Inversions for the Hua-Radon and the polarized Hua-Radon transform
aa r X i v : . [ m a t h . C A ] F e b INVERSIONS FOR THE HUA-RADON AND THE POLARIZEDHUA-RADON TRANSFORM
TEPPO MERTENS AND FRANK SOMMEN
Abstract.
The Hua-Radon and polarized Hua-Radon transform are two or-thogonal projections defined on holomorphic functions in the Lie sphere. Bothtransformations can be written as integral transforms with respect to a suitablereproducing kernel. Integrating both kernels over a Stiefel manifold yields alinear combination of zonal spherical monogenics. Using an Almansi type de-composition of holomorphic functions and reproducing properties of the zonalmonogenics, we obtain an inversion formula for both the Hua-Radon and thepolarized Hua-Radon transform.
Contents
1. Introduction 12. Preliminaries 22.1. Clifford algebras 32.2. Clifford analysis 42.3. Functions over the Lie Sphere 63. Inversion of the Hua-Radon transform 73.1. The Hua-Radon transform 73.2. Some technical results 83.3. The inversion 144. Inversion of the polarized Hua-Radon transform 164.1. The polarized Hua-Radon transform 164.2. The dual transform of the kernel 174.3. The inversion 185. Conclusions 21References 211.
Introduction
The Szeg˝o-Radon transform was defined by Sabadini and Sommen in [6] as anorthogonal projection of left monogenic functions onto a subspace ML ( τ ). TheSzeg˝o-Radon transform is a variant of the Clifford Radon transform which wasestablished in [5, 19, 20, 22]. Using techniques of Clifford analysis, a reproducingkernel for the basis of ML ( τ ) was obtained. This kernel was then used to write Date : February 19, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Holomorphic functions, Monogenic functions, Lie ball, Lie sphere,Radon-type transforms. the Szeg˝o-Radon transform as an integral transform over the m -dimensional unitsphere S m − . Finally, the dual Radon transform was defined as the integral overa Stiefel manifold of a function depending on two orthogonal unit vectors. Thisdual transform was the key to invert the Szeg˝o-Radon transform, since applyingit termwise to the kernel of the Szeg˝o-Radon transform yields a scalar multipleof the zonal spherical monogenics. This scalar coefficient was reformulated as anoperator using the Gamma operator Γ x . This has lead to an inversion formula forthe Szeg˝o-Radon transform.Sabadini and Sommen recently extended the concept of the Szeg˝o-Radon trans-form to the Lie sphere [16]. This was motivated by the fact that monogenic functionsadmit a holomorphic extension in the Lie ball, see e.g. [13, 15, 17]. Multiple kindsof mutually interrelated Radon type transforms were defined. These transformswere also reformulated as an integral transform with respect to a certain kernel,but this time the Lie sphere was considered as the integration domain.A crucial development which has not yet appeared in the literature on theseRadon type transforms is the description of their inversion. In this paper we willoutline how inversion formulas can be obtained for two important Radon typetransforms, namely the Hua–Radon and the polarized Hua–Radon transform. Asimilar analysis on yet another transform named after Hua, namely the monogenicHua–Radon transform, was conducted by the authors in [7]. We will use the tech-niques of [6, 7] also in the present setting.We start our exposition with some preliminaries on Clifford analysis. In Section3 we define Hua-Radon transform as an orthogonal projection onto a subspace ofholomorphic functions over the Lie sphere. This orthogonal projection is then writ-ten as an integral transform with respect to a reproducing kernel K τ . In Proposition3.5, we show that the zonal spherical harmonics K m,k ( x, y ) are spin-invariant, k -homogeneous and harmonic polynomials with respect to both x and y and moreoverthey are unique with this property up to a scalar multiple. Since each of the basiselements of OL ( τ ) is a null-solution of some power of the complexified Laplaceoperator ∆ z = P mj =1 ∂ z j , we first show that the dual Radon transform appliedto each of the terms of K τ is a linear combination of the zonal spherical harmon-ics. Decomposing the zonal spherical harmonics in terms of the zonal sphericalmonogenics, we obtain an inversion formula for holomorphic functions over the Liesphere in Theorem 3.16. Quintessential in this respect is the fact that holomorphicfunctions admit an Almansi decomposition [1, 2, 10, 14, 21].In Section 4 we recall the definition of the polarized Hua-Radon transform, whichis an orthogonal projection onto a subspace spanned by null-solutions of a power ofthe Dirac operator ∂ z = P mj =1 e j ∂ z j . Writing this projection as an integral trans-form with kernel L τ and using the results of Section 3, we can formulate the dualtransform of each of the terms of L τ as a linear combination of the zonal sphericalmonogenics C m,k ( x, y ). Finally, an inversion formula is determined in Theorem 4.5,relying again on the Almansi decomposition for holomorphic functions.2. Preliminaries
In this section we introduce all notations and preliminary results that will beuseful for the paper. We mostly follow the notations from [16].
NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 3
Clifford algebras.
Let R m denote the real vector space with basis ( e , e , . . . , e m ).Throughout the paper we will assume that m ≥
3. We define the real Clifford al-gebra R m as the real algebra generated by the basis elements e , e , . . . , e m whichsatisfy the following relations e j = − , j ∈ { , . . . , m } , (2.1) e j e k + e k e j = 0 , j = k. (2.2)The complex Clifford algebra C m is the complex algebra generated by e , e , . . . , e m which satisfy (2.1) and (2.2). Any element α ∈ R m (respectively α ∈ C m ), can bewritten as α = X A ⊂{ ,...,m } α A e A with α A ∈ R (or α A ∈ C ), A = { i , . . . , i ℓ } is a multi-index for which i < . . . < i ℓ ,such that e A = e i . . . e i ℓ and e ∅ = 1. A 1-vector is a linear combination of onlybasis vectors e j . We will denote the k -vector part of a real or complex Cliffordelement α by [ α ] k , i.e. [ α ] k = X A ⊂{ ,...,m }| A | = k α A e A We will be using the Hermitian conjugation, which is an automorphism on C m defined for α, β ∈ C m as ( αβ ) † = β † α † , ( α + β ) † = α † + β † , ( α A e A ) † = α A e † A ,e † j = − e j j ∈ { , . . . , m } , where α A is the complex conjugate of α A ∈ C . Note that for A = { i , . . . , i k } ⊂{ , . . . , m } , we have e † A e A = ( − k e i k . . . e i e i . . . e i k = ( − k = 1(2.3)and thus we have for any Clifford element α = P A ⊂{ ,...,m } α A e A (2.4) (cid:2) α † α (cid:3) = X A ⊂{ ,...,m } α A α A . The following result, proven in [6], will be useful.
