# Space of all topological knots (tame and wild)

Does anyone know something about the space of all topological knots (continuous maps from $S^1$) in $\mathbb R^3$ (or in some manifold)?

I found only one wonderful fact:

"Montesinos also proved that there exists a universal wild knot, i.e. every closed orientable 3-manifold is a 3- fold branched covering of $S^3$ with branched set a wild knot. This shows how rich the wild knot theory can be." (http://arxiv.org/pdf/math/0509124v1)

Added: I mean the situation same as finite-type invariant theory for tame knots - homotopy type of injective part of this space (I found nothing about it) or homotopy type of topological knots with self-intersections. So, any general-topological properties of singular knot (knot with self-intersection in this context) neighborhood is interesting too.

Added2: Ryan, thanks for clarification! Theo is right, question is "what is the correct topology on the space of topological knots for which knot theory is interesting". Usually only piecewise-smooth knots are studied.

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Your question is rather vague. Of course people know things about wild knots, there's a significant literature on them. But MO isn't the place to look for literature reviews. Why not start by picking up Rolfsen's "Knots and links"? –  Ryan Budney Oct 19 '10 at 13:47
Also, beware of the "space of all knots", wild or otherwise. If you use an obvious topology on the set of knots, then every knot can be deformed into the unknot by making the "knotted" part very small. –  Tom Goodwillie Oct 19 '10 at 13:54
The continuous maps should be injective. –  S. Carnahan Oct 19 '10 at 13:58
I am not sure the functional-analysis tag is appropriate –  Yemon Choi Oct 19 '10 at 18:07
Tom: what you've said (every wild knot is isotopic to the unknot) is a long-standing open problem (from Rolfsen's 1972 paper). Certainly, every PL knot is isotopic to the unknot, and a wild knot is isotopic to the unknot if it has a tame arc (so the "knotted" part is not the entire circle, and the deformation you mention makes sense). A wild knot conjecturally non-isotopic to the unknot is the "Bing sling". –  Sergey Melikhov May 9 '12 at 5:08

A "long topological knot" in $\mathbb R^n$ is a topological embedding $f : \mathbb R \to \mathbb R^n$ such that $f(x) = (x,0)$ for all $x \in \mathbb R \setminus (-1,1)$.

Let $K_n$ be the space of all long topological knots in $\mathbb R^n$ with the compact-open topology. Then $K_n$ is contractible. The contraction is given by

$F : [0,1] \times K_n \to K_n$ defined by

$F(t,f)(x) = (1-t)f(\frac{x}{1-t})$ provided $t \in [0,1)$ and $F(1,f)(x) = (x,0)$.

This map, $F$, is sometimes called "The Alexander Trick". See the Wikipedia Alexander Trick page for context.

In response to your edit, perhaps an interesting topology on $K_n$ could be given this way. Given $f \in K_n$ and $\epsilon > 0$ we'll say an $\epsilon$-ball about $f$ consists of all knots $\phi \circ f$ where $\phi : \mathbb R^n \to \mathbb R^n$ is a homeomorphism which agrees with the identity map outside of $D^n$, and such that $|\phi(x)-x|<\epsilon$ for all $x \in \mathbb R^n$. The topology on $K_n$ could be the topology generated by all $\epsilon$-balls about all $f \in K_n$. Presumably this kind of topology has a name?

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Thanks. It is useful clearing. The question should be transformed (I do it, see "added2") to the question about appropriate space topology. –  Nikita Kalinin Oct 19 '10 at 20:06
> In response to your edit... This topology, I think, catch only a) isotopic type of knot b) may be, homotopy type knot set same isotopic type. So, we can't speak about dicriminant, chambers ... as in Vassiliev theory. May be topology I want doesn't exist - then we need knot set between smooth and topology... In principle, how are knots in topological manifold (which doesn't have smooth structure) studied? –  Nikita Kalinin Oct 20 '10 at 7:02