Abstract
A kfan in the plane is a point x∈ℝ^{2} and k halflines starting from x. There are k angular sectors σ _{1},…,σ _{ k } between consecutive halflines. The kfan is convex if every sector is convex. A (nice) probability measure μ is equipartitioned by the kfan if μ(σ _{ i })=1/k for every sector. One of our results: Given a nice probability measure μ and a continuous function f defined on sectors, there is a convex 5fan equipartitioning μ with f(σ _{1})=f(σ _{2})=f(σ _{3}).
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1 Introduction
Let S ^{2} be the unit sphere in ℝ^{3}. A kfan on the sphere S ^{2} is formed by a point x∈S ^{2} and k≥3 great semicircles ℓ _{1},…,ℓ _{ k }, starting from x and ending at −x, listed in anticlockwise order when seen from x. The spherical sector σ _{ i } is delimited by ℓ _{ i } and ℓ _{ i+1} and its interior is disjoint from all ℓ _{ j }. The kfan on the sphere is convex, by definition, if the angle of each sector is at most π. Given a probability measure μ on S ^{2}, the kfan (x;ℓ _{1},…,ℓ _{ k }) equipartitions μ if μ(σ _{ i })=1/k for all i.
This paper is a continuation of the one by Bárány, Blagojević and Szűcs [3] which is about the following question of Nandakumar and Ramana Rao [9]. Given an integer k≥2 and a convex set K⊂ℝ^{2} of positive area does there exist a convex kpartition of K such that all pieces have the same area and the same perimeter? The case k=2 is trivial. In [3] the existence of such a partition for k=3 is proved by the following, more general theorem.
Theorem 1.1
Assume μ is a Borel probability measure on S ^{2} with μ(ℓ)=0 for all great circles ℓ, and f is a continuous function defined on the sectors in S ^{2}. Then there is a convex 3fan (x;ℓ _{1},ℓ _{2},ℓ _{3}) equipartitioning μ such that
We will explain later how this theorem settles the k=3 case of the question of Nandakumar and Ramana Rao. In this paper we prove analogous properties of continuous functions defined on the sectors of equipartitioning convex 4 and 5fans. Here are our main results.
Theorem 1.2
Assume μ is a Borel probability measure on S ^{2} with μ(ℓ)=0 for all great circles ℓ, and f is a continuous function defined on the sectors in S ^{2}. Then

(1)
there is a convex 4fan equipartitioning μ with f(σ _{1})=f(σ _{3}),f(σ _{2})=f(σ _{4}),

(2)
there is a convex 4fan equipartitioning μ with f(σ _{1})=f(σ _{2}),f(σ _{3})=f(σ _{4}),

(3)
there is a convex 5fan equipartitioning μ with f(σ _{1})=f(σ _{2})=f(σ _{3}),

(4)
there is a convex 5fan equipartitioning μ with f(σ _{1})=f(σ _{2})=f(σ _{4}).
The theorem is proved by the standard configuration space/test map method with some unusual twists. It is carried out in three steps:

The set of all equipartitioning kfans is known to be V _{2}(ℝ^{3}) the Stiefel manifold of orthogonal twoframes in ℝ^{3}. The configuration space V ^{conv} is going to be the so called convex part of the set of all equipartitioning kfans. It depends on the measure μ. Its definition and its topological properties will be established in Sect. 2 by geometric methods, similar to the ones used in [3].

Defining the suitable (ℤ_{4} and ℤ_{5}equivariant) test maps from V ^{conv} to the phase space is done in Sect. 3. Some extra care has to be exercised in case (2) of Theorem 1.2. We will show that the nonexistence of such a test map implies Theorem 1.2.

The nonexistence of such ℤ_{4} and ℤ_{5}equivariant maps is established in Theorems 5.1 and 6.1 with the help of Serre spectral sequences of Borel constructions. This is in Sects. 5 and 6.
It would be better to show that, under the conditions of Theorem 1.2, there is an equipartitioning convex 4fan resp, 5fan with f(σ _{1})=f(σ _{ j }) for all i,j. But this is too much to hope for as the following results show. The examples are in the plane ℝ^{2} but they work on S ^{2} as well (see the remark below).
Theorem 1.3
There are absolutely continuous measures in the plane μ and ν such that

(1)
there is no convex 4fan simultaneously equipartitioning μ and ν.

(2)
there is no convex 5fan equipartitioning μ such that ν(σ _{ i })=ν(σ _{ i+1})=ν(σ _{ i+2})=ν(σ _{ i+3}) for some i=1,2,3,4,5, the subscripts are taken mod5.

