Exotic Spheres Malik Obeidin May 2, 2014 1 Introduction One of the biggest shocks in the history of differential geometry and topology occured in 1956, after John Milnor published the first constructed examples of manifolds which topologically were the same, but differentiably were not. In fact, all of these manifolds are topologically the 7-sphere, but distinguished by an invariant. This bizarre result opened up new areas studying the relationship between topological, smooth, and PL manifolds, and from that point, all results had to be careful about the category in which one was working. Since then, many more examples are known, and many classified, but the original construction due to Milnor is actually quite simple and elegant. We outline his reasoning as given in the original 1956 paper [1]. For the remainder of this paper, all manifolds considered will be differentiable (of class C1), orientable, and compact, and all cohomology groups will have integer coefficients. 2 Constructing the Invariant Let M 7 be a differentiable, oriented, closed manifold satisfying H3(M 7) = H4(M 7) = 0 (1) On any such manifold, we can define a residue class λ(M 7) modulo 7, which will be the invariant which will distinguish between smooth structures on our constructed 7-spheres. By a result of Thom [2], every closed 7-manifold M 7 is the boundary of some 8-manifold B8; in particular, we can choose an ori- 8 7 7 entation ν 2 H8(B ;M ) with the property that @ν = µ, where µ 2 H7(M ) is the chosen generator for the orientation. We define a map on the free part H4(B8;M 7)=(torsion) by sending α ! hν; α2i; this is a quadratic map Zn ! Z, hence a quadratic form. Now we call τ(B8) the index of this form; that is, after diagonalizing over R, we take the number of positive terms minus the number of negative terms. This is the first of two numbers which will be used to define λ. 1 4 8 The second is given as follows: let p1 2 H (B ) be the first Pontrjagin class of the tangent bundle of B8. The assumption (1) above means that, by the long exact sequence of a pair in cohomology, the quotient map q∗ : H4(B8;M 7) ! H4(B8) is sandwiched between zero maps, and hence is an isomorphism. So, similarly to τ, we can define what Milnor calls a "Pontrjagin number" 8 ∗ −1 2 q(B ) = hν; ((q ) p1) i With these two numbers, we define λ(M 7) to be the residue class modulo 7 of 2q(B8) − τ(B8) for the following reason: Theorem 1. The residue class of λ modulo 7 does not depend on the choice of the manifold B8. 8 8 To prove this theorem, we choose two possibly different manifolds B1 , B2 with boundary M 7. Then, sewing these along the boundary M 7 gives a closed 8 8 8 8-manifold C = B1 [ B2 , possessing a differentiable structure compatible with 8 8 both B1 and B2 , and an orientation ν which is a consistent with the orientation 8 8 2 8 ν1 on B1 . We can then define q(C ) = hν; p1(C )i. The index τ(C8) defined as above was previous computed by Thom in [2] to be 1 τ(C8) = hν; (7p (C8) − p2(C8)i 45 2 1 8 where p2(C ) is the second Pontrjagin class. Then, we have that 8 8 8 2 8 2 8 45τ(C ) + q(C ) = hν; 7p2(C ) − p1(C )i + hν; p1(C )i 8 = 7hν; p2(C )i ≡ 0 (mod 7) So, reducing everything modulo 7 and multiplying by 2 gives 2q(C8) − τ(C8) ≡ 0 (mod 7) (2) But, the following lemma gives immediately the result: Lemma 1. The given manifold C8 satisfies 8 8 8 τ(C ) = τ(B1 ) − τ(B2 ) 8 8 8 q(C ) = q(B1 ) − q(B2 ) as one might guess from the construction. Combined with (2), this proves Theorem 1. The proof of Lemma 1 is not hard but moderately tedious, so we'll skip it. Note the following facts we get immediately from Theorem 1: Corollary 1. If λ(M 7) 6= 0 then M 7 is not the boundary of any 8-manifold with fourth Betti number zero. 2 Corollary 2. If λ(M 7) 6= 0 then M 7 possesses no orientation reversing self diffeomorphism. The first fact is obvious as such an 8-manifold B8 would force both τ(B8) and q(B8) to be zero. The second is also straightforward; an orientation reversing self diffeomorphism takes both q and τ (hence λ) to their negatives, though λ is diffeomorphism invariant as constructed. In the end, constructing a topological 7-sphere with