Is every knot isotopic to a PL knot? (Cancelled)

D. Rolfsen posed the following problem in 1974: Is every knot in $S^3$ isotopic (=homotopic through embeddings) to a PL knot or, equivalently, to the unknot? In particular, is the Bing sling isotopic to a PL knot?

In 2005, at the conference "Manifolds and their Mappings" (Siegen, Germany) I announced the following results on this subject.
1) There exists a 2-component link with zero linking number which is not isotopic to any PL link.
2) If $K$ is any PL knot that is disjoint from the Bing sling $B$ and links it with nonzero linking number, then the link $(B,K)$ is not isotopic to any PL link.
3) The Bing sling is not isotopic to any PL knot through knots that are intersections of nested sequences of solid tori.

The proof of (1) has been written up (and published) a couple of years ago, see arXiv:2011.01409. This argument can be called geometric; it is based on Cochran's derived invariants, which are extended to topological links by using infinite homological Seifert surfaces. A couple of weeks ago this proof was presented at this seminar (modulo some lemmas), see this video at Youtube (in English).
The proof of (2) for those Bing slings that are intersections of sufficiently rapidly decreasing nested sequences of solid tori has been written up only very recently and will be presented in this talk (modulo some lemmas). This argument can be called algebraic; it is based on the Conway polynomial, whose reduced version $\nabla_{(K_1,K_2)}/\nabla_{K_1}\nabla_{K_2}$ is extended to topological links by using their PL approximations and the notion of $n$-quasi-isotopy, whose origins can be traced to the Homma-Bryant proof of the Chernavsky–Miller codimension three approximation theorem.
Assertion (3) for Bing slings of the same type is proved similarly, and can be regarded as an excercise for those who will listen to the talk.

Connect to Zoom: https://zoom.us/j/97302991744
Access code: the Euler characteristic of the wedge of two circles
(the password is not the specified phrase but the number that it determines)

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The talk had to be cancelled because of the overtime by Y. Rudyak and some of the previous speakers.