MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}^3$?

share|cite|improve this question
I suspect one of the main motivations is pedagogical rather than theoretical. It's less of a stretch of the imagination to visualize knots in $\mathbb R^3$. Visualization in $S^3$, especially when it comes to symmetry, takes more time to digest. – Ryan Budney Apr 27 '11 at 17:28
Another question might be, why $S^3$ or $\mathbb{R}^3$? Why not some other 3-manifold? – Kevin H. Lin May 1 '11 at 6:53
@Kevin: $S^3$ is universal -- given an embedding in $S^3$, and given a manifold $M$, the connect sum of the embedding in $S^3$ with $M$ is an embedding in $M$. – Ryan Budney May 7 '11 at 18:56
@Kevin: For example, some authors have studied knots in the real projective space $\mathbb{R}P^3$: – hsp May 8 '11 at 12:01

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.

share|cite|improve this answer
Regarding your 1st paragraph, the homotopy and homology groups of both spaces are quite closely related -- the homotopy-types of the two spaces are related via a pair of fibre bundles which are readily understood. So in that sense there's not much difference. – Ryan Budney Apr 27 '11 at 17:19
I'm confused about the phrase "higher homotopies = higher bordisms". There is certainly a relationship between the two concepts, but I'm confused about why you wrote an "=" sign. – Daniel Moskovich Apr 27 '11 at 17:30
To be more precise, let $Emb(S^1,S^3)$ be the space of smooth embeddings of the circle in $S^3$, $K_{3,1}$ the space of long smooth embeddings of $\mathbb R$ in $\mathbb R^3$ and $Emb(S^1,\mathbb R^3)$ the space of embeddings of the circle in euclidean space. Then there are homotopy-equivalence $Emb(S^1,S^3) \simeq SO_4 \times_{SO_2} K_{3,1}$ and $Emb(S^1,\mathbb R^3) \simeq SO_4 \times_{SO_2} (C \rtimes K_{3,1})$. $C \rtimes K_{3,1}$ is the space of pairs consisting of a point in $K_{3,1}$ together with a point in its complement. – Ryan Budney Apr 27 '11 at 17:43
That might be a bit unclear $C \rtimes K_{3,1}$ is the set of pairs $(p,f)$ where $f \in K_{3,1}$ and $p \in \mathbb R^3 \setminus img(f)$. So it's the tautological bundle over $K_{3,1}$ with fibre the corresponding knot complement. – Ryan Budney Apr 27 '11 at 17:51
@Daniel: sorry I ment something much less precise than I guess I wrote. I only meant that these are related. @Ryan: Cool! – Theo Johnson-Freyd Apr 27 '11 at 18:05

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).

share|cite|improve this answer

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!

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
On the converse, given a knot in $\mathbb R^3$ if you apply geometrization to the complement the first step is to split off a ball summand. This gives you the corresponding knot complement in $S^3$ connect sum a ball, and then you're in the case you describe above. – Ryan Budney Apr 27 '11 at 19:54

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 $S^3$.

share|cite|improve this answer

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 $R^3$.

share|cite|improve this answer

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.

share|cite|improve this answer
There is also a homeomorphism of $R^3$ sending $T_{a,b}$ to $T_{b,a}$. – Sam Nead May 20 '11 at 21:46
I would definitely like to know it. It is not obvious to me... – user14548 May 26 '11 at 8:14
@sebastiano: Since all points of $S^3$ are equivalent, one can construct the desired homeomorphism $R^3 \to R^3$ by simply restricting the homeomorphism $S^3 \to S^3$ to the complement of a point. That said, your example can be salvaged as follows. There does not exist a homeomorphism $R^3 \to R^3$ which maps the torus containing $T_{a,b}$ to the torus containing $T_{b,a}$, and interchanges the knots. (Proof: cutting $R^3$ along the torus produces a compact and a non-compact piece.) – Dave Futer May 26 '11 at 19:58

I think it is much more convenient to work in $S^3$ when you want to compute the fundamental group of the complement of a toric knot , than in $R^3.$ In $S^3$ you use the torus containing the knot to divide $S^3$ into two $S^1 \times D^2$, and then apply Van Kampen theorem. In $R^3$ the corresponding divison of the complement of the knot is less natural, it is harder to describe and visualize. And ofcourse the fundamental group is the same if you take the knot-complement in $R^3$ or in $S^3.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.