\documentclass[12pt]{article} \usepackage{latexsym} \usepackage{amsmath} \usepackage{amsthm} \usepackage{enumerate} \textwidth=6.5in \hoffset=-.50in \textheight=8.2in \voffset -.55in\newtheorem{theorem}{Theorem}[section] \newtheorem{pr}{Proposition}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{definition}{Definition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{conjecture}{Conjecture}[section] \newtheorem{remark}{Remark}[section] \newtheorem{proposition}{Proposition}[section] \newcommand{\eproof}{{\mbox{\ }~\hfill \mbox{\large $\Box$} \par \vskip 10pt}} \newcommand{\supp}{\mbox{\rm supp\,}} \newcommand{\eps}{\varepsilon} \newcommand{\R}{{\bf R}} \newcommand{\s}{\sum_{i,j=1}^3} \newcommand{\Id}{\mbox{\it Id}} \newcommand{\zero}{^{(0)}} \newcommand{\mat}[4]{\left(\begin{array}{cc} #1 \\#3 & #4 \renewcommand{\a}{\alpha} %\newcommand{\C}{\mathbb{C}} \newcommand{\D}{\Delta} \renewcommand{\d}{\delta} \newcommand{\G}{\Gamma} \newcommand{\g}{\gamma} \newcommand{\gij}{\gamma^{ij}} \newcommand{\ident}{\mathrm{Identity}} \renewcommand{\L}{\Lambda} \renewcommand{\O}{\Omega} \newcommand{\oo}{\overline\O} \newcommand{\p}{\partial} \newcommand{\po}{\p\O} %\newcommand{\R}{\mathbb{R}} \newcommand{\rn}{\R^n} %\newcommand{\supp}{\mathrm{supp}} \renewcommand{\th}{\mathrm{th}} \newcommand{\ep} {\varepsilon} \newcommand{\pa}{} \newcommand{\CD}{} \newcommand{\F}{} \newcommand{\U}{} \newcommand{\dist}{} \newcommand{\Q}{} \newcommand{\Dr}{} \newcommand{\mathspan}{\mbox{span}} \renewcommand{\v}{} \renewcommand{\u}{} \newcommand{\uo}{} \newcommand{\ut}{} \newcommand{\cS}{} \newcommand{\cH}{\mathcal{H}} \end{array}\right)} \title{Two dimensional compact simple Riemannian manifolds are boundary distance rigid} \author{Leonid Pestov\thanks{Part of this work was done while the author\/ was visiting MSRI and the University of Washington.}\\ Institute of Computational Mathematics and Mathematical Geophysics\\Russian Academy of Sciences\\Novisibirsk 630090, Russia \and Gunther Uhlmann\thanks{Partly supported by NSF and a John Simon Guggenheim Fellowship.}\\ Department of Mathematics\\University of Washington\\Seattle, WA 98195, USA} \date{} \begin{document} \renewcommand{\theequation}{\thesection.\arabic{equation}} \maketitle \begin{abstract} We prove that knowing the lengths of geodesics joining points of the boundary of a two-dimensional, compact, simple Riemannian manifold with boundary, we can determine uniquely the Riemannian metric up to the natural obstruction. \end{abstract} \section{Introduction and statement of the results} \label{section1} \setcounter {equation}{0} Let $(M,g)$ be a compact Riemannian manifold with boundary $\partial M$. Let $d_g(x,y)$ denote the geodesic distance between $x$ and $y$. The inverse problem we address in this paper is whether we can determine the Riemannian metric $g$ knowing $d_g(x,y)$ for any $x\in\partial M$, $y\in \partial M$. This problem arose in rigidity questions in Riemannian geometry \cite{M}, \cite{C}, \cite{Gr}. For the case in which $M$ is a bounded domain of Euclidean space and the metric is conformal to the Euclidean one, this problem is known as the inverse kinematic problem which arose in Geophysics and has a long history (see for instance \cite{R} and the references cited there). The metric $g$ cannot be determined from this information alone. We have $d_{\psi^*g}=d_g$ for any diffeomorphism $\psi: M\to M$ that leaves the boundary pointwise fixed, i.e., $\psi|_{\partial M}=\Id$, where $\Id$ denotes the identity map and $\psi^*g$ is the pull-back of the metric $g$. The natural question is whether this is the only obstruction to unique identifiability of the metric. It is easy to see that this is not the case. Namely one can construct a metric $g$ and find a point $x_0$ in $M$ so that $d_g(x_0, \partial M)> \hbox{ sup }_{x,y \in \partial M} d_g(x,y)$. For such a metric, $ d_g $ is independent of a change of $ g $ in a neighborhood of $ x_0 $. The hemisphere of the round sphere is another example. Therefore it is necessary to impose some a-priori restrictions on the metric. One such restriction is to assume that the Riemannian manifold is {\bf simple}, i.e., given two points there is a unique geodesic joining the points and $\partial M$ is strictly convex. $\partial M$ is strictly convex if the second fundamental form of the boundary is positive definite in every boundary point. R.~Michel conjectured in \cite{M} that simple manifolds are boundary distance rigid that is $d_g$ determines $g$ uniquely up to an isometry which is the identity on the boundary. This is known for simple subspaces of Euclidean space (see \cite{Gr}), simple subspaces of an open hemisphere in two dimensions (see \cite{M}), simple subspaces of symmetric spaces of constant negative curvature [BCG], simple two dimensional spaces of negative curvature (see \cite{C1} or \cite{O}). In this paper we prove that simple two dimensional compact Riemannian manifolds are boundary distance rigid. More precisely we show \begin{theorem} Let $(M,g_i), i=1,2$ be a two dimensional simple Riemannian compact Riemannian manifold. Assume $$d_{g_1}(x,y)=d_{g_2}(x,y) \quad \forall (x,y)\in\partial M\times\partial M $$ then there exists a diffeomorphism $\psi: M\to M$, $\psi|_{\partial M}=Id$, so that $$g_2=\psi^*g_1.$$ \end{theorem} As it has been shown in \cite{Sh}, Theorem 1.1 follows from \begin{theorem} Let $(M,g_i), i=1,2$ be a two dimensional simple Riemannian compact Riemannian manifold. Assume $$d_{g_1}(x,y)=d_{g_2}(x,y) \quad \forall (x,y)\in\partial M\times\partial M$$ and $$g_1|_{\partial M}=g_2|_{\partial M}.$$ then there exists a diffeomorphism $\psi: M\to M$, $\psi|_{\partial M}=Id$, $\psi'|_{\partial M}=Id$, so that $$g_2=\psi^*g_1.$$ \end{theorem} We will prove Theorem 1.2. The function $d_g$ measures the travel times of geodesics joining points of the boundary. In the case that both $g_1$ and $g_2$ are conformal to the Euclidean metric $e$ (i.e., $(g_k)_{ij}= \alpha_k\delta_{ij}$, $k=1,2$ with $\delta_{ij}$ the Kr\"onecker symbol), as mentioned earlier, the problem we are considering here is known in seismology as the inverse kinematic problem. In this case, it has been proven by Mukhometov in two dimensions \cite{Mu} that if $(M,g_i), i=1,2$ is simple and $d_{g_1}=d_{g_2}$, then $g_1=g_2$. More generally the same method of proof shows that if $(M, g_i), i=1,2$ are simple compact Riemannian manifolds with boundary and they are in the same conformal class, i.e. $g_1=\alpha g_2$ for a positive function $\alpha$ and $d_{g_1}=d_{g_2}$ then $g_1=g_2$ \cite{Mu1}. In this case the diffeomorphism $\psi$ must be the identity. For related results and generalizations see \cite{B}, \cite{BG}, \cite{C}, \cite{GN}, \cite{MR}. We mention a closely related inverse problems. Suppose we have a Riemannian metric in Euclidean space which is the Euclidean metric outside a compact set. The inverse scattering problem for metrics is to determine the Riemannian metric by measuring the scattering operator (see \cite{G}). A similar obstruction occurs in this case with $\psi$ equal to the identity outside a compact set. It was proven in \cite{G} that from the wave front set of the scattering operator one can determine, under some non-trapping assumptions on the metric, the {\bf scattering relation} on the boundary of a large ball. We proceed to define in more detail the scattering relation and its relation with the boundary distance function. Let $\nu$ denote the unit-inner normal to $\partial M.$ We denote by $\Omega \left( M\right) \rightarrow M$ the unit-sphere bundle over $M$: $$\Omega(M) =\bigcup\limits_{x\in M}\Omega _{x},\quad \Omega _{x}=\{\xi \in T_{x}(M):\left| \xi \right|_g =1\}.