## FANDOM

36 Pages

Springer GTM 71.

"In this book we present the theory of Riemann surfaces and its many different facets. We begin from the most elementary aspects and try to bring the reader up to the frontier of present-day research."

# Chapter I: Riemann SurfacesEdit

## Section I.1: Definitions and ExamplesEdit

• I.1.6: Thus, we also have (since a non-vanishing holomorphic function on a disc has a logarithm) that:

From above we have

$\zeta = f(\tilde{z}) = \sum_{k=n}^\infty a_k \tilde{z}^k$

with n positive and $a_n$ nonzero. Pulling out a factor of $\tilde{z}^n$, we can rewrite this as

$\zeta = \tilde{z}^n\sum_{k=0}^\infty a_k \tilde{z}^k =: \tilde{z}^n g(\tilde{z})$

Now g converges wherever f does -- at $\tilde{z}=0$ it converges to $a_n$ and elsewhere we can just divide $f(\tilde{z})$ by $\tilde{z}^n$. Since $g(\tilde{z})$ is nonzero at $\tilde{z}=0$ and holomorphic, it's nonzero on some disc of positive radius, and consequently has a logarithm on this disc. Write $h(\tilde{z}) := \exp \left\{\frac{1}{n} \log h(\tilde{z})\right\}$ on this disc; then $g(\tilde{z}) = h(\tilde{z})^n$, so $\zeta = \tilde{z}^n h(\tilde{z})^n$ as claimed.

• I.1.6, Proposition:  The "normal form" of the mapping f given by (1.6.1) shows that this is open in N.

# Chapter III: Compact Riemann SurfacesEdit

## Section III.4 Divisors and the Riemann-Roch TheoremEdit

• III.4.8 Theorem (Riemann-Roch)

There's way too much stuff in this proof. It should be broken up into multiple pieces, and the notation should be pulled out.

Notation:

• Let M be a compact Riemann surface with canonical homology basis $\{a_1, \ldots, a_g, b_1, \ldots, b_g$, as in this detail from figure I.4:

• If $A = \sum_{j=1}^m n_j P_j \in \operatorname{Div} M$ for some Riemann surface M, write $A' := \sum_{j=1}^m (n_j+1) P_j$.
• Write $\Omega_0(A)$ for the meromorphic 1-forms with no 'a'-periods and no residues:

$\Omega_0(A) := \{ \omega \; | \; \omega \text{ a mero 1-form}, \forall j \int_{a_j} \omega = 0, \forall P \in M \operatorname{res}_P \omega = 0, (\omega) \geq A \}$.