Mixing
Show Fiber Product of Rational Elliptic Surfaces is Calabi-Yau
$begingroup$
In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.
Let $pi: S to mathbb{P}^{1}$ and $pi' : S' to mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product
$$X = S times_{mathbb{P}^{1}} S'.$$
With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).
I believe one should start by considering the obvious map induced by $pi$ and $pi'$
$$f: S times S' to mathbb{P}^{1} times mathbb{P}^{1}.$$
We can then write $X$ as the pullback of the diagonal $Delta subset mathbb{P}^{1} times mathbb{P}^{1}$,
$$X = f^{*} Delta.$$
So we can realize $X$ as a hypersurface in $S times S'$, so you should then be able to apply the adjunction formula:
$$omega_{X} = omega_{S times S'}|_{X} otimes mathcal{N}_{X/S times S'},$$
where $mathcal{N}_{X/S times S'}$ is the normal bundle of $X$ in $S times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?
ag.algebraic-geometry complex-geometry algebraic-surfaces
asked Jan 31 at 16:11
BenightedBenighted
51129
$endgroup$
add a comment |
$begingroup$
In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.
Let $pi: S to mathbb{P}^{1}$ and $pi' : S' to mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product
$$X = S times_{mathbb{P}^{1}} S'.$$
With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).
I believe one should start by considering the obvious map induced by $pi$ and $pi'$
$$f: S times S' to mathbb{P}^{1} times mathbb{P}^{1}.$$
We can then write $X$ as the pullback of the diagonal $Delta subset mathbb{P}^{1} times mathbb{P}^{1}$,
$$X = f^{*} Delta.$$
So we can realize $X$ as a hypersurface in $S times S'$, so you should then be able to apply the adjunction formula:
$$omega_{X} = omega_{S times S'}|_{X} otimes mathcal{N}_{X/S times S'},$$
where $mathcal{N}_{X/S times S'}$ is the normal bundle of $X$ in $S times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?
ag.algebraic-geometry complex-geometry algebraic-surfaces
asked Jan 31 at 16:11
BenightedBenighted
51129
$endgroup$
add a comment |
$begingroup$
In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.
Let $pi: S to mathbb{P}^{1}$ and $pi' : S' to mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product
$$X = S times_{mathbb{P}^{1}} S'.$$
With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).
I believe one should start by considering the obvious map induced by $pi$ and $pi'$
$$f: S times S' to mathbb{P}^{1} times mathbb{P}^{1}.$$
We can then write $X$ as the pullback of the diagonal $Delta subset mathbb{P}^{1} times mathbb{P}^{1}$,
$$X = f^{*} Delta.$$
So we can realize $X$ as a hypersurface in $S times S'$, so you should then be able to apply the adjunction formula:
$$omega_{X} = omega_{S times S'}|_{X} otimes mathcal{N}_{X/S times S'},$$
where $mathcal{N}_{X/S times S'}$ is the normal bundle of $X$ in $S times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?
ag.algebraic-geometry complex-geometry algebraic-surfaces
asked Jan 31 at 16:11
BenightedBenighted
51129
$endgroup$
In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.
Let $pi: S to mathbb{P}^{1}$ and $pi' : S' to mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product
$$X = S times_{mathbb{P}^{1}} S'.$$
With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).
I believe one should start by considering the obvious map induced by $pi$ and $pi'$
$$f: S times S' to mathbb{P}^{1} times mathbb{P}^{1}.$$
We can then write $X$ as the pullback of the diagonal $Delta subset mathbb{P}^{1} times mathbb{P}^{1}$,
$$X = f^{*} Delta.$$
So we can realize $X$ as a hypersurface in $S times S'$, so you should then be able to apply the adjunction formula:
$$omega_{X} = omega_{S times S'}|_{X} otimes mathcal{N}_{X/S times S'},$$
where $mathcal{N}_{X/S times S'}$ is the normal bundle of $X$ in $S times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?
ag.algebraic-geometry complex-geometry algebraic-surfaces
ag.algebraic-geometry complex-geometry algebraic-surfaces
asked Jan 31 at 16:11
BenightedBenighted
51129
asked Jan 31 at 16:11
BenightedBenighted
51129
asked Jan 31 at 16:11
BenightedBenighted
51129
asked Jan 31 at 16:11
BenightedBenighted
51129
asked Jan 31 at 16:11
BenightedBenighted
51129
51129
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
The diagonal $Delta $ is linearly equivalent to ${p}times mathbb{P}^1 +mathbb{P}^1times {p} $ for any $p$ in $mathbb{P}^1$. Therefore $X$ is the zero locus in $Stimes S'$ of a section of $L:=pi^*mathcal{O}(1) boxtimes pi'^*mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives
$omega _S=pi ^*mathcal{O}(-1) $ and $omega _{S'}=pi' ^*mathcal{O}(-1) $, therefore $omega _{Stimes S'}= L^{-1}$. Then the adjunction formula gives indeed $omega_Xcong mathcal{O}_{X}$.
