Continuous pointwise orientation

Question

Lee defines an orientation on a smooth manifold $M$ as a continuous pointwise orientation.

A pointwise orientation (that is, a choice of orientation on each tangent space of $M$) is said to be continuous if every point of $M$ is contained in the domain of a oriented local frame (= a local frame with positively oriented basis throughout the domain); we call such a domain a frame domain.

I am a bit confused about this definition. Since each manifold is locally diffeomorphic to the Euclidean space, is it not vacuous that each point is in a frame domain? Simply take the coordinate domain.

I would appreciate an example. Can you help me understand why continuity breaks for the Mobius band? There must be a point which does not belong to a frame domain, according to this definition.

Edit: In general, I have difficulties to understand why one would define a (global) orientation in this way.

Answer 1

Around a point in a manifold, there always exists a pointwise orientation and a local frame domain around that point, by using the coordinate domain. But this is not what we are interested in. We want one global pointwise orientation such that there is a local frame domain around each point.

You can't give me one global pointwise orienation on the möbius band, because the twist changes the orientation discontinuously along the "glued" sides. The möbius band is not orientable.

Answer 2

Because every manifold is locally Euclidean, you can define an orientation on a neighbourhood of any point, using a local coordinate frame, as you say. But this definition is saying something more. It's saying that you first orient every tangent space, and then you check that each point has a frame around it that agrees with this previously chosen orientation of every tangent space. So on a Möbius band, for example, there's no special nonorientable point; but if you orient every tangent space, then there must be a point where the orientation is discontinuous.

To say this with some symbols: First you have a pointwise orientation $ o \colon ( P \in M ) \mapsto ( o _ P \in \mathcal O T _ P M ) $, where $ T _ p M $ is the tangent space at $ P $ and $ \mathcal O V $ is the two-element set of orientations of a vector space $ V $. Then you check that for each point $ P $, for some neighbourhood $ U $ of $ P $, there is a continuous frame $ e \colon ( Q \in U ) \mapsto ( e _ Q \in \mathcal B T _ Q M ) $, where $ \mathcal B V $ is the set of oriented bases of a vector space $ V $; such that for each $ Q $ in $ U $, $ e _ Q $ is positively oriented relative to $ o _ Q $.

Although saying it this way depends on knowing what a continuous frame is, so I think that it's better to say that each point has a coordinate chart around it that makes this work for the coordinate frame, since the coordinate charts are more fundamental (and automatically continuous, since they define the topology). So after defining $ o $, you check that for each point $ P $, for some coordinate chart $ \phi \colon U \to \mathbb R ^ n $, for each $ Q $ in $ U $, the basis $ ( \phi _ Q ^ * ( \mathbf e _ i ) ) _ { i = 1 } ^ n $ is positively oriented relative to $ o _ Q $, where $ \phi _ Q ^ * ( \mathbf e _ i ) $ is the pullback of the $ i $th standard basis vector $ ( 0 , \ldots , 0 , 1 , 0 , \ldots , 0 ) $ of $ \mathbb R ^ n $ to $ T _ Q M $. This is what it means for $ o $ to be continuous at $ P $.

Answer 3

I elaborate a bit on Toby Bartels answer:

Clearly, a smooth atlas gives a pointwise orientation via the coordinate vector fields, however, these may not be oriented (as local frames).

More precisely, for every $p \in M$ and (centered) coordinate chart $(U_p, \phi_p)$, we can choose the orientation of $T_p M$ to be the coordinate vector fields $$\left(\frac{\partial}{\partial \phi_p^1}{\big\vert_{p}}, ..., \frac{\partial}{\partial \phi_p^m}\big\vert_{p}\right\} =: \mathcal{U}_p,$$ where $\frac{\partial}{\partial \phi_p^i} := \phi^{-1}_{p\ast}(e_i) \in \Gamma(TU_p)$ and $e_i$ is the $i$th standard coordinate vector of $\mathbb{R}^m$.

Now, fix $p \in M$. Then, for every $q \in U_p$, the ordered set of vectors $(\frac{\partial}{\partial \phi_p^1}{\big\vert_{q}}, ..., \frac{\partial}{\partial \phi_p^m}\big\vert_{q})$ defines a basis for $T_q M$. However, one should note carefully that the orientation of $T_q M$ is determined by the tuple $\mathcal{U}_q$. So, it is meaningful to ask whether these two tuple belong to the same equivalence class, or in other words, agree in orientation. Whenever they do, then $\mathcal{U}_p$ is indeed an oriented local frame for $p \in M$, but, of course, it may happen that they do not agree. Similar holds for any other point of $M$.

So, saying that a pointwise orientation is continuous means essentially that the local orientations induced by $\{\mathcal{U}_{p}\}$ "patch together".

Addendum: It turns out that each coordinate frame constitutes an oriented local frame (with respect to the point-wise orientation induced by the coordinate charts) if and only if the transition functions are orientation-preserving.

In the abstract setting, then, we allow

• any choice of orientation for each $T_p M$, rather than the one induced by the atlas,
• any local frame, rather than the coordinate vector fields.

How one goes about showing with this definition that the Möbius band is non-orientable, I still do not know.