5
$\begingroup$

Let $S$ be a regular surface covered by coordinate neighborhoods $V_1$ and $V_2$. Assume that $V_1\cap V_2$ has two connected components, $W_1$, $W_2$, and that the Jacobian of the change of coordinates is positive in $W_1$ and negative in $W_2$. Show that $S$ is non-orientable.

I know that, if a regular surface $S$, can be covered by two coordinate neighborhoods, whose intersection is connected, then the surface is orientable.

Furthermore, if $f:S\subset\mathbb{R}^3\to\mathbb{R}$ is a continuous function, in a connected surface $S$, then $f$ doesn't change of sign on $S$. Can give any hint! Thanks!

$\endgroup$
3
  • 1
    $\begingroup$ What's your definition of orientable? $\endgroup$
    – arkeet
    Oct 5, 2016 at 2:47
  • $\begingroup$ Regular surface S, is called orientable if it is possible to cover it with a family of coordinate neighborhoods. And a point $p\in S$ belongs to two neighborhoods of this family, then the change of coordinates has positive Jacobian at $p$. The choice of such a family is called and orientation of S $\endgroup$
    – MathUser
    Oct 5, 2016 at 2:51
  • 3
    $\begingroup$ Possible duplicate of Manifold is not orientable $\endgroup$ Jun 20, 2019 at 17:10

1 Answer 1

1
$\begingroup$

Rough idea: If you have a loop in a manifold, it can be subdivided such that each segment is contained in a coordinate chart, and such that the initial and final segments lie in the same chart. Then the sign of the product of the Jacobians of successive transition maps is independent of the choice of charts and subdivision.

Now if $S$ is orientable, then the sign for any loop must be positive (by using an oriented atlas). But if $p \in W_1$ and $q \in W_2$, and we consider a loop that starts at $p$, goes to $q$ within $V_1$, and returns to $p$ within $V_2$, then its sign must be negative.

I'm not entirely sure that this is correct or that there isn't a much simpler solution, but it's all I can think of for now.

$\endgroup$
1
  • $\begingroup$ What is the mathematical expression of a loop? BTW, to find a path from p to q, how to make sure S is path-connected so that such path is possible? $\endgroup$ Jul 5, 2023 at 13:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .