In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$? - MathOverflow

In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T12:26:49Z

In several textbooks on knot theory (e.g. Lickorish's, Rolfsen's) knots are considered in $\mathbb{R}^3$ or $S^3$. The reason for working in $S^3$ is sometimes given at the beginning of a text as that $S^3=\mathbb{R}^3\cup \{\infty\}$ is compact, so working there is "more convenient".

What are some specific benefits or conveniences for working in $S^3$ instead of $\mathbb{R}^3$?

I could think of one example: In knot theory we work in piecewise-linear category, so working in $S^3$ allows for considering finite triangulations of knot complements.

Another example is in studying achirality, where Smith theory for fixed points of periodic maps on the sphere is used.

Furthermore, some results in these textbooks are stated for knots in either $\mathbb{R}^3$ or $S^3$, but some other results are stated only for knots in $S^3$.

Are there results that are correct for knots in $S^3$ but not for knots in $\mathbb{R}^3$?

Answer by Mark Grant for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T13:24:04Z

The higher homotopy groups of the complement of a knot in $S^3$ are all trivial. This is not true for a knot in $\mathbb{R}^3$ (consider a $2$-sphere enclosing the image of the knot).

Answer by Daniel Moskovich for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T13:36:23Z

This is a nice question!
Knot theory is in fact knot-complement theory, and a knot complement in S³ is a compact 3-manifold, while a knot complement in R³ is an open 3-manifold. Compact (or closed) 3-manifolds are technically easier to work with than open 3-manifolds, because various chain complexes which you care about are finitely generated, and various analytic tools become available.
Ultimately, I guess that the fundamental reason that people consider knots in S³ rather than in R³ is that bordism theory is tractable for compact manifolds, but not for open manifolds. In particular, bordism groups and Wall groups for compact manifolds are of finite rank (this matters for example for Blanchfield pairings). Analytically, characteristic numbers, topological and analytic index, analytic torsion and the eta invariant are only defined in the compact case, because for open manifolds the relevant integrals can diverge, elliptic operators need not be Fredholm, the spectrum of self-adjoint operators need not be discrete, etc.
Classical knot theory usually takes place in a more low-tech world than the previous paragraph might have suggested; nevertheless, all of the above technology is lurking in the background to algebraic topological invariants associated to covering spaces (the Alexander polynomial), in particular to questions related to knot concordance.
Because knots in R³ are fairly special objects, maybe classical knot theory can somehow be made to work despite the limitations listed above. But, in the absence of more general tools, I would have no idea how to do it. I think it would be interesting to have analogues to the classical algebraic invariants in knot theory for knots in R³!

Answer by Theo Johnson-Freyd for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T14:03:37Z

Recall that the space of knots $\mathcal K(M)$ in a 3-manifold $M$ is the space of embeddings $S^1 \hookrightarrow M$ (one must think a little to topologize this space usefully). Usual know theory is concerned with studying $\pi_0(\mathcal K(M))$ — its set of connected components — because already that's interesting. But from a topologist's point of view, one should be also interested in the higher homotopy groups. Then, although $\pi_0(\mathcal K(\mathbb R^3)) = \pi_0 (\mathcal K(S^3))$, because any classes in $\pi_0$ and $\pi_1$ can be arranged to avoid the point $S^3 \smallsetminus \mathbb R^3$, it's certainly not true that the higher homotopy groups are the same. (You already know this: that $\pi_3(S^3) > 0$ means that $\pi_2(\mathcal K(S^3)) > \pi_2(\mathcal K(\mathbb R^3))$.

It's not just the topologists who are interested in these higher homotopies <edit>, which are related to </edit> higher bordisms. Physicists, whose quantum field theory connects with knot theory in a number of ways, are also interested, in programs that go by names like "categorification" and "string theory". For example, not long ago Witten released his paper proposing a physical interpretation for Khovanov homology.

Actually, that paper interestingly illustrates my point above. Namely, Witten must carefully switch between $S^3$ and $\mathbb R^3$ a few times. $S^3$, being compact, is much better suited for defining certain integrals. On the other hand, $\mathbb R^3$ has its translation-invariant framing, and since many quantum field theoretical invariants are framing dependent, it matters whether you use the translation-invariant framing (which does not extend to $S^3$) or some other one.

