Hairy balls and ham sandwiches

Transcription

1 Hairy balls and ham sandwiches Graduate Student Seminar, Carnegie Mellon University Thursday 14 th November 2013 Clive Newstead Abstract Point-set topology studies spaces up to homeomorphism. For many purposes this notion of equivalence is too strong, but by studying spaces up to homotopy equivalence we can assign algebraic invariants that allow us to prove some interesting results. One of these is the so-called hairy ball theorem, which says that any tangent vector field on a 2n-sphere must vanish somewhere. Another is the ham sandwich theorem, which says that given any collection of n objects of finite measure in n-space, there is an (n 1)-hyperplane which slices each object in half. This talk aims to give a picture of what algebraic topology is all about, without getting too bogged down in details. Contents Motivating questions 1 1 Homotopy 1 2 The fundamental group 2 3 Singular homology 3 4 Some interesting results 4 R m = R n implies m = n Brouwer s fixed point theorem Hairy ball theorem Borsuk Ulam theorem Ham sandwich theorem References 5 Please send comments and corrections to cnewstead@cmu.edu.

2 Motivating questions Recall that topological spaces X and Y are homeomorphic, written X = Y, if there exists a bijective continuous map f : X Y such that f 1 is also continuous. Some questions that motivate what we will talk about are: Is it possible that m n but R m = R n? Is it possible to comb the hairs on a coconut without creating any tufts? Is it possible to cut a ham sandwich once in such a way that the two slices of bread, and the ham inside, are each cut exactly in half? Trying to answering these questions armed only with homeomorphisms is a fruitless task. But if we introduce homotopy equivalences, we stand a better chance. 1 Homotopy Let f, g : X Y be continuous maps. A homotopy from f to g, denoted H : f g, is a continuous map H : X [0, 1] Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x X. We can think of H as being a time-indexed family of continuous maps H t : X Y such that, as time t varies from 0 to 1, the maps H t continuously deform f into g. Some properties of homotopies include: I f : f f, where I f (x, t) = f(x) for all t. If H : f g then H : g f, where H (x, t) = H(x, 1 t). If H : f g and K : g h then H K : f h, where H(x, 2t) if 0 t 1 (H K)(x, t) = 2 K(x, 2t 1) if 1 2 t 1 In particular, there is a homotopy f g is an equivalence relation on continuous maps X Y. When there is a homotopy f g we write f g. Denote by [f] the homotopy class of f, i.e. the -equivalence class of f. It will be useful for later to know that if f g then h f h g and f k g k. We say spaces X and Y are homotopy equivalent, denoted X Y, if there are continuous maps f : X Y and g : Y X such that g f id X and f g id Y It is easy to see that is an equivalence relation on topological spaces, and that if X = Y then X Y. Intuitively, X Y whenever we can continuously deform X into Y. examples. Let s look at some 1

3 Example 1.1. [0, 1] }. Define f : [0, 1] } by f(x) =, and define g : } [0, 1] by g( ) = 0. Clearly f g = id }. Conversely, g f is the constant function with value 0. Define H : id [0,1] g f by H(x, t) = tx It is clear that H is a homotopy, so g f id [0,1], and hence [0, 1] }. When a space is homotopy equivalent to } we say it is contractible. Other examples of contractible spaces include: the (open or closed) disc, R n, graph-theoretic trees,... Example 1.2. S 1 R 2 \(0, 0)}. Here S 1 denotes the unit circle (x, y) R 2 : x 2 +y 2 = 1}. To see this, let f : S 1 R 2 \ (0, 0)} be the inclusion map, and define g : R 2 \ (0, 0)} S 1 to be a retraction of R 2 \ (0, 0)} onto S 1. 1 Then g f = id, so certainly g f id. And f g id: define H((x, y), t) to be t(f g)(x, y) + (1 t)(x, y). Then H : f g id. 2 The fundamental group The first algebraic invariant we assign to our topological spaces is the fundamental group. In general the fundamental group is easy to define but difficult to compute. Given a space X, a loop in X based at x X is a path u : [0, 1] X with u(0) = u(1) = x. There is a very simple loop in any space given by c x (t) = x for all 0 t 1. Given any loop u there is another loop ū defined by ū(t) = u(1 t). We can define an operation on loops in X based at x by u(2t) if 0 t 1 u v(t) = 2 v(2t 1) if 1 2 t 1 This is painfully close to defining a group structure on the set of loops based at X. Indeed, for all loops u, v, w we have c x u u, u c x u, u (v w) (u v) w, u ū c x, ū u c x But if we quotient out by then we do get a group. That is, the elements are homotopy classes [u] for loops u based at x, the group operation is given by [u][v] = [u v], the inverse is given by [u] 1 = [ū] and the unit is given by 1 = [c x ]. This is the fundamental group of X based at x, denoted π 1 (X, x). If X is path-connected (as all our spaces will be) then, given x, y X, π 1 (X, x) = π 1 (X, y), so we just write π 1 (X). Theorem 2.1. Fix spaces X, Y and a basepoint x X. Any continuous map f : X Y induces a homomorphism f : π 1 (X, x) π 1 (Y, f(x)). Furthermore, the assignment f f is functorial, i.e. (f g) = f g and (1 X ) = 1 π1 (X,x). (In particular, if X Y then π 1 (X) = π 1 (Y ).) Proof. Define f ([u]) = [f u] and check the details. Example 2.2. If X is contractible then π 1 (X) = 1}. It s clear that π 1 ( }) = 1} since the only loop in } is c, so if X } then π 1 (X) = π 1 ( }) = 1}. 1 A concrete example of such a retraction was given in the talk. 2

4 Unfortunately, the fundamental group only allows us to answer low-dimensional analogues of our questions above. One solution might be to generalise to higher dimensions, by mapping n- spheres (instead of loops) into our spaces. Doing this we obtain homotopy groups π n (X)... but in general these are very hard to compute. For instance, we don t even know what π n (S 2 ) is for all n! 3 Singular homology The idea behind singular homology is that we probe a space X by mapping in n-dimensional spaces and looking for holes. Define the n-simplex n by n = (x 1,..., x n ) R n : x i = 1 and each x i 0} R n For each n 0 let C n (X) be the group of all formal sums, with coefficients in Z, of continuous maps n X. Given a generator σ : n X there is a natural boundary map σ = ( 1) i σ (i th face of n ) This map can be extended linearly to all of C n (X). Then im ker. The nth (singular) homology group of X is then defined to be H n (X) = ker im I don t expect anyone to get their head around this if this is their first time seeing it. The moral of singular homology is: in a suitably nice subspace X R n, if k > 0 then number of (k 1)-dimensional holes in X H k (X) = Z and number of connected components of X H 0 (X) = Z Example 3.1. Let X be any contractible space, e.g. }, [0, 1], R n. Then H n (X) = Z if n = 0 0 if n > 0. Example 3.2. The circle S 1 has one connected component and one 2-dimensional hole, so H n (S 1 Z if n = 0, 1 ) = 0 if n > 1 More generally, the k-sphere S k has one connected component and one (k + 1)-dimensional hole, so H n (S k Z if n = 0, k ) = 0 if n 0, k Theorem 3.3. Any continuous map f : X Y induces a homomorphism f : H n (X) H n (Y ) for all n. Furthermore, the assignment f f is functorial, and if X Y then H n (X) = H n (Y ) for all n. 3

