## 89.19 Infinitesimal automorphisms

Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$. Given a morphism $x' \to x$ in $\mathcal{F}$ lying over $A' \to A$, there is an induced homomorphism

$\text{Aut}_{A'}(x') \to \text{Aut}_ A(x).$

Lemma 89.16.7 says that the cokernel of this homomorphism determines whether condition (RS) on $\mathcal{F}$ passes to $\overline{\mathcal{F}}$. In this section we study the kernel of this homomorphism. We will see that it also gives a measure of how far $\mathcal{F}$ is from $\overline{\mathcal{F}}$.

Definition 89.19.1. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda$. Let $x' \to x$ be a morphism in $\mathcal{F}$ lying over $A' \to A$. The kernel

$\text{Inf}(x'/x) = \mathop{\mathrm{Ker}}(\text{Aut}_{A'}(x') \to \text{Aut}_ A(x))$

is the group of infinitesimal automorphisms of $x'$ over $x$.

Definition 89.19.2. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda$. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Assume a choice of pushforward $x_0 \to x_0'$ of $x_0$ along the map $k \to k[\epsilon ], a \mapsto a$ has been made. Then there is a unique map $x'_0 \to x_0$ such that $x_0 \to x_0' \to x_0$ is the identity on $x_0$. Then

$\text{Inf}_{x_0}(\mathcal F) = \text{Inf}(x'_0/x_0)$

is the group of infinitesimal automorphisms of $x_0$

Remark 89.19.3. Up to canonical isomorphism $\text{Inf}_{x_0}(\mathcal{F})$ does not depend on the choice of pushforward $x_0 \to x_0'$ because any two pushforwards are canonically isomorphic. Moreover, if $y_0 \in \mathcal{F}(k)$ and $x_0 \cong y_0$ in $\mathcal{F}(k)$, then $\text{Inf}_{x_0}(\mathcal{F}) \cong \text{Inf}_{y_0}(\mathcal{F})$ where the isomorphism depends (only) on the choice of an isomorphism $x_0 \to y_0$. In particular, $\text{Aut}_ k(x_0)$ acts on $\text{Inf}_{x_0}(\mathcal{F})$.

Remark 89.19.4. Assume $\mathcal{F}$ is a predeformation category. Then

1. for $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ the automorphism group $\text{Aut}_ k(x_0)$ is trivial and hence $\text{Inf}_{x_0}(\mathcal{F}) = \text{Aut}_{k[\epsilon ]}(x'_0)$, and

2. for $x_0, y_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ there is a unique isomorphism $x_0 \to y_0$ and hence a canonical identification $\text{Inf}_{x_0}(\mathcal{F}) = \text{Inf}_{y_0}(\mathcal{F})$.

Since $\mathcal{F}(k)$ is nonempty, choosing $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ and setting

$\text{Inf}(\mathcal{F}) = \text{Inf}_{x_0}(\mathcal{F})$

we get a well defined group of infinitesimal automorphisms of $\mathcal{F}$. With this notation we have $\text{Inf}(\mathcal{F}_{x_0}) = \text{Inf}_{x_0}(\mathcal{F})$. Please compare with the equality $T\mathcal{F}_{x_0} = T_{x_0}\mathcal{F}$ in Remark 89.12.5.

We will see that $\text{Inf}_{x_0}(\mathcal{F})$ has a natural $k$-vector space structure when $\mathcal{F}$ satisfies (RS). At the same time, we will see that if $\mathcal{F}$ satisfies (RS), then the infinitesimal automorphisms $\text{Inf}(x'/x)$ of a morphism $x' \to x$ lying over a small extension are governed by $\text{Inf}_{x_0}(\mathcal{F})$, where $x_0$ is a pushforward of $x$ to $\mathcal{F}(k)$. In order to do this, we introduce the automorphism functor for any object $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F})$ as follows.

Definition 89.19.5. Let $p : \mathcal{F} \to \mathcal{C}$ be a category cofibered in groupoids over an arbitrary base category $\mathcal{C}$. Assume a choice of pushforwards has been made. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F})$ and let $U = p(x)$. Let $U/\mathcal{C}$ denote the category of objects under $U$. The automorphism functor of $x$ is the functor $\mathit{Aut}(x) : U/\mathcal{C} \to \textit{Sets}$ sending an object $f : U \to V$ to $\text{Aut}_ V(f_*x)$ and sending a morphism

$\xymatrix{ V' \ar[rr] & & V\\ & U \ar[ul]^{f'} \ar[ur]_ f & }$

to the homomorphism $\text{Aut}_{V'}(f'_*x) \to \text{Aut}_ V(f_*x)$ coming from the unique morphism $f'_*x \to f_*x$ lying over $V' \to V$ and compatible with $x \to f'_*x$ and $x \to f_*x$.

