Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: symmetric elements and group(oid) actions

David Egolf (Dec 07 2024 at 16:59):

Let $\rho:G \to \mathrm{Aut}_C(S)$ be a group action, where $S$ is an object of some category $C$. If we consider the case where $S$ has "elements" (for example if $S$ is a set or a vector space), then an element $s$ is "symmetric" with respect to the action if $\rho(g)(s) = s$ for all $g \in G$. For example, if $G$ acts to rotate shapes in space about some axis, then the shapes in that space that are rotationally symmetric about that axis will be invariant under this action. As another example, if $G$ acts to permute the "variable" symbols in a polynomial, then the "symmetric polynomials" are the ones invariant under this action.

I recently learned that when $G$ is finite and $S$ is a vector space, one can construct a "symmetric" element from any element $s$ of $S$. We do this by letting each element of $G$ act on $s$ and then we add up (or "average") the results. That's really cool!

I wonder if any of these ideas can be generalized to the groupoid setting. A "groupoid action" is presumably just a functor from a groupoid $G$ to some category $C$. Some questions in this direction:

• If the objects of $C$ have elements, can we obtain a reasonable notion of a "symmetric element" with respect to a groupoid action?
• If so, can we come up with a way to "symmetrize" any element that belongs to one of the objects that $G$ is acting on?

David Egolf (Dec 07 2024 at 16:59):

More broadly, what I'm really interested in is this: Can we generalize the notion of a "symmetric" or "invariant" thing?

John Baez (Dec 07 2024 at 17:13):

This is a fun puzzle.

I find it easier to think about the case where $G$ is a group because I'm constantly thinking about invariant elements in this case. So for starters, at least, suppose we have a functor $\rho: BG \to C$ where $BG$ is the one-object category corresponding to $G$. The image of the one object in $BG$ is some object $c \in C$, and we then call $\rho$ an action of the group $G$ on the object $c$.

Next, what's an 'element' of $c$ and what's an 'invariant' element?

John Baez (Dec 07 2024 at 17:14):

As you've probably heard, a generalized element of $c$ is simply any morphism $f: a \to c$ for any object $a \in c$. I think this is a good level of generality for this puzzle - i.e., I see no good reason to reduce the level of generality.

John Baez (Dec 07 2024 at 17:22):

It's easy to say when $f$ is invariant: it's invariant when

$f = \rho(g) \circ f$

for all $g \in G$.

(Remember, for each $g \in G$ we have $\rho(g) \colon c \to c$, so we can postcompose it with $f: a \to c$ and get a new morphism $\rho(g) \circ f : a \to c$. We can check that this gives an action of $G$ on the set $\text{hom}(a,c)$... in the usual sense of a group acting on a set! Thus, my definition above merely says that $f \in \text{hom}(a,c)$ is an invariant element of this set... in the usual sense of an invariant element of a set acted on by a group!)

John Baez (Dec 07 2024 at 17:24):

Now, when can we 'average' an arbitrary element $f: a \to c$ and get an invariant element? I know how to do it when

1. our group is finite
2. our category $C$ is enriched over $\mathsf{Vect}_k$, the category of vector spaces over some field $k$ with characteristic zero.

In this case we can average $f: a \to c$ this way:

$\displaystyle{ \frac{1}{|G|} \sum_{g \in G} \rho(g) \circ f }$

and this is an invariant morphism from $a$ to $c$.

John Baez (Dec 07 2024 at 17:43):

Note I need $G$ to be finite for $|G|$ to be finite, and I need a field $k$ of characteristic zero to be sure that $1/|G|$ makes sense for for any finite group $G$.

Notice that if we were working in a field of characteristic 17, like $k = \mathbb{Z}/17\mathbb{Z}$, then all hell would break loose in the above equation if $|G|$ were divisible by 17.

