Math 225B: Differential Geometry, Homework 8

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1 Math 225B: Differential Geometry, Homewor 8 Ian Coley February 26, 204 Problem.. Find H (S S ) by induction on the number n of factors. We claim that H (T n ) ( n ). For the base case, we now that H 0 (S ) H (S ) R, which is consistent with ( ( 0) ). Assume the case of n. Then we may split T n A B, where A T n S \ {N} and B T n S \ {S}, the north and south poles of that particular circle. Then A B is two disconnected copies of T n, and A and B are homotopy equivalent to T n. Then by induction, we use the Mayer-Vietoris sequence H (A) H (B) H (A B) H (T n ) H (A) H (B) H (A B) which in our case is H (T n ) 2 H (T n ) 2 H (T n ) H (T n ) 2 H (T n ) 2. Going by dimensions, we have ( ) ( ) ( ) ( ) n n n n 2 2 H (T n ) 2 2. The image of the first map has dimension ( n ) since it is essentially projection onto one of the disconnected factors. Therefore the image of the second map in H (T n ) has dimension ( n ) (. Similarly, since the ernel of the last map has dimension n ), the image of the third map has dimension ( ) n. Therefore the total dimension of H (T n ) is ( ) n + This completes the proof. ( ) n (n )! ( )!(n )! + (n )!!(n )! (n )! + (n )(n )!!(n )! n(n )!!(n )! ( ) n. Problem.2. (a) Use the Mayer-Vietoris sequence to determine H (M \ {p}) in terms of H (M), for a connected manifold M.

2 (b) If M and N are two connected n-manifolds, let M#N be be the connected sum. Find the cohomology of M#N in terms of that of M and N. (c) Find χ for the n-holed torus. (a) Let dim M n, and assume n > so that M \ {p} is connected, since we now the cohomology of all connected -manifolds. Let U be a chart around the point {p}, so that U M \ {p} M. Then U R n, and U M \ {p} S n. Therefore we have a sequence H (S n ) H (M) H (R n ) H (M \ {p}) H (S n ) H + (M), where > 0 since we now what H 0 (M \ {p}) is since we assumed it was connected. For < n, we have 0 H (M) H (M \ {p}) 0 so H (M) H (M \ {p}). The last part of this sequence, starting at n, is 0 H n (M) H n (M \ {p}) R H n (M) H n (M \ {p}) 0 If M is nonorientable or noncompact, then H n (M) 0, so we obtain 0 H n (M) H n (M \ {p}) R 0 H n (M \ {p}) 0, so H n (M \ {p}) 0 as well and dim H n (M \ {p}) + dim H n (M). If M is compact and orientable, then H n (M) R. However M \ {p} is not compact anymore, so H n (M \ {p}) 0. Hence we obtain 0 H n (M) H n (M \ {p}) R R 0. Since R R 0 is surjective, H n (M \ {p}) R must be the zero map. Hence we again have 0 H n (M) H n (M \ {p}) 0 so H n (M) H n (M \ {p}). This completes the classification. (b) We can deconstruct the connected sum via A M\{p}, B N\{q}, and A B S n. Essentially, A is M plus part of the connecting tube, B is N plus part of the connecting tube, and A B is the tube itself. First, since the connected sum of connected manifolds is connected, H 0 (M#N) R. We have the sequence H (M \ {p}) H (N \ {q}) H (S n ) For < n, we have (from the results above) H (M#N) H (M \ {p}) H (N \ {q}) H (S n ) 0 H (M#N) H (M) H (N) 0, 2

