Mixing
Computing singularities of a surface
$begingroup$
Let $Y$ be the abelian variety $mathbb{C}/mathbb{Z}[i] times mathbb{C}/mathbb{Z}[i] $. Let $X$ be the quotient of $Y$ by action of the group generated by the map $eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by ${e, -e, eta, -eta}$ where $e$ is the identity map.
How can we show that $X$ is in fact a rational surface and has 10 singularities? I do know we have to look at the fixed points by the subgroups generated by $eta$ but I have very little idea on how to proceed. Any hints given or links to papers describing this particular construction would be greatly appreciated!
Edit: I have been told that I need to look at the fixed points of the orbit, which in this case is simply the 0 class. How can I proceed?
algebraic-geometry surfaces complex-geometry singularity
asked Jan 29 at 11:57
thedilatedthedilated
7551617
$endgroup$
add a comment |
$begingroup$
Let $Y$ be the abelian variety $mathbb{C}/mathbb{Z}[i] times mathbb{C}/mathbb{Z}[i] $. Let $X$ be the quotient of $Y$ by action of the group generated by the map $eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by ${e, -e, eta, -eta}$ where $e$ is the identity map.
How can we show that $X$ is in fact a rational surface and has 10 singularities? I do know we have to look at the fixed points by the subgroups generated by $eta$ but I have very little idea on how to proceed. Any hints given or links to papers describing this particular construction would be greatly appreciated!
Edit: I have been told that I need to look at the fixed points of the orbit, which in this case is simply the 0 class. How can I proceed?
algebraic-geometry surfaces complex-geometry singularity
asked Jan 29 at 11:57
thedilatedthedilated
7551617
$endgroup$
1
$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30
add a comment |
$begingroup$
Let $Y$ be the abelian variety $mathbb{C}/mathbb{Z}[i] times mathbb{C}/mathbb{Z}[i] $. Let $X$ be the quotient of $Y$ by action of the group generated by the map $eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by ${e, -e, eta, -eta}$ where $e$ is the identity map.
How can we show that $X$ is in fact a rational surface and has 10 singularities? I do know we have to look at the fixed points by the subgroups generated by $eta$ but I have very little idea on how to proceed. Any hints given or links to papers describing this particular construction would be greatly appreciated!
Edit: I have been told that I need to look at the fixed points of the orbit, which in this case is simply the 0 class. How can I proceed?
algebraic-geometry surfaces complex-geometry singularity
asked Jan 29 at 11:57
thedilatedthedilated
7551617
$endgroup$
Let $Y$ be the abelian variety $mathbb{C}/mathbb{Z}[i] times mathbb{C}/mathbb{Z}[i] $. Let $X$ be the quotient of $Y$ by action of the group generated by the map $eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by ${e, -e, eta, -eta}$ where $e$ is the identity map.
How can we show that $X$ is in fact a rational surface and has 10 singularities? I do know we have to look at the fixed points by the subgroups generated by $eta$ but I have very little idea on how to proceed. Any hints given or links to papers describing this particular construction would be greatly appreciated!
Edit: I have been told that I need to look at the fixed points of the orbit, which in this case is simply the 0 class. How can I proceed?
algebraic-geometry surfaces complex-geometry singularity
algebraic-geometry surfaces complex-geometry singularity
asked Jan 29 at 11:57
thedilatedthedilated
7551617
asked Jan 29 at 11:57
thedilatedthedilated
7551617
edited Feb 10 at 11:02
thedilated
asked Jan 29 at 11:57
thedilatedthedilated
7551617
asked Jan 29 at 11:57
thedilatedthedilated
7551617
asked Jan 29 at 11:57
thedilatedthedilated
7551617
7551617
1
$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30
add a comment |
1
$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30
1
1
$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us call $E=mathbb{C}/mathbb{Z}[i]$. Then $Y= Etimes E$ and $X=Y/eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.
Since $eta$ has order four, a possible stabilizer is generated either by $eta$ or $eta^2$. Hence we first look for the points fixed by $eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.
