Proving that every local frame on a connected domain has either positive orientation either negative orientation

Question

I am trying to solve Exercise 15.4 from John M. Lee's Introduction to Smooth Manifolds. Here is the statement:

Suppose M is an oriented smooth n-manifold with $n \geq 1$. Show that every local frame with a connected domain is either positively oriented or negatively oriented.

I recall that the orientation of a vector space is defined in terms of equivalence classes of ordered bases (two bases are equivalent if the determinant of the change-of-basis matrix between them is strictly positive). The pointwise orientation of a smooth manifold is a choice of orientation (in the vector space sense) for each tangent space. A manifold is said to be oriented when this pointwise orientation varies continuously, meaning that for every point, we can find a local frame $(E_1,\ldots,E_n)$ that is linearly independent in a neighborhood of p and positively oriented at each point in this neighborhood.

I think I need this result to prove the "conversely" part of Proposition 15.6, which explains how we can define an orientation from a consistently oriented atlas.

To solve the exercise, I was trying to construct a continuous map from M to {0,1}. My rough idea was that since orientation is defined in such a way that it does not change randomly in the tangent spaces of nearby points, it would be surprising for it to suddenly switch at some moment in a connected domain. Usually, in this kind of mathematical situation, finding the right continuous map into a discrete set is key to formalizing this intuition that "it can't jump." However, I must have skipped several sections on vector bundles and frames because I don't have time to study the entire chapter, so I am finding these notions difficult to work with. Could someone give me a hint?

Answer 1

For a frame $\eta = (E_1,\ldots,E_n)$ on an open $U$ let $[\eta_p] = [E_1 \mid_p,\ldots, E_n \mid_p]$ denote the induced orientation on $T_pM$. The opposite orientation on $T_pM$ is denoted by $-[\eta_p]$. It is induced by the frame $-\eta = (-E_1,\ldots,E_n)$.

Lemma. For any two frames $\eta^1, \eta^2$ on a connected $U$ either $[\eta^1_p] = [\eta^2_p] $ for all $p \in U$ or $[\eta^1_p] = -[\eta^2_p] $ for all $p \in U$.

Proof. We have $[\eta^1_p] = \sigma_p[\eta^2_p]$ where $\sigma_p$ is the sign of the determinant $\det A_p$ of the transition matrix of the linear automorphism $A_p : T_pM \to T_pM$ with $A_p(E^1_i \mid_p) = E^2_i \mid_p$. Since the functions $E_i, E'_i$ are continuous, also the function $p \mapsto A_p \in GL(n)$, the space of is invertible $(n \times n)$-matrices, is continuous. A formal proof requires to use charts $\phi : V \to V' \subset \mathbb R^n$ with $V \subset U$ which reduces the problem to frames $\alpha^1, \alpha^2: V' \to TV' = V' \times \mathbb R^n$. In that case it is easy to see that the transformation matrix depends continuously on $x \in V'$. Actually we can regard the $\alpha^i = (F^i_1,\ldots, F^i_n)$ as smooth functions $a^i : V' \to GL(n)$, where $a^i(x)$ = matrix whose $j$-th column is $F^i_j \mid_x \in \mathbb R^n$. Then the transformation matrix $A_x$ is characterized by $A_x \cdot a^1(x) = a^2(x)$, thus $A_x = a^2(x) \cdot (a^1(x))^{-1}$. Since matrix multiplication and matrix inversion are continuous, we are done.
But then also $p \mapsto \det A_p \in \mathbb R \setminus \{0\}$ is continuous, and thus $\sigma_p$ is constant.

Now let $\omega$ be an orientation (= continuous pointwise orientation) on $M$. This means that for each $p_0 \in M$ there exists an open neighborhood $U$ and a frame $\eta$ on $U$ such that either $[\eta_p] = \omega_p$ for all $p \in U$ or $-[\eta_p] = \omega_p$ for all $p \in U$. Lee calls such a frame an oriented frame (with respect to $\omega$).

Corollary. Let $\omega$ be an orientation on $M$ and $\eta$ be a frame on on open $U$. Then each $p_0 \in U$ has an open neighborhood $U_0$ contained in $U$ such that $\eta \mid_{U_0}$ is an oriented frame.

Proof. There exists a oriented frame $\eta'$ on some open neighborhood $U'$ of $p_0$. Choose a connected open neighborhood $U_0$ of $p_0$ contained in $U \cap U'$. Then $\eta \mid_{U_0}$ and $\eta' \mid_{U_0}$ are frames on $U_0$ and the Lemma applies to show that $\eta \mid_{U_0}$ is oriented (since $\eta' \mid_{U_0}$ is).

The theorem in your question says

Let $\omega$ be an orientation on $M$. The each frame $\eta$ on a connected domain $U$ is an oriented frame.

Proof. Let $A = \{ p \in M \mid [\eta_p] = \omega_p \}$ and $B = \{ p \in M \mid -[\eta_p] = \omega_p \}$. These are disjoint sets covering $U$. We shall show that both are open which implies that one of them equals $U$ and the other is empty.

So let $p_0 \in A$. The Corollary says that $p_0$ has an open neighborhood $U_0$ such that $\eta \mid_{U_0}$ is oriented. Since $[\eta_{p_0}] = \omega_{p_0}$ we see that $[\eta_p] = \omega_p$ for all $p \in U_0$. Thus $U_0 \subset A$. Hence $A$ is open.

The proof for $B$ is similar.