Recent Advances in Topological Manifolds Part III course of A. J. Casson, notes by Andrew Ranicki Cambridge, Lent 1971 Introduction A topological n-manifold is a Hausdorff space which is locally n-Euclidean (like Rn). No progress was made in their study (unlike that in PL and differentiable manifolds) until 1968 when Kirby, Siebenmann and Wall solved most questions for high dimensional manifolds (at least as much as for the PL and differentiable cases). Question: can compact n-manifolds be triangulated? Yes, if n 6 3 (Moise 1950's). This is unknown in general. However, there exist manifolds (of dimension > 5) which don't have PL structures. (They might still have triangulations in which links of simplices aren't PL spheres.) There is machinery for deciding whether manifolds of di- mension > 5 have a PL structure. Not much is known about 4-manifolds in the topological, differentiable, and PL cases. n n+1 Question 2: the generalized Sch¨onfliestheorem. Let B = x 2 R : kxk 6 1 , n n+1 n n S = x 2 R : kxk = 1 . Given an embedding f : B ! S (i.e. a 1-1 con- tinuous map), is Sn n f(Bn) =∼ Bn? No{the Alexander horned sphere. n n+1 0 n n ∼ n Let λB = x 2 R : kxk 6 λ . Question 2 : is S n f(λB ) = B (where 0 < λ < 1)? Yes. In 1960, Morton Brown, Mazur, and Morse proved the following: if g : Sn−1× n n n−1 [−1; 1] ! S is an embedding, then S ng(S ×f0g) has 2 components, D1;D2, ∼ ∼ n 0 such that D1 = D2 = B , which implies 2 as a corollary. (The proof is easier than that of PL topology). Question 3: the annulus conjecture. Let f : Bn ! Int Bn be an embedding. n 1 n ∼ n 1 n ∼ n−1 Is B n f( 2 B ) = B n 2 B (= S × I)? In 1968, Kirby, Siebenmann, and Wall proved this for n > 5. This was already known for n 6 3. The n = 4 case is still unknown. Outline of course: • Basic facts about topological manifolds • Morton Brown's theorem { the first \recent" result 1 • Kirby's trick: Homeo(M) is a topological group (with the compact-open topology). This is locally contractible: any homeomorphism h near 1 can be joined by a path in Homeo(M) to 1 • Product structure theorem: if M n is a topological manifold and M × Rk n has a PL structure, then M has a PL structure (n > 5). • sketch of proof of the annulus conjecture (complete except for deep PL theorems). 1 Basic Properties of Topological Manifolds n n n−1 Let R+ = f(x1; : : : ; xn) 2 R : xn > 0g. Identify R with n n f(x1; : : : ; xn) 2 R : xn = 0g = @R+: Definition 1.1. A (topological) n-manifold (with boundary) is a Hausdorff n space M such that each point of M has a neighborhood homeomorphic to R+. The interior of M, Int M, is the set of points in M which have neighborhoods homeomorphic to Rn. The boundary of M, @M = M n Int M. Int M is an open set in M, @M is closed in M. M is an open manifold if it is non-compact and @M = ;. M is a closed manifold if it is compact and @M = ;. Example. Any open subset of an n-manifold is an n-manifold. Let M be a connected manifold with @M = ;. If x; y 2 M then there is a homeomorphism h : M ! M with h(x) = y. Theorem 1.2 (Invariance of domain). Let U; V ⊂ Rn be subsets such that U =∼ V . Then if U is open in Rn, then so is V . Corollary 1.3. If M is an N-manifold, then @M is an (n−1)-manifold without boundary. n Proof. Suppose x 2 M and f : R+ ! M be a homeomorphism onto a neigh- borhood N of x in M. Then n−1 x 2 @M () x 2 f(R ) (1) n−1 n n−1 ∼ n If x 62 f(R ), then x 2 f(R+ n R ) = R , so x 2 Int M and x 62 @M. If x 62 @M, then x 2 Int M, i.e. there is a neighborhood U of x homeo- n n morphic to R ⊂ f(R+). So there is a neighborhood V of x which is open in M such that V ⊂ U, homeomorphic to an open set in Rn. Therefore −1 n n −1 n f (V ) ⊂ R+ ⊂ R . By theorem 1.2, f (V ) is open in R . Suppose x 62 f(Rn−1). Then f −1(x) 2 Rn−1, but then f −1(V ) can't be a neighborhood of f −1(x), so f −1(V ) is not open. This is a contradiction, therefore x 2 f(Rn−1) =) x 2 @M. n Now suppose y 2 @M. Let g : R+ ! M be a homeomorphism onto a neighborhood P of y in M. P contains an open neighborhood W of y in M. 2 Now W \@M = W \g(Rn−1) by (1). Therefore W \g(Rn−1) is a neighborhood of y in @M homeomorphic to an open set N, so y has a neighborhood in @M homeomorphic to Rn−1, as required. Corollary 1.4. If M m;N n are manifolds then M × N is an (m + n)-manifold with @(M × N) = (@M × N) [ (M × @N), i.e. Int(M × N) = Int M × Int N. m n ∼ Proof. If x 2 M × N, then x has a neighborhood homeomorphic to R+ × R+ = m+n R+ , so M × N is an (m + n)-manifold. Clearly Int M × Int N ⊂ Int(M × N). If x 2 (@M × N) [ (M × @N), then x has a neighborhood homeomorphic to m n m n m n m+n R+ × R , R × R+, or R+ × R+ { all homeomorphic to R+ by a homeomor- phism carrying x to Rm+n−1. By (1), x 2 @(M × N). Hence the result. Example. Examples of manifolds: • Rm is an m-manifold without boundary, open. • Sm is a closed m-manifold. (Stereographic projection gives neighbor- hoods.) • Bm is a compact manifold with boundary Sm−1. m m−1 • R+ is an m-manifold with boundary R . • Products of these, • CP n, orthogonal groups O(n) are manifolds. These are all differentiable manifolds. There exist topological manifolds which do not possess a differentiable structure. n k n Lemma 1.5. If X ⊂ S is homeomorphic to B , then Her(S n X) = 0 for all r 2 Z. Proof. By induction on k. The lemma is true if k = 0: Sn n fpt.g =∼ Rn. Assume true if k = l, we prove it for k = l + 1. Choose a homeomorphism l ∼ l+1 n f : B × I = B ! X, suppose α 2 Her(S n X). Take t 2 I. By induction n l hypothesis, Her(S nf(B ×ftg)) = 0. Therefore α is represented by the boundary n l of some singular chain c lying in S n f(B × ftg). There is a neighborhood Nt l n n of f(B × ftg) in S such that c lies in S n Nt. Therefore there is an open interval Jt ⊂ I containing t such that c lies in n l S n f(B × Jt). Since the unit interval is compact, we can cover by finitely many of the Jt's. Therefore there is a dissection 0 = t0 < t1 < ··· < tk = 1 such that [tp−1; tp] ⊂ some Jt. n n l Let φp;q : Her(S n X) ! Her(S n f(B × [tp; tq])) where p < q and the map is induced by inclusion. Now φp−1;p(α) = 0 for all p. Suppose inductively that φ0;i(α) = 0 starts with i = 1. By the main in- n l ductive hypothesis, Hes(S n f(B × ftig)) = 0 for s = r; r + 1. The sets n l S n f(B × [tp; tq]) are open. We have the lattice 3 n l n l S n f(B × [0; ti]) / S n f(B × ftig) O O n l n l S n f(B × [0; ti+1]) / S n f(B × [ti; ti+1]) and the corresponding Mayer-Vietoris sequence: n l 0 −! Her(S n f(B × [0; ti+1])) n l n l −! Her(S n f(B × [0; ti])) ⊕ Her(S n f(B × [ti; ti+1])) −! 0; with the maps induced by inclusion. Since φ0;i(α) = 0 and φi;i+1(α) = 0, we have φ0;i+1(α) = 0. Therefore, n φ0;k(α) = 0, i.e. α = 0 and Her(S n X) = 0 as required. Lemma 1.6. If X ⊂ Sn is homeomorphic to Sk, then ( n ∼ n−k−1 Z if r = n − k − 1, Her(S n X) = Her(S ) = 0 otherwise. Proof. By induction on k. The result is true if k = 0, for Sk n pair of points =∼ Sn−1. (???) Now assume the result holds for k = l − 1 and try to prove it for k = l. l Choose a homeomorphism f : S ! X. Let D1;D2 be northern and southern l l ∼ l−1 n hemispheres of S so that D1 [ D2 = S and D1 \ D2 = S . The sets S n X, n n S n f(Di), and S n f(D1 \ D2) are open. We have the lattice n n S n f(D1) / S n f(D1 \ D2) O O n n S n X / S n f(D2) and the Mayer-Vietoris sequence n n 0 −! Her+1(S n f(D1 \ D2)) −! Her(S n X) −! 0 n ∼ n ∼ since Her+1(S n f(D1)) = Her+1(S n f(D2)) = 0 by the previous lemma.