Lemma 2.1.
Let t, s ∈ S m − be such that h t, s i = 0 and let τ = t + is ∈ C m . Then τ † = − t + is and(i) τ τ † τ = 4 τ ,(ii) τ = ( τ † ) = 0 ,(iii) τ τ † + τ † τ = 4 . TEPPO MERTENS AND FRANK SOMMEN
Clifford analysis.
The scalar product of two 1-vectors u and v is given by h u, v i = P mj =1 u j v j and the wedge-product by u ∧ v = P i 1) denote the unit ball with center at the origin in R m , whereas the unitsphere will be denoted by S m − , i.e. S m − = (cid:8) u ∈ R m | | u | = 1 (cid:9) . The area of theunit sphere is given by A m = 2 π m/ Γ (cid:0) m (cid:1) , where Γ is the gamma function.The standard Dirac operator is given by ∂ x = m X j =1 ∂ x j . Using (2.1) and (2.2), we get that the square of the Dirac operator satisfies ∂ x = − ∆ x , where ∆ x = m P j =1 ∂ x j is the Laplace operator. The symbol of the Dirac operator ∂ x is denoted by the vector variable x = m X j =1 e j x j Definition 2.2. A function f : Ω ⊂ R m → C m which is continuously differentiablein the open set Ω is called (left) monogenic in Ω if f is in the kernel of the Diracoperator ∂ x , i.e. ∂ x f = P mj =1 e j ( ∂ x j f ( x )) = 0. The right C m -module of (left)monogenic functions in Ω is denoted by M (Ω).A function f : Ω ⊂ R m → C m which is continuously differentiable in the open setΩ is called harmonic in Ω if f is in the kernel of the Laplace operator ∆ x = − ∂ x .The right C m -module of harmonic functions in Ω is denoted by H (Ω). Remark . Note that there is also the notion of a right monogenic, i.e. a function f for which f ( x ) ∂ x = m X j =1 ( ∂ x j f ( x )) e j = 0 . The Gamma-operator Γ x is given byΓ x = − x∂ x − E x , NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 5 where E x is the Euler operator, which is defined as E x = m X j =1 x j ∂ x j . Let P ( R m ) be the space of polynomials in m variables x , . . . , x m with coefficientsin R . Using the Euler operator we can define the space P k ( R m ) of k -homogeneouspolynomials as P k ( R m ) = { P ( x ) ∈ P ( R m ) | E x P ( x ) = kP ( x ) } . which leads to the definition of the space of monogenic polynomials of degree k M k ( R m ) = ( P k ( R m ) ⊗ C m ) ∩ M ( R m )and the space of spherical harmonics of degree k as H k ( R m ) = ( P k ( R m ) ⊗ C m ) ∩ H ( R m ) . We will usually write P k , M k and H k for simplicity. Remark . The vectors x and y will denote real-valued variables, whereas z willdenote a complex-valued variable. This also applies to any of the operators definedabove. Note that if we use a complex variable z we complexify the operator. Forexample: we work with the complexified version of the Laplace operator∆ z = m X j =1 ∂ z j . If we talk about monogenic or harmonic functions in the complex case, we meannull-solutions of the complexified Dirac operator ∂ z or the complexified Laplaceoperator ∆ z . The space of polynomials of degree k monogenic with respect to ∂ z is denoted by M k ( C m ) and the corresponding space of harmonics by H k ( C m ).The space of k -homogeneous polynomials P k can be decomposed in terms ofthese monogenic polynomials and harmonic polynomials in what is known as theharmonic Fischer decomposition (see e.g. [9]):(2.6) P k ( R m ) ⊗ R m = ⌊ k ⌋ M j =0 | x | j H k − j . Moreover we can decompose H k into H k = M k ⊕ x M k − . Proposition 2.5. Let H k ( x ) ∈ H k . Then we have H k ( x ) = M k ( x ) + xM k − ( x ) ,with M k − ( x ) = − k + m − ∂ x H k ( x ) ,M k ( x ) = (cid:18) k + m − x∂ x (cid:19) H k ( x ) and M k − ∈ M k − , M k ∈ M k . TEPPO MERTENS AND FRANK SOMMEN Combining Proposition 2.5 and (2.6) leads to the monogenic Fischer decompo-sition:(2.7) P k ( R m ) ⊗ R m = k M j =0 x j M k − j . For the decomposition in (2.6), there is a projection operator in order to deter-mine each term of the decomposition (see e.g. [3]): Proposition 2.6. The projection operator of a k -homogeneous polynomial onto itsharmonic component of degree k − ℓ is given by the following operatorProj k,l = ⌊ k ⌋ − ℓ X j =0 α j,k,ℓ | x | j ∆ j + ℓx where α j,k,ℓ = ( − j ( m + k − ℓ − j + l j ! ℓ ! Γ( m + k − ℓ − j − m + k − ℓ ) . Functions over the Lie Sphere. The aim of the paper is to invert the Hua-Radon transform and the polarized Hua-Radon transform, which were defined in[16] as an orthogonal projection of functions over the Lie sphere. The definition ofthe Lie sphere can be found in e.g. [21]. Let us recall the necessary definitions andresults that we will need. Definition 2.7. The Lie ball LB (0 , 1) is defined as LB (0 , 1) = { z = x + iy ∈ C m | S x,y ⊂ B (0 , } where S x,y is the codimension 2 sphere: S x,y = { u ∈ R m | | u − x | = | y | , h u − x, y i = 0 } . Definition 2.8. The Lie sphere LS m − is given by LS m − = { e iθ ω ∈ C m | ω ∈ S m − , θ ∈ [0 , π ) } . We will be working with a space of holomorphic functions on the Lie sphere: Definition 2.9. The right C m -module containing all holomorphic functions f : LB (0 , → C m such that (cid:20)Z S m − Z π (cid:2) f ( e iθ ω ) (cid:3) † f ( e iθ ω ) dθdS ( ω ) (cid:21) < ∞ will be denoted by OL ( LB (0 , OL ( LB (0 , h f, g i OL ( LB (0 , = Z S m − Z π (cid:2) f ( e iθ ω ) (cid:3) † g ( e iθ ω ) dθdS ( ω ) . We will often write h f, g i OL for the inner product of f and g . NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 7 It was shown in [18, 21] that any f ∈ OL ( LB (0 , f ( z ) = ∞ X k =0 M k ( z ) + z ∞ X ℓ =0 N ℓ ( z )where M k ∈ M k ( C m ), N ℓ ∈ M l ( C m ). A detailed account on the Almansi decom-position can be found in [1, 2, 10, 14].3. Inversion of the Hua-Radon transform The Hua-Radon transform. Let us first recall the definition of the Hua-Radon transform which was introduced in [16].Let τ = t + is with t, s ∈ S m − , h t, s i = 0 and let f τ,k,ℓ ( z ) = h z, τ i k h z, τ † i ℓ . Note that the functions f τ,k,ℓ ( z ) are null-solutions of ∆ min { k,ℓ } +1 z .The closure of the C m -submodule of OL ( LB (0 , { f τ,k,ℓ ( z ) | k, ℓ ∈ N } will be denoted by OL ( τ ).The functions f τ,k,ℓ ( z ) form an orthogonal basis for OL ( τ ) with respect to theinner product h· , ·i OL . We have the following result from [16]: Proposition 3.1. Let τ = t + is , where t, s ∈ S m − and h t, s i = 0 . The functions f τ,k,ℓ ( z ) are such that h f τ,k,ℓ , f τ,k ′ ,ℓ ′ i OL = 0 ( k, ℓ ) = ( k ′ , ℓ ′ ) and h f τ,k,ℓ , f τ,k,ℓ i OL = 2 π m − Γ( k + l + 1)Γ( k + l + m/ . The Hua-Radon transform is defined as the orthogonal projection H τ : OL ( LB (0 , → OL ( τ ) : f Z S m − Z π K τ ( z, e − iθ ω ) f ( e iθ ω ) dθdS ( ω )with K τ ( z, e − iθ ω ) = 1 πA m ∞ X k,ℓ =0 ( − k + ℓ Γ (cid:0) k + ℓ + m (cid:1) Γ( k + ℓ + 1)Γ (cid:0) m (cid:1) a k b ℓ = 1 πA m ∞ X s =0 s X k =0 ( − s Γ (cid:0) s + m (cid:1) Γ( s + 1)Γ (cid:0) m (cid:1) a s − k b k , where a = h z, τ ih e − iθ ω, τ † i and b = h z, τ † ih e − iθ ω, τ i . The function K τ ( z, e − iθ ω ) isthe reproducing kernel of the functions f τ,k,ℓ ( uz ), i.e. Z S m − Z π K τ ( z, e − iθ ω ) f τ,k,ℓ ( e iθ ω ) dθdS ( ω ) = f τ,k,ℓ ( z ) TEPPO MERTENS AND FRANK SOMMEN Some technical results. In order to invert the Hua-Radon transform, wewill integrate its kernel K τ over a Stiefel manifold, which leads to the definition ofthe dual Radon transform: Definition 3.2. The dual Radon transform ˜ R [ F ( z, τ )] is defined as˜ R [ F ( z, τ )] = 1 A m A m − Z S m − (cid:18)Z S m − F ( z, τ ) dS ( s ) (cid:19) dS ( t ) , where S m − ⊆ S m − is the ( m − t .We will also need some results of zonal spherical harmonics and zonal sphericalmonogenics. The zonal spherical harmonics are given by, see e.g. [8] K m,k ( x, y ) = 2 k + m − m − | x | k | y | k C m − k ( t ) , where t = h x,y i| x || y | and C m − k ( t ) is a Gegenbauer polynomial. These zonal sphericalharmonics have the property of reproducing harmonic polynomials, i.e.(3.1) 1 A m Z S m − K m,j ( x, ω ) H k ( ω ) dS ( ω ) = δ j,k H x ( x )where H k ∈ H k . The zonal spherical monogenics are given by (see [8]) C m,k ( x, y ) = ( | x || y | ) k (cid:18) k + m − m − C m − k ( t ) + x ∧ y | x || y | C m k − ( t ) (cid:19) = ( | x || y | ) k (cid:18) C m k ( t ) + xy | x || y | C m k − ( t ) (cid:19) . and likewise, they are the reproducing kernel of the monogenic polynomials, i.e. for M k ∈ M k we have(3.2) 1 A m Z S m − C m,j ( x, ω ) M k ( ω ) dS ( ω ) = δ j,k M x ( x ) . Remark . Note that the reproducing properties (3.1) and (3.2) also hold whenintegrating over the Lie sphere, i.e.1 πA m Z S m − Z π K m,j ( z, e − iθ ω ) H k ( e iθ ω ) dθdS ( ω ) = δ jk H k ( z )and 1 πA m Z S m − Z π C m,j ( z, e − iθ ω ) M k ( e iθ ω ) dθdS ( ω ) = δ jk M k ( z ) , see [16, Remark 3.6].In [6] it is shown that these zonal spherical monogenics are, up to a constantfactor λ + µe ...m with λ, µ ∈ C , the unique k -homogeneous polynomials satisfyingcertain properties as mentioned in Proposition 3.4. Proposition 3.4. Let F ( x, y ) be a Clifford valued function which is a homogeneouspolynomial of degree k in both x and y . If F is left-monogenic in x and right-monogenic in y , i.e. ∂ x F = 0 = F ∂ y and if F is also spin-invariant, i.e. for σ ∈ Spin( m ) = { Q si =1 σ i | σ i ∈ S m − } we