Post History
#3: Post edited
by
WheatWizard
·
2023-11-30T01:20:38Z (11 months ago)
Fixed claim per comment.
- I'm interested in a proof of the following claim:
> If $M$ is an $n$-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](https://math.stackexchange.com/a/4168159/276060) 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.
- 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](https://math.stackexchange.com/a/4168159/276060) 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.
#2: Post edited
by
WheatWizard
·
2023-11-10T16:24:26Z (12 months ago)
- I'm interested in a proof of the following claim:
> If $M$ is an $n$-dimensional 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](https://math.stackexchange.com/a/4168159/276060) 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.
- I'm interested in a proof of the following claim:
- > If $M$ is an $n$-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](https://math.stackexchange.com/a/4168159/276060) 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.
#1: Initial revision
by
WheatWizard
·
2023-11-10T04:52:07Z (12 months ago)
Is the nth Betti number determined by orientability?
I'm interested in a proof of the following claim: > If $M$ is an $n$-dimensional 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](https://math.stackexchange.com/a/4168159/276060) 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.