This section is the continuation of Deformation Theory, Section 85.13 which we urge the reader to read first. We briefly recall the setup. We have a first order thickening $t : (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}'), \mathcal{O}_{\mathcal{B}'})$ of ringed topoi with $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$, a morphism of ringed topoi $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B})$, an $\mathcal{O}$-module $\mathcal{G}$, and a map $f^{-1}\mathcal{J} \to \mathcal{G}$ of sheaves of $f^{-1}\mathcal{O}_\mathcal {B}$-modules. We ask whether we can find the question mark fitting into the following diagram
86.22.0.1
\begin{equation} \label{cotangent-equation-to-solve-ringed-topoi} \vcenter { \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & {?} \ar[r] & \mathcal{O} \ar[r] & 0 \\ 0 \ar[r] & f^{-1}\mathcal{J} \ar[u]^ c \ar[r] & f^{-1}\mathcal{O}_{\mathcal{B}'} \ar[u] \ar[r] & f^{-1}\mathcal{O}_\mathcal {B} \ar[u] \ar[r] & 0 } } \end{equation}
and moreover how unique the solution is (if it exists). More precisely, we look for a first order thickening $i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ and a morphism of thickenings $(f, f')$ as in Deformation Theory, Equation (85.9.1.1) where $\mathop{\mathrm{Ker}}(i^\sharp )$ is identified with $\mathcal{G}$ such that $(f')^\sharp $ induces the given map $c$. We will say $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ is a solution to (86.22.0.1).
Lemma 86.22.1. In the situation above we have
There is a canonical element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {O}(L_ f, \mathcal{G})$ whose vanishing is a sufficient and necessary condition for the existence of a solution to (86.22.0.1).
If there exists a solution, then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(L_ f, \mathcal{G})$.
Given a solution $X'$, the set of automorphisms of $X'$ fitting into (86.22.0.1) is canonically isomorphic to $\mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {O}(L_ f, \mathcal{G})$.
Proof.
Via the identifications $\mathop{N\! L}\nolimits _ f = \tau _{\geq -1}L_ f$ (Lemma 86.21.4) and $H^0(L_{X/S}) = \Omega _{X/S}$ (Lemma 86.21.2) we have seen parts (2) and (3) in Deformation Theory, Lemmas 85.13.1 and 85.13.3.
Proof of (1). We will use the results of Deformation Theory, Lemma 85.13.4 without further mention. Denote
\[ p : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (*), \mathbf{Z}) \quad \text{and}\quad q : (\mathop{\mathit{Sh}}\nolimits (\mathcal{B}), \mathcal{O}_\mathcal {B}) \to (\mathop{\mathit{Sh}}\nolimits (*), \mathbf{Z}). \]
Let $\alpha \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_\mathcal {B}}(\mathop{N\! L}\nolimits _ q, \mathcal{J})$ be the element corresponding to the isomorphism class of $\mathcal{O}_{\mathcal{B}'}$. The existence of $\mathcal{O}'$ corresponds to an element $\beta \in \mathop{\mathrm{Ext}}\nolimits _\mathcal {O}^1(\mathop{N\! L}\nolimits _ p, \mathcal{G})$ which maps to the image of $\alpha $ in $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _ q, \mathcal{G})$. Note that
\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _ q, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*L_ q, \mathcal{G}) \]
and
\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _ p, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(L_ p, \mathcal{G}) \]
by Lemma 86.21.4. The distinguished triangle of Lemma 86.21.3 for $p = q \circ f$ gives rise to a long exact sequence
\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(L_ p, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*L_ q, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(L_ f, \mathcal{G}) \to \ldots \]
We obtain the result with $\xi $ the image of $\alpha $.
$\square$
Comments (2)
Comment #3285 by Eric Ahlqvist on
Comment #3372 by Johan on