EEquipartitions with Wedges and Cones
Patrick Schnider ∗ Abstract
A famous result about mass partitions is the so called
Ham-Sandwich theorem . Itstates that any d mass distributions in R d can be simultaneously bisected by a singlehyperplane. In this work, we study two related questions.The first one is, whether we can bisect more than d masses, if we allow for bisectionswith more general objects such as cones, wedges or double wedges. We answer this questionin 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 familiesof d + 1 masses each can each by simultaneously bisected by some double wedge such thatall 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 − Equipartitions of point sets and mass distributions are essential problems in combinatorialgeometry. The general goal is the following: given a number of masses in some space, we wantto find a dissection of this space into regions such that each mass is evenly distributed over allthe regions, that is, each region contains the same amount of this mass. Usually, the underlyingspace considered is a sphere or Euclidean space, and the regions are required to satisfy certainconditions, for example convexity. Arguably the most famous result about equipartitions isthe
Ham-Sandwich theorem (see e.g. [16, 19], Chapter 21 in [21]), which states that any d massdistributions in R d can be simultaneously bisected by a single hyperplane. A mass distribution µ in R d is a measure on R d such that all open subsets of R d are measurable, 0 < µ ( R d ) < ∞ and µ ( S ) = 0 for every lower-dimensional subset S of R d . The result is tight in the sense thatthere are collections of d + 1 masses in R d that cannot be simultaneously bisected by a singlehyperplane.However, the restriction that the regions of the dissection should be half-spaces is quitestrong, maybe we can bisect more masses by relaxing the conditions on the regions? This istrue in many cases. For example, any 3 masses in R can be simultaneously bisected by awedge [4]. Using double wedges, even 4 masses can be simultaneously bisected [6]. Both ofthese objects will be explained in more detail below.On the other hand, Ham-Sandwich theorem only provides a dissection of the space intotwo regions, and it is a natural question whether similar statements can be found for morethan two regions. This is indeed the case, Soberon [18] and independently Karasev, Hubardand Aronov [15] have shown that for any d masses in R d , one can find a partition of R d intoconvex sets such that each convex set contains a n -fraction of each mass. Further, B´ar´any andMatouˇsek [4] have shown that for any two masses in R , we can partition R into 3 or 4 wedgessuch that each wedge contains a - or a -fraction of each mass, respectively. ∗ Department of Computer Science, ETH Z¨urich, Switzerland. [email protected] a r X i v : . [ c s . C G ] O c t igure 1: Three masses simultaneously bisected by a 2-cone.In this paper, we generalize many of the above results into higher dimensions. More specifi-cally, we study bisections with k -cones and double wedges, as well as equipartitions with k -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 R d is defined by an apex a ∈ R d , a central axis −→ (cid:96) , which is a one-dimensional ray emanating from the apex a , and and angle α .The cone is now the set of all points that lie on a ray −→ r emanating from a such that the anglebetween −→ r and −→ (cid:96) is at most α . Note that for α = 90 ◦ , the cone is a half-space. Also, for α > ◦ , the cone is not convex, which we explicitly allow. Finally note that the complementof a cone is also a cone and that either a cone or its complement are convex. Let now H k be some k -dimensional linear subspace of R d and let π : R d → H k be the natural projection.A k -cone C is now a set π − ( C k ), where C k is a cone in H k . The apex a of C is the set π − ( a k ), where a k is the apex of C k . It has dimension d − k . Again, note that the complementof a k -cone is again a k -cone and that one of the two is convex. Also, a 2-cone is either theintersection or the union of two halfspaces, that is, a so-called wedge . Further, a d -cone is justa spherical cone. Alternatively, we could also define a k -cone by a ( d − k )-dimensional apexand a ( d − k + 1)-dimensional half-hyperplane h emanating from it. The k -cone would then bethe union of points on all ( d − k + 1)-dimensional half-hyperplanes emanating from the apexsuch that their angle with h is at most α . We say that a k -cone C simultaneously bisects themass distributions µ , . . . , µ n if µ i ( C ) = µ i ( R d ) for all i ∈ { , . . . , n } . See Figure 1 for anillustration. In Section 3 we will prove the following: Theorem 1.
