Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently reading Forster. Let $X$ be a Riemann suface, and $Y$ an open subset of $X$. Assume we have a meromorphic 1-form $\omega \in \Omega(Y \setminus \{a\})$ with a pole at $a \in Y$. One defines the order of $\omega$ at $a$ as the order of $f$ at $a$, where $\left. \omega \right|_{Y \cap U} = f \, \mathrm{d}z$ in some chart $(U,z)$ around $a$.

How exactly can I prove that this definition is independent on the choice of chart? My strategy was to try to show that different charts give rise to functions that differ by a non-zero constant, so that the respective power series have the same order. However, having done very little differential geometry, I'm not very confident working with the technicalities involved.

Thanks.

share|improve this question

1 Answer 1

Not exactly a nonzero constant, rather a non-vanishing holomorphic function.

Let $z : U_1 \subset X \to V_1 \subset \mathbb{C}$, $z': U_2 \subset X \to V_2 \subset \mathbb{C}$ be two coordinate systems involved, and $\phi :z(U_1 \cap U_2) \to z'(U_2 \cap U_1)$ sends the coordinate system of $z'$ to that of $z$. Here $\phi$ is a biholomorphic map, i.e. $\frac{d\phi}{dz'}$ is nonvanishing. Then you can check that $$fdz = (f\circ \phi) \frac{d\phi}{dz'}dz'$$ So the coefficient is off by $\frac{d\phi}{dz'}$, a nonvanishing holomorphic function.

share|improve this answer
    
I feel really dense for not getting this, but I can't wrap my head around some parts of your answer. I understand how $\frac{d\phi}{dz'}$ makes sense, as $\phi=z\circ z'^{-1}:z'(U \cap U')\rightarrow z(U \cap U')$ is a biholomorphic map of a complex variable, but how does $\frac{dz}{dz'} = \frac{d\phi}{dz'}$? In fact I'm not really sure what $\frac{dz}{dz'}$ actually represents. –  A.P. Jan 25 '13 at 23:00
    
@Silencer, sorry for my unclear notation. $\frac{dz}{dz'}$ is by definition $\frac{d\phi}{dz'}$ here. I have edited the answer to replace all of them by $\frac{d\phi}{dz'}$. –  user27126 Jan 26 '13 at 2:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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