We will be concerned with the automorphism functors of objects in a category cofibered in groupoids $\mathcal{F}$ over $\mathcal{C}_\Lambda$. If $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$, then the category $A/\mathcal{C}_\Lambda$ is nothing but the category $\mathcal{C}_ A$, i.e. the category defined in Section 89.3 where we take $\Lambda = A$ and $k = A/\mathfrak m_ A$. Hence the automorphism functor of an object $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$ is a functor $\mathit{Aut}(x) : \mathcal{C}_ A \to \textit{Sets}$.

The following lemma could be deduced from Lemma 89.16.12 by thinking about the “inertia” of a category cofibred in groupoids, see for example Stacks, Section 8.7 and Categories, Section 4.34. However, it is easier to see it directly.

Lemma 89.19.6. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$. Then $\mathit{Aut}(x): \mathcal{C}_ A \to \textit{Sets}$ satisfies (RS).

Proof. It follows that $\mathit{Aut}(x)$ satisfies (RS) from the fully faithfulness of the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ in Lemma 89.16.4. $\square$

Lemma 89.19.7. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$. Let $x_0$ be a pushforward of $x$ to $\mathcal{F}(k)$.

1. $T_{\text{id}_{x_0}} \mathit{Aut}(x)$ has a natural $k$-vector space structure such that addition agrees with composition in $T_{\text{id}_{x_0}} \mathit{Aut}(x)$. In particular, composition in $T_{\text{id}_{x_0}} \mathit{Aut}(x)$ is commutative.

2. There is a canonical isomorphism $T_{\text{id}_{x_0}} \mathit{Aut}(x) \to T_{\text{id}_{x_0}} \mathit{Aut}(x_0)$ of $k$-vector spaces.

Proof. We apply Remark 89.6.4 to the functor $\mathit{Aut}(x) : \mathcal{C}_ A \to \textit{Sets}$ and the element $\text{id}_{x_0} \in \mathit{Aut}(x)(k)$ to get a predeformation functor $F = \mathit{Aut}(x)_{\text{id}_{x_0}}$. By Lemmas 89.19.6 and 89.16.11 $F$ is a deformation functor. By definition $T_{\text{id}_{x_0}} \mathit{Aut}(x) = TF = F(k[\epsilon ])$ which has a natural $k$-vector space structure specified by Lemma 89.11.8.

Addition is defined as the composition

$F(k[\epsilon ]) \times F(k[\epsilon ]) \longrightarrow F(k[\epsilon ] \times _ k k[\epsilon ]) \longrightarrow F(k[\epsilon ])$

where the first map is the inverse of the bijection guaranteed by (RS) and the second is induced by the $k$-algebra map $k[\epsilon ] \times _ k k[\epsilon ] \to k[\epsilon ]$ which maps $(\epsilon , 0)$ and $(0, \epsilon )$ to $\epsilon$. If $A \to B$ is a ring map in $\mathcal{C}_\Lambda$, then $F(A) \to F(B)$ is a homomorphism where $F(A) = \mathit{Aut}(x)_{\text{id}_{x_0}}(A)$ and $F(B) = \mathit{Aut}(x)_{\text{id}_{x_0}}(B)$ are groups under composition. We conclude that $+ : F(k[\epsilon ]) \times F(k[\epsilon ])\to F(k[\epsilon ])$ is a homomorphism where $F(k[\epsilon ])$ is regarded as a group under composition. With $\text{id} \in F(k[\epsilon ])$ the unit element we see that $+(v, \text{id}) = +(\text{id}, v) = v$ for any $v \in F(k[\epsilon ])$ because $(\text{id}, v)$ is the pushforward of $v$ along the ring map $k[\epsilon ] \to k[\epsilon ] \times _ k k[\epsilon ]$ with $\epsilon \mapsto (\epsilon , 0)$. In general, given a group $G$ with multiplication $\circ$ and $+ : G \times G \to G$ is a homomorphism such that $+(g, 1) = +(1, g) = g$, where $1$ is the identity of $G$, then $+ = \circ$. This shows addition in the $k$-vector space structure on $F(k[\epsilon ])$ agrees with composition.

Finally, (2) is a matter of unwinding the definitions. Namely $T_{\text{id}_{x_0}} \mathit{Aut}(x)$ is the set of automorphisms $\alpha$ of the pushforward of $x$ along $A \to k \to k[\epsilon ]$ which are trivial modulo $\epsilon$. On the other hand $T_{\text{id}_{x_0}} \mathit{Aut}(x_0)$ is the set of automorphisms of the pushforward of $x_0$ along $k \to k[\epsilon ]$ which are trivial modulo $\epsilon$. Since $x_0$ is the pushforward of $x$ along $A \to k$ the result is clear. $\square$

Remark 89.19.8. We point out some basic relationships between infinitesimal automorphism groups, liftings, and tangent spaces to automorphism functors. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$. Let $x' \to x$ be a morphism lying over a ring map $A' \to A$. Then from the definitions we have an equality

$\text{Inf}(x'/x) = \text{Lift}(\text{id}_ x, A')$

where the liftings are of $\text{id}_ x$ as an object of $\mathit{Aut}(x')$. If $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ and $x'_0$ is the pushforward to $\mathcal{F}(k[\epsilon ])$, then applying this to $x'_0 \to x_0$ we get

$\text{Inf}_{x_0}(\mathcal{F}) = \text{Lift}(\text{id}_{x_0}, k[\epsilon ]) = T_{\text{id}_{x_0}} \mathit{Aut}(x_0),$

the last equality following directly from the definitions.

