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April 7, 2014

The Modular Flow on the Space of Lattices

Posted by Simon Willerton

Guest post by Bruce Bartlett

The following is the greatest math talk I’ve ever watched!

  • Etienne Ghys (with pictures and videos by Jos Leys), Knots and Dynamics, ICM Madrid 2006. [See below the fold for some links.]

Etienne Ghys A modular knot

I wasn’t actually at the ICM; I watched the online version a few years ago, and the story has haunted me ever since. Simon and I have been playing around with some of this stuff, so let me share some of my enthusiasm for it!

The story I want to tell here is how, via modular flow of lattices in the plane, certain matrices in SL(2,)\SL(2,\mathbb{Z}) give rise to knots in the 3-sphere less a trefoil knot. Despite possibly sounding quite scary, this can be easily explained in an elementary yet elegant fashion.

As promised above, here are some links related to Ghys’ ICM talk.

I’m going to focus on the last third of the talk — the modular flow on the space of lattices. That’s what produced the beautiful picture above (credit for this and other similar pics below goes to Jos Leys; the animation is Simon’s.)

Lattices in the plane

For us, a lattice is a discrete subgroup of \mathbb{C}. There are three types: the zero lattice, the degenerate lattices, and the nondegenerate lattices:

Lattices

Given a lattice LL and an integer n4n \geq 4 we can calculate a number — the Eisenstein series of the lattice: G n(L)= ωL,ω01ω n. G_{n}(L) = \sum _{\omega \in L, \omega \neq 0} \frac{1}{\omega ^{n}}. We need n3n \geq 3 for this sum to converge. For, roughly speaking, we can rearrange it as a sum over rr of the lattice points on the boundary of a square of radius rr. The number of lattice points on this boundary scales with rr, so we end up computing something like r0rr n\sum _{r \geq 0} \frac{r}{r^{n}} and so we need n3n \geq 3 to make the sum converge.

Note that G n(L)G_{n}(L) = 0 for nn odd since every term ω\omega is cancelled by the opposite term ω-\omega . So, the first two nontrivial Eisenstein series are G 4G_{4} and G 6G_{6}. We can use them to put `Eisenstein coordinates’ on the space of lattices.

Theorem: The map {lattices} 2 L (G 4(L),G 6(L)) \begin{aligned} \{ \text{lattices} \} &\rightarrow \mathbb{C}^{2} \\ L & \mapsto (G_{4} (L), \, G_{6}(L)) \end{aligned} is a bijection.

The nicest proof is in Serre’s A Course in Arithmetic, p. 89. It is a beautiful application of the Cauchy residue theorem, using the fact that G 4G_{4} and G 6G_{6} define modular forms on the upper half plane HH. (Usually, number theorists set up their lattices so that they have basis vectors 11 and τ\tau where τH\tau \in H. But I want to avoid this ‘upper half plane’ picture as far as possible, since it breaks symmetry and mystifies the geometry. The whole point of the Ghys picture is that not breaking the symmetry reveals a beautiful hidden geometry! Of course, sometimes you need the ‘upper half plane’ picture, like in the proof of the above result.)

Lemma: The degenerate lattices are the ones satisfying 20G 4 349G 6 2=020 G_{4}^{3} - 49G_{6}^{2} = 0.

Let’s prove one direction of this lemma — that the degenerate lattices do indeed satisfy this equation. To see this, we need to perform a computation. Let’s calculate G 4G_{4} and G 6G_{6} of the lattice \mathbb{Z} \subset \mathbb{C}. Well, G 4()= n01n 4=2ζ(4)=2π 490 G_{4}(\mathbb{Z}) = \sum _{n \neq 0} \frac{1}{n^{4}} = 2 \zeta (4) = 2 \frac{\pi ^{4}}{90} where we have cheated and looked up the answer on Wikipedia! Similarly, G 6()=2π 6945G_{6}(\mathbb{Z}) = 2 \frac{\pi ^{6}}{945}.

So we see that 20G 4() 349G 6() 2=020 G_{4}(\mathbb{Z})^{3} - 49 G_{6}(\mathbb{Z})^{2} = 0. Now, every degenerate lattice is of the form tt \mathbb{Z} where tt \in \mathbb{C}. Also, if we transform the lattice via LtLL \mapsto t L, then G 4t 4G 4G_{4} \mapsto t^{-4} G_{4} and G 6t 6G 6G_{6} \mapsto t^{-6} G_{6}. So the equation remains true for all the degenerate lattices, and we are done.

Corollary: The space of nondegenerate lattices in the plane of unit area is homeomorphic to the complement of the trefoil in S 3S^{3}.

The point is that given a lattice LL of unit area, we can scale it LλLL \mapsto \lambda L, λ +\lambda \in \mathbb{R}^{+} until (G 4(L),G 6(L))(G_{4}(L), G_{6}(L)) lies on the 3-sphere S 3={(z,w):|z| 2+|w| 2=1} 2S^{3} = \{ (z,w) : |z|^{2} + |w|^{2} = 1\} \subset \mathbb{C}^{2}. And the equation 20z 349w 2=020 z^{3} - 49 w^{2} = 0 intersected with S 3S^{3} cuts out a trefoil knot… because it is “something cubed plus something squared equals zero”. And the lemma above says that the nondegenerate lattices are precisely the ones which do not satisfy this equation, i.e. they represent the complement of this trefoil.

Since we have not divided out by rotations, but only by scaling, we have arrived at a 3-dimensional picture which is very different to the 2-dimensional moduli space (upper half-plane divided by SL(2,)\SL(2,\mathbb{Z})) picture familiar to a number theorist.

