Trace of a morphism of sheaves












1












$begingroup$


$underline {Background}$:Suppose $X$ be a scheme and $mathcal F$ and $mathcal G$ are two sheaves of $mathcal O_X$ modules.



Also assume that $exists ${U}$ $ a cover of $X$ by open sets such that each $mathcal F(U),mathcal G(U)$ are finitely free $mathcal O(U)$ modules.



Let $ exists phi:mathcal Ftomathcal G$ a sheaf of $mathcal O_X$ modules.



$underline {Question}$:what is the meaning of the statement "trace$phi=0$"



$underline {Guess}$:Does it mean for all open set $V$ of $X$



$phi(V):mathcal F(V)to mathcal G(V)$ has trace of the matrix of $phi(V)=0$



But this makes sense only for $V=U$ because,we only have $phi(U)$ is a morphism between 2 finitely free $mathcal O(U)$ modules,and hence its matrix w.r.to canonical bases exists.



So,I would like to know the appropriate definition of trace being $0$ and any text which mentions it clearly.



Any help from anyone is welcome.






algebraic-geometry sheaf-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Where did you see this?
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:35










  • $begingroup$
    @Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:42








  • 2




    $begingroup$
    You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:48










  • $begingroup$
    I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
    $endgroup$
    – K B Dave
    Dec 31 '18 at 5:48










  • $begingroup$
    @Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:55
















1












$begingroup$


$underline {Background}$:Suppose $X$ be a scheme and $mathcal F$ and $mathcal G$ are two sheaves of $mathcal O_X$ modules.



Also assume that $exists ${U}$ $ a cover of $X$ by open sets such that each $mathcal F(U),mathcal G(U)$ are finitely free $mathcal O(U)$ modules.



Let $ exists phi:mathcal Ftomathcal G$ a sheaf of $mathcal O_X$ modules.



$underline {Question}$:what is the meaning of the statement "trace$phi=0$"



$underline {Guess}$:Does it mean for all open set $V$ of $X$



$phi(V):mathcal F(V)to mathcal G(V)$ has trace of the matrix of $phi(V)=0$



But this makes sense only for $V=U$ because,we only have $phi(U)$ is a morphism between 2 finitely free $mathcal O(U)$ modules,and hence its matrix w.r.to canonical bases exists.



So,I would like to know the appropriate definition of trace being $0$ and any text which mentions it clearly.



Any help from anyone is welcome.






algebraic-geometry sheaf-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Where did you see this?
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:35










  • $begingroup$
    @Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:42








  • 2




    $begingroup$
    You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:48










  • $begingroup$
    I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
    $endgroup$
    – K B Dave
    Dec 31 '18 at 5:48










  • $begingroup$
    @Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:55














1












1








1





$begingroup$


$underline {Background}$:Suppose $X$ be a scheme and $mathcal F$ and $mathcal G$ are two sheaves of $mathcal O_X$ modules.



Also assume that $exists ${U}$ $ a cover of $X$ by open sets such that each $mathcal F(U),mathcal G(U)$ are finitely free $mathcal O(U)$ modules.



Let $ exists phi:mathcal Ftomathcal G$ a sheaf of $mathcal O_X$ modules.



$underline {Question}$:what is the meaning of the statement "trace$phi=0$"



$underline {Guess}$:Does it mean for all open set $V$ of $X$



$phi(V):mathcal F(V)to mathcal G(V)$ has trace of the matrix of $phi(V)=0$



But this makes sense only for $V=U$ because,we only have $phi(U)$ is a morphism between 2 finitely free $mathcal O(U)$ modules,and hence its matrix w.r.to canonical bases exists.



So,I would like to know the appropriate definition of trace being $0$ and any text which mentions it clearly.



Any help from anyone is welcome.






algebraic-geometry sheaf-theory







share|cite|improve this question









$endgroup$




$underline {Background}$:Suppose $X$ be a scheme and $mathcal F$ and $mathcal G$ are two sheaves of $mathcal O_X$ modules.