Let µ , . . . , µ d +1 be d + 1 mass distributions in R d and let ≤ k ≤ d . Then thereexists a k -cone C that simultaneously bisects µ , . . . , µ d +1 . This is tight in the sense that there is a family of d + 2 mass distributions in R d that cannotbe simultaneously bisected: place d + 1 point-like masses at the vertices of a d -dimensionalsimplex and a last point-like mass in the interior of the simplex. Assume that there is a k -cone C which simultaneously bisects all the masses and assume without loss of generality that C is convex (otherwise, just consider the complement of C ). As it simultaneously bisects allmasses, C must now contain all vertices of the simplex, so by convexity C also contains theinterior of the simplex, and thus all of the last mass. Hence C cannot simultaneously bisectall masses. Still, while we cannot hope to bisect more masses, in some cases we are able toenforce additional restrictions: for d -cones in odd dimensions, we can always enforce the apexto lie on a given line.Another relaxation of Ham-Sandwich cuts was introduced by Bereg et al. [8] and hasreceived some attention lately (see [6, 10, 13, 17]): instead of cutting with a single hyperplane,how many masses can we bisect if we cut with several hyperplanes? In this setting, the massesare distributed into two parts according to a natural 2-coloring of the induced arrangement. Wewill only consider bisections with two hyperplanes: let h and h be two (oriented) hyperplanes.Let h +1 and h − be the positive and negative side of h , respectively, and analogous for h .We define the double wedge D = ( h , h ) as the union ( h +1 ∩ h +2 ) ∪ ( h − ∩ h − ). Note that2igure 2: Three masses simultaneously bisected by a double wedge.the complement of a double wedge is again a double wedge. We say that a double wedge D simultaneously bisects the mass distributions µ , . . . , µ n if µ i ( D ) = µ i ( R d ) for all i ∈{ , . . . , n } . See Figure 2 for an illustration.Alternatively, bisections with double wedges can also be viewed as Ham-Sandwich cuts aftera projective transformation: if D = ( h , h ) is a bisecting double wedge we can find a projectivetransformation which sends h to the hyperplane at infinity. After this transformation, h is aHam-Sandwich cut of the transformed masses. In Section 4, we will prove the following: Theorem 2.
Let P , . . . , P d +1 , P , . . . , P d +1 , . . . , P dd +1 be d families each containing d +1 pointsets in R d such that their union is in general position. Then there exists a projective transfor-mation ϕ such that ϕ ( P i ) , . . . , ϕ ( P id +1 ) can be simultaneously bisected by a single hyperplanefor every i ∈ { , . . . , d } . We will actually prove a very similar statement for mass distributions, but in odd dimensionswe need some technical restrictions.Finally, the last question that we want to address is what type of equipartitions can be doneusing several wedges. More precisely, we define a k -fan in R d as a ( d −
2) dimensional hyperplane a , which we call apex , and k semi-hyperplanes emanating from it. Each k -fan partitions R d into k wedges, which can be given a cyclic order W , . . . , W k . For α = ( α . . . , α k ) with α + . . . + α k = 1 we say that a k -fan simultaneously α -equipartitions the mass distributions µ , . . . , µ n if µ i ( W j ) = α j · µ i ( R d ) for every i ∈ { , . . . , n } and j ∈ { , . . . , k } . See Figure 3 foran illustration.Equipartitions with k -fans in the plane have been extensively studied, see [2, 3, 4, 5, 7, 9,11, 22]. In Section 5 we will use the techniques from [4] to prove results for equipartitions with k -fans in higher dimensions. In particular, we will prove the following: Theorem 3.
Let p be an odd prime.1. Any (cid:98) d − p − (cid:99) +1 mass distributions in R d , where d is odd, can be simultaneously ( p . . . , p ) -equipartitioned by a p -fan;2. Any (cid:98) d +1 p − (cid:99) mass distributions in R d , where d is even, can be simultaneously ( p . . . , p ) -equipartitioned by a p -fan;3. Let ( a , . . . , a q ) ∈ N q with q < n and a + . . . + a q = p . Then any (cid:98) dp − (cid:99) + 1 massdistributions in R d , where d is odd, can be simultaneously ( a p . . . , a q p ) -equipartitioned bya q -fan;4. Let ( a , . . . , a q ) ∈ N q with q < n and a + . . . + a q = p . Then any (cid:98) d +1 p − (cid:99) mass distributionsin R d , where d is even, can be simultaneously ( a p . . . , a q p ) -equipartitioned by a q -fan. , , )-equipartitioned by a 3-fan.This directly extends the planar results in [4].All our results are proved using topological methods. In some cases, we use establishedtechniques, such as the configuration space/test map scheme (see e.g. the excellent book byMatouˇsek [16]) or degree arguments, sometimes in perhaps unusual combinations. In othercases, we use newer tools, such as a recent result about sections in canonical line bundles offlag manifolds [17], which we rephrase as an extension of the Borsuk-Ulam theorem to flagmanifolds, and a strengthening of Dold’s theorem due to Jelic [14]. In this section we will discuss two technical tools that we will use throughout the paper. Thefirst one is a generalization of the Borsuk-Ulam theorem to flag manifolds. This result wasalready used in [17], but phrased in a slightly different setting, which is why we will still provethat the version stated here actually follows from the more general statement in [17]. After thatwe give a general argument called gnonomic projection , which will allow us to only considerwedges and cones whose apex contain the origin in the following sections.
