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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.

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I do not understand what difficulty you are having with the M-V sequence in that example. –  Mariano Suárez-Alvarez Aug 27 '12 at 21:19
    
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. –  Honghao Sep 3 '12 at 16:31
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I do not understand your comment. –  Mariano Suárez-Alvarez Sep 3 '12 at 17:23

1 Answer 1

up vote 9 down vote accepted

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 $M\#_{rev}N$.

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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 $N$. –  Mariano Suárez-Alvarez Aug 28 '12 at 5:41
    
Yes, that's certainly another way to do it. I like this approach a little better, though (personal preference, I guess). –  KReiser Aug 28 '12 at 6:28
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@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? –  Honghao Sep 3 '12 at 16:27
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@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) –  KReiser Sep 3 '12 at 18:37
    
(con't) when things follow such nice rules as manifolds, lots more can be said. –  KReiser Sep 3 '12 at 18:47

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