Is every smooth simple closed curve a smooth embedding of the circle?

Suppose I have a smooth curve [itex]\gamma:[0,1] \to M[/itex], where [itex]M[/itex] is a smooth [itex]m[/itex]-dimensional manifold such that [itex]\gamma(0) = \gamma(1)[/itex], and [itex]\hat{\gamma}:=\gamma|_{[0,1)}[/itex] is an injection. Suppose further that [itex]\gamma[/itex] is an immersion; i.e., the pushforward [itex]\gamma_*[/itex] is injective at every [itex]t\in [0,1][/itex].

Claim: The image set [itex]\gamma([0,1])[/itex] is a smooth embedding of the unit circle [itex]S^1[/itex].

This seems intuitively obvious to me -- I can "bend and stretch" a circle into any "nice" closed curve (and vice versa). However, I'm having trouble proving this, even for the special case where [itex]M=\mathbb{R}^n[/itex].

Here is how I've been approaching this: If [itex]\beta:[0,1]\to \mathbb{R}^2[/itex] is a smooth parametrization of the unit circle such that [itex]\hat{\beta}:=\beta|_{[0,1)}[/itex] is an injection, [itex]\beta(0) = \beta(1)[/itex], [itex]\dot{\beta}(t) \not = 0[/itex] for all [itex]t[/itex], and [itex]\beta([0,1]) = \hat{\beta}([0,1)) = S^1[/itex], then [tex]\hat{\beta}^{-1}:S^1 \to [0,1)[/tex] and [tex]\alpha \circ \hat{\beta}^{-1}:S^1 \to C [/tex] where [itex]C:= \gamma([0,1])[/itex]. It seems that if I can show that [itex]\alpha \circ \hat{\beta}^{-1}[/itex] is smooth with an everywhere-injective pushforward, I'd be done. However, I can't figure out how to do this as [itex]\hat{\beta}^{-1}[/itex] isn't even continuous, so I don't think it is differentiable.

Any ideas? I feel like I'm missing something obvious. What little differential geometry (and topology) I know is self-taught, so maybe there is a standard trick I haven't learned.
 

Ben Niehoff

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It looks like you're running up against the fact that the circle can't be covered by a single coordinate chart. Try using two maps, that each map an open interval into M, together with a pair of transition functions that describe how the portions of the circle should be glued together.
 

WWGD

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Maybe you can argue that ## \gamma (0)= \gamma(1) :[0,1] \rightarrow M ## , that your map is a map from a quotient ##[0,1]/~ ; 0~1 ## which is homeo to ## S^1 ##. Since you have a continuous injection, the map is a bijection into its image, and it is a bijection between compact space and Hausdorff space, which is an embedding.
 
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I think the result is false. Certainly the image ##\gamma([0,1])## will be homeomorphic to the circle ##S^1##, but smoothness fails.

The reason is that you have no guarantee of smoothness in ##\gamma(0)##. The curve ##\gamma## might have a cusp in ##\gamma(0)## which means it won't be a smooth embedding of the unit circle into ##M##.

Consider for example the cardioid: http://en.wikipedia.org/wiki/Cardioid which is given by

cardioid.gif


I think it satisfies all the requirements in the OP but it is not the smooth embedding of ##S^1## in ##\mathbb{R}^2## because of the cusp in the origin.
 

WWGD

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Yes, I forgot the smoothness part and just proved the part about the map being an embedding.
 
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Yes, I forgot the smoothness part and just proved the part about the map being an embedding.
Your argument is a really nice one. It can actually be used to prove the smooth case. Not the smooth case like the OP formulates it, but if you reformulate it a bit to let ##\gamma:\mathbb{R}\rightarrow M## be a smooth periodic curve with period ##1## which is injective at ##[0,1)## and such that ##\gamma_*## is injective.

Then you can consider the quotient manifold ##\mathbb{R}/\mathbb{Z}## which is just quotienting out with respect to the Lie group ##\mathbb{Z}##. This is diffeomorphic to the circle ##S^1##. Then your argument can be generalized to this case.
 
Thanks everyone!

Your argument is a really nice one. It can actually be used to prove the smooth case. Not the smooth case like the OP formulates it, but if you reformulate it a bit to let ##\gamma:\mathbb{R}\rightarrow M## be a smooth periodic curve with period ##1## which is injective at ##[0,1)## and such that ##\gamma_*## is injective.

Then you can consider the quotient manifold ##\mathbb{R}/\mathbb{Z}## which is just quotienting out with respect to the Lie group ##\mathbb{Z}##. This is diffeomorphic to the circle ##S^1##. Then your argument can be generalized to this case.
I'm not very comfortable with Lie groups yet, but I think I understand what you're saying. Basically, I just need to require that [itex]\dot{\gamma}(1) = \dot{\gamma}(0)[/itex], right? It sounds like requiring the curve to be a periodic function ensures this, and eliminates the issues of differentiability at the endpoints of ##[0,1]##.
 
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I'm not very comfortable with Lie groups yet, but I think I understand what you're saying. Basically, I just need to require that [itex]\dot{\gamma}(1) = \dot{\gamma}(0)[/itex], right? It sounds like requiring the curve to be a periodic function ensures this, and eliminates the issues of differentiability at the endpoints of ##[0,1]##.
Yes, exactly right.
 
That was very helpful. Thank you!
 

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