Morphisms between torsion sheaves
2
$begingroup$
Let $X$ be a smooth projective variety over $mathbb{C}$, and $Y,Zsubset X$ be closed subvarieties of $X$.
Denote the embeddings of $Y$ and $Z$ into $X$ by $i_Y$ and $i_Z$ respectively.
What is the necessary and sufficient condition for $mathrm{Hom}(i_{Y*}mathcal{O}_Y, i_{Z*}mathcal{O}_Z)neq0$?
Is it that $Zsubset Y$?
Thank you!
algebraic-geometry sheaf-theory coherent-sheaves
asked Dec 19 '18 at 5:22
hsiuhsiu
112
$endgroup$
add a comment |
2
$begingroup$
Let $X$ be a smooth projective variety over $mathbb{C}$, and $Y,Zsubset X$ be closed subvarieties of $X$.
Denote the embeddings of $Y$ and $Z$ into $X$ by $i_Y$ and $i_Z$ respectively.
What is the necessary and sufficient condition for $mathrm{Hom}(i_{Y*}mathcal{O}_Y, i_{Z*}mathcal{O}_Z)neq0$?
Is it that $Zsubset Y$?
Thank you!
algebraic-geometry sheaf-theory coherent-sheaves
asked Dec 19 '18 at 5:22
hsiuhsiu
112
$endgroup$
4
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32
add a comment |
2
2
2
$begingroup$
Let $X$ be a smooth projective variety over $mathbb{C}$, and $Y,Zsubset X$ be closed subvarieties of $X$.
Denote the embeddings of $Y$ and $Z$ into $X$ by $i_Y$ and $i_Z$ respectively.
What is the necessary and sufficient condition for $mathrm{Hom}(i_{Y*}mathcal{O}_Y, i_{Z*}mathcal{O}_Z)neq0$?
Is it that $Zsubset Y$?
Thank you!
algebraic-geometry sheaf-theory coherent-sheaves
asked Dec 19 '18 at 5:22
hsiuhsiu
112
$endgroup$
Let $X$ be a smooth projective variety over $mathbb{C}$, and $Y,Zsubset X$ be closed subvarieties of $X$.
Denote the embeddings of $Y$ and $Z$ into $X$ by $i_Y$ and $i_Z$ respectively.
What is the necessary and sufficient condition for $mathrm{Hom}(i_{Y*}mathcal{O}_Y, i_{Z*}mathcal{O}_Z)neq0$?
Is it that $Zsubset Y$?
Thank you!
algebraic-geometry sheaf-theory coherent-sheaves
algebraic-geometry sheaf-theory coherent-sheaves
asked Dec 19 '18 at 5:22
hsiuhsiu
112
asked Dec 19 '18 at 5:22
hsiuhsiu
112
edited Dec 19 '18 at 22:31
hsiu
asked Dec 19 '18 at 5:22
hsiuhsiu
112
asked Dec 19 '18 at 5:22
hsiuhsiu
112
asked Dec 19 '18 at 5:22
hsiuhsiu
112
112
4
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32
add a comment |
4
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32
4
4
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32
add a comment |
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4
$begingroup$
Did you try the case of $X$ affine first?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:51
$begingroup$
Recall that if $Y=V(mathcal{I})$ then $O_Y=i^{-1}(O_X/mathcal{I})$ is a sheaf over $Y$. Hence $Hom(O_Y,O_Z)$ doesn't make much sense as these are sheaves over different spaces. But in the same spirit we can study the set $$Hom(O_X/mathcal{I},O_X/mathcal{J})$$ for $Y=V(mathcal{I})$ and $Z=V(mathcal{J})$. For an arbitrary $X$, when $mathcal{I}subseteq mathcal{J}$ (i.e, $Zsubseteq Y$) we can glue the projections $O_X(U)/mathcal{I}(U)rightarrow O_X(U)/mathcal{J}(U)$ to obtain an element in $Hom(O_X/mathcal{I},O_X/mathcal{J})$ so at least in this case the set is non empty.
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:00
$begingroup$
The other inclusion seem to be more subtle. For example take $Z=text{Spec}(mathbb{C}times mathbb{C})$ and $Y=V(0times mathbb{C})$. Then $Hom(mathbb{C}, mathbb{C}times mathbb{C})$ is not empty even though $Znotsubseteq Y$ (and $Z$, $Y$ are projective here).
$endgroup$
– yamete kudasai
Dec 19 '18 at 20:04
$begingroup$
Thanks for the answer! I have edited the question to clarify it.
$endgroup$
– hsiu
Dec 19 '18 at 22:32