Mixing
Definition of the sheaf $GL_n(mathcal{O}_X)$ of invertible $mathcal{O}_X-$linear functions
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Let $(X, mathcal{O}_X)$ be a ringed space. Is there such a thing as the sheaf of invertible linear functions $GL_n(mathcal{O}_X)$? The point is that I cannot see how to define the restriction functions: if I have an $mathcal{O}_X(U)-$linear function $mathcal{O}_X(U)^n to mathcal{O}_X(U)^n$, how does this restrict to an $mathcal{O}_X(V)-$linear function $mathcal{O}_X(V)^n to mathcal{O}_X(V)^n$ for $V subseteq U$ open?
This question came to me as I was considering equivalent definitions of a locally free sheaf $mathcal{F}$ on $X$. I think that the transition functions between the trivializations $mathcal{F}(U_i) cong mathcal{O}_{U_i}^n$ on open sets $U_i$ and $U_j$ should live in such a $GL_n(mathcal{O}_{U_i cap U_j})$, and not just in $GL_n(mathcal{O}_X(U_i cap U_j))$, otherwise I cannot see how the data of the transition functions would be enough to determine the sheaf $mathcal{F}$ up to isomorphism. I wouldn't even know what it means for the transition functions to satisfy the cocycle condition, if I cannot restrict them to a common domain.
algebraic-geometry sheaf-theory ringed-spaces
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal{O}_X)$ be a ringed space. Is there such a thing as the sheaf of invertible linear functions $GL_n(mathcal{O}_X)$? The point is that I cannot see how to define the restriction functions: if I have an $mathcal{O}_X(U)-$linear function $mathcal{O}_X(U)^n to mathcal{O}_X(U)^n$, how does this restrict to an $mathcal{O}_X(V)-$linear function $mathcal{O}_X(V)^n to mathcal{O}_X(V)^n$ for $V subseteq U$ open?
This question came to me as I was considering equivalent definitions of a locally free sheaf $mathcal{F}$ on $X$. I think that the transition functions between the trivializations $mathcal{F}(U_i) cong mathcal{O}_{U_i}^n$ on open sets $U_i$ and $U_j$ should live in such a $GL_n(mathcal{O}_{U_i cap U_j})$, and not just in $GL_n(mathcal{O}_X(U_i cap U_j))$, otherwise I cannot see how the data of the transition functions would be enough to determine the sheaf $mathcal{F}$ up to isomorphism. I wouldn't even know what it means for the transition functions to satisfy the cocycle condition, if I cannot restrict them to a common domain.
algebraic-geometry sheaf-theory ringed-spaces
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
$endgroup$
$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53
add a comment |
$begingroup$
Let $(X, mathcal{O}_X)$ be a ringed space. Is there such a thing as the sheaf of invertible linear functions $GL_n(mathcal{O}_X)$? The point is that I cannot see how to define the restriction functions: if I have an $mathcal{O}_X(U)-$linear function $mathcal{O}_X(U)^n to mathcal{O}_X(U)^n$, how does this restrict to an $mathcal{O}_X(V)-$linear function $mathcal{O}_X(V)^n to mathcal{O}_X(V)^n$ for $V subseteq U$ open?
This question came to me as I was considering equivalent definitions of a locally free sheaf $mathcal{F}$ on $X$. I think that the transition functions between the trivializations $mathcal{F}(U_i) cong mathcal{O}_{U_i}^n$ on open sets $U_i$ and $U_j$ should live in such a $GL_n(mathcal{O}_{U_i cap U_j})$, and not just in $GL_n(mathcal{O}_X(U_i cap U_j))$, otherwise I cannot see how the data of the transition functions would be enough to determine the sheaf $mathcal{F}$ up to isomorphism. I wouldn't even know what it means for the transition functions to satisfy the cocycle condition, if I cannot restrict them to a common domain.
algebraic-geometry sheaf-theory ringed-spaces
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
$endgroup$
Let $(X, mathcal{O}_X)$ be a ringed space. Is there such a thing as the sheaf of invertible linear functions $GL_n(mathcal{O}_X)$? The point is that I cannot see how to define the restriction functions: if I have an $mathcal{O}_X(U)-$linear function $mathcal{O}_X(U)^n to mathcal{O}_X(U)^n$, how does this restrict to an $mathcal{O}_X(V)-$linear function $mathcal{O}_X(V)^n to mathcal{O}_X(V)^n$ for $V subseteq U$ open?
