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.

Consider $F(x,y,z)$ a non-singular homogeneous polynomial of degree $d$. If I consider the zero locus $X=\{[x:y:z]\in\mathbb{P}^2:F(x,y,z)=0\}$

How can I define a complex structure in $X$ in order to obtain a Riemann Surface? If I do it on the three open subsets $$ U_1=\{[x:y:z]\in X:x\neq 0\} \\ U_2=\{[x:y:z]\in X:y\neq 0\} \\ U_3=\{[x:y:z]\in X:z\neq 0\} $$ then I'm defining them as $$ \begin{array}{ccccc} \phi_i:&U_i&\rightarrow&\mathbb{C}\\ &[x_1:x_2:x_3]&\mapsto &\dfrac{x_j}{x_i},& \text{ for any fixed }j\neq i \end{array} $$ but with this definition how can $\phi_i$ be a local homeomorphism? I cannot guarantee local injectivity.

share|improve this question
    
Your $F$ has to be nonsingular. Use implicit function to solve $F(x,y,z) = 0$ locally. –  user27126 Apr 26 '13 at 1:53
    
That means, if I have $\frac{\partial F}{\partial x}(x_0,y_0,z_0)\neq 0$ I can find a function $h(y,z)$ defined on a neighbourhood of $(y_0,z_0)$ such that $h(y_0,z_0)=x_0$ and $F(h(y,z),y,z)=0$ This is easily done for the other two partial derivatives as well. Now, if $x_0\neq 0$ I can define two complex charts, $\phi_1:U_1\rightarrow\mathbb{C}$ such that $\phi_1([x:y:z])=\frac{y}{x}$ or $\phi_2:U_2\rightarrow\mathbb{C}$ such that $\phi_1([x:y:z])=\frac{z}{x}$. The cases where $y_0\neq 0$ and $z_0 \neq 0$ are similar; these charts are invertible since the derivative of the implicit (...) –  Marra Apr 26 '13 at 21:03
    
(...) function is non-zero locally, therefore I can guarantee local injectivity. Can someone tell me if this is correct? –  Marra Apr 26 '13 at 21:04

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.