The Stacks project

37.9 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 $X$.

Lemma 37.9.1. Let $S$ be a scheme. Let $X \subset X'$ and $Y \subset Y'$ be two first order thickenings over $S$. Let $(a, a'), (b, b') : (X \subset X') \to (Y \subset Y')$ be two morphisms of thickenings over $S$. Assume that

  1. $a = b$, and

  2. the two maps $a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ (Morphisms, Lemma 29.31.3) 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 $\mathcal{O}_ S$-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 $\mathcal{O}_ S$-algebra maps $(a')^\sharp $ and $(b')^\sharp $ is an $\mathcal{O}_ S$-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 $\mathcal{O}_ S$-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
\begin{equation} \label{more-morphisms-equation-D} D = \text{d}_{Y/S} \circ \theta \end{equation}

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

Lemma 37.9.2. Let $S$ be a scheme. Let $(a, a') : (X \subset X') \to (Y \subset Y')$ be a morphism of first order thickenings over $S$. Let

\[ \theta : a^*\Omega _{Y/S} \to \mathcal{C}_{X/X'} \]

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 37.9.1 hold and the derivation $D$ and $\theta $ are related by Equation (

Proof. We simply set $b = a$ and we define $(b')^\sharp $ to be the map

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

where $D$ is as in Equation ( We omit the verification that $(b')^\sharp $ is a map of sheaves of $\mathcal{O}_ S$-algebras and that (1) and (2) of Lemma 37.9.1 hold. Equation ( holds by construction. $\square$

Remark 37.9.3. Assumptions and notation as in Lemma 37.9.2. 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 (

Lemma 37.9.4. Let $S$ be a scheme. Let $X \subset X'$ and $Y \subset Y'$ be first order thickenings over $S$. 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 open subscheme $U' \subset X'$ consider morphisms $a' : U' \to Y'$ such that

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

  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 $U$.

Here $U = X \cap U'$. Then the rule
\begin{equation} \label{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 $X'$.

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 are defined locally. $\square$

In the following lemma we identify sheaves on $X$ and any thickening of $X$.

Lemma 37.9.5. Same notation and assumptions as in Lemma 37.9.4. There is an action of the sheaf

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

on the sheaf ( Moreover, the action is simply transitive for any open $U' \subset X'$ over which the sheaf ( has a section.

Remark 37.9.6. A special case of Lemmas 37.9.1, 37.9.2, 37.9.4, and 37.9.5 is where $Y = Y'$. In this case the map $A$ is always zero. The sheaf of Lemma 37.9.4 is just given by the rule

\[ U' \mapsto \{ a' : U' \to Y\text{ over }S\text{ with } a'|_ U = a|_ U\} \]

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

Remark 37.9.7. Another special case of Lemmas 37.9.1, 37.9.2, 37.9.4, and 37.9.5 is where $S$ itself is a thickening $Z \subset Z' = S$ 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 37.9.4 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 37.9.8. Let $S$ be a scheme. Let $X \subset X'$ be a first order thickening over $S$. Let $Y$ be a scheme over $S$. Let $a', b' : X' \to Y$ be two morphisms over $S$ with $a = a'|_ X = b'|_ X$. This gives rise to a commutative diagram

\[ \xymatrix{ X \ar[r] \ar[d]_ a & X' \ar[d]^{(b', a')} \\ Y \ar[r]^-{\Delta _{Y/S}} & Y \times _ S Y } \]

Since the horizontal arrows are immersions with conormal sheaves $\mathcal{C}_{X/X'}$ and $\Omega _{Y/S}$, by Morphisms, Lemma 29.31.3, we obtain a map $\theta : a^*\Omega _{Y/S} \to \mathcal{C}_{X/X'}$. Then this $\theta $ and the derivation $D$ of Lemma 37.9.1 are related by Equation (

Proof. Omitted. Hint: The equality may be checked on affine opens where it comes from the following computation. If $f$ is a local section of $\mathcal{O}_ Y$, then $1 \otimes f - f \otimes 1$ is a local section of $\mathcal{C}_{Y/(Y \times _ S Y)}$ corresponding to $\text{d}_{Y/S}(f)$. It is mapped to the local section $(a')^\sharp (f) - (b')^\sharp (f) = D(f)$ of $\mathcal{C}_{X/X'}$. In other words, $\theta (\text{d}_{Y/S}(f)) = D(f)$. $\square$

For later purposes we need a result that roughly states that the construction of Lemma 37.9.2 is compatible with étale localization.

Lemma 37.9.9. Let

\[ \xymatrix{ X_1 \ar[d] & X_2 \ar[l]^ f \ar[d] \\ S_1 & S_2 \ar[l] } \]

be a commutative diagram of schemes with $X_2 \to X_1$ and $S_2 \to S_1$ étale. Then the map $c_ f : f^*\Omega _{X_1/S_1} \to \Omega _{X_2/S_2}$ of Morphisms, Lemma 29.32.8 is an isomorphism.

