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There is another question that I am trying to solve involving isotopy. I know that this question is resolved by the fact that I can untie any knot formed from a smooth embedding of $S^1$ so long as I have four dimensions.

However, I haven't found any reference to this in any of the several algebraic topology and manifold theory books that I've been reading. This includes manipulations like Whitney tricks or the like. I can't in good faith feel like I understand how to solve the question unless I can understand a proof on how any of these knots can be untangled in $\mathbb R^4$ -- for instance, while I understand that you can form and can't untangle a trefoil knot in $\mathbb R^4$, I can't properlty explain right now why there isn't some kind of "hyper trefoil" knot in $\mathbb R^4$ that also can't be untangled there (even though I am aware this can't happen).

I'm looking for books, papers, videos, or any source of information that explicitly shows how to prove this property, with whatever theorems and machinery are required.

Thanks in advance.

Nate
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    How about on this website itself? https://math.stackexchange.com/questions/1426501/why-are-all-knots-trivial-in-4d – HackR Sep 13 '24 at 19:04
  • @HackR This looks great. Thanks! – Nate Sep 13 '24 at 20:03
  • For a knot K embedded in 3-space via h : [0,1] —> R^3 with h(0) = h(1), we can temporarily remove a small interval J of K, and embed the closure of its complement via H : [eps, 1] —> R^4 in 4-space via H(s) = (h(s), s) in R^3 x R. (The remaining interval can now be replaced with a large loop connecting H(eps) to H(1), avoiding the rest of the image.) Now each 3-space R^3 x {t} in R^4 contains at most two points of the image, so it is now easy to isotop the knot to be a standard circle in R^4. – Dan Asimov Sep 13 '24 at 20:05
  • If you are working in the topological category, the claim you "know" to be true is actually false: There are wild nontrivial 1-dimensional knots in $R^n$ for all $n>2$. And "homotopy theory" tag is irrelevant here. – Moishe Kohan Sep 13 '24 at 23:50
  • @MoisheKohan In the smooth categoryi it's not totally irrelevant :). The smale hirsch implies that $\mathrm{Imm}(S^1,\mathbb{R}^4) \simeq \mathrm{Map}(S^1,S^{3})$, and an easy transversality argument shows that the map $\mathrm{Emb}(M,N) \to \mathrm{Imm}(M,N)$ is $4-2-1=1$ connected, implying the result – Andres Mejia Sep 14 '24 at 05:54
  • I mean just a stupid argument lol. – Andres Mejia Sep 14 '24 at 05:56
  • Does making it a smooth embedding help? The original problem asked for a smooth embedding (I made the change to the title). Knots aren't my forte; I wasn't aware of wild knots when I wrote this question. – Nate Sep 14 '24 at 08:26
  • Yes, assuming smoothness everything is fine. My guess is that the expected solution is by using a relative form of the Whitney embedding theorem which was likely covered in a Stanford differential topology class. – Moishe Kohan Sep 14 '24 at 16:57
  • I am closing this as a duplicate, although I suppose one has to twist the logic of duplicates to justify do this: as said by @HackR, the underlying mathematical question is duplicated on this site. – Lee Mosher Sep 15 '24 at 14:19

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