want to find a dissection of this space into regions such that each mass is evenly distributed over all the regions, that is, each region contains the same amount of this mass. Usually, the underlying space considered is a sphere or Euclidean space, and the regions are required to satisfy certain conditions, for example convexity. Arguably the most famous result about equipartitions is the Ham-Sandwich theorem (see e.g. [Matousek, StoneTukey], Chapter 21 in [Handbook]), which states that any mass distributions in can be simultaneously bisected by a single hyperplane. A mass distribution in is a measure on such that all open subsets of are measurable, and for every lower-dimensional subset of . The result is tight in the sense that there are collections of masses in that cannot be simultaneously bisected by a single hyperplane.

However, the restriction that the regions of the dissection should be half-spaces is quite strong, maybe we can bisect more masses by relaxing the conditions on the regions? This is true in many cases. For example, any 3 masses in can be simultaneously bisected by a wedge [Barany2]. Using double wedges, even 4 masses can be simultaneously bisected [pizza_cccg]. Both of these objects will be explained in more detail below.

On the other hand, Ham-Sandwich theorem only provides a dissection of the space into two regions, and it is a natural question whether similar statements can be found for more than two regions. This is indeed the case, Soberon [Soberon] and independently Karasev, Hubard and Aronov [Karasev2] have shown that for any masses in , one can find a partition of into convex sets such that each convex set contains a -fraction of each mass. Further, Bárány and Matoušek [Barany2] have shown that for any two masses in , we can partition into 3 or 4 wedges such that each wedge contains a - or a -fraction of each mass, respectively.

In this paper, we generalize many of the above results into higher dimensions. More specifically, we study bisections with -cones and double wedges, as well as equipartitions with -fans. In the following, we give an definition of these objects and an overview over our results.

Recall the definition of a cone: a (spherical) cone in is defined by an apex , a central axis , which is a one-dimensional ray emanating from the apex , and and angle . The cone is now the set of all points that lie on a ray emanating from such that the angle between and is at most . Note that for , the cone is a half-space. Also, for , the cone is not convex, which we explicitly allow. Finally note that the complement of a cone is also a cone and that either a cone or its complement are convex. Let now be some -dimensional linear subspace of and let be the natural projection. A -cone is now a set , where is a cone in . The apex of is the set , where is the apex of . It has dimension . Again, note that the complement of a -cone is again a -cone and that one of the two is convex. Also, a -cone is either the intersection or the union of two halfspaces, that is, a so-called wedge. Further, a -cone is just a spherical cone. Alternatively, we could also define a -cone by a -dimensional apex and a -dimensional half-hyperplane emanating from it. The -cone would then be the union of points on all -dimensional half-hyperplanes emanating from the apex such that their angle with is at most . We say that a -cone simultaneously bisects the mass distributions if for all . See Figure 1 for an illustration. In Section 3 we will prove the following:

###### Theorem 1.

Let be mass distributions in and let . Then there exists a -cone that simultaneously bisects .

This is tight in the sense that there is a family of mass distributions in that cannot be simultaneously bisected: place point-like masses at the vertices of a -dimensional simplex and a last point-like mass in the interior of the simplex. Assume that there is a -cone which simultaneously bisects all the masses and assume without loss of generality that is convex (otherwise, just consider the complement of ). As it simultaneously bisects all masses, must now contain all vertices of the simplex, so by convexity also contains the interior of the simplex, and thus all of the last mass. Hence cannot simultaneously bisect all masses. Still, while we cannot hope to bisect more masses, in some cases we are able to enforce additional restrictions: for

-cones in odd dimensions, we can always enforce the apex to lie on a given line.

Another relaxation of Ham-Sandwich cuts was introduced by Bereg et al. [Bereg] and has received some attention lately (see [pizza_cccg, pizza2, pizza1, HSSoCG]): instead of cutting with a single hyperplane, how many masses can we bisect if we cut with several hyperplanes? In this setting, the masses are distributed into two parts according to a natural 2-coloring of the induced arrangement. We will only consider bisections with two hyperplanes: let and be two (oriented) hyperplanes. Let and be the positive and negative side of , respectively, and analogous for . We define the double wedge as the union . Note that the complement of a double wedge is again a double wedge. We say that a double wedge simultaneously bisects the mass distributions if for all . See Figure 2 for an illustration.

Alternatively, bisections with double wedges can also be viewed as Ham-Sandwich cuts after a projective transformation: if is a bisecting double wedge we can find a projective transformation which sends to the hyperplane at infinity. After this transformation, is a Ham-Sandwich cut of the transformed masses. In Section 4, we will prove the following:

###### Theorem 2.

Let be families each containing point sets in such that their union is in general position. Then there exists a projective transformation such that can be simultaneously bisected by a single hyperplane for every .

