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Dec 4, 2010 at 18:15 comment added rita @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 $\pi_1(X-S)$.
Dec 4, 2010 at 16:29 comment added Francesco Polizzi @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.
Dec 4, 2010 at 15:59 comment added Vivek Shende @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?
Dec 3, 2010 at 21:26 history edited Francesco Polizzi CC BY-SA 2.5
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Dec 3, 2010 at 14:30 comment added rita 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.
Dec 3, 2010 at 14:15 history edited Francesco Polizzi CC BY-SA 2.5
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Dec 3, 2010 at 13:12 history answered Francesco Polizzi CC BY-SA 2.5