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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.

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

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

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