## 100.5 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 100.5.1. 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 100.3.1) 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 66.23.6. Thus $K$ is an algebraic space and $K \to T$ is locally of finite type. This proves the assertion on the relative inertia. $\square$

Remark 100.5.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. In Properties of Stacks, Remark 99.3.7 we have seen that the $2$-category of morphisms $\mathcal{Z} \to \mathcal{X}$ representable by algebraic spaces with target $\mathcal{X}$ forms a category. In this category the inertia stack of $\mathcal{X}/\mathcal{Y}$ is a group object. Recall that an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is just a pair $(x, \alpha )$ where $x$ is an object of $\mathcal{X}$ and $\alpha$ is an automorphism of $x$ in the fibre category of $\mathcal{X}$ that $x$ lives in with $f(\alpha ) = \text{id}$. The composition

$c : \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \times _\mathcal {X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$

is given by the rule on objects

$((x, \alpha ), (x', \alpha '), \beta ) \mapsto (x, \alpha \circ \beta ^{-1} \circ \alpha ' \circ \beta )$

which makes sense as $\beta : x \to x'$ is an isomorphism in the fibre category by our definition of fibre products. The neutral element $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is given by the functor $x \mapsto (x, \text{id}_ x)$. We omit the proof that the axioms of a group object hold.

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 100.5.1 that for any scheme $T$ and object $x$ of $\mathcal{X}$ over $T$ there is an exact sequence of sheaves of groups

100.5.2.1
$$\label{stacks-morphisms-equation-exact-sequence-isom} 0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \to \mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(y, y)$$

The group structure on the second and third term is the one defined in Lemma 100.3.2 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 100.5.2 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 100.5.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $Z$ be an algebraic space.

1. 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 100.5.2. We will sometimes refer to $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$ as the relative sheaf of automorphisms of $x$.

2. 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 100.5.4. 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

1. $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ is a group algebraic space over $Z$,

2. there is an exact sequence of groups

$0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \to \mathit{Isom}_\mathcal {X}(x_2, x_2) \to \mathit{Isom}_\mathcal {Y}(f \circ x_2, f \circ x_2)$
3. 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

$\xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2) \ar[d] \ar[r] & Z \ar[d]^\alpha \\ \mathit{Isom}_\mathcal {X}(x_1, x_2) \ar[r] & \mathit{Isom}_\mathcal {Y}(f \circ x_1, f \circ x_2) }$
4. 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 100.5.3. Part (2) comes from the exact sequence (100.5.2.1) 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 100.3.2. $\square$

Lemma 100.5.5. 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 100.5.6. 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 99.8.4) 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 100.5.7. 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 77.26.2. $\square$

Comment #1426 by Ariyan on

Minor typo: In the first line of the proof of Lemma 79.5.4 "is the defined as" should be "is defined as".

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