## March 2, 2009

### The Stabilizer of a Subcategory

#### Posted by David Corfield A very long time ago, John gave us a definition of a stabilizer of an object in a category on which some 2-group is acting, having had a moment of insight near the Nine Zigzag Bridge in Shanghai.

I was desperately trying to understand sub-2-groups. So, I thought: in Klein geometry, the conceptual meaning of “subgroup” is really “stabilizer of some point in a set on which a group acts”.

So, let’s take a 2-group $G$ acting on a category $X$, and let’s study the the stabilizer of some object $x$ in $X$. Whatever this stabilizer is like, maybe this should become the definition of a sub-2-group!

(Or, maybe not - there are also stabilizers of things more complicated and interesting than a mere object. But never mind! - it’s still an interesting exercise.)

Of course we need to define the stabilizer, say $Stab(x)$. There’s an obvious way to do this if you’re careful not to be evil. I’ll just sketch it.

The stabilizer $Stab(x)$ is a 2-group with the following objects and morphisms. An object of $Stab(x)$ is an object $g$ of $G$ together with an isomorphism

$a: g x \to x$

Nota bene: we’re not evilly demanding that $g x = x$; we’re specifying an isomorphism between them!

A morphism of $Stab(x)$, say from

$g, a: g x \to x$

to

$g', a': g' x \to x$

is a morphism $f: g \to g'$ in $G$ making the obvious triangle commute. Namely,

$a: g x \to x$

should equal the composite of

$f x: g x \to g' x$

and

$a': g' x \to x.$

It really looks much prettier as a triangle!

With some work one makes $Stab(x)$ into a 2-group - I didn’t check everything here, but I’m following the tao of mathematics so I’m sure everything works, even when $G$ is a weak 2-group and its action on $X$ is also weak - the general case. I also feel sure we get a 2-group homomorphism

$i: Stab(x) \to G.$

Presumably we can think of what John’s doing here as finding the stabilizer of the subcategory of $X$ composed of $x$ and its identity morphism. What I’m not so clear about is which $a$, isomorphisms between $g x$ and $x$, we’re allowed. Presumably just any such isomorphism in $X$, and not just those coming from restrictions of natural equivalences arising from the action of $G$.

But then if we look at the stabilizer of larger subcategories of $X$, what does the counterpart of $a$ look like?

Posted at March 2, 2009 11:26 AM UTC

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### Re: The Stabilizer of a Subcategory

Let me see if I have things right. Consider the category (with structure) whose objects are points on the Euclidean plane, and with a vertex group $S_2$ at each point.

There’s a 2-group acting on this whose objects are $E(2)$ and whose morphisms again make up a vertex group $S_2$ at each point.

The Baezian stabilizer of a point in the plane has as objects $O(2) \times S_2$, and for each element, $\phi$ of $O(2)$, and each $g$ and $h$ in $S_2$ there is a single arrow between $\langle \phi, g \rangle$ and $\langle \phi, h \rangle$.

So the stabilizer is equivalent to $O(2)$ with trivial morphisms, and the quotient of the original 2-group by the stabilizer is the plane with internal $S_2$ symmetry, just as we started with.

Posted by: David Corfield on March 2, 2009 11:58 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

One way to approach such questions is to get the solution in terms of general abstract nonsense and then turn the crank in any given concrete realization and work out what it looks like in detail.

Here it seems the general situation is:

we are in some higher category $C$ (of 2-groups, say) which we can safely assume to be groupoidal (no non-invertible morphisms in sight) and in $C$ we have a morphism

$f : K \to G \,.$

Now we are looking, it seems to me, for the thing of automorphisms of $f$ in the under-category $K \downarrow C$.

Suppose we have managed to say under-category $K \downarrow C$ in our context, then the object $Stab_G(f)$ in question would be the suitable weak limit

$\array{ Stab_G(f) &\to& pt \\ \downarrow && \downarrow^{f} \\ pt &\stackrel{f}{\to}& K \downarrow C } \,.$

Posted by: Urs Schreiber on March 2, 2009 4:19 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

How, if at all, does this picture relate to Mike’s?

