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Stream: learning: questions

Topic: groupoid of objects equipped with automorphism

John Baez (Dec 12 2024 at 17:58):

Suppose you have a groupoid $X$. Then there's groupoid $A(X)$ where

• an object is an object $x \in X$ together with an automorphism $g: x \to x$
• a morphism from $g: x \to x$ to $h: y \to y$ is an isomorphism $f : x \to y$ making the obvious square commute: $f g = h f$

What are some good ways to think about this construction, or interesting facts about it?

Some initial thoughts:

1. We can get a groupoid this way if $X$ is any category, but it only depends on the underlying groupoid of that category, called its [[core groupoid]], so we might as well just talk about groupoids.
2. $A(X)$ is the subcategory of the [[arrow category]] of $X$ where the objects are only endo-arrows and the morphisms are only pairs of arrows that are equal.
3. When $X$ is the one-object groupoid corresponding to a group $G$, $A(X)$ is the groupoid whose objects are group elements and whose morphisms $f: g \to h$ are group elements such that $f g f^{-1} = h$. It's the [[weak quotient]] of $G$ by itself, where it acts on itself by conjugation, so it's often called $G//G$. Since $G$ acts on itself as group automorphisms, $G//G$ is actually a [[2-group]], called the [[inner automorphism 2-group]] of $G$.

Mike Shulman (Dec 12 2024 at 18:00):

$A(X)$ is also the [[powering]] of $X$ by the circle groupoid, just as the arrow category is the powering by the interval category.

Mike Shulman (Dec 12 2024 at 18:01):

And if $X$ is a category rather than a groupoid, it sits inside the analogous powering in Cat, which has the same objects but more morphisms where $f$ isn't required to be invertible.

John Baez (Dec 12 2024 at 18:14):

Nice! In your reply, what 2-category (or if we have to go there, $\infty$-category) are you thinking of as powered by what?

John Baez (Dec 12 2024 at 18:15):

If it's Gpd powered by Gpd, do people say "powered" or "internal hom"?

Fumbling around here....

Jean-Baptiste Vienney (Dec 12 2024 at 18:16):

Maybe not very useful, but if $f:x \rightarrow x$ is an endomorphism in a category $\mathcal{C}$, then we get an endofunctor $1_f:\mathcal{C} \rightarrow \mathcal{C}$ defined by $1_f(a):=x$ for every object $a$ and $1_f(u):=f$ for every morphism $u$. If we have two endomorphisms $g:x\rightarrow x$ and $h:y \rightarrow y$, then a morphism $f:x \rightarrow y$ such that the obvious square commute: $fg=hf$ is exactly a natural transformation from $1_g$ to $1_h$.

Thus a morphism from $g:x \rightarrow x$ to $h:y \rightarrow y$ in your groupoid $X$ is exactly a natural transformation from $1_g$ to $1_h$.

It explains why these morphisms feel like natural transformations: because they are natural transformations.

Jean-Baptiste Vienney (Dec 12 2024 at 18:17):

Oops, no sorry, it doesn't work ahah (the $1_f$ doesn't preserve identities).

Mike Shulman (Dec 12 2024 at 18:18):

The general notion of power is that if we have a category $C$ enriched over $V$, there can be a power $X^K$ for $X\in C$ and $K\in V$. In your construction, $C=\rm Gpd$ and you could take either $V=\rm Cat$ or $V=\rm Gpd$. And you're right that when $C=V$, the power is the same as the internal-hom.

Mike Shulman (Dec 12 2024 at 18:19):

So maybe I should just have said that it's the functor groupoid out of the circle!

Mike Shulman (Dec 12 2024 at 18:19):

(This is the way in which the morphisms are natural transformations.)

Mike Shulman (Dec 12 2024 at 18:20):

I like to think of it as the power because then it generalizes nicely to other 2-categories, including Cat.

John Baez (Dec 12 2024 at 18:22):

I'm glad you said "powering" instead of "internal-homming" both because it opened my mind and because it sounds nicer.

James Gilles (Dec 12 2024 at 19:12):

Applications perspective: if X smells like a concrete category, then the objects of A(X) resemble homomorphic encryption schemes. Really this is true for isomorphisms in any concrete category.

David Egolf (Dec 12 2024 at 19:45):

If we start with the fundamental groupoid $X$ of some topological space, I wonder what the morphisms in $A(X)$ are intuitively describing. In $X$, each automorphism is an endpoint-preserving homotopy equivalence class of loops, and $f:x \to y$ is an endpoint-preserving homotopy equivalence class of paths from $x$ to $y$. Then $f \circ g = h \circ f$ says that doing loop $g$ and then path $f$ gives the same result (up to endpoint-preserving homotopy) as doing path $f$ first and then the loop $h$.

David Egolf (Dec 12 2024 at 19:47):

Maybe there could be a nice way to visualize this condition, in terms of some specific paths?

James Gilles (Dec 12 2024 at 19:51):

publicatioin_quality.png

John Baez (Dec 12 2024 at 20:57):

David Egolf said:

If we start with the fundamental groupoid $X$ of some topological space, I wonder what the morphisms in $A(X)$ are intuitively describing. In $X$, each automorphism is an endpoint-preserving homotopy equivalence class of loops, and $f:x \to y$ is an endpoint-preserving homotopy equivalence class of paths from $x$ to $y$. Then $f \circ g = h \circ f$ says that doing loop $g$ and then path $f$ gives the same result (up to endpoint-preserving homotopy) as doing path $f$ first and then the loop $h$.

Nice question. Maybe it's the fundamental groupoid of the $[S^1, Y]$ where $Y$ is that topological space (alas, you didn't give it a name :cry:). $[S^1, Y]$ is the topological space of continuous maps from the circle to $Y$: remember, if we use a [[convenient category of topological spaces]], it will be cartesian closed, so the space of continuous maps from $S^1$ to $Y$ gets a topology.

John Baez (Dec 12 2024 at 20:58):

$[S^1,Y]$ is usually called the [[free loop space]] of $Y$ and denoted $LY$.

John Baez (Dec 12 2024 at 21:00):

I'm not sure we have

$\Pi_1[S,T] \simeq [\Pi_1 S ,\Pi_1 T]$

where $A,B$ are nice topological spaces and $\Pi_1$ is the fundamental groupoid; if it is then I think my guess is correct.

Mike Shulman (Dec 12 2024 at 21:00):

It certainly looks like that, but I suspect that it isn't exactly the same, because there was quotienting that happened in constructing the fundamental groupoid before taking the loops.

Mike Shulman (Dec 12 2024 at 21:03):

For instance, suppose $Y$ is a simply connected space with nontrivial $\pi_2$, like the 2-sphere $S^2$. Then $\Pi_1 Y$ is trivial, and hence so is $A(\Pi_1 Y)$. But $LY$, while it may be connected, seems unlikely to have trivial $\pi_1$. The lack of basepoints makes it tricky, but intuitively $\Pi_1(LY)$ should be saying something about $\pi_2(Y)$.

John Baez (Dec 12 2024 at 21:13):

Yes, that sounds right. There may be a map one way between these two guys:

$\Pi_1[S,T] \simeq [\Pi_1 S ,\Pi_1 T]$

maybe from left to right, since you're talking about how the right side has lost information.

Luckily for me my actual examples of interest aren't coming from topology - although every groupoid comes from topology if you want it to.

Mike Shulman (Dec 12 2024 at 21:16):

Another reason for the map to go left-to-right is that the thing on the right side has a mapping-in universal property.