Also assume that $exists ${U}$ $ a cover of $X$ by open sets such that each $mathcal F(U),mathcal G(U)$ are finitely free $mathcal O(U)$ modules.



Let $ exists phi:mathcal Ftomathcal G$ a sheaf of $mathcal O_X$ modules.



$underline {Question}$:what is the meaning of the statement "trace$phi=0$"



$underline {Guess}$:Does it mean for all open set $V$ of $X$



$phi(V):mathcal F(V)to mathcal G(V)$ has trace of the matrix of $phi(V)=0$



But this makes sense only for $V=U$ because,we only have $phi(U)$ is a morphism between 2 finitely free $mathcal O(U)$ modules,and hence its matrix w.r.to canonical bases exists.



So,I would like to know the appropriate definition of trace being $0$ and any text which mentions it clearly.



Any help from anyone is welcome.






algebraic-geometry sheaf-theory




algebraic-geometry sheaf-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 5:29









HARRYHARRY

889




889












  • $begingroup$
    Where did you see this?
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:35










  • $begingroup$
    @Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:42








  • 2




    $begingroup$
    You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:48










  • $begingroup$
    I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
    $endgroup$
    – K B Dave
    Dec 31 '18 at 5:48










  • $begingroup$
    @Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:55


















  • $begingroup$
    Where did you see this?
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:35










  • $begingroup$
    @Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:42








  • 2




    $begingroup$
    You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 5:48










  • $begingroup$
    I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
    $endgroup$
    – K B Dave
    Dec 31 '18 at 5:48










  • $begingroup$
    @Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
    $endgroup$
    – HARRY
    Dec 31 '18 at 5:55
















$begingroup$
Where did you see this?
$endgroup$
– Eric Wofsey
Dec 31 '18 at 5:35




$begingroup$
Where did you see this?
$endgroup$
– Eric Wofsey
Dec 31 '18 at 5:35












$begingroup$
@Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
$endgroup$
– HARRY
Dec 31 '18 at 5:42






$begingroup$
@Eric Wolfsey ,Actually ,I have seen it in a handwritten note where $mathcal F$ is a vector bundle and $mathcal G$ is some line bundle tensor $mathcal F$ .But,I have generalized those properties in question to get a meaning of trace map being $0$ in a general situation
$endgroup$
– HARRY
Dec 31 '18 at 5:42






2




2




$begingroup$
You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
$endgroup$
– Eric Wofsey
Dec 31 '18 at 5:48




$begingroup$
You have generalized way too far! It was very very crucial that $mathcal{G}$ was a line bundle tensor $mathcal{F}$, in order for the trace being $0$ to be meaningful.
$endgroup$
– Eric Wofsey
Dec 31 '18 at 5:48












$begingroup$
I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
$endgroup$
– K B Dave
Dec 31 '18 at 5:48




$begingroup$
I'm not an expert, but a priori I wouldn't expect the notion of trace to make sense except when $mathcal{G}=mathcal{F}$, $mathcal{F}$ locally finite free, in which case I believe one globalizes the trace for finitely generated projective modules.
$endgroup$
– K B Dave
Dec 31 '18 at 5:48












$begingroup$
@Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
$endgroup$
– HARRY
Dec 31 '18 at 5:55




$begingroup$
@Eric wolfsey,even when in that special situation the problem that I mentioned in the question arises.So even in that special case how one makes sense of tr$phi=0$
$endgroup$
– HARRY
Dec 31 '18 at 5:55










1 Answer
1






active

oldest

votes


















3












$begingroup$

As you say, the trace of $phi$ is only meaningful when $mathcal{G}=mathcal{F}$. Slightly more generally, though it still makes sense to talk about the trace of $phi$ being $0$ if you have a chosen isomorphism $mathcal{G}congmathcal{F}otimesmathcal{L}$ for some line bundle $mathcal{L}$. Indeed, picking a local trivialization of $mathcal{L}$, we get local isomorphisms $mathcal{G}congmathcal{F}$ which we can use to compute the trace. Now of course the trace will depend on the trivialization of $mathcal{L}$ chosen, but changing the trivialization just multiplies the isomorphism by some invertible section of $mathcal{O}_X$, and so will not change whether the trace is $0$.



