1
$\begingroup$

I asked the following question in the mathstackexchange but I did not get an answer, probably the level of this question is not appropriate for mathoverflow, so I would like to apologize in advance if it is the case:

Let us consider the algebraic curve $$ f(z,w)=z(z-a^{-1})w^4-(z-a)^3=0 \ \ \ \ \ \ (*)$$

I want to understand the Riemann surface of this curve; strictly speaking, I want to find the genus, and neighborhood structure corresponding to singular points. One can easily see that the singular points are $0, a, a^{-1}$ and $\infty$. It seems to me that all the places $(0,\infty)$, $(a^{-1}, \infty)$, $(a, 0)$ and $(\infty, \infty)$ all have 4-cyclic neighborhoods, i.e. the neighborhoods of these places on the Riemann surface go through all four sheets. my reason for this is that , say for $(0,\infty)$, $w=\infty$ is the only solution of the equation $f(0,w)=0$. (But it may happen that there are two disjoint ,except at $(0,\infty)$, 2-cyclic neighborhoods, right\wrong ?!)

what if we had a curve such that the equation $f(0,w)=0$ has two solutions , say $w_1$ and $w_2$, how do we know $(0,w_1)$ has a 2-cyclic nighborhood (and hence $(0,w_2)$ also has two) or possibly $(0,w_1)$ has a 3-cyclic nighborhood (and hence $(0,w_2)$ also has a simple 1-cyclic neighborhood) ?

And finally how I should find the genus of the Riemann surface corresponding to (*) ?

I appreciate any help. Thanks

$\endgroup$

1 Answer 1

3
$\begingroup$

The genus $g$ of a Riemann surface is found from the Riemann-Hurwitz formula: $$2g-2=\sum(n_k-1)-2d,$$ where $d$ is the number of sheets, $n_j$ are the orders of ramification points. In your case, $d=4$, and there are $4$ ramification points, all of order $4$. So the genus is $3$.

The proof of the Riemann-Hurwitz formula is elementary: you make a triangulation of the $z$-sphere so that your ramification points are among the vertices, and pull it back on your Riemann surface. Every triangle and every edge has $d$ preimages, and every vertex over $a$ has $d-\Sigma(n_j-1)$ preimages, where the summation is over ramification points lying over $a$. (Your calculation of $n_j$ was is correct).

Then use Euler's theorem expressing the Euler characteristic in terms of the numbers of vertices, edges and triangles in a triangulation: $\chi=v-e+f$, where $v,e,f$ are the numbers of vertices, edges and triangles in a triangulation. The Euler characteristic is related to the genus by the formula $\chi=2-2g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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