1 Introduction

Let S 2 be the unit sphere in ℝ3. A k-fan on the sphere S 2 is formed by a point xS 2 and k≥3 great semicircles 1,…, k , starting from x and ending at −x, listed in anticlockwise order when seen from x. The spherical sector σ i is delimited by i and i+1 and its interior is disjoint from all j . The k-fan 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 k-fan (x; 1,…, 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⊂ℝ2 of positive area does there exist a convex k-partition 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()=0 for all great circles , and f is a continuous function defined on the sectors in S 2. Then there is a convex 3-fan (x; 1, 2, 3) equipartitioning μ such that

$$f(\sigma_{1})=f(\sigma_{2})= f(\sigma_{3}).$$
(1)

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 5-fans. Here are our main results.

Theorem 1.2

Assume μ is a Borel probability measure on S 2 with μ()=0 for all great circles , and f is a continuous function defined on the sectors in S 2. Then

1. (1)

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

2. (2)

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

3. (3)

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

4. (4)

there is a convex 5-fan 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 k-fans is known to be V 2(ℝ3) the Stiefel manifold of orthogonal two-frames in ℝ3. The configuration space V conv is going to be the so called convex part of the set of all equipartitioning k-fans. 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 (ℤ4 and ℤ5-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 non-existence of such a test map implies Theorem 1.2.

• The non-existence of such ℤ4 and ℤ5-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 4-fan resp, 5-fan 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 ℝ2 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. (1)

there is no convex 4-fan simultaneously equipartitioning μ and ν.

2. (2)

there is no convex 5-fan 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. (3)

there is no convex 4-fan 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 k-fan in the plane is formed by a point x∈ℝ2 and k halflines 1,…, 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 i and i+1. The k-fan 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 ℝ2 is embedded in ℝ3 as the tangent plane to S 2 at the point (0,0,1). Let ρ:{(x 1,x 2,x 3)∈S 2x 3>0}→ℝ2 be the central projection. The map ρ lifts any nice measure in the plane to a nice measure on the sphere. Also, a k-fan in the plane lifts to a k-fan on the sphere and a k-fan on the sphere projects to a k-fan 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()=0 for all great circles . 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 k-Fans

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()=0 for all great circles . For k≥3 consider the following family of k-fans on S 2:

$$F_{k}=\biggl\{(x;\ell_{1},\ldots,\ell_{k})\mid \mu(\sigma_{1})=\cdots=\mu(\sigma_{k})=\frac{1}{k} \biggr\}.$$

For (x; 1,…, k )∈F k , let y= 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 ( 1,…, k ), Fig. 1(B). Thus, z=x×y, where × denotes the cross product. The correspondence (x; 1,…, k )↦(x,y) induces a homeomorphism between the family of fans F k and the Stiefel manifold V 2(ℝ3). Let ℤ k =〈ε〉 be a cyclic group. There is natural free ℤ k -action on F k given by

$$\varepsilon\cdot(x;\ell_{1},\ldots,\ell_{k})=(x; \ell_{2},\ldots,\ell_{k},\ell_{1}).$$

The main objective of this section is to describe the subfamily of all convex k-fans contained in F k as a ℤ k -invariant subspace.

Let p:(F k =V 2(ℝ3))→S 2 denotes the S 1 fibration given by (x; 1,…, 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)={vS 2vz≤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 1-dimensional embedded submanifold of S 2, [5, Corollary 7.4, p. 84].

