8
$\begingroup$

First, some preliminaries:

  • Define an "LCA group" to be a locally compact Hausdorff abelian topological group.

  • Define "smooth manifold" in a way that requires Hausdorffness, but not connectedness or paracompactness. Define a "Lie group" to be a smooth manifold with smooth group operations. Note that with these definitions, any discrete topological space is a manifold, and any discrete topological group is a Lie group.

Now:

I have been told that any LCA group A has a compact subgroup K such that A/K is a Lie group.

However, I have not been able to extract this result from the literature. For some attempts, see this post to the n-Category Cafe.

Can anyone find a proof of this result, or prove it?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Paracompactness is not an issue with locally compact groups: Every $T_{0}$ locally compact group is paracompact and normal, see e.g. Hewitt-Ross, Abstract Harmonic Analysis, I, Ch. II, Thm (8.13), p.76. $\endgroup$
    – Theo Buehler
    Commented Feb 7, 2011 at 5:52
  • $\begingroup$ Thanks, Theo. I should have said "Define "smooth manifold" in a way that requires Hausdorffness, but not connectedness or second-countability". For example, I'm counting the real line with its discrete topology as a 0-dimensional Lie group. $\endgroup$
    – John Baez
    Commented Feb 8, 2011 at 6:05

2 Answers 2

Reset to default
8
$\begingroup$

Corollary 7.54 in Hoffman and Morris, The Structure of Compact Groups, seems to be what you want.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, that seems to do it - thanks! I'll comment in more detail after I get the book. $\endgroup$
    – John Baez
    Commented Feb 7, 2011 at 15:17
  • 2
    $\begingroup$ Okay, Corollary 7.54 of The Structure of Compact Groups does the job: if $A$ is an LCA group, then each neighborhood of the identity contains a compact subgroup $K$ such that $A/K \cong \mathbb{R}^m \times \mathbb{T}^n \times D$ where $D$ is a discrete abelian group. So, $A/K$ is a Lie group by my definition. Great! $\endgroup$
    – John Baez
    Commented Feb 8, 2011 at 5:40
6
$\begingroup$

I am not too familiar to Lie groups, but I think that your claim can follow from the following Theorem which is proven in Rudin:

If A is locally compact abelian group, then there exists a loc.cpct.abelian group $H$ such that $A$ is isomorphic to some $R^k \times H$, and $H$ contains an open compact subgroup $K$.

Reference: This is on page 95 in Deitmar Echterhoff "Principles of Harmonic Analysis", see also When does a LCA group not contain a (closed) infinite cyclic subgroup?.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ In Hewitt and Ross (Abstract Harmonic Analysis, Vol I), Theorem 9.8 of Chapter Two is: Every compactly-generated LCA group is isomorphic to ${\mathbb R}^n\times{\mathbb Z}^m\times K$, with $K$ compact. (An example of a non-compactly-generated group, apparently due to Hensel, is given in the next section, though most of the discussion is omitted from preview.) $\endgroup$
    – B R
    Commented Feb 7, 2011 at 4:51
  • $\begingroup$ The structure of compactly generated LCA groups is much less scary than that of general LCA groups; the hard part was finding the general result I needed. The easiest sort of non-compactly-generated LCA group is a free abelian group on infinitely many generators, given its discrete topology. But, they get worse. (By the way, for those not in the know, "compactly generated topological group" has nothing to do with the concept of "compactly generated topological space" - it means that some compact subset generates the topological group.) $\endgroup$
    – John Baez
    Commented Feb 8, 2011 at 8:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .