# What are all the group endomorphisms of the circle group?

The circle group is the unit circle in the complex plane under multiplication. What are all the group endomorphisms of this group? (I can think of $z\mapsto z^n$ for $n\in \mathbb{Z}$.)

• $\bar{z}=z^{-1}$ for $\|z\|=1$ right? – Dan Rust Dec 2 '14 at 20:23
• @DanielRust As far as the continuous ones are concerned, yes. I expect there are discontinuous ones, however. – Daniel Fischer Dec 2 '14 at 20:25
• One would think that the answer might then depend on the axiom of choice. Strange things happen with uncountably generated groups and the axiom of choice. For instance AoC implies $(\mathbb{R},+)\cong(\mathbb{C},+)$. – Dan Rust Dec 2 '14 at 20:32
• Some more "discontinuous ones": you can separate the torsion and torsion free part to obtain a decomposition of the form $\mathbb{Q}^{(\mathfrak{c})}\oplus \bigoplus_{p \in \mathbb{P}}\mathbb{Z}_{p^{\infty}}$ and you can define endomorphisms "component by component". In particular, any collection $(n_p)_{p \in \mathbb{P}},$ where each $n_p$ is a $p$-adic integer, gives rise to a unique endomorphism which fixes all elements of infinite order (to maintain some sense of "constructiveness", I will not mention the vast amount of endomorphisms of the torsion-free part). – Pavel Čoupek Dec 2 '14 at 21:55
• @DanielRust All the summands are divisible groups, i.e. injective $\mathbb{Z}$-modules, inside the unit group which is also divisible, i.e. injective $\mathbb{Z}$-module. More genrally, every divisible group has a decomposition of this form (i.e. some summands $\mathbb{Q}$ and some summands $\mathbb{Z}_{p^\infty}$). – Pavel Čoupek Dec 2 '14 at 22:10

## 1 Answer

This is a large ring. In general, recall that if $G$ is a locally compact (Hausdorff) abelian group then $\text{Hom}(G, S^1)$ is the Pontryagin dual of $G$. Hence $\text{Hom}((S^1)_{\delta}, S^1)$, where $(S^1)_{\delta}$ denotes $S^1$ with the discrete topology, is the Pontryagin dual of $(S^1)_{\delta}$. Like the Pontryagin dual of any discrete group, it must be compact, and hence must be much larger than $\mathbb{Z}$. In fact it is precisely the Bohr compactification of $\mathbb{Z}$.

(Edit: Incidentally, the statement that the Pontryagin dual of a discrete group is compact makes use of Tychonoff's theorem for compact Hausdorff spaces and so depends on a weak form of choice: Tychonoff's theorem for compact Hausdorff spaces is equivalent to the ultrafilter lemma, which is in particular known to be strictly weaker than AC.)

To get an idea of how large the Bohr compactification of $\mathbb{Z}$ has to be, some manipulations involving Pontryagin duality show that it is the free compact (Hausdorff) abelian group on one generator. In particular, there is a natural map from it to the profinite integers which I believe splits (as a map of abelian groups, at least) as described by PavelC in the comments.