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

Let $G$ be a locally compact group and $H$ be a closed subgroup of $G$. Is there any way to define a reasonable orthonormal basis for $L^2(G/H)$? By "reasonable" I mean elements of the orthonormal basis should be continuous and whose supports are compact and of bounded measure (i.e. there is a large number $M$ such that $\mu(supp f)\leq M$ for all $f$ in the basis).

At least, do you know any reference to learn about concrete orthonormal bases on $L^2(G)$, where $G$ is a locally compact group?

share|improve this question
8  
I'm afraid that, except for compact $G$'s, there is no general answer to your question. Already for $G$ non-compact abelian and $L^2(G)$, the question is very serious: after all, one of the goals of wavelet theory is to construct orthonormal bases of $L^2(\mathbb{R})$! –  Alain Valette Jan 21 '13 at 11:43
    
Thanks Alain for sharing your insight. –  Vahid Shirbisheh Jan 21 '13 at 11:49

2 Answers 2

This was merely too long for a comment.

There is a certain preferable choice in a special situation, where one can choose an orthornormal basis of a rather specific nature.

Let $H$ be cocompact in $G$, then $L^2(H \backslash G)$ decomposes into a direct sum of unitary, irreducible subspaces $V_\pi$. Now assume furthermore that $G$ admits a large compact subgroup $K$ in the sense that $Res_K V_\pi$ decomposes with finite multiplicity, then you can construct an orthonormal basis of $V_\pi$ in terms of irreducible representations of $K$. This is often done in the theory of automorphic representations, where these orthonormal vectors are then called automorphic forms (or rather finite linear combinations of them).

For example, consider $H$ to be a uniform lattice in $G=SL_2(\mathbb{R})$ and $K=SO(2)$. The vectors introduced above will then be real analytic, but have no finite support.

In general, there is no hope for a canonical basis in terms of representations. Already for $H =SL_2(\mathbb{Z}) \subset G = SL_2(\mathbb{R})$, the decomposition will involve a direct integral of irreducible representations.

share|improve this answer
    
Your suggestion looks promising (in special cases), but my original problem has an easier solution when $H\backslash G$ is compact. Anyway, I would like to learn more about the technique you suggested. Could you give some references? –  Vahid Shirbisheh Mar 14 '13 at 6:21
    
I am curious now. What is the easier solution? For the discrete decomposition, there is Deitmar Echterhoff "Principles of Harmonic Analysis" Chapter 9. The decomposition with respect to K is standard for semisimple groups over local field.Harish Chandra "Discrete series 2" first few pages will do for general Lie groups. Writing general G locally compact as projective limit of Lie groups helps to extend HC results. I also discuss related stuff in my Phd Thesis Chapter 5 ff. available on my homepage. –  Marc Palm Mar 14 '13 at 7:14
    
Thanks for references. The original problem is to show the Hecke algebra $\mathcal{H}(G,H)$ has a left regular representation on $L^2(H\backslash G)$. When $H\backslash G$ is discrete, there is a proof based on an orthonormal basis which I wanted to generalize to locally compact case. However, when $H$ is cocompact, one can use standard techniques like the Fubini theorem to show that the convolution product defines a representation. –  Vahid Shirbisheh Mar 14 '13 at 7:33

I do not see what the group has to do with the question--maybe you should redefine 'reasonable' to include the group structure. I know nothing about the groups so here is a general argument which I hope can be made to work. Fix a sequence of disjoint, open sets $A_n\subset G$ such that $\mu(A_n)\leq 1$ for all $n$, $\mu(clA_n\setminus A_n)=0$ for all $n$ and $\bigcup_n cl A_n=G$. Those sets are like cubes in $R^n$. Let $C_0(A_n)$ be the space of all continuous functions on $cl A_n$ which are zero on $cl A_n\setminus A_n$. (If $G$ is metrisable) $C_0(A_n)$ is separable and dense in $L_2(A_n;\mu)$ so do a Gram-Schmidt orthogonalisation on any sequence spaning $C_0(A_n)$ to get the orthonormal basis in $L_2(A_n)$. It consists of continuous functions that are sero on the boundary so extend to continuous on $G$ and measure of the support is $\leq 1$. Take all those bases together and it fits your requirements.

share|improve this answer
1  
I am interested in groups or homogeneous spaces, because I am looking for a suitable way to define a convolution like product and define a regular representation for certain algebras on $L^2(H\backslash G)$. I have been thinking about a similar construction as you described in your answer. But it does seem have several problems: 1. There is no general recipe to partition a general group as the union of cubes in $\mathbb{R}^n$. The Gram-Schmidt orthogonalisation process does not give us an explicit orthonormal basis to work with, but it proves the existence of an orthonormal basis. –  Vahid Shirbisheh Mar 14 '13 at 6:20

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.