4.5. Deligne-Mumford 叠

Last time we talked about separatedness. In particular, if $X$ is separated over $S$, $G$ proper over $S$, then $[X/G]$ is separated over $S$.

The next topic is to work towards the first big theorem in stacks.

定义 4.5.0.1. If $\mathfrak{X}/S$ is Artin stack, then we say it is Deligne-Mumford (DM) if there exists scheme $X$ with .

In general, its not easy to check when a stack is DM. To check this, we recall the notion of formally etale/smooth. That is, we say $X\stackrel{f}{\to }Y$ is formally etale/smooth if for all diagramswith $I^2=0$ in $A$, there exists unique arrow/exists arrow from $\mathrm{Spec}A/I$ to $Y$.

So, etale means exists unique arrow, smooth means exists arrow, and now we define formally unramified, means unique arrow.

定义 4.5.0.2. We say $X\stackrel{f}{\to }Y$ is formally unramified if for all diagramswith $I^2=0$ then $\alpha =\beta$.

定理 4.5.0.3. If $\mathfrak{X}$ is Artin stack over $S$, then $\mathfrak{X}$ is DM iff ${\Delta }_{\mathfrak{X}/S}$ is formally unramified.

We first state some corollaries, before we prove this.

推论 4.5.0.4. If $\mathfrak{X}$ is Artin stack over $S$, then $\mathfrak{X}$ is algebraic space over $S$ iff for all $x\in \mathfrak{X}(U)$, $\mathrm{Aut}(x)$ is trivial, i.e. algebraic spaces are Artin stacks with no stablizers.

证明. First, if $\mathfrak{X}$ is algebraic space, then $\mathfrak{X}$ is sheaf, i.e. fibered in sets, so no automorphisms.

Conversely, ifIf $\mathrm{Isom}(x,y)=\varnothing$, then it is a scheme and $\mathrm{Isom}(x,y)\to U$ is formally unramified. If $\mathrm{Isom}(x,y)\ne \varnothing$, then it is an $\mathrm{Aut}(x)$-torsor. That is, if we have two isomorphisms then $\alpha ,\beta$ are related by unique automorphism ${\beta }^{-1}\circ \alpha$. By assumption, $\mathrm{Aut}(x)\to U$ is the trivial group scheme, i.e. $U\stackrel{\mathrm{Id}}{\to }U$. So it is affine group scheme, and so $\mathrm{Isom}(x,y)$ is a scheme. Also, $\mathrm{Isom}(x,y)\to U$ is a monomorphism, so it is formally unramified. Therefore $\Delta$ is formally unramified and representable by schemes. Now by the big theorem above, we see there exists a etale cover.

Lastly, $\mathfrak{X}$ is a sheaf because the automorphisms are trivial and hence $\mathfrak{X}$ is algebraic space as desired.

$\square$

注 4.5.0.5. Suppose ${\Delta }_{\mathfrak{X}}$ is locally of finite presentation. Then the theorem says $\mathfrak{X}$ is DM iff for all $k=\overline{k}$ and $\mathrm{Spec}k\stackrel{x}{\to }\mathfrak{X}$, the automorphism group $\mathrm{Aut}(x)\to \mathrm{Spec}k$ is a finite group.

Why is this? We know ${\Delta }_{\mathfrak{X}}$ is formally unramified iff $\mathrm{Isom}(x,y)\to U$ is formally unramified for all $x,y\in \mathfrak{X}(U)$. However, since ${\Delta }_{\mathfrak{X}}$ is locally of finite presentation, $\mathrm{Isom}(x,y)\to U$ is locally of finite presentation. Hence, we can check formally unramified for $\mathrm{Isom}(x,y)\to U$ on geometric points, i.e. for all $k=\overline{k}$ with $\mathrm{Spec}k\stackrel{f}{\to }U$ we can check $\mathrm{Isom}(f^{\ast }x,f^{\ast }y)\to \mathrm{Spec}k$ formally unramified.

If $\mathrm{Isom}(f^{\ast }x,f^{\ast }y)=\varnothing$ there is nothing to do, else (i.e. its non-empty) it is isomorphic to $\mathrm{Aut}(x)$. But then $\mathrm{Aut}(x)\to \mathrm{Spec}k$ is locally of finite presentation, so its formally unramified if and only if etale if and only if $\mathrm{Aut}(x)$ is finite if and only if $\mathrm{Aut}(x)$ is a group.

例 4.5.0.6. From the above remark, we see $B{\mathbb{G}}_m$ is not DM as ${\mathbb{G}}_m$ is not finite. More generally, $BG$ is not DM if $G$ is a positive dimension group scheme.

