18
$\begingroup$

What can be said about a knot $K\subseteq S^3$ for which there exists a (Euclidean) polyhedral metric (aka Euclidean cone-manifold structure) on $S^3$ whose singular locus is precisely $K$? I'm particularly interested in the case where the conical angle $\alpha$ around $K$ is less than $2\pi$.

The classic example of this is the figure-8 knot. If you quotient $\mathbb E^3$ by the crystallographic group $\mathrm P2_13$, you get a Euclidean orbifold structure on $S^3$ whose singular locus is the figure-8 knot with angle $2\pi/3$. Dunbar (pp.82–86) gives a classification of all Euclidean orbifold structures on $S^3$, and this is the only one whose singular locus is a knot. Therefore, we know that if $K$ is not the figure-8 knot, then $\alpha$ cannot equal $2\pi/n$. That's a starting point, at least!

Dunbar, William D., Geometric orbifolds, Rev. Mat. Univ. Complutense Madr. 1, No. 1-3, 67-99 (1988). ZBL0655.57008.


Edit. Mednykh and Rasskazov construct a very simple family of fundamental polyhedra $\mathcal P(\theta)$ for the polyhedral manifolds $\mathcal C(\theta)$ ($S^3$ with conical angle $\theta$ along the figure-8 knot), for $\theta\in[0,4\pi/3)$. They show that, as $\theta$ increases, $\mathcal P(\theta)$ goes from hyperbolic to spherical, passing through Euclidean at $\theta=2\pi/3$. The reason I mention this is that they note (p.446):

...that [their] approach is general and can be applied to other two-bridge links and knots.

And again, they (p.448):

...describe an algorithm for the construction of the fundamental set $\mathcal P$ for the figure-eight orbifold which effectively works in every space of constant sectional curvature. With slight modifications this algorithm can also be used to construct the fundamental set of any 2-bridge link orbifold.

Mednykh, Alexander; Rasskazov, Alexey, Volumes and degeneration of cone-structures on the figure-eight knot, Tokyo J. Math. 29, No. 2, 445-464 (2006). ZBL1124.57008.


Final Edit. According to a recent paper of Mednykh, the two-bridge knots $4_1,5_2,6_1,6_2,6_3,7_2,7_3,7_4,7_5,7_6$ and $7_7$ all admit polyhedral metrics with a given conical angle $\alpha_0<\pi$. With the exception of $7_5,7_6$ and $7_7$, the value of $\alpha_0$ is given implicitly, but exactly.

Do all two-bridge knots have this property?

$\endgroup$
5
  • 3
    $\begingroup$ The following paper looks relevant to the final question: mat.uab.cat/~porti/twobridge040127.pdf Porti claims that the answer is known to be positive for hyperbolic two-bridge knots (Theorem 1.1). But it doesn't look like the answer is positive for torus knots, at least judging by what is written in Propositions 2.1 and 2.2. Porti says that in this case there is a Nil structure on the boundary of spherical ones. So I guess there is no Euclidean one $\endgroup$ – Dmitri Panov Apr 28 at 22:53
  • $\begingroup$ By googling, I found the following paper of Souto, that might be relevant to the original question: arxiv.org/pdf/math/0401003.pdf He says that on any(!) compact 3-manifold (including $S^3$) there is a hyperbolic cone metric with angles in $(2\pi-\varepsilon, 2\pi)$ with singularities along a link. I haven't looked into details but doubt a bit that the link can be connected (i.e. a knote). One might optimistically guess that some of such structures can be deformed to polyhedral ones by increasing the angle slightly to get a flat metric singular along a link. $\endgroup$ – Dmitri Panov Apr 29 at 9:39
  • $\begingroup$ Is it possible to visualize the $P2_13$ action? say, what is its fundamental domain? $\endgroup$ – Anton Petrunin May 6 at 18:35
  • 1
    $\begingroup$ One way to visualise it is to see that it's generated by two rotations of order 3 about axes $[(0,0,0),(1,1,1)]$ and $[(-1,0,0),(0,-1,1)]$. It's pretty tricky to write down a nice fundamental domain, but one is given in the paper of Mednykh and Rasskazov I cited above (see p.448, fig.2). $\endgroup$ – Tom Sharpe May 7 at 10:07
  • $\begingroup$ It helped, and I see now a simpler domain: the union of two square pyramids with a common base; each pyramid is a third of the unit cube. After identifying two pairs of triangular faces by order-3 rotations you left with a top. ball bounded by two squares; it remains to glue one square to the other. But still, I do not see why the singular locus becomes 8 after this gluing. $\endgroup$ – Anton Petrunin May 7 at 19:06
9
$\begingroup$

Here is my original answer. It is much too optimistic. See below.