The modular flow

There is an intriguing flow on the space of lattices of unit area, called the modular flow. Think of LL as sitting in 2\mathbb{R}^{2}, and then act on 2\mathbb{R}^{2} via the transformation (e t 0 0 e t), \left ( \begin{array}{cc} e^{t} & 0 \\ 0 & e^{-t} \end{array} \right ), dragging the lattice LL along for the ride. (This isn’t just some formula we pulled out the blue — geometrically this is the ‘geodesic flow on the unit tangent bundle of the modular orbifold’.)

We are looking for periodic orbits of this flow.

“Impossible!” you say. “The points of the lattice go off to infinity!” Indeed they do… but disregard the individual points. The lattice itself can ‘click’ back into its original position:

animation

How are we to find such periodic orbits? Start with an integer matrix A=(a b c d)SL(2,) A = \left ( \begin{array}{cc} a & b \\ c & d \end{array}\right ) \in \SL(2, \mathbb{Z}) and assume AA is hyperbolic, which simply means |a+d|2|a + d| \geq 2. Under these conditions, we can diagonalize AA over the reals, so we can find a real matrix PP such that PAP 1=±(e t 0 0 e t) P A P^{-1} = \pm \left ( \begin{array}{cc} e^{t} & 0 \\ 0 & e^{-t} \end{array} \right ) for some tt \in \mathbb{R}. Now set LP( 2)L \coloneqq P(\mathbb{Z}^{2}). We claim that LL is a periodic orbit of period tt. Indeed: L t =(e t 0 0 e t)P( 2) =±PA( 2) =±P( 2) =L. \begin{aligned} L_{t} &= \left ( \begin{array}{cc} e^{t} & 0 \\ 0 & e^{-t} \end{array} \right ) P (\mathbb{Z}^{2}) \\ &= \pm PA (\mathbb{Z}^{2}) \\ &= \pm P (\mathbb{Z}^{2}) \\ &= L. \end{aligned} We have just proved one direction of the following.

Theorem: The periodic orbits of the modular flow are in bijection with the conjugacy classes of hyperbolic elements in SL(2,)\SL(2, \mathbb{Z}).

These periodic orbits produce fascinating knots in the complement of the trefoil! In fact, they link with the trefoil (the locus of degenerate lattices) in fascinating ways. Here are two examples, starting with different matrices ASL(2,)A \in \SL(2, \mathbb{Z}).

animation

The trefoil is the fixed orange curve, while the periodic orbits are the red and green curves respectively.

Ghys proved the following two remarkable facts about these modular knots.

  • The linking number of a modular knot with the trefoil of degenerate lattices equals the Rademacher function of the corresponding matrix in SL(2,)\SL(2, \mathbb{Z}) (the change in phase of the Dedekind eta function).
  • The knots occuring in the modular flow are the same as those occuring in the Lorenz equations!

Who would have thought that lattices in the plane could tell the weather!!

I must say I have thought about many aspects of these closed geodesics, but it had never crossed my mind to ask which knots are produced. – Peter Sarnak

Posted at April 7, 2014 11:55 PM UTC

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Re: The Modular Flow on the Space of Lattices

That’s a very nice post Bruce.

I first learnt about the space of non-degenerate lattices up-to-homothety being the complement of the trefoil from my friend and old office-mate Jacob Mostovoy.

He has a nice, short paper on how this space is related to subsets of the circle with at most three elements.

Posted by: Simon Willerton on April 8, 2014 10:04 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Yes, I found Jacob Mostovoy’s paper a few years ago when googling these things and was fascinated. I remember having the following question:

How do we see the braid group B 3B_3 in Mostovoy’s picture?

More precisely, he mentions results of Bott and Schepin that say that exp 3S 1Δ=S 3trefoil \exp_3 S^1 - \Delta = S^3 - trefoil where exp 3S 1\exp_3 S^1 is the space of nonempy finite subsets of S 1S^1 of cardinality at most 33, and Δ=S 1exp 3S 1\Delta = S^1 \subset \exp_3 S^1 is the subspace of nonempty finite subsets of cardinality 11.

Now, π 1(S 3trefoil)=B 3\pi_1 (S^3 - \mathrm{trefoil}) = B_3, the 3-strand braid group.

And a braid on 3-strands is conceptually very similar to 3 points on a circle moving around.

This suggests there should be a nice direct way to see that π 1(exp 3S 1Δ)=B 3. \pi_1 (\exp_3 S^1 - \Delta) = B_3. How precisely does this work?

Posted by: Bruce Bartlett on April 8, 2014 2:24 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Identify left and right (dark blue edges) so that the black lines illustrate a homotopy in exp 3(S 1)exp_3 (S^1). The complement is 3-colorable in essentially 3!3! ways; now, pretend each color is a rope, and the rope you can’t see is between the other two, while the one showing-up on both edges is around the back.

Layer 1

Posted by: Jesse C. McKeown on April 8, 2014 7:12 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

… so first the builg-in svg thing outputs xml that the café doesn’t like, and then it drops a red blob!

Posted by: Jesse C. McKeown on April 8, 2014 7:13 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Maybe this picture:

is clearer?

Posted by: Jesse C. McKeown on April 8, 2014 7:42 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Thanks Jesse, this is an elegant notation. I almost get it, but not quite. I see how a path in exp 3S 1\exp_3 S^1 is the same thing as one of your “trivalent pictures” where each horizontal slice has 1,2 or 3 points. And that a path in exp 3S 1Δ\exp_3 S^1 - \Delta is the same thing as a “trivalent picture” where each horizontal slice has 2 or 3 points.

And that you can draw an alternative picture by 3 coloured regions on discs.

How do I see the 3-strand braid though? You say ” pretend each color is a rope”, can you explain a bit more?

Posted by: Bruce Bartlett on April 10, 2014 10:31 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

I think I’m going to stick with the disc picture; Each coloured region is contractible, so a homotopy of the arrangement of the three coloured regions gives a homotopy of three points in the disc/plane, while a homotopy of our three coloured regions in turn is given by a homotopy in exp 3Δexp_3 - \Delta. And vice-versa. That’s about it.