Lemma 2.1

For the fan (x; 1,…, k )=(x;σ 1,…,σ k )=(x,y), z=x×y=p(x,y), the sector σ k is not convex if

$$( h\circ p ) (x;\sigma_{1},\ldots,\sigma_{k})=h(z)< \frac{1}{k}.$$

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 k-fans:

$$(x;\sigma_{1},\ldots,\sigma_{k})\quad \text{is convex} \quad \Longleftrightarrow\quad (\forall i\geq0) ( h\circ p ) \bigl (\varepsilon^{i} ( x;\sigma_{1},\ldots,\sigma_{k} ) \bigr)\geq \frac{1}{k}$$

or

$$(x;\sigma_{1},\ldots,\sigma_{k})\quad \text{is not convex}\quad \Longleftrightarrow\quad (\exists i\geq0) ( h\circ p ) \bigl (\varepsilon^{i} ( x;\sigma_{1},\ldots,\sigma_{k} ) \bigr)< \frac{1}{k}.$$

Lemma 2.2

After a possible rotation of the measure μ, the circle

$$C=\bigl\{(e_{3},y)\in V_{2}\bigl(\mathbb{R}^{3} \bigr)\mid y\in S\bigl( \bigl( \mathrm{span}\{e_{3}\} \bigr)^{\bot}\bigr)\bigr\}\subset V_{2}\bigl( \mathbb{R}^{3}\bigr)$$

is invariant under the k -action and every point (e 3,y)∈C defines a convex k-fan.

Proof

The following result of Dolnikov [7] and Živaljević, Vrećica [10] is needed.

For nd probability measures in ℝd, there exists a (n−1)-dimensional affine subspace such that the measure of every halfspace containing this affine subspace is at least $$\frac{1}{d+2-n}$$ 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 yS((span{e 3})). Thus, the circle C is invariant under the ℤ k -action and each (e 3,y)∈C defines a convex k-fan. □

The point $$\frac{1}{k}$$ is the regular value of the function h. Thus, $$h^{-1}(\{\frac{1}{k}\})$$ is an 1-dimensional 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 yp(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(ℝ3)→V 2(ℝ3). Hence U i ,εU i ,…,ε k−1U i are solid tori and T i ,εT i ,…,ε k−1T 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. (1)

The cycle C is disjoint from all solid tori U i ,εU i ,…,ε k−1U i , i∈[m].

2. (2)

ε αT i ε βT j ≠∅ ⟹ i=j and α=β.

3. (3)

The tori U i ,εU i ,…,ε k−1U i are pairwise disjoint, i∈[m].

Proof

1. (1)

Let us assume that Cε αU i ≠∅. Since εC=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. (2)

Let α=β and ε αT i ε αT 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<βk. Without losing the generality, we can assume that α=0. Let (x; 1,…, k )∈T i ε βT j ≠∅. Then (x; 1,…, k )∈T i , ε β⋅(x; 1,…, k )=(x; kβ+1,…, kβ )∈T j and consequently σ k and σ kβ are hemispheres. This cannot be: a contradiction.

3. (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(ℝ3),ℤ)=0, every torus ε αT i splits V 2(ℝ3) into two disjoint parts. Let us assume that ε αU i ε βU i ≠∅, 0≤α<βk. Again, it is enough to consider the case α=0. Since T i ε βT i =∅ and H 2(V 2(ℝ3),ℤ)=0 then the complement V 2(ℝ3)∖(T i ε βT i ) has three components. The intersection U i ε βU i is one of these three components with the boundary T i or ε βT i or T i ε βT i . We discuss these three cases separately.

1. (a)

Let (U i ε βU i )=T i U i . Then U i ε βU 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 βU i and so T i =ε βT i , contradiction.

2. (b)

Let (U i ε βU i )=ε βT i ε βU i . Then ε βU i U i and consequently ε βU i =U i . Thus ε βT i =T i gives the contradiction.

3. (c)

Let (U i ε βU i )=T i ε βT i . Then ε βT i U i and so ε βU i is contained in U i or its complement (ε βU i )c is contained in U i . Thus either U i ε βU i =U i or U i ε βU i ⊇(ε βU i )cε βU i =V 2(ℝ3). The later is not possible since C is disjoint from both U i and ε βU i . Therefore ε βU i U i and consequently ε βU i =U i and ε βT i =T i , contradiction.

□

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≤rm. 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 $$\mathcal{O}(U_{i}):=U_{i}\cup ( \varepsilon\cdot U_{i} ) \cup\cdots\cup ( \varepsilon^{k-1}\cdot U_{i} )$$.