3 so H (M#N) H (M) H (N). For the end of the sequence, 0 H n (M#N) H n (M \ {p}) H n (N \ {q}) R H n (M#N) H n (M \ {p}) H n (N \ {q}) 0. We have two cases here. If either M or N is noncompact or nonorientable, then the connected sum M#N will also be noncompact or nonorientable, so H n (M#N) 0. Therefore we have 0 H n (M#N) H n (M \ {p}) H n (N \ {q}) R 0. Therefore dim H n (M#N) dim H n (M \ {p}) + dim H n (N \ {q}), where the dimension of the individual manifolds minus a point are calculated as in (a), depending on their top homology. If both M and N are both compact and orientable, then so is their connected sum, so H n (M#N) R. We also have H n (M \ {p}) H n (N \ {q}) 0. Putting this together, 0 H n (M#N) H n (M) H n (N) R R 0. At the end of this sequence, we have R 0 has zero image, so R R is an isomorphism. Therefore the image of the direct sum to R is zero, so we have 0 H n (M#N) H n (M) H n (N) 0. Thus dim H n (M#N) dim H n (M) + dim H n (N). This completes the classification. (c) We use the definition that χ(m) n ( ) i dim H i (M). i0 The n-holed torus is the connected sum of n tori. We proceed by induction. Let T n denote the n-holed torus. Then Assume the n case. Then χ(t ) (). χ(t n ) χ(t #T n ) 2 + H (T #T n ) since each T g is compact, orientable, and connected. In particular, we see that dim H (T n ) 2(n ). From above, dim H (T #T n ) dim H (T ) + dim H (T n ) 2 + 2(n ) 2n. Therefore χ(t n ) 2 2n, as claimed. This completes the proof. 3

4 Problem.3. (a) Find H (Möbius strip). (b) Find H (P 2 ). (c) Find H (P n ). (d) Find H (Klein bottle). (e) Find the cohomology of M#(Möbius strip) and M#(Klein bottle) if M is the n-holed torus. (a) The Möbius strip is a nonorientable connected 2-manifold. Therefore H 0 (M) R and H 2 (M) 0. For H, we apply Problem.2(a). Since M \ {p} S for an interior point p, we have H (M) H (S ) R. (b) Consider P 2 A B, where A {[x : ] : x R} and B {[x : y] : x, y R, y 0}. Then A {p} and B P S, and A B S as well. Then we have H 0 (P 2 ) R since it is connected and H 2 (P 2 ) 0 since it is nonorientable. For H (P 2 ), and in our case, 0 H (P 2 ) H (A) H (B) H (A B) 0, 0 H (P 2 ) R R 0. Since R R 0 is an isomorphism, we have the image of H (P 2 ) R is zero. Since this map is also injective, we have H (P 2 ) 0. (c) We proceed by induction on n, using the above as a guideline. We claim that { H (P n R 0 or n if n is odd ). 0 else This clearly holds for the case of n 2 above. Let A {[x 0 : : x n : ] : x i R} and B {[x 0 : : x n ] : x i R, x j 0 for j n}. Then A B P n, A B S n, A {p}, and B P n. H 0 (P n ) R is clear. For < n, we have 0 H (A B) H (P n ) H (A) H (B) 0, so H (P n ) 0. At the end of this sequence, starting at n, we have 0 H n (P n ) H n (A) H n (B) H n (A B) H n (P n ) 0. Assume first that n is odd. Then n is even, so H n (A) H n (P n ) 0. Hence we have 0 H n (P n ) 0 R H n (P n ) 0, 4

5 so H n (P n ) 0 and H n (P n ) R. If n is even, then H n (A) R, so we have 0 H n (P n ) R R H n (P n ) 0. Then R H n (P n ) 0 is surjective, so R R is an injection. Hence the image of 0 H n (P n ) R is 0, and since it is an injection, this implies that H n (P n ) 0. Hence 0 R R H n (P n ) 0 implies that H n (P n ) 0 as well. This completes the proof. (d) The Klein bottle is homotopy equivalent to P 2 #P 2. Since this space is nonorientable and connected, we have H 0 (K) R and H 2 (K) 0. From Problem.2(b), we have dim H (K) 2 dim H (P 2 \ {p}). From Problem.2(a) and Problem.3(b), we have dim H (P 2 \ {p}) + dim H (P 2 ). Therefore dim H (K) 2, so H (K) R. (e) Since both the Möbius strip M and Klein bottle K are nonorientable, we have H 0 R and H 2 0 in both cases. Now, we have dim H (T n #M) dim H (T n \ {p}) + dim H (M \ {q}). Since T n is compact and orientable, dim H (T n \ {p}) dim H (T n ) 2n Since M \ {p} S, we have dim H (M \ {q}). Hence dim H (T n #M) 2n. Now, Since K is nonorientable, we have dim H (T n #K) 2n + dim H (K \ {q}). dim H (K \ {q{) + dim H (K) 2. Therefore we have dim H (T n #K) 2n +. This completes the proof. 5