have σF ( σxσ, σyσ ) σ = F ( x, y ) . NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 9 Then there exist complex constants λ and µ such that F ( x, y ) = ( λ + µe ...m ) C m,k ( x, y ) . We have a similar result if the function is harmonic in both x and y . Proposition 3.5. Let F ( x, y ) be a k -homogeneous polynomial in x and y that isharmonic in both x and y , i.e. ∆ x [ F ] = 0 = ∆ y [ F ] . If F is also spin-invariant,i.e. for σ ∈ Spin( m ) we have σF ( σxσ, σyσ ) σ = F ( x, y ) , then there exists complex constants ρ , ρ , ν , ν ∈ C such that F ( x, y ) = ( ρ + ν e ...m ) K m,k ( x, y ) + ( ρ + ν e ...m ) C m,k ( x, y ) . Proof. Using Proposition 2.5 for both x and y we get F ( x, y ) = M k,k ( x, y ) + xM k − ,k ( x, y ) + M k,k − ( x, y ) y + xM k − ,k − ( x, y ) y where M i,j ( x, y ) is a left-monogenic, i -homogeneous function with respect to x and right-monogenic, j -homogeneous function with respect to y . Since F is spin-invariant, we get for each σ ∈ Spin( m ) F ( x, y ) = σF ( σxσ, σyσ ) σ = σM k,k ( σxσ, σyσ ) σ + xσM k − ,k ( σxσ, σyσ ) σ + σM k,k − ( σxσ, σyσ ) σy + xσM k − ,k − ( σxσ, σyσ ) σy. as σσ = 1. In [12] it was shown that ∂ x is spin-invariant and hence σM i,j ( σxσ, σyσ ) σ is monogenic. Now using the uniqueness of the decomposition in Proposition 2.5,we get σM i,j ( σxσ, σyσ ) σ = M i,j ( x, y ) i, j ∈ { k, k − } . Hence the functions M i,j only depend on | x | , | y | and h x, y i , see [6, Lemma 5.3]. Thelatter are of course invariant under the transformation x 7→ − x, y 7→ − y , whereason the other hand M i,j ( x, y ) is sent to ( − i + j M i,j ( x, y ) by this transformation.Consequently, we must have that M k,k − ( x, y ) = M k − ,k ( x, y ) = 0, and thus wefind F ( x, y ) = M k,k ( x, y ) + xM k − ,k − ( x, y ) y where M k,k and M k − ,k − are homogeneous polynomials, left-monogenic in x andright-monogenic in y and they are spin-invariant. Using Proposition 3.4, we get F ( x, y ) = α C m,k ( x, y ) + βx C m,k − ( x, y ) y = α ( | x || y | ) k (cid:18) C m k ( t ) + xy | x || y | C m k − ( t ) (cid:19) + βx ( | x || y | ) k − (cid:18) C m k − ( t ) + xy | x || y | C m k − ( t ) (cid:19) y =( α + β )( | x || y | ) k (cid:18) C m k ( t ) + xy | x || y | C m k − ( t ) (cid:19) (3.3) − β ( | x || y | ) k (cid:16) C m k ( t ) − C m k − ( t ) (cid:17) =( α + β ) C m,k ( x, y ) − β k + m − m − | x || y | ) k C m − k ( t )=( α + β ) C m,k ( x, y ) − βK m,k ( x, y ) , where α = a + b e ...n and β = a + b e ...n with a , a , b , b ∈ C and in thesecond to last line we used that (see e.g. [8]) C m k ( t ) − C m k − ( t ) = 2 k + m − m − C m − k ( t ) . (cid:3) Remark . Note that if F is a scalar valued function with the same properties asin Proposition 3.5, then F ( x, y ) = αK m,k ( x, y ) with α ∈ C . Remark . Using the equations in (3.3), we can easily see that K m,k +1 ( x, y ) = C m,k +1 ( x, y ) − x C m,k ( x, y ) y. Using Proposition 3.4 and Proposition 3.5 we will prove that the dual Radontransform of each of the terms of the kernel of the Hua-Radon transform are equalto a linear combination of the zonal spherical harmonics. Recall that when we areworking in the complex case, we are complexifying our operators. Hence it sufficesto show this equality for real variables. Afterwards, we use Remark 3.7 and thereproducing properties (3.2) to find an inversion formula.We have the following: Proposition 3.8. Suppose that F ( x, y ) is a homogeneous polynomial of degree k + ℓ in both x and y , then F ( x, y ) = ⌊ k + ℓ ⌋ X j =0 ⌊ k + ℓ ⌋ X j ′ =0 | x | j H k + ℓ − j,k + ℓ − j ′ ( x, y ) | y | j ′ where H k + ℓ − j,k + ℓ − j ′ ( x, y ) is a spherical harmonic of degree k + ℓ − j in x andof degree k + ℓ − j ′ in y .Proof. The proof follows by applying the harmonic Fischer decomposition (2.6) firstin x and then in y . (cid:3) NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 11 Remark . Let F ( x, y ) be a homogeneous polynomial of degree k + l in both x and y . If ∆ r +1 x [ F ( x, y )] = ∆ s +1 x [ F ( x, y )] = 0 and ∆ rx [ F ( x, y )] = ∆ sx [ F ( x, y )] = 0,then F ( x, y ) = r X j =0 s X j ′ =0 | x | j H k + ℓ − j,k + ℓ − j ′ ( x, y ) | y | j ′ Lemma 3.10. For j ≤ min { k, ℓ } we have ∆ jx (cid:2) h x, τ i k h x, τ † i ℓ (cid:3) = ( − j k !( k − j )! ℓ !( ℓ − j )! h x, τ i k − j h x, τ † i ℓ − j . If j > min { k, ℓ } then ∆ jx (cid:2) h x, τ i k h x, τ † i ℓ (cid:3) = 0 . Proof. Using Lemma 2.1 one can verify that∆ x (cid:2) h x, τ i k h x, τ † i ℓ (cid:3) = 2 kℓ h x, τ i k − h x, τ † i ℓ − h τ , τ † i = − kℓ h x, τ i k − h x, τ † i ℓ − . Hence repeating this process gives∆ jx (cid:2) h x, τ i k h x, τ † i ℓ (cid:3) = ( − j k !