We will actually prove a very similar statement for mass distributions, but in odd dimensions we need some technical restrictions.

Finally, the last question that we want to address is what type of equipartitions can be done using several wedges. More precisely, we define a -fan in as a dimensional hyperplane , which we call apex, and semi-hyperplanes emanating from it. Each -fan partitions into wedges, which can be given a cyclic order . For with we say that a -fan simultaneously -equipartitions the mass distributions if for every and . See Figure 3 for an illustration.

Equipartitions with -fans in the plane have been extensively studied, see [Barany4, Barany3, Barany2, Barany, Bereg2, Blagojevic2, Blagojevic, ZivFans]. In Section 5 we will use the techniques from [Barany2] to prove results for equipartitions with -fans in higher dimensions. In particular, we will prove the following:

###### Theorem 3.

Let be an odd prime.

1. Any mass distributions in , where is odd, can be simultaneously -equipartitioned by a -fan;

2. Any mass distributions in , where is even, can be simultaneously -equipartitioned by a -fan;

3. Let with and . Then any mass distributions in , where is odd, can be simultaneously -equipartitioned by a -fan;

4. Let with and . Then any mass distributions in , where is even, can be simultaneously -equipartitioned by a -fan.

This directly extends the planar results in [Barany2].

All our results are proved using topological methods. In some cases, we use established techniques, such as the configuration space/test map scheme (see e.g. the excellent book by Matoušek [Matousek]) or degree arguments, sometimes in perhaps unusual combinations. In other cases, we use newer tools, such as a recent result about sections in canonical line bundles of flag manifolds [HSSoCG], which we rephrase as an extension of the Borsuk-Ulam theorem to flag manifolds, and a strengthening of Dold’s theorem due to Jelic [Jelic].

## 2 Preliminaries

In this section we will discuss two technical tools that we will use throughout the paper. The first one is a generalization of the Borsuk-Ulam theorem to flag manifolds. This result was already used in [HSSoCG], but phrased in a slightly different setting, which is why we will still prove that the version stated here actually follows from the more general statement in [HSSoCG]. After that we give a general argument called gnonomic projection, which will allow us to only consider wedges and cones whose apex contain the origin in the following sections.

### 2.1 A Borsuk-Ulam theorem for flag manifolds

Recall the definition of a flag manifold: a flag

in a vector space

of dimension is an increasing sequence of subspaces of the form

 F={0}=V0⊂V1⊂⋯⊂Vk=V.

To each flag we can assign a signature vector of dimensions of the subspaces. A flag is a complete flag if for all (and thus ). A flag manifold is the set of all flags with the same signature vector. We denote the complete flag manifold, that is, the manifold of complete flags, by . We will consider flag manifolds with oriented lines: a flag manifold with oriented lines is a flag manifold where each flag contains a 1-dimensional subspace (that is, a line), and this line is oriented. We denote the complete flag manifold with oriented lines by . In particular, each flag manifold with oriented lines is a double cover of the underlying flag manifold . Further, reorienting the line defines a natural antipodal action.

###### Theorem 4 (Borsuk-Ulam for flag manifolds).

Let be a flag manifold with oriented lines in . Then every antipodal map has a zero.

Let us briefly mention how this implies the Borsuk-Ulam theorem: the simplest flags containing a line are those of the form . the corresponding flag manifold is the manifold of all lines through the origin in , that is, the projective space. Orienting the lines, we retrieve the manifold of all oriented lines through the origin in , which is homeomorphic to the sphere . The above theorem now says that every antipodal map from to has a zero, which is one of the versions of the Borsuk-Ulam theorem.

Note that any antipodal map can be extended to an antipodal map by concatenating it with the natural projection from to , thus it is enough to show the statement for .

In [HSSoCG] the above is phrased in terms of sections of canonical line bundles, so let us briefly recall some of the relevant terms. A vector bundle consists of a base space , a total space , and a continuous projection map . Furthermore, for each , the fiber over has the structure of a vector space over the real numbers. Finally, a vector bundle satisfies the local triviality condition, meaning that for each there is a neighborhood containing such that is homeomorphic to . A section of a vector bundle is a continuous mapping such that equals the identity map, i.e., maps each point of to its fiber. We can define a canonical bundle for each in a complete flag, which we will denote by . The bundle has a total space consisting of all pairs , where is a complete flag in and is a vector in , and a projection given by . In [HSSoCG], the following is proved:

###### Lemma 5.

Let be sections of the canonical bundle . Then there is a flag such that .

