II

# Normal operators for momentum ray transforms,
I: The inversion formula

Shubham R. Jathar  Indian Institute of Science Education and Research (IISER)
Bhopal, India<EMAIL_ADDRESS>,  Manas Kar  Indian Institute of
Science Education and Research (IISER) Bhopal, India<EMAIL_ADDRESS>,
Venkateswaran P. Krishnan  Centre for Applicable Mathematics, Tata Institute
of Fundamental Research, India<EMAIL_ADDRESS>and  Vladimir A.
Sharafutdinov  Sobolev Institute of Mathematics, 4 Koptyug Av., 630090,
Novosibirsk, Russia<EMAIL_ADDRESS>
###### Abstract.

The momentum ray transform $I_{m}^{k}$ integrates a rank $m$ symmetric tensor
field $f$ on ${{\mathbb{R}}}^{n}$ over lines with the weight $t^{k}$,
$I_{m}^{k}f(x,\xi)=\int_{-\infty}^{\infty}t^{k}\langle
f(x+t\xi),\xi^{m}\rangle\,\mathrm{d}t$. We compute the normal operator
$N_{m}^{k}=(I_{m}^{k}){}^{*}I_{m}^{k}$ and present an inversion formula
recovering a rank $m$ tensor field $f$ from the data
$(N_{m}^{0}f,\dots,N_{m}^{m}f)$.

Keywords. Ray transform, inverse problems, symmetric tensor fields, tensor
tomography, momentum ray transform.

Mathematics Subject Classification (2020): Primary 44A12, Secondary 53C65.

##  1\. Introduction

Let $\langle\cdot,\cdot\rangle$ be the standard dot product on
${{\mathbb{R}}}^{n}$ and $|\cdot|$, the corresponding norm.

For Schwartz class functions, the ray transform (also called the X-ray
transform) is defined by

(1.1) $If(x,\xi)=\int_{-\infty}^{\infty}f(x+t\xi)\,\mathrm{d}t$

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$ satisfying $|\xi|=1$
and $\langle x,\xi\rangle=0.$ For a Schwartz class symmetric $m$-tensor field
$f=(f_{i_{1}\dots i_{m}})$, the ray transform is defined by

(1.2) $\begin{split}I_{m}f(x,\xi)&=\int_{-\infty}^{\infty}f_{i_{1}\dots
i_{m}}(x+t\xi)\xi^{i_{1}}\dots\xi^{i_{m}}\,\mathrm{d}t\\\
&=\int_{-\infty}^{\infty}\langle
f(x+t\xi),\xi^{m}\rangle\,\mathrm{d}t.\end{split}$

We use the Einstein summation rule to sum from $1$ to $n$ over every repeated
index in lower and upper positions in a monomial. In particular, when $m=0$,
the definition (1.2) coincides with (1.1) and when $m=1$, (1.2) represents the
ray transform of vector fields which is also called the Doppler transform.

In the case of $m=0$, the ray transform $If$ uniquely determines a function
$f$ and there is an explicit inversion formula. However, if $m\geq 1$, the ray
transform $I_{m}$ has a nontrivial kernel. In particular, $I(\sigma\nabla
h)=0$ whenever $h$ is a smooth symmetric $(m-1)$-tensor field on
${{\mathbb{R}}}^{n}$ decaying at infinity, $\nabla$ is the total covariant
derivative, and $\sigma$ denotes the symmetrization of a tensor. A symmetric
$m$-tensor field $f$ sufficiently fast decaying at infinity can be uniquely
decomposed

$f=f^{s}+\sigma\nabla h,\quad h(x)\to 0\ \mbox{as}\ |x|\to\infty$

to the solenoidal (= divergence-free) part $f^{s}$ and potential part
$\sigma\nabla h$; see [Sha94, Theorem 2.6.2] and [PSU23, Theorem 6.4.7] for
the detailed explanation in the Euclidean case as well as in the case of
Riemannian manifolds. The solenoidal part of a symmetric $m$-tensor field $f$
can be uniquely determined from $I_{m}f$ and there is an explicit inversion
formula [Sha94, Theorem 2.12.2].

It is natural to ask: what additional information required along with $I_{m}f$
so that one could recover the entire tensor field $f$. This leads to the
notion of the momentum ray transform $I_{m}^{k}$ introduced in [Sha94, Section
2.17] by

$\displaystyle I_{m}^{k}f(x,\xi)$
$\displaystyle=\int_{-\infty}^{\infty}t^{k}f_{i_{1}\dots
i_{m}}(x+t\xi)\xi^{i_{1}}\dots\xi^{i_{m}}\,\mathrm{d}t$
$\displaystyle=\int_{-\infty}^{\infty}t^{k}\langle
f(x+t\xi),\xi^{m}\rangle\,\mathrm{d}t\quad(k=0,1,2.\dots)$

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$ satisfying $|\xi|=1$
and $\langle x,\xi\rangle=0$. In particular $I_{m}^{0}=I_{m}$.

We restrict ourselves by considering the data $(I_{m}^{0}f,\dots,I_{m}^{m}f)$
since, for $k>m$, the function $I_{m}^{k}f$ can be easily expressed through
$I_{m}^{0}f,\dots,I_{m}^{m}f$.

A rank $m$ symmetric tensor field $f$ is uniquely determined by the data
$(I_{m}^{0}f,\dots,I_{m}^{m}f)$. This was proved in [Sha94, Theorem 2.17.2].
Later this result was extended to a Helgason type support theorem for tensor
fields on a simple real analytic Riemannian manifold [AM19]. An algorithm for
recovering $f$ from the data $\left(I_{m}^{0}f,\dots,I_{m}^{m}f\right)$ is
presented in [KMSS19] as well as a stability estimate in (generalized) Sobolev
norms. A range characterization for the operator
$f\mapsto(I_{m}^{0}f,\dots,I_{m}^{m}f)$ on the Schwartz space was established
in [KMSS20].

Let us introduce the normal operator
$N_{m}^{k}=\left(I_{m}^{k}\right)^{*}I_{m}^{k}$, where
$\left(I_{m}^{k}\right)^{*}$ is the $L^{2}$-adjoint of the momentum ray
transform $I_{m}^{k}$. Since $N_{m}^{k}$ is an averaging operator, the data
$N_{m}^{k}f$ could represent a better measurement model rather than
$I^{k}_{m}f$. We present an algorithm of recovering a rank $m$ tensor field
$f$ from the data $(N_{m}^{0}f,\dots,N_{m}^{m}f)$. In terms of Fourier
transforms $\widehat{f}$ and
$(\widehat{N_{m}^{0}f},\dots,\widehat{N_{m}^{m}f})$, we derive the inversion
formula

$\widehat{f}(y)=|y|\sum\limits_{k=0}^{m}P_{m}^{k}(\widehat{N_{m}^{k}f})$

with some linear operators $P_{m}^{k}$ on the space of rank $m$ symmetric
tensor fields. Given $m$, the operators $P_{m}^{k}\ (k=0,\dots,m)$ are
calculated by explicit recurrent formulas; but the volume of calculations
grows fast with $m$. We perform the calculations for $m=1,2,3$.

The ray transform has several important applications that include X-ray
computer tomography (CT) in medical imaging when $m=0.$ In the case of $m=1$,
the ray transform is used in Doppler tomography to analyze vector fields. In
cases where $m=2$ or $m=4$, the ray transform and its variants are applied to
tomography problems in anisotropic media regarding the elasticity and Maxwell
systems, see [Sha94] and [LS09, SW12]. Recently, the momentum ray transform
has been adopted as a solution tool for the classical Calderón problem for the
bi-Laplace model and other higher-order operators [BKS23, SS23, BK23]. The
unique continuation principle for $I_{m}$ and $I^{k}_{m}$ is proved in
[AKS22]. See also [IKS23] for a related work involving a fractional momentum
operator.

### Acknowledgments

S.R.J. and V.P.K. would like to thank the Isaac Newton Institute for
Mathematical Sciences, Cambridge, UK, for support and hospitality during _Rich
and Nonlinear Tomography - a multidisciplinary approach_ in 2023 where part of
this work was done (supported by EPSRC Grant Number EP/R014604/1). S.R.J.
acknowledges the Prime Minister’s Research Fellowship (PMRF) from the
Government of India for his PhD work. M.K. was supported by MATRICS grant
(MTR/2019/001349) of SERB. The work of V.A.S. was performed according to the
Russian Government research assignment for IM SB RAS, project FWNF-2022-0006.

##  2\. Basic definitions

First of all, mostly following [Sha94, Chapter 2], we introduce some notation
and definitions concerning tensor algebra and analysis which will be used
throughout the article.

###  2.1. Tensor algebra over $\mathbb{R}^{n}$

Let $T^{m}\mathbb{R}^{n}$ be the $n^{m}$-dimensional complex vector space of
$m$-tensors on $\mathbb{R}^{n}$. In particular,
$T^{0}\mathbb{R}^{n}={{\mathbb{C}}}$ and $T^{1}\mathbb{R}^{n}={\ C}^{n}$. We
need complex tensors since we are going to use the Fourier transform. Assuming
$n$ to be fixed, the notation $T^{m}\mathbb{R}^{n}$ will be mostly abbreviated
to $T^{m}$. For a fixed orthonormal basis $(e_{1},\ldots,e_{n})$ of
$\mathbb{R}^{n}$, by $u_{i_{1}\dots i_{m}}=u^{i_{1}\dots
i_{m}}=u(e_{i_{1}},\ldots,e_{i_{m}})$ we denote coordinates (= components) of
a tensor $u\in T^{m}$ with respect to the basis. There is no difference
between covariant and contravariant tensors since we use orthonormal bases
only. Given $u\in T^{m}$ and $v\in T^{k}$, the tensor product $u\otimes v\in
T^{m+k}$ is defined by $(u\otimes v)_{i_{1}\dots i_{m+k}}=u_{i_{1}\dots
i_{m}}v_{i_{m+1}\dots i_{m+k}}.$ The standard dot product on $\mathbb{R}^{n}$
extends to $T^{m}$ by $\langle u,v\rangle=u^{i_{1}\dots
i_{m}}\overline{v_{i_{1}\dots i_{m}}}.$ Throughout the article, the Einstein
summation convention is used.

Let $S^{m}=S^{m}\mathbb{R}^{n}$ be the $\binom{n+m-1}{m}$-dimensional subspace
of $T^{m}$ consisting of symmetric tensors. The partial symmetrization
$\sigma(i_{1}\dots i_{m}):T^{m+k}\to T^{m+k}$ in the indices
$(i_{1},\dots,i_{m})$ is defined by

$\sigma(i_{1}\dots i_{m})u_{i_{1}\dots i_{m}j_{1}\dots
j_{k}}=\frac{1}{m!}\sum_{\pi\in\Pi_{m}}u_{i_{\pi(1)},\dots,i_{\pi(m)}j_{1}\dots
j_{k}},$

where the summation is performed over the group $\Pi_{m}$ of all substitutions
of the set $\\{1,\dots,m\\}$. In particular, $\sigma:T^{m}\rightarrow S^{m}$
is the symmetrization in all indices. Given $u\in S^{m}$ and $v\in S^{k}$, the
symmetric product $uv\in S^{m+k}$ is defined by $uv=\sigma(u\otimes v)$. Being
furnished with the symmetric product,
$S^{*}{{\mathbb{R}}}^{n}=\bigoplus\limits_{m=0}^{\infty}S^{m}{{\mathbb{R}}}^{n}$
becomes a commutative graded algebra that is called the algebra of symmetric
tensors over ${{\mathbb{R}}}^{n}$. The algebra $S^{*}{{\mathbb{R}}}^{n}$ is
canonically isomorphic to the algebra of polynomials on ${{\mathbb{R}}}^{n}$.
Every statement on symmetric tensors can be translated to the langauge of
polynomials, and vice versa.

Given $u\in S^{m}$, let $i_{u}:S^{k}\to S^{m+k}$ be the operator of symmetric
multiplication by $u$ and let $j_{u}:S^{m+k}\to S^{k}$ be the adjoint of
$i_{u}$. These operators are written in coordinates as

$\displaystyle\left(i_{u}v\right)_{i_{1}\ldots i_{m+k}}$
$\displaystyle=\sigma\left(i_{1}\ldots i_{m+k}\right)u_{i_{1}\ldots
i_{m}}v_{i_{m+1}\ldots i_{m+k}}$
$\displaystyle\left(j_{u}v\right)_{i_{1}\ldots i_{k}}$
$\displaystyle=v_{i_{1}\ldots i_{m+k}}u^{i_{k+1}\ldots i_{m+k}}.$

The tensor $j_{u}v$ will be also denoted by $v/u$. For the Kronecker tensor
$\delta$, the notations $i_{\delta}$ and $j_{\delta}$ will be abbreviated to
$i$ and $j$ respectively.

###  2.2. Tensor fields

Recall that the Schwartz space $\mathcal{S}\left(\mathbb{R}^{n}\right)$ is the
topological vector space consisting of $C^{\infty}$-smooth complex-valued
functions on ${{\mathbb{R}}}^{n}$ fast decaying at infinity together with all
derivatives, furnished with the standard topology. Let
$\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)=\mathcal{S}\left(\mathbb{R}^{n}\right)\otimes
S^{m}$ be the topological vector space of smooth fast decaying symmetric
$m$-tensor fields, defined on $\mathbb{R}^{n}$, whose components belong to the
Schwartz space. In Cartesian coordinates, such a tensor field is written as
$f=(f_{i_{1}\dots i_{m}})$ with coordinates (= components) $f_{i_{1}\dots
i_{m}}=f^{i_{1}\dots i_{m}}\in\mathcal{S}\left(\mathbb{R}^{n}\right)$
symmetric in all indices. We again emphasize that there is no difference
between covariant and contravariant coordinates since we use Cartesian
coordinates only.

We use the Fourier transform
$\mathcal{F}:\mathcal{S}(\mathbb{R}^{n})\rightarrow\mathcal{S}(\mathbb{R}^{n}),$
$f\mapsto\widehat{f}$ in the form (hereafter i is the imaginary unit)

$\mathcal{F}{f}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}e^{-\textsl{i}\langle
y,x\rangle}f(x)\,\mathrm{d}x.$

The Fourier transform
$\mathcal{F}:\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)\rightarrow\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$,
$f\mapsto\widehat{f}$ of symmetric tensor fields is defined component-wise:
$\widehat{f}_{i_{1}\ldots i_{m}}=\widehat{f_{i_{1}\ldots i_{m}}}$.

Besides $\mathcal{S}\left({{\mathbb{R}}}^{n};S^{m}\right)$, we use some other
spaces of tensor fields. In particular, $C^{\infty}\left(U;T^{m}\right)$ is
the space of smooth $m$-tensor fields on an open set
$U\subset{{\mathbb{R}}}^{n}$.See details in [Sha94, Section 2.1].

The $L^{2}$-product on $C_{0}^{\infty}\left({{\mathbb{R}}}^{n};T^{m}\right)$
is defined by

(2.1)
$(f,g)_{L^{2}({{\mathbb{R}}}^{n};T^{m})}=\int_{{{\mathbb{R}}}^{n}}\langle
f(x),g(x)\rangle\,dx.$

###  2.3. Inner derivative and divergence

The first order differential operator

$d:C^{\infty}({{\mathbb{R}}}^{n};S^{m})\to
C^{\infty}({{\mathbb{R}}}^{n};S^{m+1})$

defined by

$(df)_{i_{1}\dots i_{m+1}}=\sigma(i_{1}\dots i_{m+1})\frac{\partial
f_{i_{1}\dots i_{m}}}{\partial x^{i_{m+1}}}=\frac{1}{m+1}\Big{(}\frac{\partial
f_{i_{2}\dots i_{m+1}}}{\partial x^{i_{1}}}+\dots+\frac{\partial f_{i_{1}\dots
i_{m}}}{\partial x^{i_{m+1}}}\Big{)}$

is called the inner derivative.

