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\jyear

2021

[1]\fnmChao \surDing

[1]\orgdivCenter for Pure Mathematics, School of Mathematical Sciences, \orgnameAnhui University, \orgaddress \cityHefei, \postcode230061, \stateAnhui, \countryChina

2]\orgdivSchool of Mathematics, \orgnameHefei University of Technology, \orgaddress \cityHefei, \postcode230061, \stateAnhui, \countryChina

Invariance of iterated global differential operator for slice monogenic functions

[email protected]    \fnmZhenghua \surXu [email protected] * [
Abstract

In this article, we present the symmetry group of a global slice Dirac operator and its iterated ones. Further, the explicit forms of intertwining operators of the iterated global slice Dirac operator are given. At the end, we introduce a variant of the global slice Dirac operator, which allows functions considered to be defined on the whole Euclidean space. The invariance property and the intertwining operators of this variant of the global slice Dirac operator are also presented.

keywords:
global slice Dirac operator, symmetry group, intertwining operator, iterated global slice Dirac operator
pacs:
[

MSC Classification]30G35

1 Introduction

Classical Clifford analysis is a generalization of complex analysis used to study generalized Cauchy-Riemann equations (named as Dirac equations) over Euclidean spaces. The functions annihilated by the Dirac operator are called monogenic functions. These functions have many important properties just as holomorphic functions do in complex analysis. For instance, Cauchy’s theorem, Cauchy integral formula, the mean value property, Liouville’s theorem, Maximal principle, etc. More details on classical Clifford analysis can be found in Brackx ; Del .

On the other hand, there is a significant difference between complex analysis and classical Clifford analysis: it is well-known that the polynomials given in terms of the complex variable are holomorphic in complex analysis; meanwhile, powers of the paravector-valued variable are not monogenic in classical Clifford analysis. This situation was changed when the theory of slice analysis over quaternions was introduced by Gentili and Struppa in 2006 Gen2 ; Gen3 , which was motivated by an earlier work done by Cullen in Cullen . Later, Colombo, Sabadini and Struppa Co1 generalized this idea to the general Clifford algebras and introduced the concept of slice monogenic functions in Euclidean space in 2009. Gentili and Struppa Gen1 investigated slice regularity for octonions in 2010, and in the next year, the theory of slice regular functions was established on real alternative algebras by Ghiloni and Perotti Ghi1 . Further investigations on slice regularity of slice Dirac regular functions have been conducted by Ghiloni GhiJGA , Jin, Ren and Sabadini RenJin . Recently, a mean value formula for slice regular functions was introduced by Bisi and Winkelmann in Bisi . This formula leads to an important result, which says that a slice regular function over quaternions is also harmonic in certain sense.

The symmetry group of slice monogenic functions was first investigated by Colombo, Kraußhar and Sabadini Co in 2020. The authors described the group under which slice monogenic functions are taken into slice monogenic functions. Further, the authors proved a transformation formula for composing slice monogenic functions with Möbius transformations and described their conformal invariance. Recently, the conformal invariance has been investigated for generalized partial-slice monogenic functions in XS . However, the results in Co ; XS are limited in that the domains of functions considered are slice and symmetric. In this article, we continue their work by considering the symmetry group of the iterated global slice Dirac operator, which acts on functions defined on any domain. Intertwining operators for the iterated global slice Dirac operator are also provided here. In particular, the transformation formula mentioned above Co is a special case here. Note that the global slice Dirac operator involves a norm term in the denominator, which leads to some restrictions on the domain of the functions considered. Hence, at the end, we introduce a variant of the global slice Dirac operator which is well-defined on the whole Euclidean space. This operator also has the same symmetry group as the global slice Dirac operator, but the invariance property does not hold anymore for iterated ones.

This article is organized as follows. Some definitions and notation for classical Clifford analysis and slice Clifford analysis are introduced in Section 2. The intertwining operators for slice Dirac operators are introduced in Section 3. Section 4 is devoted to the intertwining operators for iterated slice Dirac operators. A variant of slice Dirac operator and its invariance property are studied in Section 5.

2 Preliminaries

In this section, we introduce some definitions and notation for classical Clifford analysis and slice Clifford analysis.

2.1 Classical Clifford analysis

Let {e1,…,em}\{e_{1},\ldots,e_{m}\} be a standard orthonormal basis for the mm-dimensional real Euclidean space ℝm\mathbb{R}^{m}. A real Clifford algebra, 𝒞​lm,\mathcal{C}l_{m}, can be generated from ℝm\mathbb{R}^{m} by considering the relationship

ei​ej+ej​ei=−2​δi​j,e_{i}e_{j}+e_{j}e_{i}=-2\delta_{ij},

where δi​j\delta_{ij} is the Kronecker delta function. Hence, an arbitrary element of the basis of the Clifford algebra can be written as eA=ej1​⋯​ejr,e_{A}=e_{j_{1}}\cdots e_{j_{r}}, where A={j1,⋯,jr}⊂{1,2,⋯,m}A=\{j_{1},\cdots,j_{r}\}\subset\{1,2,\cdots,m\} and 1≤j1<j2<⋯<jr≤m.1\leq j_{1}<j_{2}<\cdots<j_{r}\leq m. Further, for any element a∈𝒞​lma\in\mathcal{C}l_{m}, we have a=∑AaA​eA,a=\sum_{A}a_{A}e_{A}, where aA∈ℝa_{A}\in\mathbb{R}. The linear subspace of 𝒞​lm\mathcal{C}l_{m} generated by the (nk)\begin{pmatrix}n\\ k\end{pmatrix} elements of the form eA=ei1​⋯​eik,il∈{1,2,⋯,m},i1<i2<⋯<ike_{A}=e_{i_{1}}\cdots e_{i_{k}},\ i_{l}\in\{1,2,\cdots,m\},\ i_{1}<i_{2}<\cdots<i_{k}, will be denoted by 𝒞​lmk\mathcal{C}l_{m}^{k}. The elements in 𝒞​lmk\mathcal{C}l_{m}^{k} are called kk-vectors.

For a=∑AaA​eA∈𝒞​lma=\sum_{A}a_{A}e_{A}\in\mathcal{C}l_{m}, we define the reversion of aa as a~=∑AaA​eA~,\tilde{a}=\sum_{A}a_{A}\widetilde{e_{A}}, where ej1​⋯​ejr~=ejr​⋯​ej1\widetilde{e_{j_{1}}\cdots e_{j_{r}}}=e_{j_{r}}\cdots e_{j_{1}}. Also a​b~=b~​a~\widetilde{ab}=\tilde{b}\tilde{a} for a,b∈𝒞​lm.a,b\in\mathcal{C}l_{m}. We also need the Clifford conjugation defined by a¯:=∑AaA​eA¯\overline{a}:=\sum_{A}a_{A}\overline{e_{A}}, where ej1​…​ejr¯=ejr¯​…​ej1¯,ej¯=−ej,1≤j≤m,e0¯=e0=1\overline{e_{j_{1}}\ldots e_{j_{r}}}=\overline{e_{j_{r}}}\ldots\overline{e_{j_{1}}},\ \overline{e_{j}}=-e_{j},1\leq j\leq m,\ \overline{e_{0}}=e_{0}=1.

The Dirac operator in ℝm\mathbb{R}^{m} is defined to be

D𝒙:=∑i=1mei​∂xi.D_{\boldsymbol{x}}:=\sum_{i=1}^{m}e_{i}\partial_{x_{i}}.

Note D𝒙2=−Δ𝒙D_{\boldsymbol{x}}^{2}=-\Delta_{\boldsymbol{x}}, where Δ𝒙\Delta_{\boldsymbol{x}} is the Laplacian in ℝm\mathbb{R}^{m}. A 𝒞​lm\mathcal{C}l_{m}-valued function f​(𝒙)f(\boldsymbol{x}) defined on a domain UU in ℝm\mathbb{R}^{m} is left monogenic if D𝒙​f​(𝒙)=0.D_{\boldsymbol{x}}f(\boldsymbol{x})=0. Since Clifford multiplication is not commutative in general, there is a similar definition for right monogenic functions.

2.2 Slice Clifford analysis

In this subsection, we firstly introduce some definitions in slice Clifford analysis given in Co3 , and then we introduce the global slice Dirac operator studied in this article.

Recall that elements in 𝒞​lm0⊕𝒞​lm1\mathcal{C}l_{m}^{0}\oplus\mathcal{C}l_{m}^{1} are called paravectors, which can be identified as vectors in ℝm+1\mathbb{R}^{m+1}. It is easy to see that each paravector can be written as 𝒙=x0+r​𝝎\boldsymbol{x}=x_{0}+r\boldsymbol{\omega}, where r=|𝒙|r=\lvert\boldsymbol{x}\rvert and 𝝎∈𝕊m−1:={𝝎∈𝒞​lm1:|𝝎|=1}\boldsymbol{\omega}\in\mathbb{S}^{m-1}:=\{\boldsymbol{\omega}\in\mathcal{C}l_{m}^{1}:\lvert\boldsymbol{\omega}\rvert=1\}. This observation leads to the definition of slice monogenic functions on each slice in ℝm+1\mathbb{R}^{m+1} as follows.

Definition 1.

Let Ω\Omega be a domain in ℝm+1\mathbb{R}^{m+1}. A function f:Ω⟶𝒞​lmf:\ \Omega\longrightarrow\mathcal{C}l_{m} is called left slice monogenic if its restriction f𝛚f_{\boldsymbol{\omega}} to Ω𝛚=Ω∩(ℝ⊕𝛚​ℝ)\Omega_{\boldsymbol{\omega}}=\Omega\cap(\mathbb{R}\oplus\boldsymbol{\omega}\mathbb{R}) is holomorphic for all 𝛚∈𝕊m−1\boldsymbol{\omega}\in\mathbb{S}^{m-1}, i.e., it has continuous partial derivatives and satisfies (∂x0+𝛚​∂r)​f𝛚​(x0+r​𝛚)=0(\partial_{x_{0}}+\boldsymbol{\omega}\partial_{r})f_{\boldsymbol{\omega}}(x_{0}+r\boldsymbol{\omega})=0.

Remark 1.

Roughly speaking, a function ff defined on a domain Ω⊂ℝm+1\Omega\subset\mathbb{R}^{m+1} is called left slice monogenic if and only if it is holomorphic on each slice of Ω\Omega, which is the intersection of Ω\Omega and the plane spanned by {1,𝛚},𝛚∈𝕊m−1\{1,\boldsymbol{\omega}\},\ \boldsymbol{\omega}\in\mathbb{S}^{m-1}.

On one hand, the definition of slice monogenic functions was given by the holomorphicity of the restriction of functions on each slice, on the other hand, in CoG , the authors introduced a nonconstant coefficients differential operator (also called the generalized slice Cauchy-Riemman operator) given by

G=|𝒙¯|2​∂x0+𝒙¯​∑j=1mxj​∂xj,\displaystyle G=\lvert\underline{\boldsymbol{x}}\rvert^{2}\partial_{x_{0}}+\underline{\boldsymbol{x}}\sum_{j=1}^{m}x_{j}\partial_{x_{j}},

where 𝒙=x0+𝒙¯∈ℝm+1\boldsymbol{x}=x_{0}+\underline{\boldsymbol{x}}\in\mathbb{R}^{m+1}. The null solutions of GG are exactly the slice monogenic functions when the domain of the functions considered does not intersect the real line. If we consider ℝm+1=ℝ⊕ℝm\mathbb{R}^{m+1}=\mathbb{R}\oplus\mathbb{R}^{m}, then the first term ∂x0\partial_{x_{0}} corresponds to ℝ\mathbb{R} and the second term corresponds to ℝm\mathbb{R}^{m}. In this article, we generalize this generalized slice Cauchy-Riemann operator to the case of the decomposition ℝp+q=ℝp⊕ℝq\mathbb{R}^{p+q}=\mathbb{R}^{p}\oplus\mathbb{R}^{q}.

