geometric representation theory
representation, 2-representation, ∞-representation
Grothendieck group, lambda-ring, symmetric function, formal group
principal bundle, torsor, vector bundle, Atiyah Lie algebroid
Eilenberg-Moore category, algebra over an operad, actegory, crossed module
Be?linson-Bernstein localization?
Given a topological group $G$, a (continuous) $G$-set is a set $X$ equipped with a continuous group action $\mu: G \times X \to X$, where $X$ is given the discrete topology.
In the case where $G$ is a discrete group, the continuity requirement is void, and this is just a permutation representation of the discrete group $G$.
Note that since $X$ must be given the discrete topology, this behaves rather unlike topological G-spaces. In particular, a topological group does not act continuously on itself, in general. Thus this notion is not too useful when $G$ is a “usual” topology group like $SU(2)$. Instead, the topology on the group acts as a filter of subgroups (where the filter contains the open subgroups), and each element of a continuous $G$-set is required to have a “large” stabilizer.
The $G$-sets form a category, where the morphisms are the $G$-invariant maps. See category of G sets.
($G$-sets are the free coproduct completion of $G$-orbits)
Let $G \,\in\, Grp(Set)$ be a discrete group. Since every G-set $X$ decomposes as a disjoint union of transitive actions, namely of orbits of elements of $X$, the defining inclusion of the orbit category into $G Set$ exhibits the latter as the free coproduct completion of the orbit category (see also this Prop.).
Let $G$ be a topological group, and $X$ be a set with a $G$ action $\mu: G \times X \to X$. Then the action is continuous if and only if the stabilizer of each element is open.
Suppose $\mu$ is continuous. Since $X$ has the discrete topology, $\{x\}$ is an open subset of $X$. So $\mu^{-1}(\{x\})$ is open. So we know the stabilizer
is open.
Conversely, suppose each such set is open. Given any (necessarily open) subset $A \subseteq X$, its inverse image is
So it suffices to show that each $\mu^{-1}(\{a\})$ is open. We have
Thus we only have to show that for each $a, x \in X$, the set $\{g \in G: g \cdot x = a\}$ is open. If there is no such $g$, then this is empty, hence open. Otherwise, let $g_0$ be such that $g_0 \cdot x= a$. Then we have
Since $g_0$ is a homeomorphism, and $I_x$ is open, this is open. So done.
In the following examples, all groups are discrete.
A $\mathbb{Z}_2$-set is a set equipped with an involution.
Any permutation $\pi : X \to X$ gives $X$ the structure of a $\mathbb{Z}$-set, with the action of $\mathbb{Z}$ on $X$ defined by iterated composition of $\pi$ or $\pi^{-1}$.
$G$ is itself a $G$-set via the (left or right) regular representation.
A normal subgroup $N \lhd G$ defines a $G$-set by the action of conjugation.
For $G$ a finite group then Mackey functors on finite $G$-sets are equivalent to genuine G-spectra.
action: a $G$-set is a set with an action of the given group, $G$.
An early account (where the term “representation group” is used to refer to a finite set equipped with a permutation action):
For a more modern account see
Last revised on October 12, 2021 at 12:44:28. See the history of this page for a list of all contributions to it.