The Borsuk-Ulam Theorem

Transcription

1 The Borsuk-Ulam Theorem Artur Bicalho Saturnino June 2018 Abstract I am going to present the Borsuk-Ulam theorem in its historical context. After that I will give a proof using differential topology and also show some well-known corollaries. To close the talk, I will show some applications of this theorem to measure theory and combinatorics. 1 Introduction The degree of a map f : S n S n is a topological invariant that count how many time f wraps the sphere around itself. Rigorously, it can be defined as the integer satisfying f ([σ]) = (deg f)[σ] where [σ] H n (S n ) is non-zero. The study of this early topological invariant led to some remarkable results in the early 20th century, namely: Brower fixed point theorem (1912): Every map from the closed n-dimensional ball to itself has a fixed point. Hopf theorem (1927): Two maps from the sphere to itself are homotopic iff they have the same degree. Borsuk-Ulam theorem (1933): Every antipode-preserving map (i.e. a map f : S n S n such that f( x) = f(x)) has odd degree. Note that the result conjectured by Ulam and proven by Borsuk differs from the other two and from most of the results that we see in basic algebraic topology to the extend that it takes into account symmetry, in this case the antipode action of Z 2 on S n. The antipode-preserving condition might seem strong, but any map f : S n S n such that f(x) f( x) is homotopic to an antipode-preserving map. More explicitly, this homotopy can be defined by moving f(x) and f( x) away from each other on the great circle that connects this two points. 1

2 Figure 1: We can move f(x) and f( x) to antipode points tough the great circle in blue. 2 Proof We are now going to proof the Borsuk-Ulam theorem. We will assume for now that f : S n S n is smooth. Then the degree of f is given by deg f = sig(det d x f) x f 1 (y) where y S n is a regular value. The proof proceeds by induction on the dimension. Case n = 1: Let r : R S 1 be the covering map r(x) = e 2πx and let f : R R be the lift of f by r. The antipode-preserving condition implies that for all x R there is an odd number k(x) such that f(x + 1/2) = f(x) + k(x)/2. By continuity k(x) must be constant in x. Finally note that deg f = f(1) = f(1/2) + k/2 = k. Case n > 1: Assume WLOG that the north pole is a regular value of f and denote it by p. Let p 1,, p k be the preimages of p. Since this set is finite there must be a totally geodesic n 1-sphere in S n that does not intersect any of the p i s, so by pre-composing f with a rotation we may assume WLOG that none of the p i s are in the equator. Let q be the south pole and let q i be antipodal to p i. Let U be an open ball around p contained in the northern hemisphere and sufficiently small so that f 1 (U) is disjoint from the equator and has k components U 1, U k, such that p i U i and f Ui is a diffeomorphism onto its image. Let V = U, V i = U i. 2

3 Figure 2: Map f and the neighborhoods we are interested in. U and the U i s are blue, V and the V i s are red. 3

4 Figure 3: Streching U and V in an antipode-preserving way so that their complement is mapped to the equator. Let ρ : S n S n be the reflection around the equator. By moving points in S n \ {p, q} to the equator by the great circle joining each point to its image by ρ and controlling the speed at which points near the poles move we can define a smooth homotopy that preserves antipodes and brings the complement of U V to the equator. So f is smoothly homotopic to an atipode-preserving map g that brings S n \ k i=1 U i V i to the equator. Let M be the oriented n 1-manifold given by the union of the equator (with positive orientation) and the boundary of all U i s and V i s in the northern hemisphere (with negative orientation). Denote the equator by S n 1 and consider the restriction of g to M as a map to the equator. Since g M extends to a map from a compact, connected, oriented manifold that has M as its border (namely the part of S n \ k i=1 U i V i on the northern hemisphere) we have that 0 = deg g M = deg g S n 1 deg g Ui deg g Vi V i NH where NH S n is the northern hemisphere and all functions are seen as maps to the equator. Now note that by the anipode-preserving property we have deg g Ui + deg g Vi deg g Ui + deg g Ui mod 2 V i NH U i SH Since for all i = 1,, k we have that deg deg g Ui is 1 if g keeps orientation at p i and 1 otherwise we can conclude that deg g Ui + deg g Ui = deg g = deg f. U i SH 4

5 So we have that deg f deg g S n 1 mod 2 and we are done by induction. Now consider the case where f and antipode-preserving and continuous. Since smooth functions are dense in the space of functions with metric defined by the supremum of distances we can homotope f to a smooth map h such that h(x) h( x) for all x S n. It is easy to check that the homotopy bringing h to an antipode-preserving map that we have defined on the introduction is smooth. So f is homotopic to a smooth, antipode-preserving map, to which we can apply our proof. 3 Corollaries The Bursuk-Ulam theorem has some of well-know topological corollaries, a couple of which we will highlight: Corollary 1. There is no antipode-preserving map f : S n S n 1 Proof. Composing such and f with the inclusion of S n 1 on the equator of S n gives a nullhomotopic, antipode-preserving map. This contradicts Borsuk- Ulam. Corollary 2. For every map f : S n R n there is an x S n such that f(x) = f( x). Proof. If there was no such x we could define and antipode-preservig map g : S n S n 1 by letting g(x) = (f(x) f( x))/ f(x) f( x). 4 Applications to measure theory and combinatorics Surprisingly, this result in topology has applications to distant areas of math, such as measure theory and combinatoris. Starting from measure theory, using the Borsuk-Ulam thorem we will show that Ham Sandwich Theorem, which states that for every sandwich made of ham, cheese and bread there is a planar cut that simultaneously halves the ham, the cheese and the bread. More precisely: Theorem 1. Let µ 1,, µ n be finite Borel measures in R n absolutely continuous with respect to the Lebesgue measure. Then there exits a hyperplane h such that that µ i (h + ) = 1 2 µ i(r n ) where h + is the upper half space defined by h. 5

6 Figure 4: The Ham Sandwich theorem states that there is a line that simultaneously divides the set contoured by the blue curves and the set contoured by the red curves in subset with the same area. Proof. Relate each x = (x 1,, x n+1 ) to the half-space h + (x) = {(y 1,, y n ) : x 1 y x n y n x n+1 } and consider the map f : S n R n, f(x) = (µ 1 (h + (x),, µ n (h + (x))). Since f is continuous f(x) = f( x) for some x, now use the fact that h + (x) and h + ( x) are complementary half spaces. By considering the measures to be given by integration against the indicator function of small balls we get the discrete version of this theorem: Theorem 2. A set of points of n distinct colors in R n with an even number of points of each color accepts a hyperplane such that each half-space has half the points of each color. Thi theorem in turn has applications to combinatorics, such as the Necklace theorem. Before stating this theorem let us tell the story that motivates it. Suppose two thieves have stolen a necklace with n different kinds of stone, and that there are an even number of stones of each kind. The thieves want to cut the necklace in such a way that each has the same number of stones for each type of stone, the Necklace theorem states Theorem 3. Every open necklace with n different kinds of stone can be divided between two thieves using no more than n cuts Proof. The trick of the proof is to embed the necklace in R n along the curve t (t, t 2,, t n ) and apply the Ham Sandwich theorem. 6

7 Figure 5: Each half space defined by the black line has half the blue dots and half the red dots. Figure 6: The necklace above and the divided between the thieves 1 and 2 using the 3 cuts above. 7