例 4.5.0.7. A long time ago we talked about moduli space of genus $g$ curves $M_g$, where $M_g(T)$ are given by $C\stackrel{\pi }{\to }T$ with $\pi$ smooth proper and on geometric fibers it is genus $g$ curve.

When Deligne and Mumford defined DM stacks, they showed that for $g\ge 2$, $M_g$ is DM and defined a compactification ${\overline{M}}_g$. We will go through the idea of how to show $M_g$ is DM for $g\ge 2$.

We start with a curve $C$ of genus $g\ge 2$ over a ACF $k=\overline{k}$. We want $\mathrm{Aut}(C)\to \mathrm{Spec}k$ to be formally unramified (i.e. we want $\mathrm{Aut}(C)$ is finite group). So we want to show for all diagramswith $A=A^{\prime }/I$, we have $\alpha =\beta$. Now, we consider the diagramwhere we get two arrows from $C_{A^{\prime }}$ to itself ($\alpha ,\beta$) and one arrow from $C_A$ to itself ($\gamma$), where $\alpha ,\beta$ both reduce to be $\gamma$ over $A$. Now by deformation theory, “$\alpha -\beta$” is a class in $H^0(T_{C_A/A})$, where $T_{C_A/A}$ is tangent bundle. However, the degree of tangent bundles are given by $2-2g$. Hence we see for $g\ge 2$, the degree become negative, i.e. $H^0(T_{C_A/A})=0$ and hence $\alpha =\beta$, as desired.

What about $g=0,1$?

If $g=0$, then $M_0$ is just $\mathbb{P}$ over ACF $k=\overline{k}$, and over $T$ it is not. It is a Brauer-Severi variety. These are $\mathrm{Aut}({\mathbb{P}}^1)$-torsors and so $M_0=B({\mathrm{PGL}}_2)$. We know $\dim {\mathrm{PGL}}_2=3>0$, so $M_0$ is Artin, not DM. In particular, the dimension of $M_0$ is dimension of a point subtract dimension of ${\mathrm{PGL}}_2$, i.e. $\dim M_0=-3$.

Next, we consider $M_{0,3}$, the moduli space of genus $0$ curves with $3$ marked points. Let $C$ be a curve with $g=0$ and three marked points, then its isomorphic to ${\mathbb{P}}^1$ with three additional points. Thus $M_{0,3}=$points.

In terms of deformation theory, we get $\alpha -\beta$ lives in $H^0(T(-3\mathrm{pt}s))$, where $T(-3\mathrm{pt}s)$ is twisted down by three points, thus the degree is $2-2g-3<0$.

Similarly, $M_1$ is Artin, but $M_{1,1}$ is genus $1$ curves with one marked point, which is just the moduli space of elliptic curves. In particular, $\dim M_{1,1}=1$, which is exactly the $j$-invariant of elliptic curves. We get a map $M_{1,1}\to {\mathbb{A}}^1$ which sends elliptic curve $E$ to isomorphism class of $E$ (i.e. sends it to the $j$-invariant).

We give a rough picture of what this looks like:

[scale=0.5]pic/4.png

generically we have $\mathbb{Z}/2$-stablizers since $E$ has automorphisms equal multiply by $-1$. However, for $j=0$ and $j=1728$, we have more automorphisms: $\mathbb{Z}/4$ and $\mathbb{Z}/6$.

We don’t have time for the proof of the big theorme, but we go through the idea first.

If $X\to Y$ is a map of schemes, its smooth iff we can find Zariski cover $U\twoheadrightarrow X$ withwhere $n=\dim X-\dim Y$ and $F$ comes from the following: ${\Omega }_{X/Y}^1$ is locally free, we look locally on $U$ where $U_{X/Y}^1{|}_U$ is free. We choose basis $d{f}_1,...,d{f}_n$ which yields map to ${\mathbb{A}}_Y^n$ coming from $(f_1,...,f_n)$.

For us, we have smooth $X\twoheadrightarrow \mathfrak{X}$, we would like to have something like “${\Omega }_{X/\mathfrak{X}}^1$”. We look etale locally where ${\Omega }_{X/\mathfrak{X}}^1$ is free to getFormally unramified will allow us to “slice” ${\mathbb{A}}_{\mathfrak{X}}^n\to \mathfrak{X}$ to get $W\subseteq {\mathbb{A}}_{\mathfrak{X}}^n$ with etale arrow $W\to X$

Before we end, we talk about how to define ${\Omega }_{X/\mathfrak{X}}^1$.

