Lemma 115.21.3. In the situation of Remark 115.21.2 assume that $\mathcal{F}$ is flat over $U$. Then the vanishing of the class $\xi _{U'}$ is a necessary and sufficient condition for the existence of a $\mathcal{O}_{X \times _ B U'}$-module $\mathcal{F}'$ flat over $U'$ with $i^*\mathcal{F}' \cong \mathcal{F}$.

Proof (sketch). We will use the criterion of Deformation Theory, Lemma 91.11.8. We will abbreviate $\mathcal{O} = \mathcal{O}_{X \times _ B U}$ and $\mathcal{O}' = \mathcal{O}_{X \times _ B U'}$. Consider the short exact sequence

$0 \to \mathcal{I} \to \mathcal{O}_{U'} \to \mathcal{O}_ U \to 0.$

Let $\mathcal{J} \subset \mathcal{O}'$ be the quasi-coherent sheaf of ideals cutting out $X \times _ B U$. By the above we obtain an exact sequence

$\text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U) \to q^*\mathcal{I} \to \mathcal{J} \to 0$

where the $\text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U)$ is an abbreviation for

$\text{Tor}_1^{h^{-1}\mathcal{O}_ B}(p^{-1}\mathcal{O}_ X, q^{-1}\mathcal{O}_ U) \otimes _{(p^{-1}\mathcal{O}_ X\otimes _{h^{-1}\mathcal{O}_ B}q^{-1}\mathcal{O}_ U)} \mathcal{O}.$

Tensoring with $\mathcal{F}$ we obtain the exact sequence

$\mathcal{F} \otimes _\mathcal {O} \text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U) \to \mathcal{F} \otimes _\mathcal {O} q^*\mathcal{I} \to \mathcal{F} \otimes _\mathcal {O} \mathcal{J} \to 0$

(Note that the roles of the letters $\mathcal{I}$ and $\mathcal{J}$ are reversed relative to the notation in Deformation Theory, Lemma 91.11.8.) Condition (1) of the lemma is that the last map above is an isomorphism, i.e., that the first map is zero. The vanishing of this map may be checked on stalks at geometric points $\overline{z} = (\overline{x}, \overline{u}) : \mathop{\mathrm{Spec}}(k) \to X \times _ B U$. Set $R = \mathcal{O}_{B, \overline{b}}$, $A = \mathcal{O}_{X, \overline{x}}$, $B = \mathcal{O}_{U, \overline{u}}$, and $C = \mathcal{O}_{\overline{z}}$. By Cotangent, Lemma 92.28.2 and the defining triangle for $E(\mathcal{F})$ we see that

$H^{-2}(E(\mathcal{F}))_{\overline{z}} = \mathcal{F}_{\overline{z}} \otimes \text{Tor}_1^ R(A, B)$

The map $\xi _{U'}$ therefore induces a map

$\mathcal{F}_{\overline{z}} \otimes \text{Tor}_1^ R(A, B) \longrightarrow \mathcal{F}_{\overline{z}} \otimes _ B \mathcal{I}_{\overline{u}}$

We claim this map is the same as the stalk of the map described above (proof omitted; this is a purely ring theoretic statement). Thus we see that condition (1) of Deformation Theory, Lemma 91.11.8 is equivalent to the vanishing $H^{-2}(\xi _{U'}) : H^{-2}(E(\mathcal{F})) \to \mathcal{F} \otimes \mathcal{I}$.

To finish the proof we show that, assuming that condition (1) is satisfied, condition (2) is equivalent to the vanishing of $\xi _{U'}$. In the rest of the proof we write $\mathcal{F} \otimes \mathcal{I}$ to denote $\mathcal{F} \otimes _\mathcal {O} q^*\mathcal{I} = \mathcal{F} \otimes _\mathcal {O} \mathcal{J}$. A consideration of the spectral sequence

$\mathop{\mathrm{Ext}}\nolimits ^ i(H^{-j}(E(\mathcal{F})), \mathcal{F} \otimes \mathcal{I}) \Rightarrow \mathop{\mathrm{Ext}}\nolimits ^{i + j}(E(\mathcal{F}), \mathcal{F} \otimes \mathcal{I})$

using that $H^0(E(\mathcal{F})) = \mathcal{F}$ and $H^{-1}(E(\mathcal{F})) = 0$ shows that there is an exact sequence

$0 \to \mathop{\mathrm{Ext}}\nolimits ^2(\mathcal{F}, \mathcal{F} \otimes \mathcal{I}) \to \mathop{\mathrm{Ext}}\nolimits ^2(E(\mathcal{F}), \mathcal{F} \otimes \mathcal{I}) \to \mathop{\mathrm{Hom}}\nolimits (H^{-2}(E(\mathcal{F})), \mathcal{F} \otimes \mathcal{I})$

Thus our element $\xi _{U'}$ is an element of $\mathop{\mathrm{Ext}}\nolimits ^2(\mathcal{F}, \mathcal{F} \otimes \mathcal{I})$. The proof is finished by showing this element agrees with the element of Deformation Theory, Lemma 91.11.8 a verification we omit. $\square$

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