Section 91.5 (08LG): Infinitesimal deformations of modules on ringed spaces

91.5 Infinitesimal deformations of modules on ringed spaces

Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. We freely use the notation introduced in Section 91.3. Let $\mathcal{F}'$ be an $\mathcal{O}_{X'}$ -module and set $\mathcal{F} = i^*\mathcal{F}'$ . In this situation we have a short exact sequence

$0 \to \mathcal{I}\mathcal{F}' \to \mathcal{F}' \to \mathcal{F} \to 0$

of $\mathcal{O}_{X'}$ -modules. Since $\mathcal{I}^2 = 0$ the $\mathcal{O}_{X'}$ -module structure on $\mathcal{I}\mathcal{F}'$ comes from a unique $\mathcal{O}_ X$ -module structure. Thus the sequence above is an extension as in (91.4.0.1). As a special case, if $\mathcal{F}' = \mathcal{O}_{X'}$ we have $i^*\mathcal{O}_{X'} = \mathcal{O}_ X$ and $\mathcal{I}\mathcal{O}_{X'} = \mathcal{I}$ and we recover the sequence of structure sheaves

$0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$

Lemma 91.5.1. Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. Let $\mathcal{F}'$ , $\mathcal{G}'$ be $\mathcal{O}_{X'}$ -modules. Set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$ . Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$ -linear map. The set of lifts of $\varphi$ to an $\mathcal{O}_{X'}$ -linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ is, if nonempty, a principal homogeneous space under $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{I}\mathcal{G}')$ .

Proof. This is a special case of Lemma 91.4.1 but we also give a direct proof. We have short exact sequences of modules

$0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \quad \text{and}\quad 0 \to \mathcal{I}\mathcal{G}' \to \mathcal{G}' \to \mathcal{G} \to 0$

and similarly for $\mathcal{F}'$ . Since $\mathcal{I}$ has square zero the $\mathcal{O}_{X'}$ -module structure on $\mathcal{I}$ and $\mathcal{I}\mathcal{G}'$ comes from a unique $\mathcal{O}_ X$ -module structure. It follows that

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{I}\mathcal{G}') \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$

The lemma now follows from the exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G})$

see Homology, Lemma 12.5.8. $\square$

Lemma 91.5.2. Let $(f, f')$ be a morphism of first order thickenings of ringed spaces as in Situation 91.3.1. Let $\mathcal{F}'$ be an $\mathcal{O}_{X'}$ -module and set $\mathcal{F} = i^*\mathcal{F}'$ . Assume that $\mathcal{F}$ is flat over $S$ and that $(f, f')$ is a strict morphism of thickenings (Definition 91.3.2). Then the following are equivalent

$\mathcal{F}'$ is flat over $S'$ , and
the canonical map $f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ is an isomorphism.

Moreover, in this case the maps

$f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$

are isomorphisms.

Proof. The map $f^*\mathcal{J} \to \mathcal{I}$ is surjective as $(f, f')$ is a strict morphism of thickenings. Hence the final statement is a consequence of (2).

Proof of the equivalence of (1) and (2). We may check these conditions at stalks. Let $x \in X \subset X'$ be a point with image $s = f(x) \in S \subset S'$ . Set $A' = \mathcal{O}_{S', s}$ , $B' = \mathcal{O}_{X', x}$ , $A = \mathcal{O}_{S, s}$ , and $B = \mathcal{O}_{X, x}$ . Then $A = A'/J$ and $B = B'/I$ for some square zero ideals. Since $(f, f')$ is a strict morphism of thickenings we have $I = JB'$ . Let $M' = \mathcal{F}'_ x$ and $M = \mathcal{F}_ x$ . Then $M'$ is a $B'$ -module and $M$ is a $B$ -module. Since $\mathcal{F} = i^*\mathcal{F}'$ we see that the kernel of the surjection $M' \to M$ is $IM' = JM'$ . Thus we have a short exact sequence

$0 \to JM' \to M' \to M \to 0$

Using Sheaves, Lemma 6.26.4 and Modules, Lemma 17.16.1 to identify stalks of pullbacks and tensor products we see that the stalk at $x$ of the canonical map of the lemma is the map

$(J \otimes _ A B) \otimes _ B M = J \otimes _ A M = J \otimes _{A'} M' \longrightarrow JM'$

The assumption that $\mathcal{F}$ is flat over $S$ signifies that $M$ is a flat $A$ -module.

Assume (1). Flatness implies $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Lemma 10.75.8. This means $J \otimes _{A'} M' \to M'$ is injective by Algebra, Remark 10.75.9. Hence $J \otimes _ A M \to JM'$ is an isomorphism.

