Natural Topology on Automorphism Group of a Topological Group – General Topology

automorphism-groupsgn.general-topologytopological-groups

$\DeclareMathOperator\TAut{TAut}\DeclareMathOperator\Homeo{Homeo}$Let $G$ be a topological group, and let $\TAut(G)$ denote the group of topological automorphisms of $G$ under composition (i.e. the group of maps $f \colon G \to G$ that are simultaneously group automorphisms and self-homeomorphisms).

We wish to give $\TAut(G)$ a reasonable topology in the following sense:

  1. $\TAut(G)$ becomes a topological group with respect to this topology.
  2. This topology should interact with/depend on the topology of $G$ in some way, i.e. we can require that the natural action $\TAut(G) \times G \to G$ is continuous.

In the case that $G$ is compact, it is known that giving $\TAut(G)$ the compact–open topology satisfies the above conditions, where the compact–open topology has as a subbasis sets of the form $$V(C, U) = \{f \colon G \to G \mid f(C) \subseteq U\},$$ where $C, U \subseteq G$ are compact and open, respectively.

If $G$ is locally compact, we can instead give $\TAut(G)$ the $g$-topology, which has as a subbasis sets of the form $$V(K, W) = \{f \colon G \to G \mid f(K) \subseteq W\},$$ where either $K$ or $G \setminus W$ is compact.

The cases where $G$ is compact or locally compact are discussed in Dijkstra – On Homeomorphism Groups and the
Compact–Open Topology and Arens – Topologies for Homeomorphism Groups. In fact, these two papers discuss the group of self-homeomorphisms $\Homeo(X)$ for a space $X$ which is not necessarily a topological group, so the case for $G$ and $\TAut(G)$ follows from that (since $\TAut(G)$ is a subgroup of $\Homeo(G)$).

My question is, for a general topological group $G$, is there a good way to describe a topology on $\TAut(G)$ satisfying the two conditions above? We can simply say: "give $\TAut(G)$ the coarsest topology such that it becomes a topological group and the action $\TAut(G) \times G \to G$ is continuous," but I am hoping for something more explicit than that.

Best Answer

$\DeclareMathOperator\Aut{Aut}$There is a recent paper Uniformly locally bounded spaces and the group of automorphisms of a topological group by Maxime Gheysens where he among other nice things systematically investigates the topologies on $\Aut(G)$ for any topological group $G$. On every topological group there are several natural uniform structures making translations become uniformly continuous: first of all, the left and right uniform structure, their supremum (the upper uniform structure) and their infimum (the lower or Roelcke uniform structure). Now, as it turns out, on $\Aut(G)$ one can usefully consider:

  • the topology of uniform biconvergence on bounded sets with respect to the left, right, or upper uniform structure (they all give the same topology) or
  • the topology of uniform biconvergence on bounded sets with respect to the lower uniform structure.

In general, these two topologies are different, but they coincide for the so-called SIN groups (and even broader, for coarsely SIN groups), i.e. groups with admitting a basis of conjugation-invariant identity neighborhoods as well as for all locally compact groups. So the existence of two really different useful topologies on $\Aut(G)$ is purely a phenomenon in the world of “very big” groups.

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