101.5 Inertia stacks
The (relative) inertia stack of a stack in groupoids is defined in Stacks, Section 8.7. The actual construction, in the setting of fibred categories, and some of its properties is in Categories, Section 4.34.
Lemma 101.5.1. Let $\mathcal{X}$ be an algebraic stack. Then the inertia stack $\mathcal{I}_\mathcal {X}$ is an algebraic stack as well. The morphism
\[ \mathcal{I}_\mathcal {X} \longrightarrow \mathcal{X} \]
is representable by algebraic spaces and locally of finite type. More generally, let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then the relative inertia $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic stack and the morphism
\[ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{X} \]
is representable by algebraic spaces and locally of finite type.
Proof.
By Categories, Lemma 4.34.1 there are equivalences
\[ \mathcal{I}_\mathcal {X} \to \mathcal{X} \times _{\Delta , \mathcal{X} \times _ S \mathcal{X}, \Delta } \mathcal{X} \quad \text{and}\quad \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X} \times _{\Delta , \mathcal{X} \times _\mathcal {Y} \mathcal{X}, \Delta } \mathcal{X} \]
which shows that the inertia stacks are algebraic stacks. Let $T \to \mathcal{X}$ be a morphism given by the object $x$ of the fibre category of $\mathcal{X}$ over $T$. Then we get a $2$-fibre product square
\[ \xymatrix{ \mathit{Isom}_\mathcal {X}(x, x) \ar[d] \ar[r] & \mathcal{I}_\mathcal {X} \ar[d] \\ T \ar[r]^ x & \mathcal{X} } \]
This follows immediately from the definition of $\mathcal{I}_\mathcal {X}$. Since $\mathit{Isom}_\mathcal {X}(x, x)$ is always an algebraic space locally of finite type over $T$ (see Lemma 101.3.1) we conclude that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is representable by algebraic spaces and locally of finite type. Finally, for the relative inertia we get
\[ \vcenter { \xymatrix{ \mathit{Isom}_\mathcal {X}(x, x) \ar[d] & K \ar[l] \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ \mathit{Isom}_\mathcal {Y}(f(x), f(x)) & T \ar[l]_-e \ar[r]^ x & \mathcal{X} } } \]
with both squares $2$-fibre products. This follows from Categories, Lemma 4.34.3. The left vertical arrow is a morphism of algebraic spaces locally of finite type over $T$, and hence is locally of finite type, see Morphisms of Spaces, Lemma 67.23.6. Thus $K$ is an algebraic space and $K \to T$ is locally of finite type. This proves the assertion on the relative inertia.
$\square$
Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks and let $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ be its inertia stack. Let $T$ be a scheme and let $x$ be an object of $\mathcal{X}$ over $T$. Set $y = f(x)$. We have seen in the proof of Lemma 101.5.1 that for any scheme $T$ and object $x$ of $\mathcal{X}$ over $T$ there is an exact sequence of sheaves of groups
101.5.2.1
\begin{equation} \label{stacks-morphisms-equation-exact-sequence-isom} 0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \to \mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(y, y) \end{equation}
The group structure on the second and third term is the one defined in Lemma 101.3.2 and the sequence gives a meaning to the first term. Also, there is a canonical cartesian square
\[ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ T \ar[r]^ x & \mathcal{X} } \]
In fact, the group structure on $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ discussed in Remark 101.5.2 induces the group structure on $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$. This allows us to define the sheaf $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$ also for morphisms from algebraic spaces to $\mathcal{X}$. We formalize this in the following definition.
Definition 101.5.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $Z$ be an algebraic space.
Let $x : Z \to \mathcal{X}$ be a morphism. We set
\[ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) = Z \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \]
We endow it with the structure of a group algebraic space over $Z$ by pulling back the composition law discussed in Remark 101.5.2. We will sometimes refer to $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$ as the relative sheaf of automorphisms of $x$.
Let $x_1, x_2 : Z \to \mathcal{X}$ be morphisms. Set $y_ i = f \circ x_ i$. Let $\alpha : y_1 \to y_2$ be a $2$-morphism. Then $\alpha $ determines a morphism $\Delta ^\alpha : Z \to Z \times _{y_1, \mathcal{Y}, y_2} Z$ and we set
\[ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2) = (Z \times _{x_1, \mathcal{X}, x_2} Z) \times _{Z \times _{y_1, \mathcal{Y}, y_2} Z, \Delta ^\alpha } Z. \]
We will sometimes refer to $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2)$ as the relative sheaf of isomorphisms from $x_1$ to $x_2$.
If $\mathcal{Y} = \mathop{\mathrm{Spec}}(\mathbf{Z})$ or more generally when $\mathcal{Y}$ is an algebraic space, then we use the notation $\mathit{Isom}_\mathcal {X}(x, x)$ and $\mathit{Isom}_\mathcal {X}(x_1, x_2)$ and we use the terminology sheaf of automorphisms of $x$ and sheaf of isomorphisms from $x_1$ to $x_2$.