(3)
there is no convex 4fan and no t∈(0,1/3) such that μ(σ _{ i })=ν(σ _{ i })=t for three subscripts i∈{1,2,3,4}.
The first part of the theorem is the result of Bárány and Matoušek [2, Theorem 1.1.(i).(d)]. The proof of their result is repeated in Sect. 7. The same section contains the proof of the second and third parts. In all cases the construction works because the convexity condition reduces the degree of freedom by one.
Remark
Here is the short explanation on how Theorem 1.1 answers the question of Nandakumar and Ramana Rao affirmatively. A kfan in the plane is formed by a point x∈ℝ^{2} and k halflines ℓ _{1},…,ℓ _{ k }, starting from x, listed in anticlockwise order around x. There are k angular sectors σ _{1},…,σ _{ k } determined by the fan. Here σ _{ i } is the sector between halflines ℓ _{ i } and ℓ _{ i+1}. The kfan in the plane is convex if and only if each of the sectors σ _{1},…,σ _{ k } is convex.
It is easier to work with spherical fans than with planar ones mainly because S ^{2} is compact. The plane ℝ^{2} is embedded in ℝ^{3} as the tangent plane to S ^{2} at the point (0,0,1). Let ρ:{(x _{1},x _{2},x _{3})∈S ^{2}∣x _{3}>0}→ℝ^{2} be the central projection. The map ρ lifts any nice measure in the plane to a nice measure on the sphere. Also, a kfan in the plane lifts to a kfan on the sphere and a kfan on the sphere projects to a kfan in the plane, Fig. 1(A). Also, convexity of the fan is preserved under lifting and projection. Therefore any theorem about fan partitions in the plane is a consequence of a similar and more general theorem about fan partitions on the sphere S ^{2}.
Remark
In Theorem 1.2 the measure μ is required to satisfy μ(ℓ)=0 for all great circles ℓ. By a standard compactness argument it suffices to prove the theorem for a dense set of Borel probability measures.
2 Configuration Space of Equipartitioning Convex kFans
This section is taken from [3, Sections 3 and 4] with the view towards the high dimensional applications. Here we work with general k fans for all k>3 although what we have in mind k=4,5.
Let μ be an absolutely continuous (with respect to the Lebesgue measure) Borel probability measure on S ^{2} such that μ(ℓ)=0 for all great circles ℓ. For k≥3 consider the following family of kfans on S ^{2}:
For (x;ℓ _{1},…,ℓ _{ k })∈F _{ k }, let y=ℓ _{1}∩(span{x})^{⊥}∈S ^{2} and z∈(span{x})^{⊥}∩(span{y})^{⊥}∩S ^{2} be such that the base (y,z) of the linear space (span{x})^{⊥} induce the orientation given by ordering of great semicircles (ℓ _{1},…,ℓ _{ k }), Fig. 1(B). Thus, z=x×y, where × denotes the cross product. The correspondence (x;ℓ _{1},…,ℓ _{ k })↦(x,y) induces a homeomorphism between the family of fans F _{ k } and the Stiefel manifold V _{2}(ℝ^{3}). Let ℤ_{ k }=〈ε〉 be a cyclic group. There is natural free ℤ_{ k }action on F _{ k } given by
The main objective of this section is to describe the subfamily of all convex kfans contained in F _{ k } as a ℤ_{ k }invariant subspace.
Let p:(F _{ k }=V _{2}(ℝ^{3}))→S ^{2} denotes the S ^{1} fibration given by (x;ℓ _{1},…,ℓ _{ k })=(x,y)↦z=x×y and let h:S ^{2}→ℝ be the function defined by h(z)=μ(H(z)), where H(z)={v∈S ^{2}∣v⋅z≤0} is the lower hemisphere with respect to z. As shown in [3], one can assume that the composition h:S ^{2}→ℝ is a smooth map and that has a regular value at the point \(\frac{1}{k}\), i.e., \(h^{1}(\{\frac{1}{k}\})\) is an 1dimensional embedded submanifold of S ^{2}, [5, Corollary 7.4, p. 84].
Lemma 2.1
For the fan (x;ℓ _{1},…,ℓ _{ k })=(x;σ _{1},…,σ _{ k })=(x,y), z=x×y=p(x,y), the sector σ _{ k } is not convex if
Proof
Since \(\mu(\sigma_{k})=\frac{1}{k}\) and \(\mu(H(z))<\frac{1}{k}\), then σ _{ k } properly contains the hemisphere H(z) and therefore is not convex. □
Direct consequence of the previous lemma is the characterization of the (non)convex kfans:
or
Lemma 2.2
After a possible rotation of the measure μ, the circle
is invariant under the ℤ_{ k }action and every point (e _{3},y)∈C defines a convex kfan.
Proof
The following result of Dolnikov [7] and Živaljević, Vrećica [10] is needed.
For n≤d probability measures in ℝ^{d}, there exists a (n−1)dimensional affine subspace such that the measure of every halfspace containing this affine subspace is at least \(\frac{1}{d+2n}\) in every one of the k measures.
We use it with d=3 and n=2. Let the first measure be μ and the second one concentrated at the origin. Then the affine space is a line passing through the origin. We may assume, by rotating S ^{2} if necessary, that the line passes through e _{3}. Since k>3, then \(h(e_{3}\times y)\geq\frac{1}{3}>\frac{1}{k}\) for every y∈S((span{e _{3}})^{⊥}). Thus, the circle C is invariant under the ℤ_{ k }action and each (e _{3},y)∈C defines a convex kfan. □
The point \(\frac{1}{k}\) is the regular value of the function h. Thus, \(h^{1}(\{\frac{1}{k}\})\) is an 1dimensional embedded submanifold of S ^{2}, i.e., union of disjoint cycles S _{ i }, i∈[m]={1,…,m}. The image p(C) is the equator of the sphere S ^{2} and \(h(e_{3}\times y)>\frac{1}{k}\) for every y∈p(C). Therefore, every cycle S _{ i } is disjoint from the equator p(C) and so belongs to the upper or lower hemisphere.
Let Ω _{ i } denote the closed disc bounded by S _{ i }, ∂Ω _{ i }=S _{ i }, and not containing p(C). Notice that also p(C)∩Ω _{ i }=∅. Let U _{ i }=p ^{−1}(Ω _{ i }) and T _{ i }=p ^{−1}(S _{ i }). The fibrations p:U _{ i }→Ω _{ i } is the fibration over the contractible space Ω _{ i } and therefore homeomorphic to the trivial fibration. Thus U _{ i }≈S ^{1}×Ω _{ i } is a solid torus and its boundary T _{ i }≈S ^{1}×S _{ i }≈S ^{1}×S ^{1} is an ordinary torus.
The ℤ_{ k }action is given by the homeomorphism ε:V _{2}(ℝ^{3})→V _{2}(ℝ^{3}). Hence U _{ i },ε⋅U _{ i },…,ε ^{k−1}⋅U _{ i } are solid tori and T _{ i },ε⋅T _{ i },…,ε ^{k−1}⋅T _{ i } are ordinary tori for every i∈[m]. The relationships between these tori are described in the following proposition which is just the modification of [3, Claim 3.7, 3.8, 3.9].
Proposition 2.3

(1)
The cycle C is disjoint from all solid tori U _{ i },ε⋅U _{ i },…,ε ^{k−1}⋅U _{ i }, i∈[m].

(2)
ε ^{α}⋅T _{ i }∩ε ^{β}⋅T _{ j }≠∅ ⟹ i=j and α=β.

(3)
The tori U _{ i },ε⋅U _{ i },…,ε ^{k−1}⋅U _{ i } are pairwise disjoint, i∈[m].
Proof

(1)
Let us assume that C∩ε ^{α}⋅U _{ i }≠∅. Since ε⋅C=C, we have
$$ C\cap U_{i}\neq\emptyset\quad \Longrightarrow\quad p(C)\cap p ( U_{i} ) \neq\emptyset\quad \Longrightarrow\quad p(C)\cap \varOmega_{i}\neq\emptyset. $$Contradiction with definition of Ω _{ i }.