The divergence

$\mbox{div}:C^{\infty}({{\mathbb{R}}}^{n};S^{m+1})\to
C^{\infty}({{\mathbb{R}}}^{n};S^{m})$

is defined by

$(\mbox{div}\,f)_{i_{1}\dots i_{m}}=\delta^{jk}\frac{\partial f_{i_{1}\dots
i_{m}j}}{\partial x^{k}}.$

The operators $d$ and $-\mbox{div}$ are formally adjoint to each other with
respect to the $L^{2}$-product (2.1). The divergence is denoted by $\delta$ in
[Sha94]. But we will always use the notation div since some our formulas
involve the divergence and Kronecker tensor simultaneously.

###  2.4. The space ${\mathcal{S}}(T{\mathbb{S}}^{n-1})$

The Schwartz space ${\mathcal{S}}(E)$ is well defined for a smooth vector
bundle $E\to M$ over a compact manifold with the help of a finite atlas and
partition of unity subordinate to the atlas.

In particular, the Schwartz space ${\mathcal{S}}(T{\mathbb{S}}^{n-1})$ is well
defined for the tangent bundle

$T\mathbb{S}^{n-1}=\\{(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}:\langle
x,\xi\rangle=0\\}\to\mathbb{S}^{n-1},\quad(x,\xi)\mapsto\xi$

of the unit sphere $\mathbb{S}^{n-1}=\\{x\in{{\mathbb{R}}}^{n}:|x|=1\\}$.

The Fourier transform
$\mathcal{F}:\mathcal{S}\left(T\mathbb{S}^{n-1}\right)\rightarrow\mathcal{S}\left(T\mathbb{S}^{n-1}\right),\varphi\mapsto\widehat{\varphi}$
is defined by

$\mathcal{F}{\varphi}(y,\xi)=\frac{1}{(2\pi)^{(n-1)/2}}\int_{\xi^{\perp}}e^{-\textsl{i}\langle
y,x\rangle}\varphi(x,\xi)\,\mathrm{d}x,$

where $\mathrm{d}x$ is the $(n-1)$-dimensional Lebesgue measure on the
hyperplane $\xi^{\perp}=\left\\{x\in\mathbb{R}^{n};\right.$
$\langle\xi,x\rangle=0\\}$. Notice that it is the standard Fourier transform
in the $(n-1)$-dimensional variable $x$ while $\xi\in\mathbb{S}^{n-1}$ is
considered as a parameter.

The $L^{2}$-product on ${\mathcal{S}}(T{\mathbb{S}}^{n-1})$ is defined by

(2.2)
$(\varphi,\psi)_{L^{2}(T\mathbb{S}^{n-1})}=\int\limits_{\mathbb{S}^{n-1}}\int\limits_{\xi^{\bot}}\varphi(x,\xi)\overline{\psi(x,\xi)}\,\mathrm{d}x\,\mathrm{d}\xi,$

where $\mathrm{d}\xi$ is the $(n-1)$-dimensional Euclidean volume form on the
unit sphere $\mathbb{S}^{n-1}$.

###  2.5. Ray transforms

It is convenient to parameterize the family of oriented lines in
${{\mathbb{R}}}^{n}$ by points of the manifold $T\mathbb{S}^{n-1}$. Namely, a
point $(x,\xi)\in T\mathbb{S}^{n-1}$ determines the line
$\\{x+t\xi:t\in{{\mathbb{R}}}\\}$ through $x$ in the direction $\xi$.

The ray transform

$I_{m}:\mathcal{S}({{\mathbb{R}}}^{n};S^{m})\to\mathcal{S}\left(T\mathbb{S}^{n-1}\right)$

is the linear continuous operator defined by

$I_{m}f(x,\xi)=\int_{\mathbb{R}}f_{i_{1}\dots
i_{m}}(x+t\xi)\xi^{i_{1}}\dots\xi^{i_{m}}\mathrm{~{}d}t=\int_{\mathbb{R}}\left\langle
f(x+t\xi),\xi^{m}\right\rangle\mathrm{d}t.$

The ray transform is related to the Fourier transform by the important formula
[Sha94, formula 2.1.15].

(2.3)
$\widehat{I_{m}f}(y,\xi)=(2\pi)^{1/2}\langle\widehat{f}(y),\xi^{m}\rangle\quad\big{(}(y,\xi)\in
T\mathbb{S}^{n-1}\big{)}.$

For $0\leq k\leq m$, the momentum ray transform

$I_{m}^{k}:\mathcal{S}({{\mathbb{R}}}^{n};S^{m})\to\mathcal{S}\left(T\mathbb{S}^{n-1}\right)$

is the linear continuous operator defined by

(2.4) $I_{m}^{k}f(x,\xi)=\int_{\mathbb{R}}t^{k}f_{i_{1}\dots
i_{m}}(x+t\xi)\xi^{i_{1}}\dots\xi^{i_{m}}\mathrm{~{}d}t=\int_{\mathbb{R}}t^{k}\left\langle
f(x+t\xi),\xi^{m}\right\rangle\mathrm{d}t.$

The formula (2.3) is generalized as follows [KMSS19, formula (2.9)]:

$\widehat{I_{m}^{k}f}(y,\xi)=(2\pi)^{1/2}{\textsl{i}}^{k}\left\langle
d^{k}\widehat{f}(y),\xi^{m+k}\right\rangle\quad\big{(}(y,\xi)\in
T\mathbb{S}^{n-1}\big{)}.$

As we will see later, $I_{m}^{k}$ should be considered together with lower
degree operators $I_{m}^{0},\dots,I_{m}^{k-1}$, i.e., the collection
$(I_{m}^{0}f,\dots,I_{m}^{k}f)$ represents more convenient information about
$f$ than $I_{m}^{k}f$ alone.

###  2.6. Normal operators

The formal adjoint of the ray transform $I_{m}$ with respect to
$L^{2}$-products (2.1) and (2.2)

$I_{m}^{*}:\mathcal{S}\left(T\mathbb{S}^{n-1}\right)\rightarrow
C^{\infty}\left(\mathbb{R}^{n};S^{m}\right)$

is expressed by

$\left(I_{m}^{*}\varphi\right)_{i_{1}\dots
i_{m}}(x)=\int_{\mathbb{S}^{n-1}}\xi_{i_{1}}\dots\xi_{i_{m}}\varphi\big{(}x-\langle
x,\xi\rangle\xi,\xi\big{)}\,\mathrm{d}\xi.$

We emphasize that, for $\varphi\in\mathcal{S}(T\mathbb{S}^{n-1})$, the tensor
field $I_{m}^{*}\varphi$ does not need fast decay at infinity.

Similarly, the formal $L^{2}$-adjoint of the momentum ray transform
$I_{m}^{k}$

$\left(I_{m}^{k}\right)^{*}:\mathcal{S}\left(T\mathbb{S}^{n-1}\right)\rightarrow
C^{\infty}\left(\mathbb{R}^{n};S^{m}\right)$

is expressed by

(2.5) $\big{(}(I_{m}^{k})^{*}\varphi\big{)}_{i_{1}\ldots
i_{m}}(x)=\int_{\mathbb{S}^{n-1}}\langle
x,\xi\rangle^{k}\xi_{i_{1}}\dots\xi_{i_{m}}\varphi\big{(}x-\langle
x,\xi\rangle\xi,\xi\big{)}\,\mathrm{d}\xi.$

Let

$N_{m}=I_{m}^{*}I_{m}:\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)\rightarrow
C^{\infty}\left(\mathbb{R}^{n};S^{m}\right)$

be the normal operator for the ray transform $I_{m}$. Similarly, let

$N_{m}^{k}=(I_{m}^{k})^{*}I_{m}^{k}:\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)\rightarrow
C^{\infty}\left(\mathbb{R}^{n};S^{m}\right)$

be the normal operator for the momentum ray transform $I_{m}^{k}$.

Given $f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$, the tensor field
$N_{m}^{k}f$ does not grow too fast at infinity, i.e., the estimate

$|N_{m}^{k}f(x)|\leq C(1+|x|)^{N}$

holds with some constants $C$ and $N$. In particular, $N_{m}^{k}f$ can be
considered as a tempered tensor field-distribution, i.e.,
$N_{m}^{k}f\in{\mathcal{S}}^{\prime}\left(\mathbb{R}^{n};S^{m}\right)$. Hence
the Fourier transform
$\widehat{N_{m}^{k}f}\in{\mathcal{S}}^{\prime}\left(\mathbb{R}^{n};S^{m}\right)$
is well defined at least in the distribution sense. We will show that, for
$f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$, the restriction of
$\widehat{N_{m}^{k}f}$ to ${{\mathbb{R}}}^{n}\setminus\\{0\\}$ belongs to
$C^{\infty}\left(\mathbb{R}^{n}\setminus\\{0\\};S^{m}\right)$.

The operator $N_{m}$ was computed in [Sha94, formula 2.11.3] where the
notation $\mu^{m}$ was used instead of $I_{m}^{*}$. In Section 4, we will
derive a similar formula for $N_{m}^{k}$. In [AKS22], a similar expression for
the normal operator is considered to study the unique continuation principle
for momentum ray transforms.

##  3\. Main results

We start with inversion formulas for vector fields and for second rank
symmetric tensor fields.

######  Theorem 3.1.

A vector field $f\in\mathcal{S}\left(\mathbb{R}^{n};{{\mathbb{C}}}^{n}\right)\
(n\geq 2)$ is recovered from the data $(N_{1}^{0}f,N_{1}^{1}f)$ by the
inversion formula

(3.1)
$f(x)=\frac{2^{n/2-1}\Gamma\Big{(}\frac{n+1}{2}\Big{)}}{\sqrt{\pi}}(-\Delta)^{1/2}\Big{[}N_{1}^{0}f-\frac{1}{n\\!-\\!1}\,d\,j_{x}\,N_{1}^{0}f-\frac{1}{n\\!-\\!1}\,\,d\,\mbox{\rm
div}\,N_{1}^{1}f\Big{]}.$

where $\Gamma$ is Euler’s Gamma function and the operator $(-\Delta)^{1/2}$ is
defined with the help of the Fourier transform by
$|y|{\mathcal{F}}={\mathcal{F}}(-\Delta)^{1/2}$. The vector field in brackets
belongs to the domain of $(-\Delta)^{1/2}$.

######  Theorem 3.2.

A tensor field $f\in\mathcal{S}\left(\mathbb{R}^{n};S^{2}\right)$ is recovered
from the data $(N_{2}^{0}f,N_{2}^{1}f,N_{2}^{2}f)$ by the inversion formula

(3.2) $\displaystyle
f(x)=\frac{2^{n/2-1}\Gamma\Big{(}\frac{n+3}{2}\Big{)}}{\sqrt{\pi}}(-\Delta)^{1/2}\bigg{[}$
$\displaystyle N_{2}^{0}f-\frac{1}{n+1}\,ij\,N_{2}^{0}f$
$\displaystyle-\frac{2}{n+1}\,d\Big{(}j_{x}N_{2}^{0}f+\mbox{\rm
div}\,N_{2}^{1}f\Big{)}$
$\displaystyle+\frac{1}{(n\\!-\\!1)(n\\!+\\!1)}\,d^{2}\Big{(}j_{x}^{2}\,N_{2}^{0}f-2\,j_{x}\,\mbox{\rm
div}\,N_{2}^{1}f+\frac{1}{2}\,\mbox{\rm div}^{2}\,N_{2}^{2}f\Big{)}\bigg{]}.$

The tensor field in brackets belongs to the domain of $(-\Delta)^{1/2}$.

We use the definition

$(2l+1)!!=1\cdot 3\cdots(2l+1),\quad(-1)!!=1.$

For tensor fields of arbitrary rank, our result is as follows.

######  Theorem 3.3.

Given integers $m\geq 0$ and $n\geq 2$, the Fourier transform of a tensor
field $f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$ is recovered from the
data
$(\widehat{N_{m}^{0}f},j_{y}\widehat{N_{m}^{0}f},\dots,j_{y}^{m}\widehat{N_{m}^{m}f})$
by the algorithm consisting of three steps.

1\. Compute tensor fields $F^{(m,k)}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m-k})\ (0\leq k\leq m)$ by

(3.3)
$F^{(m,k)}(y)=\frac{c_{m,n}}{k!}\,j_{y}^{k}\,\widehat{N_{m}^{k}f}(y),\quad
c_{m,n}=\pi^{-1/2}\,(2m-1)!!\,2^{m+n/2-2}\,\Gamma\Big{(}\frac{2m\\!+\\!n\\!-\\!1}{2}\Big{)}.$

2\. Compute tensor fields $H^{(m,k)}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m-k})\ (0\leq k\leq m)$ by

(3.4)
$H^{(m,k)}=\frac{(2m-2k-1)!!(n+2m-2k-3)!!}{(2m-1)!!(n+2m-3)!!}\sum\limits_{p=0}^{k}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{k-p}\,F^{(m,p)}.$

3\. Recover $\widehat{f}$ by the formula

(3.5) $\displaystyle\widehat{f}(y)$
$\displaystyle=\frac{|y|}{m!}\,\bigg{[}\frac{1}{(2m\\!-\\!1)!!}\,H^{(m,0)}(y)$
$\displaystyle\qquad+\sum\limits_{k=1}^{m}\frac{(-1)^{k}}{(2m\\!-\\!2k\\!-\\!1)!!}\binom{m}{k}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\frac{(-1)^{p}}{2^{p}}\binom{m-k}{p}\,i^{p}i_{y}^{k-p}j_{y}^{p}H^{(m,k)}(y)\bigg{]}.$

Of course, (3.3)–(3.5) can be combined to give a formula that expresses the
rank $m$ symmetric tensor field $\widehat{f}$ through
$(\widehat{N_{m}^{0}f},j_{y}\widehat{N_{m}^{0}f},\dots,j_{y}^{m}\widehat{N_{m}^{m}f})$
that does not involve $F^{(m,k)}$ and $H^{(m,k)}$. We will present the latter
formula for $m=1,2,3$ in the last section. The proof of Theorems 3.1 and 3.2
is also presented in the last section. Formulas (3.1) and (3.2) are obtained
from (3.5) just by applying the inverse Fourier transform; nevertheless, some
commutator formulas for the Fourier transform and operators participating in
(3.5) should be used.

The rest of the article is mostly devoted to the proof of Theorem 3.3. Now, we
discuss the scheme of the proof.

We introduce the tensor fields

(3.6) $A^{(m,k)}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{2m-k})\quad(0\leq k\leq m)$

by

(3.7) $A^{(m,k)}(y)=d^{2m-k}|y|^{2m-2k-1}.$

These tensor fields play an important role in all our arguments.