We will now show how Lemma 5 implies Theorem 4.

###### Proof of Theorem 4.

Let be antipodal. Write as , where each is an antipodal map. In particular, each defines a section in . Let be the zero section in . Thus, are sections in and thus by Lemma 5 there is a flag on which the coincide. As is the zero section, this means that for all and thus , which is what we wanted to show. ∎

### 2.2 Gnonomic projection

Gnonomic projection is a projection of the upper hemisphere of a (unit) sphere to its tangent space at the north pole. It works as follows: for some point on , let be the line through and the origin. The projection of is then defined as the intersection of and . Note that this is a bijection from the (open) upper hemisphere to the tangent space. Gnonomic projection maps great circles to lines. More generally, we say that a great -circle on a sphere is the intersection on with a -dimensional linear subspace (i.e., a -flat containing the origin). In particular, gnonomic projection then maps great -circles to -flats.

Using gnonomic projection, we can show the following:

###### Lemma 6 (Gnonomic projection).

Assume that any mass distributions in can be simultaneously partitioned by a -cone (or double wedge or -fan) whose apex contains the origin. Then any mass distributions in can be simultaneously partitioned by a -cone (or double wedge or -fan).

###### Proof.

We will only prove the statement for -cones, as the proof is analogous for the other objects. Let be mass distributions in . Use the inverse of gnonomic projection to map to the upper hemisphere of . By our assumption there exists a -cone whose apex contains the origin that simultaneously partitions . Let be the intersection of with the upper hemisphere. From the alternative definition of -cones it is not hard to see that the image of under the gnonomic projection is a -cone in which simultaneously bisects . ∎

In the following, we will only prove statements about -cones, double wedges and fans whose apexes contain the origin. The general results then follow from the above lemma.

## 3 k-cones

In this section, we will use the Borsuk-Ulam theorem for flag manifolds to show the existence of bisections with -cones.

###### Theorem 7.

Let be mass distributions in , let be a point and let . Then there exists a -cone whose apex contains and that simultaneously bisects .

###### Proof.

Without loss of generality, let be the origin. Let be the flag manifold with oriented lines defined by the flags , where has dimension 1 and has dimension . Each defines a unique -cone that bisects the total mass and whose projection to is a cone with central axis and apex . For , define , where denotes the complement of . Then is a map from to . Further, as every bisects the total mass, so does the complement . In particular, is the unique -cone that we get when switching the orientation of . Thus, is antipodal, and by Theorem 4 it has a zero. Let be the -cone defined by this zero. By the definition of we have that simultaneously bisects . By construction, also bisects the total mass, thus, it must also bisect . ∎

Theorem 1 now follows by gnonomic projection.

See 1

As noted in the introduction, these results are tight with respect to the number of masses that are bisected. However, in some cases, we can enforce additional restrictions without sacrificing a mass.

###### Theorem 8.

Let be mass distributions in , where is odd and let be a line. Then there exists a -cone whose apex lies on that simultaneously bisects .

###### Proof.

Without loss of generality, let be the -axis. Place somewhere on . We will move along from to . Consider all directed lines through . Each such line defines a unique -cone bisecting the total mass. For each , define , where again denotes the complement of . Thus, for each choice of , we get a map . We claim that for some this map has a zero. Assume for the sake of contradiction that none of the maps have a zero. Then we can normalize them to get maps . In particular, each such map has a degree. Further, all of the maps are antipodal, implying that their degree is odd, and thus non-zero. Finally, note that , and thus . (Here we require that is even.) In particular, as the degrees are non-zero, and have different degrees. But moving along from to defines a homotopy from to , which is a contradiction. Thus, there exists some such that has a zero and analogous to above this zero defines a -cone that simultaneously bisects . ∎

## 4 Double Wedges

Before proving Theorem 2, we will prove a more general statement about bisections of mass distributions with double wedges. Let us first explain how bisections with double wedges can be regarded as Ham-Sandwich cuts after a projective transformation: Let be a mass distribution in and let be a double wedge that bisects . Use gnonomic projection to map and to the upper hemisphere of . Now, antipodally copy and to the lower hemisphere. Note that both and are oriented -dimensional great circles on , so we can extend them to oriented hyperplanes through the origin in , which we denote as and , respectively. Also, we will denote the defined measure on by . Note now that and that bisects . Further, the above is invariant under rotations of the sphere, thus we can rotate the sphere until is one of the two orientations of the hyperplane . Using gnonomic projection to map the upper hemisphere to , we get a projective transformation of with the property that is the sphere at infinity and that bisects . Thus, we get the following:

###### Lemma 9.

Let be mass distributions in and let be a double wedge which simultaneously bisects . Then there is a projective transformation of with the property that is the sphere at infinity and that simultaneously bisects .