Assume (2). Then $J \otimes _{A'} M' \to M'$ is injective. Hence $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Remark 10.75.9. Hence $M'$ is flat over $A'$ by Algebra, Lemma 10.99.8. $\square$

Lemma 91.5.3. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}'$ , $\mathcal{G}'$ be $\mathcal{O}_{X'}$ -modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$ . Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$ -linear map. Assume that $\mathcal{G}'$ is flat over $S'$ and that $(f, f')$ is a strict morphism of thickenings. The set of lifts of $\varphi$ to an $\mathcal{O}_{X'}$ -linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ is, if nonempty, a principal homogeneous space under

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J})$

Proof. Combine Lemmas 91.5.1 and 91.5.2. $\square$

Lemma 91.5.4. Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. Let $\mathcal{F}'$ , $\mathcal{G}'$ be $\mathcal{O}_{X'}$ -modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$ . Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$ -linear map. There exists an element

$o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')$

whose vanishing is a necessary and sufficient condition for the existence of a lift of $\varphi$ to an $\mathcal{O}_{X'}$ -linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ .

Proof. It is clear from the proof of Lemma 91.5.1 that the vanishing of the boundary of $\varphi$ via the map

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}')$

is a necessary and sufficient condition for the existence of a lift. We conclude as

$\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')$

the adjointness of $i_* = Ri_*$ and $Li^*$ on the derived category (Cohomology, Lemma 20.28.1). $\square$

Lemma 91.5.5. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}'$ , $\mathcal{G}'$ be $\mathcal{O}_{X'}$ -modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$ . Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$ -linear map. Assume that $\mathcal{F}'$ and $\mathcal{G}'$ are flat over $S'$ and that $(f, f')$ is a strict morphism of thickenings. There exists an element

$o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J})$

whose vanishing is a necessary and sufficient condition for the existence of a lift of $\varphi$ to an $\mathcal{O}_{X'}$ -linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ .

First proof. This follows from Lemma 91.5.4 as we claim that under the assumptions of the lemma we have

$\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J})$

Namely, we have $\mathcal{I}\mathcal{G}' = \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J}$ by Lemma 91.5.2. On the other hand, observe that

$H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{O}_ X)$

(local computation omitted). Using the short exact sequence

$0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$

we see that this $\text{Tor}_1$ is computed by the kernel of the map $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ which is zero by the final assertion of Lemma 91.5.2. Thus $\tau _{\geq -1}Li^*\mathcal{F}' = \mathcal{F}$ . On the other hand, we have

$\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\tau _{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')$

by the dual of Derived Categories, Lemma 13.16.1. $\square$

Second proof. We can apply Lemma 91.4.2 as follows. Note that $\mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ and $\mathcal{L} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{G}$ by Lemma 91.5.2, that $c_{\mathcal{F}'} = 1 \otimes 1$ and $c_{\mathcal{G}'} = 1 \otimes 1$ and taking $\psi = 1 \otimes \varphi$ the diagram of the lemma commutes. Thus $o(\varphi ) = o(\varphi , 1 \otimes \varphi )$ works. $\square$

Lemma 91.5.6. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}$ be an $\mathcal{O}_ X$ -module. Assume $(f, f')$ is a strict morphism of thickenings and $\mathcal{F}$ flat over $S$ . If there exists a pair $(\mathcal{F}', \alpha )$ consisting of an $\mathcal{O}_{X'}$ -module $\mathcal{F}'$ flat over $S'$ and an isomorphism $\alpha : i^*\mathcal{F}' \to \mathcal{F}$ , then the set of isomorphism classes of such pairs is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}( \mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F})$ .

Proof. If we assume there exists one such module, then the canonical map

$f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$

is an isomorphism by Lemma 91.5.2. Apply Lemma 91.4.3 with $\mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ and $c = 1$ . By Lemma 91.5.2 the corresponding extensions $\mathcal{F}'$ are all flat over $S'$ . $\square$

Lemma 91.5.7. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}$ be an $\mathcal{O}_ X$ -module. Assume $(f, f')$ is a strict morphism of thickenings and $\mathcal{F}$ flat over $S$ . There exists an $\mathcal{O}_{X'}$ -module $\mathcal{F}'$ flat over $S'$ with $i^*\mathcal{F}' \cong \mathcal{F}$ , if and only if

the canonical map $f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ is an isomorphism, and
the class $o(\mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}, 1) \in \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}( \mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F})$ of Lemma 91.4.4 is zero.

Proof. This follows immediately from the characterization of $\mathcal{O}_{X'}$ -modules flat over $S'$ of Lemma 91.5.2 and Lemma 91.4.4. $\square$

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91.5 Infinitesimal deformations of modules on ringed spaces

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