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, αβ, such that ε αU k(i)ε βU k(j)≠∅. Without losing the generality we can assume that α=0. There are two separate cases:

1. (1)

Let β=0. Then

2. (2)

Let β≠0. Since T i ε βT k(j)=∅ we see that the complement V 2(ℝ3)∖(T i ε βT j ) has three components. One of them is U i ε βU j with boundary either T j or ε βT j or T i ε βT j . We discuss all three possibilities:

1. (a)

Let (U i ε βU j )=T i U i . Then U i ε βU j and consequently the orbit $$\mathcal{O}(U_{i})$$ is contained in the orbit $$\mathcal{O}(U_{j})$$.

2. (b)

Let (U i ε βU j )=ε βT j ε βU k(j). Then ε βU j U i and consequently the orbit $$\mathcal{O}(U_{j})$$ is contained in the orbit $$\mathcal{O}(U_{i})$$.

3. (c)

Let (U j ε βU j )=T i ε βT j . Consequently T i ε βU j and ε βT j U i . Therefore ε βU j U i or (ε βU j )cU i . Since (ε βU j )cU i implies that U i ε βU j =V 2(ℝ3), and this is not possible, we conclude that ε βU j U i and consequently the orbit $$\mathcal{O}(U_{j})$$ is contained in the orbit $$\mathcal{O}(U_{i})$$.

□

Consider the following subset of the family of all equipartitioning k-fans on the sphere S 2:

$$V^{\mathrm{conv}}=V_{2}\bigl(\mathbb{R}^{3}\bigr)\backslash \biggl( \bigcup_{i\in [r]}\bigcup _{\alpha\in\{0,\ldots,k-1\}}\varepsilon^{\alpha}\cdot U_i \biggr) .$$

The previous results imply that

1. (1)

ε αU i , for all i∈[r] and α∈{0,…,k−1}, are pairwise disjoint closed solid tori,

2. (2)

every (x; 1,…, k )=(x,y)∈V conv is a convex k-fan,

3. (3)

CV conv and V conv are ℤ k -invariant subspaces of V 2(ℝ3).

Therefore the set V conv will be called the convex part of V 2(ℝ3). Notice that, as Fig. 2 indicates, there might be some convex k-fans 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 5-fans 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(ℝ3) and

$$A_{4}= \biggl( \bigcup_{i\in\lbrack r_{4}]}\bigcup _{\alpha\in \{0,1,2,3\}}\varepsilon^{\alpha}\cdot U_{k(i)} \biggr)\quad\text{and}\quad A_{5}= \biggl( \bigcup _{i\in\lbrack r_{5}]}\bigcup_{\alpha\in \{0,1,2,3,4\}} \varepsilon^{\alpha}\cdot U_{k(i)} \biggr) .$$

Notice that both spaces A 4 and A 5 are homotopy equivalent to disjoint unions of 1-dimensional spheres.

Some Real ℤ4 and ℤ5-Representations

Let ℝ4 be a real ℤ4 -representation equipped with the following ℤ4-action ε⋅(x 1,x 2,x 3,x 4)=(x 2,x 3,x 4,x 1). The subspaces

are ℤ4-invariant subspace or real ℤ4-representations. It is not hard to prove that there is an isomorphism of real ℤ4-representations W 4 UV.

Similarly, consider ℝ5 as a real ℤ5 -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

$$W_{5}=\bigl\{(x_{1},x_{2},x_{3},x_{4},x_{5}) \mid x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0 \bigr\}\subset\mathbb{R}^{5}$$

is ℤ5-invariant and therefore a real subrepresentation.

