Serre duality explicitly on curves

Question

Consider a Riemann surface $X$, with genus and marking so that a suitable moduli space exists. It's a well-known fact that the tangent space to that moduli space at $X$ (in other words, the space of first-order deformations of $X$) is dual to the space $\Omega_X^{\otimes 2}(X)$ of holomorphic quadratic differentials on $X$. This fact goes back at least to Teichmuller, who used quadratic differentials to explicitly deform the Cauchy-Riemann operator and get a new complex structure.

A more algebraic approach proceeds by noting that this tangent space equals the space of lifts of $X\to \mathbb C$ to $\mathbb C[\epsilon]/\epsilon^2$. (See Harris and Morrison Moduli of Curves.) For a reason I'm not sure of, such a lift is Zariski-locally trivial (i.e. isomorphic to $U\times_\mathbb{C} \mathbb C[\epsilon]/\epsilon^2$, and so can be described by a cover of $C$, together with gluing automorphisms on the overlaps $$\mathscr O_X(U_i\cap U_j)[\epsilon]/\epsilon^2\to \mathscr O_X(U_i\cap U_j)[\epsilon]/\epsilon^2$$ satisfying a cocycle condition.

An $R$-automorphism of the dual numbers over $R$ is the same thing as a derivation of $R$, so we have a system of derivations satisfying a cocycle condition. In other words, a first-order deformation gives a 1-Cech cocycle taking values in the tangent bundle, and by Serre duality $H^1(X,T_X)$ is dual to the space of quadratic differentials.

I'd really like to see a transparent way to turn a smooth curve $\mathfrak{X}\to\mathbb C[\epsilon]/\epsilon^2$ into a quadratic differential on the special fiber, and while the first step of the above is pretty explicit, Serre duality is less so.

Is there, in general or in this specific case (a smooth curve, Riemann surface, or even just for the tangent bundle) an explicit way to write down a pairing on Cech cocycles that descends to Serre duality?

In particular, can we explicitly describe a pairing between the set of first-order deformations and the quadratic differentials?

Answer 1

For any curious parties, here is one possible answer to this question (without proof - it follows from comments in Harris and Morrison Moduli of Curves).

Consider a deformation $\mathfrak{X}$ of $X$ and a quadratic differential $q$. $q$ defines a map $T_X \to \Omega_X$ by multiplication.

1. The inner product $\left\langle\mathfrak{X},q\right\rangle$ is equal to zero if and only if this lifts to a map $T_\mathfrak{X}\to\Omega_X$ which restricts to multiplication-by-$q$ on $T_X\subseteq T_\mathfrak{X}$.
2. Otherwise, the pushout of the $T_X\subseteq T_\mathfrak{X}$ along $q$ gives a diagram: $\require{AMScd}$ \begin{CD} T_X @>>> T_{\mathfrak{X}} \\ @VVV @VVV \\ \Omega_X @>{\left\langle\mathfrak{X},q\right\rangle^{-1}}>> \mathscr{O}_{X\times X}/\Delta^2 \end{CD} where the lower map is the usual inclusion, but scaled by the inverse of the inner product. This gives an explicit calculation of the pairing, but not of the isomorphism between the deformation space and the space of quadratic differentials - that'd likely require analyzing the Weil-Petersson product of quadratic differentials.

Finally, my confusion about local triviality: the point is that near a smooth point of $X$, any such extension is trivial, which is certainly intuitive since it is topoloigically and even analytically trivial. In fact, this follows from the fact that the sheaf of Kahler differentials is free, so the local ext group $\text{Ext}^1_{\mathscr{O}_{X,x}}(\Omega_{X,x},\mathscr{O_X,x})$ vanishes. In the singular case, there is a discrepancy between the space of locally trivial deformations (which are parametrized by $H^1(X,\mathscr O_X)$ as in the smooth case) and the space of all deformations. The quotient of the space of deformations by the subspace of locally trivial ones is isomorphic to the direct sum of the local Ext-groups at the singularities.