We don’t realy know what to do, hence the first thing is to descent. Thus, consider the following diagramWe have ${\Omega }_{Y/X}^1={\Omega }_{{\pi }^{\prime }}^1$ and we get canonical isomorphism $p^{\ast }{\Omega }_{Y/X}^1\cong {\Omega }_{Z/Y}^1\cong q^{\ast }{\Omega }_{Y/X}^1$. Thus it satisfies the cocycle condition. Thus by descent of coherent sheaves, we get ${\Omega }_{X/\mathfrak{X}}^1$ such that ${\pi }^{\ast }{\Omega }_{X/\mathfrak{X}}^1\cong {\Omega }_{Y/X}^1$ where $Y/X$ is via the map ${\pi }^{\prime }:Y\to X$.

Moreover, ${\Omega }_{X/\mathfrak{X}}^1$ is locally free sheaf on $X$ because ${\pi }^{\ast }{\Omega }_{X/\mathfrak{X}}^1\cong {\Omega }_{Y/X}^1$ is.

In addition, we get ${\Omega }_{X/S}^1\to {\Omega }_{X/\mathfrak{X}}^1$. To show this, use descent:Thus we get ${\Omega }_{X/S}^1\to {\Omega }_{X/\mathfrak{X}}^1$.

We note that, for Artin stacks, this is usually not surjective. But it is surjective for DM stacks.

Last time we defined DM stacks to be Artin stack $\mathfrak{X}$ such that it has etale cover by a scheme.

定理 4.5.0.8. $\mathfrak{X}$ is DM iff ${\Delta }_{\mathfrak{X}/S}$ is formally unramified.

Last time, for smooth $X\twoheadrightarrow \mathfrak{X}$ we defined ${\Omega }_{X/\mathfrak{X}}^1$ coherent locally free. We also mentioned the idea of the proof, which is to look at where ${\Omega }_{X/\mathfrak{X}}$ is free, then we getThen using ${\Delta }_{\mathfrak{X}/S}$ is formally unramified, we will “slice” $g$ until it becomes relative dim $0$, i.e. etale.

Now we start the proof.

证明. We start with the easy direction. Assume $\mathfrak{X}$ is DM. Choose etale cover $X\twoheadrightarrow \mathfrak{X}$. Consider the following diagramThe goal is to show $b$ is formally unramified. Let $X{\times }_{S}X\to X$ be the projection, then we get the following diagramHowever, note $\pi \circ b$ is etale, hence $b$ is unramified, and hence $\Delta$ is unramified as desired.

Conversely, suppose $\Delta$ is formally unramified. Let $k=\overline{k}$ be a ACF, $y\in \mathfrak{X}(k)$. Choose $p:X\stackrel{sm}{\to }\mathfrak{X}$ with $X$ affine, such thatHowever, since $k$ is ACF, we get a section $x^{\prime }$:Let $k_0$ be the residue field of image of $x^{\prime }$ in $S$ and let $k_0^{sep}$ be the separable closure of $k_0$.

Our goal is to show etale locally on $S$ and $X$, we will find $W\subseteq X$ closed such that $W\stackrel{et}{\to }\mathfrak{X}$ with $W_y\ne \varnothing$. Then we are done as ${\coprod }_y{W}_y\twoheadrightarrow \mathfrak{X}$ is etale.

Now we get the diagramwhich is the same as the following diagram

Now let ${\Omega }_{{\pi }^{\prime }}^{\prime }:={\Omega }_{Z/X}^1={\pi }^{\ast }{\Omega }_{X/\mathfrak{X}}^1$. Last time we also constructed ${\Omega }_{X/S}^1\to {\Omega }_{X/\mathfrak{X}}^1$. This is usually not surjective, but we will show it is the case for DM stacks. To show it is surjective, by descent, it is enough to show surjective after applying ${\pi }^{\ast }$. We see $\Delta$ is formally unramified implies $(\pi ,{\pi }^{\prime })$ is formally unramified, and hence$$(\pi ,{\pi }^{\prime }{)}^{\ast }{\Omega }_{X{\times }_{S}X/S}={\pi }^{\ast }{\Omega }_{X/S}^1\oplus ({\pi }^{\prime }{)}^{\ast }{\Omega }_{X/S}\twoheadrightarrow {\Omega }_{Z/S}$$This givesSince we have that $0$ map, it gives us that ${\pi }^{\ast }\varphi$ is surjective, and hence $\varphi$ is surjective.

