DDOMAINS WITH RADICAL-POLYNOMIAL X -RAYTRANSFORM MARK AGRANOVSKY
Abstract.
Let K be a compact convex body in R n . For any affine line L, denote (cid:98) χ K ( L ) = (cid:82) L χ K ( x ) dl ( x ) , where dl is the arc length measure, the X -raytransform of the characteristic function χ K , i.e., the length of the chord K ∩ L. We prove that if K is bounded by a C ∞ real algebraic hypersurface ∂K andthe X -ray transform (cid:98) χ K ( L ) behaves, under small parallel translations of theline L to the distance t, as the m -th root of a polynomial of t , for some fixed m ∈ N , then ∂K is an ellipsoid. Introduction
This article is devoted to characterization bodies in R n in integral-geometricterms and is motivated by study of so called polynomially integrable bodies. Letus explain this relation.Given a bounded domain K ⊂ R n , denote A K ( ξ, t ) , ξ ∈ R n , | ξ | = 1 , t ∈ R , thesectional volume function , which equals to the ( n − K by the affine hyperplane {(cid:104) ξ, x (cid:105) = t } . Here (cid:104) , (cid:105) is the innerproduct in R n . In other words, A K ( ξ, t ) is the Radon transform of the characteristic function χ K : A K ( ξ, t ) = (cid:90) (cid:104) ξ,x (cid:105) = t χ K ( x ) dvol n − ( x ) = vol n − (cid:0) K ∩ {(cid:104) ξ, x (cid:105) = t } (cid:1) . The body K is called polynomially integrable [1], if A K ( ξ, t ) is a polynomial withrespect to t so long as the above cross-section is non-empty.It was proved in [9] (see also [1], [2]) that the only polynomially integrabledomains with C ∞ boundary are solid ellipsoids in odd-dimensional spaces. Inparticular, the sectional volume function A K ( ξ, t ) is never a polynomial in t when n is even. Nevertheless, for any ellipsoidal domain E ⊂ R k , the squared sectionalvolume function A E ( ξ, t ) does polynomially depend on t. Thus, in any dimension,the sectional volume functions of ellipsoids are either polynomials or radicals ofpolynomials with respect to t. We conjecture that this property fully characterizesellipsoids, disregarding the parity of the dimension of the space:
Conjecture 1.1.
Let K ⊂ R n be a compact body with C ∞ boundary ∂K. Supposethat for some m ∈ N the m -th power A mK ( ξ, t ) of the sectional volume function is apolynomial in t , whenever A K ( ξ, t ) (cid:54) = 0 . Then ∂K is an ellipsoid. Mathematics Subject Classification.
Primary 44A12; Secondarily 51M25; keywords: X -ray transform, chords, volumes, algebraic hypersurface, ellipsoids. a r X i v : . [ m a t h . M G ] F e b M. AGRANOVSKY If ∂K is an ellipsoid, then, in the case n is odd, the function A K ( ξ, t ) is apolynomial with respect to t of degree n − , i.e., the condition is satisfied with m = 1 , while if n is even, then A K ( ξ, t ) is the square root of a polynomial of degree n − m = 2 . Remark 1.2.
