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Suppose $G$ is a compact topological group. We can construct the profinite completion of $G$; let's call this $\Gamma$.

My questions are:

1) Assuming that we know nothing about the (original) topology of $G$ other than that it is compact, is there anything we can say linking the topology of $G$ to the topology of $\Gamma$?

I assume the answer to this question is "no" since it seems to me that we usually regard $G$ as an abstract group when thinking about constructing its profinite completion.

2) If not (and, like I said, I assume the answer to (1) is "no") is there a way to construct a profinite completion (or something like this) of $G$ that takes into account the topology we already have?

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1 Answer 1

up vote 4 down vote accepted

In principle, the answer to question (1) should be "no".

As for question (2), it seems that a basic construction would be to form a "profinite completion" of the topological group $G$ by taking the inverse limit of all quotients $G/N$, where $N$ ranges over all finite-index open normal subgroups of $G$. The resulting completion $\hat{G}$ will have the property that the canonical function $G\to\hat{G}$ is continuous, and it should be somehow universal with respect to this property.

For example, let $\mathbb{Z}_2^\infty$ be the direct sum of infinitely many copies of $\mathbb{Z}_2$, which is a subspace of the infinite product $\mathbb{Z}_2^\omega$. Then $\mathbb{Z}_2^\infty$ has many finite-index subgroups that are not open (e.g. the subgroup of elements with an even number of $1$'s), but it seems to me that every finite-index open subgroup should be the intersection of an open subgroup of $\mathbb{Z}_2^\omega$ with $\mathbb{Z}_2^\infty$. Then the resulting topological profinite completion of $\mathbb{Z}_2^\infty$ ought to be $\mathbb{Z}_2^\omega$.

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Using $\mathbb{Z}_2$ for the group of two elements is potentially confusing, especially when talking about profinite groups. Some people will think you mean the additive group of the ring of $2$-adic integers. –  Pete L. Clark Sep 30 '10 at 9:13
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P.S.: the answer is otherwise right on the money. There are two kinds of profinite completion: as a discrete group (i.e., take the system of all finite index subgroups) and as a topological group (i.e., take the system of all open finite index subgroups). A profinite group is necessarily complete in the second sense, but not necesarily the first: this is called "strongly complete". A big recent theorem is that any topologically finitely generated profinite group is strongly complete. –  Pete L. Clark Sep 30 '10 at 9:16

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