# Existence of an isotopy in Riemannian manifold

Let $$(M,g)$$ be a Riemannian manifold, and $$p,q\in M$$ be two fixed points. We assume $$p,q$$ are close enough. Say, we assume $$p$$ and $$q$$ are in the same normal coordinate chart. It is clear that there is some diffeomorphism $$F:M\to M$$ so that $$F(p)=q$$ and $$F$$ equals the identity outside a small neighborhood of $$p$$ and $$q$$. For example, we take $$F=\exp(X)$$ for some vector field $$X$$ which flows from $$p$$ to $$q$$ and which is supported near $$p$$ and $$q$$.

Question: Now, suppose $$F'$$ is another such diffeomorphism ($$F'(p)=q$$ and equal to the identity outside a small neighborhood of $$p$$ and $$q$$ too). Can we find an isotopy of diffeomorphisms $$F_s,s\in [0,1]$$ so that $$F_0=F$$, $$F_1=F'$$ and for each $$s$$ we have $$F_s(p)=q$$? Moreover, can we require $$F_s=\exp(X_s)$$ for a smooth family of vector fields $$X_s,s\in[0,1]$$?

I guess the answer should be positive. For example, in the special case $$p=q$$ and $$F'=\mathrm{id}$$; write $$F=\exp(X)$$ and then $$X_p=0$$. Thus, $$F_s=\exp(sX)$$ suffices. My attempt in general is that if $$F'=\exp(X')$$ then we may first take $$G_s=\exp((1-s)X+sX')$$. But we cannot guarantee $$G_s(p)=q$$. Maybe we can do some modification.

• Normally, when a function is said to be supported on a set, it means the function is zero outside that set. So it's not really the right way to express what you mean. I think it would be better if you say that $F$ and $F'$ are equal to the identity outside a neighborhood of $p$ and $q$. – Deane Yang Mar 30 at 22:42
• Another quibble: You should be more precise about what you mean by $p$ and $q$ being close enough, because any two points in a Riemannian manifold lie inside a geodesic ball. – Deane Yang Mar 30 at 22:43
• @DeaneYang Thanks for the comment, and I have edited my post according to what you suggest. – Hang Mar 30 at 22:52

For the first question, which concerns just the smooth category without reference to metrics, the answer depends on the dimension of $$M$$. Before explaining this a small clarification is needed. By "a small neighborhood of $$p$$ and $$q$$" you probably mean a ball containing $$p$$ and $$q$$, otherwise one could take the neighborhood to consist of disjoint balls about $$p$$ and $$q$$, and then there would be no chance of finding a diffeomorphism $$F$$ with $$F(p)=q$$ and with $$F$$ the identity outside the neighborhood.

Let $$Diff(D^n)$$ be the group of diffeomorphisms $$F:D^n\to D^n$$ which are the identity in a small neighborhood $$N$$ of the boundary of $$D^n$$, with the $$C^\infty$$ topology on $$Diff(D^n)$$. The set $$\pi_0Diff(D^n)$$ of path components of $$Diff(D^n)$$ is the same as the set of isotopy classes of diffeomorphisms in $$Diff(D^n)$$ with the understanding that isotopies $$F_s$$ restrict to the identity on $$N$$ for all $$s$$. You are interested in the subspace $$Diff(D^n;p,q)$$ consisting of diffeomorphisms taking $$p$$ to $$q$$. It is not hard to see that every $$F\in Diff(D^n)$$ can be isotoped to take $$p$$ to $$q$$, and also that any isotopy $$F_s$$ between diffeomorphisms $$F_0,F_1$$ taking $$p$$ to $$q$$ can be deformed, fixing $$F_0$$ and $$F_1$$, to an isotopy with $$F_s(p)=q$$ for all $$s$$. In other words the natural map $$\pi_0Diff(D^n;p,q)\to \pi_0Diff(D^n)$$ is a bijection. (In fact this holds for all higher homotopy groups as well.)

