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$\DeclareMathOperator\Ext{Ext}$Let $C \in \bar{\mathcal{M}}_g$ be a nodal curve. It is a classical result that the tangent space of $\bar{\mathcal{M}}_g$ at $C$ is given by \begin{align*} T_C \bar{\mathcal{M}}_g= \Ext^1(\Omega_C, \mathcal{O}_C) \end{align*}

If $C$ is in the boundary of $\bar{\mathcal{M}}_g$ one can also consider $C$ as an element of the boundary strata $\mathcal{M}^\Gamma$ consisting of those curves with same dual graph or topological type as $C$. If we assume that $C$ (or more precisely $\Gamma$, but this is not relevant here) has no automorphisms, we can conclude that \begin{align*} T_C \mathcal{M}^\Gamma = \Ext^1(\Omega_{\tilde{C}}(D), \mathcal{O}_\tilde{C}) \end{align*} where $D$ is the reduced divisor supported at the preimages of the nodes under the normalization $\nu: \tilde{C} \to C$ as any deformation is given by a product of deformations of the connected components of the normalization. The twist is necessary as we have to keep track of the preimages of nodes as "marked" points.

Now the immersion $\mathcal{M}^\Gamma \to \bar{\mathcal{M}}_g$ induces an injection on tangent spaces $T_C \mathcal{M}^\Gamma \to T_C \bar{\mathcal{M}}_g$. My problem is, that I do not see a canonical map \begin{align*} \Ext^1(\Omega_{\tilde{C}}(D), \mathcal{O}_\tilde{C}) \to \Ext^1(\Omega_C, \mathcal{O}_C) \end{align*} as at some point morphisms go in the wrong direction, so something has to be wrong. I would be happy if someone could explain how one can describe this inclusion of tangent spaces as algebraic as possible.

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    $\begingroup$ You are twisting by $-D$ where you should twist by $+D$: consider the case where $\widetilde{C}$ is a disjoint union of two elliptic curves and $C$ has arithmetic genus $2$. $\endgroup$
    – Jason Starr
    Commented Dec 20, 2023 at 21:44
  • $\begingroup$ Thank you for pointing this out. Now I have a morphism $\Omega_C \to \nu_* \Omega_\tilde{C}(D)$ but still no morphism $\nu_* \mathcal{O}_\tilde{C} \to \mathcal{O}_C$, so I do not see yet why your proposed example is easier than the general case. $\endgroup$
    – Matthias
    Commented Dec 20, 2023 at 23:22
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    $\begingroup$ Since $\Omega_{\widetilde{C}}$ is locally free, $\text{Ext}^1_{\mathcal{O}_{\widetilde{C}}}(\Omega_{\widetilde{C}}(D),\mathcal{O}_{\widetilde{C}})$ is canonically isomorphic to $H^1(\widetilde{C},T_{\widetilde{C}}(-D))$. Since pushforward by the finite (thus affine) morphism $\nu$ is exact on the category of quasi-coherent sheaves, this is canonically isomorphic to $H^1(C,\nu_*T_{\widetilde{C}}(-D))$. Similarly $\text{Ext}^1_{\mathcal{O}_C}(\Omega_C,\mathcal{O}_C)$ is canonically isomorphic to $\mathbb{H}^1(C,\mathbf{R}\textit{Hom}_{\mathcal{O}_C}(\Omega_C,\mathcal{O}_C))$ . . . $\endgroup$
    – Jason Starr
    Commented Dec 21, 2023 at 0:25
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    $\begingroup$ . . . So it suffices to give a morphism in the derived category from $\nu_*T_{\widetilde{C}}(-D)[0]$ to $\mathbb{R}\textit{Hom}_{\mathcal{O}_C}(\Omega_C,\mathcal{O}_C)$. Both of these complexes are canonically quasi-isomorphic when pulled back to $U=C\setminus\nu(D)$. By adjunction, both of these map to the pushforwards from $U$ of their pullbacks to $U$, and these are canonically isomorphic. The claim is that there is a unique map of complexes that commutes with these canonical maps. That claim can be checked locally, e.g., for the coordinate axes in $\mathbb{A}^2$. $\endgroup$
    – Jason Starr
    Commented Dec 21, 2023 at 0:29

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