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Are there two diffeomorphic smoothly embedded homology 3-spheres $M_1^3, M_2^3 \subset S^4$ that have diffeomorphic complements but such that $M_1$ and $M_2$ are not isotopic? I would be interested in examples in other 4-manifolds as well. I am more interested in examples where $M_1$ and $M_2$ are not smoothly isotopic (I imagine there is a difference between smooth and topological isotopy of 3-manifolds in 4-manifolds, I guess I'd love to hear about references to that as well, specifically in $S^4$).

On a more general note, what are some general ways people tend to tell apart isotopy classes of codimension one submanifolds?

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  • $\begingroup$ In my upcomming paper I proved some of these kind of results on "exotic 3-manfiolds" in 4-manifolds (i.e two copies are topologically isotopic but not smoothly). But those 4-manifolds may have very large $b_2$. Also in my best knowledge (after emailing to a few experts), these kind of problems are not yet wel-studied. $\endgroup$
    – Anubhav Mukherjee
    Commented Dec 29, 2021 at 2:53
  • $\begingroup$ There aren't many tools to distinguish isotopy classes of codimension $1$ submanifolds of $S^4$. In the case of $S^1 \times D^3$ we have an invariant we call "the $W_3$ invariant" that can distinguish some isotopy classes. This invariant has variants that apply to some submanifolds of $S^4$, but we have not completed a computation (yet). But it gives one possibly non-trivial invariant. $\endgroup$
    – Ryan Budney
    Commented Dec 29, 2021 at 3:03
  • $\begingroup$ Also one of my collaborator told me that Watanabe had developed some tools for $S^1\times D^3$ (maybe using theta graph techniques). I am not sure if those are similar to Budney--Gabai's invariant. We developed some gauge theoretic tools to distinguish isotopy classes. At this point I cannot tell the details as the paper is not yet in arXiv. $\endgroup$
    – Anubhav Mukherjee
    Commented Dec 29, 2021 at 3:35
  • $\begingroup$ @AnubhavMukherjee There likely is some relation, although at present Watanabe's techniques are rational homotopy techniques, while ours can also give torsion invariants. For example, the Tom Farrell embeddings $D^{n-1} \to S^1 \times D^{n-1}$ for $n \geq 6$ appear to be detectable with our invariants. This last point should appear in our 2nd paper, which will hopefully be released soon. $\endgroup$
    – Ryan Budney
    Commented Dec 29, 2021 at 6:22
  • $\begingroup$ @RyanBudney I have a few questions. 1) In your techniques can you tell that embedded balls are topological isotopic? 2) Do you know what happened when you stabilize $S^1\times D^3$, i.e connected sum with $S^2\times S^2$? $\endgroup$
    – Anubhav Mukherjee
    Commented Dec 29, 2021 at 16:06

1 Answer 1

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There is an example of an Akbulut cork, proved to be a strong cork by Dai-Hedden-Mallick, $W= W(0,1)$ whose boundary $M$ is $+1$ surgery on the Stevedore knot, and has a mapping class group of order 4 generated by two involutions $S$ and $T$. If one doubles $W$ along $M$, then one gets $S^4=-W\cup W$, and hence an embedding of $M$ into $S^4$, since $W$ is a Mazur manifold (described first by Akbulut-Kirby). (Proposition 1 of Mazur proves that any such manifold has double diffeomorphic to $S^4$)

involutions

Akbulut also shows that $-W \cup_S W$ obtained by gluing two copies by $S$ is diffeomorphic to $S^4$, giving another embedding of $M$ into $S^4$, and same for $-W\cup_T W$. Suppose the two embeddings are isotopic, in particular there is a diffeomorphism taking one embedding to the other. Then this diffeomorphism takes the $W$ on each side to each other. We may think of one $W$ as fixed, and the diffeomorphism extends over the other $W$. But the other two copies of $W$ are glued by the identity and $S$ respectively, so we see that the involution $S$ extends to a diffeomorphism of $W$, a contradiction. Thus we have two copies of $M$ embedded in $S^4$ which are not isotopic, and which have diffeomorphic complements.

One comment: if the isotopy takes the $W$ on one side to the other copy of $W$ in such a way that the diffeomorphism is not isotopic to the identity on $M$, then it must be the involution $ST$ in the mapping class group of $M$, since $S$ and $T$ don’t extend. Then the two mapping classes will differ by $T$, which still doesn’t extend over the other copy of $W$.

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  • $\begingroup$ Dear Professor @Ian Agol, Here what's confusing me the most... If we know that both of the copies are "diffeomorphic" to S^4, then we will have a self diffeomorphism from $f: S^4\to S^4$. Now knowing that two copies of 3-manifolds are not isotopic, doesn't that mean $f$ is not smoothly isotopic to identity and thus contradicting Smale conjecture? $\endgroup$
    – Anubhav Mukherjee
    Commented Feb 9, 2022 at 15:54
  • $\begingroup$ @AnubhavMukherjee I don’t get the logic, why is it not isotopic to the identity? $\endgroup$
    – Ian Agol
    Commented Feb 9, 2022 at 16:02
  • $\begingroup$ Let's say $S_1 = -W\cup_S W$ and $S_2= -W\cup_T W$. Then Both $S_1,S_2$ are diffeomorphic to $S^4$. So maybe it is not clear that how these two different identifications can give a self-diffeomorphism on $S^4$? Also is it clear that it's an "exotic" embedding, that is two copies of $Y$ are top isotopic? Maybe this is where I am getting confused. $\endgroup$
    – Anubhav Mukherjee
    Commented Feb 9, 2022 at 16:10
  • $\begingroup$ @AnubhavMukherjee If by $Y$ you mean what I am calling $M$, then the point of the answer is to show that the two copies are not isotopic. One manifold was shown to be S^4 by Mazur, one by Akbulut. So the manifolds are diffeomorphic, but not in a way that takes one copy of M to the other. $\endgroup$
    – Ian Agol
    Commented Feb 12, 2022 at 5:32
  • $\begingroup$ @AnubhavMukherjee Also, I think that Akbulut shows that the involutions $S$ and $T$ extend to a homeomorphism of $W$, so there is a homeomorphism sending one embedded M to the other. If the space of orientation preserving homeos is connected, then the two embeddings are topologically isotopic. But I don’t know if this is the case. $\endgroup$
    – Ian Agol
    Commented Feb 12, 2022 at 5:39

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