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For a link $L\subset S^3$ and two Seifert surfaces (edit: a better name would be slice surfaces as the comments below 1 2 point out) with minimal genus $S_1,S_2\subset B^4$, I have the following questions:

  1. Is it true that the surfaces are isotopic?
  2. And if the knots are algebraic?
  3. I am interested in the Hopf link, so the surface is a cylinder, is it true if the surfaces have genus zero?

I know of the existence of the paper Livingston - Surfaces bounding the unlink and Using a 4th dimension to make Seifert surfaces isotopic? but it answers a particular case and I hope that there are updates since 1980 (does not seem so from the citations).

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  • $\begingroup$ Google "Kakimizu complex." $\endgroup$
    – Moishe Kohan
    Commented Jan 19, 2023 at 14:11
  • $\begingroup$ @MoisheKohan Despite the use of the term Seifert surface OP seems to be asking about slice surfaces, as they write $S_1, S_2 \subset B^4$. $\endgroup$
    – mme
    Commented Jan 19, 2023 at 15:09
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    $\begingroup$ OP: What do you mean to assume about $S_1, S_2$? Are these merely surfaces embedded in the 4-ball for which $\partial (B^4, S_i) = (S^3, L)$? (I would call these slice surfaces, in analogy to slice disks.) Or do you mean to assume that these start their life as Seifert surfaces $S_i \subset S^3$ for which you have pushed $S_i \setminus L$ into the interior of the 4-ball? $\endgroup$
    – mme
    Commented Jan 19, 2023 at 15:11
  • $\begingroup$ @mme: I see, I should have read beyond the title... $\endgroup$
    – Moishe Kohan
    Commented Jan 19, 2023 at 16:31
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    $\begingroup$ By the way, even if the question were about "usual" Seifert surfaces, I think that the downvote was a bit hasty. The answer is well-known to specialists, but I don't think that this alone makes the question unsuitable for MO. $\endgroup$
    – Marco Golla
    Commented Jan 19, 2023 at 17:06

1 Answer 1

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The question is very loaded and the question would almost require a survey...

Anyway, the answer to your questions is mostly no. Let $S \subset S^4$ be a 2-knot (i.e. an embedded 2-sphere), $p \in S$ a point, and remove a small ball around $p$. The complement of $S$ in the complement of the ball is a slice disc $D$ for the unknot, and the fundamental group of the complement of $D$ is isomorphic to the complement of the fundamental group of $S^4\setminus S$, which can be non-trivial. (It is possible that $D$ is trivial if and only if $S$ is, but I don't want to make the claim without having properly checked). So, any $2$-knot with non-cyclic fundamental group gives you a slice disc for the unknot which is not isotopic to the standard one.

You can plant the construction into any minimal-genus Seifert surface, and probably you get the same construction for any knot. (Perhaps using the Alexander polynomial is a good way to show that you get different things, without worrying too much about Seifert--van Kampen.)

Now, you might ask: what about slice surfaces whose complement has cyclic fundamental group? Then the answer depends on the category. For locally-flat slice discs, Freedman showed that they are topologically isotopic. There's a nice preprint of Conway, Piccirillo, and Powell where they treat higher-genus surfaces and they give a complete classification in terms of algebro-topological data (Blanchfield forms). In the smooth category, things get pretty weird: Kyle Hayden has examples of slice discs that are topologically isotopic but not smoothly isotopic (see here) and of pushed-in Seifert surfaces that stay non-isotopic in $B^4$ (note that these automatically have cyclic fundamental group).

Back on the subject of pushed-in Seifert surfaces, Hayden, Kim, M. Miller, Park, and Sundberg proved that they can be topologically non-isotopic and topologically isotopic but not smoothly isotopic (see here and the Quanta Magazine article on their result).

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    $\begingroup$ This is a helpful overview! A side note: my understanding is that the Hayden-Kim-Miller-Park-Sundberg paper is not just about the smooth category: they first give a counterexample in the topological category (where the obstruction comes from the intersection form on the double cover, which I'm assuming fits into the Conway-Piccirillo-Powell framework) and then they use it to get a counterexample in the smooth category that isn't a counterexample in the topological category. $\endgroup$
    – Alison Miller
    Commented Jan 20, 2023 at 17:42
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    $\begingroup$ (Before anyone gets confused, I should note that I'm the number theorist Alison Beth Miller who likes Seifert/Blanchfield pairings, not the low-dimensional topological Allison N. Miller who knows this subject better than I do, not to mention that I'm not Maggie Miller of the Hayden-Kim-Miller-Park-Sundberg collaboration.) $\endgroup$
    – Alison Miller
    Commented Jan 20, 2023 at 17:46
  • $\begingroup$ Thanks Alison, you're right. The last sentence is indeed incorrect. I'll fix it now. $\endgroup$
    – Marco Golla
    Commented Jan 20, 2023 at 20:06

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