If Conjecture is true then the similar version using k -dimensionalaffine cross-sections, ≤ k ≤ n − is fixed, is true. It immediately follows byapplying Conjecture to intersections of K with k +1 -dimensional affine hyperplanes. In Theorem 2.1 of this article, we confirm Conjecture for n = 2 and under apriori assumption of algebraicity of the boundary ∂K. Applying this result to two-dimensional sections yields a characterization of n -dimensional ellipsoids in termsof the chord length function (cid:98) χ K ( L + tξ ) , i.e., to the situation corresponding inRemark 1.2 to arbitrary n and k = 1 . Let us start with discussion of the basic, two-dimensional, case. When n = 2 thenthe hyperplanes are affine straight lines L ξ,t = R · ξ ⊥ + tξ = { x ∈ R : (cid:104) x, ξ (cid:105) = t } and the sectional volume function A K ( ξ, t ) boils down to the chord length function (1.1) A K ( ξ, t ) = (cid:98) χ K ( L ξ,t ) = length of the chord K ∩ L ξ,t . We want to characterize those domains K for which (cid:98) χ K ( ξ, t ) is an algebraicfunction of a simple form, namely, is a radical of a polynomial in t. There is a relation of the question under discussion with a well known Newton’sLemma about ovals (see [4]). It says that the area cut off a planar domain K withsmooth boundary by a straight line is never algebraic function of the parametersof the secant line. The area V ± K ( ξ, t ) = area (cid:0) K ∩ {(cid:104) x, ξ (cid:105) ≷ t } (cid:1) of a portion of K on one side of the line ( the solid volume function ) is just the primitive functionof the chord length function. Therefore, if A K ( ξ, t ) = m (cid:112) P ξ ( t ) where P ξ ( t ) is apolynomial in t then(1.2) V K ( ξ, t ) = (cid:90) m (cid:113) P ξ ( t ) dt is an Abelian integral.Thus, Newton’s lemma says that the solid area function V K of a domain K in the plane is always transcendental, Theorem 2.1 specifies that among thosetranscendental functions, Abelian integrals (1.2) characterize ellipses.Multi-dimensional generalization of Newton’s Lemma are related to Arnold con-jecture about algebraically integrable domains in R n (see [3],[4],[10]).2. Main result
Let K be a compact (connected) domain in R n . Given an affine line L ⊂ R n anda unit vector ξ ∈ R n , we will call the function µ L,ξ ( t ) = (cid:99) χ K ( L + tξ ) = length of K ∩ ( L + tξ ) , t ∈ R , the chord length function .We will be considering domains K which boundary Γ = ∂K is a semi-algebraic curve. This means that Γ is a connected component of the zero locus of a polynomial Q with real coefficients. The polynomial Q is assumed irreducible over the field C . Theorem 2.1.
Let K ⊂ R n , n ≥ , be a compact convex domains with C ∞ semi-algebraic boundary ∂K. Suppose for any fixed affine line
L, L ∩ intK (cid:54) = ∅ , and unit OMAINS WITH RADICAL-POLYNOMIAL X -RAY TRANSFORM 3 vector ξ, the chord length function has for small t the form µ L,ξ ( t ) = m (cid:113) P L,ξ ( t ) , where m ∈ N and P L,ξ ( t ) a polynomial with respect to t : P L,ξ ( t ) = N (cid:88) j =1 a j ( L, ξ ) t j . Then ∂K is an ellipsoid (and therefore a posteriori m can be taken 2 ). Remark 2.2.
The convexity of the domains in Theorem 2.1 can be derived fromthe main condition for the chord length function (see [1] ) and hence Theorem 2.1is valid without assumption of convexity. However, for the sake of simplicity of theexposition, we a priori assume the domain K to be convex. Theorem 2.1 is, in a certain sense, similar to Theorem 2 from [8] which statesthat if there exists a function f with the X-ray transform identically one then thedomain is a ball. 3. Outline of the proof of Theorem 2.1
