MathOverflow is a question and answer site for professional mathematicians. 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 am looking for an algorithm to obtain the period matrix tau= (tau1 & tau12\ tau12 & tau2) from the Igusa invariants of a genus two curve. More precisely:

I consider a family of genus two curves given in hyperlliptic form y^2 = f(x,z1, z2, z3). Here, f is a polynomial of degree six in x and z1, z2, z3 are moduli parametrizing the family. Now I can compute the Igusa invariants I2(z1, z2, z3), ... I10(z1, z2, z3). But what I am interested in is the dependence tau(z1, z2, z3). I know that the absolute Igusa invariants x1, x2, x3 can be written in terms of Siegel modular forms of genus two, but it seems difficult to invert these relations.

Thanks a lot for your help in advance.


share|cite|improve this question
I don't know if I understand your question properly. I think you know the Thomae Theorem that expresses the 4th powers of theta constants in terms of the tableaux invariants of the 6 points on the line, don't you? – IMeasy Jan 31 '14 at 12:44
No to be honest I don't know it. Can you recommend me a good reference? So to perhaps reformulate my question: In the case of an elliptic curve, I can compute the J-function from the Weierstrass normal form and use the Fourier expansion in exp(2 pi i tau) of the J-function to compute the modulus tau in terms of the parameters parametrizing my curve. I am looking for an analog on in case of genus two curves. – Maximilian Feb 1 '14 at 13:15

It's par 5-6-7 of chapter eight of Asterisque 165 by Dolgachev and Ortland. About your question on EC, it seems reasonable, you should check especially par 6. Let us know what you find out!

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.