Posted by: David Corfield on March 3, 2009 10:33 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

I thought about this for a bit, and maybe I’m just being stupid, but I don’t even understand how what Urs wrote answers the question. Where is the object $X$ that $G$ is acting on? What is $K$?

Posted by: Mike Shulman on March 3, 2009 4:15 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

What I described was supposed to be the full automorphism 2-group of a category fixing a subcategory.

So my $G$ was your $X$ and my $K$ your $\{x\}$ and Mike’s $Y$. My $f$ is Mike’s $i$.

So let me say it again, with different symbols:

let $X$ be a category and $i : Y \to X$ a morphism, which can be regarded as an object of $Y \downarrow Cat$.

Notice that the endomorphism 2-monoid of $X$ itself is the weak pullback

$\array{ AUT(X) &\to& pt \\ \downarrow && \downarrow^X \\ pt &\stackrel{X}{\to}& Cat } \,.$

Similarly for the automorphism 2-group with $Cat$ replaced by the core of $Cat$.

Analogously, the weak pullback

$\array{ Stab(i) &\to& pt \\ \downarrow && \downarrow^i \\ pt &\stackrel{i}{\to}& Y\downarrow Cat }$

computes endo/auto–morphisms of $X$ that weakly fix $Y$.

Here, too, an object is a 2-cell

$\array{ && Y \\ & {}^{i}\swarrow &\Rightarrow& \searrow^i \\ X &&\to && X }$

etc.

Posted by: Urs Schreiber on March 3, 2009 4:49 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

You have to be careful here. The 3-categorical pullback

$\array{Aut(X) & \to & 1\\ \downarrow & & \downarrow ^X\\ 1 & \underset{X}{\to} & Cat}$

is the automorphism 2-group of the category $X$, meaning the auto-equivalences and isomorphisms between them. It doesn’t change if you replace $Cat$ by its core. The 3-categorical comma object

$\array{Core(End(X)) & \to & 1\\ \downarrow & \Downarrow & \downarrow^X\\ 1 & \underset{X}{\to} & Cat}$

is the core of the endomorphism 2-monoid of $X$: it consists of endofunctors $X\to X$ and isomorphisms between them. Generally, comma objects in $n$-categories for $n\gt 2$ seem not to be quite right; see for instance the bottom of this page. This is currently tripping me up in my attempts to define 3-toposes.

If you want to recover the full endomorphism 2-monoid $End(X)$ of endofunctors $X\to X$ and all transformations between them, then you have to consider the universal 2-category equipped with a lax natural transformation

$\array{End(X) & \to & 1\\ \downarrow & \Downarrow_{lax} & \downarrow ^X\\ 1 & \underset{X}{\to} & Cat}$

which doesn’t live in a 3-category but rather in the biclosed non-symmetric monoidal category $(2Cat,\otimes_{Gray,lax})$ where $\otimes_{Gray,lax}$ is the lax version of the Gray tensor product.

I’m still thinking about whether&why your definition of the stabilizer is the same as mine. They’ve definitely got the same intuition behind them.

Posted by: Mike Shulman on March 3, 2009 10:15 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

For amplification:

taking endomorphisms means forming loops:

for $X$ an object of an $\infty$-category $C$ we have generally

$End_C(X) := \Omega_{X} C \,,$

where on the right we have the based (at $X$) loop space object, which is the weak pullback

$\array{ \Omega_X(C) &\to& pt \\ \downarrow && \downarrow^X \\ pt &\stackrel{X}{\to}& C }$

(you know this – but for the record I mention that this is described for instance in the examples at $n$Lab: homotopy limit )

And endomorphisms fixing a map $i : Y \to X$ are endomorphisms of $i$ in $Y \downarrow C$.

Posted by: Urs Schreiber on March 3, 2009 5:05 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Let $G$ be an $n$-group acting on an object $X$ of an $n$-category $C$ (such as a 2-group acting on a category = object of the 2-category $Cat$), and let $i:Y\to X$ be a morphism (thought of as the “inclusion of a subobject”). Then I would define the stabilizer of $i$ under the action of $G$ to be the ($n$-categorical) pullback $\array{Stab(i) && \to && 1\\ \downarrow &&&&\downarrow i\\ G & \to & Aut_C(X) & \underset{-\circ i}{\to} & C(Y,X)}.$ in $(n-1)Cat$. Thus, an object of $Stab(i)$ consists of a $g\in G$ and a 2-cell isomorphism $g i \cong i$, a morphism of $Stab(i)$ consists of $h:g\cong g'$ in $G$ and a triangle commuting up to a 3-cell isomorphism, and so on.

