The Stacks project

89.20 Applications

We collect some results on deformation categories we will use later.

Lemma 89.20.1. Let $f : \mathcal{H} \to \mathcal{F}$ and $g : \mathcal{G} \to \mathcal{F}$ be $1$-morphisms of deformation categories. Then

  1. $\mathcal{W} = \mathcal{H} \times _\mathcal {F} \mathcal{G}$ is a deformation category, and

  2. we have a $6$-term exact sequence of vector spaces

    \[ 0 \to \text{Inf}(\mathcal{W}) \to \text{Inf}(\mathcal{H}) \oplus \text{Inf}(\mathcal{G}) \to \text{Inf}(\mathcal{F}) \to T\mathcal{W} \to T\mathcal{H} \oplus T\mathcal{G} \to T\mathcal{F} \]

Proof. Part (1) follows from Lemma 89.16.12 and the fact that $\mathcal{W}(k)$ is the fibre product of two setoids with a unique isomorphism class over a setoid with a unique isomorphism class.

Part (2). Let $w_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{W}(k))$ and let $x_0, y_0, z_0$ be the image of $w_0$ in $\mathcal{F}, \mathcal{H}, \mathcal{G}$. Then $\text{Inf}(\mathcal{W}) = \text{Inf}_{w_0}(\mathcal{W})$ and simlarly for $\mathcal{H}$, $\mathcal{G}$, and $\mathcal{F}$, see Remark 89.19.4. We apply Lemmas 89.12.4 and 89.19.10 to get all the linear maps except for the “boundary map” $\delta : \text{Inf}_{x_0}(\mathcal{F}) \to T\mathcal{W}$. We will insert suitable signs later.

Construction of $\delta $. Choose a pushforward $w_0 \to w'_0$ along $k \to k[\epsilon ]$. Denote $x'_0, y'_0, z'_0$ the images of $w'_0$ in $\mathcal{F}, \mathcal{H}, \mathcal{G}$. In particular we obtain isomorphisms $b' : f(y'_0) \to x'_0$ and $c' : x'_0 \to g(z'_0)$. Denote $b : f(y_0) \to x_0$ and $c : x_0 \to g(z_0)$ the pushforwards along $k[\epsilon ] \to k$. Observe that this means $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b')$ and $w_0 = (k, y_0, z_0, c \circ b)$ in terms of the explicit form of the fibre product of categories, see Remarks 89.5.2 (13). Given $\alpha : x'_0 \to x'_0$ we set $\delta (\alpha ) = (k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ which is indeed an object of $\mathcal{W}$ over $k[\epsilon ]$ and comes with a morphism $(k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b') \to w_0$ over $k[\epsilon ] \to k$ as $\alpha $ pushes forward to the identity over $k$. More generally, for any $k$-vector space $V$ we can define a map

\[ \text{Lift}(\text{id}_{x_0}, k[V]) \longrightarrow \text{Lift}(w_0, k[V]) \]

using exactly the same formulae. This construction is functorial in the vector space $V$ (details omitted). Hence $\delta $ is $k$-linear by an application of Lemma 89.11.5.

Having constructed these maps it is straightforward to show the sequence is exact. Injectivity of the first map comes from the fact that $f \times g : \mathcal{W} \to \mathcal{H} \times \mathcal{G}$ is faithful. If $(\beta , \gamma ) \in \text{Inf}_{y_0}(\mathcal{H}) \oplus \text{Inf}_{z_0}(\mathcal{G})$ map to the same element of $\text{Inf}_{x_0}(\mathcal{F})$ then $(\beta , \gamma )$ defines an automorphism of $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b')$ whence exactness at the second spot. If $\alpha $ as above gives the trivial deformation $(k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ of $w_0$, then the isomorphism $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b') \to (k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ produces a pair $(\beta , \gamma )$ which is a preimage of $\alpha $. If $w = (k[\epsilon ], y, z, \phi )$ is a deformation of $w_0$ such that $y'_0 \cong y$ and $z \cong z'_0$ then the map

\[ f(y'_0) \to f(y) \xrightarrow {\phi } g(z) \to g(z'_0) \]

is an $\alpha $ which maps to $w$ under $\delta $. Finally, if $y$ and $z$ are deformations of $y_0$ and $z_0$ and there exists an isomorphism $\phi : f(y) \to g(z)$ of deformations of $f(y_0) = x_0 = g(z_0)$ then we get a preimage $w = (k[\epsilon ], y, z, \phi )$ of $(x, y)$ in $T\mathcal{W}$. This finishes the proof. $\square$

Lemma 89.20.2. Let $\mathcal{H}_1 \to \mathcal{G}$, $\mathcal{H}_2 \to \mathcal{G}$, and $\mathcal{G} \to \mathcal{F}$ be maps of categories cofibred in groupoids over $\mathcal{C}_\Lambda $. Assume

  1. $\mathcal{F}$ and $\mathcal{G}$ are deformation categories,

  2. $T\mathcal{G} \to T\mathcal{F}$ is injective, and

  3. $\text{Inf}(\mathcal{G}) \to \text{Inf}(\mathcal{F})$ is surjective.

