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Homework 3∗ I 3b: Let Σ be a closed surface with χ(Σ) ≤ 1. Show that any Morse function on Σ must have at least |χ(Σ)| + 2 critical points. For a Morse function f , let crit(f ) = criteven (f )∪critodd (f ), where even/odd denotes the parity of the index of a critical point. Then 3a claims that for any manifold, χ(Σ) = | criteven (f )| − | critodd (f )|. For any even-dimensional closed manifold, | criteven (f )| > 2 always, since f must have a maximum and a minimum. Now compare with the equation | crit(f )| = | criteven (f )| + | critodd (f )|. If χ(Σ) = 1, then we see that | crit(f )| > 3, and if χ(Σ) 6 0 we get that | crit(f )| > |χ(Σ)| + 4. B c: Let M be a closed orientable 3-manifold, and let f : M → R be a Morse function with four critical points, which has distinct critical values. Show that there is an embedded circle S 1 ⊆ M , so that M \ S 1 ∼ = S 1 × R2 . You may assume two facts without proof: the only boundaryless surface which is homotopy equivalent to a point is R2 , and every orientation preserving diffeomorphism of R2 is isotopic to the identity. (Hint: If c ∈ R has two critical values below it and two above it, what is the topology of f −1 ((−∞, c])?. Use cases.) Let m− ∈ M be the minimum of f , m+ the maximum, and p, q the critical points with f (p) < c < f (q). Since χ(M ) = 0, we know that ind(p) is opposite of the parity of ind(q). This reduces the problem into 6 cases: • Case max : ind(p) = 2 and ind(q) = 3. We show in this case M ∼ = S3. • Case min/max : ind(p) = 0 and ind(q) = 3. We show in this case M ∼ = S3 q S3. • Case none: ind(p) = 3 and ind(q) = 2. We show that this is impossible. • Case max/min: ind(p) = 3 and ind(q) = 0. We show in this case M ∼ = S3 q S3. • Case 2/1 : ind(p) = 2 and ind(q) = 1. We show in this case M ∼ = S2 × S1. • Case 1/2 : ind(p) = 1 and ind(q) = 2. There are infinitely many diffeomorphism classes of manifolds which admit such a Morse function, so we prove the statement directly. The other two remaining cases have two local minima. By replacing f with −f we see that these are already covered by Case max and Case none. ∗ Due November 13th 1 2 In every case, the manifolds f −1 (−∞, f (p)−ε] and f −1 [f (q)+ε, ∞) are both diffeomorphic to the closed ball B 3 . Indeed, f −1 (−∞, f (m− ) + ε] is diffeomorphic to B 3 by the standard neighborhood of critical points, and f −1 [f (m− ) + ε, f (p) − ε] has no critical points and is therefore diffeomorphic to S 2 × I. Since S 2 has a unique orientation preserving diffeomorphism up to isotopy, we see that f −1 (∞, f (p) − ε] ∼ = B3. = B 3 ∪S 2 S 2 × I ∼ We show that, if ind(p) = 2, then f −1 (−∞, c] ∼ = S 2 × I. We know that −1 3 2 f (−∞, c] deformation retracts onto B ∪ D , but in fact we know more: that f −1 (−∞, c] is diffeomorphic to B 3 ∪ U , where U is a small closed neighborhood of D2 . U ∼ = S 1 × I. Up to isotopy, there is a unique = D2 × I, so U ∩ ∂B 3 ∼ 1 2 ∼ embedding of S × I into S = ∂B 3 (Jordan curve theorem). Therefore the diffeomorphism type of the gluing is unique. Since S 2 × I is such a gluing (taking S 1 × I as a neighborhood of the equator) this completes the claim. Case max : Since M = f −1 (−∞, c] ∪f −1 (c) f −1 [c, ∞) ∼ = S 2 × I ∪∂ (B 3 q B 3 ), and since there is a unique orientation preserving diffeomorphism of S 2 , the only