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Suppose $C$ is a complex projective curve (or a compact $1$-dimensional connected reduced complex space). If $C$ is smooth, then its module of differentials $\Omega^1_C$ is locally free. If $C$ is a singular prestable curve (i.e. its singularities are at worst ordinary nodes), then $\Omega^1_C$ is coherent, but not a locally free sheaf.

Can we find a canonical finite complex of vector bundles $E^\bullet$ on $C$ resolving $\Omega^1_C$?

It's easy to find such a canonical resolution locally near a node. Let $p\in C$ be a node and assume that the two branches meeting at $C$ are $A,B$. Then, we may embed $C$ locally into $A\times B$ as $A\times\{p\}\cup\{p\}\times B$. This is the zero locus of a tautologically defined section $s$ of the bundle $L =\mathcal O_A(p)\boxtimes\mathcal O_B(p)$. Then, the conormal sequence $0\to L^\vee|_C\to\Omega^1_{A\times B}|_C\to\Omega^1_C\to0$ is a (local) resolution by a two term complex of vector bundles. I don't see any obvious way to globalize this construction though.

My main motivation is to set up a Dolbeault type complex to better understand the vector space $\text{Ext}^1(\Omega^1_C,\mathcal O_C)$ which is the space of first order deformations of $C$. More generally, I also want to use this type of Dolbeault complex to understand the deformations of (possibly nodal) pseudo-holomorphic maps in almost complex manifolds.

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    $\begingroup$ Here is a first thought. If it had a finite resolution as you write (canonical or otherwise), if $C$ is irreducible and sheaf of differential forms is torsion free, then it would be locally free by Auslanser-Buchsbaum. This will solve a long standing conjecture known as Berger's conjecture. So, the answer is most likely to be unknown in your case. $\endgroup$
    – Mohan
    Commented Jul 29, 2020 at 20:23
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    $\begingroup$ Embed the prestable curve $C$ as a Cartier divisor in a surface $S$. Then the resolution is $\mathcal{O}_S(-\underline{C})|_C \to\Omega_S|_C$. $\endgroup$
    – Jason Starr
    Commented Jul 29, 2020 at 21:54
  • $\begingroup$ ... "surface" --> "smooth surface". $\endgroup$
    – Jason Starr
    Commented Jul 29, 2020 at 22:31
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    $\begingroup$ If $C$ is prestable then its dualizing sheaf is a line bundle; if $C$ is stable this bundle is ample so you can use it to embed in projective space. Now project from a well chosen linear space to obtain an immersion in $\mathbb P^2$, and blow up $\mathbb P^2$ at those singularities which arose from projection (as opposed to those which are part of the abstract pre-stable curve). Now you have an embedding of $C$ in a rational surface. How is that? $\endgroup$
    – Tabes Bridges
    Commented Jul 30, 2020 at 2:50
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    $\begingroup$ Yes, this is pretty explicit. In the unstable case, I think we can just stabilize by adding marked points $p_1,\ldots,p_n$ and then take the ample line bundle $\omega_C(p_1 + \cdots + p_n)$ instead of $\omega_C$. $\endgroup$
    – Mohan Swaminathan
    Commented Jul 30, 2020 at 3:12

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