For a link $L\subset S^3$ and two *Seifert surfaces* (edit: a better name would be slice surfaces as the comments below [1](https://mathoverflow.net/questions/438878/are-two-slice-surfaces-with-minimal-genus-isotopic#comment1131877_438878) [2](https://mathoverflow.net/questions/438878/are-two-slice-surfaces-with-minimal-genus-isotopic#comment1131880_438878) 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](https://doi.org/10.1307/mmj/1029002727) and [Using a 4th dimension to make Seifert surfaces isotopic?](https://mathoverflow.net/questions/376600/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).



Are two slice surfaces with minimal genus isotopic?
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](https://arxiv.org/abs/2205.12774) 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](https://arxiv.org/abs/2003.13681)) 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](https://arxiv.org/abs/2205.15283) and the Quanta Magazine [article](https://www.quantamagazine.org/special-surfaces-remain-distinct-in-four-dimensions-20220616/) on their result).


