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

[Only a Mirage]

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

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

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 M=\mathbb{R}^n.

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

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]

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.

 

[micromass]

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

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]

Yes, I forgot the smoothness part and just proved the part about the map being an embedding.

 

[micromass]

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.

 

[Only a Mirage]

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 \dot{\gamma}(1) = \dot{\gamma}(0), right? It sounds like requiring the curve to be a periodic function ensures this, and eliminates the issues of differentiability at the endpoints of $[0,1]$.

 

[micromass]

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 \dot{\gamma}(1) = \dot{\gamma}(0), 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.

 

[Only a Mirage]

That was very helpful. Thank you!