( k − j )! ℓ !( ℓ − j )! h x, τ i k − j h x, τ † i ℓ − j . If j > min { k, ℓ } then it suffices to see that∆ x (cid:2) h x, τ i k − j (cid:3) = ( k − j )( k − j − h x, τ i k − j − m X i =1 τ i = ( k − j )( k − j − h x, τ i k − j − ( − τ )= 0where we used Lemma 2.1. Analogously we get ∆ x (cid:2) h x, τ † i ℓ − j (cid:3) = 0. (cid:3) Corollary 3.11. Let K m,j be the zonal spherical harmonic of degree j and let k ≥ l . Then for τ = t + is , there exists constants θ j,k,l such that A m A m − Z S m − Z S m − h x, τ i k h x, τ † i ℓ h y, τ i ℓ h y, τ † i k dS ( s ) dS ( t )= ℓ X j =0 θ j,k,ℓ | x | j K m,k + ℓ − j ( x, y ) | y | j . Proof. If we define the integral on the left-hand side in Corollary 3.11 as L k,ℓ ( x, y ),then we see that this integral is homogeneous of degree k + ℓ in both x and y . UsingLemma 3.10 we see that L k,ℓ is a solution of ∆ ℓ +1 [ F ] = 0 and not of ∆ ℓ [ F ] = 0.Hence applying Proposition 3.8 yields L k,ℓ ( x, y ) = ℓ X j =0 ℓ X j ′ =0 | x | j H k + ℓ − j,k + ℓ − j ′ ( x, y ) | y | j ′ . If we interchange x and y , one can easily see that L k,ℓ ( y, x ) = (cid:2) L k,ℓ ( x, y ) (cid:3) † andtherefore ℓ X j =0 ℓ X j ′ =0 | y | j H k + ℓ − j,k + ℓ − j ′ ( y, x ) | x | j ′ = ℓ X j =0 ℓ X j ′ =0 (cid:16) | x | j H k + ℓ − j,k + ℓ − j ′ ( x, y ) | y | j ′ (cid:17) † = ℓ X j =0 ℓ X j ′ =0 | y | j (cid:2) H k + ℓ − j,k + ℓ − j ′ ( x, y ) (cid:3) † | x | j ′ . By uniqueness of the Fischer decomposition, we get H k + ℓ − j,k + l − j ′ ( y, x ) = (cid:2) H k + ℓ − j,k + l − j ′ ( x, y ) (cid:3) † . But the left-hand side is a polynomial of degree k + ℓ − j in y and of degree k + ℓ − j ′ in x , whereas the right-hand side is a polynomial of degree k + ℓ − j ′ in y and of degree k + ℓ − j in x . Consequently, H k + ℓ − j,k + ℓ − j ′ ( y, x ) = 0 whenever j = j ′ . Hence the remaining decomposition is L k,ℓ ( x, y ) = ℓ X j =0 | x | j H k + ℓ − j,k + ℓ − j ( x, y ) | y | j . Furthermore, let σ ∈ Spin( m ). First of all note that Spin( m ) / {− , } ≃ SO ( m )and σxσ is just a rotation of the vector x , see [11, 12]. Hence σL k,ℓ ( σxσ, σyσ ) σ = Z S m − Z S m − h x, στ σ i k h x, στ † σ i ℓ h y, στ σ i ℓ h y, στ † σ i k dS ( s ) dS ( t )= Z S m − Z S m − h x, τ i k h x, τ † i ℓ h y, τ i ℓ h y, τ † i k dS ( s ) dS ( t ) . Hence L k,ℓ ( x, y ) is Spin-invariant, which implies that H k + ℓ − j,k + ℓ − j ( x, y ) is Spin-invariant for each j . Note that the integral L k,l is scalar valued. This means that,using Remark 3.6, we can find constants θ j,k,ℓ ∈ C such that H k + l − j,k + ℓ − j ( x, y ) = θ j,k,ℓ K m,k + ℓ − j ( x, y ) . (cid:3) Remark . Note that if ℓ ≥ k in Corollary 3.11, then the result stays the same,but the upper limit of the sum will be k instead of ℓ .The following result from [6] will be useful to calculate θ n,k,ℓ . Lemma 3.13. Let ω ∈ S m − and m ≥ , then A m A m − Z S m − Z S m − h ω, τ i j h ω, τ † i j dS ( s ) dS ( t ) = ( − j A m − πA m Γ( m − j + 1)Γ( m + j )= ( − j Γ( m )Γ( j + 1)Γ( m + j ) . NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 13 Proposition 3.14. The coefficients θ n,k,ℓ are given by: θ n,k,ℓ = ( − k + ℓ Γ( m − (cid:0) m + k − n − (cid:1) Γ (cid:0) m + ℓ − n − (cid:1) ( m − ℓ − n )!Γ (cid:0) m + k + ℓ − n (cid:1) Γ( m + k + ℓ − n − (cid:18) k ! ℓ ! n ! (cid:19) ( k + ℓ − n )!( k − n )! . Proof. Let L k,ℓ denote the left-hand side of the equation in Corollary 3.11 andassume ℓ ≤ k . By Lemma 3.10, applying ∆ jx with j ≤ ℓ yields∆ jx [ L k,ℓ ] =( − j k !( k − j )! l !( ℓ − j )! 1 A m A m − × Z S m − Z S m − h x, τ i k − j h x, τ † i ℓ − j h y, τ i ℓ h y, τ † i k dS ( s ) dS ( t )and ∆ jx (cid:2) L k,l ( x, y ) (cid:3) = 0 when j > l . Hence by Proposition 2.6 we have l − n X j =0 α j,k + ℓ,n | x | j ∆ j + nx (cid:2) L k,ℓ ( x, y ) (cid:3) = θ n,k,ℓ K m,k + ℓ − n ( x, y ) | y | n . As θ n,k,ℓ is the same for all x, y , it suffices to consider the case where x = y = ω ∈ S m − . Using the fact that, see e.g. [4], C m − r (1) = Γ( m − r )Γ( m − r + 1)we get ℓ − n X j =0 α j,k + ℓ,n | x | j ∆ j + nx (cid:2) L k,ℓ ( x, y ) (cid:3) | x = y = ω = θ n,k,ℓ k + ℓ − n ) + m − m − m − k + ℓ − n )Γ( m − k + ℓ − n + 1) . Now using Lemma 3.13 we get ℓ − n X j =0 α j,k + l,n | x | j ∆ j + nx (cid:2) L k,ℓ ( x, y ) (cid:3) | x = y = ω = ℓ − n X j =0 ( − j ( m + k + ℓ − n − j + n j ! n ! Γ( m + k + ℓ − n − j − m + k + ℓ − n ) × ( − j + n k ! ℓ !