The truncated moment problem on reducible cubic curves I: Parabolic and Circular type relations

In this article we study the bivariate truncated moment problem (TMP) of degree $2k$ on reducible cubic curves. First we show that every such TMP is equivalent after applying an affine linear transformation to one of 8 canonical forms of the curve. The case of the union of three parallel lines was solved in 2022 by the second author, while the degree 6 cases in 2017 by the first author. Second we characterize in terms of concrete numerical conditions the existence of the solution to the TMP on two of the remaining cases concretely, i.e., a union of a line and a circle $y(ay+x^2+y^2)=0, a\in \mathbb{R}\setminus \{0\}$, and a union of a line and a parabola $y(x-y^2)=0$. In both cases we also determine the number of atoms in a minimal representing measure.

If such a measure exists, we say that β (2k) has a representing measure supported on K and µ is its K-representing measure (K-rm).
Let R[x, y] ≤k := {p ∈ R[x, y] : deg p ≤ k} stand for the set of real polynomials in variables x, y of total degree at most k.For every p(x, y) = i,j a i,j x i y j ∈ R[x, y] ≤k we define its evaluation p(X, Y ) on the columns of the matrix M(k) by replacing each capitalized monomial X i Y j in p(X, Y ) = i,j a i,j X i Y j by the column of M(k), indexed by this monomial.Then p(X, Y ) is a vector from the linear span of the columns of M(k).If this vector is the zero one, i.e., all coordinates are equal to 0, then we say p is a column relation of M(k).A column relation p is nontrivial, if p ̸ ≡ 0. We denote by Z(p) := {(x, y) ∈ R 2 : p(x, y) = 0}, the zero set of p.We say that the matrix M(k) is recursively generated (rg) if for p, q, pq ∈ R[x, y] ≤k such that p is a column relation of M(k), it follows that pq is also a column relation of M(k).The matrix M(k) is p-pure, if the only column relations of M(k) are those determined recursively by p.We call a sequence β p-pure, if M(k) is p-pure.
A concrete solution to the TMP is a set of necessary and sufficient conditions for the existence of a K-representing measure µ, that can be tested in numerical examples.Among necessary conditions, M(k) must be positive semidefinite (psd) and rg [CF04,Fia95], and by [CF96], if the support supp(µ) of µ is a subset of Z(p) for a polynomial p ∈ R[x, y] ≤k , then p is a column relation of M(k).The bivariate K-TMP is concretely solved in the following cases: (1) K = Z(p) for a polynomial p with 1 ≤ deg p ≤ 2. Assume that deg p = 2.By applying an affine linear transformation it suffices to consider one of the canonical cases: x 2 +y 2 = 1, y = x 2 , xy = 1, xy = 0, y 2 = y.The case x 2 +y 2 = 1 is equivalent to the univariate trigonometric moment problem, solved in [CF02].The other four cases were tackled in [CF02,CF04,CF05,Fia15] by applying the far-reaching flat extension theorem (FET) [CF96, Theorem 7.10] (see also [CF05b,Theorem 2.19] and [Lau05] for an alternative proof), which states that β (2k) admits a (rank M(k))-atomic rm if and only if M(k) is psd and admits a rank-preserving extension to a moment matrix M(k +1).For an alternative approach with shorter proofs compared to the original ones by reducing the problem to the univariate setting see [BZ21, Section 6] (for xy = 0), [Zal22a] (for y 2 = y), [Zal22b] (for xy = 1) and [Zal23] (for y = x 2 ).
For deg p = 1 the solution is [CF08, Proposition 3.11] and uses the FET, but can be also derived in the univariate setting (see [Zal23,Remark 3.3
(5) M(3) satisfies symmetric cubic column relations which can only cause extremal moment problems.In order to satisfy the variety condition, another symmetric column relation must exist, and the solution was obtained by checking consistency [CY14].
The solutions to the K-TMP, which are not concrete in the sense of definition from the previous paragraph, are known in the cases K = Z(y − q(x)) and K = Z(yq(x) − 1), where q ∈ R[x].[Fia11, Section 6] gives a solution in terms of the bound on the degree m for which the existence of a positive extension M(k + m) of M(k) is equivalent to the existence of a rm.In [Zal23] the bound on m is improved to m = deg q − 1 for curves of the form y = q(x), deg q ≥ 3, and to m = ℓ + 1 for curves of the form yx ℓ = 1, ℓ ∈ N \ {1}.
References to some classical work on the TMP are monographs [Akh65, AK62, KN77], while for a recent development in the area we refer a reader to [Sch17].Special cases of the TMP have also been considered in [Kim14, Ble15, Fia17, DS18, BF20, Kim21], while [Nie14] considers subspaces of the polynomial algebra and [CGIK+] the TMP for commutative R-algebras.
The motivation for this paper was to solve the TMP concretely on some reducible cubic curves, other than the case of three parallel lines solved in [Zal22a].Applying an affine linear transformation we show that every such TMP is equivalent to the TMP on one of 8 canonical cases of reducible cubics of the form yc(x, y) = 0, where c ∈ R[x, y], deg c = 2.In this article we solve the TMP for the cases c(x, y) = ay + x 2 + y 2 , a ∈ R \ {0}, and c(x, y) = x − y 2 , which we call the circular and the parabolic type, respectively.The main idea is to characterize the existence of a decomposition of β into the sum β (ℓ) + β (c) , where β (ℓ) = {β (ℓ) i,j } i,j∈Z + , 0≤i+j≤2k and β (c) = {β (c) i,j } i,j∈Z + , 0≤i+j≤2k admit a R-rm and a Z(c)-rm, respectively.Due to the form of the cubic yc(x, y) = 0, it turns out that all but two moments of β (ℓ) and β (c) are not already fixed by the original sequence, i.e., β 2k,0 in the parabolic type case.Then, by an involved analysis, the characterization of the existence of a decomposition β = β (ℓ) + β (c) can be done in both cases.We also characterize the number of atoms in a minimal representing measure, i.e., a measure with the minimal number of atoms in the support.
1.1.Readers Guide.The paper is organized as follows.In Section 2 we present some preliminary results needed to establish the main results of the paper.In Section 3 we show that to solve the TMP on every reducible cubic curve it is enough to consider 8 canonical type relations (see Proposition 3.1).In Section 4 we present the general procedure for solving the TMP on all but one of the canonical types and prove some results that apply to them.Then in Sections 5 and 6 we specialize to the circular and the parabolic type relations and solve them concretely (see Theorems 5.1 and 6.1).In both cases we show, by numerical examples, that there are pure sequences β (6) with a psd M(3) but without a rm (see Examples 5.3 and 6.3).