Recall the definition of a flag manifold : a flag F in a vector space V of dimension d is anincreasing sequence of subspaces of the form F = { } = V ⊂ V ⊂ · · · ⊂ V k = V. To each flag we can assign a signature vector of dimensions of the subspaces. A flag is a complete flag if dim V i = i for all i (and thus k = n ). A flag manifold F is the set of all flagswith the same signature vector. We denote the complete flag manifold, that is, the manifoldof complete flags, by ˜ V n,n . We will consider flag manifolds with oriented lines : a flag manifoldwith oriented lines −→F 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 orientedlines by −→ V n,n . In particular, each flag manifold with oriented lines −→F is a double cover of theunderlying flag manifold F . Further, reorienting the line defines a natural antipodal action. Theorem 4 (Borsuk-Ulam for flag manifolds) . Let −→F be a flag manifold with oriented linesin R d +1 . Then every antipodal map f : −→F → R d has a zero. Let us briefly mention how this implies the Borsuk-Ulam theorem: the simplest flags con-taining a line are those of the form F = { } ⊂ (cid:96) ⊂ R d +1 . the corresponding flag manifold is4he manifold of all lines through the origin in R d +1 , that is, the projective space. Orientingthe lines, we retrieve the manifold of all oriented lines through the origin in R d +1 , which ishomeomorphic to the sphere S d . The above theorem now says that every antipodal map from S d to R d has a zero, which is one of the versions of the Borsuk-Ulam theorem.Note that any antipodal map f : −→F → R d can be extended to an antipodal map g : −→ V d +1 ,d +1 → R d by concatenating it with the natural projection from −→ V d +1 ,d +1 to −→F , thus itis enough to show the statement for −→ V d +1 ,d +1 .In [17] the above is phrased in terms of sections of canonical line bundles, so let us brieflyrecall some of the relevant terms. A vector bundle consists of a base space B , a total space E ,and a continuous projection map π : E (cid:55)→ B . Furthermore, for each b ∈ B , the fiber π − ( b )over b has the structure of a vector space over the real numbers. Finally, a vector bundlesatisfies the local triviality condition , meaning that for each b ∈ B there is a neighborhood U ⊂ B containing p such that π − ( U ) is homeomorphic to U × R d . A section of a vectorbundle is a continuous mapping s : B (cid:55)→ E such that πs equals the identity map, i.e., s mapseach point of B to its fiber. We can define a canonical bundle for each V i in a complete flag,which we will denote by ϑ di . The bundle ϑ di has a total space E consisting of all pairs ( F, v ),where F is a complete flag in R d and v is a vector in V i , and a projection π : E (cid:55)→ ˜ V n,n givenby π (( F, v )) = F . In [17], the following is proved: Lemma 5.
Let s , . . . , s m +1 be m + 1 sections of the canonical bundle ϑ m + ll . Then there is aflag F ∈ ˜ V m + l,m + l such that s ( F ) = . . . = s m +1 ( F ) . We will now show how Lemma 5 implies Theorem 4.
Proof of Theorem 4.
Let f : −→ V d +1 ,d +1 → R d be antipodal. Write f as ( f , . . . , f d ), where each f i : −→ V d +1 ,d +1 → R is an antipodal map. In particular, each f i defines a section s i in ϑ d +11 .Let s be the zero section in ϑ d +11 . Thus, s , s , . . . , s d are d + 1 sections in ϑ d +11 and thus byLemma 5 there is a flag F on which the coincide. As s is the zero section, this means that f i ( F ) = 0 for all i ∈ { , . . . , d } and thus f ( F ) = 0, which is what we wanted to show. Gnonomic projection is a projection π of the upper hemisphere S + of a (unit) sphere to itstangent space T at the north pole. It works as follows: for some point p on S + , let (cid:96) ( p ) bethe line through p and the origin. The projection π ( p ) of p is then defined as the intersectionof (cid:96) ( p ) and T . Note that this is a bijection from the (open) upper hemisphere to the tangentspace. Gnonomic projection maps great circles to lines. More generally, we say that a great k -circle on a sphere S d is the intersection on S d with a k + 1-dimensional linear subspace(i.e., a ( k + 1)-flat containing the origin). In particular, gnonomic projection then maps great k -circles to k -flats.Using gnonomic projection, we can show the following: Lemma 6 (Gnonomic projection) . Assume that any m mass distributions in R d +1 can besimultaneously partitioned by a k -cone (or double wedge or q -fan) whose apex contains theorigin. Then any m mass distributions in R d can be simultaneously partitioned by a k -cone(or double wedge or q -fan).Proof. We will only prove the statement for k -cones, as the proof is analogous for the otherobjects. Let µ , . . . , µ m be mass distributions in R d . Use the inverse π − of gnonomic projectionto map µ , . . . , µ m to the upper hemisphere of S d ⊆ R d +1 . By our assumption there exists a k -cone C whose apex contains the origin that simultaneously partitions π − ( µ ) , . . . , π − ( µ m ).Let C S be the intersection of C with the upper hemisphere. From the alternative definition of k -cones it is not hard to see that the image π ( C S ) of C S under the gnonomic projection is a k -cone in R d which simultaneously bisects µ , . . . , µ m .In the following, we will only prove statements about k -cones, double wedges and fanswhose apexes contain the origin. The general results then follow from the above lemma.5 k -cones In this section, we will use the Borsuk-Ulam theorem for flag manifolds to show the existenceof bisections with k -cones. Theorem 7.