Answer by Sam Nead for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T15:34:52Z

Here is a way to say it directly in the language of three-manifolds. By Alexander's theorem, both $R^3$ and $S^3$ are irreducible. However, as Mark indicates, a knot complement in $R^3$ is reducible. In fact, a knot complement in $R^3$ decomposes as the connect sum of a copy of $R^3$ and... the knot complement in $S^3, which is irreducible.

Answer by Bruno Martelli for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-27T19:48:00Z

One may freely pass from $S^3$ to $\mathbb R^3$ just by adding/removing a point; however $S^3$ is nicer when using 3-dimensional topology techniques, in particular geometrization.

For instance, the complement of a "generic" (in some sense) knot in $S^3$ should admit a complete hyperbolic metric of finite volume, in which case it is called a hyperbolic knot. Such a metric is unique by (a version of) Mostow rigidity theorem and is hence a rigid nice geometric object assigned to the knot. The topology of the knot complement determines the knot by Gordon-Luecke theorem, hence such a rigid object is actually all you need to determine the knot. The Epstein-Penner decomposition is a canonical combinatorial decomposition of the complement of a hyperbolic knot (into ideal polyhedra) which may be used to determine algorithmically whether two hyperbolic knots are isotopic or not.

To use all these very powerful geometric theorems and techniques you need to look at the knot complement in $S^3$, not in $\mathbb R^3$. A knot complement in $\mathbb R^3$ cannot have any complete hyperbolic metric of finite volume.

Answer by Dave Futer for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-04-28T21:02:53Z

In the same spirit as Bruno's answer (which focuses on hyperbolic geometry), let me mention a purely topological result. By the Lickorish-Wallace theorem, every closed, orientable 3-manifold can be obtained using Dehn surgery on a link in $S^3$. This result is central in the study of 3-manifolds, and simply doesn't hold true if one only considers knots in $R^3$.

Answer by sebastiano-peotta for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-05-07T10:06:58Z

A very down-to-earth but nice example (suggested by Kauffman himself during a seminar) of why working in $S^3$ can be useful. Let $T_{a,b}$, with $a,b$ coprime integers, be the the torus knot with $a$ horizontal windings and $b$ vertical windings. $T_{a,b}$ is equivalent to $T_{b,a}$ and this can be proved by constructing a homeomorphism $S^{3} \to S^{3}$ which sends one knot into the other. I'll leave the explicit construction for your entertainment.

Answer by Kevin Walker for In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?

2011-05-20T16:45:08Z

You asked about the advantages of links in $S^3$ over links in $\mathbb R^3$ (or $B^3$, which I prefer). Here's an advantage of $B^3$ over $S^3$: Khovanov homology is a (functorial) invariant of links in $B^3$, but so far as I know there is no known proof that it is an invariant of links in $S^3$.

More specifically, the standard literature on Khovanov homology constructs a functor from this category

Objects: links in $B^3$
Morphisms: Isotopy classes of isotopies in $B^3\times I$ (or more generally isotopy classes of cobordisms in $B^3\times I\;$).

to the category of bigraded chain complexes and homotopy classes of chain maps. The proof involves defining a chain map for each Reidemeister move and elementary cobordism, and then checking that these generators satisfy various "movie move" relations.

If you want to prove an analogous theorem for links in $S^3$, then there is an additional "global movie move" to check. It corresponds to the case where a 1-parameter family of isotopies in $S^3\times I$ (i.e. a second order isotopy) transversely intersects {south pole of $S^3$} $\times I$. I don't think anyone knows how to prove invariance under this global movie move. Perhaps it's not even true. (Scott Morrison and I announced a proof a couple of years ago, but we later discovered a sign issue. We're currently writing up a $\mathbb Z/2$ version with Chris Douglas. If anyone knows how to prove/disprove the relation with $\mathbb Z$ coefficients, please speak up!)

Morally, the reason for this difficulty is that there is no nice projection of links in $S^3$ to link diagrams in $S^2$. The obvious attempt at a map from $S^3$ to $S^2$ is not well-defined at the north and south pole of $S^3$, and this ambiguity actually matters for categorical/functorial link invariants.

In summary, if you are studying categorical link invariants (like Khovanov homology) in terms of planar projections, you are dealing in links in $B^3$, not $S^3$.