(2)
Let α=β and ε ^{α}⋅T _{ i }∩ε ^{α}⋅T _{ j }≠∅. Then
$$ T_{i}\cap T_{j}\neq\emptyset \Longrightarrow\ p(T_{i})\cap p(T_{j})\neq\emptyset\quad \Longrightarrow\quad S_{i}\cap S_{j}\neq\emptyset\quad \Longrightarrow\quad i=j. $$Let 0≤α<β≤k. Without losing the generality, we can assume that α=0. Let (x;ℓ _{1},…,ℓ _{ k })∈T _{ i }∩ε ^{β}⋅T _{ j }≠∅. Then (x;ℓ _{1},…,ℓ _{ k })∈T _{ i }, ε ^{−β}⋅(x;ℓ _{1},…,ℓ _{ k })=(x;ℓ _{ k−β+1},…,ℓ _{ k−β })∈T _{ j } and consequently σ _{ k } and σ _{ k−β } are hemispheres. This cannot be: a contradiction.

(3)
In this part of the proof we use the Generalized Jordan Curve theorem [5, Corollary 8.8, p. 353]. Since H _{1}(V _{2}(ℝ^{3}),ℤ)=0, every torus ε ^{α}⋅T _{ i } splits V _{2}(ℝ^{3}) into two disjoint parts. Let us assume that ε ^{α}⋅U _{ i }∩ε ^{β}⋅U _{ i }≠∅, 0≤α<β≤k. Again, it is enough to consider the case α=0. Since T _{ i }∩ε ^{β}⋅T _{ i }=∅ and H _{2}(V _{2}(ℝ^{3}),ℤ)=0 then the complement V _{2}(ℝ^{3})∖(T _{ i }∪ε ^{β}⋅T _{ i }) has three components. The intersection U _{ i }∩ε ^{β}⋅U _{ i } is one of these three components with the boundary T _{ i } or ε ^{β}⋅T _{ i } or T _{ i }∪ε ^{β}⋅T _{ i }. We discuss these three cases separately.

(a)
Let ∂(U _{ i }∩ε ^{β}⋅U _{ i })=T _{ i }⊂U _{ i }. Then U _{ i }⊆ε ^{β}⋅U _{ i } and consequently
$$ U_{i}\subseteq\varepsilon^{\beta}\cdot U_{i} \subseteq\varepsilon^{2\beta}\cdot U_{i}\subseteq\cdots \subseteq U_{i}. $$Thus, U _{ i }=ε ^{β}⋅U _{ i } and so T _{ i }=ε ^{β}⋅T _{ i }, contradiction.

(b)
Let ∂(U _{ i }∩ε ^{β}⋅U _{ i })=ε ^{β}⋅T _{ i }⊂ε ^{β}⋅U _{ i }. Then ε ^{β}⋅U _{ i }⊆U _{ i } and consequently ε ^{β}⋅U _{ i }=U _{ i }. Thus ε ^{β}⋅T _{ i }=T _{ i } gives the contradiction.

(c)
Let ∂(U _{ i }∩ε ^{β}⋅U _{ i })=T _{ i }∪ε ^{β}⋅T _{ i }. Then ε ^{β}⋅T _{ i }⊆U _{ i } and so ε ^{β}⋅U _{ i } is contained in U _{ i } or its complement (ε ^{β}⋅U _{ i })^{c} is contained in U _{ i }. Thus either U _{ i }∪ε ^{β}⋅U _{ i }=U _{ i } or U _{ i }∪ε ^{β}⋅U _{ i }⊇(ε ^{β}⋅U _{ i })^{c}∪ε ^{β}⋅U _{ i }=V _{2}(ℝ^{3}). The later is not possible since C is disjoint from both U _{ i } and ε ^{β}⋅U _{ i }. Therefore ε ^{β}⋅U _{ i }⊆U _{ i } and consequently ε ^{β}⋅U _{ i }=U _{ i } and ε ^{β}⋅T _{ i }=T _{ i }, contradiction.

(a)
□
Since the cycles S _{ i }, i∈[m] are pairwise disjoint (Fig. 2), the discs Ω _{ i } and Ω _{ j } are either disjoint or one is contained in the other. Consider the discs Ω _{ i } that are not contained in any other disc Ω _{ j }. They are the maximal elements among the Ω _{ i } with the respect to inclusion. For simpler writing we assume that these disks are Ω _{ i } with i∈[r] where, of course, 1≤r≤m. Consequently, the related U _{ i }, i∈[r], are also maximal between U _{ i } with the respect to inclusion. Let us denote the ℤ_{ k } orbit of U _{ i } by \(\mathcal{O}(U_{i}):=U_{i}\cup ( \varepsilon\cdot U_{i} ) \cup\cdots\cup ( \varepsilon^{k1}\cdot U_{i} ) \).
Lemma 2.4
For distinct i,j∈[r], the orbits \(\mathcal{O}(U_{i})\) and \(\mathcal {O}(U_{j})\) are either disjoint or one is contained in the other.
Proof
Let \(\mathcal{O}(U_{i})\cap\mathcal{O}(U_{j})\neq\emptyset\). Then there are α and β, α≤β, such that ε ^{α}⋅U _{ k(i)}∩ε ^{β}⋅U _{ k(j)}≠∅. Without losing the generality we can assume that α=0. There are two separate cases:

(1)
Let β=0. Then

(2)
Let β≠0. Since T _{ i }∩ε ^{β}⋅T _{ k(j)}=∅ we see that the complement V _{2}(ℝ^{3})∖(T _{ i }∪ε ^{β}⋅T _{ j }) has three components. One of them is U _{ i }∩ε ^{β}⋅U _{ j } with boundary either T _{ j } or ε ^{β}⋅T _{ j } or T _{ i }∪ε ^{β}⋅T _{ j }. We discuss all three possibilities:

(a)
Let ∂(U _{ i }∩ε ^{β}⋅U _{ j })=T _{ i }⊂U _{ i }. Then U _{ i }⊆ε ^{β}⋅U _{ j } and consequently the orbit \(\mathcal{O}(U_{i})\) is contained in the orbit \(\mathcal{O}(U_{j})\).

(b)
Let ∂(U _{ i }∩ε ^{β}⋅U _{ j })=ε ^{β}⋅T _{ j }⊂ε ^{β}⋅U _{ k(j)}. Then ε ^{β}⋅U _{ j }⊆U _{ i } and consequently the orbit \(\mathcal{O}(U_{j})\) is contained in the orbit \(\mathcal{O}(U_{i})\).

