Orientation of 1-dim manifold

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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?
 

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  • #2
quasar987
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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.
 
  • #3
lavinia
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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.
 
  • #4
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Minor nitpick: every connected one-dimensional topological manifold without boundary is R or S1.
 
  • #5
Hurkyl
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"Manifold" is a special kind of "manifold with boundary", not the other way around. A "manifold with boundary" that has a non-empty boundary is, in fact, not a manifold at all. (Of course, if you are considering "manifolds with boundary", you might decide to rename "manifold" to "manifold without boundary" to avoid this trap of language)

The connected bit is relevant though. (Also, we need second-countable, although some people opt to include that in their definition of manifold)
 

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