Let U⊂ℝp+qU\subset\mathbb{R}^{p+q} be a domain and let f:U⟶𝒞​lp+qf:\ U\longrightarrow\mathcal{C}l_{p+q} be a real differential function. In this paper, we consider functions f:ℝp+q⟶𝒞​lp+qf:\mathbb{R}^{p+q}\longrightarrow\mathcal{C}l_{p+q}. Given 𝒙∈ℝp+q\boldsymbol{x}\in\mathbb{R}^{p+q}, we rewrite it as

𝒙=∑i=1p+qeixi=∑i=1peixi+∑i=p+1p+qeixi=:𝒙p+𝒙q∈ℝp⊕ℝq.\boldsymbol{x}=\sum_{i=1}^{p+q}e_{i}x_{i}=\sum_{i=1}^{p}e_{i}x_{i}+\sum_{i=p+1}^{p+q}e_{i}x_{i}=:\boldsymbol{x}_{p}+\boldsymbol{x}_{q}\in\mathbb{R}^{p}\oplus\mathbb{R}^{q}.

Similarly, we denote the Dirac operator and the Euler operator as follows.

D𝒙=∑i=1p+qei∂xi=∑i=1pei∂xi+∑i=p+1p+qei∂xi=:D𝒙p+D𝒙q,\displaystyle D_{\boldsymbol{x}}=\sum_{i=1}^{p+q}e_{i}\partial_{x_{i}}=\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\sum_{i=p+1}^{p+q}e_{i}\partial_{x_{i}}=:D_{\boldsymbol{x}_{p}}+D_{\boldsymbol{x}_{q}},
𝔼𝒙=∑i=1p+qxi∂xi=∑i=1pxi∂xi+∑i=p+1p+qxi∂xi=:𝔼𝒙p+𝔼𝒙q.\displaystyle\mathbb{E}_{\boldsymbol{x}}=\sum_{i=1}^{p+q}x_{i}\partial_{x_{i}}=\sum_{i=1}^{p}x_{i}\partial_{x_{i}}+\sum_{i=p+1}^{p+q}x_{i}\partial_{x_{i}}=:\mathbb{E}_{\boldsymbol{x}_{p}}+\mathbb{E}_{\boldsymbol{x}_{q}}.

The (global) slice Dirac operator considered here is defined as

G𝒙=D𝒙p+𝒙q|𝒙q|2​𝔼𝒙q=∑i=1pei​∂xi+𝒙q|𝒙q|2​∑i=p+1p+qxi​∂xi.\displaystyle G_{\boldsymbol{x}}=D_{\boldsymbol{x}_{p}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\rvert^{2}}\mathbb{E}_{\boldsymbol{x}_{q}}=\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\rvert^{2}}\sum_{i=p+1}^{p+q}x_{i}\partial_{x_{i}}.

Note that when p=1p=1, the differential operator G𝒙G_{\boldsymbol{x}} is not exactly the differential operator GG given above. But, the two operators are closely connected and more details on the connections can be found at the end of Section 3 below.

3 Symmetry group for the global slice Dirac operator

For a domain UU in ℝm\mathbb{R}^{m}, a diffeomorphism ϕ:U⟶ℝm\phi:U\longrightarrow\mathbb{R}^{m} is said to be conformal if, for each 𝒙∈U\boldsymbol{x}\in U and each 𝐮,𝐯∈T​U𝒙\mathbf{u,v}\in TU_{\boldsymbol{x}}, the angle between 𝐮\mathbf{u} and 𝐯\mathbf{v} is preserved under the corresponding differential at 𝒙\boldsymbol{x}, d​ϕ𝒙d\phi_{\boldsymbol{x}}. For m≥3m\geq 3, a theorem of Liouville tells us that the only conformal transformations are Möbius transformations. Ahlfors and Vahlen showed that any Möbius transformation on ℝm∪{∞}\mathbb{R}^{m}\cup\{\infty\} can be expressed as y=(a​𝒙+b)​(c​𝒙+d)−1y=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1} with a,b,c,d∈𝒞​lma,\ b,\ c,\ d\in\mathcal{C}l_{m} satisfying the following conditions Lou :

1.a,b,c,d​a​r​e​a​l​l​p​r​o​d​u​c​t​s​o​f​v​e​c​t​o​r​s​i​n​ℝm.\displaystyle 1.\ a,\ b,\ c,\ d\ are\ all\ products\ of\ vectors\ in\ \mathbb{R}^{m}.
2.a​b~,c​d~,b~​c,d~​a∈ℝm.\displaystyle 2.\ a\tilde{b},\ c\tilde{d},\ \tilde{b}c,\ \tilde{d}a\in\mathbb{R}^{m}.
3.a​d~−b​c~=±1.\displaystyle 3.\ a\tilde{d}-b\tilde{c}=\pm 1.

Since y=(a​𝒙+b)​(c​𝒙+d)−1=a​c−1+(b−a​c−1​d)​(c​𝒙+d)−1y=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}=ac^{-1}+(b-ac^{-1}d)(c\boldsymbol{x}+d)^{-1}, a conformal transformation can be decomposed as compositions of translation, dilation, reflection and inversion. This gives an Iwasawa decomposition for Möbius transformations. See Li for more details. Further, if we rewrite a Möbius transformation (a​𝒙+b)​(c​𝒙+d)−1(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1} in terms of a 2×22\times 2 Clifford valued matrix (abcd)\begin{pmatrix}a&b\\ c&d\end{pmatrix}, then the set of all these matrices is known as the special Ahlfors-Vahlen group, denoted by S​A​V​(ℝm)SAV(\mathbb{R}^{m}). The four basic transformations in the Iwasawa decomposition correspond to the following four types of Ahlfors-Vahlen matrices.

  1. 1.

    Translation: (1𝒃01)\begin{pmatrix}1&\boldsymbol{b}\\ 0&1\end{pmatrix}, where b∈ℝmb\in\mathbb{R}^{m}, inducing Möbius transformations 𝒚=𝒙+𝒃\boldsymbol{y}=\boldsymbol{x}+\boldsymbol{b}.

  2. 2.

    Dilation: (λ00λ−1)\begin{pmatrix}\lambda&0\\ 0&\lambda^{-1}\end{pmatrix}. where λ∈ℝ\{0}\lambda\in\mathbb{R}\backslash\{0\}, inducing Möbius transformations 𝒚=λ2​𝒙\boldsymbol{y}=\lambda^{2}\boldsymbol{x}.

  3. 3.

    Reflection: (𝒂00−𝒂−1)\begin{pmatrix}\boldsymbol{a}&0\\ 0&-\boldsymbol{a}^{-1}\end{pmatrix}, where 𝒂∈𝕊m−1\boldsymbol{a}\in\mathbb{S}^{m-1}, inducing Möbius transformations 𝒚=𝒂​𝒙​𝒂\boldsymbol{y}=\boldsymbol{a}\boldsymbol{x}\boldsymbol{a}.

  4. 4.

    Inversion: (01−10)\begin{pmatrix}0&1\\ -1&0\end{pmatrix}, inducing Möbius transformations 𝒚=−𝒙−1\boldsymbol{y}=-\boldsymbol{x}^{-1}.

Now, let us define a subgroup of the Möbius group on ℝp+q\mathbb{R}^{p+q} as follows.

G​R​A​V​(ℝp+q)=⟨(1𝒃01),(λ00λ−1),(𝒂00𝒂−1),(01−10)⟩,\displaystyle GRAV(\mathbb{R}^{p+q})=\left\langle\begin{pmatrix}1&\boldsymbol{b}\\ 0&1\end{pmatrix},\begin{pmatrix}\lambda&0\\ 0&\lambda^{-1}\end{pmatrix},\begin{pmatrix}\boldsymbol{a}&0\\ 0&\boldsymbol{a}^{-1}\end{pmatrix},\begin{pmatrix}0&1\\ -1&0\end{pmatrix}\right\rangle, (1)

where 𝒃∈ℝp,𝒂∈𝕊q−1\boldsymbol{b}\in\mathbb{R}^{p},\ \boldsymbol{a}\in\mathbb{S}^{q-1} and λ∈ℝ\{0}\lambda\in\mathbb{R}\backslash\{0\}. When p=1p=1, this is the symmetry group for slice monogenic functions, see Co for more details. Here, we claim that G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}) is the symmetry group of the global slice Dirac operator G𝒙G_{\boldsymbol{x}}, in other words, the space of null solutions to the slice Dirac operator is invariant with respect to the transformations in G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}). This result can be justified immediately by the intertwining operators of G𝒙G_{\boldsymbol{x}} under transformations in G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}) given below.

Theorem 1.

Let 𝐲=φ​(𝐱)=(a​𝐱+b)​(c​𝐱+d)−1∈G​R​A​V​(ℝp+q)\boldsymbol{y}=\varphi(\boldsymbol{x})=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}\in GRAV(\mathbb{R}^{p+q}), and let f∈C1​(U)f\in C^{1}(U), where UU is a domain in ℝp+q\ℝp\mathbb{R}^{p+q}\backslash\mathbb{R}^{p}. Then, we have

G𝒚​f​(𝒚)=(c​𝒙+d|c​𝒙+d|p+3)−1​G𝒙​c​𝒙+d~|c​𝒙+d|p+1​f​((a​𝒙+b)​(c​𝒙+d)−1).\displaystyle G_{\boldsymbol{y}}f(\boldsymbol{y})=\bigg{(}\frac{c\boldsymbol{x}+d}{\lvert c\boldsymbol{x}+d\rvert^{p+3}}\bigg{)}^{-1}G_{\boldsymbol{x}}\frac{\widetilde{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\rvert^{p+1}}f((a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}). (2)
Remark 2.

The two terms (c​𝐱+d|c​𝐱+d|p+3)−1\bigg{(}\displaystyle\frac{c\boldsymbol{x}+d}{\lvert c\boldsymbol{x}+d\rvert^{p+3}}\bigg{)}^{-1} and c​𝐱+d~|c​𝐱+d|p+1\displaystyle\frac{\widetilde{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\rvert^{p+1}} are usually called the weight functions of G𝐱G_{\boldsymbol{x}} under the transformations in G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}).

Proof: According to the Iwasawa decomposition, we only need to show that (3) is true for the four basic transformations respectively. More specifically, we consider the following four transformations.

  1. 1.

    𝒚=φ​(𝒙)=𝒙+𝒃,𝒃∈ℝp\boldsymbol{y}=\varphi(\boldsymbol{x})=\boldsymbol{x}+\boldsymbol{b},\ \boldsymbol{b}\in\mathbb{R}^{p}, in this case, a=1,c=0,d=1a=1,\ c=0,\ d=1.

