Embedding abelian varieties into projective spaces of small dimension
$begingroup$
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
asked Jan 21 at 8:58
KimKim
484314
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add a comment |
$begingroup$
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
asked Jan 21 at 8:58
KimKim
484314
$endgroup$
add a comment |
$begingroup$
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
asked Jan 21 at 8:58
KimKim
484314
$endgroup$
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
asked Jan 21 at 8:58
KimKim
484314
asked Jan 21 at 8:58
KimKim
484314
asked Jan 21 at 8:58
KimKim
484314
asked Jan 21 at 8:58
KimKim
484314
asked Jan 21 at 8:58
KimKim
484314
484314
add a comment |
add a comment |
1 Answer
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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).
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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).
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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).
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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).
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
$endgroup$
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).
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
edited Jan 21 at 11:55
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
answered Jan 21 at 10:10
Mere ScribeMere Scribe
6721819
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
add a comment |
$begingroup$
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
add a comment |
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