Imposing that $-(a+bi) equiv a+bi mod{mathbb{Z}[i]}$ it follows that $$a+bi mod{mathbb{Z}[i]}in left{0, frac{1}{2}, frac{i}{2}, frac{1}{2}+frac{i}{2} right} = Ksubset E $$
Then the points with nontrivial stabilizer by $eta$ are in $Ktimes K$. This gives us ten orbits
$$
left{left(0,0right)right},
left{left(0,frac{1}{2}right),left(0,frac{i}{2}right)right},
left{left(0,frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2},0right),left(frac{i}{2},0right)right},
left{left(frac{1}{2},frac{1}{2}right),left(frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2},frac{i}{2}right),left(frac{i}{2},frac{1}{2}right)right},
left{left(frac{1}{2},frac{1}{2}+frac{i}{2}right),left(frac{i}{2},frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},0right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}right),left(frac{1}{2}+frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}+frac{i}{2}right)right}.
$$
We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= Etimes E$ we have -- by Proposition 2.2 here --
$$
q(S) = g(E/eta|_E) + g(E/eta|_E) = 0.
$$
and using a Kunneth type formula for coherent sheaves we have
$$
P_2 = h^0(S,K_S^{otimes 2}) = h^0(Etimes E, K_E^{otimes 2} boxtimes K_E^{otimes 2} )^eta = h^0(E, K_E^{otimes 2})^{eta|_E}cdot h^0(E, K_E^{otimes 2})^{eta|_E} = 0.
$$
Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.
Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.
Let me know if there is any point that is not clear.
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
$endgroup$
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let us call $E=mathbb{C}/mathbb{Z}[i]$. Then $Y= Etimes E$ and $X=Y/eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.
Since $eta$ has order four, a possible stabilizer is generated either by $eta$ or $eta^2$. Hence we first look for the points fixed by $eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.
Imposing that $-(a+bi) equiv a+bi mod{mathbb{Z}[i]}$ it follows that $$a+bi mod{mathbb{Z}[i]}in left{0, frac{1}{2}, frac{i}{2}, frac{1}{2}+frac{i}{2} right} = Ksubset E $$
Then the points with nontrivial stabilizer by $eta$ are in $Ktimes K$. This gives us ten orbits
$$
left{left(0,0right)right},
left{left(0,frac{1}{2}right),left(0,frac{i}{2}right)right},
left{left(0,frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2},0right),left(frac{i}{2},0right)right},
left{left(frac{1}{2},frac{1}{2}right),left(frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2},frac{i}{2}right),left(frac{i}{2},frac{1}{2}right)right},
left{left(frac{1}{2},frac{1}{2}+frac{i}{2}right),left(frac{i}{2},frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},0right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}right),left(frac{1}{2}+frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}+frac{i}{2}right)right}.
$$
We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= Etimes E$ we have -- by Proposition 2.2 here --
$$
q(S) = g(E/eta|_E) + g(E/eta|_E) = 0.
$$
and using a Kunneth type formula for coherent sheaves we have
$$
P_2 = h^0(S,K_S^{otimes 2}) = h^0(Etimes E, K_E^{otimes 2} boxtimes K_E^{otimes 2} )^eta = h^0(E, K_E^{otimes 2})^{eta|_E}cdot h^0(E, K_E^{otimes 2})^{eta|_E} = 0.
$$
Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.
Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.
Let me know if there is any point that is not clear.
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
$endgroup$
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
add a comment |
$begingroup$
Let us call $E=mathbb{C}/mathbb{Z}[i]$. Then $Y= Etimes E$ and $X=Y/eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.
Since $eta$ has order four, a possible stabilizer is generated either by $eta$ or $eta^2$. Hence we first look for the points fixed by $eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.
Imposing that $-(a+bi) equiv a+bi mod{mathbb{Z}[i]}$ it follows that $$a+bi mod{mathbb{Z}[i]}in left{0, frac{1}{2}, frac{i}{2}, frac{1}{2}+frac{i}{2} right} = Ksubset E $$
Then the points with nontrivial stabilizer by $eta$ are in $Ktimes K$. This gives us ten orbits
$$
left{left(0,0right)right},
left{left(0,frac{1}{2}right),left(0,frac{i}{2}right)right},
left{left(0,frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2},0right),left(frac{i}{2},0right)right},
left{left(frac{1}{2},frac{1}{2}right),left(frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2},frac{i}{2}right),left(frac{i}{2},frac{1}{2}right)right},
left{left(frac{1}{2},frac{1}{2}+frac{i}{2}right),left(frac{i}{2},frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},0right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}right),left(frac{1}{2}+frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}+frac{i}{2}right)right}.
$$
We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= Etimes E$ we have -- by Proposition 2.2 here --
$$
q(S) = g(E/eta|_E) + g(E/eta|_E) = 0.
$$
and using a Kunneth type formula for coherent sheaves we have
$$
P_2 = h^0(S,K_S^{otimes 2}) = h^0(Etimes E, K_E^{otimes 2} boxtimes K_E^{otimes 2} )^eta = h^0(E, K_E^{otimes 2})^{eta|_E}cdot h^0(E, K_E^{otimes 2})^{eta|_E} = 0.
$$
Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.
Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.
Let me know if there is any point that is not clear.
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
$endgroup$
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
add a comment |
$begingroup$
Let us call $E=mathbb{C}/mathbb{Z}[i]$. Then $Y= Etimes E$ and $X=Y/eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.
Since $eta$ has order four, a possible stabilizer is generated either by $eta$ or $eta^2$. Hence we first look for the points fixed by $eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.
Imposing that $-(a+bi) equiv a+bi mod{mathbb{Z}[i]}$ it follows that $$a+bi mod{mathbb{Z}[i]}in left{0, frac{1}{2}, frac{i}{2}, frac{1}{2}+frac{i}{2} right} = Ksubset E $$
Then the points with nontrivial stabilizer by $eta$ are in $Ktimes K$. This gives us ten orbits
$$
left{left(0,0right)right},
left{left(0,frac{1}{2}right),left(0,frac{i}{2}right)right},
left{left(0,frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2},0right),left(frac{i}{2},0right)right},
left{left(frac{1}{2},frac{1}{2}right),left(frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2},frac{i}{2}right),left(frac{i}{2},frac{1}{2}right)right},
left{left(frac{1}{2},frac{1}{2}+frac{i}{2}right),left(frac{i}{2},frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},0right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}right),left(frac{1}{2}+frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}+frac{i}{2}right)right}.
$$
We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= Etimes E$ we have -- by Proposition 2.2 here --
$$
q(S) = g(E/eta|_E) + g(E/eta|_E) = 0.
$$
and using a Kunneth type formula for coherent sheaves we have
$$
P_2 = h^0(S,K_S^{otimes 2}) = h^0(Etimes E, K_E^{otimes 2} boxtimes K_E^{otimes 2} )^eta = h^0(E, K_E^{otimes 2})^{eta|_E}cdot h^0(E, K_E^{otimes 2})^{eta|_E} = 0.
$$
Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.
Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.
Let me know if there is any point that is not clear.
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
$endgroup$
Let us call $E=mathbb{C}/mathbb{Z}[i]$. Then $Y= Etimes E$ and $X=Y/eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.
Since $eta$ has order four, a possible stabilizer is generated either by $eta$ or $eta^2$. Hence we first look for the points fixed by $eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.
Imposing that $-(a+bi) equiv a+bi mod{mathbb{Z}[i]}$ it follows that $$a+bi mod{mathbb{Z}[i]}in left{0, frac{1}{2}, frac{i}{2}, frac{1}{2}+frac{i}{2} right} = Ksubset E $$
Then the points with nontrivial stabilizer by $eta$ are in $Ktimes K$. This gives us ten orbits
$$
left{left(0,0right)right},
left{left(0,frac{1}{2}right),left(0,frac{i}{2}right)right},
left{left(0,frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2},0right),left(frac{i}{2},0right)right},
left{left(frac{1}{2},frac{1}{2}right),left(frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2},frac{i}{2}right),left(frac{i}{2},frac{1}{2}right)right},
left{left(frac{1}{2},frac{1}{2}+frac{i}{2}right),left(frac{i}{2},frac{1}{2}+frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},0right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}right),left(frac{1}{2}+frac{i}{2},frac{i}{2}right)right},
left{left(frac{1}{2}+frac{i}{2},frac{1}{2}+frac{i}{2}right)right}.
$$
We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= Etimes E$ we have -- by Proposition 2.2 here --
$$
q(S) = g(E/eta|_E) + g(E/eta|_E) = 0.
$$
and using a Kunneth type formula for coherent sheaves we have
$$
P_2 = h^0(S,K_S^{otimes 2}) = h^0(Etimes E, K_E^{otimes 2} boxtimes K_E^{otimes 2} )^eta = h^0(E, K_E^{otimes 2})^{eta|_E}cdot h^0(E, K_E^{otimes 2})^{eta|_E} = 0.
$$
Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.
Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.
Let me know if there is any point that is not clear.
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
edited Feb 26 at 17:28
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
answered Feb 23 at 15:16
Alan MunizAlan Muniz
2,61311030
2,61311030
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
add a comment |
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general.
$endgroup$
– thedilated
Mar 4 at 11:19
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
$begingroup$
I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y to X$.
$endgroup$
– Alan Muniz
Mar 5 at 19:18
add a comment |
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$begingroup$
From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$.
$endgroup$
– Alan Muniz
Feb 22 at 19:12
$begingroup$
From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup.
$endgroup$
– thedilated
Feb 23 at 4:30