Posted by: Mike Shulman on March 2, 2009 8:08 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

That looks good!

So now can we choose a topological category for $X$ with a nice continuous transitive action of $G$, and a shape $Y$ a subcategory of $X$, so that the category of cosets $G/Stab(i)$ (perhaps //) is equivalent to the functor category $Fun(Y, X)$?

One thing that’s given me some concern is that $G$ acting on the morphisms of $X$ can’t be transitive, having to send identity maps to other such.

Posted by: David Corfield on March 3, 2009 8:56 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Posted by: Mike Shulman on March 3, 2009 4:33 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

I’m trying to get us back on the chase of Klein 2-geometry. Here I’m trying to categorify homogeneous spaces. Things like the Euclidean group acting on the plane with the stabilizer of a point being $O(2)$ and the coset space $E(2)/O(2)$ being isomorphic to the plane.

Posted by: David Corfield on March 3, 2009 4:50 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

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 $H\to G$.

Posted by: Mike Shulman on March 4, 2009 9:06 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Wow, excellent!

What happens in the case of Lie groups? In your description at the 1-categorical level, you required $K$ to be regular. The category of manifolds isn’t regular, is it? But Klein geometry works there.

A Klein Geometry requires a Lie group and closed Lie subgroup. Can we just ask for a closed Lie sub-2-group at the 2-level?

Do we have a Cartan’s 2-theorem: every closed sub-2-group of a Lie 2-group is a Lie sub-2-group?

Posted by: David Corfield on March 5, 2009 9:13 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

No, of course the category of manifolds isn’t regular or exact, and doesn’t even have finite limits. But looking at the argument I gave, to show that a transitive $G$-object is the quotient of $G$ by its stabilizer, I think all you really need is that the action map $act_x:G\to X$ has a kernel pair of which it is the quotient. For that, it’s probably good enough for $act_x$ to be a surjective submersion. Presumably that always happens when the action is transitive on underlying sets.

Going in the other direction, to make a homogenous space $G/H$ you just need the equivalence relation $G\times H \to G\times G$ to be effective. I’d guess that $H$ being closed is exactly what you need to make that true, probably partly because of Cartan’s theorem.

I’ll leave the other questions to the Lie 2-group theorists. But one way to avoid them entirely would be to work in the 2-category of stacks on the category of manifolds, which is certainly regular and exact.

Posted by: Mike Shulman on March 5, 2009 4:28 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

What’s a nice nontrivial example of an $\infty$-group?

Posted by: David Corfield on March 5, 2009 10:38 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

…setting up an equivalence between transitive actions of G and “$\infty$-subgroups.” But now, an “$\infty$-subgroup” just means any $\infty$-group homomorphism $H \to G$.

So take the fundamental $\infty$-groups of the 2-sphere and the 3-sphere. Each of these is an $\infty$-subgroup of the other (induced by inclusion and Hopf fibration)? Are either of $\pi_{\infty}(S^3)//\pi_{\infty}(S^2)$ or $\pi_{\infty}(S^2)//\pi_{\infty}(S^3)$ interesting? I would imagine that the former is an $\infty$-groupoid equivalent to the fundamental $\infty$-groupoid of the circle.

Or do we have to be more careful working in Top?

Posted by: David Corfield on March 5, 2009 12:14 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Well, whatever $\pi_\infty(S^3)// \pi_\infty(S^2)$ is, it’s a transitive $\pi_\infty(S^3)$-space. Do you have an action of $\pi_\infty(S^3)$ on $S^1$ in mind?

Classical homotopy theorists would write $\pi_\infty(S^3)// \pi_\infty(S^2)$ as a bar construction $B(*,\Omega S^2,\Omega S^3)$. I feel like there may be more to say here, but I haven’t thought of it yet.

Posted by: Mike Shulman on March 5, 2009 5:11 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Okay, got it.

