Locally Compact Hausdorff Abelian Groups
Posted by John Baez
This is a followup to my post on Pontryagin duality. In the comments to that, Yves de Cornulier said:
It’s well known (and not trivial) that any locally compact abelian group $A$ has a compact subgroup $K$ such that $A/K$ is a Lie group.
This sent me back to the books. I learned some stuff I should have known a long time ago. Let me tell you about it.
The following is some learned commentary on Theorem 25 in here:
- Sidney A. Morris, Pontryagin Duality and the Structure of Locally Compact Abelian Groups, London Math. Soc. Lecture Notes 29, Cambridge U. Press, 1977.
The theorem is supposed to shed light on the structure of abelian topological groups that are locally compact and Hausdorff — the groups that allow for a massive generalization of the Fourier transform called Pontryagin duality.
Morris calls these LCA groups. Lots of people just call ‘em locally compact abelian groups, and hide “Hausdorff” in the fine print.
Anyway, here’s the theorem:
Principal Structure Theorem for LCA Groups: If $G$ is an LCA group, $G$ has an open subgroup that’s isomorphic (as a topological group) to $\mathbb{R}^n \times K$ for some finite $n$ and some compact abelian group $K$.
The meaning of this theorem is a bit obscure at first. It’s instantly striking how $\mathbb{R}^n$ shows up out of the blue, starting from hypotheses that don’t involve the real numbers! But the theorem doesn’t deliver what you naively want, namely a classification of LCA groups.
If you’re hoping for a classification, maybe it’s because normal mathematicians only know a few easy examples of LCA groups, namely:
- the real line $\mathbb{R}$
- the circle $S^1$
- the integers $\mathbb{Z}$
- finite cyclic groups $\mathbb{Z}/n$
and products of these — possibly infinite products, but with only finitely many factors of $\mathbb{R}$, since an infinite product of copies of $\mathbb{R}$ isn’t locally compact.
For easy examples like this, the Principal Structure Theorem is trivial, since the whole group is a product
$G = \mathbb{R}^n \times K \times D$
where $K$ is compact and $D$ is discrete. Why? Just take $\mathbb{R}^n$ to be the product of all your copies of $\mathbb{R}$, take $D$ to be the product of all your copies of $\mathbb{Z}$, and stuff all the rest into your $K$. The subgroup $\mathbb{R}^n \times K$ will be open in $G$, so the theorem holds.
There are, however, much weirder LCA groups! It’s hopeless to classify them! It’s hard to even understand some of them! Spend a few hours trying to visualize the Bohr compactification of the real line. You can do it, but here’s the last guy who succeeded:
So, before telling you about another weird example, let me point out some spinoffs of the Principal Structure Theorem.
First of all, an obvious corollary:
Theorem: If $G$ is a connected LCA group, it’s isomorphic (as a topological group) to $\mathrm{R}^n \times K$, with $K$ compact and connected.
The point is that if $G$ is connected, any open subgroup has to be all of $G$.
This has a further corollary that Todd Trimble pointed out to me:
Theorem: If $G$ is an LCA group, there is a short exact sequence of groups $0 \to G_0 \to G \to \pi_0(G) \to 0$ where the connected component of the identity of $G$, denoted $G_0$, is isomorphic (as a topological group) to $\mathbb{R}^n \times K$, where $K$ is compact and connected.
Now this is very nice. However, if you aren’t paying careful attention, you may be lulled by the easy examples into believing this:
False Theorem: If $G$ is an LCA group, it’s isomorphic (as a topological group) to $\mathbb{R}^n \times K \times \pi_0(G)$ with $K$ compact and connected.
This is wrong! In fact, it’s contradicted by this result of Morris:
Theorem: Not every LCA group $G$ is isomorphic (as a topological group) to $\mathbb{R}^n \times K \times D$ with $K$ compact and $D$ discrete.
