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Timeline for Unknotted $S^{n-2}$ in $S^n$

Current License: CC BY-SA 4.0

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Jun 21, 2021 at 20:27 comment added Ryan Budney 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.
Jun 21, 2021 at 20:12 comment added Ryan Budney @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. . .
Jun 21, 2021 at 17:09 comment added Danny Ruberman 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.
Jun 20, 2021 at 18:54 comment added Ryan Budney 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.
Jun 20, 2021 at 10:22 comment added Victor 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?
Jun 20, 2021 at 10:17 vote accept Victor
Jun 19, 2021 at 19:07 history answered Ryan Budney CC BY-SA 4.0