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I know two definitions of an orientation of a smooth n-manifold $M$:

1) A continuous pointwise orientation for $M$.

2) A continuous choice of generators for the groups $H_n(M,M-\{x\})=\mathbb{Z}$.

Why are these two definitions equivalent? In other words, why is a choice of basis of $\mathbb{R}^n$ equivalent to a choice of generator of $H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})=\mathbb{Z}$?

See comments for precise definitions.

Thanks!

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    $\begingroup$ The first definition you have given is called a parallelization, not an orientation. The two concepts are not equivalent. See en.wikipedia.org/wiki/Parallelizable. $\endgroup$
    – Jim Belk
    Commented Jun 7, 2011 at 1:29
  • $\begingroup$ @Jim Belk, I disagree, a pointwise choice of basis for each tangent space is continuous if every point of $M$ is in the domain of an oriented local frame. If there is a global frame then the manifold is parallelizable. $\endgroup$
    – Manuel
    Commented Jun 7, 2011 at 1:46
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    $\begingroup$ @Manuel I do not agree. What is your precise definition of a continuous pointwise choice of basis for the tangent space? $\endgroup$
    – Jim Belk
    Commented Jun 7, 2011 at 2:14
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    $\begingroup$ A pointwise orientation on $M$ is a choice of orientation of each tangent space. A local frame over a nieghborhood $U$ in $M$ is an ordered n-tuple $(e_1,...,e_n)$ of vector fields over $U$ such that $(e_1(p),...,e_n(p))$ is a basis for $T_pM$ for each $p\in U$. A local frame over $U$ is oriented if $(e_1(p),...,e_n(p))$ is positively oriented for each $p\in U$. As I said above, a pointwise orientation on $M$ is continuous if every point of $M$ is in the domain of an oriented local frame. I will edit my question so there is no ambiguity. $\endgroup$
    – Manuel
    Commented Jun 7, 2011 at 2:28
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    $\begingroup$ See this answer to a very similar MO question: mathoverflow.net/questions/10966/… $\endgroup$
    – Matt E
    Commented Jun 7, 2011 at 2:41

2 Answers 2

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Recall that an element of $H_n(M,M-\{x\})$ is an equivalence class of singular $n$-chains, where the boundary of any chain in the class lies entirely in $M-\{x\}$. In particular, any generator of $H_n(M,M-\{x\})$ has a representative consisting of a single singular $n$-simplex $\sigma\colon \Delta^n\to M$, whose boundary lies in $M-\{x\}$. Moreover, the map $\sigma$ can be chosen to be a differentiable embedding. (Think of $\sigma$ as an oriented simplex in $M$ that contains $x$.)

Now, the domain $\Delta^n$ of $\sigma$ is the standard $n$-simplex, which has a canonical orientation as a subspace of $\mathbb{R^n}$. Since $\sigma$ is differentiable, we can push this orientation forward via the derivative of $\sigma$ onto the image of $\sigma$ in $M$. This gives a pointwise orientation on a neighborhood of $x$.

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  • $\begingroup$ Will the converse follow from the existence of solutions for ODE's? Given an oriented local frame on a neigborhood $U\subset M$ of $x$ we want to find a smooth non zero cycle $\sigma: \Delta^n \rightarrow M$ such that $\partial \sigma$ lies in $M-\{x\}$ the pushforward of the orientation of $\Delta^n$ is the one determined by the frame... $\endgroup$
    – Manuel
    Commented Jun 7, 2011 at 19:27
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Observe that in (1) there is no difference between using tangent and cotangent bundle, and in (2) one can use $H^n$ instead of $H_n$.

Now, the equivalence becomes especially clear if in (2) one uses de Rham cohomology (instead of, say, singular).

Indeed, (1) is just existence of a (non-vanishing) section $\omega$ for $\Lambda^{top} T^*M$. So $\omega$ is a differential form, and for any $x\in U$ one can take a function $f_U$ that is 1 near $x$ and 0 outside of $U$ — and $\omega\cdot f_U$ is a generator of $H^n_{dR,c}(U)=H^n(M,M-\{x\})$. And using partitions of unit it's not hard to go in the opposite direction (i.e. reconstruct $\omega$ from local orientations).

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  • $\begingroup$ I am bit confused with the dimensions of the cohomology/homology groups. Do you mean $H^n_{dR}(U)=H_n(M,M-\{x\})$ where the first group is the n-th de rham compactly supported cohomology group? If so, what is this isomorphism? $\endgroup$
    – Manuel
    Commented Jun 7, 2011 at 23:33
  • $\begingroup$ @Manuel Yes, exactly. It's an isomorphism of Poincare duality, if you will (but it's Poincare duality just for $S^n$: $H^n_{dR,c}(U)=\tilde H^n(S^n)=\tilde H_n(S^n)=H_n(M,M-\{x\})$). $\endgroup$
    – Grigory M
    Commented Jun 8, 2011 at 6:50
  • $\begingroup$ Wouldn't Poincare duality give an iso $H^n(S^n)=H_0(S^n)$ sending a cell to its dual cell? $\endgroup$
    – Manuel
    Commented Jun 8, 2011 at 12:27
  • $\begingroup$ Well, right, one shouldn't call it Poincare duality — it's just duality by universal coefficients. And the statement implicit in my answer is that one can use $H^n$ (but not $H^0$!) instead of $H_n$ in the definition (2). $\endgroup$
    – Grigory M
    Commented Jun 8, 2011 at 12:34
  • $\begingroup$ Right, you mean the isomorphism given by $H^n(S^n)=\text{Hom}(H_n(S^n),\mathbb{Z})$ $\endgroup$
    – Manuel
    Commented Jun 8, 2011 at 16:08

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