Test Space and Test Map for 4-Fans

Let f and $$g:V_{4}^{\mathrm{conv}}\rightarrow\mathbb{R}$$ be continuous function on the sectors of the convex 4-fan. 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

(2)

Proposition 3.1

1. (1)

If there is no4-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. (2)

If there is no4-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. (1)

If there is no ℤ4-equivariant map $$V_{4}^{\mathrm{conv}}\rightarrow W_{4}\backslash T_{1}$$, then there exists a convex 4-fan 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 ℤ4-equivariant map $$V_{4}^{\mathrm{conv}}\rightarrow ( W_{4}\oplus W_{4} ) \backslash T_{2}$$, then there is a convex 4-fan 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. (2)

The existence of ℤ4-homotopies,

and (1) imply the claim (2).

□

Test Space and Test Map for 5-Fans

Let $$h:V_{5}^{\mathrm{conv}}\rightarrow\mathbb{R}$$ be a continuous function on the sectors of 5-fans. 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}⊆ℝ5.

There are two test spaces T 3 and T 4 we are interested in. They are unions of the minimal ℤ5-invariant arrangements $$\mathcal{A}_{3}$$ and $$\mathcal{A}_{4}$$ containing the linear subspace L 3W 5 and L 4W 5, respectively, given by

(3)
(4)

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. (1)

If there is no5-equivariant map $$V_{5}^{\mathrm{conv}}\rightarrow W_{5}\backslash T_{3}$$, then part 3 of Theorem 1.2 holds.

2. (2)

If there is no5-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[ℤ4] and $${\mathbb {F}}_{5}[{\mathbb {Z}}_{5}]$$-module, respectively. This will turn out to be an important step in the proof of the non-existence of the appropriate ℤ4 and ℤ5-equivariant maps, Sects. 5 and 6.

4.1 Cohomology of $$V_{4}^{\mathrm{conv}}$$

We establish the following isomorphisms of ℤ[ℤ4]-modules:

$$H^{0}\bigl(V_{4}^{\mathrm{conv}};\mathbb{Z}\bigr)=\mathbb{Z}\quad \text{and}\quad H^{1}\bigl(V_{4}^{\mathrm{conv}};\mathbb{Z}\bigr)\cong \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}}.$$
(5)

Proposition 4.1

The cohomology with thecoefficients of the pair (B 4,A 4) is given by

H^{i}(B_{4},A_{4};\mathbb{Z})\cong \left \{ \begin{array}{l@{\quad }l} ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus k}/_{(1+\varepsilon+\varepsilon ^{2}+\varepsilon^{3})^{\oplus k}\mathbb{Z}}\\ \noalign{\vspace*{3pt}} \quad{} \cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus(k-1)}\oplus ( \mathbb{Z}[\mathbb{Z}_{4}]/_{(1+\varepsilon+\varepsilon^{2}+\varepsilon^{3})\mathbb{Z}} ), & i=1, \\ M, & i=2, \\ \mathbb{Z}, & i=3, \\ 0 , & \text{\textit{otherwise}}.\end{array} \right .

where ℤ[ℤ4]-module M is a part of the following exact sequence of ℤ[ℤ4]-modules:

$$0\longrightarrow\bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus k} \longrightarrow M\longrightarrow\mathbb{Z}_{2}\longrightarrow0.$$
(6)

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

$$\varPhi_{0}(a)= \bigl( a+\varepsilon\cdot a+\varepsilon^2 \cdot a+\varepsilon^3\cdot a \bigr) \oplus\cdots\oplus\bigl( a+ \varepsilon\cdot a+\varepsilon^2\cdot a+\varepsilon^3\cdot a \bigr) .$$

Thus Φ 0 is an injection. Since H 1(B 4)=0, there is a short exact sequence

$$0\longrightarrow\mathbb{Z}\overset{\varPhi_{0}}{\longrightarrow} \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}}\longrightarrow H^{1}(B_{4},A_{4})\longrightarrow0$$

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)=ℤ2 we obtain an exact sequence

$$0\longrightarrow\bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}} \longrightarrow H^{2}(B_{4},A_{4})\longrightarrow \mathbb{Z}_{2}\longrightarrow0.$$