$$ $\Omega(M)$ is a $(2\mbox{ dim }M-1)$-dimensional compact manifold with boundary, which can be written as the union $\partial \Omega \left( M\right) =\partial _{+}\Omega \left( M\right) \cup \partial _{-}\Omega \left( M\right) $ $$\partial _{\pm }\Omega \left( M\right) =\{(x,\xi )\in \partial \Omega \left( M\right) ,\;\pm \,(\nu \left( x\right) ,\xi )\geq 0\;\}.$$ The manifold of inner vectors $\partial _{+}\Omega \left( M\right) $ and outer vectors $\partial _{-}\Omega \left( M\right) $ intersect at the set of tangent vectors $$\partial _{0}\Omega \left( M\right) =\{(x,\xi )\in \partial \Omega \left( M\right),\quad (\nu \left( x\right) ,\xi )=0\;\}.$$ Let $(M,g)$ be an n-dimensional compact manifold with boundary. We say that $(M,g)$ is {\bf non-trapping} if each maximal geodesic is finite. Let $(M,g)$ be non-trapping and the boundary $\partial M$ is strictly convex. Denote by $\tau(x,\xi)$ the length of the geodesic $\gamma(x,\xi,t), t\geq 0$, starting at the point $x$ in the direction $\xi \in \Omega_{x}$. These function is smooth on $\Omega(M)\setminus \partial_{0}\Omega(M)$. The function $\tau^{0}=\tau|_{\partial\Omega(M)}$ is equal zero on $\partial_{-}\Omega(M)$ and is smooth on $\partial_{+}\Omega(M)$. Its odd part with respect to $\xi$ $$\tau _{-}^{0}(x,\xi )=\frac{1}{2}\left( \tau ^{0}(x,\xi )-\tau ^{0}\left( x,-\xi \right) \right)$$ is the smooth function. \begin{definition} Let $(M,g)$ be non-trapping with strictly convex boundary. The scattering relation $\alpha :\partial \Omega \left( M\right) \rightarrow \partial \Omega \left( M\right)$ is defined by $$\alpha (x,\xi )= (\gamma (x,\xi ,2\tau _{-}^{0}(x,\xi )),\dot{\gamma}(x,\xi ,2\tau _{-}^{0}(x,\xi ))). $$ \end{definition} The scattering relation is a diffeomorphism $\partial \Omega \left( M\right) \rightarrow \partial \Omega \left( M\right).$ Notice that $\alpha |_{\partial _{+}\Omega \left( M\right) }:\partial _{+}\Omega \left( M\right) \rightarrow \partial _{-}\Omega \left( M\right) ,$ $\alpha |_{\partial _{-}\Omega \left( M\right) }:\partial _{-}\Omega \left( M\right) \rightarrow \partial _{+}\Omega \left( M\right) $ are diffeomorphisms as well. Obviously, $\alpha $ is an involution, $\alpha ^{2}=id$ and $\partial _{0}\Omega \left( M\right) $ is the hypersurface of its fixed\ points,\ $\alpha (x,\xi )=(x,\xi ),\;(x,\xi )\in \partial _{0}\Omega \left( M\right) .$ A natural inverse problem is whether the scattering relation determines the metric $g$ up to an isometry which is the identity on the boundary. In the case that $(M,g)$ is a simple manifold, and we know the metric at the boundary, knowing the scattering relation is equivalent to knowing the boundary distance function (\cite{M}). We show in this paper that if we know the scattering relation we can determine the Dirichlet-to-Neumann (DN) map associated to the Laplace-Beltrami operator of the metric. We proceed to define the DN map. Let $(M,g)$ be a compact Riemannian manifold with boundary. The Laplace-Beltrami operator associated to the metric $g$ is given in local coordinates by $$ \Delta_ g u=\frac{1}{\sqrt{\det g}}\sum^n_{i,j=1}\frac{\partial}{\partial x_i}\left(\sqrt{\det g} g^{ij} \frac{\partial u}{\partial x_j}\right)$$ where $(g^{ij})$ is the inverse of the metric $g$. Let us consider the Dirichlet problem $$ \Delta_ g u = 0\hbox{ on }M,\quad u\Big|_{\partial M} = f.$$ We define the DN map in this case by $$\Lambda_ g(f)=(\nu,\nabla u|_{\partial M})$$ The inverse problem is to recover $g$ from $\Lambda_g.$ In the two dimensional case the Laplace-Beltrami operator is conformally invariant. More precisely \[ \Delta_{\beta g}=\frac{1}{\beta}\Delta_g \] for any function $\beta$, $\beta\ne 0$. Therefore we have that for $n=2$ \[ \Lambda_{\beta(\psi^\ast g)}=\Lambda_{g} \] for any non-zero $\beta$ satisfying $\beta|_{\partial M}=1.$ Therefore the best that one can do in two dimensions is to show that we can determine the conformal class of the metric $g$ up to an isometry which is the identity on the boundary That this is the case is a result proven in [LeU] for simple metrics and for general connected two dimensional Riemannian manifolds with boundary in [LaU]. In this paper we prove: \begin{theorem} Let $(M,g_i), i=1,2$ be compact, simple two dimensional Riemannian manifolds with boundary. Assume that $\alpha_{g_1}=\alpha_{g_2}.$ Then $\Lambda_{g_1}= \Lambda_{g_2}$. \end{theorem} The proof of Theorem 1.2 is reduced then to the proof of Theorem 1.3. In fact from Theorem 1.3 and the result of [LaU] we get that we can determine the conformal class of the metric up to an isometry which is the identity on the boundary. Now by Mukhometov's result we have that the conformal factor must be one proving that the metrics are isometric via a diffeomorphism which is the identity at the boundary. In other words $d_{g_1}=d_{g_2}$ implies that $\alpha_{g_1}=\alpha_{g_2}$. By Theorem 1.3 $\Lambda_{g_1}=\Lambda_{g_2}.$ By the result of [LeU], [LaU], there exists a diffeomorphism $\psi: M\longrightarrow M$, $\psi|_{\partial M}=\mbox{Identity}$ and a function $\beta\ne 0, \beta|_{\partial M}=\mbox{identity}$ such that $g_1= \beta \psi^{\ast} g_2.$ By Mukhometov's theorem $\beta=1$ showing that $g_1=\psi^{\ast} g_2$ proving Theorem 1.2. and Theorem 1.1. The proof of Theorem 1.3 consists in showing that from the scattering relation we can determine the traces at the boundary of conjugate harmonic functions, which is equivalent information to knowing the DN map associated to the Laplace-Beltrami operator. The steps to accomplish this are outlined below. It relies in a connection between the Hilbert transform and geodesic flow. We embed $(M,g)$ into a compact Riemannian manifold $(S,g)$ with no boundary. Let $\varphi_{t}$ be the geodesic flow on $\Omega(S)$ and $\mathcal{H}=\frac{d}{dt}\varphi_{t}|_{t=0}$ be the geodesic vector field. Introduce the map $\psi:\Omega(M)\rightarrow \partial_{-}\Omega(M)$ defined by \[ \psi(x,\xi)=\varphi_{\tau(x,\xi)}(x,\xi), \quad (x,\xi)\in \Omega(M). \] The solution of the boundary value problem for the transport equation $$\mathcal{H}u=0,\quad u|_{\partial _{+}\Omega (M)}=w$$ can be written in the form $$u=w_{\psi}=w\circ\alpha\circ \psi.$$ Let $u^{f}$ be the solution of the boundary value problem \[ \mathcal{H}u=-f,\quad u|_{\partial _{-}\Omega(M)}=0, \] which we can write as \[ u^{f}(x,\xi)=\int\limits_{0}^{\tau(x,\xi)}f(\varphi_{t}(x,\xi))dt,\quad (x,\xi)\in \Omega(M). \] In particular \[ \mathcal{H}\tau=-1. \] The trace \[ If=u^{f}|_{\partial _{+}\Omega(M)} \] is called {\bf the geodesic X-ray transform} of the function $f$. By the fundamental theorem of calculus we have \begin{equation} I\mathcal{H}f=(f\circ\alpha-f) |_{\partial _{+} \Omega(M)}. \end{equation} In what follows we will consider the operator $I$ acting only on functions that do not depend on $\xi$, unless otherwise indicated. Let $L^{2}_{\mu}(\partial_{+}\Omega(M))$ is the real Hilbert space, with scalar product given by \[ (u,v)_{L^{2}_{\mu}(\partial_{+}\Omega(M))}=\int_{\partial_{+}\Omega(M)}\mu uvd\Sigma,\quad \mu=(\xi,\nu). \] Here the measure $d\Sigma= d(\partial M)\wedge d\Omega_x$ where $d(\partial M)$ is the induced volume form on the boundary by the standard measure on $M$ and $$d\Omega_x=\sum_{k=1}^{n} (-1)^{k+1} \xi^k d\xi^1\wedge...\wedge \hat{d\xi^k}\wedge...d\xi^n.$$ As usual the scalar product in $L^{2}(M)$ is defined by $$(u,v)=\int_{M}uv\sqrt{detg}dx.