On $operatorname{End}(mathcal F)$ of a locally free sheaf of finite rank












0














$underline{Background}$: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.










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  • 2




    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.
    – Mohan
    Dec 1 at 14:28










  • @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?
    – HARRY
    Dec 1 at 14:34










  • Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
    – Mohan
    Dec 1 at 17:39
















0














$underline{Background}$: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.










share|cite|improve this question




















  • 2




    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.
    – Mohan
    Dec 1 at 14:28










  • @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?
    – HARRY
    Dec 1 at 14:34










  • Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
    – Mohan
    Dec 1 at 17:39














0












0








0







$underline{Background}$: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.










share|cite|improve this question















$underline{Background}$: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






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edited Dec 1 at 11:35

























asked Dec 1 at 6:59









HARRY

629




629








  • 2




    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.
    – Mohan
    Dec 1 at 14:28










  • @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?
    – HARRY
    Dec 1 at 14:34










  • Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
    – Mohan
    Dec 1 at 17:39














  • 2




    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.
    – Mohan
    Dec 1 at 14:28










  • @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?
    – HARRY
    Dec 1 at 14:34










  • Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
    – Mohan
    Dec 1 at 17:39








2




2




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.
– Mohan
Dec 1 at 14:28




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.
– Mohan
Dec 1 at 14:28












@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?
– HARRY
Dec 1 at 14:34




@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?
– HARRY
Dec 1 at 14:34












Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
– Mohan
Dec 1 at 17:39




Yes, it is correct. To get a splitting, of course, you have to divide by the rank of the vector bundle, hence characteristic zero.
– Mohan
Dec 1 at 17:39










1 Answer
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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 $r^2$.






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    1 Answer
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    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 $r^2$.






    share|cite|improve this answer


























      3














      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 $r^2$.






      share|cite|improve this answer
























        3












        3








        3






        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 $r^2$.






        share|cite|improve this answer












        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 $r^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 10:04









        Roland

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