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.

`$\{\pm n\}$`

in all stalks. – algori Oct 27 '11 at 0:18