Via Galois theory, when $A$ is connected, spaces with an action of $\pi_\infty(A)$ can be identified with fibrations over $A$. The $\pi_\infty(A)$-space corresponding to a fibration is, guess what, its fiber.

Now given any map $f:A\to C$, we can pull back fibrations over $C$ to fibrations over $A$; this corresponds to restricting an action of $\pi_\infty(C)$ to an action of $\pi_\infty(A)$. We can also compose with $f$ to turn a fibration over $A$ into a fibration over $C$. (Well, it won’t necessarily be a fibration, but we can just replace it by one.) This corresponds to the $(\infty,1)$-categorical left adjoint to restriction along $\pi_\infty(f)$, which we might call (Kan) “extension” or “induction;” a homotopy theorist would write it as a bar construction $B(\Omega C, \Omega A, -)$. (There’s also a right adjoint, but we won’t need it.) A careful study of the relationships between these equivalences and transition functors can be found in my paper Parametrized spaces model locally constant homotopy sheaves (arXiv).

Now, the quotient $\pi_\infty(C)//\pi_\infty(A)$, written by the homotopy-theorist as $B(\star, \Omega A, \Omega C)$, is constructed by starting with $\pi_\infty(C)$ as a $\pi_\infty(C)$-space, restricting it to a $\pi_\infty(A)$-space, then quotienting by the $\pi_\infty(A)$-action, i.e. extending forward along the projection $\pi_\infty(A)\to \star$ to the trivial $\infty$-group.

Translating over into the world of fibrations, the fibration corresponding to $\pi_\infty(C)$ as a $\pi_\infty(C)$-space is the path fibration $P C \to C$, where $P C$ is the space of paths starting at the basepoint $c_0\in C$, and the map $P C \to C$ projects to the endpoint of the path. The fiber of this map, over the basepoint, is the loop space $\Omega C$ of paths starting and ending at the basepoint, so that makes sense. We now pull back this fibration along $f:A\to C$, and what we get is the space of pairs $(a,\gamma)$ where $a\in A$ and $\gamma$ is a path in $C$ from $f(a)$ to the basepoint $c_0$; in other words, it’s just the homotopy fiber of $f$ over the basepoint, equipped with its obvious projection to $A$. Next we compose with the projection $A\to \star$ from $A$ to a point, which doesn’t change the space but just forgets the projection to $A$. Now we have a fibration over $\star$, i.e. just a space. We then have to transform this fibration over $\star$ into a space with an action by $\pi_\infty(\star)=\star$ by taking its fiber; but of course this does nothing.

In conclusion, for any map $f:A\to C$ of connected based spaces, the $(\infty,1)$-categorical quotient $\pi_\infty(C)//\pi_\infty(A)$ is just the homotopy fiber of $f$. And if $f$ is itself a fibration, this is equivalent to its ordinary fiber. The $\infty$-group $\pi_\infty(C)$ acts on this in the standard way, by transporting along paths in $C$; the action is transitive because $A$ is connected.

So, if we start with the Hopf fibration $S^3\to S^2$, then the quotient $\pi_\infty(S^2)//\pi_\infty(S^3)$ is just the fiber $S^1$. I read what you wrote as saying that $\pi_\infty(S^3)//\pi_\infty(S^2)$ should be $S^1$, but maybe this is actually what you meant; if so, you’re right! The other one $\pi_\infty(S^3)//\pi_\infty(S^2)$ is the homotopy fiber of the inclusion $S^2\to S^3$, whatever that is.

In the other direction, suppose I start with a connected based space $C$ and a space $X$ having a transitive action of $\pi_\infty(C)$, and I want to find a map $f:A\to C$ of connected based spaces such that $X\simeq \pi_\infty(C)//\pi_\infty(A)$, or equivalently that $X$ is the homotopy fiber of $f$. To use the above approach, I should start by translating $X$ into a fibration over $C$—but then I’m already done! This is a fibration with fiber $X$, and its total space is connected since the action is transitive.

We can sort of give an explicit description of that total space: if $X$ has an action of $\Omega C$, we can make the bar construction $B(\star,\Omega C, X)$. This isn’t exactly a fibration over $C$, but it comes with a map to $B(\star, \Omega C, \star) = B\Omega C$, which is equivalent to $C$ (since $C$ is connected), and its fiber is indeed $X$. Presumably we could trace through the equivalences and figure out why this looks like the “stabilizer” of a point in $X$.

