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?
ag.algebraic-geometry at.algebraic-topology fundamental-group covering-spaces etale-covers
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add a comment |
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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?
ag.algebraic-geometry at.algebraic-topology fundamental-group covering-spaces etale-covers
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2
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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.
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– Francesco Polizzi
21 hours ago
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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...
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– Francesco Polizzi
21 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago
add a comment |
$begingroup$
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?
ag.algebraic-geometry at.algebraic-topology fundamental-group covering-spaces etale-covers
$endgroup$
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
ag.algebraic-geometry at.algebraic-topology fundamental-group covering-spaces etale-covers
asked 22 hours ago
Daniel LoughranDaniel Loughran
11.1k22672
11.1k22672
2
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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.
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– Francesco Polizzi
21 hours ago
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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...
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– Francesco Polizzi
21 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago
add a comment |
2
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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.
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– Francesco Polizzi
21 hours ago
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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...
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– Francesco Polizzi
21 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago
2
2
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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.
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– Francesco Polizzi
21 hours ago
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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.
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– 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...
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– Francesco Polizzi
21 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago
add a comment |
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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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.
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– Daniel Loughran
14 hours ago
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@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 $B$.
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– Francesco Polizzi
14 hours ago
add a comment |
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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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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
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.
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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.
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– Daniel Loughran
14 hours ago
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@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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 12 hours ago
answered 13 hours ago
Piotr AchingerPiotr Achinger
8,59212854
8,59212854
add a comment |
add a comment |
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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.
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– Francesco Polizzi
21 hours ago
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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...
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– Francesco Polizzi
21 hours ago
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@Francesco Polizzi: Are you able to provide an answer with an explicit counter-example?
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– Daniel Loughran
20 hours ago
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@DanielLoughran: I will try
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– Francesco Polizzi
19 hours ago