Lattices
Lattices in the Plane and the Trefoil Knot
Steve Trettel

The space of lattices in $\mathbb{C}$ is an open 4manifold  however if we identify lattices up to scaling, the associated deformation space is three dimensional: the trefoil knot complement. This identification allows us to understand the Seifert fibration on the trefoil complement visually (it corresponds to rotation on the space of lattices), and to identify the base space (by looking for especially symmetric lattices). We also can gain fruitful information in the other direction, for example two natural flows on the space of lattices correspond to the geodesic and horocyclic flows on the base orbifold of the fibration.
The Space of Lattices
A lattice in a Lie group is a discrete subgroup of finite covolume.
Of course the case we are interested in here is particularly simple: when the group
in question is the additive group ℝ2 (which we will think of as $\mathbb{C}$) lattices
are just infinite grids of dots in the plane. To get our hands on something concrete,
we can parameterize the space of lattices (nonuniquely) by linearly independent
vectors in $\mathbb{C}$: given two such vectors v,w we can produce the lattice
$\Lambda(v,w)=\mathbb{Z}v\oplus\mathbb{Z}w$.
Its easy to see precisely how this fails to be non unique:
a lattice has many bases and all of them represent the same point.
$$\Lambda(v,w)=\Lambda(w,v)=\Lambda(w+v,v)=\Lambda(2w+v,wv)=\cdots$$
In fact, given a basis $(v,w)$ the rest of the bases for the lattice are given by
the $GL(2,\mathbb{Z})$ orbit of this pair  so we may identify the space of
lattices with pairs of linearly independent vectors up to this equivalence.
But as a pair of linearly independent vectors is a basis for \mathbb{R}^2$ and so gives
rise to an invertible transformation (a matrix in $GL(2,\mathbb{R})$, we can identify
the space of lattices with
$$\mathcal{L}=\frac{GL(2,\mathbb{R})}{GL(2,\mathbb{Z})}$$
We can use this identification to give the space of lattices a topology, and this topology behaves as we would expect: nearby pairs of vectors generate lattices which look similar, and which we now declare to be nearby each other in the space of lattices (we declare the above map to be a homemorphism)
Drawing a line segment in $GL(2,\mathbb{R})$ between two elements of $GL(2,\mathbb{Z})$ lets us see how the quotient space is indeed the natrual structure  this clearly projects to a closed loop in $\mathcal{L}$
Some other information we can immediately reap from this identification is the dimensionality: $GL(2,\mathbb{R})$ is four dimensional and the space of lattices is a quotient of this by a discrete subgroup  thus is four dimensional. This can also be seen intuitively as given a lattice, choosing any basis we can form nearby lattices by moving either of the basis vectors around inside of a little tiny disk  giving four local degrees of freedom.
Unit Covolume Lattices
Of course, we’d really like to understand this space in more detail than just its dimensionality  is there any other structure we can exploit? The space of lattices admits an action of ℝ+ by scaling, and so the space of lattices decomposes into a bunch of lines, each containing one lattice of unit covolume.
Thinking a little more we can see that this decomposition is actually a product:
$\mathcal{L}\cong\mathbb{R}_+\times\mathbb{L}_1$$
where $\mathcal{L}_1$ is the space of unit covolume lattices.
Thus, all the interesting topology is contained in $\mathcal{L}_1$, and this space is
three dimensional! But what threemanifold is it?
Eisenstein Series
To answer this question, we are going to try and find a second description of the space of lattices (it turns out our current identification with a quotient space makes it a little difficult to see the global topology)  and to do so we will consider the following functions on the space of lattices, called Eisenstein series:
$$G_k(\Lambda)=\sum_{\omega\neq 0}\frac{1}{\omega^{2k}}$$
These functions come from the world of modular forms, but the important point for us here is that these functions do a particularly good job of distinguishing lattices: in fact, just taking the first two $k=2,3$ we can construct a function
$$\mathcal{G}=(G_2,G_3)\colon\mathcal{L}\to\mathbb{C}^2$$
which turns out to be injective! Thus, we can identify the space of lattices with their image under this map. Even better: a pair of complex numbers $(z,w)$ actually comes from a lattice (ie $(z,w)=(G_2(\Lambda),G_3(\Lambda))$) so long as $(60z)^3−27(140w)^2\neq0$. Thus we can identify $\mathcal{L}$ with the subset of $\mathbb{C}^2$ satisfying this equation. After a linear change of variables $(\xi,eta)=(60z,140\sqrt{27}w)$ we may express this cleanly as
$$\mathcal{L}={(\xi,\eta)\in\mathbb{C}^2\mid \xi^3\neq \eta^2}$$
Now, how does the action of ℝ+ by scaling carry over to this new model? Let $t\Lambda$ be the scaling of $\Lambda$ by $t>0$. Then
$$G_2(t\Lambda)=\sum_{\omega\neq 0}\frac{1}{t\omega}^4=\frac{1}{t^4}G_2(\Lambda)$$
and similarly for $G_3$. Thus, the action of the positive reals on our new coordinates is as follows:
$$t.(\xi,\eta)=\left(\frac{\xi}{t^4},\frac{\eta}{t^6}\right)$$
And so our equivalence classes are curves in $\mathbb{C}^2$ of the form ${(t^4\xi,t^6\eta)}$. Each of these curves passes through a unique point on the three sphere and contains a unique covolume one lattice. Thus we may identify the space of unit covolume lattices with the subset of the three sphere off of the curve $\xi^3=\eta^2$.
