Finite etale covers of products of curves












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Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $mathbb{C}$.




Let $C_1, C_2 subset mathbb{P}^1$ be non-empty open subsets and $f: X to C_1 times C_2$ a non-trivial finite etale cover. Does there exist $iin {1,2}$ such that the composition $X to C_1 times C_2 to C_i$ has non-connected fibres?











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  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    20 hours ago












  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    19 hours ago
















5












$begingroup$


Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $mathbb{C}$.




Let $C_1, C_2 subset mathbb{P}^1$ be non-empty open subsets and $f: X to C_1 times C_2$ a non-trivial finite etale cover. Does there exist $iin {1,2}$ such that the composition $X to C_1 times C_2 to C_i$ has non-connected fibres?











share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    20 hours ago












  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    19 hours ago














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$begingroup$


Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $mathbb{C}$.




Let $C_1, C_2 subset mathbb{P}^1$ be non-empty open subsets and $f: X to C_1 times C_2$ a non-trivial finite etale cover. Does there exist $iin {1,2}$ such that the composition $X to C_1 times C_2 to C_i$ has non-connected fibres?











share|cite|improve this question









$endgroup$




Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $mathbb{C}$.




Let $C_1, C_2 subset mathbb{P}^1$ be non-empty open subsets and $f: X to C_1 times C_2$ a non-trivial finite etale cover. Does there exist $iin {1,2}$ such that the composition $X to C_1 times C_2 to C_i$ has non-connected fibres?








ag.algebraic-geometry at.algebraic-topology fundamental-group covering-spaces etale-covers






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asked 22 hours ago









Daniel LoughranDaniel Loughran

11.1k22672




11.1k22672








  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    20 hours ago












  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    19 hours ago














  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
    $endgroup$
    – Francesco Polizzi
    21 hours ago












  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    20 hours ago












  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    19 hours ago








2




2




$begingroup$
I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
$endgroup$
– Francesco Polizzi
21 hours ago






$begingroup$
I do not think so, at least in general. Think of a double cover $bar{X}$ of $mathbb{P}^1 times mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbb{P}^1$ minus two points). The general fibres of the composition $bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$ are smooth double cover of $mathbb{P}^1$ branched at two points, hence they are again isomorphic to $mathbb{P}^1$, and so the fibres of your composition are isomorphic to $mathbb{P}^1$ minus the ramification, i.e. $mathbb{P}^1$ minus two points.
$endgroup$
– Francesco Polizzi
21 hours ago














$begingroup$
If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
$endgroup$
– Francesco Polizzi
21 hours ago






$begingroup$
If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on...
$endgroup$
– Francesco Polizzi
21 hours ago














$begingroup$
@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
$endgroup$
– Daniel Loughran
20 hours ago






$begingroup$
@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
$endgroup$
– Daniel Loughran
20 hours ago














$begingroup$
@DanielLoughran: I will try
$endgroup$
– Francesco Polizzi
19 hours ago




$begingroup$
@DanielLoughran: I will try
$endgroup$
– Francesco Polizzi
19 hours ago










2 Answers
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The answer is no, at least in general, as shown by the following counterexample.



Take a double cover $bar{f} colon bar{X} to mathbb{P}^1 times mathbb{P}^1$, branched over a reducible
curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).



Such a cover exists because $B$ is $2$-divisible in $mathrm{Pic}(mathbb{P}^1 times mathbb{P}^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbb{P}^1$ - {two points}.



If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbb{C} times (mathbb{C}-{0, , 1})^2$ is
$$z^2 = xy(x-1)(y-1), quad f(z, (x, ,y)) = (x,, y).$$



It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $mathbb{P}^1$ (double cover of $mathbb{P}^1$ branched at two points).



Then both compositions $$X to C_1 times C_2 to C_i$$ have connected fibres, the general one being isomorphic to the $mathbb{P}^1$ above minus the ramification, i.e. $mathbb{P}^1$ minus two points, that is clearly connected.



In the same vein, choosing as $B subset mathbb{P}^1 times mathbb{P}^1$ a divisor of type $$B = sum_{i=1}^{2g+2} L_i + sum_{i=1}^{2g+2} M_i,$$
both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $Sigma_g$ of genus $g$, and so both compositions $$X to C_1 times C_2 to C_i$$ (here each $C_i$ is $mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $Sigma_g$ minus $2g+2$ distinct points.






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$endgroup$













  • $begingroup$
    Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    14 hours ago












  • $begingroup$
    @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    14 hours ago



















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The question already has a beautiful answer, but here's a different point of view which you may find helpful.



Let $F_i = pi_1(C_i, x_i)$, which is a free group on $#(mathbf{P}^1setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = pi_1(C_1times C_2, x_1times x_2) = F_1times F_2$.



