User agol - MathOverflow

Answer by Agol for When do two positive braids represent the same link?

2013-05-16T23:15:02Z

I'll make some comments on this question.

First of all, two positive braids which are conjugate (equivalently represented by the same labelled oriented 2-component link when taken with the braid axis) will be equivalent by a sequence of positive braid type III Reidemeister moves. This follows from Garside's solution to the conjugacy problem in the braid group (see also Theorem 9.4.2 by Thurston).

For closures of 3-strand braids, the complete classification was given by Birman-Menasco. They show that most 3-strand braids which represent the same link are conjugate, except for braids which are braid index 1 or 2, and a special class of braids $s_1^ps_2^qs_1^rs_2$, where $p,q,r \geq 2$ (restricting to the positive case) which are equivalent to $s_1^ps_2s_1^rs_2^q$. One sees directly from their classification that the braid index 1 and 2 cases have a unique positive representative up to conjugacy. For the exceptional case, they show (or refer to a result of Fein which is given on p. 100 of Birman's book) that these exceptional cases are conjugate after a single stabilization. In fact, one may check directly that the stabilization may be taken to be positive, so this answers your question in the case of pairs of positive 3-strand braids. Also, note that these two examples are related by flypes, and the equivalence by one positive stabilization holds for any pair of braids which are locally related by the same type of flype given in Figure 1.2 (but which are not necessarily conjugate).

If the answer to your question is true, then I think it might give an efficient algorithm to test if two positive braids are equivalent. For positive braids, the Seifert genus is determined by the braid index and number of crossings. When one does several positive Markov stabilizations, one sees that in the positive conjugacy class of the stabilized braid, most braid generators will occur only once. So I think one ought to be able to get a bound on the number of stabilizations needed, which would then lead to an algorithm to tell them apart. One could try to apply the techniques of Birman-Menasco (after Bennequin) to attempt to understand your conjecture.

Answer by Agol for Gap between first two nonzero Laplacian eigenvalues on closed compact surface?

2013-05-02T18:26:26Z

The question is a bit vague, but I'm assuming you're asking for an upper bound on the gap between the smallest non-zero eigenvalue and the second smallest non-zero eigenvalue (ignoring multiplicity), let's call this number $E$, taking the supremum over all closed hyperbolic surfaces. If $\lambda_j$ is the $j$th eigenvalue, counted with multiplicity, then Otal has shown that $\lambda_{2g-2}>\frac14$. Moreover, Buser has shown that this estimate is sharp: one can find for any $\epsilon >0$ a surface with $\lambda_{2g-3}<\epsilon$. Now, I believe that in these Buser examples, one could actually arrange that $\lambda_1=\lambda_2=\cdots=\lambda_{2g-3}$, by adjusting the moduli (recall moduli space has dimension $3g-3$). If so, this would show that $E\geq \frac14$.

On the other hand, a result of Besson implies the maximal multiplicity of $\lambda_1$ is $4g+3$, so we know that $E\leq \lambda_{4g+4}$. There is some universal bound on $\lambda_{4g+4}$, since by the Margulis Lemma, there exists $R>0$ so that on a Riemann surface of genus $g$, one can find $4g+5$ disjointly embedded disks of radius $R$ (there is such an $R$ for any linear function of $g$). By Corollary 4.65 of Gallot-Hulin-Lafontaine, one has $\lambda_{4g+4} \leq \lambda_1^D(D_{R})$, where $\lambda_1^D(D_R)$ is the first Dirichlet eigenvalue on a hyperbolic disk of radius $R$. Thus, $E\leq \lambda_1^D(D_R)$. One ought to be able to get explicit bounds on $R$ and $\lambda_1^D(D_R)$ using computations of Margulis constants to get an estimate of $E$.

It is conjectured by Colbois and Colin de Verdiere that the maximal multiplicity of $\lambda_1$ equals the chromatic number of a genus $g$ surface, which grows like $\sqrt{g}$. Given Huber's result that $\sup \lambda_1 \to \frac14$ as $g\to \infty$, this would imply that $E_g$ is asymptotic to $\frac14$, where $E_g$ is the maximal difference between these eigenvalues for genus $g$ hyperbolic surfaces.