Posted by: Mike Shulman on March 5, 2009 6:22 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

It may seem weird that we get the quotient $\pi_\infty(C)//\pi_\infty(A)$ by taking a fiber of $f:A\to C$, since fibers are dual to quotients. At least, it seems weird to me! But we can reassure ourselves by dropping back to low dimensions.

Let $G$ be an ordinary discrete 1-group and let $i:H\hookrightarrow G$ be a subgroup. Now the claim is that we can recover $G/H$ as the essential fiber of the map $B i:B H \to B G$ of groupoids. Of course, now I’m using $B G$ for the groupoid with one object $\star$ and $B G (\star,\star) = G$.

Well, in general, for a map $f:A\to C$ of groupoids, an object of the essential fiber over $c_0\in C$ is an object $a\in A$ and an isomorphism $f(a)\cong c_0$, and a morphism in the essential fiber is a morphism $a\to a'$ in $A$ making the obvious triangle commute in $C$. For $B i:B H \to B G$, this means that an object of the essential fiber over $\star\in B G$ is the unique object $\star \in B H$ together with an isomorphism $B i(\star) \cong \star$—in other words, an element of $G$. And a morphism in the essential fiber from $g_1$ to $g_2$ will be a morphism $\star\to\star$ in $B H$—that is, an element of $H$—such that $g_1 h = g_2$ in $G$. Thus, the essential fiber is equivalent to the discrete set $G/H$, which is what we wanted.

This still feels like magic to me.

Posted by: Mike Shulman on March 5, 2009 7:07 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Let $G$ be an ordinary discrete 1-group and let $i : H \hookrightarrow G$ be a subgroup. Now the claim is that we can recover $G/H$ as the essential fiber of the map $\mathbf{B} i : \mathbf{B}H \to \mathbf{B}G$ of groupoids.

Yes, this is a nice fact. Let me just remark that we have discussed this here from time to time in slightly different but equivalent language:

As described at homotopy limit and at generalized universal bundle one way to compute the essential fiber of a morphism $g : X \to \mathbf{B}G$ by an ordinary pullback is to pull back the universal $G$-bundle $\mathbf{E}_{pt} G$ in its $\infty$-groupoid incarnation.

Dually, for $S$ any pointed object and $\rho : \mathbf{B}H \to S$ a morphism of pointed objects, the homotopy fiber is the pullback of $\mathbf{E}_{pt} S$ and computes the action groupoid $pt_S//H$ of $H$ acting on the point of $S$, as described there.

The situation you just mentioned is the combination of these two cases: that where we have a morphism $\mathbf{B}H \to \mathbf{B}G$ of groups:

here the homotopy fiber is both:

a) the $G$-bundle associated to the universal $H$-bundle by this morphism in its $\infty$-groupoid incarnaton;

- the action groupoid $G//H$.

Posted by: Urs Schreiber on March 5, 2009 8:53 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Very neat! I didn’t see those connections, but of course you’re right. It still feels like magic to me that we compute a quotient by computing a fiber, but at least now I see that it’s an instance of a familiar general phenomenon.

Posted by: Mike Shulman on March 5, 2009 11:40 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

It still feels like magic to me that we compute a quotient by computing a fiber, but at least now I see that it’s an instance of a familiar general phenomenon.

Yes, indeed. Just for the record, some may recall that a while ago I had been wondering about this in the context of not $\infty$-groupoids but their infinitesimal version, $L_\infty$-algebroids.

I wanted to get a good definition of $\infty$-action of one $L_\infty$-algebroid $g$ on another $h$.

The direct way to do this would be to define something like an $L_\infty$-algebroid $end(h)$ of $h$ and then look at $L_\infty$-morphisms $g \to end(h)$.

But it turns out that defining $end(h)$ in the right way is at best painful. Well, I wouldn’t be shocked if somebody presents me a nice solution, but in all the discussion we had about this here no really good general such seems to have surfaced.

So at some point I decided to go the other route and instead of defining the action directly, define the fibration that it should induce, i.e. the “action $L_\infty$-algebroid” and let the action $L_\infty$-algebroid define the action instead of the other way round.

