4.4. 代数叠

Today we are going to define algebraic stacks.

定义 4.4.0.1. Let $P$ be property of morphisms which is local on the source for the etale topology, local on the target for the sm topology, and stable under base change. Let $f:\mathfrak{X}\to \mathfrak{Y}$ be a representable map of stacks over $S$. We say $\mathbf{f}$ has $\mathbf{P}$ if for all schemes $T$ and diagramwe have $g$ has property $P$, where we know ${\mathfrak{X}}_T$ is algebraic space.

Recall that, if $g:X\to T$ where $T$ is scheme and $X$ is algebraic space, then we say $g$ has $P$ if there exists etale covering $\tilde{X}$ withsuch that $g\pi$ has $P$.

例 4.4.0.2. $P$ could be etale, smooth, relative dim $d$, affine, finite, closed, immersion, open immersion, surjection, and so on. We note this list is smaller than the list for fppf descent as we require local on the source.

定义 4.4.0.3. A stack $\mathfrak{X}/S$ is an algebraic/Artin stack if:

1.	${\Delta }_{\mathfrak{X}}:\mathfrak{X}\to \mathfrak{X}{\times }_S\mathfrak{X}$ is representable,
2.	There exists scheme $U$ so that

We note ${\Delta }_{\mathfrak{X}}$ is representable implies for all algebraic spaces $X$, the maps $X\to \mathfrak{X}$ are representable. So, $U\to \mathfrak{X}$ being smooth surjection is well-defined.

定义 4.4.0.4. A morphism of stacks over $S$, say $\mathfrak{X}\to \mathfrak{Y}$, is defined as an element of ${\mathrm{Hom}}_S(\mathfrak{X},\mathfrak{Y})$, i.e. they are morphisms of fibered categories over $S$.

引理 4.4.0.5. Let $\mathfrak{X}$ be stack over $S$. Then ${\Delta }_{\mathfrak{X}}$ is representable iff for all $S$-schemes $U,V$, $U\in \mathfrak{X}(U)$ and $V\in \mathfrak{X}(V)$ withwe have $\mathrm{Isom}(p_U^{\ast }U,p_V^{\ast }V)$ is an algebraic space. This is also equivalent to: for all $U,V\in \mathfrak{X}(U)$, $\mathrm{Isom}(U,V)$ is an algebraic space.

证明. Equivalence of last $2$ conditions: for all $U,V\in \mathfrak{X}(U)$,so we see $\mathrm{Isom}(p_U^{\ast }U,p_U^{\ast }V)$ is algebraic space imply $\mathrm{Isom}(U,V)$ is algebraic space. Conversely, $\mathrm{Isom}(p_U^{\ast }U,p_V^{\ast }V)$ is the special case with $U^{\prime }=U\times V$, $U^{\prime }=p_U^{\ast }U$, $V^{\prime }=p_V^{\ast }V$.

Next, consider diagramJust like the proof for algebraic space, we see we getSo, $\mathfrak{Y}(T)$ giveswhere $\xi :f(u)\stackrel{\sim }{\to }g(v)$. Then ${\pi }_{1}f={\pi }_g$ implies $T\stackrel{h}{\to }U{\times }_{S}V$ and hence $h^{\ast }{p}_U^{\ast }u=f^{\ast }(u)\underset{\xi }{\overset{\sim }{\to }}{g}^{\ast }v=h^{\ast }{p}_V^{\ast }v$. Thus we see $\xi \in \mathrm{Isom}(p_U^{\ast }u,p_V^{\ast }v)(T)$ which concludes the proof.

$\square$

The next goal is to define one of the most important example of stacks, namely quotient stacks.

例 4.4.0.6. Consider $\mathbb{Z}/2\mathbb{Z}$ acts on ${\mathbb{A}}^2$, i.e. $\mathbb{Z}/2\mathbb{Z}\circlearrowleft {\mathbb{A}}^2$ via $(x,y)\mapsto (-x,-y)$. In particular, we see ${\mathbb{A}}^2/(\mathbb{Z}/2\mathbb{Z}):=\mathrm{Spec}k[x,y{]}^{\mathbb{Z}/2\mathbb{Z}}$ where the ring $k[x,y{]}^{\mathbb{Z}/2\mathbb{Z}}=\{f(x,y):f(x,y)=f(-x,-y)\}=k[x^2,xy,y^2]$ is the ring of invariant of the $\mathbb{Z}/2\mathbb{Z}$ action. We also ntoe $k[x^2,xy,y^2]=k[a,b,c]/(ac-b^2)$ which is given by ${\mathbb{A}}^2\to {\mathbb{A}}^2/(\mathbb{Z}/2\mathbb{Z})$. Thus we get ${\mathbb{A}}^2\to {\mathbb{A}}^2/(\mathbb{Z}/2\mathbb{Z})$ is sort of like the $xy$-plane map to the cone defined by $ac-b^2$. But this is bad, because $V(ac-b^2)$ is singular.