(c)
Let ∂(U _{ j }∩ε ^{β}⋅U _{ j })=T _{ i }∪ε ^{β}⋅T _{ j }. Consequently T _{ i }⊂ε ^{β}⋅U _{ j } and ε ^{β}⋅T _{ j }⊂U _{ i }. Therefore ε ^{β}⋅U _{ j } ⊆U _{ i } or (ε ^{β}⋅U _{ j })^{c}⊆U _{ i }. Since (ε ^{β}⋅U _{ j })^{c}⊆U _{ i } implies that U _{ i }∪ε ^{β}⋅U _{ j }=V _{2}(ℝ^{3}), and this is not possible, we conclude that ε ^{β}⋅U _{ j }⊆U _{ i } and consequently the orbit \(\mathcal{O}(U_{j})\) is contained in the orbit \(\mathcal{O}(U_{i})\).

(a)
□
Consider the following subset of the family of all equipartitioning kfans on the sphere S ^{2}:
The previous results imply that

(1)
ε ^{α}⋅U _{ i }, for all i∈[r] and α∈{0,…,k−1}, are pairwise disjoint closed solid tori,

(2)
every (x;ℓ _{1},…,ℓ _{ k })=(x,y)∈V ^{conv} is a convex kfan,

(3)
C⊂V ^{conv} and V ^{conv} are ℤ_{ k }invariant subspaces of V _{2}(ℝ^{3}).
Therefore the set V ^{conv} will be called the convex part of V _{2}(ℝ^{3}). Notice that, as Fig. 2 indicates, there might be some convex kfans that are not contained in the convex part V ^{conv}.
3 Test Maps
In this section we describe four similar test map schemes associated with the parts of Theorem 1.2. Let ℤ_{ k }=〈ε〉 denotes the usual cyclic group of order k.
Configuration Spaces
Consider as the configuration spaces the spaces of equipartitioning convex 4 and 5fans described in the previous section. Let us denote these spaces by \(V_{4}^{\mathrm {conv}}=B_{4}\backslash A_{4}\) and \(V_{5}^{\mathrm{conv}}=B_{5}\backslash A_{5}\), where B _{4}=B _{5}=V _{2}(ℝ^{3}) and
Notice that both spaces A _{4} and A _{5} are homotopy equivalent to disjoint unions of 1dimensional spheres.
Some Real ℤ_{4} and ℤ_{5}Representations
Let ℝ^{4} be a real ℤ_{4} representation equipped with the following ℤ_{4}action ε⋅(x _{1},x _{2},x _{3},x _{4})=(x _{2},x _{3},x _{4},x _{1}). The subspaces
are ℤ_{4}invariant subspace or real ℤ_{4}representations. It is not hard to prove that there is an isomorphism of real ℤ_{4}representations W _{4}≅_{ℝ} U⊕V.
Similarly, consider ℝ^{5} as a real ℤ_{5} representation via the action ε⋅(x _{1},x _{2},x _{3},x _{4},x _{5})=(x _{2},x _{3},x _{4},x _{5},x _{1}). The subspace
is ℤ_{5}invariant and therefore a real subrepresentation.
Test Space and Test Map for 4Fans
Let f and \(g:V_{4}^{\mathrm{conv}}\rightarrow\mathbb{R}\) be continuous function on the sectors of the convex 4fan. Consider two test maps \(\tau_{1}:V_{4}^{\mathrm{conv}}\rightarrow W_{4}\) and \(\tau _{2}:V_{4}^{\mathrm{conv}}\rightarrow W_{4}\oplus W_{4}\) given by
where Δ_{ f }=f(σ _{1})+f(σ _{2})+f(σ _{3})+f(σ _{4}) and
where, similarly, Δ_{ g }=g(σ _{1})+g(σ _{2})+g(σ _{3})+g(σ _{4}). Having in mind that for g one can take for example function f ^{2}.
There are two test spaces of interest
Proposition 3.1

(1)
If there is no ℤ_{4}equivariant map \(V_{4}^{\mathrm{conv}}\rightarrow W_{4}\backslash T_{1}\) and \(V_{4}^{\mathrm{conv}}\rightarrow ( W_{4}\oplus W_{4} ) \backslash T_{2}\), then parts 1 and 2 of Theorem 1.2 hold.

(2)
If there is no ℤ_{4}equivariant map \(V_{4}^{\mathrm{conv}}\rightarrow S(V)\) and \(V_{4}^{\mathrm{conv}}\rightarrow S ( U\oplus U ) \), then parts 1 and 2 of Theorem 1.2 hold.
Proof

(1)
If there is no ℤ_{4}equivariant map \(V_{4}^{\mathrm{conv}}\rightarrow W_{4}\backslash T_{1}\), then there exists a convex 4fan with sectors σ _{1},σ _{2},σ _{3},σ _{4} such that τ _{1}(σ _{1},σ _{2},σ _{3},σ _{4})∩T _{1}≠∅ and consequently
$$ f ( \sigma_{1} ) =f ( \sigma_{3} )\quad\text{and}\quad f ( \sigma_{2} ) =f ( \sigma_{4} ). $$If there is no ℤ_{4}equivariant map \(V_{4}^{\mathrm{conv}}\rightarrow ( W_{4}\oplus W_{4} ) \backslash T_{2}\), then there is a convex 4fan with sectors σ _{1},σ _{2},σ _{3},σ _{4} such that τ _{2}(σ _{1},σ _{2},σ _{3},σ _{4})∩T _{1}≠∅. Taking for g=f ^{2} we get
$$ f ( \sigma_{1} ) +f ( \sigma_{3} ) =f ( \sigma_{2} ) +f ( \sigma_{4} )\quad\text{and}\quad f ( \sigma_{1} )^{2}+f^{2} ( \sigma_{3} ) =f^{2} ( \sigma_{2} ) +f^{2} ( \sigma_{4} ) . $$This implies that either f(σ _{1})=f(σ _{2}), f(σ _{3})=f(σ _{4}) or f(σ _{1})=f(σ _{4}), f(σ _{2})=f(σ _{3}).

