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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.

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    $\begingroup$ 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 $q$. $\endgroup$ – Deane Yang Mar 30 at 22:42
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    $\begingroup$ 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. $\endgroup$ – Deane Yang Mar 30 at 22:43
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    $\begingroup$ @DeaneYang Thanks for the comment, and I have edited my post according to what you suggest. $\endgroup$ – Hang Mar 30 at 22:52
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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – Deane Yang Mar 30 at 22:15
  • $\begingroup$ @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. $\endgroup$ – Hang Mar 30 at 22:41

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