Non-(affine line)s

This is my collection of examples of algebraic spaces and stacks that look something like $\AA^1$.

Non-(affine line)s 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 tangent direction. An example in the introduction of Knutson 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: $[V(f(y)-x^k)/(\ZZ/k)]$.

$\AA^1_\RR$ with a complex origin

Knutson 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 $\CC$.

A smooth non-(affine line) stack

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) 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.

maybe throw in some Artin stack examples like $[\AA^2/_{(1\ -1)}\GG_m]$