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Let $M$ be an oriented, compact, connected $n$-dimensional smooth manifold with boundary. Then, is it true that $n$-th singular homology of M, that is $H_n(M)$, is vanish? I can't make counterexamples for this statement, but I don't have the slightest idea for proof.

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migrated from mathoverflow.net Dec 12 '13 at 9:55

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    $\begingroup$ Hint: use a collar to show that $M$ is homotopic to a noncompact $n$-manifold without boundary. $\endgroup$ – Dan Petersen Dec 12 '13 at 9:27
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Just to complete the details of Dan's hint. The top homology of a non-compact manifold vanishes, e.g. by Poincare duality $H_n(M) \cong H^0_c(M)$ where $H^*_c(M)$ is cohomology with compact support. But $H^0_c(M)$ is just the kernel of $d : C^0_c(M) \to C^1_c(M)$. Therefore $C^0_c(M)$ consists of constant functions and because $M$ is non-compact the only such function is the zero function.

Alternatively, one can see the result directly from Lefschetz duality, a generalization of Poincare duality. This gives $H_n(M) \cong H^0(M,N)$ where $N$ is the boundary of $M$. But $H^0(M,N) \cong \tilde H^0(M/N) = 0$.

I'm sure there are many other and probably simpler ways to see this. But the upshot is that non-compact manifolds and manifolds with boundary are simpler from the point of view of homotopy theory than their closed cousins (just thinking of ${\bf R}^n$ which has homototy type of a point).

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  • $\begingroup$ Thank you a lot! Your answer help me understand all! $\endgroup$ – user115322 Jan 10 '14 at 14:10
  • $\begingroup$ In fact, that H^n(M)=0 for the deRham cohomology for noncompact manifolds is proved in Spivak (Vol1, chapter 8), but he uses a "Problem 20" in Chapter 8 that I think it is not correct (a manifold with two ends and infinite topology should be a counterexample). Do you know an elementary proof (e.g., no Poincare duality or simplicial homology) of this without using the Problem 20? $\endgroup$ – Luis A. Florit Oct 20 '14 at 16:11
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The top dimensional cycle of an orientable triangulated manifold without boundary is an oriented sum of its n-simplices. Each n-simplex shares each of its n-1-faces with exactly one other n-simplex and these two n-1 faces cancel each other out under the boundary operator. One can show that this cycle is unique up to orientation.

With a manifold with boundary the n-1 simplices on the boundary are not paired off and can not cancel out. Thus any cycle must exclude them. But if one removes the n-simplices that contain them then none of their interior n-1 faces can be cancelled. Continue removing n-simplices until there is nothing left.

This shows that the top homology is zero.

A good reference is Spanier's Algebraic Topology

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  • $\begingroup$ Thanks! This answer is very interesting! $\endgroup$ – user115322 Jan 10 '14 at 14:12
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You can prove that $H_n(M, \mathfrak d M)\cong \mathbb Z$ in essentially the same way as the special case $\mathfrak d M=\emptyset$, and also that for $\mathfrak d M\ne\emptyset$ the connecting homomorphism $\mathfrak d\colon H_n(M, \mathfrak d M)\to H_{n-1}(\mathfrak d M)$ is an isomorphism. It takes a relative orientation cycle to an orientation cycle of the boundary. The result now follows from looking at the long exact sequence $$0=H_n(\mathfrak d M)\to H_n(M)\to H_n(M, \mathfrak d M)\xrightarrow{\cong} H_{n-1}(\mathfrak d M).$$

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