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(I'm trying to learn logarithmic geometry, and I'm extremely confused about something very basic in log deformation theory. It's very likely my question is nonsense, but I don't get it.)

Let's pick for instance a reducible nodal curve $C$ over an algebraically closed field (characteristic 0, if it matters), $C = C_1 \cup C_2$ with $C_i$ smooth and irreducible meeting at a point $x$. Let $Def^1(C) = \mathrm{Ext}^1(\Omega_C,{\mathscr O}_C)$ be the usual space of first order deformations. (When I write $\Omega_C$ I always mean the non-logarithmic one.)

If I understand correctly, there is a canonical way to think of $C$ as a log curve $(C,{\mathcal M}_C)$ (I'm talking about the "unmarked" version with characteristic $\overline{\mathcal M}_C = {\mathbb Z}_x$) over the standard logarithmic point. Let $Def^1_{log}(C)$ be the space of first order deformations of $(C,{\mathcal M}_C)$,

EDIT: by which I mean the following: let $B_0$ be the standard logarithmic point, and $B$ given by $\underline{B} = \mathrm{Spec} \mathop{} \mathbb{C}[\epsilon]/(\epsilon^2)$, with its logarithmic structure obtained by pulling back the log structure on ${\mathbb A}^1$ relative to the divisor $0$ along $\underline{B} \to {\mathbb A}^1 = \mathrm{Spec} \mathop{} {\mathbb C}[t]$ corresponding to $t \mapsto \epsilon$; then $Def^1_{log}(C)$ is the space of (integral, log smooth) deformations of $(C,{\mathcal M}_C)$ over $B$.

According to a general theorem (EDIT: which applies in the setup of the edit above), this should be an affine space for $H^1(C,T^{log}_C) = H^1(C,\omega_C^\vee)$, where $\omega_C$ is the dualizing sheaf of the usual curve.

The question. I'm confused about the relation between these spaces of deformations. I think there should be a map $$ \psi:Def^1_{log}(C) \to Def^1(C) $$ which just forgets the logarithmic structure, but I can't begin to comprehend what this map looks like. (The image of this map doesn't contain $0$, right?) Is it an affine map? If yes, which one?

Some thoughts. There is a canonical map $\Omega_C \to \omega_C$, which induces $\phi: H^1(C,\omega_C^\vee) = \mathrm{Ext}^1(\omega_C,{\mathcal O}_C) \to {\mathrm{Ext}}^1(\Omega_C,{\mathcal O}_C)$. A wild guess: maybe $\psi$ is a "translate" of $\phi$? Then what's the distinguished element of $Coker(\phi)$?

Bonus: What happens for more complicated nodal curves?

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  • $\begingroup$ Why doesn't the image of $\psi$ contain $0$? I think the element $0$ should correspond to the trivial deformation and the trivial deformation of the log scheme $(C, M_C)$ maps to the trivial deformation of $C$ when you forget $M_C$. $\endgroup$ Commented Mar 30, 2022 at 3:44
  • $\begingroup$ @ Dori Bejleri: You're right I didn't define properly the log deformations I'm talking about. You're supposed to fix the log structure on the thickened base scheme in advance, and if you take the one which allows "smoothings", I think that one disallows the trivial deformation. Edited the question. (I'm still not sure I know what I'm talking about, but at least note that the general theorem about 1st order log deformations says "torsor" or something equivalent -- if the trivial deformation was allowed, it wouldn't say that). $\endgroup$
    – aiz89
    Commented Mar 30, 2022 at 11:41

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