Computing the homology and cohomology of connected sum

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.

• I do not understand what difficulty you are having with the M-V sequence in that example. Aug 27, 2012 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. Sep 3, 2012 at 16:31
• I do not understand your comment. Sep 3, 2012 at 17:23

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 $$M\#_{rev}N$$.

Addendum: It has been pointed out in the comments that the arguments about what happens with the maps between the various copies of $$\Bbb Z$$ are not quite complete. The key gap is showing that the map from $$\widetilde{H}_n(M\vee N)\to \widetilde{H}_{n-1}(S^{n-1})$$ is surjective when at least one of $$M,N$$ are orientable. The fix is reasonable, and is presented below to make this answer as complete as possible.

The sequence for a good pair gives that the map $$H_n(M\# N,S^{n-1})\to H_{n-1}(S^{n-1})$$ is given by taking the boundary of relative chains. Without loss of generality, $$M$$ is orientable, so consider the relative class in $$H_n(M\# N,S^{n-1})$$ which is the fundamental class of $$M$$ minus the disc we delete to perform the connected sum operation. This has boundary exactly the fundamental class of $$S^{n-1}$$ (up to possibly a sign change for orientation issues) by the definition of how we chose $$S^{n-1}$$ to glue along. As we identified $$H_n(M\# N,S^{n-1}) \cong \widetilde{H}_n(M\vee N) \cong \widetilde{H}_n(M)\oplus \widetilde{H}_n(N)$$, we see that both the map $$\Bbb Z\oplus\Bbb Z\to \Bbb Z$$ and $$\Bbb Z\oplus 0\to \Bbb Z$$ are surjective. Thus the conclusion about $$\widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N)$$ being an isomorphism when at least one of $$M,N$$ is orientable holds.

• 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 $N$. Aug 28, 2012 at 5:41
• Yes, that's certainly another way to do it. I like this approach a little better, though (personal preference, I guess). Aug 28, 2012 at 6:28
• @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? Sep 3, 2012 at 16:27
• @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) Sep 3, 2012 at 18:37
• @KReiser: The answer is still wrong as written: You are implicitly assuming that at least one of the manifolds $M, N$ is compact. If both are noncompact, then $H_{n-1}(S^{n-1})$ injects in $H_{n-1}(M\# N)$. Jan 4, 2020 at 23:39