Definition of stabilizer of a point in a stack

Question

I was reading in Adeel Khan's ''A Modern Introduction to Algebraic Stacks'' and came across the definition of the stabilizer of an $R$-valued point $x$ of a stack $X$. My question will be about that the definition does not quite make intuitive sense to me, but let me first make some definitions. The eventual questions are bold-faced.

I look at stacks as $(2,1)$-sheaves $\mathrm{Sch}^\mathrm{op}_\mathrm{fppf}\to\mathsf{Grpd}$ in the fppf-topology, where $\mathsf{Grpd}$ is the $(2,1)$-category of groupoids. Then, given an $R$-valued point $x\colon\mathrm{Spec}\,R\to X$ in $X(R)$, we define the stabilizer of $x$ in $X$ as the pullback (in the $(2,1)$-category of stacks) $\require{AMScd}$ \begin{CD} {\mathrm{St}_X(x)} @>>> {\mathrm{Spec}\,R}\\ @V V V @VV x V\\ {\mathrm{Spec}\,R} @>>x> X \end{CD} We can also define the intertia stack $\mathcal{I}_X$ of $X$ by the pullback $\require{AMScd}$ \begin{CD} {\mathcal{I}_X} @>>> X\\ @V V V @VV \Delta V\\ X @>>\Delta> X\times X \end{CD} where $\Delta\colon X\to X\times X$ is the diagonal.

Because $\mathrm{Spec}\,R$ is a set-valued stack, I can find that for any ring $R'$, the groupoid $\mathrm{St}_X(x)(R')$ is actually the set of triples $(y,z,\varphi)$ with $y,z\colon R\to R'$ maps of rings, and $\varphi\colon x|_y\to x|_z$ an isomorphism in $X(R')$.

Now comes my question: why do we call this a stabilizer of $x$ in $X$? I would expect that a stabilizer consists of all the automorphisms of $x$, in some sense. More precisely, I would expect the $R'$-valued points of the stabilizer to consist precisely of a groupoid with objects $x|_y$ for all $y\colon R\to R'$, and morphisms the automorphisms of these $x|_y$ in $X(R')$. This is not what $\mathrm{St}_X(x)$ does: we get not only automorphisms, but just general isomorphisms between different restrictions of $x$, and we only have a set, and not a groupoid. All in all, it seems like it is not completely resembling a stabilizer. Does anyone have an explanation why this is then the correct notion of the stabilizer of $x$? Why do we want to look at isomorphisms between different restrictions of $x$?

When I compute the $R'$-valued points of $\mathcal{I}_X$, I find that $\mathcal{I}_X(R')$ is equivalent to a groupoid consisting of objects $(y,\alpha)$, with $y\in X(R')$ and $\alpha\colon y\to y$ a morphism in $X(R')$. The morphisms $(y,\alpha)\to(y',\alpha')$ in this groupoid are morphisms $\zeta\colon y\to y'$ in $X(R')$ such that $\alpha=\zeta^{-1}\alpha'\zeta$. So $\mathcal{I}_X(R')$ has the automorphisms of $R'$-valued points of $X$ as objects, and their conjugations as morphisms. This seems much closer to what I would want from a stabilizer, but this is of course not the stabilizer of a single point.

We can remedy this by defining for $x\colon\mathrm{Spec}\,R\to X$ a stack $F_x$ by a pullback $\require{AMScd}$ \begin{CD} {F_x} @>>> {\mathcal{I}_X}\\ @V V V @VV \pi V\\ {\mathrm{Spec}\,R} @>>x> X \end{CD} where $\pi\colon\mathcal{I}_X\to X$ maps an object $(y,\alpha)$ to $y$ and a morphism $\zeta\colon(y,\alpha)\to(y',\alpha')$ to $\zeta\colon y\to y'$.

I can show that for a ring $R'$ the groupoid $F_x(R')$ is actually equivalent to the set consisting of pairs $(y,\omega)$, for $y\colon R\to R'$ and $\omega\colon x|_y\to x|_y$ in $X(R')$. In other words, $F_x$ collects all automorphisms of restrictions of $x$ to $R'$ into a set, but forgets about the conjugations (which is not weird, given its definition). Now, this seems more like a stabilizer to me than $\mathrm{St}_X(x)$, where we also allow isomorphisms between different restrictions of $x$. Why is this not the definition of the stabilizer of $x$?

As an aside, via the definition of $\mathrm{St}_X(x)$ we have a pullback $\require{AMScd}$ \begin{CD} {\mathrm{St}_X(x)} @>>> {\mathrm{Spec}\,R}\\ @V V V @VV (x,\mathrm{id}_x) V\\ {F_x} @>>> {\mathcal{I}_X} \end{CD} So we can recover $\mathrm{St}_X(x)$ from $F_x$ by ''further restriction'' along the identity automorphism of $x$. Because this is a $(2,1)$-pullback, it is of course not purely a restriction, but it seems still like $F_x$ is a slightly more powerful stack.

Answer 1

It comes from thinking of groupoids as quotients of group actions. Before thinking about stacks, you should first think about what happens in the 2-category of groupoids – which anyway is the same as the 2-category of stacks on the point.

