Topology Hmwk 1 All problems are from Allen Hatcher Algebraic Topology (online) ch 3.3

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1 Topology Hmwk 1 All problems are from Allen Hatcher Algebraic Topology (online) ch 3.3 Andrew Ma April 7, Show that Q[x, y]/(x 3, y 3, x 2 y 2 ) is not the rational cohomology ring of any closed orientable 195-manifold. I am presenting the proof by grad student Brandon and a person he worked with. Proof. Let α = x and β = y the dimension of each generator in the cohomology ring. We have the following list of terms with their corresponding dimensions x y xy x 2 y 2 x 2 y xy 2 α β α + β 2α 2β 2α + β α + 2β By orientability of M we have H 195 (M; Q) = Q [H, Thm 3.26 section 3.3] for a 195- manifold M = H 195 (M; Q) = 0. Thus either x 2 y or xy 2 is a generator in H 195 (M; Q) = 2α + β or α + 2β = 195. Wlog 2α + β = 195 = β is odd. However by anticommutativity this would imply y y = y y = y = 0. This is a contradiction because y 2 is clearly not 0 in this ring This is a non-rigorous example that I found from math stack exchange - I had attempted doing a rigorous example using variants of a the line with two origins, but nothing was as clear as this example - and interesting note: by the usual definitions of orientable manifolds, all 1 dimension manifolds are orientable. Proof. Consider a quotient Q of R where x x for x > 1 (this map essentially folds the ends of the real line together with a loop remaining). Using the usual sense of an orientation i.e. a consistent way to traverse along the 1 dimensional space, it follows that Q is not an orientable manifold. This is because 1

2 assigning an orientation to Im (1, ) induces an orientation on [ 1, 1] but for the orientation to remain consistent about -1 this forces a reverse orientation on Im (1, ). Hence it Q cannot be oriented Claim 1. Every cover space of a manifold M is a manifold Proof. Let p : M M be a covering space of a manifold M where I assume that this cover is surjective. By definition of a cover space [H, section 1.3 page 56] there exists {U α an open cover of M s.t. each U α lifts to disjoint sets in {U αi M each homeomorphic to U α. By the definition of a manifold [H, section 3.3 page 231] each point x M has a nbhd. homeomorphic to R n. Then by intersecting these nbhds with the {U α we may assume without loss of generality that the {U α are each sets homeomorphic to R n and they lift to disjoint open sets in M. Next, given x M say p( x) U α M. Then U α lifts homemorphically to a nbhd say U α1. It follows that x U α1 = Uα = R n. Claim 2. Every cover space of an orientable manifold M is an orientable manifold. Proof. Let p : M M be a cover space of M. By the previous claim we know that that M is manifold so I will show just the orientablity is inherited in M. By the homological definition of orientability [H, section 3.3, page 234] for each point x M there is a map x µ x H n (M x). From this define an orientation map on M by f : x µ p( x) H n (M x) = H n ( M x) Finally I will show that this map will satisfy the consistency condition. Given x M consider p( x) = x.by assumption f is a consistent orientation so there exists a ball x B s.t. y B, µ y are the images of one generator µ B H n (M B). By taking B of small enough radius we may assume that B is in an open set that lifts homeomorphically to M. Hence there exists a B = B and it follows from the definition of the map that ỹ B µỹ H n M B) = H n (M B) and hence are all images of a single generator While it is true that M, N manifolds = M N is a manifold, the converse is not true. From a post on math stack exchange The dogbone space, D, by R.H. Bing (1957) is an example of a non-manifold that has the property D R = R 4. Hence for this problem I will assume that M, N and M N are all manifolds. 2

3 Claim 3. If M, N orientable then M N is orientable. Proof. Assuming M, N are orientable there exist maps Hence we may construct a map f : x µ x H m (M x) x M g : y µ y H n (zn y) y N f g : (x, y) µ x µ y H m (M x) H n (N y) = H m+n (M N (x, y)) (x, y) M N where the isomorphism is by the Kunneth formula and the fact that for a k-manifold the homology groups higher than k are trivial [H, Theorem 3.26 section 3.3, page 236]. All that is left to check is that this orientation is consistent on the product space. Given a point (x, y) M N we may apply the consistency of orientation f to say there exists nbhd x B 1 where p B 1 f (p) = µ p = aµ B1 i.e. it is a multiple of generator µ B1 H m (M B 1 ). Similarly there is a nbhd y B 2 with µ q = bµ B2. Then (p, q) B 1 B 2 f g(p, q) = µ p µ q = ab(µ B1 µ B2 ) where is µ B1 µ B2 is the generator of H m+n H m+n (M N B 1 B 2 ) by the Kunneth formula. I will try doing an alternative method of proof because I think this might actually be a faster proof. Claim 4. If M, N are not both orientable manifolds then manifold M N is not orientable. Proof. By Theorem 3.26 on page 236 [H, section 3.3] a k-dimensional manifold K has H k (K) = Z or 0 depending on if it is orientable or not. Hence, wlog, if M is not an orientable m-dimensional manifold then H m (M) = 0 and then for M N, H m+n (M N) = H m (M) H n (N) = 0 by the Kunneth formula and hence is not orientable I got help on this problem from Evan Dummmit s online solutions. I hadn t previously considered mapping everything to the surface of a ball. I had previously been trying to produce mappings from within the region of the ball - this was a flawed idea. Really the clear thing to do is to map to boundary of the circle. I may want to consider removing this useless comment. Proof. Since M is a manifold there exists an n-dimensional ball B which is contained in M. Then consider the quotient map f : M/M B = S n since f sends M to the surface of the B. This induces the map f : H n (M) H n (M/M B) = H n (S n ) Next I will show the degree f = 1. Notice that H n (M/M B) = H n (M, M B) I actually don t know how to justify this. I know this is true for good pairs, but I assume that 3

4 it s true here too Then by the natural map provided by the local consistency condition for orientable manifolds [H, section 3.3 page 234] H n (M B) H n (M x), the generator µ b µ x. Because f factors into this map, if f ([M]) = d[µ B ] where d = ±1 then this would imply [M] will be mapped to a multiple of µ x, which is a not a generator and would contradict that [M] is a fundamental class. Hence deg f = ±1. Finally, by switching the choice of orientation on S n if needed, we can have deg f = I looked up a hint online because I forgot that H n (X Y) = H n (X) H n (Y) by Mayer- Vietoris Proof. Take n-balls B 1,... B k = B where Bi M and B N. Then since M/( k B i ) = k S n we have H n (M k B i ) = H n ( k S n ) = k H n (S n ). Thus we have the induced map Since this map factors through H n (M) f : k i H n(s n ) H n (N) f H n (N B) H n (N) and [M] is a fundamental class we conclude that f : k i H n(s n ) H n (N) [M]... [M] ɛ 1 [N]... ɛ k [N] (This reasoning is similar to that done in problem 3.3.7) I got help from this strange blog on topology. Professor Kent even posted this his answer to this problem on the blog. Claim 5. If g h then there exists a degree 1 map from M g to M h Proof. Consider M g as the 4g-gon with edges labeled a i, b i for i = 1 to g in the following order [a 1, b 1 ][a 2, b 2 ]... [a h, b h ]... [a g, b g ]. Take the map f to be the quotient map M g M g /[a h+1, b h+1 ]... [a g, b g ] that quotients some of the edges of the 4g-gon together. Since M g /[a h+1, b h+1 ]... [a g, b g ] = M h and f preserves the two-cell it must be a degree 1 map. Claim 6. If g < h then there does not exist a degree 1 map from M g to M h 4