The idea of the proof is as follows.First of all, it suffices to prove Theorem 2.1 with n = 2 , then the statementfor arbitrary n follows by considering two-dimensional affine cross-sections. In thecase n = 2 , the chord function µ L,ξ ( t ) turns to µ L,ξ ( t ) = A K ( ξ, t ) if we take L = ξ ⊥ = { x ∈ R : (cid:104) ξ, x (cid:105) = 0 } . Also,we will use notation P ( ξ, t ) instead of P ξ ( t ) . The key point is to determine the degree of the polynomial t → P ( ξ, t ) . First ofall, we want to obtain an upper bound for the degree. For this purpose it sufficesto understand the order of growth of P ( ξ, t ) as t → ∞ . Since the information aboutvalues of the polynomial P ( ξ, t ) for large real t, when the line {(cid:104) ξ, x (cid:105) = t } becomesdisjoint from K, is unavailable, we extend P ( ξ, t ) for complex t. At this point, we use algebraicity of the boundary curve ∂K . This curve has anatural complexification, which is a complex algebraic curve in C . This allows usto construct, in Section 4, analytic extension of the chord length function A K ( ξ, t )to complex values t ∈ C . Determining the growth of the analytic extension alongregular pathes going to ∞ delivers the upper bound deg t P ≤ m (Section 5).The lower bound for deg t P ( ξ, t ) (Section 6) follows much easier, from vanishingthe chord length function A K ( ξ, t ) on tangent lines to ∂K. We show that at Morsepoints the order of vanishing is and hence P ( ξ, t ) = A mK ( ξ, t ) vanishes at tangentlines to the order m . Since there are two tangent lines with the same normal vector ξ, we conclude that deg t P ≥ m. In Section 7, we finish the proof of Theorem 2.1. Together with the upper bound,this implies deg t P = m and hence all zeros of P ( ξ, t ) are delivered by tangentlines. Knowing zeros allows us to reconstruct the polynomial P ( ξ, t ) up to a factordepending on ξ, and express the chord length function A K ( ξ, t ) via the supportingfunction of K. Then the range conditions (the first three power moments) for X -ray transform applied to the function A K imply that the supporting function of K coincides with the supporting function of an ellipse. M. AGRANOVSKY Analytic continuation of the chord length function
Let n = 2 . Let K be a domain satisfying the conditions of Theorem 2.1. Weassume that the boundary Γ = ∂K is a non-singular real semi-algebraic curve,which means that there is a real irreducible polynomial Q ( x , x ) such that Q ( x ) = 0 , x = ( x , x ) ∈ Γand ∇ Q ( x ) (cid:54) = 0 , x ∈ Γ . Extend polynomial Q to the complex space C and denote Γ C the complexalgebraic curve Γ C = { z = ( z , z ) ∈ C : Q ( z ) = 0 } . The domain K is regarded as a set in the real subspace R = { z ∈ C : Imz = Imz = 0 } , so that ∂K ⊂ Γ C ∩ R . Given a real unit vector ξ ∈ R and t ∈ C denote the complex affine line in C : X ( ξ, t ) = { z ∈ C : (cid:104) ξ, z (cid:105) = ξ z + ξ z = t } . Lemma 4.1.
Fix a unit vector ξ ∈ S . There is a finite set Z ⊂ R such thatis t ∈ R \ Z and the real affine line {(cid:104) ξ , x (cid:105) = t } intersects transversally thecurve Γ = ∂K at two points a and b the following is true. There is a path T ⊂ C , joining t and ∞ , an open connected neighborhood U ⊂ C of T and two holomorphicmappings F a , F b : U → C such that(i) F a ( t ) , F b ( t ) ∈ X ( ξ , t ) ∩ Γ C for all t ∈ U. (ii) If t ∈ U ∩ R then F a ( t ) , F b ( t ) ∈ Γ . (iii) F a ( t ) = a, F b ( t ) = b. Proof
