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quasar987

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## Main Question or Discussion Point

*Question: Let M, N be orientable closed n-manifolds. Knowing that the nth homology group of an orientable closed n-manifolds is isomorphic to*

**Z**, compute the homology of M#N (the connected sum of M and N).I was thinking of using the Mayer-Vietoris sequence with A=M u (a little bit of N), B=N u (a little bit of M) such that

(1) A n B deformation retracts onto the "circle"

**S**^{n-1} where M and N are glued together,

(2) A deformation retracts onto M\

**B**^n,

(3) B deformation retracts onto N\

**B**^n,

with

**B**^n (something homeomorphic to) an open n-ball.

This way, the M-V sequence reads

[tex]\ldots\rightarrow \widetilde{H}_{i+1}(M\# N)\rightarrow \widetilde{H}_i(\mathbb{S}^{n-1})\rightarrow \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_i(N\backslash \mathbb{B}^n)\rightarrow \widetilde{H}_i(M\# N)\rightarrow \widetilde{H}_{i-1}(\mathbb{S}^{n-1})\rightarrow\ldots[/tex]

So [tex]\widetilde{H}_i(M\# N)\cong \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_i(N\backslash \mathbb{B}^n)[/tex] as soon as [itex]i\neq n-1, n[/itex]. But M#N is itself a closed n-manifold so we know that [tex]\widetilde{H}_n(M\# N)\cong\mathbb{Z}[/tex].

Now if I could just figure out what the [tex]\widetilde{H}_i(M\backslash \mathbb{B}^n)[/tex] and [tex] \widetilde{H}_i(N\backslash \mathbb{B}^n)[/tex] are...

The M-V sequence for the obvious decomposition of M into M\

**B**^n and

**D**^n with

**D**^n an n-disk covering

**B**^n gives

[tex]\ldots\rightarrow \widetilde{H}_{i+1}(M)\rightarrow \widetilde{H}_i(\mathbb{S}^{n-1})\rightarrow \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus 0\rightarrow \widetilde{H}_i(M)\rightarrow \widetilde{H}_{i-1}(\mathbb{S}^{n-1})\rightarrow\ldots[/tex]

So [tex]\widetilde{H}_i(M\backslash \mathbb{B}^n)\cong \widetilde{H}_i(M)[/tex] as soon as [itex]i\neq n-1, n[/itex], and similarly for N\

**B**^n.

Hence, [tex]\widetilde{H}_i(M\# N)\cong \widetilde{H}_i(M)\oplus \widetilde{H}_i (N)[/tex] for all [itex]i\neq n-1,n[/itex] and [tex]\widetilde{H}_n(M\# N)\cong \mathbb{Z}[/tex].

But what about [tex]\widetilde{H}_{n-1}(M\# N)[/tex] ??