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91.23 Deformations of ringed topoi and the cotangent complex

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
\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 ( 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 (

Lemma 91.23.1. In the situation above we have

  1. 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 (

  2. If there exists a solution, then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(L_ f, \mathcal{G})$.

  3. Given a solution $X'$, the set of automorphisms of $X'$ fitting into ( 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

The notation in diagram ( differs from the definitions above it.

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