Another way to see it is that a morphism $mathcal{F}tomathcal{F}otimesmathcal{L}$ is equivalent to a morphism $mathcal{L}^veetomathcal{F}otimesmathcal{F}^vee$. There is a canonical "trace" map $mathcal{F}otimesmathcal{F}^veetomathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $mathcal{L}^veetomathcal{O}_X$ (or equivalently, a section of $mathcal{L}$) which we can call the "trace" of $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:06












  • $begingroup$
    Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:24








  • 1




    $begingroup$
    Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 8:30











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1 Answer
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1 Answer
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active

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active

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active

oldest

votes









3












$begingroup$

As you say, the trace of $phi$ is only meaningful when $mathcal{G}=mathcal{F}$. Slightly more generally, though it still makes sense to talk about the trace of $phi$ being $0$ if you have a chosen isomorphism $mathcal{G}congmathcal{F}otimesmathcal{L}$ for some line bundle $mathcal{L}$. Indeed, picking a local trivialization of $mathcal{L}$, we get local isomorphisms $mathcal{G}congmathcal{F}$ which we can use to compute the trace. Now of course the trace will depend on the trivialization of $mathcal{L}$ chosen, but changing the trivialization just multiplies the isomorphism by some invertible section of $mathcal{O}_X$, and so will not change whether the trace is $0$.



Another way to see it is that a morphism $mathcal{F}tomathcal{F}otimesmathcal{L}$ is equivalent to a morphism $mathcal{L}^veetomathcal{F}otimesmathcal{F}^vee$. There is a canonical "trace" map $mathcal{F}otimesmathcal{F}^veetomathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $mathcal{L}^veetomathcal{O}_X$ (or equivalently, a section of $mathcal{L}$) which we can call the "trace" of $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:06












  • $begingroup$
    Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:24








  • 1




    $begingroup$
    Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 8:30
















3












$begingroup$

As you say, the trace of $phi$ is only meaningful when $mathcal{G}=mathcal{F}$. Slightly more generally, though it still makes sense to talk about the trace of $phi$ being $0$ if you have a chosen isomorphism $mathcal{G}congmathcal{F}otimesmathcal{L}$ for some line bundle $mathcal{L}$. Indeed, picking a local trivialization of $mathcal{L}$, we get local isomorphisms $mathcal{G}congmathcal{F}$ which we can use to compute the trace. Now of course the trace will depend on the trivialization of $mathcal{L}$ chosen, but changing the trivialization just multiplies the isomorphism by some invertible section of $mathcal{O}_X$, and so will not change whether the trace is $0$.



Another way to see it is that a morphism $mathcal{F}tomathcal{F}otimesmathcal{L}$ is equivalent to a morphism $mathcal{L}^veetomathcal{F}otimesmathcal{F}^vee$. There is a canonical "trace" map $mathcal{F}otimesmathcal{F}^veetomathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $mathcal{L}^veetomathcal{O}_X$ (or equivalently, a section of $mathcal{L}$) which we can call the "trace" of $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:06












  • $begingroup$
    Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:24








  • 1




    $begingroup$
    Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 8:30














3












3








3





$begingroup$

As you say, the trace of $phi$ is only meaningful when $mathcal{G}=mathcal{F}$. Slightly more generally, though it still makes sense to talk about the trace of $phi$ being $0$ if you have a chosen isomorphism $mathcal{G}congmathcal{F}otimesmathcal{L}$ for some line bundle $mathcal{L}$. Indeed, picking a local trivialization of $mathcal{L}$, we get local isomorphisms $mathcal{G}congmathcal{F}$ which we can use to compute the trace. Now of course the trace will depend on the trivialization of $mathcal{L}$ chosen, but changing the trivialization just multiplies the isomorphism by some invertible section of $mathcal{O}_X$, and so will not change whether the trace is $0$.