A finite etale cover of $C_i$ or $C_1times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.



Let $sigma_i colon C_ito C_1times C_2$ be the section $sigma_1(x) = (x, x_2)$, $sigma_2(x) = (x_1, x)$. Then for a finite etale cover $Xto C_1times C_2$, the composition $Xto C_1times C_2to C_i$ has connected fibres if and only if the pull-back of $X$ along $sigma_{2-i}$ is connected.



So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples.



The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.






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    2 Answers
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    2 Answers
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    6












    $begingroup$

    The answer is no, at least in general, as shown by the following counterexample.



    Take a double cover $bar{f} colon bar{X} to mathbb{P}^1 times mathbb{P}^1$, branched over a reducible
    curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).



    Such a cover exists because $B$ is $2$-divisible in $mathrm{Pic}(mathbb{P}^1 times mathbb{P}^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbb{P}^1$ - {two points}.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbb{C} times (mathbb{C}-{0, , 1})^2$ is
    $$z^2 = xy(x-1)(y-1), quad f(z, (x, ,y)) = (x,, y).$$



    It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $mathbb{P}^1$ (double cover of $mathbb{P}^1$ branched at two points).



    Then both compositions $$X to C_1 times C_2 to C_i$$ have connected fibres, the general one being isomorphic to the $mathbb{P}^1$ above minus the ramification, i.e. $mathbb{P}^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbb{P}^1 times mathbb{P}^1$ a divisor of type $$B = sum_{i=1}^{2g+2} L_i + sum_{i=1}^{2g+2} M_i,$$
    both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $Sigma_g$ of genus $g$, and so both compositions $$X to C_1 times C_2 to C_i$$ (here each $C_i$ is $mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $Sigma_g$ minus $2g+2$ distinct points.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      14 hours ago












    • $begingroup$
      @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      14 hours ago
















    6












    $begingroup$

    The answer is no, at least in general, as shown by the following counterexample.



    Take a double cover $bar{f} colon bar{X} to mathbb{P}^1 times mathbb{P}^1$, branched over a reducible
    curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).



    Such a cover exists because $B$ is $2$-divisible in $mathrm{Pic}(mathbb{P}^1 times mathbb{P}^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbb{P}^1$ - {two points}.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbb{C} times (mathbb{C}-{0, , 1})^2$ is
    $$z^2 = xy(x-1)(y-1), quad f(z, (x, ,y)) = (x,, y).$$



    It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $mathbb{P}^1$ (double cover of $mathbb{P}^1$ branched at two points).



    Then both compositions $$X to C_1 times C_2 to C_i$$ have connected fibres, the general one being isomorphic to the $mathbb{P}^1$ above minus the ramification, i.e. $mathbb{P}^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbb{P}^1 times mathbb{P}^1$ a divisor of type $$B = sum_{i=1}^{2g+2} L_i + sum_{i=1}^{2g+2} M_i,$$
    both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $Sigma_g$ of genus $g$, and so both compositions $$X to C_1 times C_2 to C_i$$ (here each $C_i$ is $mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $Sigma_g$ minus $2g+2$ distinct points.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      14 hours ago












    • $begingroup$
      @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      14 hours ago














    6












    6








    6





    $begingroup$

    The answer is no, at least in general, as shown by the following counterexample.



    Take a double cover $bar{f} colon bar{X} to mathbb{P}^1 times mathbb{P}^1$, branched over a reducible
    curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).



    Such a cover exists because $B$ is $2$-divisible in $mathrm{Pic}(mathbb{P}^1 times mathbb{P}^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbb{P}^1$ - {two points}.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbb{C} times (mathbb{C}-{0, , 1})^2$ is
    $$z^2 = xy(x-1)(y-1), quad f(z, (x, ,y)) = (x,, y).$$



    It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $mathbb{P}^1$ (double cover of $mathbb{P}^1$ branched at two points).



    Then both compositions $$X to C_1 times C_2 to C_i$$ have connected fibres, the general one being isomorphic to the $mathbb{P}^1$ above minus the ramification, i.e. $mathbb{P}^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbb{P}^1 times mathbb{P}^1$ a divisor of type $$B = sum_{i=1}^{2g+2} L_i + sum_{i=1}^{2g+2} M_i,$$
    both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $Sigma_g$ of genus $g$, and so both compositions $$X to C_1 times C_2 to C_i$$ (here each $C_i$ is $mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $Sigma_g$ minus $2g+2$ distinct points.






    share|cite|improve this answer











    $endgroup$



    The answer is no, at least in general, as shown by the following counterexample.



    Take a double cover $bar{f} colon bar{X} to mathbb{P}^1 times mathbb{P}^1$, branched over a reducible
    curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).



