On $operatorname{End}(mathcal F)$ of a locally free sheaf of finite rank
$begingroup$
$underline{Background}$:Let,$X$ be a scheme(for simplicity we can assume it to be noetherian).Let $mathcal F$ be a locally free sheaf of rank $r$ on $X$.Let, us also assume that {$operatorname{Spec}A_i:iin I$} is its trivializing open sets,i.e
we have $forall iin I$ , $mathcal F_{big|operatorname{Spec}cA_i}cong{mathcal O_{operatorname{Spec}cA_i^{oplus r}}}cong{oplus_{substack r}}tilde{A_i}$
$underline{Aim}$:I am trying to give a $O_X$ module morphism from $operatorname{End}(mathcal F)=mathcal Hom(mathcal F,mathcal F)$ to $O_X$.
$underline{Attempt}$:$1)$ we can give maps $f(operatorname{Spec}A_i):mathcal Hom (F_{big|operatorname{Spec}A_i},mathcal F_{big|operatorname{Spec}A_i}) to mathcal O(operatorname{Spec}A_i)=A_i$,
which is equivalent to give a morphism $f(operatorname{Spec}A_i):mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})to A_i$
Let,$gin mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})$
$Rightarrow g(operatorname{Spec}A_i):oplus_{substack r}{A_i}to oplus_{substack r}{A_i}$ is an $A_i$ module morphism.
Since $oplus_{substack r}{A_i}$ is a free $A_i$ module of rank r, let {$e_1,.....e_r$} be a basis(its canonical basis say).
Then let the matrix(w.r.to that basis) corresponding to $g(operatorname{Spec}A_i)$ is $mathcal M_g$.
Then we define $f(operatorname{Spec}A_i)(g)=operatorname{trace}(mathcal M_g)in{A_i}$
$underline{Question}$: Even if we assume that $f(operatorname{Spec}A_i)$s are valid $A_i$-module morphism how can we say that $exists phi :operatorname{End}(mathcal F)to {mathcal O_X}$ such that $phi(operatorname{Spec}A_i)$ is same as $f(operatorname{Spec}A_i)$ ?
Because we know that we can do that if we have {$operatorname{Spec}A_i:iin I$} forms a base of $X$ and $f(operatorname{Spec}A_i)$s are morphisms of sheaves on that base.but we only have {$operatorname{Spec}A_i:iin I$} forms a cover of $X$ ,not necessarily a base.
$underline{Attempt(2)}$:So to avoid the base problem if we define maps at all affine opens {$operatorname{Spec}A$}, then we have $g(operatorname{Spec}A):mathcal F(operatorname{Spec}A)to{mathcal F(operatorname{Spec}A)}$.
Now ,we have $mathcal F$ is coherent on $X$,so $ mathcal F_{big|operatorname{Spec}A}$s are coherent on $operatorname{Spec}A$.But that does not necessarily mean that $mathcal F(operatorname{Spec}A)$s are finitely generated $A$ module,so we do not necessarily have corresponding matrix and hence its trace.
$underline{Question}$:How do we define $f(operatorname{Spec}A)$ in this case?
$underline{Question}$(not directly related to previous discussion)}:For $mathcal F$ a rank $r$ locally free sheaf is $operatorname{End}(mathcal F)$ always a locally free sheaf of rank r?
Any help from anyone is welcome.
algebraic-geometry sheaf-theory vector-bundles
$endgroup$
add a comment |
$begingroup$
$underline{Background}$:Let,$X$ be a scheme(for simplicity we can assume it to be noetherian).Let $mathcal F$ be a locally free sheaf of rank $r$ on $X$.Let, us also assume that {$operatorname{Spec}A_i:iin I$} is its trivializing open sets,i.e
we have $forall iin I$ , $mathcal F_{big|operatorname{Spec}cA_i}cong{mathcal O_{operatorname{Spec}cA_i^{oplus r}}}cong{oplus_{substack r}}tilde{A_i}$
$underline{Aim}$:I am trying to give a $O_X$ module morphism from $operatorname{End}(mathcal F)=mathcal Hom(mathcal F,mathcal F)$ to $O_X$.
