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I am looking for examples for a certain kind of topological groups:

Definition: A topological group G is called syndetically separated if for every compact subset $K \subseteq G \setminus \{1\}$ there exists a syndetic closed normal subgroup $H \unlhd G$ such that $H \cap K = \emptyset$ (a subgroup $H \leq G$ is called syndetic if there exists a compact subset $C \subseteq G$ such that $CH = G$).

Of course, every compact topological group is syndetically separated, but I am looking for non-compact examples.

Note that, in case of a locally compact Hausdorff topological group $G$, a subgroup $H \leq G$ is syndetic iff the quotient space $G/H$ is compact. In particular, I am interested in examples of non-compact locally compact Hausdorff topological groups.

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Discrete groups with no torsion elements, or more general non-compact groups with no compact subgroups. Why are these groups interesting? –  Marc Palm Apr 8 '13 at 11:42
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Marc - are you sure? If I understood the definition correctly then a syndetic subgroup in the discrete case is just finite-index, so a discete group is syndetically separated if and only if it's residually finite. Or did I misunderstand? In either case there are lots of discrete examples, and the OP hasn't explained whether or not he's interested in these. –  HJRW Apr 8 '13 at 13:06
    
Okay, I misread subset as subgroup:( –  Marc Palm Apr 8 '13 at 14:15
    
In fact, I am not so much interested in these. I am more interested in non-discrete examples. –  Niemi Apr 8 '13 at 15:17

1 Answer 1

The definition is a bit misleading as syndetic subgroups are not assumed to be normal, but you require them to be normal in defining "syndetically separated". So let me change the definition, removing the word "normal" in the definition of "syndetically separated", turning your original definition into "normally syndetically separated".

1) If $G$ is a linear Lie group with a cocompact lattice, then it is syndetically separated (in my sense). Indeed, if $\Gamma$ is a cocompact lattice, then $\Gamma$ is residually finite, in the sense that the intersection of all finite index subgroups is trivial. If $K$ is a compact subset of $G-\{1\}$, then $K\cap\Gamma$ is finite and it follows that some finite index subgroup $\Gamma_1$ of $\Gamma$ satisfies $\Gamma_1\cap K=\emptyset$.

2) The simplest noncompact example of such a group $G$ is just the additive group $\mathbf{R}$ of real numbers and more generally $\mathbf{R}^n$. These groups are therefore normally syndetically separated (this answers your question). Also, residually finite discrete abelian groups are normally syndetically separated, e.g. $\mathbf{Z}[1/p]$.

3) Other examples satisfying (1) are linear connected semisimple Lie groups, as well as $p$-adic analogues. On the other hand, if $G$ is a nontrivial connected Lie group whose Lie algebra does not have any quotient abelian or compact simple (e.g. $G$ is simple and noncompact, e.g. $\text{SL}_2(\mathbf{R}))$, then the only normal cocompact (=syndetic) subgroup of $G$ is $G$ itself and therefore $G$ is not normally syndetically separated.

4) More generally, a connected Lie group $G$ is normally syndetically separated iff its Lie algebra is product of an abelian one and a compact semisimple one. Equivalently, $G$, is the direct product (up to finite covering) of a compact semisimple Lie group and an abelian Lie group. Indeed, if $\mathfrak{n}$ is the intersection of the kernels of all homomorphism from $\mathfrak{g}$ to abelian or compact simple Lie algebras, then it can be shown that the Lie subgroup $N$ it generates is closed and equal to the intersection of all kernels of homomorphisms from $G$ to compact Lie groups (I skip the proof). Example: the Heinsenberg group is not normally syndetically separated.

5) In view of (1), it is natural to wonder about the existence of a connected Lie group which is not syndetically separated (in my sense). An example is the affine ``ax+b" group $\mathbf{R}\rtimes\mathbf{R}$. Indeed it is not hard to show that any proper closed cocompact subgroup (not assumed normal!) has the form, in the above decomposition $\mathbf{R}\rtimes \lambda\mathbf{Z}$.

6) $\mathbf{Q}_p^n$ for $n\ge 1$ is not syndetically separated (it has no proper cocompact subgroup).

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I am not very familiar with Lie groups, so it would be very helpful and kind if you could give me a short answer to the following questions: 1) Could you explain why the cocompact lattice $\Gamma$ has to be residually finite? 2) For which classes of Lie groups is the existence of a cocompact lattice guaranteed? 3) Is is true that in particular every abelian Lie group is syndetically separated? –  Niemi Apr 9 '13 at 9:17
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@Sebastian 1) $\Gamma$ is finitely generated, as is any cocompact lattice in a connected Lie group $G$. This holds because it acts properly cocompactly on a connected manifold (namely $G$, with the action by left translations). On the other hand, Malcev's theorem is that any finitely generated linear group is residually finite. Combining, it follows that any cocompact lattice of a connected Lie group is residually finite. –  YCor Apr 9 '13 at 16:49
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@Sebastian 2) It's complicated. Sufficient conditions: $G$ abelian, $G$ semisimple. Necessary condition: $G$ unimodular. For a simply connected nilpotent Lie group (that's always unimodular), a necessary and sufficient condition is that the Lie algebra be $\mathbf{Q}$-defined, i.e. isomorphic to $\mathfrak{g}\otimes_\mathbf{Q}\mathbf{R}$ for some Lie algebra $\mathfrak{g}$ over $\mathbf{Q}$. Also, $\text{SL}_2(\mathbf{R})\ltimes\mathbf{R}^2$ has no cocompact lattices. Reference: Raghtnathan's book for most of this. –  YCor Apr 9 '13 at 16:53
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@Sebastian. In 2) I assumed $G$ connected. So the answer is yes for (3) if you assume $G$ connected or more generally with finitely many components. But false in general (since arbitrary discrete groups are particular instances of Lie groups). –  YCor Apr 9 '13 at 16:56

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