    It is easy to observe that ∂yi=∂xi\partial_{y_{i}}=\partial_{x_{i}} and 𝒚q=𝒙q\boldsymbol{y}_{q}=\boldsymbol{x}_{q}. Then, we immediately have

    G𝒚=∑i=1pei​∂yi+𝒚q|𝒚q|2​∑i=p+1p+qyi​∂yi=G𝒙.\displaystyle G_{\boldsymbol{y}}=\sum_{i=1}^{p}e_{i}\partial_{y_{i}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\rvert^{2}}\sum_{i=p+1}^{p+q}y_{i}\partial_{y_{i}}=G_{\boldsymbol{x}}.
  2. 2.

    𝒚=φ​(𝒙)=λ2​𝒙,λ∈ℝ\{0}\boldsymbol{y}=\varphi(\boldsymbol{x})=\lambda^{2}\boldsymbol{x},\ \lambda\in\mathbb{R}\backslash\{0\}, in this case, a=λ=d−1,b=c=0a=\lambda=d^{-1},\ b=c=0.

    It is also easy to see that ∂yi=λ−2​∂xi\partial_{y_{i}}=\lambda^{-2}\partial_{x_{i}}, and these give us

    G𝒚=∑i=1pei​∂yi+𝒚q|𝒚q|2​∑i=p+1p+qyi​∂yi=λ−2​G𝒙.\displaystyle G_{\boldsymbol{y}}=\sum_{i=1}^{p}e_{i}\partial_{y_{i}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\rvert^{2}}\sum_{i=p+1}^{p+q}y_{i}\partial_{y_{i}}=\lambda^{-2}G_{\boldsymbol{x}}.
  3. 3.

    𝒚=φ​(𝒙)=𝒂​𝒙​𝒂−1,𝒂∈𝕊q−1\boldsymbol{y}=\varphi(\boldsymbol{x})=\boldsymbol{a}\boldsymbol{x}\boldsymbol{a}^{-1},\ \boldsymbol{a}\in\mathbb{S}^{q-1}, which gives us that 𝒚p=−𝒙p\boldsymbol{y}_{p}=-\boldsymbol{x}_{p} and 𝒚q=𝒂​𝒙q​𝒂−1\boldsymbol{y}_{q}=\boldsymbol{a}\boldsymbol{x}_{q}\boldsymbol{a}^{-1}. In this case, b=c=0,d=𝒂−1b=c=0,\ d=\boldsymbol{a}^{-1}.

    Thus, we have

    G𝒚=\displaystyle G_{\boldsymbol{y}}= D𝒚p+𝒚q|𝒚q|2​𝔼𝒚q=−D𝒙p+𝒂​𝒙q​𝒂−1|𝒙q|2​𝔼𝒚q\displaystyle D_{\boldsymbol{y}_{p}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\rvert^{2}}\mathbb{E}_{\boldsymbol{y}_{q}}=-D_{\boldsymbol{x}_{p}}+\frac{\boldsymbol{a}\boldsymbol{x}_{q}\boldsymbol{a}^{-1}}{\lvert\boldsymbol{x}_{q}\rvert^{2}}\mathbb{E}_{\boldsymbol{y}_{q}}
    =\displaystyle= 𝒂​(D𝒙p+𝒙q|𝒙q|2​𝔼𝒙q)​𝒂−1=𝒂​G𝒙​𝒂−1.\displaystyle\boldsymbol{a}\bigg{(}D_{\boldsymbol{x}_{p}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\rvert^{2}}\mathbb{E}_{\boldsymbol{x}_{q}}\bigg{)}\boldsymbol{a}^{-1}=\boldsymbol{a}G_{\boldsymbol{x}}\boldsymbol{a}^{-1}.

    To see the equation above is in the form of (3), we only need to see that 𝒂−1~=𝒂\widetilde{\boldsymbol{a}^{-1}}=\boldsymbol{a} and |𝒂|=1\lvert\boldsymbol{a}\rvert=1 for 𝒂∈𝕊q−1\boldsymbol{a}\in\mathbb{S}^{q-1}.

  4. 4.

    𝒚=φ​(𝒙)=−𝒙−1=𝒙|𝒙|2\boldsymbol{y}=\varphi(\boldsymbol{x})=-\boldsymbol{x}^{-1}=\displaystyle\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{2}}, in this case, a=d=0,b=−c=1a=d=0,\ b=-c=1. In other words, we need to show that

    (𝒙|𝒙|p+3)−1​G𝒙​𝒙|𝒙|p+1=G𝒚.\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3}}\bigg{)}^{-1}G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}=G_{\boldsymbol{y}}.

    First, we notice that yi=xi|𝒙|2y_{i}=\displaystyle\frac{x_{i}}{\lvert\boldsymbol{x}\lvert^{2}}, which gives us that

    ∂xj\displaystyle\partial_{x_{j}} =∑k=1p+q∂yk∂xj​∂yk=∑k=1,k≠jp+q−2​xj​xk|𝒙|4​∂yk+(1|𝒙|2−2​xj2|𝒙|4)\displaystyle=\sum_{k=1}^{p+q}\frac{\partial y_{k}}{\partial x_{j}}\partial_{y_{k}}=\sum_{k=1,k\neq j}^{p+q}\frac{-2x_{j}x_{k}}{\lvert\boldsymbol{x}\lvert^{4}}\partial_{y_{k}}+\bigg{(}\frac{1}{\lvert\boldsymbol{x}\lvert^{2}}-\frac{2x_{j}^{2}}{\lvert\boldsymbol{x}\lvert^{4}}\bigg{)}
    =\displaystyle= 1|𝒙|2∂yj−∑k=1p+q2yjyk∂yk=|𝒚|2∂yj−2yj𝔼𝒚.\displaystyle\frac{1}{\lvert\boldsymbol{x}\lvert^{2}}\partial_{y_{j}}-\sum_{k=1}^{p+q}2y_{j}y_{k}\partial_{y_{k}}=\lvert\boldsymbol{y}\lvert^{2}\partial_{y_{j}}-2y_{j}\mathbb{E}_{\boldsymbol{y}}.

    This leads to

    G𝒙=∑i=1pei​∂xi+𝒙q|𝒙q|2​𝔼𝒙q\displaystyle G_{\boldsymbol{x}}=\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\mathbb{E}_{\boldsymbol{x}_{q}}
    =\displaystyle= ∑i=1pei(|𝒚|2∂yj−2yj𝔼𝒚)+𝒚q|𝒚q|2|𝒚|2∑i=p+1p+qyi|𝒚|2(|𝒚|2∂yj−2yj𝔼𝒚)\displaystyle\sum_{i=1}^{p}e_{i}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}\partial_{y_{j}}-2y_{j}\mathbb{E}_{\boldsymbol{y}}\bigg{)}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\sum_{i=p+1}^{p+q}\frac{y_{i}}{\lvert\boldsymbol{y}\lvert^{2}}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}\partial_{y_{j}}-2y_{j}\mathbb{E}_{\boldsymbol{y}}\bigg{)}
    =\displaystyle= |𝒚|2∑i=1pei∂yi−2𝒚p𝔼𝒚+𝒚q|𝒚q|2|𝒚|2(𝔼𝒚q−2|𝒚q|2|𝒚|2𝔼𝒚)\displaystyle\lvert\boldsymbol{y}\lvert^{2}\sum_{i=1}^{p}e_{i}\partial_{y_{i}}-2\boldsymbol{y}_{p}\mathbb{E}_{\boldsymbol{y}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\bigg{(}\mathbb{E}_{\boldsymbol{y}_{q}}-2\frac{\lvert\boldsymbol{y}_{q}\lvert^{2}}{\lvert\boldsymbol{y}\lvert^{2}}\mathbb{E}_{\boldsymbol{y}}\bigg{)}
    =\displaystyle= |𝒚|2D𝒚p−2𝒚𝔼𝒚+𝒚q|𝒚q|2|𝒚|2𝔼𝒚q=|𝒚|2G𝒚−2𝒚𝔼𝒚.\displaystyle\lvert\boldsymbol{y}\lvert^{2}D_{\boldsymbol{y}_{p}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\mathbb{E}_{\boldsymbol{y}_{q}}=\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}.

    Now, we calculate

    G𝒙𝒙|𝒙|p+1=(|𝒚|2G𝒚−2𝒚𝔼𝒚)𝒚|𝒚|p−1\displaystyle G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}=(\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}})\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}
    =\displaystyle= |𝒚|2∑i=1pei(ei|𝒚|p−1+(p−1)yi𝒚|𝒚|p−3+𝒚|𝒚|p−1∂yi)−2p𝒚2|𝒚|p−1\displaystyle\lvert\boldsymbol{y}\lvert^{2}\sum_{i=1}^{p}e_{i}\bigg{(}e_{i}\lvert\boldsymbol{y}\lvert^{p-1}+(p-1)y_{i}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-3}+\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}\partial_{y_{i}}\bigg{)}-2p\boldsymbol{y}^{2}\lvert\boldsymbol{y}\lvert^{p-1}
    +2|𝒚|p+1𝔼𝒚+𝒚q|𝒚q|2|𝒚|2∑i=p+1p+qyi(ei|𝒚|p−1+(p−1)yi𝒚|𝒚|p−3+𝒚|𝒚|p−1∂yi)\displaystyle+2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\sum_{i=p+1}^{p+q}y_{i}\bigg{(}e_{i}\lvert\boldsymbol{y}\lvert^{p-1}+(p-1)y_{i}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-3}+\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}\partial_{y_{i}}\bigg{)}
    =\displaystyle= |𝒚|2(−p|𝒚|p−1+(p−1)𝒚p𝒚|𝒚|p−3+∑i=1p(−𝒚ei−2yi)|𝒚|p−1∂yi)+2p|𝒚|p+1\displaystyle\lvert\boldsymbol{y}\lvert^{2}\bigg{(}-p\lvert\boldsymbol{y}\lvert^{p-1}+(p-1)\boldsymbol{y}_{p}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-3}+\sum_{i=1}^{p}(-\boldsymbol{y}e_{i}-2y_{i})\lvert\boldsymbol{y}\lvert^{p-1}\partial_{y_{i}}\bigg{)}+2p\lvert\boldsymbol{y}\lvert^{p+1}
    +2|𝒚|p+1𝔼𝒚+𝒚q|𝒚q|2|𝒚|2(𝒚q|𝒚|p−1+(p−1)|𝒚q|2𝒚|𝒚|p−3+𝒚|𝒚|p−1𝔼𝒚q)\displaystyle+2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\bigg{(}\boldsymbol{y}_{q}\lvert\boldsymbol{y}\lvert^{p-1}+(p-1)\lvert\boldsymbol{y}_{q}\lvert^{2}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-3}+\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}\mathbb{E}_{\boldsymbol{y}_{q}}\bigg{)}
    =\displaystyle= −p|𝒚|p+1+(p−1)𝒚p𝒚|𝒚|p−1−𝒚|𝒚|p+1D𝒚p−2|𝒚|p+1𝔼𝒚p+2|𝒚|p+1𝔼𝒚\displaystyle-p\lvert\boldsymbol{y}\lvert^{p+1}+(p-1)\boldsymbol{y}_{p}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}-\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p+1}D_{\boldsymbol{y}_{p}}-2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}_{p}}+2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}}
    −|𝒚|p+1+(p−1)𝒚q𝒚|𝒚|p−1+𝒚q​𝒚|𝒚q|2|𝒚|p+1𝔼𝒚q\displaystyle-\lvert\boldsymbol{y}\lvert^{p+1}+(p-1)\boldsymbol{y}_{q}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p-1}+\frac{\boldsymbol{y}_{q}\boldsymbol{y}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}_{q}}
    =\displaystyle= −𝒚|𝒚|p+1D𝒚p−2|𝒚|p+1𝔼𝒚p+2|𝒚|p+1𝔼𝒚+−𝒚𝒚q−2|𝒚q|2|𝒚q|2|𝒚|p+1𝔼𝒚q\displaystyle-\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p+1}D_{\boldsymbol{y}_{p}}-2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}_{p}}+2\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}}+\frac{-\boldsymbol{y}\boldsymbol{y}_{q}-2\lvert\boldsymbol{y}_{q}\lvert^{2}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{p+1}\mathbb{E}_{\boldsymbol{y}_{q}}
    =\displaystyle= −𝒚|𝒚|p+1G𝒚.\displaystyle-\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p+1}G_{\boldsymbol{y}}.

