Example 96.24.1. Let $\Lambda , S, W, \mathcal{F}$ be as in Example 96.22.3. Assume that $W \to S$ is proper and $\mathcal{F}$ coherent. By Cohomology of Schemes, Remark 30.22.2 there exists a finite complex of finite projective $\Lambda $-modules $N^\bullet $ which universally computes the cohomology of $\mathcal{F}$. In particular the obstruction spaces from Example 96.22.3 are $\mathcal{O}_ x(M) = H^1(N^\bullet \otimes _\Lambda M)$. Hence with $K^\bullet = N^\bullet \otimes _\Lambda A[-1]$ we see that $\mathcal{O}_ x(M) = H^2(K^\bullet \otimes _ A^\mathbf {L} M)$.

## 96.24 A dual notion

Let $(x, A' \to A)$ be a deformation situation for a given category $\mathcal{X}$ fibred in groupoids over a locally Noetherian scheme $S$. Assume $\mathcal{X}$ has an obstruction theory, see Definition 96.22.1. In practice one often has a complex $K^\bullet $ of $A$-modules and isomorphisms of functors

In this section we formalize this a little bit and show how this leads to a verification of openness of versality in some cases.

Situation 96.24.2. Let $S$ be a locally Noetherian scheme. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume that $\mathcal{X}$ has (RS*) so that we can speak of the functor $T_ x(-)$, see Lemma 96.21.2. Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme of finite type over $S$ which maps into an affine open $\mathop{\mathrm{Spec}}(\Lambda )$. Let $x$ be an object of $\mathcal{X}$ over $U$. Assume we are given

a complex of $A$-modules $K^\bullet $,

a transformation of functors $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$,

for every deformation situation $(x, A' \to A)$ with kernel $I = \mathop{\mathrm{Ker}}(A' \to A)$ an element $o_ x(A') \in H^2(K^\bullet \otimes _ A^\mathbf {L} I)$

satisfying the following (minimal) conditions

the transformation $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$ is an isomorphism,

given a morphism $(x, A'' \to A) \to (x, A' \to A)$ of deformation situations the element $o_ x(A')$ maps to the element $o_ x(A'')$ via the map $H^2(K^\bullet \otimes _ A^\mathbf {L} I) \to H^2(K^\bullet \otimes _ A^\mathbf {L} I')$ where $I' = \mathop{\mathrm{Ker}}(A'' \to A)$, and

$x$ lifts to an object over $\mathop{\mathrm{Spec}}(A')$ if and only if $o_ x(A') = 0$.

It is possible to incorporate infinitesimal automorphisms as well, but we refrain from doing so in order to get the sharpest possible result.

In Situation 96.24.2 an important role will be played by $K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits _{A/\Lambda }$. Suppose we are given an element $\xi \in H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits _{A/\Lambda })$. Then (1) for any surjection $A' \to A$ of $\Lambda $-algebras with kernel $I$ of square zero the canonical map $\mathop{N\! L}\nolimits _{A/\Lambda } \to \mathop{N\! L}\nolimits _{A/A'} = I[1]$ sends $\xi $ to an element $\xi _{A'} \in H^2(K^\bullet \otimes _ A^\mathbf {L} I)$ and (2) the map $\mathop{N\! L}\nolimits _{A/\Lambda } \to \Omega _{A/\Lambda }$ sends $\xi $ to an element $\xi _{can}$ of $H^1(K^\bullet \otimes _ A^\mathbf {L} \Omega _{A/\Lambda })$.

Lemma 96.24.3. In Situation 96.24.2. Assume furthermore that

given a short exact sequence of deformation situations as in Remark 96.21.5 and a lift $x'_2 \in \text{Lift}(x, A_2')$ then $o_ x(A_3') \in H^2(K^\bullet \otimes _ A^\mathbf {L} I_3)$ equals $\partial \theta $ where $\theta \in H^1(K^\bullet \otimes _ A^\mathbf {L} I_1)$ is the element corresponding to $x'_2|_{\mathop{\mathrm{Spec}}(A_1')}$ via $A_1' = A[I_1]$ and the given map $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$.

In this case there exists an element $\xi \in H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits _{A/\Lambda })$ such that

for every deformation situation $(x, A' \to A)$ we have $\xi _{A'} = o_ x(A')$, and

$\xi _{can}$ matches the canonical element of Remark 96.21.8 via the given transformation $T_ x(-) \to H^1(K^\bullet \otimes _ A^\mathbf {L} -)$.