This is actually a well-known problem in group representation theory: if we're studying the representations of a finite group over a field of characteristic $p$, it becomes a lot harder when the order of our group is divisible by $p$. Some very smart people have burnt out a lot of brain cells dealing with this.

John Baez (Dec 07 2024 at 17:45):

Finally I'll say there's a more general averaging trick we often use when $G$ is not a finite group but instead a compact topological group (like the group of rotations of $\mathbb{R}^n$). Essentially, "compact" is still pretty close to "finite". This makes the representation theory of compact topological groups delightfully nice compared to the noncompact ones.

Amar Hadzihasanovic (Dec 07 2024 at 17:48):

Isn't the "unnormalised" sum also invariant? What role does normalisation play here?

John Baez (Dec 07 2024 at 18:06):

The unnormalized sum is also invariant, and sometimes that's good enough, but often we want a splitting of the inclusion of invariant elements in all elements, and that's what the normalization gives.

David Egolf (Dec 07 2024 at 18:32):

Awesome! That makes a lot of sense!

Let me check that post-composition indeed gives an action on the set $\mathrm{hom}(a,c)$ of $a$-shaped generalized elements of $c$. I'll write the result of acting on $f:a \to c$ by $g \in G$ as $g.f$. Then we need $(gh).f = g.(h.f)$ and $e_G.f = f$ for all $f:a \to c$ and all $g,h \in G$. (Here $e_G$ is the identity element of $G$).

Since $\rho:BG \to C$ is a functor, we have $\rho(e_G) = \rho(1_{BG}) = 1_{\rho(BG)} = 1_c$, so that $e_G.f$ amounts to post-composition by the identity morphism of $c$. Thus, $e_G.f = f$ as desired.

We also need $(gh).f = g.(h.f))$. This amounts to needing $\rho(gh) \circ f = \rho(g) \circ (\rho(h) \circ f)$. Since $\rho$ is a functor and composition of morphisms is associative, the two sides of this equation are equal.

David Egolf (Dec 07 2024 at 18:36):

I suspect that this approach can be used to talk about an action on subobjects of a given shape. Instead of taking all $a$-shaped generalized elements, we could restrict attention to monomorphic $a$-shaped generalized elements. Post composition by an isomorphism should send a monomorphism to a monomorphism.

If that works, then we can talk about "symmetric/invariant" subobjects, which seems pretty cool! And, in the right setting, I suspect we can "symmetrize" subobjects using the procedure outlined above.

John Baez (Dec 07 2024 at 18:50):

All that sounds right, except that I'm scared that symmetrizing a monomorphism may not give a monomorphism. Why would it?

David Egolf (Dec 07 2024 at 19:34):

Ah, good point! I suspect that in general a monomorphism wouldn't necessarily symmetrize to a monomorphism. So what I suggested above about symmetrizing a subobject probably doesn't work in general.

David Egolf (Dec 07 2024 at 19:41):

The problem is that the sum of two monomorphisms is probably not a monomorphism in general. I'm trying to cook up an example of this, currently.

David Egolf (Dec 07 2024 at 19:43):

Ah, chatGPT helpfully suggests the maps $f,g:\mathbb{R} \to \mathbb{R}$ where $f(x)=x$ and $g(x)=-x$ for all $x \in\mathbb{R}$. Then $f$ and $g$ are both injective - and hence monomorphisms in $\mathsf{Set}$ - but their sum is the zero function, which is very non-injective.

David Egolf (Dec 07 2024 at 19:52):

Ideally, though, I'd like to find an example in a category that's enriched over $\mathsf{Vect}_\mathbb{k}$ for some field $\mathbb{k}$ of characteristics zero. However, I think we can view these $f$ and $g$ as morphisms in $\mathsf{Vect}_\mathbb{R}$, which I think is enriched over $\mathsf{Vect}_\mathbb{R}$.