5 Proof. There are two ways to prove this. The first way is to apply the problem from Hatcher in which it states that a degree 1 map induces a surjective map on the first fundamental groups. However by examining the fundamental groups of M g and M h for g < h we see that such a surjection is not possible. In the second method, we will use cup products. It is straight forward to compute the cohomology groups of these spaces and doing so gets H 1 (M g ; Z) = Z 2g, H 1 (M h ; Z) = Z 2h. Any map f : M g M h must induce the map f : Z 2h Z 2g and since g < h, ker f =. Let α = 0 ker f, α will have infinte order, and by Corollary 3.39 on page 250 [H, section 3.3] there exists a β H 1 (M h ; Z) s.t. α β generates H 2 (M h ; Z). This will also generate H 2 (M h ; Z) since H 2 (M h ; Z) = H 2 (M h ; Z) by universal coefficients. By similar reasoning, the image f (α β) will generate H 2 (M g ; Z). However, by naturality of the cup product f (α β) = f (α) f (β) = 0 and so any map f : M g M h sends the fundamental class of H 2 (M g ; Z) to zero and therefore is degree Given a 3-manifold M with H 1 (M; Z) = Z r F for a finite group F. Claim 7. If M is orientable then H 2 (M; Z) = Z r Proof. By Poincare Duality on page 245 we have H 1 (M; Z) = H 2 (M; Z) [H, Thm 3.35], section 3.3]. By Universal coefficients formula we have H 1 (M; Z) = Ext(H 0 (M)) hom(h 1 (M; Z); Z) = 0 Z r because H 0 (M) = Z by connectedness = Z r Claim 8. If M is nonorientable then H 2 (M; Z) = Z r 1 Z 2 Proof. Duality doesn t hold for nonorientable manifolds so I ll use Euler characteristic. We have the following betti numbers or ranks rank H 0 (M; Z) = 1 rank H 1 (M; Z) = r rank H 0 (M; Z) = x rank H 0 (M; Z) = 0 where x is to be determined Note that the last rank is due to M being nonorientable (as discussed on page 236)[H, Thm. 3.26, section 3.3]. Since the Euler characteristic of a closed manifold of odd dimesions is 0 [H, Cor 3.37, section 3.3] we have that 1 r + x = 0 = x = r 1. Finally, the torsion part of H 2 (M; Z) must be Z 2 [H, Cor 3.28, section 3.3] hence we conclude that H 2 (M; Z) = Z r 1 Z 2. 5

6 To construct this example I made needed to use the following solution by Hal Sodofsky for the idea of taking a space with a group action to get the F term (this will make sense in the context below. sadofsky/636/hw10-solutions.pdf Interestingly, from searching around online, in the same idea as this problem there exists a way to prime factor 3-manifolds. Claim 9. There exists a closed, connected, orientable 3-manifold with H 1 (M; Z) = Z r F Proof. Following the hint I will be using exercise 6 without proof. To start I will produce a space, X, that has H 1 (X) = Z. Take X = S 1 S 2, without proof I assume that this is a closed connected manifold. By direct computation of the homology groups using the Kunneth formula H 3 (X) = Z, hence X is an orientable manifold. Similarly one may compute that H 1 (X) = Z. Next I will produce a space Y n s.t. H 1 (Y n ) = Z n. Take S 3 with a Z n group action by rotation about one axis of S 3. Without proof I assume that this is satisfactory group action on a space for which I may apply Proposition 1.40 from page 72 [H, section 1.3]. Let Y n = S 3 /Z n and by proposition 1.40 Now, if F = Z a1 Z an then let Z n = π 1 (Y n ) = H 1 (Y n ) S = X#. {{.. #X #Y a1... #Y an r-times Then by exercise 6 H 1 (S) = H 1 (M; Z) = Z r F. Finally I assume without proving that the connected sum of orientable manifolds is an orientable manifold. (One can t apply exercise 6 to say this but it seems intuitively true). Here I also borrowed from Hal Sodofsky s idea that N = S 1 RP 2 to get this to work but I couldn t prove this as the problem stated. Claim 10. There exists a closed, connected, nonorientable 3-manifold with H 1 (M; Z) = Z r F where F = Z 2 Z a1 Z an. Proof. I believe that the manifold N describe in the hint is S 1 RP 2. I assume without proof that N is a closed connected manifold. By computation using Kunneth formula H 1 (N) = Z Z 2 and H 3 (N) = 0. Hence N is nonorientable and by taking S = X#. {{.. #X #N#Y a1... #Y an r 1 times by exercise 6 H 1 (S) = Z r F. Additionally I assume without proof that S is nonorientable because N is nonorientable and used in the connected sum. 6

7 Proof. If H k 1 (M; Z) is torsion free then by the Universal coefficients theorem H k (M; Z) = hom(h k (M); Z). Next by Poincare Duality [H, Thm 3.30] H k (M; Z) = H k (M; Z). Together we conclude H k (M; Z) = hom(h k (M); Z) and it follows that H k (M; Z) cannot have a torsion part since a torsion term would not appear in hom(h k (M); Z). 7