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.

  • $\begingroup$ "any other orientation $\hat{\mathcal O}$ of $M$ would have $\hat{\mathcal O_p}=\mathcal O_p$ or $\hat{\mathcal O_p}=\tilde{\mathcal O_p}$", why ? It seems you're already know that if we have a third orientation it must coincide with one of the two given $\endgroup$ Jan 1, 2021 at 14:28

2 Answers 2


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., $\mathcal{O}_1 = \mathcal{O}_2$.

  • $\begingroup$ Thank-you Sou, I like your proof :) why your answer here math.stackexchange.com/questions/3152728/… disappeared? $\endgroup$
    – Minato
    Mar 19, 2019 at 18:13
  • $\begingroup$ I just feel uncomfortable to post a full answer of this problem in MSE. $\endgroup$ Mar 20, 2019 at 5:13

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.


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