Applying a suitable rotation in R , we can assume for simplicity that ξ = (0 , . OMAINS WITH RADICAL-POLYNOMIAL X -RAY TRANSFORM 5 In this case, X ( ξ, t ) is the complex line { z = t } in C and a = ( a , t ) , b = ( b , t ) . The condition z ∈ X ( ξ , t ) ∩ Γ C translates in this case as Q ( z , t ) = 0 . Then Lemma 4.1 asserts, in fact, that if we consider the projection π : Γ C → C , π ( z , t ) = t then there is a path T ⊂ C , joining t and ∞ and a neighborhood U of T such thatthe holomorphic mapping π possesses two holomorphic sections F a , F b : U → Γ C of π over the set U with the initial conditions (ii), (iii) . The existence of suchsections follows from the path lifting property (see, e.g., [5], Proposition 11.6) ofcovering maps and from the fact that the projection π is a covering outside of thefinite set of poles and ramification points.Let us give more extended analytic arguments, for the sake of self-sufficiency.Represent the polynomial Q, defining the complex algebraic curve Γ C , in the form Q ( z , z ) = q ( z ) + q ( z ) z + · · · + q M ( z ) z M , where q j are polynomials of one complex variable and q M (cid:54)≡ . Consider the discriminant D ( t ) = Disc z Q t of the polynomial Q t ( z ) := Q ( z , t ) : D ( t ) = q M ( t ) (cid:89) i The monodromy theorem [7], Thm. 16.2, implies that the algebraic equation(4.1) so completely solvable. This means that there is no monodromy on t ∈ T andthere exist M + 1 continuous functions f ( t ) , · · · , f M ( t ) on T, satisfying equation(4.1): Q t ( f j ( t )) = 0 , j = 0 , ..., M. Since all the roots are simple, we have f i ( t ) (cid:54) = f j ( t ) for all t ∈ T and i (cid:54) = j. Let us explain this point in more details. Consider the spaces E = { ( λ , · · · λ M ) ∈ C M +1 : λ i (cid:54) = λ j , i, j = 0 , ..., M, } and B = { ( p , ..., p M − ) ∈ C M : p + p z + ... + p M − z M − + z M has no multiple roots } . Define the mapping π : E → B as follows: π ( λ ) , λ ∈ E is the vector p = ( p , ..., p M − ) of the coefficients of themonic polynomial with the roots λ j , i.e., M (cid:88) j =0 p j λ j = M (cid:89) j =0 ( λ − λ j ) , p M = 1 . The roots λ j of the polynomial in the right hand side are symmetric functions ofthe coefficients p j and by Implicit Function Theorem, π is a ( M + 1)!-covering map.Let λ (0) = ( λ (0)0 , · · · , λ (0) M ) ∈ E be the roots of the polynomial Q t ( z ) = Q ( z , t ) , i.e., π ( λ (0) ) = (cid:0) p ( t ) , ..., p M − ( t ) (cid:1) . The mapping g ( t ) := (cid:0) p ( t ) , ..., p M − ( t ) (cid:1) ∈ B, t ∈ T, where p j ( t ) are coefficients (4.2) of the polynomial Q t defines a path g : T → B in the base space B. The path lifting property of covering mappings (see, e.g.,[5],Proposition 11.6; [7],Theorem 16.2) says that there is a lifting path f = ( f , · · · , f M ) : T → E such that π ◦ f = g and f ( t ) = λ (0) . Then the functions f j ( t ) , j = 0 , · · · , M define the above claimed continuous family of roots of the polynomials Q t . The point a = ( a , t ) , b = ( b , t ) satisfy Q ( a , t ) = Q ( b , t ) = 0 . Therefore,there are two branches, say, f j ( t ) , f j ( t ) which take at t the values a , b , corre-spondingly. Denote f i = f a , f j = f b . Then we have(1) Q ( f a ( t ) , t ) = Q t ( f a ( t )) = 0 , t ∈ T, (2) Q ( f b ( t ) , t ) = Q t ( f b ( t )) = 0 , t ∈ T, (3) f a ( t ) = a , f b ( t ) = b . For any fixed t ∈ C we have Q ( f a ( t ) , t ) = 0 , ∂Q∂z ( f a ( t ) , t ) (cid:54) = 0 , OMAINS WITH RADICAL-POLYNOMIAL X -RAY TRANSFORM 7 because f a ( t ) is a simple root. The same is true for f b ( t ) . By Implicit FunctionTheorem, there is a complex neighborhood U at of t and a complex neighborhood V f a ( t ) of f a ( t ) such that for any s ∈ U at there exists a unique w := f a ( s ) ∈ V f a ( t ) such that Q ( w, s ) = 0 , and the function w = f a ( s ) is holomorphic in U at . Thus, given t ∈ T the function f a ( t ) extends to the neighborhood U at as aholomorphic function. The union U a = ∪ t ∈ T U a t constitutes an open connected setcontaining the path T. The function f a ( t ) , t ∈ T extends to U a as a holomorphicfunction. The extensions satisfies the same polynomial equation Q ( f a ( t ) , t ) = 0 , t ∈ U a . Similarly, we construct an open set U b and the holomorphic function f b in U b with analogous properties.Now set(4.3) U = U a ∩ U b ,F a ( t ) = ( f a ( t ) , t ) , F b ( t ) = ( f b ( t ) , t ) . Check properties (i)-(iii). The mapping F a : U → C is holomorphic in U, andby construction Q ( F a ( t )) = Q ( f a ( t ) , t ) = 0 for all t ∈ U. Also, since ξ = (0 , , wehave (cid:104) ξ, F a ( t ) (cid:105) = t, hence F a ( t ) ∈ Γ C ∩ X ( ξ, t ) . Furthermore, F a ( t ) = ( f a ( t ) , t ) = ( a , t ) = a. For t ∈ U ∩ R near t , the straight line x = t intersects ∂K at a point a t = ( a ,t , t ) close to a. Then Q ( a ,t , t ) = 0 for t in a neighborhood of t , i.e. the polynomial Q vanishes on anopen subarc of the real-analytic curve ∂K. By the uniqueness theorem for holo-morphic functions, the identity holds for all real t ∈ U. Since the root is unique,we conclude that a t = F a ( t ) and thus F a ( t ) ∈ ∂K ∩ U. Hence the set U and themapping F a satisfy all properties (i)-(iii) . Similarly, we proceed with the mapping F b . Lemma is proved. Lemma 4.2. Assume, as in Theorem 2.1, A mK ( ξ, t ) = P ( ξ, t ) , where P ( ξ, t ) isa polynomial in t. Fix a unit vector ξ ∈ S . Let the straight line (cid:104) ξ , x (cid:105) = t intersects ∂K at the points a and b and t / ∈ Z , where Z is the finite exceptionalset from Lemma 4.1. Construct the open set U ⊂ C and the holomorphic mappings F a , F b : U → C as in Lemma 4.1. Then (4.4) P ( ξ , t ) = (cid:0) (cid:104) ξ ⊥ , F a ( t ) − F b ( t ) (cid:105) (cid:1) m , t ∈ U. where ξ ⊥ = a − b | a − b | is the vector orthogonal to ξ . Proof By Lemma 4.1, (i), (ii) , when t ∈ U ∩ R then the segment [ F a ( t ) , F b ( t )]is just the chord X ( ξ , t ) ∩ K. The length of the chord is A K ( ξ, t ) = | F a ( t ) − F b ( t ) | = (cid:104) F a ( t ) − F b ( t ) , F a ( t ) − F b ( t ) | F a ( t ) − F b ( t ) | (cid:105) . Also, (cid:104) F a ( t ) − F b ( t ) , ξ (cid:105) = t − t = 0 and hence the second factor in the inner productis a unit vector orthogonal to ξ and does not depend on t. By taking t = t , wefind, due to F a ( t ) = a, F b ( t ) = b, that this vector is ξ ⊥ = a − b | a − b | . Then P ( ξ , t ) = A m ( ξ , t ) = (cid:0) (cid:104) F a ( t ) − F b ( t ) , ξ ⊥ (cid:105) (cid:1) m for t ∈ U ∩ R . Since P ( ξ, t )and the function in the right hand side of the equality are holomorphic in t ∈ U, and the set U is open and connected, the equality is satisfied for all t ∈ U by theuniqueness theorem. Lemma is proved. M. AGRANOVSKY Upper bound of the degree of the polynomial P ( ξ, t )As before, n = 2 . We fix ξ ∈ S and t ∈ R such that the line {(cid:104) ξ , x (cid:105) meetsΓ = ∂K at the points a, b and t does not belong to the finite exceptional set Z inLemma 4.1. Let the path T ⊂ C and the open set U ⊂ C , T ⊂ U, be as in Lemma4.1.In order to obtain the upper bound for the degree of the polynomial P ( ξ , t ) , itsuffices to estimate the order of its growth as t → ∞ . According to the representa-tion (4.4), the problem is reduced to understanding the behaviour of the mappings F a ( t ) , F b ( t ) as t → ∞ . Let Q = Q + · · · + Q N be the decomposition of the polynomial Q, defining the complexified boundary Γ C , into homogeneous polynomials Q j , degQ j = j, j = 1 , ..., N = degQ. The leading homogeneous polynomial Q N ( z , z ) of two complex variables