The n-Category Café

I’m not sure when I’ll have the time to read all the stuff you guys have been doing with Klein 2-geometry. But this particular question sounds like a job for regularity and exactness.

Let me start by rephrasing the 1-categorical version. Let $G$ be an internal group object in a regular category $K$, and let $G$ act on the object $X\in K$ via a map $act:G\times X\to X$. Let’s say this action is transitive if

1. $X$ has a global element $x:1\to X$, and
2. the map $act_x:G\overset{(1_G,x)}{\to} G\times X \overset{act}{\to} X$, which sends $g$ to $g\cdot x$, is a regular epi.

It’s not unreasonable to ask that $X$ have a global element, since any homogeneous space $G/H$ will have one, namely the image of the identity $e:1\to G$. Here if $H$ is a sub-group-object of $G$, then $G/H$ is the coequalizer $G\times H \;\rightrightarrows\; G$, on which $G$ clearly acts transitively in the usual way. This coequalizer will exist if we also assume that $K$ is exact, since $G\times H \;\rightrightarrows\; G$ is an equivalence relation.

Now, if $G$ acts transitively on $X$, then the regular epi $act_x:G\to X$, like any regular epi in a regular category, is the quotient of its kernel pair. Intuitively, this kernel pair $K\to G\times G$ consists of pairs $(g_1,g_2)$ such that $g_1 x = g_2 x$. It’s easy to see that in fact we have $K\cong G\times Stab(x)$, where $Stab(x)$ is the pullback of $act_x:G\to X$ along $x:1\to X$; the isomorphism takes $(g_1,g_2)$ to $(g_1,g_2 g_1^{-1})$. Thus the quotient of $K$ is precisely the quotient $G/Stab(x)$, so we have $G/Stab(x)\cong X$ whenever $G$ acts transitively on $X$. And with a little more work we can show that transitive actions of $G$ are equivalent to subgroups, at least if $K$ is exact so that we have quotients. (In fact, $K$ being exact means that equivalence relations on an object $G$ are equivalent to regular epis out of $G$, so all that remains is to identify those regular epis coming from transitive actions with those equivalence relations coming from subgroups.)

Now let’s push it up a notch. Let $G$ be an internal group object in a regular 2-category $K$ (you can think of $K=Cat$ or $Gpd$). Thus, we have morphisms $mult:G\times G\to G$, $id:1\to G$, and $inv:G\to G$ satisfying the usual axioms up to coherent isomorphism. Clearly any group object in $Gpd$ is a 2-group in the usual sense. The converse follows because we can construct the functor $inv:G\to G$ by choosing, for each object $g\in G$, an adjoint inverse $inv(g)$, and for a morphism $\gamma:g\to g'$ defining $inv(\gamma)$ to be the composite $$inv(g) \overset{\cong}{\to} inv(g) \cdot g' \cdot inv(g') \overset{1\cdot \gamma^{-1} \cdot 1}{\to} inv(g)\cdot g \cdot inv(g') \overset{\cong}{\to} inv(g').$$ Thus, 2-groups are the same as group objects in $Gpd$. I’m pretty sure they are also the same as group objects in $Cat$ as well; you can reverse the above argument to show that given the functor $inv$, the category $G$ must be a groupoid. Have 2-group-theorists written this down already?

Moving on, suppose our group object $G$ acts on the object $X\in K$, and let’s say that this action is transitive if

1. $X$ has a global element $x:1\to X$, and
2. The morphism $act_x:G\to X$ is eso.

Note that we’re only requiring the action to be transitive on objects. “Transitivity on arrows” turns out to be unnecessary, “because” the “2-inclusion” of a 2-subgroup is only required to be faithful, not full. In particular, if $H\to G$ is a faithful 2-group homomorphism, then the (weak) quotient $G//H$ has a transitive action of $G$ in this sense, but it might not be transitive on arrows. As an extreme example, if $G=1$ is the trivial (2-)group and $H$ is any 1-group, considered as a 2-group with only identity morphisms, then $H\to G$ is faithful and its quotient $G//H$ is $B H$; and $G$ certainly doesn’t act transitively on the arrows of $B H$.

Now if $G$ acts transitively on $X$, then the eso $act_x:G\to X$, like any eso in a regular 2-category, is the quotient of its kernel. This kernel is the comma object

$$\array{K & \overset{}{\to} & G\\ \downarrow & \Downarrow & \downarrow act_x\\ G& \underset{act_x}{\to} & X}$$

In particular, in $Cat$, an object of $K$ is a triple $(g_1,g_2,\alpha)$ where $\alpha: g_1 x \to g_2 x$ is not necessarily invertible! Still, using the fact that $G$ is a group object, we can identify $K$ with the product $G\times Stab_2(x)$ where now an object of $Stab_2(x)$ is an object $g\in G$ equipped with a not-necessarily-invertible morphism $x\to g x$, and a morphism in $Stab_2(x)$ is a morphism $g_1\to g_2$ in $G$ making the obvious triangle commute. Then $X$ being the quotient of $K$ means that we have a lax codescent diagram

$$\array{ G\times Stab_2(x) \times Stab_2(x)& \underoverset{\to}{\to}{\to}& G\times Stab_2(x) & \underoverset{\to}{\to}{\leftarrow} & G & \to & X. }$$

Thus, in a certain sense, we have $X\simeq G//Stab_2(x)$, but it doesn’t look quite like the usual weak quotient of $G$ by the action of $Stab_2(x)$. Rather, it’s a sort of lax quotient in which we only glue in a morphism, rather than an isomorphism, from $g$ to $g h$ for $h\in Stab_2(x)$. This is connected with the fact that $Stab_2(x)$ is not necessarily itself a 2-group! It’s a monoidal groupoid, but its objects need not be invertible. So perhaps for the purposes of Klein 2-geometry, a “2-subgroup” of a 2-group need not be itself a 2-group?

However, if $X$ is itself a groupoid, then the above comma object becomes a pullback, $Stab_2(x)$ becomes a 2-group, and the lax quotient becomes an ordinary (weak) quotient. So in the case of a 2-group acting on a groupoid, things seem to work out just as we would expect, but if we act on a category then some laxness creeps in. In either case, I guess that if $K$ is also exact, then the appropriate type of “2-subgroups” can be shown to be equivalent to transitive actions.

Finally, we can guess a generalization of this to $\infty$-groups acting on $\infty$-groupoids. The action of $G$ on $X$ should be transitive if $X$ has a point $x:1\to X$ and the map $act_x:G\to X$ is surjective on $\pi_0$, and in this case we should have $X\simeq G//Stab_\infty(x)$, setting up an equivalence between transitive actions of $G$ and “$\infty$-subgroups.” But now, an “$\infty$-subgroup” just means any $\infty$-group homomorphism $H\to G$.