# Bohr compactification as a topological compactification

Let $G$ be a locally compact Hausdorff group. Denote its Bohr compactification by $bG$.

Despite group structure, $G$ has several (Hausdorff) compactifications that, in a sense, the smallest one is the one-point compactification, and the largest one is the Stone-Čech compactification.

Is $bG$ is isomorphic to one of the (topological) compactifications of $G$ as a topological space? In case of a positive answer, to which one? I don't know the answer even in the case of $G=\mathbb{R}$.

• Maybe first work on $b\mathbb Z$ and figure out whether the natural image of $\mathbb Z$ in there has the discrete topology. Dec 18, 2017 at 19:03
• Bohr compactification (of a not already-compact group) is not a "topological" compactification. Dec 18, 2017 at 19:09
• Thanks. I don't know the topology of $b \mathbb{Z}$, good point. @GeraldEdgar
– XIE
Dec 18, 2017 at 19:20
• Would you define "topological compactification"? I'm not aware of any difference between "compactification" and "topological compactification". Does the latter mean that the subspace carries the compactification map induces a homeomorphism onto its image?
– YCor
Dec 18, 2017 at 20:14
• @HamedPourmohammad, $G$ is always dense in $bG$ by definition. The problem is that it might not be open in (or might not inject into) it. Dec 21, 2017 at 20:09

As Francois Ziegler answered, for a locally compact group $G$, the Bohr compactification of $G$ is a compactification in the usual sense iff $G$ is compact. This is true with no further restrictions.

Recall that the forgetful functor from compact groups to topological groups has a left adjoint functor, denoted here $b$. For any topological group $G$, the identity in $\text{Hom}(bG,bG)$ corresponds to an element of $\text{Hom}(G,bG)$ called the unit, $u:G\to bG$. It is easy to check that $u(G)$ is dense in $bG$ (this is usually seen directly from the construction of $b$, but also follows from the abstract nonsense). The homomorphism $u:G\to bG$ is known as the Bohr compactification of $G$. This terminology is a bit unfortunate, as $u$ is not a compactification in the sense usually used in topology, unless $G$ was compact to begin with (in fact, $u$ is rarely injective - groups for which $u$ is injective are called "maximally almost periodic").

By a "compactification in the usual sense" of a locally compact space $X$ one means a continuous map $f:X\to Y$ where $f(X)$ is dense in $Y$ and $f:X\to f(X)$ is a homeomorphism. In that case $f(X)$ is necessarily open in $Y$ (this an easy exercise).

In case $X$ and $Y$ were groups and $f$ a group homomorphism, $f(X)$ was also closed, as any open subgroup is closed, thus $f(X)=Y$ by density. So $X$, being homeomorphic to $Y$, must have been compact to begin with.

More generally, let me note that any locally compact subgroup of any topological group is necessarily closed (and if the ambient group is locally compact, a subgroup is locally compact iff it is closed).

• Yes; this may be the argument Rudin had in mind, it’s also in van Douwen (1990, pp. 75-76), still assuming local compactness: “it is known that $b$ is a homeomorphism iff $G$ is totally bounded. (In particular, $b$ never is a homeomorphism if $G$ is a noncompact locally compact group. This can also be proved directly: A locally compact group never is a dense proper subgroup of a topological group, since locally compact dense subspaces are open, and since open subgroups are closed.)” Dec 21, 2017 at 12:34
• @FrancoisZiegler by "no restrictions" I really meant "no restrictions apart of local compactness" (and having in mind the your restrction to the abelian case), surely a non locally compact space cannot have a "compactification in the usual sense" as above. I'll edit. Thanks. Dec 21, 2017 at 12:36

The answer is no: unless $G$ is compact, its subspace topology inside $bG$ (called its Bohr topology) is much weaker than its own. So $\iota:G\to bG$, while continuous with dense image, is not an embedding (homeomorphism onto $\iota(G)$), hence not a compactification in the sense of topologists.

Rudin (1962, pp. 30-31) notes this without proof. Later Katznelson (1973) gave a method to show that very “thin” subsets of $G$ are dense in the Bohr topology: e.g. $\mathbf N$ inside $\mathbf Z$, a parabola inside $\smash{\mathbf R^2}$, etc. For more details and references see (with apologies for the plug) this paper, esp. Lemma 1(4).

Note: I have restricted my answer to abelian $G$; $\iota$ can be defined for any topological group, but your question only makes sense when $\iota$ is injective, i.e. for “maximally almost-periodic” $G$, and these are very nearly the compact $\times$ abelian ones: see Dixmier (1977, §§16.1, 16.4).

• "[maximally almost periodic] are very nearly the compact $\times$ abelian ones" is not true. For instance it includes all residually finite discrete groups.
– YCor
Dec 21, 2017 at 11:31
• @YCor right... my weasel words “very nearly” were meant to cover that; Dixmier’s precise statement is that connected ones are compact $\times\,\mathbf R^n$, then I don’t know if disconnected ones have since been classified. Dec 21, 2017 at 12:12
• In the totally disconnected case, the discrete case is clear-cut (MAP = residually finite). In general there's a result of Caprace and Monod (2011) in the direction that, roughly, there are not much more than discrete ones: if $G$ is totally disconnected, MAP and compactly generated, then $G$ has a basis of neighborhoods of 1 made up of compact normal open subgroups. In particular $G$ is compact-by-discrete. In general, it already says that every compactly generated subgroup satisfies this condition, which is quite restrictive. ArXiv ref: arxiv.org/abs/0811.4101
– YCor
Dec 21, 2017 at 12:45
• Disscusion in Rudin was interesting. But as you mentioned, Dixmier has proved in his book that in what he called 'injectable groups' have the desird property.
– XIE
Dec 23, 2017 at 14:50
• I could ask the question in the case of injectable groups
– XIE
Dec 23, 2017 at 14:50