This is described in p. 9, section Actions and their homotopy quotients of On $\infty$-Lie, more details from p. 21 on (from a talk I once gave).

Later it turned out that the same definition, but in in rather different language, was given by Jonathan Block. I talk about that in the entry Block on $L_\infty$-module categories.

(With Jim Stasheff and Hisham Sati we have an almost finished article where we apply this kind of thing to $L_\infty$-connections to reproduce various formulas known in String theory. Much to the infuration of at least Jim I kept postponing finishing this article because I felt I needed to better understand the homotopy-theoretic abstract nonsense in which it lives. It seems I am getting closer…)

Posted by: Urs Schreiber on March 6, 2009 1:33 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Oh yes, the wrong way round.

You’d kinda want $\pi_{\infty}(S^3)//\pi_{\infty}(S^2)$ to be an ‘inverse’ to $S^1$.

Posted by: David Corfield on March 5, 2009 7:46 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Sorry to keep adding more while probably no one else has yet digested what I’m saying, but this is too nice not to share. I just realized that the equivalence between $G$-actions and spaces over $B G$ is not limited to classical homotopy theory; it’s true in any exact $(\infty,1)$-category.

In an exact 1-category $K$, regular epis are automatically descent morphisms, meaning that $X\mapsto K/X$ is a stack for them. That means that for an equivalence relation $R\to A\times A$, objects over the quotient $A/R$ are equivalent to objects over $A$ with an “action” by $R$, which amounts to being constant along equivalence classes.

Now, the same should be true in an exact $(\infty,1)$-category, for the quotient of a groupoid. But recall what I said here about the construction of $B G$ from a group object $G$ in any exact $(\infty,1)$-category. That means that $1\to B G$ is a descent morphism, so that objects over $B G$ can be identified with objects over $1$ (that is, just objects) having an action by $G$. But this is precisely what we wanted.

That means that my argument above that identifies the quotient $G//H$ with the fiber of $B H \to B G$ should also work in any exact $(\infty,1)$-category. In particular, it should work in the $(\infty,1)$-category of stacks of $\infty$-groupoids on the category of manifolds.

Posted by: Mike Shulman on March 5, 2009 8:27 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

That means that my argument above that identifies the quotient $G//H$ with the fiber of $\mathbf{B}H \to \mathbf{B}G$ should also work in any exact $(\infty,1)$-category. In particular, it should work in the $(\infty,1)$-category of stacks of $\infty$-groupoids on the category of manifolds.

Compare section 2.2 of

J. F. Jardine, Cocycle categories

which has a proof of this kind of statement for $\infty$-stacks with $\infty$-stacks modeled by simplicial presheaves.

Posted by: Urs Schreiber on March 5, 2009 9:03 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Excellent! I also like seeing it as a consequence of exactness, though.

Posted by: Mike Shulman on March 6, 2009 1:39 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

What’s a nice nontrivial example of an $\infty$-group?

For a topological space $X$, let $\Pi X$ be its fundamental $\infty$-groupoid. For any point $x \in X$ we have the $\infty$-group $\Pi (X) (x,x)$ of automorphisms of $x$.

This $\Pi(X)(x,x)$ is an $\infty$-group which subsumes all the ordinary homotopy groups $\pi_n(X,x)$ of $X$ at $x$, together with all their actions and interrelations on each other.

This is the canonical example but also a bit tautological: equivalently one could say: an $\infty$-groups is a connected topological space.

More generally: for any $\infty$-groupoid $G$ and every object $a \in G$ the thing $G(a,a)$ is an $\infty$-group.

That, too, is a bit tautological, of course.

Posted by: Urs Schreiber on March 5, 2009 2:31 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Old-fashioned homotopy theorists would call $\Pi(X)(x,x)$ the loop space $\Omega X$ of the based space $(X,x)$. And in fact, every example is of this form: given any $\infty$-group $G$ there is a space $B G$ such that $G\simeq \Omega B G$. This is a very classical theorem in homotopy theory, although classical homotopy theorists would say “grouplike $A_\infty$-space” rather than “$\infty$-group.”

