24
$\begingroup$

Suppose $M$ and $N$ are two connected oriented smooth manifolds of dimension $n$. Conventionally, people use $M\#N$ to denote the connecte sum of the two. (The connected sum is constructed from deleting an $n$-ball from each manifold and glueing the boundary.) I think there should be a general approach of finding the homology and cohomology groups of $M\#N$ from the homology and cohomology groups of $M$ amd $N$. However, using Mayer-Vietoris sequence becomes quite complicated whent the spaces are complicated. For example: 1) $M=S^1 \times S^3$, $N= \mathbb{CP}^2$. The difficulty comes from understanding the map $H^i(U\cap V)\rightarrow H^{i+1}(M\#N)$.

If we define the $M\#_{rev}N$ to be the connected sum, however, glueing the $n$-ball in the reversed order, then we can ask the same problem. the case $n=2$ is easier to think of the picture when the orientation is reversed. It is also easier since the glueing is "fixed" in the sense there is only one parameter discribing the boundary of $n$-ball, aka, $S^1$. What happens when $n$ goes higher?

I would appreciate it if someone could explain this general question with examples ($M\#N$ and $M\#_{rev}N$, where $M=S^1 \times S^3$, $N= \mathbb{CP}^2$) in the context.

  • 1
    I do not understand what difficulty you are having with the M-V sequence in that example.2012-08-27
  • 0
    Dear @MarianoSuárez-Alvarez, I think I can compute the homology of $M$ retracts a $n$-ball using relative homology. However, $M\#N$ is a union of $M-D_n$ and $N-D_n$, where a ball is removed other than collapsed. I could not really compute the homology of $U$ and $V$ (in your comments) from that of $M$ and $N$. If this is done, the M-V is easier.2012-09-03
  • 1
    I do not understand your comment.2012-09-03

1 Answers 1

34

The procedure for finding homology and cohomology of the spaces in question is a neat little trick. From here on out, I'll just treat the homology case, but the cohomology follows from the same arguments. Collapse the $S^{n-1}$ you're gluing along to a point- this turns $M\# N$ into $M\vee N$. Since $(M\# N, S^{n-1})$ is a good pair, the homology can be identified with the relative homology of the pair $(M\# N,S^{n-1})$. From this, we get the following long exact sequence:

$$\cdots\to \widetilde{H_i}(S^{n-1})\to \widetilde{H_i}(M\# N) \to \widetilde{H_i}(M\vee N)\to\cdots$$

By a simple Mayer-Vietoris argument, we have that $\widetilde{H}_i(M\vee N)\cong \widetilde{H}_i(M)\oplus \widetilde{H}_i(N)$. Since $\widetilde{H_i}(S^{n-1})$ is zero except for $i=n-1$, we have automatically that $H_i(M\# N)\cong H_i(M\vee N)\cong H_i(M)\oplus H_i(N)$ for $i\neq n-1,n$. The only interesting case is as follows:

$$0\to \widetilde{H_n}(M\# N)\to \widetilde{H_n}(M\vee N) \to \widetilde{H}_{n-1}(S^{n-1})\to \widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N) \to 0$$

Now, we start getting into some casework depending on whether none, one, or both of $M,N$ are orientable. In the case that both are orientable, the above sequence turns into

$$0\to \mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z} \to \widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N) \to 0$$

as their connected sum is also orientable. From this, we see that $\widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N)$ must be an isomorphism.

If just one of $M,N$ is orientable, then their connected sum is non-orientable, and the following happens:

$$0\to 0 \to \mathbb{Z}\oplus0 \to \mathbb{Z} \to \widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N) \to 0$$

in which case we still have that that $\widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N)$ must be an isomorphism.

If neither of $M,N$ are orientable, then their connected sum is non-orientable, in which case the long exact sequence does the following:

$$0\to 0 \to 0 \to \mathbb{Z} \to \widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N) \to 0$$

and thus $\widetilde{H}_{n-1}(M\# N)$ is an extension of $\widetilde{H}_{n-1}(M\vee N)$ by $\mathbb{Z}$. To figure out what extension it is, one needs to inspect the map $S^{n-1}\to M\# N$ and the corresponding map on homology. Nothing too surprising can happen- $H^{n-1}(M\# N)$ is the direct sum of a free abelian group and a finite abelian group.

Note that during this argument, it was never necessary to talk about the orientation of the gluing- so $M\# N$ and $M\#_{rev}N$ have the same homology/cohomology. No description of $S^{n-1}$ was ever used except for it having reduced homology only in degree $n-1$, so the process does not care very much about what dimension your manifolds are.

Now, for the example where $M=S^1\times S^3$ and $N=\mathbb{C}P^2$. $M$ has homology $H_0\cong H_1\cong H_3\cong H_4\cong \mathbb{Z}$ and all other groups zero, while $N$ has homology $H_0\cong H_2\cong H_4\cong \mathbb{Z}$ and all other groups zero. Using the procedure above, we have that the homology of $M\# N$ is as follows: $H_0\cong H_1\cong H_2\cong H_3\cong H_4\cong \mathbb{Z}$. The result is the same for $M\#_{rev}N$.

