Consider the following generalization of residual finiteness to topological groups.

A locally compact Hausdorff group $G$ is called residually compact if for every compact $K \subseteq G$ there is a normal cocompact lattice $\Lambda \subseteq G$ such that the projection $G \to G/\Lambda$ is injective on $K$.

Notice that this clashes with the normal definition of a residual property, but I don't know a better name.

Q1: Has this property been studied under another name?

Up to now we know the following. If $G$ contains a normal cocompact lattice which is residually finite, then the group is residually compact.

On the other hand a residually compact group embeds (algebraically) into a compact group, hence every finitely generated lattice in $G$ is residually finite by Malcev's theorem and the fact that the irreducible linear presentations of a compact group are finite dimensional. If $G$ is compactly generated, then every lattice is finitely generated. In the compactly generated case, residual finiteness is thus the same as containing a residually finite normal cocompact lattice.

Q2: Is there an example of a residually compact topological group that does not contain a residually finite normal lattice?

  • $\begingroup$ What is a "normal lattice"? Is it the same as a "lattice"? $\endgroup$
    – Misha
    Oct 27, 2015 at 2:21
  • $\begingroup$ @Misha With a "normal lattice" L in G I mean a normal subgroup that is a lattice (discrete subgroup such that G/L has finite measure). But I would also be interested in an example with "normal" being dropped. $\endgroup$ Oct 27, 2015 at 8:21
  • 2
    $\begingroup$ If you remove the normality assumption on the lattice (which quite relaxes the definition), it is called "residually systolic" in arxiv.org/abs/1403.5295, Definition 6.1. $\endgroup$
    – YCor
    Oct 27, 2015 at 8:29
  • $\begingroup$ A discrete cocompact normal subgroup $\Lambda$ is already a lot to ask for. For instance, if your group is compactly generated, then any such $\Lambda$ must have open centraliser, so if $G$ is connected then $\Lambda$ would be central, and if $G$ is totally disconnected you'd have a finite index subgroup of the form $\Lambda \times U$ where $U$ is a compact open subgroup. $\endgroup$
    – Colin Reid
    Oct 27, 2015 at 12:00
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    $\begingroup$ For an arbitrary compactly generated locally compact group $G$, it implies the existence of a compact normal subgroup $W$ such that the quotient $G/W$ is a isomorphic to an extension with kernel $\mathbf{R}^k$ and discrete quotient, whose action on $\mathbf{R}^k$ goes through $\mathrm{GL}_k(\mathbf{Z})$. Without the compact generation assumption, it still holds that if $W$ is the maximal compact normal subgroup in $G^\circ$, then $(G/W)^\circ=G^\circ/W$ is isomorphic to $\mathbf{R}^k$ for some $k$ and that the conjugation action preserves a lattice in this $\mathbf{R}^k$. $\endgroup$
    – YCor
    Oct 27, 2015 at 12:42

1 Answer 1


Here's an answer to the mathematical part of the question (namely Q2).

An example is the group of adeles. Start from $\mathbf{A'}$ defined as the inverse image of $\bigoplus_p\mathbf{Q}_p/\mathbf{Z}_p$ in $\prod\mathbf{Q}_p$, with $\prod\mathbf{Z}_p$ prescribed to be a compact open subgroup (the sum being over all primes); then $\mathbf{A}=\mathbf{R}\oplus\mathbf{A}'$. This is a locally compact ring with $\mathbf{Q}$ standing as a discrete cocompact subring through the diagonal embedding (let $Q$ denote its image).

Then $\mathbf{A}$ is "residually compact". Indeed, if we pick a compact subset, then there exists an integer $n\ge 0$, a finite set of primes $I$ with complement $J$ such that this compact subset is contained in the compact subset $$K_{I,n}=[-n,n]\times\bigoplus_{p\in I}p^{-n}\mathbf{Z}_p\oplus\prod_{p\in J}\mathbf{Z}_p.$$ Define $m=\prod_{p\in I}p^n$. The intersection $Q\cap K_{I,n}$ consists of those rationals of the form $m^{-1}k$ with $k\in\mathbf{Z}$ with real absolute value $|m^{-1}k|\le n$. So if we define $\phi$ as the topological group automorphism which the identity on $\mathbf{A}'$ and multiplies the real component by, say, $2mn$, then $\phi(Q)\cap K_{I,n}=\{0\}$. So $\mathbf{A}$ is "residually compact".

On the other hand, $\mathbf{A}$ has no residually finite lattice. Indeed, let $R$ be a lattice. The image of $R$ in $\bigoplus\mathbf{Q}_p/\mathbf{Z}_p$ has finite index, and since the latter group has no proper finite index subgroup, it is all of $\bigoplus\mathbf{Q}_p/\mathbf{Z}_p\simeq\mathbf{Q}/\mathbf{Z}$. The kernel of this projection is $R_1=R\cap (\mathbf{R}\oplus\prod\mathbf{Z}_p)$ and is a cocompact lattice in the open subgroup $\mathbf{R}\oplus\prod\mathbf{Z}_p$, which is torsion-free and compactly generated. In particular $R_1\cap\prod\mathbf{Z}_p$ is trivial and the projection of $R_1$ in $\mathbf{R}$ is injective and its image is a cocompact lattice in $\mathbf{R}$ (discreteness follows from compactness of $\prod\mathbf{Z}_p$). So $R_1$ is an infinite cyclic group. Now let us show that $R$ is divisible by any $p$ (this will imply that $R$ is not residually finite). Let $x\in R$. Then since $R/R_1$ is divisible, we can write $x=py+z$ with $z\in R_1$ and $y\in R$. Now let $M$ be the inverse image in $R$ of the subgroup of order $p$ of $R/R_1$. Then $M$ is torsion-free and contains $R_1$ with index $p$. Hence every element of $R_1$ has a $p$-root in $M$, hence in $R$. So $z=pz'$ with $z'\in R$, thus $x\in pR$. Thus $pR=R$ and $R$ is divisible, hence not residually finite (and actually, necessarily is isomorphic to $\mathbf{Q}$).

  • $\begingroup$ Thanks a lot for this extensive answer. It took me a while to digest this and I am still struggling to understand some bits. Is "in this open subgroup" referring to $R \oplus \prod Z_p$? $\endgroup$ Nov 30, 2015 at 9:43
  • $\begingroup$ OK, I edited to make this more precise. $\endgroup$
    – YCor
    Nov 30, 2015 at 12:36

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