It’s also a special case of the statement that the $(\infty,1)$-category of spaces (= $\infty$-groupoids) is exact: an $\infty$-group $G$ is a group object in spaces, hence $G \;\rightrightarrows\; 1$ is a groupoid object, and so (by exactness) it is the kernel of some map $1\to B G$, i.e. we have a pullback $\array{G & \overset{}{\to} & 1\\ \downarrow && \downarrow\\ 1& \underset{}{\to} & B G}$ which says exactly that $G\simeq \Omega B G$.

Posted by: Mike Shulman on March 5, 2009 4:35 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

So is the lesson that infinitely categorified Klein geometry (at least the ($\infty$, 1) variety) is just homotopy theory for a pair of connected spaces with base points and a mapping between them?

Double cosets are next on the agenda.

Posted by: David Corfield on March 5, 2009 4:50 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Yes, I think so. That shouldn’t really be a surprise; most $(\infty,1)$-category theory is just a new language for homotopy theory. Although, like any new language, it does suggest new ways of looking at old things. I doubt that homotopy theorists ever thought of a map between connected based spaces as analogous to a Klein geometry.

Posted by: Mike Shulman on March 5, 2009 5:27 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

In light of what I said above, it might be more accurate to say that Klein $(\infty,1)$-geometry is the homotopy theory of pairs of connected based spaces and fibrations between them—the corresponding homogeneous space then being the fiber of the fibration.

Of course, things get more complicated when you want to add in a smooth structure. To do that in homotopy theory, you’d probably have to think about, say, simplicial (pre)sheaves on the category of manifolds.

Posted by: Mike Shulman on March 5, 2009 8:15 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

David wrote:

infinitely categorified Klein geometry […] is just homotopy theory

Mike replied

That shouldn’t really be a surprise; most $(\infty,1)$-category theory is just a new language for homotopy theory

Maybe a remark on this point:

Saying that

Working in the $(\infty,1)$-category of $(\infty,0)$-categories (= $\infty$-groupoids) is nothing but working with topological spaces.

is the analog of saying

Working in the $1$-category of 0-categories (= sets) is nothing but doing set theory.

In both cases one can ask: So what’s the point? Why do we talk about ($\infty$-)categories and not just about sets and spaces.

The answer is both cases is: because the category theory allows us to consider parameterized sets and parameterized spaces: we have the Yoneda lemma and (pre)sheaves of sets and (pre)sheaves spaces, hence (Grothendieck) topoi and (Rezk-Lurie) $\infty$-topoi.

This allows us to do things that set theorists and homotopy theorists can not do with plain sets and plain topological spaces: add extra structure, for instance smooth structure, by considering presheaves over suitable test objects.

So a central point is that there are (presheaf) $\infty$-topoi. Whence the title of Lurie’s book, I suppose.

(E.g. his remarks at the beginning of section 5.1).

Posted by: Urs Schreiber on March 6, 2009 11:06 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Yes!

Of course, set theorists and homotopy theorists do study parametrized things and add extra structure. Set theorists call it “forcing.” Homotopy theorists do “equivariant homotopy theory” and “parametrized homotopy theory” and talk about simplicial (pre)sheaves. The merits of the categorical approach are that we have a unified framework in which to do all of these things, and that when we phrase classical things correctly in the categorical language, we don’t have to re-prove them in all the other situations. And probably others.

Posted by: Mike Shulman on March 7, 2009 4:08 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Ahem

Just note that the link at that comment to the ‘notes’ has moved to here, due to departmental server migration.

Posted by: David Roberts on March 11, 2009 1:38 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

As an added inducement to get you onboard, Mike, with a categorified Erlanger Programme we can categorify Tarski’s What are Logical Notions?, using what Todd was telling us about here and here.

Posted by: David Corfield on March 3, 2009 4:57 PM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

I’m glad we’re taking up this theme again. I’m pretty tired, so all I have to say tonight is: “a moment of insight near the Nine Zigzag Bridge in Shanghai” sounds pretty cool.

Cooler than if you also know I was in a Starbucks packed with tourists trying to recover from the heat.

Posted by: John Baez on March 3, 2009 6:45 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

What sort of object do we expect the stabiliser to be? Certainly the stabiliser of a point is a sub-2-group (i.e. a faithful functor to the group), but can we allow more general things? Here is a silly idea that comes from just copying my own private definition of the stabiliser of a point, replacing the point with something more general.