possibility if M ∼ = S3. Case min/max : Here M = f −1 (−∞, c] ∪f −1 (c) f −1 [c, ∞) ∼ = (B 3 q B 3 ) ∪∂ 3 3 ∼ 3 3 (B q B ) = S q S . Case none: f −1 (−∞, f (p) + ε] = f −1 (−∞, f (p) − ε] ∪f −1 (f (p)−ε) f −1 [f (p) − ε, f (p) + ε] ∼ = B 3 ∪S 2 B 3 ∼ = S 3 . Therefore f −1 (f (p) + ε) = ∂f −1 (−∞, f (p) + ε] ∼ = 3 −1 ∂S = ∅. Thus f [f (p) + ε, ∞) is a closed manifold, but f has no minimum on this manifold. Case max/min: Just as in the previous case we see that f −1 (−∞, f (p)+ε] ∼ = 3 S . But now f −1 [f (p) + ε, ∞) is a closed manifold, and f has two critical points on it, a minimum and a maximum. Therefore M ∼ = S3 q S3. Case 2/1 : By the claim above the cases we know f −1 (−∞, c]congS 2 × I. −1 f [c, ∞) ∼ = S 2 × I as well, since we can replace f with −f and apply the same claim. Therefore M ∼ = S 2 × I ∪S 2 qS 2 S 2 × I ∼ = S 2 × (I ∪∂ I) = S 2 × S 1 . −1 Case 1/2 : We first show that f (−∞, c] ∼ = S 1 × D2 . We know that −1 3 f (−∞, c] deformation retracts onto B ∪ I, but more than that, we know that it is diffeomorphic to B 3 ∪ U , where U is a small closed neighborhood of I ⊆ M . Therefore U ∼ = I × D2 , so U ∩ ∂B 3 ∼ = D2 q D2 . Since up to isotopy there is a unique way to embed two disks into S 2 in an orientation preserving manner, we see that this gluing is unique. Since S 1 ×D2 is such a gluing (taking D2 q D2 to be neighborhoods of the poles), we completed the claim. But then also f −1 [c, ∞) ∼ = S 1 × D2 , since the critical point data is the same after being turned upside down. Therefore M ∼ = S 1 ×D2 ∪T 2 S 1 ×D2 . Removing 1 the curve S ×{0} in one of the factors leaves M \S 1 ∼ = S 1 ×D2 ∪T 2 S 1 ×(D2 \{0}). 2 Any diffeomorphism of T extends to a diffeomorphism of S 1 × D2 \ {0} = T 2 ×[0, 1) (obviously). Therefore M \S 1 ∼ = S 1 ×D2 ∪T 2 T 2 ×[0, 1) ∼ = S 1 ×Int(D2 ). 3 It remains to reach the conclusion for the manifolds S and S 2 × S 1 . For 3 S = {(z, w) ∈ C2 ; |z|2 + |w|2 = 1}, we can let S 1 = {w = 0} and then S 3 \ S 1 = {|z|2 < 1; |w|2 = 1 − |z|2 } ∼ = S 1 × Int(D2 ). For the case S 2 × S 1 1 simply remove {point} × S . B d: Give an example of a closed 3-manifold M , so that any Morse function on M has at least 6 critical points. 3 Let M = T 3 . Since χ(T 3 ) = 0 we know that any Morse function has an even number of critical points, so it suffices to show that T 3 \ S 1 is not diffeomorphic to S 1 × R2 for any S 1 ⊆ T 3 . In T 3 , let α = {point} × {point} × S 1 , β = S 1 × {point} × {point}, B = {point}×T 2 , and A = T 2 ×{point}. Then [α]·[A] = ±1, [α]·[B] = 0, [β]·[A] = 0, and [β] · [B] = ±1. Since intersection product is a homotopy invariant, it follows that neither α nor β is homotopic to a constant, and they are not homotopic to each other. Furthermore the only maps f : S 1 → α, g : S 1 → β so that f and g are homotopic are degree zero, since [f ] · [A] = ± deg(f ) and etc. (we drop composition with the inclusion from the notation). Now let S 1 ⊆ T 3 be an arbitrary embedding. We can assume that α and β are disjoint from S 1 by Sard’s theorem. In S 1 × R2 , every map h : S 1 → S 1 × R2 is homotopic to a map e h : S 1 → S 1 × {0} (just compose with the deformation retraction). Since α and β are never homotopic in T 3 even after composing with a function S 1 → S 1 , they cannot be homotopic in T 3 \ S 1 either, so we see that T 3 \ S 1 is not diffeomorphic to S 1 × R2 .