Every connected orientable smooth manifold has exactly two orientations, Lee Proposition 15.9

Question

The proof of Proposition 15.9 from John Lee's book "Introduction to Smooth Manifolds" is left as an exercise.

Here is the statement:

Let $M$ be a connected, orientable, smooth manifold with or without boundary. Then $M$ has exactly two orientations. If two orientations of $M$ agree at one point, they are equal.

Here is my argument

Let $\mathcal{O}=\{\mathcal{O}_p:p\in M\}$ and $\tilde{\mathcal{O}}=\{\tilde{\mathcal{O}}_p:p\in M\}$ be orientations for $M$.

Let $\mathcal {C}=\{p\in M:\mathcal O_p=\tilde{\mathcal O}_p\}$, and let $p\in \mathcal C$.

By definition there exist $(E_i:U\to TM)$ and $(\tilde E_i:\tilde U\to TM)$ (continuous) local frames such that $p\in U\cap \tilde U$ and $(E_i)$ is positively oriented with respect to $\mathcal O$ and $(\tilde E_i)$ is positively oriented with respect to $\tilde {\mathcal O}$.

We can suppose $U\subseteq \tilde U$ and $U$ connected. Since $\mathcal O_p=\tilde{\mathcal O}_p$ we have that the ordered basis $(E_1|_p,\dots,E_n|_p)$ and $(\tilde E_1|_p,\dots,\tilde E_n|_p)$ are consistently oriented, meaning that the transition matrix $A(p)=(A_i^j(p))$ has positive determinant.

The map det$_A:U\to \mathbb{R}, q \mapsto$det$A(q)$ is continuous (where $A(q)$ is the transition matrix between the ordered basis $(E_1|_q,\dots,E_n|_q)$ and $(\tilde E_1|_q,\dots,\tilde E_n|_q)$) and since $U$ is connected and det$A(p)>0$ we have that det$_A$ is always positive on $U$. This implies that $U\subseteq \mathcal C$, and thus $\mathcal C$ is open in $M$.

Analogously we show that $M-\mathcal C$ is open in $M$.

Since $M$ is connected we have $\mathcal C=\emptyset$ or $\mathcal C=M$. In the second case we have $\mathcal O=\tilde{\mathcal O}$. In the first case $\mathcal O$ and $\tilde{\mathcal O}$ are two distinct orientations of $M$ and any other orientation $\hat{\mathcal O}$ of $M$ would have $\hat{\mathcal O_p}=\mathcal O_p$ or $\hat{\mathcal O_p}=\tilde{\mathcal O_p}$, and thus we would have $\hat{\mathcal O}=\mathcal O$ or $\hat{\mathcal O}=\tilde{\mathcal O}$. $\qquad\square$

To be precise I should also prove the existence of two distinct orientations. But if $\mathcal O$ is the orientation which exists by hypothesis, then $-\mathcal O$ is another orientation. So we have at least two (distinct) orientations of $M$.

Please let me know if my proof is correct and if it can be shortened/ simplified.

Answer 1

I think that is correct. One can also do that by considering orientation forms as follows.

Let $(M,\mathcal{O})$ be the connected smooth oriented manifold. By hypothesis, there exists a nonvanishing $n$-form $\omega$ on $M$ such that $\omega$ is positively oriented at each point. Any other choice of orientation $\widetilde{\mathcal{O}}$ for $M$ will induce a nonvanishing $n$-form $\widetilde{\omega}$ on $M$. Since $\omega = f \widetilde{\omega}$ for some smooth function $f : M \to \mathbb{R}$ such that $f(p) \neq 0$ for all $p \in M$, then by the connectedness of $M$, the image $f(M) \subseteq \mathbb{R}$ is a connected subset which does not contain $0$. I.e., the function $f$ either always positive or negative on $M$. If $f$ is positive, then $\omega_p$ and $\widetilde{\omega}_p$ determine the same orientation at $T_pM$ for each $p \in M$. Hence $\widetilde{\mathcal{O}} = \mathcal{O}$. So lets assume that $f$ is negative. By the similar argument $\omega_p$ and $\widetilde{\omega}_p$ determine different orientation at $T_pM$ for each $p \in M$. So $\mathcal{O}$ is different than $\widetilde{\mathcal{O}}$. By similar argument as above, any other orientation $\mathcal{O}'$ for $M$ will induce a non-vanishing $n$-form $\omega'$ such that $\omega' = g \omega$ for either $g : M \to \mathbb{R}$ is positive or negative function. If $g$ is positive, then $\mathcal{O}' = \mathcal{O}$. If $g$ is negative then $\mathcal{O}'$ different than $\mathcal{O}$ and then the product $gf$ is a positive function. Hence above relation $\omega' =g \omega= gf \widetilde{\omega}$ implies that $\mathcal{O} = \widetilde{\mathcal{O}}$. This proves that there are exactly two orientation on $M$.

Suppose $\mathcal{O}_1$ and $\mathcal{O}_2$ are orientation for $M$ and $\omega_1$ and $\omega_2$ are the orientation forms for $\mathcal{O}_1$ and $\mathcal{O}_2$ respectively. Let $\omega_1 = f \omega_2$. If they agree at a point $p \in M$, then the orientation forms is positive multiples of each other at $p$. This implies that $f$ is a positive function. Hence $\omega_1$ and $\omega_2$ determines the same orientation for each point in $M$, i.e., $\mathcal{O}_1 = \mathcal{O}_2$.

Answer 2

I think there's a much short solution as follows: From the converse of proposition 15.5, given two orientiations $O_1$ and $O_2$ on $M$, we can find two nonvanishing $n$-forms $\omega_1, \omega_2$ corresponding to $O_1$ and $O_2$ respectively. By changing $\omega_2$ with $-\omega_2$ if necessary, we can assume they both define the same orientation on a point $p \in M$.

From a previous exercise in Lee, for any other point $q \in M$, we can find a smooth curve $\gamma : [0,1] \mapsto M$ joining $p$ to $q$. Since $M$ is oriented, using the orientation $O_1$ we get the open set $\bigwedge^+ T^* M \subseteq \bigwedge T^* M$ consisting of positively oriented $n$-covectors. Let $\Psi: M \mapsto \bigwedge T^* M$ be the evaluation map, i.e sending $s \in M$ to the value of $\omega_2$ at $s$. Note that $\bigwedge^+ T^*M$ and $\bigwedge^- T^*M$ are dwo disjoint open sets, and $\Psi(M) \subset \bigwedge^+ T^*M \cup \bigwedge^- T^*M$. Then $(\Psi \circ \gamma)^{-1}(\bigwedge^+ T^*M)$ and $(\Psi \circ \gamma)^{-1}(\bigwedge^- T^*M)$ are two disjoint open sets whose union is $[0,1]$; by the connectness of $[0,1]$ the latter is nonempty since $0$ lies in the first open set.

Thus $\omega_2$ is a section of $\bigwedge^+ T^*M$, as is $\omega_1$. By Proposition 15.3 last line, these two define the same orientation.