In the next section, we compute the normal operators $N_{m}^{k}\ (0\leq k\leq
m)$ and prove that a tensor field
$f\in{\mathcal{S}}({{\mathbb{R}}}^{n};S^{m})$ satisfies
$A^{(m,0)}/(d^{k}\widehat{f})=F^{(m,k)}$, where $F^{(m,k)}$ is defined by
(3.3). To avoid proliferation of the $\widehat{\cdot}$ symbol, we denote
$g(y)=\widehat{f}(y)$ and write the latter equation in the form

(3.8) $A^{(m,0)}/(d^{k}g)=F^{(m,k)}\quad(0\leq k\leq m).$

Given the data $(F^{(m,0)},\dots,F^{(m,m)})$, we consider (3.8) as a system of
linear equations for the unknown tensor field $g$.

The first equation of the system (3.8)

$A^{(m,0)}(y)/g(y)=F^{(m,0)}(y)\quad(y\in{{\mathbb{R}}}^{n}\setminus\\{0\\})$

is a pure algebraic equation. More precisely, being written in coordinates, it
constitutes a system of linear algebraic equations in the components of the
tensor field $g(y)$ with coefficients depending on $y$. The system was
considered in [Sha94, Theorem 2.12.1] where the tensor field
$\varepsilon^{m}(y)=\frac{|y|}{((2m-1)!!)^{2}}\,d^{2m}|y|^{2m-1}$ was used
instead of $A^{(m,0)}$. It allows to determine the tangential part of the
tensor field $g$ which corresponds to the solenoidal part of
$f={\mathcal{F}}^{-1}g$ (see [Sha94, Section 2.6] for the definition of the
tangential part).

The second equation of the system (3.8), $A^{(m,1)}/(dg)=F^{(m,1)}$,
constitutes a system of linear first order PDEs in components of the tensor
field $g$, the third equation constitutes a system of linear second order
PDEs, and so on.

At first sight, the following statement may seem incredible. The system (3.8)
can be reduced to the purely algebraic system

(3.9) $A^{(m,k)}/g=H^{(m,k)}\quad(0\leq k\leq m)$

with right-hand side defined by (3.4). The reduction is presented in Section
5. The precise sense of the reduction is expressed by Proposition 5.3 below;
see also the paragraph after Proposition 5.3.

Some consistency conditions should be imposed on right-hand side $H^{(m,k)}$
for solvability of the system (3.9). In the case of a general $m$, it is not
easy to write down the consistency conditions explicitly. Fortunately, we do
not need to know the consistency conditions; in our setting, the system (3.9)
has a solution by Propositions 4.3 and 5.3 presented below. If the system
(3.9) has a solution, then the solution is unique and is expressed by (3.5)
with $\widehat{f}=g$. This fact is proved in Section 6.

##  4\. Normal operator

We start with the proof of (2.5). For
$f\in{\mathcal{S}}({{\mathbb{R}}}^{n};S^{m})$ and
$\varphi\in{\mathcal{S}}(T{\mathbb{S}}^{n-1})$,

(4.1) $\displaystyle(I_{m}^{k}f,\varphi)_{L^{2}(T{\mathbb{S}}^{n-1})}$
$\displaystyle=\int\limits_{{\mathbb{S}}^{n-1}}\int\limits_{\xi^{\bot}}(I_{m}^{k}f)(x,\xi)\,\overline{\varphi(x,\xi)}\,{\mathrm{d}}x\,{\mathrm{d}}\xi$
$\displaystyle=\int\limits_{{\mathbb{S}}^{n-1}}\int\limits_{\xi^{\bot}}\int\limits_{-\infty}^{\infty}t^{k}\langle
f(x^{\prime}+t\xi),\xi^{m}\rangle\,\overline{\varphi(x^{\prime},\xi)}\,{\mathrm{d}}t\,{\mathrm{d}}x^{\prime}\,{\mathrm{d}}\xi.$

We transform the inner integral by the change $x=x^{\prime}+t\xi$ of
integration variables

$\displaystyle\int\limits_{\xi^{\bot}}\int\limits_{-\infty}^{\infty}t^{k}\langle
f(x^{\prime}+t\xi),\xi^{m}\rangle\,\overline{\varphi(x^{\prime},\xi)}\,{\mathrm{d}}t\,{\mathrm{d}}x^{\prime}$
$\displaystyle=\int\limits_{{{\mathbb{R}}}^{n}}\langle x,\xi\rangle^{k}\langle
f(x),\xi^{m}\rangle\,\overline{\varphi(x-\langle
x,\xi\rangle\xi,\xi)}\,{\mathrm{d}}x$
$\displaystyle=\int\limits_{{{\mathbb{R}}}^{n}}f^{i_{1}\dots
i_{m}}(x)\,\xi_{i_{1}}\dots\xi_{i_{m}}\,\overline{\varphi(x-\langle
x,\xi\rangle\xi,\xi)}\,{\mathrm{d}}x.$

Substituting this expression into (4.1), we obtain

$\displaystyle(I_{m}^{k}f,\varphi)_{L^{2}(T{\mathbb{S}}^{n-1})}$
$\displaystyle=\int\limits_{{{\mathbb{R}}}^{n}}f^{i_{1}\dots
i_{m}}(x)\,\overline{\int\limits_{{\mathbb{S}}^{n-1}}\xi_{i_{1}}\dots\xi_{i_{m}}\,\varphi(x-\langle
x,\xi\rangle\xi,\xi)\,{\mathrm{d}}\xi}\,{\mathrm{d}}x$
$\displaystyle=(f,(I_{m}^{k})^{*}\varphi)_{L^{2}({{\mathbb{R}}}^{n};S^{m})}.$

This proves (2.5).

Recall that $N_{m}^{k}=(I_{m}^{k})^{*}I_{m}^{k}$ is the normal operator for
the momentum ray transform.

######  Proposition 4.1.

Let $0\leq k\leq m$ and $n\geq 2$. For a tensor field
$f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$,

(4.2) $(N_{m}^{k}f)_{i_{1}\dots
i_{m}}(x)=2\sum_{l=0}^{k}\binom{k}{l}(x^{k+l}f)^{j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}\ast\frac{(x^{2m+k+l})_{i_{1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}}{|x|^{2m+2l+n-1}},$

where $\ast$ denotes the convolution.

The right-hand side of (4.2) needs the following comment. For
$x\in{{\mathbb{R}}}^{n}$, according to our definition of the symmetric
product, $x^{k+l}\in S^{k+l}$ with coordinates $(x^{k+l})^{p_{1}\dots
p_{k+l}}=x^{p_{1}}\dots x^{p_{k+l}}$. Therefore, for $f\in S^{m}$,

$(x^{k+l}f)^{j_{1}\dots j_{m}p_{1}\dots p_{k+l}}=\sigma(j_{1}\dots
j_{m}p_{1}\dots p_{k+l})(x^{p_{1}}\dots x^{p_{k+l}}f^{j_{1}\dots j_{m}}).$

Before proving Proposition 4.1, we observe that it implies some regularity of
the tensor field $N_{m}^{k}f$. Indeed, the first factor
$(x^{k+l}f)^{j_{1}\dots j_{m}p_{1}\dots p_{k+l}}$ on the right-hand side of
(4.2) belongs to ${\mathcal{S}}({{\mathbb{R}}}^{n})$. The second factor is a
function locally summable over ${{\mathbb{R}}}^{n}$ and bounded for $|x|\geq
1$. Hence the second factor can be considered as an element of the space
${\mathcal{S}}^{\prime}({{\mathbb{R}}}^{n})$ of tempered distributions. As is
well known [Vla79], for $u\in{\mathcal{S}}({{\mathbb{R}}}^{n})$ and
$v\in{\mathcal{S}}^{\prime}({{\mathbb{R}}}^{n})$, the convolution $u\ast v$ is
defined and belongs to the space of smooth functions whose every derivative
increases at most as a polynomial at infinity. In this case, the standard
formula is valid: $\widehat{u\ast v}=\widehat{u}\widehat{v}$. Thus, we can
state that

$N_{m}^{k}:\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)\to
C^{\infty}\left(\mathbb{R}^{n};S^{m}\right)$

is a continuous operator.

To prove Proposition 4.1 we need the following

######  Lemma 4.2.

Let $k\geq 0$ be an integer, $0\neq a\in{\mathbb{R}}$, and $b\in{\mathbb{R}}$.
Then

$\displaystyle\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\frac{(a^{2}+b)^{2k-l}}{a^{2k-2l}}=\sum_{l=0}^{k}\binom{k}{l}\frac{b^{k+l}}{a^{2l}}.$

###### Proof.

By the binomial formula,

$\displaystyle\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\frac{(a^{2}+b)^{2k-l}}{a^{2k-2l}}$
$\displaystyle=\frac{1}{a^{2k}}\sum_{l=0}^{k}\binom{k}{l}(-a^{2})^{l}(a^{2}+b)^{2k-l}$
$\displaystyle=\frac{1}{a^{2k}}\sum_{l=0}^{k}\binom{k}{l}(-a^{2}+b-b)^{l}(a^{2}+b)^{2k-l}$
$\displaystyle=\frac{(a^{2}+b)^{k}}{a^{2k}}\sum_{l=0}^{k}\binom{k}{l}(-a^{2}+b-b)^{l}(a^{2}+b)^{k-l}$
$\displaystyle=\frac{b^{k}\left(a^{2}+b\right)^{k}}{a^{2k}}=\sum_{l=0}^{k}\binom{k}{l}\frac{b^{k+l}}{a^{2l}}.$

∎

###### Proof of Proposition 4.1.

Using (2.4) and (2.5), we first compute

$\displaystyle\left(N_{m}^{k}f\right)_{i_{1}\ldots i_{m}}(x)$
$\displaystyle=\left(I_{m}^{k}\right)_{i_{1}\dots i_{m}}^{*}I_{m}^{k}f(x)$
$\displaystyle=\int_{\mathbb{S}^{n-1}}\langle
x,\xi\rangle^{k}\xi_{i_{1}}\ldots\xi_{i_{m}}(I_{m}^{k}f)(x-\langle
x,\xi\rangle\xi,\xi)\,\mathrm{d}\xi$
$\displaystyle=\int_{\mathbb{S}^{n-1}}\int_{{\mathbb{R}}}t^{k}\langle
x,\xi\rangle^{k}f^{j_{1}\ldots j_{m}}(x-\langle
x,\xi\rangle\xi+t\xi)\xi_{j_{1}}\dots\xi_{j_{m}}\xi_{i_{1}}\ldots\xi_{i_{m}}\,\mathrm{d}t\mathrm{d}\xi$
$\displaystyle=2\int_{\mathbb{S}^{n-1}}\int_{0}^{\infty}t^{k}\langle
x,\xi\rangle^{k}f^{j_{1}\ldots j_{m}}(x-\langle
x,\xi\rangle\xi+t\xi)(\xi^{2m})_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}\,\mathrm{d}t\mathrm{d}\xi.$

Replacing $t-\langle x,\xi\rangle$ by $t$ in the last integral, we have

$\displaystyle\left(N_{m}^{k}f\right)_{i_{1}\ldots i_{m}}(x)$
$\displaystyle=2\int_{\mathbb{S}^{n-1}}\int_{0}^{\infty}\big{(}t+\langle
x,\xi\rangle\big{)}^{k}\langle x,\xi\rangle^{k}f^{j_{1}\ldots
j_{m}}(x+t\xi)(\xi^{2m})_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}\,\mathrm{d}t\mathrm{d}{\xi}$
$\displaystyle=2\sum_{l=0}^{k}\binom{k}{l}\int_{\mathbb{S}^{n-1}}\int_{0}^{\infty}t^{l}\langle
x,\xi\rangle^{2k-l}f^{j_{1}\ldots j_{m}}(x+t\xi)(\xi^{2m})_{i_{1}\dots
i_{m}j_{1}\dots j_{m}}\,\mathrm{d}t\mathrm{d}{\xi}.$

Changing integration variables by

$x+t\xi=z,\quad
t=|z-x|,\quad\xi=\frac{z-x}{|z-x|},\quad\mathrm{d}t\mathrm{d}\xi=|z-x|^{1-n}\mathrm{d}z,$

we obtain

$\left(N_{m}^{k}f\right)_{i_{1}\ldots
i_{m}}(x)=2\sum_{l=0}^{k}\binom{k}{l}\int_{\mathbb{R}^{n}}\langle
x,z\\!-\\!x\rangle^{2k-l}\big{(}(z\\!-\\!x)^{2m}\big{)}_{i_{1}\ldots
i_{m}j_{1}\ldots j_{m}}\frac{f^{j_{1}\ldots
j_{m}}(z)}{|z-x|^{2m+2k-2l+n-1}}\,\mathrm{d}z.$

Let us write this in the form

$\displaystyle\left(N_{m}^{k}f\right)_{i_{1}\ldots i_{m}}(x)$
$\displaystyle=2\int_{\mathbb{R}^{n}}\bigg{[}\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\frac{\langle
x,x-z\rangle^{2k-l}}{|x-z|^{2k-2l}}\bigg{]}\big{(}(z\\!-\\!x)^{2m}\big{)}_{i_{1}\ldots
i_{m}j_{1}\ldots j_{m}}\frac{f^{j_{1}\ldots
j_{m}}(z)}{|z-x|^{2m+n-1}}\,\mathrm{d}z.$

By Lemma 4.2 with $a=|x-z|$ and $b=\langle z,x-z\rangle$,

$\displaystyle\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\frac{\langle
x,x-z\rangle^{2k-l}}{|x-z|^{2k-2l}}$
$\displaystyle=\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\frac{\big{(}|x-z|^{2}+\langle
z,x-z\rangle\big{)}^{2k-l}}{|x-z|^{2k-2l}}$
$\displaystyle=\sum_{l=0}^{k}\binom{k}{l}\frac{\langle
z,x-z\rangle^{k+l}}{|x-z|^{2l}}=\sum_{l=0}^{k}\binom{k}{l}\frac{\langle
z,x-z\rangle^{k+l}}{|z-x|^{2l}}.$

Substitute this expression into the previous formula

$\left(N_{m}^{k}f\right)_{i_{1}\ldots
i_{m}}(x)=2\sum_{l=0}^{k}\binom{k}{l}\int_{\mathbb{R}^{n}}\langle
z,x-z\rangle^{k+l}\big{(}(x\\!-\\!z)^{2m}\big{)}_{i_{1}\ldots i_{m}j_{1}\ldots
j_{m}}\frac{f^{j_{1}\ldots j_{m}}(z)}{|x-z|^{2m+2l+n-1}}\,\mathrm{d}z.$

Then we represent the first factor of the integrand as follows

$\langle z,x-z\rangle^{k+l}=(z^{k+l})^{p_{1}\dots
p_{k+l}}\big{(}(x-z)^{k+l}\big{)}_{p_{1}\dots p_{k+l}}.$

Substituting this expression into the previous formula, we write the result in
the form

$\displaystyle\left(N_{m}^{k}f\right)_{i_{1}\ldots i_{m}}(x)$
$\displaystyle=2\sum_{l=0}^{k}\binom{k}{l}\int_{\mathbb{R}^{n}}\big{(}z^{k+l}\otimes
f(z)\big{)}^{p_{1}\dots p_{k+l}j_{1}\dots
j_{m}}\frac{\big{(}(x\\!-\\!z)^{2m+k+l}\big{)}_{i_{1}\ldots i_{m}j_{1}\ldots
j_{m}p_{1}\dots p_{k+l}}}{|x-z|^{2m+2l+n-1}}\,\mathrm{d}z.$

We can replace $\big{(}z^{k+l}\otimes f(z)\big{)}^{p_{1}\dots
p_{k+l}j_{1}\dots j_{m}}$ with $\big{(}z^{k+l}f(z)\big{)}^{j_{1}\dots
j_{m}p_{1}\dots p_{k+l}}$ since the second factor in the integrand is
symmetric in all indices. Hence

$\displaystyle\left(N_{m}^{k}f\right)_{i_{1}\ldots i_{m}}(x)$
$\displaystyle=2\sum_{l=0}^{k}\binom{k}{l}\int_{\mathbb{R}^{n}}\big{(}z^{k+l}f(z)\big{)}^{j_{1}\dots
j_{m}p_{1}\dots p_{k+l}}\frac{\big{(}(x\\!-\\!z)^{2m+k+l}\big{)}_{i_{1}\ldots
i_{m}j_{1}\ldots j_{m}p_{1}\dots p_{k+l}}}{|x-z|^{2m+2l+n-1}}\,\mathrm{d}z.$

Every integral on the right-hand side is the convolution of
$\big{(}x^{k+l}f\big{)}^{j_{1}\dots j_{m}p_{1}\dots p_{k+l}}$ with
$\frac{(x^{2m+k+l})_{i_{1}\ldots i_{m}j_{1}\ldots j_{m}p_{1}\dots
p_{k+l}}}{|x|^{2m+2l+n-1}}$. We thus arrive at (4.2). ∎

We use the abbreviated notation $\partial_{i_{1}\dots
i_{k}}=\frac{\partial^{k}}{\partial y^{i_{1}}\dots\partial y^{i_{k}}}$ for
partial derivatives. Recall that indices can be written either in lower
position or in upper position. In particular, $\partial^{i_{1}\dots
i_{k}}=\partial_{i_{1}\dots i_{k}}$. Recall that $j_{y}:S^{m}\to S^{m-1}$ is
the operator of contraction with $y$, see Subsection 2.1 where the operator
$j_{u}$ is defined. For $0\leq k\leq m$, tensor fields $A^{(m,k)}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{2m-k})$ are defined by (3.7)
or in coordinates

$A^{(m,k)}_{i_{1}\dots i_{2m-k}}=\partial_{i_{1}\dots i_{2m-k}}|y|^{2m-2k-1}.$

Proposition 4.1 can be equivalently written in terms of Fourier transforms
$\widehat{f}$ and $\widehat{N_{m}^{k}f}$.