We havewhere we call the arrow ${\mathcal{O}}_X\to {\Omega }_{X/\mathfrak{X}}^1$ $d$ as well. Locally, $d:{\mathcal{O}}_X\to {\Omega }_{X/S}^1$ has image generates, so the same is true for $d:{\mathcal{O}}_X\to {\Omega }_{X/\mathfrak{X}}^1$.

Since ${\Omega }_{X/\mathfrak{X}}^1$ is locally free, so we need to look etale locally where ${\Omega }_{X/\mathfrak{X}}^1$ is free and there exists $f_1,...,f_r\in \Gamma ({\mathcal{O}}_U)$ such that $d{f}_1,...,d{f}_r$ is a basis for ${\Omega }_{X/\mathfrak{X}}^1{|}_U$.

Shrinking $X$, we may assume $U=X$. Let $f_i\in \Gamma ({\mathcal{O}}_X)$ that $d{f}_i$ generates ${\Omega }_{X/\mathfrak{X}}^1$, so we get $F:X\stackrel{(p,f_1,...,f_r)}{\to }\mathfrak{X}{\times }_S{\mathbb{A}}_S^r$. Then,and ${\dim }_{x^{\prime }}(X)={\dim }_{x^{\prime }}({\mathbb{A}}_{\mathfrak{X}}^r)$ because ${\Omega }_{X/\mathfrak{X}}^{\prime }$ is free for rank $r$.

$F$ is smooth map, representable and relative dim $0$ in neighbourhood of $x^{\prime }$, so it is etale in neighbourhood of $x^{\prime }$. Shrink to assume $F$ is etale. Then $F_y:X_y\stackrel{et}{\to }{\mathbb{A}}_k^r$. Then $F_y$ is etale implies the image is open. Let $f\in k[t_1,...,t_r]$ such that $\varnothing \ne D(f)\subseteq F_y(X_y)$. Over $k_0^{sep}$, there exists $a_1,...,a_r\in k_0^{sep}$ such that $f(a_1,...,a_r)\ne 0$.

In particular, we see Now, we import a fact, which is that, we can take an etale neighbourhood $S^{\prime }\to S$ of $x^{\prime }$ such that there exists closed $E$ with $E\subseteq {\mathbb{A}}_{S^{\prime }}^r$ and diagramthen there exists unique arrow $\mathrm{Spec}k\to S^{\prime }$ with $Q\in E$, i.e. we get the following diagramTherefore, we seeTherefore, $W_y\ne \varnothing$ by construction and henceNow $W\to \mathfrak{X}$ is our desired map.

$\square$

推论 4.5.0.9. If $G$ is a finite group, then $\mathfrak{X}=[X/G]$ DM.

证明. We just need $\mathrm{Aut}$ groups are finite over $k=\overline{k}$. Over $k=\overline{k}$, $G$-torsors are trivial, so our point isand $\mathrm{Aut}$ group isSo $\gamma (1)=g$. Then $\gamma (h)=h\circ \gamma (1)=hg$ and hence we getIn other word, $\mathrm{Aut}$ is stablizer of $\alpha (1)$. Hence we getand so $\mathrm{Aut}$ is finite as desired.

$\square$

At the start of the course, we talked about $5$ general points determine conic. We did this through geometry on moduli space. In particular, moduli space of singular conics is exactly ${\mathbb{P}}^5$ (if we drop singular, then this is not true!).

Even if we only interested in smooth curves, i.e. $M_g$. To do intersection theory, we need a compactification $\overline{M_g}$. So, we need a notion of properness. To do this, we need quasi-coherent sheaves on $\mathfrak{X}$.

定理 4.5.0.10 (Eisenbud-Harris). It is impossible to write down a general $g\ge 24$ curve.

This uses intersection theory on $\overline{M_g}$. Note here we are asking for general $g\ge 24$ curve. To see what this means, we consider the example of elliptic curve. For that, we know the short form for elliptic curve is given by $y^2=x^3+ax+b$. Thus it is the same as a dominant rational map ${\mathbb{A}}_{a,b}^2⇢M_{1,1}$, i.e. ${\mathbb{A}}_{a,b}^2\backslash (\text{discriminant}=0)\to M_{1,1}$. They showed $M_g$ is of general type.

There is also another similar problem, where we look at ${\mathcal{A}}_g$, the moduli space of dim $g$ abelian varities. Then we know ${\mathcal{A}}_{\ge 7}$ is of general type, ${\mathcal{A}}_{\le 5}$ is , and ${\mathcal{A}}_6$ is unknown.