iscompletely reducible over C : Q ( z , z ) = const J (cid:89) j =1 ( A j z + B j z ) m j , J (cid:88) j =1 m j = N = degQ. Lemma 5.1. Suppose that α B j − β A j (cid:54) = 0 for any j = 1 , · · · , J, where ξ =( α , β ) . Then F a ( t ) t , F b ( t ) t are bounded for t ∈ U. Proof The zero set Q − N (0) of the homogeneous polynomial Q N in C consist of J complex lines A j z + B j z = 0 , j = 1 , ..., J, counting multiplicities. The conditionfor ξ means that the affine complex line X ( ξ , 0) is none of those. Again, it wouldbe convenient to apply rotation in the real plane R and make ξ = (0 , . Then wehave α = 0 , β = 1 and A j (cid:54) = 0 for all j = 1 , ..., J. Let z ∈ Q − (0) ∩ X ( ξ , t ) . It means that z = t and Q ( z ) = Q ( z , t ) = 0 , or, thesame, Q + tQ ( z t , 1) + ... + t N − Q N − ( z t , 1) + t N Q N ( z t , 1) = 0 . Dividing both sides by t N yields(5.1) Ψ( w, s ) := s N Q + s N − Q ( w, 1) + · · · + sQ N − ( w, 1) + Q N ( w, 1) = 0 , where w, s and z , t are related by w = z t , s = 1 t . Denote w j = − B j A j the root of the polynomial Q N ( w, , of multiplicity m j . Consider the logarithmic residue r j ( s ) = 12 πi (cid:90) | w − w j | = ε Ψ (cid:48) w ( w, s )Ψ( w, s ) dw, where ε is so small that w j is the only root of Ψ( w, 0) = Q N ( w, 1) in the disc | w − w j | ≤ ε. Then r j (0) = m j . OMAINS WITH RADICAL-POLYNOMIAL X -RAY TRANSFORM 9 Thus, the total sum is r (0) + · · · + r J (0) = m + · · · + m J = N. By continuity, for every j = 1 , ..., J and sufficiently small ε > δ j > | s | < δ j , Ψ( w, s ) (cid:54) = 0 when | w − w j | = ε. Then the function r j ( s ) is continuous in | s | < δ j and integer-valued, therefore r ( s ) + · · · + r J ( s ) = N for | s | < δ = min { δ j , ..., δ J } . . This means that for | s | < δ, all N roots, countingmultiplicities, of the polynomial w → Ψ( w, s ) are located in ε -neighborhood of theset of zeros of the polynomial Ψ( w, 0) = Q N ( w, . Going back to the variable z = tw and the function Q ( z , t ) we conclude thatif | t | > | δ | , then | z t − w j | < ε for some root w j of Q N ( w, . Since we consider the situation ξ = (0 , , the condition (i) in Lemma 4.1 meansthat F a ( t ) , F b ( t ) have the form F a ( t ) = (cid:0) f a ( t ) , t (cid:1) , F b ( t ) = (cid:0) f b ( t ) , t (cid:1) , t ∈ U. Now, by the construction in Lemma 4.1, if t ∈ U then Q t ( z ) = Q ( z , t ) hasonly simple roots Λ( t ) = { λ ( t ) , ..., λ M ( t ) } . Therefore, when | t | > δ then λ j ( t ) t arein an ε -neighborhood of a root of Q N ( w, . Among the collection Λ( t ) of the rootsof Q t , there are two branches, say, λ i ( t ) , λ j ( t ) , determined by the initial conditions λ i ( t ) = a, λ j ( t ) = b. They must coincide, correspondingly, with f a ( t ) and f b ( t )and hence (cid:12)(cid:12)(cid:12)(cid:12) f a ( t ) t − w i (cid:12)(cid:12)(cid:12)(cid:12) < ε, (cid:12)(cid:12)(cid:12)(cid:12) f b ( t ) t − w j (cid:12)(cid:12)(cid:12)(cid:12) < ε, t ∈ U, | t | > δ . Since the set of roots w i is finite, we conclude that f a ( t ) t , f b ( t ) t , t ∈ U, are bounded.Lemma is proved. Lemma 5.2. For all ξ ∈ S but finite set of those, deg t P ( ξ, t ) is at most m. Proof The set of vectors ξ which do not satisfy the condition of Lemma 5.1is finite. Let ξ = ξ be not such a vector and let t , a, b, be as in Lemma 4.1. ByLemma 4.2, A mK ( ξ, t ) = P ( ξ, t ) = (cid:0) (cid:104) ξ ⊥ , F a ( t ) − F b ( t ) (cid:105) (cid:1) m . Then by Lemma 5.1 P ( ξ,t ) t m is bounded, as t → ∞ , t ∈ U. Therefore, deg t P ( ξ, t ) ≤ m. The zeros and exact degree of P ( ξ, t )In this section n = 2 . Given ξ ∈ S , denote ρ + ( ξ ) = max x ∈ K (cid:104) ξ, x (cid:105) , ρ − ( ξ ) = min x ∈ K (cid:104) ξ, x (cid:105) . Denote M ± ( ξ ) ∈ ∂K the points where the maximum and minimum are attained: ρ + ( ξ ) = (cid:104) ξ, M + ( ξ ) (cid:105) , ρ − ( ξ ) = (cid:104) ξ, M − ( ξ ) (cid:105) . The lines (cid:104) ξ, x (cid:105) = ρ ± ( ξ ) are supporting lines to ∂K at the points M ± ( ξ ) and thevectors ± ξ are the unit outward normal vectors ± ξ = ν M ± ( ξ ) to the curve ∂K atthe points M ± ( ξ ) , correspondingly. Lemma 6.1. The polynomial P has the form P ( ξ, t ) = c ( ξ ) (cid:0) ρ + ( ξ ) − t (cid:1) m (cid:0) t − ρ − ( ξ ) (cid:1) m , c ( ξ ) > . Proof By continuity, it suffices to establish the representation for almost all ξ. Since ∂K is a non-singular algebraic curve, it is real-analytic. Then all points,except finite number of those, are Morse points. Therefore, only for a finite vectors ξ, the points M ± ( ξ ) are non-Morse. Choose ξ such that this is the case.Using rotation and translation, we can assume that ξ = (0 , 1) and M ( ξ ) =(0 , . The outward normal vector at M − ( ξ ) is − ξ = (0 , − 1) and ρ − ( ξ ) = 0 . Ina neighborhood of the point M − ( ξ ) = (0 , 0) the curve Γ can be represented as agraph x = ϕ ( x ) of a real-analytic function, with ϕ (cid:48) (0 , 0) = 0 , ϕ (cid:48)(cid:48) (0 , (cid:54) = 0 : x = 12 ϕ (cid:48)(cid:48) (0 , x + o ( x ) , x → . Then the length of the chord x = t is A K ( ξ, t ) = 2 (cid:113) tϕ (cid:48)(cid:48) (0 , + o ( √ t ) , t → . It shows that the length of the chord, obtained by the parallel translation of atangent line to the distance t , behaves at √ t. This yields that if M ± ( ξ ) ∈ Γ areMorse points then A K ( ξ, t ) = const ( t − ρ − ( ξ )) + o (( t − ρ − ( ξ )) , t → ρ − ( ξ ) . Similarly, A K ( ξ, t ) = const ( ρ + ( ξ ) − t ) + o (( ρ + ( ξ ) − t ) , t → ρ + ( ξ ) . Thus, the polynomial P ( ξ, t ) = A mK ( ξ, t ) vanishes, to the order m , at t = ρ + ( ξ ) and t = ρ − ( ξ ) . This means, firstly, that m is even and, secondarily, that deg t P ( ξ, t ) ≥ m. According to the remark at he beginning of the proof, it holds for all ξ ∈ S exceptfor a finite set. Together with Lemma 5.2 it proves that deg t P ( ξ, t ) = m for all ξ except for a finite set, and for those ξ we have P ( ξ, t ) = c ( ξ ) (cid:0) ρ + ( ξ ) − t (cid:1) m (cid:0) t − ρ − ( ξ ) (cid:1) m . By continuity, P ( ξ, t ) has the claimed representation for all ξ ∈ S . Proof of Theorem 2.1 First of all, as it has been mentioned before, it suffices to prove Theorem 2.1 for n = 2 . Indeed, each transversal intersection of K with two-dimensional affine planeproduces a domain in this plane satisfying all the conditions of Theorem 2.1. If wecould conclude that all such two-dimensional cross-sections are bounded by ellipsesthen the entire body K is bounded by an ellipsoid.Thus, we assume that n = 2 . Then we follow the arguments from [1]. Therepresentation of the polynomial P ( ξ, t ) given by Lemma 6.1 yields(7.1) A K ( ξ, t ) = m (cid:112) P ( ξ, t ) = d ( ξ ) (cid:0) ρ + ( ξ ) − t (cid:1) (cid:0) t − ρ − ( ξ ) (cid:1) , where d ( ξ ) = m (cid:112) c ( ξ ) . Representation (7.1) holds whenever t ∈ [ ρ − ( ξ ) , ρ + ( ξ )] , oth-erwise A K ( ξ, t ) = 0 . The next step is applying the range conditions for Radon transform [6]. Function A K ( ξ, t ) is the Radon transform of the characteristic function χ K and hence satisfythe moment conditions. Namely, the moments(7.2) M k ( ξ ) = ρ + ( ξ ) (cid:90) ρ − ( ξ ) A K ( ξ, t ) t k dt must be restriction to the unit circle S of a homogeneous polynomial of degree k. Notice, that M ( ξ ) = area K = const > . OMAINS WITH RADICAL-POLYNOMIAL X -RAY TRANSFORM 11 Then substituting (7.2) into (7.1) yields: M k ( ξ ) = d ( ξ ) ρ + ( ξ ) (cid:90) ρ − ( ξ ) (cid:0) ρ + ( ξ ) − t (cid:1) (cid:0) t − ρ − ( ξ ) (cid:1) t k dt. Denote(7.3) B ( ξ ) = 12 (cid:0) ρ + ( ξ ) + ρ − ( ξ ) (cid:1) ,C ( ξ ) = 12 (cid:0) ρ + ( ξ ) − ρ − ( ξ ) (cid:1) . and perform the change of variables in the integral s = t − B ( ξ ) . Since (cid:0) ρ + ( ξ ) − t (cid:1)(cid:0) t − ρ − ( ξ ) (cid:1) = (cid:0) C ( ξ ) − s (cid:1) , substitution in the integral yields: M k ( ξ ) = d ( ξ ) C ( ξ ) (cid:90) − C ( ξ ) (cid:0) C ( ξ ) − s (cid:1) (cid:0) s + B ( ξ ) (cid:1) k ds. Finally, the change of variable s = C ( ξ ) v leads to M k ( ξ ) = G ( ξ ) (cid:90) − (1 − v ) (cid:0) C ( ξ ) v + B ( ξ ) (cid:1) k dv where G ( ξ ) = d ( ξ ) C ( ξ ) . Let us write the first three moments:(7.4) M ( ξ ) = G ( ξ ) α ,M ( ξ ) = G ( ξ ) (cid:0) C ( ξ ) α + B ( ξ ) α (cid:1) ,M ( ξ ) = G ( ξ ) (cid:0) C ( ξ ) α + 2 C ( ξ ) B ( ξ ) α + B ( ξ ) α (cid:1) , where α k = (cid:82) − (1 − v ) v k dv. Since α = 0 we have B ( ξ ) = M ( ξ ) M ( ξ ) . By the range conditions, M ( ξ ) = const, G ( ξ ) = const on S and M ( ξ ) extendsfrom S as a homogeneous linear polynomial. Therefore, B ( ξ ) is the restriction tothe unit circle of a linear form: B ( ξ ) = (cid:104) ξ, b (cid:105) , where b is a fixed vector.Since M ( ξ ) extends from S as a homogeneous quadratic polynomial, G ( ξ ) isconstant and B ( ξ ) is a linear form, the third equality in (7.4) implies that C ( ξ ) is the restriction of a (strictly positive) quadratic form. Applying a suitable rotationwe can reduce C ( ξ ) to the form(7.5) C ( ξ ) = c ξ + c ξ , c , c > . Now, by translating K by the vector b, we can make B ( ξ ) = 0 for all ξ. Indeed,the supporting functions ρ ± transform, under the translation by the vector b, asfollows: ρ + ( ξ ) → ρ + ( ξ ) − (cid:104) ξ, b (cid:105) , ρ − ( ξ ) → ρ − ( ξ ) − (cid:104) ξ, b (cid:105) . Then (7.3) shows that B ( ξ ) transforms to B ( ξ ) − (cid:104) ξ, b (cid:105) = 0 . Thus, we can apply the translation K by the vector b and assume that B ( ξ ) = 0 . Then from (7.3) we have ρ + ( ξ ) = − ρ − ( ξ ) and C ( ξ ) = ρ + ( ξ ) . From (7.5), we obtain ρ + ( ξ ) = (cid:113) c x + c x . Thus, the supporting function ρ + ( ξ ) of the body K coincides with the supportingfunction of the ellipse E = { x c + x c = 1 } . Thus, we conclude that ∂K = E. Theorem 2.1 is proved. References [1] M. Agranovsky, On polynomially integrable domains in Euclidean spaces, in ComplexAnalysis and Dynamical Systems,Trends in Mathematics, Birkhauser, Springer, 2018;arXiv.1701.05551v1.[2] M. Agranovsky, On algebraically integrable domains , Contemp. Math., vol.733, 33-44;arXiv.1705.06063v2.[3] V. I. Arnold, Arnold’s Problem, 2nd edition , Springer-Verlag, Belin, 2004.[4] V.I. Arnold, V.A. Vassiliev, Newton’s Principia read 300 years later , Notices AMS, 36:9(1989), 1148-1154.[5] W. Fulton, Algebraic Topology: A First Course , Springer Sci, 1995.[6] S. Helgason, Groups and Geometric Analysis , Acad.Press, 1984.[7] Sze-Tsen Hu, Homotopy Theory , Academic Press, 1959.[8] J.Ilmavirta, G. P. Paternain, Functions of constant geodesic X-ray transform , Inverse Prob-lems, vol. 35, 6, 2019; arXiv.1702.00429[9] A. Koldobsky, A. Merkurjev, V. Yaskin, On polynomially integrable convex bodies , Advancesin Math., vol. 320, 2017, 876-886; arXiv.1702.00429[10] V. A. Vassiliev, Newton’s lemma XXVIII on integrable ovals in higher dimensions and re-flection groups , Bull. Lond. Math. Soc., 47:2 (2015), 290-300. Bar-Ilan University and Holon Institute of Technology; Israel. E-mail address: [email protected] Bar-Ilan University and Holon Institute of Technology Email address ::