Posted by: Jesse C. McKeown on April 10, 2014 3:59 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Jesse, I still don’t get it. How do I translate from a three strand braid to a homotopy of coloured regions in the disc? What do the generators correspond to?

Posted by: Simon Willerton on April 10, 2014 4:32 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Hello, Simon; I don’t know how to say it more precisely, but I’ve drawn two more pictures trying to illustrate more how I think this thing should work; the colouring in the picture is just to hint at what corresponds to what (unless one wants to talk of the braid groupoid with objects permutations of 33 things…)

Homotopies running vertically; braid picture on the left, slices of the disc picture in the middle, unrolled the circle on the right.

They’re still wobbly, but I hope the pictures help some?

Posted by: Jesse C. McKeown on April 10, 2014 10:43 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

This is a fascinating perspective.

Posted by: Bruce Bartlett on April 11, 2014 10:11 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Oddly enough, I actually did see this talk in person in 2006. I agree that it was excellent.

Posted by: Ben Webster on April 8, 2014 10:33 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Here’s some stuff I wrote about certain aspects of this business:

May 20, 2006

This Week’s Finds in Mathematical Physics (Week 233)

On Tuesday I’m supposed to talk with Lee Smolin about an idea he’s been working on with Fotini Markopoulou and Sundance Bilson-Thompson. This idea relates the elementary particles in one generation of the Standard Model to certain 3-strand framed braids:

1) Sundance O. Bilson-Thompson, A topological model of composite preons, available as hep-ph/0503213.

2) Sundance O. Bilson-Thompson, Fotini Markopoulou, and Lee Smolin, Quantum gravity and the Standard Model, hep-th/0603022. It’s a very speculative idea: they’ve found some interesting relations, but nobody knows if these are coincidental or not.

Luckily, one of my hobbies is collecting mysterious relationships between basic mathematical objects and trying to figure out what’s going on. So, I already happen to know a bunch of weird facts about 3-strand braids. I figure I’ll tell Smolin about this stuff. But if you don’t mind, I’ll practice on you!

So, today I’ll try to tell a story connecting the 3-strand braid group, the trefoil knot, rational tangles, the groups SL(2,Z) and PSL(2,Z), conformal field theory, and Monstrous Moonshine.

I’ve talked about some of these things before, but now I’ll introduce some new puzzle pieces, which come from two places:

3) Imre Tuba and Hans Wenzl, Representations of the braid group B3 and of SL(2,Z), available as math.RT/9912013. 4) Terry Gannon, The algebraic meaning of genus-zero, available as math.NT/0512248.

You could call it "a tale of two groups".

On the one hand, the 3-strand braid group has generators

     |   |   |
      \ /    |
A  =   /     |
      / \    |
     |   |   |

and

     |   |   |   
     |    \ /    
B =  |     /    
     |    / \  
     |   |   |

and the only relation is

ABA = BAB

otherwise known as the "third Reidemeister move":

|   |  |         |  |   | 
 \ /   |         |   \ /
  /    |         |    / 
 / \   |         |   / \
|   \ /           \ /   |  
|    /      =      /    | 
|   / \           / \   | 
 \ /   |         |   \ / 
  /    |         |    /
 / \   |         |   / \  

On other hand, the group SL(2,Z) consists of 2×2 integer matrices with determinant 1. It’s important in number theory, complex analysis, string theory and other branches of pure mathematics. I’ve described some of its charms in "week125", "week229" and elsewhere.

These groups look pretty different at first. But, there’s a homomorphism from B3 onto SL(2,Z)! It goes like this:

      
        1   1
A |->  
        0   1
        1   0
B |->  
       -1   1

Both these matrices describe "shears" in the plane. You may enjoy drawing these shears and visualizing the equation ABA = BAB in these terms. I did.

I would like to understand this better… and here are some clues.

The center of B3 is generated by the element (AB)3. This element corresponds to a "full twist". In other words, it’s the braid you get by hanging 3 strings from the ceiling, grabbing them all with one hand at the bottom, and giving them a full 360-degree twist:

|   |  |  
 \ /   | 
  /    |       A
 / \   |
|   \ /    
|    /         B 
|   / \   
 \ /   | 
  /    |       A
 / \   |
|   \ /    
|    /         B 
|   / \   
 \ /   | 
  /    |       A
 / \   |
|   \ /    
|    /         B 
|   / \   
|  |   |

This full twist gets sent to -1 in SL(2,Z):

 
             -1   0 
(AB)^3 |->  
              0  -1

So, the double twist gets sent to the identity:

             1   0 
(AB)^6 |->  
             0   1

In fact, Tuba and Wenzl say the double twist generates the kernel of our homomorphism from B3 to SL(2,Z). So, SL(2,Z) is isomorphic to the group of 3-strand braids modulo double twists!

This reminds me of spinors… since you have to twist an electron around twice to get its wavefunction back to where it started. And indeed, SL(2,Z) is a subgroup of SL(2,C), which is the double cover of the Lorentz group. So, 3-strand braids indeed act on the state space of a spin-1/2 particle, with double twists acting trivially!

(For more on this, check out Trautman’s work on "Pythagorean spinors" in "week196". There’s also a version where we use integers mod 7, described in "week219".)

If instead we take 3-strand braids modulo full twists, we get the so-called "modular group":

PSL(2,Z) = SL(2,Z)/{+-1}

Now, SL(2,Z) is famous for being the "mapping class group" of the torus - that is, the group of orientation-preserving diffeomorphisms, modulo diffeomorphisms connected to the identity. Similary, PSL(2,Z) is famous for acting on the rational numbers together with a point at infinity by means of fractional linear transformations:

         az + b
z |->   -------
         cz + d

where a,b,c,d are integers and ad-bc = 1. The group PSL(2,Z) also acts on certain 2-strand tangles called "rational tangles". In "week229", I told a nice story I heard from Michael Hutchings, explaining how these three facts fit together in a neat package.

