## 100.6 Higher diagonals

Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. In this situation it makes sense to consider not only the diagonal

\[ \Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \]

but also the diagonal of the diagonal, i.e., the morphism

\[ \Delta _{\Delta _ f} : \mathcal{X} \longrightarrow \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Y} \mathcal{X})} \mathcal{X} \]

Because of this we sometimes use the following terminology. We denote $\Delta _{f, 0} = f$ the *zeroth diagonal*, we denote $\Delta _{f, 1} = \Delta _ f$ the *first diagonal*, and we denote $\Delta _{f, 2} = \Delta _{\Delta _ f}$ the *second diagonal*. Note that $\Delta _{f, 1}$ is representable by algebraic spaces and locally of finite type, see Lemma 100.3.3. Hence $\Delta _{f, 2}$ is representable, a monomorphism, locally of finite type, separated, and locally quasi-finite, see Lemma 100.3.4.

We can describe the second diagonal using the relative inertia stack. Namely, the fibre product $\mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Y} \mathcal{X})} \mathcal{X}$ is equivalent to the relative inertia stack $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by Categories, Lemma 4.34.1. Moreover, via this identification the second diagonal becomes the *neutral section*

\[ \Delta _{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \]

of the relative inertia stack. By analogy with what happens for groupoids in algebraic spaces (Groupoids in Spaces, Lemma 77.29.2) we have the following equivalences.

Lemma 100.6.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

The following are equivalent

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is separated,

$\Delta _{f, 1} = \Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is separated, and

$\Delta _{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion.

The following are equivalent

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is quasi-separated,

$\Delta _{f, 1} = \Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is quasi-separated, and

$\Delta _{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a quasi-compact.

The following are equivalent

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is locally separated,

$\Delta _{f, 1} = \Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is locally separated, and

$\Delta _{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an immersion.

The following are equivalent

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is unramified,

$f$ is DM.

The following are equivalent

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is locally quasi-finite,

$f$ is quasi-DM.

**Proof.**
Proof of (1), (2), and (3). Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $G = U \times _\mathcal {X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic space over $U$ (Lemma 100.5.1). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark 100.5.2. Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is surjective and smooth as a base change of $U \to \mathcal{X}$. Finally, the base change of $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is the identity $U \to G$ of $G/U$. Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma 77.6.1. Since $\Delta _{f, 2}$ is the diagonal of $\Delta _ f$ we have (b) $\Leftrightarrow $ (c) by definition.

Proof of (4) and (5). Recall that (4)(b) means $\Delta _ f$ is unramified and (5)(b) means that $\Delta _ f$ is locally quasi-finite. Choose a scheme $Z$ and a morphism $a : Z \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. Then $a = (x_1, x_2, \alpha )$ where $x_ i : Z \to \mathcal{X}$ and $\alpha : f \circ x_1 \to f \circ x_2$ is a $2$-morphism. Recall that

\[ \vcenter { \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2) \ar[d] \ar[r] & Z \ar[d] \\ \mathcal{X} \ar[r]^{\Delta _ f} & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \ar[d] \ar[r] & Z \ar[d]^{x_2} \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} } } \]

are cartesian squares. By Lemma 100.5.4 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 $Z$. Thus the equivalences in (4) and (5) follow from Groupoids in Spaces, Lemma 77.9.5.
$\square$

Lemma 100.6.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

the morphism $f$ is representable by algebraic spaces,

the second diagonal of $f$ is an isomorphism,

the group stack $ \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is trivial over $\mathcal X$, and

for a scheme $T$ and a morphism $x : T \to \mathcal{X}$ the kernel of $\mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x))$ is trivial.

**Proof.**
We first prove the equivalence of (1) and (2). Namely, $f$ is representable by algebraic spaces if and only if $f$ is faithful, see Algebraic Stacks, Lemma 93.15.2. On the other hand, $f$ is faithful if and only if for every object $x$ of $\mathcal{X}$ over a scheme $T$ the functor $f$ induces an injection $\mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x))$, which happens if and only if the kernel $K$ is trivial, which happens if and only if $e : T \to K$ is an isomorphism for every $x : T \to \mathcal{X}$. Since $K = T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma 100.5.1 we have seen that the kernel in (4) corresponds to the fibre product $T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ over $T$.
$\square$

This lemma leads to the following hierarchy for morphisms of algebraic stacks.

Lemma 100.6.3. A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is

a monomorphism if and only if $\Delta _{f, 1}$ is an isomorphism, and

representable by algebraic spaces if and only if $\Delta _{f, 1}$ is a monomorphism.

Moreover, the second diagonal $\Delta _{f, 2}$ is always a monomorphism.

**Proof.**
Recall from Properties of Stacks, Lemma 99.8.4 that a morphism of algebraic stacks is a monomorphism if and only if its diagonal is an isomorphism of stacks. Thus Lemma 100.6.2 can be rephrased as saying that a morphism is representable by algebraic spaces if the diagonal is a monomorphism. In particular, it shows that condition (3) of Lemma 100.3.4 is actually an if and only if, i.e., a morphism of algebraic stacks is representable by algebraic spaces if and only if its diagonal is a monomorphism.
$\square$

Lemma 100.6.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then

$\Delta _{f, 1}$ separated $\Leftrightarrow $ $\Delta _{f, 2}$ closed immersion $\Leftrightarrow $ $\Delta _{f, 2}$ proper $\Leftrightarrow $ $\Delta _{f, 2}$ universally closed,

$\Delta _{f, 1}$ quasi-separated $\Leftrightarrow $ $\Delta _{f, 2}$ finite type $\Leftrightarrow $ $\Delta _{f, 2}$ quasi-compact, and

$\Delta _{f, 1}$ locally separated $\Leftrightarrow $ $\Delta _{f, 2}$ immersion.

**Proof.**
Follows from Lemmas 100.3.5, 100.3.6, and 100.3.7 applied to $\Delta _{f, 1}$.
$\square$

The following lemma is kind of cute and it may suggest a generalization of these conditions to higher algebraic stacks.

Lemma 100.6.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then

$f$ is separated if and only if $\Delta _{f, 1}$ and $\Delta _{f, 2}$ are universally closed, and

$f$ is quasi-separated if and only if $\Delta _{f, 1}$ and $\Delta _{f, 2}$ are quasi-compact.

$f$ is quasi-DM if and only if $\Delta _{f, 1}$ and $\Delta _{f, 2}$ are locally quasi-finite.

$f$ is DM if and only if $\Delta _{f, 1}$ and $\Delta _{f, 2}$ are unramified.

**Proof.**
Proof of (1). Assume that $\Delta _{f, 2}$ and $\Delta _{f, 1}$ are universally closed. Then $\Delta _{f, 1}$ is separated and universally closed by Lemma 100.6.4. By Morphisms of Spaces, Lemma 66.9.7 and Algebraic Stacks, Lemma 93.10.9 we see that $\Delta _{f, 1}$ is quasi-compact. Hence it is quasi-compact, separated, universally closed and locally of finite type (by Lemma 100.3.3) so proper. This proves “$\Leftarrow $” of (1). The proof of the implication in the other direction is omitted.

Proof of (2). This follows immediately from Lemma 100.6.4.

Proof of (3). This follows from the fact that $\Delta _{f, 2}$ is always locally quasi-finite by Lemma 100.3.4 applied to $\Delta _ f = \Delta _{f, 1}$.

Proof of (4). This follows from the fact that $\Delta _{f, 2}$ is always unramified as Lemma 100.3.4 applied to $\Delta _ f = \Delta _{f, 1}$ shows that $\Delta _{f, 2}$ is locally of finite type and a monomorphism. See More on Morphisms of Spaces, Lemma 75.14.8.
$\square$

Lemma 100.6.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a separated (resp. quasi-separated, resp. quasi-DM, resp. DM) morphism of algebraic stacks. Then

given algebraic spaces $T_ i$, $i = 1, 2$ and morphisms $x_ i : T_ i \to \mathcal{X}$, with $y_ i = f \circ x_ i$ the morphism

\[ T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times _{y_1, \mathcal{Y}, y_2} T_2 \]

is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified),

given an algebraic space $T$ and morphisms $x_ i : T \to \mathcal{X}$, $i = 1, 2$, with $y_ i = f \circ x_ i$ the morphism

\[ \mathit{Isom}_\mathcal {X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal {Y}(y_1, y_2) \]

is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

**Proof.**
Proof of (1). Observe that the diagram

\[ \xymatrix{ T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times _{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

is cartesian. Hence this follows from the fact that $f$ is separated (resp. quasi-separated, resp. quasi-DM, resp. DM) if and only if the diagonal is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof of (2). This is true because

\[ \mathit{Isom}_\mathcal {X}(x_1, x_2) = (T \times _{x_1, \mathcal{X}, x_2} T) \times _{T \times T, \Delta _ T} T \]

hence the morphism in (2) is a base change of the morphism in (1).
$\square$

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