$$ The operator $I$ is a bounded operator from $L^{2}(M)$ into $L^{2}_{\mu}(\partial_{+}\Omega(M))$. The adjoint $I^{\ast}:L^{2}_{\mu}(\partial_{+}\Omega(M))\rightarrow L^{2}(M)$ is given by \[ I^{\ast}w(x)=\int_{\Omega_{x}}w_{\psi}(x,\xi)d\Omega_{x}, \] We will study the solvability of equation $I^{\ast}w=h$ with smooth right hand side. Let $w\in C^{\infty}(\partial_{+}\Omega(M))$. Then the function $w_{\psi}$ will not be smooth on $\Omega(M)$ in general. We have that $w_{\psi}\in C^{\infty} (\Omega(M)\setminus\partial_{0}\Omega(M))$. We give below necessary and sufficient conditions for smoothness of $w_{\psi}$ on $\Omega(M)$. We introduce the operators of even and odd continuation with respect to $\alpha $: \[ A_{\pm }w(x,\xi )=w(x,\xi ),\;\;\;\;\;\; (x,\xi )\in \partial _{+}\Omega \left( M\right), \] \[ A_{\pm }w(x,\xi )=\pm \left( \alpha ^{\ast }w\right) (x,\xi ), \;\;(x,\xi )\in \partial _{-}\Omega \left( M\right). \] The scattering relation preserves the measure $|(\xi, \nu)| d\Sigma$ and therefore the operators $A_{\pm}:L^2_{\mu}(\partial_{+}\Omega(M))\rightarrow L_{\left| \mu \right| }^{2}\left( \partial\Omega \left( M\right) \right)$ are bounded, where $L_{\left| \mu \right| }^{2}\left(\partial \Omega \left( M\right) \right) $ is real Hilbert space with scalar product \[ \left( u,v\right) _{L_{\left| \mu \right| }^{2}\left(\partial \Omega \left( M\right) \right) }=\int\limits_{\partial \Omega \left( M\right) }\left| \mu \right| uvd\Sigma,\quad \mu=(\xi,\nu). \] The adjoint of $A_{\pm}$ is a bounded operator $A^*_{\pm}:L_{\left| \mu \right| }^{2}\left(\partial \Omega \left( M\right) \right)\rightarrow L^2_{\mu}(\partial_{+}\Omega (M))$ given by \[ A_{\pm}^{\ast}u=(u\pm u\circ \alpha)|_{\partial_{+}\Omega(M)}. \] Using $A_{-}^{\ast}$ formula (1.1) can be written in the form \begin{equation} I\mathcal{H}f=-A_{-}^{\ast}f^{0}, \quad f^{0}=f|_{\partial \Omega (M)}. \end{equation} The space $C_{\alpha }^{\infty }\left( \partial _{+}\Omega \left( M\right) \right) $ is defined by \[ C_{\alpha }^{\infty }\left( \partial _{+}\Omega \left( M\right) \right) =\{w\in C^{\infty } \left( \partial _{+}\Omega \left( M\right) \right) :w_{\psi }\in C^{\infty }\left( \Omega \left( M\right) \right) \}. \] We have the following characterization of the space of smooth solutions of the transport equation \begin{lemma} $$C_{\alpha}^{\infty}(\partial_{+}\Omega(M))= \{w\in C^{\infty}(\partial_{+}\Omega(M)):A_{+}w\in C^{\infty}(\partial\Omega(M))\}.$$ \end{lemma} Now we can state the main theorem for solvability for $I^{\ast}$. \begin{theorem} Let $(M,g)$ be a simple, compact two dimensional Riemannian manifold with boundary. Then the operator $I^{\ast}:C_{\alpha}^{\infty}(\partial_{+}\Omega(M))\rightarrow C^{\infty}(M)$ is onto. \end{theorem} Now we define the Hilbert transform: \begin{equation} Hu(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{1+(\xi,\eta)}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta),\quad \xi\in \Omega_{x}, \end{equation} where the integral is understood as a principle value integral. Here $\perp $\ means\ a $90^o$ degree rotation.\ In coordinates $(\xi _{\perp })_{i}=\varepsilon _{ij}\xi ^{j},$\ \vspace{1pt}where \[ \varepsilon =\sqrt{\det g}\begin{pmatrix} _{0} & _{1} \\ _{-1} & _{0} \end{pmatrix}. \] The Hilbert transform $H$ transforms even (respectively odd) functions with respect to $\xi$ to even (respectively odd) ones. If $H_{+}$ (respectively $H_{-})$ is the even (respectively odd) part of the operator $H$: \[ H_{+}u(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{(\xi,\eta)}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta), \] \[ Hu_{-}(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{1}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta) \] and $u_{+}, u_{-}$ are the even and odd parts of the function $u$, then $H_{+}u=Hu_{+}, H_{-}u=Hu_{-}$. \smallskip We introduce the notation $\mathcal{H}_{\perp}=(\xi_{\perp},\nabla)=-(\xi,\nabla_{\perp})$, where $\nabla_{\perp}=\varepsilon\nabla$ and $\nabla$ is the covariant derivative with respect to the metric $g$. The following commutator formula for the geodesic vector field and the Hilbert transform is very important in our approach. \begin{theorem} Let $(M,g)$ be a two dimensional Riemannian manifold. For any smooth function $u$ on $\Omega(M)$ we have the identity \begin{equation} [H,\mathcal{H}]u=\mathcal{H}_{\perp}u_{0}+(\mathcal{H}_{\perp}u)_{0} \end{equation} where \[ u_{0}(x)=\frac{1}{2\pi}\int_{\Omega_{x}}u(x,\xi)d\Omega_{x} \] is the average value. \end{theorem} \smallskip Now we can prove the Theorem 1.3. \bigskip Separating the odd and even parts with respect to $\xi$ in (1.4) we obtain the identities: \[ H_{+}\mathcal{H}u-\mathcal{H}H_{-}u=(\mathcal{H}_{\perp}u)_{0},\quad H_{-}\mathcal{H}u-\mathcal{H}H_{+}u={\mathcal{H}}_{\perp}u_{0}. \] Let $(M,g)$ be a non-trapping strictly convex manifold. Take $u=w_{\psi},w\in C_{\alpha}^{\infty}(\partial_{+}(\Omega))$. Then $$2\pi \mathcal{H}H_{+}w_{\psi}=-\mathcal{H}_{\perp}I^{\ast}w$$ and using the formula (1.2) we conclude \begin{equation} 2\pi A_{-}^{\ast}H_{+}A_{+}w=I\mathcal{H}_{\perp}I^{\ast}w, \end{equation} since $w_{\psi }|_{\partial \Omega (M)}=A_{+}w$. Let $(h,h_{\ast})$ be a pair of conjugate harmonic functions on $M$, \[ \nabla h=\nabla_{\perp}h_{\ast},\quad \nabla h_{\ast}=-\nabla_{\perp}h. \] Notice, that $\delta\nabla=\triangle$ is the Laplace-Beltrami operator and $\delta\nabla_{\perp}=0$. Let $I^{\ast}w=h$. Since $I\mathcal{H}_\perp h=I\mathcal{H} h_{\ast}=-A^{\ast}_{-}h^{0}_{\ast}$, where $h^{0}_{\ast}=h_{\ast}|_{\partial M}$, we obtain from (1.5) \begin{equation} 2\pi A_{-}^{\ast}H_{+}A_{+}w=-A^{\ast}_{-}h_{\ast}^{0}. \end{equation} The following theorem gives the key to obtain the DN map from the scattering relation. \smallskip \begin{theorem} Let $M$ be a 2-dimensional simple manifold. Let $w\in C^{\infty}_{\alpha}(\partial_{+}\Omega(M))$ and $h_{\ast}$ is harmonic continuation of function $h_{\ast}^{0}$. Then the equation (1.6) holds iff the functions $h=I^{\ast}w$ and $h_{\ast}$ are conjugate harmonic functions. \end{theorem} \begin{proof} The necessity has already been established. Using (1.2) and (1.5) the equality (1.6) can be written in the form \[ I\mathcal{H}_{\perp}h=I\mathcal{H}q, \] where $q$ is an arbitrary smooth continuation onto $M$ of the function $h_{\ast}^{0}$ and $h=I^{\ast}w$. Thus, the ray transform of the vector field $\nabla q+\nabla _{\perp }h$ equal $0$. Consequently, this field is potential ([An]), that is, $\nabla q+\nabla _{\perp }h=\nabla p$ and $\ p|_{\partial M}=0.$ Then the functions $h$ and $h_{\ast}=q-p$ are conjugate harmonic functions and $h_{\ast}|_{\partial M}=h^{0}_{\ast}$. We have finished the proof of the main theorem. \end{proof} \smallskip In summary we have the following procedure to obtain the DN map from the scattering relation. For an arbitrary given smooth function $h_{\ast}^{0}$ on $\partial M$ we find a solution $w\in C^{\infty}_{\alpha}(\partial_{+}\Omega(M))$ of the equation (1.6). Then the functions $h^{0}=2\pi (A_{+}w)_{0}$ (notice, that $2\pi (A_{+}w)_{0}=I^{\ast}w|_{\partial M} $) and $h_{\ast}^{0}$ are the traces of conjugate harmonic functions. It gives the map \[ h_{\ast}^{0}\rightarrow (\nu_{\perp},\nabla^{\parallel}h^{0})=(\nu,\nabla h_{\ast})|_{\partial M}. \] which is the DN map. This proves Theorem 1.3. For the definition of $\nabla^{\parallel}$ see section 2. A brief outline of the paper is as follows. In section 2 we collect some facts and definition we need later. In section 3 we study the solvability of $I^{\ast}w=h$ on Sobolev spaces and prove Theorem 1.4. In section 4 we make a detailed study of the scattering relation and prove Lemma 1.1. In section 5 we prove Theorem 1.5. % ------------------------------------------------------------------------ \section{\bf Preliminaries and Notation} \setcounter {equation}{0} Here we will give some definitions and formulas, that we need in what follows. For further references see \cite{E}, \cite{J}, \cite{K}, \cite{Sh}. Let $\pi:T(M)\rightarrow M$ be the tangent bundle over an $n$-dimensional Riemannian manifold $(M,g)$. We will denote points of the manifold $T(M)$ by pairs $(x,\xi)$. The connection map $K:T(T(M))\rightarrow T(M)$ is defined by its local representation $$K(x,\xi,y,\eta)=(x,\eta+\Gamma(x)(y,\xi)),\quad(\Gamma(x)(y,\xi))^{i}=\Gamma^{i}_{jk}(x)y^{j}\xi^{k}, i=1,...,n,$$ where $\Gamma^{i}_{jk}$ are the Christoffel symbols of the metric $g$ $$\Gamma^{i}_{jk}=\frac{1}{2}g^{il}(\frac{\partial g_{jl}}{\partial x^{k}}+\frac{\partial g_{kl}}{\partial x^{j}} -\frac{\partial g_{jk}}{\partial x^{l}}).