2
$\begingroup$

I am studying a particular class of real Riemann surfaces; these are branched covers of the complex plane or Riemann sphere where all of the branch points lie on the real line. I am wondering if there are any known restrictions on the period matrices of such surfaces. For example, are they pure imaginary in this case? I am quite sure that they are for 2 branches (hyperelliptic curves).

$\endgroup$
  • 1
    $\begingroup$ What do you mean by period matrix? If $C$ is a compact Riemann surface of genus $g$, you can always normalize with respect to a symplectic basis for $H_1(C,\mathbb{Z})$ so that the period matrix takes the form $\Omega = (I\ \ Z)$ where $I$ is the $g\times g$ identity matrix and $Z$ is symmetric with positive-definite imaginary part. If you are referring just to the matrix $Z$, then you might want to look at the paper "Period Matrices of Real Riemann Surfaces and Fundamental Domains" by Giavedoni. $\endgroup$ – Kevin Jul 7 '12 at 15:58
  • 1
    $\begingroup$ Isn't the condition that $Z$ is purely imaginary equivalent to the existence of a direct sum decomposition $H_1(C,\mathbf{Z})=H_1^+(C,\mathbf{Z}) \oplus H_1^-(C,\mathbf{Z})$ ? $\endgroup$ – François Brunault Jul 7 '12 at 21:25
  • $\begingroup$ You might then be able to construct explicitly such an adapted basis in the case of curves $y^2=f(x)$ with $f$ having only real roots. $\endgroup$ – François Brunault Jul 7 '12 at 21:26
  • $\begingroup$ Kevin -- Indeed I mean the matrix $Z$. Thanks for that reference (I have also found an old paper by Gross and Harris, "Real Algebraic Curves", which has some relevant results). Francois -- What do the two factors of that decomposition refer to explicitly? (For the curve $y^2 = f(z)$, I can indeed show that $Z$ is pure imaginary by using a standard basis for holomorphic differentials and appropriate canonical homology basis). $\endgroup$ – Albion Lawrence Jul 9 '12 at 21:05
  • $\begingroup$ @Albion Lawrence : $H_1^\pm$ refers to the $\pm 1$-eigenspace with respect to complex conjugation acting on $C$. $\endgroup$ – François Brunault Jul 10 '12 at 6:22

Your Answer

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

Browse other questions tagged or ask your own question.