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Say we have a topological group $G$. It's easy to see that if $\cdot: G \times G \rightarrow G$ is uniformly continuous (with respect to either the right or left uniformity), then $G$ must have equivalent uniform structures. I figure the converse is probably false, on the basis that otherwise I would have seen this mentioned somewhere. But I can't think of an example, because I don't know many examples of topological groups, and all the examples I can think of with equivalent uniformities are built out of compact, abelian, or discrete groups, all of which do have uniformly continuous multiplication! Can anyone give me a counterexample? Or are these indeed equivalent?

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Exercise 1.8.c on p.79 of Arhangel'skii-Tkachenko, Topological groups and related structures asks the reader to prove that uniform continuity of the multiplication mapping is equivalent to the group being balanced (having equivalent left and right uniformities). –  t.b. Jul 20 '11 at 19:23
...huh. That's a surprise. You should post that as an answer so I can accept it and consider this closed. –  Harry Altman Jul 20 '11 at 19:42
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1 Answer 1

up vote 3 down vote accepted

Disclaimer: I haven't done the exercise carefully myself (and I'm not really in the mood to), but on Harry's request I'm posting it as an answer.

According to Exercise 1.8.c on page 79 of Arhangel'skii-Tkachenko, Topological groups and related structures, the following are equivalent for a topological group (uniformly continuous means uniformly continuous with respect to both the left and the right uniform structures):

  1. The multiplication map $G^{2} \to G$ is uniformly continuous.
  2. The multiplication map $G^{n} \to G$ is uniformly continuous for all $n \geq 2$.
  3. The group is balanced in the sense that the left and right uniformities are equivalent.

Since you're interested in $3 \implies 1$ that's good enough (provided that it is true).

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I would say that there is not much more to it than the fact that inversion is uniformly continuous on a balanced group. –  t.b. Jul 20 '11 at 20:04
Yeah, I feel silly - this is actually easy, but I miscalculated when trying to prove it and figured it probably wasn't true. –  Harry Altman Jul 20 '11 at 20:13
@Harry: Well, this happens to all of us :) Once we're convinced of the wrong track, we often need a nudge from outside... Glad I could help in that respect. –  t.b. Jul 20 '11 at 20:19
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