7

For a group $G$ there are always two actions of $G$ on itself: $g*h=gh$ and $g\star h=ghg^{-1}$. As objects in the category $G$-$Set$, the former is the free $G$-set on a singleton set. When $G$ is considered as a one-object category, this same $G$-action is also the unique representable functor on $G$. Is there any similar way to understand the action by conjugation in categorical terms?

Ittay Weiss
  • 81,796
  • @ArnaudD. Or possibly $h^{-1}gh$. Depends on whether you want the action acting from the right or from the left. At any rate the way it's written in the OP is off. – Arthur Mar 13 '18 at 14:23
  • 1
    Too busy to write an answer right now but this might be a starting point. Also worth noting that $g \mapsto hgh^{-1}$ is a left action, so is an object of the functor category $[G, \mathbf{Set}]$, and $g \mapsto h^{-1}gh$ is a right action, so is an object of the functor category $[G^{\mathrm{op}}, \mathbf{Set}]$. – Clive Newstead Mar 13 '18 at 14:24
  • @CliveNewstead it's an interesting link - thanks. It's not quite the adjoint action of $G$ on $G$ though. If you get the time to expand a bit on your comment, that will be greatly appreciated. – Ittay Weiss Mar 13 '18 at 14:51
  • https://math.stackexchange.com/questions/2662005/how-much-of-a-group-g-is-determined-by-the-category-of-g-sets Also has a further characterization of the regular $G$-set as the unique transitive one with an epimorphism to any other transitive one. I have no idea what sort of description there might be for the conjugation action. – Tobias Kildetoft Mar 13 '18 at 19:41
  • $G$ can be identified with the set of $G$-morphisms on the regular $G$-set $G\to G$ (as the image of $1$ determines the morphism), then, according to the commented nLab page, the internal hom $(G,G)$ will be just this conjugation action. In other words, it is the natural $G$ action arising on this homset of $G$-morphisms. – Berci Mar 13 '18 at 23:32
  • @Berci This isn't quite right, see my answer. – Kevin Carlson Mar 14 '18 at 05:26

3 Answers3

4

One way to see this rather abstractly is as the diagonal of the Yoneda bimodule $G\times G^{\mathrm{op}}\to \mathbf{Set}$. That is, compose the $G$-biset $(g,h)*k=gkh$ with the diagonal homomorphism/functor $G\to G\times G^{\mathrm{op}}$, $g\mapsto (g,g^{-1})$. Under this formulation we find a similar conjugation action of any groupoid, or even any dagger-category, on itself, or generalizing in a different direction, on any bi-$G$-set.

The question was raised in the comments of constructing this action as the internal hom of $G$ in $G-\mathbf{Set}$. This is trickier than it first appears, however. Let's calculate $G^G$ from first principles. We have isomorphisms of underlying sets: $G^G\cong G-\mathbf{Set}(G,G^G)\cong G-\mathbf{Set}(G\times G,G)\cong \mathbf{Set}(G,G)$. So the internal hom is actually the set of all functions $G\to G$, with the conjugation action, $f:G\to G\mapsto g*f, g*f(g_1)=gf(g^{-1}g_1)$. The odd thing is that we can't construct the conjugation action this way. Indeed, the equivariant maps between any two $G$-sets are precisely the fixed points of the action! Or perhaps this isn't odd-we certainly don't want to start claiming we have conjugation actions on arbitrary $G$-sets. One way to make something closer to this work is to transfer the conjugation action along the natural isomorphism between $G-\mathbf{Set}(G,G)$ and $G^{\mathrm{op}}-\mathbf{Set}(G,G)$, which more or less brings us back to the suggestion of the previous paragraph.

Kevin Carlson
  • 54,515
  • 1
    Thanks Kevin. I'm wondering if there is a way to avoid explicitly using the dagger structure. So, as a follow up, is there a categorical extension of Yoneda that identifies, among the functors $C\to Set$ a class $W$ wider than the representable ones in such a way that: 1) when $C$ happens to be a group, the action by conjugation is found in $W$; and 2) when $C$ happens to have a dagger structure, the Yoneda diagonal action is present in $W$. – Ittay Weiss Mar 15 '18 at 10:17
2

Associated to any action of a group $G$ on a set $X$ is its action groupoid, which you can think of as the homotopy quotient of $X$ by the action of $G$, which I'll write $X//G$. The unique map $X \to \bullet$ induces a map

$$X//G \to \bullet//G$$

where $\bullet//G$, otherwise known as $BG$, is the groupoid with one object with automorphism group $G$. This construction extends to an equivalence of categories between the category of $G$-sets and the category of groupoids equipped with a covering map to $BG$ in a suitable sense.

So, to understand the conjugation action of $G$ on itself we can equivalently attempt to understand the action groupoid $G//G$, together with its map to $BG$. The action groupoid $G//G$ can in turn be understood as the free loop space

$$G//G \cong L(BG) \cong [S^1, BG] \cong [B\mathbb{Z}, BG]$$

of $BG$, and the map to $BG$ can be understood as induced by applying $[-, BG]$ to any map $\bullet \to S^1$.

Note that $G$ acting on itself by left multiplication appears as the point $\bullet$, together with any map $\bullet \to BG$.

Qiaochu Yuan
  • 468,795
-1

The way Galois came up with the operation of conjugation is that he asked himself how does a group act on its subgroups. The only subgroup closed under the actions by multiplication (be it on the left or on the right) is the whole group. The nontrivial equivariant subobjects of the group acting on itself are precisely the normal subgroups, acted on by conjugation.

dusko
  • 1