In the following we will now prove results about bisections of different families with different double wedges which still share one of the hyperplanes.

###### Lemma 10.

Let be families each containing mass distributions in , where is odd. Then there exists an oriented hyperplane and double wedges , , whose apexes all contain the origin and such that simultaneously bisects .

###### Proof.

The space of pairs of oriented hyperplanes in containing the origin is . For some mass distribution , we can thus define a function by , where is the double wedge defined by and . Note that and that if and only if bisects . Further, for each we get a function .

Let us now fix some and consider the family . Assume that cannot be simultaneously bisected by a double wedge defined by and some other hyperplane through the origin. In particular, defining a function as above for each mass yields a map which has no zero. Thus, after normalizing, we get a map . In particular, this map has a degree. As we have , this degree is odd and thus non-zero. Further, varying again, we note that , and thus, as is even, we have . In particular, any path from to defines a homotopy between two maps of different degree, which is a contradiction. Thus, along every path from to we encounter a hyperplane such that has a zero. In particular, this partitions into regions where has a zero and where it does not. Let be the region where has a zero. We note that all regions are antipodal (i.e., has a zero if and only if does) and no connected component of contains two antipodal points. Hence, we can define a map as follows: for each , set . Further, for each , set , where denotes the distance from to . Note that is continuous and and if and only if has a zero.

We can do this for all families to get an antipodal map . By the Borsuk-Ulam theorem, this map has a zero. This zero gives us a hyperplane through the origin which, by construction, has the property that for each family of masses there exists another hyperplane through the origin such that simultaneously bisects . ∎

Using the same lifting argument as for the proof of Theorem 1, we get the following:

###### Corollary 11.

Let be families each containing mass distributions in , where is even. Then there exists an oriented hyperplane and double wedges , , such that simultaneously bisects .

Note that for the argument we require that the considered sphere has even dimension. This means, that if we want to prove Lemma 10 for even dimensions, we have to use different arguments. In the following we try to do this, but at the expense that we will only be able to ’almost’ bisect the masses. More precisely, we say that a double wedge -bisects a mass if . Similarly we say that simultaneously -bisects if it -bisects for every . In the following we will show Lemma 10 for even dimensions, with ’bisect’ replaced by ’-bisect’. For this we first need an auxiliary lemma.

###### Lemma 12.

Let and let be families each containing mass distributions in , where is odd. Then there exists an oriented hyperplane containing the -axis, a point at distance to the -axis and double wedges , , whose apexes all contain the origin and such that simultaneously bisects and contains .

The proof is very similar to the proof of Lemma 10.

###### Proof.

The space of pairs of oriented hyperplanes in with containing the -axis and containing the origin is . Let us again fix some and consider the family . As above, we can define functions which give rise to a function , which has a zero if and only if can be simultaneously bisected by a double wedge using . Further, for each we can define in a continuous fashion a point which lies in the positive side of and on the upper hemisphere of , and which has distance to the -axis. Define now , where denotes the distance from to and if lies in the double wedge and otherwise. Note that is continuous, if and only if is on , and . Thus, together with the functions defined by the masses, we get a map , which has a zero if and only if can be simultaneously bisected by a double wedge using and an passing through . Again, assuming this map has no zero, we get a map , which, because of the antipodality condition, has odd degree. Again, we have , and thus along every path from to we encounter a hyperplane such that has a zero. As above, this partitions the sphere into antipodal regions, the only difference being that this time we only consider the sphere . In particular, the Borsuk-Ulam theorem now gives us a hyperplane containing the -axis which, by construction, has the property that for each family of masses there exists another hyperplane through the origin such that simultaneously bisects . Further, the hyperplane also defines a point at distance to the -axis with the property that each contains . ∎

After gnonomic projection, we thus get a double wedges that simultaneously bisect the masses and such that contains the origin and the distance from is at most . If we now translate to contain the origin, by continuity we get that there is some such that simultaneously -bisects the masses . In particular, for every we can choose in the above lemma such that after gnonomic projection we get the following:

###### Corollary 13.

Let .

1. Let be families each containing mass distributions in , where is even. Then there exists an oriented hyperplane containing the origin and double wedges , , whose apexes all contain the origin and such that simultaneously -bisects .

2. Let be families each containing mass distributions in , where is odd. Then there exists an oriented hyperplane and double wedges , , such that simultaneously -bisects .