Another way to see it is that a morphism $mathcal{F}tomathcal{F}otimesmathcal{L}$ is equivalent to a morphism $mathcal{L}^veetomathcal{F}otimesmathcal{F}^vee$. There is a canonical "trace" map $mathcal{F}otimesmathcal{F}^veetomathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $mathcal{L}^veetomathcal{O}_X$ (or equivalently, a section of $mathcal{L}$) which we can call the "trace" of $phi$.






share|cite|improve this answer









$endgroup$



As you say, the trace of $phi$ is only meaningful when $mathcal{G}=mathcal{F}$. Slightly more generally, though it still makes sense to talk about the trace of $phi$ being $0$ if you have a chosen isomorphism $mathcal{G}congmathcal{F}otimesmathcal{L}$ for some line bundle $mathcal{L}$. Indeed, picking a local trivialization of $mathcal{L}$, we get local isomorphisms $mathcal{G}congmathcal{F}$ which we can use to compute the trace. Now of course the trace will depend on the trivialization of $mathcal{L}$ chosen, but changing the trivialization just multiplies the isomorphism by some invertible section of $mathcal{O}_X$, and so will not change whether the trace is $0$.



Another way to see it is that a morphism $mathcal{F}tomathcal{F}otimesmathcal{L}$ is equivalent to a morphism $mathcal{L}^veetomathcal{F}otimesmathcal{F}^vee$. There is a canonical "trace" map $mathcal{F}otimesmathcal{F}^veetomathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $mathcal{L}^veetomathcal{O}_X$ (or equivalently, a section of $mathcal{L}$) which we can call the "trace" of $phi$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 6:01









Eric WofseyEric Wofsey

182k12209337




182k12209337












  • $begingroup$
    @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:06












  • $begingroup$
    Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:24








  • 1




    $begingroup$
    Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 8:30


















  • $begingroup$
    @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:06












  • $begingroup$
    Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
    $endgroup$
    – HARRY
    Dec 31 '18 at 8:24








  • 1




    $begingroup$
    Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
    $endgroup$
    – Eric Wofsey
    Dec 31 '18 at 8:30
















$begingroup$
@ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
$endgroup$
– HARRY
Dec 31 '18 at 8:06






$begingroup$
@ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $phi|_U:mathcal{F}|_Utomathcal {F}|_U$ with $U$ local trivialization of $mathcal L$ $Rightarrow phi|_Uinmathcal End(mathcal F)(U))$ $Rightarrow tr(U)(phi|_U)in mathcal O(U)$ and we mean by tr$phi=0$ that all $ tr(U)(phi|_U)=0$ for all member of that trivialization.So it does not depend upon the trivialization of $mathcal F$
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– HARRY
Dec 31 '18 at 8:06














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Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
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– HARRY
Dec 31 '18 at 8:24






$begingroup$
Wolfsey,can you explain why it does not depend upon trivilizations of $mathcal L$
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– HARRY
Dec 31 '18 at 8:24






1




1




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Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
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– Eric Wofsey
Dec 31 '18 at 8:30




$begingroup$
Changing the trivialization of $mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $mathcal{G}|_Utomathcal{F}|_U$ by some invertible element $ainmathcal{O}(U)$. So, when we turn $phi|_U$ into a matrix, we just multiply that matrix by the scalar $a$, which multiplies its trace by $a$. Since $a$ is invertible, this does not change whether the trace is $0$.
$endgroup$
– Eric Wofsey
Dec 31 '18 at 8:30


















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