Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?

Question

The posting of this question was suggested by Yemon Choi: see Discrete cyclic subgroup.. The question is not mine; it's just a rephrasing of Discrete cyclic subgroup.

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EDIT 4. This post claims that the answer is No in general. I hope it's correct. END OF EDIT 4.

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By page 110 of Weil's book L'intégration dans les groupes topologiques et ses applications, the answer is No in the abelian case.

I know almost nothing about locally compact groups. The question might be very easy for experts, and perhaps even for laymen. In the unlikely event the question is difficult, here is a particular case:

Let G be a non-compact connected Lie group. Does G admit a discrete infinite cyclic subgroup?

EDIT 1. I think that, by known results about lattices, the answer is No for semisimple Lie groups. Thanks for correcting me if I'm wrong, or (even better) for providing precise statements and references. Again, plenty of MathOverflowers know this stuff much better than I. I'm making it a Community Wiki. END OF EDIT 1.

EDIT 2. I scanned a few pages of Weil's L'intégration dans les groupes topologiques et ses applications and of Raghunathan's Discrete subgroups of Lie groups, and highlighted some statements. The highlighted statements from Raghunathan's book imply that the answer to the main question is No for semisimple Lie groups. (Of course, there might be more elementary arguments.)

[On page 100 of Raghunathan's book (one of the scanned pages) one reads "As will be seen later ... any lattice in a connected Lie group is finitely generated". Unfortunately, I haven't been able to find where, in the sequel of the book, this is proved. If somebody could indicate the appropriated page (and even scan it), it would be great!] END OF EDIT 2.

EDIT 3. Keivan Karai's answer convinced me that there are elementary arguments showing the negativity of the answer to the main question for semisimple Lie groups. END OF EDIT 3.

Answer 1

The answer to the title question is No, that is

A locally compact group has a compact open subgroup or a discrete infinite cyclic subgroup.

[Keivan Karai helped me a lot, without being responsible for the possible mistakes in this post.]

Step 1. The case of a connected noncompact Lie group.

We claim that a connected noncompact Lie group $G$ has a discrete infinite cyclic subgroup.

Arguing by contradiction, assume $G$ is a counterexample of smallest dimension, and let $Z$ be its center.

Suppose $Ad(G)$ is compact. Then $Z$ is noncompact. If $\dim Z=0$, then $G$ is a noncompact connected covering of the connected compact group $Ad(G)$, which is impossible. Thus, $\dim Z\ge1$. Let $Z_1$ be the connected component of $Z$. If $Z_1$ were compact, $G/Z_1$ would be a counterexample of smaller dimension. If $Z_1$ were noncompact, it would contain a discrete infinite cyclic subgroup, again a contradiction.

Thus, $Ad(G)$ is noncompact. Replacing $G$ with $Ad(G)$, we can suppose $G\subset GL_n(\mathbb R)$.

Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $exp(\mathbb Z x)$ of $GL(V)$ is not an infinite discrete subgroup.

Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ is nonzero and satisfies (C), the same holds for $ad(x)$, implying $Killing(x,x) < 0$. If it were so for all nonzero $x$ in $\mathfrak g$, then $\mathfrak g$, and thus $G$, would be compact.

Step 2. The general case.

Let $G$ be our locally compact group, $C$ its connected component, and recall the following facts (see [1], [2], and references therein):

(1) If $G$ is totally disconnected, then $G$ contains a compact open subgroup.

(2) $C$ is a normal subgroup of $G$, and $G/C$ is totally disconnected.

(3) If $G$ is connected, then every neighborhood of 1 contains a compact normal subgroup $K$ such that $G/K$ is a connected Lie group.

Assume $G$ has no compact open subgroups.

We must show that $G$ has a discrete infinite cyclic subgroup.

As $C$ is noncompact by (1) and (2), we can assume $G=C$, that is $G$ is connected. Then (3) implies that $G$ contains a compact normal subgroup $K$ such that $G/K$ is a connected noncompact Lie group, and we can assume $K=1$, that is $G$ is a connected noncompact Lie group, and the conclusion follows from Step 1.

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[1] Willis, G. The structure of totally disconnected, locally compact groups. Math. Ann. 300 (1994), no. 2, 341-363.

http://gdz.sub.uni-goettingen.de/en/dms/load/img/?IDDOC=167209

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[2] Willis, G. Totally disconnected, nilpotent, locally compact groups. Bull. Austral. Math. Soc. 55 (1997), no. 1, 143-146.

http://journals.cambridge.org/action/displayFulltext?type=1&fid=4856560&jid=BAZ&volumeId=55&issueId=01&aid=4856552

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Answer 2

Hi--

Please refer this link : http://www.math.uiuc.edu/documenta/vol-06/03.pdf

Answer 3

Here is an argument for the semi-simple case: First, suppose that the center of $G$ is trivial. Then $G$ (via adjoint representation) can be considered as a subgroup of $GL(n)$. Consider a neighborhood $U$ of $0$ in the Lie algebra on which the exponential map is a diffeomorphism. If there is any $g \in \exp(U)$ with an eigenvalue which is off the unit circle in $\mathbb C$ then $\langle g \rangle$ is discrete. Otherwise the eigenvalue of every $X \in U$ will be in $\sqrt{-1} \mathbb{R}$, which then would be the case for all $X \in Lie(G)$. Now, it is easy to see that this can only happen for non-compact semi-simple groups, for instance because the Lie algebra must have a copy of $sl_2$. (Jacobson-Morozov). Now, the argument would work in general semi-simple case because, as long as $G/Z(G)$ is not compact, a pre-image of the generator of a non-discrete cyclic subgroup, generates such a subgroup and the universal cover of a semi-simple compact Lie group is also compact.

Answer 4

This is somehow related. A simply-connected Lie group $G$ has uncountably many infinite cyclic subgroups. To see this simply observe that the exponential map at the identity gives a group homomorphism $\mathbb{R}\to G$ defined by $t\to\operatorname{exp}(tX)$ for each $X\in\mathfrak{g}$. This homorpphism is injective because $G$ is simply connected. Then resticting it to $\mathbb{Z}\subset\mathbb{R}$ gives an infinite cyclic subgroup of $G$ for each vector $X$.

Maybe this was obvious but I just realised it, found it surprising, and tought this could be a good spot to leave it.

Also I think this argument could be used to answer the easy case: if $G$ is non-compact then at least one geodesic is non-compact, hence the map is surjectve and the claim follows.