Is it possible to explicitly describe the Bohr compactification of $\mathbb Z$? This is equivalent to describing all the group homomorphisms $\mathbb R/\mathbb Z \to \mathbb R/\mathbb Z$ including the discrete ones.

Keith Conrad has written up an explicit construction of the set of all homomorphisms $\mathbb Q \to \mathbb T$ and this can be used to find the Bohr Compactification of $\mathbb R$ as follows. The dual group of $\mathbb R$ is just $\mathbb R$ again. So $b\mathbb R$ is the set of all group homomorphisms $\mathbb R \to \mathbb T$ under pointwise convergence. To see what theese look like, choose a Hamel basis $B$ and write $\mathbb R = \oplus_{b \in B} \mathbb Q$. Each group homomorphism factors into an homomorphism $\mathbb Q \to \mathbb T$ for each element of $B$.

However the construction given by Conrad is so complicated there is no chance I would have come up with it myself. So I doubt I will be able to describe $b \mathbb Z$ either.

Is there a known description?


1 Answer 1


The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $((\mathbb{Q}/\mathbb{Z})_\mathrm{disc})^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\tfrac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

  • $\begingroup$ Neat! I think I can see why $\mathbb T \cong \mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$. But is it hard to prove that $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime \cong \prod_p \mathbb{Z}_p$ ? $\endgroup$
    – Daron
    Commented May 24 at 5:34
  • $\begingroup$ The product of p-adics is just the profinite completion of integers. $Hom(-, \Bbb Q / \Bbb Z)$ is an anti-equivalence between categories of artinian and noetherian abelian groups, which restricts to an auto-anti-equivalence of finite abelian groups. $\Bbb Q / \Bbb Z$ is the colimit of $\Bbb Z/n$, so dual of it is the limit $\hat{\Bbb Z}$. (Those groups themselves are not artinian/noetherian, but their p-localisations are.) $\endgroup$
    – Denis T
    Commented May 24 at 14:02
  • $\begingroup$ @DenisT Yes indeed, $\hat{\mathbb{Z}}$ can also be thought of as the algebraic double dual: $$\hat{\mathbb{Z}} \cong \hom(\hom(\mathbb{Z}, \mathbb{Q}/\mathbb{Z}), \mathbb{Q}/\mathbb{Z}).$$ One pleasant thing about $\hat{\mathbb{Z}} \cong \hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ is that the ring multiplication on $\hat{\mathbb{Z}}$ comes from the composition of endomorphisms $\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. $\endgroup$
    – Todd Trimble
    Commented May 24 at 15:25

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