Non-(affine line)s: Difference between revisions

From stacky wiki
 
Line 27: Line 27:
'''Remark.''' Matsushima's theorem (Theorem 12.15 of Alper<ref>Alper, [http://arxiv.org/abs/0804.2242 Good moduli spaces for Artin stacks]</ref>) says that a subgroup of a linearly reductive group is linearly reductive if and only if the quotient space is affine. Since $G$ is not affine, this shows that $H$ is not linearly reductive. In particular, this shows that linear reductivity of a relative group cannot be checked on fibers.
'''Remark.''' Matsushima's theorem (Theorem 12.15 of Alper<ref>Alper, [http://arxiv.org/abs/0804.2242 Good moduli spaces for Artin stacks]</ref>) says that a subgroup of a linearly reductive group is linearly reductive if and only if the quotient space is affine. Since $G$ is not affine, this shows that $H$ is not linearly reductive. In particular, this shows that linear reductivity of a relative group cannot be checked on fibers.


Now $B_{\AA^1}G$ is a smooth DM stack with a single stacky point with residual gerbe $B(\ZZ/2)$. However, it has ''non-separated diagonal'', so it is isomorphic to the usual $[\AA^1/(\ZZ/2)]$, with the action given by negation of the coordinate.
Now $B_{\AA^1}G$ is a smooth DM stack with a single stacky point with residual gerbe $B(\ZZ/2)$. However, it has ''non-separated diagonal'', so it is not isomorphic to the usual $[\AA^1/(\ZZ/2)]$, with the action given by negation of the coordinate.


= Some smooth non-(affine line) Artin stacks =
= Some smooth non-(affine line) Artin stacks =

Latest revision as of 16:19, 11 January 2021

This is my collection of examples of algebraic spaces and stacks that look something like $\AA^1$. If you like the affine line with a doubled origin, and the stack $[\AA^1/\mu_2]$, you've found the right place.

Non-(affine line) spaces with various singularities at the origin

Gluing two copies of $\AA^1$ along the complement of the origin gives the usual line with a doubled origin. An example in the introduction of Knutson[1] is the "affine line with a doubled tangent direction", obtained by taking two intersecting lines $C=V(y^2-x^2)$, considering the $\ZZ/2$-action which negates $y$, "removing the action at the origin," and taking the algebraic space quotient. More precisely, take the algebraic space quotient by the relation $R=C\sqcup C'\rightrightarrows C$, where $C'$ is the complement of the origin in $C$, and the two maps $C'\to C$ are given by the inclusion and the inclusion followed by negating $x$. The result looks like a line, but with a "doubled tangent direction" at the origin since it has an étale cover by two intersecting lines.

An alternative description of this example is to let $C$ be $V(y^2-x^2)$ with a doubled origin, and let $\ZZ/2$ act by negating $x$ and swapping the two origins. Then the affine line with a doubled tangent direction is the algebraic space $C/(\ZZ/2)$.

Example. Generalizing the above example, we can consider the action of $\ZZ/2$ on $V(y^n-x^2)$ with a doubled origin. The space quotient is an affine line with a higher (tac)node (if $n$ is even) or higher cusp (if $n$ is odd) at the origin. More generally, any singularity of the form $f(y)-x^k$ can appear on a tweaked copy of the affine line by "$k$-folding up" the points where $x=0$ and acting by $\ZZ/k$. (Assuming we're working over a base where $\ZZ/k$ is isomorphic to $\mu_k$)

Example. All the above singularities can appear on Deligne-Mumford stacks with coarse space $\AA^1$. Just don't introduce nonseparatedness: $[V(f(y)-x^k)/(\ZZ/k)]$.

Example (space or stack with a non-lci singularity at the origin). Let $C$ be the union of the coordinate axes in $\AA^3$, and let $\ZZ/3$ act by cyclic permutation of the three coordinates. Then $[C/(\ZZ/3)]$ (resp. the algebraic space obtained by tripling the origin in $C$ before quotienting) is a non-(affine line) stack (resp. algebraic space). The interesting thing about this example is that the singularity is not a local complete intersection singularity.

