Tetrahedra on deformed spheres and integral group cohomology

arXiv:0808.3841v1 [math.AT] 28 Aug 2008 Tetrahedra on deformed spheres and integral group cohomology Pavle V. M. Blagojevi´c∗ Mathematiˇcki Institut Knez Michailova 35/1 11001 Beograd, Serbia pavleb@mi.sanu.ac.yu G¨unter M. Ziegler∗∗ Inst. Mathematics, MA 6-2 TU Berlin D-10623 Berlin, Germany ziegler@math.tu-berlin.de August 28, 2008 Abstract We show that for every injective continuous map f : S2 →R3 there are four distinct points in the image of f such that the convex hull is a tetrahedron with the property that two opposite edges have the same length and the other four edges are also of equal length. This result represents a partial result for the topological Borsuk problem for R3. Our proof of the geometrical claim, via Fadell– Husseini index theory, provides an instance where arguments based on group cohomology with integer coefficients yield results that cannot be accessed using only field coefficients. 1 Introduction The motivation for the study of the existence of particular types of tetrahedra on deformed 2-spheres is twofold. The topological Borsuk problem, as considered in [7], along with the square peg problem [6] inspire the search for possible polytopes with nice metric properties whose vertices lie on the continuous images of spheres. Beyond their intrinsic interest, these problems can be used as testing grounds for tools from equivariant topology, e.g. for comparing the strength of Fadell–Husseini index theory with ring resp. field coefficients. Figure 1: D8-invariant tetrahedra on deformed sphere S2 The following theorem will be proved through the use of Fadell–Husseini index theory with coefficients in the ring Z. It is also going to be demonstrated that Fadell–Husseini index theory with coefficients in field F2 has no power in this instance (Section 4.1). ∗Supported by the grant 144018 of the Serbian Ministry of Science and Technological development ∗∗Partially supported by the German Research Foundation DFG 1 Theorem 1.1. Let f : S2 →R3 an injective continuous map. Then its image contains vertices of a tetrahedron that has the symmetry group D8 of a square. That is, there are four distinct points ξ1, ξ2, ξ3 and ξ4 on S2 such that d(f(ξ1), f(ξ2)) = d(f(ξ2), f(ξ3)) = d(f(ξ3), f(ξ4)) = d(f(ξ4), f(ξ1)) and d(f(ξ1), f(ξ3)) = d(f(ξ2), f(ξ4)). Remark 1.2. The proof is not going to use any properties of R3 except that it is a metric space. Thus in the statement of the theorem, R3 can be replaced by any metric space (M, d). Remark 1.3. Unfortunately, the methods used for the proof of Theorem 1.1 do not imply any conclusion when applied to the square peg problem (see Section 4.2). On the other hand, if the square peg problem could be solved for the continuous Jordan curves, then it would imply the result of Theorem 1.1. 2 Introducing the equivariant question Let f : S2 →R3 be an injective continuous map. Denote by D8 the symmetry group of a square, that is, the 8-element dihedral group D8 = ⟨ω, j | ω4 = j2 = 1, ωj = jω3⟩. A few D8-representations. The vector spaces U4 = {(x1, x2, x3, x4) ∈R4 | x1 + x2 + x3 + x4 = 0}, U2 = {(x1, x2) ∈R2 | x1 + x2 = 0} are D8-representations with actions given by (a) for (x1, x2, x3, x4) ∈U4: ω · (x1, x2, x3, x4) = (x2, x3, x4, x1), j · (x1, x2, x3, x4) = (x3, x2, x1, x4), (b) for (x1, x2) ∈U2 : ω · (x1, x2) = (x2, x1), j · (x1, x2) = (x2, x1), The configuration space. Let X = S2 × S2 × S2 × S2 and let Y be the subspace given by Y =  (x, y, x, y) | x, y ∈S2 ≈S2 × S2. The configuration space to be considered is the space Ω:= X\Y. Let a D8-action on X be induced by ω · (ξ1, ξ2, ξ3, ξ4) = (ξ2, ξ3, ξ4, ξ1), j · (ξ1, ξ2, ξ3, ξ4) = (ξ4, ξ3, ξ2, ξ1), for (ξ1, ξ2, ξ3, ξ4) ∈X. 2 A test map. Let τ : Ω→U4 × U2 be a map defined for (ξ1, ξ2, ξ3, ξ4) ∈X by τ(ξ1, ξ2, ξ3, ξ4) = (d12 −∆ 4 , d23 −∆ 4 , d34 −∆ 4 , d41 −∆ 4 ) × (d13 −Φ 2 , d24 −Φ 2 ) (1) where dij := d(f (ξi) , f (ξj)) and ∆= d12 + d23 + d34 + d14, Φ = d13 + d24. With the D8-actions introduced above the test map τ is D8-equivariant. The following proposition connects our set-up with the tetrahedron problem. Proposition 2.1. If there is no D8-equivariant map α : Ω→(U4 × U2)\({0} × {0}) (2) then Theorem 1.1 follows. Proof. If there is no D8-equivariant map Ω→(U4×U2)\({0}×{0}), then for every continuous embedding f : S2 →R3 there is a point ξ = (ξ1, ξ2, ξ3, ξ4) ∈Ω= X\Y such that τ(ξ1, ξ2, ξ3, ξ4) = (0, 0) ∈U4 × U2. (3) From (3) we conclude that d12 = d23 = d34 = d14 = ∆ 4 and d13 = d24 = Φ 2 . (4) It only remains to prove that all four points are different. Since (ξ1, ξ2, ξ3, ξ4) /∈Y we have ξ1 ̸= ξ3 or ξ2 ̸= ξ4. By symmetry we may assume that ξ1 ̸= ξ3. The map f is injective, therefore f(ξ1) ̸= f(ξ3) and consequently d13 ̸= 0. Now d13 ̸= 0 ⇒ d24 ̸= 0 ⇒ f(ξ1) ̸= f(ξ3), f(ξ2) ̸= f(ξ4) ⇒ ξ1 ̸= ξ3, ξ2 ̸= ξ4. Let us assume, without loss of generality, that ξ1 = ξ2. Then d12 = d23 = d34 = d14 = 0, which implies that d13 ≤d12 + d23 = 0. This yield a contradiction to d13 ̸= 0. Thus ξ1 ̸= ξ2. By Proposition 2.1, Theorem 1.1 is a consequence of

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