# Is $SO_2$ an amenable group?

In S. Wagon's "The Banach-Tarski Paradox," amenable groups are defined on p. 12 as follows:

[amenable] groups bear a left-invariant, finitely additive measure of total measure one that is defined on all subsets.

He defines $SO_2$ to be the group of rotations of the unit circle, which he has used to show that $S_1$, the unit circle, is $SO_2$-paradoxical (as an analogue to the usual non-measurable set defined in the interval $[0,1)$ ). I am taking measure theory this term, but am not sure how to assign a measure to subsets of $SO_2$. Thus, I am not really sure where to start in showing whether or not $SO_2$ is an amenable group.

When I look at the Wikipedia entry about amenable groups, I'm unable to make much more sense of the definition in the context of the material.

Is $SO_2$, the group of rotations of the unit circle, an amenable group? If not, why (so that I may build an intuition for these objects)?

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For groups with a topology, like $SO_2$, one usually uses a slightly different definition. You will find it in Wikipedia.

Now, it is easy to see that a compact group is amenable, and $SO_2$ is compact.

If you insist on ignoring the topology of $SO_2$, though, you are still left with an abelian group. And all discrete abelian groups are amenable---but this is harder to see.

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I'm curious: In what sense is it easier to see that compact groups are amenable than that discrete abelian groups are amenable? – t.b. Aug 11 '11 at 22:25
@Theo: using the Wikipedia definition, you'd integrate with respect to the Haar measure. For $SO_2$ that's the $SO_2$-invariant Lebesgue measure. – Ryan Budney Aug 11 '11 at 22:42
@Ryan: of course, $SO_2$ is fine and very easy. However, amenability of abelian groups is obtained in a blow using Kakutani-Markov, while I can think of no way to prove amenability for compact groups that is similarly easy. – t.b. Aug 11 '11 at 22:45
One might guess that the OP had in mind the "analogous" case of $SO_3$, whose non-amenability leads to the Banach-Tarski paradox. – GEdgar Aug 11 '11 at 23:12

An amusing point is that the proof that there is a measure on the Abelian group of rotations is generally done using the Hahn-Banach Theorem, and that requires the Axiom of Choice. So this means that the proof that a Banach-Tarski Paradox does NOT exist in the plane requires the same Axiom of Choice that yields the paradox in 3-space.

But in fact one can eliminate (almost all of) AC from the proof that there is no BTP in the Euclidean plane.

Stan Wagon

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Hahn-Banach doesn't require the full force of AC; you can prove it (so I am told) with the ultrafilter lemma. – Qiaochu Yuan Aug 11 '11 at 22:12

According to Terry Tao's notes here and here it is amenable.

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+1 for the resources! I had not come across this reference to B-T. – Rachel Mar 8 '11 at 4:27