89.10 Schlessinger's conditions

In the following we often consider fibre products $A_1 \times _ A A_2$ of rings in the category $\mathcal{C}_\Lambda$. We have seen in Example 89.3.7 that such a fibre product may not always be an object of $\mathcal{C}_\Lambda$. However, in virtually all cases below one of the two maps $A_ i \to A$ is surjective and $A_1 \times _ A A_2$ will be an object of $\mathcal{C}_\Lambda$ by Lemma 89.3.8. We will use this result without further mention.

We denote by $k[\epsilon ]$ the ring of dual numbers over $k$. More generally, for a $k$-vector space $V$, we denote by $k[V]$ the $k$-algebra whose underlying vector space is $k \oplus V$ and whose multiplication is given by $(a, v) \cdot (a', v') = (aa', av' + a'v)$. When $V = k$, $k[V]$ is the ring of dual numbers over $k$. For any finite dimensional $k$-vector space $V$ the ring $k[V]$ is in $\mathcal{C}_\Lambda$.

Definition 89.10.1. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda$. We define conditions (S1) and (S2) on $\mathcal{F}$ as follows:

1. Every diagram in $\mathcal{F}$

$\vcenter { \xymatrix{ & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & A_2 \ar[d] \\ A_1 \ar[r] & A } }$

in $\mathcal{C}_\Lambda$ with $A_2 \to A$ surjective can be completed to a commutative diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d] \\ A_1 \ar[r] & A. } }$
2. The condition of (S1) holds for diagrams in $\mathcal{F}$ lying over a diagram in $\mathcal{C}_\Lambda$ of the form

$\xymatrix{ & k[\epsilon ] \ar[d] \\ A \ar[r] & k. }$

Moreover, if we have two commutative diagrams in $\mathcal{F}$

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

then there exists a morphism $b : y \to y'$ in $\mathcal{F}(A \times _ k k[\epsilon ])$ such that $a = a' \circ b$.

We can partly explain the meaning of conditions (S1) and (S2) in terms of fibre categories. Suppose that $f_1 : A_1 \to A$ and $f_2 : A_2 \to A$ are ring maps in $\mathcal{C}_\Lambda$ with $f_2$ surjective. Denote $p_ i : A_1 \times _ A A_2 \to A_ i$ the projection maps. Assume a choice of pushforwards for $\mathcal{F}$ has been made. Then the commutative diagram of rings translates into a $2$-commutative diagram

$\xymatrix{ \mathcal{F}(A_1 \times _ A A_2) \ar[r]_-{p_{2, *}} \ar[d]_{p_{1, *}} & \mathcal{F}(A_2) \ar[d]^{f_{2, *}} \\ \mathcal{F}(A_1) \ar[r]^{f_{1, *}} & \mathcal{F}(A) }$

of fibre categories whence a functor

89.10.1.1
$$\label{formal-defos-equation-compare} \mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$$

into the $2$-fibre product of categories. Condition (S1) requires that this functor be essentially surjective. The first part of condition (S2) requires that this functor be a essentially surjective if $f_2$ equals the map $k[\epsilon ] \to k$. Moreover in this case, the second part of (S2) implies that two objects which become isomorphic in the target are isomorphic in the source (but it is not equivalent to this statement). The advantage of stating the conditions as in the definition is that no choices have to be made.

Lemma 89.10.2. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda$. Then $\mathcal{F}$ satisfies (S1) if the condition of (S1) is assumed to hold only when $A_2 \to A$ is a small extension.

Proof. Proof omitted. Hints: apply Lemma 89.3.3 and use induction similar to the proof of Lemma 89.8.2. $\square$

Remark 89.10.3. When $\mathcal{F}$ is cofibered in sets, conditions (S1) and (S2) are exactly conditions (H1) and (H2) from Schlessinger's paper [Sch]. Namely, for a functor $F: \mathcal{C}_\Lambda \to \textit{Sets}$, conditions (S1) and (S2) state:

1. If $A_1 \to A$ and $A_2 \to A$ are maps in $\mathcal{C}_\Lambda$ with $A_2 \to A$ surjective, then the induced map $F(A_1 \times _ A A_2) \to F(A_1) \times _{F(A)} F(A_2)$ is surjective.

2. If $A \to k$ is a map in $\mathcal{C}_\Lambda$, then the induced map $F(A \times _ k k[\epsilon ]) \to F(A) \times _{F(k)} F(k[\epsilon ])$ is bijective.