By a standard argument (see e.g. [Matousek]), a bisection partition result for mass distributions also implies the analogous result for point sets in general position. Further, for point sets in general position, we can choose small enough to get an actual bisection. Thus, Theorem 2 now follows from Lemma 9, Corollary 11 and the second part of Corollary 13.

See 2

Let us mention that this result is tight with respect to the number of families that are bisected: Consider families each containing point sets in where each point set consists of many points that are very close together. Place these point sets in such a way that no hyperplane passes through of them. Look at one family of point sets. if are to simultaneously bisect the point sets in this family, each point set must be transversed by either or , or both. In particular, as can pass through at most point sets, must pass through at least one point set. This is true for all families, which means that must pass through at least point sets in total, which cannot happen by our construction of the point sets.

For larger families, a similar argument shows that at most families each containing point sets can have a Ham-Sandwich cut after a common projective transformation. We conjecture, that this is always possible, even for general mass distributions:

###### Conjecture 1.

Let be families each containing mass distributions in . Then there exists a projective transformation such that can be simultaneously bisected by a single hyperplane for every .

Note that for , this would imply a Ham-Sandwich cut after a projective transformation or, equivalently, a bisection with a double wedge, for a single family of mass distributions. Such a bisection is known to exist for any which is a power of 2, see [pizza1].

## 5 Fans

Similar to the previous sections, we will again first prove all results with the apex containing the origin. We will call such a fan a fan through the origin. For technical reasons, we again distinguish whether the dimension is even or odd. The general results will then again follow from a lifting argument. Our proofs are very similar to those in [Barany2].

###### Lemma 14.

Let be prime and let be odd.

1. Any mass distributions in can be simultaneously -equipartitioned by a -fan through the origin;

2. Let with and . Then any mass distributions in can be simultaneously -equipartitioned by a -fan through the origin.

###### Proof.

We start with equipartitions. Assume without loss of generality that for each mass distribution we have . Let . Consider the Stiefel manifold of all pairs

. To each we assign a -fan as follows: Let by the linear subspace spanned by and let be the canonical projection. The apex of the -fan is then . Further, note that defines an orientation on , so we can consider a ray on rotating in clockwise direction. Start this rotation at , and let be the (unique) ray such that the area between and is the projection of a wedge which contains exactly a -fraction of the total mass. Analogously, let be the (unique) ray such that the area between and is the projection of a wedge which contains exactly a -fraction of the total mass. This construction thus continuously defines a -fan through the origin for each . Further note that there is a natural action on , defined by , i.e., by turning by one sector.

For a mass distribution , we introduce a test map by

 fi(x,y):=(μi(W1)−1p,μi(W2)−1p,…,μi(Wp)−1p).

Note that the image of is contained in the hyperplane of dimension and that implies that -equipartitions . In particular, if for all , then simultaneously equipartitions , and thus, as equipartitions the total mass by construction, it also equipartitions . We thus want to show that all test maps have a common zero. To this end, we note that the action on induces a action on by . Further, as is prime, this action is free on . Thus, if we assume that the test maps do not have a common zero, they induce a -map . We will now show that there is no such map.

To this end, we first note that the dimension of is and that, after normalizing, induces a map . We will use the following strengthening of Dold’s theorem due to Jelic [Jelic]:

###### Theorem 15 ([Jelic], Thm. 2.1).

Let be a finite group acting freely on a cell -complex of dimension at most , and let be a -space. Let be acommutative ring with unit such that and for . Then there is no -equivariant map .

Setting , we get that for all , see [tomDieck], Theorem III 2.5. Further, for odd, we have for , , and all other cohomology groups are trivial (see [Borel], Prop. 10.1). From the universal coefficients theorem we thus get for , and otherwise. As

 (p−1)k−1=(p−1)⌊2d−3p−1⌋−1≤2d−3−1=2d−4,

we can thus apply Theorem 15 with , where is , to show that , and thus , cannot exist. This finishes the proof for the equipartitions.

As for the more general partitions, we let . We take the same configuration space and test maps, only that we now have more possible solutions to exclude from . In particular, let . Further let . We now want to show that there is no -equivariant map . By Lemma 6.1 in [Barany2], would induce a map , where is a -dimensional manifold on which acts freely. As

 (p−1)k−2=(p−1)⌊2d−2p−1⌋−2≤2d−2−2=2d−4,

the non-existence of again follows from Theorem 15. ∎

When is even, we cannot use the above method, as any Cohomology of has too many non-trivial groups. However, we can again use lifting to the upper hemisphere of

and use some degrees of freedom to enforce that the apex of the equipartitioning fan contains the

-axis. Projecting back to then gives us an equipartitioning fan through the origin. For this we need the following lemma: