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Consider a locally profinite group $G$, i.e. a locally compact, totally disconnected topological group. Suppose it admits an open maximal compact subgroup named $K$. It is known that $G$ admits as a neighborhood basis of the identity element a collection $\{ K_i \}$ of open compact subgroups, but what can we say about the countability of this family?

In the basic examples I know, $GL_1(\mathbb{Q}_p)$ and $GL_2(\mathbb{Q}_p)$, the family ${K_i}$ is in fact countable, is it always true? Any reference is greatly appreciated, thanks.

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First, since you're only speaking of neighborhoods of identity, you may as well assume G=K is compact. Anyway, a Hausdorff topological group is metrizable if and only if it has a countable basis of neighborhoods of the identity (that is the Birkhoff-Kakutani theorem)- so the criterion you're looking for is simply the metrisability of the group. –  Julien Melleray May 26 '12 at 9:42
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it seems to me that a product over an uncountable set, something like $\prod_{x\in \mathbb R} \mathbb F_p$ will not have a countable basis. –  vytas May 26 '12 at 9:58
    
the terminology "locally profinite" is in contradiction with the mainstream use of "locally" in topological group theory. Its natural meaning would be: every compact subset is contained in a compact open subgroup. Many people deal with totally disconnected LC-groups (LC= locally compact) and they're only referred as "locally profinite" by a few people, mainly around Langlands's theory. –  Yves Cornulier May 26 '12 at 13:38
    
Number of mathscinet/google references: "locally profinite group": 11/4220 "totally disconnected locally compact group": 60/22700; "locally compact totally disconnected group" 23/16300. Wikipedia contains some nontrivial general nontrivial facts (Willis' theory) about totally disconnected LC-groups, inside the page "totally disconnected groups", but also redirects to a page "locally profinite groups" which essentially contains nothing. I hope the latter page will be renamed but I'm not technically qualified to do this. –  Yves Cornulier May 26 '12 at 13:54
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Check out Casselman's notes. They are quite enlightening, and just came up yesterday for another question.

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