Lemma 101.5.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $Z$ be an algebraic space and let $x_ i : Z \to \mathcal{X}$, $i = 1, 2$ be morphisms. Then
$\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ is a group algebraic space over $Z$,
there is an exact sequence of groups
\[ 0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \to \mathit{Isom}_\mathcal {X}(x_2, x_2) \to \mathit{Isom}_\mathcal {Y}(f \circ x_2, f \circ x_2) \]
there is a map of algebraic spaces $ \mathit{Isom}_\mathcal {X}(x_1, x_2) \to \mathit{Isom}_\mathcal {Y}(f \circ x_1, f \circ x_2) $ such that for any $2$-morphism $\alpha : f \circ x_1 \to f \circ x_2$ we obtain a cartesian diagram
\[ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2) \ar[d] \ar[r] & Z \ar[d]^\alpha \\ \mathit{Isom}_\mathcal {X}(x_1, x_2) \ar[r] & \mathit{Isom}_\mathcal {Y}(f \circ x_1, f \circ x_2) } \]
for any $2$-morphism $\alpha : f \circ x_1 \to f \circ x_2$ the algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2)$ is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ over $Z$.
Proof.
Part (1) follows from Definition 101.5.3. Part (2) comes from the exact sequence (101.5.2.1) étale locally on $Z$. Part (3) can be seen by unwinding the definitions. Locally on $Z$ in the étale topology part (4) reduces to part (2) of Lemma 101.3.2.
$\square$
Lemma 101.5.5. Let $\pi : \mathcal{X} \to \mathcal{Y}$ and $f : \mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks. Set $\mathcal{X}' = \mathcal{X} \times _\mathcal {Y} \mathcal{Y}'$. Then both squares in the diagram
\[ \xymatrix{ \mathcal{I}_{\mathcal{X}'/\mathcal{Y}'} \ar[r] \ar[d]_{ \text{Categories, Equation}\ (04Z4) } & \mathcal{X}' \ar[r]_{\pi '} \ar[d] & \mathcal{Y}' \ar[d]^ f \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r]^\pi & \mathcal{Y} } \]
are fibre product squares.
Proof.
The inertia stack $\mathcal{I}_{\mathcal{X}'/\mathcal{Y}'}$ is defined as the category of pairs $(x', \alpha ')$ where $x'$ is an object of $\mathcal{X}'$ and $\alpha '$ is an automorphism of $x'$ with $\pi '(\alpha ') = \text{id}$, see Categories, Section 4.34. Suppose that $x'$ lies over the scheme $U$ and maps to the object $x$ of $\mathcal{X}$. By the construction of the $2$-fibre product in Categories, Lemma 4.32.3 we see that $x' = (U, x, y', \beta )$ where $y'$ is an object of $\mathcal{Y}'$ over $U$ and $\beta $ is an isomorphism $\beta : \pi (x) \to f(y')$ in the fibre category of $\mathcal{Y}$ over $U$. By the very construction of the $2$-fibre product the automorphism $\alpha '$ is a pair $(\alpha , \gamma )$ where $\alpha $ is an automorphism of $x$ over $U$ and $\gamma $ is an automorphism of $y'$ over $U$ such that $\alpha $ and $\gamma $ are compatible via $\beta $. The condition $\pi '(\alpha ') = \text{id}$ signifies that $\gamma = \text{id}$ whereupon the condition that $\alpha , \beta , \gamma $ are compatible is exactly the condition $\pi (\alpha ) = \text{id}$, i.e., means exactly that $(x, \alpha )$ is an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. In this way we see that the left square is a fibre product square (some details omitted).
$\square$
Lemma 101.5.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Then the diagram
\[ \xymatrix{ \mathcal{I}_\mathcal {X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{I}_\mathcal {Y} \ar[r] & \mathcal{Y} } \]
is a fibre product square.
Proof.
This follows immediately from the fact that $f$ is fully faithful (see Properties of Stacks, Lemma 100.8.4) and the definition of the inertia in Categories, Section 4.34. Namely, an object of $\mathcal{I}_\mathcal {X}$ over a scheme $T$ is the same thing as a pair $(x, \alpha )$ consisting of an object $x$ of $\mathcal{X}$ over $T$ and a morphism $\alpha : x \to x$ in the fibre category of $\mathcal{X}$ over $T$. As $f$ is fully faithful we see that $\alpha $ is the same thing as a morphism $\beta : f(x) \to f(x)$ in the fibre category of $\mathcal{Y}$ over $T$. Hence we can think of objects of $\mathcal{I}_\mathcal {X}$ over $T$ as triples $((y, \beta ), x, \gamma )$ where $y$ is an object of $\mathcal{Y}$ over $T$, $\beta : y \to y$ in $\mathcal{Y}_ T$ and $\gamma : y \to f(x)$ is an isomorphism over $T$, i.e., an object of $\mathcal{I}_\mathcal {Y} \times _\mathcal {Y} \mathcal{X}$ over $T$.
$\square$
Lemma 101.5.7. Let $\mathcal{X}$ be an algebraic stack. Let $[U/R] \to \mathcal{X}$ be a presentation. Let $G/U$ be the stabilizer group algebraic space associated to the groupoid $(U, R, s, t, c)$. Then
\[ \xymatrix{ G \ar[d] \ar[r] & U \ar[d] \\ \mathcal{I}_\mathcal {X} \ar[r] & \mathcal{X} } \]
is a fibre product diagram.
Proof.
Immediate from Groupoids in Spaces, Lemma 78.26.2.
$\square$
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