David Egolf (Dec 07 2024 at 20:24):

I'm still curious regarding whether we can generalize this stuff to groupoid actions. My intuition is that a groupoid contains multiple "kinds" of transformations that we can perform on objects of interest. But I want to talk about invariant elements... so maybe we can consider functors $F:G \to C$ where each object of the groupoid $G$ is mapped to the same object of the category $C$ by $F$.

David Egolf (Dec 07 2024 at 20:26):

Let's assume that each object of the groupoid $G$ is mapped to $c \in C$ by $F$. Then an $a$-shaped generalized element of $c$ is a morphism $f:a \to c$. For any morphism $g \in G$, we have $F(g):c \to c$ so we can still consider acting on generalized elements by post composition: $f \mapsto F(g) \circ f$.

An invariant element $f:a \to c$ is one such that $F(g) \circ f = f$ for all $g \in G$.

David Egolf (Dec 07 2024 at 20:30):

Given an arbitrary generalized element $f:a \to c$, and assuming that $C$ is enriched over $\mathsf{Vect}_\mathbb{k}$ for $\mathbb{k}$ a field of characteristic zero, can we make an invariant element from $f$ using a recipe similar to that described above?

David Egolf (Dec 07 2024 at 20:34):

$F(g') \circ (\sum_{g \in G}F(g) \circ f) = \sum_{g \in G}F(g') \circ F(g) \circ f$ because composition is bilinear in $C$. I would next like to rewrite $F(g') \circ F(g)$ as $F(g \circ g')$ but that's not possible in general because $g$ and $g'$ might not be composable in $G$.

David Egolf (Dec 07 2024 at 20:35):

So, I guess the question is, do we have $\sum_{g \in G}F(g') \circ F(g) \circ f = \sum_{g \in G}F(g) \circ f$ for any $g' \in G$ and any $f:a \to c$ in $C$?

David Egolf (Dec 07 2024 at 20:39):

I suspect that contemplating a small example will help make this clearer.

John Baez (Dec 09 2024 at 00:51):

David Egolf said:

I'm still curious regarding whether we can generalize this stuff to groupoid actions. My intuition is that a groupoid contains multiple "kinds" of transformations that we can perform on objects of interest. But I want to talk about invariant elements... so maybe we can consider functors $F:G \to C$ where each object of the groupoid $G$ is mapped to the same object of the category $C$ by $F$.

I find that unnatural. Why? I have to psychoanalyze myself.

• First, it's not something that I've seen happen, at least not consciously. It's easy to map a one-object groupoid $G$ to a category $C$ in a way where each object of $G$ gets mapped to the same object of $C$ :upside_down:, and that's what we call a group action. But it's less likely, when you map a multi-object groupoid to $C$, that all the objects get mapped to the same one!

• Second, I feel the answer to this question is "not usually":

So, I guess the question is, do we have $\sum_{g \in G}F(g') \circ F(g) \circ f = \sum_{g \in G}F(g) \circ f$ for any $g' \in G$ and any $f:a \to c$ in $C$?

• Third, say you map a groupoid $G$ to a category $C$ via a functor $F: G \to C$ and all the objects of $G$ get mapped to $c \in C$. Then $F$ should factor through some one-object groupoid $G'$ formed from $C$ by glomming together all the objects of $G$. I haven't thought about that construction before, but I feel should exist: it should be some sort of colimit. If this true, then $F$ equals some composite
  $G \xrightarrow{Q} G' \xrightarrow{F'} C$
  so we have an action of some group on $C$, namely the group corresponding to the 1-object groupoid $G'$ If so, the subject you're proposing to study gets reduced to the study of group actions and the study of how we form the groupoid $G'$.

David Egolf (Dec 09 2024 at 19:35):

John Baez said:

David Egolf said:

I'm still curious regarding whether we can generalize this stuff to groupoid actions. My intuition is that a groupoid contains multiple "kinds" of transformations that we can perform on objects of interest. But I want to talk about invariant elements... so maybe we can consider functors $F:G \to C$ where each object of the groupoid $G$ is mapped to the same object of the category $C$ by $F$.