$underline{Attempt}$:$1)$ we can give maps $f(operatorname{Spec}A_i):mathcal Hom (F_{big|operatorname{Spec}A_i},mathcal F_{big|operatorname{Spec}A_i}) to mathcal O(operatorname{Spec}A_i)=A_i$,
which is equivalent to give a morphism $f(operatorname{Spec}A_i):mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})to A_i$
Let,$gin mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})$
$Rightarrow g(operatorname{Spec}A_i):oplus_{substack r}{A_i}to oplus_{substack r}{A_i}$ is an $A_i$ module morphism.
Since $oplus_{substack r}{A_i}$ is a free $A_i$ module of rank r, let {$e_1,.....e_r$} be a basis(its canonical basis say).
Then let the matrix(w.r.to that basis) corresponding to $g(operatorname{Spec}A_i)$ is $mathcal M_g$.
Then we define $f(operatorname{Spec}A_i)(g)=operatorname{trace}(mathcal M_g)in{A_i}$
$underline{Question}$: Even if we assume that $f(operatorname{Spec}A_i)$s are valid $A_i$-module morphism how can we say that $exists phi :operatorname{End}(mathcal F)to {mathcal O_X}$ such that $phi(operatorname{Spec}A_i)$ is same as $f(operatorname{Spec}A_i)$ ?
Because we know that we can do that if we have {$operatorname{Spec}A_i:iin I$} forms a base of $X$ and $f(operatorname{Spec}A_i)$s are morphisms of sheaves on that base.but we only have {$operatorname{Spec}A_i:iin I$} forms a cover of $X$ ,not necessarily a base.
$underline{Attempt(2)}$:So to avoid the base problem if we define maps at all affine opens {$operatorname{Spec}A$}, then we have $g(operatorname{Spec}A):mathcal F(operatorname{Spec}A)to{mathcal F(operatorname{Spec}A)}$.
Now ,we have $mathcal F$ is coherent on $X$,so $ mathcal F_{big|operatorname{Spec}A}$s are coherent on $operatorname{Spec}A$.But that does not necessarily mean that $mathcal F(operatorname{Spec}A)$s are finitely generated $A$ module,so we do not necessarily have corresponding matrix and hence its trace.
$underline{Question}$:How do we define $f(operatorname{Spec}A)$ in this case?
$underline{Question}$(not directly related to previous discussion)}:For $mathcal F$ a rank $r$ locally free sheaf is $operatorname{End}(mathcal F)$ always a locally free sheaf of rank r?
Any help from anyone is welcome.
algebraic-geometry sheaf-theory vector-bundles
$endgroup$
2
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39
add a comment |
$begingroup$
$underline{Background}$:Let,$X$ be a scheme(for simplicity we can assume it to be noetherian).Let $mathcal F$ be a locally free sheaf of rank $r$ on $X$.Let, us also assume that {$operatorname{Spec}A_i:iin I$} is its trivializing open sets,i.e
we have $forall iin I$ , $mathcal F_{big|operatorname{Spec}cA_i}cong{mathcal O_{operatorname{Spec}cA_i^{oplus r}}}cong{oplus_{substack r}}tilde{A_i}$
$underline{Aim}$:I am trying to give a $O_X$ module morphism from $operatorname{End}(mathcal F)=mathcal Hom(mathcal F,mathcal F)$ to $O_X$.
$underline{Attempt}$:$1)$ we can give maps $f(operatorname{Spec}A_i):mathcal Hom (F_{big|operatorname{Spec}A_i},mathcal F_{big|operatorname{Spec}A_i}) to mathcal O(operatorname{Spec}A_i)=A_i$,
which is equivalent to give a morphism $f(operatorname{Spec}A_i):mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})to A_i$
Let,$gin mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})$
$Rightarrow g(operatorname{Spec}A_i):oplus_{substack r}{A_i}to oplus_{substack r}{A_i}$ is an $A_i$ module morphism.
Since $oplus_{substack r}{A_i}$ is a free $A_i$ module of rank r, let {$e_1,.....e_r$} be a basis(its canonical basis say).
Then let the matrix(w.r.to that basis) corresponding to $g(operatorname{Spec}A_i)$ is $mathcal M_g$.
Then we define $f(operatorname{Spec}A_i)(g)=operatorname{trace}(mathcal M_g)in{A_i}$
$underline{Question}$: Even if we assume that $f(operatorname{Spec}A_i)$s are valid $A_i$-module morphism how can we say that $exists phi :operatorname{End}(mathcal F)to {mathcal O_X}$ such that $phi(operatorname{Spec}A_i)$ is same as $f(operatorname{Spec}A_i)$ ?