Let µ , . . . , µ d be d mass distributions in R d , let p ∈ R d be a point and let ≤ k ≤ d . Then there exists a k -cone C whose apex contains p and that simultaneously bisects µ , . . . , µ d .Proof. Without loss of generality, let p be the origin. Let −→F be the flag manifold with orientedlines defined by the flags (0 , −→ (cid:96) , V k , R d ), where −→ (cid:96) has dimension 1 and V k has dimension k .Each (0 , −→ (cid:96) , V k , R d ) defines a unique k -cone that bisects the total mass µ + . . . + µ d and whoseprojection to V k is a cone C with central axis −→ (cid:96) and apex p . For i ∈ { , . . . , d − } , define f i := µ i ( C ) − µ i ( C ), where C denotes the complement of C . Then f := ( f , . . . , f d − ) is amap from −→F to R d − . Further, as every C bisects the total mass, so does the complement C .In particular, C is the unique k -cone that we get when switching the orientation of −→ (cid:96) . Thus, f is antipodal, and by Theorem 4 it has a zero. Let C be the k -cone defined by this zero. Bythe definition of f we have that C simultaneously bisects µ , . . . , µ d − . By construction, C also bisects the total mass, thus, it must also bisect µ d .Theorem 1 now follows by gnonomic projection. Theorem 1.
Let µ , . . . , µ d +1 be d + 1 mass distributions in R d and let ≤ k ≤ d . Then thereexists a k -cone C that simultaneously bisects µ , . . . , µ d +1 . As noted in the introduction, these results are tight with respect to the number of massesthat are bisected. However, in some cases, we can enforce additional restrictions withoutsacrificing a mass.
Theorem 8.
Let µ , . . . , µ d +1 be d + 1 mass distributions in R d , where d is odd and let g be a line. Then there exists a d -cone C whose apex a lies on g that simultaneously bisects µ , . . . , µ d +1 .Proof. Without loss of generality, let g be the x d -axis. Place a somewhere on g . We will move a along g from −∞ to + ∞ . Consider all directed lines through a . Each such line −→ (cid:96) defines aunique d -cone C bisecting the total mass. For each i ∈ { , . . . , d } , define f i := µ i ( C ) − µ i ( C ),where C again denotes the complement of C . Thus, for each choice of a , we get a map f a = ( f , . . . , f d ) : S d − → R d . We claim that for some a this map has a zero. Assume forthe sake of contradiction that none of the maps have a zero. Then we can normalize themto get maps f a : S d − → S d − . In particular, each such map has a degree. Further, all ofthe maps are antipodal, implying that their degree is odd, and thus non-zero. Finally, notethat f −∞ = − f + ∞ , and thus deg( f −∞ ) = − deg( f + ∞ ). (Here we require that d − f −∞ and f + ∞ have different degrees. But moving a along g from −∞ to + ∞ defines a homotopy from f −∞ to f + ∞ , which is a contradiction.Thus, there exists some a such that f a has a zero and analogous to above this zero defines a d -cone C that simultaneously bisects µ , . . . , µ d +1 . Before proving Theorem 2, we will prove a more general statement about bisections of massdistributions with double wedges. Let us first explain how bisections with double wedges can beregarded as Ham-Sandwich cuts after a projective transformation: Let µ be a mass distributionin R d and let D = ( h , h ) be a double wedge that bisects µ . Use gnonomic projection to map µ and D to the upper hemisphere of S d ⊆ R d +1 . Now, antipodally copy µ and D to the lowerhemisphere. Note that both h and h are oriented ( d − S d , sowe can extend them to oriented hyperplanes through the origin in R d +1 , which we denote as6 and H , respectively. Also, we will denote the defined measure on S d by µ S . Note nowthat µ S ( S d ) = 2 µ ( R d ) and that ( H , H ) bisects µ S . Further, the above is invariant underrotations of the sphere, thus we can rotate the sphere until H is one of the two orientationsof the hyperplane x d = 0. Using gnonomic projection to map the upper hemisphere to R d , weget a projective transformation ϕ of R d with the property that ϕ ( h ) is the sphere at infinityand that ϕ ( h ) bisects ϕ ( µ ). Thus, we get the following: Lemma 9.
Let µ , . . . , µ k be mass distributions in R d and let D = ( h , h ) be a double wedgewhich simultaneously bisects µ , . . . , µ k . Then there is a projective transformation ϕ of R d with the property that ϕ ( h ) is the sphere at infinity and that ϕ ( h ) simultaneously bisects ϕ ( µ ) , . . . , ϕ ( µ k ) . In the following we will now prove results about bisections of different families with differentdouble wedges which still share one of the hyperplanes.
Lemma 10.