    Such a cover exists because $B$ is $2$-divisible in $mathrm{Pic}(mathbb{P}^1 times mathbb{P}^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbb{P}^1$ - {two points}.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbb{C} times (mathbb{C}-{0, , 1})^2$ is
    $$z^2 = xy(x-1)(y-1), quad f(z, (x, ,y)) = (x,, y).$$



    It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $mathbb{P}^1$ (double cover of $mathbb{P}^1$ branched at two points).



    Then both compositions $$X to C_1 times C_2 to C_i$$ have connected fibres, the general one being isomorphic to the $mathbb{P}^1$ above minus the ramification, i.e. $mathbb{P}^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbb{P}^1 times mathbb{P}^1$ a divisor of type $$B = sum_{i=1}^{2g+2} L_i + sum_{i=1}^{2g+2} M_i,$$
    both compositions $$bar{X} to mathbb{P}^1 times mathbb{P}^1 to mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $Sigma_g$ of genus $g$, and so both compositions $$X to C_1 times C_2 to C_i$$ (here each $C_i$ is $mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $Sigma_g$ minus $2g+2$ distinct points.







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    edited 18 hours ago

























    answered 19 hours ago









    Francesco PolizziFrancesco Polizzi

    48.7k3130212




    48.7k3130212












    • $begingroup$
      Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      14 hours ago












    • $begingroup$
      @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      14 hours ago


















    • $begingroup$
      Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      14 hours ago












    • $begingroup$
      @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      14 hours ago
















    $begingroup$
    Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    14 hours ago






    $begingroup$
    Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $mathbb{P}^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    14 hours ago














    $begingroup$
    @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    14 hours ago




    $begingroup$
    @DanielLoughran: You are welcome. Note that $bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    14 hours ago











    5












    $begingroup$

    The question already has a beautiful answer, but here's a different point of view which you may find helpful.



    Let $F_i = pi_1(C_i, x_i)$, which is a free group on $#(mathbf{P}^1setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = pi_1(C_1times C_2, x_1times x_2) = F_1times F_2$.



    A finite etale cover of $C_i$ or $C_1times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.



    Let $sigma_i colon C_ito C_1times C_2$ be the section $sigma_1(x) = (x, x_2)$, $sigma_2(x) = (x_1, x)$. Then for a finite etale cover $Xto C_1times C_2$, the composition $Xto C_1times C_2to C_i$ has connected fibres if and only if the pull-back of $X$ along $sigma_{2-i}$ is connected.



    So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples.



    The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      The question already has a beautiful answer, but here's a different point of view which you may find helpful.



      Let $F_i = pi_1(C_i, x_i)$, which is a free group on $#(mathbf{P}^1setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = pi_1(C_1times C_2, x_1times x_2) = F_1times F_2$.



      A finite etale cover of $C_i$ or $C_1times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.



      Let $sigma_i colon C_ito C_1times C_2$ be the section $sigma_1(x) = (x, x_2)$, $sigma_2(x) = (x_1, x)$. Then for a finite etale cover $Xto C_1times C_2$, the composition $Xto C_1times C_2to C_i$ has connected fibres if and only if the pull-back of $X$ along $sigma_{2-i}$ is connected.



      So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples.



      The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        The question already has a beautiful answer, but here's a different point of view which you may find helpful.



        Let $F_i = pi_1(C_i, x_i)$, which is a free group on $#(mathbf{P}^1setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = pi_1(C_1times C_2, x_1times x_2) = F_1times F_2$.



        A finite etale cover of $C_i$ or $C_1times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.



        Let $sigma_i colon C_ito C_1times C_2$ be the section $sigma_1(x) = (x, x_2)$, $sigma_2(x) = (x_1, x)$. Then for a finite etale cover $Xto C_1times C_2$, the composition $Xto C_1times C_2to C_i$ has connected fibres if and only if the pull-back of $X$ along $sigma_{2-i}$ is connected.



        So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples.



        The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.






        share|cite|improve this answer











        $endgroup$



        The question already has a beautiful answer, but here's a different point of view which you may find helpful.



        Let $F_i = pi_1(C_i, x_i)$, which is a free group on $#(mathbf{P}^1setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = pi_1(C_1times C_2, x_1times x_2) = F_1times F_2$.



        A finite etale cover of $C_i$ or $C_1times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.



        Let $sigma_i colon C_ito C_1times C_2$ be the section $sigma_1(x) = (x, x_2)$, $sigma_2(x) = (x_1, x)$. Then for a finite etale cover $Xto C_1times C_2$, the composition $Xto C_1times C_2to C_i$ has connected fibres if and only if the pull-back of $X$ along $sigma_{2-i}$ is connected.



        So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$? It is very easy to construct such examples.



        The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 12 hours ago

























        answered 13 hours ago









        Piotr AchingerPiotr Achinger

        8,59212854




        8,59212854






























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