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Suppose that $M$ is an $n-$ dimensional smooth manifold and if there exists an atlas consisting of only two charts $(U,x)$ and $(V,y)$. If $U\cap V$ is connected, then $M$ is orientable.

To prove this, I must show that the transition map $x\circ y^{-1}: y(U\cap V)\to x(U\cap V)$ has a positive Jacobian determinant. That is, if $x=(x_1,...,x_n), y=(y_1,...,y_n)$, then the determinant of the matrix $(\partial x_i/\partial{y_j})_{i,j}$ is positive. But I am not sure how to show that.

Since $U\cap V$ is connected and $\det$ is continuous, it suffices to show that the determinant is positive only at a point $p\in U\cap V$.

Consider the standard orientation given by $dr_1\wedge \cdots\wedge dr_n$ on $\mathbb R^n$. Let's pull these back to $U$ and $V$:

$\begin{align}x^\ast (dr_1\wedge \cdots\wedge dr_n)&=d(r_1\circ x)\wedge\cdots\wedge d(r_n\circ x)\\ &=dx_1\wedge\cdots\wedge dx_n\\ &=\det\left((\partial x_i/\partial{y_j})_{i,j}\right)dy_1\wedge\cdots\wedge dy_n\\ &=\det\left((\partial x_i/\partial{y_j})_{i,j}\right)y^\ast (dr_1\wedge \cdots\wedge dr_n)\end{align}$

If $x,y$ are orientation preserving then the determinant is $>0$ and we are done.

If $x,y$ are both orientation reversing, then also the determinant is $>0$ and we are done.

But what if $x$ is orientation preserving and $y$ is not? In this situation, the determinant could be negative. How to contradict this?

Koro
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  • Not marking as a duplicate, but https://math.stackexchange.com/questions/1403649/manifold-is-not-orientable is closely related. – Andrew D. Hwang Mar 11 '24 at 17:35

1 Answers1

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You can’t get a contradiction. Note you’re asked to prove orientability, not that the specific $\{(U,x),(V,y)\}$ is an oriented atlas (which is a false statement). So, you should instead replace one of the coordinate functions, say $y^1$ with $-y^1$ if you get a negative determinant; then the new pair will have transition with positive determinant.

Also, your ‘proof’ doesn’t really make sense. You can’t speak of $x,y$ being orientation preserving/reversing unless you have an orientation on $M$ already. So, you should rephrase your argument:

If $\det \frac{\partial x}{\partial y}>0$, then we’re done.

Otherwise, consider the chart $(z^1,\dots, z^n)=(-y^1,y^2,\dots, y^n)$ (to be super precise, one should first check this is a chart which is compatible with the atlas given by $\{(U,x),(V,y)\}$, i.e belongs to the maximal atlas… but this is almost obvious). Then, $\det\frac{\partial x}{\partial z}>0$, so the atlas $\{(U,x),(V,z)\}$ provides an orientation.

peek-a-boo
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  • Thanks a lot, very helpful :-). I understand that one can't talk about orientability of $x$ because M was not given to be oriented. But pullback form is still valid. If I define the orientation on U by the pullback of the standard orientation, then x is by default orientation preserving. Similarly $y$ is orientation preserving. So the determinant is positive. What's wrong in this? – Koro Mar 09 '24 at 03:48
  • @Koro sure, you could transport structure to make $x$ orientation-preserving. But really, note that this is referring to orientations on $U$ and $V$ respectively, rather than $M$, so it’s a little awkward to mix all these notions up in a question asking about orientability of $M$ – peek-a-boo Mar 09 '24 at 03:53
  • Never mind. I understood now. Transporting the form is possible, fine. But the question in the post would still remain: on the intersection, the pullback forms by x and y need not be of the same sign. – Koro Mar 09 '24 at 04:01