Because we know that we can do that if we have {$operatorname{Spec}A_i:iin I$} forms a base of $X$ and $f(operatorname{Spec}A_i)$s are morphisms of sheaves on that base.but we only have {$operatorname{Spec}A_i:iin I$} forms a cover of $X$ ,not necessarily a base.
$underline{Attempt(2)}$:So to avoid the base problem if we define maps at all affine opens {$operatorname{Spec}A$}, then we have $g(operatorname{Spec}A):mathcal F(operatorname{Spec}A)to{mathcal F(operatorname{Spec}A)}$.
Now ,we have $mathcal F$ is coherent on $X$,so $ mathcal F_{big|operatorname{Spec}A}$s are coherent on $operatorname{Spec}A$.But that does not necessarily mean that $mathcal F(operatorname{Spec}A)$s are finitely generated $A$ module,so we do not necessarily have corresponding matrix and hence its trace.
$underline{Question}$:How do we define $f(operatorname{Spec}A)$ in this case?
$underline{Question}$(not directly related to previous discussion)}:For $mathcal F$ a rank $r$ locally free sheaf is $operatorname{End}(mathcal F)$ always a locally free sheaf of rank r?
Any help from anyone is welcome.
algebraic-geometry sheaf-theory vector-bundles
$endgroup$
$underline{Background}$:Let,$X$ be a scheme(for simplicity we can assume it to be noetherian).Let $mathcal F$ be a locally free sheaf of rank $r$ on $X$.Let, us also assume that {$operatorname{Spec}A_i:iin I$} is its trivializing open sets,i.e
we have $forall iin I$ , $mathcal F_{big|operatorname{Spec}cA_i}cong{mathcal O_{operatorname{Spec}cA_i^{oplus r}}}cong{oplus_{substack r}}tilde{A_i}$
$underline{Aim}$:I am trying to give a $O_X$ module morphism from $operatorname{End}(mathcal F)=mathcal Hom(mathcal F,mathcal F)$ to $O_X$.
$underline{Attempt}$:$1)$ we can give maps $f(operatorname{Spec}A_i):mathcal Hom (F_{big|operatorname{Spec}A_i},mathcal F_{big|operatorname{Spec}A_i}) to mathcal O(operatorname{Spec}A_i)=A_i$,
which is equivalent to give a morphism $f(operatorname{Spec}A_i):mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})to A_i$
Let,$gin mathcal Hom(oplus_{substack r}tilde{A_i},oplus_{substack r}tilde{A_i})$
$Rightarrow g(operatorname{Spec}A_i):oplus_{substack r}{A_i}to oplus_{substack r}{A_i}$ is an $A_i$ module morphism.
Since $oplus_{substack r}{A_i}$ is a free $A_i$ module of rank r, let {$e_1,.....e_r$} be a basis(its canonical basis say).
Then let the matrix(w.r.to that basis) corresponding to $g(operatorname{Spec}A_i)$ is $mathcal M_g$.
Then we define $f(operatorname{Spec}A_i)(g)=operatorname{trace}(mathcal M_g)in{A_i}$
$underline{Question}$: Even if we assume that $f(operatorname{Spec}A_i)$s are valid $A_i$-module morphism how can we say that $exists phi :operatorname{End}(mathcal F)to {mathcal O_X}$ such that $phi(operatorname{Spec}A_i)$ is same as $f(operatorname{Spec}A_i)$ ?
Because we know that we can do that if we have {$operatorname{Spec}A_i:iin I$} forms a base of $X$ and $f(operatorname{Spec}A_i)$s are morphisms of sheaves on that base.but we only have {$operatorname{Spec}A_i:iin I$} forms a cover of $X$ ,not necessarily a base.
$underline{Attempt(2)}$:So to avoid the base problem if we define maps at all affine opens {$operatorname{Spec}A$}, then we have $g(operatorname{Spec}A):mathcal F(operatorname{Spec}A)to{mathcal F(operatorname{Spec}A)}$.
Now ,we have $mathcal F$ is coherent on $X$,so $ mathcal F_{big|operatorname{Spec}A}$s are coherent on $operatorname{Spec}A$.But that does not necessarily mean that $mathcal F(operatorname{Spec}A)$s are finitely generated $A$ module,so we do not necessarily have corresponding matrix and hence its trace.
