# On the definition of a meromorphic differential on a Riemann surface

In the book "Complex Algebraic Curves", Frances Kirwan gives the following definition of a meromorphic differential on a Riemann surface.

Definition. Let $\{\phi_\alpha:U_\alpha\rightarrow V_\alpha: \alpha \in A\}$ be a holomorphic atlas on a Riemann surface $S$. Then a meromorphic differential $\eta$ on $S$ is given by a collection $$\{\eta_\alpha: V_\alpha \rightarrow \mathbb{C}\cup\{\infty\}: \alpha \in A\}$$ of meromorphic functions on the open subsets $V_\alpha$ of $\mathbb{C}$ such that if $\alpha,\beta \in A$ and $u\in U_\alpha\cap U_\beta$ then $$\eta_\alpha(\phi_\alpha(u))=\eta_\beta(\phi_\beta(u))(\phi_\beta \circ \phi_\alpha^{-1})'(\phi_\alpha(u)).$$

For two meromorphic functions $f$ and $g$ on $S$, the differential $fdg$ on $S$ is defined by $fdg=\eta$, where $\eta_\alpha=(f\circ \phi_\alpha^{-1})(g\circ\phi_\alpha^{-1})'$.

Further, the following remark is given:

If $\eta$ and $\zeta$ are meromorphic differentials according to this definition and $\zeta$ is not identically zero on any connected component of $S$ then the ratios $\eta_\alpha/\zeta_\alpha$ define meromorphic functions on the open subsets of $V_\alpha$ of $\mathbb{C}$ satisfying $$\frac{\eta_\alpha(\phi_\alpha(u))}{\zeta_\alpha(\phi_\alpha(u))}=\frac{\eta_\beta(\phi_\beta(u))}{\zeta_\beta(\phi_\beta(u))}$$ for all $u\in U_\alpha$; or equivalently $\eta=f\zeta$. Therefore to show that every meromorphic differential $\eta$ in the sense of definition 6.6 is a meromorphic differential of the form $fdg$ it is enough to show that there is at least one nonconstant meromorphic function $g$ on every Riemann suraface

Here are my questions:

1. Why the differential $fdg$ is well-defined, i.e., why $(f\circ \phi_\alpha^{-1})(g\circ\phi_\alpha^{-1})'(\phi_\alpha(u))=(f\circ \phi_\beta^{-1})(g\circ\phi_\beta^{-1})'(\phi_\beta(u))(\phi_\beta \circ \phi_\alpha^{-1})'(\phi_\alpha(u))$ for $u\in U_\alpha\cap U_\beta$ and $\alpha,\beta \in A$?
2. Why the equality $\frac{\eta_\alpha(\phi_\alpha(u))}{\zeta_\alpha(\phi_\alpha(u))}=\frac{\eta_\beta(\phi_\beta(u))}{\zeta_\beta(\phi_\beta(u))}$ is the same as $\eta=f\zeta$, and what is $f$ in the end? If it is some function, on which the notation hints, my mind still refuses to understand how a set (formally, $\eta$ is a set, isn't it?) can be equal to another set times a function.
3. How the existence of a meromorphic functioon $g$ on every Riemann surface implies that every meromorphic differential on $S$ in the sence of the given definition is of the form $fdg$?
4. Finally, an "ideological question": is it ideologically correct to think of meromorphic differentials on Riemann surfaces according to this definition? I like this definition because on the one hand it requires no knowledge about cotangent bundles and its sections for example, and on the other hand it is quite rigorous (unlike the definition which says that a differential is "something that can be written as $fdz$"). And by the way, is "differential" the same as "differential 1-form"?
• maybe you should look at the Riemann surface $\mathbb{C} \setminus \mathbb{Z}$, whose field of meromorphic functions are simply the $1$ periodic meromorphic functions. so $\eta_\alpha(\phi_\alpha(u))=\eta_\beta(\phi_\beta(u))(\phi_\beta \circ \phi_\alpha^{-1})'(\phi_\alpha(u))$ is really a constraint such that $\eta \$ (a meromorphic function/differential on some subset of $\mathbb{C}$) can be thought as a meromorphic function/differential on $S$, i.e. its analytic continuation coincides with itself in every charts of $S$ – reuns Aug 31 '16 at 18:01

1. This just follows from the chain rule, using the fact that$$f \circ (\phi_\alpha)^{-1} = (f \circ \phi_\beta^{-1})(\phi_\beta \circ \phi_\alpha^{-1})$$and similarly for $g$.
2. The remark is not very well-expressed. The equality says that there is a well-defined meromorphic function $f$ whose restriction coincides with $\eta_\alpha(\phi_\alpha(u))$ where this is defined, for any choice of $\alpha$, and then $\eta = f\zeta$ for this $f$. Conversely, if $\eta = f\zeta$ for any meromorphic function $f$, then the equality must hold. Here the products are pointwise products — just like multiplying polynomial functions to get another polynomial function.
3. If $g$ is a meromorphic function which is not constant, then $dg$ is a meromorphic differential which is not identically zero, and so taking $\zeta = dg$ we deduce that any meromorphic differential $\eta$ can be expressed as$$\eta = f\zeta = f\,dg$$for some meromorphic function $f$.
4. Yes, this is fine, and "differential" is the same as "differential $1$-form".