######  Proposition 4.3.

Let $0\leq k\leq m$ and $n\geq 2$. For
$f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$, the equation (3.8) holds
with $g=\hat{f}$ and $F^{(m,k)}$ defined by (3.3).

###### Proof.

Applying the Fourier transform to the equality (4.2), we obtain

$\widehat{N_{m}^{k}f}_{i_{1}\dots
i_{m}}=2\sum_{l=0}^{k}\binom{k}{l}\big{(}\widehat{x^{k+l}f}\big{)}^{j_{1}\dots
j_{m}p_{1}\dots p_{k+l}}{\mathcal{F}}\Big{[}\frac{(x^{2m+k+l})_{i_{1}\dots
i_{m}j_{1}\dots j_{m}p_{1}\dots p_{k+l}}}{|x|^{2m+2l+n-1}}\Big{]}.$

Using the standard properties of the Fourier transform [Hör83, Lemma 7.1.2]

$\widehat{x_{j}f}=\textsl{i}\,\partial_{j}\widehat{f},\quad\widehat{\partial_{j}f}=\textsl{i}\,y_{j}\widehat{f},$

we transform our formula to the form

$\displaystyle\widehat{N_{m}^{k}f}_{i_{1}\dots
i_{m}}=2(-1)^{m\\!+\\!k}\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}$
$\displaystyle\sigma(j_{1}\dots j_{m}p_{1}\dots
p_{k+l})\big{(}\partial^{p_{1}\dots p_{k+l}}{\widehat{f}}^{j_{1}\dots
j_{m}}\big{)}$ $\displaystyle\times\partial_{i_{1}\dots i_{m}j_{1}\dots
j_{m}p_{1}\dots p_{k+l}}{\mathcal{F}}\Big{[}|x|^{-2m-2l-n+1}\Big{]}.$

Here we can omit the symmetrization $\sigma(j_{1}\dots j_{m}p_{1}\dots
p_{k+l})$ since the second factor
$\partial_{i_{1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}{\mathcal{F}}\Big{[}|x|^{-2m-2l-n+1}\Big{]}$ is symmetric in all
indices. Thus,

(4.3) $\widehat{N_{m}^{k}f}_{i_{1}\dots
i_{m}}=2(-1)^{m\\!+\\!k}\sum_{l=0}^{k}(-1)^{l}\binom{k}{l}\big{(}\partial^{p_{1}\dots
p_{k+l}}{\widehat{f}}^{j_{1}\dots j_{m}}\big{)}\partial_{i_{1}\dots
i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}{\mathcal{F}}\Big{[}|x|^{-2m-2l-n+1}\Big{]}.$

Let ${\mathcal{S}}^{\prime}({{\mathbb{R}}}^{n})$ be the space of tempered
distributions. Recall that $\lambda\mapsto|x|^{\lambda}$ is the meromorphic
${\mathcal{S}}^{\prime}({{\mathbb{R}}}^{n})$-valued function of
$\lambda\in{\mathbb{C}}$ with simple poles at points $-n,-n-2,-n-4,\dots$. The
Fourier transform of $|x|^{\lambda}$ is expressed by

$\displaystyle{\mathcal{F}}[|x|^{\lambda}]$
$\displaystyle=\frac{2^{\lambda+n/2}\Gamma\Big{(}\frac{\lambda+n}{2}\Big{)}}{\Gamma(-\lambda/2)}|y|^{-\lambda-n}\quad(\lambda,-\lambda-n\notin
2{{\mathbb{Z}}}^{+}),$ $\displaystyle{\mathcal{F}}[|x|^{2k}]$
$\displaystyle=(2\pi)^{n/2}(-\Delta)^{k}\delta\quad(k\in{{\mathbb{Z}}}^{+}),$

where $\delta$ is the Dirac function. In particular,

${\mathcal{F}}\Big{[}|x|^{-2m-2l-n+1}\Big{]}=\frac{\Gamma\Big{(}\frac{1-2m-2l}{2}\Big{)}}{2^{2m+2l+n/2-1}\Gamma\Big{(}\frac{2m+2l+n-1}{2}\Big{)}}\,|y|^{2m+2l-1}.$

Substitute this value into (4.3)

(4.4) $\displaystyle\widehat{N_{m}^{k}f}_{i_{1}\dots
i_{m}}(y)=\frac{(-1)^{m\\!+\\!k}}{2^{2m\\!+\\!n/2\\!-\\!2}}\sum_{l=0}^{k}\binom{k}{l}\frac{(-1)^{l}\Gamma\Big{(}\frac{1\\!-\\!2m\\!-\\!2l}{2}\Big{)}}{2^{2l}\Gamma\Big{(}\frac{2m\\!+\\!2l\\!+\\!n\\!-\\!1}{2}\Big{)}}$
$\displaystyle\Big{(}\partial_{i_{1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}|y|^{2m\\!+\\!2l\\!-\\!1}\Big{)}$
$\displaystyle\times\partial^{p_{1}\dots p_{k+l}}{\widehat{f}}^{j_{1}\dots
j_{m}}(y).$

Let us contract the equation (4.4) with $y^{i_{1}}\dots y^{i_{k}}$, i.e.,
multiply the equation by $y^{i_{1}}\dots y^{i_{k}}$ and perform the summation
over indices $i_{1}\dots i_{k}$

(4.5) $\displaystyle\big{(}j_{y}^{k}$
$\displaystyle\widehat{N^{k}_{m}f}\big{)}_{i_{k+1}\dots
i_{m}}(y)=\frac{(-1)^{m\\!+\\!k}}{2^{2m\\!+\\!n/2\\!-\\!2}}\sum_{l=0}^{k}\binom{k}{l}\frac{(-1)^{l}\Gamma\Big{(}\frac{1\\!-\\!2m\\!-\\!2l}{2}\Big{)}}{2^{2l}\Gamma\Big{(}\frac{2m\\!+\\!2l\\!+\\!n\\!-\\!1}{2}\Big{)}}$
$\displaystyle\times\bigg{[}y^{i_{1}}\dots y^{i_{k}}\partial_{i_{1}\cdots
i_{k}}\Big{(}\partial_{i_{k+1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}|y|^{2m\\!+\\!2l\\!-\\!1}\Big{)}\bigg{]}\partial^{p_{1}\dots
p_{k+l}}{\widehat{f}}^{j_{1}\dots j_{m}}(y).$

On the right-hand side, all summands corresponding to $l>0$ are equal to zero.
Indeed,
$\partial_{i_{k+1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}|y|^{2m\\!+\\!2l\\!-\\!1}$ is the positively homogeneous function of
degree $l-1$. By the Euler equation for homogeneous functions,

${y^{i_{1}}\dots y^{i_{k}}\partial_{i_{1}\cdots
i_{k}}}\Big{(}\partial_{i_{k+1}\dots i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k+l}}|y|^{2m\\!+\\!2l\\!-\\!1}\Big{)}=\left\\{\begin{array}[]{ll}(-1)^{k}k!|y|^{2m-1}&\mbox{if}\
l=0,\\\ 0&\mbox{if}\ l>0.\end{array}\right.$

The formula (4.5) becomes

(4.6) $\big{(}j_{y}^{k}\,\widehat{N^{k}_{m}f}\big{)}_{i_{k+1}\dots
i_{m}}(y)=\frac{(-1)^{m}k!\Gamma\Big{(}\frac{1\\!-\\!2m}{2}\Big{)}}{2^{2m\\!+\\!n/2\\!-\\!2}\Gamma\Big{(}\frac{2m\\!+\\!n\\!-\\!1}{2}\Big{)}}\Big{(}\partial_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k}}|y|^{2m\\!-\\!1}\Big{)}\partial^{p_{1}\dots
p_{k}}{\widehat{f}}^{j_{1}\dots j_{m}}(y).$

This is equivalent to (3.8). ∎

######  Lemma 4.4.

Let $0\leq k\leq m$ and $n\geq 2$. Then
$j_{y}^{k+1}\,\widehat{N^{k}_{m}f}(y)=0$ for any tensor field
$f\in\mathcal{S}\left(\mathbb{R}^{n};S^{m}\right)$ and for any
$y\in{{\mathbb{R}}}^{n}$.

###### Proof.

The statement trivially holds in the case of $k=m$. In the case of $k<m$ we
apply the operator $j_{y}$ to the equality (4.6)

$\big{(}j_{y}^{k+1}\,\widehat{N^{k}_{m}f}\big{)}_{i_{k+2}\dots
i_{m}}(y)=\frac{1}{C_{m}^{k}}\Big{[}\big{(}y^{i_{k+1}}\partial_{i_{k+1}}\big{)}\Big{(}\partial_{i_{k+2}\dots
i_{m}j_{1}\dots j_{m}p_{1}\dots
p_{k}}|y|^{2m\\!-\\!1}\Big{)}\Big{]}\partial^{p_{1}\dots
p_{k}}{\widehat{f}}^{j_{1}\dots j_{m}}(y),$

where
$C_{m}^{k}=(-1)^{m}k!\Gamma\big{(}\frac{1\\!-\\!2m}{2}\big{)}/\Big{(}2^{2m\\!+\\!n/2\\!-\\!2}\Gamma\big{(}\frac{2m\\!+\\!n\\!-\\!1}{2}\big{)}\Big{)}$.
The expression in brackets is equal to zero since $\partial_{i_{k+2}\dots
i_{m}j_{1}\dots j_{m}p_{1}\dots p_{k}}|y|^{2m\\!-\\!1}$ is a positively
homogeneous function of zero degree. ∎

From now on we can forget the momentum ray transform. The rest of the article
is devoted to investigation of the system (3.8).

Lemma 4.4 implies that right-hand sides of equations (3.8) satisfy

(4.7) $j_{y}\,F^{(m,k)}(y)=0\quad(0\leq k\leq m).$

Thus, equalities (4.7) constitute necessary conditions for existence of a
solution $g\in{\mathcal{S}}({{\mathbb{R}}}^{n};S^{m})$ to the system (3.8).
Most probably, equalities (4.7) are necessary and sufficient consistency
conditions for the system (3.8), but this fact is not proved.

##  5\. Reduction of the system (3.8) to an algebraic system

Tensor fields $A^{(m,k)}\in
C^{\infty}({\mathbb{R}}^{n}\setminus\\{0\\};S^{2m-k})\ (0\leq k\leq m)$ are
defined by (3.7). There exist two important relations between these tensor
fields.

######  Lemma 5.1.

The following equalities are valid:

(5.1) $j_{y}A^{(m,k)}=-kA^{(m-1,k-1)}\quad(0\leq k\leq m),$ (5.2) $\mbox{\rm
div}\,A^{(m,k)}=(2m-2k-1)(n+2m-2k-3)A^{(m,k+1)}\quad(0\leq k\leq m).$

###### Proof.

Applying the operator $j_{y}$ to the equality (3.7), we have

$j_{y}A^{(m,k)}=j_{y}d^{2m-k}|y|^{2m-2k-1}.$

With the help of the operator $\langle
y,\partial\rangle=y^{j}\frac{\partial}{\partial y^{j}}$, the latter formula
can be written as

(5.3) $j_{y}A^{(m,k)}=\langle y,\partial\rangle\,d^{2m-k-1}|y|^{2m-2k-1}.$

The tensor field $d^{2m-k-1}|y|^{2m-2k-1}$ is positively homogeneous of degree
$-k$. By the Euler equation for homogeneous functions,

$\langle
y,\partial\rangle\,d^{2m-k-1}|y|^{2m-2k-1}=-k\,d^{2m-k-1}|y|^{2m-2k-1}.$

Substituting this expression into (5.3), we obtain

$j_{y}A^{(m,k)}=-k\,d^{2m-k-1}|y|^{2m-2k-1}.$

By (3.6), the right-hand side of this formula is equal to $-kA^{(m-1,k-1)}$.
This proves (5.1).

Let us write (3.7) in the coordinate form

$A^{(m,k)}_{i_{k+1}\dots i_{2m}}=\partial_{i_{k+1}\dots i_{2m}}|y|^{2m-2k-1}.$

Differentiate this equality

$\frac{\partial A^{(m,k)}_{i_{k+1}\dots i_{2m}}}{\partial
y^{j}}=\partial_{ji_{k+1}\dots i_{2m}}|y|^{2m-2k-1}.$

From this

$\displaystyle(\mbox{\rm div}\,A^{(m,k)})_{i_{k+2}\dots i_{2m}}$
$\displaystyle=\delta^{jl}\,\frac{\partial A^{(m,k)}_{li_{k+2}\dots
i_{2m}}}{\partial y^{j}}=\delta^{jl}\,\partial_{jli_{k+2}\dots
i_{2m}}|y|^{2m-2k-1}$ $\displaystyle=\partial_{i_{k+2}\dots
i_{2m}}(\Delta|y|^{2m-2k-1}).$

This can be written in the coordinate-free form

(5.4) $\mbox{\rm div}\,A^{(m,k)}=d^{2m-k-1}(\Delta|y|^{2m-2k-1}).$

Using the obvious formula

$\Delta|y|^{\alpha}=\alpha(\alpha+n-2)|y|^{\alpha-2},$

we obtain

$\Delta|y|^{2m-2k-1}=(2m-2k-1)(n+2m-2k-3)|y|^{2m-2k-3}.$

Substituting this expression into (5.4), we have

$\mbox{\rm div}\,A^{(m,k)}=(2m-2k-1)(n+2m-2k-3)\,d^{2m-k-1}|y|^{2m-2k-3}.$

By (3.7),

$d^{2m-k-1}|y|^{2m-2k-3}=A^{(m,k+1)}.$

Two last formulas imply (5.2). ∎

From (5.2), one easily proves by induction on $k$

(5.5) $\mbox{\rm
div}^{k}\,A^{(m,0)}=\frac{(2m-1)!!(n+2m-3)!!}{(2m-2k-1)!!(n+2m-2k-3)!!}\,A^{(m,k)}\quad(0\leq
k\leq m).$

We reproduce the system (3.8)

(5.6) $A^{(m,0)}/(d^{l}g)=F^{(m,l)}\quad(l=0,1,\dots,m).$

Here $F^{(m,l)}\in C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m-l})\
(0\leq l\leq m)$ are arbitrary tensor fields belonging to the kernel of
$j_{y}$.