Finally, the fact that H 3(B 4)=ℤ gives the exact sequence

$$0\longrightarrow H^{3}(B_{4},A_{4}) \longrightarrow\mathbb{Z}\longrightarrow0$$

and the isomorphism H 3(B 4,A 4)≅ℤ. □

Corollary 4.2

H_{i}(B_{4}\backslash A_{4};{\mathbb {Z}})\cong \left \{ \begin{array}{l@{\quad }l} ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}/_{(1+\varepsilon +\varepsilon^{2}+\varepsilon^{3})^{\oplus r_{4}}\mathbb{Z}}\\ \noalign{\vspace*{3pt}} \quad{}\cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus(r_{4}-1)}\oplus ( \mathbb{Z}[ \mathbb{Z}_{4}]/_{(1+\varepsilon+\varepsilon^{2}+\varepsilon^{3})\mathbb{Z}} ) , & i=2, \\ M , & i=1, \\ \mathbb{Z} , & i=0 ,\\ 0 , & \text{\textit{otherwise}}. \end{array} \right .

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 4A 4 with the cohomology of the pair (B 4,A 4), i.e.,

$$H_{i}(B_{4}\backslash A_{4};{\mathbb {Z}})\cong H^{3-i}(B_4,A_4;{\mathbb {Z}}).$$

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[ℤ4]-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 ℤ4-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[ℤ4]-modules,

the exact sequence transforms into

$$0\longrightarrow\mathrm{Hom}(M,\mathbb{Z})\longrightarrow\bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}} \longrightarrow \mathbb{Z}_{2} \longrightarrow\mathrm{Ext}(M,\mathbb{Z}) \longrightarrow0 .$$

First notice that rank(Hom(M,ℤ))≤4r 4 and therefore

$$\mathrm{rank} \bigl( \mathrm{Hom}(M,\mathbb{Z}) \bigr) =\mathrm {rank} \bigl( \mathrm{Hom}\bigl(\mathrm{Free}(M),\mathbb{Z}\bigr)\bigr) = \mathrm{rank} \bigl( \mathrm{Free}(M) \bigr) =4r_{4}.$$

Since the exact sequence (6) gives an inclusion of ℤ[ℤ4]-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[ℤ4]-modules we can conclude that $$\mathrm{Free}(M) \cong ( \mathbb{Z}[\mathbb{Z}_{4}] )^{\oplus r_{4}}$$. Thus we have an isomorphism of ℤ[ℤ4]-modules

$$\mathrm{Hom}(M,\mathbb{Z})\cong\mathrm{Hom}\bigl(\mathrm{Free}(M), \mathbb{Z}\bigr)\cong\mathrm{Hom}\bigl( \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}},\mathbb{Z}\bigr)\cong \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}}.$$

□

Finally, we have to verify the isomorphisms (5) of ℤ[ℤ4]-modules.

Corollary 4.4

H 0(B 4A 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 4A 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 4A 4;ℤ),ℤ)=Ext(ℤ,ℤ)=0, the exact sequence gives the isomorphism

$$H^{1}(B_{4}\backslash A_{4};\mathbb{Z})\cong \mathrm{Hom} \bigl( H_{1}(B_{4}\backslash A_{4}; \mathbb{Z}),\mathbb{Z} \bigr) =\mathrm{Hom}(M,\mathbb{Z})\cong \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}}.$$

□

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:

$$\everymath{\displaystyle} \begin{array}{l} H^{0} \bigl(V_{5}^{\mathrm{conv}};\mathbb{F}_{5}\bigr)=\mathbb{F}_{5} \quad \text{and}\quad H^{1} \bigl(V_{5}^{\mathrm{conv}};\mathbb{F}_{5}\bigr)\cong \bigl( \mathbb{F}_{5} [\mathbb{Z}_{5}] \bigr)^{\oplus r_{5}}.\end{array}$$

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,

$$H^{1}\bigl(V_{5}^{\mathrm{conv}};\mathbb{F}_{5} \bigr)\cong H_{2}(B_{5},A_{5}; \mathbb{F}_{5})\cong H_{1}(A_{5}; \mathbb{F}_{5})\cong\bigl( \mathbb{F}_{5} [\mathbb{Z}_{5}] \bigr)^{\oplus r_{5}}.$$

