The Stacks project

90.15 Miniversal formal objects and tangent spaces

The general notion of minimality introduced in Definition 90.14.4 can sometimes be deduced from the behaviour on tangent spaces. Let $\xi $ be a formal object of the predeformation category $\mathcal{F}$ and let $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F}$ be the corresponding morphism. Then we can consider the following the condition
\begin{equation} \label{formal-defos-equation-bijective} d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F} \text{ is bijective} \end{equation}

and the condition
\begin{equation} \label{formal-defos-equation-bijective-orbits} d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F} \text{ is bijective on }\text{Der}_\Lambda (k, k)\text{-orbits.} \end{equation}

Here we are using the identification $T\underline{R}|_{\mathcal{C}_\Lambda } = \text{Der}_\Lambda (R, k)$ of Example 90.11.11 and the action ( of derivations on the tangent spaces. If $k' \subset k$ is separable, then $\text{Der}_\Lambda (k, k) = 0$ and the two conditions are equivalent. It turns out that, in the presence of condition (S2) a versal formal object is minimal if and only if $\underline{\xi }$ satisfies ( Moreover, if $\underline{\xi }$ satisfies (, then $\mathcal{F}$ satisfies (S2).

Lemma 90.15.1. Let $\mathcal{F}$ be a predeformation category. Let $\xi $ be a versal formal object of $\mathcal{F}$ such that ( holds. Then $\xi $ is a minimal versal formal object. In particular, such $\xi $ are unique up to isomorphism.

Proof. If $\xi $ is not minimal, then there exists a morphism $\xi ' \to \xi $ lying over $R' \to R$ such that $R = R'[[x_1, \ldots , x_ n]]$ with $n > 0$, see Lemma 90.14.5. Thus $d\underline{\xi }$ factors as

\[ \text{Der}_\Lambda (R, k) \to \text{Der}_\Lambda (R', k) \to T\mathcal{F} \]

and we see that ( cannot hold because $D : f \mapsto \partial /\partial x_1(f) \bmod \mathfrak m_ R$ is an element of the kernel of the first arrow which is not in the image of $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$. $\square$

Lemma 90.15.2. Let $\mathcal{F}$ be a predeformation category. Let $\xi $ be a versal formal object of $\mathcal{F}$ such that ( holds. Then

  1. $\mathcal{F}$ satisfies (S1).

  2. $\mathcal{F}$ satisfies (S2).

  3. $\dim _ k T\mathcal{F}$ is finite.

Proof. Condition (S1) holds by Lemma 90.13.1. The first part of (S2) holds since (S1) holds. Let

\[ \vcenter { \xymatrix{ y \ar[r]_ c \ar[d]_ a & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } } \]

be diagrams as in the second part of (S2). As above we can find morphisms $b : \xi \to y$ and $b' : \xi \to y'$ such that

\[ \xymatrix{ \xi \ar[r]^{b'} \ar[d]_ b & y' \ar[d]^{a'} \\ y \ar[r]^{a} & x } \]

commutes. Let $p : \mathcal{F} \to \mathcal{C}_\Lambda $ denote the structure morphism. Say $\widehat{p}(\xi ) = R$, i.e., $\xi $ lies over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. We see that the pushforward of $\xi $ via $p(c) \circ p(b)$ is $x_\epsilon $ and that the pushforward of $\xi $ via $p(c') \circ p(b')$ is $x_\epsilon $. Since $\xi $ satisfies (, we see that $p(c) \circ p(b) = p(c') \circ p(b')$ as maps $R \to k[\epsilon ]$. Hence $p(b) = p(b')$ as maps from $R \to A \times _ k k[\epsilon ]$. Thus we see that $y$ and $y'$ are isomorphic to the pushforward of $\xi $ along this map and we get a unique morphism $y \to y'$ over $A \times _ k k[\epsilon ]$ compatible with $b$ and $b'$ as desired.

Finally, by Example 90.11.11 we see $\dim _ k T\mathcal{F} = \dim _ k T\underline{R}|_{\mathcal{C}_\Lambda }$ is finite. $\square$

Example 90.15.3. There exist predeformation categories which have a versal formal object satisfying ( but which do not satisfy (S2). A quick example is to take $F = \underline{k[\epsilon ]}/G$ where $G \subset \text{Aut}_{\mathcal{C}_\Lambda }(k[\epsilon ])$ is a finite nontrivial subgroup. Namely, the map $\underline{k[\epsilon ]} \to F$ is smooth, but the tangent space of $F$ does not have a natural $k$-vector space structure (as it is a quotient of a $k$-vector space by a finite group).

