In what sense is an orientation continuous?

Question

I have been reading from Lee's Introduction to Smooth Manifolds and I have gotten to the section on how an orientation on a manifold is defined.

First Lee defines an orientation on a vector space $V$ by considering the set of all possible ordered bases for $V$, call this set $B$. Thus if $(v_1, \ldots v_n)\in B$ then $(v_1, \ldots, v_n)$ is an ordered set such that $V = \text{span}(v_1, \ldots ,v_n)$ and the vectors are linearly independent. We then consider two bases $(v_1, \ldots v_n)$ and $(w_1, \ldots, w_n)$ to be equivalent if there exists an invertible matrix $A$ such $v_i = (A)_i^j w_j$ and $\text{det}(A) > 0$. This creates an equivalence on $B$ of which there are two equivalence classes and an orientation on $V$ is a choice of either of the two equivalence classes.

Now for a manifold $M$ we defined a point wise orientation as follows. For all $p\in M$, we can select an orientation on $T_pM$ since $T_pM$ is a vector space. Next on some open set $U\subset M$, if we have a local frame $(E_1, \ldots E_n)$ so that $(\forall p\in U)\, \text{span}(E_1|_p, \ldots, E_n|_p) = T_pM$ such that $(E_1|_p, \ldots, E_n|_p)$ is positively oriented on $T_pM$ with respect to the point wise orientation on $M$, then we say that $(E_1,\ldots, E_n)$ is positively oriented on $U$.

Finally we say $M$ is oriented if for all $p\in M$ we may find an open set $U$ containing $p$ and a positively oriented local frame on $U$. Further the orientation is said to be continuous in this case.

My question is about the last part. In what sense is the orientation continuous? I cannot see how the point wise orientation is a weaker assumption than the definition of a local frame being positively oriented. I understand that the vector fields in the local frame are continuous maps from $U\subset M$ to $TM$, but I don't how this tells us anything about how the orientation evolves on $M$. I saw the answer on Definition of pointwise continuous orientation of smooth manifolds which elaborates on how for each $p\in M$ we can find a map $f:U\to GL(n,\mathbb{R})$ where $f(p)$ is a transition matrix from our local frame to any fixed representative of the orientation. In the answer, the author posts that $f$ is a continuous map and I do not see how that parts follows either.

Answer 1

To see in what sense a continuous orientation for $M$ is continuous, first prove the following easy topological lemma.

Lemma. If $f\colon X\to D$ is a function on a topological space $X$ valued in a discrete space $D$, then the function $f$ is locally constant if and only if $f$ is continuous.

We will use the lemma for $D = \{1,-1\}$. Now, by the definition of a continuous orientation, first of all, we have a pointwise orientation $\mathcal O$ such that each point of the manifold $M$ is contained in some open subset $U$ on which we have a local frame $(E_1,\dots,E_n)$ satisfying $[(E_1,\dots,E_n)] = \mathcal O$ on $U$. By considering the determinant $\det(E_1,\dots,E_n)\in\mathbb R\smallsetminus\{0\}$ of the matrix whose columns are the vectors $E_1,\dots,E_n$, and recalling the continuity of the local frame $(E_1,\dots,E_n)$, you should check that, possibly after shrinking $U$, either for all $p\in U$, $\det(E_1,\dots,E_n)<0$ or for all $p\in U$, $\det(E_1,\dots,E_n)>0$. Thus, we see that our orientation $\mathcal O$ may be regarded as a locally constant function $M\to \{1,-1\}$ according to the sign of $\det(E_1,\dots,E_n)$ for whatever our local frame $(E_1,\dots,E_n)$ is. Since $\mathcal O$ is locally constant, the orientation is continuous when regarded as a map $M\to\{1,-1\}$, which justifies the terminology.

Here is the conclusion, summarized. A pointwise orientation allows the orientation $\mathcal O$ to flip-flop arbitrarily between "nearby" tangent spaces, whereas a continuous orientation is a pointwise orientation that is also locally constant as a consequence of the continuity of the map $p\mapsto \det(E_1(p),\dots, E_n(p))$.

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Added: To compute the determinants $\det(E_1,\dots,E_n)$, you can shrink $U$ to be a coordinate patch, and then express $E_j = a^i_j\partial_i$ in terms of coordinate vector fields. Now, the functions $p\mapsto a^i_j(p)$ are continuous by the assumption that $(E_1,\dots,E_n)$ is a local frame, and we are concretely taking the determinant of the matrix $(a^i_j)$.

Note. We also recall here that a local frame $(E_1,\dots,E_n)$ in the author's convention is by definition an $n$-tuple of continuous vector fields that span the tangent space at each point. Continuity can be understood in various equivalent senses, the most concrete of which probably being the one in terms of coordinates with respect to some coordinate vector fields.

Answer 2

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 that a pointwise orientation $\omega$ on $M$ is compatible with a frame $(E_i)$ over an open $U \subset M$. He expresses the same fact by saying that $(E_i)$ is positively oriented on $U$. 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 choose an orientaton $\omega_p$ of $T_pM$ and declare it to be the positive orientation. This is of course an arbitrary act.

Lee then says that $M$ is oriented (with orientation $\omega$) if for all $p\in M$ we can find an open set $U$ containing $p$ and a positively oriented local frame on $U$. In our terminology this means that for all $p\in M$ we may find an open set $U$ containing $p$ and a frame $(E_i)$ over $U$ such that $\omega$ is compatible with $(E_i)$.

Finally the orientation $\omega$ is said to be continuous in this case.

And that is the question: Why is the word continuous used in that case?

Formally we can use each frame $\eta = (E_i)$ to define $$(\eta,\omega)^* : U \to \{+1,-1\}, \eta^*(p) = \begin{cases} +1 & [\eta_p] = \omega_p \\ -1 & [\eta_p] = -\omega_p \end{cases}$$ Here $\{+1,-1\}$ is regarded as a topological space with the discrete topology. The continuity of $(\eta,\omega)^*$ does not agree with our above concept of $\omega$ being compatible with $\eta$. In the latter case $\eta^*$ is constant with value $+1$ (in particular continuous), but if $(\eta,\omega)^*$ is continuous, we can only conclude that it is constant on each connected component of $U$. However, if $(\eta,\omega)^*$ is continuous, we can easily construct a frame $\eta'$ over $U$ such that $\omega$ is compatible with $\eta'$. 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 $\eta_p = - \omega_p$.

Therefore we get the following two results:

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.