Talk:Bohr compactification

message: Yes, that's all very interesting, but what actually is it and what is it for? --Phil | Talk 13:26, Sep 3, 2004 (UTC)
: The article says what it is. Though more on the relationship touched on at the end could be expanded. Dysprosia 13:33, 3 Sep 2004 (UTC)
replies: text:  I just put something in to start the article.  Later I will add more stuff. It used to be a redirect to Pontryagin duality which is completely misleading. CSTAR 13:51, 3 Sep 2004 (UTC)

text: Just browse a bit. See compactification. Anyway, Wikipedia is not the best way to teach yourself mathematics - it aims to be a reference. Wikibooks aims to be more oriented to education. Dysprosia 00:24, 4 Sep 2004 (UTC)
replies: text: And though she's too modest to say so herhimself, Dysposia has produced a couple of quite readable math Wikibooks. -- orthogonal 09:09, 5 Sep 2004 (UTC)
replies: text:  How sweet of himher to say! Dysprosia 10:05, 5 Sep 2004 (UTC)
thread_title: [Untitled]

message: The Bohr compactification is universal in the comma category of mappings into compact Hausdorff spaces.  For instance see Dixmier 16.1.2.  Note that in that reference
* Compact means compact Hausdorff. 
* Quasi-compact means compact  without any separation property.
Under the current definition of Bohr compactification (which does not require the spaces to be Hausdorff), theorems based on application of the Peter-Weyl theory are false! The Bohr cmpactification may not have a Haar measure.  Why was this condition removed? 
CSTAR 17:05, 20 Sep 2004 (UTC)
replies: text: After my wikibreak, I tend to agree with that.  Note that the non-Hausdorff definition is not really all that different, and should probably still be mentioned on Bohr topology (which is a non-Hausdorff topology on G such that the completion of its Hausdorff quotient is the (Hausdorff) Bohr compactification.  All quasicompact groups have Haar measures, obviously, since all measurable sets correspond to measurable subsets of the Hausdorff quotient, so I'm still a bit confused about that.  Prumpf 09:38, 21 Sep 2004 (UTC)
replies: text:  Agreed quasi-compact groups are extensions of compact groups by the closure of the identity.  It seems reasonable that all this is well-defined, e.g. that the boring technical details, such as the closure of the identity is a normal subgroup in the non-Hausdorff case etc., etc, all work out.  I don't have a handy reference and don't feel like checking the general topology. Thus Haar measure on quasi-compact groups can indeed be defined in terms of Haar measure on the quotient. When I refered to Haar measure I meant in the sense of Borel measure on a locally compact Hausdorff group.  But in any case, I think we now clearly have some agreement.  Yes by all means, mention the Bohr topology.CSTAR 14:24, 21 Sep 2004 (UTC)
replies: text: A compactification of the group G is a pair (X; φ), such that X is a compact group,φ ∈ Hom(G; X), and ran(φ) is dense in X.  As usual, "compact group" means "compact Hausdorff topological group".
* Joan E. Hart and Kenneth Kunen, Bohr Compactifications of Non-Abelian Groups,  May 23, 2002
CSTAR 01:34, 21 Sep 2004 (UTC)
thread_title:  Hausdorff 

message: The theorem is said to be "a direct application of the Tychonoff theorem". From this, I understand that one should consider an embedding of G into a product of compact groups H such that there is a morphism G -> H, each of these morphisms being a coordinate of the product. Can someone clarify on the logical problems? That is, why is it that one can chose a SET of compact groups such that the corresponding embedding is enough to reach all compact groups. Maybe scanning all quotients of G with different toplogies will do it? Can someone add a line to that stament to clarify?
: There is a standard argument based on the Peter-Weyl theory for compact groups. Using it one can show  that any homomorphism of a topological $G$ into a compact group factors through a product of finite dimensional unitary groups. As a result, in defining the Bohr compactification, one can consider continuous homomorphisms from $G$ into finite dimensional unitary groups. It is fairly easy to show that this can be reduced to a product over a set. 
:There may be simpler proofs which do not rely on the Peter-Weyl theory along the lines you suggest. That actually seems quite plausible.--CSTAR (talk) 22:09, 15 July 2008 (UTC)
thread_title:  Proof