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Section 2.3 Riemann Hurwitz Formula (Sachi)

The genus is invariant under changing triangulation.

In particular there are at least two distinct ways of thinking about genus for Riemann surfaces \(R\)

  1. \begin{equation*} \chi(R) = V -E + F = 2-2g \end{equation*}
  2. The dimension of the space of holomorphic differentials on \(R\text{.}\)

Goal: given \(R\) calculate genus

\begin{equation*} y^2 = (x+1)(x-1)(x+2)(x-2) \end{equation*}

so in an ad hoc way

\begin{equation*} y = \sqrt{(x+1)(x-1)(x+2)(x-2)} \end{equation*}

when \(x\) is not a root of the above we have two distinct values for \(y\text{,}\) we can imagine two copies of \(\CC\) sitting above each other and then square root will land in both copies. We have to make branch cuts between the roots and glue along these to account for the fact that going around a small loop surrounding a root will change the sign of our square root. We end up with something looking like a torus here.

Here we examined the value where there were not enough preimages when we plugged in a value for \(x\text{.}\) The idea is to project to \(x\text{,}\) and understand the number of preimages.

\begin{equation*} P(x,y) = y^n + p_{n-1}(x) y^{n-1} + \cdots + p_0(x) \end{equation*}

an irreducible polynomial.

\begin{equation*} R= \{(x,y) : P(x,y) = 0\}\text{.} \end{equation*}

If we fix \(x_0 \in \PP^1 \CC\) we can analyse how many \(y\) values lie over this \(x\text{.}\) If we have fixed our coefficients we expect \(n\) solutions in \(y\) over \(\CC\text{,}\) i.e. points \((x_0,y)\in R\text{.}\)

For some values of \(x_0\) this will not be true, there will be fewer \(y\)-values, this occurs when we have a multiple root. This happens precisely when the discriminant of this polynomial vanishes, the discriminant is a polynomial and so has finitely many roots.

Definition 2.3.2. Branch points.

Let \(\pi\colon R \to \PP^1 \CC\text{.}\) We say \(x_0\) is a branch point if there are fewer than \(n\) distinct \(y\)-values above \(x\text{.}\) Then define the total branching index

\begin{equation*} b = \sum_{x\in \PP^1 \CC} (\deg(\pi) - \# \pi\inv (x))\text{.} \end{equation*}

In Sachi's notes.

Of Claim 2.3.3.

Triangulate \(R\) so that every face lies in some small coordinate neighborhood s.t.

\begin{equation*} \pi \colon R\to \PP^1 \CC \end{equation*}

is given by \(w \mapsto w^m\text{,}\) s.t. every edge, all branch points are vertices. This ensures that each face edge and vertex has \(n = \deg (\pi)\) preimages (except branch points). Then accounting for branch points we have \(\deg(\pi) - \# \pi\inv (x_0)\) preimages.

Example 2.3.5.

\(P(x,y)\) plane curve, classically have

\begin{equation*} g = \frac{(d-1)(d-2)}{2} \end{equation*}

\(\PP^2 = \{ \lb x:y:z \rb\}\) and \((\PP^2)^* = \lb a:b:c \rb\text{,}\) lines in \(\PP^2\)

\begin{equation*} ax + by + cz = 0 \end{equation*}

and we have lines \(\leftrightarrow\) points. We have \(C^*\) the dual curve in \(\PP^2\) cut out by the tangent lines \(t_Q\) for \(Q \in C\text{.}\) Claim \(\deg C^* = (d-1)d\text{.}\)

Want

\begin{equation*} R:\{P(x,y) = 0\} \xrightarrow\pi \PP^1\CC \end{equation*}

compute \(b\text{.}\) In other words, if we fix an arbitrary point \(Q\in C\) then there are \(d(d-1)\) lines through \(Q\) which are tangent to \(C\text{.}\) Projecting to the \(x\)-coordinate \(\iff\) family of lines through a point at \(\infty\) \(\iff\) \(\ast\) line in \((\PP^2)^*\text{.}\) We have a new question: How many points does this line intersect (up to multiplicity). By bezout \(\iff \deg C^*\text{.}\)

