Let $M$ be a differentiable manifold of dimension $n$. First I give two definitions of Orientability.

The first definition should coincide with what is given in most differential topology text books, for instance Warner's book.

Orientability using differential forms: There exists a nowhere vanishing differential form $\omega$ of degree $n$ on $M$.

The second one is from Greenberg and Harper, "Algebraic Topology". This is the "fundamental class" approach. Let $x$ be a point on $X$, and let $R$ be a commutative ring and in the following the homologies are with coefficients in $R$.

Local orientability: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H_n(X, X-x)$.

By a simple application of Excision, it is seen that the above homology module is indeed isomorphic to $R$. We can also so arrange a neighborhood around every point that this local orientation can be "continued to a neighborhood" and is "coherent". Forgive me for being imprecise here; the detailed lemmas are in the reference given above. With this background in mind, we define:

A Global $R$-orientation of $X$ consists of: 1. A family $U_i$ of open sets covering of $X$, 2. For each $i$, a local orientation $\alpha_i \in H_n(X, X -U_i)$ of along $X$, such that a "compatibility condition" holds.

Here again I am imprecise about the compatibility condition; please check in the reference given above for details. I mean this basically as a question for those who already know both the definitions, as fully writing down the second definition would take 2-3 pages with all the necessary lemmas.

Also we define "orientation" to be a such a global choice.

Now the question:

How do the two definitions, the first one using differential forms, and the second one using homology, match?

Of course, to match we have to take $\mathbb{Z}$ to be the base ring for homology. A related question is about the meaning of orientability and orientation when we take a base ring other than $\mathbb{Z}$. It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring to be $\mathbb{Z}/5\mathbb{Z}$?

Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?

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    $\begingroup$ In case you're unfamiliar with the notion, there is a definition of orientability of a manifold (or vector bundle) with respect to an arbitrary (extraordinary) homology theory. Moreover, orientable manifolds satisfy Poincare duality for this (co)homology theory. The book by Rudyak on Thom Spectra, Orientability and Cobordism does a pretty nice job describing these ideas. $\endgroup$ – Ryan Budney Jan 21 '10 at 0:14
  • $\begingroup$ Related question on Math.SE $\endgroup$ – Grigory M Jan 11 '14 at 20:28

If $X$ is a differentiable manifold, so that both notions are defined, then they coincide.

The ``patching'' of local orientations that you describe can be expressed more formally as follows: there is a locally constant sheaf $\omega_R$ of $R$-modules on $X$ whose stalk at a point is $H^n(X,X\setminus\{x\}; R).$ Of course, $\omega_R = R\otimes_{\mathbb Z} \omega_{\mathbb Z}$.

This sheaf is called the orientation sheaf, and appears in the formulation of Poincare duality for not-necessarily orientable manifolds. It is not the case that any section of this sheaf gives an orientation. (For example, we always have the zero section.) I think the usual definition would be something like a section which generates each stalk.

I will now work just with $\mathbb Z$ coefficients, and write $\omega = \omega_{\mathbb Z}$.

Since the stalks of $\omega$ are free of rank one over $\mathbb Z$, to patch them together you end up giving a 1-cocyle with values in $GL_1({\mathbb Z}) = \{\pm 1\}.$ Thus underlying $\omega$ there is a more elemental sheaf, a locally constant sheaf that is a principal bundle for $\{\pm 1\}$. Equivalently, such a thing is just a degree two (not necessarily connected) covering space of $X$, and it is precisely the orientation double cover of $X$.

Now giving a section of $\omega$ that generates each stalk, i.e. giving an orientation of $X$, is precisely the same as giving a section of the orientation double cover (and so $X$ is orientable, i.e. admits an orientation, precisely when the orientation double cover is disconnected).

Instead of cutting down from a locally constant rank 1 sheaf over $\mathbb Z$ to just a double cover, we could also build up to get some bigger sheaves.

