# Can an exotic diffeomorphism of the 4-ball change the isotopy class of an embedded surface?

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?

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.
• Thanks, that's a simpler argument than what I had in mind (at least in the $B^4$ case). – Kevin Walker Jun 19 '19 at 22:42
• @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 $x\times I$). – Ian Agol Jun 19 '19 at 23:01