Lemma 89.19.9. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Then $\text{Inf}_{x_0}(\mathcal{F})$ is equal as a set to $T_{\text{id}_{x_0}} \mathit{Aut}(x_0)$, and so has a natural $k$-vector space structure such that addition agrees with composition of automorphisms.

Proof. The equality of sets is as in the end of Remark 89.19.8 and the statement about the vector space structure follows from Lemma 89.19.7. $\square$

Lemma 89.19.10. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of categories cofibred in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Then $\varphi$ induces a $k$-linear map $\text{Inf}_{x_0}(\mathcal{F}) \to \text{Inf}_{\varphi (x_0)}(\mathcal{G})$.

Proof. It is clear that $\varphi$ induces a morphism from $\mathit{Aut}(x_0) \to \mathit{Aut}(\varphi (x_0))$ which maps the identity to the identity. Hence this follows from the result for tangent spaces, see Lemma 89.12.4. $\square$

Lemma 89.19.11. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x' \to x$ be a morphism lying over a surjective ring map $A' \to A$ with kernel $I$ annihilated by $\mathfrak m_{A'}$. Let $x_0$ be a pushforward of $x$ to $\mathcal{F}(k)$. Then $\text{Inf}(x'/x)$ has a free and transitive action by $T_{\text{id}_{x_0}} \mathit{Aut}(x') \otimes _ k I = \text{Inf}_{x_0}(\mathcal{F}) \otimes _ k I$.

Proof. This is just the analogue of Lemma 89.17.5 in the setting of automorphism sheaves. To be precise, we apply Remark 89.6.4 to the functor $\mathit{Aut}(x') : \mathcal{C}_{A'} \to \textit{Sets}$ and the element $\text{id}_{x_0} \in \mathit{Aut}(x)(k)$ to get a predeformation functor $F = \mathit{Aut}(x')_{\text{id}_{x_0}}$. By Lemmas 89.19.6 and 89.16.11 $F$ is a deformation functor. Hence Lemma 89.17.5 gives a free and transitive action of $TF \otimes _ k I$ on $\text{Lift}(\text{id}_ x, A')$, because as $\text{Lift}(\text{id}_ x, A')$ is a group it is always nonempty. Note that we have equalities of vector spaces

$TF = T_{\text{id}_{x_0}} \mathit{Aut}(x') \otimes _ k I = \text{Inf}_{x_0}(\mathcal{F}) \otimes _ k I$

by Lemma 89.19.7. The equality $\text{Inf}(x'/x) = \text{Lift}(\text{id}_ x, A')$ of Remark 89.19.8 finishes the proof. $\square$

Lemma 89.19.12. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x' \to x$ be a morphism in $\mathcal{F}$ lying over a surjective ring map. Let $x_0$ be a pushforward of $x$ to $\mathcal{F}(k)$. If $\text{Inf}_{x_0}(\mathcal{F}) = 0$ then $\text{Inf}(x'/x) = 0$.

Lemma 89.19.13. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Then $\text{Inf}_{x_0}(\mathcal{F}) = 0$ if and only if the natural morphism $\mathcal{F}_{x_0} \to \overline{\mathcal{F}_{x_0}}$ of categories cofibered in groupoids is an equivalence.

Proof. The morphism $\mathcal{F}_{x_0} \to \overline{\mathcal{F}_{x_0}}$ is an equivalence if and only if $\mathcal{F}_{x_0}$ is fibered in setoids, cf. Categories, Section 4.39 (a setoid is by definition a groupoid in which the only automorphism of any object is the identity). We prove that $\text{Inf}_{x_0}(\mathcal{F}) = 0$ if and only if this condition holds for $\mathcal{F}_{x_0}$. Obviously if $\mathcal{F}_{x_0}$ is fibered in setoids then $\text{Inf}_{x_0}(\mathcal{F}) = 0$. Conversely assume $\text{Inf}_{x_0}(\mathcal{F}) = 0$. Let $A$ be an object of $\mathcal{C}_\Lambda$. Then by Lemma 89.19.12, $\text{Inf}(x/x_0) = 0$ for any object $x \to x_0$ of $\mathcal{F}_{x_0}(A)$. Since by definition $\text{Inf}(x/x_0)$ equals the group of automorphisms of $x \to x_0$ in $\mathcal{F}_{x_0}(A)$, this proves $\mathcal{F}_{x_0}(A)$ is a setoid. $\square$

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