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.

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Ben Niehoff

Gold Member
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.

• 1 person

WWGD

Gold Member
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

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

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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.

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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.

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Only a Mirage

That was very helpful. Thank you!

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