# Kuznetsov trace formula, orthogonality of Bessel functions

Sorry if this is a vague question. I remember from my younger days that before proving his trace formula, Kuznetsov had a pretty result on orthogonality of Bessel functions. The formulas that I am going to write are WRONG, but the whole point is that I would like to remember the correct one (or a pointer to a reference).

If $J_{\nu}$ denotes the $\nu$th $J$-Bessel function, then if $n$ and $m$ are positive integers, then $<J_n,J_m>=0$ when $n\ne m$ for the scalar product $\int_0^\infty f(x)g(x)dx/x$ (I know this is wrong, let me continue). Thus, as for Fourier series, one can think of expanding reasonable functions into linear combinations of $J_n$ by computing the scalar product to get the coefficients. This does not work because the $J_n$ are not complete, contrary to the $e^{inx}$. Thus, in addition, Kuznetsov adds the "continuous spectrum" functions which are $J_{i\nu}-J_{-i\nu}$ for $\nu\in\Bbb R$, and shows that these are orthonormal, orthogonal to the $J_n$, and now form a complete set. I know that this is very weak form of the Selberg trace formula, but could someone give me the correct formula ?

The Bessel functions $J_\ell$ for $\ell\geq 1$ odd are pairwise orthogonal on the positive axis with respect to the measure $dx/x$. They correspond to the holomorphic spectrum (of various even weights $\ell+1$) of $L^2(\Gamma\backslash H)$. The orthogonal complement of the span of these $J_\ell$'s is continuously (and orthogonally) spanned by the functions $J_{2it}-J_{-2it}$ with $t>0$. This corresponds to the (weight zero and tempered) Maass and Eisenstein spectrum of $L^2(\Gamma\backslash H)$ (of various Laplace eigenvalues $1/4+t^2$). For more details, I recommend Sections 9.3-9.4 in Iwaniec: Introduction to the spectral theory of automorphic forms.
• A small comment: In fact, the Arthur-Selberg trace formula associated to $G$ is, essentially, equivalent to the Jacquet's relative trace formula associated to $(G\times G, \Delta_G)$, where $\Delta_G\hookrightarrow G\times G$ is the diagonal embedding. So the Arthur-Selberg trace formula can be viewed as a special case of the relative trace formula. – Subhajit Jana Nov 20 '18 at 15:03