Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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 $\tilde M \to \mathcal O_n$.

share|improve this answer
    
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
2  
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
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.