  • 1
    There is no need to collapse the sphere, no? You can just consider the open cover of the connected sum given by the two open sets which result in removing slightly smaller closed balls from $M$ and $N$.2012-08-28
  • 0
    Yes, that's certainly another way to do it. I like this approach a little better, though (personal preference, I guess).2012-08-28
  • 1
    @KReiser. Sorry I left someday reading the reduced Homology before I get to understand this solution. One place I got little confusion is the first exact sequence, which might be $\cdots\to H_i(S^{n-1})\to H_i(M\#N)\to \widetilde{H_i}(M\vee N)\to\cdots$. The last tilde comes from the isomorphism $H_i(M\#N, S^{n-1})\cong H_i(M\#N/S^{n-1})\cong \widetilde{H_i}(M\vee N)$. It this correct? I would also like to ask a question about orientation. I think I can get the top homology of oriented smooth manifold is $\mathbb{Z}$, but how is that true for a general topological space?2012-09-03
  • 3
    @Honghao, The first exact sequence is correct as stated and can be referenced as theorem 2.14 in Hatcher. It is important to note that $\widetilde{H}_i(M\# N/S^{n-1})\cong \widetilde{H}_i(M\vee N)$ can be very false if $i=n-1,n$ (this is the whole reason why we need to talk about the exact sequence!). In regards to your last question, connected sum is an operation which only works for manifolds, so we're not losing any generality by talking about orientable/non-orientable manifolds here. In general, the situation for an arbitrary topological space and orientation is hopeless, but (con't)2012-09-03
  • 0
    (con't) when things follow such nice rules as manifolds, lots more can be said.2012-09-03
  • 0
    @KReiser. Thanks for your reply. What I meant by $M\# N/S^{n-1}$ was the quotient topology which is homeomorphic to $M\vee N$ and we should have the isomorphism in homology. Could you please review the equation again? I would also like to ask how do you think if we approach the question by removing a disc from $M$ and $N$(That is how we get $U$ and $V$ in M-V sequence). How can we get the (co)homology of $U$ and $V$?2012-09-05
  • 0
    The original post is correct as stated, and the explanation in your comment is essentially right, except you need to be careful with which homology is reduced and which isn't. The correct formula is for a good pair, $H_i(A,B)\cong \widetilde{H}_i(A/B)$ As far as using a small removed disc, just go through the M-V sequence and recognize that since $U$ and $V$ are $n$-discs, they are contractible and thus have trivial (co)homology outside dimension 0.2012-09-05
  • 0
    The Klein Bottle $K$ is the connect sum of two projective planes $P$. So your answer in the third case gives >$\mathbb{Z}\oplus\mathbb{Z}_2=\tilde{H}_{1}(K)=\tilde{H}_1(P)\oplus \tilde{H}_1(P)\oplus\mathbb{Z}=\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}$ What am I missing?2013-08-28
  • 0
    A typo on my part. If one has a short exact sequence $0\to A\to B\to C\to 0$, $B$ is termed an extension of $C$ by $A$ and the isomorphism classes of such are given by $\mathrm{Ext}(C,A)$. The post originally had this flipped. To answer your specific question, the short exact sequence in question ought to be $0\to \mathbb{Z}\stackrel{(2,0)}{\rightarrow} \mathbb{Z}\oplus\mathbb{Z}/2 \to \mathbb{Z}/2\oplus\mathbb{Z}/2$. In the 2 non-orientables case, one has to inspect the map $S^{n-1}\to M\# N$ and the corresponding map on homology. So the picture isn't quite as rosy as I painted.2013-08-28
  • 0
    If you have the time, could you explain how you found what the short exact sequence in reference to my question was? Specifically the $(2,0)$ map?2013-08-28
  • 0
    I drew the polygonal representation of a connect sum of two real projective planes, drew in the gluing circle, and then cut/pasted to get the usual diagram for a Klein bottle. Tracking the curve I drew earlier gave that it is homotopic to 2 times the generator of the $\mathbb{Z}$ summand of the homology. I'm not completely satisfied with this approach in terms of teaching, but it's how I attacked it.2013-08-30
  • 0
    @KReiser sorry for upping an old post, but I don't see the reason behind the fact that $$\widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N)$ must be an isomorphism. I tried searching on other questions, but all of the people gave this iso as a "triviality" :( Could you add some details? thanks in advance!2015-04-11
  • 0
    Compute the map. The generator of $\widetilde{H}_{n-1}(M\# N)$ is the funamental class, which is mapped to the sum of the fundamental classes of $M$ and $N$ included in $M\wedge N$, so the map $\mathbb{Z}\to \mathbb{Z}\times\mathbb{Z}$ is $1\mapsto (1,1)$. The quotient of $\mathbb{Z}\times\mathbb{Z}$ by $\mathbb{Z}$ embedded in this way is $\mathbb{Z}$, so you can break up the long exact sequence into a direct sum of $0\to\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\to 0$ and the relevant homology groups.2015-04-13
  • 0
    @KReiser: However, this is still not enough. You need to show that the map $\tilde{H_n}(M \vee N) \to \tilde{H}_{n-1}(S^{n-1})$ is surjective. The description of the map you gave only computes the kernel of the map I mentioned.2018-01-18