# On action of sheaf of symmetric algebra

$$\underline {Background}$$:Let,$$X$$ be a projective scheme over a field $$K$$.

Let, $$\phi:\mathcal F\to\mathcal F\otimes\mathcal L$$ be a morphism of $$\mathcal O_X$$ modules for some vector bundle $$\mathcal F$$ of rank $$n$$ and for some line bundle $$\mathcal L$$.

Let us denote $$T$$:=total space of line bundle $$\mathcal L$$=relative spectrum of(Sym($$\mathcal L^*$$))

By universal property we have a morphism $$p:T\to X$$

$$\underline {Question}$$:(1) What is the meaning of the statement "$$\phi:\mathcal F\to\mathcal F\otimes\mathcal L$$ corresponds to an action of Sym($$\mathcal L^*$$) on the sheaf $$\mathcal F$$ "

(I mean what is the precise meaning of action of one sheaf over another?)

(2) How do we construct a coherent sheaf of modules over $$T$$ out of this action?

Any help from anyone is welcome.



Well, we can define the morphism $$\tau: \Sym(\calL^*)\to \shHom(\calF,\calF)$$ on open affine sets $$U$$, and on degree one elements of $$\Sym(\calL^*)$$, $$\alpha\in\Sym(\calL^*)(U)_1$$, define $$\tau_U(\alpha) = \mu_U\circ (1\otimes\alpha)\circ \phi_U$$, where $$\mu:\calF\otimes \calO_X\to \calF$$ is the usual natural isomorphism.

You do have to check that the images of degree one elements commute in $$\shHom(\calF,\calF)$$ to make sure that the morphism is well defined. This isn't hard to check though.


Anyway, in our particular case, the concrete construction goes like this.

Let $$p:T\to X$$ be the structure map. Fix $$\{U_i\}$$ an open cover of $$X$$. $$p\newcommand\inv{^{-1}}\inv(U_i)=\Spec (\Sym(\calL^*)(U_i))$$, and $$\calF(U_i)$$ is conveniently a module over $$\Sym(\calL^*)(U_i)$$, so it defines a quasicoherent sheaf on the spectrum, and the restriction maps coming from the sheaf structure of $$\calF$$ induce gluing maps for the corresponding modules on $$T$$, allowing the sheaves corresponding to $$\calF(U_i)$$ to glue together into a quasicoherent sheaf on $$T$$.

Then you just need to check that, let's call it $$\tilde{\calF}$$, is in fact coherent, not just quasicoherent. However, this follows since $$K$$ is noetherian, so $$X$$ is noetherian, so $$T$$ is noetherian, hence the fact that $$\calF$$ is finitely generated over $$\Sym(\calL^*)$$ (which contains $$\calO_X$$) implies that it is coherent.