# Thursday January 16th

## Correspondence

**Recall:**
Start with a translation surface with cone points with angles $2\pi n i$.
This yields a Riemann surface $\Sigma$ and a holomorphic $1\dash$form $\omega$ with zeros of order $n_i -1$.

Given a square fundamental domain, there is an order 4 automorphism given by rotating 90 degrees.
In charts, this is multiplication by $i$ and possibly a translation, which is a holomorphic map $f: \Sigma \selfmap$.
We then have $f^*(dz) = d(f(z)) = idz$, so $dz$ is an eigenvector for $f^*$.

An elliptic curve can be specified by $y^2 = f_4(x)$ for a degree 4 polynomial, so we can obtain it as a double branched cover of $S^2$.
(i.e. glue along the slits joining pairs of roots)

![Image](figures/2020-01-22-21:56.png)\

Take $E : y^2 = x^4 - 1$, this is the only elliptic curve with an order 4 automorphism.
In coordinates, this is generated by $(x, y) \mapsto (-ix ,y)$.
So $\omega = c \frac{dx}{y}$, i.e. $dx/y$ up to scaling, and $f^*(dx/y) = i \frac{dx}{y}$.
What is the constant $c$?

Take closed cycles on $E$ given by $\alpha, \beta$ (see diagram), then by FTC $\int_\alpha \omega = z \mid_0^1 = 1$.
Negation is 4 fixed points on the elliptic curve, the 2-torsion points.

![Image](figures/2020-01-22-21:55.png)\

![Image](figures/2020-01-22-21:57.png)\

We can then compute

\[
I = \int_\alpha  \frac{dx}{y} = \int_1^2 \frac{dx}{\sqrt{x^4 - 1}} 
.\]

and since $2c I = 1$, this uniquely determines $c$.

> Note that this can be numerically evaluated, but this is an elliptic integral with (possibly) no elementary antiderivative.

Consider the decagon with sides identified.
We get a complex structure $(\Sigma, \omega)$ with cone angles $4\pi$ and $dz$ has 2 zeros of order 1 (namely the two cone points).

![Image](figures/2020-01-22-21:58.png)\

What is the genus?
The degree of the canonical bundle is 
$$
g(\Sigma) = \deg K_\Sigma = 2g - 2 = \sum_{p\in \Sigma}\ord_p \omega
$$ 
and thus $g = 2$.

**Fact:**
Every genus 2 curve is a double branched cover of $\PP^1$ branched over 6 points.

> Use Riemann-Roch.

Consider automorphisms that preserve the decagon.
Rotation by $\pi/10$ swaps the two cone points, to take rotation by $-2\pi/5$ (inserting the negative to account for pullbacks).
Then $f^* \omega = \zeta_5 \omega$, where again we just write locally $z \mapsvia{f} \zeta_5\inv + c \implies f^* dz = \zeta_5 dz$.

Consider points of order $5$ on $\PP^1$, we can take $\zeta_5^k$ and $\infty$.

![Image](figures/2020-01-22-21:58.png)\

This corresponds to the curves $y^2 = x^5 - 1$, with an automorphism $(x, y) \mapsto (\zeta_5 x, y)$.
This is the only genus 2 curve with an order 5 automorphism.

**Fact:**
The space of sections of the canonical $H^0(K_{\mathcal E}) = \CC^g$.

We can write a basis for the space of 1-forms: $\frac{dx}{y}$ and $(1-x) \frac{dx}{y}$.
Alternatively, $\omega = (a+bx) \frac{dx}{y}$ where $a, b \in \CC$.
What are the zeros of $\omega$?
$V(\omega) = V(a + bx)$ where if $b=0$ it's $\infty$.
Because this has to preserve the order 5 symmetric and map cone points to themselves, this forces $\omega = bx \frac{dx}{y}$, which has exactly two zeros: $(x, y) = (0, \pm i)$.

We can also consider the doubled pentagon, which has only one point.
This has an automorphism given by rotating each pentagon by $1/5$, it has cone angle $6\pi$, and $\omega$ has a double zero at the cone point.
Since there is only *one* genus 2 curve with order 5 automorphisms, this yields the previous Riemann surface but a distinct translation surface and a distinct form.

