Any 1-d compact connect submanifold of R3 is diffeomorphic to a circle?

Question

It's problem 2.2.1 in Guillemin's Differential Topology. I can't understand the statement. Why is it true? Why the submanifold can't diffeomorphic to an interval? And for the question, can the submanifold be deformed into a circle within $\mathbb R^3$?

Answer 1

Fix a Riemannian metric on the manifold $M$ (it is in $\mathbb{R}^3$, so you can take the induced metric. However, every manifold admits a Riemannian metric, so the following argument is general) and take a geodesic $\gamma$ emanating from a point $p$. Since the manifold is compact, the geodesic is defined for all real numbers. By the local form of submersions, $\gamma$ is an open map. A local chart argument using the fact that the charts are on segments shows that $\gamma$ is also closed, hence $\gamma$ is surjective since the manifold is connected. We also have that $\gamma$ is not injective. If it were, being a bijective open map, it would be a homeomorphism, which is absurd.

Now, we then have $t_1<t_2$ such that $\gamma(t_1)=\gamma(t_2)$. Let $\gamma(t_1)$ be our new point $q$, and consider the geodesic $\zeta$ emanating now from $q$ with same velocity with which $\gamma$ passes through $q$. Since it is only a shift from our starting geodesics $\gamma$, it is also continuous and surjective. Let $T$ be the smallest positive number such that $\zeta(T)=\zeta(0)$. It is then clear that $\zeta(a+nT)=\zeta(a)$ for all $n \in \mathbb{N}$ by uniqueness of geodesics. To see that it also holds for $n \in \mathbb{Z}$, notice that $\alpha(t)=\zeta(T-t)$ is a geodesic such that $\alpha(T)=\alpha(0),$ hence $\alpha(2T)=\zeta(-T)=\zeta(0)$ and then we have that $\zeta(a+nT)=\zeta(a)$ is also true for the integers.

It then follows that $\zeta$ induces a map $\hat{\zeta}:S^1 \to M$ which is smooth. It must be surjective, since $\zeta$ was. It is injective because if it was not, then $T$ would not have been the smallest number such that $\zeta(T)=\zeta(0)$ (we would be able to have $0<s_1<s_2<T$ such that $\zeta(s_1)=\zeta(s_2)$, and then "shifting" to the origin $\zeta(0)=\zeta(s_1-s_1)$). It follows that $\hat{\zeta}$ is a diffeomorphism.

Answer 2

It's true because there's a proof.

An open interval is not compact, and a closed interval is not a 1-d sub manifold.

Finally, there exist examples which cannot be deformed to a circle, such as the trefoil knot; here I assume by "deformed" you mean "isotopic".