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January 10

Knot equivalence

Our knot theory article gives two definitions, asserted to be equivalent, for when one knot is equivalent to another. Both of these involve auto-homeomorphisms of ${𝐑}^3$.

To me this just kind of feels -- heavy. To move one trefoil knot to another, verifying their equivalence, I apparently have to account for all the rest of space. Is this actually necessary?

Concretely, suppose I define knot equivalence as follows. If $K_0$ and $K_1$ are two knots, defined as continuous maps from [0, 1] into ${𝐑}^3$ that are injective except that $K_i(0)=K_i(1)$ ($i=0,1$), then say $K_0$ and $K_1$ are equivalent if there is a continuous map $f$ from $[0,1]\times [0,1]$ into ${𝐑}^3$ such that:

• the map $x\mapsto f(x,0)$ equals $K_0$
• the map $x\mapsto f(x,1)$ equals $K_1$
• for $0<t<1$, the map $x\mapsto f(x,t)$ is injective except that $f(0,t)=f(1,t)$

Is this notion of equivalence the same as the one in the article? --Trovatore (talk) 07:01, 10 January 2025 (UTC)

Since no $K_t$ intersects itself, my intuition tells me there ought to be a lower bound on how close it can come to self-intersection. More precisely, the boundary of the Minkowski sum $K_t+B_{\delta }$ of $K_t$ and a constant ball $B_{\delta }$ with a sufficiently small (Template:Nowrap) radius $\delta$ should be a torus. Proving this formally may not be easy, but a proof will establish the required ambient isotopy.

If your conjecture is indeed correct – I'm hedging my bet because counterexamples in topology can be quite counterintuitive – it would be amazing if your markedly simpler characterization is not found in the literature.  --Lambiam 10:20, 10 January 2025 (UTC)

I think knot (lol). I believe any nontrivial not is "injectively homotopic" to the unknot. Start with your favorite knot and stretch one arc while compressing the rest so that in the limit the compressed part tends to a point. This can be done injectively. Tito Omburo (talk) 12:03, 10 January 2025 (UTC)

Yeah, that sounds right. Thanks. --Trovatore (talk) 18:22, 10 January 2025 (UTC)

With the R³ embedding definition you can definitely say that if the complements are not homeomorphic then the knots are inequivalent. That's a useful property since you may be able to show that two knots are different without having to go into the details of knot theory, in fact you can show the trefoil knot is not the unknot with a little algebraic topology. In practice, two knots are equivalent if their diagrams can be transformed into each other by a sequence of Reidemeister moves, so the problem is really to find an intuitive definition that's equivalent. As Lambiam pointed out, topology can be counterintuitive, and perhaps "injectively homotopic" = "homotopic" is an example of that. --RDBury (talk) 00:40, 11 January 2025 (UTC)

Think yourself lucky. Maybe. When I had a look at knot theory a while ago the definition was in terms of a finite set of straight lines and Reidemeister moves, and things like infinite knots were an extension. And then there were the extensions like knotting a sphere in 4d. NadVolum (talk) 14:50, 10 January 2025 (UTC)