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.
Best Answer
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 $p\mapsto \det(E_1(p),\dots, E_n(p))$.
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.