(2)
The existence of ℤ_{4}homotopies,
and (1) imply the claim (2).
□
Test Space and Test Map for 5Fans
Let \(h:V_{5}^{\mathrm{conv}}\rightarrow\mathbb{R}\) be a continuous function on the sectors of 5fans. Consider the test map \(\tau _{3}:V_{5}^{\mathrm{conv}}\rightarrow W_{5}\) given by
where Δ_{ f }=f(σ _{1})+f(σ _{2})+f(σ _{3})+f(σ _{4})+f(σ _{5}). Here W _{5}={(x _{1},…,x _{5})x _{1}+⋯+x _{5}=0}⊆ℝ^{5}.
There are two test spaces T _{3} and T _{4} we are interested in. They are unions of the minimal ℤ_{5}invariant arrangements \(\mathcal{A}_{3}\) and \(\mathcal{A}_{4}\) containing the linear subspace L _{3}⊂W _{5} and L _{4}⊂W _{5}, respectively, given by
The intersection posets of the arrangements \(\mathcal{A}_{3}\) and \(\mathcal{A}_{4}\) are isomorphic, Fig. 3.
The basic property of the test map scheme follows directly.
Proposition 3.2

(1)
If there is no ℤ_{5}equivariant map \(V_{5}^{\mathrm{conv}}\rightarrow W_{5}\backslash T_{3}\), then part 3 of Theorem 1.2 holds.

(2)
If there is no ℤ_{5}equivariant map \(V_{5}^{\mathrm{conv}}\rightarrow W_{5}\backslash T_{4}\), then part 4 of Theorem 1.2 holds.
4 Cohomology of the Configuration Spaces as an R[ℤ_{ n }]Module
In this section we study the cohomology of the configuration spaces \(V_{4}^{\mathrm{conv}}\) and \(V_{5}^{\mathrm{conv}}\) as ℤ[ℤ_{4}] and \({\mathbb {F}}_{5}[{\mathbb {Z}}_{5}]\)module, respectively. This will turn out to be an important step in the proof of the nonexistence of the appropriate ℤ_{4} and ℤ_{5}equivariant maps, Sects. 5 and 6.
4.1 Cohomology of \(V_{4}^{\mathrm{conv}}\)
We establish the following isomorphisms of ℤ[ℤ_{4}]modules:
Proposition 4.1
The cohomology with the ℤ coefficients of the pair (B _{4},A _{4}) is given by
where ℤ[ℤ_{4}]module M is a part of the following exact sequence of ℤ[ℤ_{4}]modules:
Proof
The pare (B _{4},A _{4}) generates the following long exact sequence in cohomology with ℤ coefficients:
We know that H ^{0}(B _{4})=ℤ, \(H^{0}(A_{4})= ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\) and
Thus Φ _{0} is an injection. Since H ^{1}(B _{4})=0, there is a short exact sequence
and therefore
From the fact that \(H^{1}(A_{4})= ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\), H ^{2}(A _{4})=0 and H ^{2}(B _{4})=ℤ_{2} we obtain an exact sequence
Finally, the fact that H ^{3}(B _{4})=ℤ gives the exact sequence
and the isomorphism H ^{3}(B _{4},A _{4})≅ℤ. □
Corollary 4.2
Proof
The Poincaré–Lefschetz duality [8, Theorem 70.2, p. 415] applied on the compact manifold B _{4} relates the homology of the difference B _{4}∖A _{4} with the cohomology of the pair (B _{4},A _{4}), i.e.,
Now the claim follows directly from the previous proposition. □
Proposition 4.3
\(\mathrm{Hom}(M,\mathbb{Z})\cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\).
Proof
The ℤ[ℤ_{4}]module M seen as an abelian group can be decomposed into the direct sum of the free and the torsion part, M=Free(M)⊕Torsion(M). This is decomposition is a ℤ_{4}invariant. Then Hom(M,ℤ)≅Hom(Free(M),ℤ)≅Free(M) and therefore Hom(M,ℤ) is a free abelian group. The exact sequence (6) implies that rank(Hom(M,ℤ))≥4r _{4}. Application of the Hom functor on the same exact sequence (6) yields the exact sequence
Since, as ℤ[ℤ_{4}]modules,
the exact sequence transforms into
First notice that rank(Hom(M,ℤ))≤4r _{4} and therefore
Since the exact sequence (6) gives an inclusion of ℤ[ℤ_{4}]modules \(( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\longrightarrow \mathrm{Free}(M)\), and \(( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\) is the direct sum of the free ℤ[ℤ_{4}]modules we can conclude that \(\mathrm{Free}(M) \cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\). Thus we have an isomorphism of ℤ[ℤ_{4}]modules
□
Finally, we have to verify the isomorphisms (5) of ℤ[ℤ_{4}]modules.
Corollary 4.4
H ^{0}(B _{4}∖A _{4};ℤ)=ℤ and \(H^{1}(B_{4}\backslash A_{4};\mathbb{Z})\cong\mathrm{Hom}(M,\mathbb{Z})\cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}\).
Proof
The complement B _{4}∖A _{4} is connected. Therefore the cohomology in dimension zero is ℤ. The Universal coefficient theorem applied for the first cohomology gives the exact sequence
Since Ext(H _{0}(B _{4}∖A _{4};ℤ),ℤ)=Ext(ℤ,ℤ)=0, the exact sequence gives the isomorphism
□
4.2 Cohomology of \(V_{5}^{\mathrm{conv}}\)
Like in [3, Sect. 6], we establish the following isomorphisms of \(\mathbb{F}_{5}[{\mathbb {Z}}_{5}]\)modules:
Proposition 4.5
\(H^{0}(V_{5}^{\mathrm{conv}};\mathbb {F}_{5})=\mathbb{F}_{5}\) and \(H^{1}(V_{5}^{\mathrm{conv}};\mathbb{F}_{5})=\bigoplus _{i=1}^{r_{5}}\mathbb{F}_{5}[\mathbb{Z}_{5}]\).
Proof
Since the complement \(V_{5}^{\mathrm{conv}}=B_{5}\backslash A_{5}\) is connected, the first claim easily follows. The second claim follows from Poincaré–Lefschetz duality [8, Theorem 70.2, p. 415] and the homology exact sequence of the pair (B _{5},A _{5}) since \(H_{1}(B_{5};\mathbb{F}_{5})=H_{2}(B_{5};\mathbb{F}_{5})=0\). Indeed,
□
5 Nonexistence of the Test Map, Proof of Theorem 1.2(1)–(2)
The first two parts of Theorem 1.2, via Proposition 3.1, are direct consequences of the following theorem.
Theorem 5.1
There is no ℤ_{4}equivariant map