Understanding $\xi^3=\eta^2$
We are getting close! We now have a very concrete description of our space:
$$\mathcal{L}_1=\mathbb{C}^2\smallsetminus{\xi^3=\eta^2}$$
All that remains is to understand what the solution set looks like inside the 3sphere, and this turns out to be quite tractable. Indeed, if (ξ,η)∈𝕊3 then $\xi^2+\eta^2=1$ and if in addition $\xi^3=\eta^2$ then $\xi^2+\xi3=1$, which has a unique solution for $\xi$
$$\xi=\frac{ 2 + 2^{2/3}\sqrt[3]{25  3 \sqrt{69}}+ 2^{2/3}\sqrt[3]{25 + 3 \sqrt{69}}}{6}\approx 0.7548776:=R_\xi$$
But now as $\xi$ is constant, we have that $\eta=\sqrt{1\xi^2}:=R_\eta$ is constant as well, so each lies on a circle making $(\xi,\eta)$ lie on a torus in $\mathbb{S}^3$ with radii $(R_\xi,R_\eta)$.
Thus, for some values of $\theta,\psi$ we may write
$$(\xi,\eta)=\left(R_\xi e^{i\theta},R_\eta e^{i\psi}\right)$$
and again utilizing the condition $\xi^3=\eta^2$ we can see that $e^{i3\theta}=e^{i2\psi}$. This means we may write $\theta=\frac{2}{3}\psi$ and so all of our solutions lie along a curve of slope $2/3$ on the torus  but this is a trefoil knot!
Its easy to see that each point on this curve satisfies $\xi^3=\eta^2$, and so we have identified the space of unit covolume lattices: it is homeomorphic to the trefoil knot complement!
Seifert Fibration
This allows us to identify a lattice with a distinct point in the trefoil knot complement, and likewise a path in the space of lattices with a path in the trefoil complement. For instance, below is some random path in the space of lattices together with its image in the trefoil complement.
One natural path associated to any lattice is that of rotation: the lattices eiθΛ give a circle of lattices through Λ, and these trace out a loop in the complement of the trefoil knot.
In fact, rotation provides a circle action on the entire space of lattices  and the orbits under this action decompose $\mathcal{L}_1$ into a disjoint union of circles, giving it the structure of a Seifert fiber space! Because lattices are symmetric under negation (Λ=−Λ), these circles have length at most π. However, some lattices are more symmetric than others, coming back to themselves under rotation in less than a full π turn  these correspond to the exceptional Seifert fibers, around which the generic ones twist.
A generic lattice returns to tiself after a π rotation. The first exceptional case, a square lattice, returns to itself after only rotation by π/2, and the the other exeception, the hexagonal lattice, returns after $\pi/3$. Thus, the two circles of rotations traced out by these lattices are of length 1/2 and 1/3 as long as a generic circle respectively. As these are the only lattices with extra symmetries, these are the only exceptional fibers in the Seifert fibration.
That is, rotation on the space of lattices corresponds to the (2,3) Seifert fibration of the three sphere  the two exceptional orbits correspond to the two exceptional lattices, and the degenerate lattices correspond to the trefoil knot which ha been removed.
The base space of this fibration is then not the entire two sphere, but the two sphere minus a point (corresponding to the rotations of degenerate lattices)  a disk. In fact, if you have seen the moduli space of Euclidean tori, this particular disk should be familiar to you: a Euclidean structure on a torus (up to scaling) corresponds to a rotation class of unitcovolume lattices in the plane. The traditional description of this space is ℍ2/SL(2,ℤ)  the $(2,3,\infinty)$ triangle group orbifold, which is a punctured sphere with two cone points. The cone points correspond to the square and hexagonal lattice rotation classes! We will not go into this here (it is covered in the next article) but if this orbifold inherits its hyperbolic structure from ℍ2 then the trefoil complement is actually diffeomorphic to its unit tangent bundle! This allows us to think of paths in the space of lattices not only as paths in the trefoil complement, but also as unit speed paths on the this disk with cone points.