We can write $\alpha = a \frac{dx}{y}$.

**Proposition:**
One Riemann surface has many translation structures, and the space of such structures is the space of 1-forms.

*Proof:*
Pick a chart $w: U \to \CC$ avoiding the zeros of $\omega$.
Then $\omega = f(z) ~dz$ in this chart, and we want to compose with a biholomorphism $x$ to obtain a new chart $z$ such that $\omega = dz$ in this chart.
We can solve $dz = e = f(w) ~dw$ and $\frac{dz}{dw} = f(w)$ and by integrating we get $z(w) = \int_{p_0}^w f(w_0) dw_0$.
In this chart, $\omega = dz$, and we can correct near zeros by finding charts such that $\omega = z^n dz$ where $n$ is the order of the zero.

![Image](figures/2020-01-22-21:59.png)\

What does this buy us?
We get a translation structure by considering transitions, and $\omega = dz_1 = dz_2 \implies z_1 = z_2 + c$, which is exactly a translation structure.
Thus every $(\Sigma, \omega)$ has a translation structure for which $\omega = dz$ in local polygonal charts.

![Image](figures/2020-01-22-22:00.png)\

**Theorem:**
There is a bijection 

\[
\correspond{\text{Translation surfaces with cone angles $2\pi n i$ }} \iff
\correspond{\text{$(\Sigma, \omega)$ a Riemann surface with holomorphic}\\ \text{1-forms with zeros of order $n_i - 1$}}
.\]

What do half-translations correspond to?
Note that $dz \mapsto -dz$, so we don't get a well-defined holomorphic 1-form.

![Image](figures/2020-01-22-22:01.png)\

The fix? 
$(dz)^2$ is some well-defined object. 
What is it?

The set $\theset{f(z) (dz)^2 = g(w)(dw)^2}$ corresponds with line bundles with transition functions given by $(dz/dw)^2$.

Thus the correspondence is 

\[
\correspond{\text{Half translation surfaces with cone angle $\pi n_i$}}
\iff
\correspond{\text{Riemann surfaces with $q$ a section of $K_\Sigma^{\tensor 2}$ }}
.\]

I.e., these correspond with sections of the second tensor power of the canonical bundle of $\Sigma$.

A defining property of $q = (dz)^2$ for half-translation charts $z$: we can measure the order of the zero by going to charts, finding a chart to $\CC$ (see image) e.g. $w = z^{2/3}$, and then defining 
$$
\omega = dz = d(w^{3/2}) = 3/2 w^{1/2} dw
$$

Then $q = w^2 = (dz)^2$ is well-defined and equal to $\frac{9}{4} w (dw)^2$ in the local chart $w$. 

In this case, we get points that are *poles* of order 1 for the quadratic differential (sections of $K^{\tensor 2})$.

> Note: 1-forms are referred to as "abelian differentials" in the literature.

We know that $K_{\PP^1} = \mathcal O(-2)$ and $K_{\PP^1}^{\tensor 2} = \mathcal O(-4)$.

## Moduli Spaces

**Definition:**
$\mathcal{H}(k_1, \cdots, k_n) = \theset{ \Sigma \text{ with abelian differential}  \div \omega = \sum k p_i}$ where the $k_i$ record the orders of zeros of $\omega$.

> Second condition on divisor records zeros.

Similarly, define $Q(k_1, \cdots, k_n) = \theset{\cdots \suchthat \text{quadratic differential}}$ where now $k_i \geq -1, \neq 0$.

These moduli spaces are called *strata of abelian/quadratic differentials*.

Consider $M_g$, the moduli space of genus $g$ surfaces.
There is a vector bundle (the Hodge bundle) over $M_g$ where the fiber over $[c]$ is $H^0(K_C)$ with strata given by $Q$.

This is a bundle of rank $g$, and by Riemann-Roch, ? $3g-3$ and is the (co?)tangent bundle of something.

There is an $\SL(2, \RR)$ action on any stratum.
How to define -- use isomorphism with translation structures (see image) by just applying any such form $\gamma$ to the entire plane and gluing polygons in the same way.
This gives a non-holomorphic action on any stratum.

> Next time: to special case of square tiled surfaces.

Tags: #projects/research  #AG/moduli-spaces