Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

By real curve, we mean a Riemann surface $X$ together with an anti-holomorphic involution $\sigma : X\rightarrow X$. Let $S$ be a finite subset of $X$. For each point $x\in S$, we associate a positive integer $m_x\geq 2$. Then by Riemann existence theorem, there exists a universal covering $\pi : Y\rightarrow X$ such that $S$ is the branch locus of $\pi$ and $m_x$ is ramification index of $\pi$ over $x\in S$.

Question:- Is it true that the involution $\sigma$ can be lift to $Y$ making it real curve?

share|improve this question
add comment

1 Answer

In general, the property of being real is not preserved by finite coverings, not even by Galois ones.

For instance, take $X= \mathbb{P}^1$, which is a real curve with the anti-holomorphic involution $\sigma(z) = \bar{z}$.

Now every elliptic curve $Y$ is a double cover of $\mathbb{P}^1$ branched in four points, but not all elliptic curves have a real structure.

More precisely, $\sigma$ can be lifted to $Y$ if and only if the double cover $Y \to \mathbb{P}^1$ admits an affine equation of the form

$w^2=(z-a)(z-\bar{a})(z-b)(z-\bar{b}) \quad a,b \in \mathbb{C}$.

In this case there are exactly two liftings, namely

$(z,w) \to (\bar{z}, \bar{w}) \quad $ and $ \quad (z,w) \to (\bar{z}, -\bar{w})$.

share|improve this answer
1  
In general you can lift the involution to a double cover of $P^1$ if $S$ is invariant for $\sigma$. I think one can work out a similar condition for more general covers of $P^1$, but you have to give not only the branching order but the monodromy at each point of S, as one actually does in the Riemann construction. –  rita Dec 3 '10 at 14:30
    
@rita: it seems to me that you would at least have to require that the monodromy can be made $\sigma$-equivariant. Is the point that this can always be done for double covers? Also, what does it precisely mean to make the monodromy $\sigma$-equivariant, anyway? –  Vivek Shende Dec 4 '10 at 15:59
    
@Vivek: If $S$ is invariant for $\sigma$, then $\sigma$ restricts to an anti-holomorphic involution $\sigma \colon X-S \to X-S$, which induces an isomorphism $\sigma_* \colon \pi_1(X-S) \to \pi_1(X-S)$. Therefore a necessary condition for lifting $\sigma$ to $Y$ is that the monodromy representation $p \colon \pi_1(X-S) \to S_n$ satisfies $p \circ \sigma_*=p$, which is the $\sigma$-equivariance condition you are looking for. I suspect that this condition is also sufficient; at any rate, it is always satisfied for double covers. –  Francesco Polizzi Dec 4 '10 at 16:29
    
@Francesco: thanks for clarifying my comment. Since the monodromy rep. is defined (I think) only up to inner automorphisms of $S_n$, my impression is that it might be enough to ask that $p$ and $p\circ \sigma_*$ differ by an inner automorphism of $S_n$. I would also pick a real base point for $\pi_1(X-S)$. –  rita Dec 4 '10 at 18:15
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.