5 4 Some interesting results R m = R n implies m = n We now have all the power we need to prove the following theorem. Theorem 4.1. If R m = R n then m = n. Proof. Suppose f : R m = R n. Then R m \ 0} = R n \ f(0)}. Now R m \ 0} = S m 1 and R n \ f(0)} = S n 1, and so H r (S m 1 ) = H r (S n 1 ) for all r Z. In particular so m 1 = n 1 and m = n. Z = H m 1 (S m 1 ) = H m 1 (S n 1 ) Brouwer s fixed point theorem Homology yields an impressively short proof of Brouwer s fixed point theorem. Theorem 4.2 (Brouwer s fixed point theorem). Every continuous function g : B n B n has a fixed point. Proof. Suppose f : B n B n has no fixed points. Define g : B n S n 1 = B n by letting g(x) be the (unique) point on the boundary of B n which lies on the ray from f(x) to x. Continuity of f implies continuity of g, and moreover g(x) = x for all x B n = S n 1. Let i : S n 1 B n be the inclusion map. Then S n 1 i B n g S n 1 is the identity map. Now H n 1 (S n 1 ) = Z and H n 1 (B n ) = 0. By functoriality of homology, g i = id Z. But i is the zero map, so g i = 0... a contradiction. Corollary 4.3. If a map of the United States is laid on the ground in the United States, some point on the map is directly over the point it refers to. Hairy ball theorem The proof of the hairy ball theorem answers another of the motivating questions of the talk. Intuitively we can think of the hairs on an 2n-sphere being vectors in (2n + 1)-space; if they re combed without any tufts then this means they all lie tangent to the n-sphere and are nonzero. Theorem 4.4 (Hairy ball theorem). Let f : S 2n R 2n+1 be a continuous map such that f(x) x for all x S 2n. Then there exists x S 2n with f(x) = 0. Proof. Suppose f : S 2n R 2n+1 is continuous and f(x) x for all x S 2n, and that f(x) 0 for any x. Let u x : [0, 1] S 2n be the path from x to x along the great semicircle in the direction of f(x). Let a : S 2n S 2n be the antipodal map. Then a = ( 1) 2n+1 id Z = id Z. But there is a homotopy H : id S 2n a defined by H(x, t) = u x (t), so id Z = id Z... a contradiction. Corollary 4.5. Somewhere in the world, the wind isn t blowing. 4

6 Proof. Model the surface of the earth as S 2 and the speed and direction of wind at a point x by a vector v x in R 3. On the assumption that x v x is continuous, there exists x S 2 such that v x = 0. (Presumably x Pittsburgh.) Borsuk Ulam theorem The Borsuk Ulam theorem is a seemingly innocuous result which says that any continuous maps from the n-sphere to n-space maps some pair of antipodal points of the n-space to the same place. Theorem 4.6 (Borsuk Ulam). Let f : S n R n be a continuous map. Then there exists x S n such that f(x) = f( x). Proof. Let f : S n R n and suppose f maps all pairs of antipodal points to distinct points. Define g : S n S n 1 by f(x) f( x) g(x) = f(x) f( x) Then g( x) = g(x) for all x S n. Let i : S n 1 S n be an inclusion. Then i g = id S n 1, so (i g) = id Z. But i g c, so (i g) = id Z... a contradiction. Corollary 4.7. There are two opposite points in the world with the same temperature and humidity. Proof. Model the surface of the earth as S 2. At a point x on the surface of the earth, let T x be the temperature and H x be the humidity. The assignment x (H x, T x ) is presumably continuous, so by the Borsuk Ulam theorem, there exists x with H x = H x and T x = T x. Ham sandwich theorem Finally we have a very nice result that follows from the Borsuk Ulam theorem. Theorem 4.8. Given any collection of n measurable subsets X 1,..., X n R n, there is an (n 1)-hyperplane which splits each X i into two pieces of equal measure. Proof. For each point x S n 1 there is a (canonical) (n 1)-hyperplane Π x perpendicular to x which bisects A n. Give the Π x some uniform orientation. Define f : S n 1 R n 1 by f(x) = (V 1 (x), V 2 (x),..., V n 1 (x)) where V i (x) is the measure of X i on the positive side of Π x. Then f is continuous, so by the Borsuk Ulam theorem there exists x S n 1 with f(x) = f( x). But Π x = Π x with opposite orientations, so V i ( x) = µ(x i ) V i (x). Hence V i (x) = µ(x i ) V i (x), so V i (x) = 1 2 µ(x i). So Π x bisects each X i. Corollary 4.9. There is a way of cutting any ham sandwich in such a way that the two slices of bread, and the ham inside, are each cut exactly in half. 5