Let µ , . . . , µ d , µ , . . . , µ d , . . . , µ d − d be d − families each containing d massdistributions in R d , where d is odd. Then there exists an oriented hyperplane h and d − double wedges D i = ( h , h i ) , i ∈ { , . . . , d − } , whose apexes all contain the origin and suchthat D i simultaneously bisects µ i , . . . , µ id .Proof. The space of pairs ( h , h ) of oriented hyperplanes in R d containing the origin is S d − × S d − . For some mass distribution µ , we can thus define a function f : S d − × S d − → R by f ( h , h ) := µ ( D ) − µ ( D ), where D is the double wedge defined by h and h . Note that f ( − h , h ) = f ( h , − h ) = − f ( h , h ) and that f ( h , h ) = 0 if and only if D bisects µ .Further, for each h ∈ S d − we get a function f h := f ( h , · ) : S d − → R .Let us now fix some h and consider the family µ , . . . , µ d . Assume that µ , . . . , µ d cannotbe simultaneously bisected by a double wedge defined by h and some other hyperplane throughthe origin. In particular, defining a function as above for each mass yields a map g h : S d − → R d which has no zero. Thus, after normalizing, we get a map g h : S d − → S d − . In particular,this map has a degree. As we have g h ( − h ) = − g h ( h ), this degree is odd and thus non-zero.Further, varying h again, we note that g − h ( h ) = − g h ( h ), and thus, as d − g − h ) = − deg( g h ). In particular, any path from − h to h defines a homotopybetween two maps of different degree, which is a contradiction. Thus, along every path from − h to h we encounter a hyperplane h ∗ such that g h ∗ has a zero. In particular, this partitions S d − into regions where g h has a zero and where it does not. Let Z ⊆ S d − be the regionwhere g h has a zero. We note that all regions are antipodal (i.e., g h has a zero if and onlyif g − h does) and no connected component of S d − \ Z contains two antipodal points. Hence,we can define a map t : S d − → R as follows: for each h ∈ Z , set t ( h ) = 0. Further, foreach h ∈ S d − \ Z , set t ( h ) = deg( g h ) · d ( h , Z ), where d ( h , Z ) denotes the distance from h to Z . Note that t is continuous and t ( − h ) = − t ( h ) and t ( h ) = 0 if and only if g h has a zero.We can do this for all families to get an antipodal map t := ( t , . . . , t d − ) : S d − → R d − .By the Borsuk-Ulam theorem, this map has a zero. This zero gives us a hyperplane h throughthe origin which, by construction, has the property that for each family of masses µ i , . . . , µ id there exists another hyperplane h through the origin such that ( h , h ) simultaneously bisects µ i , . . . , µ id .Using the same lifting argument as for the proof of Theorem 1, we get the following: Corollary 11.
Let µ , . . . , µ d +1 , µ , . . . , µ d +1 , . . . , µ dd +1 be d families each containing d + 1 mass distributions in R d , where d is even. Then there exists an oriented hyperplane h and d double wedges D i = ( h , h i ) , i ∈ { , . . . , d } , such that D i simultaneously bisects µ i , . . . , µ id +1 . Note that for the argument deg( g − h ) = − deg( g h ) we require that the considered spherehas even dimension. This means, that if we want to prove Lemma 10 for even dimensions, wehave to use different arguments. In the following we try to do this, but at the expense thatwe will only be able to ’almost’ bisect the masses. More precisely, we say that a double wedge7 = ( h , h ) ε -bisects a mass µ if | µ ( D ) − µ ( D ) | < ε . Similarly we say that D simultaneously ε -bisects µ , . . . , µ k if it ε -bisects µ i for every i ∈ { , . . . , k } . In the following we will showLemma 10 for even dimensions, with ’bisect’ replaced by ’ ε -bisect’. For this we first need anauxiliary lemma. Lemma 12.
Let δ > and let µ , . . . , µ d − , µ , . . . , µ d − , . . . , µ d − d − be d − families eachcontaining d − mass distributions in R d , where d is odd. Then there exists an orientedhyperplane h containing the x d -axis, a point p at distance δ to the x d -axis and d − doublewedges D i = ( h , h i ) , i ∈ { , . . . , d − } , whose apexes all contain the origin and such that D i simultaneously bisects µ i , . . . , µ id − and h i contains p . The proof is very similar to the proof of Lemma 10.
Proof.
The space of pairs ( h , h ) of oriented hyperplanes in R d with h containing the x d -axis and h containing the origin is S d − × S d − . Let us again fix some h and consider thefamily µ , . . . , µ d − . As above, we can define d − g h : S d − → R d − , which has a zero if and only if µ , . . . , µ d − can be simultaneously bisectedby a double wedge using h . Further, for each h we can define in a continuous fashion apoint p which lies in the positive side of h and on the upper hemisphere of S d − ⊆ R d ,and which has distance δ to the x d -axis. Define now d h := I p ∈ D · d ( p, h ), where d ( p, h )denotes the distance from p to h and I p ∈ D = 1 if p lies in the double wedge D = ( h , h )and I p ∈ D = − d h is continuous, d h = 0 if and only if p is on h , and d h ( − h ) = − d h ( h ). Thus, together with the d − g h : S d − → R d , which has a zero if and only if µ , . . . , µ d − can be simultaneouslybisected by a double wedge using h and an h passing through p . Again, assuming this maphas no zero, we get a map g h : S d − → S d − , which, because of the antipodality condition,has odd degree. Again, we have g − h ( h ) = − g h ( h ), and thus along every path from − h to h we encounter a hyperplane h ∗ such that g h ∗ has a zero. As above, this partitions the sphereinto antipodal regions, the only difference being that this time we only consider the sphere S d − . In particular, the Borsuk-Ulam theorem now gives us a hyperplane h containing the x d -axis which, by construction, has the property that for each family of masses µ i , . . . , µ id − there exists another hyperplane h i through the origin such that ( h , h i ) simultaneously bisects µ i , . . . , µ id − . Further, the hyperplane h also defines a point p at distance δ to the x d -axiswith the property that each h i contains p .After gnonomic projection, we thus get a double wedges ( h , h i ) that simultaneously bisectthe masses µ i , . . . , µ id and such that h contains the origin and the distance from h i is atmost δ . If we now translate h i to contain the origin, by continuity we get that there is some ε > h , h i ) simultaneously ε -bisects the masses µ i , . . . , µ id . In particular, for every ε > δ > Corollary 13.
