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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.

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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 $K$).

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    $\begingroup$ 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. $\endgroup$ – Arkadij Apr 12 '20 at 11:46
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    $\begingroup$ For me a resolution is a quasiisomorphism, no matter which direction it goes. $\endgroup$ – Sasha Apr 12 '20 at 12:33
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    $\begingroup$ I have changed the question to be more explicit now. It is important to me that it goes the way that I have mentioned, otherwise it is trivial as your answer shows. $\endgroup$ – Arkadij Apr 12 '20 at 12:36
  • $\begingroup$ 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. $\endgroup$ – Arkadij Apr 12 '20 at 15:59
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    $\begingroup$ Because the projective dimension of a locally free sheaf on a Cartier divisor is 1. $\endgroup$ – Sasha Apr 12 '20 at 16:01

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