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 ?

• See (2.15) and following of iopscience.iop.org/article/10.1070/SM1981v039n03ABEH001518 See also Section 16.4 of Iwaniec-Kowalski. Jul 9, 2018 at 19:30
• Perhaps you're thinking of the Sears--Titchmarsh formula? You can find that in an appendix to Iwaniec's spectral theory book that GH referred to. Jul 11, 2018 at 4:20

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.

The Bruggeman-Kuznetsov formula is not a weak form of the Selberg trace formula, in fact in many situations it is a more refined (or more suitable) tool than the Selberg trace formula. It can be interpreted as a relative trace formula, see e.g. the paper of Knightly and Li in Acta Arithmetica.

• 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. Nov 20, 2018 at 15:03

I was acquainted with Nikolay Vasil'evich Kuznetsov while worked in Vladivostok, 1990s. And he was very kind to mee, too. He tought me that many asymptotics for Bessel functions are not valid, many from Erdelyi's book on asymptotics, and some formulas for integral transforms. And as you I do not remember all his lessons, unfortunately. But I think the answer to your question is in this paper or this one.

Fortunately they are digitized officially now by the Steklov Institute project on mathnet.

• There is a book of Kuznetsov in Russian on his formula. Jul 9, 2018 at 20:58