## 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.

1. The following are equivalent

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

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

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

2. The following are equivalent

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

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

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

3. The following are equivalent

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

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

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

4. The following are equivalent

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

2. $f$ is DM.

5. The following are equivalent

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

2. $f$ 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:

1. the morphism $f$ is representable by algebraic spaces,

2. the second diagonal of $f$ is an isomorphism,

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

4. 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

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

2. 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

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

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

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

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

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

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

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

4. $f$ 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

1. 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),

2. 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). $\square$

Comment #1407 by Ariyan Javanpeykar on

Would it be possible to split off (3) of Lemma 79.6.2? I would suggest finishing with (2) and adding an extra sentence "Moreover, the second diagonal [..] is always a monomorphism".

Comment #1415 by on

@#1407: Do you really think that this makes the statement easier to read? I thought it was fun how the numbering suggests a hierarchy of results...

Comment #1417 by jojo on

I think Ariyan was trying to say that the 3) is not well phrased and doesn't belong in the list. I agree with him and his suggestion of correction.

Comment #1418 by Ariyan on

I also think it is cool how the numbering suggests a hierarchy. To keep this hierarchy and make the Lemma well-phrased, would it be possble to add "and" after "is a monomorphism," in 79.6.2.(2)?

Comment #1419 by jojo on

Honnestly I think your first suggestion was better Ariyan, I'm not sure if the second one would really make more sense than what is written right now. But maybe i'm missing something.

Comment #1431 by Ariyan on

Let $f:X\to Y$ be a morphism of algebraic stacks (as considered throughout this tag.) Is the group stack $K\to X$ defined as the kernel of $I_X\to I_Y$ given any special attention elsewhere in the Stacks Project? It occurs in this Tag because the second diagonal $\Delta_{f,2}$ can be identified with the identity section of $K\to X$. Would it be useful to record that the representability of $f$ is equivalent to $K\to X$ being the trivial group (i.e., the second diagonal being an isomorphism as in Lemma 79.6.1)?

Comment #1432 by on

As requested in #1407 and #1417, I made the change as suggested. See here.

Comment #1433 by on

@#1431: Yes, we could add this to the first lemma. Feel free to send us an edited version of the chapter doing so.

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