Orthonormal basis for $L^2(G/H)$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:15:31Z http://mathoverflow.net/feeds/question/119456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h Orthonormal basis for $L^2(G/H)$. Vahid Shirbisheh 2013-01-21T10:59:05Z 2013-03-13T14:26:05Z <p>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).</p> <p>At least, do you know any reference to learn about concrete orthonormal bases on $L^2(G)$, where $G$ is a locally compact group?</p> http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h/124411#124411 Answer by PW for Orthonormal basis for $L^2(G/H)$. PW 2013-03-13T13:29:35Z 2013-03-13T13:29:35Z <p>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.</p> http://mathoverflow.net/questions/119456/orthonormal-basis-for-l2g-h/124420#124420 Answer by Marc Palm for Orthonormal basis for $L^2(G/H)$. Marc Palm 2013-03-13T14:26:05Z 2013-03-13T14:26:05Z <p><em>This was merely too long for a comment.</em></p> <p>There is a certain preferable choice in a special situation, where one can choose an orthornormal basis of a rather specific nature.</p> <p>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).</p> <p>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.</p> <p>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.</p>