Then $\mathcal{H}_1 \times _\mathcal {G} \mathcal{H}_2 \to \mathcal{H}_1 \times _\mathcal {F} \mathcal{H}_2$ is smooth.

Proof. Denote $p_ i : \mathcal{H}_ i \to \mathcal{G}$ and $q : \mathcal{G} \to \mathcal{F}$ be the given maps. Let $A' \to A$ be a small extension in $\mathcal{C}_\Lambda $. An object of $\mathcal{H}_1 \times _\mathcal {F} \mathcal{H}_2$ over $A'$ is a triple $(x'_1, x'_2, a')$ where $x'_ i$ is an object of $\mathcal{H}_ i$ over $A'$ and $a' : q(p_1(x'_1)) \to q(p_2(x'_2))$ is a morphism of the fibre category of $\mathcal{F}$ over $A'$. By pushforward along $A' \to A$ we get $(x_1, x_2, a)$. Lifting this to an object of $\mathcal{H}_1 \times _\mathcal {G} \mathcal{H}_2$ over $A$ means finding a morphism $b : p_1(x_1) \to p_2(x_2)$ over $A$ with $q(b) = a$. Thus we have to show that we can lift $b$ to a morphism $b' : p_1(x'_1) \to p_2(x'_2)$ whose image under $q$ is $a'$.

Observe that we can think of

\[ p_1(x'_1) \to p_1(x_1) \xrightarrow {b} p_2(x_2) \quad \text{and}\quad p_2(x'_2) \to p_2(x_2) \]

as two objects of $\textit{Lift}(p_2(x_2), A' \to A)$. The functor $q$ sends these objects to the two objects

\[ q(p_1(x'_1)) \to q(p_1(x_1)) \xrightarrow {b} q(p_2(x_2)) \quad \text{and}\quad q(p_2(x'_2)) \to q(p_2(x_2)) \]

of $\textit{Lift}(q(p_2(x_2)), A' \to A)$ which are isomorphic using the map $a' : q(p_1(x'_1)) \to q(p_2(x'_2))$. On the other hand, the functor

\[ q : \textit{Lift}(p_2(x_2), A' \to A) \to \textit{Lift}(q(p_2(x_2)), A' \to A) \]

defines a injection on isomorphism classes by Lemma 89.17.5 and our assumption on tangent spaces. Thus we see that there is a morphism $b' : p_1(x_1') \to p_2(x'_2)$ whose pushforward to $A$ is $b$. However, we may need to adjust our choice of $b'$ to achieve $q(b') = a'$. For this it suffices to see that $q : \text{Inf}(p_2(x'_2)/p_2(x_2)) \to \text{Inf}(q(p_2(x'_2))/q(p_2(x_2)))$ is surjective. This follows from our assumption on infinitesimal automorphisms and Lemma 89.19.11. $\square$

Lemma 89.20.3. Let $f : \mathcal{F} \to \mathcal{G}$ be a map of deformation categories. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ with image $y_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{G}(k))$. If

  1. the map $T\mathcal{F} \to T\mathcal{G}$ is surjective, and

  2. for every small extension $A' \to A$ in $\mathcal{C}_\Lambda $ and $x \in \mathcal{F}(A)$ with image $y \in \mathcal{G}(A)$ if there is a lift of $y$ to $A'$, then there is a lift of $x$ to $A'$,

then $\mathcal{F} \to \mathcal{G}$ is smooth (and vice versa).

Proof. Let $A' \to A$ be a small extension. Let $x \in \mathcal{F}(A)$. Let $y' \to f(x)$ be a morphism in $\mathcal{G}$ over $A' \to A$. Consider the functor $\text{Lift}(A', x) \to \text{Lift}(A', f(x))$ induced by $f$. We have to show that there exists an object $x' \to x$ of $\text{Lift}(A', x)$ mapping to $y' \to f(x)$, see Lemma 89.8.2. By condition (2) we know that $\text{Lift}(A', x)$ is not the empty category. By condition (2) and Lemma 89.17.5 we conlude that the map on isomorphism classes is surjective as desired. $\square$

Lemma 89.20.4. Let $\mathcal{F} \to \mathcal{G} \to \mathcal{H}$ be maps of categories cofibred in groupoids over $\mathcal{C}_\Lambda $. If

  1. $\mathcal{F}$, $\mathcal{G}$ are deformation categories

  2. the map $T\mathcal{F} \to T\mathcal{G}$ is surjective, and

  3. $\mathcal{F} \to \mathcal{H}$ is smooth.

Then $\mathcal{F} \to \mathcal{G}$ is smooth.

Proof. Let $A' \to A$ be a small extension in $\mathcal{C}_\Lambda $ and let $x \in \mathcal{F}(A)$ with image $y \in \mathcal{G}(A)$. Assume there is a lift $y' \in \mathcal{G}(A')$. According to Lemma 89.20.3 all we have to do is check that $x$ has a lift too. Take the image $z' \in \mathcal{H}(A')$ of $y'$. Since $\mathcal{F} \to \mathcal{H}$ is smooth, there is an $x' \in \mathcal{F}(A')$ mapping to both $x \in \mathcal{F}(A)$ and $z' \in \mathcal{H}(A')$, see Definition 89.8.1. This finishes the proof. $\square$


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