Embedding abelian varieties into projective spaces of small dimension












17












$begingroup$


Given a (complex) abelian variety $A$ of a fixed dimension $g$, let $d(A)$ be the dimension of the smallest complex projective space it embeds into.



Is $d(A)$ uniform over all abelian varieties of a fixed $g$? Or are there special ones that embed into even smaller projective spaces?



Can $d(A)$ be computed explicitly? I am particularly interested in the case $g = 2$.






ag.algebraic-geometry abelian-varieties line-bundles







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$endgroup$

















    17












    $begingroup$


    Given a (complex) abelian variety $A$ of a fixed dimension $g$, let $d(A)$ be the dimension of the smallest complex projective space it embeds into.



    Is $d(A)$ uniform over all abelian varieties of a fixed $g$? Or are there special ones that embed into even smaller projective spaces?



    Can $d(A)$ be computed explicitly? I am particularly interested in the case $g = 2$.






    ag.algebraic-geometry abelian-varieties line-bundles







    share|cite|improve this question









    $endgroup$















      17












      17








      17


      3



      $begingroup$


      Given a (complex) abelian variety $A$ of a fixed dimension $g$, let $d(A)$ be the dimension of the smallest complex projective space it embeds into.



      Is $d(A)$ uniform over all abelian varieties of a fixed $g$? Or are there special ones that embed into even smaller projective spaces?



      Can $d(A)$ be computed explicitly? I am particularly interested in the case $g = 2$.






      ag.algebraic-geometry abelian-varieties line-bundles







      share|cite|improve this question









      $endgroup$




      Given a (complex) abelian variety $A$ of a fixed dimension $g$, let $d(A)$ be the dimension of the smallest complex projective space it embeds into.



      Is $d(A)$ uniform over all abelian varieties of a fixed $g$? Or are there special ones that embed into even smaller projective spaces?



      Can $d(A)$ be computed explicitly? I am particularly interested in the case $g = 2$.






      ag.algebraic-geometry abelian-varieties line-bundles




      ag.algebraic-geometry abelian-varieties line-bundles






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 at 8:58









      KimKim

      484314




      484314






















          1 Answer
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          27












          $begingroup$

          Recall that any smooth projective variety of dimension $g$ embeds into $mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $mathbf{P}^{2g}$) that the degree of $A$ in $mathbf{P}^{2g}$ is given by ${2g+1choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$.
          Of course elliptic curves embed into $mathbf{P}^2$. One cannot embed abelian surfaces into $mathbf{P}^3$, and if an abelian surface embeds into $mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $Ccdot D=3$, then $J(C)$ embeds into $mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $mathbf{P}^4$ is projectively equivalent to one of theirs).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
            $endgroup$
            – Kim
            Jan 21 at 23:32












          • $begingroup$
            @Kim Possibly, but not to me.
            $endgroup$
            – Mere Scribe
            Jan 22 at 10:29











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          1 Answer
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          active

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          27












          $begingroup$

          Recall that any smooth projective variety of dimension $g$ embeds into $mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $mathbf{P}^{2g}$) that the degree of $A$ in $mathbf{P}^{2g}$ is given by ${2g+1choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$.
          Of course elliptic curves embed into $mathbf{P}^2$. One cannot embed abelian surfaces into $mathbf{P}^3$, and if an abelian surface embeds into $mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $Ccdot D=3$, then $J(C)$ embeds into $mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $mathbf{P}^4$ is projectively equivalent to one of theirs).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
            $endgroup$
            – Kim
            Jan 21 at 23:32












          • $begingroup$
            @Kim Possibly, but not to me.
            $endgroup$
            – Mere Scribe
            Jan 22 at 10:29
















          27












          $begingroup$

          Recall that any smooth projective variety of dimension $g$ embeds into $mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $mathbf{P}^{2g}$) that the degree of $A$ in $mathbf{P}^{2g}$ is given by ${2g+1choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$.
          Of course elliptic curves embed into $mathbf{P}^2$. One cannot embed abelian surfaces into $mathbf{P}^3$, and if an abelian surface embeds into $mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $Ccdot D=3$, then $J(C)$ embeds into $mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $mathbf{P}^4$ is projectively equivalent to one of theirs).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
            $endgroup$
            – Kim
            Jan 21 at 23:32












          • $begingroup$
            @Kim Possibly, but not to me.
            $endgroup$
            – Mere Scribe
            Jan 22 at 10:29














          27












          27








          27





          $begingroup$

          Recall that any smooth projective variety of dimension $g$ embeds into $mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $mathbf{P}^{2g}$) that the degree of $A$ in $mathbf{P}^{2g}$ is given by ${2g+1choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$.
          Of course elliptic curves embed into $mathbf{P}^2$. One cannot embed abelian surfaces into $mathbf{P}^3$, and if an abelian surface embeds into $mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $Ccdot D=3$, then $J(C)$ embeds into $mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $mathbf{P}^4$ is projectively equivalent to one of theirs).






          share|cite|improve this answer











          $endgroup$



          Recall that any smooth projective variety of dimension $g$ embeds into $mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $mathbf{P}^{2g}$) that the degree of $A$ in $mathbf{P}^{2g}$ is given by ${2g+1choose g}$, and notes that the Riemann-Roch theorem implies that the degree has to be divisible by $g!$. This is possible only if $g=1$ or $g=2$.
          Of course elliptic curves embed into $mathbf{P}^2$. One cannot embed abelian surfaces into $mathbf{P}^3$, and if an abelian surface embeds into $mathbf{P}^4$, then its degree is $10$. Abelian surfaces with this property exist (Mumford-Horrocks surfaces). This is all folklore. Maybe less known is the fact that Comessatti proved in 1919 (see this paper of Lange for a modern account) that for some curves of genus $2$ the Jacobian embeds into $mathbf{P}^4$. More precisely: if $C$ is a curve of genus $2$ and $J(C)$ contains a curve $D$ with self-intersection $2$ and $Ccdot D=3$, then $J(C)$ embeds into $mathbf{P}^4$ (with embedding given by $|C+D|$). But these should also be of the Mumford-Horrocks type (Theorem 5.2 in the paper of Mumford and Horrocks says that any abelian surface in $mathbf{P}^4$ is projectively equivalent to one of theirs).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 21 at 11:55

























          answered Jan 21 at 10:10









          Mere ScribeMere Scribe

          6721819




          6721819












          • $begingroup$
            Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
            $endgroup$
            – Kim
            Jan 21 at 23:32












          • $begingroup$
            @Kim Possibly, but not to me.
            $endgroup$
            – Mere Scribe
            Jan 22 at 10:29


















          • $begingroup$
            Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
            $endgroup$
            – Kim
            Jan 21 at 23:32












          • $begingroup$
            @Kim Possibly, but not to me.
            $endgroup$
            – Mere Scribe
            Jan 22 at 10:29
















          $begingroup$
          Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
          $endgroup$
          – Kim
          Jan 21 at 23:32






          $begingroup$
          Is it known what is the smallest degree that the image of $mathbf{A}$ in $mathbf{P}^{2g+1}$ can be?
          $endgroup$
          – Kim
          Jan 21 at 23:32














          $begingroup$
          @Kim Possibly, but not to me.
          $endgroup$
          – Mere Scribe
          Jan 22 at 10:29




          $begingroup$
          @Kim Possibly, but not to me.
          $endgroup$
          – Mere Scribe
          Jan 22 at 10:29


















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