# Map to a given vector bundle from a split vector bundle

Let $$X$$ be a connected scheme, smooth and proper over $$\mathbb{C}$$. Let $$F$$ be a locally free $$\mathcal{O}_X$$-module of finite rank $$r>1$$. Suppose on a non-empty affine open $$U\subset X$$ whose complement is irreducible we have an isomorphism of $$\mathcal{O}_X$$-modules $$f:\mathcal{O}^{\oplus r}_X|_{U}\rightarrow F|_{U}$$.

Does there necessarily exist a locally free $$\mathcal{O}_X$$-module $$L$$ of rank $$1$$ with an identification $$\mathcal{O}_X|_U\rightarrow L|_U$$ such that there is a morphism of $$\mathcal{O}_X$$-modules $$g:L^{\oplus r}\rightarrow F$$ with $$g|_{U}=f$$?

• Try the tautological bundle over the $r$th Grassmanian: I don't think it admits any nontrivial maps from line bundles. Apr 27 '19 at 10:23

Yes. Slightly more precisely, we have the following result.

Proposition: Let $$X$$ be a normal, integral, finite-type scheme over a field. Let $$F$$ be a locally free $$\mathcal O_X$$-module of finite rank. Let $$U\subseteq X$$ be a nonempty open subset. Let $$r$$ be a nonnegative integer. Let $$\phi : \mathcal O_U^{\oplus r}\to F|_U$$ be an $$\mathcal O_U$$-linear map. Then there exists an effective Weil divisor $$D$$ supported on $$X\setminus U$$ and an $$\mathcal O_X$$-linear map $$\bar \phi : \mathcal O_X(-D)^{\oplus r} \to F$$ such that $$\bar\phi|_U = \phi$$. Furthermore, if $$X$$ is $$\mathbb Q$$-factorial, then the divisor $$D$$ can be chosen to be Cartier.

Proof: Given an effective Weil divisor $$D$$ on $$X$$ with support on $$X\setminus U$$, there exists at most one $$\mathcal O_X$$-linear map $$\phi(D) : \mathcal O_X(-D)^{\oplus r} \to F$$ such that $$\phi(D)|_U = \phi$$. If $$\phi(D)$$ exists and $$E$$ is an effective Weil divisor supported on $$X\setminus U$$, then $$\phi(D+E)$$ exists, since we can take $$\phi(D+E)$$ to be the restriction of $$\phi(D)$$ to $$\mathcal O_X(-D-E)^{\oplus r}\subseteq \mathcal O_X(-D)^{\oplus r}$$. In particular, if $$\phi(D)$$ exists, then $$\phi(nD)$$ exists for all positive integers $$n$$. The last statement of the proposition follows.

To find an effective Weil divisor $$D$$ on $$X$$ supported on $$X\setminus U$$ such that $$\phi(D)$$ exists, we may work locally on $$X$$. Indeed, let $$X=\bigcup_{i\in I} V_i$$ be a finite Zariski-open cover. Suppose that, for each $$i\in I$$, there exists an effective Weil divisor $$D_i$$ on $$V_i$$ with support on $$V_i\setminus U$$ and an $$\mathcal O_{V_i}$$-linear map $$\bar\phi_i : \mathcal O_{V_i}(-D_i)^{\oplus r}\to F|_{V_i}$$ such that $$\bar\phi_i|_{V_i\cap U} = \phi_{V_i\cap U}$$. For each $$i\in I$$, let $$\overline D_i\subseteq X$$ be the closure of $$D_i \subseteq V_i$$. Let $$D= \sum_{i\in I} \overline D_i$$. Then $$\mathcal O_X(-D)|_{V_i} \subseteq \mathcal O_{V_i}(-D_i)$$ for all $$i\in i$$, and there exists a unique map $$\bar\phi : \mathcal O_X(-D)^{\oplus r} \to F$$ such that $$\bar\phi|_{V_i} = \bar\phi_i|\mathcal O_{V_i}(-D\cap V_i)^{\oplus r}$$ for all $$i\in I$$.

Thus we may assume that $$F=\mathcal O_X^{\oplus f}$$. Then the $$\mathcal O_U$$-linear map $$\phi : \mathcal O_U^{\oplus r}\to F|_U$$ is given by multiplication by a matrix $$[\phi_{ij}]$$ of rational functions on $$X$$. Let $$D_1,\dotsc,D_m$$ denote the irreducible components of $$X\setminus U$$ that have codimension 1 in $$X$$. For each $$l=1,\dotsc, m$$, $$i=1,\dotsc,f$$ and $$j=1,\dotsc,r$$, let $$n_l(i,j)\ge 0$$ denote the order of the pole that the rational function $$\phi_{ij}$$ has at $$D_l$$. For each $$l=1,\dotsc, m$$, let $$n_l := \max_{i,j} n_l(i,j)$$. Let $$D= \sum_{i=1}^m n_l D_l$$. Then $$\phi(D)$$ exists. QED

• are you sure normality, as opposed to factoriality, is enough?
– user138661
Apr 28 '19 at 3:43
• @schematic_boi You are right, that is not clear from the argument I wrote. I edited my answer, adding the assumption that X is Q-factorial and replacing a divisor by a multiple. Apr 28 '19 at 9:27