[Math] (Closures of sets of) operations in topological groups.

Question

Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$.

Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure?

Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$ with respect to the topology of pointwise convergence. However, I do not know how the induced subspace topology on $F$ can be described. (It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology, but this does not tell anything about the topology of pointwise convergence.)

Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference?

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Edit:

I originally stated the example differently, namely

Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology.

However, in this example, the topology of pointwise convergence on $F$ does not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so the second statement of the example is just not true.

Answer 1

Here is what I think happens in the category of compact (Hausdorff) groups. I know it is true in the category of profinite groups and I assume the argument carries over. First of all I believe the closure in the compact-open topology and the pointwise convergence topology are the same. The closure should be described this way.

Let $C$ be the free compact group on 1-generator $a$. To each element $\nu$ of $C$ and compact group $G$, we get a cts map $\nu_G\colon G\to G$ as follows. Let $x\in G$. By the universal property of $C$ there is a unique cts map $C\to G$ sending $a$ to $x$. Define $\nu_G(x)$ to be the image of $\nu$ under this map. This gives a cts mapping $C\to Cts(G,G)$ (natural in $G$) in the compact open topology. The closure of the family $F$ of mappings in the OP's question is the image of $C$ in $Cts(G,G)$.

To make this clearer, the mapping on $G$ associated to $a^n$ with $n\in \mathbb Z$ is $x\mapsto x^n$.

Edit. My intuition above was from the profinite case. In the profinite setting the family of mappings $x\to x^n$ with $n\geq 1$ is equicontinuous. Indeed if $G$ is profinite, then it has a fundamental system of entourages for its uniformity consisting of partitions into cosets of an open normal subgroup. Now if $x,y$ are in the same coset of an open normal subgroup $N$ then $x^nN=y^nN$ for all $n\geq 1$. This gives equicontinuity in the sense of uniform spaces (in fact all these maps are contractive in the uniform sense). It follows that for profinite groups the closure in the compact-open topology is the pointwise closure and hence compact. The discussion above using free pro-cyclic groups works then in the profinite setting.

In the general setting of compact groups I was probably naive. Doubling on a circle is chaotic and not contractive nor are its powers equicontinuous. So most likely what I wrote above doesn't work in general.

However do note that all maps of raising to a negative power are pointwise limits of positive powers. So the OP's statement for the circle case about the topology being discrete is incorrect.