Orthonormal basis for $L^2(G/H)$. - MathOverflow

Orthonormal basis for $L^2(G/H)$.

2013-01-21T10:59:05Z

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?

Answer by PW for Orthonormal basis for $L^2(G/H)$.

2013-03-13T13:29:35Z

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.

Answer by Marc Palm for Orthonormal basis for $L^2(G/H)$.

2013-03-13T14:26:05Z

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.