For example, there is the sheaf $\mathcal{C}_X^{\infty}$ of smooth functions on $X$. We can form the tensor product $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega,$ to get a locally free sheaf of rank one over ${\mathcal C}^{\infty}$, or equivalently, the sheaf of sections of a line bundle on $X$. This is precisely the line bundle of top-dimensional forms on $X$.

If we give a section of $\omega$ giving rise to an orientation of $X$, call it $\sigma$, then we certainly get a nowhere-zero section of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega$, namely $1\otimes\sigma$.

On the other hand, if we have a nowhere zero section of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega$, then locally (say on the the members of some cover $\{U_i\}$ of $X$ by open balls) it has the form $f_i\otimes\sigma_i,$ where $f_i$ is a nowhere zero real-valued function on $U_i$ and $\sigma_i$ is a generator of $\omega_{| U_i}.$

Since $f_i$ is nowhere zero, it is either always positive or always negative; write $\epsilon_i$ to denote its sign. It is then easy to see that sections $\epsilon_i\sigma_i$ of $\omega$ glue together to give a section $\sigma$ of $X$ that provides an orientation.

One also sees that two different nowhere-zero volume forms will give rise to the same orientation if and only if their ratio is an everywhere positive function.

This reconciles the two notions.

  • $\begingroup$ What is the point of defining orientations over rings other than $\mathbb{Z}$? Is there any useful consequence, except that anything is orientable over $\mathbb{Z}/2\mathbb{Z}$? $\endgroup$ – Anweshi Jan 6 '10 at 21:57
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    $\begingroup$ One answer, relating more to the sheaf $\omega_R$ then to the (slightly) more limited notion of orientations themselves, is that for lots of reasons one wants to be able to work with (co)homology with coefficients in an arbitrary ring $R$, and then having the sheaf $\omega_R$ is important, since it gives Poincare duality. More succintly: on an orientable manifold, choosing an orientation gives Poincare duality. Since one wants to have Poincare duality with arbitrary coefficients, one needs the notion of orientation with arbitrary coefficients. $\endgroup$ – Emerton Jan 6 '10 at 22:07
  • $\begingroup$ Important special case: If your manifold happens to be non-orientable (or you just don't know), you also switch to $R=\mathbb{F}_2$ and apply Poincare duality. Because of $\mathbb{F}_2^\times = \{1\}$ there is always a unique $\mathbb{F}_2$-orientation. $\endgroup$ – Johannes Hahn Jan 28 '16 at 21:15
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    $\begingroup$ Also: How can one see that $\mathcal{C}_X^\infty \otimes \omega$ corresponds to the line bundle $\bigwedge^{top} T^\ast M$ and not some other line bundle? $\endgroup$ – Johannes Hahn Jan 28 '16 at 21:20

Your main question was answered by Emerton. Regarding other notions of orientability, there's many. A popular one is the obstruction-theoretic approach:

1) A manifold $M$ is orientable if the tangent bundle $TM$ admits a trivialization when restricted to a $1$-skeleton of a CW-decomposition of $M$. An orientation of $M$ is taken to be a (homotopy class of) trivialization of $TM_{|M^0}$ that extends over $M^1$.

2) [Corrected to take into account Chris's comment] You can restate definition 1 in a way that avoids skeleta. A popular one is to define the associated orthogonal (principal) bundle to $TM$, lets call it $O(TM)$. This is the bundle over $M$ whose fibers over points $p \in M$ is the linear isomorphisms between $\mathbb R^m$ and $T_pM$. Then $M$ is orientable if every loop $S^1 \to M$ lifts to a loop $S^1 \to O(TM)$.

3) There's a small variant on these ideas called the "orientation cover", this is a 2-sheeted covering space of $M$, and it is connected if and only if $M$ is non-orientable. This has the additional assumption that $M$ is connected.

4) Another variant on this comes from bundle classifying-space machinery. Every vector bundle has a classifying map $M \to B(GL_m)$, and $GL_m$ has a subgroup of positive-determinant matrices, call it $GL^+_m$. $M$ is orientable if and only if the classifying map $M \to BGL_m$ lifts to a map $M \to BGL^+_m$, and an orientation is a homotopy-class of such lifts (flexible enough to allow homotopy of the original classifying map).

