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Let us consider the group of continuous functions $C(S^1, S^1)$ from the circle to itself with the compact open topology. Does it have a chance to contain a dense cyclic subgroup?

Unfortunately it is not compact so we cannot apply the standard argument about compact, connected groups.

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  • $\begingroup$ No: if $f$ is not injective, then $\mathbf{Z}f$ is not dense because its closure is contained in $\{g:g(x)=g(y)\}$ for some $x\neq y$. If $f$ is injective, then it is surjective, and hence vanishes, hence its closure is contained in $\{g:g(x)=0\}$ for some $x$. $\endgroup$
    – YCor
    Commented May 3, 2019 at 21:58
  • $\begingroup$ @YCor, I guess $g(x) = 0$ should be $g(x) = 1$, since we're talking about functions to the circle. $\endgroup$
    – LSpice
    Commented May 3, 2019 at 22:05
  • $\begingroup$ @LSpice I was thinking of $S^1=\mathbf{R}/\mathbf{Z}$ (also justifying the notation $\mathbf{Z}f$), but you're right if you think of $S^1$ as the unit circle in the complex numbers. $\endgroup$
    – YCor
    Commented May 3, 2019 at 22:07
  • $\begingroup$ Also $C(X,S^1)$ is scarcely monothetic, even for pointwise convergence over $X$ (i.e., in the Tychonov topology of $(S^1)^X$). Indeed, if $\mathbf{Z}f$ is dense, then clearly $f$ takes no value in the set of torsion elements of the circle, and in particular $f$ is constant on every connected component, which, say for $X$ metrizable and not totally disconnected, is an obstruction to density. $\endgroup$
    – YCor
    Commented May 3, 2019 at 22:14

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No. Actually, for every, say metrizable space $X$ that is not totally disconnected, $C(X,S^1)$ is not even monothetic for the pointwise convergence topology (that induced by inclusion in $(S^1)^X$ with the Tychonov topology). Indeed, let $C$ be a connected component of $X$ not reduced to a point.

If $f$ is constant on $C$, then so is every pointwise limit point of $\mathbf{Z}f$ (I'm writing $S^1=\mathbf{R}/\mathbf{Z}$ additively), so $\mathbf{Z}f$ is not dense. If $f$ is not constant on $C$, then there exists $x\in C$ such that $nf(x)=0$, and this property is shared by every pointwise limit point of $\mathbf{Z}f$. In both cases we conclude that $\mathbf{Z}f$ is not dense.

However, for $X$ Cantor (or more generally compact metrizable totally disconnected), I claim that $C(X,S^1)$ is monothetic, by showing that the set $W$ of $f$ such that $\mathbf{Z}f$ is dense, is $G_\delta$-dense.

Indeed, there is a countable basis of the topology $\mathcal{D}$, such that every $U\in\mathcal{D}$ is given as follows: a number $d=d(U)$, a clopen partition $X=K_1\sqcup \dots K_d$, open intervals $I_1,\dots,I_d$, such that $U$ is the set of $f$ such that $f(K_j)\subset I_j$ for all $j$.

For $U\in\mathcal{D}$, define $U^\sharp=\{f:\exists n\in\mathbf{Z}:nf\in U\}$: as a union of open subsets, $U^\sharp$ is open.

Then $W=\bigcap_{U\in\mathcal{D}}U^\sharp$ is $G_\delta$. Hence, to show that it is dense, it is enough to show that $U^\sharp$ is open for every $U$. Indeed, fix $U$ as above, $\varepsilon$ and $f$. Then there exists a clopen partition $(L_k)$, such that each $L_k$ is contained in some $K_j$, and a function $g$ constant (say equal to $g_j$) on each $L_j$, such that $\|f-g\|_\infty\le\varepsilon$. We can suppose that the $g_j\in\mathbf{R}/\mathbf{Z}$ form a $\mathbf{Z}$-free family. Then a density argument in finite-dimensional tori shows that $ng$ belongs to $U$ for some $u$, that is, $g\in U^\sharp$. This shows density.

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