Let $G$ be a 2-group, $i:Y \to X$ a sub-thingie of the groupoid $X$, which has an action $a:X \times G \to X$. I guess we would want $i$ to be faithful.

Consider the weak pullback $i\downarrow (a\big|_Y)$ of $i$ along the composite $a\big|_Y:Y \times G \stackrel{i\times id}{\to} X \times G \stackrel{a}{\to} X.$

This admits the structure of a double groupoid where one of the directions is as weak as the 2-group (so strict 2-group gives a double groupoid, and a weak 2-group gives a pseudo double groupoid). There is a functor $j$ from $i\downarrow (a\big|_Y)$ to $G$ given by the composite of the canonical arrow to $Y\times G$ and projection to $G$. I think that this double groupoid has connections (at least in the strict case) so is the ‘same as’ a 2-groupoid (and bigroupoid in the non-strict case, once we know that this would even work. Has it been done anywhere in the literature?) In the case that $Y=\ast$, the double groupoid reduces to a 2-group.

How much $j$ could be thought of as a sub-2-group of $G$ is debatable, but some homotopy-theoretic considerations may be able to guide these ill-formed thoughts into something sensible.

Posted by: David Roberts on March 4, 2009 1:32 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

I don’t understand. The pullback you mention presumably takes place in $Gpd$, so it will be a groupoid. Where does the double groupoid come from?

Posted by: Mike Shulman on March 5, 2009 3:00 AM | Permalink | Reply to this

### Re: The Stabilizer of a Subcategory

Apologies for the delay, public holiday here in Australia, plus the one day a week I’m not at uni. All of what is below may be less than optimal to the problem at hand, but the structure seems interesting.

Where does the double groupoid come from?

The pullback admits the structure of a double groupoid as follows:

Recall: we have a faithful functor $i:Y \to X$ between strict $G$-groupoids for $G$ a 2-group ($G$ strict for now, as well as $i$, so it is really equivariant). The weak pullback has as objects

$(y;\gamma;y',g) \in Y_0\times X_1 \times Y_0 \times G_0$

such that $\gamma:i(y) \to i(y')\cdot g$. Morphisms between $(y_1;\gamma_1;y_1',g_1)$ and $(y_2;\gamma_2;y_2',g_2)$ are triples $(a;a',h) \in Y_1 \times Y_1 \times G_1$ such that $\gamma_2 \circ i(a) = i(a')\cdot h \circ \gamma_1$ It helps now to draw the commuting square I could not (Oh when will I learn to do diagrams in here?), with the $a$’s vertical.

The objects of the pullback groupoid become horizontal arrows, we let the objects of the double groupoid be just the objects of $Y$ and the arrows of the pullback groupoid become the squares in the double groupoid.

Then vertical composition of squares is just composition in the pullback groupoid (componentwise composition of $(a;a',h)$’s). The horizontal composition is akin to semidirect product multiplication.

The horizontal composition of $(y;\gamma;y',g)$ and $(y';\delta;y'',k)$ is $(y;\delta\cdot g \circ \gamma;y'',kg).$ That this is associative uses strictness of $G$, $i$ and the action. I haven’t thought seriously about precisely to say what thing I get for weak $G$ etc. A pseudo double groupoid maybe.

It may be worth noting that the vertical category in this double category is just $Y$.

There is a functor to $G$, which is considered as a double category as follows: There is only one object, the vertical arrows are trivial, the horizontal arrows are the objects of $G$, and the squares are the arrows of $G$. Clearly this is the 2-groupoid with one object corresponding to $G$ considered as a double groupoid in the usual way.

My claim above that this has connections is obviously wrong, as it isn’t edge symmetric.

Interesting. That double groupoid is sort of a “$G$-equivariant kernel” of $i:Y\to X$. In particular, although it doesn’t have connections, it does have companions and conjoints for vertical arrows (which are, of course, the same in a double groupoid). And if $Y=X$, its quotient (aka its reflection into groupoids, left adjoint to forming the double groupoid of commutative squares in a groupoid) is the weak quotient $X//G$.