# Start with a topological group, take the meet of the two uniformities, and take the topology. Is the result again a topological group? [xpost from math.SE]

And what else can be said, if so?

In more detail: Say $(G,\mathscr{T})$ is a topological group. It has a left uniformity $\mathscr{L}$ and a right uniformity $\mathscr{R}$. (It also has a two-sided uniformity $\mathscr{U}$, which is the join of the two.)

Now, uniformities on a given set form a complete lattice, so we can also consider the meet of the two, $\mathscr{V}$. However, the meet of two uniformities that yield the same topology does not necessarily again yield the same topology, so it's possible that $\mathscr{T}'$, the topology coming from $\mathscr{V}$, is coarser than our original topology $\mathscr{T}$.

(Obviously, this does not happen if the group is balanced, i.e. $\mathscr{L}=\mathscr{R}$; it also does not happen if $\mathscr{T}$ is locally compact, since the meet of two uniformities yielding the same locally compact topology does again yield the same topology. Actually, I don't know an actual case where this does happen, so I guess a first question I can ask is, are there any actual examples of this?)

So my question is, is $(G,\mathscr{T}')$ again a topological group? Obviously inversion is continuous, since $\mathscr{V}$ makes inversion uniformly continuous, but it's not clear what would happen with multiplication.

If it is a topological group, then we can ask things like, how does $\mathscr{V}$ compare to $\mathscr{L}'$, $\mathscr{R}'$, $\mathscr{U}'$, and $\mathscr{V'}$? (Well, obviously it's coarser than the last of these.) And considering $\mathscr{T} \mapsto \mathscr{T}'$ as an operation on group topologies on $G$, what happens when we iterate it? When we iterate it transfinitely?

• What is a good example of a nonabelian, non-locallycompact, non-balanced group to try this out on? Linear transformations on some Banach space with some appropriate topology? – Gerald Edgar Oct 19 '12 at 14:28
• Are you sure that you chose the correct lattice of uniformities? I think the uniformities (including left and right uniformities) which produce the same topology, form a lattice. The two-sided uniformity is the join of the left and right uniformities in this lattice. The same for the meet of them. So the meet of the left and right uniformities will produce the same topology as the left an right one. – nano - merge me Sep 4 '15 at 2:48
• No, that's not the case. In general the meet $\mathscr{W}$ of two uniformities $\mathscr{U}$ and $\mathscr{V}$ that yield the same topology $\mathscr{T}$ does not necessarily again yield $\mathscr{T}$. So if we were to try to take the meet of $\mathscr{U}$ and $\mathscr{V}$ inside the set of uniformities yielding $\mathscr{T}$, it would still have to be contained in both $\mathscr{U}$ and $\mathscr{V}$ and thus inside $\mathscr{W}$, and so it would yield a topology contained in that from $\mathscr{W}$, which is a strict subset of $\mathscr{T}$. So that subset is not a lattice. – Harry Altman Sep 4 '15 at 18:32
• @HarryAltman: So you must prove "strict subsetness". But as far as I remember from uniform space theory if $(\mathcal D_i)_{i\in I}$ is a collection of uniformities on a set $X$ producing the same topology $\mathcal T$, then the smallest uniformity containing $\bigcup_{i\in I} \mathcal D_i$ exists and produces $\mathcal T$. So the set of all $\mathcal T$-compatible uniformities on $X$ forms a complete lattice, because any subset of it has a supremum. Remember that the fine uniformity of a topology is defined to be the greatest uniformity producing that topology. – nano - merge me Sep 5 '15 at 1:22
• Yes, your statement about joins is is correct. But meets of uniformities producing the same topology do not in general produce the same topology. However, your inference from joins to meets doesn't work, because the subset isn't actually complete with respect to joins -- you ignored the possibility that $I$ is empty. There is not, in general, a coarsest uniformity producing a given topology. IIRC, this happens iff the space is locally compact -- so in that case it would work as you say, with it being a complete lattice. Otherwise it needn't even be a lattice. – Harry Altman Sep 5 '15 at 21:24

This is most definitely not my field of expertise (so be kind!), but Section 1.8 of the book "Topological Groups and Related Structures, an introduction to topological algebra" by Arhangel'skii and Tkachenko deals with these sorts of questions.

The book is available online at SpringerLink if you have access. Theorem 1.8.15 deals with something called the Roelcke uniformity, and if I'm reading the result correctly, proves that it is compatible with the topology on the group, and also the finest uniformity on the group coarser than the left and right uniformities.

I hope this is helpful! I can edit with better information after my colleague Vladimir Uspenskij gets out of class, as he's an expert on this.

Edit: I asked Uspenskij about this, and his quote was something like "In general, the meet of two uniformities is something horrible, but in topological groups we get the nice Roelcke uniformity."

The meet of the left- and right- uniformities is called the Roelcke uniformity, as Todd Eisworth mentions. The topology it generates is the original topology (the same is true for the join of the two uniformities). One way to see it is as follows: a fundamental system of entourages for the Roelcke uniformity is given by sets of the form $\{(x,y) \colon y \in VxV \}$, for $V$ a neighborood of the neutral element. If $U$ is an open neighborhood of some $g \in G$, then by joint continuity of the group operations there exists an open $V$ containing the neutral element and such that $VgV \subseteq U$, which shows that $U$ is open for the topology induced by the Roelcke uniformity. So the topology induced by the Roelcke uniformity is finer than the original one, and it is clearly coarser.

• If someone knows how to format the sets properly (so that curly brackets appear), please edit my answer! – Julien Melleray Oct 19 '12 at 20:40
• To have curly brackets on MO, you sometimes have to put two backslash symbols before the $\{$. – Mikael de la Salle Oct 19 '12 at 20:45
• Ordinarily, "\lbrace" and "\rbrace" should work. – Lubin Oct 20 '12 at 13:54
• Huh, that's much simpler than I anticipated. Thank you. – Harry Altman Oct 20 '12 at 21:34