PRELIMINARIES
We write R n×m for the set of n × m real matrices.For a matrix M we call the linear span of its columns a column space and denote it by C(M ).The set of real symmetric matrices of size n will be denoted by S n .For a matrix A ∈ S n the notation A ≻ 0 (resp.A ⪰ 0) means A is positive definite (pd) (resp.positive semidefinite (psd)).We write 0 t 1 ,t 2 for a t 1 × t 2 matrix with only zero entries and 0 t = 0 t,t for short, where t 1 , t 2 , t ∈ N. The notation E (ℓ) i,j , ℓ ∈ N, stands for the usual ℓ × ℓ coordinate matrix with the only nonzero entry at the position (i, j), which is equal to 1.
In the rest of this section let k ∈ N and β ≡ β (2k) = {β i,j } i,j∈Z + , 0≤i+j≤2k be a bivariate sequence of degree 2k.

2.1.
consisting of the rows indexed by the elements of Q 1 and the columns indexed by the elements of Q 2 .In case 2.2.Affine linear transformations.The existence of representing measures is invariant under invertible affine linear transformations of the form where p = i,j∈Z + , 0≤i+j≤2k a i,j x i y j .

Generalized Schur complements. Let
be a real matrix where where A † (resp.D † ) stands for the Moore-Penrose inverse of A (resp.D).
The following lemma will be frequently used in the proofs.
Lemma 2.1.Let n, m ∈ N and where A ∈ S n , B ∈ R n×m and C ∈ S m .If rank M = rank A, then the matrix equation where W ∈ R n×m , is solvable and the solutions are precisely the solutions of the matrix equation Proof.
The assumption rank M = rank A implies that for some W ∈ R n×m .So the equation (2.2) is solvable.In particular, AW = B.It remains to prove that any solution W to AW = B is also a solution to (2.3).Note that all the solutions of the equation where each column of Z ∈ R n×m is an arbitrary vector from ker A. So W satisfiying (2.3) is also of the form A † B + Z 0 for some Z 0 ∈ R n×m with columns belonging to ker A. We have that where we used the fact that each column of B belongs to C(A) and ker(A) ⊥ = C(A).Replacing W with any W of the form (2.4) in the calculation (2.5) gives the same result, which proves the statement of the proposition.□ The following theorem is a characterization of psd 2 × 2 block matrices.(2) If M ⪰ 0, then

