$\underline {Background}$: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 F\to\mathcal 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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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 $\mathcal{G}\cong\mathcal{F}\otimes\mathcal{L}$ for some line bundle $\mathcal{L}$. Indeed, picking a local trivialization of $\mathcal{L}$, we get local isomorphisms $\mathcal{G}\cong\mathcal{F}$ 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 $\mathcal{F}\to\mathcal{F}\otimes\mathcal{L}$ is equivalent to a morphism $\mathcal{L}^\vee\to\mathcal{F}\otimes\mathcal{F}^\vee$. There is a canonical "trace" map $\mathcal{F}\otimes\mathcal{F}^\vee\to\mathcal{O}_X$ (just the evaluation pairing) and so we can compose to get a morphism $\mathcal{L}^\vee\to\mathcal{O}_X$ (or equivalently, a section of $\mathcal{L}$) which we can call the "trace" of $\phi$.

  • $\begingroup$ @ Eric Wolfsey,If I understand you correctly then are you trying to say the following? $\phi|_U:\mathcal{F}|_U\to\mathcal {F}|_U$ with $U$ local trivialization of $\mathcal L$ $\Rightarrow \phi|_U\in\mathcal End(\mathcal F)(U))$ $\Rightarrow tr(U)(\phi|_U)\in \mathcal O(U)$ 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
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    $\begingroup$ Changing the trivialization of $\mathcal{L}|_U$ just amounts to multiplying the chosen isomorphism $\mathcal{G}|_U\to\mathcal{F}|_U$ by some invertible element $a\in\mathcal{O}(U)$. 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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