Answer by Agol for Isotopy classes on the disk and mapping tori

2013-04-27T00:44:28Z

As I said in the comment above, knots in $D^2\times S^1$ which have a Dehn filling giving $D^2\times S^1$ were classified by Gabai and Berge. Gabai proved that knots in $D^2\times S^1$ giving back $D^2\times S^1$ are either cables or 1-bridge braids (note that the original statement was given before the knot complement problem, so Gabai states that the knot could also lie in a ball, but now this is known to not occur). I didn't check, but I believe the cable case will not work: even though there are many solid torus Dehn fillings, the link complement has an infinite image of the mapping class group in the mapping class group of the boundary, so I believe these all give equivalent cablings (although I didn't check this).

For 1-bridge braids in a solid torus with solid torus surgery, Gabai gave a partial classification, and Berge gave a complete classification. Gabai shows that the other 1-bridge braid will have the same winding number (Corollary 3.3), so the same number of strands. For some of these examples, Dehn filling gives back a 1-bridge braid in the solid torus of the same type. But Berge shows that most examples will give back a different braid.

To be explicit, I took the braid from Figure 8 of Gabai's paper, which is a braid on 10 strands. I input this into SnapPy, which shows that the braid is hyperbolic, and $(63,1)$ Dehn filling on cusp $0$ gives a manifold with fundamental group $\mathbb{Z}$, which therefore must be $D^2\times S^1$. Moreover, the symmetry group is $\mathbb{Z}/2^2$, and there is a $\mathbb{Z}/2$ subgroup which preserves the cusps, and acts as an elliptic involution, therefore preserving slopes. So the two 1-bridge braids must be inequivalent, but have the same complement.

Answer by Agol for Fibered knot with periodic homological monodromy

2013-04-23T20:57:29Z

My previous answer was flawed: the example I came up with was not unlinked.

The fibered knots with periodic monodromy are the torus knots. If one has a separating multicurve on the Seifert surface of a torus knot which is an unlink in $S^3$ and has zero framing, then Dehn twisting about it will yield fibered knots with periodic homological monodromy.

However, on a torus knot Seifert surface, such a multicurve cannot exist. The easiest case is for a separating curve on the Seifert surface. This cannot be unknotted, since this would give a lower 4-ball genus for the torus knot (in fact, the knot is quasi-positive).

If one has a separating multicurve which has framing zero, then for a non-separating component which is non-separating, one can surger the Seifert surface in the 4-ball to get a smaller genus surface, again a contradiction.

So this gives some positive evidence for your conjecture. One could attempt to analyze Giroux's classification of fibered links by Murasugi summing with Hopf bands, and see how the monodromy changes.

Answer by Agol for Are virtual cubulated groups cubulated?

2013-04-17T22:02:57Z

A paper of Caprace-Muhlherr (pdf) characterizes the Coxeter groups that act cocompactly on the CAT(0) cube complex associated to the group by Niblo-Reeves. It seems possible that this gives a characterization of the Coxeter groups which may act cocompactly on any CAT(0) cube complex, although I think this may still be open.

Answer by Agol for The existence of meromorphic functions on Riemann surfaces

2013-03-26T21:22:23Z

(Deleted incorrect suggestion).

I think one can use uniformization and the construction of automorphic functions on the universal cover to produce meromorphic functions. A google search for these terms found these notes.

Answer by Agol for Seifert genus of the lift of a knot in its cyclic branched covers.

2013-03-26T02:02:59Z

A non-trivial (and much more general) result of Gabai implies that $g_n(K)=g_1(K)$ for all $n$. This is encapsulated in the phrase ``Gromov norm equals Thurston norm". Roughly, the Gromov norm represents the minimal genus of an immersed Seifert surface, whereas the Thurston norm represents the minimal genus embedded Seifert surface. It follows that the minimal genus embedded Seifert surface realizes the minimal genus over all immersed Seifert surfaces.

Gabai's proof is interesting and enlightening. He constructs a taut foliation of the knot complement with a minimal genus Seifert surface as a leaf, making use of the theory of taut sutured manifolds that he develops in the papers. The euler class of this foliation gives a lower bound on the genus of immersed Seifert surfaces, and realized by the embedded Seifert surface.

Answer by Agol for Existence of different knots in $RP^3$ having the equivalent liftings in $S^3$

2013-03-13T18:51:44Z

If either knot is hyperbolic, this is not possible.

