## 76.17 Infinitesimal deformations of maps

In this section we explain how a derivation can be used to infinitesimally move a map. Throughout this section we use that a sheaf on a thickening $X'$ of $X$ can be seen as a sheaf on $X$, see Equations (76.9.1.1) and (76.9.1.2).

Lemma 76.17.1. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $X \subset X'$ and $Y \subset Y'$ be two first order thickenings of algebraic spaces over $B$. Let $(a, a'), (b, b') : (X \subset X') \to (Y \subset Y')$ be two morphisms of thickenings over $B$. Assume that

1. $a = b$, and

2. the two maps $a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ (Lemma 76.5.3) are equal.

Then the map $(a')^\sharp - (b')^\sharp$ factors as

$\mathcal{O}_{Y'} \to \mathcal{O}_ Y \xrightarrow {D} a_*\mathcal{C}_{X/X'} \to a_*\mathcal{O}_{X'}$

where $D$ is an $\mathcal{O}_ B$-derivation.

Proof. Instead of working on $Y$ we work on $X$. The advantage is that the pullback functor $a^{-1}$ is exact. Using (1) and (2) we obtain a commutative diagram with exact rows

$\xymatrix{ 0 \ar[r] & \mathcal{C}_{X/X'} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & a^{-1}\mathcal{C}_{Y/Y'} \ar[r] \ar[u] & a^{-1}\mathcal{O}_{Y'} \ar[r] \ar@<1ex>[u]^{(a')^\sharp } \ar@<-1ex>[u]_{(b')^\sharp } & a^{-1}\mathcal{O}_ Y \ar[r] \ar[u] & 0 }$

Now it is a general fact that in such a situation the difference of the $\mathcal{O}_ B$-algebra maps $(a')^\sharp$ and $(b')^\sharp$ is an $\mathcal{O}_ B$-derivation from $a^{-1}\mathcal{O}_ Y$ to $\mathcal{C}_{X/X'}$. By adjointness of the functors $a^{-1}$ and $a_*$ this is the same thing as an $\mathcal{O}_ B$-derivation from $\mathcal{O}_ Y$ into $a_*\mathcal{C}_{X/X'}$. Some details omitted. $\square$

Note that in the situation of the lemma above we may write $D$ as

76.17.1.1
$$\label{spaces-more-morphisms-equation-D} D = \text{d}_{Y/B} \circ \theta$$

where $\theta$ is an $\mathcal{O}_ Y$-linear map $\theta : \Omega _{Y/B} \to a_*\mathcal{C}_{X/X'}$. Of course, then by adjunction again we may view $\theta$ as an $\mathcal{O}_ X$-linear map $\theta : a^*\Omega _{Y/B} \to \mathcal{C}_{X/X'}$.

Lemma 76.17.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(a, a') : (X \subset X') \to (Y \subset Y')$ be a morphism of first order thickenings over $B$. Let

$\theta : a^*\Omega _{Y/B} \to \mathcal{C}_{X/X'}$

be an $\mathcal{O}_ X$-linear map. Then there exists a unique morphism of pairs $(b, b') : (X \subset X') \to (Y \subset Y')$ such that (1) and (2) of Lemma 76.17.1 hold and the derivation $D$ and $\theta$ are related by Equation (76.17.1.1).

Proof. Consider the map

$\alpha = (a')^\sharp + D : a^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'}$

where $D$ is as in Equation (76.17.1.1). As $D$ is an $\mathcal{O}_ B$-derivation it follows that $\alpha$ is a map of sheaves of $\mathcal{O}_ B$-algebras. By construction we have $i_ X^\sharp \circ \alpha = a^\sharp \circ i_ Y^\sharp$ where $i_ X : X \to X'$ and $i_ Y : Y \to Y'$ are the given closed immersions. By Lemma 76.9.2 we obtain a unique morphism $(a, b') : (X \subset X') \to (Y \subset Y')$ of thickenings over $B$ such that $\alpha = (b')^\sharp$. Setting $b = a$ we win. $\square$

Remark 76.17.3. Assumptions and notation as in Lemma 76.17.2. The action of a local section $\theta$ on $a'$ is sometimes indicated by $\theta \cdot a'$. Note that this means nothing else than the fact that $(a')^\sharp$ and $(\theta \cdot a')^\sharp$ differ by a derivation $D$ which is related to $\theta$ by Equation (76.17.1.1).

Lemma 76.17.4. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $X \subset X'$ and $Y \subset Y'$ be first order thickenings over $B$. Assume given a morphism $a : X \to Y$ and a map $A : a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ of $\mathcal{O}_ X$-modules. For an object $U'$ of $(X')_{spaces, {\acute{e}tale}}$ with $U = X \times _{X'} U'$ consider morphisms $a' : U' \to Y'$ such that

1. $a'$ is a morphism over $B$,

2. $a'|_ U = a|_ U$, and

3. the induced map $a^*\mathcal{C}_{Y/Y'}|_ U \to \mathcal{C}_{X/X'}|_ U$ is the restriction of $A$ to $U$.

Then the rule

76.17.4.1
$$\label{spaces-more-morphisms-equation-sheaf} U' \mapsto \{ a' : U' \to Y'\text{ such that (1), (2), (3) hold.}\}$$

defines a sheaf of sets on $(X')_{spaces, {\acute{e}tale}}$.