Anyhow, those are a few. There's of course more since all these ideas admit perturbations in various directions. For example, another small variant would be that the 1st Stiefel-Whitney class is trivial. One advantage to approaches (1), (2), (4) is that any of them are natural lead-in to other notions of orientation, like $spin$ or $spin^c$ structures.

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    $\begingroup$ Just a small correction to number (2). The principal bundle you want doesn't have fibers which are GL(T_pM), but rather the frames (i.e. collections of bases of T_pM). These are the same as the invertible homomorphisms from a fixed R^n to T_pM. This has a clear O(n)-action. The bundle you describe doesn't have a clear O(n)-action. But even if it did, it would be a trivial principal bundle since it has a section (the identity map from T_pM to itself). $\endgroup$ – Chris Schommer-Pries Jan 20 '10 at 12:58
  • $\begingroup$ Ah, right, I should have said the fibre over $T_pM$ is $Iso(\mathbb R^m, T_pM)$ provided $M$ is $m$ dimensional. $\endgroup$ – Ryan Budney Jan 20 '10 at 16:19

Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?

Another notion of orientability is the existence of an atlas whose transition functions have derivatives with everywhere positive determinant. This gives a clear cut way, along with the Cauchy-Riemann equations, of showing that every complex manifold (say, for simplicity, a Riemann surface) is orientable.

  • $\begingroup$ One problem with this definition is that it needs significant tweaking to work for manifolds with boundary -- and there's only one manifold with boundary for which it fails! $[0,1]$. $\endgroup$ – Ryan Budney Jan 6 '10 at 23:24

Speaking personally, I was not really comfortable with the notion of orientation until I understood the notion for vector bundles, so I will tell you about that.

Given a vector bundle $\pi: E \to B$, first select an orientation for each fiber $\pi^{-1}(b)$. The bundle will be oriented is you made these choices in a coherent manner and the following two are equivalent notions of 'coherent'.

1) For every point $b$ in $B$ has a neighborhood $N$ such that there are sections $s_1, \ldots, s_r: N \to E$ such that for all $n \in N$: {$s_1(n), \ldots, s_r(n)$} is an oriented basis for the fiber $\pi^{-1}(n)$.

2) Every point $b$ in $B$ is in a vector bundle chart $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that $\phi(n,\cdot): n \times \mathbb{R}^r \to \pi^{-1}(n)$ is orientation preserving.

Forgetting about picking an orientation for each fiber ahead of time, being orientable is also equivalent to:

3) You can cover $B$ with vector bundle charts $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that for any two $\phi$ and $\psi$ the linear isomorphism $n \times \mathbb{R}^r \stackrel{\phi(n,\cdot)}{\to} \pi^{-1}(n) \stackrel{\psi(n,\cdot)^{-1}}{\to} n \times \mathbb{R}^r$ is orientation preserving.

4) There is a nonzero section of the line bundle $\wedge^rE \to B$.

Now a manifold $M$ being orientable is equivalent to its tangent bundle being orientable. Given what has been said, the quickest way to see this is to note that a nonzero n-form on $M$ is by definition a nonzero section of the bundle $\wedge^n (T^\*M)$. (Note: $T^\*M$ is orientable iff $TM$ is orientable since they are isomorphic as bundles by picking a Riemannian metric.)

The canonical example of a nonorientable bundle is the Mobius bundle which is the line bundle over the circle whose total space looks like a Mobius band. In terms of 1) this bundle is not orientable since if you pick a nonzero section (vector) at a point and try to extend to the whole circle, by the time you get back to where you started your vector is now pointing the other way.

  • $\begingroup$ 3) is also the generalization of Justin Curry's comment about having transition maps whose derivatives have positive determinant. $\endgroup$ – user1835 Jan 20 '10 at 7:05

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