Archive/Tilings/Tilings VRG]]

# Thursday January 23rd

## Counting Square Tiled Surfaces 
Square tiled surfaces in $\mch(k)$ with $d$ squares correspond to degree $d$ branched covers of the identification square, branched over the origin, with profile (?).

To count square-tiled surfaces: label squares, look at inverse images of $\ast$ by $\theset{1 ,\cdots, d}$.
Consider the monodromy [representation](representation) $\rho: \pi_1( \TT \setminus \theset{0}, \ast ) \to S_d$ where $\sigma = \rho(\alpha) = (1)(23)$ and $\tau = \rho(\beta) = (12)(3)$.
We compute ramification orders by considering the commutators $[\alpha, \beta]$.
Then $\rho([\alpha, \beta] )$ has cycle type $(1, 1, \cdots 1, 1+k, \cdots, 1+ k_n)$.
Note that $[(23), (12)]$ is a 3 cycle.

![](Archive/Tilings/sections/figures/2020-01-23-14_32.png)
![](Archive/Tilings/sections/figures/2020-01-23-14_32.png)

**Conclusion:**
The number of square-tiled surfaces in $\mch(k)$ with $d$ squares is exactly 
$$
\frac{1}{d!} \abs{\theset{\sigma, \tau \in S_d \suchthat [\sigma, \tau] \in C_{1, \cdots, 1, 1+k, \cdots, 1+ k_n}}}
.$$

Note that the division is due to the artificial labeling of squares.

**Main theorem from "Branched Covers of Torus" Paper:**
The generating function 
$$
f_\kappa(q)\definedas \sum_{d\geq 1} \size\theset{\text{Square tiled, $d$ squares in $\mch(k)$}} q^d
$$ 
is a **modular form**.

Follows from taking $q = e^{2\pi i \tau}$ which is holomorphic on $\HH$ the upper half-plane, satisfying a transformation rule with respect to $\tau \mapsto -1/\tau$, which is a finite-dimensional space.

> Actual: quasimodular mixed form.

The weights are bounded by $\abs \kappa + \ell(\kappa)$.

Concretely, $f_\kappa \in \QQ[E_2, E_4, E_8]$ where $E_k(q) = \text{const} + \sum_{d \geq 1} \sigma_{k-1}(d) q^d$, where $\sigma_{k-1}(d) = \sum_{e\mid d} e^{k-1}$.
This is the ring of [quasimodular forms](quasimodular%20forms).

- $1$ is weight 0,
- $E_2$ is weight 2,
- $E_2^2, E_4$ are weight 4,
- $E_2^3, E_2 E_4, E_6$ are weight 6, etc

*Example:*
Take $\kappa = \theset{2} \iff \mch(2)$, then $\abs \kappa + \ell(\kappa) = 3$ and $f_{\theset z}(q) = c_1 + c_2 E_2(q)$.
In this case $[q^1] = [q^2] = 0$.

Note that surfaces in $\mch(2)$ have 1 vertex of cone angle $6\pi$ and all others of angle $2\pi$, corresponding to an abelian differential with a single zero of order 2.

A special type of square-tiled surfaces: 1 cylinder surfaces, where $\rho(\alpha)$ is a full length cycle.

This is in $\mch(3, 1)$, and corresponding surface $(\Sigma, \omega)$, which is a holomorphic 1-form with a triple zero and a single zero.
By [Riemann-Hurwitz](Riemann-Hurwitz), $2g-2 = \deg \omega = 3+1 \implies g = 3$.

> Note: the genus here difficult to compute otherwise!


**Main Result of 1-Cylinder Surface Paper:**
1-cylinder surfaces have roughly a $1/d$ proportion in all square tiled surfaces, where $d = \dim \mch(\kappa)$.

Recall that we can get a square tiled surface from any unicursal curve:

![](Archive/Tilings/sections/figures/2020-01-23-14:41.png)

Note that these aren't always translation surfaces:

![](Archive/Tilings/sections/figures/2020-01-23-14_43.png)

This has transition maps that looks like $z \mapsto i^k z + z_0 = w$ and thus $dz = i^k dw$, so $dz^k$ is the well-defined object here.

Recall

- $(\Sigma, dz) \iff$ translation surfaces
- $(\Sigma, (dz)^2) \iff \frac 1 2 \dash$translation surfaces
- $(\Sigma, (dz)^4) \iff \frac 1 4\dash$translation surfaces
  - I.e. a Riemann surface with a section of $K_\eps^{\tensor 4}$. 


Can consider a *tricursal* curve instead (a curve that requires lifting the pen 3 times).
Taking the dual complex yields a cube. 

![](Archive/Tilings/sections/figures/2020-01-23-14_52.png)

This has charts $w = z^{4/3}$ and thus $(dz)^4 = w\inv (dw)^4$.
Let $\mch(\kappa)$ be the quartic differentials with $dv \omega = \sigma \kappa_i p_i$.
Then the cube is in $\mch_4(-1, \cdots -1)$ with $8$ copies of $-1$.

This gives cone angles $n \frac{2\pi}{4}$ and the order of the zero/pole is $n-4$.

This example is in $\mch_4(-3, -3, -2)$.

**Proposition:**
The generating functions for square-tiled surfaces $\mch_4(\kappa)$ is now a quasimodular form for $\Gamma_1(4)$.

## Open Questions

**Question (can find numerical evidence?):**
How can we count these in terms of the symmetric group? 
Analogous result to proportion result earlier?
Can try to lift square example, but admits no map from a torus -- instead, quotient square by $\ZZ/4\ZZ$ and take fundamental domain.
What kind of branching do these covers have?

Every center of a square is branched of order 4.
Every center of an edge is branched of order 2.
The ramification order of a vertex is its valence, divided by the number of squares meeting at that vertex.
The degree of the covering map is $4n$ where $n$ is the number of squares.

Identify the fundamental domain with $\PP^1$,
We get a monodromy representation:

\[
\rho: \pi_1(\PP^1 \setminus(0, 1,\infty), \ast) \to S_{4d} \\
\gamma_0 \mapsto \rho(\gamma_0)
\]

Note that $\gamma_0 \gamma_1 \gamma_\infty = 1$.

![](Archive/Tilings/sections/figures/2020-01-23-15_13.png)s

It then follows that this has cycle type $(4, \cdots ,4)$.

So the number of square tiled surfaces in $\mch_4(\kappa)$ is given by

\[
\frac{1}{(4d)!} = \size \theset{ (\sigma_0, \sigma_1, \sigma_\infty) \suchthat \sigma_0 \in C_{4, \cdots, 4} (d), \sigma_1 \in C_{2, \cdots ,2} (2d), \sigma_{\infty} \in C_{4, \cdots ,4, 4+k, \cdots, 4 + k_n}, \sigma_0 \sigma_1 \sigma_\infty = 1}
.\]

Would be nice to figure out what the proportionality constant here is.