$$ The linear map $K(x,\xi)=K|_{(x,\xi)}:T_{(x,\xi)}(T(M))\rightarrow T_{x}M$ defines the horizontal subspace $H_{(x,\xi)}=KerK(x,\xi)$. It can be identified with the tangent space $T_{x}(M)$ using the operator $$J^{h}_{(x,\xi)}=(\pi'(x,\xi)|_{H_{(x,\xi)}})^{-1}:T_{x}(M)\rightarrow H_{(x,\xi)}.$$ The vertical space $V_{(x,\xi)}=Ker\pi'(x,\xi)$ can also identified with $T_{x}(M)$ by using the operator $$J^{v}_{(x,\xi)}=(K(x,\xi)|_{V_{(x,\xi)}})^{-1}:T_{x}(M)\rightarrow V_{(x,\xi)}.$$ The tangent space $T_{(x,\xi)}(T(M))$ is the direct sum of the horizontal and vertical subspaces, $T_{(x,\xi)}(T(M))=H_{(x,\xi)}\oplus V_{(x,\xi)}$. An arbitrary vector $X\in T _{(x,\xi)}(T(M))$ can be uniquely decomposed in the form $$X=J^{h}_{(x,\xi)}X_{h}+J^{v}_{(x,\xi)}X_{v},$$ where $$X_{h}=\pi'(x,\xi)X,\quad X_{v}=K(x,\xi)X.$$ We will call $X_{h},X_{v}$ the horizontal and vertical components of the vector $X$ and use the notation $X=(X_{h},X_{v})$. If in local coordinates $X=(X^{1},...,X^{2n})$ then $X_{h},X_{v}$ is given by $$X_{h}^{i}=X^{i},\quad X_{v}^{i}=X^{i+n}+\Gamma^{i}_{jk}(x)X^{j}\xi^{k},\quad i=1,...,n.$$ The horizontal and vertical components $Y^{h},Y^{v}\in T^{\ast}_{x}(M)$ of a covector $Y\in T^{\ast} _{(x,\xi)}(T(M)), Y=(Y^{h},Y^{v})$ are given by $$\langle X,Y\rangle=(X_{h},Y^{h})+(X_{v},Y^{v}).$$ In local coordinates we have $$Y^{h}_{j}=Y_{j}-\Gamma^{i}_{jk}(x)Y_{i+n}\xi^{k},\quad Y^{v}_{j}=Y_{j+n},\quad j=1,...,n.$$ \smallskip Let $N$ be a smooth manifold and $f:T(M)\rightarrow N$ a smooth map. Then the derivative $f'(x,\xi):T_{(x,\xi)}(T(M))\rightarrow T_{f(x,\xi)}(N)$ defines the horizontal $\nabla_{h} f(x,\xi)$ and vertical $\nabla_{v} f(x,\xi)$ derivatives: \[ \nabla_{h} f(x,\xi)=f'(x,\xi)\circ J^{h}_{(x,\xi)}:T_{x}(M)\rightarrow T_{f(x,\xi)}(N), \] \[ \nabla_{v} f(x,\xi)=f'(x,\xi)\circ J^{v}_{(x,\xi)}:T_{x}(M)\rightarrow T_{f(x,\xi)}(N). \] We have then that \begin{equation} f'(x,\xi)X=(\nabla_{h}f(x,\xi),X_{h})+(\nabla_{v}f(x,\xi),X_{v}) \end{equation} In local coordinates \[ \nabla_{h j} f^{(\alpha)}(x,\xi)=(\frac{\partial}{\partial x^{j}}- \Gamma^{i}_{jk}(x)\xi^{k}\frac{\partial}{\partial \xi^{i}})f^{(\alpha)}(x,\xi), \] \[ \nabla_{v j} f^{(\alpha)}(x,\xi)=\frac{\partial}{\partial \xi^{j}}f^{(\alpha)}(x,\xi),\quad\alpha=1,...,\mbox{dim}N. \] \bigskip Let $e:\Sigma\rightarrow T(M)$ be a smooth embedding and $p:T(M)\rightarrow \Sigma$ denotes a normal projection, i.e. $p e=id.$ For a point $(x,\xi)\in e(\Sigma)$, $\mbox{ Ker }p'(x,\xi)=N(\Sigma)\bigcap T_{(x,\xi)}(T(M))$, where $N(\Sigma)$ is the normal bundle of $\Sigma$. \begin{definition} The horizontal and vertical derivatives $\nabla_{h}^{\parallel}, \nabla_{v}^{\parallel}$ on $\Sigma$ are the tangent components of $\nabla_{h}$ and $\nabla_{v}$ with respect to $\Sigma$: \[ \nabla^{\parallel}_{h}f=\nabla_{h}(f\circ p)|_{\Sigma},\quad \nabla_{v}^{\parallel}f=\nabla_{v}(f\circ p)|_{\Sigma}. \] \end{definition} This definition is obviously independent of the choice of the normal projection. From the identity $f=(f\circ p)\circ e$ and (2.1) we have for any vector $X\in T_{(x,\xi)}(T(\Sigma)),$ by identifying $X$ with its embedding $e'(x,\xi)X$, that \[ f'(x,\xi)X=(f\circ p)'(x,\xi)X=(\nabla_{h}(f\circ p)(x,\xi),X_{h})+(\nabla_{v}(f\circ p)(x,\xi),X_{v}). \] Therefore \[ f'(x,\xi)X=(\nabla^{\parallel}_{h}f(x,\xi),X_{h})+(\nabla^{\parallel}_{v}f(x,\xi),X_{v}). \] Let us give an equivalent definition. Denote by $i( x,\xi)$ the isomorphism \[ i\left( x,\xi \right) :T_{\left( x,\xi \right) }\left( T\left( M\right) \right) \rightarrow T_{x}(M)\times T_{x}\left( M\right), \mbox{ defined by } i\left( x,\xi \right) X=(X_{h},X_{v}) \] Let $i_{\Sigma }\left( x,\xi \right) $ be its restriction on $T_{\left( x,\xi \right) }\left( \Sigma \right) ,\;i_{\Sigma }\left( x,\xi \right) =i\left( x,\xi \right) |_{T_{\left( x,\xi \right) }\left( \Sigma \right) }$. \ Denote by $T_{x}^{\left( i\right) }\left( M\right) =pr_{i}\;Ran\,i_{\Sigma }\left( x,\xi \right) \subset T_{x}(M)\times T_{x}\left( M\right) ,\;i=1,2,$\ the projections to the first and second component of the range of $i_\Sigma.$ Then \[\nabla _{h}^{\parallel }f\left( x,\xi \right) =\nabla _{h}\tilde{f}\left( x,\xi \right) |_{T_{x}^{\left( 1\right) }\left( M\right) },\;\;\;\nabla _{v}^{\parallel }f\left( x,\xi \right) =\nabla _{v}\tilde{f}\left( x,\xi \right) |_{T_{x}^{\left( 2\right) }\left( M\right) }, \] where $\tilde{f}:T\left( M\right) \rightarrow N$\ is an arbitrary smooth continuation of $f$. \smallskip \noindent{\bf Example 1} Let us consider the special case of $\Sigma=\Omega(M)$ the unit sphere bundle. Then $p(x,\xi)=(x,\xi/|\xi|)$ is a normal projection. For simplicity of notation in the case of the manifold $\Omega(M)$ we will use notation $\partial$ instead of $\nabla_{v}^{\parallel}$. Then $$\partial=(\frac{\partial}{\partial\xi}-\xi(\xi,\frac{\partial}{\partial\xi}))|_{|\xi|=1}.$$ Clearly, $(\xi,\partial)=0$. Notice, that the horizontal derivative is tangent to the submanifold $\Omega(M)$ since $\nabla_{h}|\xi|=0$. Because of this we keep the original notation for horizontal derivative on $\Omega(M)$ and moreover simplify this notation to just $\nabla$. Thus, in the case of the manifold $\Omega(M)$ we will use notation $\nabla$ and $\partial$ for the horizontal and vertical derivatives. We will also use $\nabla$ for horizontal derivative on $T(M)$. Notice relation \[ T_{(x,\xi)}(\Omega(M))=\{X\in T_{(x,\xi)}(T(M)):(\xi,X_{v})=0\}. \] \smallskip \noindent{\bf Example 2} Let $\Gamma$ be a smooth hypersurface in $M$ with normal $\nu$ and $\Sigma$ its lift to $\Omega(M)$, $\Sigma=\pi^{-1}_{1}(\Gamma)=\{(x,\xi):x\in \Gamma, \xi\in\Omega_{x}(M)\}$, where $\pi_{1}=\pi|_{\Omega(M)}$. Then \begin{equation} T_{(x,\xi)}(\pi^{-1}_{1}(\Gamma))=\{X\in T_{(x,\xi)}(T(M)):(\nu(x),X_{h})=0,(\xi,X_{v})=0\} \end{equation} It means that $T_{x}^{\left( 1\right) }\left( M\right) =\{\eta \in T_{x}\left( M\right) :\left( \eta ,\nu \left( x\right) \right) =0\},\;T_{x}^{\left( 2\right) }(M)=\{\eta \in T_{x}\left( M\right) :\left( \xi ,\eta \right) =0\}$. We have the same vertical derivative as in the case of $\Omega \left( M\right) .$\ The horizontal derivative for this case will be denoted by $\nabla ^{\parallel },$ so that \[ \nabla^{\parallel}=(\nabla-\nu(\nu,\nabla))|_{\pi^{-1}_{1}(\Gamma)}. \] Clearly, $(\nu,\nabla^{\parallel})=0$. We now state a similar definition of vertical and horizontal derivatives on a submanifold for semibasic tensor fields. We recommend chapter 3 of [Sh] for more details. Let $T_{s}^{r}(M)$ denote the bundle of tensor fields of degree $(r,s)$ on $M.$ A section of this bundle is called a tensor field of degree $(r,s)$. Let $\pi _{s}^{r}:T_{s}^{r}(M)\rightarrow M$ be the projection. A\ fiber map $u:T(M)\rightarrow T_{s}^{r}(TM)$, i.e. $\pi _{s}^{r}\circ u=\pi $ is called a semibasic tensor field of degree $(r,s)$ on the manifold $T(M)$. Denote by $\xi $ the semibasic vector field given by the identity map $T(M)\rightarrow T(M)$. An arbitrary tensor field $u$ of degree $(r,s)$ on the manifold $M$, i.e. section $u:M\rightarrow T_{s}^{r}(M)$ defines by the formula $u\circ \pi $ a semibasic tensor field (since $\pi _{s}^{r}\circ (u\circ \pi )=(\pi _{s}^{r}\circ u)\circ \pi =id\circ \pi =\pi $). The map $u\rightarrow u\circ \pi $ identifies tensor fields on $M$ and $\xi $-constant semibasic tensor fields on $T(M)$. A semibasic tensor field on the submanifold $\Sigma \subset T(M)$ is defined as the fiber map $u:\Sigma \rightarrow T_{s}^{r}(M)$, i.e. $\pi _{s}^{r}\circ u=\pi |_{\Sigma }$. If $u$ is semibasic tensor field on $T(M)$, then its restriction $u|_{\Sigma }$ is a semibasic tensor field on $\Sigma $. Using the metric $g$ we can identify the bundle $T_{s}^{r}(M)$ with $T_{0}^{r+s}(M)$ and the bundle $T_{r+s}^{0}(M)$ with $T_{r+s}^{\ast }(M).