Is the symmetric product of an abelian variety a CY variety?
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Let $n>1$ be a positive integer and let $A$ be an abelian variety over $mathbb{C}$. Then the symmetric product $S^n(A)$ is a normal projective variety over $mathbb{C}$ with Kodaira dimension zero (see for instance https://arxiv.org/pdf/math/0006107.pdf).
Let $A(n)to S^n(A)$ be a resolution of singularities. Then, up to finite etale cover, $A(n)$ is a product of hyperkaehler varieties, an abelian variety, and simply connected strict Calabi-Yau varieties. (This should follow from the Beauville-Bogomolov decomposition theorem. Or does this require an additional hypothesis on $A(n)$.)
I am wondering how the decomposition of $A(n)$ looks like as $n$ grows. Is it always a strict Calabi-Yau variety? Could it be that $A(n)$ is an abelian variety in fact?
I am looking for examples and would appreciate any comments.
ag.algebraic-geometry complex-geometry abelian-varieties
$endgroup$
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$begingroup$
Let $n>1$ be a positive integer and let $A$ be an abelian variety over $mathbb{C}$. Then the symmetric product $S^n(A)$ is a normal projective variety over $mathbb{C}$ with Kodaira dimension zero (see for instance https://arxiv.org/pdf/math/0006107.pdf).
Let $A(n)to S^n(A)$ be a resolution of singularities. Then, up to finite etale cover, $A(n)$ is a product of hyperkaehler varieties, an abelian variety, and simply connected strict Calabi-Yau varieties. (This should follow from the Beauville-Bogomolov decomposition theorem. Or does this require an additional hypothesis on $A(n)$.)
I am wondering how the decomposition of $A(n)$ looks like as $n$ grows. Is it always a strict Calabi-Yau variety? Could it be that $A(n)$ is an abelian variety in fact?
I am looking for examples and would appreciate any comments.
ag.algebraic-geometry complex-geometry abelian-varieties
$endgroup$
add a comment |
$begingroup$
Let $n>1$ be a positive integer and let $A$ be an abelian variety over $mathbb{C}$. Then the symmetric product $S^n(A)$ is a normal projective variety over $mathbb{C}$ with Kodaira dimension zero (see for instance https://arxiv.org/pdf/math/0006107.pdf).
Let $A(n)to S^n(A)$ be a resolution of singularities. Then, up to finite etale cover, $A(n)$ is a product of hyperkaehler varieties, an abelian variety, and simply connected strict Calabi-Yau varieties. (This should follow from the Beauville-Bogomolov decomposition theorem. Or does this require an additional hypothesis on $A(n)$.)
I am wondering how the decomposition of $A(n)$ looks like as $n$ grows. Is it always a strict Calabi-Yau variety? Could it be that $A(n)$ is an abelian variety in fact?
I am looking for examples and would appreciate any comments.
ag.algebraic-geometry complex-geometry abelian-varieties
$endgroup$
Let $n>1$ be a positive integer and let $A$ be an abelian variety over $mathbb{C}$. Then the symmetric product $S^n(A)$ is a normal projective variety over $mathbb{C}$ with Kodaira dimension zero (see for instance https://arxiv.org/pdf/math/0006107.pdf).
Let $A(n)to S^n(A)$ be a resolution of singularities. Then, up to finite etale cover, $A(n)$ is a product of hyperkaehler varieties, an abelian variety, and simply connected strict Calabi-Yau varieties. (This should follow from the Beauville-Bogomolov decomposition theorem. Or does this require an additional hypothesis on $A(n)$.)
I am wondering how the decomposition of $A(n)$ looks like as $n$ grows. Is it always a strict Calabi-Yau variety? Could it be that $A(n)$ is an abelian variety in fact?
I am looking for examples and would appreciate any comments.
ag.algebraic-geometry complex-geometry abelian-varieties
ag.algebraic-geometry complex-geometry abelian-varieties
asked 7 hours ago
HarryHarry
1533
1533
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2 Answers
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$begingroup$
When $dim A = 1$, $S^nA$ is a $mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-infty$.
When $dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map
$$
A^{[n]} to A
$$
(summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.
$endgroup$
add a comment |
$begingroup$
You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $dim A =2$, the story is as Sasha described. For $dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $mathbb{C}^{2d}/{pm 1}$ and for $d>1$ these do not admit crepant resolutions.
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2 Answers
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2 Answers
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active
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$begingroup$
When $dim A = 1$, $S^nA$ is a $mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-infty$.
When $dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map
$$
A^{[n]} to A
$$
(summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.
$endgroup$
add a comment |
$begingroup$
When $dim A = 1$, $S^nA$ is a $mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-infty$.
When $dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map
$$
A^{[n]} to A
$$
(summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.
$endgroup$
add a comment |
$begingroup$
When $dim A = 1$, $S^nA$ is a $mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-infty$.
When $dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map
$$
A^{[n]} to A
$$
(summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.
$endgroup$
When $dim A = 1$, $S^nA$ is a $mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-infty$.
When $dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map
$$
A^{[n]} to A
$$
(summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.
answered 7 hours ago
SashaSasha
20.5k22655
20.5k22655
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$begingroup$
You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $dim A =2$, the story is as Sasha described. For $dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $mathbb{C}^{2d}/{pm 1}$ and for $d>1$ these do not admit crepant resolutions.
$endgroup$
add a comment |
$begingroup$
You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $dim A =2$, the story is as Sasha described. For $dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $mathbb{C}^{2d}/{pm 1}$ and for $d>1$ these do not admit crepant resolutions.
$endgroup$
add a comment |
$begingroup$
You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $dim A =2$, the story is as Sasha described. For $dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $mathbb{C}^{2d}/{pm 1}$ and for $d>1$ these do not admit crepant resolutions.
$endgroup$
You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $dim A =2$, the story is as Sasha described. For $dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $mathbb{C}^{2d}/{pm 1}$ and for $d>1$ these do not admit crepant resolutions.
answered 6 hours ago
Jim BryanJim Bryan
4,96421937
4,96421937
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