Low Dimensional Topology

May 17, 2013

An old corker on the unknotting of knots

Filed under: Knot theory — Ryan Budney @ 11:13 am
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I imagine many readers of this blog are familiar with the fact that you can knot a circle in 3-space, but not in 4-space.    If you enjoy thinking about why that is true, please read on!

Think of euclidean 3-space, \mathbb R^3 as a linear subspace of euclidean 4-space, \mathbb R^3 \equiv \mathbb R^3 \times \{0\} \subset \mathbb R^4.  So if you have a knotted circle in 3-space, you can consider it as an embedded circle in 4-space.  And you can unknot it! I think one of the simplest explanations of of this would be the idea to push the knot up into the 4-th dimension every time a strand is close to being an overcrossing (in a planar diagram).   At this stage you could in effect change the crossing to be anything you want, after you’re done modifying the crossings, you could push the knot back into 3-space to get a different knot. 

But there’s a more uniform version of this construction.  I think I first learned of it from Rolfsen’s textbook.  It sits most naturally in a slightly different formalism. 

\mathcal K_n = \{ f : \mathbb R \to \mathbb R^n is a smooth embedding and  f(x) = (x,0) whenever |x| > 1 \}.

is called the space of long knots in \mathbb R^n.  There is a natural inclusion map i : \mathcal K_n \to \mathcal K_{n+1}.  I claim it is a null homotopic map. 

The idea is pretty simple. When you perform the inclusion, there is a new direction in \mathbb R^{n+1} orthogonal to the original \mathbb R^n \times \{0\}. Let’s call the unit vector in that direction e_{n+1}. The idea is to take a little bump function \beta : \mathbb R \to \mathbb R, and add \beta e_{n+1} to i(f) \in \mathcal K_{n+1}. Choose \beta so that it is strictly increasing along the interval [-1,1], and then have it decrease to zero quickly near 1.

There is a straight-line isotopy from i(f) to i(f) + \beta e_{n+1}, and also from i(f) + \beta e_{n+1} to i(s) + \beta where s \in \mathcal K_n is the standard inclusion, i.e. s(x) = (x,0) always. Similarly there is a straight-line isotopy from i(s) + \beta to i(s). Assemble these three maps together and you have your null-homotopy of \mathcal K_n \to \mathcal K_{n+1}.

Once you see this null-homotopy, notice that there actually two such null-homotopies! If instead of choosing \beta to be increasing along [-1,1], we could have chosen it to be decreasing along [-1,1]. If you assemble those two null-homotopies together, you get a map \mathcal K_n \to \Omega \mathcal K_{n+1}, or equivalently \Sigma \mathcal K_n \to \mathcal K_{n+1}.

My question for people here is, is this map null-homotopic?

I don’t know the answer to this question. From the perspective of the Vassiliev spectral sequence, or the Goodwillie embedding calculus, this is a difficult map to understand. You can check that this map is zero on rational homology and rational homotopy groups. Not enough is known about the torsion in homotopy or homology to say what’s going on there — in a way that’s part of why I love this question. It’s also possible one of you will notice there’s a naive null-homotopy of this map. I just don’t see it.

Edit: One further note. There can’t be too naive a null-homotopy of the map \Sigma \mathcal K_n \to \mathcal K_{n+1}. This is because if you evaluate your knots f \in \mathcal K_n at a finite collection of points in \mathbb R, this map restricts to the “Freudenthal suspension map” \Sigma C_n \mathbb R^n \to C_n \mathbb R^{n+1}. This map is known to be an isomorphism on the first non-trivial homology (and homotopy) group (see for example Fred Cohen’s dissertation). So in particular, given a finite set of points in \mathbb R there are restriction maps \mathcal K_n \to C_n \mathbb R^n. These maps to the configuration spaces are themselves null-homotopic, so they do not inform on whether or not \Sigma \mathcal K_n \to \mathcal K_{n+1} is null-homotopic, but they do put restrictions on the nature of any such null-homotopy, were it to exist.

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2 Comments »

  1. I very much like this question! An answer either way would be really interesting.

    Comment by dmoskovich — May 25, 2013 @ 10:34 am | Reply

  2. I’m glad you like it. It’s sort of like a homotopy-theoretic analogue of the question of if a knot is doubly-slice.

    Comment by Ryan Budney — May 25, 2013 @ 6:30 pm | Reply


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