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Among the smooth 1-manifolds (with or without boundary) which embed into $\mathbb{R}^2$, which ones can be represented by a single parametrization $z = (x,y) = f(t)$, for $t \in I$, where $I$ is an interval (not necessarily open or closed), and $f$ is smooth (i.e. infinitely differentiable)?

The reason I ask is that in my reading, I've come across two definitions for "line integral" (which I'm sure turn out to be equivalent):

The first is the standard definition given in most multivariable calculus and complex analysis classes, which relies on defining a "curve" as a (sufficiently nice) function (or image set) $z = \gamma(t)$. That is: $\int_\gamma f(z) dz = \int_a^b f(g(t))g'(t) dt$.

The second is the more high-powered definition involving 1-manifolds and differential 1-forms.

So my question is not really about whether or not these two definitions are equivalent per se, but rather about how much generality is lost by looking at only the "special" 1-manifolds which admit representations as $z = \gamma(t)$.

(My own thoughts: (1) Do these 1-manifolds end up being the 1-manifolds with an atlas consisting of one chart? I'm thinking not, because the circle cannot be given an atlas with one chart, yet can still be parametrized by a single function. (2) Does the Implicit Function Theorem have any role to play?)

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Every compact connected 1-manifold is diffeomorphic to either $S^1$ or $[0,1]$. So your "special" 1-manifolds are precisely connected 1-manifolds. –  Ryan Budney Sep 15 '10 at 6:59
So every connected 1-manifold (not necessarily compact) in $\mathbb{R}^2$ can be given the form $(x,y) = f(t)$ for some smooth function $f$? Am I understanding you correctly? –  Jesse Madnick Sep 15 '10 at 7:48
The OP didn't say "compact" so there's also $(0,1)$ and $[0,1)$. Anyway, if I am understanding your terminology correctly: yes, this means that every embedded $1$-manifold in the plane is parameterized by a smooth function. –  Pete L. Clark Sep 15 '10 at 11:12
So, every connected 1-manifold (with or without boundary) (not necessarily compact) is diffeomorphic to one of: $(0,1)$, $[0,1)$, $[0,1]$, and $\mathbb{S}^1$, is that right? OK, so then you just take that diffeomorphism as the parametrization, I see. Thank you! –  Jesse Madnick Sep 15 '10 at 18:24
@Ryan Budney Should one or more of these comments become an "answer" ? –  isomorphismes May 7 '11 at 21:31

1 Answer 1

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I believe this question has been answered:

several users commented on the fact that "every connected 1-manifold (with or without boundary) (not necessarily compact) is diffeomorphic to one of: $(0,1)$, $[0,1)$, $[0,1]$, and $\mathbb{S}^1$", and Ryan Budney noted that "The proof that 1-manifolds are all of this form isn't particularly hard, either. The idea is to orient a component then take the unit speed vector field on that component and use it to parametrize the component by arc length. You have to argue that such a parametrization is onto. Once you have that, the domain of the parametrization has to be a 1-dimensional submanifold of $\mathbb R$ which puts the problem into the land of single-variable calculus/analysis."

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