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1 Introduction

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1 Introduction 1 Introduction The homotopy theory we will discuss in these lectures has historical origins in very concrete geometric problems. We will begin, in this lecture, by de- scribing some of those problems and their solutions. For the most part these solutions date from work by Frank Adams in the 1960's, but the problems themselves are much older. The central problem we are concerned with involves the construction of interesting sets of tangent vector fields on the (n − 1)-sphere n−1 n S = fx 2 R : kxk = 1g : This embedding of the sphere into Euclidean space provides us with a con- crete way to visualize tangent vectors: A tangent vector at x is a vector n n−1 v 2 R such that x · v = 0. The tangent space to S at a point x is the n n−1 orthogonal complement in R of the line through x. Write TxS , or just Tx, for this vector space. The identification of the tangent space at a point with a subspace of n R provides us with an inner product on each tangent space. This inner product varies continuously with the point on the sphere; it is a \metric" on the tangent bundle. In particular each tangent vector v has a length jjvjj. A vector field on Sn−1 is a continuous section of the tangent bundle. In prosaic terms, this is a continuous function n−1 n v : S ! R such that x · v(x) = 0 for all x 2 Sn−1 : A first question is whether there exists a nowhere vanishing vector field. If we have a nonwhere vanishing vector field, we may normalize it by dividing by its length (which is a continuous function on Sn−1) to obtain a unit tangent vector field: a map v : Sn−1 ! Sn−1 such that x · v(x) = 0 for all x 2 Sn−1. There are advantages to replacing the open condition of being everywhere nonzero with the closed condition of being everywhere of length 1, just as there are advantages to thinking about the unit sphere Sn−1 in n place of the noncompact set of nonzero vectors in R . The answer to this question is very well known. Proposition 1.1. Sn−1 admits a unit tangent vector field if and only if n is even. 1 Proof. Suppose first that n is even; say n = 2m. Then Sn−1 can be thought m of as the unit vectors in C , and v(x) = ix satisfies jjv(x)jj = 1 and x·v(x) = 0 for all x 2 Sn−1. This gives us a nonvanishing vector field on each odd sphere. However, when n is odd there are none. Such a v(x) would give a ho- motopy between the identity and the antipodal map α(x) = −x: ht(x) = (cos πt)x + (sin πt)v(x) : We can get a contradiction by using the degree, i.e. the effect of a self-map of n−1 n−1 ∼ S on the (n − 1)-dimensional integral homology group Hn−1(S ; Z) = Z. The degree of the identity is 1, so by homotopy invariance deg α = 1. n But α is a composite of n reflections in R , each of degree −1, so deg α = n (−1) = −1, giving a contradiction. A more refined question now arises: for a given positive integer k, can n−1 we produce a sequence (v1; v2; : : : ; vk−1) of (k −1) vector fields on S that are everywhere linearly independent? How big can we make k and still get an affirmative answer to this question? Once again, we can replace this by an equivalent compact question, this time using the Gram-Schmidt process (which continuous, so we can apply it fiber-wise): find a sequence v1; v2; : : : ; vk−1 of everywhere orthonormal vector fields on Sn−1. And, once again, in certain cases there are rather obvious constructions. If n = 4m, for example, we can regard Sn−1 as the unit vectors in quater- nionic m-space, and then v1(x) = ix ; v2(x) = jx ; v3(x) = ijx provide three orthonormal vector fields. Similarly, the \octonion" multipli- 8 7 cation on R provides seven everywhere independent vector fields on S . In n−1 n−1 n−1 these cases, the vector fields provide an isomorphism S ×R ! TS of vector bundles; that is, a trivialization of the tangent bundle or a par- allelization of the sphere. In fact, Kervaire proved that these are the only spheres that are parallelizable. While these questions are clearly interesting in their own right, we shall see, in this course, that they are equivalent to fundamental