First, consider the non-null homologous case. Let $K_1,K_2\subset \mathbb{RP}^3$ be two knots, such that their preimages $K_1',K_2'\subset S^3$ are isotopic to the hyperbolic knot $K$ (Remark: if $K_1$ is hyperbolic, then so is $K$ and therefore $K_2$ by the geometrization theorem). The covering translation induces involutions $\iota_{1,2}:S^3 \to S^3$ such that $\iota_{i}(K)=K$. Restricting to the hyperbolic space $S^3\backslash K$, we get involutions $\iota_i:S^3\backslash K \to S^3\backslash K$, $i=1,2$, which are isotopic to (fixed-point free) hyperbolic isometries. The hyperbolic isometry restricts to the torus $T=\partial\mathcal{N}(K)$ as an isometry (e.g. taking a horotorus representative). Each isometry is determined by its action on $T$. There are 3 possible fixed-point free involutions of a torus. Let $\mu, \lambda\subset T$ be representatives of the meridian and longitude. The involutions rotate around $\mu$ (preserving $\mu$), or rotate around $\lambda$, or rotate around $\mu\lambda$. The rotation around $\mu$ is not possible, since this would extend to an involution of $S^3$ with fixed-point set $K$, which is impossible by the Smith conjecture. Thus, $\iota_1$ and $\iota_2$ must induce the other two involutions. But then $\iota_1\circ\iota_2$ must be the forbidden involution, a contradiction. So there is a unique involution preserving $S^3\backslash K$, and therefore at most one isotopy class of knot in $\mathbb{RP}^3$.

In the null homologous case, there are knots $K_1,K_2\subset \mathbb{RP}^3$ such that the preimage is isotopic to a two-component link $L$. There are two fixed-point free involutions of $S^3$ preserving $L$ and exchanging its components, and isotopic to isometries of the hyperbolic metric on $S^3\backslash L$ which exchanges the two cusps. Thus, they generate a dihedral group in the isometries of the hyperbolic knot complement, which must be finite order. Since both isometries preserve the meridians of $L$, this may be extended to a dihedral group action on $S^3$, generated by two fixed point free involutions. However, any such (smooth) action is conjugate to a group of isometries of $S^3$ by the orbifold theorem, which gives a contradiction, since the only fixed-point free involution is the antipodal map.

I think the general case should follow in a similar fashion, but one would have to consider how the involutions permute the JSJ decomposition.

Answer by Agol for Applications of n-dimensional crystallographic groups

2013-03-11T00:12:23Z

They occur as cusps cross-sections of non-uniform hyperbolic lattices of one higher dimension. For example, they are useful in the classification of minimal volume lattices.

Answer by Agol for How do you prove that every curve of constant width is convex?

2013-03-09T00:30:04Z

First, observe that for a closed convex region $R$ of constant width $w$, there can be no interval $[a,b]\subset \partial R$. For take the support line $l$ to $R$ containing $[a,b]\subset l$, and let $c$ be a point in the other parallel support line for $R$ realizing the width $w$. Then the height of the triangle $abc$ is $w = d(c,l)$. However, the distance $w<\max\{d(a,c),d(b,c)\}$ , and therefore the width of $R$ will be greater that the maximum of these two distances by choosing the support lines perpendicular to the longer side, contradicting constant width.

Now, suppose $K$ is a constant width region, and let $R$ be the convex hull of $K$. If $K \neq R$, then there exists an interval $[a,b]\subset \partial R-\partial K$, a contradiction.

Answer by Agol for Can you characterize the group of transformations of knot diagrams which preserve the knot embedding?

2013-03-07T20:02:31Z

If you consider knots made of piecewise linear segments, then one can say a couple of things. First, crossings will be between various segments, and for any pair of segments, one could compute the possible projections for any given direction. This will give a quadrilateral of projection directions where the segments cross. If you're interested in generic projections, then one could compute the non-generic phenomena. A Reidemeister 3 move occurs when there's a triple of segments which project to a triple crossing. The set of lines through 3 skew segments forms part of a hyperbolic paraboloid or hyperboloid. So the Gauss image will be part of a quadratic curve in $S^2$. There are some other possible non-generic phenomena, such as the planes containing adjacent segments, or vertices and segments (the boundaries of the quadrilaterals). In any case, once you've computed these, $S^2$ gets partitioned up into regions in which a generic projection occurs, and in principle one could compute each projection for each region. In fact, this is one way to prove that the Reidemeister moves suffice for PL equivalence of knots.