answered Jan 31 at 16:42
abxabx
24k34885
$endgroup$
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
add a comment |
$begingroup$
$Stimes_{mathbb{P}^1}S'$ is a complete intersection in $mathbb{P}^1times mathbb{P}^2times mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
$endgroup$
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The diagonal $Delta $ is linearly equivalent to ${p}times mathbb{P}^1 +mathbb{P}^1times {p} $ for any $p$ in $mathbb{P}^1$. Therefore $X$ is the zero locus in $Stimes S'$ of a section of $L:=pi^*mathcal{O}(1) boxtimes pi'^*mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives
$omega _S=pi ^*mathcal{O}(-1) $ and $omega _{S'}=pi' ^*mathcal{O}(-1) $, therefore $omega _{Stimes S'}= L^{-1}$. Then the adjunction formula gives indeed $omega_Xcong mathcal{O}_{X}$.
answered Jan 31 at 16:42
abxabx
24k34885
$endgroup$
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
add a comment |
$begingroup$
The diagonal $Delta $ is linearly equivalent to ${p}times mathbb{P}^1 +mathbb{P}^1times {p} $ for any $p$ in $mathbb{P}^1$. Therefore $X$ is the zero locus in $Stimes S'$ of a section of $L:=pi^*mathcal{O}(1) boxtimes pi'^*mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives
$omega _S=pi ^*mathcal{O}(-1) $ and $omega _{S'}=pi' ^*mathcal{O}(-1) $, therefore $omega _{Stimes S'}= L^{-1}$. Then the adjunction formula gives indeed $omega_Xcong mathcal{O}_{X}$.
answered Jan 31 at 16:42
abxabx
24k34885
$endgroup$
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
add a comment |
$begingroup$
The diagonal $Delta $ is linearly equivalent to ${p}times mathbb{P}^1 +mathbb{P}^1times {p} $ for any $p$ in $mathbb{P}^1$. Therefore $X$ is the zero locus in $Stimes S'$ of a section of $L:=pi^*mathcal{O}(1) boxtimes pi'^*mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives
$omega _S=pi ^*mathcal{O}(-1) $ and $omega _{S'}=pi' ^*mathcal{O}(-1) $, therefore $omega _{Stimes S'}= L^{-1}$. Then the adjunction formula gives indeed $omega_Xcong mathcal{O}_{X}$.
answered Jan 31 at 16:42
abxabx
24k34885
$endgroup$
The diagonal $Delta $ is linearly equivalent to ${p}times mathbb{P}^1 +mathbb{P}^1times {p} $ for any $p$ in $mathbb{P}^1$. Therefore $X$ is the zero locus in $Stimes S'$ of a section of $L:=pi^*mathcal{O}(1) boxtimes pi'^*mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives
$omega _S=pi ^*mathcal{O}(-1) $ and $omega _{S'}=pi' ^*mathcal{O}(-1) $, therefore $omega _{Stimes S'}= L^{-1}$. Then the adjunction formula gives indeed $omega_Xcong mathcal{O}_{X}$.
answered Jan 31 at 16:42
abxabx
24k34885
answered Jan 31 at 16:42
abxabx
24k34885
answered Jan 31 at 16:42
abxabx
24k34885
answered Jan 31 at 16:42
abxabx
24k34885
24k34885
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
add a comment |
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
$begingroup$
Thanks a lot, that nearly clears everything up for me. In your expressions for $omega_{S}$ and $omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau.
$endgroup$
– Benighted
Jan 31 at 17:15
1
1
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
$begingroup$
I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $omega _{S}cong mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$endgroup$
– abx
Jan 31 at 17:49
add a comment |
$begingroup$
$Stimes_{mathbb{P}^1}S'$ is a complete intersection in $mathbb{P}^1times mathbb{P}^2times mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
$endgroup$
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
add a comment |
$begingroup$
$Stimes_{mathbb{P}^1}S'$ is a complete intersection in $mathbb{P}^1times mathbb{P}^2times mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
$endgroup$
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
add a comment |
$begingroup$
$Stimes_{mathbb{P}^1}S'$ is a complete intersection in $mathbb{P}^1times mathbb{P}^2times mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
$endgroup$
$Stimes_{mathbb{P}^1}S'$ is a complete intersection in $mathbb{P}^1times mathbb{P}^2times mathbb{P}^2$: it is given by two equations of degree $(1,3,0)$ and $(1,0,3)$. The canonical bundle is trivial by the adjunction formula.
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
answered Jan 31 at 17:46
Jim BryanJim Bryan
5,03921937
5,03921937
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
add a comment |
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
$begingroup$
I see, my argument only applies directly in the generic case. @abx gives an argument which applies to all relatively minimal rational elliptic surfaces with a section.
$endgroup$
– Jim Bryan
Jan 31 at 17:54
add a comment |
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