The injectivity of the map $F(A \times _ k k[\epsilon ]) \to F(A) \times _{F(k)} F(k[\epsilon ])$ comes from the second part of condition (S2) and the fact that morphisms are identities.

Lemma 89.10.4. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda$. If $\mathcal{F}$ satisfies (S2), then the condition of (S2) also holds when $k[\epsilon ]$ is replaced by $k[V]$ for any finite dimensional $k$-vector space $V$.

Proof. In the case that $\mathcal{F}$ is cofibred in sets, i.e., corresponds to a functor $F : \mathcal{C}_\Lambda \to \textit{Sets}$ this follows from the description of (S2) for $F$ in Remark 89.10.3 and the fact that $k[V] \cong k[\epsilon ] \times _ k \ldots \times _ k k[\epsilon ]$ with $\dim _ k V$ factors. The case of functors is what we will use in the rest of this chapter.

We prove the general case by induction on $\dim (V)$. If $\dim (V) = 1$, then $k[V] \cong k[\epsilon ]$ and the result holds by assumption. If $\dim (V) > 1$ we write $V = V' \oplus k\epsilon$. Pick a diagram

$\vcenter { \xymatrix{ & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & k[V] \ar[d] \\ A \ar[r] & k } }$

Choose a morphism $x_ V \to x_{V'}$ lying over $k[V] \to k[V']$ and a morphism $x_ V \to x_\epsilon$ lying over $k[V] \to k[\epsilon ]$. Note that the morphism $x_ V \to x_0$ factors as $x_ V \to x_{V'} \to x_0$ and as $x_ V \to x_\epsilon \to x_0$. By induction hypothesis we can find a diagram

$\vcenter { \xymatrix{ y' \ar[d] \ar[r] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V'] \ar[d] \ar[r] & k[V'] \ar[d] \\ A \ar[r] & k } }$

This gives us a commutative diagram

$\vcenter { \xymatrix{ & x_\epsilon \ar[d] \\ y' \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & k[\epsilon ] \ar[d] \\ A \times _ k k[V'] \ar[r] & k } }$

Hence by (S2) we get a commutative diagram

$\vcenter { \xymatrix{ y \ar[d] \ar[r] & x_\epsilon \ar[d] \\ y' \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ (A \times _ k k[V']) \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \times _ k k[V'] \ar[r] & k } }$

Note that $(A \times _ k k[V']) \times _ k k[\epsilon ] = A \times _ k k[V' \oplus k\epsilon ] = A \times _ k k[V]$. We claim that $y$ fits into the correct commutative diagram. To see this we let $y \to y_ V$ be a morphism lying over $A \times _ k k[V] \to k[V]$. We can factor the morphisms $y \to y' \to x_{V'}$ and $y \to x_\epsilon$ through the morphism $y \to y_ V$ (by the axioms of categories cofibred in groupoids). Hence we see that both $y_ V$ and $x_ V$ fit into commutative diagrams

$\vcenter { \xymatrix{ y_ V \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x_{V'} \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ x_ V \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x_{V'} \ar[r] & x_0 } }$

and hence by the second part of (S2) there exists an isomorphism $y_ V \to x_ V$ compatible with $y_ V \to x_{V'}$ and $x_ V \to x_{V'}$ and in particular compatible with the maps to $x_0$. The composition $y \to y_ V \to x_ V$ then fits into the required commutative diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V] \ar[d] \ar[r] & k[V] \ar[d] \\ A \ar[r] & k } }$

In this way we see that the first part of $(S2)$ holds with $k[\epsilon ]$ replaced by $k[V]$.

To prove the second part suppose given two commutative diagrams

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V] \ar[d] \ar[r] & k[V] \ar[d] \\ A \ar[r] & k } }$

We will use the morphisms $x_ V \to x_{V'} \to x_0$ and $x_ V \to x_\epsilon \to x_0$ introduced in the first paragraph of the proof. Choose morphisms $y \to y_{V'}$ and $y' \to y'_{V'}$ lying over $A \times _ k k[V] \to A \times _ k k[V']$. The axioms of a cofibred category imply we can find commutative diagrams

$\vcenter { \xymatrix{ y_{V'} \ar[r] \ar[d] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y'_{V'} \ar[r] \ar[d] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V'] \ar[d] \ar[r] & k[V'] \ar[d] \\ A \ar[r] & k } }$

By induction hypothesis we obtain an isomorphism $b : y_{V'} \to y'_{V'}$ compatible with the morphisms $y_{V'} \to x$ and $y'_{V'} \to x$, in particular compatible with the morphisms to $x_0$. Then we have commutative diagrams