$underline{Question}$:How do we define $f(operatorname{Spec}A)$ in this case?
$underline{Question}$(not directly related to previous discussion)}:For $mathcal F$ a rank $r$ locally free sheaf is $operatorname{End}(mathcal F)$ always a locally free sheaf of rank r?
Any help from anyone is welcome.
algebraic-geometry sheaf-theory vector-bundles
algebraic-geometry sheaf-theory vector-bundles
edited Dec 1 '18 at 11:35
HARRY
asked Dec 1 '18 at 6:59
HARRYHARRY
889
889
2
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39
add a comment |
2
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39
2
2
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For your first question : no you don't need to define a morphism of sheaves $f:mathcal{F}tomathcal{G}$ on a basis. It is enough to define it on a cover ${U_i}$. However, you need to check that this agree on the two-fold intersections $U_icap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $mathcal{F}$ is locally isomorphic to $mathcal{O}_X^r$, so $End(mathcal{F})$ is locally isomorphic to $End(mathcal{O}_X^r)$ which is the sheaf of $rtimes r$-matrices. Hence $End(mathcal{F})$ is locally free of rank $r^2$.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021079%2fon-operatornameend-mathcal-f-of-a-locally-free-sheaf-of-finite-rank%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For your first question : no you don't need to define a morphism of sheaves $f:mathcal{F}tomathcal{G}$ on a basis. It is enough to define it on a cover ${U_i}$. However, you need to check that this agree on the two-fold intersections $U_icap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $mathcal{F}$ is locally isomorphic to $mathcal{O}_X^r$, so $End(mathcal{F})$ is locally isomorphic to $End(mathcal{O}_X^r)$ which is the sheaf of $rtimes r$-matrices. Hence $End(mathcal{F})$ is locally free of rank $r^2$.
$endgroup$
add a comment |
$begingroup$
For your first question : no you don't need to define a morphism of sheaves $f:mathcal{F}tomathcal{G}$ on a basis. It is enough to define it on a cover ${U_i}$. However, you need to check that this agree on the two-fold intersections $U_icap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $mathcal{F}$ is locally isomorphic to $mathcal{O}_X^r$, so $End(mathcal{F})$ is locally isomorphic to $End(mathcal{O}_X^r)$ which is the sheaf of $rtimes r$-matrices. Hence $End(mathcal{F})$ is locally free of rank $r^2$.
$endgroup$
add a comment |
$begingroup$
For your first question : no you don't need to define a morphism of sheaves $f:mathcal{F}tomathcal{G}$ on a basis. It is enough to define it on a cover ${U_i}$. However, you need to check that this agree on the two-fold intersections $U_icap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $mathcal{F}$ is locally isomorphic to $mathcal{O}_X^r$, so $End(mathcal{F})$ is locally isomorphic to $End(mathcal{O}_X^r)$ which is the sheaf of $rtimes r$-matrices. Hence $End(mathcal{F})$ is locally free of rank $r^2$.
$endgroup$
For your first question : no you don't need to define a morphism of sheaves $f:mathcal{F}tomathcal{G}$ on a basis. It is enough to define it on a cover ${U_i}$. However, you need to check that this agree on the two-fold intersections $U_icap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $mathcal{F}$ is locally isomorphic to $mathcal{O}_X^r$, so $End(mathcal{F})$ is locally isomorphic to $End(mathcal{O}_X^r)$ which is the sheaf of $rtimes r$-matrices. Hence $End(mathcal{F})$ is locally free of rank $r^2$.
answered Dec 1 '18 at 10:04
RolandRoland
7,0041913
7,0041913
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021079%2fon-operatornameend-mathcal-f-of-a-locally-free-sheaf-of-finite-rank%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
The natural map is from $mathcal{O}$ to $mathcal{End}(mathcal{F})$, sending 1 to the identity. This splits, at least in characteristic zero, by the trace map going the other way.
$endgroup$
– Mohan
Dec 1 '18 at 14:28
$begingroup$
@Mohan ,Is the trace map going from $End(mathcal F)$ to $O_X$ same as the map that I have tried to construct in Attempt (1) by using trace of the corresponding matrix ?is this why it is called the trace map?
$endgroup$
– HARRY
Dec 1 '18 at 14:34
$begingroup$
Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
$endgroup$
– Mohan
Dec 1 '18 at 17:39