( k − j − n )( ℓ − j − n )! ( − k + ℓ − j − n Γ( m )Γ( k + ℓ − j − n + 1)Γ( m + k + ℓ − j − n )= ( m + k + ℓ − n − m )( − k + ℓ k ! ℓ ! n !Γ( m + k + ℓ − n ) × ℓ − n X j =0 ( − j Γ( m + k + ℓ − n − j − k + ℓ − j − n + 1) j !Γ( m + k + ℓ − j − n )( k − j − n )!( ℓ − j − n )! . (3.4)The summation (3.4) is actually a multiple of a hypergeometric function F : ℓ − n X j =0 ( − j Γ( m + k + ℓ − n − j − k + ℓ − j − n + 1) j !Γ( m + k + ℓ − j − n )( k − j − n )!( ℓ − j − n )! =Γ( m + k + ℓ − n − k + ℓ − n + 1)Γ( m + k + ℓ − n )( k − n )!( ℓ − n )! F ([ − p, a, b ] , [ c, d ]; 1) where p = ℓ − n,a = − m − k − ℓ + n + 1 ,b = − k + n,c = − k − ℓ + n,d = − m − k − ℓ + 2 n + 2 , = a + b − c + 1 − p. Using the Pfaff-Saalsch¨utz Balanced sum, see [4], we find F ([ − p, a, b ] , [ c, d ]; 1) = ( c − a ) p ( c − b ) p ( c ) p ( c − a − b ) p = ( m − ℓ − n ( − ℓ ) ℓ − n ( − k − ℓ + n ) ℓ − n ( m + k − n − ℓ − n = Γ( m + ℓ − n − m − 1) Γ( m + k − n − m + k + ℓ − n − ℓ ! n ! k !( k + ℓ − n )! . Now combining everything yields the result stated in the theorem. The case k ≤ ℓ is the same as interchanging k ↔ ℓ in the above calculations. Note that due to thesymmetry of θ n,k,l with respect to k and ℓ , the result does not change. (cid:3) The inversion. We now have the necessary tools to find an inversion formulafor the Hua-Radon transform. Theorem 3.15. Let f ∈ OL ( LB (0 , be of the form f ( z ) = z n H t − n ( z ) where H t − n ∈ H t − n ( C m ) is a spherical harmonic of degree t − n ≥ . Then ˜ R [ H τ [ f ]]( z ) = ϕ t,n f ( z ) with ϕ t,n = Γ( t + m )Γ( m − t − n )! Γ( t − n + m − t !Γ( m + t − n ) Γ( m + t − n − × F ([ n + 1 , n + 1 , n − t, m − , [ n − t, n − t, n − t − m . Proof. We can write the kernel of the Hua-Radon transform as K τ ( z, e − iθ ω ) = 1 πA m ∞ X s =0 s X k =0 ( − s Γ( s + m )Γ( m )Γ( s + 1) a s − k b k where a = h z, τ ih e − iθ ω, τ † i and b = h z, τ † ih e − iθ ω, τ i . Thus using Corollary 3.11 wehave NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 15 ˜ R [ H τ [ f ]]( z ) = 1 πA m A m − ∞ X s =0 s X k =0 ( − s Γ( s + m )Γ( m )Γ( s + 1) × Z S m − Z S m − Z S m − Z π a s − k b k ( e iθ ω ) n H t − n ( e iθ ω ) dθdS ( ω ) dS ( s ) dS ( t )= 1 πA m ∞ X s =0 s X k =0 min { k,s − k } X n ′ =0 θ n ′ ,s − k,k ( − s Γ( s + m )Γ( m )Γ( s + 1)(3.5) × Z S m − Z π z n ′ K m,s − n ′ ( z, e − iθ ω )( e − iθ ω ) n ′ ( e iθ ω ) n H t − n ( e iθ ω ) dθdS ( ω ) . If we now use the homogeneity of the zonal spherical harmonics and of H t − n , wecan rewrite the last integral in (3.5) as follows (cid:18)Z π e iθ ( t − s ) dθ (cid:19) Z S m − z n ′ K m,s − n ′ ( z, ω )( ω ) n ′ + n ) H t − n ( ω ) dS ( ω )= (cid:18)Z π e iθ ( t − s ) dθ (cid:19) Z S m − ( − n ′ + n z n ′ K m,s − n ′ ( z, ω ) H t − n ( ω ) dS ( ω ) . Therefore, if we use the reproducing property of the zonal spherical harmonic, weget that the above integral vanishes except when s − n ′ = t − n . But this meansthat 2 n − n ′ = t − s and thus, by looking at the integral over θ , we get thatthe only non-vanishing integral occurs when n ′ = n and consequently s = t . As n = n ′ ≤ min { k, t − k } , we have k ≥ n and k ≤ t − n . So now (3.5) becomes˜ R [ H τ [ f ]]( z ) = f ( z ) t − n X k = n θ n,t − k,k ( − t Γ( t + m )Γ( m )Γ( t + 1)= f ( z ) ψ t,n t − n X k = n ξ n,t − k,k where ψ t,n = Γ( t + m )Γ( m ) t ! ( m − m − t − n + 1)4Γ( m + t − n ) Γ( m + t − n − n !) ξ n,t − k,k = (( t − k )!) ( k !) Γ( t − k − n + m − k − n + m − k − n )!( t − k − n )! . We can rewrite this last summation as a multiple of a hypergeometric function: t − n X k = n ξ n,t − k,k = ( t − n )! n ! Γ( t − n + m − m − t − n )! × F ([ n + 1 , n + 1 , n − t, m − , [ n − t, n − t, n − t − m . Hence if we combine everything we get ϕ t,n = Γ( t + m )Γ( m − t − n )! Γ( t − n + m − t !Γ( m + t − n ) Γ( m + t − n − × F ([ n + 1 , n + 1 , n − t, m − , [ n − t, n − t, n − t − m . (cid:3) We can now use the fact that M t ( z ) and zM t − ( z ) are both spherical harmon-ics, where M t ( z ) ∈ M t ( C m ) , M t − ( z ) ∈ M t − ( C m ) are spherical monogenics, torewrite the result of Theorem 3.15 as˜ R [ H τ [ f ]]( z ) = ϕ t,n f ( z ) f ( z ) = z n M t − n ( z ) , (3.6) ˜ R [ H τ [ h ]]( z ) = ϕ t,n h ( z ) h ( z ) = z n +1 M t − n − ( z ) . (3.7)In doing so we can use E z (cid:2) z j M t − j ( z ) (cid:3) = tz j M t − j ( z ) , Γ z (cid:2) z n M t − n ( z ) (cid:3) = (2 n − t ) z n M t − n ( z ) , (3.8) Γ z (cid:2) z n +1 M t − n − ( z ) (cid:3) = ( t − n + m − z n +1 M t − n − ( z ) . Thus we can write ϕ t,n = ϕ E z , Γ z + E z in case (3.6) and ϕ t,n = ϕ E z , E z − Γ z + m − in