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ y'_{V'} \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r] \ar[d] & x_\epsilon \ar[d] \\ y'_{V'} \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } }$

where the morphism $y \to y'_{V'}$ is the composition $y \to y_{V'} \xrightarrow {b} y'_{V'}$ and where the morphisms $y \to x_\epsilon$ and $y' \to x_\epsilon$ are the compositions of the maps $y \to x_ V$ and $y' \to x_ V$ with the morphism $x_ V \to x_\epsilon$. Then the second part of (S2) guarantees the existence of an isomorphism $y \to y'$ compatible with the maps to $y'_{V'}$, in particular compatible with the maps to $x$ (because $b$ was compatible with the maps to $x$). $\square$

Lemma 89.10.5. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$.

1. If $\mathcal{F}$ satisfies (S1), then so does $\overline{\mathcal{F}}$.

2. If $\mathcal{F}$ satisfies (S2), then so does $\overline{\mathcal{F}}$ provided at least one of the following conditions is satisfied

1. $\mathcal{F}$ is a predeformation category,

2. the category $\mathcal{F}(k)$ is a set or a setoid, or

3. for any morphism $x_\epsilon \to x_0$ of $\mathcal{F}$ lying over $k[\epsilon ] \to k$ the pushforward map $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective.

Proof. Assume $\mathcal{F}$ has (S1). Suppose we have ring maps $f_ i : A_ i \to A$ in $\mathcal{C}_\Lambda$ with $f_2$ surjective. Let $x_ i \in \mathcal{F}(A_ i)$ such that the pushforwards $f_{1, *}(x_1)$ and $f_{2, *}(x_2)$ are isomorphic. Then we can denote $x$ an object of $\mathcal{F}$ over $A$ isomorphic to both of these and we obtain a diagram as in (S1). Hence we find an object $y$ of $\mathcal{F}$ over $A_1 \times _ A A_2$ whose pushforward to $A_1$, resp. $A_2$ is isomorphic to $x_1$, resp. $x_2$. In this way we see that (S1) holds for $\overline{\mathcal{F}}$.

Assume $\mathcal{F}$ has (S2). The first part of (S2) for $\overline{\mathcal{F}}$ follows as in the argument above. The second part of (S2) for $\overline{\mathcal{F}}$ signifies that the map

$\overline{\mathcal{F}}(A \times _ k k[\epsilon ]) \to \overline{\mathcal{F}}(A) \times _{\overline{\mathcal{F}}(k)} \overline{\mathcal{F}}(k[\epsilon ])$

is injective for any ring $A$ in $\mathcal{C}_\Lambda$. Suppose that $y, y' \in \mathcal{F}(A \times _ k k[\epsilon ])$. Using the axioms of cofibred categories we can choose commutative diagrams

$\vcenter { \xymatrix{ y \ar[r]_ c \ar[d]_ a & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x'_\epsilon \ar[d]^{e'} \\ x' \ar[r]^{d'} & x'_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } }$

Assume that there exist isomorphisms $\alpha : x \to x'$ in $\mathcal{F}(A)$ and $\beta : x_\epsilon \to x'_\epsilon$ in $\mathcal{F}(k[\epsilon ])$. This also means there exists an isomorphism $\gamma : x_0 \to x'_0$ compatible with $\alpha$. To prove (S2) for $\overline{\mathcal{F}}$ we have to show that there exists an isomorphism $y \to y'$ in $\mathcal{F}(A \times _ k k[\epsilon ])$. By (S2) for $\mathcal{F}$ such a morphism will exist if we can choose the isomorphisms $\alpha$ and $\beta$ and $\gamma$ such that

$\xymatrix{ x \ar[d]^\alpha \ar[r] & x_0 \ar[d]^\gamma & x_\epsilon \ar[d]^\beta \ar[l]^ e \\ x' \ar[r] & x'_0 & x'_\epsilon \ar[l]_{e'} }$

is commutative (because then we can replace $x$ by $x'$ and $x_\epsilon$ by $x'_\epsilon$ in the previous displayed diagram). The left hand square commutes by our choice of $\gamma$. We can factor $e' \circ \beta$ as $\gamma ' \circ e$ for some second map $\gamma ' : x_0 \to x'_0$. Now the question is whether we can arrange it so that $\gamma = \gamma '$? This is clear if $\mathcal{F}(k)$ is a set, or a setoid. Moreover, if $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective, then we can adjust the choice of $\beta$ by precomposing with an automorphism of $x_\epsilon$ whose image is $\gamma ^{-1} \circ \gamma '$ to make things work. $\square$

Lemma 89.10.6. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Let $\mathcal{F}_{x_0}$ be the category cofibred in groupoids over $\mathcal{C}_\Lambda$ constructed in Remark 89.6.4.

