Are nearby simple closed geodesics ambient isotopic?

Let $(M,g)$ be a closed Riemannian manifold. I want to show that

there exists a sufficiently small $\delta > 0$ such that if $\gamma_1: S^1 \to M$ and $\gamma_2: S^1 \to M$ are simple closed geodesics with $d(\gamma_1, \gamma_2) = \sup\ \{d(\gamma_1(\theta), \gamma_2(\theta)) : \theta \in S^1\} < \delta$, then they are ambient isotopic.

There are easy counterexamples if you drop the simple condition. I have attempted a proof as follows: Since $M$ is closed, there is some $\epsilon$ such that every point in $M$ has a totally normal neighborhood of radius $\epsilon$. As long as $\delta < \epsilon$, there is a unique shortest geodesic connecting $\gamma_1(\theta)$ to $\gamma_2(\theta)$ for each $\theta$. Geodesic flow along these connecting segments takes $\gamma_1$ to $\gamma_2$. But a priori, two of those connecting segments might intersect such that the flow is not an isotopy. I suspect that something about being geodesics prevents this, since in a neighborhood nearby geodesics should look something like parallel lines oriented in the same direction, but I haven't managed to come up with an argument.

Can this proof be finished, possibly with much smaller $\delta$? Is this statement even true?

• I am little confused about the involement of Riemannian metric. In general if you have two smooth curves in a manifold and they are isotopic (not ambient) then under fairly general hypothesis you can extend the isotopy to an ambient isotopy. This is known an isotopy extension theorem. See Differential topology book of Hirsch. But this is a statement about differential manifold. You don't need Riemannian structure. – tessellation Apr 15 '17 at 6:53
• Yes, the ambient part isn't really the issue. Showing that nearby geodesics are isotopic is the issue. Clearly you can draw arbitrary close by (or just arbitrarily small) loops that are not isotopic, say if one is simple and the other is not. But it feels like that behavior should be ruled out by being geodesics, because geodesics that stay close together should look locally like parallel lines. – Nathaniel Mayer Apr 15 '17 at 16:43
• Do you really need the condition that they are simple and geodesic, or you only need that they are simple. – Anubhav Mukherjee Dec 18 '17 at 22:40
• Also if we have a metric with set of singulartity as cantor set, then it might possible that arbitrary close geodesic of a simple geodesic may not be simple. Now I wonder whether I can prove that any two arbitrary closed simple cuvers are isotopic in a compact manifold or not. – Anubhav Mukherjee Dec 18 '17 at 22:42
• Hmm, I’m not sure. I had thought that the geodesic assumption was needed, but I don’t have an example to prove it. Your last statement can’t be true without at least assuming they’re in the same homotopy class – Nathaniel Mayer Dec 19 '17 at 0:07

Fix a geodesic loop $\gamma$ Define a tubular neighborhood $U$ s.t. $$U =\{ x| d(x,\gamma )<\delta \}$$ where $\delta < \frac{{\rm Inj}\ M}{3}$ and ${\rm Inj}\ M$ is an injectivity radius of $M$. Hence any geodesic of length $< 3\delta$ is minimizing.

Assume that another geodesic loop $\gamma_2$ is in $U$

And define $f$ to be a function on $U$ by $$f(x)=d(x,\gamma )$$

Since $\gamma$ is a geodesic so gradient vector field $X:=\nabla f$ is well-defined. If $F$ is a local flow of $X$, then $F$ is an isotopy sending $\gamma_2$ onto $\gamma$ :

Assume that $$\gamma_2(s_1)=F(\varepsilon,\gamma_2(s_2)),\ s_1<s_2$$

Here ${\rm length}\ \gamma_2|[s_1,s_2]\leq 2\delta$ so that it is minimizing. In further $c(t):=F(t,\gamma_2 (s_2))$ is geodesic so that $\gamma_2$ goes out $U$. So it is a contradiction.

• Why should such an $f$ be an isotopy? Your construction doesn't seem to use the fact that the $\gamma_i$ are geodesics, which is certainly necessary in dimensions $\ge 3$ (otherwise we can tie a very small knot in one of the curves). – Anthony Carapetis Apr 17 '17 at 9:27
• Thank you for your comment. (1) Any two geodesics $c_t$ do not intersect. I know that isotopy is homotopy path s.t. paths starting at any two points do not intersect. (2) I see. You are right. We must assume that $\gamma_i$ are geodesics. – HK Lee Apr 17 '17 at 9:45
• Why don't the $c_t$ intersect? – Anthony Carapetis Apr 17 '17 at 10:08
• I rewrite answer in different approach. – HK Lee Apr 19 '17 at 5:15