But now let’s combine those facts with the stuff I just said! Since PSL(2,Z) acts on rational tangles, and there’s a homomorphism from B3 to PSL(2,Z), 3-strand braids must act on rational tangles. How does that go?

There’s an obvious guess, or two, or three, or four, but let’s just work it out.

I just said that the 3-strand braid A gets mapped to this shear:

 
        1   1
A |->  
        0   1

In "week229" I said what this shear does to a rational tangle. It gives it a 180 degree twist at the bottom, like this:

  
  |   |                |   |
  |   |                |   |
  |   |                |   |
 -------              -------
 |  T  |   |---->     |  T  |        
 -------              -------
  |   |                 \ /
  |   |                  / 
  |   |                 / \

Next, Tuba and Wenzl point out that

 
                  0   1
ABA = BAB |-> 
                 -1   0

which is a quarter turn. From "week229" you can see how this quarter turn acts on a rational tangle:

  
  |   |                       |     | 
  |   |              ____     |     |
  |   |             /     \   |     |
 -------           |     -------    |
 |  T  |   |---->  |     |  T  |    |    
 -------           |     -------    |
  |   |            |      |   \____/
  |   |            |      |    
  |   |            |      |  

It gives it a quarter turn!

From these facts, we can figure out what the braid B does to a rational tangle. So, let me do the calculation.

Scribble, scribble, curse and scribble…. Eureka!

Since we know what A does, and what ABA does, we can figure out what B must do. But, to make the answer look cute, I needed a sneaky fact about rational tangles, which is that A also acts like this:

  
  |   |                 \ / 
  |   |                  /
  |   |                 / \ 
 -------              -------
 |  T  |   |---->     |  T  |        
 -------              -------
  |   |                |   |
  |   |                |   | 
  |   |                |   |

This is proved in Goldman and Kauffman’s paper cited in "week228". With the help of this, I can show B acts like this:

 
  |   |              |          |
  |   |              |   ___    |
  |   |              |  /   \   |
 -------             | /    -------   
 |  T  |   |---->     \     |  T  |  
 -------             / \    ------- 
  |   |             |   \___/   | 
  |   |             |           | 
  |   |             |           |

And this is great! It means our action of 3-strand braids on rational tangles is really easy to describe. Just take your tangle and let the upper left strand dangle down:

           |
   ____    |               
  /    \   |              
 |    -------           
 |    |  T  |   
 |    -------  
 |     |   |  
 |     |   | 
 |     |   |

To let a 3-strand braid act on this, just attach it to the bottom of the picture!

(That’s why there were four obvious guesses about this would work: one can easily imagine four variations on this trick, depending on which strand is the "odd man out" - here it’s the upper right. It’s just a matter of convention which we use, but my conventions give this.)

In fact, even the group of 4-strand braids acts on rational tangles in an obvious way, but the 3-strand braid group is enough for now.

Let me summarize. The 3-strand braid group B3 acts on rational tangles in an obvious way. The subgroup that acts trivially is precisely the center of B3, generated by the full twist. Using stuff from "week229", it follows that the quotient of B3 by its center acts on the projectivized rational homology of the torus. We thus get a topological explanation of why B3 mod its center is PSL(2,Z).

But there’s more.

For starters, the 3-strand braid group is also the fundamental group of S3 minus the trefoil knot!

And, S3 minus the trefoil knot is secretly the same as SL(2,R)/SL(2,Z)!

In fact, Terry Gannon writes that the 3-strand braid group can be regarded as "the universal central extension of the modular group, and the universal symmetry of its modular forms". I’m not completely sure what that means, but here’s part of what it means.

Just as PSL(2,C) is the Lorentz group in 4d spacetime, PSL(2,R) is the Lorentz group in 3d spacetime. This group has a double cover SL(2,R), which shows up when you study spinors. But, it also has a universal cover, which shows up when you study anyons. The universal cover has infinitely many sheets. And up in this universal cover, sitting over the subgroup SL(2,Z), we get… the 3-strand braid group!

In math jargon, we have this commutative diagram where the rows are short exact sequences:

    1 ----> Z -----> B^3 -------> SL(2,Z) ----> 1
            |        |             |          
            |        |             |         
            v        v             v        
    1 ----> Z ---> SL(2,R)~ ---> SL(2,R) ----> 1

Here SL(2,R)~ is the universal cover of SL(2,R). Since π1(SL(2,R)) = Z, this is a Z-fold cover. You can describe this cover explicitly using the Maslov index, which is a formula that actually computes an integer for any loop in SL(2,R), or indeed any symplectic group.

But anyway, fiddling around with this diagram and the long exact sequence of homotopy groups for a fibration, you can show that indeed:

π1(SL(2,R)/SL(2,Z)) = B3.

This also follows from the fact that SL(2,R)/SL(2,Z) looks like S3 minus a trefoil.

Gannon believes that number theorists should think about all this stuff, since he thinks it’s lurking behind that weird network of ideas called Monstrous Moonshine (see "week66").

And here’s the basic reason why. I’ll try to get this right….

Any rational conformal field theory has a "chiral algebra" A which acts on the left-moving states. Mathematicians call this sort of thing a "vertex operator algebra". A representation of this on some vector space V is a space of states for the circle in some "sector" of our theory. Let’s pick some state v in V. Then we can define a "one-point function" where we take a Riemann surface with little disk cut out and insert this state on the boundary. This is a number, essentially the amplitude for a string in the give state to evolve like this Riemann surface says.