Answer by Agol for Hyperbolic groups with infinitely generated commutator subgroups

2013-03-06T18:12:10Z

No, consider a hyperbolic 3-manifold $M$ with $b_1(M)=2$, and all faces of the Thurston norm fibered. Then there are only finitely many surjections to $\mathbb{Z}$ with infinitely generated kernel, corresponding to the (projective classes of the) vertices of the Thurston norm ball.

For an explicit example, consider the Whitehead link, which has all fibered faces. This is not a hyperbolic group, but one may perform orbifold Dehn filling along the longitudes to get a closed hyperbolic orbifold with this property (since the linking number is zero).

"The" random tree

2009-11-18T19:21:48Z

One time I heard a talk about "the" random tree. This tree has one vertex for each natural number, and the edges are constructed probabilistically. Connect vertex 2 to vertex 1. Connect vertex 3 to vertex 1 or 2 with probability 1/2. Connect vertex n+1 to exactly one of vertices 1, ..., n with equal probability (1/n). This procedure will construct an infinite tree. The theorem is that with probability 1, any tree constructed this way will be the same (up to permutation of the vertices).

My question is, does anyone know of a reference for this result? What is the automorphism group of this tree? Can anyone draw a picture of it?

I don't have any reason for knowing about this, just curiosity, and I wasn't able to turn up anything with a (not too extensive) internet/mathscinet search.

Answer by Agol for A malnormal embedding theorem?

2013-02-28T20:25:56Z

This appears to be an open question: see Remark 5.23 of Sapir's paper, which attributes this as a question to Denis Osin.

Answer by Agol for fundamental groups of surfaces

2013-02-26T03:21:57Z

Poincar\'e duality groups of dimension 2 are surface groups.

Answer by Agol for Property of lattices in Lie groups

2013-02-24T07:16:35Z

This is false for uniform lattices in rank one semi simple Lie groups and large $n$ by a result of Ivanov and Olshanskii, which implies that the normal subgroup generated by $n$th powers is infinite index for certain large $n$.

Answer by Agol for Is a knotted trivalent graph determined by its set of unzips?

2013-02-23T16:34:57Z

A framed theta graph (ribbon surface) is really a thrice puncture sphere embedded in $S^3$ (pants). A stronger question is whether one may have two pants whose boundaries give the same 3-component link? In fact, suppose the interiors of the two pants are disjointly embedded in the link complement. The union is a genus 2 surface. Conversely, given a genus 2 surface, one may choose a pants decomposition of it to get such a pair. I claim that if one takes a genus 2 surface bounding an incompressible surface to one side (so the other side is compressible), a generic pants decomp will give inequivalent theta graphs. One may assume the link is hyperbolic without symmetries. If they were equivalent, there would be an isometry taking one pants to the other. This isometry would be finite order, giving a symmetry of the link complement, a contradiction. It remains to find such a surface and pants decomp, but I'm sure any sufficiently complicated asymmetric example will work.

Answer by Agol for Non-tame 3-manifolds covered by the Euclidean space

2013-02-22T23:44:22Z

I believe the original such examples (with fundamental group $\mathbb{Z}$) were due to Scott and Tucker.

Answer by Agol for Is a knotted trivalent graph determined by its set of unzips?

2013-02-21T21:31:29Z

As a special case, consider a (framed) $\theta$ graph, such that one zipping contains an unknotted component and another zipping is a split link, then the $\theta$ graph is isotopic to a planar one. This follows from a theorem of Scharlemann, which states that if a band sum of knots is trivial, then the knots form a trivial link.