    Hence, we have that

    (𝒙|𝒙|p+3)−1G𝒙𝒙|𝒙|p+1=−𝒚|𝒚|−p−3𝒚|𝒚|p+1G𝒚=G𝒚,\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3}}\bigg{)}^{-1}G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}=-\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{-p-3}\boldsymbol{y}\lvert\boldsymbol{y}\lvert^{p+1}G_{\boldsymbol{y}}=G_{\boldsymbol{y}},

which completes the proof.

Remark 3.

When p=1,q=np=1,\ q=n, the slice Dirac operator is given by

G𝒙=e1​∂𝒙1+𝒙n|𝒙n|2​∑i=2n+1xi​∂xi.G_{\boldsymbol{x}}=e_{1}\partial_{\boldsymbol{x}_{1}}+\frac{\boldsymbol{x}_{n}}{\lvert\boldsymbol{x}_{n}\lvert^{2}}\sum_{i=2}^{n+1}x_{i}\partial_{x_{i}}.

One notices that this operator is slightly different from the global slice Dirac operator introduced in CoG

G=|𝒙n|2∂𝒙0+𝒙n∑i=1nxi∂xi,\displaystyle{G}=\lvert\boldsymbol{x}_{n}\lvert^{2}\partial_{\boldsymbol{x}_{0}}+\boldsymbol{x}_{n}\sum_{i=1}^{n}x_{i}\partial_{x_{i}},

which acts on functions defined on ℝ⊕ℝn\mathbb{R}\oplus\mathbb{R}^{n}. However, one can rewrite

G𝒙=e1|𝒙n|2(|𝒙n|2∂𝒙1−∑i=2n+1(e1ei)xi∑i=1nxi∂xi),\displaystyle G_{\boldsymbol{x}}=\frac{e_{1}}{\lvert\boldsymbol{x}_{n}\lvert^{2}}\bigg{(}\lvert\boldsymbol{x}_{n}\lvert^{2}\partial_{\boldsymbol{x}_{1}}-\sum_{i=2}^{n+1}(e_{1}e_{i})x_{i}\sum_{i=1}^{n}x_{i}\partial_{x_{i}}\bigg{)},

and let ei†=−e1​eie_{i}^{\dagger}=-e_{1}e_{i}, so one can see that the operator in the parentheses above is G{G}. This suggests that with a similar argument as above, one can obtain a similar result for G{G} as follows. This result also justifies the invariance of the slice monogenic operator given in (CoG, , Theorem 3.1).

Corollary 2.

Let 𝐲=φ​(𝐱)=(a​𝐱+b)​(c​𝐱+d)−1∈G​R​A​V​(ℝ⊕ℝp)\boldsymbol{y}=\varphi(\boldsymbol{x})=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}\in GRAV(\mathbb{R}\oplus\mathbb{R}^{p}), and f∈C1​(U)f\in C^{1}(U), where UU is a domain in ℝ⊕ℝp\mathbb{R}\oplus\mathbb{R}^{p}. Then, we have

G​f​(𝒚)=(c​𝒙+d)−1​G​c​𝒙+d¯|c𝒙+d|2​f​((a​𝒙+b)​(c​𝒙+d)−1).\displaystyle Gf(\boldsymbol{y})=\big{(}c\boldsymbol{x}+d\big{)}^{-1}G\frac{\overline{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\lvert^{2}}f((a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}). (3)

It is worth pointing out that the GG operators on the left and right sides above are the GG operators with respect to 𝒚\boldsymbol{y} and 𝒙\boldsymbol{x} respectively.

4 Intertwining operators for iterated global slice Dirac operators

In this section, we show that the iterated slice Dirac operator G𝒙lG_{\boldsymbol{x}}^{l} also has G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}) as its symmetry group. However, we prove this for ll odd and even separately, since the intertwining operators for the odd and even cases are different.

The reason that we use G𝒙G_{\boldsymbol{x}} instead of GG to construct the iterated slice Dirac operator is the following: GG plays the same role as the generalized Dirac operator D0=∂x0+∑i=1nei​∂xiD_{0}=\partial_{x_{0}}+\sum_{i=1}^{n}e_{i}\partial_{x_{i}} does in classical Clifford analysis. It is well-known that D0kD_{0}^{k} is not conformally invariant anymore when k>1k>1. In contrast, the kkth power of the Dirac operator is also conformally invariant for k>1k>1. More details can be found in Peetre . The same phenomenon happens here for GG and G𝒙G_{\boldsymbol{x}}.

Theorem 3.

Let 𝐲=φ​(𝐱)=(a​𝐱+b)​(c​𝐱+d)−1∈G​R​A​V​(ℝp+q)\boldsymbol{y}=\varphi(\boldsymbol{x})=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}\in GRAV(\mathbb{R}^{p+q}), and let ff be a sufficiently smooth function over a domain U⊂ℝp+q\ℝpU\subset\mathbb{R}^{p+q}\backslash\mathbb{R}^{p}. Then, we have

G𝒚l​f​(𝒚)=(c​𝒙+d|c𝒙+d|p+2+l)−1​G𝒙l​c​𝒙+d~|c𝒙+d|p+2−l​f​((a​𝒙+b)​(c​𝒙+d)−1),l​o​d​d.\displaystyle G_{\boldsymbol{y}}^{l}f(\boldsymbol{y})=\bigg{(}\frac{c\boldsymbol{x}+d}{\lvert c\boldsymbol{x}+d\lvert^{p+2+l}}\bigg{)}^{-1}G^{l}_{\boldsymbol{x}}\frac{\widetilde{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\lvert^{p+2-l}}f((a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}),\quad l\ odd.

The strategy is similar as in the proof of the previous theorem and it is a straightforward check that the theorem above is true when φ​(𝒙)\varphi(\boldsymbol{x}) is a translation, dilation or reflection. Hence, we only need to prove it is also true for the inversion. More specifically, we need to show that

(𝒙|𝒙|p+2+l)−1​G𝒙l​𝒙|𝒙|p+2−l=G𝒚l\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+2+l}}\bigg{)}^{-1}G_{\boldsymbol{x}}^{l}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+2-l}}=G_{\boldsymbol{y}}^{l} (4)

for 𝒚=−𝒙−1∈ℝp+q\boldsymbol{y}=-\boldsymbol{x}^{-1}\in\mathbb{R}^{p+q}. To show this, we need the following technical lemma. It is worth pointing out that the functions on the left sides of the equations below are actually considered as multiplication operators.

Lemma 4.

Let 𝐲=−𝐱−1∈ℝp+q\boldsymbol{y}=-\boldsymbol{x}^{-1}\in\mathbb{R}^{p+q}, then we have

(a).G𝒙2​𝒙|𝒙|m=−m​(m−p−1)​𝒙|𝒙|m+2+2​(m−p−1)​𝒙|𝒙|m+2​𝔼𝒙\displaystyle(a).\ G_{\boldsymbol{x}}^{2}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}=-m(m-p-1)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+2(m-p-1)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}\mathbb{E}_{\boldsymbol{x}}
−2|𝒙|−m−2G𝒚+𝒙|𝒙|m+4G𝒚2,\displaystyle\quad\quad\quad\quad\quad\quad\quad\ \ -2\lvert\boldsymbol{x}\lvert^{-m-2}G_{\boldsymbol{y}}+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+4}}G_{\boldsymbol{y}}^{2},
(b).G𝒚2​k−2​𝒚=−(2​k−2)​G𝒚2​k−3+𝒚​G𝒚2​k−2,\displaystyle(b).\ G_{\boldsymbol{y}}^{2k-2}\boldsymbol{y}=-(2k-2)G_{\boldsymbol{y}}^{2k-3}+\boldsymbol{y}G_{\boldsymbol{y}}^{2k-2},
(c).G𝒚2​k−1​𝒚=−(p+2​k−1+2​𝔼𝒚)​G𝒚2​k−2−𝒚​G𝒚2​k−1,\displaystyle(c).\ G_{\boldsymbol{y}}^{2k-1}\boldsymbol{y}=-(p+2k-1+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2k-2}-\boldsymbol{y}G_{\boldsymbol{y}}^{2k-1},
(d).G𝒚2​k−2|𝒚|2=−2(k−1)(p+2k−3+2𝔼𝒚)G𝒚2​k−4+|𝒚|2G𝒚2​k−2.\displaystyle(d).\ G_{\boldsymbol{y}}^{2k-2}\lvert\boldsymbol{y}\lvert^{2}=-2(k-1)(p+2k-3+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2k-4}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}.
(e).𝔼𝒙=−𝔼𝒚.\displaystyle(e).\ \mathbb{E}_{\boldsymbol{x}}=-\mathbb{E}_{\boldsymbol{y}}.

Proof: (a). This is indeed a straightforward calculation as follows. Firstly, we calculate

G𝒙|𝒙|−m=(∑i=1pei∂xi+𝒙q|𝒙q|2∑i=p+1p+qxi∂xi)|𝒙|−m\displaystyle G_{\boldsymbol{x}}\lvert\boldsymbol{x}\lvert^{-m}=\bigg{(}\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\sum_{i=p+1}^{p+q}x_{i}\partial_{x_{i}}\bigg{)}\lvert\boldsymbol{x}\lvert^{-m}
=\displaystyle= ∑i=1pei(−mxi|𝒙|−m−2+|𝒙|−m∂xi)+𝒙q|𝒙q|2∑i=p+1p+qxi(−mxi|𝒙|−m−2+|𝒙|−m∂xi)\displaystyle\sum_{i=1}^{p}e_{i}(-mx_{i}\lvert\boldsymbol{x}\lvert^{-m-2}+\lvert\boldsymbol{x}\lvert^{-m}\partial_{x_{i}})+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\sum_{i=p+1}^{p+q}x_{i}(-mx_{i}\lvert\boldsymbol{x}\lvert^{-m-2}+\lvert\boldsymbol{x}\lvert^{-m}\partial_{x_{i}})
=\displaystyle= −m𝒙p|𝒙|m+2+|𝒙|−mD𝒙p+𝒙q|𝒙q|2(−m|𝒙q|2|𝒙|−m−2+|𝒙|−m𝔼𝒙q)\displaystyle-m\frac{\boldsymbol{x}_{p}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\lvert\boldsymbol{x}\lvert^{-m}D_{\boldsymbol{x}_{p}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}(-m\lvert\boldsymbol{x}_{q}\lvert^{2}\lvert\boldsymbol{x}\lvert^{-m-2}+\lvert\boldsymbol{x}\lvert^{-m}\mathbb{E}_{\boldsymbol{x}_{q}})
=\displaystyle= −m𝒙|𝒙|m+2+|𝒙|−mG𝒙.\displaystyle-m\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\lvert\boldsymbol{x}\lvert^{-m}G_{\boldsymbol{x}}.

