Talk:Pontryagin duality

message: It would be nice to add the dual of the set of p-adic numbers in the section Examples [march, 10, 2009].
:Done. Ringspectrum (talk) 08:42, 10 March 2009 (UTC)
thread_title: Older comments

message: Can someone give some examples that show where Pontryagin duality can be used to prove something or understand something better? —The preceding unsigned comment was added by 217.189.228.213 (talk) 08:49, 3 May 2007 (UTC)
thread_title:  What is it good for? 

message: It is analogous to the dual vector space of a finite-dimensional vector space: a vector space V and its dual vector space V * are not naturally isomorphic
This is unclear, but seems to say that a finite-dimensional vector space is not isomorphic to its dual.  That's not right, is it? 72.75.67.226 (talk) 01:53, 3 October 2009 (UTC)
thread_title:  Dual of finite-dimensional vector space 

message: I think one should require the the topological group is also Hausdorff space.  —Preceding unsigned comment added by 132.64.102.21 (talk) 13:12, 24 November 2009 (UTC)
thread_title:  Topological Groups 

message: The article's first para gives a suitably gentle description of where Pontryagin duality might be useful, how, and why. But I don't anywhere find anything approaching a definition. The overall feeling I get from this article is similar to the effect I'd expect if the article on lion consisted of this text:
:"Lions are native to Africa. The males have manes. Their complex behavior, including apex predation, gives them a central role in the culture of many indigenous social groups.  See also: circus."
(Ummm, OK, but shouldn't we mention, for instance, that lion is a species (Panthera leo) of large cat?)
Can somebody please mend this shortcoming. I'd do it myself if I knew enough about the subject matter.—PaulTanenbaum (talk) 16:18, 1 March 2010 (UTC)
thread_title:  What '''is''' it? 

message: Excuse me, this propositions is a little bit vague to me:
Such a theory cannot exist in the same form for non-commutative groups G, since in that case the appropriate dual object G^ of isomorphism classes of representations cannot only contain one-dimensional representations, and will fail to be a group.
Is it possible to reformulate it is such a way that it could be presented as a theorem?
And if you don't mind, I would insert here a little review of non-commutative generalizations.Eozhik (talk) 08:39, 10 November 2012 (UTC)
replies: text: Seems to be all that statement says is, for non-abelian groups, one cannot multiply irreducible unitary representations to get another one unlike the characters in the abelian case. That said, clarification/elaboration is always good. The simplest nonabelian case is listed in See Also:Peter-Weyl theorem. Maybe the main body of the article can say something about that. Mct mht (talk) 10:12, 10 November 2012 (UTC)
replies: text:  But if we do not see how to multiply representations, I suppose, this doesn't mean that such a multiplication does'n t exist... Perhaps one can define an operation of multiplication such that the space of all irreducible unitary representrations becomes a group, is it possible? Eozhik (talk) 14:02, 10 November 2012 (UTC)
replies: text: In a very naive sense, and assuming choice, yes. The axiom of choice implies that every nonempty set carries some group structure, so we can pick one essentially at random for each of our sets of representations. Of course, this will not have any nice properties; the interesting question is whether there is a functor between appropriate categories that we can define. Honestly, the statement above confuses me, too: what isomorphism classes are we taking? where did the notion of dimension come in? I suspect that there is not such a functor; perhaps one could prove this by taking a small (i.e. finite) nonabelian group and doing some casework. In particular, functoriality (which is really what we want) would require our multiplication on characters to satisfy (χf)•(ρf) = (χ•ρ)f where the dot is our supposed multiplication and juxtaposition is computation, for every homomorphism f : A -> A, A being the nonabelian group under consideration. Cyrapas (talk) 01:47, 21 January 2014 (UTC)

text: I interpreted that sentence to mean that in a more abstract treatment of G^, we would find that its elements are precisely the 1-dimensional representations (up to isomorphism) and the group operation is the tensor product. This has a chance of being invertible for one-dimensional representations, but since dimensions multiply under tensoring, it has no chance of being invertible in general, so we would generally just have a monoid. Still, the classical definition of the Brauer group for CSA's naively has the same issue--tensoring only gives a monoid structure--which gets resolved by weakening the notion of equivalence slightly. The quoted text is vague about why one should never expect such a theory to exist, and it's also vague about its own abstract underpinnings. I like it, I just wish it were expanded, clarified, and referenced. (I know nothing about this subject.) 50.132.4.93 (talk) 22:20, 17 November 2015 (UTC)
replies: text:  I must say that I don't like this vagueness. It seems to me it would be more honest to write that in the noncommutative case the classical construction stops working for various reasons, in particular, because characters stop separating points, irreducible representations cease to be one-dimensional, it is not clear how to introduce multiplication on the dual object if it is supposed to be the set of irreducible unitary representatioons, and it is even not clear whether the set of irrducuble unitary representations is indeed a good choice for the status of the dual object for the group. So the task of constructing duality in this case requires complete rethinking. Would anybody mind if I write something like this? Eozhik (talk) 14:54, 1 March 2019 (UTC)

text:  I'll make some edits this week-end, if there will be questions, we can roll back the text and discuss. Eozhik (talk) 06:04, 2 March 2019 (UTC)
thread_title:  About non-commutative theories 

message: In the section about the duality theorem and the canonical map, the canonical map is defined, but it is not shown to be an isomorphism. This, to me, seems a rather important step, since in other contexts (e.g. infinite-dimensional vector spaces) analogous canonical maps can fail to be isomorphisms. Presumably this proof uses the fact that the groups are "small" because they are locally compact. Does anyone know it? Cyrapas (talk) 01:52, 21 January 2014 (UTC)
replies: text: Not sure what you want in a Wiki article--a sketch of a proof? The explicit map was probably only included because it's so easy to define. Any textbook presentation of this material will include a proof that it's in fact an isomorphism. Of course "locally compact" is standing in for "finite-dimensional" here. 50.132.4.93 (talk) 22:57, 17 November 2015 (UTC)
thread_title:  Invertibility of canonical map 

message: What are those patterns of colored hyperbolic tiles supposed to represent? 166.170.23.122 (talk) 04:57, 15 January 2016 (UTC)
replies: text: Does :Commons:File:2-adic_integers_with_dual_colorings.svg answer your question? Melchoir (talk) 04:59, 15 January 2016 (UTC)
thread_title:  Meaning of picture? 

message: I must say that I don't understand the meaning of this phrase in the text: 
but this diverges from the connection with harmonic analysis, which needs to tackle the question of the Plancherel measure on \widehat{G}. I would like to ask its author to clarify this. The duality theories for non commutative groups are constructed. What is the reason to mention the Plancherel measure in this connection? Eozhik (talk) 10:52, 24 February 2019 (UTC)
thread_title:  Phrase about Plancherel measure 

message: The section Pontrjagin duality and the Fourier transform has a subsection titled Fourier transform and Fourier inversion formula for L1-functions.
In that subsection, this sentence appears:
"As an example, suppose G=\R^n, so we can think about \widehat{G} as \R^n by the pairing (\mathbf{v}, \mathbf{w}) \mapsto e^{i \mathbf{v}\cdot \mathbf{w}}.
But this is the first use of the word "pairing" in the article, so the reader does not know why or how one is supposed to think about \widehat{G} by this pairing.
It would be very helpful if the article elaborated on how one is supposed to think of \widehat{G} by this pairing, and why.50.205.142.50 (talk) 18:40, 25 March 2020 (UTC)
thread_title:  Pairing???