When is the orientation presheaf a sheaf?

Question

The orientation presheaf of a topological $n$-manifold $X$ is $$U \mapsto H_n(X, X-U)$$

The manifold $X$ is orientable iff there exists a global section which is a generator of each stalk (I believe).

How is this related to the condition that the orientation presheaf be a sheaf? What does it mean morally for the orientation presheaf to be a sheaf? (I'm not asking for a restatement of the definition...I'm comfortable with that.)

Answer 1

Because of the excision theorem, you can check the sheaf property on a neighborhood of a point homeomorphic to $\mathbb R^n$, in which case it is trivial using the covering space of $R$-orientations of a topological manifold (where $R$ is a general commutative unital ring ; in our case, use $R = \mathbb Z$). A very detailed explanation of the construction of this two-sheeted cover can be found in Hatcher, page 234-235. On $\mathbb R^n$, this covering space is trivial, i.e. equal to $X \times \mathbb Z$, which is to be expected since $\mathbb R^n$ is orientable, hence $\mathbb Z$-orientable. Translating the language of covering spaces into the language of sheaves, one must realize that an element $\sigma \in H_n(X,X \setminus U)$ is the same as a section $\sigma : U \to U_{\mathbb Z}$ where $\sigma(x)$ generates the $\mathbb Z$-module $H_n(X|x) = H_n(X, X \setminus \{x\})$ (using the language of Hatcher), since points in $U_{\mathbb Z}$ are stalks of the sheaf $(-) \mapsto H_n(X,X \setminus (-))$. Therefore, saying that the associated covering space is trivial ($\simeq X \times \mathbb Z$) is the same as saying that the sheaf is free of rank $1$. This gives locally freeness, hence $U \mapsto H_n(X,X \setminus U)$ is a sheaf of abelian groups on $X$.

Now that you have locally freeness, the orientability question becomes obvious ; an orientation is an element $\sigma \in H_n(X) = H_n(X,X \setminus X)$ such that the corresponding element $\sigma_x \in H_n(X,X\setminus \{x\})$ is a generator of that $\mathbb Z$-module. Translating in the language of covering spaces, this is a global section of the covering map $\pi_X : X_{\mathbb Z} \to X$, say $\sigma : X \to X_{\mathbb Z}$ such that $\sigma_x$ generates the $\mathbb Z$-module $\pi_X^{-1}(\{x\})$. Translating in the language of sheaves, this amounts to saying that the orientation sheaf is free sheaf of abelian groups of rank $1$, i.e. the constant sheaf equal to $\mathbb Z$. (The translation is not necessary, but I feel it is helpful to see a section as an actual section of a projection map, which gives a good sense of the word "section".)

To read about the details, I guess Hatcher does a pretty good job of explaining how the maps work but it doesn't mention sheaves. If you know of a good book detailing the use of sheaves in algebraic topology, I'd love to hear your comments on that ; the only one I know is S. Ramanan's Global Calculus, which in my opinion contains a lot of good ideas but is (a tiny bit) sketchy in its presentation.

Hope that helps,

Answer 2

It is always a sheaf, as it is the sheaf of sections of a surjective map, the covering map from the canonical double cover.

Answer 3

I think it is important that $X$ is compact.

Otherwise, we can let $X = \mathbb{N}$, with the discrete topology. Call your presheaf $F$, so $F(U) = H_n(X, X \setminus U)$. Then $F(X) = H_n(X, X \setminus X) = \oplus_{X} \mathbb{Z}$. However, $F(x_i) = H_n(X,X \setminus x_i) = \mathbb{Z}$, so if we could glue then we would get global sections to be $F(X) = F(\bigcup x_i) = \Pi F(x_i) = \Pi_{\mathbb{X}} \mathbb{Z}$.

Of course, you can make similar counter examples in different dimensions, e.g. by taking a countable disjoint union of spheres.

Another example is $\mathbb{R}^n$ itself. This is orientable (in the sense of having a way to choose locally consistent generators of $H_n(\mathbb{R}^n, \mathbb{R}^n \setminus x)$), so the stalkwise sheafification has $\mathbb{Z}$ global sections. But $F(\mathbb{R}^n) = H_n(\mathbb{R}^n, \emptyset) = 0$. If $F$ was a sheaf, it would be isomorphic to its sheafification; but this cannot be, because their global sections have different ranks.

I guess the point is that $H_n(X)$ is an inherently compact thing, since chains always have compact support.

I guess the other answers are correct provided one has the compactness assumption. I don't know, maybe I'm totally wrong. Maybe everyone is implicitely thinking of Borel-Moore homology?