Now, we check

G𝒙​𝒙|𝒙|m=(∑i=1pei​∂xi+𝒙q|𝒙q|2​∑i=p+1p+qxi​∂xi)​𝒙|𝒙|m\displaystyle G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}=\bigg{(}\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\sum_{i=p+1}^{p+q}x_{i}\partial_{x_{i}}\bigg{)}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}
=\displaystyle= ∑i=1pei​(ei|𝒙|m−m​xi​𝒙|𝒙|m+2+𝒙​∂xi|𝒙|m)+𝒙q|𝒙q|2​∑i=p+1p+qxi​(ei|𝒙|m−m​xi​𝒙|𝒙|m+2+𝒙​∂xi|𝒙|m)\displaystyle\sum_{i=1}^{p}e_{i}\bigg{(}\frac{e_{i}}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{mx_{i}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\frac{\boldsymbol{x}\partial_{x_{i}}}{\lvert\boldsymbol{x}\lvert^{m}}\bigg{)}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\sum_{i=p+1}^{p+q}x_{i}\bigg{(}\frac{e_{i}}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{mx_{i}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\frac{\boldsymbol{x}\partial_{x_{i}}}{\lvert\boldsymbol{x}\lvert^{m}}\bigg{)}
=\displaystyle= −p|𝒙|m−m​𝒙p​𝒙|𝒙|m+2+∑i=1p−𝒙​ei−2​xi|𝒙|m​∂xi+𝒙q|𝒙q|2​(𝒙q|𝒙|m−m|𝒙q|2𝒙|𝒙|m+2+𝒙​𝔼𝒙q|𝒙|m)\displaystyle\frac{-p}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{m\boldsymbol{x}_{p}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\sum_{i=1}^{p}\frac{-\boldsymbol{x}e_{i}-2x_{i}}{\lvert\boldsymbol{x}\lvert^{m}}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\bigg{(}\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{m\lvert\boldsymbol{x}_{q}\lvert^{2}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\frac{\boldsymbol{x}\mathbb{E}_{\boldsymbol{x}_{q}}}{\lvert\boldsymbol{x}\lvert^{m}}\bigg{)}
=\displaystyle= −p|𝒙|m−m​𝒙p​𝒙|𝒙|m+2−𝒙​D𝒙p|𝒙|m−2​𝔼𝒙p|𝒙|m−1|𝒙|m−m​𝒙q​𝒙|𝒙|m+2−𝒙​𝒙q​𝔼𝒙q|𝒙q|2|𝒙|m−2​𝔼𝒙q|𝒙|m\displaystyle\frac{-p}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{m\boldsymbol{x}_{p}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}-\frac{\boldsymbol{x}D_{\boldsymbol{x}_{p}}}{\lvert\boldsymbol{x}\lvert^{m}}-2\frac{\mathbb{E}_{\boldsymbol{x}_{p}}}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{1}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{m\boldsymbol{x}_{q}\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}-\frac{\boldsymbol{x}\boldsymbol{x}_{q}\mathbb{E}_{\boldsymbol{x}_{q}}}{\lvert\boldsymbol{x}_{q}\lvert^{2}\lvert\boldsymbol{x}\lvert^{m}}-\frac{2\mathbb{E}_{\boldsymbol{x}_{q}}}{\lvert\boldsymbol{x}\lvert^{m}}
=\displaystyle= (m−p−1)|𝒙|−m−𝒙|𝒙|mD𝒙p−2𝔼𝒙p|𝒙|m−−𝒙q𝒙−2|𝒙q|2|𝒙q|2|𝒙|m𝔼𝒙q−2|𝒙|m𝔼𝒙q\displaystyle(m-p-1)\lvert\boldsymbol{x}\lvert^{-m}-\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}D_{\boldsymbol{x}_{p}}-2\frac{\mathbb{E}_{\boldsymbol{x}_{p}}}{\lvert\boldsymbol{x}\lvert^{m}}-\frac{-\boldsymbol{x}_{q}\boldsymbol{x}-2\lvert\boldsymbol{x}_{q}\lvert^{2}}{\lvert\boldsymbol{x}_{q}\lvert^{2}\lvert\boldsymbol{x}\lvert^{m}}\mathbb{E}_{\boldsymbol{x}_{q}}-\frac{2}{\lvert\boldsymbol{x}\lvert^{m}}\mathbb{E}_{\boldsymbol{x}_{q}}
=\displaystyle= |𝒙|−m(m−p−1−2𝔼𝒙)−𝒙|𝒙|mG𝒙.\displaystyle\lvert\boldsymbol{x}\lvert^{-m}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}G_{\boldsymbol{x}}.

Hence,

G𝒙2𝒙|𝒙|m=G𝒙(|𝒙|−m(m−p−1−2𝔼𝒙)−𝒙|𝒙|mG𝒙)\displaystyle G_{\boldsymbol{x}}^{2}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}=G_{\boldsymbol{x}}\bigg{(}\lvert\boldsymbol{x}\lvert^{-m}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}G_{\boldsymbol{x}}\bigg{)}
=\displaystyle= G𝒙|𝒙|−m(m−p−1−2𝔼𝒙)−G𝒙𝒙|𝒙|mG𝒙\displaystyle G_{\boldsymbol{x}}\lvert\boldsymbol{x}\lvert^{-m}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}G_{\boldsymbol{x}}
=\displaystyle= (−m​𝒙|𝒙|m+2+|𝒙|−mG𝒙)(m−p−1−2𝔼𝒙)−(m−p−1−2​𝔼𝒙)​G𝒙|𝒙|m+𝒙​G𝒙2|𝒙|m\displaystyle\bigg{(}\frac{-m\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+\lvert\boldsymbol{x}\lvert^{-m}G_{\boldsymbol{x}}\bigg{)}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-\frac{(m-p-1-2\mathbb{E}_{\boldsymbol{x}})G_{\boldsymbol{x}}}{\lvert\boldsymbol{x}\lvert^{m}}+\frac{\boldsymbol{x}G_{\boldsymbol{x}}^{2}}{\lvert\boldsymbol{x}\lvert^{m}}
=\displaystyle= −m𝒙|𝒙|m+2(m−p−1−2𝔼𝒙)−2|𝒙|−mG𝒙+𝒙|𝒙|mG𝒙2\displaystyle-m\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-2\lvert\boldsymbol{x}\lvert^{-m}G_{\boldsymbol{x}}+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}G_{\boldsymbol{x}}^{2}
=\displaystyle= −m𝒙|𝒙|m+2(m−p−1−2𝔼𝒙)−2|𝒙|−m(|𝒚|2G𝒚−2𝒚𝔼𝒚)\displaystyle-m\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-2\lvert\boldsymbol{x}\lvert^{-m}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\bigg{)}
+𝒙|𝒙|m(|𝒚|2G𝒚−2𝒚𝔼𝒚)(|𝒚|2G𝒚−2𝒚𝔼𝒚)\displaystyle+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\bigg{)}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\bigg{)}
=\displaystyle= −m𝒙|𝒙|m+2(m−p−1−2𝔼𝒙)−2|𝒙|−m(|𝒚|2G𝒚−2𝒚𝔼𝒚)\displaystyle-m\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}(m-p-1-2\mathbb{E}_{\boldsymbol{x}})-2\lvert\boldsymbol{x}\lvert^{-m}\bigg{(}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\bigg{)}
+𝒙|𝒙|m[|𝒚|2(2𝒚+|𝒚|2G𝒚)G𝒚−2|𝒚|2(−p−1−2𝔼𝒚−𝒚G𝒚)𝔼𝒚−2𝒚𝔼𝒚|𝒚|2G𝒚\displaystyle+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m}}\bigg{[}\lvert\boldsymbol{y}\lvert^{2}(2\boldsymbol{y}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}})G_{\boldsymbol{y}}-2\lvert\boldsymbol{y}\lvert^{2}(-p-1-2\mathbb{E}_{\boldsymbol{y}}-\boldsymbol{y}G_{\boldsymbol{y}})\mathbb{E}_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}
+4𝒚2(𝔼𝒚+1)𝔼𝒚]\displaystyle+4\boldsymbol{y}^{2}(\mathbb{E}_{\boldsymbol{y}}+1)\mathbb{E}_{\boldsymbol{y}}\bigg{]}
=\displaystyle= −m(m−p−1)𝒙|𝒙|m+2+2(m−p−1)𝒙|𝒙|m+2𝔼𝒙−2|𝒙|−m−2G𝒚+𝒙|𝒙|m+4G𝒚2.\displaystyle-m(m-p-1)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}+2(m-p-1)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+2}}\mathbb{E}_{\boldsymbol{x}}-2\lvert\boldsymbol{x}\lvert^{-m-2}G_{\boldsymbol{y}}+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{m+4}}G_{\boldsymbol{y}}^{2}.

The identity (b) can be easily proved by induction and (c) can be proved immediately by applying G𝒚G_{\boldsymbol{y}} to (b). Now, we prove the identity (d).

G𝒚2​k−2|𝒚|2=G𝒚2​k−3(2𝒚+|𝒚|2G𝒚)=2G𝒚2​k−3𝒚+G𝒚2​k−4(2𝒚+|𝒚|2G𝒚)G𝒚=⋯\displaystyle G_{\boldsymbol{y}}^{2k-2}\lvert\boldsymbol{y}\lvert^{2}=G_{\boldsymbol{y}}^{2k-3}(2\boldsymbol{y}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}})=2G_{\boldsymbol{y}}^{2k-3}\boldsymbol{y}+G_{\boldsymbol{y}}^{2k-4}(2\boldsymbol{y}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}})G_{\boldsymbol{y}}=\cdots
=\displaystyle= 2G𝒚2​k−3𝒚+2G𝒚2​k−4𝒚G𝒚+⋯+2𝒚G𝒚2​k−3+|𝒚|2G𝒚2​k−2\displaystyle 2G_{\boldsymbol{y}}^{2k-3}\boldsymbol{y}+2G_{\boldsymbol{y}}^{2k-4}\boldsymbol{y}G_{\boldsymbol{y}}+\cdots+2\boldsymbol{y}G_{\boldsymbol{y}}^{2k-3}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}
=\displaystyle= ∑j=12​k−22G𝒚2​k−2−j𝒚G𝒚j−1+|𝒚|2G𝒚2​k−2\displaystyle\sum_{j=1}^{2k-2}2G_{\boldsymbol{y}}^{2k-2-j}\boldsymbol{y}G_{\boldsymbol{y}}^{j-1}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}
=\displaystyle= ∑n=1k−1G𝒚2​k−2−2​n𝒚G𝒚2​n−1+∑n=1k−1G𝒚2​k−1−2​n𝒚G𝒚2​n−2+|𝒚|2G𝒚2​k−2\displaystyle\sum_{n=1}^{k-1}G_{\boldsymbol{y}}^{2k-2-2n}\boldsymbol{y}G_{\boldsymbol{y}}^{2n-1}+\sum_{n=1}^{k-1}G_{\boldsymbol{y}}^{2k-1-2n}\boldsymbol{y}G_{\boldsymbol{y}}^{2n-2}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}
=\displaystyle= ∑n=1k−1(−(2​k−2​n−2)​G𝒚2​k−2​n−3+𝒚​G𝒚2​k−2​n−2)​G𝒚2​n−1\displaystyle\sum_{n=1}^{k-1}\bigg{(}-(2k-2n-2)G_{\boldsymbol{y}}^{2k-2n-3}+\boldsymbol{y}G_{\boldsymbol{y}}^{2k-2n-2}\bigg{)}G_{\boldsymbol{y}}^{2n-1}
+∑n=1k−1(−(p+2k−2n−1+2𝔼𝒚)G𝒚2​k−2​n−2−𝒚G𝒚2​k−2​n−1)G𝒚2​n−2+|𝒚|2G𝒚2​k−2\displaystyle+\sum_{n=1}^{k-1}\bigg{(}-(p+2k-2n-1+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2k-2n-2}-\boldsymbol{y}G_{\boldsymbol{y}}^{2k-2n-1}\bigg{)}G_{\boldsymbol{y}}^{2n-2}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}
=\displaystyle= −2(k−1)(p+2k−3+2𝔼𝒚)G𝒚2​k−4+|𝒚|2G𝒚2​k−2.\displaystyle-2(k-1)(p+2k-3+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2k-4}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}.

