|
5
|
|
edited Aug 2 2010 at 7:50 |
My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take
$X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).
EXAMPLE 1.
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,
$B=B_1 + B_2$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points;
set $p_{ij}:=B_i \cap B_j$.
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_3$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.
In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate covers" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; any each $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.
|
|
|
|
|
4
|
|
edited Aug 2 2010 at 7:37 |
My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take
$X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).
EXAMPLE 1.
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,
$B=B_1 + B_2$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points;
set $p_{ij}:=B_i \cap B_j$.
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_2$, B_3$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.
In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate coveringscovers" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; any $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.
|
|
|
|
|
3
|
|
edited Aug 1 2010 at 23:08 |
My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take
$X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).
EXAMPLE 1.
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,
$B=B_1 + B_2$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points;
set $p_{ij}:=B_i \cap B_j$.
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.
In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate coverings" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; any $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.
|
|
|
|
2
|
|
edited Aug 1 2010 at 22:57 |
My answer does not intend to completely solve your problem, but I just wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take
$X=\mathbb{P}^2$ (if you want your a map over $\mathbb{A}^1$ with fibre $\mathbb{P}^1$, take an affine chart and just blow-up a point).
EXAMPLE 1.
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,
$B=B_1 + B_2$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise at in three distinct points;
set $p_{ij}:=B_i \cap B_j$.
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.
In this case $Y$ is smoothand , in fact it is isomorphic to $\mathbb{P}^2$; however\mathbb{P}^2$. However, there are three "intermediate coverings" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivialsubgroups non-trivial subgroups of $G$; any $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.
|
|
|
|
|
1
|
|
answered Aug 1 2010 at 22:51 |
My answer does not intend to completely solve your problem, but I just wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take
$X=\mathbb{P}^2$ (if you want your map over $\mathbb{A}^1$ with fibre $\mathbb{P}^1$, take an affine chart and blow-up a point).
EXAMPLE 1.
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,
$B=B_1 + B_2$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise at three distinct points;
set $p_{ij}:=B_i \cap B_j$.
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.
In this case $Y$ is smooth and isomorphic to $\mathbb{P}^2$; however, there are three "intermediate coverings" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivialsubgroups of $G$; any $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.
|
|
|
