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?


1 Answer 1


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})$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – rita
    Dec 3, 2010 at 14:30
  • $\begingroup$ @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? $\endgroup$ Dec 4, 2010 at 15:59
  • $\begingroup$ @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. $\endgroup$ Dec 4, 2010 at 16:29
  • $\begingroup$ @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 $\pi_1(X-S)$. $\endgroup$
    – rita
    Dec 4, 2010 at 18:15

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