I've just came across this question, which gives us a great tool for showing that smooth manifold is non-orientable. Namely

Thm. If $M$ is a smooth manifold and there are two charts $(U_a,\phi_a),(U_b,\phi_b),$ such that $U_a,U_b$ are connected, $U_a\cap U_b\neq\emptyset$ and transformation function $\phi_{ab}$ neither preserves nor reverses the orientation, then $M$ is non-orientable.

Andrew D. Hwang gave an excellent answer. But still wonders me, if converse statement holds. I.e.

Question. If $M$ is non-orientable, can we find two such charts $(U_a,\phi_a),(U_b,\phi_b)?$

Again Andrew D. Hwang suggested that it holds and gave me the following hint

@Fallen: Yes, the converse is true. Intuitively, if M is non-orientable, there exists a loop (closed curve) around which "the orientation reverses". Thicken the loop into a tube and cover it with two "solid cylinders" Ua and Ub. There are details to check, and this sketch may not be the easiest strategy to make rigorous. (E.g., if no such charts exist, start with an arbitrary atlas and reverse chart orientations as needed to get an atlas of compatible charts. Again, though, there are details to check....).

I only managed to do it assuming that $M$ is connected and second countable.

Proof. Assume that for any two charts $(U_a,\phi_a),(U_b,\phi_b),$ such that $U_a,U_b$ are connected and $U_a\cap U_b\neq\emptyset$ transformation function $\phi_{ab}$ preserves or reverses orientation.

We want to show that $M$ is orientable.

Let $\{(V_i,\psi_i)\}_{i}$ be a countable atlas such that $V_i$ are connected. From that we want to construct a new atlas $\{(V_i,\overline{\psi}_i)\}_{i}$ which is oriented. Construction will be inductive. $\overline{\psi}_1:=\psi_1. \overline{\psi}_i:=\psi_i$ or $\overline{\psi}_i:=\psi^*_i$ (here if $\psi_i=(x^1,\dots,x^n),$ then $\psi^*_i=(x^1,\dots,-x^n)$) to satisfy condition $$\forall_{j<i}\overline{\psi}_{ji}\hspace{5pt}\text{preserves orientation,}$$

where $\overline{\psi}_{ji}$ is transformation function from $\overline{\psi}_j$ to $\overline{\psi}_i.$

But the obvious question is, if such procedure exists. Suppose it fails at some $i.$ Hence there are $j,k<i,$ such that transformation from $\overline{\psi_j}$ to $\psi_i$ preserves orientation and transformation from $\overline{\psi_k}$ to $\psi_i$ reverses. Consider a curve $\gamma$ starting in $V_i\cap V_j$ and ending in $V_i\cap V_k$ and its neighberhoud $U_a$ such that it sits in $V_p$ for $p\leqslant i$ and it meets $V_i$ only in $V_i\cap V_j$ and $V_i\cap V_k.$ Let additionaly $U_a$ be such that $(U_a,\phi_a)$ is a chart for $\phi_a$ "glued" from $\overline{\psi_p}$ for $p<i.$ Set $(U_b,\phi_b):=(V_i,\psi_i).$ As a result $\phi_{ab}$ preserves and reverses an orientation. Hence contradiction with the assumption from the beggining. Hence procedure cannot fail at $i.$ So $M$ is orientable.

These reasoning can be captured in a picture enter image description here

I ask about some simplification of the proof. Now it is too constructive and descriptive for me, especially construction of $\phi_a.$ Do you have any ideas how to prove it simpler?

  • 1
    $\begingroup$ To most of us, all manifolds are second countable. Connectedness shouldn't be relevant: Just work with a connected component. $\endgroup$ – Ted Shifrin Aug 22 '15 at 15:59
  • $\begingroup$ @TedShifrin Yes, connectedness is not the issue. Second countable makes the difference. To hold my reasoning would require transfinite induction and still I will be unable to create a path. I belive construction of $\phi_a$ requires paracompactness. $\endgroup$ – Fallen Apart Aug 22 '15 at 16:02
  • 3
    $\begingroup$ Right. Once again. I don't know any mathematician who works with manifolds (topological or smooth) who doesn't put second countable or paracompact in the definition. Partitions of unity are essential for work in topology/geometry. $\endgroup$ – Ted Shifrin Aug 22 '15 at 16:08
  • $\begingroup$ @TedShifrin You are right. But, it would be nice, if those conditions would be redundant. Especially, when the proof in one way doesn't need them. $\endgroup$ – Fallen Apart Aug 22 '15 at 16:12
  • $\begingroup$ It's entirely possible they're not equivalent for non-second-countable manifolds. Those things can be terrible. I would have trouble cooking up an example. $\endgroup$ – user98602 Aug 23 '15 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.