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Let us consider a closed Riemann surface $\Sigma_b$ of genus $B$, and let $\Delta \subset \Sigma_b \times \Sigma_b$ be the diagonal.

If $G$ is a finite group, then any group epimorphism $$\varphi \colon \pi_1( \Sigma_b \times \Sigma_b - \Delta) \to G$$ induces, by Grauert-Remmert Extension Theorem, the existence of a compact complex manifold $X$ (actually, a complex projective surface), endowed with a Galois cover $$\pi \colon X \to \Sigma_b \times \Sigma_b$$ branched at most over $\Delta$.

Let us now denote by $a$ the involutory automorphism of $\Sigma_b \times \Sigma_b$ given by $a(x, \, y)=(y, \, x)$; it leaves $\Delta$ (pointwise) invariant, so we may ask the following

Question. Under which conditions on $\varphi$ the automorphism $a \colon \Sigma_b \times \Sigma_b \to \Sigma_b \times \Sigma_b $ lifts to an automorphism $\bar{a} \colon X \to X$?

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  • $\begingroup$ Are you claiming that $X$ is smooth? I think it is constructed as the normalisation of $C \times C$ in the étale Galois cover $U \to (C\times C)\setminus\Delta$, so wouldn't it a priori only be normal? (And conversely, any normal surface with a finite Galois map to $C \times C$ is generically a finite étale Galois cover, right?) Or is there something special about this situation that makes it smooth? $\endgroup$ Commented Aug 14, 2020 at 19:37
  • $\begingroup$ If $X \to Y$ is a (branched) analytic covering of complex analytic spaces, with $Y$ smooth, the singularities of $X$ lie above the singularities of the branch locus. In our situation, since $\Delta$ is smooth, the same is true for $X$. Of course, the existence of a complex structure on the closure of $U$ is strongly non-trivial (this is ensured by Grauert-Remmert, indeed). $\endgroup$ Commented Aug 14, 2020 at 19:43
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    $\begingroup$ Ah, you're right; in the type of example I had in mind the branch locus is singular (I was thinking of a cone $V(z^2-xy)$ projecting onto the $(x,y)$-plane, by way of toy example). $\endgroup$ Commented Aug 14, 2020 at 19:50
  • $\begingroup$ Right. In this example of double cover, the projective branch locus is given by two lines in $\mathbb{P}^2$, and the conical singularity lies over their cross point. $\endgroup$ Commented Aug 14, 2020 at 19:58

1 Answer 1

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In this post I will work in the greatest generality I can think of. In particular, fundamental group means étale fundamental group, but for varieties over $\mathbf C$ the same argument carries through using the topological fundamental group instead.

Lemma. Let $X$ and $Y$ be separated normal integral schemes, let $f \colon Y \to X$ be a finite and finitely presented separable Galois cover with group $G$, and let $a \colon X \to X$ be an automorphism. Let $U \subseteq X$ be the dense open locus where $f$ is étale, let $V = f^{-1}(U)$, let $\bar y \to V$ be a geometric point with image $\bar x \to U$, and let $\phi \colon \pi_1(U,\bar x) \twoheadrightarrow G$ be the surjection corresponding to the $G$-cover $V \to U$. Then the following are equivalent:

  1. There exists an automorphism $b \colon Y \to Y$ lifting $a$;
  2. There exists a dominant rational map $b \colon Y \to Y$ lifting $a$;
  3. The isomorphism $a$ takes $U$ to itself, and the pullback $V' \to U$ of $V \to U$ along $a$ is isomorphic to $V \to U$ (as étale $G$-covers of $U$);
  4. For any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$, the subgroups $\ker \phi$ and $\ker(\phi a_* \gamma_*)$ of $\pi_1(U,\bar x)$ are conjugate (see proof for precise statement).

Moreover, the set of such lifts is a $G$-bitorsor via pre- and post-composition of deck transformations.

Proof. For (1) $\Leftrightarrow$ (2), note that a dominant rational lift $b$ is automatically an automorphism. Indeed, given a commutative diagram $$\begin{array}{ccc}Y & \stackrel{b}\dashrightarrow & Y \\ \downarrow & & \downarrow \\ X & \underset a\to & X,\!\end{array}\tag{1}\label{1}$$ multiplicativity of function field degrees shows that $b$ is birational. Since $Y$ is the integral closure of $X$ in $K(Y)$, we conclude that $b$ is an isomorphism since normalisation is a functor.

Thus for (2) $\Leftrightarrow$ (3), we know that $b$ gives an isomorphism $V \to V$ lifting $a|_U \colon U \to U$. This is exactly the same thing as an isomorphism $V \to V'$ over $U$, where $V'$ is the pullback $$\begin{array}{ccc}V' & \to & V \\ \downarrow & & \downarrow \\ U & \stackrel a\to & U.\!\end{array}$$ Finally, for (3) $\Leftrightarrow$ (4), we note that the cover $V' \to U$ corresponds to the surjection $$\pi_1(U,a^*\bar x) \stackrel{a_*}\to \pi_1(U,\bar x) \stackrel \phi\twoheadrightarrow G.$$ Any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$ gives an identification \begin{align*} \gamma_* \colon \pi_1(U,\bar x) &\stackrel\sim\longrightarrow \pi_1(U,a^* \bar x)\\ [\alpha] &\longmapsto [\gamma^{-1}] \cdot [\alpha] \cdot [\gamma], \end{align*} well-defined up to conjugation. Under this identification, the surjection $\pi_1(U,a^*\bar x) \twoheadrightarrow G$ above corresponds to the surjection $\pi_1(U,\bar x) \twoheadrightarrow G$ given by $\phi a_* \gamma_*$. The induced cover is isomorphic to the cover $V \to U$ given by $\phi$ if and only if the kernels are conjugate (see e.g. [Munkres, Thm. 79.4] in the topological setting), proving (3) $\Leftrightarrow$ (4).

The final statement follows for example because $\operatorname{Isom}_X(Y,Y')$ is naturally a $G$-bitorsor, as $G$ agrees with both $\operatorname{Aut}_X(Y)$ and $\operatorname{Aut}_X(Y')$. (See also this post for a general discussion of Galois covers of normal schemes.) $\square$


References.

[Munkres] J. R. Munkres, Topology (second edition). Pearson, 2018.

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  • $\begingroup$ Thank you for the nice answer. I have a couple of remarks: (1) In the ramified case one does not have (as far I can see) an epimorphism $\pi_1(X) \to G$, but an epimorphism $\pi_1(X-B) \to G$, where $B$ is the branch locus. Are you sure that your reduction to the étale case is correct in this respect, or am I missing something? (2) Do you have a reference for the result on "Galois theory of the fundamental group" that you quote in the proof of equivalence $(1)-(3)$? $\endgroup$ Commented Aug 15, 2020 at 7:15
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    $\begingroup$ You're absolutely right, and in looking up good references I realised that my criterion wasn't quite right because I ignored basepoints. $\endgroup$ Commented Aug 15, 2020 at 20:13

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