The last equation comes from a straightforward calculation as follows.

𝔼𝒙=∑i=1p+qxi​∂xi=∑i=1p+qyi|𝒚|2​∑j=1p+q∂yj∂xi​∂yj=∑i=1p+qyi|𝒚|2​∑j=1p+q(δi​j|𝒙|2−2​xi​xj|𝒙|4)​∂yj\displaystyle\mathbb{E}_{\boldsymbol{x}}=\sum_{i=1}^{p+q}x_{i}\partial_{x_{i}}=\sum_{i=1}^{p+q}\frac{y_{i}}{\lvert\boldsymbol{y}\lvert^{2}}\sum_{j=1}^{p+q}\frac{\partial y_{j}}{\partial_{x_{i}}}\partial_{y_{j}}=\sum_{i=1}^{p+q}\frac{y_{i}}{\lvert\boldsymbol{y}\lvert^{2}}\sum_{j=1}^{p+q}\bigg{(}\frac{\delta_{ij}}{\lvert\boldsymbol{x}\lvert^{2}}-\frac{2x_{i}x_{j}}{\lvert\boldsymbol{x}\lvert^{4}}\bigg{)}\partial_{y_{j}}
=\displaystyle= ∑i=1p+qyi|𝒚|2​∂yj|𝒙|2−2​∑i,j=1p+qyi|𝒚|2​yi​yj​∂yj=−𝔼𝒚.\displaystyle\sum_{i=1}^{p+q}\frac{y_{i}}{\lvert\boldsymbol{y}\lvert^{2}}\frac{\partial_{y_{j}}}{\lvert\boldsymbol{x}\lvert^{2}}-2\sum_{i,j=1}^{p+q}\frac{y_{i}}{\lvert\boldsymbol{y}\lvert^{2}}y_{i}y_{j}\partial_{y_{j}}=-\mathbb{E}_{\boldsymbol{y}}.

Now, we can prove the identity (4).

Proof: We prove (4) by induction. Firstly, it is true for l=1l=1, see Theorem 1. Next, we assume that it is true for l=2​k−1l=2k-1, i.e,

(𝒙|𝒙|p+1+2​k)−1​G𝒙2​k−1​𝒙|𝒙|p+3−2​k=G𝒚2​k−1.\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1+2k}}\bigg{)}^{-1}G_{\boldsymbol{x}}^{2k-1}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3-2k}}=G_{\boldsymbol{y}}^{2k-1}.

We need to show that it is also true for l=2​k+1l=2k+1 as follows.

(𝒙|𝒙|p+3+2​k)−1​G𝒙2​k+1​𝒙|𝒙|p+1−2​k=(𝒙|𝒙|p+3+2​k)−1​G𝒙2​k−1​G𝒙2​𝒙|𝒙|p+1−2​k\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3+2k}}\bigg{)}^{-1}G_{\boldsymbol{x}}^{2k+1}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1-2k}}=\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3+2k}}\bigg{)}^{-1}G_{\boldsymbol{x}}^{2k-1}G_{\boldsymbol{x}}^{2}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1-2k}}
=\displaystyle= (𝒙|𝒙|p+3+2​k)−1G𝒙2​k−1(2k(p+1−2k)𝒙|𝒙|p+3−2​k−4k𝒙|𝒙|p+3−2​k𝔼𝒙\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3+2k}}\bigg{)}^{-1}G_{\boldsymbol{x}}^{2k-1}\bigg{(}2k(p+1-2k)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3-2k}}-4k\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3-2k}}\mathbb{E}_{\boldsymbol{x}}
−2|𝒙|2​k−p−3G𝒚+𝒙|𝒙|p+5−2​kG𝒚2).\displaystyle-2\lvert\boldsymbol{x}\lvert^{2k-p-3}G_{\boldsymbol{y}}+\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+5-2k}}G_{\boldsymbol{y}}^{2}\bigg{)}.

The last equation comes from the identity (a) of Lemma 4. Now, with the assumption for l=2​k−1l=2k-1, the equation above becomes

(𝒙|𝒙|p+3+2​k)−1[2k(p+1−2k)𝒙|𝒙|p+1+2​kG𝒚2​k−1+4k𝒙|𝒙|p+1+2​kG𝒚2​k−1𝔼𝒚\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3+2k}}\bigg{)}^{-1}\bigg{[}2k(p+1-2k)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1+2k}}G_{\boldsymbol{y}}^{2k-1}+4k\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1+2k}}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}
−2G𝒙2​k−1𝒙|𝒙|p+3−2​k(−𝒙|𝒙|2)G𝒚+G𝒙2​k−1𝒙|𝒙|p+3−2​k1|𝒙|2G𝒚2]\displaystyle-2G_{\boldsymbol{x}}^{2k-1}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3-2k}}\bigg{(}\frac{-\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{2}}\bigg{)}G_{\boldsymbol{y}}+G_{\boldsymbol{x}}^{2k-1}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3-2k}}\frac{1}{\lvert\boldsymbol{x}\lvert^{2}}G_{\boldsymbol{y}}^{2}\bigg{]}
=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2G𝒚2​k−1𝔼𝒚+2|𝒙|2G𝒚2​k−1𝒚G𝒚+|𝒙|2G𝒚2​k−1|𝒚|2G𝒚2\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}+2\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\boldsymbol{y}G_{\boldsymbol{y}}+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2}
=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2G𝒚2​k−1𝔼𝒚+2|𝒙|2G𝒚2​k−2G𝒚𝒚G𝒚\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}+2\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}G_{\boldsymbol{y}}\boldsymbol{y}G_{\boldsymbol{y}}
+|𝒙|2G𝒚2​k−2G𝒚2|𝒚|2G𝒚2\displaystyle+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}G_{\boldsymbol{y}}^{2}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2}
=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2G𝒚2​k−1𝔼𝒚\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}
+2|𝒙|2G𝒚2​k−2(−(p+1+2𝔼𝒚)−𝒚G𝒚)G𝒚+|𝒙|2G𝒚2​k−2(2𝒚+|𝒚|2G𝒚)G𝒚2\displaystyle+2\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}\bigg{(}-(p+1+2\mathbb{E}_{\boldsymbol{y}})-\boldsymbol{y}G_{\boldsymbol{y}}\bigg{)}G_{\boldsymbol{y}}+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}(2\boldsymbol{y}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2}
=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2G𝒚2​k−1𝔼𝒚−2|𝒙|2G𝒚2​k−2(p+1+2𝔼𝒚)G𝒚\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}-2\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}(p+1+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}
+|𝒙|2G𝒚2​k−2|𝒚|2G𝒚3.\displaystyle+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{3}.

Now, we apply the identity (d) in Lemma 4 to the last term above to obtain

=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2G𝒚2​k−1𝔼𝒚−2|𝒙|2G𝒚2​k−2(p+1+2𝔼𝒚)G𝒚\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}\mathbb{E}_{\boldsymbol{y}}-2\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-2}(p+1+2\mathbb{E}_{\boldsymbol{y}})G_{\boldsymbol{y}}
+|𝒙|2(−2(k−1)(p+2k+2𝔼𝒚−3)G𝒚2​k−1+|𝒚|2G𝒚2​k+1)\displaystyle+\lvert\boldsymbol{x}\lvert^{2}\bigg{(}-2(k-1)(p+2k+2\mathbb{E}_{\boldsymbol{y}}-3)G_{\boldsymbol{y}}^{2k-1}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k+1}\bigg{)}
=\displaystyle= 2k(p+1−2k)|𝒙|2G𝒚2​k−1+4k|𝒙|2(𝔼𝒚+2k−1)G𝒚2​k−1\displaystyle 2k(p+1-2k)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k-1}+4k\lvert\boldsymbol{x}\lvert^{2}(\mathbb{E}_{\boldsymbol{y}}+2k-1)G_{\boldsymbol{y}}^{2k-1}
−2|𝒙|2(p+2𝔼𝒚+4k−3)G𝒚2​k−1−2(k−1)|𝒙|2(p+2k+2𝔼𝒚−3)G𝒚2​k−1+G𝒚2​k+1\displaystyle-2\lvert\boldsymbol{x}\lvert^{2}(p+2\mathbb{E}_{\boldsymbol{y}}+4k-3)G_{\boldsymbol{y}}^{2k-1}-2(k-1)\lvert\boldsymbol{x}\lvert^{2}(p+2k+2\mathbb{E}_{\boldsymbol{y}}-3)G_{\boldsymbol{y}}^{2k-1}+G_{\boldsymbol{y}}^{2k+1}
=\displaystyle= G𝒚2​k+1.\displaystyle G_{\boldsymbol{y}}^{2k+1}.

Hence, the identity (4) is also true for l=2​k+1l=2k+1, which completes the proof of Theorem 4.

Remark 4.

We can easily see that the following intertwining operators for the ll odd cases

(c​𝒙+d|c𝒙+d|p+2+l)−1,c​𝒙+d~|c𝒙+d|p+2−l\displaystyle\bigg{(}\frac{c\boldsymbol{x}+d}{\lvert c\boldsymbol{x}+d\lvert^{p+2+l}}\bigg{)}^{-1},\ \frac{\widetilde{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\lvert^{p+2-l}}

are two vectors. In contrast, for the ll even cases, we can adapt the strategy used above to see that G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}) is also the symmetric group but with scalar intertwining operators. More specifically,

Theorem 5.

Let 𝐲=φ​(𝐱)=(a​𝐱+b)​(c​𝐱+d)−1∈G​R​A​V​(ℝp+q)\boldsymbol{y}=\varphi(\boldsymbol{x})=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}\in GRAV(\mathbb{R}^{p+q}), and let ff be a sufficiently smooth function over a domain U⊂ℝp+q\ℝpU\subset\mathbb{R}^{p+q}\backslash\mathbb{R}^{p}. Then, we have

G𝒚lf(𝒚)=|c𝒙+d|p+1+lG𝒙l|c𝒙+d|−p−1+lf((a𝒙+b)(c𝒙+d)−1),leven.\displaystyle G_{\boldsymbol{y}}^{l}f(\boldsymbol{y})=\lvert c\boldsymbol{x}+d\lvert^{p+1+l}G^{l}_{\boldsymbol{x}}\lvert c\boldsymbol{x}+d\lvert^{-p-1+l}f((a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}),\quad l\ even.