nonzero λ will give the exotic sphere which we are looking for. We will be able to show the constructed manifolds are homeomorphic to S7 by using the following result of Morse theory, (a version of) the Reeb Sphere Theorem: Theorem 2. Let M n be a closed manifold. Suppose there exists a differentiable function f : M n ! R such that f has only two critical points, both nondegener- ate (i.e., the Hessian is nonsingular). Then, M n is homeomorphic to the sphere Sn. 3 Constructing the 7-Manifolds Consider the isomorphism classes of S3 bundles over S4, with structure group SO(4). In fact, it can be shown that this set of isomorphism classes is in one- ∼ to-one correspondence with the group π3(SO(4)) = Z ⊕ Z, with an explicit isomorphism given as follows: for any pair (h; j) 2 Z ⊕ Z, define a map fhj : S3 ! SO(4) by the formula h j fhj(u) · v = u vu where v is an element of R4. The multiplication between u and v is quaternion multiplication. 3 So, let ξhj be the S bundle corresponding the (h; j) and fhj. Then, if we denote the standard orientation ι 2 H4(S4), we have the following formula: Lemma 2. The Pontrjagin class p1(ξhj) is equal to ±2(h − j). It is easy to see that the Pontrjagin class p1(ξhj) is linear as a function of h and j, and it is well known that it is independent of the orientation on the 3 fiber. But, reversing the orientation of S switches ξhj to ξ−h−j, leaving only linear functions of the form c(h − j) for some c to be determined later. Given any odd k 2 Z, we have a unique (h; j) such that the equations 7 h + j = 1 and h − j = k are satisfied. For any such k, we define Mk to be the total space of ξhj, which is clearly a topological manifold. In fact, we will show 7 Mk has a natural smooth structure and orientation, and is in fact topologically a sphere. 7 Lemma 3. Mk has a natural smooth structure and orientation, and a smooth 7 function f : Mk ! R with exactly two nondegenerate critical points. 7 So, by applying Theorem 2, Mk is homeomorphic to the standard 7-sphere. 3 Proof. Take coordinate charts in the base S4 given by the complements of the north and south poles, which are each identified with R4 by stereographic pro- jection. By elementary geometry, a point corresponding to u 2 R4 in one chart 0 u is given by u = kuk2 in the other. Over each of these charts, the bundle is necessarily trivial, so we describe 7 4 3 the total space Mk by taking two copies of R × S and identifying the subsets (R4 − 0) × S3 with the diffeomorphism u uhvuj (u; v) ! (u0; v0) = ( ; ) kuk2 kuk which gives us back the bundle ξhj. Again here we have exploited the quaternion multiplication on R4, and fixed a smooth structure on the total space by starting with smooth manifolds and identifying via diffeomorphisms. Making the change of coordinates (u0; v0) by (u00; v0), where u00 = u0(v0)−1, 7 we can now define the function f : Mk ! R by Re(v) Re(u00) f(x) = 1 = 1 (1 + kuk2) 2 (1 + ku00k2) 2 which can be easily verified to have the desired property. The critical points will be at (u; v) = (0; ±1) and will be non-degenerate, completing the proof of the lemma. 7 So, Theorem 2 now concludes that Mk is in fact homeomorphic to a sphere. To prove the existence of manifolds homeomorphic but not diffeomorphic to 7 7 S , we need only to find our previous invariant λ(Mk ), which is now defined, 3 7 4 7 7 since H (Mk ) = H (Mk ) = 0 as Mk is homeomorphic to the sphere. In fact, we have the following: 7 2 Lemma 4. The invariant λ(Mk ) is the residue class of k − 1 modulo 7. 7 4 Proof. With any 3-sphere bundle Mk ! S we can extend the total space to a 8 4 8 4-cell bundle ρk : Bk ! S , where Bk is a differentiable manifold with boundary 7 4 8 ∗ Mk . The fourth cohomology H (Bk) is generated by the pullback α = ρk(ι). Choosing proper orientations, we can fix hν; ((q∗)−1α)2i = 1, hence we will have 7 the index τ(Bk) = 1. 7 The tangent bundle TBk will decompose as a direct sum of the bundle of vectors tangent to the fiber and the bundle of vectors normal to the fiber. The ∗ first bundle is simply the pullback ρk(ξhj), so by naturality and the proof of ∗ Lemma 2 has the Pontrjagin class p1 = ρk(c(h − j)ι) = ckα, as k = h − j.