questions about the structure of unstable homotopy theory, especially the behavior of the homotopy groups of spheres. For a start, notice that they can be phrased in terms of standard ques- tions about the existence of sections of certain fiber bundles. To define these 2 n fiber bundles, recall that a k-frame (or an orthogonal k-frame) in R is a sequence (v1; : : : ; vk) of mutually orthogonal unit vectors. Regarding them as column vectors, they can be assembled into an n × k matrix v with the T property that the product v v is the k × k identity matrix Ik. Definition 1.2. The Stiefel manifold Vk(E) of an inner product space E is the set of all k-frames in E, topologized as a subset of the product Ek. n If E = R with it standard inner product, we will write Vn;k in place of n Vk(R ). The Stiefel manifold is a compact manifold. It comes equipped with a smooth map n−1 π : Vn;k ! S n−1 sending the k-frame (v0; v1; : : : ; vk−1) to v0 2 S . The rest of the k-frame, (v1; : : : ; vk−1), can be viewed as giving a (k − 1)-frame in the tangent space −1 to the sphere at v0. That is, π (v0) = Vk−1(Tv0 ). n−1 Lemma 1.3. The map π : Vn;k ! S sending v to v0 is the projection map of a fiber bundle. Proof. This means that any point v0 in the base has a neighborhood W over which the projection is isomorphic to a product projection. In this case we can take for W the upper open hemisphere centered at v0, W = 0 n−1 0 −1 fv0 2 S : v0 · v0 > 0g. Then define a map κ : π W ! Vk−1(Tv0 ) 0 −1 0 0 as follows. Let v 2 π W . The sequence (v0; v1; : : : ; vk−1) is still linearly independent, so we can apply the Gram-Schmidt process to it to obtain a 0 sequence (v0; v1; : : : ; vk−1). Define κ(v ) = (v1; : : : ; vk−1). Then −1 (π; κ): π W ! W × Vk−1(Tv0 ) is a homeomorphism compatible with the projections to W . So the projection map always has local cross-sections; the sphere (or any (n − 1)-manifold) admits local (k − 1)-frames as long as k ≤ n. A global cross-section is the same thing as a tangential (k − 1)-frame field (or briefly a (k − 1)-frame) on Sn−1. Finding obstructions to the existence of a cross- section of a fiber bundle is one of a small collection of standard problems in homotopy theory, and we will bring a number of homotopy-theoretic tools to bear on this particular sectioning problem. We may ask how large k can be, given n. The answer is given in terms of a function ρ(n), defined as follows. Let ν(n) be the largest integer such that 2ν divides n. Then write ν = ν(n) as ν = 4b + c, 0 ≤ c ≤ 3, and set ρ(n) = 8b + 2c : 3 Theorem 1.4 (Hurwitz, Radon, Eckmann; Adams). There exist ρ(n) − 1 linearly independent vector fields on Sn−1 (Hurwitz-Radon-Eckmann) and no more (Adams). Here's a table of a few values of this strange function ρ(n). Since ρ(n) depends only on ν(n), we tabulate its values in terms of ν. ν 0 1 2 3 4 5 6 7 ··· 2ν 1 2 4 8 16 32 64 128 ··· ρ(n) 1 2 4 8 9 10 12 16 ··· Thus ρ(n) is roughly twice the exponent of 2 in n; it quickly falls behind n. In fact the only cases in which ρ(n) = n are n = 1; 2; 4; 8; the only paralleliz- able spheres are S0;S1;S3;S7. The sphere S127 admits fifteen everywhere independent vector fields and no more. There are two steps to proving Theorem 1.4: 1. Construct enough vector fields. It turns out that linear algebra { in the form of representations of Clifford algebras { produces a (ρ(n)−1)- frame on Sn−1. 2. Show that no other method can produce more. This is much harder, and was the first major victory for K-theory in topology, the occasion for Frank Adam's introduction of the operations named after him. Let's focus for the moment on the parallelizable case. There are many statements that are equivalent to saying that Sn−1 is parallelizable. If Sn−1 is a Lie group, we can pick a basis for the tangent space at the identity element and translate it around the group; this gives a different way of thinking of the parallelization we got above by thinking of S1 and 3 S as unit vectors in C and H. The 7-sphere is not a Lie group, but it does posess a smooth (though non-associative) multiplication, which can be used to parallelize it.
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