# Explicit locally free resolution of a perfect complex $E\oplus F\to (E\oplus F)\otimes \mathcal{O}_X(D)\to (E\otimes \mathcal{O}_X(D))|_D$

Let $$X$$ be a smooth projective variety over $$\mathbb{C}$$. Let $$E,F\to X$$ be 2 holomorphic vector bundles and $$D\hookrightarrow X$$ be a smooth divisor. Denote by $$\mathcal{O}_X(D)$$ the line bundle associated to the divisor and let $$s: \mathcal{O}_X\to \mathcal{O}_X(D)$$ be a section, such that $$s^{-1}(0) = D$$. Assume also that we have an isomorphism $$E|_D\cong F|_D$$ of the restrictions to $$D$$.

Consider the following perfect complex $$\mathcal{E}^\bullet$$ on $$X$$: $$\begin{equation*} E\oplus F\xrightarrow{ \begin{pmatrix} \text{id}\otimes s& 0\\ 0&\text{id}\otimes s \end{pmatrix}} (E\oplus F)\otimes \mathcal{O}_X(D)\xrightarrow{ \begin{pmatrix} \rho_E&\rho_F \end{pmatrix} } (E\otimes \mathcal{O}_X(D))|_D\,, \end{equation*}$$ where $$\rho_{(-)}$$ corresponds to restriction of sections to $$D$$, and we use the isomorphism $$F|_D\cong E|_D$$.

Now it is well known that there exists a locally free resolution of this complex on $$X$$. However, is it possible to write it down explicitly from what we know?

Edit: By a locally free resolution I mean a complex of vector bundles $$L^\bullet$$ with a quasi-isomorphism $$L^\bullet\to \mathcal{E}^\bullet$$. It is important to me that I have such a map, as I would like to use it to construct a differential operator.

Projections to the second summands define a morphism from that complex to the complex $$F \stackrel{s}\to F(D)\tag{*}$$ of locally free sheaves. The cone of this morphism is the complex $$0 \to E \stackrel{s}\to E(D) \stackrel{\rho_E}\to E(D)\vert_D \to 0$$ which is acyclic. Therefore, $$(*)$$ is a locally free resolution of the original complex.
EDIT. Alternatively, let $$K = \mathrm{Ker}\Big((\rho_E, \rho_F) \colon E(D) \oplus F(D) \to E(D)\vert_D\Big).$$ Then $$K$$ is locally free, the morphism $$(s, s) \colon E \oplus F \to E(D) \oplus F(D)$$ factors through $$K$$, and the complex $$E \oplus F \to K$$ is quasiisomorphic to the original one (via the morphism, which is identical on $$E \oplus F$$ and is the natural embedding on $$K$$).
• For me a resolutions is a complex of vector bundle $L^\bullet$ with a map $L^\bullet \to \mathcal{E}^\bullet$ which is a quasi-isomorphism. The map goes the wrong way in your case. Apr 12, 2020 at 11:46
• Thank you for the edit Sasha. Could you please add some argument explaining why $K$ is locally free? I am struggling to see why that is true. Apr 12, 2020 at 15:59