(i)
\(V_{4}^{\mathrm{conv}}\rightarrow S(V)\),

(ii)
\(V_{4}^{\mathrm{conv}}\rightarrow S ( U\oplus U ) \).
Proof
The proof is obtained by studying the morphism of Serre spectral sequences associated with the Borel constructions of B _{4}∖A _{4}, S(V) and S(U⊕U). We denote the cohomology of the group ℤ_{4} with ℤ coefficients by H ^{∗}(ℤ_{4};ℤ). It is well known that
where degT=2.
The Serre Spectral Sequence of \(V_{4}^{\mathrm{conv}}\times_{_{\mathbb{Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\)
The E _{2}term of the sequence is given by \(E_{2}^{p,q}=H^{p}(\mathbb {Z}_{4},H^{q}(V_{4}^{\mathrm{conv}},\mathbb{Z}))\). For q=1, from Corollary 4.4 and [6, Example 2, p. 58] we find that the first row is
Since the differentials in the spectral sequence are H ^{∗}(ℤ_{4};ℤ)module maps, we have \(d_{2}^{0,1}=0\). This means, in particular, that \(T,2T\in H^{2}(\mathbb{Z}_{4};\mathbb{Z})=E_{2}^{2,0}\) survive to the E _{∞}term.
The Serre Spectral Sequence of \(S(V)\times_{_{\mathbb{Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\)
The E _{2}term of the sequence is given by
In general, the coefficients should be twisted, but the ℤ_{4} action on S(V) is orientation preserving, hence the coefficients are untwisted. The action of ℤ_{4} on S(V)≈S ^{1} is free and therefore
The spectral sequence converges to \(H^{\ast}(S(V)\times_{\mathbb {Z}_{4}}\mathrm{E}\mathbb{Z}_{4};\mathbb{Z})\) and therefore in the E _{∞}term everything in positions p+q>1 must vanish. Since our spectral sequence has only two nonzero rows and the only possibly nonzero differential is d _{2} it follows that \(d_{2}(1\otimes L)=T\in H^{2}(\mathbb{Z}_{4};\mathbb{Z})=E_{2}^{2,0}\). Here L∈H ^{1}(S(V);ℤ) denotes a generator. Therefore, the element \(T\in H^{2}(\mathbb{Z}_{4}; \mathbb{Z})=E_{2}^{2,0}\) vanishes in the E _{3}term.
The Serre Spectral Sequence of \(S(U\oplus U)\times_{{\mathbb{Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\)
The representation V is the 1dimensional complex representation of ℤ_{4} induced by 1↦e ^{iπ/2}. Then U⊕U≅V⊗_{ℂ} V. Following [1, Sect. 8, p. 271 and Appendix, p. 285] we deduce the first Chern class of the ℤ_{4}representation U⊕U
There by [4, Proposition 3.11] we know that in the E _{2}term of the Serre spectral sequence associated to \(S(U\oplus U)\times _{_{\mathbb{Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\) the second (0,1)differential is given by \(d_{2}(1\otimes L)=2T\in H^{2}(\mathbb{Z}_{4};\mathbb{Z})=E_{2}^{2,0}\). Here again L∈H ^{1}(S(U⊕U);ℤ) denotes the generator. Thus the element \(2T\in H^{2}(\mathbb{Z}_{4};\mathbb{Z})=E_{2}^{2,0}\) vanishes in the E _{3}term.
The Nonexistence of Both ℤ_{4}Equivariant Maps
Assume that in both cases there exists a ℤ_{4}equivariant,

(i)
\(f:V_{4}^{\mathrm{conv}}\rightarrow S(V)\),

(ii)
\(g:V_{4}^{\mathrm{conv}}\rightarrow S(U\oplus U)\).
Then f and g induce maps between

Borel constructions, \(V_{4}^{\mathrm{conv}}\times_{\mathbb {Z}_{4}}\mathrm{E}\mathbb{Z}_{4}\rightarrow S(V)\times_{_{\mathbb {Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\) and \(V_{4}^{\mathrm{conv}}\times_{\mathbb {Z}_{4}}\mathrm{E}\mathbb{Z}_{4}\rightarrow S(U\oplus U)\times_{_{\mathbb {Z}_{4}}}\mathrm{E}\mathbb{Z}_{4}\),

