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] - MathOverflow

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]

2012-10-19T03:12:54Z

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?

Answer by Todd Eisworth for 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]

2012-10-19T16:22:58Z

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."

Answer by Julien Melleray for 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]

2012-10-19T20:38:47Z

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.