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 (91.23.0.1).

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 (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. $\square$

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