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sniffnoyEdit Oct 20: Rechecking things, much of this may be wrong?? I am confused.
Here's a math question I've thought of but don't really want to think about. I want to ask this on MO or Math.SE, but, well, I don't actually have a precise question. I could ask anyway, I suppose, but first I thought I'd work out my thoughts here.
Say we have a topological group G -- let's explicitly name the topology, call it T. So, G has a left uniformity and a right uniformity (call them L and R). (Terminology: If these are the same, we say the group is balanced. Abelian groups are and compact groups are balanced, e.g.) It also has a two-sided uniformity (call it U) which is the join of the two. The left uniformity makes left-translation uniformly continuous, but it only makes right-translation uniformly continuous if the group is balanced (in which case multiplication itself becomes uniformly continuous); obviously the same is true on the other side. More obviously, neither makes inversion uniformly continuous unless the group is balanced. The two-sided uniformity doesn't make either sort uniformly continuous unless the group is balanced, but it does make inversion uniformly continuous.
Now, uniformities on a set, like topologies, form a complete lattice. However, taking the topology given by a uniformity, while certainly monotonic, is not a homomorphism of complete lattices -- it preserves joins, but not meets. So if we take the meet of the left and right uniformities (call it V), we get some fourth uniformity, but it might not give us back the original group topology, making it possibly kind of dumb in general. Obviously if the group is balanced we get the same thing back. Also if the group is locally compact; if we have a bunch of uniformities yielding the same locally compact topology, then so does their meet. I think also if the group embeds in one that's locally compact, but I didn't bother working out the details there (obviously there I am using the particular uniformities on the group!). So actually finding one where it's different may be hard; there go most of my ideas for making topological groups! (I can't really claim to know much about them.)
So then the question becomes -- firstly, with this new topology, call it T', is G necessarily still a topological group? (Aha! So there's a precise question I can ask. I forgot that this was a prerequisite to the coming less precise questions.) If it is, how does V (the uniformity we used to make T') compare to the various uniformities coming from the fact that (G,T') is a topological group? (I.e. L', R', U' -- obviously it's finer than V'.) Does this uniformity make either sort of translation uniformly continuous? Obviously it makes inversion uniformly continuous, so it does at least make it continuous, which is part of the way towards showing that you have a topological group. And -- the big, imprecise question I originally wanted to ask -- what happens when we iterate this operation? (Consider it as an operation on topologies on G.) Perhaps iterate transfinitely? Since T' is always coarser than T, iterating transfinitely makes sense.
...yeah, I think you can see why I'd prefer to ask someone else about these questions.
Edit: Eh, why the hell not. Asked on Math.SE.
Edit Oct 20: And, answered. The meet of the two is called the Roelcke uniformity and it gives the same topology back; it actually has a pretty simple description (see the link). And it not only makes inversion uniformly continuous, it makes both left and right translation uniformly continuous! Don't know about multiplication as a whole, though.
-Harry
Here's a math question I've thought of but don't really want to think about. I want to ask this on MO or Math.SE, but, well, I don't actually have a precise question. I could ask anyway, I suppose, but first I thought I'd work out my thoughts here.
Say we have a topological group G -- let's explicitly name the topology, call it T. So, G has a left uniformity and a right uniformity (call them L and R). (Terminology: If these are the same, we say the group is balanced. Abelian groups are and compact groups are balanced, e.g.) It also has a two-sided uniformity (call it U) which is the join of the two. The left uniformity makes left-translation uniformly continuous, but it only makes right-translation uniformly continuous if the group is balanced (in which case multiplication itself becomes uniformly continuous); obviously the same is true on the other side. More obviously, neither makes inversion uniformly continuous unless the group is balanced. The two-sided uniformity doesn't make either sort uniformly continuous unless the group is balanced, but it does make inversion uniformly continuous.
Now, uniformities on a set, like topologies, form a complete lattice. However, taking the topology given by a uniformity, while certainly monotonic, is not a homomorphism of complete lattices -- it preserves joins, but not meets. So if we take the meet of the left and right uniformities (call it V), we get some fourth uniformity, but it might not give us back the original group topology, making it possibly kind of dumb in general. Obviously if the group is balanced we get the same thing back. Also if the group is locally compact; if we have a bunch of uniformities yielding the same locally compact topology, then so does their meet. I think also if the group embeds in one that's locally compact, but I didn't bother working out the details there (obviously there I am using the particular uniformities on the group!). So actually finding one where it's different may be hard; there go most of my ideas for making topological groups! (I can't really claim to know much about them.)
So then the question becomes -- firstly, with this new topology, call it T', is G necessarily still a topological group? (Aha! So there's a precise question I can ask. I forgot that this was a prerequisite to the coming less precise questions.) If it is, how does V (the uniformity we used to make T') compare to the various uniformities coming from the fact that (G,T') is a topological group? (I.e. L', R', U' -- obviously it's finer than V'.) Does this uniformity make either sort of translation uniformly continuous? Obviously it makes inversion uniformly continuous, so it does at least make it continuous, which is part of the way towards showing that you have a topological group. And -- the big, imprecise question I originally wanted to ask -- what happens when we iterate this operation? (Consider it as an operation on topologies on G.) Perhaps iterate transfinitely? Since T' is always coarser than T, iterating transfinitely makes sense.
...yeah, I think you can see why I'd prefer to ask someone else about these questions.
Edit: Eh, why the hell not. Asked on Math.SE.
Edit Oct 20: And, answered. The meet of the two is called the Roelcke uniformity and it gives the same topology back; it actually has a pretty simple description (see the link). And it not only makes inversion uniformly continuous, it makes both left and right translation uniformly continuous! Don't know about multiplication as a whole, though.
-Harry