That’s why we want stakcs, where we replace the origin by the point $\mathbb{Z}/2\mathbb{Z}$, i.e. we get a “quotient stack” $[{\mathbb{A}}^2/(\mathbb{Z}/2\mathbb{Z})]=:\mathfrak{X}$.

In particular, we get a diagramwhere $\tilde{X}$ is the minimal resolution of $X$, and $\mathfrak{X}$ is the canonical stack we discussed above, and $X$ is any surface with mild singularity.

In physics, we get McKay correspondence that comparing $\mathfrak{X}$ and $\tilde{X}$. We see a lot of interesting math about comparing the two, and it also relates to deriving categories.

Before we jump to definition, we give one or two words about the idea. Say we have group scheme $G\circlearrowleft X$ over $S$. Then we get $X\to [X/G]$ and what we want is to have the arrow $X\to [X/G]$ to be a $G$-torsor.

We don’t realy know what $X\to [X/G]$ is $G$-torsor means, thus we want to pullback and getThis is going to be our definition.

定义 4.4.0.7. Continue the above set-up, for any scheme $T$ the category $[X/G](T)$ is defined as follows. The objects areThe morphisms are $(T^{\prime },P^{\prime },{\pi }^{\prime })\to (T,P,\pi )$ is given bywhere

We let $\mathfrak{X}=[X/G]$. Why is $\mathfrak{X}$ a stack? We know $G$-torsors are sheaves, so we have descent for sheaves. Then descent as sheaf with $G$-action we see $G$-action is $G\times P\to P$ map of sheaves, so those descent as well. We see $G\circlearrowleft P$ is torsor if $G\times p\stackrel{\sim }{\to }P\times P$ given by $(g,p)\mapsto (p,gp)$ and we can check this isomorphism locally.

Why is ${\Delta }_{\mathfrak{X}}$ representable? Let $(P_1,{\pi }_1)$, $(P_2,{\pi }_2)$ be over $T$. Let$$I=\mathrm{Isom}((P_1,{\pi }_1),(P_2,{\pi }_2))$$We want to show $I$ is algebraic space.

First, we claim (its an exercise!) that if $\mathcal{F}\to W$ where $W$ is scheme and $\mathcal{F}$ a sheaf, then we can check $\mathcal{F}$ is algebraic space etale locally on $W$.

Thus, to check $I$ is algebraic space, we can make etale base change on $T$ so that $(P_i,{\pi }_i)$ are trivial torsors. Thus now we havewhere $\xi \in I$. However, note if we have $\xi (1)=g$, then for any $h$ we get $\xi (h)=\xi (h\cdot 1)=h\cdot \xi (1)=hg$ and hence $\xi$ is right multiplication by $g$. Thus we see ${\pi }_1(1)={\pi }_2(g)$. Thus we see $I$ has a very simple description:In particular, since $G_T,X_T,X_T{\times }_T{X}_T$ are all schemes, hence $I$ is a scheme, as desired.

Why is $\mathfrak{X}$ has smooth cover by scheme?

We see we getwhere $q$ is defined asWhy is $q$ a smooth surjection? Well, we getwhere $G=U\to P$ is given by $1\mapsto p$. So, we get map $I\stackrel{\sim }{\to }P$ given by $(U\stackrel{f}{\to }P)\mapsto f(1)$.

The point is that we getand since $P\to T$ is smooth, $q$ is smooth surjection as desired. Thus, we showed, every torsor is the pullback of the torsor $X\to [X/G]$!

例 4.4.0.8. Let $X=S$ and $G\circlearrowleft S$ be the trivial action. Then we let $BG:=[X/G]$ and hence we getHowever, note the action is trivial, thus $G$-equivariant $P\to S$ is just any arrow $P\to S$. Hence we see $(BG)(T)$ is just $G$-torsors on $T$.

例 4.4.0.9. Let $G={\mathbb{G}}_m={\mathrm{GL}}_1$, then $B{\mathbb{G}}_m$ is just line bundles, and $B{\mathrm{GL}}_n$ is vector bundles.

定义 4.4.0.10. We say $\mathfrak{X}$ is quotient stack if $\mathfrak{X}\cong [X/G]$ for some $X,G$.