Lemma 90.15.4. Let $\mathcal{F}$ be a predeformation category satisfying (S2) which has a versal formal object. Then its minimal versal formal object satisfies (

Proof. Let $\xi $ be a minimal versal formal object for $\mathcal{F}$, see Lemma 90.14.5. Say $\xi $ lies over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. In order to parse ( we point out that $T\mathcal{F}$ has a natural $k$-vector space structure (see Lemma 90.12.2), that $d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F}$ is linear (see Lemma 90.12.4), and that the action of $\text{Der}_\Lambda (k, k)$ is given by addition (see Lemma 90.12.6). Consider the diagram

\[ \xymatrix{ & \mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k) \\ K \ar[r] & \text{Der}_\Lambda (R, k) \ar[r]^{d\underline{\xi }} \ar[u] & T\mathcal{F} \\ & \text{Der}_\Lambda (k, k) \ar[u] \ar[ru] } \]

The vector space $K$ is the kernel of $d\underline{\xi }$. Note that the middle column is exact in the middle as it is dual to the sequence ( If ( fails, then we can find a nonzero element $D \in K$ which does not map to zero in $\mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k)$. This means there exists an $t \in \mathfrak m_ R$ such that $D(t) = 1$. Set $R' = \{ a \in R \mid D(a) = 0\} $. As $D$ is a derivation this is a subring of $R$. Since $D(t) = 1$ we see that $R' \to k$ is surjective (compare with the proof of Lemma 90.3.12). Note that $\mathfrak m_{R'} = \mathop{\mathrm{Ker}}(D : \mathfrak m_ R \to k)$ is an ideal of $R$ and $\mathfrak m_ R^2 \subset \mathfrak m_{R'}$. Hence

\[ \mathfrak m_ R/\mathfrak m_ R^2 = \mathfrak m_{R'}/\mathfrak m_ R^2 + k\overline{t} \]

which implies that the map

\[ R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \to R/\mathfrak m_ R^2 \]

sending $\epsilon $ to $\overline{t}$ is an isomorphism. In particular there is a map $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$.

Let $\xi \to y$ be a morphism lying over $R \to R/\mathfrak m_ R^2$. Let $y \to x$ be a morphism lying over $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$. Let $y \to x_\epsilon $ be a morphism lying over $R/\mathfrak m_ R^2 \to k[\epsilon ]$. Let $x_0$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k$. By the axioms of a category cofibred in groupoids we obtain a commutative diagram

\[ \vcenter { \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ R'/\mathfrak m_ R^2 \ar[r] & k. } } \]

Because $D \in K$ we see that $x_\epsilon $ is isomorphic to $0 \in \mathcal{F}(k[\epsilon ])$, i.e., $x_\epsilon $ is the pushforward of $x_0$ via $k \to k[\epsilon ], a \mapsto a$. Hence by Lemma 90.10.7 we see that there exists a morphism $x \to y$. Since $\text{length}_\Lambda (R'/\mathfrak m_ R^2) < \text{length}_\Lambda (R/\mathfrak m_ R^2)$ the corresponding ring map $R'/\mathfrak m_ R^2 \to R/\mathfrak m_ R^2$ is not surjective. This contradicts the minimality of $\xi /R$, see part (1) of Lemma 90.14.2. This contradiction shows that such a $D$ cannot exist, hence we win. $\square$

Theorem 90.15.5. Let $\mathcal{F}$ be a predeformation category. Consider the following conditions

  1. $\mathcal{F}$ has a minimal versal formal object satisfying (,

  2. $\mathcal{F}$ has a minimal versal formal object satisfying (,

  3. the following conditions hold:

    1. $\mathcal{F}$ satisfies (S1).

    2. $\mathcal{F}$ satisfies (S2).

    3. $\dim _ k T\mathcal{F}$ is finite.

We always have

\[ (1) \Rightarrow (3) \Rightarrow (2). \]

If $k' \subset k$ is separable, then all three are equivalent.

Proof. Lemma 90.15.2 shows that (1) $\Rightarrow $ (3). Lemmas 90.13.4 and 90.15.4 show that (3) $\Rightarrow $ (2). If $k' \subset k$ is separable then $\text{Der}_\Lambda (k, k) = 0$ and we see that ( $=$ (, i.e., (1) is the same as (2).

