Let $W$ be $B^4$ or $S^3 \times I$. Let $Y$ be a properly embedded surface in $W$. Let $f : W \to W$ be a diffeomorphism which is the identity near $\partial W$. Very little is known about $\pi_0(\mbox{Diff($W$) rel $\partial W$})$ (see page 4 of this survey by Hatcher), so $f$ might not be isotopic to the identity (or, in the $S^3\times I$ case, isotopic to a standard twist diffeomorphism). Nevertheless, I believe it is true that $f(Y)$ is isotopic rel boundary to $Y$. In other words, exotic diffeomorphisms cannot do anything exotic to embedded surfaces.

The proof I have in mind (based on conversations with an expert) uses a "push-pull" type argument (and, in the $S^3\times I$ case, straightening along an arc) to concentrate any potential exoticness of $f$ in the neighborhood of one or two points. We can take these points to be far from $Y$, and it follows that $f(Y)$ is isotopic to $Y$.

Rather than write down the details of this proof, I would prefer to just cite a published result. So my question is:

Is there a citable reference for the above claim that $f(Y)$ is isotopic rel boundary to $Y$? Is there a citable reference for the claim that diffeomorphisms of simple 4-manifolds (like $B^4$ or $S^3 \times I$) can be "tamed" in the complement of a finite number of points?


1 Answer 1


Isn't this relatively obvious for $W=B^4$ (i.e. it only took me several hours to realize it was trivial)? Isotope $Y$ to $Y'$ by an isotopy $g_t$ into a small collar neighborhood of $\partial W$ (which we can do by general position), then $f(Y')=Y'$ since you've assumed $f$ is the identity in a neighborhood of $\partial W$. But also $f(Y)$ is isotopic to $f(Y')=Y'$ by the conjugate isotopy $fg_tf^{-1}$, hence $f(Y)$ is isotopic to $Y$.

For $S^3\times I$ one can probably also achieve this by the lightbulb trick, but I haven't thought it through carefully.

  • $\begingroup$ Thanks, that's a simpler argument than what I had in mind (at least in the $B^4$ case). $\endgroup$ Commented Jun 19, 2019 at 22:42
  • $\begingroup$ @KevinWalker For the $W=S^3\times I$ case, all that one needs to observe is that $\pi_0 Diff(W,\partial W)$ is surjected by $\pi_0 Diff(W, \partial W \cup x\times I)$. This follows because any two arcs connecting $S^3 \times 0$ and $S^3\times 1$ are isotopic (again by general position). Now up to isotopy one may assume that the diffeomorphism is the identity in a neighborhood of $\partial W \cup x\times I$, and hence one is reduced to the ball case (we can assume that the surface misses $x\times I$). $\endgroup$
    – Ian Agol
    Commented Jun 19, 2019 at 23:01
  • $\begingroup$ There is a slight issue of trivializing in a neighborhood of the arc. If it represents a nontrivial element in π_1 ( SO(4)), then rotate once to trivialize. $\endgroup$
    – Ian Agol
    Commented Jun 20, 2019 at 17:13

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