Last time we talked about quotient stacks $\mathfrak{X}=[X/G]$ where $G$ is a group scheme over $S$ and $G\circlearrowleft X$ with $X$ a $S$-scheme.

We also showed that $X\to [X/G]$ is universal, in the sense that if we have $T\to [X/G]$ then we get the following diagram

We also talked about examples. In particular, ${\mathbb{P}}^n=[{\mathbb{A}}^{n+1}\backslash 0/{\mathbb{G}}_m]$ and more generally, if $G\circlearrowleft X$ is a free action, i.e. all stablizers are trivial, then $[X/G]$ is an algebraic space that is exactly $X/G$. For example, if $X=\mathrm{Spec}A$ then $[X/G]=X/G=\mathrm{Spec}(A^G)$.

命题 4.4.0.11. If $\mathfrak{X},\mathfrak{Y},\mathfrak{Z}$ are Artin stacks withthen $\mathfrak{X}{\times }_{\mathfrak{Z}}\mathfrak{Y}$ is an Artin stack.

Next, we are going to define a very useful stack, called the inertia stack.

定义 4.4.0.12. For $\mathfrak{X}$, we define the inertia stack $I_{\mathfrak{X}}=I\mathfrak{X}$ to be the pullback

The point of this is that, suppose we havethen a lift of the arrow $x$ is equivalent to giving $x^{\prime }\in \mathfrak{X}(T)$ and $\xi :(x,x)\stackrel{\sim }{\to }(x^{\prime },x^{\prime })$, i.e. ${\xi }_1,{\xi }_2:x\stackrel{\sim }{\to }{x}^{\prime }$. In particular, we get that for $\sigma \in \mathrm{Aut}(x)$, we have $\sigma ={\xi }_1^{-1}\circ {\xi }_2:x\stackrel{\sim }{\to }x$.

A different point of $I\mathfrak{X}$ is $(x,x)\underset{(\mathrm{Id},\sigma )}{\overset{\sim }{\to }}(x,x)$. This new point is isomorphic to the old point: So, we seeIn other word, we get and $I\mathfrak{X}\to X$ is a relative group algebraic space, but it is normally not flat!

The next topic is properties for stacks and morphisms.

定义 4.4.0.13. Let $P$ be a property that is local for smooth topology. Then we say $\mathfrak{X}$ has $P$ if there exists smooth cover $X\twoheadrightarrow \mathfrak{X}$ with $X$ scheme, such that $X$ has $P$.

例 4.4.0.14. $P$ could be local Noetherian, regular, of finite type over $S$, of finite presentation over $S$.

引理 4.4.0.15. If $P$ is local for smooth topology, and $\mathfrak{X}$ has $P$, and for $Y$ a scheme we have $Y\stackrel{sm}{\to }\mathfrak{X}$ then $Y$ has $P$.

证明. Considerwhere we assume $X$ has P. But $X$ has $P$ implies $Z$ has $P$ and hence $Y$ has $P$ as desired.

$\square$

注 4.4.0.16. The proof shows that if $Y\to \mathfrak{X}$ is a morphism, then smooth locally on $Y$ $Y\to \mathfrak{X}$ factors through $X\to \mathfrak{X}$.

定义 4.4.0.17. If $f:\mathfrak{X}\to \mathfrak{Y}$ is morphism of Artin stacks, then a chart for $\mathbf{f}$ is a diagramIf $P$ is property of morphisms stable under base change, local on source and target for smooth topology, then we say $\mathbf{f}$ has $\mathbf{P}$ if $g$ has $P$. In this case we also say that this chart $\mathbf{g}$ has $\mathbf{P}$.

例 4.4.0.18. $P$ could be smooth, locally of finite presentation, surjective, etc.

例 4.4.0.19. If we are given quotient stack $[X/G]$ over $S$, andThen $\mathfrak{X}/S$ is smooth iff $X/S$ is smooth. For example, we see $[{\mathbb{A}}^2/(\mathbb{Z}/2)]$ is smooth because ${\mathbb{A}}^2$ is smooth. On the other hand, ${\mathbb{A}}^2/(\mathbb{Z}/2)$ is singular as its equal $\mathrm{Spec}k[x,y{]}^{\mathbb{Z}/2}$.

命题 4.4.0.20. The morphism $f:\mathfrak{X}\to \mathfrak{Y}$ has $P$ iff every chart has $P$.

证明. We start with a chartThen we get another chartNow we want that: $X\to Y$ has $P$ iff $X^{\prime }\to Y^{\prime }$ has $P$.

