*Note:
These are notes on an online graduate course in stacks by Jarod Alper in Spring 2021. As such, any errors or inaccuracies are almost certainly my own.
*

dzackgarza@gmail.com

Last updated: 2021-09-06

References:

- Course website: https://sites.math.washington.edu/~jarod/math582C.html
- Gómez 99: Expository article on algebraic stacks

Stated goal of the course: prove that the moduli space \(\mkern 1.5mu\overline{\mkern-1.5mu{ \mathcal{M}_{g} }\mkern-1.5mu}\mkern 1.5mu\) of stable curves (for \(g\geq 2\)) is a smooth, proper, irreducible Deligne-Mumford stack of dimension \(3g-3\). Moreover, it admits a projective coarse moduli space.

In the process we’ll define **algebraic spaces** and **stacks**.

Prerequisites:

- Schemes
- Existence of Hilbert schemes
- Artin approximation
- Resolution of singularities for surfaces
- Deformation theory

Last time: functors, sheaves on sites, descent, and Artin approximation. Today: groupoids and stacks.

Recall that a **site** \(\mathsf{S}\) is a category such that for all \(U\in {\operatorname{Ob}}(\mathsf{S})\), there exists a set \({\mathsf{Cov}}(U) \mathrel{\vcenter{:}}=\left\{{U_i \to U}\right\}_{i\in I}\) (a *covering family*) such that

- \(\operatorname{id}_U \in {\mathsf{Cov}}(U)\),
- \({\mathsf{Cov}}(U)\) is closed under composition.
- \({\mathsf{Cov}}(U)\) is closed under pullbacks:

Take \(\mathsf{S} \mathrel{\vcenter{:}}={\mathsf{Sch}}_{\text{Ét}}\) to be the big étale site: the category of all schemes, with covering families given by étale morphisms \(\left\{{U_i\to U}\right\}_{i\in I}\) such that \(\displaystyle\coprod_i U_i \twoheadrightarrow U\). Note that there is a special covering family given by *surjective* etale morphisms.

Let \(\mathsf{C}\) be a category (e.g. \(\mathsf{C} \mathrel{\vcenter{:}}={\mathsf{Set}}\)) and recall that a *presheaf* on a category \(\mathsf{S}\) is a contravariant functor \(\mathsf{S}\to \mathsf{C}\).

A \(\mathsf{C}{\hbox{-}}\)valued **sheaf** on a site \(\mathsf{S}\) is a presheaf
\begin{align*}
{\mathcal{F}}:\mathsf{S} \to \mathsf{C}
\end{align*}
such that for all \(U_i, U_j\in {\mathsf{Cov}}(U)\), the following equalizer diagram is exact in \(\mathsf{C}\)

Show that a presheaf \(F\) is a sheaf on \({\mathsf{Sch}}_\text{Ét}\) iff

- \(F\) is a sheaf on \({\mathsf{Sch}}_{\mathrm{Zar}}\) and
- For all etale surjections \(U' \twoheadrightarrow_{\text{ét}} U\) of affines, the equalizer diagram is exact.

For \(X\in {\mathsf{Sch}}\), the presheaf \begin{align*} h_X \mathrel{\vcenter{:}}=\operatorname{Mor}({-}, X): {\mathsf{Sch}}\to {\mathsf{Set}} \end{align*} is a sheaf on \({\mathsf{Sch}}_{\text{Ét}}\).

We’ll often consider *moduli functors*: functors \(F: {\mathsf{Sch}}\to {\mathsf{Set}}\) where \(F(S)\) is a family of objects over \(S\). Then \(F\) will be a sheaf iff families glue uniquely in the étale topology, and representability of such functors will imply they are sheaves.

Consider the following moduli functor:

This is *not* representable by a scheme and not a sheaf.

Why care about representability? Suppose there were a scheme \(M\), so \begin{align*} F_{{\mathsf{Alg}}}(S) \simeq \operatorname{Mor}(S, M) .\end{align*} Then taking \(\operatorname{id}_M \in \operatorname{Mor}(M, M)\) should yield a universal family \({\mathcal{U}}\to M\):

Then the points of \(M\) would correspond to isomorphism classes of curves, and every family of curves would be a pullback of this.

For any \(S\in{\mathsf{Sch}}\) and a family \({\mathcal{C}}\xrightarrow{f} S\), the fiber \(f^{-1}(s)\in{\mathcal{C}}\) is a curve for any \(s\in S\), so one could define a map \begin{align*} g: S &\to M \\ s &\mapsto [s] ,\end{align*} where we send a curve to its isomorphism class. Then \({\mathcal{C}}\) would fit into a pullback diagram:

If \(S\) was itself a curve, then \(g: S\to M\) would be a path in \(M\) deforming a base curve.