Extension principle.
Proposition 2.3.Let A ∈ S n be positive semidefinite, Q a subset of the set {1, . . ., n} and A| Q the restriction of A to the rows and columns from the set Q.
, where v is a vector with the only nonzero entries in the rows from Q and such that the restriction v| Q to the rows from Q equals to v.
Proof.See [Fia95, Proposition 2.4] or [Zal22a, Lemma 2.4] for an alternative proof.□ 2.5.Partially positive semidefinite matrices and their completions.A partial matrix A = (a i,j ) n i,j=1 is a matrix of real numbers a i,j ∈ R, where some of the entries are not specified.A partial symmetric matrix A = (a i,j ) n i,j=1 is partially positive semidefinite (ppsd) (resp.partially positive definite (ppd)) if the following two conditions hold: (1) a i,j is specified if and only if a j,i is specified and a i,j = a j,i .
(2) All fully specified principal minors of A are psd (resp.pd).
For n ∈ N write [n] := {1, 2, . . ., n}.We denote by A Q 1 ,Q 2 the submatrix of A ∈ R n×n consisting of the rows indexed by the elements of Q 1 ⊆ [n] and the columns indexed by the elements of It is well-known that a ppsd matrix A(x) of the form as in Lemma 2.4 below admits a psd completion (This follows from the fact that the corresponding graph is chordal, see e.g., [GJSW84,Dan92,BW11]). Since we will need an additional information about the rank of the completion A(x 0 ) and the explicit interval of all possible x 0 for our results, we give a proof of Lemma 2.4 based on the use of generalized Schur complements.
Lemma 2.4.Let A(x) be a partially positive semidefinite symmetric matrix of size n × n with the missing entries in the positions (i, j) and (j, i), (iii) The following statements are equivalent: (a) Proof.We write Let P be a permutation matrix, which changes the order of columns to 1, 2, . . ., i − 1, i + 1, . . ., j − 1, j + 1, . . ., n, i, j. Then Lemma 2.4 with the missing entries in the positions (n − 1, n) and (n, n − 1) was proved in [Zal21, Lemma 2.11] using computations with generalized Schur complements under one additional assumption: (2.7) Here we explain why the assumption (2.7) can be removed from [ (2.8) where However, the formula (2.8) has been generalized [CHM74, Theorem 4] to noninvertible A 1 , A 2 , where all Schur complements are the generalized ones, under the conditions: (2.9) b x T ∈ C(A 2 ) and a ∈ C(A 1 ).
So if we show that the conditions (2.9) hold, the same proof as in [Zal21, Lemma 2.11] can be applied in the case A 1 is singular.From A 2 (resp.
The assumption for every x ∈ R, which concludes the proof of (2.9).□ 2.6.Hamburger TMP.Let k ∈ N.For v = (v 0 , . . ., v 2k ) ∈ R 2k+1 we define the corresponding Hankel matrix as (2.10) We denote by v j := (v j+ℓ ) k ℓ=0 the (j + 1)-th column of A v , 0 ≤ j ≤ k, i.e., As in [CF91], the rank of v, denoted by rank v, is defined by For m ≤ k we denote the upper left-hand corner (v i+j ) m i,j=0 ∈ S m+1 of A v of size m + 1 by A v (m).A sequence v is called positively recursively generated (prg) if for r = rank v the following two conditions hold: • A v (r − 1) ≻ 0.
the equality (2.12) holds for j = r, . . ., 2k.The solution to the R-TMP is the following.
(2) There exists a (rank A v )-atomic R-representing measure for β.
(3) A v is positive semidefinite and one of the following holds: ) v is positively recursively generated.
2.7.TMP on the unit circle.The solution to the Z(x 2 + y 2 − 1)-TMP is the following.