Proof: The proof is similar to the odd case. The same argument as in the odd case shows us the invariance with respect to translation, dilation and rotation. Here, we only show the invariance under inversion. More specifically, we show that

G𝒚l=|𝒙|p+1+lG𝒙l|𝒙|−p−1+l,leven,\displaystyle G_{\boldsymbol{y}}^{l}=\lvert\boldsymbol{x}\lvert^{p+1+l}G^{l}_{\boldsymbol{x}}\lvert\boldsymbol{x}\lvert^{-p-1+l},\ l\ even, (5)

where 𝒚=−𝒙−1\boldsymbol{y}=-\boldsymbol{x}^{-1}. We also prove this by induction. When l=2l=2, using the identities obtained in the proof of Lemma 4 (a)(a), we have

G𝒙2|𝒙|1−p=G𝒙((1−p)𝒙|𝒙|p+1+|𝒙|1−pG𝒙)\displaystyle G_{\boldsymbol{x}}^{2}\lvert\boldsymbol{x}\lvert^{1-p}=G_{\boldsymbol{x}}\bigg{(}(1-p)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}+\lvert\boldsymbol{x}\lvert^{1-p}G_{\boldsymbol{x}}\bigg{)}
=\displaystyle= (1−p)(−2|𝒙|−1−p𝔼𝒙−𝒙|𝒙|p+1G𝒙)+((1−p)𝒙|𝒙|p+1+|𝒙|1−pG𝒙)G𝒙\displaystyle(1-p)\bigg{(}-2\lvert\boldsymbol{x}\lvert^{-1-p}\mathbb{E}_{\boldsymbol{x}}-\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}G_{\boldsymbol{x}}\bigg{)}+\bigg{(}(1-p)\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}+\lvert\boldsymbol{x}\lvert^{1-p}G_{\boldsymbol{x}}\bigg{)}G_{\boldsymbol{x}}
=\displaystyle= 2(p−1)|𝒙|−1−p𝔼𝒙+|𝒙|1−pG𝒙2.\displaystyle 2(p-1)\lvert\boldsymbol{x}\lvert^{-1-p}\mathbb{E}_{\boldsymbol{x}}+\lvert\boldsymbol{x}\lvert^{1-p}G_{\boldsymbol{x}}^{2}.

Further, we have

G𝒙=∑i=1pei(1|𝒙|2∂yi−2yi𝔼𝒚)+𝒚q|𝒚q|2|𝒚|2(𝔼𝒚q−2|𝒚q|2|𝒚|2𝔼𝒚)=|𝒚|2G𝒚−2𝒚𝔼𝒚,\displaystyle G_{\boldsymbol{x}}=\sum_{i=1}^{p}e_{i}\bigg{(}\frac{1}{\lvert\boldsymbol{x}\lvert^{2}}\partial_{y_{i}}-2y_{i}\mathbb{E}_{\boldsymbol{y}}\bigg{)}+\frac{\boldsymbol{y}_{q}}{\lvert\boldsymbol{y}_{q}\lvert^{2}}\lvert\boldsymbol{y}\lvert^{2}\bigg{(}\mathbb{E}_{\boldsymbol{y}_{q}}-2\frac{\lvert\boldsymbol{y}_{q}\lvert^{2}}{\lvert\boldsymbol{y}\lvert^{2}}\mathbb{E}_{\boldsymbol{y}}\bigg{)}=\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}},

and

G𝒙2=(|𝒚|2G𝒚−2𝒚𝔼𝒚)(|𝒚|2G𝒚−2𝒚𝔼𝒚)\displaystyle G_{\boldsymbol{x}}^{2}=(\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}})(\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}})
=\displaystyle= |𝒚|2(2𝒚+|𝒚|2G𝒚)G𝒚−2|𝒚|2(−(p+1+2𝔼𝒚)−𝒚G𝒚)𝔼𝒚−2𝒚𝔼𝒚|𝒚|2G𝒚\displaystyle\lvert\boldsymbol{y}\lvert^{2}(2\boldsymbol{y}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}})G_{\boldsymbol{y}}-2\lvert\boldsymbol{y}\lvert^{2}(-(p+1+2\mathbb{E}_{\boldsymbol{y}})-\boldsymbol{y}G_{\boldsymbol{y}})\mathbb{E}_{\boldsymbol{y}}-2\boldsymbol{y}\mathbb{E}_{\boldsymbol{y}}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}
−4|𝒚|2(𝔼𝒚+1)𝔼𝒚\displaystyle-4\lvert\boldsymbol{y}\lvert^{2}(\mathbb{E}_{\boldsymbol{y}}+1)\mathbb{E}_{\boldsymbol{y}}
=\displaystyle= 2(p−1)|𝒚|2𝔼𝒚+|𝒚|4G𝒚2.\displaystyle 2(p-1)\lvert\boldsymbol{y}\lvert^{2}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{y}\lvert^{4}G_{\boldsymbol{y}}^{2}.

Hence, when l=2l=2, we have

|𝒙|p+3G𝒙2|𝒙|1−p=|𝒚|−p−3(2(p−1)|𝒙|−1−p𝔼𝒙+|𝒙|1−pG𝒙2)\displaystyle\lvert\boldsymbol{x}\lvert^{p+3}G_{\boldsymbol{x}}^{2}\lvert\boldsymbol{x}\lvert^{1-p}=\lvert\boldsymbol{y}\lvert^{-p-3}\bigg{(}2(p-1)\lvert\boldsymbol{x}\lvert^{-1-p}\mathbb{E}_{\boldsymbol{x}}+\lvert\boldsymbol{x}\lvert^{1-p}G_{\boldsymbol{x}}^{2}\bigg{)}
=\displaystyle= |𝒚|−p−3(2(1−p)|𝒚|1+p𝔼𝒚+|𝒚|p−1(2(p−1)|𝒚|2𝔼𝒚+|𝒚|4G𝒚2))=G𝒚2.\displaystyle\lvert\boldsymbol{y}\lvert^{-p-3}\bigg{(}2(1-p)\lvert\boldsymbol{y}\lvert^{1+p}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{y}\lvert^{p-1}(2(p-1)\lvert\boldsymbol{y}\lvert^{2}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{y}\lvert^{4}G_{\boldsymbol{y}}^{2})\bigg{)}=G_{\boldsymbol{y}}^{2}.

Next, we assume that (5) is true for l=2​kl=2k, i.e.,

G𝒚2​k=|𝒙|p+1+2​kG𝒙2​k|𝒙|−p−1+2​k.\displaystyle G_{\boldsymbol{y}}^{2k}=\lvert\boldsymbol{x}\lvert^{p+1+2k}G^{2k}_{\boldsymbol{x}}\lvert\boldsymbol{x}\lvert^{-p-1+2k}. (6)

For l=2​k+2l=2k+2, we have

|𝒙|p+2​k+3G𝒙2​k+2|𝒙|−p+2​k+1=|𝒙|p+2​k+3G𝒙2​kG𝒙2|𝒙|−p−1+2​k+2\displaystyle\lvert\boldsymbol{x}\lvert^{p+2k+3}G^{2k+2}_{\boldsymbol{x}}\lvert\boldsymbol{x}\lvert^{-p+2k+1}=\lvert\boldsymbol{x}\lvert^{p+2k+3}G_{\boldsymbol{x}}^{2k}G_{\boldsymbol{x}}^{2}\lvert\boldsymbol{x}\lvert^{-p-1+2k+2}
=\displaystyle= |𝒙|p+2​k+3[G𝒙2​k((−p+2k+1)|𝒙|−p+2​k−1(−2k−2𝔼𝒙)+|𝒙|−p+2​k+1G𝒙2)]\displaystyle\lvert\boldsymbol{x}\lvert^{p+2k+3}\bigg{[}G_{\boldsymbol{x}}^{2k}\bigg{(}(-p+2k+1)\lvert\boldsymbol{x}\lvert^{-p+2k-1}(-2k-2\mathbb{E}_{\boldsymbol{x}})+\lvert\boldsymbol{x}\lvert^{-p+2k+1}G_{\boldsymbol{x}}^{2}\bigg{)}\bigg{]}
=\displaystyle= |𝒙|p+2​k+3[(−p+2k+1)G𝒙2​k|𝒙|−p+2​k−1(−2k−2𝔼𝒙)+G𝒙2​k|𝒙|−p+2​k−1|𝒙|2G𝒙2].\displaystyle\lvert\boldsymbol{x}\lvert^{p+2k+3}\bigg{[}(-p+2k+1)G_{\boldsymbol{x}}^{2k}\lvert\boldsymbol{x}\lvert^{-p+2k-1}(-2k-2\mathbb{E}_{\boldsymbol{x}})+G_{\boldsymbol{x}}^{2k}\lvert\boldsymbol{x}\lvert^{-p+2k-1}\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{x}}^{2}\bigg{]}.

Now, we apply our assumption for l=2​kl=2k to obtain

=\displaystyle= |𝒙|p+2​k+3[(−p+2k+1)|𝒙|−p−1−2​kG𝒚2​k(−2k−2𝔼𝒙)+|𝒙|−p−1−2​kG𝒚2​k|𝒙|2G𝒙2]\displaystyle\lvert\boldsymbol{x}\lvert^{p+2k+3}\bigg{[}(-p+2k+1)\lvert\boldsymbol{x}\lvert^{-p-1-2k}G_{\boldsymbol{y}}^{2k}(-2k-2\mathbb{E}_{\boldsymbol{x}})+\lvert\boldsymbol{x}\lvert^{-p-1-2k}G_{\boldsymbol{y}}^{2k}\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{x}}^{2}\bigg{]}
=\displaystyle= (−p+2k+1)|𝒙|2G𝒚2​k(−2k+2𝔼𝒚)+|𝒙|2G𝒚2​k|𝒚|−2(2(p−1)|𝒚|2𝔼𝒚+|𝒚|4G𝒚2)\displaystyle(-p+2k+1)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}(-2k+2\mathbb{E}_{\boldsymbol{y}})+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\lvert\boldsymbol{y}\lvert^{-2}\bigg{(}2(p-1)\lvert\boldsymbol{y}\lvert^{2}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{y}\lvert^{4}G_{\boldsymbol{y}}^{2}\bigg{)}
=\displaystyle= (−p+2k+1)|𝒙|2G𝒚2​k(−2k+2𝔼𝒚)+2(p−1)|𝒙|2G𝒚2​k𝔼𝒚+|𝒙|2G𝒚2​k|𝒚|2G𝒚2\displaystyle(-p+2k+1)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}(-2k+2\mathbb{E}_{\boldsymbol{y}})+2(p-1)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2}
=\displaystyle= 2k(p−2k−1)|𝒙|2G𝒚2​k+4k|𝒙|2G𝒚2​k𝔼𝒚+|𝒙|2G𝒚2​k|𝒚|2G𝒚2.\displaystyle 2k(p-2k-1)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\mathbb{E}_{\boldsymbol{y}}+\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2}.

