Finite etale covers of products of curves Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Deligne-Simpson problem in the symmetric groupAn algebraic vector bundle is trivialized by open sets. How many does one need?Is the following morphism etaleGood covers on complex algebraic varieties with normal crossings singularitiesisogeny and congruence subgroupExample: Principal G bundle that is not Zariski locally trivial, G not finite and G simply connectedPerfectoid universal coversUniformization over finite fields?How to test a divisor is or not the fixed part of a linear system on algebraic surfaceWhy are modular curves non-trivial covers of the $j$-line

Finite etale covers of products of curves



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Deligne-Simpson problem in the symmetric groupAn algebraic vector bundle is trivialized by open sets. How many does one need?Is the following morphism etaleGood covers on complex algebraic varieties with normal crossings singularitiesisogeny and congruence subgroupExample: Principal G bundle that is not Zariski locally trivial, G not finite and G simply connectedPerfectoid universal coversUniformization over finite fields?How to test a divisor is or not the fixed part of a linear system on algebraic surfaceWhy are modular curves non-trivial covers of the $j$-line










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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 $mathbbC$.




Let $C_1, C_2 subset mathbbP^1$ 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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$endgroup$







  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 9:01











  • $begingroup$
    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
    Apr 10 at 9:08











  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    Apr 10 at 9:58











  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 10:44















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 $mathbbC$.




Let $C_1, C_2 subset mathbbP^1$ 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 $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 9:01











  • $begingroup$
    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
    Apr 10 at 9:08











  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    Apr 10 at 9:58











  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 10:44













5












5








5


2



$begingroup$


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 $mathbbC$.




Let $C_1, C_2 subset mathbbP^1$ 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 $mathbbC$.




Let $C_1, C_2 subset mathbbP^1$ 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 Apr 10 at 8:11









Daniel LoughranDaniel Loughran

11.2k22672




11.2k22672







  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 9:01











  • $begingroup$
    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
    Apr 10 at 9:08











  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    Apr 10 at 9:58











  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 10:44












  • 2




    $begingroup$
    I do not think so, at least in general. Think of a double cover $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 9:01











  • $begingroup$
    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
    Apr 10 at 9:08











  • $begingroup$
    @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
    $endgroup$
    – Daniel Loughran
    Apr 10 at 9:58











  • $begingroup$
    @DanielLoughran: I will try
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 10:44







2




2




$begingroup$
I do not think so, at least in general. Think of a double cover $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
$endgroup$
– Francesco Polizzi
Apr 10 at 9:01





$begingroup$
I do not think so, at least in general. Think of a double cover $barX$ of $mathbbP^1 times mathbbP^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=mathbbP^1$ minus two points). The general fibres of the composition $barX to mathbbP^1 times mathbbP^1 to mathbbP^1$ are smooth double cover of $mathbbP^1$ branched at two points, hence they are again isomorphic to $mathbbP^1$, and so the fibres of your composition are isomorphic to $mathbbP^1$ minus the ramification, i.e. $mathbbP^1$ minus two points.
$endgroup$
– Francesco Polizzi
Apr 10 at 9:01













$begingroup$
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
Apr 10 at 9:08





$begingroup$
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
Apr 10 at 9:08













$begingroup$
@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
$endgroup$
– Daniel Loughran
Apr 10 at 9:58





$begingroup$
@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
$endgroup$
– Daniel Loughran
Apr 10 at 9:58













$begingroup$
@DanielLoughran: I will try
$endgroup$
– Francesco Polizzi
Apr 10 at 10:44




$begingroup$
@DanielLoughran: I will try
$endgroup$
– Francesco Polizzi
Apr 10 at 10:44










2 Answers
2






active

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



Take a double cover $barf colon barX to mathbbP^1 times mathbbP^1$, 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 $mathrmPic(mathbbP^1 times mathbbP^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbbP^1$ - two points.



If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbbC times (mathbbC-0, , 1)^2$ 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 $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ have connected fibres, the general one being isomorphic to $mathbbP^1$ (double cover of $mathbbP^1$ 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 $mathbbP^1$ above minus the ramification, i.e. $mathbbP^1$ minus two points, that is clearly connected.



