Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In measure theoretic language there is a notion of matched pair of locally compact (l.c.) groups due to Baaj-Skandalis-Vaes. A pair $(G_{1}, G_{2})$ is called a matched pair of l.c. groups if there exists a l.c. group $G$ such that, $G_{1}$, $G_{2}$ are closed subgroups of $G$, intersection of $G_{1}$ and $G_{2}$ is trivial inside $G$ and the complement of $G_{1}G_{2}$ is of measure zero in $G$. Does there exists any notion which replace the last condition by some topological criterion?

share|improve this question
    
What do you mean by “topological”? Without using the group structure? Only regarding $G_1G_2$ as a subspace of $G$? –  The User Jul 20 '13 at 19:23
    
I think it means topological as opposed to measure-theoretic. –  MTS Jul 20 '13 at 19:24
    
@MTS “of measure zero with respect to a Haar measure” only depends on the topological-group-structure. It seems to be a plausible interpretation of “topological criterion” to take this as a valid answer. –  The User Jul 20 '13 at 19:50
    
@TheUser yes, of course you're right. –  MTS Jul 20 '13 at 19:51
    
I think that this "matched" criterion is somewhat inspired by Furstenberg's disjointness definition in dynamics (especially in light of MTS answer below), which has interpretation both in measure theory (Ergodic theory) and in topological dynamics (topology), although the interpretations gives you some different outcomes in each category... –  Asaf Jul 20 '13 at 20:45

1 Answer 1

up vote 0 down vote accepted

I suppose you could ask that the complement of $G_1G_2$ is nowhere dense, or more generally a meagre set. But whether this notion is appropriate or not really depends on what application you have in mind.

Also, unless I am missing something, isn't every pair of locally compact groups is a matched pair? Just take $G = G_1 \times G_2$.

share|improve this answer
    
Thanks for your comment. Obviously direct product provides the trivial matching. But I think nowhere dense is a good candidate for that. –  Sutanu Jul 20 '13 at 19:34
    
Why isn’t it trivial in the context of Baaj-Skandalis-Vaes? –  The User Jul 20 '13 at 19:51
1  
The point is not that this notion admits trivial cases, but that it admits highly non trivial ones. The emphasis is not on whether a given pair is a matched pair but whether a group $G$ can be reconstructed from a matched pair... –  Nicola Ciccoli Jul 21 '13 at 9:02
    
@NicolaCiccoli Yes, I wouldn't suspect Baaj-Skandalis-Vaes to come up with a trivial notion! It just seems a funny way to phrase the definition - it makes more sense if $G$ is the focus of the definition rather than $G_1$ and $G_2$. –  MTS Jul 21 '13 at 11:40
    
Previously Baaj and Skandalis considered the map $(g_{1},g_{2})\to g_{1}g_{2}$ to be a homeomorphism. The Definition mentioned above was one step generalization. As per my understanding the goal was is to carry forward Drinfeld's bicrossed product construction in operator algebraic framework. General theory of Double crossed product for quantum groups in operator algebraic setting is due to Baaj-Vaes. One of the delicate parts of their construction was to give an explicit description of the $C^{*}$-algebra of the double crossed product. –  Sutanu Jul 21 '13 at 12:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.