Let’s see how he shows this. He exhibits a counterexample that provides a tiny window into the world of weird LCA groups.
Let $G$ be a countable product of copies of $\mathbb{Z}/4$. With its product topology, $G$ is compact. But Morris will give it a sneakier topology!
$G$ has a subgroup $H$ consisting of a countable product of copies of $\mathbb{Z}/2$, one copy sitting inside each copy of $\mathbb{Z}/4$.
Morris puts the product topology on $H$, making it a compact totally disconnected topological group. Totally disconnected means that each point is its own connected component. There are lots of spaces that are totally disconnected but not discrete: the rational numbers are one, and this $H$ is another. But this $H$ is also compact, thanks to Tychonoff’s Theorem.
Next he puts a sneaky topology on $G$. He chooses a base of open neighborhoods of the identity in $G$ that consists of all open sets in $H$ containing the identity.
The sneaky topology is a lot finer than the product topology! The sequence
$(1,0,0,0,...)$ $(0,1,0,0,...)$ $(0,0,1,0,...)$
never gets into $H$, so it doesn’t converge to $0 \in G$ in the sneaky topology. It would in the product topology.
He claims that now $G$ is a totally disconnected locally compact abelian group with $H$ as an open subgroup. I guess all those statements are obvious if you think about each one for a minute!
Now, the Principal Structure Theorem says $G$ has an open subgroup isomorphic (as a topological group) to $\mathbb{R}^n \times K$ with $K$ compact.
Of course $n = 0$, so let’s forget about the $\mathbb{R}^n$ stuff.
So the theorem says: $G$ has an open subgroup $K$ that’s compact.
I suppose $H$ itself is such an open subgroup! The identity is not such a subgroup
Then he says: suppose $G$ were a compact group $K$ times a discrete group $D$:
$G = K \times D$
Then we get a contradiction. Since $G$ is not compact and $K$ is, $D$ must be infinite. But this is impossible because:
1) every infinite subgroup of $G$ has infinitely many elements in $H$
and
2) every discrete subgroup of $H$ is finite.
Think about it.
But what’s the point? We’ll he’s basically saying we’ve got an LCA group $G$ that’s not of the form $\mathbb{R}^n \times K \times D$ with $K$ compact and $D$ discrete.
So, it sure as heck ain’t of the form $\mathbb{R}^n \times K \times \pi_0(G)$ with $K$ compact and connected!
I haven’t reached the point of talking about Yves de Cornulier’s claim, and there’s a lot more fun stuff to say. For example, despite the impossibility of classifying all LCA groups, we can classify them if we throw on some other conditions.
But, this is about as much as anyone could be expected to read in one sitting! The takeaway point is: if you know and love the classification of finite abelian groups, and you know and love some topology, or maybe Fourier transforms, you should get to know Pontryagin duality, and learn a bit about LCA groups. Some of them are very familiar, and others are quite scary — but they sit at a nice intermediate spot between ‘too simple to be interesting’ and ‘too complicated to comprehend’. E. H Gombrich said it well:
Aesthetic delight lies somewhere between boredom and confusion.
Re: Locally Compact Hausdorff Abelian Groups
On the minor point of the “Hausdorff” question. On Wikipedia (or in Hewitt and Ross, if you can find a copy of this old book) you’ll find that a topological group which is T_0 is already Hasudorff. Now, a T_0 space is one with a very weak separation property: given any two distinct points, there is an open set containing precisely one of them. So a topological group which isn’t Hausdorff is already very badly behaved. I think this is why the Hausdorff condition gets pushed aside.
That, and most people these days take “Hausdorff” as being necessary for a space to be locally compact! But this is one of those points (like: “Is 0 a natural number?”) which we could argue about forever, to no particularly conclusion…
Have you looked at the book “Locally compact groups” by Markus Stroppel? I didn’t like it very much, but it’s a new book, concentrating on the Abelian case, and so might be of interest. The MathSciNet reference is MR2226087.