□

5 Non-existence 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 no4-equivariant map

1. (i)

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

2. (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 4A 4, S(V) and S(UU). We denote the cohomology of the group ℤ4 with ℤ coefficients by H (ℤ4;ℤ). It is well known that

$$H^{\ast}(\mathbb{Z}_{4};\mathbb{Z})=\mathbb{Z}[T]/_{\langle 4T\rangle},$$

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

$$E_{2}^{p,1}=H^{p}\bigl(\mathbb{Z}_{4}; \bigl( \mathbb{Z}[\mathbb{Z}_{4}] \bigr)^{\oplus r_{4}}\bigr)=\left \{ \begin{array}{l@{\quad}l} \mathbb{Z}^{\oplus r_{4}}, & p=0, \\ 0, & p\neq0. \end{array} \right .$$

Since the differentials in the spectral sequence are H (ℤ4;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 ℤ4 action on S(V) is orientation preserving, hence the coefficients are untwisted. The action of ℤ4 on S(V)≈S 1 is free and therefore

$$S(V)\times_{\mathbb{Z}_{4}}\mathrm{E}\mathbb{Z}_{4}\simeq S^{1}/\mathbb{Z}_{4}\quad \Rightarrow\quad H^{i}\bigl(S(V)\times_{\mathbb{Z}_{4}}\mathrm{E}\mathbb{Z}_{4};\mathbb{Z}\bigr)=0\quad\text{for}\ i>1.$$

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 non-zero rows and the only possibly non-zero 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 LH 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 1-dimensional complex representation of ℤ4 induced by 1↦e /2. Then UUV V. Following [1, Sect. 8, p. 271 and Appendix, p. 285] we deduce the first Chern class of the ℤ4-representation UU

$$c_{1}(U\oplus U)=c_{1}(V\otimes_{\mathbb{C}}V)=c_{1}(V)+c_{1}(V)=T+T=2T \in H^{2}(\mathbb{Z}_{4};\mathbb{Z}).$$

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 LH 1(S(UU);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 Non-existence of Both ℤ4-Equivariant Maps

Assume that in both cases there exists a ℤ4-equivariant,

1. (i)

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

2. (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 0-row 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 TH 2(ℤ4;ℤ) and 2TH 2(ℤ4;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 ℤ4-equivariant maps in both cases:

$$V_{4}^{\mathrm{conv}}\rightarrow S(V) ,\qquad V_{4}^{\mathrm{conv}}\rightarrow S(U\oplus U).$$

The theorem is proved.  □

6 Non-existence 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 non-existence theorem.

Theorem 6.1

There is no5-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 5T 3. The cohomology ring of the group ℤ5 with $$\mathbb{F}_{5}$$ coefficients will be denoted by $$H^{\ast}(\mathbb{Z}_{5};\mathbb{F}_{5})$$. It is known that

$$H^{\ast}(\mathbb{Z}_{5};\mathbb{F}_{5})= \mathbb{F}_{5}[t]\otimes\bigl( \mathbb{F}_{5}[e]/e^{2} \bigr),$$

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

$$E_{2}^{p,1}=H^{p}\bigl(\mathbb{Z}_{5}; \bigl( \mathbb{F}_{5}[\mathbb{Z}_{5}] \bigr)^{\oplus r_{5}}\bigr)=\left \{ \begin{array}{l@{\quad}l} \mathbb{F}_{5}^{\oplus r_{5}} , & p=0, \\ 0 , & p\neq0. \end{array} \right .$$

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 5T j with $$\mathbb{F}_{5}$$ coefficients. According to Goresky–MacPherson formula

$$\tilde{H}^{i}(W_{5}\backslash T_j; \mathbb{F}_{5})\cong\bigoplus_{p\in P_{\mathcal{A}_j}} \tilde{H}_{2-i-\dim p} \bigl(\Delta\bigl(( P_{\mathcal{A}_j})_{<p} \bigr);\mathbb{F}_{5} \bigr).$$