######  Proposition 5.2.

If a tensor field $g\in C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m})$
satisfies (5.6), then

(5.7) $(\mbox{\rm
div}^{k}A^{(m,0)})/(d^{l}g)=(-1)^{k}\sum\limits_{p=0}^{k}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p}\,F^{(m,k+l-p)}\quad(0\leq k\leq m,0\leq l\leq m-k).$

###### Proof.

We prove (5.7) by induction on $k$. For $k=0$, (5.7) coincides with (5.6).
Assume (5.7) to be valid for some $k$. Apply the operator div to the equation
(5.7)

(5.8) $\mbox{\rm div}\Big{(}(\mbox{\rm
div}^{k}A^{(m,0)})/(d^{l}g)\Big{)}=(-1)^{k}\sum\limits_{p}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p+1}\,F^{(m,k+l-p)}.$

We assume binomial coefficients to be defined for all integers $k$ and $p$
under the agreement:

(5.9) $\binom{k}{p}=0\quad\mbox{if either}\ k<0\ \mbox{or}\ p<0\ \mbox{or}\
k<p.$

Due to the agreement, we can assume the summation to be performed over all
integers $p$ in (5.8) and formulas below.

The equality

$\mbox{\rm div}(u/v)=(\mbox{\rm div}\,u)/v+u/(dv)$

is valid for any two tensor fields. It is easily proved on the base of
definitions of the operators $d$ and $\rm div$. With the help of this
equality, we write (5.8) in the form

$(\mbox{\rm div}^{k+1}A^{(m,0)})/(d^{l}g)+(\mbox{\rm
div}^{k}A^{(m,0)})/(d^{l+1}g)=(-1)^{k}\sum\limits_{p}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p+1}\,F^{(m,k+l-p)}.$

By the induction hypothesis,

$(\mbox{\rm
div}^{k}A^{(m,0)})/(d^{l+1}g)=(-1)^{k}\sum\limits_{p}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p}\,F^{(m,k+l-p+1)}.$

Substituting this expression into the previous formula, we write the result in
the form

$\displaystyle(\mbox{\rm div}^{k+1}A^{(m,0)})/(d^{l}g)$
$\displaystyle=(-1)^{k}\sum\limits_{p}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p+1}\,F^{(m,k+l-p)}$
$\displaystyle\quad+(-1)^{k+1}\sum\limits_{p}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p}\,F^{(m,k+l-p+1)}.$

Changing the summation variable of the first sum as $p:=p-1$, we obtain

$(\mbox{\rm
div}^{k+1}A^{(m,0)})/(d^{l}g)=(-1)^{k+1}\sum\limits_{p}(-1)^{p}\bigg{[}\binom{k}{p-1}+\binom{k}{p}\bigg{]}\mbox{\rm
div}^{p}\,F^{(m,k+l-p+1)}.$

By the Pascal triangle equality, $\binom{k}{p-1}+\binom{k}{p}=\binom{k+1}{p}$.
Substituting this expression into the last formula, we arrive at (5.7) for
$k:=k+1$. ∎

Setting $l=0$ in (5.7), we obtain

$(\mbox{\rm
div}^{k}A^{(m,0)})/g=(-1)^{k}\sum\limits_{p=0}^{k}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p}\,F^{(m,k-p)}\quad(0\leq k\leq m).$

Substituting the value (5.5) of $\mbox{\rm div}^{k}A^{(m,0)}$, we arrive at
the equation

$A^{(m,k)}/g=(-1)^{k}\frac{(2m-2k-1)!!(n+2m-2k-3)!!}{(2m-1)!!(n+2m-3)!!}\sum\limits_{p=0}^{k}(-1)^{p}\binom{k}{p}\,\mbox{\rm
div}^{p}\,F^{(m,k-p)}.$

We have thus proved

######  Proposition 5.3.

If a tensor field $g\in C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m})$
satisfies (3.8), then it also solves the system (3.9) with right-hand sides
defined by (3.4).

We emphasize that (3.9) is a system of linear algebraic equations in
coordinates of the unknown tensor field $g$. Of course the system (3.9) is not
equivalent to (3.8). Proposition 5.3 states that (3.8) implies (3.9) but not
vice versa. Nevertheless, we will see that $g$ can be uniquely recovered from
(3.9). In this sense the system (3.8) is reduced to the algebraic system
(3.9).

##  6\. Solution of the system (3.9)

The following statement completes the proof of Theorem 3.3.

######  Proposition 6.1.

If the system (3.9) is solvable, then the solution $g=\widehat{f}$ is unique
and is expressed by the formula (3.5).

The proof of Proposition 6.1 is not easy. The main part of the proof is
contained in the following two lemmas.

######  Lemma 6.2.

Given a tensor field $g\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m+1})$, let us fix a
Cartesian coordinate system on ${{\mathbb{R}}}^{n}$, fix a value of the index
$i_{m+1}$ and introduce the tensor field $\tilde{g}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m})$ by

(6.1) $\tilde{g}_{i_{1}\dots i_{m}}=g_{i_{1}\dots i_{m}i_{m+1}}.$

Let us also introduce the vector field $\tilde{\delta}$ by

$\tilde{\delta}_{i}=\delta_{ii_{m+1}}.$

Then, for every $0\leq k\leq m$,

(6.2) $\displaystyle(A^{(m,k)}/\tilde{g})_{i_{k+1}\dots i_{m}}$
$\displaystyle=\frac{1}{m\\!+\\!1}\Big{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}$ $\displaystyle-y_{i_{m+1}}\,(A^{(m+1,k+1)}/g)_{i_{k+1}\dots
i_{m}}-(m\\!-\\!k)\,\Big{(}\tilde{\delta}\,(A^{(m,k)}/g)\Big{)}_{i_{k+1}\dots
i_{m}}\Big{]}.$

###### Proof.

The identity

$\partial_{i_{1}\dots i_{p}}(|y|^{\alpha}y_{k})=y_{k}\,\partial_{i_{1}\dots
i_{p}}|y|^{\alpha}+p\,\sigma(i_{1}\dots
i_{p})(\delta_{i_{1}k}\,\partial_{i_{2}\dots i_{p}}|y|^{\alpha})$

holds for any integer $p\geq 0$, any real $\alpha$ and any $1\leq k\leq n$. It
is easily proved by induction on $p$. With the help of this identity, we
obtain

$\displaystyle\partial_{i_{k+1}\dots i_{m+1}j_{1}\dots
j_{m+1}}|y|^{2m-2k+1}=(2m\\!-\\!2k\\!+\\!1)\partial_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m+1}}(|y|^{2m-2k-1}y_{i_{m+1}})$
$\displaystyle=(2m\\!-\\!2k\\!+\\!1)\bigg{[}y_{i_{m+1}}\,\partial_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m+1}}|y|^{2m-2k-1}$
$\displaystyle+(2m\\!-\\!k\\!+\\!1)\sigma(i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1})\Big{(}\delta_{i_{m+1}j_{m+1}}\,\partial_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m}}|y|^{2m-2k-1}\Big{)}\bigg{]}.$

Expanding the symmetrization $\sigma(i_{k+1}\dots i_{m}j_{1}\dots j_{m+1})$
with respect to the index $j_{m+1}$ (see [Sha94, Lemma 2.4.1]), we write this
in the form

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}$
$\displaystyle\partial_{i_{k+1}\dots i_{m+1}j_{1}\dots
j_{m+1}}|y|^{2m-2k+1}=y_{i_{m+1}}\,\partial_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}|y|^{2m-2k-1}$ $\displaystyle+\sigma(i_{k+1}\dots i_{m}j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}j_{m+1}}\,\partial_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m}}|y|^{2m-2k-1}$
$\displaystyle+(2m\\!-\\!k)\delta_{i_{m+1}j_{1}}\,\partial_{i_{k+1}\dots
i_{m}j_{2}\dots j_{m+1}}|y|^{2m-2k-1}\Big{)}.$

Since the tensor $\delta_{i_{m+1}j_{m+1}}\,\partial_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}|y|^{2m-2k-1}$ is symmetric in
$i_{k+1},\dots,i_{m},j_{1},\dots,j_{m}$, the formula can be written as
follows:

(6.3) $\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\partial_{i_{k+1}\dots
i_{m+1}j_{1}\dots j_{m+1}}|y|^{2m-2k+1}=y_{i_{m+1}}\,\partial_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m+1}}|y|^{2m-2k-1}$
$\displaystyle+\delta_{i_{m+1}j_{m+1}}\,\partial_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m}}|y|^{2m-2k-1}$ $\displaystyle+(2m\\!-\\!k)\sigma(i_{k+1}\dots
i_{m}j_{1}\dots j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,\partial_{i_{k+1}\dots
i_{m}j_{2}\dots j_{m+1}}|y|^{2m-2k-1}\Big{)}.$

By (3.7),

$\displaystyle\partial_{i_{k+1}\dots i_{m}j_{1}\dots j_{m+1}}|y|^{2m-2k-1}$
$\displaystyle=A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots j_{m+1}},$
$\displaystyle\partial_{i_{k+1}\dots i_{m}j_{1}\dots j_{m}}|y|^{2m-2k-1}$
$\displaystyle=A^{(m,k)}_{i_{k+1}\dots i_{m}j_{1}\dots j_{m}},$
$\displaystyle\partial_{i_{k+1}\dots i_{m}j_{2}\dots j_{m+1}}|y|^{2m-2k-1}$
$\displaystyle=A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots j_{m+1}}.$

Substitute these expressions into (6.3)

(6.4) $\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\partial_{i_{k+1}\dots
i_{m+1}j_{1}\dots
j_{m+1}}|y|^{2m-2k+1}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}$ $\displaystyle+\delta_{i_{m+1}j_{m+1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}$ $\displaystyle+(2m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m}j_{1}\dots j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{2}\dots j_{m+1}}\Big{)}.$

The equality

(6.5) $\displaystyle\sigma(i_{k+1}\dots i_{m}j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots
j_{m+1}}\Big{)}$ $\displaystyle=\frac{1}{2m\\!-\\!k}\sigma(i_{k+1}\dots
i_{m})\sigma(j_{1}\dots
j_{m})\Big{[}(m\\!-\\!k)\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m-1}j_{1}\dots j_{m+1}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+m\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{2}\dots j_{m+1}}\Big{]}$

is easily proved on the base of the only fact: the tensor $A^{(m,k)}$ is
symmetric. Formally speaking, the first term
$(m\\!-\\!k)\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}$ in brackets makes sense for $k\leq m-2$ only. Nevertheless, the
formula (6.5) holds for $k=m-1$ if we assume that $A^{(m,k)}_{i_{m}\dots
i_{m-1}j_{1}\dots j_{m+1}}=A^{(m,k)}_{j_{1}\dots j_{m+1}}$. In the case of
$k=m$, the first term in brackets is equal to zero because of the factor
$(m-k)$. Thus, the formula (6.5) holds for all $0\leq k\leq m$.

With the help of (6.5), the formula (6.4) becomes

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\partial_{i_{k+1}\dots
i_{m+1}j_{1}\dots
j_{m+1}}|y|^{2m-2k+1}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}$ $\displaystyle+\delta_{i_{m+1}j_{m+1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}$ $\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}\Big{)}$ $\displaystyle+m\,\sigma(i_{k+1}\dots i_{m})\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots
j_{m+1}}\Big{)}.$

On the last line, the symmetrization $\sigma(i_{k+1}\dots i_{m})$ can be
deleted since the tensor $A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots j_{m+1}}$ is
symmetric in these indices. The formula simplifies to the following one:

(6.6) $\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\partial_{i_{k+1}\dots
i_{m+1}j_{1}\dots
j_{m+1}}|y|^{2m-2k+1}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}$ $\displaystyle+\delta_{i_{m+1}j_{m+1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}$ $\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}\Big{)}$ $\displaystyle+m\,\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots
j_{m+1}}\Big{)}.$

Formulas (6.3) and (6.6) imply

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}=\bigg{[}y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}$ $\displaystyle+\delta_{i_{m+1}j_{m+1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}$ $\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}\Big{)}$ $\displaystyle+m\,\sigma(j_{1}\dots
j_{m})\Big{(}\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{2}\dots
j_{m+1}}\Big{)}\bigg{]}g^{j_{1}\dots j_{m+1}}.$

The symmetrization $\sigma(j_{1}\dots j_{m})$ can be deleted after opening
brackets since the tensor $g^{j_{1}\dots j_{m+1}}$ is symmetric in these
indices. We thus obtain

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}$
$\displaystyle(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}$
$\displaystyle+\delta_{i_{m+1}j_{m+1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}g^{j_{1}\dots j_{m+1}}$
$\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}\Big{)}$
$\displaystyle+m\,\delta_{i_{m+1}j_{1}}\,A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{2}\dots j_{m+1}}g^{j_{1}\dots j_{m+1}}.$

Implementing the contraction with the Kronecker tensor in second and last
lines, we obtain

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}$
$\displaystyle(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}$ $\displaystyle+A^{(m,k)}_{i_{k+1}\dots
i_{m}j_{1}\dots j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}$
$\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}\Big{)}$
$\displaystyle+m\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}.$

In the last line, we have replaced the summation indices $j_{2},\dots,j_{m+1}$
with $j_{1},\dots,j_{m}$. We see now that second and last lines contain
similar terms. Grouping this terms, we write the formula as follows:

(6.7) $\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}$
$\displaystyle(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}=y_{i_{m+1}}\,A^{(m+1,k+1)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}$
$\displaystyle+(m\\!-\\!k)\,\sigma(i_{k+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,k)}_{i_{k+1}\dots i_{m-1}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}\Big{)}$
$\displaystyle+(m+1)\,A^{(m,k)}_{i_{k+1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}.$

Recall that the value of the index $i_{m+1}$ is fixed and
$\tilde{g}^{j_{1}\dots j_{m}}=g_{i_{m+1}}^{j_{1}\dots j_{m}}$. The formula
(6.7) can be written as

$\displaystyle\frac{1}{2m\\!-\\!2k\\!+\\!1}$
$\displaystyle(A^{(m+1,k)}/g)_{i_{k+1}\dots
i_{m+1}}=y_{i_{m+1}}\,(A^{(m+1,k+1)}/g)_{i_{k+1}\dots i_{m}}$
$\displaystyle+(m\\!-\\!k)\,\Big{(}\tilde{\delta}\,(A^{(m,k)}/g)\Big{)}_{i_{k+1}\dots
i_{m}}+(m+1)\,(A^{(m,k)}/\tilde{g})_{i_{k+1}\dots i_{m}}.$