I find that unnatural. Why? I have to psychoanalyze myself.

• First, it's not something that I've seen happen, at least not consciously. It's easy to map a one-object groupoid $G$ to a category $C$ in a way where each object of $G$ gets mapped to the same object of $C$ :upside_down:, and that's what we call a group action. But it's less likely, when you map a multi-object groupoid to $C$, that all the objects get mapped to the same one!

That makes sense! I admit I don't have a compelling example in mind. But I had vaguely hoped it could be an interesting way to study several related group actions. I could imagine a situation where multiple groups act on a single thing. Two group elements from different groups can't be composed, but what those group elements do can be composed.

David Egolf (Dec 09 2024 at 19:39):

John Baez said:

• Third, say you map a groupoid $G$ to a category $C$ via a functor $F: G \to C$ and all the objects of $G$ get mapped to $c \in C$. Then $F$ should factor through some one-object groupoid $G'$ formed from $C$ by glomming together all the objects of $G$. I haven't thought about that construction before, but I feel should exist: it should be some sort of colimit.

That makes some intuitive sense to me. I could imagine "freely combining" several groups to form a bigger one, where the composition of the elements from the different groups is presumably pretty boring.

David Egolf (Dec 09 2024 at 19:44):

Let me see if I can figure out how to build a corresponding one-object groupoid $G'$ in the case where our acting groupoid $G$ just consists of two groups (with no morphisms between them).

David Egolf (Dec 09 2024 at 19:55):

(...I wonder if this $G'$ is "universal" in the sense that it can uniquely factor any groupoid action of the form under consideration. We can consider the full subcategory of the coslice category $G/\mathsf{Cat}$ where we take only the objects that send every object of $G$ to the same object. Then we can ask if this category has an initial object.)

David Egolf (Dec 09 2024 at 20:31):

I think (but haven't carefully checked) that in this case we form the one object groupoid $G'$ by taking the coproduct (in $\mathsf{Grp}$) of the two groups in $G$ and then viewing that as a one object groupoid. Presumably things would get more complicated if we have morphisms between distinct objects in our groupoid.

John Baez (Dec 09 2024 at 21:20):

David Egolf said:

I could imagine a situation where multiple groups act on a single thing. Two group elements from different groups can't be composed, but what those group elements do can be composed.

For several groups to act on the same thing, is the same as for their coproduct to act on that thing. So, we never need to think about more than one group acting on something if we don't want to. But maybe we want to.

The coproduct of groups is, among ordinary mortals, usually called their free product, and for two groups it's written $G \ast H$. It's widely used when we study groups presented by generators and relations. If we take generators and relations for a group $G$ and generators and relations for a group $H$ and we "lump them together", we get generators and relations for $G \ast H$.

John Baez (Dec 09 2024 at 21:21):

David Egolf said:

Let me see if I can figure out how to build a corresponding one-object groupoid $G'$ in the case where our acting groupoid $G$ just consists of two groups (with no morphisms between them).

David Egolf (Dec 09 2024 at 22:17):

Regarding having multiple (related) groups act at once - I'm reminded of the paper on Schur functors you wrote with Moeller and Trimble recently.

That would probably an interesting paper for me to work on reading parts of, if I wished to better understand groupoid actions...

John Baez (Dec 09 2024 at 23:04):

We study functors from the groupoid of finite sets and bijections to the category of finite-dimensional vector spaces. These are two of the best categories in the world, so a huge amount is known about such functors, but it hadn't all been phrased in terms of category theory and made maximally elegant!

John Baez (Dec 09 2024 at 23:05):

(I doubt we made it maximally elegant, but we pushed in that direction.)

Joe Moeller (Dec 10 2024 at 01:21):

I'll say that you should really read our first paper as a trio before the most recent one.

Joe Moeller (Dec 10 2024 at 01:21):

And these are really three or four papers crammed into two, so read the first half of the first one first.