Proof (Matt emerton) Consider a point on \(C\) in \(\PP^2\) such that no tangent line to the curve at \(\infty\) passes through it. Move this point to the origin. If we write

\begin{equation*} P(x,y) = f_d + f_{d-1} + \cdots + f_0 \end{equation*}

then

\begin{equation*} (f_d, f_{d-1}) = 1 \end{equation*}

suppose they share a linear factor:

\begin{equation*} 0 = (f_d)_x x + (f_d)_y y + f_{d-1}\text{,} \end{equation*}

then this defines a line through the origin. (Because this gives an equation of an asymptote, this is a contradiction).

\begin{equation*} f_d + f_{d-1} + \cdots + f_0 = 0 \end{equation*}
\begin{equation*} d f_d + (d-1) f_{d-1} + \cdots + f_1 = 0 \end{equation*}
\begin{equation*} \implies \end{equation*}
\begin{equation*} \begin{cases} f_d + f_{d-1} + \cdots + f_0 = 0 \\ f_{d-1} + 2f_{d-2} + \cdots + (d-1) f_1 = 0\end{cases}\text{.} \end{equation*}

Now these have \(d(d-1)\) common solutions. \(C^*\) has degree \(d(d-1)\) so \(b = d(d-1)\text{.}\) Riemann-Hurwitz implies

\begin{equation*} \chi(R) = 2\deg \pi - d(d-1) \end{equation*}
\begin{equation*} \chi(R) = 2d - d(d-1) \end{equation*}

so

\begin{equation*} g= \frac{(d-1)(d-2)}{2}\text{.} \end{equation*}
A 3-fold equivalence of categories.

Amazing synthesis.

  1. Analysis: Compact connected riemann surfaces.
  2. Algebra: Field extensions \(K/\CC\) where \(K\) is finitely generated of transcendence degree 1 over \(\CC\text{.}\)
  3. Geometry: Complete nonsingular irreducible algebraic curves in \(\PP^n\text{.}\)

3) curve \(\to\) 2) field extension. Over \(C\) all rational functions \(\frac{P(x)}{Q(x)}\) \(\deg P= \deg Q\text{,}\) \(P,Q \colon C\to \CC \cup \{\infty\}\text{.}\)

3) \(\to\) 1) take complex structure induced by \(\PP^n\text{.}\)

1) \(\to\) 2) associated field of meromorphic functions on \(X\text{.}\)

1) \(\to\) 3) Any curve which is holomorphic has an embedding into \(\PP^n\) (Riemann-Roch).

2) \(\to\) 1) \(K/\CC\) consider valuation rings \(R\) such that \(K\supseteq R\supseteq \CC\text{.}\)

Example 2.3.6.

\(g =0\text{,}\) \(\PP^1 \CC\) \(\CC(t)\text{,}\) \(\CC\cup \{\infty\}\text{.}\)

Example 2.3.7.

\(g =1\text{,}\) elliptic curves, \(f(x,y,z)\) smooth plane cubic, \(f= 0\text{,}\) \(\CC(\sqrt{f(x)}, x)\text{.}\)

\begin{equation*} \CC/\Lambda \to \PP^2 \end{equation*}
\begin{equation*} z\mapsto (z,\wp(z), \wp'(z)) \end{equation*}
\begin{equation*} z\not\in \Lambda \end{equation*}

backwards

\begin{equation*} (x,y) \mapsto \int_{(x_0, y_0)}^{(x,y)} \frac{\diff x}{y} \end{equation*}
Riemann-Hurwitz (generally).

There's nothing that doesn't generalise about the previous proof.

\begin{equation*} f\colon \PP^1 \CC \to S \end{equation*}
\begin{equation*} \chi(\PP^1 \CC) = \deg f \chi(S) - b \end{equation*}
\begin{equation*} 2 = (+)\cdot (-) - b\text{.} \end{equation*}
\begin{equation*} x^n +y^ n + z^n = 0 \end{equation*}

is not solvable in non-constant polynomials for \(n \gt 2\text{.}\)

\begin{equation*} E = \CC/ \ZZ + \ZZ i \end{equation*}

multiplication by \(i\) rotates \(x \mapsto xi\) let \(x \sim xi\text{.}\) If we mod out by \(\sim\) to get \(E/\sim\) this is still a Riemann surface and the quotient map

\begin{equation*} f \colon E \to E / \sim \end{equation*}

is nice, compute the branch points of order 4 and order 2.

\(X\) compact Riemann surface of \(g\ge 2\) then there are at most \(84(g-1)\) automorphisms of \(X\text{.}\)

Klein quartic

\begin{equation*} x^3 y + y^3 z + z^3 x = 0 \end{equation*}

has 168 automorphisms and is genus 3.