TMP ON REDUCIBLE CUBICS -CASE REDUCTION
In this section we show that to solve the TMP on reducible cubic curves it suffices, after applying an affine linear transformation, to solve the TMP on 8 canonical forms of curves.
Proposition 3.1.Let k ∈ R and β := β (2k) = (β i,j ) i,j∈Z + ,i+j≤2k .Assume M(k; β) does not satisfy any nontrivial column relation between columns indexed by monomials of degree at most 2, but it satisfies a column relation p(X, Y ) = 0, where p ∈ R[x, y] is a reducible polynomial with deg p = 3.If β admits a representing measure, then there exists an invertible affine linear transformation ϕ of the form (2.1) such that the moment matrix ϕ M(k; β) satisfies a column relation q(x, y) = 0, where q has one of the following forms: Parallel lines type: q(x, y) = y(a Circular type: q(x, y) = y(ay + x 2 + y 2 ), a ∈ R \ {0}.Parabolic type: q(x, y) = y(x − y 2 ).
Intersecting lines type: q(x, y) = yx(y + 1), Mixed type: q(x, y) = y(1 Remark 3.2.The name of the types of the form q in Proposition 3.1 comes from the type of the conic q(x,y) y = 0.The conic x + y + axy = 0, a ∈ R \ {0}, is a hyperbola, since the discriminant a 2 is positive.Similarly, the conic ay + x 2 − y 2 = 0, a ∈ R, is a hyperbola, since its discriminant is equal to 4. Clearly, the conic ay + x 2 + y 2 = 0, a ∈ R, is a circle with the center (0, − a 2 ) and radius a 2 .Now we prove Proposition 3.1.
Proof of Proposition 3.1.Since p(x, y) is reducible, it is of the form p = p 1 p 2 , where Without loss of generality we can assume that a 2 ̸ = 0, since otherwise we apply the alt (x, y) → (y, x) to exchange the roles of x and y.Since a 2 ̸ = 0, the alt

is invertible and hence:
A sequence ϕ 1 (β) has a moment matrix ϕ 1 M(k; β) satisfying the column relation We separate two cases according to the value of c 3 .
Case 1: c 3 = 0.In this case (3.1) is equal to A sequence ϕ 1 (β) has a moment matrix ϕ 1 M(k; β) satisfying the column relation Since by assumption β and hence ϕ 1 (β) admit a rm, supported on Hence, also M(k; β) satisfies a nontrivial column relation between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition.Therefore (c 0 , c 1 , c 2 ) ̸ = (0, 0, 0).
Case 1.1: c 0 ̸ = 0. Dividing the relation in (3.2) by c 0 , we get: The quadratic equation 1+ c 2 y + c 5 y 2 = 0 must have two different real nonzero solutions, otherwise Z(y(1 + c 2 x + c 5 y)) is a union of two parallel lines.Then it follows by [CF96] that there is a nontrivial column relation in M(k; β) between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition.So we have the parallel lines type relation from the proposition.
Case 1.1.2.1.2:c2 ̸ = 0. We apply the alt Further on, the relation in (3.9) is equivalent to Finally, applying the alt satisfying the parabolic type relation in the proposition.
Case 1.2: c 0 = 0.In this case (3.2) is equivalent to: A sequence ϕ 1 (β) has a moment matrix ϕ 1 M(k; β) satisfying the column relation Assume that c 1 = 0. Since by assumption β and hence ϕ 1 (β) admits a rm, supported on Hence, also M(k; β) satisfies a nontrivial column relation between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition.Hence, c 1 ̸ = 0. Applying the alt (x, y) → (c 1 x, y) to ϕ 1 (β), we obtain a sequence with the moment matrix satisfying the column relation of the form (3.6) and we can proceed as in the Case 1.1.2above.
Applying the alt satisfying the mixed type relation in the proposition.
Case 2.2.2: c0 ̸ = 0. Dividing the relation in (3.18) with c0 , it becomes equal to the relation in (3.16) from the Case 2.1.2,so we can proceed as above.□