1. If $\mathcal{F}$ satisfies (S1), then so does $\mathcal{F}_{x_0}$.

2. If $\mathcal{F}$ satisfies (S2), then so does $\mathcal{F}_{x_0}$.

Proof. Any diagram as in Definition 89.10.1 in $\mathcal{F}_{x_0}$ gives rise to a diagram in $\mathcal{F}$ and the output of condition (S1) or (S2) for this diagram in $\mathcal{F}$ can be viewed as an output for $\mathcal{F}_{x_0}$ as well. $\square$

Lemma 89.10.7. Let $p: \mathcal{F} \to \mathcal{C}_\Lambda$ be a category cofibered in groupoids. Consider a diagram of $\mathcal{F}$

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

in $\mathcal{C}_\Lambda$. Assume $\mathcal{F}$ satisfies (S2). Then there exists a morphism $s : x \to y$ with $a \circ s = \text{id}_ x$ if and only if there exists a morphism $s_\epsilon : x \to x_\epsilon$ with $e \circ s_\epsilon = d$.

Proof. The “only if” direction is clear. Conversely, assume there exists a morphism $s_\epsilon : x \to x_\epsilon$ with $e \circ s_\epsilon = d$. Note that $p(s_\epsilon ) : A \to k[\epsilon ]$ is a ring map compatible with the map $A \to k$. Hence we obtain

$\sigma = (\text{id}_ A, p(s_\epsilon )) : A \to A \times _ k k[\epsilon ].$

Choose a pushforward $x \to \sigma _*x$. By construction we can factor $s_\epsilon$ as $x \to \sigma _*x \to x_\epsilon$. Moreover, as $\sigma$ is a section of $A \times _ k k[\epsilon ] \to A$, we get a morphism $\sigma _*x \to x$ such that $x \to \sigma _*x \to x$ is $\text{id}_ x$. Because $e \circ s_\epsilon = d$ we find that the diagram

$\xymatrix{ \sigma _*x \ar[r] \ar[d] & x_\epsilon \ar[d]_ e \\ x \ar[r]^ d & x_0 }$

is commutative. Hence by (S2) we obtain a morphism $\sigma _*x \to y$ such that $\sigma _*x \to y \to x$ is the given map $\sigma _*x \to x$. The solution to the problem is now to take $a : x \to y$ equal to the composition $x \to \sigma _*x \to y$. $\square$

Lemma 89.10.8. Consider a commutative diagram in a predeformation category $\mathcal{F}$

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_2 \ar[d]^{a_2} \\ x_1 \ar[r]^{a_1} & x } } \quad \text{lying over} \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d]^{f_2} \\ A_1 \ar[r]^{f_1} & A } }$

in $\mathcal{C}_\Lambda$ where $f_2 : A_2 \to A$ is a small extension. Assume there is a map $h : A_1 \to A_2$ such that $f_2 = f_1 \circ h$. Let $I = \mathop{\mathrm{Ker}}(f_2)$. Consider the ring map

$g : A_1 \times _ A A_2 \longrightarrow k[I] = k \oplus I, \quad (u, v) \longmapsto \overline{u} \oplus (v - h(u))$

Choose a pushforward $y \to g_*y$. Assume $\mathcal{F}$ satisfies (S2). If there exists a morphism $x_1 \to g_*y$, then there exists a morphism $b: x_1 \to x_2$ such that $a_1 = a_2 \circ b$.

Proof. Note that $\text{id}_{A_1} \times g : A_1 \times _ A A_2 \to A_1 \times _ k k[I]$ is an isomorphism and that $k[I] \cong k[\epsilon ]$. Hence we have a diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & g_*y \ar[d] \\ x_1 \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A_1 \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ A_1 \ar[r] & k. } }$

where $x_0$ is an object of $\mathcal{F}$ lying over $k$ (every object of $\mathcal{F}$ has a unique morphism to $x_0$, see discussion following Definition 89.6.2). If we have a morphism $x_1 \to g_*y$ then Lemma 89.10.7 provides us with a section $s : x_1 \to y$ of the map $y \to x_1$. Composing this with the map $y \to x_2$ we obtain $b : x_1 \to x_2$ which has the property that $a_1 = a_2 \circ b$ because the diagram of the lemma commutes and because $s$ is a section. $\square$

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