relative Grassmannian

Question

When one considers the relative Grassmannian $\mathrm{Grass}(r,\mathcal{E})$ for a locally free sheaf of finite rank on some smooth projective variety $X$. Now consider an invertible sheaf $\mathcal{L}$ on $X$; Do we have $\mathrm{Grass}(r,\mathcal{E})\simeq \mathrm{Grass}(r,\mathcal{E}\otimes \mathcal{L})$?

Answer 1

This is best seen by the functorial definition of the Grassmannian (see EGA I, §9).

Let $X$ be an arbitrary(!) base scheme and $E$ be an arbitrary(!) quasi-coherent module on $X$. If $p : T \to X$ is an $X$-scheme, then $\mathrm{Grass}(r,E)(T)$ equals the set of (isomorphism classes of) pairs $(F,h)$, where $F$ is locally free of rank $r$ on $T$ and $h : p^*(E) \to F$ is an epimorphism of quasi-coherent module on $T$. If $L$ is an invertible module on $X$, then we may tensor $h$ with the invertible module $p^*(L)$ on $T$ to obtain an epimorphism $h : p^*(E \otimes L) \to F \otimes p^*(L)$. Notice that $F \otimes p^*(L)$ is still locally free of rank $r$. This establishes a natural map $$\mathrm{Grass}(r,E)(T) \to \mathrm{Grass}(r,E \otimes L)(T).$$ It is invertible $-$ the inverse map just tensors with $p^*(L^{\otimes -1})$. By the Yoneda-Lemma, this induces an isomorphism of $X$-schemes $$\mathrm{Grass}(r,E) \to \mathrm{Grass}(r,E \otimes L).$$ More generally, if $L$ is locally free of rank $s$, there is a natural map of $X$-schemes $$\mathrm{Grass}(r,E) \to \mathrm{Grass}(r \cdot s,E \otimes L).$$