# When are two resolutions of a coherent sheaf homotopic

Let $$\mathcal{F}$$ be a coherent sheaf on a projective manifold $$X$$. It is well known that one can construct a resolution of $$\mathcal{F}$$ by holomorphic vector bundles (locally free sheaves).

Are two such resolutions homotopic? Any reference would be much appreciated.

• Crossposted at MSE at the same time it was posted here. Please see meta for site rules about cross-posting. Nov 15, 2021 at 3:28

If this were true, then any short exact sequence of vector bundles would split. Indeed, if $$0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0$$ is a short exact sequence of vector bundles, then both \begin{align*} K^\bullet = \cdots \to 0 \to \mathscr E_1 \to \mathscr E_2 \to 0 \to \cdots \end{align*} and $$L^\bullet = \mathscr E_3[0]$$ are resolutions of $$\mathscr E_3$$. If $$g \colon L^\bullet \to K^\bullet$$ is a homotopy equivalence (or even a quasi-isomorphism!), then the map $$g^0 \colon \mathscr E_3 \to \mathscr E_2$$ induces an isomorphism $$\phi \colon \mathscr E_3 \stackrel\sim\to H^0(K^\bullet) = \mathscr E_3.$$ Then $$g^0 \circ \phi^{-1} \colon \mathscr E_3 \to \mathscr E_2$$ is a splitting of $$0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0$$. $$\square$$
(On the other hand, there does always exist a quasi-isomorphism $$K^\bullet \to L^\bullet$$ in this case, just not in the other direction.)
An example of a short exact sequence of vector bundles that doesn't split is the Koszul sequence $$0 \to \mathcal O_{\mathbf P^1}(-2) \stackrel{\left(\begin{smallmatrix}-y \\ x\end{smallmatrix}\right)}\longrightarrow \mathcal O_{\mathbf P^1}(-1) \oplus \mathcal O_{\mathbf P^1}(-1) \stackrel{\left(x\ \ y\right)}\longrightarrow \mathcal O_{\mathbf P^1} \to 0$$ on $$X = \mathbf P^1$$. Indeed, $$\operatorname{Hom}(\mathcal O_{\mathbf P^1}, \mathcal O_{\mathbf P^1}(-1) \oplus \mathcal O_{\mathbf P^1}(-1)) = 0$$.
No. The simplest example is given by the following two resolutions of the structure sheaf of a point $$P \in \mathbb{P}^1$$: $$0 \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{\mathbb{P}^1} \to \mathcal{O}_{P} \to 0$$ and $$0 \to \mathcal{O}_{\mathbb{P}^1}(-2) \to \mathcal{O}_{\mathbb{P}^1}(-1) \to \mathcal{O}_{P} \to 0$$ (the second is obtained from the first by a twist).