case(3.7) so that it becomes an operator. Hence we have ϕ − E z , Γ z + E z ˜ R [ H τ [ g ]]( z ) = g ( z ) g ( z ) = z n M t − n ( z ) ,ϕ − E z , E z − Γ z + m − ˜ R [ H τ [ h ]]( z ) = h ( z ) h ( z ) = z n +1 M t − n − ( z ) . Since the Hua-Radon transform is defined for holomorphic functions, we can use(2.8), so that we can invert it on any holomorphic function. Theorem 3.16. Let f ( z ) = P ∞ k =0 M k ( z ) + z P ∞ ℓ =0 N ℓ ( z ) ∈ OL ( LB (0 , , with M k ∈ M k ( C m ) , N ℓ ∈ M ℓ ( C m ) , then f ( z ) = ϕ − E z , Γ z + E z ˜ R " H τ " ∞ X k =0 M k ( z ′ ) ( z )+ ϕ − E z , E z − Γ z + m − ˜ R " H τ " z ′ ∞ X ℓ =0 N ℓ ( z ′ ) ( z ) . Inversion of the polarized Hua-Radon transform The polarized Hua-Radon transform. In this section we determine theinversion of the polarized Hua-Radon transform which was defined in [16] as follows.Let τ = t + is with t, s ∈ S m − , h t, s i = 0 and let ψ τ, r,k ( z ) = τ h z, τ i r + k h z, τ † i r = τ f τ,r + k,r ( z ) ,ψ τ, r +1 ,k ( z ) = τ † τ h z, τ i r + k +1 h z, τ † i r = τ † τ f τ,r + k +1 ,r ( z ) . One can show that the functions ψ τ,α,k ( z ) are null-solutions of ∂ α +1 z (see [16]).For s ∈ N , we define the right C m -submodule M s ( τ ) of OL ( LB (0 , { ψ τ,s,k ( z ) | k ∈ N } . The spaces M s ( τ )are orthogonal with respect to h· , ·i OL . The space M ( τ ) is defined as the directorthogonal sum ⊕ ∞ s =0 M s ( τ ) (see [16]). NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 17 Remark . The space M ( τ ) was constructed in such a way that we can decompose OL ( τ ) into OL ( τ ) = M ( τ ) ⊕ M ( τ † )The polarized Hua-Radon transform is defined as the orthogonal projection R Hτ : OL ( LB (0 , → M ( τ ) : f Z S m − Z π L τ ( z, e − iθ ω ) f ( e iθ ω ) dθdS ( ω )with L τ ( z, e − iθ ω ) given by1 πA m ∞ X s =0 ∞ X k =0 ( − k Γ( k + 2 s + m )Γ( k + 2 s + 1)Γ( m ) a k + s b s (4.1) − πA m τ † τ ∞ X s =0 Γ(2 s + m )Γ(2 s + 1)Γ( m ) a s b s (4.2)where a = h z, τ ih e − iθ ω, τ † i , b = h z, τ † ih e − iθ ω, τ i . The functions ψ τ,α,k ( z ) arereproduced by the kernel L τ ( z, e − iθ ω ).4.2. The dual transform of the kernel. In order to invert the polarized Hua-Radon transform, we will integrate its kernel over a Stiefel manifold aiming toget a linear combination of the zonal spherical monogenics. Recall that we aretrying to use the reproducing properties of the zonal spherical monogenics to findan inversion formula for the polarized Hua-Radon transform. These reproducingproperties also hold when working over the Lie sphere, see Remark 3.3. Since this isjust a complexified version of the zonal spherical monogenics, it suffices to considerthe kernel evaluated in real variables and show equality in the real case.In order to evaluate the action of the dual Radon transform of (4.1), we can use theCorollary 3.11, obtained for the Hua-Radon transform in Section 3, and Proposition2.5. For (4.2), we have for a term a s b s (4.3) 1 A m A m − Z S m − Z S m − h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k τ † τ dS ( s ) dS ( t )where S m − is the unit sphere perpendicular to t . Now consider h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k τ † τ as a function f of the variable s . As τ † τ = 2 − its , we can write f ( s ) = f ( s )+ f ( s ),where f ( s ) = 2 h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k is the scalar part of f ( s ) and f ( s ) = − i h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k ts is its bivector part. We have the following: f ( − s ) = − i h x, t − is i k h x, − t − is i k h y, t − is i k h y, − t − is i k t ( − s )= ( − k +2 i h x, − t + is i k h x, t + is i k h y, − t + is i k h y, t + is i k ts = − f ( s ) . Hence f ( s ) is an odd function, therefore Z S m − f ( s ) dS ( s ) = 0 . Consequently we have1 A m A m − Z S m − Z S m − h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k τ † τ dS ( s ) dS ( t )= 2 A m A m − Z S m − Z S m − h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k dS ( s ) dS ( t )= 2 L k,k ( x, y )with L k,k ( x, y ) the integral defined in Corollary 3.11. Thus for (4.2) we have thefollowing result: Proposition 4.2. The dual Radon transform of a term of (4.2) is a linear combi-nation of the zonal spherical monogenics, i.e. A m A m − Z S m − Z S m − h x, τ i k h x, τ † i k h y, τ i k h y, τ † i k τ † τ dS ( s ) dS ( t )= k X j =0 ϑ j,k x j C m, k − j ( x, y ) y j where ϑ j,k = 2 θ j,k,k = ( m − k !) (2 k − n )!