Let ε > .1. Let µ , . . . , µ d , µ , . . . , µ d , . . . , µ d − d be d − families each containing d mass distributionsin R d , where d is even. Then there exists an oriented hyperplane h containing the originand d − double wedges D i = ( h , h i ) , i ∈ { , . . . , d − } , whose apexes all contain theorigin and such that D i simultaneously ε -bisects µ i , . . . , µ id .2. Let µ , . . . , µ d +1 , µ , . . . , µ d +1 , . . . , µ dd +1 be d families each containing d + 1 mass distri-butions in R d , where d is odd. Then there exists an oriented hyperplane h and d doublewedges D i = ( h , h i ) , i ∈ { , . . . , d } , such that D i simultaneously ε -bisects µ i , . . . , µ id +1 . By a standard argument (see e.g. [16]), a bisection partition result for mass distributionsalso implies the analogous result for point sets in general position. Further, for point sets ingeneral position, we can choose ε small enough to get an actual bisection. Thus, Theorem 2now follows from Lemma 9, Corollary 11 and the second part of Corollary 13.8 heorem 2. Let P , . . . , P d +1 , P , . . . , P d +1 , . . . , P dd +1 be d families each containing d +1 pointsets in R d such that their union is in general position. Then there exists a projective transfor-mation ϕ such that ϕ ( P i ) , . . . , ϕ ( P id +1 ) can be simultaneously bisected by a single hyperplanefor every i ∈ { , . . . , d } . Let us mention that this result is tight with respect to the number of families that arebisected: Consider d + 1 families each containing d + 1 point sets in R d where each point setconsists of many points that are very close together. Place these point sets in such a way thatno hyperplane passes through d + 1 of them. Look at one family of d + 1 point sets. if ( h , h )are to simultaneously bisect the point sets in this family, each point set must be transversedby either h or h , or both. In particular, as h can pass through at most d point sets, h mustpass through at least one point set. This is true for all d + 1 families, which means that h must pass through at least d + 1 point sets in total, which cannot happen by our constructionof the point sets.For larger families, a similar argument shows that at most (cid:98) dk (cid:99) families each containing d + k point sets can have a Ham-Sandwich cut after a common projective transformation. Weconjecture, that this is always possible, even for general mass distributions: Conjecture 1.
Let µ , . . . , µ d + k , µ , . . . , µ d + k , . . . , µ md + k be m = (cid:98) dk (cid:99) families each contain-ing d + k mass distributions in R d . Then there exists a projective transformation ϕ suchthat ϕ ( µ i ) , . . . , ϕ ( µ id +1 ) can be simultaneously bisected by a single hyperplane for every i ∈{ , . . . , m } . Note that for k = d , this would imply a Ham-Sandwich cut after a projective transformationor, equivalently, a bisection with a double wedge, for a single family of 2 d mass distributions.Such a bisection is known to exist for any d which is a power of 2, see [13]. Similar to the previous sections, we will again first prove all results with the apex containingthe origin. We will call such a fan a fan through the origin . For technical reasons, we againdistinguish whether the dimension is even or odd. The general results will then again followfrom a lifting argument. Our proofs are very similar to those in [4].
Lemma 14.