Recall that a **groupoid** is a category where every morphism is an isomorphism. Morphisms of groupoids are functors, and isomorphisms of groupoids are equivalences of categories.

A basic example is the category of sets where \begin{align*} \operatorname{Mor}(A, B) \mathrel{\vcenter{:}}= \begin{cases} \operatorname{id}_A & A=B \\ \emptyset & \text{else}. \end{cases} \end{align*}

A similar construction: for any set \(\Sigma\), one can form a groupoid \({\mathcal{C}}_\Sigma\):

- Object: Elements \(x\in \Sigma\).
- Morphisms: \(\operatorname{id}_x\)

Define a category \({ \mathcal{M}_{g} }({\mathbb{C}})\):

- Objects: smooth projective curves over \({\mathbb{C}}\) of genus \(g\).
- Morphisms: \begin{align*} \operatorname{Mor}(C, C') = \mathop{\mathrm{Isom}}_{{\mathsf{Sch}}_{/ {{\mathbb{C}}}} }(C, C') \subseteq \operatorname{Mor}_{{\mathsf{Sch}}_{/ {{\mathbb{C}}}} }(C, C') .\end{align*}

Groupoids are equivalent iff they are equivalent as categories. The following is an example of mapping the quotient groupoid \([C_2/C_4]\) to \({\mathsf{B}}C_2\):

If a groupoid \({\mathfrak{X}}\) is equivalent to \(\mathsf{C}_{\Sigma}\) for any \(\Sigma \in {\mathsf{Set}}\), we say \({\mathfrak{X}}\) is **equivalent to a set**. For example, the following groupoid is equivalent to a 2-element set:

For \(G\curvearrowright\Sigma\) a group acting on any set, define the **quotient groupoid** \([\Sigma/G]\) in the following way:

- Objects: \(x\in \Sigma\), i.e. one object for each element of the set \(\Sigma\).
- Morphisms: \(\operatorname{Mor}(x, x') = \left\{{g\in G {~\mathrel{\Big|}~}gx' = x}\right\}\).

Show that \([\Sigma/G]\) is equivalent to a set iff \(G\curvearrowright\Sigma\) is a free action.

For \(\Sigma = \left\{{{\operatorname{pt}}}\right\}\), we obtain \begin{align*} {\mathsf{B}}G \mathrel{\vcenter{:}}=[{\operatorname{pt}}/ G] ,\end{align*} where there is one object \({\operatorname{pt}}\) and \(\operatorname{Mor}({\operatorname{pt}}, {\operatorname{pt}}) = G\).

Define \({\mathsf{FinSet}}\) to be the category of finite sets where the morphisms are set bijections. Then \({\mathsf{FinSet}}= \displaystyle\coprod_{n\in {\mathbb{Z}}_{\geq 0}} {\mathsf{B}}S_n\) for \(S_n\) the symmetric group.

For \(C, D' \to D\) morphisms of groupoids, we can construct their **fiber product** as the cartesian diagram:

It can be constructed as the following category:

\begin{align*} {\operatorname{Ob}}(C{ \underset{\scriptscriptstyle {D} }{\times} } D') \mathrel{\vcenter{:}}= \left\{ \begin{array}{l} (c, d', \alpha) \end{array} \middle\vert \begin{array}{l} c\in C, d'\in D', \\ \\ \alpha: f(c) \xrightarrow{\sim} g(d') \end{array} \right\} \end{align*}

Describe the universal property of the pullback in the 2-category of groupoids.

\(G\) regarded as a groupoid is the pullback over inclusions of points into \({\mathsf{B}}G\):

Let \(G\curvearrowright\Sigma\) and \(x\in \Sigma\), and let \(Gx\) be the orbit and \(G_x\) be the stabilizer. Then there is a morphism of groupoids \(f \in \operatorname{Mor}({\mathsf{B}}G_x, [\Sigma/G])\) inducing a pullback:

Motivation: to specify a moduli functor, we’ll need the data of

- Families over \(S\),
- How to pull back families under morphisms, and
*How*objects are isomorphic.

As a first attempt, we might try to define a 2-functor \(F: {\mathsf{Sch}}\to {\mathsf{Grpd}}\) between 2-categories, where the latter is the category of groupoids. For this, we need the following data:

- For all \(S\in {\mathsf{Sch}}\), an assignment of a groupoid \(F(S)\),
- For all morphisms \(f\in \operatorname{Mor}_{{\mathsf{Sch}}}(S, T)\), an assignment of morphisms of groupoids \begin{align*} f^* \in \operatorname{Mor}_{{\mathsf{Grpd}}}(F(T), F(S)) .\end{align*}
- For compositions of morphisms of schemes \(S \xrightarrow{f} T \xrightarrow{g} U\), an isomorphism of functors \begin{align*} \psi_{fg}: g^* \circ f^* \xrightarrow{\sim} (g \circ f)^* .\end{align*}
- Compatibility of these isomorphisms on chains of compositions \(S \to T \to U \to V \to \cdots\).
^{1}

This is a lot of data to track, so instead we’ll construct a large category \({\mathfrak{X}}\) that encodes all of this, along with a fibration

Here \(S \in {\mathsf{Sch}}\) and \(F(S) \in {\mathsf{Grpd}}\), so the “fibers” above \(S\) are groupoids.