In the same vein, choosing as $B subset mathbbP^1 times mathbbP^1$ a divisor of type $$B = sum_i=1^2g+2 L_i + sum_i=1^2g+2 M_i,$$
both compositions $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ 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 $mathbbP^1$ 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 $mathbbP^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    Apr 10 at 15:55











  • $begingroup$
    @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 16:02


















7












$begingroup$

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 $#(mathbfP^1setminus C_i) - 1$ 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
    2






    active

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    active

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    active

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    6












    $begingroup$

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



    Take a double cover $barf colon barX to mathbbP^1 times mathbbP^1$, 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 $mathrmPic(mathbbP^1 times mathbbP^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbbP^1$ - two points.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbbC times (mathbbC-0, , 1)^2$ 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 $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ have connected fibres, the general one being isomorphic to $mathbbP^1$ (double cover of $mathbbP^1$ 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 $mathbbP^1$ above minus the ramification, i.e. $mathbbP^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbbP^1 times mathbbP^1$ a divisor of type $$B = sum_i=1^2g+2 L_i + sum_i=1^2g+2 M_i,$$
    both compositions $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ 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 $mathbbP^1$ 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 $mathbbP^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      Apr 10 at 15:55











    • $begingroup$
      @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      Apr 10 at 16:02















    6












    $begingroup$

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



    Take a double cover $barf colon barX to mathbbP^1 times mathbbP^1$, 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 $mathrmPic(mathbbP^1 times mathbbP^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbbP^1$ - two points.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbbC times (mathbbC-0, , 1)^2$ 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 $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ have connected fibres, the general one being isomorphic to $mathbbP^1$ (double cover of $mathbbP^1$ 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 $mathbbP^1$ above minus the ramification, i.e. $mathbbP^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbbP^1 times mathbbP^1$ a divisor of type $$B = sum_i=1^2g+2 L_i + sum_i=1^2g+2 M_i,$$
    both compositions $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ 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 $mathbbP^1$ 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 $mathbbP^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      Apr 10 at 15:55











    • $begingroup$
      @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      Apr 10 at 16:02













    6












    6








    6





    $begingroup$

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



    Take a double cover $barf colon barX to mathbbP^1 times mathbbP^1$, 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 $mathrmPic(mathbbP^1 times mathbbP^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbbP^1$ - two points.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbbC times (mathbbC-0, , 1)^2$ 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 $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ have connected fibres, the general one being isomorphic to $mathbbP^1$ (double cover of $mathbbP^1$ 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 $mathbbP^1$ above minus the ramification, i.e. $mathbbP^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbbP^1 times mathbbP^1$ a divisor of type $$B = sum_i=1^2g+2 L_i + sum_i=1^2g+2 M_i,$$
    both compositions $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ 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 $mathbbP^1$ 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 $barf colon barX to mathbbP^1 times mathbbP^1$, 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 $mathrmPic(mathbbP^1 times mathbbP^1)$, and corresponds to an étale cover $f colon X to C_1 times C_2$, where each $C_i$ is $mathbbP^1$ - two points.



    If these points are (say) $0$ and $1$ in both factors, then the equation for $X subset mathbbC times (mathbbC-0, , 1)^2$ 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 $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ have connected fibres, the general one being isomorphic to $mathbbP^1$ (double cover of $mathbbP^1$ 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 $mathbbP^1$ above minus the ramification, i.e. $mathbbP^1$ minus two points, that is clearly connected.



    In the same vein, choosing as $B subset mathbbP^1 times mathbbP^1$ a divisor of type $$B = sum_i=1^2g+2 L_i + sum_i=1^2g+2 M_i,$$
    both compositions $$barX to mathbbP^1 times mathbbP^1 to mathbbP^1$$ 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 $mathbbP^1$ 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














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 10 at 12:23

























    answered Apr 10 at 11:13









    Francesco PolizziFrancesco Polizzi

    48.7k3132214




    48.7k3132214











    • $begingroup$
      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 $mathbbP^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      Apr 10 at 15:55











    • $begingroup$
      @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      Apr 10 at 16:02
















    • $begingroup$
      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 $mathbbP^1$ are conic bundles on the surface.
      $endgroup$
      – Daniel Loughran
      Apr 10 at 15:55











    • $begingroup$
      @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
      $endgroup$
      – Francesco Polizzi
      Apr 10 at 16:02















    $begingroup$
    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 $mathbbP^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    Apr 10 at 15:55





    $begingroup$
    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 $mathbbP^1$ are conic bundles on the surface.
    $endgroup$
    – Daniel Loughran
    Apr 10 at 15:55













    $begingroup$
    @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 16:02




    $begingroup$
    @DanielLoughran: You are welcome. Note that $barX$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$.
    $endgroup$
    – Francesco Polizzi
    Apr 10 at 16:02











    7












    $begingroup$

    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 $#(mathbfP^1setminus C_i) - 1$ 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











    $endgroup$

















      7












      $begingroup$

      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 $#(mathbfP^1setminus C_i) - 1$ 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











      $endgroup$















        7












        7








        7





        $begingroup$

        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 $#(mathbfP^1setminus C_i) - 1$ 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 $#(mathbfP^1setminus C_i) - 1$ 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 Apr 10 at 17:48

























        answered Apr 10 at 16:57









        Piotr AchingerPiotr Achinger

        8,62212854




        8,62212854



























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