Let p be prime and let d be odd.1. Any (cid:98) d − p − (cid:99) + 1 mass distributions in R d can be simultaneously ( p . . . , p ) -equipartitionedby a p -fan through the origin;2. Let ( a , . . . , a q ) ∈ N q with q < n and a + . . . + a q = p . Then any (cid:98) d − p − (cid:99) + 1 massdistributions in R d can be simultaneously ( a p . . . , a q p ) -equipartitioned by a q -fan throughthe origin.Proof. We start with equipartitions. Assume without loss of generality that for each massdistribution µ i we have µ i ( R ) = 1. Let k := (cid:98) d − p − (cid:99) . Consider the Stiefel manifold V ( R d )of all pairs ( x, y ) of orthonormal vectors in R d . To each ( x, y ) ∈ V ( R d ) we assign a p -fan F ( x, y ) as follows: Let h by the linear subspace spanned by ( x, y ) and let π : R d → h be thecanonical projection. The apex of the p -fan F ( x, y ) is then π − (0). Further, note that ( x, y )defines an orientation on h , so we can consider a ray on h rotating in clockwise direction. Startthis rotation at x , and let r be the (unique) ray such that the area between x and r is theprojection of a wedge W which contains exactly a p -fraction of the total mass. Analogously,let r i be the (unique) ray such that the area between r i − and r i is the projection of a wedge W i which contains exactly a p -fraction of the total mass. This construction thus continuouslydefines a p -fan F ( x, y ) through the origin for each ( x, y ) ∈ V ( R d ). Further note that there isa natural Z p action on V ( R d ), defined by ( W , W , . . . , W p ) (cid:55)→ ( W p , W , . . . , W p − ), i.e., byturning by one sector. 9or a mass distribution µ i , i ∈ { , . . . , k } we introduce a test map f i ( x, y ) : V ( R d ) → R p by f i ( x, y ) := ( µ i ( W ) − p , µ i ( W ) − p , . . . , µ i ( W p ) − p ) . Note that the image of f i is contained in the hyperplane Z = { y ∈ R (cid:112) : y + y + . . . + y p = 0 } ofdimension p − f i ( x, y ) = 0 implies that F ( x, y ) ( p . . . , p )-equipartitions µ i . In partic-ular, if f i ( x, y ) = 0 for all i ∈ { , . . . , k } , then F ( x, y ) simultaneously equipartitions µ , . . . , µ k ,and thus, as F ( x, y ) equipartitions the total mass by construction, it also equipartitions µ k +1 .We thus want to show that all test maps have a common zero. To this end, we note that the Z p action on V ( R d ) induces a Z p action ν on Z by ν ( y , y , . . . , y p ) = ( y p , y , . . . , y p − ). Further,as p is prime, this action is free on Z \ { } . Thus, if we assume that the test maps do not havea common zero, they induce a Z p -map f : V ( R d ) → Z k \ { } . We will now show that there isno such map.To this end, we first note that the dimension of Z k is ( p − k and that, after normalizing, f induces a map f (cid:48) : V ( R d ) → S ( p − k − . We will use the following strengthening of Dold’stheorem due to Jelic [14]: Theorem 15 ([14], Thm. 2.1) . Let G be a finite group acting freely on a cell G -complex Y of dimension at most n , and let X be a G -space. Let R be acommutative ring with unit suchthat H n +1 ( BG ; R ) (cid:54) = 0 and ˜ H k ( X ; R ) = 0 for ≤ k ≤ n . Then there is no G -equivariant map g : X → Y . Setting G = R = Z p , we get that H i ( B Z p ; Z p ) (cid:54) = 0 for all k , see [20], Theorem III 2.5.Further, for d odd, we have H i ( V ( R d ); Z ) ∼ = Z for i ∈ { , d − } , H d − ( V ( R d ); Z ) ∼ = Z , andall other cohomology groups are trivial (see [12], Prop. 10.1). From the universal coefficientstheorem we thus get H i ( V ( R d ); Z p ) ∼ = Z p for i ∈ { , d − } , and H i ( V ( R d ); Z p ) = 0 otherwise.As ( p − k − p − (cid:98) d − p − (cid:99) − ≤ d − − d − , we can thus apply Theorem 15 with n = 2 d −
4, where Y is S ( p − k − , to show that f (cid:48) , andthus f , cannot exist. This finishes the proof for the equipartitions.As for the more general partitions, we let k := (cid:98) d − p − (cid:99) . We take the same configuration spaceand test maps, only that we now have more possible solutions to exclude from Z k . In particular,let L i := { y ∈ R p : y + . . . y a = 0 , y a +1 + . . . y a + a = 0 , . . . , y a + ... + a q − +1 + . . . + y p = 0 } .Further let L i := { L i , ν ( L i ) , ν ( L i ) , . . . , ν p − ( L i ) } . We now want to show that there is no Z p -equivariant map f : V ( R d ) → Z k \ (cid:83) ≤ i ≤ k L i . By Lemma 6.1 in [4], f would induce a map f (cid:48) : V ( R d ) → M , where M is a (( p − k − Z p acts freely.As ( p − k − p − (cid:98) d − p − (cid:99) − ≤ d − − d − , the non-existence of f again follows from Theorem 15.When d is even, we cannot use the above method, as any Cohomology of V ( R d ) hastoo many non-trivial groups. However, we can again use lifting to the upper hemisphere of S d ⊆ R d +1 and use some degrees of freedom to enforce that the apex of the equipartitioningfan contains the x d +1 -axis. Projecting back to R d then gives us an equipartitioning fan throughthe origin. For this we need the following lemma: Lemma 16.
