In what sense is an orientation continuous?

Lee considers (local) frames over open subsets $U \subset M$. These are $n$-tuples of smooth vector fields $E_i : U \to TM \mid_U$ such that $(E_1\mid_p,\ldots, E_n\mid_p)$ forms a basis of $T_pM$ for each $p \in U$. Let us call $U$ a frame domain if there exists a frame over $U$. Coordinate domains are always frame domains, but in general there are more frame domains than coordinate domains. For example, if $TM$ is trivial (e.g. for $M = S^1$), then $M$ is a frame domain though in general not a coordinate domain.

Now consider a pointwise orientation on $M$, i.e. a family $\omega$ of orientations $\omega_p$ of $T_pM$, $p \in M$. As Lee writes

By itself, this is not a very useful concept, because the orientations of nearby points may have no relation to each other.

We can use frames $(E_i)$ over open $U \subset M$ to relate all orientations $\omega_p$ with $p \in U$. Let us say that $\omega$ is compatible with the frame $(E_i)$ if the basis $(E_1\mid_p,\ldots, E_n\mid_p)$ of $T_pM$ represents the orientation $\omega_p$ for all $p \in U$. Of course this property depends on the choice of the frame. In fact, for each frame domain $U$ there always exist frames $(E_i)$ over $U$ such that $\omega$ is not compatible with $(E_i)$: Either there does not exist any frame $(E_i)$ such that $\omega$ is compatible with $(E_i)$, or such a frame $(E_i)$ exists. In the latter case we can consider the frame $(E'_i) = (-E_1,E_2,\ldots,E_n)$ and see that $(E'_i\mid_p)$ always represents the opposite orientation $-\omega_p$ in which case $\omega$ is not compatible with $(E'_i)$.

Lee does not use our above terminology. He expresses the same fact by saying that $(E_i)$ is positively oriented. In my opinion this wording may be misleading because it suggests that there exists something like a positive orientation on $T_pM$ in an absolute sense. What Lee really means is this: For each $p \in M$ we have chosen an orientation $\omega_p$ of $T_pM$ and declare it to be the positive orientation. This is of course an arbitrary act. I think it would be better to say that $(E_i)$ is positively oriented with respect to the given pointwise orientation $\omega$.

Lee calls a frame $(E_i)$ negatively oriented if the basis $(E_1\mid_p,\ldots, E_n\mid_p)$ of $T_pM$ represents the opposite orientation $-\omega_p$. Moreover, a frame is called oriented if it is positively or negatively oriented.

Lee then defines a pointwise orientation $\omega$ to be continuous if for each $p\in M$ there exist an open neighborhood $U$ of $p$ and an oriented frame over $U$. Of course this is equivalent to requiring that for each $p\in M$ there exist an open neighborhood $U$ of $p$ and a positively oriented frame over $U$. In fact, if we have a negatively oriented frame $(E_i)$, then $(E'_i) = (-E_1, E_2,\ldots, E_n)$ is a positively oriented frame.

In our terminology this means that for all $p\in M$ there an open neighborhood $U$ of $p$ and a frame $(E_i)$ over $U$ such that $\omega$ is compatible with $(E_i)$.

And here is the question: Why does Lee use the phrase continuous pointwise orientation?

Formally we can use each frame $\eta = (E_i)$ to define $$(\eta,\omega)^* : U \to \{+1,-1\}, (\eta,\omega)^*(p) = \begin{cases} +1 & [\eta_p] = \omega_p \\ -1 & [\eta_p] = -\omega_p \end{cases}$$ Here $\{+1,-1\}$ is regarded as a topological space with the discrete topology.

Let us call $\eta$ a constant resp. continuous frame with respect to $\omega$ if $(\eta,\omega)^*$ is constant resp. continuous. Clearly constant frames are continuous, and continuous frames over a connected $U$ are constant.

Therefore a pointwise orientation $\omega$ is continuous in the sense of Lee iff for each $p\in M$ there exists a constant frame over some open neighborhood $U$ of $p$. Such a frame is in particular continuous, and certainly this is why the phrase "continuous pointwise orientation" is used. One could alternatively call a pointwise orientation to be constant if for each $p\in M$ there exists a constant frame over some open neighborhood $U$ of $p$, but perhaps the use of the word "constant" would be a bit irritating in this context.

Lee's definition is clearly equivalent to this:

A pointwise orientation $\omega$ is continuous if for all $p\in M$ there exist a connected open neighborhood $U$ of $p$ and a continuous frame over $U$.

What does it mean that a frame $\eta$ over an arbitrary open $U$ is continuous? Since each manifold $M$ is locally connected, all connected components of $U$ are open (see e.g. here). Hence $\eta$ is continuous on $U$ iff $\eta$ is continuous on each connected component of $U$. The latter means that $\eta$ is constant on each connected component of $U$.