$ A semibasic tensor field $u:T(M)\rightarrow T_{m}^{0}(TM)$ is\ naturally identified with the polynomial of degree $m$ in $\eta$, $u_{(m)}:T(M)\times T^{m}(M)\rightarrow R,$ $u_{m}\left( x,\xi ,\eta _{\left( 1\right) },...,\eta _{\left( m\right) }\right) =u_{i_{1}...i_{m}}\left( x,\xi \right) \eta _{\left( 1\right) }^{i_{1}}...\eta _{\left( m\right) }^{i_{m}},$ where the fibers of the vector bundle $\ T(M)\times T^{m}(M)=T^{m+1}(M)$ are given by $$T_{x}^{m+1}(M)=\underset{m+1}{\text{ }\underbrace{T_{x}\left( M\right) \times ...\times T_{x}\left( M\right) }}.$$ The horizontal derivative $\nabla _{h}u_{\left( m\right) }$ and vertical derivative $\nabla _{v}u_{\left( m\right) }$ are the fiber maps $T(M)\times T^{m}(M)\rightarrow T^{\ast }(M)=T_{1}^{0}(M)$,\ and \ therefore are semibasic tensor fields on $T(M)\times T^{m}(M)$ of degree 1. The corresponding polynomials $(\nabla _{h}u_{\left( m\right) })_{\left( 1\right) },(\nabla _{v}u_{\left( m\right) })_{\left( 1\right) }$ on $T(M)\times T^{m+1}(M)$\ define semibasic tensor fields $\nabla u,\;\nabla _{\xi }u$\ of degree $(m+1)$, which we call the horizontal and vertical derivatives of $u.$ Thus, by definition \[ (\nabla u)_{(m+1)}= (\nabla _{h}u_{\left( m\right) })_{\left( 1\right) },\;(\nabla _{\xi }u)_{(m+1)}= (\nabla _{v}u_{\left( m\right) })_{\left( 1\right)}. \] Straightforward calculations give the representation in local coordinates: \[ \left( \nabla u\right) _{i_{1}...i_{m+1}}=\tilde{\nabla}_{i_{m+1}}u_{i_{1}...i_{m}}- \Gamma _{i_{\left( m+1\right) }k}^{j}\xi ^{k}\frac{\partial u_{i_{1}...i_{m}}}{\partial \xi ^{j}},\quad (\nabla _{\xi }u)_{i_{1}...i_{m+1}}= \frac{\partial u_{i_{1}...i_{m}}}{\partial \xi ^{i_{m+1}}}, \] where $\tilde{\nabla}$ denotes the usual covariant derivative on the manifold $(M,g).$ Notice, that for $\xi $-constant tensor fields, $\nabla u=\tilde{\nabla}u$ and since we identify $\xi $-constant semibasic tensor fields with tensor fields on $M$, we will use one notation $\nabla $\ for covariant and horizontal derivatives. As in the case of maps, we define tangent derivatives of semibasic tensor fields on the submanifold $\Sigma \subset T(M)$: \[ \nabla ^{\parallel }u=\nabla (u\circ p)|_{\Sigma },\quad \nabla _{\xi }^{\parallel }u=\nabla _{\xi }(u\circ p)|_{\Sigma }. \] In the case $\Sigma =\Omega (M)$ we keep the notation $\nabla $ instead of $\nabla ^{\parallel }$ and use $\partial $ instead $\nabla _{\xi }^{\parallel }$. We mention the following formulas, [Sh] : \begin{equation} \nabla g=0,\quad \nabla \xi =0,\quad \partial _{j}\xi ^{i}= \delta _{j}^{i}-\xi ^{i}\xi _{j}, \end{equation} \begin{equation} \lbrack \nabla ,\partial \rbrack =0,\quad \lbrack \partial _{i},\partial _{j}\rbrack =\xi _{i}\partial _{j}-\xi _{j}\partial _{i}, \end{equation} \begin{equation} \lbrack \nabla _{i},\nabla _{j}\rbrack u=-R_{qij}^{p}\partial _{p}u, \end{equation} where $R$ is the curvature tensor. In the last formula $u$ is a scalar. In the case that $\Sigma =\pi _{1}^{-1}\left( \Gamma \right) $ we will use the same\ notations $\nabla ^{\parallel },\partial $ as in the case of maps. \section{\bf The geodesic X-ray transform} \setcounter {equation}{0} In this section we study the solvability of the equation $I^*w=h$ and prove Theorem 1.4. \begin{lemma} Let $V$ be an open set of a Riemannian manifold $(M,g).$ We can define the ray transform as before. Then the normal operator $I^{\ast}I$ is an elliptic pseudodifferential operator of order $-1$ on $V$ with principal symbol $c_{n}\left| \xi \right| ^{-1}$ where $c_{n}$ is a constant. \end{lemma} \begin{proof} It is easy to see, that \begin{equation} \left( I^{\ast }If\right) \left( x\right) = \int\limits_{\Omega _{x}}d\Omega _{x}\int\limits_{-\tau \left( x,-\xi \right) }^{\tau \left( x,\xi \right) }f\left( \gamma \left( x,\xi ,t\right) \right) dt= 2\int\limits_{\Omega _{x}}d\Omega _{x}\int\limits_{0}^{\tau \left( x,\xi \right) }f\left( \gamma \left( x,\xi ,t\right) \right) dt. \end{equation} Before we continue we make a remark concerning notation. We have used up to know the notation $\gamma (x,\xi ,t)$ for a geodesic. But it is known \cite{J} , that a geodesic depends smoothly on the point $x$ and vector $\xi t\in T_{x}(M).$ Therefore in what follows we will also use sometimes the notation $\gamma (x,\xi t)$ for a geodesic. Since the manifold $M$ is simple and any small enough neighborhood $U$ (in $\left( S,g\right) $) is also simple (an open domain is simple if its closure is simple). \ For any point $x\in U$\ \ there is an open domain $D_{x}^{U}\subset T_{x}\left( U\right) $ such that exponential map $exp_{x}:D_{x}^{U}\rightarrow U,\;exp_{x}\eta =\gamma (x,\eta )$\ is a diffeomorphism onto $U.$ Let $D_{x},\;x\in M$ \ be the inverse image of $M$, then $exp_{x}(D_{x})=M$ and $exp_{x}|_{D_{x}}:D_{x}\rightarrow M$ is a diffeomorphism. Now we change variables in (3.1), $y=\gamma (x,\xi t).$ Then $t=d_g\left( x,y\right)$ and \[ (I^{\ast }If)\left( x\right) = \int\limits_{M}K\left( x,y\right) \,f\left( y\right) dy, \] where \[ K\left( x,y\right) =2\frac{\det \left( exp_{x}^{-1}\right) ^{\prime }\left( x,y\right) \sqrt{\det g\left( x\right) }}{d_g^{n-1}\left( x,y\right)}. \] Notice, that since \begin{equation} \gamma (x,\eta )=x+\eta +O(\left| \eta \right| ^{2}), \end{equation} it follows, that the Jacobian matrix of the exponential map is $1$ at 0, and then $\det (exp_{x}^{-1\prime }\left)( x,x\right) =1/\det \left( exp_{x}\right) ^{\prime }\left( x,0\right) =1$. >From (3.2) we also conclude that \[ d^{2}\left( x,y\right) =G_{ij}\left( x,y\right) \left( x-y\right) ^{i}\left( x-y\right) ^{j},\;\;G_{ij}\left( x,x\right) = g_{ij}\left( x\right) ,\;\;G_{ij}\in C^{\infty }\left( M\times M\right) \] Therefore the kernel of $I^*I$ can be written in the form \[ K\left( x,y\right) =\frac{2\det \left( exp_{x}^{-1}\right) ^{\prime }\left( x,y\right) \sqrt{\det g\left( x\right) }}{\left( G_{ij}\left( x,y\right) \left( x-y\right) ^{i}\left( x-y\right) ^{j}\right) ^{\left( n-1\right) /2}}. \] Thus \ the kernel $K$ has at the diagonal $x=y$ \ a singularity of type $\left| x-y\right| ^{-n+1}.$ \ The kernel \[ K_{0}^{\,}\left( x,y\right) =\frac{2\sqrt{\det g\left( x\right) }}{\left( g_{ij}\left( x\right) \left( x-y\right) ^{i}\left( x-y\right) ^{j}\right) ^{\left( n-1\right) /2}} \] has the same singularity. Clearly, the difference $K-K_{0}$ has a singularity of type $\left| x-y\right| ^{-n+2}.$ Therefore the principal symbols of both operators coincide.\ The principal symbol of the integral operator, corresponding to the kernel $K_{0}$ coincide with its full symbol and is easily\ calculated.\ As a result \[ \sigma \left( I^{\ast }I\right) (x,\xi )= 2\sqrt{\det g\left( x\right) }\int \frac{e^{- i(y,\xi )}}{\left( g_{ij}\left( x\right) y^{i}y^{j}\right) ^{\left( n-1\right) /2}}dy= c_{n}\left| \xi \right| ^{-1}. \] \end{proof} Let $r_{M}$ denotes the restriction from $S$ onto $M.$ \begin{theorem} Let $U$ be a simple neighborhood of the simple manifold $M$. Then for any function $h\in H^{s}\left( M\right) ,\,s\geq 0\;$ there exists function $f\in H^{s-1}\left( U\right) ,\;r_{M}I^{\ast }If=h.$ \end{theorem} \begin{proof} Let $(M,g)$ be simple and\ embedded into a compact Riemannian manifold $\left( S,g\right) \;$ without boundary, of the same dimension. Choose a\ finite atlas\ of $S$, which consist of simple open sets $U_{k}$ with coordinate maps $ \kappa _{k}:U_{k}\rightarrow R^{n}$. Let $\left\{ \varphi _{k}\right\} $ be the subordinated partition of unity:\; $\varphi _{k}\geq 0,\;supp\varphi _{k}\subset U_{k},\;\sum \varphi _{k}=1$. We assume without loss of generality that $M\subset U_1$ and $\varphi _{1}|_{M}=1$. We consider the operators $I_{k},$ $I_{k}^{\ast }$ for the domain $U_{k},$ and the pseudodifferential operator on $\left( S,g\right) $ \[ Pf=\sum\limits_{k}\varphi _{k}\left( I_{k}I_{k}^{\ast }\right) \left( f|_{U_{k}}\right) ,\;\;f\in D^{\prime }\left( X\right). \] Every operator $I_{k}I_{k}^{\ast }:C_{0}^{\infty }\left( U_{k}\right) \rightarrow C^{\infty }\left( U_{k}\right) $ is an elliptic pseudodifferential operator of order $-1$ with principal symbol $c_{n}\left| \xi \right| ^{-1},\;\xi \in T\left( U_{k}\right) .