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

$$H^{i}(W_{5}\backslash T_j; \mathbb{F}_{5})\cong \left \{ \begin{array}{l@{\quad}l} \mathbb{F}_{5} ,& i=0, \\ \mathbb{F}_{5}[\mathbb{Z}_{5}]\oplus\mathbb{F}_{5}[\mathbb {Z}_{5}]\oplus \mathbb{F}_{5} , & i=1 ,\\ 0 , & i\neq0,1 . \end{array} \right .$$

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 ℤ5 on W 5T j is free and therefore

$$( W_{5}\backslash T_j ) \times_{_{\mathbb{Z}_{5}}} \mathrm{E}\mathbb{Z}_{5}\simeq( W_{5}\backslash T_j )/\mathbb{Z}_{5}\quad \Rightarrow\quad H^{i}\bigl( ( W_{5}\backslash T_j ) \times_{_{\mathbb{Z}_{5}}}\mathrm{E}\mathbb{Z}_{5};\mathbb{F}_{5}\bigr)=0\quad \text{for}\ i>1.$$

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 non-zero rows and the only possibly non-zero 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

$$x\in H^{0}(\mathbb{Z}_{5};\mathbb{F}_{5}) \subset H^{0}\bigl(\mathbb{Z}_{5};\mathbb{F}_{5}[\mathbb{Z}_{5}]\bigr)\oplus H^{0} \bigl(\mathbb{Z}_{5};\mathbb{F}_{5}[\mathbb{Z}_{5}]\bigr)\oplus H^{0}(\mathbb{Z}_{5}; \mathbb{F}_{5})=E_{2}^{0,1}$$

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 Non-existence of ℤ5-Equivariant Maps

Assume that there exists a ℤ5-equivariant map f:B 5A 5W 5T 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 0-row 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,

\everymath{\displaystyle} \begin{array}{l} E_{2}^{2,0}(W_{5}\backslash T_{3};\mathbb{F}_{5})\ni t\ \overset{E_{2}^{2,0}(f)}{\longmapsto}\ t\in E_{2}^{2,0} ( B_{5}\backslash A_{5}; \mathbb{F}_{5} )\\ \noalign{\vspace*{3pt}} E_{3}^{2,0}(W_{5}\backslash T_{3};\mathbb{F}_{5})\ni0\ \overset{E_{3}^{2,0}(f)}{\longmapsto}\ t\in E_{3}^{3,0} ( B_{5}\backslash A_{5}; \mathbb{F}_{5} ) . \end{array}

The image of zero cannot be different from zero, thus we have reached a contradiction. There is no ℤ5-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 ℝ2 such that no convex 4-fan 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 4-fan α-partitioning both measures. Then three consecutive rays intersect both Q and T and so the center of the 4-fan 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 4-fan 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 μ-measure 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:

1. (i)

no line intersects the graph of F h in more than three points,

2. (ii)

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

3. (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 5-fan 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 5-fan 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

$$6he^x=3-3(1-h)e^{-x-y}=2-2(1-h)e^{-x-2y}=6(1-h)e^{-x-3y}.$$

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

$$6he^x=3he^{x+y}=2-2(1-h)e^{-x-2y}=6(1-h)e^{-x-3y}.$$

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 ℝ2 such that there is no t∈(0,1/3) and no convex 4-fan in ℝ2 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 4-fan 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 4-fan 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 yx=λt, zy=λt and F(y)−F(x)=t, F(z)−F(y)=t, or yx=λt, zy=λ(1−3t) and F(y)−F(x)=t, F(z)−F(y)=1−3t, or yx=λ(1−3t), zy=λt and F(y)−F(x)=1−3t, F(z)−F(y)=t with a suitable positive λ. In all three cases

$$\frac{F(y)-F(x)}{y-x}=\frac{F(z)-F(y)}{z-y}=\frac{1}{\lambda} .$$

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.