In fact, instead of chopping out a little disk it’s nice to just remove a point - a "puncture", they call it. But, we get an ambiguous answer unless we pick coordinates at this point, or in the lingo of complex analysis, a choice of "uniforming parameter". Then our one-point function becomes a function on the moduli space of Riemann surfaces equipped with a puncture and a choice of uniformizing parameter.

If we didn’t have this uniformizing parameter to worry about, we’d just have the moduli space of tori equipped with a marked point, which is nothing but the usual moduli space of elliptic curves,

H/PSL(2,Z)

where H is the complex upper halfplane. Then our one-point function would have nice transformation properties under PSL(2,Z).

But, with this uniformizing parameter to worry about, our one-point function only has nice transformation properties under B3. This is somehow supposed to be related to how B3 is the "universal central extension" of PSL(2,Z): in conformal field theory, all sorts of naive symmetries hold only up to a phase, so you have to replace various groups by central extensions thereof… and here that’s what’s happening to PSL(2,Z)!

That last paragraph was pretty vague. If I’m going to understand this better, either someone has to help me or I’ve got to read something like this:

5) Yongchang Zhu, Modular invariance of characters of vertex operator algebras, J. Amer. Math. Soc 9 (1996), 237-302. Also available at http://www.ams.org/jams/1996-9-01/S0894-0347-96-00182-8/home.html

But I shouldn’t need any conformal field theory to see how the moduli space of punctured tori with uniformizing parameter is related to the 3-strand braid group! I bet this moduli space is X/B3 for some space X, or something like that. There’s something simple at the bottom of all this, I’m sure.

Addenda: Another relation between the trefoil and the punctured torus: the trefoil has genus 1, meaning that it bounds a torus minus a disc embedded in R3. You can see this in the lecture "Genus and knot sum" in this course on knot theory:

6) Brian Sanderson, The knot theory MA3F2 page, http://www.maths.warwick.ac.uk/~bjs/MA3F2-page.html

This course also has material on rational tangles.

The fact that B3 is a central extension of PSL(2,Z) by Z, and the quantum-mechanical interpretation of a central extension in terms of phases, plays an important role here:

7) R. Voituriez, Random walks on the braid group B3 and magnetic translations in hyperbolic geometry, Nucl. Phys. B621 (2002), 675-688. Also available as http://arxiv.org/abs/math-ph/0103008.

Among other things, he points out that the homomorphism B3 → SL(2,Z) described above is the "Burau representation" of B3 evaluated at t = 1. In general, the Burau representation of B3 is given by:

        t   1
A |->  
        0   1
        1   0
B |->  
       -t   t

(Conventions differ, and this may not be the best, but it’s the one he uses.) The Burau representation can also be used to define a knot invariant called the Alexander polynomial. I believe that with some work, one can use this to explain why Conway could calculate the rational number associated to a rational tangle in terms of the ratio of Alexander polynomials of two links associated to it, called its "numerator" and "denominator". In fact he computed this ratio of polynomials and then evaluate it at a special value of t - presumably the same special value we’re seeing here (modulo differences in convention).

Another issue: I wrote

For starters, the 3-strand braid group is also the fundamental group of S3 minus the trefoil knot!

And, S3 minus the trefoil knot is secretly the same as SL(2,R)/SL(2,Z)!

The first one is pretty easy to see; you start with the "Wirtinger presentation" of the fundamental group of S3 minus a trefoil, and show by a fun little calculation that this isomorphic to the braid group on 3 strands. A more conceptual proof would be very nice, though. (See "week261" for such a proof - and much more on all this stuff.)

What about the second one? Why is S3 minus the trefoil knot diffeomorphic to SL(2,R)/SL(2,Z)? Terry Gannon says so in his paper above, but doesn’t say why. Some people asked about this, and eventually some people found some explanations. First of all, there’s a proof on page 84 of this book:

8) John Milnor, Introduction to Algebraic K-theory, Annals of Math. Studies 72, Princeton U. Press, Princeton, New Jersey, 1971.

Milnor credits it to Quillen. Joe Christy summarizes it below. I can’t tell if this proof is essentially the same as another sketched below by Swiatoslaw Gal, which exhibits a diffeomorphism using functions called the Eisenstein series g2 and g3. They are probably quite similar arguments.

Joe Christy writes:

I wouldn’t be surprised if this was known to Seifert in the 30’s, though I can’t lay my hands on Seifert & Threfall at the moment to check. Likewise for Hirzebruch, Brieskorn, Pham & Milnor in the 60’s in relation to singularities of complex hypersurfaces and exotic spheres. When I was learning topology in the 80’s it was considered a warm up case of Thurston’s Geometrization Program - the trefoil knot complement has PSL(2,R) geometric structure.

In any case, peruse Milnor’s Annals of Math Studies for concrete references. There is a (typically) elegant proof on p.84 of "Introduction to Algebraic K-theory" [study 72], which Milnor credits to Quillen. It contains the missing piece of John’s argument: introducing the Weierstrass P-function and remarking that the differential equation that it satisfies gives the diffeomorphism to S3-trefoil as the boundary of the pair (discriminant of diff-eq, C2 = (P,P’)-space).

This point of view grows out of some observations of Zariski, fleshed out in "Singular Points of Complex Hypersurfaces" [study 61]. The geometric viewpoint is made explicit in the paper "On the Brieskorn Manifolds M(p,q,r)" in "Knots, Groups, and 3-manifolds" [study 84].

It is also related to the intermediate case between the classical Platonic solids and John’s favorite Platonic surface - the Klein quartic. By way of a hint, look to relate the trefoil, qua torus knot, the seven-vertex triangulation of the torus, and the dual hexagonal tiling of a (flat) Clifford torus in S3. Joe

Swiatoslaw Gal writes:

In fact the isomorphism is a part of the modular theory:

Looking for f: GL(2,R)/SL(2,Z) → C2 - {x2=y3}

(there is an obvious action of R+ on both sides: M |→ tM for M in GL(2,R), x |→ t6 x, y |→ t4 y, and the quotient is what we want).

