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The Hopf link is an embedding of two circles in $\mathbb{R}^3$ which cannot be separated without intersecting each other. If we embed this in $\mathbb{R}^4$, however, it is easy to see that one can separate them.

Is it possible, however, to embed two circles in $\mathbb{R}^4$ in such a way that you will not be able to separate them without intersecting them?

To be extremely precise, set $X=S^1\times 2$. Is there an injective, continuous (preferably smooth or even analytic) function $$f:X\to \mathbb{R}^4,$$ which admits no continuous function $$H:X\times [0,1]\to \mathbb{R}^4,$$ for which

  1. For every $t\in [0,1]$, the map $H_t$ defined by $H_t(x)=H(x,t)$ is still injective,
  2. The two subsets $H_1(S^1\times \{0\})$ and $H_1(S^1\times\{1\})$ can be separated by a hyperplane.

(I hope I got the formulation correctly; if anyone thinks I made a mistake with it, please point it out).

Remark: observe that if $f$ is nice enough then both knots it is composed of can be unknotted. Otherwise they may be wild knots, however. If there is no answer where $f$ is nice, I'd prefer at least for each of the knots to be trivial, if that is possible. See here for more info.

Cronus
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  • @ThomasAndrews Your first remark is of course correct, I corrected this. Your second remark seems to me to be correct, if we assume $f$ is smooth, although I am not entirely sure the homotopy contracting the other loop to a point can be made injective at any given point in time... is this clear? – Cronus Sep 21 '21 at 16:03
  • You don’t need $f$ smooth, just injective and continuous. That’s why I didn’t say to a point, just to “near a point.” And any time you can contract a simple loop to a point, you can do so such that every intervening loop is a simple loop. But I’m working on a proof of that - it might need some geometry on $Y,$ like $Y$ being a manifold. – Thomas Andrews Sep 21 '21 at 16:42
  • @ThomasAndrews There are definitely closed simple loops in $\Bbb{R}^4$ whose complement is not simply connected. – Cronus Sep 21 '21 at 16:56
  • I’d like to see an example. We don’t study knots in four dimensions precisely because there are no interesting nots in four dimensions. – Thomas Andrews Sep 21 '21 at 17:00
  • @ThomasAndrews You can see the example in the question I linked to. People usually restrict to the case of "tame" knots, which indeed are all trivial in dimension 4 or higher, so people rarely talk about knots in that case – Cronus Sep 21 '21 at 17:03
  • Okay. Although I don’t quite understand why $\pi_1(\mathbb R^n\setminus K)\neq 1$ just because $\pi_{1}(\mathbb R^n\setminus C)\neq 1.$ Reducing the set reduces the possible homotopies, but it also reduces the possible loops. But at the very least, this show that if $f_1$ is a nice loop, there is no $f_2,$ nice or merely a simple loop, to provide a counter example. – Thomas Andrews Sep 21 '21 at 17:18
  • @thomasandrews Yes, I agree, so this solves my problem in the case one assumes $f$ is nice (although I'm not sure I understood the details of your solution). In any case, thank you. – Cronus Sep 21 '21 at 17:20
  • I think there is a misunderstanding as to what an embedding is here. You cannot get a wild knot or a long knot if $f:S^1 \to \mathbb{R}^n$ is an embedding. This is why we use embeddings, specifically to exclude these situations. – N. Owad Sep 21 '21 at 18:33
  • @N.Owad I am aware of that, that's why I gave a very precise formulation of the question. It seems that there is no such $f$ if one assumed it to be an embedding, or even smooth (Thomas Andrews gave a sketch proof for that). But there is if you do not. – Cronus Sep 21 '21 at 18:37
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    You are making a "standard mistake" defining an isotopy of a link. – Moishe Kohan Sep 21 '21 at 21:16
  • @MoisheKohan which is? – Cronus Sep 21 '21 at 21:17
  • I see now that usually a much stronger condition is required for links to be trivial. But is it really a mistake? I mean - it is a definition... – Cronus Sep 21 '21 at 21:21
  • @Cronus One can show that every link in 3d is trivial with your definition. The standard definition is to require an ambient isotopy or to work in a smooth or PL category. I suggest you read the discussion in any textbook in the subject. – Moishe Kohan Sep 21 '21 at 22:03
  • @MoisheKohan I am pretty sure you're wrong - there's no way the Hopf Link is trivial according to this definition... – Cronus Sep 22 '21 at 08:48
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    @Cronus: You are right, all knots are trivial but (some) links are not (with your definition). The linking number is indeed constant under such deformation, it even suffices to ensure that the components stay disjoint. – Moishe Kohan Sep 22 '21 at 09:15
  • @MoisheKohan Yes. Do you think that according to the "real" definition, all links are trivial in four dimensions? – Cronus Sep 22 '21 at 11:29
  • I am quite sure that for topological links there are but a proof would require some work. As for smooth/PL links, they are all trivial. – Moishe Kohan Sep 22 '21 at 14:21

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