equivariant cohomologies,

associated Serre spectral sequences,
$$ E_{r}^{p,q}(f):E_{r}^{p,q}\bigl(S(V); \mathbb{Z}\bigr)\rightarrow E_{r}^{p,q} \bigl( V_{4}^{\mathrm{conv}};\mathbb{Z} \bigr) $$and
$$ E_{r}^{p,q}(g):E_{r}^{p,q}\bigl(S(U \oplus U);\mathbb{Z}\bigr)\rightarrow E_{r}^{p,q} \bigl( V_{4}^{\mathrm{conv}};\mathbb{Z} \bigr) $$such that in the 0row of the E _{2}term
$$ E_{2}^{p,0}(f): \bigl( E_{2}^{p,0} \bigl(S(V);\mathbb{Z}\bigr)=H^{p}(\mathbb{Z}_{4};\mathbb{Z}) \bigr) \rightarrow\bigl( E_{2}^{p,0} \bigl( V_{4}^{\mathrm{conv}};\mathbb{Z} \bigr) =H^{p}( \mathbb{Z}_{4};\mathbb{Z}) \bigr) $$and
$$ E_{2}^{p,0}(g): \bigl( E_{2}^{p,0} \bigl(S(U\oplus U);\mathbb{Z}\bigr)=H^{p}(\mathbb{Z}_{4}; \mathbb{Z}) \bigr) \rightarrow\bigl( E_{2}^{p,0} \bigl( V_{4}^{\mathrm{conv}};\mathbb{Z} \bigr) =H^{p}( \mathbb{Z}_{4};\mathbb{Z}) \bigr) $$are identity maps.
The contradiction is obtained by tracking the behavior of the \(E_{r}^{2,0}(f)\) and \(E_{r}^{2,0}(g)\) images of T∈H ^{2}(ℤ_{4};ℤ) and 2T∈H ^{2}(ℤ_{4};ℤ) as r grows from 2 to 3. Explicitly,
and
Since the image of zero cannot be different from zero we have reached a contradiction. Thus, there are no ℤ_{4}equivariant maps in both cases:
The theorem is proved. □
6 Nonexistence of the Test Map, Proof of Theorem 1.2(3)–(4)
We conclude the proof of Theorem 1.2, using Proposition 3.2, by showing the following nonexistence theorem.
Theorem 6.1
There is no ℤ_{5}equivariant map \(V_{5}^{\mathrm{conv}}\rightarrow W_{5}\backslash T_{j}\), where j∈{3,4}.
Proof
Again we study the morphism of Serre spectral sequences associated with the Borel constructions of \(V_{5}^{\mathrm{conv}}\) and W _{5}∖T _{3}. The cohomology ring of the group ℤ_{5} with \(\mathbb{F}_{5}\) coefficients will be denoted by \(H^{\ast}(\mathbb{Z}_{5};\mathbb{F}_{5})\). It is known that
where degt=2, dege=1.
The Serre Spectral Sequence of \(V_{5}^{\mathrm{conv}}\times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z}_{5}\)
The E _{2}term of the sequence is given by \(E_{2}^{p,q}=H^{p}(\mathbb {Z}_{5},H^{q}(V_{5}^{\mathrm{conv}},\mathbb{F}_{5}))\). For q=1, from the Proposition 4.5 and [6, Example 2, p. 58] we have
The differentials in the spectral sequence are \(H^{\ast}(\mathbb {Z}_{5};\mathbb{F}_{5})\)module maps. Therefore \(d_{2}^{0,1}=0\). In particular, \(\alpha t\in H^{2}(\mathbb{Z}_{5};\mathbb{F}_{5})=E_{2}^{2,0}\) survive to the E _{∞}term for all \(\alpha\in\mathbb{F}_{5}\backslash\{0\}\).
The Serre Spectral Sequence of \(( W_{5}\backslash T_{j} ) \times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z}_{5}\)
First, we need to understand the cohomology of W _{5}∖T _{ j } with \(\mathbb{F}_{5}\) coefficients. According to Goresky–MacPherson formula
Here \(P_{\mathcal{A}_{j}}\) is an intersection poset of the arrangement \(\mathcal{A}_{j}\). The intersection posets \(P_{\mathcal{A}_{3}}\) and \(P_{\mathcal{A}_{4}}\) are isomorphic. Since the cohomology of the arrangement complement is completely determined by the intersection poset, we do not need the distinguish between the test spaces T _{3} and T _{4}.
From Hasse diagram of the poset \(P_{\mathcal{A}_{j}}\), Fig. 3, we have
Thus the E _{2}term of the Serre spectral sequence of \(( W_{5}\backslash T_{j} ) \times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z} _{5}\) is
The action of ℤ_{5} on W _{5}∖T _{ j } is free and therefore
The spectral sequence converges to \(H^{\ast}( ( W_{5}\backslash T_{j} ) \times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z}_{5};\mathbb{F} _{5})\) and so in the E _{∞}term everything for p+q>1 must vanish. Since our spectral sequence has only two nonzero rows and the only possibly nonzero differential is d _{2} it follows that \(d_{2}(x)=t\in H^{2}(\mathbb{Z}_{5};\mathbb{F}_{5})=E_{2}^{2,0}\). Here
denotes a suitably chosen generator. Thus the element \(t\in H^{2}(\mathbb{Z}_{4};\mathbb{Z})=E_{2}^{2,0}\) vanishes in the E _{3}term.
The Nonexistence of ℤ_{5}Equivariant Maps
Assume that there exists a ℤ_{5}equivariant map f:B _{5}∖A _{5}→W _{5}∖T _{ j }. Then f induces the maps between

Borel constructions, \((B_{5}\backslash A_{5})\times_{\mathbb {Z}_{5}}\mathrm{E}\mathbb{Z}_{5}\rightarrow ( W_{5}\backslash T_{j} ) \times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z}_{5}\),

equivariant cohomologies, \(f^{\ast}:H_{\mathbb{F}_{5}} ( W_{5}\backslash T_{j};\mathbb{F}_{5} ) \rightarrow H_{\mathbb{Z}_{5}} ( B_{5}\backslash A_{5};\mathbb{F}_{5} ) \), and

associated Serre spectral sequences,
$$ E_{r}^{p,q}(f):E_{r}^{p,q}(W_{5} \backslash T_j;\mathbb{F}_{5})\rightarrow E_{r}^{p,q} ( B_{5}\backslash A_{5}; \mathbb{F}_{5} ) $$such that on the 0row of the E _{2}term
is the identity map.
The contradiction is obtained by tracking the image of \(t\in H^{2}(\mathbb{Z}_{5};\mathbb{F}_{5})\) mapped by \(E_{r}^{2,0}(f)\) as r grows from 2 to 3. Explicitly,
The image of zero cannot be different from zero, thus we have reached a contradiction. There is no ℤ_{5}equivariant map \(V_{5}^{\mathrm{conv}}\rightarrow W_{5}\backslash T_{j}\) and the theorem is proved. □
7 Counter Examples, Proof of Theorem 1.3
7.1 Proof of Theorem 1.3(1)
We will prove more, namely, that given α _{ i }>0 (i=1,2,3,4) with \(\sum_{1}^{4}\alpha_{i}=1\), there are two probability measures μ and ν on ℝ^{2} such that no convex 4fan satisfies the conditions μ(σ _{ i })=ν(σ _{ i })=α _{ i } for all i=1,2,3,4.
This construction is from [2, Theorem 1.1.(i).(d)]. Let Q resp. T be the segment [(−2,0),(2,0)] and [(−1,1),(1,1)], and let ν be the uniform (probability) measure on Q. Also, let μ be the uniform (probability) measure on T for the time being. It will be modified soon. Assume there is a convex 4fan αpartitioning both measures. Then three consecutive rays intersect both Q and T and so the center of the 4fan cannot lie between the lines containing Q and T. It cannot be below the line containing T as otherwise one sector would meet Q in an interval too short to have the prescribed ν measure. The only way to make the 4fan convex is that there are three downward rays and the fourth ray points upward. The three downward rays split Q, resp. T into four intervals of ν and μmeasure α _{ i },α _{ i+1},α _{ i+2},α _{ i+3} in this order for some i=1,2,3,4 (the subscripts are meant modulo 4). Thus the lengths of these intervals are 4α _{ i },4α _{ i+1},4α _{ i+2},4α _{ i+3} on Q and 2α _{ i },2α _{ i+1},2α _{ i+2},2α _{ i+3} on T. So given α, the three downward rays, together with the center, are uniquely determined by the index i specifying that the starting interval is of length 4α _{ i } on Q. Let (z _{ i },1) be the point where the middle downward ray intersects T. This is four points corresponding to the four possible cases. Now we modify the measure μ a little. We move a small mass of μ from the left of (z _{ i },1) to the right, for each i=1,2,3,4. Each moving takes place in a very small neighborhood of (z _{ i },1). This changes only the position of the middle downward ray (in the modified measure μ), and the new ray will not pass through the intersection of the other two. We need to check that the four modifications are compatible. This is clearly the case when all the z _{ i } are distinct if the mass that has been moved is close enough to the corresponding (z _{ i },1). If two or more z _{ i } coincide, then the modification for one i will do for the others as well.
7.2 Proof of Theorem 1.3(2)
This construction is similar to the previous one. This time μ is the uniform measure on the interval T=[(−1,1),(1,1)], but Q, the support of ν=ν _{ h }, is the whole x axis and the distribution function of ν _{ h }, F=F _{ h }, which depends on a parameter h∈(0,1), is given explicitly as
Note that F _{ h } is concave resp. convex on [0,∞) and (−∞,0]. The following properties of F _{ h } are easily checked:

(i)
no line intersects the graph of F _{ h } in more than three points,

(ii)
no line intersects the graph of the convex (concave) part of F _{ h } in more than two points,

(iii)
F _{ h } is symmetric, in the sense that F _{1−h }(−x)=1−F _{ h }(x) for all h and x.
We are going to show that, for some h∈(0,1), the measures μ and ν _{ h } satisfy the requirements.
Assume that this is false, that is, for each h∈(0,1) there is a convex 5fan equipartitioning μ and ν(σ _{ i })=ν(σ _{ i+1})=ν(σ _{ i+2})=ν(σ _{ i+3})>0. As we have seen before, the center of the 5fan cannot lie between Q and T. Consequently four consecutive rays intersect T at points (−0.6,1),(−0.2,1),(0.2,1),(0.6,1) and then intersect Q at points x,x+y,x+2y,x+3y, say. These four points split Q into five intervals \(I_{1}=(\infty,x),\;I_{2}=(x,x+y),\;I_{3}=(x+y,x+2y),\;I_{4}=(x+2y,x+3y), \;I_{5}=(x+3y,\infty)\). Because of symmetry (iii) it suffices to consider three cases:
 Case 1:

when ν _{ h }(I _{1})=ν _{ h }(I _{2})=ν _{ h }(I _{3})=ν _{ h }(I _{4}),
 Case 2:

when ν _{ h }(I _{1})=ν _{ h }(I _{2})=ν _{ h }(I _{3})=ν _{ h }(I _{5}),
 Case 3:

when ν _{ h }(I _{1})=ν _{ h }(I _{2})=ν _{ h }(I _{4})=ν _{ h }(I _{5}).
We show that there is a small h _{0}>0 such that all three cases fail for h∈(0,h _{0}) and for h∈(1−h _{0},1). This is needed because of symmetry.
 Case 1:

This case is the simplest: the points (x+iy,F(x+iy)), i=0,1,2,3 are on the same line contradicting property (i).
 Case 2:

Now x<0 as otherwise the points (x+iy,F(x+iy)), i=0,1,2 would be on the same line contradicting property (ii). Similarly x+2y>0. The conditions say that 2F _{ h }(x)=F _{ h }(x+y), 3F _{ h }(x)=F _{ h }(x+2y) and F _{ h }(x)=1−F _{ h }(x+3y).
If 0∈(x,x+y], then we have
$$6he^x=33(1h)e^{xy}=22(1h)e^{x2y}=6(1h)e^{x3y}. $$Here the middle equation fails to hold when h is close to 1. When h is close to 0, then x+y and x+2y have to be close to 0; consequently x+3y is also close to 0. But then F _{ h }(x) is close to 0 and 1−F _{ h }(x+3y) is close to 1 so they cannot be equal.
If 0∈(x+y,x+2y), then we have
$$6he^x=3he^{x+y}=22(1h)e^{x2y}=6(1h)e^{x3y}. $$The first equation shows that e ^{y}=2. Then the last equation fails to hold when h is close to 1. We also have he ^{x}=(1−h)e ^{−x}/8, or 8he ^{2x}=1−h which cannot hold when h is close to 0.
 Case 3:

Again, x<0 and x+3y>0 follow from (ii). By (iii) it suffices to consider the case 0∈[x+y,x+3y]. Then 2F _{ h }(x)=F _{ h }(x+y) implies, again, that y=log2. Then, just as before, F _{ h }(x)=1−F _{ h }(x+3y) gives 8he ^{2x}=1−h. This cannot hold for h close to 0. When h is close to 1, then x→−∞ and x+3y>0 is not possible since y=ln2.
7.3 Proof of Theorem 1.3(3)
We construct two probability measures μ and ν on ℝ^{2} such that there is no t∈(0,1/3) and no convex 4fan in ℝ^{2} satisfy the conditions μ(σ _{ i })=ν(σ _{ i })=t for three consecutive subscripts.
This is similar to the example in Sect. 7.1. T is the same as there, μ is the uniform measure on T, and Q is again the interval [(−2,0),(2,0)]. But this time the measure ν has a continuous distribution function F(x), defined on x∈[−2,2]. We assume that F(x) is a strictly concave function with F(−2)=0 and F(2)=1 (of course). This implies that no line intersects the graph of F in more than two points. Assume there is t>0 and a convex 4fan with μ(σ _{ i })=ν(σ _{ i })=t for three subscripts i. Then for the fourth subscript j, μ(σ _{ j })=ν(σ _{ j })=1−3t.
As we have seen above, the center of the 4fan cannot be between the lines of T and Q. Consequently three consecutive rays intersect both Q and T. Let x,y,z be the intersection points of these rays with Q in this order from left to right. The conditions on μ and ν imply that either y−x=λt, z−y=λt and F(y)−F(x)=t, F(z)−F(y)=t, or y−x=λt, z−y=λ(1−3t) and F(y)−F(x)=t, F(z)−F(y)=1−3t, or y−x=λ(1−3t), z−y=λt and F(y)−F(x)=1−3t, F(z)−F(y)=t with a suitable positive λ. In all three cases
So the points (x,F(x)),(y,F(y)),(z,F(z)) from the graph of F are on the same line, contrary to the assumption of concavity of F.
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Acknowledgements
I. Bárány was partially supported by ERC Advanced Research Grant no 267165 (DISCONV), and by Hungarian National Research Grants No K84767 and NK78439.
The research of P. Blagojević leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP7/20072013) / ERC Grant agreement no. 247029SDModels. Also supported by the grant ON 174008 of the Serbian Ministry of Education and Science.
A.D. Blagojević was supported by the grant ON 174008 of the Serbian Ministry of Education and Science.
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Bárány, I., Blagojević, P. & Dimitrijević Blagojević, A. Functions, Measures, and Equipartitioning Convex kFans. Discrete Comput Geom 49, 382–401 (2013). https://doi.org/10.1007/s0045401294678
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DOI: https://doi.org/10.1007/s0045401294678