This is equivalent to (6.2). ∎

######  Lemma 6.3.

For a tensor field $g\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m})$, the following identity
holds:

(6.8)
$A^{(m,0)}/g=m!(2m-1)!!\,|y|^{-1}g+\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m,k)}/g),$

where the coefficients $\beta(m,k,p)$ are uniquely determined by the recurrent
formulas

(6.9) $\displaystyle\tilde{\beta}(m\\!+\\!1,k,p)$
$\displaystyle=\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\beta(m,k,p)-\frac{k\\!-\\!p}{k}\,\beta(m,k\\!-\\!1,p)$
$\displaystyle+\frac{m\\!-\\!k\\!-\\!p\\!+\\!2}{\textstyle
k}\,\beta(m,k\\!-\\!1,p\\!-\\!1)$

and

(6.10)
$\beta(m+1,k,p)=\left\\{\begin{array}[]{ll}(2m\\!+\\!1)\big{(}\tilde{\beta}(m+1,k,p)+1\big{)}&\mbox{\rm
if}\ (k,p)=(1,0),\\\\[8.0pt]
(2m\\!+\\!1)\big{(}\tilde{\beta}(m+1,k,p)-m\big{)}&\mbox{\rm if}\
(k,p)=(1,1),\\\\[8.0pt] (2m\\!+\\!1)\tilde{\beta}(m+1,k,p)&\mbox{\rm
otherwise}\end{array}\right.$

under the agreement

(6.11) $\beta(m,k,p)=0\quad\mbox{\rm if either}\ k=0\ \mbox{\rm or}\ k>m\
\mbox{\rm or}\ p<0\ \mbox{\rm or}\ p>\min(k,m-k).$

###### Proof.

The proof is going by induction on $m$. For $m=0$, the sum on the right-hand
side of (6.8) is absent and the formula holds since $A^{(m,0)}=|y|^{-1}$.
Assume (6.8) to be valid for some $m\geq 0$ and let $g\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m+1})$. We fix a value of the
index $i_{m+1}$ and introduce the tensor field $\tilde{g}\in
C^{\infty}({{\mathbb{R}}}^{n}\setminus\\{0\\};S^{m})$ by (6.1). By the
induction hypothesis, the formula (6.8) holds for $\tilde{g}$. Let us write
the formula in coordinates

(6.12) $\displaystyle A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}=m!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots
i_{m+1}}$ $\displaystyle+\sigma(i_{1}\dots
i_{m})\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\,y^{l_{1}}\dots
y^{l_{p}}\times$
$\displaystyle\qquad\qquad\qquad\times\delta_{i_{1}i_{2}}\dots\delta_{i_{2p-1}i_{2p}}y_{i_{2p+1}}\dots
y_{i_{k+p}}A^{(m,k)}_{i_{k+p+1}\dots i_{m}l_{1}\dots l_{p}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}.$

By Lemma 6.2,

(6.13) $\displaystyle A^{(m,k)}_{i_{k+p+1}\dots i_{m}l_{1}\dots
l_{p}j_{1}\dots j_{m}}g_{i_{m+1}}^{j_{1}\dots
j_{m}}=(A^{(m,k)}/\tilde{g})_{i_{k+p+1}\dots i_{m}l_{1}\dots l_{p}}$
$\displaystyle=\frac{1}{m\\!+\\!1}\Big{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(A^{(m+1,k)}/g)_{i_{k+p+1}\dots
i_{m+1}l_{1}\dots l_{p}}-y_{i_{m+1}}\,(A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}$
$\displaystyle-(m\\!-\\!k)\,\Big{(}\tilde{\delta}\,(A^{(m,k)}/g)\Big{)}_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}\Big{]}.$

In the case of $k=m$, the last term on the right-hand side of (6.13) is equal
to zero. In the case of $k<m$, we transform the last term on the right-hand
side of (6.13) with the help of (5.1) as follows:

(6.14)
$\displaystyle\Big{(}\tilde{\delta}\,(A^{(m,k)}/g)\Big{)}_{i_{k+p+1}\dots
i_{m}l_{1}\dots
l_{p}}=-\frac{1}{k+1}\Big{(}\tilde{\delta}(j_{y}A^{(m+1,k+1)}/g)\Big{)}_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}$ $\displaystyle=-\frac{1}{k+1}\sigma(i_{k+p+1}\dots
i_{m}l_{1}\dots
l_{p})\Big{(}\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}\Big{)}.$

The equality

$\displaystyle\sigma(i_{k+p+1}\dots i_{m}$ $\displaystyle l_{1}\dots
l_{p})\Big{(}\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}\Big{)}$
$\displaystyle=\frac{m\\!-\\!k\\!-\\!p}{m\\!-\\!k}\,\sigma(i_{k+p+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}\Big{)}$
$\displaystyle+\frac{p}{m\\!-\\!k}\,\sigma(l_{1}\dots
l_{p})\Big{(}\delta_{i_{m+1}l_{1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{2}\dots l_{p}}\Big{)}$

holds since $j_{y}A^{(m+1,k+1)}/g$ is a symmetric tensor. With the help of
this, the formula (6.14) takes the form

(6.15)
$\displaystyle\Big{(}\tilde{\delta}\,(A^{(m,k)}/g)\Big{)}_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}$
$\displaystyle=-\frac{m\\!-\\!k\\!-\\!p}{(m\\!-\\!k)(k\\!+\\!1)}\,\sigma(i_{k+p+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}\Big{)}$
$\displaystyle-\frac{p}{(m\\!-\\!k)(k\\!+\\!1)}\,\sigma(l_{1}\dots
l_{p})\Big{(}\delta_{i_{m+1}l_{1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{2}\dots l_{p}}\Big{)}.$

Replacing the last term on the right-hand side of (6.13) with its value
(6.15), we obtain

(6.16) $\displaystyle A^{(m,k)}_{i_{k+p+1}\dots i_{m}l_{1}\dots
l_{p}j_{1}\dots j_{m}}g_{i_{m+1}}^{j_{1}\dots
j_{m}}=(A^{(m,k)}/\tilde{g})_{i_{k+p+1}\dots i_{m}l_{1}\dots l_{p}}$
$\displaystyle=\frac{1}{m\\!+\\!1}\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(A^{(m+1,k)}/g)_{i_{k+p+1}\dots
i_{m+1}l_{1}\dots l_{p}}-y_{i_{m+1}}\,(A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}$
$\displaystyle\qquad\qquad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\,\sigma(i_{k+p+1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}\Big{)}$
$\displaystyle\qquad\qquad+\frac{p}{k\\!+\\!1}\,\sigma(l_{1}\dots
l_{p})\Big{(}\delta_{i_{m+1}l_{1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{2}\dots l_{p}}\Big{)}\bigg{]}.$

It is not quite obvious now that two last lines on the right-hand side of
(6.16) are equal to zero in the case of $k=m$. Nevertheless, in the case of
$k=m$ we are interested in (6.16) for $p=0$ only, as is seen from (6.12). For
$k=m$ and $p=0$, the last two lines on the right-hand side of (6.16) are equal
to zero.

We substitute the expression (6.16) into (6.12). After the substitution, the
symmetrization $\sigma(i_{k+p+1}\dots i_{m})$ can be omitted because of the
presence of the “larger” symmetrization $\sigma(i_{1}\dots i_{m})$. The
symmetrization $\sigma(l_{1}\dots l_{p})$ can be also omitted because of the
presence of the factor $y^{l_{1}}\dots y^{l_{p}}$. We thus obtain

$\displaystyle A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}=m!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots
i_{m+1}}$ $\displaystyle+\frac{1}{m\\!+\\!1}\,\sigma(i_{1}\dots
i_{m})\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\,\delta_{i_{1}i_{2}}\dots\delta_{i_{2p-1}i_{2p}}\,y_{i_{2p+1}}\dots
y_{i_{k+p}}\,y^{l_{1}}\dots y^{l_{p}}\times$
$\displaystyle\times\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(A^{(m+1,k)}/g)_{i_{k+p+1}\dots
i_{m+1}l_{1}\dots l_{p}}-y_{i_{m+1}}\,(A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{1}\dots l_{p}}$
$\displaystyle\qquad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\,\delta_{i_{m+1}i_{k+p+1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}l_{1}\dots l_{p}}$
$\displaystyle\qquad+\frac{p}{k\\!+\\!1}\,\delta_{i_{m+1}l_{1}}(j_{y}A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}l_{2}\dots l_{p}}\bigg{]}.$

After pulling the factor $y^{l_{1}}\dots y^{l_{p}}$ inside brackets, this
becomes

$\displaystyle A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}=m!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots
i_{m+1}}$ $\displaystyle+\frac{1}{m\\!+\\!1}\,\sigma(i_{1}\dots
i_{m})\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\,\delta_{i_{1}i_{2}}\dots\delta_{i_{2p-1}i_{2p}}\,y_{i_{2p+1}}\dots
y_{i_{k+p}}\times$
$\displaystyle\times\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(j_{y}^{p}\,A^{(m+1,k)}/g)_{i_{k+p+1}\dots
i_{m+1}}-y_{i_{m+1}}\,(j_{y}^{p}\,A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots i_{m}}$
$\displaystyle\qquad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\,\delta_{i_{m+1}i_{k+p+1}}(j_{y}^{p+1}\,A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}}$
$\displaystyle\qquad+\frac{p}{k\\!+\\!1}\,y_{i_{m+1}}(j_{y}^{p}\,A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}}\bigg{]}.$

Observe that second and last terms in brackets differ by coefficients only.
After grouping these terms, the formula becomes

$\displaystyle A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}=m!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots
i_{m+1}}$ $\displaystyle+\frac{1}{m\\!+\\!1}\,\sigma(i_{1}\dots
i_{m})\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\,\delta_{i_{1}i_{2}}\dots\delta_{i_{2p-1}i_{2p}}\,y_{i_{2p+1}}\dots
y_{i_{k+p}}\times$
$\displaystyle\times\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,(j_{y}^{p}\,A^{(m+1,k)}/g)_{i_{k+p+1}\dots
i_{m+1}}-\frac{k\\!-\\!p\\!+\\!1}{k\\!+\\!1}\,y_{i_{m+1}}\,(j_{y}^{p}\,A^{(m+1,k+1)}/g)_{i_{k+p+1}\dots
i_{m}}$
$\displaystyle\qquad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\,\delta_{i_{m+1}i_{k+p+1}}(j_{y}^{p+1}\,A^{(m+1,k+1)}/g)_{i_{k+p+2}\dots
i_{m}}\bigg{]}.$

Next, we pull the factor $\delta_{i_{2p-1}i_{2p}}\,y_{i_{2p+1}}\dots
y_{i_{k+p}}$ inside brackets

(6.17) $\displaystyle A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}=m!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots
i_{m+1}}$
$\displaystyle+\frac{1}{m\\!+\\!1}\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\Big{(}i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)\Big{)}_{i_{1}\dots
i_{m+1}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad-\frac{k\\!-\\!p\\!+\\!1}{k\\!+\\!1}\,y_{i_{m+1}}\Big{(}i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k+1)}/g)\Big{)}_{i_{1}\dots
i_{m}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\Big{(}i_{\tilde{\delta}}\,i_{y}^{k-p}i^{p}j_{y}^{p+1}(A^{(m+1,k+1)}/g)\Big{)}_{i_{1}\dots
i_{m}}\bigg{]}.$

Next, we apply Lemma 6.2 with $k=0$. More precisely, we reproduce the formula
(6.7) from the proof of the lemma for $k=0$

$\displaystyle(A^{(m+1,0)}/g)_{i_{1}\dots
i_{m+1}}=(m\\!+\\!1)(2m\\!+\\!1)\,A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}$
$\displaystyle+(2m\\!+\\!1)y_{i_{m+1}}\,A^{(m+1,1)}_{i_{1}\dots
i_{m}j_{1}\dots j_{m+1}}g^{j_{1}\dots j_{m+1}}$
$\displaystyle+m(2m\\!+\\!1)\,\sigma(i_{1}\dots
i_{m})\Big{(}\delta_{i_{m+1}i_{m}}\,A^{(m,0)}_{i_{1}\dots i_{m-1}j_{1}\dots
j_{m+1}}g^{j_{1}\dots j_{m+1}}\Big{)}$ .

This can be written in the form

$\displaystyle(A^{(m+1,0)}$ $\displaystyle/g)_{i_{1}\dots
i_{m+1}}=(m\\!+\\!1)(2m\\!+\\!1)\,A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}$
$\displaystyle+(2m\\!+\\!1)y_{i_{m+1}}\,(A^{(m+1,1)}/g)_{i_{1}\dots
i_{m}}+m(2m\\!+\\!1)\Big{(}i_{\tilde{\delta}}(A^{(m,0)}/g)\Big{)}_{i_{1}\dots
i_{m}}.$

By (5.1), $A^{(m,0)}=-j_{y}A^{(m+1,1)}$. Substituting this expression into the
last line of the previous formula, we obtain

(6.18) $\displaystyle(A^{(m+1,0)}/g)_{i_{1}\dots
i_{m+1}}=(m\\!+\\!1)(2m\\!+\\!1)\,A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}$
$\displaystyle+(2m\\!+\\!1)y_{i_{m+1}}\,(A^{(m+1,1)}/g)_{i_{1}\dots
i_{m}}-m(2m\\!+\\!1)\Big{(}i_{\tilde{\delta}}\,j_{y}(A^{(m+1,1)}/g)\Big{)}_{i_{1}\dots
i_{m}}.$

Now, we replace the first term $A^{(m,0)}_{i_{1}\dots i_{m}j_{1}\dots
j_{m}}g_{i_{m+1}}^{j_{1}\dots j_{m}}$ on the right-hand side of (6.18) with
its expression (6.17)

(6.19) $\displaystyle\frac{1}{2m\\!+\\!1}(A^{(m+1,0)}/g)_{i_{1}\dots
i_{m+1}}=(m\\!+\\!1)!(2m-1)!!\,|y|^{-1}g_{i_{1}\dots i_{m+1}}$
$\displaystyle+\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\Big{(}i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)\Big{)}_{i_{1}\dots
i_{m+1}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad-\frac{k\\!-\\!p\\!+\\!1}{k\\!+\\!1}\,y_{i_{m+1}}\Big{(}i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k+1)}/g)\Big{)}_{i_{1}\dots
i_{m}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\Big{(}i_{\tilde{\delta}}\,i_{y}^{k-p}i^{p}j_{y}^{p+1}(A^{(m+1,k+1)}/g)\Big{)}_{i_{1}\dots
i_{m}}\bigg{]}$ $\displaystyle+y_{i_{m+1}}\,\,(A^{(m+1,1)}/g)_{i_{1}\dots
i_{m}}-m\Big{(}i_{\tilde{\delta}}\,j_{y}(A^{(m+1,1)}/g)\Big{)}_{i_{1}\dots
i_{m}}.$

From now on, we again let $i_{m+1}$ be an arbitrary index. We apply the
symmetrization $\sigma(i_{1}\dots i_{m+1})$ to the equation (6.19). The
operator $i_{\tilde{\delta}}$ becomes $i$ after the symmetrization and the
result can be written in the coordinate-free form (recall that operators
$i_{y}$ and $i$ commute)