An alternative proof of (3) $\Rightarrow $ (1) in the classical case is to add a few words to the proof of Lemma 90.13.4 to see that one can right away construct a versal object which satisfies ( in this case. This avoids the use of Lemma 90.13.4 in the classical case. Details omitted. $\square$

Remark 90.15.6. Let $F : \mathcal{C}_\Lambda \to \textit{Sets}$ be a predeformation functor satisfying (S1) and (S2) and $\dim _ k TF < \infty $. Recall that these conditions correspond to the conditions (H1), (H2), and (H3) from Schlessinger's paper, see Remark 90.13.5. Now, in the classical case (or if $k' \subset k$ is separable) following Schlessinger we introduce the notion of a hull: a hull is a versal formal object $\xi \in \widehat{F}(R)$ such that $d\underline{\xi } : T\underline{R}|_{\mathcal{C}_\Lambda } \to TF$ is an isomorphism, i.e., ( holds. Thus Theorem 90.15.5 tells us

\[ (H1) + (H2) + (H3) \Rightarrow \text{ there exists a hull} \]

in the classical case. In other words, our theorem recovers Schlessinger's theorem on the existence of hulls.

Remark 90.15.7. Let $\mathcal{F}$ be a predeformation category. Recall that $\mathcal{F} \to \overline{\mathcal{F}}$ is smooth, see Remark 90.8.5. Hence if $\xi \in \widehat{\mathcal{F}}(R)$ is a versal formal object, then the composition

\[ \underline{R}|_{\mathcal{C}_\Lambda } \longrightarrow \mathcal{F} \longrightarrow \overline{\mathcal{F}} \]

is smooth (Lemma 90.8.7) and we conclude that the image $\overline{\xi }$ of $\xi $ in $\overline{\mathcal{F}}$ is a versal formal object. If ( holds, then $\overline{\xi }$ induces an isomorphism $T\underline{R}|_{\mathcal{C}_\Lambda } \to T\overline{\mathcal{F}}$ because $\mathcal{F} \to \overline{\mathcal{F}}$ identifies tangent spaces. Hence in this case $\overline{\xi }$ is a hull for $\overline{\mathcal{F}}$, see Remark 90.15.6. By Theorem 90.15.5 we can always find such a $\xi $ if $k' \subset k$ is separable and $\mathcal{F}$ is a predeformation category satisfying (S1), (S2), and $\dim _ k T\mathcal{F} < \infty $.

Example 90.15.8. In Lemma 90.9.5 we constructed objects $R \in \widehat{\mathcal{C}}_\Lambda $ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth and such that

\[ H_1(L_{k/\Lambda }) = \mathfrak m_ R/\mathfrak m_ R^2 \quad \text{and}\quad \Omega _{R/\Lambda } \otimes _ R k = \Omega _{k/\Lambda } \]

Let us reinterpret this using the theorem above. Namely, consider $\mathcal{F} = \mathcal{C}_\Lambda $ as a category cofibred in groupoids over itself (using the identity functor). Then $\mathcal{F}$ is a predeformation category, satisfies (S1) and (S2), and we have $T\mathcal{F} = 0$. Thus $\mathcal{F}$ satisfies condition (3) of Theorem 90.15.5. The theorem implies that (2) holds, i.e., we can find a minimal versal formal object $\xi \in \widehat{\mathcal{F}}(S)$ over some $S \in \widehat{\mathcal{C}}_\Lambda $ satisfying ( Lemma 90.9.3 shows that $\Lambda \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology (because $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F} = \mathcal{C}_\Lambda $ is smooth). Now condition ( tells us that $\text{Der}_\Lambda (S, k) \to 0$ is bijective on $\text{Der}_\Lambda (k, k)$-orbits. This means the injection $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (S, k)$ is also surjective. In other words, we have $\Omega _{S/\Lambda } \otimes _ S k = \Omega _{k/\Lambda }$. Since $\Lambda \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology, we can apply More on Algebra, Lemma 15.40.4 to conclude the exact sequence ( turns into a pair of identifications

\[ H_1(L_{k/\Lambda }) = \mathfrak m_ S/\mathfrak m_ S^2 \quad \text{and}\quad \Omega _{S/\Lambda } \otimes _ S k = \Omega _{k/\Lambda } \]

Reading the argument backwards, we find that the $R$ constructed in Lemma 90.9.5 carries a minimal versal object. By the uniqueness of minimal versal objects (Lemma 90.14.5) we also conclude $R \cong S$, i.e., the two constructions give the same answer.

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