Trace of a morphism of sheaves

$$\underline {Background}$$: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.

• Where did you see this? – Eric Wofsey Dec 31 '18 at 5:35
• @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 – HARRY Dec 31 '18 at 5:42
• 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. – Eric Wofsey Dec 31 '18 at 5:48
• 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. – K B Dave Dec 31 '18 at 5:48
• @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$\phi=0$ – 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 $$\phi$$.
• @ 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 $\mathcal F$ – HARRY Dec 31 '18 at 8:06
• Wolfsey,can you explain why it does not depend upon trivilizations of $\mathcal L$ – HARRY Dec 31 '18 at 8:24
• 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 $0$. – Eric Wofsey Dec 31 '18 at 8:30