(6.20)
$\displaystyle\frac{1}{2m\\!+\\!1}\,A^{(m+1,0)}/g=(m\\!+\\!1)!(2m-1)!!\,|y|^{-1}\,g$
$\displaystyle+\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\beta(m,k,p)\bigg{[}\frac{1}{2m\\!-\\!2k\\!+\\!1}\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle\qquad\qquad-\frac{k\\!-\\!p\\!+\\!1}{k\\!+\\!1}\,i_{y}^{k-p+1}i^{p}j_{y}^{p}(A^{(m+1,k+1)}/g)+\frac{m\\!-\\!k\\!-\\!p}{k\\!+\\!1}\,i_{y}^{k-p}i^{p+1}j_{y}^{p+1}(A^{(m+1,k+1)}/g)\bigg{]}$
$\displaystyle+i_{y}\,(A^{(m+1,1)}/g)-m\,ij_{y}\,(A^{(m+1,1)}/g).$

Let us write (6.20) in the form

$\displaystyle\frac{1}{2m\\!+\\!1}\,A^{(m+1,0)}/g=(m\\!+\\!1)!(2m-1)!!\,|y|^{-1}\,g$
$\displaystyle+\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\beta(m,k,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle-\sum\limits_{k^{\prime}=1}^{m}\sum\limits_{p=0}^{\min(k^{\prime},m-k^{\prime})}\\!\\!\frac{k^{\prime}\\!-\\!p\\!+\\!1}{k^{\prime}\\!+\\!1}\,\beta(m,k^{\prime},p)\,i_{y}^{k^{\prime}-p+1}i^{p}j_{y}^{p}(A^{(m+1,k^{\prime}+1)}/g)$
$\displaystyle+\sum\limits_{k^{\prime}=1}^{m}\sum\limits_{p^{\prime}=0}^{\min(k^{\prime},m-k^{\prime})}\\!\\!\frac{m\\!-\\!k^{\prime}\\!-\\!p^{\prime}}{k^{\prime}\\!+\\!1}\,\beta(m,k^{\prime},p^{\prime})\,i_{y}^{k^{\prime}-p^{\prime}}i^{p^{\prime}+1}j_{y}^{p^{\prime}+1}(A^{(m+1,k^{\prime}+1)}/g)$
$\displaystyle+i_{y}\,(A^{(m+1,1)}/g)-m\,ij_{y}\,(A^{(m+1,1)}/g).$

We change summation variables by $k^{\prime}=k-1$ in the second sum and by
$k^{\prime}=k-1,p^{\prime}=p-1$ in the third sum. The formula becomes

(6.21)
$\displaystyle\frac{1}{2m\\!+\\!1}\,A^{(m+1,0)}/g=(m\\!+\\!1)!(2m-1)!!\,|y|^{-1}\,g$
$\displaystyle+\sum\limits_{k=1}^{m}\sum\limits_{p=0}^{\min(k,m-k)}\\!\\!\frac{1}{2m\\!-\\!2k\\!+\\!1}\,\beta(m,k,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle-\sum\limits_{k=2}^{m+1}\sum\limits_{p=0}^{\min(k-1,m-k+1)}\frac{k\\!-\\!p}{k}\,\beta(m,k-1,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle+\sum\limits_{k=2}^{m+1}\sum\limits_{p=1}^{\min(k-1,m-k+1)+1}\frac{m\\!-\\!k\\!-\\!p\\!+\\!2}{k}\,\beta(m,k-1,p-1)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle+i_{y}\,(A^{(m+1,1)}/g)-m\,ij_{y}\,(A^{(m+1,1)}/g).$

We are going to equate summation limits in three sums on the right-hand side
of (6.21) in order to unite the sums. Then we are going to involve two terms
on the last line of (6.21) into the same sum. This needs some logical and
arithmetic analysis.

In the first sum on the right-hand side of (6.21), the summation over $k$ can
be extended to $1\leq k\leq m+1$ since $\beta(m,m+1,p)=0$ by the agreement
(6.11). Let us demonstrate that the summation over $p$ can be extended to
$0\leq p\leq\min(k,m-k+1)$. Indeed, $\min(k,m-k)=\min(k,m-k+1)$ if $k\leq
m-k$. If $k>m-k$, then there appears one extra term corresponding to $p=m-k+1$
in the first sum. But $\beta(m,k,m-k+1)=0$ by the agreement (6.11). Thus,
summation limits of the first sum can be replaced with

(6.22) $1\leq k\leq m+1,\quad 0\leq p\leq\min(k,m-k+1).$

In the second sum on the right-hand side of (6.21), the summation over $k$ can
be extended to $1\leq k\leq m+1$ since $\beta(m,0,p)=0$ by the agreement
(6.11). Let us demonstrate that the summation over $p$ can be extended to
$0\leq p\leq\min(k,m-k+1)$. Indeed, $\min(k-1,m-k)=\min(k,m-k+1)$ if
$k>m-k+1$. If $k\leq m-k+1$, then there appears one extra term corresponding
to $p=k$ in the second sum. But $\beta(m,k-1,k)=0$ by the agreement (6.11).
Thus, summation limits of the second sum can be replaced with (6.22).

In the third sum on the right-hand side of (6.21), the summation over $k$ can
be extended to $1\leq k\leq m+1$ since $\beta(m,0,p-1)=0$ by the agreement
(6.11). The lower summation limit over $p$ can be replaced with zero since
$\beta(m,k-1,-1)=0$ by the agreement (6.11). Let us demonstrate that the upper
summation limit over $p$ can be replaced with $\min(k,m-k+1)$. Indeed,
$\min(k-1,m-k+1)+1=\min(k,m-k+1)$ if either $2k<m+2$ or $2k>m+2$. The only
critical case is $2k=m+2$ when $\min(k-1,m-k+1)+1=m-k+2$ and
$\min(k,m-k+1)=m-k+1$. We are going to loose the term corresponding to
$p=m-k+2$ after the replacement. But this term is equal to zero due to the
presence of the factor $\frac{m\\!-\\!k\\!-\\!p\\!+\\!2}{k}$.

Thus, summation limits can be replaced with (6.22) in all sums on the right-
hand side of (6.21). After the replacement, we unite three sums and write
(6.21) in the form

(6.23) $\displaystyle A^{(m+1,0)}/g=(m\\!+\\!1)!(2m\\!+\\!1)!!\,|y|^{-1}\,g$
$\displaystyle+(2m\\!+\\!1)\sum\limits_{k=1}^{m+1}\sum\limits_{p=0}^{\min(k,m-k+1)}\tilde{\beta}(m+1,k,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g)$
$\displaystyle+(2m\\!+\\!1)\,i_{y}\,(A^{(m+1,1)}/g)-m(2m\\!+\\!1)\,ij_{y}\,(A^{(m+1,1)}/g),$

where $\tilde{\beta}(m+1,k,p)$ is defined by (6.9).

Finally, we have to include two terms on the last line of (6.23) into the sum.
The term $i_{y}\,(A^{(m+1,1)}/g)$ corresponds to $(k,p)=(1,0)$ and the term
$ij_{y}\,(A^{(m+1,1)}/g)$ corresponds to $(k,p)=(1,1)$. Therefore we define
$\beta(m+1,k,p)$ by (6.10). The formula (6.23) becomes now

$\displaystyle A^{(m+1,0)}/g$
$\displaystyle=(m\\!+\\!1)!(2m\\!+\\!1)!!\,|y|^{-1}\,g$
$\displaystyle+\sum\limits_{k=1}^{m+1}\sum\limits_{p=0}^{\min(k,m-k+1)}\beta(m+1,k,p)\,i_{y}^{k-p}i^{p}j_{y}^{p}(A^{(m+1,k)}/g).$

This coincides with (6.8) for $m:=m+1$. ∎

###### Proof of Proposition 6.1.

Coefficients $\beta(m,k,p)$ are determined by pretty complicated recurrent
formulas (6.9)–(6.11). Nevertheless, the coefficients can be expressed by the
explicit formula

(6.24)
$\beta(m,k,p)=\left\\{\begin{array}[]{l}(-1)^{k+p+1}\frac{\textstyle(2m-1)!!}{\textstyle(2m-2k-1)!!}\,2^{-p}\binom{\textstyle
m}{\textstyle k}\binom{\textstyle{m-k}}{\textstyle p}\quad\mbox{for}\
k>0,\\\\[8.0pt] 0\quad\mbox{for}\ k\leq 0.\end{array}\right.$

Indeed, being defined by (6.9), $\beta(m,k,p)$ satisfy (6.11) under the
agreement (5.9). Formulas (6.9)–(6.10) can be equivalently written in the form

(6.25) $\beta(m+1,1,0)=(2m+1)\Big{(}\frac{1}{2m-1}\,\beta(m,1,0)+1\Big{)},$
(6.26) $\beta(m+1,1,1)=(2m+1)\Big{(}\frac{1}{2m-1}\,\beta(m,1,1)-m\Big{)},$
(6.27)
$\displaystyle\beta(m+1,k,p)=(2m+1)\bigg{[}\frac{1}{2m-2k+1}\,\beta(m,k,p)-\frac{k-p}{k}\,\beta(m,k-1,p)$
$\displaystyle+\frac{m-k-p+2}{k}\,\beta(m,k-1,p-1)\bigg{]}\quad\big{(}(k,p)\neq(1,0),(k,p)\neq(1,1)\big{)}.$

Unlike (6.9)–(6.10), formulas (6.25)–(6.27) do not involve
$\tilde{\beta}(m+1,k,p)$. One easily proves that equations (6.25)–(6.27) are
satisfied by values (6.24). We express $g$ from (6.8). Substituting the value
(6.24) of $\beta(m,k,p)$ into the expression, we arrive at the formula (3.5)
with $\hat{f}=g$. Substituting the value (6.24) into (6.8) we arrive at the
formula (3.5) with $\widehat{f}=g$. This completes the proof of Proposition
6.1 as well as of Theorem 3.3. ∎

##  7\. Proof of Theorems 3.1 and 3.2

###  7.1. Vector fields

In the case of $m=1$, the formula (3.5) looks as follows:

(7.1) $\widehat{f}=|y|\big{(}H^{(1,0)}-i_{y}H^{(1,1)}\big{)}.$

By (3.4),

$H^{(1,0)}=F^{(1,0)},\quad
H^{(1,1)}=\frac{1}{n-1}\,\Big{(}\mbox{div}\,F^{(1,0)}-F^{(1,1)}\Big{)}$

and by (3.3),

$F^{(1,0)}=c_{1,n}\,\widehat{N_{1}^{0}f},\quad
F^{(1,1)}=c_{1,n}\,j_{y}\widehat{N_{1}^{1}f},$

where

$c_{1,n}=-\frac{2^{n/2}\Gamma\big{(}\frac{n+1}{2}\big{)}}{\Gamma\big{(}-\frac{1}{2}\big{)}}=\frac{2^{n/2-1}\Gamma\big{(}\frac{n+1}{2}\big{)}}{\sqrt{\pi}}.$

From this

$H^{(1,0)}=c_{1,n}\,\widehat{N_{1}^{0}f},\quad
H^{(1,1)}=\frac{c_{1,n}}{n-1}\,\Big{(}\mbox{div}\,\widehat{N_{1}^{0}f}-j_{y}\widehat{N_{1}^{1}f}\Big{)}.$

Substitute these expressions into (7.1)

(7.2)
$\widehat{f}=c_{1,n}\,|y|\Big{(}\widehat{N_{1}^{0}f}-\frac{1}{n-1}\,i_{y}\,\mbox{div}\,\widehat{N_{1}^{0}f}+\frac{1}{n-1}\,i_{y}j_{y}\,\widehat{N_{1}^{1}f}\Big{)}.$

We apply the inverse Fourier transform to the formula (7.2) and use the
commutator formulas

(7.3)
${\mathcal{F}}^{-1}|y|=(-\Delta)^{1/2}{\mathcal{F}}^{-1},\quad{\mathcal{F}}^{-1}i_{y}j_{y}=-d\,\mbox{div}\,{\mathcal{F}}^{-1},\quad{\mathcal{F}}^{-1}i_{y}\,\mbox{div}=d\,j_{x}\,{\mathcal{F}}^{-1}.$

In this way we obtain

$f=c_{1,n}(-\Delta)^{1/2}\Big{(}N_{1}^{0}f-\frac{1}{n\\!-\\!1}\,d\,j_{x}\,N_{1}^{0}f-\frac{1}{n\\!-\\!1}\,\,d\,\mbox{div}\,N_{1}^{1}f\Big{)}.$

This completes the proof of Theorem 3.1.

###  7.2. Second rank tensor fields

In the case of $m=2$, the formula (3.5) looks as follows:

(7.4)
$\widehat{f}=\frac{1}{6}\,|y|\Big{(}H^{(2,0)}-6\,i_{y}H^{(2,1)}+3\,ij_{y}H^{(2,1)}+3\,i_{y}^{2}H^{(2,2)}\Big{)}.$

By (3.4),

$\displaystyle H^{(2,0)}$ $\displaystyle=F^{(2,0)},\quad
H^{(2,1)}=\frac{1}{3(n+1)}\,\Big{(}\mbox{div}\,F^{(2,0)}-F^{(2,1)}\Big{)},$
$\displaystyle H^{(2,2)}$
$\displaystyle=\frac{1}{3(n-1)(n+1)}\,\Big{(}\mbox{div}^{2}\,F^{(2,0)}-2\,\mbox{div}\,F^{(2,1)}+F^{(2,2)}\Big{)}$

and by (3.3),

$F^{(2,0)}=c_{2,n}\,\widehat{N_{2}^{0}f},\quad
F^{(2,1)}=c_{2,n}\,j_{y}\widehat{N_{2}^{1}f},\quad
F^{(2,2)}=\frac{1}{2}\,c_{2,n}\,j_{y}^{2}\widehat{N_{2}^{2}f},$

where

(7.5)
$c_{2,n}=3\,\frac{2^{n/2}\Gamma\big{(}\frac{n+3}{2}\big{)}}{\sqrt{\pi}}.$

From this

$\displaystyle H^{(2,0)}$ $\displaystyle=c_{2,n}\,\widehat{N_{2}^{0}f},\quad
H^{(2,1)}=\frac{c_{2,n}}{3(n+1)}\,\Big{(}\mbox{div}\,\widehat{N_{2}^{0}f}-j_{y}\widehat{N_{2}^{1}f}\Big{)},$
$\displaystyle H^{(2,2)}$
$\displaystyle=\frac{c_{2,n}}{3(n-1)(n+1)}\,\Big{(}\mbox{div}^{2}\,\widehat{N_{2}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{2}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{2}^{2}f}\Big{)}.$

Substitute these expressions into (7.4)

$\displaystyle\widehat{f}=\frac{c_{2,n}}{6}\,|y|\bigg{[}$
$\displaystyle\widehat{N_{2}^{0}f}-\frac{2}{n+1}\,i_{y}\Big{(}\mbox{div}\,\widehat{N_{2}^{0}f}-j_{y}\widehat{N_{2}^{1}f}\Big{)}$
$\displaystyle+\frac{1}{n+1}\,i\Big{(}j_{y}\,\mbox{div}\,\widehat{N_{2}^{0}f}-j_{y}^{2}\widehat{N_{2}^{1}f}\Big{)}$
$\displaystyle+\frac{1}{(n-1)(n+1)}\,i_{y}^{2}\Big{(}\mbox{div}^{2}\,\widehat{N_{2}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{2}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{2}^{2}f}\Big{)}\bigg{]}.$

