Mixing
Definition of $mathcal{O}_X(n)$
1
$begingroup$
Let $S$ be a graded ring, and let $X = Proj S$, we define the sheaf $mathcal{O}_X(n)$ to be $S(n)^sim$. Here can you explain what $S(n)$ is?
algebraic-geometry sheaf-theory
asked Jan 29 at 21:18
KatherineKatherine
438310
$endgroup$
add a comment |
1
$begingroup$
Let $S$ be a graded ring, and let $X = Proj S$, we define the sheaf $mathcal{O}_X(n)$ to be $S(n)^sim$. Here can you explain what $S(n)$ is?
algebraic-geometry sheaf-theory
asked Jan 29 at 21:18
KatherineKatherine
438310
$endgroup$
2
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
2
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03
add a comment |
1
1
1
$begingroup$
Let $S$ be a graded ring, and let $X = Proj S$, we define the sheaf $mathcal{O}_X(n)$ to be $S(n)^sim$. Here can you explain what $S(n)$ is?
algebraic-geometry sheaf-theory
asked Jan 29 at 21:18
KatherineKatherine
438310
$endgroup$
Let $S$ be a graded ring, and let $X = Proj S$, we define the sheaf $mathcal{O}_X(n)$ to be $S(n)^sim$. Here can you explain what $S(n)$ is?
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Jan 29 at 21:18
KatherineKatherine
438310
asked Jan 29 at 21:18
KatherineKatherine
438310
asked Jan 29 at 21:18
KatherineKatherine
438310
asked Jan 29 at 21:18
KatherineKatherine
438310
asked Jan 29 at 21:18
KatherineKatherine
438310
438310
2
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
2
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03
add a comment |
2
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
2
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03
2
2
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
2
2
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03
add a comment |
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2
$begingroup$
The answer is a bit boring: $S(n)$ has the same $S$-module structure as $S$ but the graduation now is different. What was of degree $d$ before know is of degree $d-n$, i.e, $S(n)_{d}=S_{d+n}$. For example you can see the Stack project
$endgroup$
– yamete kudasai
Jan 29 at 21:24
$begingroup$
@Yesterdaywasdramatic Is there a motivation for that?
$endgroup$
– Katherine
Jan 29 at 23:00
2
$begingroup$
Of course but that is a more subtle question and there are different ways to answer it. One possible answer is that one can prove that $mathcal{O}_X(1)$ is a locally free sheaf of rank one and this sheaves are really nice. These are the analogue of the line bundles in differential geometry. The set of line bundles of $X$ is so important that it has its own name: $Pic(X)$. It has a group structure (the picard group) and one of the central problems is to understand this group. For example for $X=mathbb{P}^n$ the group is cyclic and $mathcal{O}_X(1)$ is the generator.
$endgroup$
– yamete kudasai
Jan 30 at 1:03