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.

  • 1
    $\begingroup$ 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 '19 at 22:42
  • 1
    $\begingroup$ 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 '19 at 22:43
  • 1
    $\begingroup$ @DeaneYang Thanks for the comment, and I have edited my post according to what you suggest. $\endgroup$ – Hang Mar 30 '19 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.

  • 2
    $\begingroup$ 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 '19 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 '19 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.