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







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    5












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked 7 hours ago









      HarryHarry

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          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.






          share|cite|improve this answer









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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.






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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.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  SashaSasha

                  20.5k22655




                  20.5k22655























                      2












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 6 hours ago









                          Jim BryanJim Bryan

                          4,96421937




                          4,96421937






























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