GL(2,R)/SL(2,Z) is a space of lattices in C. Such a lattice L has classical invariants

g2(L) = 60 sum_{z in L’} z-4,

and

g3(L) = 140 sum_{z in L’} z-6, where L’=L-{0}

The modular theory asserts that:

1. For every pair (g2,g3) there exist a lattice L such that g2(L) = g2 and g3(L) = g3 provided that g23 is not equal to 27g32. 2. Such a lattice is unique. Best, S. R. Gal

The quantity g23 - 27 g32 is called the "discriminant" of the lattice L, and vanishes as the lattice squashes down to being degenerate, i.e. a discrete subgroup of C with one rather than two generators.

Posted by: John Baez on April 9, 2014 4:26 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Yes I remember reading your TWF material on this stuff and there are so many strands I want to understand better.

In particular, your deeper explanation from week 261 for why the fundamental group of the complement of the trefoil is the 3-strand braid group is extremely beautiful and I’m letting it sink in. By the way, your link to week 261 is broken in your post.

Rational tangles I’m also trying to absorb!

For others: John referred to some papers by Gannon in the TWF posts above. The nicest reference is Gannon’s book which was published after the TWF posts, Moonshine beyond the monster. In particular the sections of the book numbered “Braided 0-7”.

Posted by: Bruce Bartlett on April 10, 2014 11:01 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Rational tangles are just certain tangles with 2 strands coming in and 2 strands coming out. They’re the tangles built starting from this one, called "zero"

  |   |
  |   |
  |   |
  |   |

and repeatedly doing two operations. The first is a twisting operation that we call "adding one":

  |   |                |   |
  |   |                |   |
  |   |                |   |
 -------              -------
 |  T  |   |---->     |  T  |        =  "T + 1"
 -------              -------
  |   |                 \ /
  |   |                  / 
  |   |                 / \

where the box labelled "T" stands for any tangle we’ve built so far. The second is a rotation that we call "negative reciprocal":

  |   |             |     |     
  |   |             |     |    ____
  |   |             |     |   /    \
 -------            |    -------    |
 |  T  |   |---->   |    |  T  |    |    =  "-1/T"
 -------            |    -------    |
  |   |              \___/   |      |
  |   |                      |      |
  |   |                      |      |

Using these tricks we can try to assign a rational number to any rational tangle. The shocking theorem is that this number is indeed well-defined, and in fact a complete invariant of rational tangles!

However, my post above is an attempt to demystify this fact. The first step is to grab one strand on top of your rational tangle and bend it down so now the tangle has 1 strand coming in and 3 coming out. Then the two operations described above are two generators of the 3-strand braid group B 3B_3! Thus, by definition, the 3-strand braid group B 3B_3 acts transitively on rational tangles.

But here’s the great thing: the ‘full twist’ element of B 3B_3 acts trivially on rational tangles. The ‘full twist’ is the braid you get by hanging 3 strings from the ceiling, grabbing them all with one hand at the bottom, and giving them a 360-degree twist.

What’s so great about this? On the one hand, it’s completely obvious. But on the other hand, B 3B_3 mod the full twist is the modular group PSL(2,)PSL(2,\mathbb{Z}). I show how that works…

So, PSL(2,)PSL(2,\mathbb{Z}) acts on rational tangles!

Even better, it turns out there’s a bijection between rational tangles and rational numbers (together with \infty) making the action of PSL(2,)PSL(2,\mathbb{Z}) on rational tangles isomorphic to its action on rational numbers:

zaz+bcz+d z \mapsto \frac{a z + b}{c z + d}

So, the connection between the modular group and the trefoil knot is part of a bigger picture which includes — among many other delights — a connection between the modular group and 3-strand braids!

But the fundamental group of the complement of the trefoil knot is the 3-strand braid group B 3B_3, so it all fits together… in a way I understood much better when I was writing about this stuff.

Posted by: John Baez on April 10, 2014 3:08 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

That’s really very nice, John. I just made a link to this comment from the nLab page on continued fractions, which gave a taste of this rational tangle business.

Noam Zeilberger asked a question about this at the nForum; maybe you’d like to have a look?

Posted by: Todd Trimble on April 10, 2014 6:02 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

You get a slightly different perspective on PSL(2,)PSL\left(2,\mathbb{Z}\right) and rational tangles by

1) gluing the ends of the tangle to the plane 2\mathbb{R}^2:

a) glue the ends of one strand to the points (0,0)\left(0,0\right) and (0,12)\left(0,\frac{1}{2}\right)

b) glue the ends of the other strand to the points (12,0)\left(\frac{1}{2},0\right) and (12,12)\left(\frac{1}{2},\frac{1}{2}\right)

2) Quotient the plane by 2\mathbb{Z}^2 to get a torus with the tangle glued to it.

3) Act on the plane with elements of PSL(2,)PSL\left(2,\mathbb{Z}\right), hence acting on the torus, hence acting on the tangle.

You have to stick the torus into 33-space in a funny way so that it looks like a cylinder (i.e. slice it open, but still consider it a torus for the purpose of identifying points) with the ends of the tangle lying at the corners of a square and the strands of the tangle inside the cylinder.

The rational number associated with the tangle is then the image of 01\frac{0}{1} under the fractional linear transformation given by the corresponding element of PSL(2,)PSL\left(2,\mathbb{Z}\right) … I think. It’s been many years since I thought about this so I’m a little vague about the details, but I think this picture also made it a little clearer why the rational tangles were in bijection with the rational numbers.