Answer by Agol for Hyperbolic exceptional fillings of cusped hyperbolic 3-manifolds

2013-02-21T05:34:58Z

I'll embellish on Richard's answer. Experimentally, it seems that most one-cusped hyperbolic manifolds have at most 10 exceptional Dehn fillings in the sense you consider, i.e. points in the complement of Dehn surgery space. It is conjectured that Dehn surgery space is star-like, so that one can deform any hyperbolic metric with core geodesic by decreasing the cone angle monotonically to zero, and without affecting the singular structure. Lackenby and Meyerhoff have shown that there are at most 10 non-hyperbolic Dehn fillings, but they are not able to show that these are deformations of the complete metric in the sense of Thurston. However, their method of proof gives a bit more. The core of the Dehn filling is homotopically non-trivial, and the kernel of the Dehn filling map on $\pi_1$ is a free group (this is true if the core is geodesic, since the fundamental group of the complement of a collection of geodesics in hyperbolic space is free). I'm not sure if they prove this in the paper, but it follows from the proof the 6-theorem. The number of negatively curved fillings is also much smaller than 60 from the $2\pi$-theorem. There's hope that the cross-curvature flow could flow these negatively curved metrics to the hyperbolic metric. If one could do this in the cone-manifold context, then this might enable one to obtain a sharpening of Hodgson-Kerckhoff.

Answer by Agol for group of diffeomorphisms of a manifold

2013-02-08T17:42:50Z

One can approach the study of diffeomorphism groups from many perspectives: topology, geometry, differential equations, and dynamics. I'll mention a few results that I'm aware of, giving links to literature surveys on different topics.

There is a short exact sequence $$Diff_0(M)\to Diff(M)\to MCG(M),$$ where $Diff_0(M)$ is the subgroup of diffeomorphisms isotopic to the identity.

One can regard $MCG(M)=\pi_0(Diff(M))$. There is a huge literature studying $MCG(M)$, especially when $M$ is a surface. One question that has been answered for closed surfaces is that there is no section $Diff(M)\leftarrow MCG(M)$. I'm not sure what's known about the higher-dimensional version of this question.

Topologists study the homotopy type of $Diff(M)$, which breaks down into computing $MCG(M)$ and the homotopy type of $Diff_0(M)$. Hatcher has a survey on the homotopy type of $Diff(M)$. This has more-or-less been completely resolved in dimensions $\leq 3$, but is quite complex for general higher dimensional manifolds.

It is known that $Diff_0(M)$ is simple for closed manifolds by a result of Thurston. A general strategy then for understanding the group structure of $Diff_0(M)$ is to understand its subgroups. One aspect of this is the Zimmer program, to understand homomorphisms $\Lambda\to Diff_0(M)$, where $\Lambda$ is a higher rank lattice. Another aspect is to consider homomorphisms between diffeomorphism groups for different manifolds.

There are some results on dynamics of diffeomorphisms with relation to the diffeomorphism group. There is a huge literature on the dynamics of individual diffeomorphisms, but I think this is orthogonal to your question.

Answer by Agol for Largest hyperbolic disk embeddable in Euclidean 3-space?

2009-11-01T21:10:57Z

I didn't see the exact answer to your question in the Borisenko paper, since section 2.4 only seems to address immersions of subsets of ℍ² into ℝ³. However, a perturbation of the pseudosphere, Dini's surface, which is an isometrically embedded horodisk, seems to do the trick since it contains arbitrarily large disks in the hyperbolic plane. See Dini's Surface at the Geometry Center.

Answer by Agol for open problems in Seiberg-Witten Theory on 4-Manifolds

2013-02-05T05:00:53Z

It might be useful to generalize a theorem of Donaldson and Sullivan, that the Donaldson invariants are defined for quasi-conformal 4-manifolds, to the category of Seiberg-Witten invariants. More generally, one would like to know which smooth invariants of 4-manifolds are defined for quasi-conformal 4-manifolds.

Answer by Agol for Locally finite groups of finite rank and bounded exponent

2013-02-04T22:13:49Z

I'll expand on Derek Holt's comment, which answers your question. Suppose one has a group $G$ of the type you describe, so that finitely generated subgroups are generated by $r$ elements and have exponent $n$. Consider a finitely generated subgroup $K< G$. By the restricted Burnside problem, there is a universal constant $R(r,n)$ such that $|K|\leq R(r,n)$. Now, choose the largest size subgroup $K< G$ which is finitely generated. Since $K$ is finite and $G$ is infinite, there exists $g\in G-K$ such that $K < \langle K, g\rangle <G$ is finitely generated, so $\langle K, g\rangle$ must be finite. But since $|K|$ is maximal, we have $K=\langle K,g\rangle$, so $g\in K$, a contradiction.

Answer by Agol for When is a Baumslag-Solitar group linear?