By Lemma 4.4, $j_{y}^{2}\widehat{N_{2}^{1}f}=0$, and the latter formula is
simplified to the following one:

(7.6) $\displaystyle\widehat{f}=\frac{c_{2,n}}{6}\,|y|\bigg{[}$
$\displaystyle\widehat{N_{2}^{0}f}+\frac{1}{n+1}\,ij_{y}\,\mbox{div}\,\widehat{N_{2}^{0}f}$
$\displaystyle-\frac{2}{n+1}\,i_{y}\Big{(}\mbox{div}\,\widehat{N_{2}^{0}f}-j_{y}\widehat{N_{2}^{1}f}\Big{)}$
$\displaystyle+\frac{1}{(n-1)(n+1)}\,i_{y}^{2}\Big{(}\mbox{div}^{2}\,\widehat{N_{2}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{2}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{2}^{2}f}\Big{)}\bigg{]}.$

The second term on the right-hand side of (7.6) can be simplified. Indeed, the
commutator formula

(7.7) $j_{y}\,\mbox{div}=\mbox{div}\,j_{y}-j$

is proved by an easy calculation in coordinates. By this formula,

$j_{y}\,\mbox{div}\,\widehat{N_{2}^{0}f}=\mbox{div}\,j_{y}\widehat{N_{2}^{0}f}-j\widehat{N_{2}^{0}f}.$

By Lemma 4.4, $j_{y}\widehat{N_{2}^{0}f}=0$, and the latter formula gives
$j_{y}\,\mbox{div}\,\widehat{N_{2}^{0}f}=-j\widehat{N_{2}^{0}f}$. Substitute
this value into (7.6)

(7.8) $\displaystyle\widehat{f}=\frac{c_{2,n}}{6}\,|y|\bigg{[}$
$\displaystyle\widehat{N_{2}^{0}f}-\frac{1}{n+1}\,ij\,\widehat{N_{2}^{0}f}$
$\displaystyle-\frac{2}{n+1}\Big{(}i_{y}\,\mbox{div}\,\widehat{N_{2}^{0}f}-i_{y}\,j_{y}\widehat{N_{2}^{1}f}\Big{)}$
$\displaystyle+\frac{1}{(n-1)(n+1)}\Big{(}i_{y}^{2}\,\mbox{div}^{2}\,\widehat{N_{2}^{0}f}-2\,i_{y}^{2}\,\mbox{div}\,j_{y}\,\widehat{N_{2}^{1}f}+\frac{1}{2}\,i_{y}^{2}\,j_{y}^{2}\widehat{N_{2}^{2}f}\Big{)}\bigg{]}.$

We apply the inverse Fourier transform to the formula (7.8) and use the
commutator formulas (7.3) as well as

$\displaystyle{\mathcal{F}}^{-1}i_{y}^{2}=-d^{2}{\mathcal{F}}^{-1},\quad{\mathcal{F}}^{-1}\,\mbox{div}^{2}=-j_{x}^{2}{\mathcal{F}}^{-1},\quad{\mathcal{F}}^{-1}j_{y}d=\mbox{div}\,i_{x}{\mathcal{F}}^{-1},\quad{\mathcal{F}}^{-1}\,\mbox{div}\,j_{y}=j_{x}\,\mbox{div}\,{\mathcal{F}}^{-1}.$

In this way we obtain

$\displaystyle f(x)=\frac{c_{2,n}}{6}\,(-\Delta)^{1/2}\bigg{[}$ $\displaystyle
N_{2}^{0}f-\frac{1}{n+1}\,ij\,N_{2}^{0}f$
$\displaystyle-\frac{2}{n+1}\,d\Big{(}j_{x}\,N_{2}^{0}f+\mbox{div}\,N_{2}^{1}f\Big{)}$
$\displaystyle+\frac{1}{(n\\!-\\!1)(n\\!+\\!1)}\,d^{2}\Big{(}j_{x}^{2}N_{2}^{0}f-2\,j_{x}\,\mbox{div}\,N_{2}^{1}f+\frac{1}{2}\,\mbox{div}^{2}\,N_{2}^{2}f\Big{)}\bigg{]}.$

Substituting the value (7.5) of $c_{2,n}$, we arrive at (3.2). This completes
the proof of Theorem 3.2.

###  7.3. Third rank tensor fields

In the case of $m=3$, the formula (3.5) looks as follows:

(7.9)
$\widehat{f}=\frac{|y|}{90}\Big{(}H^{(3,0)}-15\,i_{y}H^{(3,1)}+15\,ij_{y}H^{(3,1)}+45\,i_{y}^{2}H^{(3,2)}-\frac{45}{2}\,ii_{y}j_{y}H^{(3,2)}-15\,i_{y}^{3}H^{(3,3)}\Big{)}.$

By (3.4),

$\displaystyle H^{(3,0)}$ $\displaystyle=F^{(3,0)},\quad
H^{(3,1)}=\frac{1}{5(n+3)}\,\Big{(}\mbox{div}\,F^{(3,0)}-F^{(3,1)}\Big{)},$
$\displaystyle H^{(3,2)}$
$\displaystyle=\frac{1}{15(n\\!+\\!1)(n\\!+\\!3)}\,\Big{(}\mbox{div}^{2}\,F^{(3,0)}-2\,\mbox{div}\,F^{(3,1)}+F^{(3,2)}\Big{)},$
$\displaystyle H^{(3,3)}$
$\displaystyle=\frac{1}{15(n\\!-\\!1)(n\\!+\\!1)(n\\!+\\!3)}\,\Big{(}\mbox{div}^{3}\,F^{(3,0)}-3\,\mbox{div}^{2}\,F^{(3,1)}+3\,\mbox{div}\,F^{(3,2)}-F^{(3,3)}\Big{)},$

and by (3.3),

$F^{(3,0)}=c_{3,n}\,\widehat{N_{3}^{0}f},\quad
F^{(3,1)}=c_{3,n}\,j_{y}\widehat{N_{3}^{1}f},\quad
F^{(3,2)}=\frac{1}{2}\,c_{3,n}\,j_{y}^{2}\widehat{N_{3}^{2}f},\quad
F^{(3,3)}=\frac{1}{6}\,c_{3,n}\,j_{y}^{3}\widehat{N_{3}^{2}f},$

where

(7.10) $c_{3,n}=15\,\pi^{-1/2}\,2^{n/2+1}\,\Gamma\Big{(}\frac{n+5}{2}\Big{)}.$

From this

$\displaystyle H^{(3,0)}$ $\displaystyle=c_{3,n}\,\widehat{N_{3}^{0}f},\quad
H^{(3,1)}=\frac{c_{3,n}}{5(n+3)}\,\Big{(}\mbox{div}\,\widehat{N_{3}^{0}f}-j_{y}\widehat{N_{3}^{1}f}\Big{)},$
$\displaystyle H^{(3,2)}$
$\displaystyle=\frac{c_{3,n}}{15(n\\!+\\!1)(n\\!+\\!3)}\,\Big{(}\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{3}^{2}f}\Big{)},$
$\displaystyle H^{(3,3)}$
$\displaystyle=\frac{c_{3,n}}{15(n\\!-\\!1)(n\\!+\\!1)(n\\!+\\!3)}\,\Big{(}\mbox{div}^{3}\,\widehat{N_{3}^{0}f}-3\,\mbox{div}^{2}\,j_{y}\,\widehat{N_{3}^{1}f}+\frac{3}{2}\,\mbox{div}\,j_{y}^{2}\widehat{N_{3}^{2}f}-\frac{1}{6}\,j_{y}^{3}\widehat{N_{3}^{3}f}\Big{)}.$

Substitute these expressions into (7.9)

$\displaystyle\widehat{f}=\frac{c_{3,n}}{90}\,|y|\bigg{[}$
$\displaystyle\widehat{N_{3}^{0}f}-\frac{3}{n+3}\,i_{y}\Big{(}\mbox{div}\,\widehat{N_{3}^{0}f}-j_{y}\widehat{N_{3}^{1}f}\Big{)}$
$\displaystyle+\frac{3}{n+3}\,i\Big{(}j_{y}\,\mbox{div}\,\widehat{N_{3}^{0}f}-j_{y}^{2}\widehat{N_{3}^{1}f}\Big{)}$
$\displaystyle+\frac{3}{(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{2}\Big{(}\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{3}^{2}f}\Big{)}$
$\displaystyle-\frac{3}{2(n\\!+\\!1)(n\\!+\\!3)}\,ii_{y}\Big{(}j_{y}\,\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,j_{y}\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}+\frac{1}{2}\,j_{y}^{3}\widehat{N_{3}^{2}f}\Big{)}$
$\displaystyle-\frac{1}{(n\\!-\\!1)(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{3}\Big{(}\mbox{div}^{3}\,\widehat{N_{3}^{0}f}-3\,\mbox{div}^{2}\,j_{y}\widehat{N_{3}^{1}f}+\frac{3}{2}\,\mbox{div}\,j_{y}^{2}\widehat{N_{3}^{2}f}-\frac{1}{6}\,j_{y}^{3}\widehat{N_{3}^{3}f}\Big{)}\bigg{]}.$

By Lemma 4.4, $j_{y}^{2}\widehat{N_{3}^{1}f}=0$ and
$j_{y}^{3}\widehat{N_{3}^{2}f}=0$. The previous formula is simplified to the
following one:

(7.11) $\displaystyle\widehat{f}=\frac{c_{3,n}}{90}\,|y|\bigg{[}$
$\displaystyle\widehat{N_{3}^{0}f}+\frac{3}{n+3}\,ij_{y}\,\mbox{div}\,\widehat{N_{3}^{0}f}-\frac{3}{n+3}\,i_{y}\Big{(}\mbox{div}\,\widehat{N_{3}^{0}f}-j_{y}\widehat{N_{3}^{1}f}\Big{)}$
$\displaystyle+\frac{3}{(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{2}\Big{(}\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{3}^{2}f}\Big{)}$
$\displaystyle-\frac{3}{2(n\\!+\\!1)(n\\!+\\!3)}\,ii_{y}\Big{(}j_{y}\,\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,j_{y}\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}\Big{)}$
$\displaystyle-\frac{1}{(n\\!-\\!1)(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{3}\Big{(}\mbox{div}^{3}\,\widehat{N_{3}^{0}f}-3\,\mbox{div}^{2}\,j_{y}\widehat{N_{3}^{1}f}+\frac{3}{2}\,\mbox{div}\,j_{y}^{2}\widehat{N_{3}^{2}f}-\frac{1}{6}\,j_{y}^{3}\widehat{N_{3}^{3}f}\Big{)}\bigg{]}.$

At least three terms on the right-hand side of (7.11) can be simplified.
Indeed, using Lemma 4.4 and the commutator formula (7.7), we transform

(7.12)
$j_{y}\,\mbox{div}\,\widehat{N_{3}^{0}f}=\mbox{div}\,j_{y}\,\widehat{N_{3}^{0}f}-j\,\widehat{N_{3}^{0}f}=-j\,\widehat{N_{3}^{0}f}.$

Quite similarly,

(7.13)
$j_{y}\,\mbox{div}\,j_{y}\,\widehat{N_{3}^{1}f}=(\mbox{div}\,j_{y}-j)j_{y}\,\widehat{N_{3}^{1}f}=\mbox{div}\,j_{y}^{2}\,\widehat{N_{3}^{1}f}-jj_{y}\,\widehat{N_{3}^{1}f}=-jj_{y}\,\widehat{N_{3}^{1}f}=-j_{y}j\,\widehat{N_{3}^{1}f}.$

We have used that operators $j$ and $j_{y}$ commute. Let us also transform the
term containing $j_{y}\,\mbox{div}^{2}$

$\displaystyle j_{y}\,\mbox{div}^{2}\,\widehat{N_{3}^{0}f}$
$\displaystyle=(j_{y}\,\mbox{div})\mbox{div}\,\widehat{N_{3}^{0}f}=(\mbox{div}\,j_{y}-j)\mbox{div}\,\widehat{N_{3}^{0}f}$
$\displaystyle=\mbox{div}(j_{y}\,\mbox{div})\widehat{N_{3}^{0}f}-j\,\mbox{div}\,\widehat{N_{3}^{0}f}=\mbox{div}(\mbox{div}\,j_{y}-j)\widehat{N_{3}^{0}f}-j\,\mbox{div}\,\widehat{N_{3}^{0}f}$
$\displaystyle=\mbox{div}^{2}\,j_{y}\,\widehat{N_{3}^{0}f}-\mbox{div}\,j\,\widehat{N_{3}^{0}f}-j\,\mbox{div}\,\widehat{N_{3}^{0}f}=-\mbox{div}\,j\,\widehat{N_{3}^{0}f}-j\,\mbox{div}\,\widehat{N_{3}^{0}f}.$

Using that $j$ and div commute [DS10], we get

(7.14)
$j_{y}\,\mbox{div}^{2}\,\widehat{N_{3}^{0}f}=-2\,j\,\mbox{div}\,\widehat{N_{3}^{0}f}.$

Substituting expressions (7.12)–(7.14) and the value (7.10) of the constant
$c_{3,n}$ into (7.11), we obtain the inversion formula recovering the Fourier
transform of a tensor field $f\in{\mathcal{S}}({{\mathbb{R}}}^{n};S^{3})\
(n\geq 2)$ through the data
$(\widehat{N_{3}^{0}f},j_{y}\widehat{N_{3}^{1}f},j_{y}^{2}\widehat{N_{3}^{2}f},j_{y}^{3}\widehat{N_{3}^{3}f})$

$\displaystyle\widehat{f}$
$\displaystyle=\frac{2^{n/2}\,\Gamma\big{(}\frac{n+5}{2}\big{)}}{3\sqrt{\pi}}\,|y|\bigg{[}\widehat{N_{3}^{0}f}-\frac{3}{n+3}\,ij\,\widehat{N_{3}^{0}f}$
$\displaystyle-\frac{3}{(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}\Big{(}(n\\!+\\!1)\,\mbox{div}\,\widehat{N_{3}^{0}f}-\,ij\,\mbox{div}\,\widehat{N_{3}^{0}f}-(n\\!+\\!1)\,j_{y}\widehat{N_{3}^{1}f}+\,ij\,j_{y}\widehat{N_{3}^{1}f}\Big{)}$
$\displaystyle+\frac{3}{(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{2}\Big{(}\mbox{div}^{2}\,\widehat{N_{3}^{0}f}-2\,\mbox{div}\,j_{y}\widehat{N_{3}^{1}f}+\frac{1}{2}\,j_{y}^{2}\widehat{N_{3}^{2}f}\Big{)}$
$\displaystyle-\frac{1}{(n\\!-\\!1)(n\\!+\\!1)(n\\!+\\!3)}\,i_{y}^{3}\Big{(}\mbox{div}^{3}\,\widehat{N_{3}^{0}f}-3\,\mbox{div}^{2}\,j_{y}\widehat{N_{3}^{1}f}+\frac{3}{2}\,\mbox{div}\,j_{y}^{2}\widehat{N_{3}^{2}f}-\frac{1}{6}\,j_{y}^{3}\widehat{N_{3}^{3}f}\Big{)}\bigg{]}.$

The same approach can be used for deriving the inversion formula for
$m=4,5,\dots$ The length of the formula grows with $m$ as well as the volume
of calculations.

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