Proof. We recall that an étale morphism $U \to V$ is a smooth morphism with $\Omega _{U/V} = 0$. Using this we see that Morphisms, Lemma 29.32.9 implies $\Omega _{X_2/S_2} = \Omega _{X_2/S_1}$ and Morphisms, Lemma 29.34.16 implies that the map $f^*\Omega _{X_1/S_1} \to \Omega _{X_2/S_1}$ (for the morphism $f$ seen as a morphism over $S_1$) is an isomorphism. Hence the lemma follows. $\square$

Lemma 37.9.10. Consider a commutative diagram of first order thickenings

\[ \vcenter { \xymatrix{ (T_2 \subset T_2') \ar[d]_{(h, h')} \ar[rr]_{(a_2, a_2')} & & (X_2 \subset X_2') \ar[d]^{(f, f')} \\ (T_1 \subset T_1') \ar[rr]^{(a_1, a_1')} & & (X_1 \subset X_1') } } \quad \begin{matrix} \text{and a commutative} \\ \text{diagram of schemes} \end{matrix} \quad \vcenter { \xymatrix{ X_2' \ar[r] \ar[d] & S_2 \ar[d] \\ X_1' \ar[r] & S_1 } } \]

with $X_2 \to X_1$ and $S_2 \to S_1$ étale. For any $\mathcal{O}_{T_1}$-linear map $\theta _1 : a_1^*\Omega _{X_1/S_1} \to \mathcal{C}_{T_1/T'_1}$ let $\theta _2$ be the composition

\[ \xymatrix{ a_2^*\Omega _{X_2/S_2} \ar@{=}[r] & h^*a_1^*\Omega _{X_1/S_1} \ar[r]^-{h^*\theta _1} & h^*\mathcal{C}_{T_1/T'_1} \ar[r] & \mathcal{C}_{T_2/T'_2} } \]

(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 37.9.3.

Proof. The equality sign comes from the identification $f^*\Omega _{X_1/S_1} = \Omega _{X_2/S_2}$ of Lemma 37.9.9. 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 affine opens. Hence we may assume the schemes in the initial big diagram are affine. Thus we obtain commutative diagrams

\[ \vcenter { \xymatrix{ (B'_2, I_2) & & (A'_2, J_2) \ar[ll]^{a_2'} \\ (B'_1, I_1) \ar[u]^{h'} & & (A'_1, J_1) \ar[ll]_{a_1'} \ar[u]_{f'} } } \quad \text{and}\quad \vcenter { \xymatrix{ A'_2 & & R_2 \ar[ll] \\ A'_1 \ar[u] & & R_1 \ar[ll] \ar[u] } } \]

The notation signifies that $I_1, I_2, J_1, J_2$ are ideals of square zero and maps of pairs are ring maps sending ideals into ideals. Set $A_1 = A'_1/J_1$, $A_2 = A'_2/J_2$, $B_1 = B'_1/I_1$, and $B_2 = B'_2/I_2$. We are given that

\[ A_2 \otimes _{A_1} \Omega _{A_1/R_1} \longrightarrow \Omega _{A_2/R_2} \]

is an isomorphism. Then $\theta _1 : B_1 \otimes _{A_1} \Omega _{A_1/R_1} \to I_1$ is $B_1$-linear. This gives an $R_1$-derivation $D_1 = \theta _1 \circ \text{d}_{A_1/R_1} : A_1 \to I_1$. In a similar way we see that $\theta _2 : B_2 \otimes _{A_2} \Omega _{A_2/R_2} \to I_2$ gives rise to a $R_2$-derivation $D_2 = \theta _2 \circ \text{d}_{A_2/R_2} : A_2 \to I_2$. The construction of $\theta _2$ implies the following compatibility between $\theta _1$ and $\theta _2$: for every $x \in A_1$ we have

\[ h'(D_1(x)) = D_2(f'(x)) \]

as elements of $I_2$. We may view $D_1$ as a map $A'_1 \to B'_1$ using $A'_1 \to A_1 \xrightarrow {D_1} I_1 \to B_1$ similarly we may view $D_2$ as a map $A'_2 \to B'_2$. Then the displayed equality holds for $x \in A'_1$. By the construction of the action in Lemma 37.9.2 and Remark 37.9.3 we know that $\theta _1 \cdot a_1'$ corresponds to the ring map $a_1' + D_1 : A'_1 \to B'_1$ and $\theta _2 \cdot a_2'$ corresponds to the ring map $a_2' + D_2 : A'_2 \to B'_2$. By the displayed equality we obtain that $h' \circ (a_1' + D_1) = (a_2' + D_2) \circ f'$ as desired. $\square$

Remark 37.9.11. Lemma 37.9.10 can be improved in the following way. Suppose that we have commutative diagrams as in Lemma 37.9.10 but we do not assume that $X_2 \to X_1$ and $S_2 \to S_1$ are étale. Next, suppose we have $\theta _1 : a_1^*\Omega _{X_1/S_1} \to \mathcal{C}_{T_1/T'_1}$ and $\theta _2 : a_2^*\Omega _{X_2/S_2} \to \mathcal{C}_{T_2/T'_2}$ such that

\[ \xymatrix{ f_*\mathcal{O}_{X_2} \ar[rr]_{f_*D_2} & & f_*a_{2, *}\mathcal{C}_{T_2/T_2'} \\ \mathcal{O}_{X_1} \ar[rr]^{D_1} \ar[u]^{f^\sharp } & & a_{1, *}\mathcal{C}_{T_1/T_1'} \ar[u]_{\text{induced by }(h')^\sharp } } \]

is commutative where $D_ i$ corresponds to $\theta _ i$ as in Equation ( Then we have the conclusion of Lemma 37.9.10. The importance of the condition that both $X_2 \to X_1$ and $S_2 \to S_1$ are étale is that it allows us to construct a $\theta _2$ from $\theta _1$.

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