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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?

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.

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    $\begingroup$ Do you identify functions equal everywhere, even if $n$ are different? E.g. in a group of order two, e.g. in $\{0,1\}^X$ for any $X$, you have $f_{2n}=f_0$ for all $n$, so $F=\{f_0,f_1\}$. $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 6, 2012 at 13:20
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    $\begingroup$ And could you please explain how you prove your statement for the unit circle (I hope that's what you mean by the "group associated to one-dimensional sphere"). Because to me it seems that the topology on $F$ is not at all discrete and that its closure equals to the set of functions which are arbitrary at irrational points (assume that the circle is parameterized by $[0,1)$), and at rational points $f(x)\in \{nx: n\in \mathbb Z\}$. $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 6, 2012 at 13:42
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    $\begingroup$ @Yulia: If G is abelian then each $f_n$ is a homomorphism; so if $f_{n_\alpha}$ limits to $f$ pointwise, then $f(st) = \lim f_{n_\alpha}(st) = \lim f_{n_\alpha}(s) f_{n_\alpha}(t) = f(s) f(t)$ by joint continuity of the group product. Thus $F$ certainly consists of (continuous) homomorphisms. So if G is the circle group, we already see that F is a subset of the functions $x\mapsto x^n$, where now $n$ might be negative. However, I think you can get negative $n$, and so F is not closed (but I don't see a good argument for this right now). $\endgroup$
    – Matthew Daws
    Commented Jan 6, 2012 at 15:52
  • $\begingroup$ The thing with $\mathbb{N}$ is a typo. Of course, $n$ is also allowed to be negative. I have corrected that above. As for the frist comment given by Yulia, I do not know how to give a better answer than "Yes, two identical functions are the same function". The set $F$ is simply the collection of all functions $x \mapsto x^n$ where $n$ ranges over $\mathbb{Z}$. If two function $f_{n1}$,$f_{n2}$ among this set happen to the identical (i.e., $x^{n_1}=x^{n_2}$ for all $x \in X$), then they are of course the same element of the set $F$. So, yes, in your example it would be $F = \{f_0,f_1\}$. $\endgroup$
    – Niemi
    Commented Jan 6, 2012 at 16:24
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    $\begingroup$ @Matthew: thank you, closedness clear now! Note that in the question $n$ is allowed to be negative. In fact, to conclude with the circle, it remains to separate $f_0$ from all the other $f_n$. And this has to do with simultaneous rational approximations, the area I have only a vague notion of. Is the following true (or its negation): $\forall\epsilon>0 \forall t_1\dots t_m\in[0,1)$ $\exists n\in \mathbb Z$: $\{nt_j\}<\epsilon$ for all $j$, where $\{x\}$ is the fractional part of $x$ (or its distance to the closest integer)? $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 6, 2012 at 16:48

1 Answer 1

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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.

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  • $\begingroup$ Thank you for your answer. However, I do have some comments on what you say. For a better overview (and due to the character limits for comments), I will poste them in separate comments. $\endgroup$
    – Niemi
    Commented Jan 7, 2012 at 11:41
  • $\begingroup$ 1. There is no reason why the closure of $F$ must be compact. Note that $C(G,G)$ does not need to be closed in $G^G$. $\endgroup$
    – Niemi
    Commented Jan 7, 2012 at 11:41
  • $\begingroup$ 2. I think you are wrong with your assumption that the compact-open topology and the topology of pointwise convergence coincide. In fact, I seriously doubt that this happens in the profinite case. $\endgroup$
    – Niemi
    Commented Jan 7, 2012 at 11:41
  • $\begingroup$ 3. In your construction, there is no argument that substantiates that the maps $\nu_{G} \colon G \to G$ are continuous. So, what you have described is just the closure of $F$ in $G^G$. However, the point of my question was that I would like to know the closure in $C(G,G)$ in a more concrete manner than just stating that it is the intersection of the closure of $F$ in $G^G$ intersected with $C(G,G)$. $\endgroup$
    – Niemi
    Commented Jan 7, 2012 at 11:42
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    $\begingroup$ Of course these operations are cts in the profinite case since they are for finite groups and by naturality they commute with projective limits. Look in the book of Ribes and Zalesskii where they have a section on raising elements of a profinite group to a power from $\widehat Z$. See also the paper of Almeida in transactions of the AMS on dynamics of implicit operations where a more general setting is cts. I think all this works more generally for compact Hausdorff groups but there are details I have no time to check. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 7, 2012 at 13:17

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