Construction of a line bundle from a class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ as $\mathcal{O}_X^{\times}$-Torsor

Question

Let $X$ be a complex compact manifold, and write $\mathcal{O}_X$ for the sheaf of holomorphic functions on $X$. Let $\mathcal{O}_X^{\times}$ be the subsheaf consisting of holomorphic functions. These are both sheaves of abelian groups. If we identify $H^1(X, \mathcal{O}_X^{\times})$ with the Picard group $\text{Pic}(X)$, then we can regard classes of $H^1(X, \mathcal{O}_X^{\times})$ geometrically as iso classes of line bundles on $X$.

Surely, one way to obtain from an abstract class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ the line bundle back is via the Cech cocycle data containing the whole glueing information about the associated line bundle up to isomorphism.

Recently I saw a different construction associating to an abstract class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ a representing line bundle and I would like to learn more about the background of this construction and why this gives finally a line bundle which coincides with that one obtained from Cech cycle. Does it has a name under which it can be looked up in literature? What is it's geometrical origin? Is there some construction from (differential) topology motivating it?

Here the construction: We take some injective resolution $1 \to \mathcal{O}_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1 \to ... $ Then a $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ is represented by construction by a section $\alpha \in H^0(X, \mathcal{I}^1)$ becoming zero in $H^0(X, \mathcal{I}^2)$.

We consider the bundle $\mathcal{M}$ obtained as fibre product

$$ \require{AMScd} \begin{CD} \mathcal{M} @>{} >> \mathbb{Z}_X \\ @VVV @VV{\alpha}V \\ \mathcal{I}^0 @>{}>> \mathcal{I}^1 \end{CD} $$

where $\mathbb{Z}_X \to \mathcal{I}^1$ is induced by $\alpha$ via $1 \vert _U \to \alpha \vert _U $. Since $ \mathcal{I}^0 $ contains $\mathcal{M}$ and $\mathcal{O}_X^{\times}$ as abelian subsheaves, there is a action $\mathcal{O}_X^{\times} \times \mathcal{M} \to \mathcal{M}$ by $\mathcal{O}_X^{\times}$, which restricts on $\mathcal{O}_X^{\times}$ to the usual multiplication $\mathcal{O}_X^{\times} \times \mathcal{O}_X^{\times} \to \mathcal{O}_X^{\times}$.
It's well defined because for $U \subset X$ the sections $\mathcal{M}(U)$ are given as sections in $\mathcal{I}^0(U)$ mapping to $\alpha \vert _U$ and $\mathcal{O}_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1$ is exact.

Then we obtain an invertible sheaf $\mathcal{L}$ as $(\mathcal{M} \times \mathcal{O}_X)/\mathcal{O}_X^{\times}$ by moding out the diagonal action.

The claim is that this coincides with the line bundle obtained from class $[\alpha]$ via Cech cycle. Why? The construction reminds me on algebraic analoga of topological associated bundle. The aim behind the construction is obviously to exchange the fiber $\mathcal{O}_X^{\times}$ by the "right" fiber $\mathcal{O}_X$ giving the resulting object the desired structure of a line bundle. (This fiber replacing business is essentially for what the construction of associated bundle good for. Morally one wants to keep the same base space & in certain sense the same twisting behaviour, but replace the fiber)

But I not see from this construction why this one gives up to isomorphism the same line bundle one obtains from the Cech cocycle data.

Answer 1

Let $\mathcal M$ be the sub-sheaf of $\mathcal I^0$ of sections mapping to (the restrictions of) $\alpha$. (I suppose that's what the OP has intended anyways.) That's easily seen to be an $\mathcal{O}_X^\times$-torsor. By Stacks Project Tag 0A6G, if $X=\bigcup_i U_i$ is an open cover trivialising this torsor, say via sections $s_i\in\mathcal M(U_i)$, then the Čech cocycle defined by the $s_{ij}\in\mathcal O_{U_{ij}}^\times$ such that $s_{ij}(s_i|_{U_{ij}})=s_j|_{U_{ij}}$ does indeed represent $[\alpha]\in H^1(\mathcal O_X^\times)$.

Moreover, let $\mathcal L$ be the quotient sheaf of the action of $\mathcal O_X^\times$ on $\mathcal M\times \mathcal O_X$ which acts on sections as $s.(m,z)=(sm,s^{-1}z)$ – that's not quite the diagonal action, but exactly what you should be familiar with from the associated bundle construction. $\mathcal L$ is trivialised over the same open cover by the classes $[s_i,1]$ of the sections $(s_i,1)\in \mathcal M(U_i)\times\mathcal O_X(U_i)$ and by construction, $s_{ij}[s_i,1]=[s_i,s_{ij}]=[s_{ij}s_i,1]=[s_j,1]$, i.e., the cocycle yields the transition functions of the associated line bundle.

By the way, the construction works quite generally, cf. Stacks Project Tag 040E.