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I wonder is it still an open question that a smooth sphere $\Sigma^{2}\subset S^4$ is unknotted in $S^4$ iff its complement is homotopy equivalent to $S^1$? If it is an open question, how is it related to other known conjectures in 4D?

I know for all the other $n$ this has been settled by Levine 1965 "Unknotting spheres in codimension 2" and Wall 1965 "Unknotting tori in codimension one and spheres in codimension two", see relatedly Status of a conjecture of C.T.C. Wall?.

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My understanding is this remains an open problem in the smooth category.

I believe there have been a few claims of proofs of this statement in the literature over the years, but as far as I know none of these arguments have been robust. As I believe you are aware, in the topological category this was done by Mike Freedman.

If you jazz up the conjecture a little you could turn it into a recognition principle for $S^1 \times D^3$, and that could in turn be turned into a proof of the smooth 4-dimensional Poincare conjecture. Specifically, say you have a homotopy 4-sphere. Remove a small unknotted $S^2$ (i.e. in some embedded $D^4$ in your homotopy 4-sphere), then its exterior presumably would satisfy the recognition principle for $S^1 \times D^3$. From this you could argue the homotopy $4$-sphere is the standard smooth $S^4$ by filling in $D^2 \times S^2$.

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  • $\begingroup$ No, I was not aware about the topological case resolved by Freedman. Which paper is it? As far as I understand there are 2 versions of the 4D Poincare conjecture: "small" and "large" depending on whether $\Sigma^4\setminus D^4$ embeds or does not embed in $R^4$. The small version is the Schoenflies problem. One would get a recognition principle for a small $S^1\times D^3$, thus the proof of the Shoenflis conjecture? $\endgroup$
    – Victor
    Commented Jun 20, 2021 at 10:22
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    $\begingroup$ I believe it first appears in his 1983 paper "The disk theorem for four-dimensional manifolds" Proc. ICM. But it's also in the Freedman-Quinn book. $\endgroup$
    – Ryan Budney
    Commented Jun 20, 2021 at 18:54
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    $\begingroup$ There are two notions of unknotted that one can consider. The first is that there is a diffeomorphism from $S^4$ to $S^4$ taking an embedded $2$-sphere $K$ onto the standard $S^2$. This is what Ryan addresses. The second is that $K$ is isotopic to $S^2$. This is also true, and follows by Quinn/Perron's 1986 result that pseudoisotopy implies topological isotopy. Both versions are still open in the smooth case. $\endgroup$
    – Danny Ruberman
    Commented Jun 21, 2021 at 17:09
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    $\begingroup$ @DannyRuberman: I think one can show that these two notions are equivalent, provided you demand the diffeomorphism of $S^4$ is orientation-preserving. One direction is that smooth isotopy implies the diffeomorphism by isotopy extension. The reverse argument uses the proof of the Cerf decomposition $Diff(S^n) \simeq O_{n+1} \times Diff(D^n)$. The idea is if you have an orientation preserving diffeomorphism of $S^n$ sending one knot to the other, you can linearize that diffeomorphism on any hemi-sphere you like. The linear part is an element of $SO_{n+1}$ which is connected. . . $\endgroup$
    – Ryan Budney
    Commented Jun 21, 2021 at 20:12
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    $\begingroup$ So if you do a preliminary shrinking of the knot to lie in a hemi-sphere, the linearization process together with the path to the identity in $𝑆𝑂_{𝑛+1}$ gives you the isotopy of one knot to the other. i.e. for any proper submanifold $M$ of $𝑆^𝑛$, the restriction fibration $𝐷𝑖𝑓𝑓(𝑆^n)\to πΈπ‘šπ‘(𝑀,𝑆^n)$ factors (up to homotopy) through $𝑂_{n+1} \to πΈπ‘šπ‘(𝑀,𝑆^n)$, the exotic diffeomorphisms of discs are a null family for this map. $\endgroup$
    – Ryan Budney
    Commented Jun 21, 2021 at 20:27

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