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I would like to know what the singular relative cohomology $H^2(M,\partial M;\mathbb{Z})$ of a smooth connected surface with boundary $M$ is. In the orientable case I did the following: The zero-th singular homology $H_0(M;\mathbb{Z})$ is zero since the surface is connected. Then, by Lefschetz duality, $H^2(M,\partial M;\mathbb{Z})$ is also zero. In the non-orientable case, I tried considering the orientable double cover $\tilde{M}$. By the previous argument we get $H^2(\tilde{M},\partial \tilde{M};\mathbb{Z})=0$. However I don't know how to relate this to the cohomology of the surface. Can you help me?

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The relative long exact sequence gives us the exactness of $H^1(M) \to H^1(\partial M) \to H^2(M,\partial M) \to H^2(M)$. The last term is zero ($M$ deformation retracts onto a 1-complex; to see this, think about the picture of closed manifolds as polygons mod edge identifications). Call $M'$ the manifold you get when you glue in discs to each of the boundary circles; we have a Mayer-Vietoris sequence $H^1(M) \to H^1(\partial M) \to H^2(M') \to H^2(M) = 0;$ and for a non-orientable manifold, $H^2(M') = \Bbb Z/2$ by the universal coefficient theorem. So $H^2(M,\partial M) = \Bbb Z/2$.

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  • $\begingroup$ Thank you very much for your answer! I am having trouble understanding some aspects of it (and probably in general). I read that the top cohomology determines an orientation on the manifold, so in the orientable case, shouldn't the cohomology be non-zero, contrary to what I obtained first? And in the non-orientable case, shouldn't it be zero? I am really confused about this. $\endgroup$ – Sak Jun 16 '16 at 17:45
  • $\begingroup$ @Sak $H^n(M) = \Bbb Z$ if and only if $M$ is closed and orientable, and an isomorphism determines an orientation. Here your manifold is neither closed nor orientable. In the non-closed case, replace that with $H^n(M,\partial M)$. It's still not $\Bbb Z$ here; it's $\Bbb Z/2$. $\endgroup$ – user98602 Jun 16 '16 at 20:30
  • $\begingroup$ So if I am understanding correctly the theorem for manifolds with boundary would be that if $M$ is non-orientable then $H^n(M,\partial M)=0$? Could you suggest a reference where I can read this result? Thank you very much $\endgroup$ – Sak Jun 16 '16 at 20:44
  • $\begingroup$ @Sak That's clearly false, since we just proved it's $\Bbb Z/2$ for all compact non-orientable surfaces. The theorem you want is "$H^n(M,\partial M) \cong \Bbb Z$ if and only if $M$ is compact and orientable." Perhaps you'd feel more comfortable working with homology instead of cohomology, where it is true that $H_n(M,\partial M) \neq 0$ iff $M$ is compact and orientable (indeed, when it's nonzero it's automatically $\Bbb Z$.) $\endgroup$ – user98602 Jun 16 '16 at 20:46
  • $\begingroup$ I am trying to prove that fiber bundles over a smooth surface $M$ with boundary with fibre a circle always have sections. I thought about using a theorem of obstruction theory which says that they do, whenever $H^2(M,\partial M; \pi_1(S^1))=0$. Supposedly this should be true but now I am confused since you proved that this is not the case. $\endgroup$ – Sak Jun 16 '16 at 20:53

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