**Proof.**
Choose a $\alpha : \Lambda [x_1, \ldots , x_ n] \to A$ with kernel $J$. Write $P = \Lambda [x_1, \ldots , x_ n]$. In the rest of this proof we work with

which is permissible by Algebra, Lemma 10.134.2 and More on Algebra, Lemma 15.58.1. Consider the element $o_ x(P/J^2) \in H^2(K^\bullet \otimes _ A^\mathbf {L} J/J^2)$ and consider the quotient

where $J/J^2$ is embedded diagonally. Note that $C \to A$ is a surjection with kernel $\bigoplus A\text{d}x_ i$. Moreover there is a section $A \to C$ to $C \to A$ given by mapping the class of $f \in P$ to the class of $(f, \text{d}f)$ in the pushout. For later use, denote $x_ C$ the pullback of $x$ along the corresponding morphism $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$. Thus we see that $o_ x(C) = 0$. We conclude that $o_ x(P/J^2)$ maps to zero in $H^2(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$. It follows that there exists some element $\xi \in H^1(K^\bullet \otimes _ A^\mathbf {L} \mathop{N\! L}\nolimits (\alpha ))$ mapping to $o_ x(P/J^2)$.

Note that for any deformation situation $(x, A' \to A)$ there exists a $\Lambda $-algebra map $P/J^2 \to A'$ compatible with the augmentations to $A$. Hence the element $\xi $ satisfies the first property of the lemma by construction and property (ii) of Situation 96.24.2.

Note that our choice of $\xi $ was well defined up to the choice of an element of $H^1(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$. We will show that after modifying $\xi $ by an element of the aforementioned group we can arrange it so that the second assertion of the lemma is true. Let $C' \subset C$ be the image of $P/J^2$ under the $\Lambda $-algebra map $P/J^2 \to C$ (inclusion of first factor). Observe that $\mathop{\mathrm{Ker}}(C' \to A) = \mathop{\mathrm{Im}}(J/J^2 \to \bigoplus A\text{d}x_ i)$. Set $\overline{C} = A[\Omega _{A/\Lambda }]$. The map $P/J^2 \times \bigoplus A \text{d}x_ i \to \overline{C}$, $(f, \sum f_ i \text{d}x_ i) \mapsto (f \bmod J, \sum f_ i \text{d}x_ i)$ factors through a surjective map $C \to \overline{C}$. Then

is a short exact sequence of deformation situations. The associated splitting $\overline{C} = A[\Omega _{A/\Lambda }]$ (from Remark 96.21.5) equals the given splitting above. Moreover, the section $A \to C$ composed with the map $C \to \overline{C}$ is the map $(1, \text{d}) : A \to A[\Omega _{A/\Lambda }]$ of Remark 96.21.8. Thus $x_ C$ restricts to the canonical element $x_{can}$ of $T_ x(\Omega _{A/\Lambda }) = \text{Lift}(x, A[\Omega _{A/\Lambda }])$. By condition (iv) we conclude that $o_ x(P/J^2)$ maps to $\partial x_{can}$ in

By construction $\xi $ maps to $o_ x(P/J^2)$. It follows that $x_{can}$ and $\xi _{can}$ map to the same element in the displayed group which means (by the long exact cohomology sequence) that they differ by an element of $H^1(K^\bullet \otimes _ A^\mathbf {L} \bigoplus A\text{d}x_ i)$ as desired. $\square$

Lemma 96.24.4. In Situation 96.24.2 assume that (iv) of Lemma 96.24.3 holds and that $K^\bullet $ is a perfect object of $D(A)$. In this case, if $x$ is versal at a closed point $u_0 \in U$ then there exists an open neighbourhood $u_0 \in U' \subset U$ such that $x$ is versal at every finite type point of $U'$.

**Proof.**
We may assume that $K^\bullet $ is a finite complex of finite projective $A$-modules. Thus the derived tensor product with $K^\bullet $ is the same as simply tensoring with $K^\bullet $. Let $E^\bullet $ be the dual perfect complex to $K^\bullet $, see More on Algebra, Lemma 15.73.15. (So $E^ n = \mathop{\mathrm{Hom}}\nolimits _ A(K^{-n}, A)$ with differentials the transpose of the differentials of $K^\bullet $.) Let $E \in D^{-}(A)$ denote the object represented by the complex $E^\bullet [-1]$. Let $\xi \in H^1(\text{Tot}(K^\bullet \otimes _ A \mathop{N\! L}\nolimits _{A/\Lambda }))$ be the element constructed in Lemma 96.24.3 and denote $\xi : E = E^\bullet [-1] \to \mathop{N\! L}\nolimits _{A/\Lambda }$ the corresponding map (loc.cit.). We claim that the pair $(E, \xi )$ satisfies all the assumptions of Lemma 96.23.4 which finishes the proof.

Namely, assumption (i) of Lemma 96.23.4 follows from conclusion (1) of Lemma 96.24.3 and the fact that $H^2(K^\bullet \otimes _ A^\mathbf {L} -) = \mathop{\mathrm{Ext}}\nolimits ^1(E, -)$ by loc.cit. Assumption (ii) of Lemma 96.23.4 follows from conclusion (2) of Lemma 96.24.3 and the fact that $H^1(K^\bullet \otimes _ A^\mathbf {L} -) = \mathop{\mathrm{Ext}}\nolimits ^0(E, -)$ by loc.cit. Assumption (iii) of Lemma 96.23.4 is clear. $\square$

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