Therefore, if there exists a continuous frame $\eta$ over $U$, then there also exists a positively oriented frame $\eta'$ over $U$. Since $\eta$ is constant on each connected component of $U$, we can simply switch $\eta \mid_p = (E_i \mid_p)$ to $-\eta \mid_p = (-E_1 \mid_p, E_2 \mid_p,\ldots, E_n \mid_p)$ on each connected component of $U$ on which $\eta_p = - \omega_p$.

We conclude that the following is also a valid definition:

A pointwise orientation $\omega$ is continuous if for all $p\in M$ there exist an open neighborhood $U$ of $p$ and a continuous frame over $U$.

Let us close with

Theorem. Let $\omega$ be a pointwise orientation on $M$. The following are equivalent:

1. $\omega$ is continuous.

2. All local frames on $M$ with a connected frame domain are continuous (or, equivalently, constant).

3. All local frames on $M$ are continuous.

Proof. 2. $\implies$ 1. is obvious because each $p \in M$ lies in a connected frame domain.

1. $\implies$ 2. follows from the theorem in Proving that every local frame on a connected domain has either positive orientation either negative orientation.

2. $\Longleftrightarrow$ 3. is clear from the above considerations (look at the components of $U$).

Therefore we get the following:

1. Pick any frame domain $U$ and any frame $\eta$ over $U$. Then $\omega$ is compatible with some frame over $U$ iff $(\eta,\omega)^*$ is continuous. Therefore we can reasonably define that $\omega$ is continuous over a frame domain.

2. If $M$ is covered by frame domains over which $\omega$ is continuous, it makes sense to say that $\omega$ is (globally) continuous.

There are already two comprehensive answers. Both deal with certain maps into the discrete space with two element. It is hard to say what Lee "really" wanted to express by the word continuous, but I think there is an obvious interpretation.

A pointwise orientation of $M$ is a family $\omega = (\omega_p)_{p \in M}$ of orientations $\omega_p$ of the tangent spaces $T_pM$. Such a family can be completely erratic. Intuitively, an orientation of $M$ should be a pointwise orientation such that $\omega_p$ does not make leaps when moving on $M$. But the tangent spaces are pairwise disjoint, so what does this mean?

The tangent spaces are aggregated in the tangent bundle $TM$ and this allows to make precise what "not making a leap" means.

Each $\omega_p$ is represented by a frame $\eta_p = (e_{1p}, \ldots, e_{np})$ in $T_pM$. Now we can continuously move $\eta_p$ using a local trivialization $\phi : TM \mid_U \to U \times \mathbb R^n$, where $U$ is an open neighborhood of $p$ : If $q \in U$, then we move $\eta_p$ to $\eta_q = (e_{1q}, \ldots, e_{nq})$ by shifting $e_{ip} \in T_pM$ to $\hat e_{ip} \in \mathbb R^n$ via $\phi$ and then shifting $\hat e_{ip}$ to $e_{iq} \in T_qM$ via $\phi^{-1}$. For each path $u : [0,1] \to U$ starting at $p$ this construction gives a (continuous!) path in $TM \mid_U$ starting at $e_{ip}$ and ending at $e_{iu(1)}$.

Technically this results in a local frame $E = (E_1, \ldots, E_n)$ over $U$ such that $E \mid_p = (E_1 \mid_p, \ldots, E_n \mid_p) = \eta_p$. This makes precise what it means to move the given frame $\eta_p$ in $T_pM$ continuously to a frame in $T_qM$, $q \in U$.

It is then reasonable to say that the pointwise orientation $\omega$ is continuous on $U$ if there exists a local frame $E$ over $U$ such that $[E \mid_p] = \omega_p$ and $[E \mid_q] = \omega_q$ for all $p, q \in M$.

If $M$ is covered by open $U$ on which $\omega$ is continuous, then we call $\omega$ continuous.

Let us close with an example of a non-continuous pointwise orientation on $M = \mathbb R$.

The $T_p\mathbb R$ can be naturally identified with $\mathbb R$, thus each orientation of $T_p\mathbb R$ can be identified with one of the numbers $\pm 1$. Indeed, both numbers give a frame in $\mathbb R$, and each frame is equivalent to one of these two frames.

Now let $D \subset \mathbb R$ a subset $\ne \emptyset, \mathbb R$. A pointwise orientation $\omega$ of $\mathbb R$ is given by $\omega_p = +1$ for $p \in D$ and $\omega_p = -1$ else. It is easy to check that it is not continuous, the problem being the boundary points of $D$. Note that $\omega$ can be extremely erratic, e.g. for $D =\mathbb Q$.