Let $G$ be a topological group. $G$ comes equipped with a left (resp. right) uniformity $\mathscr{L}$ (resp. $\mathscr{R}$) which can be characterized as the coarsest uniformity which is compatible with the topology and which makes $x \mapsto gx$ (resp. $x \mapsto xg$) a uniformly continuous map $G \to G$ for all $g \in G$.

Edit: My question is now just:

Is there necessarily a uniformity on $G$ compatible with the topology which makes all left and right multiplication maps uniformly continuous? Bonus points if multiplication $G \times G \to G$ (using the product uniformity on $G \times G$) is uniformly continuous or inversion is continuous.

As Harry Altman points out, there must be (as for any uniformizable space) a finest uniformity $\mathscr{U}$ on $G$ compatible with the topology. Since the uniformities on $G$ form a (complete) lattice there is also a coarsest uniformity $\mathscr{V}$ refining both $\mathscr{L}$ and $\mathscr{R}$. Any uniformity which answers my question must sit between $\mathscr{V}$ and $\mathscr{U}$. Such a uniformity is automatically compatible with the topology since it will sit between, say, $\mathscr{L}$ and $\mathscr{U}$ which are compatible with the topology.


I'm adding another answer based on the answer to this question over on MathOverflow (thanks to Todd Eisworth and Julien Melleray), and some other things.

The answer to your modified question is yes.

Given a topological groups, one can take the meet of its left and right uniformity to get the Roelcke uniformity. Even though meets of uniformities are nasty in general, in this case the result is quite nice and we get the original topology back. The Roelcke uniformity can be described quite simply as the uniformity generated by the entourages $\{ (x,y): x\in VyV\}$ for $V$ a neighborhood of the origin. And, in fact, the Roelcke uniformity makes both left and right translation uniformly continuous, as well as inversion, thus answering your question.

I don't know if or to what extent the Roelcke uniformity makes the multiplication map as a whole uniformly continuous, but it does work with both sorts of translations (and inversion), as you wanted.

(By contrast, if you take the join of the two uniformities as I originally suggested, to get the two-sided uniformity, while this does make inversion uniformly continous, it doesn't make left-translation or right-translation uniformly continuous unless the group was balanced to begin with (i.e. the left and right uniform structures were the same). This is despite the fact that in general joins of uniformities are much nicer than meets of uniformities.) This paragraph is crossed out because see the comments. I think it actually does make both of them uniformly continuous? A basis for this uniformity is the entourages $\{(x,y): x^{-1}y, xy^{-1}\in V\}$ for $V$ a neighborhood of the origin.

A good source for this stuff seems to be Topological Groups and Related Structures by Arhangel'skii and Tkachenko.

  • $\begingroup$ OK, rechecking things, I find myself confused. It looks to me like actually the left uniformity does make right-multiplication continuous, and vice versa? Because if an entourage comes from a nbd $U$, then its preimage under right-multiplication by $g$ comes from the nbd $gUg^{-1}$. So we didn't need to do all this? Something seems fishy here. I'm confused. $\endgroup$ – Harry Altman Oct 21 '12 at 0:19
  • $\begingroup$ And the two-sided uniformity appears to work too? So you could actually go with any of the four uniformities? (Using two-sided or Roelcke if you want inversion to be uniformly continuous as well.) Not sure how I got this wrong before. (Or if I've got it wrong now...) $\endgroup$ – Harry Altman Oct 21 '12 at 0:28

In fact it's true given any uniformizable topological space, there is a unique finest uniformity on it. As for your question about there being coarsest common refinement, since the collection of uniformities on a set forms a lattice (in fact, a complete one) there's certainly a coarsest possible refinement of the two - the only question is whether, if all the uniformities you started out with yield the same topology, their coarsest-possible-common-refinement must also. But since this resulting uniformity is sandwiched between your original uniformities, and an even finer uniformity with the same topology, it must yield the original topology as well. For proofs of all this, I'll just refer you to General Topology by Willard...

So yes, there will be a coarsest possible common refinement of the two, and it will still yield the same topology. What properties it will actually have with regard to multiplication and inversion, I have no idea.

Edit: I may as well add some more information on the already-mentioned uniformities. Obviously, if we have the left uniformity, and right translation is uniformly continuous (or vice versa), the group must have equivalent uniformities. Less obviously, considering the two-sided uniformity (the aforementioned coarsest one containing both left and right, which is generated by sets $\{(x,y):xy^{-1}\in U, x^{-1}y\in U\}$ for neighborhoods of the identity U), if left translation is uniformly continuous on that, once again the uniformities must be the same (this is a quick computation using the above characterization of it; I'll omit it unless you really want to see it).

So the coarsest one containing both won't work unless they were already the same. Now, if the group is locally compact, there'll also be a finest one contained in both, but I have no idea as to the properties of this. (Note that there's a finest uniformity contained in both regardless of whether the group is locally compact, but if it's not locally compact I don't think there's any guarantee that you'll get back the topology you started with, which makes it not very helpful.)

  • $\begingroup$ Thanks for your reply! I didn't realize the uniformities on a set formed a complete lattice. I think I remember convincing myself at some point that the intersection or union of even two uniformities didn't need to be a uniformity and I might have given up hope at that point... I guess the idea is to generate a uniformity from the intersecion/union? $\endgroup$ – Mike F Mar 31 '11 at 5:48
  • $\begingroup$ I hope it's not bad manners if I edit the question/title to account for what I've learned from your answer. $\endgroup$ – Mike F Mar 31 '11 at 6:02
  • $\begingroup$ I don't know about intersections, but the unions are not. You have to extend them to get back to a uniformity. $\endgroup$ – dfeuer Nov 28 '11 at 3:01
  • $\begingroup$ Intersections neither. But since it's complete way, it's complete the other way. $\endgroup$ – Harry Altman Nov 28 '11 at 4:37

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