# Orientation of 1-dim manifold

I have the result that any 1 dim topological manifold is either R or S1. And I have the fact that every 1-dim topological manifold is orientable in the sense of orientation on simplices.

i want to get that any 1-dim manifold (smooth) is orientable, where orientability is given by the existence of a nowhere vanishing 1-form. Since i know my manifold is either the real line or the circle, does the section s:M --> T∗M that takes each point p to the differential dx at p work as the 1-form i need?

quasar987
Homework Helper
Gold Member
For the real line, this argument works, because R is covered by a single coordinate chart. But the circle cannot be covered by a single coordinate chart (why?), so dx only defines a 1-form locally on some open subset of the circle.

So what you try to do for the circle is to find a collection of charts U_1,...,U_n such that dx^i coincides with dx^j on U_i n U_j. This way, the 1-form a defined by "a=dx^i (on U_i)", is a well defined nowhere vanishing 1-form.

Hint: use two angular coordinate charts.

lavinia
Gold Member
I have the result that any 1 dim topological manifold is either R or S1. And I have the fact that every 1-dim topological manifold is orientable in the sense of orientation on simplices.

i want to get that any 1-dim manifold (smooth) is orientable, where orientability is given by the existence of a nowhere vanishing 1-form. Since i know my manifold is either the real line or the circle, does the section s:M --> T∗M that takes each point p to the differential dx at p work as the 1-form i need?

dx works on the line

dtheta works on the circle.

dx is not everywhere non-zero on the circle.

Minor nitpick: every connected one-dimensional topological manifold without boundary is R or S1.

Hurkyl
Staff Emeritus