Let \(p:{\mathfrak{X}}\to \mathsf{C}\) be a functor between two 1-categories, so we have the following data:

Then \({\mathfrak{X}}, p\) define a **prestack** over \(\mathsf{C}\) iff

- Pullbacks exist: for \(S \xrightarrow{f} T\), there exists a (not necessarily unique) map \(f^*b\), sometimes denoted \({ \left.{{b}} \right|_{{f}} }\), yielding a cartesian square:

- A universal property making \({\mathfrak{X}}\) a
*fibered category*: every arrow in \({\mathfrak{X}}\) is a pullback, so there are always lifts of the following form:

An alternative definition: a prestack is a category *fibered in groupoids*.

We often conflate \({\mathfrak{X}}\) and the functor \({\mathfrak{X}}\xrightarrow{p} S\), and don’t spell out the composition law in \({\mathfrak{X}}\). Moreover, we write \(f^*b\) or \({ \left.{{b}} \right|_{{f}} }\) for a *choice* of a pullback.

For \(p: {\mathfrak{X}}\to \mathsf{C}\) a functor and \(S\in {\operatorname{Ob}}(\mathsf{C})\) any fixed object, the associated **fiber category over \(S\)**, denoted \({\mathfrak{X}}(S)\), is the subcategory of \({\mathfrak{X}}\) defined by:

- Objects: \(a\in {\operatorname{Ob}}({\mathfrak{X}})\) such that \(a \xrightarrow{p} S\),
- Morphisms: \(\operatorname{Mor}(a, a')\) are morphisms \(f\in \operatorname{Mor}_{{\mathfrak{X}}}(a, a')\) over \(\operatorname{id}_S\):

We can now equivalently define presheaves as categories fibered in sets.

Show that if \({\mathfrak{X}}\to \mathsf{C}\) is a prestack, then for all \(S\in \mathsf{C}\), all maps in \({\mathfrak{X}}(S)\) are invertible. Conclude that the fiber categories \({\mathfrak{X}}(S)\) are all groupoids.

Every presheaf forms a prestack. Let \(F \in \underset{ \mathsf{pre} } {\mathsf{Sh} }({\mathsf{Sch}}, {\mathsf{Set}})\) be a presheaf of sets, and define \({\mathfrak{X}}_F\) as the following category:

- Objects: Pairs \((S, a \in F(S))\) where \(S\in {\mathsf{Sch}}\) and \(F(s) \in {\mathsf{Set}}\).
- Morphisms: \begin{align*} \operatorname{Mor}( (S, a), (T, b) ) \mathrel{\vcenter{:}}=\left\{{ S \xrightarrow{f} T {~\mathrel{\Big|}~}a = f^* b}\right\} .\end{align*}

Note that we’ll often conflate \(F\) and \({\mathfrak{X}}_F\). This yields the fibration

For \(X\in {\mathsf{Sch}}\), take its Yoneda functor \(h_X: {\mathsf{Sch}}\to {\mathsf{Set}}\). Then define the category \({\mathfrak{X}}_X\):

- Objects: Morphisms \(S\to X\) of schemes.
- Morphisms: \(\operatorname{Mor}(S\to X, T\to X)\) are morphisms over \(X\):

This yields the fibration

Define \({ \mathcal{M}_{g} }\) as the following category:

- Objects: families \({\mathcal{C}}\to S\) of smooth genus \(g\) curves,
- Morphisms: \(\operatorname{Mor}({\mathcal{C}}\to S, {\mathcal{C}}'\to S')\): cartesian squares

This yields a fibration

For \(C\) a smooth connected projective curve over \(k\) a field, define \({\mathsf{Bun}}(C)\) as the following category:

- Objects: pairs \((S, F)\) where \(F\) is a vector bundle over \(C\times S\).
- Morphisms: \begin{align*} \operatorname{Mor}((S, F), (S', F')) = \left\{ \begin{array}{l} f\in \operatorname{Mor}_{{\mathsf{Sch}}}(S, S') \\ \text{and a chosen isomorphism} \\ \alpha: (f\times \operatorname{id})^* \circ F' \xrightarrow{\sim} F \end{array} \right\} .\end{align*}

A technical point: the choice of pushforward here is not necessarily canonical. However, as part of the data, one can take morphisms \(F' \to (f\times\operatorname{id})_* \circ F\) such that the adjunction yields an isomorphism.

