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The orientation sheaf of an $n$-manifold $M$ is $\mathcal{O}_n=Sheaf(U\mapsto H_n(M,M-U;\mathbb{Z}))$, with stalks given by $(\mathcal{O}_n)_x = lim H_n(M,M-U)=H_n(M,M-x)=\mathbb{Z}$ (the limit is over neighborhoods $U$ containing $x\in M$).

Suppose $M$ is orientable and closed, so that $H_n(M)=\mathbb{Z}$. Then $\mathcal{O}_n=M\times\mathbb{Z}$ is a constant sheaf.

Suppose now that $M$ is nonorientable. Then we can take a neighborhood $N$ of each point $x\in M$ so that $\mathcal{O}_n|_N$ is constant. Thus $\mathcal{O}_n$ is a locally constant sheaf.

The orientable double cover of $M$ is $\tilde{M}=\lbrace(x,\mu_x)\rbrace$ where $\mu_x$ is a local orientation at $x\in M$ (a choice of generator of the local homology). If $M$ is nonorientable then $\tilde{M}$ is connected, and if $M$ is orientable then $\tilde{M}\approx M\times\mathbb{Z}_2$. In either case, there is an embedding $\tilde{M}\hookrightarrow \mathcal{O}_n$.

Is there anything else I can say about this enlargement from $\tilde{M}$ to $\mathcal{O}_n$ ? Does one encode more information than the other?

If we instead use $\mathbb{R}$-coefficients, I believe (but can easily be wrong) that $\tilde{M}$ is a deformation retract of $\mathcal{O}_n$ for nonorientable $M$, and is a deformation retract of $\mathcal{O}_n-\Gamma^0$ for orientable $M$ (where $\Gamma^0$ is the zero section of this real line bundle). In other words, $\tilde{M}$ and $\mathcal{O}_n$ are the same up to homotopy (over $\mathbb{R}$-coefficients), and one does not encode more information than the other.

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It's not clear to me what your question is. $\mathcal O_n$ can be viewed as an induced bundle from the orientation cover. $\mathcal O_n \simeq \tilde M \times_{\mathbb Z_2} \mathbb Z$. So there's nothing really new going on with it. –  Ryan Budney Oct 26 '11 at 22:44
I would think that they are equivalent in general. You've already shown one direction. Now suppose that you have $\pi:\tilde M\to M$. You can recover $O_n$ as the anti-invariant part of $pi_*\mathbb{Z}$ under the action of the Galois or covering group $\mathbb{Z}/2$. Or am I overlooking something? –  Donu Arapura Oct 26 '11 at 22:46
@Ryan, could you elaborate a little more on that homotopy equivalence? @Donu, I am not sure about that procedure... are you saying it gives you a map $\mathcal{O}_n\rightarrow\tilde{M}$ which when composed with the embedding is homotopic to the identity? –  Chris Gerig Oct 26 '11 at 23:02
Chris, no, but perhaps it's not important. Ryan's answer should be sufficient. –  Donu Arapura Oct 26 '11 at 23:21
Here is another description of $\mathcal{O}_n$: (the \'etale space of) $\mathcal{O}_n$ is homeomorphic to the disjoint union of $M$ (this corresponds to the zero section) and a countable number of copies of $\tilde M$; the $n$-th copy is formed by taking the union of $\{\pm n\}$ in all stalks. –  algori Oct 27 '11 at 0:18

1 Answer 1

up vote 3 down vote accepted

An elaboration.

$\mathcal O_n$ is a bundle over $M$ with fiber $\mathbb Z$. There is an action of the integers on it, because the integers act on homology. Earlier I said this was a principal bundle, I was too tired! The action is of course not free. In particular, this bundle $\mathcal O_n \to M$ has a section.

Given a manifold $M$, its orientation cover is a $\mathbb Z_2$-principal bundle. The action of $\mathbb Z_2$ is given by reversal of orientations.

So the product $\tilde M \times \mathbb Z$ has an action of $\mathbb Z_2$ given by

$$t.(x,n) = (t.x, (-1)^tn)$$

where $t.x$ is the action of $\mathbb Z_2$ on $\tilde M$.

$\tilde M \times_{\mathbb Z_2} \mathbb Z$ is the quotient of $\tilde M \times \mathbb Z$ by the action of $\mathbb Z_2$. By design, it is a $\mathbb Z$-principal bundle -- you can project onto $\tilde M / \mathbb Z_2 \equiv M$ -- and by design it is isomorphic to $\mathcal O_n$. The map $\tilde M \times_{\mathbb Z_2} \mathbb Z \to \mathcal O_n$ is induced by your inclusion $\tilde M \to \mathcal O_n$.

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The action of $\mathbb Z$ on the fibers of $\mathcal O_n$ is... multiplication? –  Mariano Suárez-Alvarez Oct 26 '11 at 23:34
Whatever the action of $\mathbb Z$ is on a homology group is called, that's what the action of $\mathbb Z$ on $\mathcal O_n$ is. The fiber over a point of $M$ is $H_n(M,M\setminus \{x\})$, so if you want to call the action multiplication, that's alright with me. –  Ryan Budney Oct 26 '11 at 23:40
My point is: there are two sensible actions of $\mathbb Z$ on a group like $H_n(M,M\setminus x)$: either by multiplication (and then the bundle is not principal) or by translation, but then there is no global action of the whole of $\mathcal O_n$, because you need to decide in which direction $1$ will transltate things to, and this can only be done coherently if the manifold is orientable. –  Mariano Suárez-Alvarez Oct 26 '11 at 23:42
I think that the most one ca say is that $\mathcal O_n$ is a bundle with structure group $\mathbb Z_2$, and that you can restrict the structure group to $\mathbb Z$ if $M$ is orientable (you will not get a principal bundle upon restriction, in any case) –  Mariano Suárez-Alvarez Oct 26 '11 at 23:43
You're right the action I'm describing isn't free. So I shouldn't have used the word principal bundle. In particular $\mathcal O_n$ always has a section. –  Ryan Budney Oct 26 '11 at 23:50

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