$\newcommand{\F}{\mathcal{F}}$ Let $\mathcal{F}$ be a $G$-equivariant sheaf over a smooth projective $G$-variety $X$, generated by global sections $s_1,...s_n$. One can construct the first locally free sheaf $\F_1$ in the resolution of $\F$. The sheaf is $\F_1=\Gamma(X, \F)$. It is defined as $\Gamma(X, \F)(U)=\rho_{U,X} \Gamma(X,\F)$.

(I am trying to simplify a construction given in Ginzburg's Representation Theory and Complex Geometry page 241)

Since this sheaf is supposed to be locally free, what is the vector bundle over $X$ such that the sheaf of sections is $\F_1$? In other words what is the vector bundle that corresponds to $\F_1$?

My first guess would have been $\epsilon^n$. This has the same global sections. But unfortunately, it has different sections over any open. In fact the space of sections over any open set in $X$ is going to be infinite dimensional whereas $\F_1$ has a finite dimensional (in fact with dimension less than $n$) space of sections over any open.


1 Answer 1


If $X$ is a variety over a field $k$, then one may take ${\cal F}_1 := \Gamma(X,{\cal F}) \otimes_k {\cal O}_X$. One gets the corresponding vector bundle from it using the standard natural construction, as in EGA, for example, and it would be isomorphic to a sum of a finite number of copies of the trivial vector bundle of rank 1.

It would be good to reserve the notation $\Gamma(X,{\cal F})$ for the $k$-vector space of global sections of $\cal F$, and to adopt a different notation for your sheaf.

Normally, the phrase locally free sheaf is interpreted to mean locally free sheaf of ${\cal O}_X$-Modules, not locally free sheaf of $k$-modules, which is what your definition seems to provide. Your proposed definition, if tensored over $k$ with ${\cal O}_X$, might work, and would be a little smaller than the standard choice above when $X$ is not connected, which is perhaps what you had in mind.

Notice that on that page in the book of Ginzburg none of the tensor product symbols $\otimes$ are decorated with the ring: some are over $k$ and some are over ${\cal O}_X$; it's up to the reader to provide the correct interpretation.

If $n$ is the dimension of $\Gamma(X,{\cal F})$ over $k$, then $k^n$ is isomorphic to $\Gamma(X,{\cal F})$, but the isomorphism is not canonical, since it depends on a choice of basis, so it is best to avoid identifying the two vector spaces.

Also, be careful about identifying $\Gamma(X,{\cal O}_X)$ with $k$, since the map $k \to \Gamma(X,{\cal O}_X)$ may not be an isomorphism.

It is best to refer to a resolution of $\cal F$, rather than the resolution of $\cal F$, since there are many.


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