Let \(X_{/ {S}} \in {\mathsf{Grp}}{\mathsf{Sch}}\) where \(G\curvearrowright X\). Then define a category \([X/G]^{\mathsf{pre}}\):

- Objects: Morphisms over \(\operatorname{id}_S\):

- Morphisms:

\begin{align*} \operatorname{Mor}(T\to X, T'\to X) \mathrel{\vcenter{:}}= \left\{ \begin{array}{l} T\to T' \end{array} \,\, \middle\vert \begin{array}{l} (T \to T' \to X ) = g(T \to X) \\ g\in G(T) \\ G(T) \curvearrowright X(T) \end{array} \right\} .\end{align*}

A group scheme can alternatively be thought of as a functor with a factorization through \({\mathsf{Grp}}\).

Show that for \(T\in {\mathsf{Sch}}\), there is an equivalence \begin{align*} [X/G]^{\mathsf{pre}}(T) \xrightarrow{\sim} [X(T) / G(T)] ,\end{align*} where the left-hand side is a fibered category over \(T\) and the right-hand side is a quotient groupoid.

A **morphism of prestacks** is a functor \({\mathfrak{X}}\xrightarrow{f} {\mathfrak{X}}'\) such that there is a (strictly) commutative triangle

Here we require a strict equality \(p_X(a) = p_Y(f(a))\) for any \(a\in {\mathfrak{X}}\)

A **2-morphism** \(\alpha\) between morphisms \(f, g\) is a natural transformation:

such that for all \(a\in {\mathfrak{X}}\), the following triangle \(\alpha_a\in \operatorname{Mor}_{{\mathfrak{X}}'}(f(a), g(a))\) is a morphisms over \(\operatorname{id}_S\) for any \(S\in \mathsf{C}\):

We define a category \(\operatorname{Mor}({\mathfrak{X}}, {\mathfrak{X}}')\) by:

- Objects: morphisms of prestacks.
- Morphisms: 2-morphisms of prestacks.

Show that \(\operatorname{Mor}({\mathfrak{X}}, {\mathfrak{X}}')\) is a groupoid.

A diagram is **2-commutative** iff there exists a 2-morphism \(\alpha: g \circ f' \xrightarrow{\sim} f\circ g'\) which is an isomorphism:

An **isomorphism** of prestacks is a 1-isomorphism of prestacks \(f: {\mathfrak{X}}\to {\mathfrak{X}}'\) along with 2-isomorphisms \(g\circ f \xrightarrow{\sim} \operatorname{id}_{{\mathfrak{X}}}\) and \(f\circ g \xrightarrow{\sim} \operatorname{id}_{{\mathfrak{X}}'}\).

Show that \({\mathfrak{X}}\to {\mathfrak{X}}'\) is an isomorphism iff \({\mathfrak{X}}(S) \xrightarrow{\sim} {\mathfrak{X}}'(S)\) is an isomorphism on all fibers.

If \({\mathfrak{X}}\in {\underset{ \mathsf{pre} } {\mathsf{St} } } {}_{/ {\mathsf{C}}}\) is a prestack over \(\mathsf{C}\), then for any \(S\in {\operatorname{Ob}}(\mathsf{C})\), there is an equivalence of categories induced by the following functor: \begin{align*} \operatorname{Mor}(S, {\mathfrak{X}}) & \xrightarrow{\sim} {\mathfrak{X}}(S) \\ f &\mapsto f_S(\operatorname{id}_S ) .\end{align*}

For \(S\in {\mathsf{Sch}}\), view \(S\) as a prestack and consider a morphism \(f:S\to {\mathfrak{X}}\). How is this specified? For all \(T\in {\mathsf{Sch}}\), the objects of \(S_{/ {T}}\) are morphisms \begin{align*} f_T: \operatorname{Mor}(T, S) \to {\mathfrak{X}}(T) \end{align*} and if \(T=S\) this sends \(\operatorname{id}_S\) to \(f_S(\operatorname{id}_S)\in {\mathfrak{X}}(S)\).

What is the inverse? For \(a\in {\mathfrak{X}}(S)\) and for each \(T \xrightarrow{g} S\), **choose** a pullback \(g^* a\). Then define \(f: S \to {\mathfrak{X}}\) by
\begin{align*}
f_T: \operatorname{Mor}(T, S) &\to {\mathfrak{X}}(T) \\
g &\mapsto g^* a
.\end{align*}

Define what this equivalence should do on morphisms.

Next time: fiber products of prestacks.

This leads to the notion of

**lax**or**pseudofunctors**.↩︎