A few experiments with Snappy very, very strongly suggest that the $5_2$ knot is the locus of a euclidean cone manifold structure on $S^3$ with cone angle $10\pi / 13$.

They also very strongly suggest that the $(-2, 3, 7)$ pretzel knot either occurs as a locus with angle $\pi$ or something very strange happens as the cone deformation from infinity ascends to $\pi$. Perhaps the knot "bumps" into itself just as the angle gets to $\pi$?

So, in an abundance of optimism I am going to guess that "most" hyperbolic knots have such a structure, with cone angle strictly less than $\pi$. This is motivated by the cone deformation machinery of Hodgson and Kerckhoff.

To be a bit more concrete, I'll conjecture that hyperbolic alternating knots with all twist regions of size one do this. These hypotheses are aimed at the $(-2, 3, 7)$ pretzel which is both non-alternating and which has has big twist regions in many (all?) "efficient" projections.

Edit: More experiments suggest that the $6_2$ knot has a cone manifold structure with angle a shade over $0.8 \pi$.


After Dmitri's comment, I worked a bit harder... I looked at the first one hundred "census knots". 19 of them are two-bridge knots (four occur among the first five census knots, explaining my results above...). 66 of them have their branched double cover being hyperbolic.

So, for the 19 two-bridge knots there is a definite chance of having the desired euclidean cone-manifold structure. For the other 66 there is no chance (at least, with angle at most $2\pi$).

That leaves 15 knots out of the one hundred. I am going to guess that they are three-bridge, each with their branched double cover being a Seifert fibered space. I don't have much evidence for this but:

  • For each, the branched double cover has fundamental groups of rank two (so the Heegaard genus is probably two).
  • For each, the transition (from hyperbolic cone-manifold to strangeness) happens at cone angle $\pi$.

As a final comment - Dmitri listed geometric three-manifolds with metrics of non-negative curvature (spherical space forms, $S^2 \times S^1$, Euclidean space forms). Most of these have a Heegaard splitting of genus one or two. So they have a (canonical!) "hyperelliptic" involution. The quotient will be the three-sphere, and the locus of fixed points will descend to give a three-bridge knot or link.

So perhaps (???) a guess at an answer to the original question is: "two-bridge knots generally, and some very very special three-bridge knots".


Actual final comment: After quickly skimming various bits of Hodgson's thesis "Degeneration and regeneration of hyperbolic structures on three-manifolds" and the introduction to Porti's paper "Regenerating hyperbolic and spherical cone structures from Euclidean ones" I tried the following experiments.

In Snappy I started to use "M.high_precision()" to compute volumes using quad doubles. Using finite differences, experiments suggest that

  1. the volume degeneration for the figure eight knot (euclidean limit) is quadratic and
  2. the volume degeneration for the (-2, 3, 7) pretzel (PSL limit) is linear.
  3. out of the 15 "interesting" census knots above, 12 have linear volume degeneration. The knots $9_{42}$, $8_{20}$, $10_{125}$ do something ... else.

This is getting too delicate for my simple mind. So I am going to go do something else. Good luck!

