This section is the continuation of Deformation Theory, Section 90.13 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
91.23.0.1
\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 (90.9.1.1) 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 (91.23.0.1).
Lemma 91.23.1. 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 (91.23.0.1).
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 (91.23.0.1) 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 91.22.4) and $H^0(L_ f) = \Omega _ f$ (Lemma 91.22.2) we have seen parts (2) and (3) in Deformation Theory, Lemmas 90.13.1 and 90.13.3.
Proof of (1). To match notation with Deformation Theory, Section 90.13 we will write $\mathop{N\! L}\nolimits _ f = \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}$ and $L_ f = L_{\mathcal{O}/\mathcal{O}_\mathcal {B}}$ and similarly for the morphisms $t$ and $t \circ f$. By Deformation Theory, Lemma 90.13.8 there exists an element
\[ \xi ' \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*\mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \]
such that a solution exists if and only if this element is in the image of the map
\[ \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \]
The distinguished triangle of Lemma 91.22.3 for $f$ and $t$ gives rise to a long exact sequence
\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G}) \]
Hence taking $\xi $ the image of $\xi '$ works.
$\square$
Comments (2)
Comment #3285 by Eric Ahlqvist on
Comment #3372 by Johan on