Even more hazily, I seem to recall that if you retain the full ratio of Alexander polynomials instead of reducing to \mathbb{Q}, you can extract a sort of qq-deformed version of PSL(2,)PSL\left(2,\mathbb{Z}\right), which at nnth roots of unity gives the symmetry group of the tiling with Schläfli symbol {n,3}\left\{n,3\right\} (in S 2S^2, E 2E^2 or H 2H^2 depending on nn).

Hmm … if you replace

TT+1T\rightarrow T+1 with Tq(T+1)T\rightarrow q\left(T+1\right)

and

T1TT\rightarrow\displaystyle\frac{-1}{T} with TqTT\rightarrow\displaystyle\frac{-q}{T} (which is still an involution!) …

I think that does it …

Posted by: Tim Silverman on April 18, 2014 10:17 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

In regard to your mentioning that some of the knots in Modular Flow are Lorenz knots (slight rewording)it has been said that the figure eight knot is not a Lorenz knot. Do you know the current status of this as there have been conjectures otherwise. See ‘Knots in Lorenz Equations’ by Jonas Bergman.

Posted by: Mark A. Thomas on April 10, 2014 12:12 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

The theorem of Ghys is that Lorenz knots and knots in the modular flow coincide, not just that one is a subset of the other.

My understanding is that the figure-eight knot is known not to be a Lorenz knot, eg. Figure 7(c) from Dana Lyon’s article.

So, I don’t understand why Bergman’s thesis searches for a figure-eight knot in the Lorenz equations! Maybe there is a subtlety to do with the parameters in the Lorenz equations.

Posted by: Bruce Bartlett on April 10, 2014 11:19 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

By the way, some years ago my AIMS essay student Ihechukwu Chinyere made a youtube animation of the “modular fibration” (z,w)z 3z 3w 2(z,w) \mapsto \frac{z^3}{z^3-w^2}. This is the representation of S 3S^3 as a S 1S^1 fibration over S 2S^2 whose fibers represent the lattices with the same G 4G_4 and G 6G_6. Each fiber is a trefoil.

It is the analogue of the Hopf fibration (z,w)zw(z,w) \mapsto \frac{z}{w}, and the code was adapted from Niles Johnson’s youtube animation of the Hopf fibration.

Posted by: Bruce Bartlett on April 10, 2014 11:27 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

These are some nice animations!

How does the modular fibration relate to the Hopf fibration homotopically – Is the former the square or the cube of the latter in π 3(S 2)\pi_3(S^2)?

This is the representation of S 3S^3 as a S 1S^1 fibration over S 2S^2 whose fibers represent the lattices with the same G 4G_4 and G 6G_6.

So in this picture, what are (z,w)(z,w)? Are they lattice generators? And is the map (z,w)z 3z 3w 2(z,w) \mapsto \frac{z^3}{z^3 - w^2} somehow a shortcut to calculating G 4G_4 and G 6G_6?

Posted by: Tim Campion on April 10, 2014 3:07 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Sorry, this was supposed to be a reply to Bruce’s last comment.

Posted by: Tim Campion on April 10, 2014 3:08 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

We are thinking of S 3 2S^3 \subset \mathbb{C}^2 as the pairs (z,w)(z,w) with |z| 2+|w| 2=1|z|^2 + |w|^2 = 1, and S 2=S^2 = \mathbb{C} \union \infty under stereographic projection from the north pole. So (putting in some scalar factors for precision) the map

(1)(z,w)20z 320z 349w 2 (z,w) \mapsto \frac{20 z^3}{20z^3 - 49w^2}

can be interpreted as a map j:S 3S 2j :S^3 \rightarrow S^2. The fiber is generically a trefoil in S 3S^3, except for the fibers above 00 \in \mathbb{C} (the south pole of S 2S^2) and the fiber above 11 \in \mathbb{C} (the east pole of S 2S^2). These fibers are z=0z=0 and w=0w=0 respectively in S 3S^3, and hence are the unknot. So this is not a principal S 1S^1-bundle since it is not locally trivial, but it is a Seifert fibration.

Topologically, the numbers “20” and “49” are of no significance. But we might choose to put them in to make contact with lattices. We can interpret z=G 4(L)z = G_4(L) and w=G 6(L)w = G_6(L), the Eisenstein series of a lattice. Then the quantity 20z 349w 220z^3 - 49w^2 measures the discriminant of the elliptic curve associated to the lattice. And the map j:S 3S 2j : S^3 \rightarrow S^2 is now interpreted as the jj-invariant of the elliptic curve corresponding to the lattice LL. So the fiber j 1)(p)j^{-1})(p) corresponds to all lattices of unit area with the same jj-invariant. The special fibers now have the following interpretations: j 1()j^{-1} (\infty) are the degenerate lattices, j 1(0)j^{-1}(0) are the hexagonal lattices, and j 1(1)j^{-1}(1) are the square lattices.

More details in Ihechukwu’s essay.

> How does the modular fibration relate to the Hopf fibration homotopically – Is the former the square or the cube of the latter in π 3(S 2)\pi_3(S^2)?

I was once curious about this, but I think I convinced myself, possibly erroneously, that the map j:S 3S 2j \colon S^3 \rightarrow S^2 has Hopf invariant 1, because the inverse image of two fibers j 1(p)j^{-1}(p) and j 1(q)j^{-1}(q) have linking number 1. I believe this characterizes the Hopf invariant of maps S 3S 2S^3 \rightarrow S^2? Have to check up on this.

