Is the symmetric product of an abelian variety a CY variety?Is the generic deformation of a symplectic...



Is the symmetric product of an abelian variety a CY variety?


Is the generic deformation of a symplectic variety affine?Log resolutions on surfaces and 3-folds in characteristic paxiomatizing the abelian part of the topological fundamental groupoid functor on algebraic varietiesDo all varieties have only finitely many etale covers of fixed degreeModuli space of log Calabi-Yau varieties exists?K-stability on Fano fibrationConstraints on the base of an elliptically fibered Calabi-Yau threefoldTotal space of canonical bundle on toric del pezzo surfaceKähler-Ricci flow approach for Beauville-Bogomolov type decomposition?Extending Beauville-Bogomolov orthogonal decomposition from variety to scheme













3












$begingroup$


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









$endgroup$

















    3












    $begingroup$


    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









    $endgroup$















      3












      3








      3





      $begingroup$


      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      HarryHarry

      1433




      1433






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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









          $endgroup$





















            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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "504"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324031%2fis-the-symmetric-product-of-an-abelian-variety-a-cy-variety%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 2 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 1 hour ago









                          Jim BryanJim Bryan

                          4,96421937




                          4,96421937






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to MathOverflow!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324031%2fis-the-symmetric-product-of-an-abelian-variety-a-cy-variety%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Szabolcs (Ungheria) Altri progetti | Menu di navigazione48°10′14.56″N 21°29′33.14″E /...

                              Image
                              Comuni della provincia di Szabolcs-Szatmár-Bereg Ungheria2001provincia di Szabolcs-Szatmár-Bereg Szabolcs comune Localizzazione Stato   Ungheria Regione Grande Pianura Settentrionale Provincia Szabolcs-Szatmár-Bereg Territorio Coordinate 48°10′14.56″N 21°29′33.14″E  /  48.17071°N 21.49254°E 48.17071; 21.49254  ( Szabolcs ) Coordinate: 48°10′14.56″N 21°29′33.14″E  /  48.17071°N 21.49254°E 48.17071; 21.49254  ( Szabolcs ) Superficie 5,88 km² Abitanti 425 (2001) Densità 72,28 ab./km² Altre informazioni Cod. postale 4467 Prefisso 42 Fuso orario UTC+1 Codice KSH 19169 Cartografia Szabolcs Sito istituzionale Modifica dati su Wikidata  · Manuale Szabolcs è un comune dell'Ungheria di 425 abitanti (dati 2001). È situato nella provincia di Szabolcs-Szatmár-Bereg. Altri progetti | Altri progetti Wikimedia Commons Wikimedia Commons contiene immagini o altri file su Szabolcs .mw-parser-output
                              Read more

                              Discografia di Klaus Schulze Indice Album in studio | Album dal vivo | Singoli | Antologie | Colonne...

                              Image
                              Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune
                              Read more

                              How to make inet_server_addr() return localhost in spite of ::1/128RETURN NEXT in Postgres FunctionConnect to...

                              Image
                              Potential client has a problematic employee I can't work with Can the "Friends" spell be used without making the target hostile? Does it take energy to move something in a circle? Why are carbons of Inositol chiral centers? Count repetitions of an array Which RAF squadrons and aircraft types took part in the bombing of Berlin on the 25th of August 1940? Memory usage: #define vs. static const for uint8_t Broad Strokes - missing letter riddle Why does 0.-5 evaluate to -5? Why is 'diphthong' pronounced the way it is? Why did Luke use his left hand to shoot?
                              Read more