$ Then $P$ is an elliptic\ pseudodifferential operator with principal symbol $c_{n}\left| \xi \right| ^{-1},$ $\xi \in T\left( S\right) $ and, therefore, is a Fredholm operator from $H^{s}(S) $ into $H^{s+1}(S).$ We have that $\mbox{ Ker }P$ has finite dimension, $\mbox{ Ran }P$ is closed and has finite codimension.\ Notice, that $P^{\ast }=P$ (more precisely if $P^{s}=P:H^{s}\left( S\right) \rightarrow H^{s+1}\left( S\right) ,$ then $\left( P_{s}\right) ^{\ast }=P_{-s-1}$)$.$ \smallskip For arbitrary $s\geq 0$\ the operator $r_{M}:H^{s}\left( S\right) \rightarrow H^{s}\left( M\right) $ is bounded and $r_{M}(H^{s}\left( S\right) )=H^{s}\left( M\right) .$ Then the range of $r_{M}P:H^{s}\left( S\right) \rightarrow H^{s+1}\left( M\right) ,\;s\geq -1$ is closed. \smallskip Since $M$ is only covered by $U_{1}$ and $\varphi _{1}|_{M}=1$ we have that $r_{M}Pf=r_{M}I_{1}^{\ast }I_{1}\left( f|_{U_{1}}\right) $. Thus, the\ range of the \ operator$\;r_{M}I_{1}^{\ast }I_{1}:H^{s}\left( U_{1}\right) \rightarrow H^{s+1}\left( M\right) ,\;s\geq -1$ is closed. Now to prove the\ solvability of the\ equation \[ r_{M}I_{1}^{\ast }I_{1}f=h\in H^{s+1}\left( M\right) ,\;s\geq -1 \] in $H^{s}\left( U_{1}\right) $\ is sufficient to show that the kernel of the adjoint $\left( r_{M}I_{1}^{\ast }I_{1}\right) ^{\ast }:\left( H^{\left( s+1\right) }\left( M\right) \right) ^{\ast }\rightarrow \left( H^{s}\left( U_{1}\right) \right) ^{\ast }$ is zero. \smallskip Let $\left\langle ,\right\rangle _{M}$\ and $\left\langle ,\right\rangle $\ be\ dualities between $H^{s}(M)$ and $\left( H^{s}\right) ^{\ast }(M)$ or $H^{s}(S)$ and $H^{-s}(S)$ respectively. The dual\ space $\left( H^{s}\left( M\right) \right) ^{\ast },\;s\geq 0$ can be identified with the subspace of $H^{-s}\left( S\right) :$ $$\left( H^{s}\left( M\right) \right) ^{\ast }=H^{-s}\left( M\right) = \left\{ u\in H^{-s}\left( S\right) :{supp}u\subset M\right\} .$$ For any $f\in H^{s}\left( U_{1}\right) ,\;u\in H^{-(1+s)}\left( M\right) $ we have $$\langle r_{M}I_{1}^{\ast }I_{1}f,u\rangle _{M}= \langle P_{s}\tilde{f},u\rangle =\langle\tilde{f},P_{-s-1}u\rangle ,$$ where $\tilde{f}$\ is an arbitrary continuation of $f$ on the manifold $S.$ On the other hand $$\left\langle r_{M}I_{1}^{\ast }I_{1}f,u\right\rangle _{M}= \left\langle f,\left( r_{M}I_{1}^{\ast }I_{1}\right) ^{\ast }u\right\rangle _{M}.$$ Since $\tilde{f}$ is arbitrary, then equality $\langle \tilde{f},P_{-s-1}u\rangle =\langle f,\left( r_{M}I_{1}^{\ast }I_{1}\right) ^{\ast }u\rangle _{M}$ implies $\left( r_{M}I_{1}^{\ast }I_{1}\right) ^{\ast }=r_{U_{1}}P_{-s-1}=r_{U_{1}}I_{1}^{\ast }I_{1}.$ Because of ellipticity the equality $r_{U_{1}}Pu =0$ implies smoothness $u |_{U_{1}},$ and then $u \in H^{-s-1}\left( M\right) $ implies of $u \in C_{0}^{\infty }( U_{1})$. Since $r_{U_{1}}Pu =I_{1}^{\ast }I_{1}u ,$ then $I_{1}^{\ast }I_{1}u =0\Longrightarrow I_{1}u =0\Longrightarrow u =0.$ \end{proof} Now we are ready to prove Theorem 1.4. \begin{proof} Let $I,I_{1}$ be the geodesic X-ray transforms on $M$ and $U_{1}$ respectively. From Theorem $3.1$ it follows that for any $h\in C^{\infty }(M)$ there exists\ $f\in C^{\infty }(\overline{U})$, such that $r_{M}I_{1}^{\ast }I_{1}f=h.$ Then $u^{f}\in C^{\infty }(\Omega (\dot{N}))$ where $\dot{N}$ denotes the interior of $N$. Let $w=2u_{+}^{f}|_{\partial _{+}\Omega \left( M \right)},$ where $u_{+}^{f}$ is the even part with respect to $\xi$. Then it easy to see, that $w_{\psi }=2u_{+}^{f}|_{\Omega(M)}$ and $I^{\ast }w=h$. The function $w\in C_{\alpha }^{\infty }(\partial _{+}\Omega (M))$ since $w_{\psi }\in C^{\infty }(\Omega (M))$. \end{proof} \bigskip \section{\bf Scattering relation and folds} \setcounter {equation}{0} In this section we prove Lemma 1.1. As indicated before we embed $(M,g)$ into a compact manifold $(S,g)$ with no boundary. Let $(N,g)$ be an arbitrary neighborhood in $(S,g)$ of the manifold $(M,g)$, such that any geodesic $\gamma(x,\xi,t),(x,\xi)\in \Omega(N)$ intersects the boundary $\partial N$ transversally. Then the length of the geodesic ray $\tau$ is a smooth function on $\Omega(\dot{N})$ and the map $\psi:\partial\Omega(M)\rightarrow\partial_{+}\Omega(N)$, defined by \begin{equation} \phi(x,\xi)=\varphi_{\tau(x,\xi)}(x,\xi),\quad (x,\xi)\in\partial\Omega(M), \end{equation} is smooth as well. Moreover it turns out $\phi$ is a fold map with fold $\partial_{0}\Omega(M)$. This fact will be proved in the next Theorem. Once this is proven Lemma 1.1 follows from [H] Theorem C.4.4. From the assumption $A_{+}w\in C^{\infty}(\partial\Omega(M))$ we deduce the existence of a smooth function $v$ on a neighborhood of the range $\phi(\partial\Omega(M))$ such that $w=v\circ\phi$. Consider function $w_{\psi }=w\circ \alpha \circ \psi .$ Change notation $\psi $\ to $\psi _{M},$ keeping $w_{\psi }.$ Denote by\ $\psi _{N}$\ the map, analogical to $\psi _{M},$ $$\psi _{N}\left( x,\xi \right) = \varphi _{\tau \left( x,\xi \right) }\left( x,\xi \right) , \;\;\left( x,\xi \right) \in \Omega \left( N\right) .$$ Then $w_{\psi }=v\vspace{1pt}\circ \phi \circ \alpha \circ \psi _{M}.$ It easy to see, that $\phi \circ \alpha \circ \psi _{M}=\psi _{N}|_{\Omega (M)}.$ Since\ the map\ $\psi _{N}$\ is smooth on $\Omega \left( M\right) ,$ then $w_{\psi }\in C^{\infty }\left( \Omega \left( M\right) \right) $, i.e. $w\in C_{\alpha }^{\infty }( \partial _{+}\Omega (M)) .$ Thus Lemma 1.1 is proven once we show that $\phi$ is a fold. \begin{theorem} Let $(M,g)$ be a strictly convex, non-trapping manifold and $N$ an arbitrary neighborhood of $M$, such, that any geodesic $\gamma(x,\xi, t),(x,\xi)\in \Omega(\dot{N})$ intersects the boundary $\partial N$ transversally. Then the map $\phi$, defined by (4.1) is a fold with fold $\partial_{0}\Omega(M)$. \end{theorem} First we recall the definition of a Whitney fold. \begin{definition} Let $M,N$ be $C^\infty$ manifolds of the same dimension and let $f:M\longrightarrow N$ be a $C^\infty$ map with $f(m)=n.$ The function $f$ is a Whitney fold (with fold L) at $m$ if $f$ drops rank by one simply at $m$, so that $\{ x; df(x) \mbox { is singular } \}$ is a smooth hypersurface near $m$ and $\mbox{ker}(df(m))$ is transverse to $T_{m}L.$ \end{definition} We rewrite the definition of a fold below in a form that we will use. Let $\gamma:(a,b)\rightarrow M, 0\in(a,b)$ be an arbitrary curve in M. Then $f$ induces the curve $\gamma_{1}:(a,b)\rightarrow N$ in the manifold $N$, $\gamma_{1}=f\circ\gamma$. The derivative $f'(m)$ of the map $f$ at the point $m$ maps the tangent vector $\xi=\dot{\gamma}(0)\in T_{m}(M)$ into the tangent vector $\eta=\dot{\gamma}_{1}(0)=f'(m)\xi\in T_{f(m}(N)$. The (acceleration) vectors $\ddot{\gamma}(0), \ddot{\gamma}_{1}(0)$ belong to the spaces $T_{\xi}(T_{m}(M))$ and $T_{\eta}(T_{f(m)}(N))$ respectively, which are tangent to the vector spaces $T_{m}(M)$ and $T_{f(m)}(N)$ and may be identified with them. Fix a point $m\in M$ and nonzero vector $\xi\in Kerf'(m)$. Let us consider the map $L_{\xi}:T_{m}\rightarrow T_{f(m)},\quad L_{\xi}(\ddot{\gamma}(0))=\ddot{\gamma}_{1}(0)$. A calculation in local coordinates gives that the range of the map $L_{\xi}$ is an affine subspaces in $T_{f(m)}(N), RanL_{\xi}=a(\xi)+Ranf'(m)$, where $a(\xi)$ is a vector, depending on $\xi$. Moreover the function $$Hessf(m)(\xi,Y)=\langle L_{\xi}(X),Y\rangle, Y\in (Ranf'(m))^{\perp}$$ does not depend on $X$. Here $\langle,\rangle$ corresponds to the duality of the spaces $T_{n}(N)$ and $T^{\ast}_{n}(N)$. It follows immediately from the that the definition of a fold is equivalent to \begin{definition} The map $f$ is a fold at the point $m$ if \begin{enumerate}[(i)] \item $\mbox{ dimKer }f'(m)=1.