2013-01-24T03:11:28Z

Presumably you've consulted the Wikipedia page on the Baumslag-Solitar group, which states that $BS(m,n)$ is not residually finite (and therefore not linear) if $|m|\neq |n|$ and $|m|>1, |n|>1$. This leaves the case that $|m|=|n|$. In this case, one can show that the group is the fundamental group of a compact Seifert-fibered 3-manifold, which is known to be linear.

Consider $BS(m,\pm m)$. Take a solid torus, and two parallel annuli in the boundary whose cores run $|m|$ times around the core of the solid torus. Attach an annulus $\times I$ to these two annuli, with opposite orientation for $BS(m,-m)$, to get a Seifert-fibered 3-manifold with fundamental group $BS(m,\pm m)$. For example, $BS(1,-1)$ is the fundamental group of the Klein bottle, which when thickened up is an interval bundle over a Klein bottle.

A Seifert 3-manifold with boundary has a finite-sheeted cover which is a product $S^1\times \Sigma^2$. The fundamental group is $\mathbb{Z}\times F$, where $F$ is a free group, and thus this group is linear. Take the induced representation to get a linear representation of the original 3-manifold group.

Answer by Agol for Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?

2013-01-23T03:43:49Z

Revision: For $n>6$, there is no embedding of $\mathcal{S}_n \hookrightarrow \mathcal{S}_{n+1}$ .

First, recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$ , where $Mod(S_{0,n})$ is the (orientation preserving) mapping class group of the $n$-punctured sphere.

Inside $Mod(S_{0,n})$, there is a subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$, which is a rotation of order $n-2$ of $S^2$, and fixes the north and south poles. The $n$ punctures include the north and south poles and one orbit of size $n-2$. The preimage of this group in $\mathcal{S}_n$ is isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ or to $\mathbb{Z}/2(n-2)\mathbb{Z}$ (of course if $n$ is odd, these are isomorphic). Then we get a corresponding subgroup of $\mathcal{S}_{n+1}$ from the injection. Project this group to $Mod(S_{0,n+1})$ via the map $\mathcal{S}_{n+1}\to Mod(S_{0,n+1})$ . The kernel of this projection will have size at most $2$. By the Nielsen Realization Theorem, any finite subgroup of $Mod(S_{0,n+1})$ must preserve a complete hyperbolic metric of finite area on $S_{0,n+1}$, and in particular by uniformization extends to a finite group of conformal automorphisms of $S^2$ which permutes $n+1$ marked points. The finite subgroups of $PSL_2(\mathbb{C})$ lie inside a conjugate of $SO(3)$, so the image is an abelian subgroup of $SO(3)$. The only abelian subgroups of $SO(3)$ are cyclic (or $\mathbb{Z}/2\mathbb{Z}^2$), so the image is either $\mathbb{Z}/(n-2)\mathbb{Z}$ or $\mathbb{Z}/2(n-2)\mathbb{Z}$ (in which case we may take an index 2 subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$; here we need $n>6$ to conclude that the image is not $\mathbb{Z}/2^2$). However, for $n>5$, there is no subgroup of $SO(3)$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ which permutes $n+1$ points, and therefore there is no such subgroup of $Mod(S_{0,n+1})$, a contradiction. To see this, note that a cyclic group of rotations of $S^2$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ has two fixed points, and every other orbit of size $n-2$. Thus, there must be some $k$ and $e$ such that there are $k$ orbits of size $n-2$, and $e$ orbits of size $1$, where $e\leq 2$. If $k\leq 1$, then we get $n+1=k(n-2)+e \leq n$, a contradiction. If $k\geq 2$, then $n+1=k(n-2)+e \geq 2(n-2)$, so $n\leq 5$, a contradiction.

Answer by Agol for Hyperbolic Brunnian links and rectangular cusp shapes

2013-01-19T01:35:57Z

Any two-component two-bridge link is Brunnian, but most of them do not have rectangular cusp shapes. For a description of the geometry of 2-bridge link complements, you may consult this appendix by David Futer.

Answer by Agol for Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements?

