Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that the surface $y^2-(z-a_1)\ldots(z-a_n)$ is a Riemann Surface (that is the Riemann surface of $\sqrt{P(z)}$ with $P(z)=(z-a_1)\ldots(z-a_n)$) of genus $g$ and that $g=\mathrm{dim}(\Omega(X))$, with $\Omega(X)$ the holomorphics 1-forms. In "Lectures on Riemann Surfaces" (17.15), Forster says that one has the basis $(\omega_j)_{j=1,\ldots,g}$ where $\omega_j=\frac{z^{j-1}dz}{\sqrt{P(z)}}$.

My problem is that when I prove the independence of this sequence I didn't use that the indices stopped at $g$ that is I had an infinite basis: if we have $\sum_j \lambda_j\omega_j=0$ then $\sum_j \lambda_j z^{j-1}=0$ and so with a Vandermonde determinant I conclude that the $\lambda_j=0$.

Where is my mistake?

share|cite|improve this question
up vote 2 down vote accepted

You overlook the fact that from $j > g$, $\omega_j$ is no longer holomorphic, particularly at the point(s) with $z = \infty$ :

Suppose $n$ is even so that the surface has two points above $z = \infty$ (corresponding to $y = \pm z^{n/2}$). Then $g = n/2-1$. Pick $q = 1/z $. $q$ is a uniformizer at those two points, and $dq = -dz/z^2$. So $\omega_j / dq = - z^{j+1}/y = \pm z^{j+1-n/2} = \pm q^{-j-1+n/2}$. This has a pole at $z = \infty$ if and only if $-j-1+n/2 < 0$, which means $j > n/2-1 = g$.

If $n$ is odd, $g = (n+1)/2-1$, and there is a branch point at $z = \infty$. A uniformizer is now $q = y /z^{(n+1)/2} \sim 1/\sqrt z$. Then $z \sim q^{-2}$ has a pole of order $2$, $y \sim q^{-n}$ has a pole of order $n$, $dq/dz \sim z^{-3/2} \sim q^3$ has a zero of order $3$, and $\omega_j/dq = (z^{j-1}/y) (dz/dq) \sim q^{-2(j-1)+n-3} = q^{-2j+n-1}$. This has a pole if and only if $-2j+n-1 < 0$, which means $j > (n-1)/2 = g$.

share|cite|improve this answer

Your Answer


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.