$ \item $Hessf(m)(\xi,Y)\neq0.$ \end{enumerate} \end{definition} \begin{remark} Condition 4.2 (i) implies that $\mbox{ dimCoker }f'(m)=1$ and $\mbox{ Ran }L_{\xi}$ is an hyperplane. and condition 4.2(ii) implies that the hyperplane $RanL_{\xi}$ does not pass trough the origin. \end{remark} The following proposition is immediate. \begin{proposition} Let $f:M\rightarrow N$ be a smooth map, $\dim M=\dim N$. Let $e:M_{0}\rightarrow M$ be an hypersurface and the map $\ f\circ e:M_{0}\rightarrow f\left( M_{0}\right) $ is non-singular. Then $\dim Ker\,f^{\prime }\left( x\right) \leq 1$ for any point $x\in M_{0}$. \end{proposition} Now we start proving Theorem 4.1. \begin{proof} We first prove Condition 4.2(i). The smooth map $\phi|_{\partial_{0}\Omega(M)}:\partial_{0}\Omega(M)\rightarrow \phi(\partial_{0}\Omega(M))$ is bijective and therefore $\mbox{ dimKer }\phi'(x_{0},\xi_{0})\leq 1$ for any point $(x_{0},\xi_{0})\in \partial_{0}\Omega(M)$. We prove that the vector $X_{0}\in T_{(x_{0},\xi_{0})}(\partial\Omega(M)), X_{0 h }=\xi_{0}, X_{0 v}=0$, where $X_{0 h},X_{0 v}$ are its horizontal and vertical components (see section 2) belongs to $\mbox{ Ker }\phi'(x_{0},\xi_{0})$. Represent $\phi=\psi_{N}\circ e$, where $e:\partial\Omega(M)\rightarrow \Omega(N)$ is the embedding. Then $\phi'(x,\xi)X=\psi_{N}'(x,\xi)X$ (we identify here the vector $X$ with its embedding $e'(x,\xi)X\in T_{(x,\xi)}(\partial\Omega(M))$). Then $X_{0}\in Ker\,\phi ^{\prime }(x_{0},\xi _{0})$ iff $X_{0}\in T_{(x_{0},\xi _{0})}( \partial \Omega ( M) )\bigcap Ker \psi _{N} ^{\prime }(x_{0},\xi _{0}).$ The fact that $X_{0}\in T_{(x_{0},\xi _{0})}\left( \partial \Omega \left( M\right) \right) $ follows from equality $\left( \nu \left( x_{0}\right) ,\xi _{0}\right) =0$ (see (2.2)). The equality \ $\psi _{N}^{\prime }(x_{0},\xi _{0})X_{0}=(\nabla \psi _{N}(x_{0},\xi _{0}),\xi _{0})=0$ is obvious, since for any point $(x,\xi)\in \Omega(N)$ the point $\psi_{N}(x,\xi)$ is left fixed by geodesic flow and $(\nabla\psi _{N}(x,\xi),\xi)=\mathcal{H}\psi_{N}(x,\xi)=0$. Thus, the vector $X_{0}=(\xi_{0},0)\in \mbox{ Ker }\phi'(x_{0},\xi_{0})$ and the first condition in the definition of a fold has been verified. Let us prove the second condition in the definition of a fold by contradiction. Assume that $\mbox{ Hess }\phi(x_{0},\xi_{0})(X_{0},Y^{0})=0$, where $Y^{0}\in Ker(\phi'(x_{0},\xi_{0}))^\ast$. We will show then $Y^{0}=0$. Denote $(y,\eta)=\phi(x,\xi)$. Then in local coordinates we have for the covector $Y^{0}=(Y^{0}_{1},...,Y^{0}_{2n})$: \begin{equation} Y_{i}^{0}\nabla _{k}^{\parallel }y^{i}(x_{0},\xi_{0} )+ Y_{n+i}^{0}\nabla _{k}^{\parallel }\eta ^{i}(x_{0},\xi_{0} )=0, \end{equation} \begin{equation} Y_{i}^{0}\partial_{k}y^{i}(x_{0},\xi_{0} )+ Y_{n+i}^{0}\partial_{k}\eta ^{i}(x_{0},\xi_{0})=0. \end{equation} We write down these equations using the horizontal and vertical parts of the covector $Y^{0}$ (see section 2) and the Jacobi fields $A_{(k)}\left( x_{0},\xi _{0},t\right) ,\;B_{\left( k\right) }\left( x_{0},\xi _{0},t\right) ,\;k=1,...,n$\ on the geodesic $\gamma (x_{0},\xi _{0},t):$ \[ A_{(k)}=\nabla _{k}^{\parallel }\gamma +\dot{\gamma}\nabla _{k}^{\parallel }\tau ,\quad B_{k}=\partial _{k}\gamma+\dot{\gamma}\partial _{k}\tau. \] Then the equations (4.2),(4.3) can be written in the form \[ \lbrack (A_{(k)},Y_{h}^{0})+(DA_{(k)},Y_{v}^{0})\rbrack |_{t=\tau }=0, \] \[ \lbrack (B_{(k)},Y_{h}^{0})+(DB_{(k)},Y_{v}^{0})\rbrack |_{t=\tau }=0, \] where $D$ means covariant derivative along geodesic. Without lost of generality one can take $Y^{0}_{v}=Z^c|_{t=\tau },$\ $Y^{0}_{h}=-DZ^c|_{t=\tau }$, where $Z^c$ is some Jacobi field on $\gamma(x_0,\xi_0, t)$. It is known that for any Jacobi fields $X,Y$, the Wronskian $\{X,Y\}=(DX,Y)-(X,DY)$ is constant. Then we get \[ \{Z^{c},A_{(k)}\}|_{t=0}=0,\quad \{Z^{c},B_{(k)}\}|_{t=0}=0. \] The initial data is given by \[ A_{(k)}^{j}|_{t=0}=\delta _{k}^{j}-\nu _{0}^{i}\nu _{0k}+\xi _{0}^{j}\nabla _{k}^{\parallel }\tau ,\quad DA_{(k)}|_{t=0}=0, \] \[ B_{(k)}|_{t=0}=\xi _{0}\partial _{k}\tau ,\quad (DB_{(k)})^{j}|_{t=0}=\delta _{k}^{j}-\xi _{0}^{j}\xi _{0k}, \] where $\nu_{0}=\nu(x_{0}).$ We have \[ Z^c|_{t=0}=a\xi_{0}+c\partial\tau,\quad DZ^c|_{t=0}=b\nu_{0}-c\nabla^{\parallel}\tau, \] where $a,b,c$ are some constants, depending on the point $(x_{0},\xi_{0})$. From these general solutions we find one satisfying the condition that $Y^{0}\in T_{\phi(x,\xi)}(\partial\Omega(N))$, or the conditions (2.2), \begin{equation} (Y^{0}_{h},\nu_{N}(y))=-(DZ^c|_{t=\tau},\nu_{N}(y))=0, \end{equation} \begin{equation} (Y^{0}_{v},\eta)=(Z^c|_{t=\tau},\eta)=0, \end{equation} where $\nu_{N}$ is the normal to $\partial N$. Consider at first the condition (4.5). Using the Jacobi field $J=(\tau-t)\dot{\gamma}$ it can be rewritten in the form \[ \{Z^c,J\}|_{t=\tau}=\{Z^c,J\}|_{t=0}=0, \] and, therefore, $a=-c\tau$. Thus, \[ Z^c|_{t=0}=c(\partial\tau-\xi_{0}\tau),\quad DZ^c|_{t=0}=b\nu_{0}-c\nabla^{\parallel}\tau. \] Now consider our hypothesis that $\mbox{ Hess }\phi_{(x_{0},\xi_{0})}(X_{0},Y^{0})=0$. A straightforward calculation shows that in local coordinates we have \[ \mbox{ Hess }\phi_{(x_{0},\xi_{0})}(X_{0},Y^{0})=-B(x_{0},\xi_{0})\{Z,P\}|_{t=\tau}= -B(x_{0},\xi_{0})\{Z^{c},P\}|_{t=0}, \] where $B=-\xi^{i}\xi^{j}\nabla^{\parallel}_{i}\nu_{j}$ is the second fundamental form of the boundary $\partial M$ and $$P=(\nu,\nabla\gamma)+(\nu ,\nabla \tau )\dot{\gamma}$$ is a Jacobi field. Since $B>0$, our hypothesis gives that $\{Z,P\}|_{t=0}=0$. Because of the initial data $P(x_{0},\xi_{0},0)=\nu_{0}+(\nu_{0} ,\nabla \tau )\xi, \quad DP(x_{0},\xi_{0},0)=0$ we obtain $b=-c(\nu_{0},\nabla\tau)$ and \[ Z^c|_{t=0}=c(\partial\tau-\xi_{0}\tau),\quad DZ^c|_{t=0}=-c\nabla\tau. \] Now we prove that for any point $( x,\xi) \in \Omega ( \dot{N}) $ the vector field \begin{equation} Z\left( x,\xi ,t\right) =(\partial \tau -\xi \tau )\left( \varphi _{t}\left( x,\xi \right) \right) ,\;t\in \lbrack -\tau \left( x,-\xi \right) ,\tau \left( x,\xi \right) \rbrack \end{equation} on the geodesic $\gamma \left( x,\xi ,t\right) $ is a Jacobi field and \begin{equation} DZ\left( x,\xi ,t\right) =\left( \nabla \tau \right) \left( \varphi _{t}\left( x,\xi \right) \right). \end{equation} Applying to the equation $\mathcal{H}\tau =-1$ the operator $\mathcal{H}\partial $ and using (2.3)-(2.5) we get $\mathcal{H}^{2}X+RX=0,\;\;\left( RX\right) ^{i}(x,\xi )=R_{jkl}^{i}\left( x\right) \xi ^{j}X^{k}\xi ^{l},$ where the semibasic vector field $X=\partial \tau -\xi \tau .$ The operators $\mathcal H$ and $D$ are related by: \begin{equation} \left( \mathcal{H} X\right)\circ \varphi _{t} =D(X\circ \varphi _{t}) , \end{equation} where $X$ is an arbitrary semibasic vector field. Then (4.6) is a Jacobi field. Since from $\mathcal{H}\tau =-1$ it follows $\mathcal{H}( \partial \tau -\xi \tau ) =-\nabla \tau ,$ we obtain (4.7) from (4.8). Thus, $Z^{c}( x_{0},\xi _{0},t) =cZ( x_{0},\xi _{0},t) $ and condition (4.4) gives \begin{equation} 0=(DZ|_{t=\tau },\nu_{N} (y))=-c(\nabla \tau ( y,\eta ) ,\nu _{N}(y)). \end{equation} Now we show that \begin{equation} \nabla \tau |_{\partial _{-}\Omega \left( N\right) }=-\frac{\nu _{N}}{\left( \nu _{N},\xi \right) }. \end{equation} Let $h$\ be a\ smooth\ function on $N$, such that $h|_{\partial M}=0,\;\nabla h|_{M}\neq 0.$ Then since $h(x,\xi ,\tau (x,\xi ))=0,\;(x,\xi )\in \Omega (N)$, we have that \[ \nu _{i}^{N}(y)\nabla \gamma _{(k)}^{i}(x,\xi ,\tau (x,\xi ))+ \left( \nu _{N}\left( y\right) ,\eta \right) \nabla _{k}\tau (x,\xi )=0, \] where $\left( y,\eta \right) =\varphi _{\tau \left( x,\xi \right) }(x,\xi )\in \partial _{-}\Omega (N)$. Change here $(x,\xi )$ to $\varphi _{t}(x,\xi ),\;t\in \left[ 0,\tau \left( x,\xi \right) \right] .