2013-01-09T23:48:04Z

The answer to your Question 1. is that there are not rank 2 subgroups of $PSL_2(\mathbb{Z})$, but there are rank 3 subgroups. This follows from a result of Wohlfahrt. He shows in Theorem 5 that any non-congruence subgroup $\Gamma < PSL_2(\mathbb{Z})$ has index $\geq 7$. Misha points out in a comment to his answer that a finite coarea fuchsian group of rank $N$ has $Area(\mathbb{H}^2/\Gamma)\leq 2\pi(N-1)$. Therefore in the case $\Gamma$ is non-congruence, $Area(\mathbb{H}^2/\Gamma)=[PSL_2(\mathbb{Z}):\Gamma] Area(\mathbb{H}^2/PSL_2(\mathbb{Z})) \geq 7\pi/3$. Thus, $N=rank(\Gamma) \geq 3$. Misha's claim (in the non-uniform case) follows from an application of the free product decomposition of non-uniform Fuchsian group into cyclic groups, Grushko's theorem, and Gauss-Bonnet.

Wohlfahrt also gives an example of a non-congruence subgroup of index $7$ in $PSL_2(\mathbb{Z})$. On p. 531, he points out that there exists an index $7$ subgroup with two cusps. One determines that the group has genus $0$, and two cone points of orders $2$ and $3$ respectively. This group is therefore rank $3$, isomorphic to $\mathbb{Z}\ast \mathbb{Z}/2\ast\mathbb{Z}/3$. I didn't check, but I think this also lifts to a rank 3 subgroup of $SL_2(\mathbb{Z})$.

Answer by Agol for A Problem about spherical transformation (circle mapping)

2012-12-31T02:01:30Z

Take two circles $C_1, C_2$ intersecting in two points $p_1, p_2$. These are contained in a unique 2-sphere $S$. One way to see this is to assume $p_1=\infty$, $S^n-p_1=\mathbb{R}^n$ (considering $S^n$ as the one-point compactification of $\mathbb{R}^n$). Then the circles $C_1, C_2$ correspond to two lines intersecting in a point $p_2$, which span a unique 2-plane in $\mathbb{R}^n$ corresponding to the sphere $S$.

The images $f(C_i)\subset D_i$, where $D_i$ are circles, and $D_1\cap D_2 = \{f(p_1),f(p_2)\}$ . Let $T$ be the unique 2-sphere containing $D_1\cup D_2$. Then I claim $f(S)\subset T$. Normalize it by post-composition with a Moebius transformation so that $f(p_1)=\infty$, then any line intersecting $(C_1\cup C_2)-\infty$ in two points distinct from $p_2$ will map to a line intersecting the lines $D_1\cup D_2-\infty$ in two points, and thus is contained in $T$. So $f(S)\subset T$ since every point of $S$ is contained in a line intersecting $D_1, D_2$ in points distinct from $f(p_2)$.

One can perform a similar inductive argument to show that higher-dimensional spheres are preserved. Take $k$ circles intersecting two points, say $0, \infty$, which span a $k$-plane $K$ in $\mathbb{R}^n$. Then any point on $K$ is contained in a $k-1$-plane $J$ intersecting the $k$ lines in points different from the origin. Then the image is contained in a similar configuration by induction, so the $k$-spheres are taken to $k$-spheres.

Answer by Agol for Explicit homeomorphism between Thurston's compactification of Teichmuller space and the closed disc

2012-12-27T17:50:36Z

One natural attempt to compactify Teichmuller space is by the visual sphere of the Teichuller metric. However, Anna Lenzhen showed that there are Teichmuller geodesics which do not limit to $PMF$ (in fact, I think it was known before by Kerckhoff that the visual compactification is not Thurston's compactification).

However, it was shown by Cormac Walsh that if one takes Thurston's Lipschitz (asymmetric) metric on Teichmuller space, and take the horofunction compactification of this metric, one gets Thurston's compactification of Teichmuller space. In fact, he shows in Corollary 1.1 that every geodesic in the Lipschitz metric converges in the forward direction to a point in Thurston's boundary. I think this gives a new proof that Thurston's compactification gives a ball.

As Misha points out, it's not clear that the horofunction compactification is a ball.

