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
a = b, and
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
\begin{equation} \label{spaces-more-morphisms-equation-D} D = \text{d}_{Y/B} \circ \theta \end{equation}
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
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
a' is a morphism over B,
a'|_ U = a|_ U, and
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
\begin{equation} \label{spaces-more-morphisms-equation-sheaf} U' \mapsto \{ a' : U' \to Y'\text{ such that (1), (2), (3) hold.}\} \end{equation}
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.
Proof.
This is a combination of Lemmas 76.17.1, 76.17.2, and 76.17.4.
\square
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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