$ Then the point $\left( y,\eta \right) $ does not change. Using the initial data $\nabla \gamma _{(k)}^{i}(x,\xi ,0)=\delta _{k}^{i}$ we get in the limit $\nu _{N}(y)+(\nu _{N}(y),\eta )\nabla \tau (y,\eta )=0.$ Thus, (4.10) is proved and we conclude from (4.9) that $c=0.$ Thus, $\mbox{ Hess }\phi_{(x_{0},\xi_{0})}(X_{0},Y^{0})=0$ iff $\mbox{ Ker }(\phi'(x_{0},\xi_{0}))^\ast=\{0\}$. We have finished the proof of the Theorem. \end{proof} \bigskip \section{\bf The Hilbert transform and geodesic flow} \setcounter {equation}{0} \bigskip In this section we prove Theorem 1.5 in the introduction. %Let $(M,g)$ be orient 2-dimensional manifold. For any vector $a\in %T_{x}$ denote by $a_{\perp}$ the vector rotated on right angle in %some fixed direction, such that $(a,a_{\perp})=0, |a|=|a_{\perp}|, %(a_{\perp})_{\perp}=-a$. For definition assume that in local %coordinates $(a_{\perp})_{i}=\varepsilon _{ij}a^{j}$, where %$\varepsilon$ is volume tensor %$$\varepsilon=\sqrt{detg}\left(% %\begin{array}{cc} % 0 & 1 \\ % -1 & 0 \\ %\end{array}% %\right)$$ Notice, that $\nabla_{\perp}\xi=0$, because of %$\nabla\varepsilon=0$. \ %Define the integral transform : %$$Hu(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{1+(\xi,\eta)}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta),\quad %\xi\in \Omega_{x},$$ where integral is in principle value. If one %use angle variables then it easy to see that it is a regular %Hilbert transform. Let $H$ be the Hilbert transform as defined in (1.3) %We summarize here some well know properties of the Hilbert transform. %\begin{itemize} % \item $H:C^{\infty}(\Omega_{x}))\rightarrow C^{\infty}(\Omega_{x}), %L^{2}(\Omega_{x}))\rightarrow L^{2}(\Omega_{x})$, % \item $Hu=0$ if $u$ does %not depend on $\xi$, % \item $(Hu,v)=-(u,Hv)$, where $(,)$ is scalar %```product in $L^{2}(\Omega_{x}),$ % \item $H^{2}u=-u+u_{0}$, where $u_{0}$ %is the average value over the circle $\Omega_{x}$, %$$u_{0}(x)=\frac{1}{2\pi}\int_{\Omega_{x}}u(x,\xi)d\Omega_{x}.$$ %\end{itemize} We have that $H$ is a unitary operator in the space $L^{2}_{0}(\Omega_{x})=\{u\in L^{2}(\Omega_{x}): u_{0}=0\}$, $$(u,v)=(Hu,Hv),\quad \forall u,v\in L^{2}_{0}(\Omega_{x}),$$ $$H^{2}(u)=-u,\quad \forall u\in L^{2}_{0}(\Omega_{x}).$$ Clearly, all these properties remain the same if we change $\Omega_{x}$ to $\Omega(M)$. \ %Notice, that Hilbert transform $H$ transforms even (odd) functions %with respect to $\xi$ to even (odd) ones. If $H_{+(-)}$ is even %(odd) part of operator $H$: %$$H_{+}u(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{(\xi,\eta)}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta),$$ %$$Hu_{-}(x,\xi)=\frac{1}{2\pi}\int_{\Omega_{x}}\frac{1}{(\xi_{\perp},\eta)}u(x,\eta)d\Omega_{x}(\eta)$$ %and $u_{+}, u_{-}$ are even and part of function $u$, then %$H_{+}u=Hu_{+}, H_{-}u=Hu_{-}$. %\smallskip %Introduce notations %$H_{\perp}=(\xi_{\perp},\nabla)=-(\xi,\nabla_{\perp})$, where %$\nabla_{\perp}=\varepsilon\nabla$. In what follows the formula of %commutation for geodesic flow and Hilbert transform has impotent %value. %\bigskip %Lemma 5.1 \emph {For any smooth function $u$ on $\Omega(M)$ the %$$[H,H_{g}]u=H_{\perp}u_{0}+(H_{\perp}u)_{0}.$$ %holds}. In order to prove Theorem 1.4 we need the following commutator formula which is valid for Riemannian manifolds of any dimension \begin{lemma} Let $u$ be a smooth function on the manifold $\Omega ^{2}(M)=\bigcup_{x\in M}\Omega^{2}_{x},\quad \Omega^{2}_{x}=\{(x,\xi,\eta):\xi,\eta\in \Omega_{x}\}.$ Then \begin{equation} \nabla \int\limits_{\Omega _{x}}u(x,\xi ,\eta) d\Omega _{x}\left( \eta \right) = \int\limits_{\Omega _{x}}\nabla^{(2)} u(x,\xi ,\eta ) d\Omega _{x}\left( \eta \right), \end{equation} where $\nabla^{(2)}$ under the integral sign in (5.1) denotes the horizontal derivative on $\Omega ^{2}(M)$, \end{lemma} \[ \nabla^{(2)}_{j} u(x,\xi ,\eta )=(\frac{\partial}{\partial x^{j}}-\Gamma^{i}_{jk}\xi^{k}\partial_{i(\xi)}-\Gamma^{i}_{jk}\eta^{k}\partial_{i(\eta)})u(x,\xi ,\eta ). \] Notice that the horizontal tangential derivative can be defined on $T(M)\times T(M)$ in a similar fashion to the case of $T(M)$ in section 2. \begin{proof} Let $\varphi \in C_{0}^{\infty }\left( \R^{+}\right) $ be arbitrary function. We define the function $v$ on $T^2\left( M\right) $ by \[ v(x,\xi ,\eta )=\varphi \left( \left| \eta \right| \right) u(x,\xi /\left| \xi \right| ,\eta /\left| \eta \right| ) \] Let us consider the integral \[ S(x,\xi )=\int\limits_{T_{x}(M)}v(x,\xi ,\eta )dT_{x}\left( \eta \right) . \] Identifying $T_{x}(M)$ with $R^{n}$ we have \[ S(x,\xi )=\int\limits_{R^{n}}v(x,\xi ,\eta )\sqrt{\det g\left( x\right) }d\eta . \] Then \[ \nabla _{j}S=\frac{\partial S}{\partial x^{j}}- \Gamma _{jk}^{i}\xi ^{k}\frac{\partial S}{\partial \xi ^{i}}= \int\limits_{R^{n}}(\frac{\partial v}{\partial x^{j}}- \Gamma _{jk}^{i}\xi ^{k}\frac{\partial v}{\partial \xi ^{i}})\sqrt{\det g}d\eta +\int\limits_{R^{n}}v\frac{\partial \ln \sqrt{\det g\left( x\right) }}{\partial x^{j}}\sqrt{\det g}d\eta. \] Since $\partial \ln \sqrt{\det g}/dx^{j}=\Gamma _{jk}^{k}$ we rewrite the last integral in the form $$\int\limits_{R^{n}}v\frac{\partial }{\partial \eta ^{k}} \left( \Gamma _{jl}^{k}\eta ^{l}\right) \sqrt{\det g}d\eta .$$ Then $$\nabla _{j}S=\int\limits_{R^{n}}(\frac{\partial v}{\partial x^{j}}- \Gamma _{jk}^{i}\xi ^{k}\frac{\partial v}{\partial \xi ^{i}}- \Gamma _{jl}^{k}\eta ^{l}\frac{\partial v}{\partial \eta ^{k}})\sqrt{\det g}d\eta.$$ Since $$(\frac{\partial }{\partial x^{j}}- \Gamma _{jk}^{i}\xi ^{k}\frac{\partial }{\partial \xi ^{i}}- \Gamma _{jl}^{k}\eta ^{l}\frac{\partial }{\partial \eta ^{k}})\left| \eta \right| =0,$$ then after changing to spherical coordinates we obtain \begin{equation} \nabla S(x,\xi )=\int\limits_{0}^{\infty } \varphi \left( t\right) t^{n-1}dt \int\limits_{\Omega _{x}}\nabla u(x,\xi ,\eta ) d\Omega _{x}\left( \eta \right) . \end{equation} Now $S$ in spherical coordinates is given by \begin{equation} S(x,\xi )=\int\limits_{0}^{\infty }\varphi \left( t\right) t^{n-1}dt \int\limits_{\Omega _{x}}u(x,\xi ,\eta ) d\Omega _{x}\left( \eta \right) \end{equation} We conclude (5.1) using (5.2),(5.3). \end{proof} \bigskip Now we prove Theorem 1.4. \begin{proof} A straightforward calculation gives that $$\nabla \frac{1+\left( \xi ,\eta \right) }{\left( \xi _{\perp },\eta \right) }=0$$ and therefore we have \[ \nabla H u(x,\xi )=\frac{1}{2\pi }\int\limits_{\Omega _{x}}\frac{1+\left( \xi ,\eta \right) }{\left( \xi _{\perp },\eta \right) }\nabla u(x,\eta )d\Omega _{x}\left( \eta \right). \] For any pair vectors $\xi ,\,\eta \in \Omega _{x}$ we have $$\eta=(\xi ,\eta )\xi +(\xi _{\perp },\eta )\xi _{\perp }, \eta _{\perp }=-(\xi _{\perp },\eta )\xi +(\xi ,\eta )\xi _{\perp },\quad \left( \xi ,\eta \right) ^{2}+\left( \xi _{\perp },\eta \right) ^{2}= 1.$$ Then $$\eta \frac{1+\left( \xi ,\eta \right) }{\left( \xi _{\perp },\eta \right) }=\xi \frac{\left( \xi ,\eta \right) +\left( \xi ,\eta \right) ^{2}}{\left( \xi _{\perp },\eta \right) }+\xi _{\perp }(1+(\xi ,\eta ))=$$ $$=\xi \frac{\left( \xi ,\eta \right) +1}{\left( \xi _{\perp },\eta \right) }-\xi \left( \xi _{\perp },\eta \right) +\xi _{\perp }(\xi ,\eta )+\xi _{\perp }=\xi \frac{1+\left( \xi ,\eta \right) }{\left( \xi _{\perp },\eta \right) }+\xi _{\perp }+\eta _{\perp }.$$ Thus \[ H\mathcal{H}u=\mathcal{H}Hu+\mathcal{H}_{\perp }u_{0}+(\mathcal{H}_{\perp }u)_{0} \] and Theorem 1.4 is proved. \end{proof} \bigskip %===================================================================== %===================================================================== % REFERENCES %===================================================================== %===================================================================== %\small \begin{thebibliography}{Mu-R} \frenchspacing \bibitem[An]{An} {\sc Yu. 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