1
$\begingroup$

Is it true or false? Let $M$ a manifold of class $C^{\infty}$ not orientable and $p \in M$, then $M - \{p\}$ is orientable.

I believe this to be false, but I do not know a counterexample

$\endgroup$
3
$\begingroup$

If $M$ is the (boundaryless) Mobius strip and $p$ is a point that is not on the central loop, then $M\setminus\{p\}$ is not orientable, because the central loop has still got a neighbourhood which is diffeomorphic to the whole Mobius strip.

(actually, it's the same even if you pick $p$ on the central loop, though the diffeomorphism may be less obvious at first glance)

$\endgroup$
2
$\begingroup$

This is never true. That is, $M$ is orientable iff $M-\{p\}$ is orientable. This follows from the following characterization, which does not "see" discrete subsets:

Proposition. $M$ is not orientable iff there exists a finite sequence of charts $(U_i, \phi_i)_{1 \leq i \leq n}$ with $U_i$ connected, $n \geq 3$ and indices $\bmod n$ such that:

  • $U_i \cap U_{i+1} \neq \varnothing$ for all $i$
  • $\det(D(\phi_i \circ \phi_{i+1}^{-1}))>0$ for all $i$ except one
$\endgroup$
  • $\begingroup$ There is possibility of doing this is also normal fields, correct? $\endgroup$ – Croos Jun 29 '18 at 19:02
  • $\begingroup$ What do you mean with 'doing this' and 'normal fields'? $\endgroup$ – rabota Jun 29 '18 at 21:09

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.