Verify proof for continuity of pointwise orientation

Question

This has been asked here, but there is no answer.

Prove that $[(X_1,...,X_n)]$ on a manifold $M$ is continuous if and only if every point $p$ in $M$ has a coordinate neighborhood $(U,\phi) = (U,x^1,...,x^n)$ such that for all $ q \in U$, the differential $ \phi_{ \ast,q}: T_qM \to T_{f(q)}\mathbb{R} \simeq \mathbb{R}^n $ carries the orientation of $T_qM$ to the stander orientation of $\mathbb{R}^n $ in the following sense: $(\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1},..., \frac{\partial}{\partial r^n})$

• This is an exercise from An Introduction to Manifolds by Loring W. Tu, but are there some mistakes? I think:

  1. $\color{red}{\phi(q)}$ and not $f(q)$ in "$T_qM \to T_{f(q)}\mathbb{R}$"

  2. $(\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1}\color{red}{\mid_{\phi(q)}},..., \frac{\partial}{\partial r^n}\color{red}{\mid_{\phi(q)}})$

Here is my proof: What is the correct proof if I am wrong?

$\rightarrow$

I'm not sure continuity is needed here. A pointwise orientation is a frame and so $X_{1,q},..., X_{n,q}$ is an ordered basis for $T_qU \cong T_qM$, by Remark 8.2. The $(U,\phi)$ in question is the same as the one in Lemma 21.4. Since $\phi$ is a diffeomorphism by Proposition 6.10, $\phi_{ \ast}$ is an isomorphism of vector spaces by Corollary 8.6 and so carries basis element to basis element.

• Perhaps this is wrong because I am supposed

  1. to deduce that locally $X_{1,q} = \frac{\partial}{\partial x^n}\mid_{q}$, perhaps with Proposition 8.9, something like what was done in the proofs of Theorem 21.5 or Theorem 21.10

  2. to somehow use Proposition 8.8.

  3. to somehow use to the $dx^1 \wedge ... \wedge dx^n (X_1, ..., X_n) > 0$ on $U$ from Lemma 21.4

$\leftarrow$

Again, without continuity assumed (all the more relevant because continuity is what we're trying to prove this time), a pointwise orientation is a frame and so $X_{1,q},..., X_{n,q}$ is an ordered basis for $T_qU \cong T_qM$, by Remark 8.2.

Again, $\phi$ is a diffeomorphism by Proposition 6.10, $\phi_{ \ast}$ is an isomorphism of vector spaces by Corollary 8.6 and so carries basis element to basis element.

The $\frac{\partial}{\partial r^1}\color{red}{\mid_{\phi(q)}},..., \frac{\partial}{\partial r^n}\color{red}{\mid_{\phi(q)}}$ are smooth and so by assumption of equivalence, the $\phi_{ \ast}X_{1,q},..., \phi_{ \ast}X_{n,q}$ are smooth. Since $\phi_{ \ast}$ is an isomorphism, I guess the $(X_{1,q},...,X_{n,q})$ are smooth and thus continuous.

• The last part might be wrong because I assume smoothness is preserved under a vector space isomorphism.

Answer 1

A pointwise orientation is not a frame but rather a equivalence class of of (possible discontinous) frames, where two frames are equivalent iff at each point the corresponding ordered bases are equivalent.

Furthermore if a pointwise orientation is continous on $U$ then this just means one can find a frame which is continous on $U$ and represents the pointwise orientation. So for example even if $(\frac{\partial}{\partial r^1},\dots, \frac{\partial}{\partial r^n})$ is continous and $(\frac{\partial}{\partial r^1},\dots, \frac{\partial}{\partial r^n})\sim (\phi_{ \ast}X_{1},\dots, \phi_{ \ast}X_n)$ on $\phi(U)$ this does not necessarily imply that $(\phi_{ \ast}X_{1},\dots, \phi_{ \ast}X_n)$ is continous.

But i think the approach of using Lemma 21.4 is good:

Set $\beta=dr^1\wedge\dots\wedge dr^n$. Then

1. $\beta(\frac{\partial}{\partial r^1}|_{\phi(q)},\dots, \frac{\partial}{\partial r^n}|_{\phi(q)})=1$
2. $\beta(\phi_{ \ast}X_{1,q},\dots, \phi_{ \ast}X_{n,q})=dx^1\wedge\dots\wedge dx^n(X_{1,q},\dots,X_{n,q})$

Now by 1., 2. and the corollary of Lemma 21.1 $(\phi_{ \ast}X_{1,q},\dots, \phi_{ \ast}X_{n,q}) \sim (\frac{\partial}{\partial r^1}|_{\phi(q)},\dots, \frac{\partial}{\partial r^n}|_{\phi(q)})$ is equivalent to $dx^1\wedge\dots\wedge dx^n(X_{1,q},\dots,X_{n,q})>0$ and hence we can apply Lemma 21.4 to prove the desired equivalence.