$\DeclareMathOperator\TAut{TAut}\DeclareMathOperator\Homeo{Homeo}$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](https://www.cs.vu.nl/%7Edijkstra/research/papers/2005compactopen.pdf) and [Arens - Topologies for Homeomorphism Groups](https://doi.org/10.2307/2371787). 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.



Is there a natural topology on the automorphism group of a topological group?
$\DeclareMathOperator\Aut{Aut}$There is a recent paper [Uniformly locally bounded spaces and the group of automorphisms of a topological group](https://arxiv.org/abs/2001.00548) 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.