Now take pullbacks of the squares of the two sides, we getand we also get natural arrows from ${\mathfrak{Z}}^{\prime \prime }\to Y^{\prime \prime }$. Viz we getThus it is good enough to show $X\to Y$ has $P$ iff $X^{\prime \prime }\to Y^{\prime \prime }$ has $P$, i.e. it is good enough to handle the case when $Y^{\prime }\to \mathfrak{Y}$ factors as $Y^{\prime }\twoheadrightarrow Y\twoheadrightarrow \mathfrak{Y}$.

First, consider the case $X^{\prime }=X{\times }_{\mathfrak{Z}}{\mathfrak{Z}}^{\prime }$ and we get diagramwhere $g^{\prime }$ is pullback of $g$ under $Y^{\prime }\twoheadrightarrow Y$. We see $P$ is smooth local on the target so $g$ has $P$ iff $g^{\prime }$ has $P$.

Now, we just need to compare two charts with the same $Y$. To see this, we note we have the following diagramand we see $g^{\prime }$ has $P$ iff $g^{\prime }{\pi }^{\prime }=g\pi$ has $P$ iff $g$ has $P$. This concludes the proof.

$\square$

Next, we consider separatedness.

命题 4.4.0.21. Consider the diagramof stacks with $g$ representable. Then $h$ is representable iff $f$ is representable.

证明. If $f$ is representable, $h$ is representable by definition. If $h$ is representable, given $Z\stackrel{\alpha }{\to }\mathfrak{Y}$ we want $Z{\times }_{\mathfrak{Y}}\mathfrak{X}$ is an algebraic space. Thus we get the following diagramwhere the bottom square is also Cartesian. We note $X,Y$ are algebraic spaces because $h$, $g$ are representable, respectively. Then, note we get a section $\beta :Z\to Y$ defined by $\alpha$ and hence we obtain the diagramNow we see $\mathfrak{X}{\times }_{\mathfrak{Y},\alpha }Z=X{\times }_{Y,\beta }Z$ is an algebraic space as $X,Y,Z$ are all algebraic spaces.

$\square$

命题 4.4.0.22. Let $\mathfrak{X},\mathfrak{Y}$ be Artin stacks over $S$ and $f:\mathfrak{X}\to \mathfrak{Y}$, then ${\Delta }_{\mathfrak{X}/\mathfrak{Y}}$ is representable.

证明. We have the following diagramTHen ${\Delta }_{\mathfrak{Y}}$ repable implies $g$ is repable. Hence ${\Delta }_{\mathfrak{X}}$ repable implies, by the proposition above, that ${\Delta }_{\mathfrak{X}/\mathfrak{Y}}$ is representable as desired.

$\square$

定义 4.4.0.23. We say $f:\mathfrak{X}\to \mathfrak{Y}$ is separated if ${\Delta }_{\mathfrak{X}/\mathfrak{Y}}$ is proper.

注 4.4.0.24. For stacks, $\Delta$ keeps track of $\mathrm{Isom}$ or $\mathrm{Aut}$ which is a group so that is is rarely a closed immersion.

注 4.4.0.25. FOr $X,Y$ schemes, ${\Delta }_{X/Y}$ is always an immersion, so they are separated and of finite type. Thus ${\Delta }_{X/Y}$ is proper iff ${\Delta }_{X/Y}$ universally closed but base changes of immersion is immersion and hence $\Delta$ is universally closed iff its closed.

Thus, $X\to Y$ separated in the usual definition (${\Delta }_{X/Y}$ closed) iff its separated as stacks.

例 4.4.0.26. Let $\mathfrak{X}=[X/G]$, then we see we getwhere $X{\times }_{\mathfrak{X}}X$ is the $\mathrm{Aut}$ of universal torsor. Hence we seewhere $G\times X\cong X{\times }_{\mathfrak{X}}X$ because we has a section $X\to X{\times }_{\mathfrak{X}}X$. Thus, we see the above diagram’s arrows are given bySo, we seewhere $\Gamma$ is the graph of the action. Therefore, we see $\mathfrak{X}$ is separated iff $\Gamma$ is proper. By definition of $G\circlearrowleft X$ is called proper if $\Gamma$ is proper.

例 4.4.0.27. If $X$ is separated and $G$ is proper (e.g. $G$ is finite) thenand hence $\Gamma$ is proper, i.e. if $X$ is separated, $G$ proper (e.g. finite), then $[X/G]$ is separated.

例 4.4.0.28. If $A$ is Abelian variety, then $BA$ is separated.