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.

Is it true that for any non-orientable Riemannian surface of genus 2 there exists Conformal mapping of degree two to a projective plane?

Also, is the following argument works? Given any Riemann surface $M$ of genus two with a free involition $\sigma$ we can find a meromorphic $f$ function with the only pole of second order (existence of Weierstrass point). Then we consider the function $f +\sigma^*f$, it is a conformal mapping to a Riemann sphere of degree 4 and it descends to a factor $M/\sigma$. As a result we have a conformal mapping of order 2 of any non-orientable to a Riemann sphere. I have some suspicions about the result, but do not seem to find an error.

share|improve this question
    
What's your definition of a conformal mapping, especially in view of the fact that your maps are branched covers? If it means a map preserving nonoriented angles away from branch-points, then the answer is negative: Such map does not exist (generically) once $\chi(M)<-2$. –  studiosus Oct 19 '13 at 21:41
add comment

1 Answer

Judging from your proposed solution, when you refer to a "non-orientable Riemann surface $M$ of genus 2" what you have in mind is an orientable genus 2 surface equipped with a fixed-point-free involution $\tau$ which reverses orientation. If this is the case, one can argue as follows. Let $J$ be the hyperelliptic involution of $M$. It is known that $\tau$ and $J$ necessarily commute. Therefore $\tau$ descends to the quotient of $M$ by $J$, i.e., the Riemann sphere. The quotient of the Riemann sphere by $\tau$ gives the real projective plane. All maps involved are locally holomorphic or anti-holomorphic, giving the result. A more detailed discussion may be found here.

share|improve this answer
add comment

Your Answer

 
discard

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.