Specifically, take a group $G$ and a (left) $G$-set $X$ and define $X // G$ to be the groupoid whose objects are the elements of $X$ and whose morphisms $x' \to x$ are the elements $g \in G$ such that $g \cdot x' = x$. Then the automorphism group of an object in $X // G$ is precisely the stabiliser group of the corresponding element of $X$.

In fact, every groupoid is the quotient of a set by a groupoid action (albeit in a tautological way). Thus, we may think of the automorphism group of an object in a groupoid as being the stabiliser group of that object. But what are "objects" and "automorphism groups" when we have an object in a general 2-category rather than the 2-category of groupoids?

Changing notation now, let $X$ be an object in a 2-category $\mathcal{K}$. When $\mathcal{K} = \textbf{Grpd}$, we can identify objects in $X$ with morphisms $1 \to X$, where $1$ is the terminal object. For general $\mathcal{K}$, we usually do not have "enough" morphisms $1 \to X$ to detect everything about $X$, so we are forced to generalise to "variable" objects by replacing $1$ with arbitrary (not fixed!) objects in $\mathcal{K}$. (In other words, a "variable" object in $X$ is any morphism in $\mathcal{K}$ with codomain $X$.)

So much for objects in $X$. What about morphisms in $X$? When $\mathcal{K} = \textbf{Grpd}$, given objects $x$ and $x'$ in $X$, we have the following bipullback diagram: $$\require{AMScd} \begin{CD} X (x', x) @>>> 1 \\ @VVV ⬃ @VV{x'}V \\ 1 @>>{x}> X \end{CD}$$ (If you have never seen this before, it is a very instructive exercise to work it out!) This has an obvious generalisation: given "variable" objects $x : T \to X$ and $x' : T' \to X$, we can think about the bipullback of $x$ and $x'$: $$\begin{CD} P_X (x', x) @>>> T' \\ @VVV ⬃ @VV{x'}V \\ T @>>{x}> X \end{CD}$$ This will be an object equipped with a projection to $T$ and another projection to $T'$. This agrees with your first definition of "stabiliser" when $T = T'$ and $x = x'$. Unfortunately (and this is related to your confusion) this construction does not work well when $T$ is not the terminal object: in general, the two projections are different!

Let us try again. When $\mathcal{K} = \textbf{Grpd}$, there is another bipullback diagram: $$\begin{CD} X (x', x) @>>> X \\ @VVV \cong @VV{\Delta}V \\ 1 @>>{\langle x', x \rangle}> X \times X \end{CD}$$ (If you construct the bipullback explicitly as an iso-comma, you will get a different but equivalent groupoid, namely the one whose objects are tuples $(y, f', f)$, where $y$ is an object in $X$ and $f' : y \to x'$ and $f : y \to x$ are morphisms in $X$, and whose morphisms are morphisms in $X$ making the two obvious triangles commute. Every object in that groupoid is uniquely isomorphic to one of the form $(x', \textrm{id}_{x'}, g)$, so it is indeed equivalent to the set $X (x', x)$.) This generalises better: given "variable" objects $x, x' : T \to X$ with the same domain, we can pull back $\Delta : X \to X \times X$ along $\langle x', x \rangle : T \to X \times X$ to get an object over $T$: $$\begin{CD} X^\cong (x', x) @>>> X \\ @VVV \cong @VV{\Delta}V \\ T @>>{\langle x', x \rangle}> X \times X \end{CD}$$

This agrees with your second definition of "stabiliser" when $x = x'$, and also agrees with your definition of "inertia" when $T = X$ and $x = x' = \textrm{id}_X$.

I would say that both constructions have their uses, but the second one is probably more deserving of the name "stabiliser group" than the first one. Focusing on the case where $x = x'$, the two projections $P_X (x, x) \rightrightarrows T$ are actually the domain and codomain maps of an internal groupoid in $\mathcal{K}$. That is, the first construction can be continued to define an internal groupoid in $\mathcal{K}$ whose object of objects is $T$ and whose object of morphisms is $P_X (x, x)$. In this light, the fact that the two projections are different is a feature, not a bug! One use of this construction is to exhibit $X$ as the quotient of $T$ by a groupoid action: in this case we need $x : T \to X$ to be essentially surjective. (Note that $x : T \to X$ does not need to be full – in fact, $T$ can be "discrete"!)

On the other hand, the second construction can be continued to define an internal group in $\mathcal{K}_{/ T}$. The two constructions are related: the automorphism group of the generic object in the groupoid $P_X (x, x) \rightrightarrows T$ is $X^\cong (x, x)$. So I would say which one is "correct" depends on what you want to use it for.

Answer 2

Your $F_x$ indeed agrees with the stabilizer as in Alper's note in this link. Look at Section 3.2.2 on page 107. Alper defines the stabilizer of a point $x: Spec(R)\to X$ as the fiber product of $(x,x): Spec(R)\to X\times X$ and $\Delta: X\to X\times X,$ which is exactly how you defined $F_x$.