# Does a degeneration always have a larger-dimensional automorphism group?

Suppose $\newcommand{\X}{\mathcal{X}}\X$ is an algebraic stack over a field $k$, $\xi$ is a $k$-point which has another $k$-point $x$ in its closure ($x$ is an isotrivial degeneration of $\xi$). Must the dimension of $Aut(x)$ be larger than the dimension of $Aut(\xi)$?

If $\X=[X/G]$, where $G$ is an algebraic group over $k$, then the answer is yes. An isotrivial degeneration corresponds to a $G$-orbit with another $G$-orbit in its closure. The closure of a $G$-orbit can only contain smaller dimensional orbits, and the dimension of the orbit is complementary to the dimension of the stabilizer of a point in that orbit, so when one orbit degenerates to another, there is always a jump in the dimension of the stabilizer.

Almost every algebraic stack I can think of is étale-locally of the form $[X/G]$, and étale maps preserve the dimensions of automorphism groups. This suggests that the answer is probably "yes."

## Example

Consider the 1-parameter family of $2\times 2$ matrices $\begin{pmatrix}1&t\\ 0&1\end{pmatrix}$. If we are studying matrices up to conjugation, then away from $t=0$, the family is (isomorphic to) the constant family $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}$. However, at $t=0$ you get a different Jordan type. So we say that this family is an isotrivial degeneration of Jordan types $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\rightsquigarrow \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$. When you pass to the more degenerate Jordan type, the automorphism group (i.e. the group of ways in which the matrix is self-conjugate) jumps from something 2-dimensional to something 4-dimensional.

## Motivation

It sometimes happens that you want to determine all degeneration relations among a collection of points in an algebraic stack. For example, you may be trying to determine if some map is weakly proper. If the answer to this question is "yes," then you can rule out certain degenerations by looking at dimensions of automorphism groups.

See Weakly proper moduli stacks of curves by Alper, Smyth, and van der Wyck.

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Note that the dimension of the automorphism group cannot jump down because of upper semi-continuity of the fiber dimension of the inertia (see mathoverflow.net/questions/193/…). The question is whether the dimension must jump up in the case of an isotrivial degeneration. –  Anton Geraschenko Mar 12 '11 at 22:36
Here's an example that might not be etale locally isomorphic to $[X/G],$ where $G$ is over $k.$ Simply take $G$ to be a group scheme over $X,$ a non-trivial family of algebraic groups, e.g. the universal elliptic curve over the $j$-line, acting on $X$ trivially, and then form the quotient $[X/G].$ (This comment doesn't help answering the question though.) –  shenghao Mar 13 '11 at 1:09
I think that in the Lie case, the answer should be "no". I will leave this as a comment, because I don't know enough about algebraic stacks. Consider the following Lie groupoid (= $C^\infty$ Artin stack, I think) $[X/G]$, where $X=\mathbb R^2$, $G=\mathbb R$, and the action integrates the vector field $(\dot x,\dot y) = (1,\sin y)$. Then the $x$-axis is a closed point; the curve $y = \frac\pi2 + \arctan x$ is a point with the $x$-axis in its closure; both these points have trivial automorphism group. –  Theo Johnson-Freyd Mar 13 '11 at 1:36
@shenghao: Thanks for the example. The examples I can think of that aren't locally $[X/G]$ are also quotients by relative groups. My favorites are open (and non-closed) subgroups of finite groups over $\mathbb A^1$, as well as their quotients. I'm in awe of how much milage I get out of these counterexamples. See, for example, mathoverflow.net/questions/1467/… –  Anton Geraschenko Mar 13 '11 at 4:01
@Theo: I agree that all points in your example have trivial stabilizers, but I don't think there are any closure relations. All the $\mathbb R$-orbits are closed in your example! In fact, the map $\mathbb R\to [X/G]$ given by $y\mapsto (0,y)$ is an isomorphism of stacks since it hits every orbit once and induces isomorphisms on stabilizers. I would be happy if somebody left an answer with a proof or counterexample for the Lie case. –  Anton Geraschenko Mar 13 '11 at 4:06
You can represent $\xi$ and $x$ by a morphism $\varphi: S\to\mathcal{X}$, where $S$ is a normal affine $k$-scheme of finite type with a point $s\in S(k)$, such that $\varphi(s)=x$ and $\varphi_{\mid U}$ factors through $\xi$, where $U:=S\setminus\{s\}$. On $S\times S$, consider the two objects $\varphi_1$ and $\varphi_2$ of $\mathcal{X}$ obtained by pullback via the projections. Put $Y:=\underline{\mathrm{Isom}}_{\mathcal{X}}(\varphi_1,\varphi_2)$. This is an algebraic space of finite type over $S\times S$, and a pseudo-torsor under the group we are interested in. The assumptions on $x$ and $\xi$ imply that the image of $f:Y\to S\times S$ is $(U\times U)\cup\{(s,s)\}$. Hence $f$ is not open, and therefore not equidimensional by EGA IV (14.4.4) since $S\times S$ is geometrically unibranch. Hopefully, this implies the result on the group but we have to be careful; for instance our group may have constant dimension but extra components at $x$, which would account for the lack of equidimensionality at some points of the fibre at $(s,s)$ (but not all?).
$Y$ isn't actually a pseudo-torsor under the group we're interested in, right? It's a pseudo-torsor under the pullback of that group from $S$ to $S\times S$ along a projection. And the group we're interested in is the pullback of $Y$ along the diagonal of $S$. If $s$ is not a $k$-point, then it's possible that $\{s\}\times\{s\}$ has many components, so the group could have many dimensions, but if $s$ is a $k$-point, how could there be a failure of equidimensionality? We have that $\{s\}\times\{s\}\cong Spec(k)$, over which a group must be equidimensional. Am I missing something? –  Anton Geraschenko Mar 15 '11 at 6:34
Consider in $\mathbb{A}^2$ the union of the $x$-axis and the point $(0,1)$. Via the $x$-projection, it has a unique structure of group scheme over $\mathbb{A}^1$ (the fiber at the origin being $\mathbb{Z}/2\mathbb{Z}$). This $\mathbb{A}^1$-scheme is not equidimensional at the point $(0,1)$, although all its fibers have the same dimension. So, the condition "$Y$ is not equidimensional" does not imply that the dimension jumps. But using the triviality along the diagonal, you might be able to prove that the group is equidimensional at the origin over $(s,s)$, which should be enough. –  Laurent Moret-Bailly Mar 18 '11 at 13:13