If $$\pi_0Diff(D^n)=0$$, i.e., there is a single isotopy class, then the the answer to the first question is Yes. It is known that $$\pi_0Diff(D^n)=0$$ for $$n=1$$ (elementary), $$n=2$$ (Smale), and $$n=3$$ (Cerf). For $$n=4$$ it is still unknown whether $$\pi_0Diff(D^n)$$ is $$0$$ or not. For $$n\geq 5$$ it is known that $$\pi_0Diff(D^n)$$ is isomorphic to the group of exotic spheres of dimension $$n+1$$ (via the h-cobordism theorem and Cerf's pseudoisotopy theorem). This group is known to be $$0$$ for $$n=5,11,55,60$$ and it is known to be nonzero for all other $$n\leq 125$$ and all other even $$n$$, using hard calculations in the stable homotopy groups of spheres.

An embedding $$D^n \subset M$$ induces maps $$\pi_0Diff(D^n)\to\pi_0Diff(M)$$ and $$\pi_0Diff(D^n;p,q)\to\pi_0Diff(M;p,q)$$ by extending diffeomorphsms via the identity outside $$D^n$$. In these terms the question becomes whether the image of the map $$\pi_0Diff(D^n;p,q)\to\pi_0Diff(M;p,q)$$ is $$0$$. If $$M$$ is simply connected then it is not hard to show that the map $$\pi_0Diff(M;p,q)\to \pi_0Diff(M)$$ is a bijection, so in these cases the question boils down to whether the image of the map $$\pi_0Diff(D^n)\to\pi_0Diff(M)$$ is $$0$$. The answer is certainly Yes if $$\pi_0Diff(D^n)=0$$, so the answer is Yes for $$M$$ simply connected of dimension $$n=1,2,3,5,11,55,60$$. In the special case that $$M$$ is the sphere $$S^n$$ there is a classical elementary argument showing that the map $$\pi_0Diff(D^n)\to\pi_0Diff(S^n)$$ is injective (also on all higher homotopy groups), so this provides a No answer for $$M=S^n$$ for all other $$n\leq 125$$ and all even $$n\geq 126$$. For other manifolds $$M$$ the question of injectivity of $$\pi_0Diff(D^n)\to\pi_0Diff(M)$$ was studied in the heyday of differential topology but I can't recall any specific results or references at the moment.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $$F$$ preserves orientation but $$F'$$ reverses orientation, then there is no isotopy between $$F$$ and $$F'$$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $$M=\mathbb{S}^1\times\mathbb{S}^1$$ and $$F:M\to M$$ to be the identity. Let $$F':M\to M$$, $$F'(z_1,z_2)=(z_2,\overline{z}_1)$$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $$y$$-axis to make sure that $$F'$$ preserves orientation). Fix $$p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$$. Then $$F(p)=F'(p)=q$$. But there is no isotopy that would deform $$F'$$ to $$F$$, because $$F$$ and $$F'$$ are not homotopic. Indeed, $$F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$$ but $$F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$$ give two different elements in $$\pi_1(M)$$.

2. The answer is yes if $$F$$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $$F:\mathbb{B}^n\to\mathbb{B}^n$$ is identity near the boundary, $$p=0$$ and $$q\in \mathbb{B}^n$$. According to our assumptions $$F':\mathbb{B}^n\to\mathbb{B}^n$$ be a diffeomorphism which is identity near the boundary and $$F'(0)=q$$. Then $$(F')^{-1}\circ F$$ maps $$0$$ to $$0$$. And $$[0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s}))$$ gives an isotopy between $$F$$ when $$s=1$$ and $$F'$$ when $$s=0$$. Moreover $$p=0$$ is mapped to $$q$$ for all $$s\in [0,1]$$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $$s=0$$.

3. There are orientation preserving diffeomorphisms of $$\mathbb{S}^6$$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $$7$$-spheres are obtained by gluing together two balls $$\mathbb{B}^7$$ along a diffeomorphism $$F:\mathbb{S}^6\to\mathbb{S}^6$$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $$F$$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $$\mathbb{S}^6$$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

• I think the question assumes that $F$ and $F'$ are the identity outside a small neighborhood of $p$ and $q$. I think the answer is yes in that case. – Deane Yang Mar 30 at 22:15
• @Piotr Thank you for this excellent example and this gives some good insight. But I indeed require some additional conditions there. I am sorry my statement is probably not clear enough, and I have edited it now. – Hang Mar 30 at 22:41