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Q&A

Comments on Is the nth Betti number determined by orientability?

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Is the nth Betti number determined by orientability?

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I'm interested in a proof of the following claim:

If $M$ is a connected $n$-dimensional compact manifold then the $n$th Betti number, $\beta_n(M) = 1$ if $M$ is orientable and $\beta_n(M) = 0$ otherwise.

This claim seems true since it basically says that orientable manifolds have some sort of "inside" while non-orientable manifolds don't. However it seems like a basic enough claim that I would expect to see it somewhere, and I've had trouble finding anywhere making this claim.

I did find this math stackexchange answer which uses a version of this claim, however the user does not prove it, which would seem to indicate it is elementary.

I only understand homology groups at a surface level, so I would appreciate answers that stick closely to first principles. I hope that because this appears elementary that will be possible.

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1 comment thread

You probably mean to specify that M is a *connected* n-dimensional compact manifold; the nth Betti nu... (2 comments)
You probably mean to specify that M is a *connected* n-dimensional compact manifold; the nth Betti nu...
r~~‭ wrote 11 months ago

You probably mean to specify that M is a connected n-dimensional compact manifold; the nth Betti number is in fact the number of connected components if M is orientable.

WheatWizard‭ wrote 11 months ago

Yes. Thank you. It's easy to forget about connected and compact. :)