Proof. Denote $\mathcal{F}$ the rule of the lemma. The restriction mapping $\mathcal{F}(U') \to \mathcal{F}(V')$ for $V' \subset U' \subset X'$ of $\mathcal{F}$ is really the restriction map $a' \mapsto a'|_{V'}$. With this definition in place it is clear that $\mathcal{F}$ is a sheaf since morphisms of algebraic spaces satisfy étale descent, see Descent on Spaces, Lemma 74.7.2. $\square$

Lemma 76.17.5. Same notation and assumptions as in Lemma 76.17.4. We identify sheaves on $X$ and $X'$ via (76.9.1.1). There is an action of the sheaf

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a^*\Omega _{Y/B}, \mathcal{C}_{X/X'})$

on the sheaf (76.17.4.1). Moreover, the action is simply transitive for any object $U'$ of $(X')_{spaces, {\acute{e}tale}}$ over which the sheaf (76.17.4.1) has a section.

Remark 76.17.6. A special case of Lemmas 76.17.1, 76.17.2, 76.17.4, and 76.17.5 is where $Y = Y'$. In this case the map $A$ is always zero. The sheaf of Lemma 76.17.4 is just given by the rule

$U' \mapsto \{ a' : U' \to Y\text{ over }B\text{ with } a'|_ U = a|_ U\}$

and we act on this by the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a^*\Omega _{Y/B}, \mathcal{C}_{X/X'})$.

Remark 76.17.7. Another special case of Lemmas 76.17.1, 76.17.2, 76.17.4, and 76.17.5 is where $B$ itself is a thickening $Z \subset Z' = B$ and $Y = Z \times _{Z'} Y'$. Picture

$\xymatrix{ (X \subset X') \ar@{..>}[rr]_{(a, ?)} \ar[rd]_{(g, g')} & & (Y \subset Y') \ar[ld]^{(h, h')} \\ & (Z \subset Z') }$

In this case the map $A : a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ is determined by $a$: the map $h^*\mathcal{C}_{Z/Z'} \to \mathcal{C}_{Y/Y'}$ is surjective (because we assumed $Y = Z \times _{Z'} Y'$), hence the pullback $g^*\mathcal{C}_{Z/Z'} = a^*h^*\mathcal{C}_{Z/Z'} \to a^*\mathcal{C}_{Y/Y'}$ is surjective, and the composition $g^*\mathcal{C}_{Z/Z'} \to a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ has to be the canonical map induced by $g'$. Thus the sheaf of Lemma 76.17.4 is just given by the rule

$U' \mapsto \{ a' : U' \to Y'\text{ over }Z'\text{ with } a'|_ U = a|_ U\}$

and we act on this by the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a^*\Omega _{Y/Z}, \mathcal{C}_{X/X'})$.

Lemma 76.17.8. Let $S$ be a scheme. Consider a commutative diagram of first order thickenings

$\vcenter { \xymatrix{ (T_2 \subset T_2') \ar[d]_{(h, h')} \ar[rr]_{(a_2, a_2')} & & (X_2 \subset X_2') \ar[d]^{(f, f')} \\ (T_1 \subset T_1') \ar[rr]^{(a_1, a_1')} & & (X_1 \subset X_1') } } \quad \begin{matrix} \text{and a commutative} \\ \text{diagram} \end{matrix} \quad \vcenter { \xymatrix{ X_2' \ar[r] \ar[d] & B_2 \ar[d] \\ X_1' \ar[r] & B_1 } }$

of algebraic spaces over $S$ with $X_2 \to X_1$ and $B_2 \to B_1$ étale. For any $\mathcal{O}_{T_1}$-linear map $\theta _1 : a_1^*\Omega _{X_1/B_1} \to \mathcal{C}_{T_1/T'_1}$ let $\theta _2$ be the composition

$\xymatrix{ a_2^*\Omega _{X_2/B_2} \ar@{=}[r] & h^*a_1^*\Omega _{X_1/B_1} \ar[r]^-{h^*\theta _1} & h^*\mathcal{C}_{T_1/T'_1} \ar[r] & \mathcal{C}_{T_2/T'_2} }$

(equality sign is explained in the proof). Then the diagram

$\xymatrix{ T_2' \ar[rr]_{\theta _2 \cdot a_2'} \ar[d] & & X'_2 \ar[d] \\ T_1' \ar[rr]^{\theta _1 \cdot a_1'} & & X'_1 }$

commutes where the actions $\theta _2 \cdot a_2'$ and $\theta _1 \cdot a_1'$ are as in Remark 76.17.3.

Proof. The equality sign comes from the identification $f^*\Omega _{X_1/S_1} = \Omega _{X_2/S_2}$ we get as the construction of the sheaf of differentials is compatible with étale localization (both on source and target), see Lemma 76.7.3. Namely, using this we have $a_2^*\Omega _{X_2/S_2} = a_2^*f^*\Omega _{X_1/S_1} = h^*a_1^*\Omega _{X_1/S_1}$ because $f \circ a_2 = a_1 \circ h$. Having said this, the commutativity of the diagram may be checked on étale locally. Thus we may assume $T'_ i$, $X'_ i$, $B_2$, and $B_1$ are schemes and in this case the lemma follows from More on Morphisms, Lemma 37.9.10. Alternative proof: using Lemma 76.9.2 it suffices to show a certain diagram of sheaves of rings on $X_1'$ is commutative; then argue exactly as in the proof of the aforementioned More on Morphisms, Lemma 37.9.10 to see that this is indeed the case. $\square$

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