$\endgroup$
10
  • 1
    $\begingroup$ Thanks for this, Sam. What 'experiments' do you mean, exactly? Also, I'm afraid I don't know what you mean by 'twist regions'. $\endgroup$ – Tom Sharpe Apr 27 at 9:53
  • 5
    $\begingroup$ Sam, for each knot you can take a double cover. If what you say were true you would get a Euclidean metric with cone angle $<2\pi$ along the knot on the cover. Then the cover is non-negatively curved in the sense of Alexandrov. But there are very few non-negatively curved 3-manifolds ($S^3/\Gamma, S^2\times S, T^3...$). Certainly, for majority of hyperbolic knots the double covers are not manifolds of this type. So if there is a flat metric, it is not with angle $<\pi$. Maybe on the contrary this can be used to prove that such metrics are rare? $\endgroup$ – Dmitri Panov Apr 27 at 22:36
  • 1
    $\begingroup$ Ah - good point! Let's see - the branched double cover of a two-bridge knot is a lens space. (The bridge sphere is covered by the Heegaard torus.) This then "explains" the "good" examples above. The branched double cover of the $(-2, 3, 7)$ pretzel is a Seifert fibered space with PSL geometry; this is not non-negatively curved. $\endgroup$ – Sam Nead Apr 28 at 8:43
  • 2
    $\begingroup$ Tom - Yes. But it is not a very good test as the transitional geometry at volume zero could be some other Seifert fibered geometry. As Dmitri points out, this technique is actually more useful to rule out examples rather than rule them in. But (?) see my "actual final edit" - I think there is further information hiding in the higher derivatives of the volume degeneration function. $\endgroup$ – Sam Nead Apr 28 at 11:17
  • 2
    $\begingroup$ Sam - just for your interest, your approximations for $5_2$ and $6_2$ were very close! See the paper of Mednykh in the final edit to my question. $\endgroup$ – Tom Sharpe Apr 28 at 16:21
9
$\begingroup$

The ones that have a Euclidean cone metric with cone angle $\leq \pi$ should only be the spherical Montesinos knots (including 2-bridge knots, excluding the (2,n) torus knots). This should follow from the proof of the orbifold theorem, but one would need to search through the proof to sort out the logic. As indicated in the comments, Porti hinted at this in his paper on 2-bridge links. If there’s a spherical metric, then the $\pi$ orbifold with knot as the singular locus is spherical, and the representation into $SO(4)$ can be deformed to decrease the cone angle until it collapses to a Euclidean cone metric. Actually, the proof proceeds in the other direction: a hyperbolic metric is deformed to increase the cone angle until it collapses to a (rescaled) Euclidean cone metric. Conversely, if there’s a metric with cone angle $\leq\pi$, then the double branched cover admits an metric of non-negative Alexandrov curvature, and therefore must be modeled on $S^3$, $S^2\times R$, or be Euclidean. One can rule out the last two cases in the case of knots (Dunbar classified the knots whose $\pi$ orbifold admits exceptional geometries).

For the general case with cone angle $> \pi$, I think this is wide open. There are some restrictions though. The case of cone angle $\geq 2\pi$ is impossible, since this corresponds to a CAT(0) metric on the 3-sphere which doesn't exist.

Similarly, if the cone angle is $> \pi$, the double branched cover will admit a CAT(0) metric. This is only possible if it is hyperbolic. If there’s an essential torus that meets the branch locus, then it can be isotoped to be a flat plane invariant under the involution, and there’s an essential Conway sphere for the knot, and the cone angle would be $= \pi$, a contradiction. An essential sphere cannot exist, since there is a CAT(0) metric so $\pi_2$ vanishes. If the torus does not meet the singular locus, then the knot was a satellite. But then the torus may be isotoped to be totally geodesic (again by the flat plane theorem). The pattern knot would then admit a flat metric with totally geodesic torus boundary. But the only such manifold is the twisted I-bundle over a Klein bottle, which is not a knot complement. So one should restrict to knots whose double cover is either spherical (and the cone angle is $\leq \pi$) or hyperbolic (and the cone angle is betweeen $\pi$ and $2\pi$). The knot will be prime and not satellite and have no essential Conway sphere. Most arborescent knots, such as the (-2,3,7) pretzel considered in Sam Nead’s answer, will be ruled out.

A possible obstruction would come from the representation of $\pi_1(S^3-K)$ to $Isom^+(\mathbb{E}^3)\cong \mathbb{R}^3 \rtimes SO(3)$. If the action of this group has no global fixed point (so is not conjugate into $SO(3)$), then the projection of this representation into $SO(3)$ gives a representation with non-trivial cohomology with coefficients in $\mathbb{R}^3$ twisted by the representation. Any knot $K$ will have many $SO(3)$ representations of $\pi_1(S^3-K)$, but one could try to look for ones for which this cohomology always vanishes. For a Euclidean cone structure, I would suspect that this representation does not have a global fixed point, since otherwise the developing map would have precompact image, which shouldn't happen for a Euclidean cone structure (a geodesic in a generic direction should miss the cone locus by general position and hence have non-compact developing map).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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