Another approach was given by Mike Wolf, who gave a compactification in terms of harmonic maps, and showed that this is equivalent to Thurston's compactification (Theorem 4.1 of the paper). Wolf shows that given a Riemann surface $\sigma \in \mathcal{T}_g$, there is a unique harmonic map to any other Riemann surface $\rho \in \mathcal{T}_g$ which has an associated quadratic differential $\Phi(\sigma,\rho) dz^2 \in QD(\sigma)$ ($QD(\sigma)$ is naturally a linear space homeomorphic to $\mathbb{R}^{6g-6}$). Wolf shows that this is a continuous bijection between $\mathcal{T}_g$ and $QD(\sigma)$, and shows that the compactification of $QD(\sigma)$ by rays is homeomorphic to Thurston's compactification $\overline{\mathcal{T}_g}$ in Theorem 4.1. I skimmed through the proof, and as far as I can tell the proof of the homeomorphism does not appeal to the fact that Thurston's compactification is a ball, so I think this might give another proof that it is a ball.

Comment by Agol

2013-05-17T19:02:14Z

There may be some insight gained from the transfer map from the homology of the Sylow subgroups. Maybe there's some general structure theory for Sylow subgroups of simple groups that shows that they have small Schur multiplier at each prime.

Comment by Agol

2013-05-15T14:49:26Z

Here's a paper addressing finite quotients of the $(2,3,7)$ triangle group, but the techniques should apply to any hyperbolic triangle group. link.springer.com/article/10.1007%2FBF02784148

Comment by Agol

2013-05-14T15:55:34Z

What you are seeking could be regarded as an analogue of the Tait flyping conjecture for positive braids.

Comment by Agol

2013-05-04T17:44:09Z

Isn't there a bijection between transitive actions on discrete sets and conjugacy classes of subgroups, given by point stabilizers and coset action respectively? I assume there are infinitely many conjugacy classes of subgroups, which would seem to indicate that the actions are inequivalent.

Comment by Agol

2013-05-02T19:15:04Z

There can be no universal lower bound, since the multiplicity of $\lambda_1$ can vary over moduli space, and thus there are surfaces with the difference arbitrarily close to $0$. That is why in my answer I interpret your question as asking for an upper bound, which I show exists for hyperbolic surfaces. The reason to restrict to the hyperbolic case is that one cannot probably not say much for general Riemannian metrics on surfaces.

Comment by Agol

2013-05-02T05:27:54Z

The smallest non-zero eigenvalue can have multiplicity $>1$, so in this case is the gap zero? Or are you counting without multiplicity?

Comment by Agol

2013-04-26T20:43:16Z

@ Lee: agreed, but the number of fiberings in a given face of Thurston norm ball which are planar surfaces of the same number of punctures is finite (if the manifold is hyperbolic), and there's a chance that all of these fiberings are related by isometries of the manifold, so give the same braid. In fact, this happens for the unique hyperbolic 1-bridge braid in $D^2\times S^1$ which has two Dehn fillings which are $D^2\times S^1$, since the symmetry group permutes the 3 fillings. But I think Berge shows which pairs of 1-bridge braids have the same complement, if I interpret his paper right.

Comment by Agol

2013-04-26T05:26:50Z

I believe this is answered by results of Gabai and Berge. Namely, there are 1-bridge braids in $D^2\times S^1$ which have surgeries which are 1-bridge braids. Some of these are equivalent braids, but I think Berge has shown that most are not. ams.org/mathscinet-getitem?mr=1093862, ams.org/mathscinet-getitem?mr=1082933

Comment by Agol

2013-04-24T22:44:14Z

Incidentally, grid position (or arc index) behaves better with respect to connect sums, since the arc index is additive ($-1$). But the proof of this (by Dynnikov) uses 3-dimensions.

Comment by Agol

2013-04-24T10:38:40Z

This will be true for surfaces (n=2).

Comment by Agol

2013-04-24T00:10:30Z

The old answer claimed to have found such a link on the Seifert surface of a torus knot, which is clearly impossible by the new answer. I think you can look at old versions if you like.

Comment by Agol

2013-04-22T20:39:48Z

this is probably more appropriate for math.stackexchange.com

Comment by Agol

2013-04-19T16:59:22Z

I guess I mean that the Hruska-Wise result is not new (2012) for hyperbolic groups, but is probably new in the relatively hyperbolic setting.

Comment by Agol

2013-04-19T16:56:47Z

I think you can just use the hyperplane stabilizers of the cubulation of $N$ to cubulate $G$ a la Bergeron-Wise.

Comment by Agol

2013-04-17T21:04:18Z

$S^1\times S^2$ works too.