Comments on Is the nth Betti number determined by orientability?
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Is the nth Betti number determined by orientability?
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.
Post
If you are an algebraic topologist then you might take $\beta_n(M)=1$ as the definition of orientable.
So to answer your question, we first have to decide on a different definition of "orientable" than $\beta_n(M)=1.$ Usually the alternative notions of orientability are calculus based, about either the tangent bundle having a chart with compatible determinants of transition functions, or equivalently that $\Lambda^n(M),$ also known as the determinant bundle is trivial.
The following are equivalent:
- $H^n(M,\mathbb{Z})\cong\mathbb{Z}$ (call the generator $[M]$).
- The $n$th Betti number is one.
- The bundle of local homology groups $H^n(M,M\setminus{x})$ is trivial
- The tangent bundle $TM$ is orientable.
- The determinant bundle $\Lambda^n(M)$ is trivial.
Any of these can be taken as the definition of orientable. Note that the homological definitions depend on the coefficients used. I am assuming $\mathbb{Z}$ coefficients throughout. If you instead use $\mathbb{Z}/2$ coefficients then all manifolds are orientable. (So this is the right homology to use when you need a Poincaré duality that works for (classically) nonorientable manifolds)
The equivalence of 1,2, and 3 can be found in algebraic topology textbooks (eg Hatcher). Actually 1 and 2 are just different notations saying the same thing. But the idea of the proof of $1\to 3$ is that a principal bundle is trivial iff it has a section. $[M]$ restricts to an element of every local homology group, and we take that as our section. We just have to show it is continuous. Since the fiber is discrete, continuous means locally constant. We have isomorphisms between $H^n(M,M\setminus x)\cong H^n(M,M\setminus U) \cong H^n(M,M\setminus y),$ where $U$ is a neighborhood containing $x$ and $y$.
And the equivalence of 4 and 5 can be seen in textbooks on smooth manifolds (eg Lee). What might be harder to find in your textbook is a discussion showing the equivalence of the algebraic definition and the smooth definition. So to see $3\to 4$, suppose a point $x$ is in two neighborhoods $U$ and $V$ with charts $\Phi$ and $\Psi$. The key point is that the transition function between $H^n(M,M\setminus U)$ and $H^n(M,M\setminus V)$ will be plus one (resp. minus one) iff the transition function $\Phi\circ\Psi^{-1}$ is orientation preserving (resp. reversing) which is determined by the sign of the Jacobian determinant of $\Phi\circ\Psi^{-1}$.
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