3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 $n=3$. $\endgroup$ – Yilong Zhang Apr 1 at 1:14
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '13 at 18:22
  • $\begingroup$ Oops, sorry, Gustavo. I didn't think carefully. I'm editing. $\endgroup$ – Ted Shifrin May 11 '13 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 '13 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.