$\AA^1_\RR$ with a complex origin and various bug eyes

Knutson[1] explains the following procedure for doing an "étale extension along a closed subscheme." Suppose $X\to Y$ is an étale morphism, $Z\subseteq Y$ is a closed subscheme, and $R=X\times_Y X$. Then $Y=X/R$. Since $X\to Y$ is \'etale, $R$ is the disjoint union of the diagonal and some other stuff: $R=X\sqcup R_0$. We can then remove the part of the relation responsible for gluing together points in the fiber over $Z$ by replacing $R_0$ by $R_0'=R_0\times_Y (Y\smallsetminus Z)$. Then $R'=X\sqcup R_0'$ is an étale relation on $X$. The algebraic space quotient $Y'=X/R'$ has a morphism to $Y$ which is an isomorphism over the complement of $Z$, but $Y'\times_Y Z\cong X\times_Y Z$.

Applying this to the case $Y=\AA^1_\RR$, $X=\AA^1_\CC$, and $Z=\{0\}$, we get an algebraic space $Y'$ which looks like $\AA^1_\RR$, except the residue field at the origin is $\CC$.

It's also possible to do non-étale extensions along closed subschemes. For example, the squaring map $\AA^1\to \AA^1$ is étale away from the origin. We can modify the induced relation on $\AA^1$ to make it étale: consider $\AA^1\sqcup (\AA^1\setminus 0)\rightrightarrows \AA^1$, where the first copy is the diagonal, and the second copy is $x\mapsto x$ and $x\mapsto -x$. The quotient is the "bug-eyed" cover of $\AA^1$. An alternative description: let $\ZZ/2$ act on the non-separated line by $x\mapsto -x$ and switching the two origins, and consider the quotient. The same trick can be done with any of the non-(affine line)s in the previous section.

A smooth non-(affine line) DM stack with non-separated diagonal

Consider the relative group $(\ZZ/2)\times \AA^1$ over $\AA^1$. Then $H=\AA^1\sqcup (\AA^1\smallsetminus \{0\})$ is an open subgroup. The quotient $G=(\ZZ/2\times \AA^1)/H$ is the affine line with a doubled origin, regarded as a group over $\AA^1$.

Remark. Matsushima's theorem (Theorem 12.15 of Alper[2]) says that a subgroup of a linearly reductive group is linearly reductive if and only if the quotient space is affine. Since $G$ is not affine, this shows that $H$ is not linearly reductive. In particular, this shows that linear reductivity of a relative group cannot be checked on fibers.

Now $B_{\AA^1}G$ is a smooth DM stack with a single stacky point with residual gerbe $B(\ZZ/2)$. However, it has non-separated diagonal, so it is not isomorphic to the usual $[\AA^1/(\ZZ/2)]$, with the action given by negation of the coordinate.

Some smooth non-(affine line) Artin stacks

Consider the action of $\GG_m$ on $\AA^2$ given by $t\cdot (x,y)=(tx,t^{-1}y)$. Then $[\AA^2/\GG_m]$ (which I would denote $[\AA^2/_{\begin{pmatrix}1& -1\end{pmatrix}}\GG_m]$) has good moduli space $\AA^1$, but contains a dense open copy of the non-separated line (namely $[(\AA^2\smallsetminus \{0\})/_{\begin{pmatrix}1& -1\end{pmatrix}}\GG_m]$). The non-separatedness disappears in the coarse space because there is a $B\GG_m$ which both origins specialize to.

By considering a the weight $\begin{pmatrix}a& -b\end{pmatrix}$ action of $\GG_m$ on $\AA^2$ for positive integers $a$ and $b$ (instead of the weight $\begin{pmatrix}1& -1\end{pmatrix}$ action), we get a similar stack, but where the two origins have stabilizers $\mu_a$ and $\mu_b$.

In Kresch's Flattening stratification and the stack of partial stabilisations of prestable curves, he builds a very interesting stack: the quotient of $\X=[(\AA^2\smallsetminus \{0\})/_{\begin{pmatrix}1& -1\end{pmatrix}}\GG_m]$ by the étale relation $\AA^1\rightrightarrows \X$ given by the two open immersions. The quotient is a smooth stack with a dense open substack isomorphic to $\AA^1$, whose complement is a closed substack of codimension 2. Interestingly, this stack has no non-trivial vector bundles, since sections of a vector bundle extend across codimension 2 on a normal stack. When any vector bundle is restricted to the copy of $\AA^1$, it has trivializing sections which extend to trivialize the whole bundle. Therefore, this stack is not a quotient stack!