Using Lemma 4 (d)(d), we have

=\displaystyle= 2k(p−2k−1)|𝒙|2G𝒚2​k+4k|𝒙|2G𝒚2​k𝔼𝒚\displaystyle 2k(p-2k-1)\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\mathbb{E}_{\boldsymbol{y}}
+|𝒙|2(−2k(p+2k−1+2E𝒚)G𝒚2​k−2+|𝒚|2G𝒚2​k)G𝒚2\displaystyle+\lvert\boldsymbol{x}\lvert^{2}\bigg{(}-2k(p+2k-1+2E_{\boldsymbol{y}})G_{\boldsymbol{y}}^{2k-2}+\lvert\boldsymbol{y}\lvert^{2}G_{\boldsymbol{y}}^{2k}\bigg{)}G_{\boldsymbol{y}}^{2}
=\displaystyle= −8k2|𝒙|2G𝒚2​k+4k|𝒙|2G𝒚2​k𝔼𝒚−4k|𝒙|2G𝒚2​k(𝔼𝒚−2k)+G𝒚2​k+2\displaystyle-8k^{2}\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}+4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}\mathbb{E}_{\boldsymbol{y}}-4k\lvert\boldsymbol{x}\lvert^{2}G_{\boldsymbol{y}}^{2k}(\mathbb{E}_{\boldsymbol{y}}-2k)+G_{\boldsymbol{y}}^{2k+2}
=\displaystyle= G𝒚2​k+2,\displaystyle G_{\boldsymbol{y}}^{2k+2},

which proves that the equation (5) is also true for l=2​k+2l=2k+2, and this completes the proof.

5 A variant of the slice Dirac operator

Recall that the slice Dirac operator is defined as

G𝒙=D𝒙p+𝒙q|𝒙q|2​𝔼𝒙q=∑i=1pei​∂xi+𝒙q|𝒙q|2​∑i=p+1p+qxi​∂xi.\displaystyle G_{\boldsymbol{x}}=D_{\boldsymbol{x}_{p}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\mathbb{E}_{\boldsymbol{x}_{q}}=\sum_{i=1}^{p}e_{i}\partial_{x_{i}}+\frac{\boldsymbol{x}_{q}}{\lvert\boldsymbol{x}_{q}\lvert^{2}}\sum_{i=p+1}^{p+q}x_{i}\partial_{x_{i}}.

Apparently, this operator is not well-defined on the whole Euclidean space ℝp+q\mathbb{R}^{p+q} since it requires that |𝒙q|≠0\lvert\boldsymbol{x}_{q}\lvert\neq 0. In other words, the domain of the functions considered can not intersect ℝp\mathbb{R}^{p}. Hence, it is reasonable to consider a variant of the slice Dirac operator given as G𝒙†=|𝒙q|2D𝒙p+𝒙q𝔼𝒙q.G^{\dagger}_{\boldsymbol{x}}=\lvert\boldsymbol{x}_{q}\lvert^{2}D_{\boldsymbol{x}_{p}}+\boldsymbol{x}_{q}\mathbb{E}_{\boldsymbol{x}_{q}}. This operator also has an invariance property with respect to the group G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}). More specifically, we have

Theorem 6.

Let 𝐲=φ​(𝐱)=(a​𝐱+b)​(c​𝐱+d)−1∈G​R​A​V​(ℝp+q)\boldsymbol{y}=\varphi(\boldsymbol{x})=(a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}\in GRAV(\mathbb{R}^{p+q}), and let f∈C1​(U)f\in C^{1}(U), where UU is a domain in ℝp+q\mathbb{R}^{p+q}. Then, we have

G𝒚†​f​(𝒚)=(c​𝒙+d|c𝒙+d|p−1)−1​G𝒙†​c​𝒙+d~|c𝒙+d|p+1​f​((a​𝒙+b)​(c​𝒙+d)−1).\displaystyle G^{\dagger}_{\boldsymbol{y}}f(\boldsymbol{y})=\bigg{(}\frac{c\boldsymbol{x}+d}{\lvert c\boldsymbol{x}+d\lvert^{p-1}}\bigg{)}^{-1}G^{\dagger}_{\boldsymbol{x}}\frac{\widetilde{c\boldsymbol{x}+d}}{\lvert c\boldsymbol{x}+d\lvert^{p+1}}f((a\boldsymbol{x}+b)(c\boldsymbol{x}+d)^{-1}). (7)

Proof: The strategy is similar to that used in Theorem 1. We only need to show (7) is true for the four basic transformations respectively. Since G𝒙†=|𝒙q|2G𝒙G^{\dagger}_{\boldsymbol{x}}=\lvert\boldsymbol{x}_{q}\lvert^{2}G_{\boldsymbol{x}}, we can applied the result obtained in Theorem 1 here. More specifically,

  1. 1.

    𝒚=φ​(𝒙)=𝒙+b,b∈ℝp\boldsymbol{y}=\varphi(\boldsymbol{x})=\boldsymbol{x}+b,\ b\in\mathbb{R}^{p}, in this case, a=1,c=0,d=1a=1,\ c=0,\ d=1.

    Since |𝒚q|2=|𝒙q|2\lvert\boldsymbol{y}_{q}\lvert^{2}=\lvert\boldsymbol{x}_{q}\lvert^{2} under this type of translations and G𝒚=G𝒙G_{\boldsymbol{y}}=G_{\boldsymbol{x}}, we have

    G𝒚†=|𝒚q|2G𝒚=|𝒙q|2G𝒙=G𝒙†.\displaystyle G^{\dagger}_{\boldsymbol{y}}=\lvert\boldsymbol{y}_{q}\lvert^{2}G_{\boldsymbol{y}}=\lvert\boldsymbol{x}_{q}\lvert^{2}G_{\boldsymbol{x}}=G^{\dagger}_{\boldsymbol{x}}.
  2. 2.

    𝒚=φ​(𝒙)=a​𝒙​a−1,a∈𝕊q−1\boldsymbol{y}=\varphi(\boldsymbol{x})=a\boldsymbol{x}a^{-1},\ a\in\mathbb{S}^{q-1}, which gives us that 𝒚p=−𝒙p\boldsymbol{y}_{p}=-\boldsymbol{x}_{p} and 𝒚q=a​𝒙q​a−1\boldsymbol{y}_{q}=a\boldsymbol{x}_{q}a^{-1}. In this case, b=c=0,d=a−1b=c=0,\ d=a^{-1}. Since |𝒚q|2=|𝒙q|2\lvert\boldsymbol{y}_{q}\lvert^{2}=\lvert\boldsymbol{x}_{q}\lvert^{2} under this type of reflection and G𝒚=a​G𝒙​a−1G_{\boldsymbol{y}}=aG_{\boldsymbol{x}}a^{-1}, we have

    G𝒚†\displaystyle G^{\dagger}_{\boldsymbol{y}} =|𝒚q|2G𝒚=|𝒙q|2aG𝒙a−1=aG𝒙†a−1.\displaystyle=\lvert\boldsymbol{y}_{q}\lvert^{2}G_{\boldsymbol{y}}=\lvert\boldsymbol{x}_{q}\lvert^{2}aG_{\boldsymbol{x}}a^{-1}=aG^{\dagger}_{\boldsymbol{x}}a^{-1}.

    To see the equation above is in the form of (3), we only need to see that a−1~=a\widetilde{a^{-1}}=a and |a|=1\lvert a\lvert=1 for a∈𝕊q−1a\in\mathbb{S}^{q-1}.

  3. 3.

    𝒚=φ​(𝒙)=λ2​𝒙,λ∈ℝ\{0}\boldsymbol{y}=\varphi(\boldsymbol{x})=\lambda^{2}\boldsymbol{x},\ \lambda\in\mathbb{R}\backslash\{0\}, in this case, a=λ=d−1,b=c=0a=\lambda=d^{-1},\ b=c=0.

    It is also easy to see that ∂yi=λ−2​∂xi\partial_{y_{i}}=\lambda^{-2}\partial_{x_{i}}, and these give us

    G𝒚†=|𝒚q|2∑i=1pei∂yi+𝒚q∑i=p+1p+qyi∂yi=λ2G𝒙†.\displaystyle G^{\dagger}_{\boldsymbol{y}}=\lvert\boldsymbol{y}_{q}\lvert^{2}\sum_{i=1}^{p}e_{i}\partial_{y_{i}}+\boldsymbol{y}_{q}\sum_{i=p+1}^{p+q}y_{i}\partial_{y_{i}}=\lambda^{2}G^{\dagger}_{\boldsymbol{x}}.
  4. 4.

    𝒚=φ​(𝒙)=−𝒙−1=𝒙|𝒙|2\boldsymbol{y}=\varphi(\boldsymbol{x})=-\boldsymbol{x}^{-1}=\displaystyle\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{2}}, in this case, a=d=0,b=−c=1a=d=0,\ b=-c=1. In other words, we need to show that

    (𝒙|𝒙|p−1)−1​G𝒙†​𝒙|𝒙|p+1=G𝒚†.\displaystyle\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p-1}}\bigg{)}^{-1}G^{\dagger}_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}=G^{\dagger}_{\boldsymbol{y}}.

    First, we notice that G𝒙†=|𝒙q|2G𝒙G^{\dagger}_{\boldsymbol{x}}=\lvert\boldsymbol{x}_{q}\lvert^{2}G_{\boldsymbol{x}}, hence with the results in Theorem 1, we have that

    G𝒚†=|𝒚q|2G𝒚=|𝒚q|2(𝒙|𝒙|p+3)−1G𝒙𝒙|𝒙|p+1=|𝒙q|2|𝒙|4(𝒙|𝒙|p+3)−1G𝒙𝒙|𝒙|p+1\displaystyle G^{\dagger}_{\boldsymbol{y}}=\lvert\boldsymbol{y}_{q}\lvert^{2}G_{\boldsymbol{y}}=\lvert\boldsymbol{y}_{q}\lvert^{2}\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3}}\bigg{)}^{-1}G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}=\frac{\lvert\boldsymbol{x}_{q}\lvert^{2}}{\lvert\boldsymbol{x}\lvert^{4}}\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+3}}\bigg{)}^{-1}G_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}}
    =(𝒙|𝒙|p−1)−1​G𝒙†​𝒙|𝒙|p+1,\displaystyle=\bigg{(}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p-1}}\bigg{)}^{-1}G^{\dagger}_{\boldsymbol{x}}\frac{\boldsymbol{x}}{\lvert\boldsymbol{x}\lvert^{p+1}},

    which completes the proof.

Remark 5.

From the proof above, we can see the reason that G𝐱†G_{\boldsymbol{x}}^{\dagger} is invariant under the transformations in G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}) is the following: since G𝐱†=|𝐱q|2G𝐱G_{\boldsymbol{x}}^{\dagger}=\lvert\boldsymbol{x}_{q}\lvert^{2}G_{\boldsymbol{x}} and G𝐱G_{\boldsymbol{x}} is invariant under the transformations in G​R​A​V​(ℝp+q)GRAV(\mathbb{R}^{p+q}), hence we only need to modify the left weight function of G𝐱G_{\boldsymbol{x}} given in Theorem 1 regarding the term |𝐱q|2\lvert\boldsymbol{x}_{q}\lvert^{2}. However, this also indicates that (G𝐱†)l,l>1(G_{\boldsymbol{x}}^{\dagger})^{l},\ l>1 does not preserve this invariant property anymore because of the interaction between G𝐱G_{\boldsymbol{x}} and |𝐱q|2\lvert\boldsymbol{x}_{q}\lvert^{2} when we consider the iterated cases.

\bmhead

Acknowledgments

The authors are grateful to the referee for helpful comments. Chao Ding is supported by the National Natural Science Foundation (NNSF) of China (No. 12271001) and the Natural Science Foundation of Anhui Province (No. 2308085MA03). Zhenghua Xu is supported by the National Natural Science Foundation (NNSF) of China (No. 11801125) and the Natural Science Foundation of Anhui Province (No. 2308085MA04).

Declarations

No potential conflict of interest was reported by the authors.

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