This question came to me as I was considering equivalent definitions of a locally free sheaf $mathcal{F}$ on $X$. I think that the transition functions between the trivializations $mathcal{F}(U_i) cong mathcal{O}_{U_i}^n$ on open sets $U_i$ and $U_j$ should live in such a $GL_n(mathcal{O}_{U_i cap U_j})$, and not just in $GL_n(mathcal{O}_X(U_i cap U_j))$, otherwise I cannot see how the data of the transition functions would be enough to determine the sheaf $mathcal{F}$ up to isomorphism. I wouldn't even know what it means for the transition functions to satisfy the cocycle condition, if I cannot restrict them to a common domain.
algebraic-geometry sheaf-theory ringed-spaces
algebraic-geometry sheaf-theory ringed-spaces
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
edited Jan 31 at 15:53
57Jimmy
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
asked Jan 31 at 14:24
57Jimmy57Jimmy
3,538422
3,538422
$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53
add a comment |
$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53
$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53
add a comment |
1 Answer
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Well, in a naive way, if we have a map of rings $A to B$, then there is an induced map $GL_n(A) to GL_n(B)$ defined coordinate wise. In this case $A = O_X(U), B = O_X(V).$
Less naively, given a $A$ linear map $f: A^n to A^n$, we get an induced map by taking tensor products $f: A^n otimes_A B to A^n otimes_A B = B^n to B^n$.
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
$endgroup$
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
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– ant
Jan 31 at 14:42
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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$begingroup$
Well, in a naive way, if we have a map of rings $A to B$, then there is an induced map $GL_n(A) to GL_n(B)$ defined coordinate wise. In this case $A = O_X(U), B = O_X(V).$
Less naively, given a $A$ linear map $f: A^n to A^n$, we get an induced map by taking tensor products $f: A^n otimes_A B to A^n otimes_A B = B^n to B^n$.
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
$endgroup$
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
add a comment |
$begingroup$
Well, in a naive way, if we have a map of rings $A to B$, then there is an induced map $GL_n(A) to GL_n(B)$ defined coordinate wise. In this case $A = O_X(U), B = O_X(V).$
Less naively, given a $A$ linear map $f: A^n to A^n$, we get an induced map by taking tensor products $f: A^n otimes_A B to A^n otimes_A B = B^n to B^n$.
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
$endgroup$
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
add a comment |
$begingroup$
Well, in a naive way, if we have a map of rings $A to B$, then there is an induced map $GL_n(A) to GL_n(B)$ defined coordinate wise. In this case $A = O_X(U), B = O_X(V).$
Less naively, given a $A$ linear map $f: A^n to A^n$, we get an induced map by taking tensor products $f: A^n otimes_A B to A^n otimes_A B = B^n to B^n$.
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
$endgroup$
Well, in a naive way, if we have a map of rings $A to B$, then there is an induced map $GL_n(A) to GL_n(B)$ defined coordinate wise. In this case $A = O_X(U), B = O_X(V).$
Less naively, given a $A$ linear map $f: A^n to A^n$, we get an induced map by taking tensor products $f: A^n otimes_A B to A^n otimes_A B = B^n to B^n$.
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
answered Jan 31 at 14:37
AsvinAsvin
3,43821432
3,43821432
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
add a comment |
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
$begingroup$
The naive map is defined term-wise which is algebraic. Furthermore $GL_n$ is a representable (i.e. contravariant) functor on $text{Sch}/k$ which yields a (covariant) corepresentable functor on $text{Alg}-k$ by taking opposite categories
$endgroup$
– ant
Jan 31 at 14:42
add a comment |
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$begingroup$
Just a nitpick : $GL_n(mathcal{O}_X)$ is not an $mathcal{O}_X$-module, just a sheaf of (non-commutative) groups.
$endgroup$
– Roland
Jan 31 at 15:10
$begingroup$
@Roland Thanks, it's actually an important remark! :) I've edited the question and title
$endgroup$
– 57Jimmy
Jan 31 at 15:53