Γ( m − (cid:0) m + k − n − (cid:1) k − n )!) ( n !) Γ (cid:0) m + 2 k − n (cid:1) Γ(2 k − n + m − ϑ j +1 ,k = − ϑ j,k . The inversion. Finally, we can use the reproducing properties of the zonalspherical monogenics to find an inversion formula. Proposition 4.3. Let M t − n ∈ M t − n ( C m ) , t, n ∈ N and t − n ≥ . Then ˜ R [ R Hτ [ z n M t − n ( z )] = ρ t, n z n M t − n ( z ) where ρ t, n = ( P ℓs = n ν ℓ − s,s θ n, ℓ − s,s − ν ,ℓ ϑ n,ℓ if t = 2 ℓ even P ℓs = n ν ℓ +1 − s,s θ n, ℓ +1 − s,s if t = 2 ℓ + 1 odd with ν k,s = ( − k Γ ( k +2 s + m ) Γ( k +2 s +1)Γ ( m ) . NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 19 Proof. We have˜ R [ R Hτ [ z n M t − n ( z )] = ˜ R (cid:20)Z π Z S m − L τ ( z, e iθ ω )( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ (cid:21) = ˜ R "Z π Z S m − πA m ∞ X s =0 ∞ X k =0 ν k,s a k + s b s ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ − πA m τ † τ ∞ X s =0 ν ,s a s b s ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ = Z π Z S m − πA m ∞ X s =0 ∞ X k =0 ν k,s ˜ R (cid:2) a k + s b s (cid:3) ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ (4.4) − πA m ∞ X s =0 ν ,s ˜ R (cid:20) a s b s τ † τ (cid:21) ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ (4.5)where a = h z, τ ih e − iθ ω, τ † i , b = h z, τ † ih e − iθ ω, τ i . Thus upon applying Corollary3.11 and Proposition 3.5, we get˜ R (cid:2) a k + s b s (cid:3) = s X n ′ =0 θ n ′ ,k + s,s z n ′ (cid:2) C m, s + k − n ′ ( z, e − iθ ω ) − z C m, s + k − n ′ − ( z, e − iθ ω ) e − iθ ω (cid:3) (cid:0) e − iθ ω (cid:1) n ′ . Hence the terms of (4.4) become Z π Z S m − z n ′ C m, s + k − n ′ ( z, e − iθ ω ) (cid:0) e − iθ ω (cid:1) n ′ ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ − Z π Z S m − z n ′ +1 C m, s + k − n ′ − ( z, e − iθ ω ) (cid:0) e − iθ ω (cid:1) n ′ +1 (cid:0) e iθ ω (cid:1) n M t − n ( e iθ ω ) dS ( ω ) dθ =( − n ′ + n z n ′ Z π e iθ ( t − s − k ) dθ Z S m − C m, s + k − n ′ ( z, ω ) M t − n ( ω ) dS ( ω )(4.6) − ( − n ′ + n z n ′ +1 Z π e iθ ( t − s − k ) dθ Z S m − C m, s + k − n ′ − ( z, ω ) ωM t − n ( ω ) dS ( ω ) . (4.7)If we now use the reproducing properties, (4.7) will vanish and the only situationin which (4.6) will not vanish is when 2 s + k − n ′ = t − n . But now Z π e iθ ( t − s − k ) dθ = Z π e iθ (2 n − n ′ ) dθ = (cid:26) n = n ′ , π if n = n ′ .This implies that the only nontrivial case will be when n ′ = n and 2 s + k = t ,resulting in: Z π Z S m − πA m ∞ X s =0 ∞ X k =0 ν k,s ˜ R (cid:2) a k + s b s (cid:3) ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ = z n M t − n ( z ) ⌊ t ⌋ X s = n ν t − s,s θ n,t − s,s . Now looking at (4.5), we can use Proposition 4.2 which yields˜ R (cid:20) a s b s τ † τ (cid:21) = 14 s X n ′ =0 ϑ n ′ ,s z n ′ C m, s − n ′ ( z, e − iθ ω ) (cid:0) e − iθ ω (cid:1) n ′ In complete analogy to (4.4), we see that n ′ = 2 n and 2 s = t , i.e. (4.5) will onlybe non-zero if t = 2 ℓ is even and in this case we have − πA m ∞ X s =0 ν ,s ˜ R (cid:20) a s b s τ † τ (cid:21) ( e iθ ω ) n M t − n ( e iθ ω ) dS ( ω ) dθ = − ν ,ℓ ϑ n,ℓ z n M t − n ( z ) . (cid:3) Proposition 4.4. Let M t − n − ∈ M t − n − ( C m ) , t, n ∈ N and t − n − ≥ .Then ˜ R [ R Hτ [ z n +1 M t − n − ( z )] = ρ t, n +1 z n +1 M t − n − ( z ) where ρ t, n +1 = ( P ℓs =0 ν ℓ − s,s θ n, ℓ − s,s − ν ,ℓ ϑ n,ℓ if t = 2 ℓ even P ℓs =0 ν ℓ +1 − s,s θ n, ℓ +1 − s,s if t = 2 ℓ + 1 oddwith ν k,s = ( − k Γ ( k +2 s + m ) Γ( k +2 s +1)Γ ( m ) .Proof. The proof is similar to the proof of Proposition 4.3. (cid:3) The last step in the inversion is to write the coefficients ρ t, n and ρ t, n +1 asoperators. We can do this using the Euler operator E z , and the Gamma operatorΓ z , using (3.8) on page 16. Thus we can write ρ t, n = ρ E z , Γ z + E z and ρ t, n +1 = ρ E z , − Γ z + E z + m − and hence the inversion of the polarized Hua-Radon transform isgiven by: z n M t − n ( z ) = ρ − E z , Γ z + E z ˜ R h R Hτ [ z n M t − n ( z )] i ( z ) z n +1 M t − n − ( z ) = ρ − E z , − Γ z + E z + m − ˜ R h R Hτ [ z n +11 M t − n − ( z )] i ( z )Analogously to the Hua-Radon transform, we can use (2.8), so that we can invertany holomorphic function. NVERSIONS FOR THE HUA-RADON AND POLARIZED HUA-RADON TRANSFORM 21 Theorem 4.5. Let f ( z ) = P ∞ k =0 M k ( z ) + z P ∞ ℓ =0 N ℓ ( z ) ∈ OL ( LB (0 , , with M k ∈ M k ( C m ) , N ℓ ∈ M ℓ ( C m ) , then f ( z ) = ρ − E z , Γ z + E z ˜ R " R τ " ∞ X k =0 M k ( z ′ ) ( z )+ ρ − E z , E z − Γ z + m − ˜ R " R τ " z ′ ∞ X ℓ =0 N ℓ ( z ′ ) ( z ) . Conclusions In this paper we have studied the Hua-Radon and polarized Hua-Radon trans-form and their inversions. 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