Let's suppose that $X$ is a Riemann Surface, $\omega$ is a meromorphic 1-form in $X$ and let $p$ be a pole of $\omega$ of order $M$. I want to show that the residue of $\omega$ at $p$, defined by $$ Res_p(\omega )=\dfrac{1}{2\pi i}\int_\gamma\omega $$ (here $\gamma$ is a closed $C^\infty$ path with $p\in Int(\gamma )$) does not depend of the chart over $X$. For this we will suppose that $\gamma$ is contained on the domain of two charts, $\phi_z:U_z\rightarrow V_z\subset\mathbb{C}$ and $\phi_w:U_w\rightarrow V_w\subset\mathbb{C}$, that is, $\gamma\subset U_z\cap U_w$. Here we'll soppuse tht both charts are centered at $p$ (that is, $\phi_z(p)=\phi_w(p)=0$.

(Please be careful to not be confused with the uses of $\omega$ and $w$!)

What I want is to show that the Residue is invariant under coordinate (or chart) changing.

This is my resolution for this (there is a proof for this in Otto's Lectures on Riemann Surfaces; it is proved as a corolary of the invariance of the integral of a 1-form over a path in Miranda's Algebraic Curves and Riemann Surfaces and in Farkas' Riemann Surfaces it is said that it can be done manipulating power series - which is the way I'm trying to do).

I can assume that in the coordinate $z$, $\omega$ is written as $\omega=f(z)dz$ (if not, then there is another local chart $\psi$ such that after coordinate changing $\omega$ has this form and then we lose no generality). We can then write $z=T(w)$ locally, where $T$ is holomorphic and $T(0)=0$.

Using Laurent Series, we have $$ f(z)=\sum_{n=-M}^{\infty}c_nz^n $$ near $p$. Note that $Res_p(\omega)=c_{-1}.$ On the $w$ coordinate, then $$ \omega=f(T(w))T'(w)dw $$ and also $$T(w)=\sum_{i=1}^{\infty}a_iw^i$$ therefore $$ T'(w)=\sum_{i=1}^\infty a_iiw^{i-1} $$

$\color{blue}{\text{This is the part where comes my dilemma}}$. I must obviously have $T(w)T(w)^{-1}=1$. Describing $T(w)^{-1}=\sum_{i=1}^\infty b_i w^{k_i}$, and working with the series, we have: $$ \left(\sum_{i=1}^\infty a_i w^i\right)\left(\sum_{i=1}^\infty b_i w^{k_i}\right)=1 $$ $$ a_1b_1xx^{k_1}+[...]=1\Rightarrow k_1=-1\text{ and }b_1=a_1^{-1} $$ and the rest of the terms will cancel themselves. Here $[...]$ are terms of higher/lower order.

Now, substituting the series in $\omega$, $$ \omega=\left[\sum_{n=-M}^\infty c_n \left(\sum_{i=1}^\infty a_iw^i\right)^{-n}\right]\left[\sum_{n=1}^\infty na_nw^{n-1}\right] $$ $$ =\left[ [...]+c_{-1}\left(\sum_{i=1}^\infty a_iw^i\right)^{-1}+[...]\right]\left[ a_1+[...]\right] $$ $$ =\left[ [...]+c_{-1}(a_1^{-1}w^{-1}+[...])+[...]\right]\left[ a_1+[...]\right] $$ $$ =[...]+c_1a_1^{-1}a_1w^{-1}+[...] $$ Since $Res_p(\omega)=c_1$, the invariance is proved.

Can someone tell me if the series manipulation of the part in blue is correct? If not, a hint of how can I do it? Thanks!

  • $\begingroup$ The proof of Otto reduces to the case when $f$ has a simple pole, where it is easy to show independence of coordinates. The idea is to write $f=g+c_{-1}z^{-1}$ where $g$ has no $z^{-1}$ terms. Now $g$ has "antiderivative" function $h$, and it is shown that the differential form $dh$ has zero residue independent of coordinate. Otherwise, as Ted mentioned, you have to expand $z^{-n}dz$ in $w$-coordinate and show that the coefficient of $w^{-1}dw$ is zero, but this is hard even for $n=3$. $\endgroup$
    – AG learner
    Apr 1, 2019 at 1:14

1 Answer 1


The flaw in this computation is that there are many cross-terms that could give you a $w^{-1}$ contribution. For example, what happened to $c_{-2}a_1^{-2}\cdot 2a_2$? The $[...]$ technique works with either higher or lower, but not both at once :)

What I guess you need, if you want to pursue this approach, is to check that each of the merormorphic $1$-forms $\dfrac{dw}{w^j}$, $j=2,\dots,M$, has $0$ residue. This can certainly be done by brute force.

  • $\begingroup$ I don't think it is; when doing the last substitution, I will have to work with $\dfrac{1}{T(w)}=T(w)^{-1}$, not the inverse... I think. $\endgroup$
    – Marra
    May 11, 2013 at 18:22
  • $\begingroup$ Oops, sorry, Gustavo. I didn't think carefully. I'm editing. $\endgroup$ May 11, 2013 at 19:46
  • $\begingroup$ Yes, you are correct... even when the "$[...]$" to the left are finite, I will have some other elements adding to the coefficient of $w^{-1}$. Thanks! $\endgroup$
    – Marra
    May 11, 2013 at 19:54

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