Let F np be the space of all p -fans through the origin in R n , endowed with thenatural Z p -action. Then there exists a Z p -equivariant map z : F np → R p − such that z ( F ) = 0 if and only if the apex of F contains the x n -axis.Proof. Let P be the north pole of the sphere S n − , that is, the point (0 , . . . , , ∈ R n . Let F be a p -fan through the origin O whose wedges W , . . . , W p have angles α , . . . , α p . Let a be the apex of F and let h be the orthogonal complement of a . Project F and P to h . Thisdefines a two-dimensional p -fan, which we again denote by F , with apex O and a point, which10 P P S S W W s α = 1 s α > s α > s α = 0Figure 4: The segment S and the fraction s i /α i for two different positions of P .we again denote by P . We will again denote the wedges of F by W , . . . , W p , and they stillhave angles α , . . . , α p . We now construct a ( p − z which is Z p -equivariantand for which z ( F ) = 0 if and only if P = O . As the projection is continuous, such a functionimplies the lemma.Fix some small ε > D := min( d ( P, O ) , − ε ), where d ( P, O ) denotes the distancefrom O to P . Assume that O (cid:54) = P . Consider the circle C with radius d ( P, O ) and center O .On this circle, draw circular segments of length (1 − D ) π to both sides of P . (See Figure 4 foran illustration). This defines a circular segment S . Note that as d ( P, O ) converges to 0, thecircular segment S converges to a full circle. Let s i denote the length of the circular segment S ∩ W i . Define g ( W i ) := s i α i · p · ( p (cid:88) j =1 ( s j α j · p )) − . Note that (cid:80) pi =1 g ( W i ) = 1 and that, assuming O (cid:54) = P , there is always a wedge W i for which s i α i <
1. Further note that g ( W i ) converges to p when d ( P, O ) converges to 0. Thus, we cancontinuously extend g by setting g ( W i ) := p for d ( P, O ) = 0.Let now z ( F ) := ( p − g ( W ) , . . . , p − g ( W p )). Note that z carries a natural Z p -action.Further note that the image of z is contained in the hyperplane { y ∈ R d : y + . . . + y p = 0 } ,i.e., we can view z as a map to R p − . Finally, it follows from the construction that z ( F ) = 0if and only if P = O .Using the above lemma, we can now prove the following: Lemma 17.
Let p be prime and let d be even. Let ( a , . . . , a q ) ∈ N q and a + . . . + a q = p .Then any (cid:98) d − p − (cid:99) mass distributions in R d can be simultaneously ( a p . . . , a q p ) -equipartitionedby a q -fan through the origin. Note that here we do not distinguish between equipartitions and general partitions. Thereason for this is that in odd dimensions, we used Lemma 6.1 in [4] to reduce the dimension ofthe target space. However, this lemma requires that the linear span of the excluded solution11s the whole target space, which is now not the case anymore, as for the function z we onlyexclude the origin of R p . Otherwise, the proof is essentially the same as above, so we will onlysketch the main differences. Proof.
Lift the mass distributions to the upper hemisphere of S d ⊆ R d +1 . We will showthat there is a q -fan whose apex contains the x d +1 -axis which simultaneously ( a p . . . , a q p )-equipartitions all mass distributions. After projecting back, we then get the desired result.Let k := (cid:98) d − p − (cid:99) −
1. Consider the same configuration space and test maps as in theproof of Lemma 14. Further, add the function z from Lemma 16 as an additional test map.Assuming there is no required q -fan, we get a Z p -equivariant map f (cid:48) : V ( R d +1 ) → S l , where l = ( p − k + ( p − − p − k + 1) −
1. As( p − k + 1) − p − (cid:98) d − p − (cid:99) − ≤ d − − d + 1) − , such a map cannot exist by Jelic’s extension of Dold’s theorem.Analogously to the previous sections, Theorem 3 now follows again from lifting to the upperhemisphere. We have shown several results about partitions of mass distributions with cones and wedges.We have proven many of our results to be tight, but not Theorem 3. Thus, it remains an openquestion whether more masses could be simultaneously partitioned using q -fans. Also, all ourresults about fans require the parameter p to be an odd prime. It is an interesting questionwhether similar statements hold when p is not prime.As for partitions with double wedges, or, equivalently, Ham-Sandwich cuts after projectivetransformations, in Cojecture 1 we conjecture that given m = (cid:98) dk (cid:99) families each containing d + k mass distributions in R d , there exists a projective transformation such that every familycan be simultaneously bisected by a single hyperplane after the transformation. The methodsused in this paper seem not to suffice to prove this conjecture.Finally, from a computational point of view, whenever the existence of an object is known,the next interesting question is the complexity of finding it. While there is some work aboutpartitions with wedges in the plane (see [1, 7]), to the authors knowledge there is nothingknown in higher dimensions. References [1] Oswin Aichholzer, Nieves Atienza, Ruy Fabila-Monroy, Pablo Perez-Lantero, Jose M Dıaz-B´a˜nez, David Flores-Pe˜naloza, Birgit Vogtenhuber, and Jorge Urrutia. Balanced islandsin two colored point sets in the plane. arXiv preprint arXiv:1510.01819 , 2015.[2] Imre B´ar´any, Pavle Blagojevi´c, and Aleksandra Dimitrijevi´c Blagojevi´c. Functions, mea-sures, and equipartitioning convex k-fans.
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