Posted by: Bruce Bartlett on April 10, 2014 7:22 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Okay, I think I see. A fiber of the modular fibration (considered as a map 2\mathbb{C}^2 \to \mathbb{C}) is a class of lattices with the same j-invariant. In turn, two lattices / elliptic curves have the same j-invariant if they can be obtained from one another by multiplication by a complex number – i.e. rotation and homothety: a similarity transformation. This we can see because G k= ωΛω kG_k =\sum_{\omega \in \Lambda} \omega^{-k} scales as λ k\lambda^{-k} when the lattice is multiplied by λ ×\lambda \in \mathbb{C}^{\times}, so j=20G 4 320G 4 349G 6 2j = \frac{20 G_4^3}{20 G_4^3 - 49 G_6^2} is invariant under multiplication by λ\lambda.

So a fiber of the modular fibration is a similarity class of lattices. This also means that the corresponding elliptic curves – the toruses you get when you quotient \mathbb{C} by the lattices – are isomorphic.

This scaling behavior also makes it a little strange to consider the domain of the modular fibration to be S 3={(z,w)|z| 2+|w| 2=1}S^3 = \{(z,w) \mid |z|^2 + |w|^2 = 1\}. Perhaps it would be a little more natural to consider the domain to be {(z,w)|z| 3+|w| 2=1}\{(z,w) \mid |z|^3 + |w|^2 = 1\}, which is still topologcially S 3S^3. That way a fiber of the modular fibration consists of lattices which are rotations of each other, rather than a certain S 1S^1 obtained by some mix of rotations and homotheties.

Posted by: Tim Campion on April 10, 2014 11:50 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Perhaps it would be a little more natural to consider the domain to be {(z,w)||z| 3+|w| 2=1}\{(z,w) | |z|^3+|w|^2=1\}, which is still topologcially S 3S^3.

Yes, I think you’re right. This is the approach John takes in TWF261. [John - your link to TWF261 is still broken in your post].

Of course, if we want to get a picture of this fibration in R 3R^3, we will need to get into the standard S 3S^3 and then stereographically project.

Posted by: Bruce Bartlett on April 11, 2014 10:23 AM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Of course, if we want to get a picture of this fibration in R 3R^3, we will need to get into the standard S 3S^3 and then stereographically project.

It seems to me that since the modified sphere is still (the boundary of) a convex shape, we can perform stereographic projection on it directly. Of course we’ll have different formulas for the projection. Maybe they’re harder to work with than the usual ones?

Posted by: Tim Campion on April 11, 2014 2:54 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Let’s see. We need to parameterize something like

(1)(z,w) 2:|z| 3+|w| 2=1. (z,w) \in \mathbb{C}^2 : |z|^3 + |w|^2 = 1.

We can write

(2)z=p(s)e iϕ 1,w=q(s)e iϕ 2 z = p(s) e^{i \phi_1}, w = q(s) e^{i \phi_2}

where the functions p(s)p(s) and q(s)q(s) parameterize the the elliptic curve in 2\mathbb{R}^2,

(3)x 3+y 2=1,x0,y0. x^3 + y^2 = 1, x \geq 0, y \geq 0.

I have written these functions as pp and qq suggestively, in the hope that one could somehow use the Weierstrass elliptic functions of the original lattice LL to do this!

However, a naive attempt to do that doesn’t seem to work, for the following reason. Let LL be our lattice, and G 4(L)G_4(L) and G 6(L)G_6(L) the “Eisenstein numbers” of the lattice (defined as in the post… no numbers in front). Let p(s)p(s) be the associated Weierstrass function,

(4)p(s)=1s 2+ ω0(1zω 21ω 2), p(s) = \frac{1}{s^2} + \sum_{\omega \neq 0} (\frac{1}{z-\omega}^2 - \frac{1}{\omega}^2),

and q(s):=p(s)q(s) := p'(s). Then the functions pp and qq parameterize the curve

(5)q(s) 2=4p 3(s)60G 4p 2(s)140G 6 q(s)^2 = 4p^3(s) - 60 G_4 p^2(s) -140 G_6

which has a cross-term and is not of the form

(6)p 3+q 2=1. p^3 + q^2 = 1.
Posted by: Bruce Bartlett on April 11, 2014 7:06 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

Bruce said:

John - your link to TWF261 is still broken in your post

Fixed.

Posted by: Simon Willerton on April 11, 2014 7:45 PM | Permalink | Reply to this

Re: The Modular Flow on the Space of Lattices

In my post, I skipped over the `Great Dictionary’. By that, I mean the equivalence Ghys mentions in his talk (pg 26 of the slides) between:

  • periodic orbits (i.e. knots) in the modular flow
  • conjugacy classes of hyperbolic elements in PSL(2,)PSL(2, \mathbb{Z})
  • ideal classes in quadratic fields
  • indefinite integral quadratic forms in two variables
  • continuous fractions

This is clearly a long and beautiful story, well known to some of you much better than me. I’d like to point out here a cool fact about the “indefinite integral quadratic forms” picture (reference: Lemmermeyer).

Each closed orbit in the modular flow corresponds to an indefinite integral quadratic form in two variables in two variables xx and yy, that is an integral polynomial of the form

(1)Ax 2+Bxy+Cy 2 Ax^2 + Bxy + Cy^2

whose discriminant D:=B 24ACD := B^2 - 4AC is positive. That’s a cool fact by itself: a vivid picture of a quadratic form as a knot! But more is true. From the length (the number tt above) of the closed orbit, one can read off the discriminant DD of the corresponding quadratic form. And so the number of closed orbits with the same length is the class number h(D) — the number of equivalence classes of quadratic forms with discriminant DD.

This `dynamical systems’ view of indefinite quadratic forms is very useful. You can construct a zeta function as in TTWF216 and apply the Selberg trace formula, and hopefully prove something in number theory. This is exactly what Peter Sarnak did in Class numbers of indefinite binary quadratic forms. He used the dynamical systems description of the space of quadratic forms to derive an asymptotic formula for the class number h(D)h(D).

Posted by: Bruce Bartlett on April 10, 2014 10:25 PM | Permalink | Reply to this

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