SOLVING THE TMP ON CANONICAL REDUCIBLE CUBIC CURVES
Let β = {β i } i∈Z 2 + ,|i|≤2k be a sequence of degree 2k, k ∈ N, and the set of rows and columns of the moment matrix M(k; β) in the degree-lexicographic order.Let be a polynomial of degree 3 in one of the canonical forms from Proposition 3.1.Hence, c(x, y) a polynomial of degree 2. β will have a Z(p)-rm if and only if it can be decomposed as where has a representing measure on y = 0, has a representing measure on the conic c(x, y) = 0, and the sum in (4.3) is a component-wise sum.On the level of moment matrices, (4.3) is equivalent to Note that if β has a Z(p)-rm, then the matrix M(k; β) satisfies the relation p(X, Y ) = 0 and it must be rg, i.e., (4.5) We write ⃗ X (0,k) := (1 , X, . . ., X k ).Let T ⊆ C be a subset, such that the columns from T span the column space C(M(k; β)) and P is a permutation matrix such that moment matrix M(k; β) := P M(k; β)P T has rows and columns indexed in the order ⃗ X (0,k) , T \ ⃗ X (0,k) , C \ ( ⃗ X (0,k) ∪ T ). (4.6) In this new order of rows and columns, (4.4) becomes equivalent to We write By the form of the atoms, we know that M(k; β (ℓ) ) and M(k; β (c) ) will be of the forms for some Hankel matrix A ∈ S k+1 .Define matrix functions F : S k+1 → S(k+1)(k+2) 2 and H : S k+1 → S k+1 by (1) The sequence with the moment matrix F(A) has a Z(c)-representing measure.
(2) The sequence with the moment matrix H(A) has a R-representing measure.
Proof.First we prove the implication (⇒).If β has a Z(p)-rm µ, then µ is supported on the union of the line y = 0 and the conic c(x, y) = 0. Since the moment matrix, generated by the measure supported on y = 0, can be nonzero only when restricted to the columns and rows indexed by ⃗ X (0,k) , it follows that the moment matrix generated by the restriction µ| {c=0} (resp.µ| {y=0} ) of the measure µ to the conic c(x, y) = 0 (resp.line y = 0), is of the form F(A) (resp.H(A) ⊕ 0k(k+1) 2 ) for some Hankel matrix A ∈ S k+1 .
It remains to establish the implication (⇐).Let M (c) (k) (resp.M (ℓ) (k)) be the moment matrix generated by the measure µ 1 (resp.µ 2 ) supported on Z(c) (resp.y = 0) such that respectively, where P is as in (4.6).The equalities (4.12) imply that M(k; β) = M (c) (k) + M (ℓ) (k; β).Since the measure µ 1 + µ 2 is supported on the curve Z(c) ∪ {y = 0} = Z(p), the implication (⇐) holds.□ Lemma 4.2.Assume the notation above and let the sequence ∈ S k+1 be a Hankel matrix such that F(A) admits a Z(c)-representing measure and H(A) admits a R-representing measure.Let c(x, y) be of the form c(x, y) = a 00 + a 10 x + a 20 x 2 + a 01 y + a 02 y 2 + a 11 xy with a ij ∈ R and exactly one of the coefficients a 00 , a 10 , a 20 is nonzero.(4.13) If: (1) a 00 ̸ = 0, then (2) a 10 ̸ = 0, then (3) a 20 ̸ = 0, then Proof.By Lemma 4.1, F(A) has a Z(c)-rm for some Hankel matrix A ∈ S k+1 .Hence, F(A) satisfies the rg relations X i Y j c(X, Y ) = 0 for i, j ∈ Z + , i + j ≤ k − 2. Let us assume that a 00 ̸ = 0 and a 10 = a 20 = 0.In particular, F(A) satisfies the relations Observing the rows 1 , X, . . ., X k of F(A), the relations (4.14) imply that (4.15) Using the forms of M(k; β) and F(A) (see (4.8) and (4.10)), it follows that β , where k ≥ 3, be a sequence of degree 2k.Assume that M(k; β) is positive semidefinite and satisfies the column relations (4.5).Then: (1) F(A) ⪰ 0 for some A ∈ S k+1 if and only if (4) We have that Proof.By the equivalence between (1a) and (1b) of Theorem 2.2 used for (M, (4.16) and (4.17) where Using the equivalence between (1a) and (1b) of Theorem 2.2 again for the pairs (M, A) = (F(A), A) and (M, A) = ( F(A) ⃗ X (0,k) ∪T , A), it follows that (4.18) Since F(A) ⪰ 0 implies, in particular, that F(A) ⃗ X (0,k) ∪T ⪰ 0, (4.18) implies that (4.19) Proof of Claim.By (4.18) and (4.19), it suffices to prove that F( A min ) ⪰ 0. By definition of T and the relations for some matrix W, A 22 (A 23 ) T X = (A 12 ) T (A 13 ) T for some matrix X.We have that where I is the identity matrix of the same size as A 22 and we used (4.22) in the second equality.This proves the Claim.■ Using (4.17  (1) First compute A min := A 12 (A 22 ) † A 12 .By Lemma 4.3.(3),there is one entry of A min , which might need to be changed to obtain a Hankel structure.Namely, in the notation (4.13), if: (a) a 00 ̸ = 0, then the value of (A min ) k,k must be made equal to (A min ) k−1,k+1 .(b) a 10 ̸ = 0, then the value of (A min ) 1,k+1 must be made equal to (A min ) 2,k .(c) a 20 ̸ = 0, then the value of (A min ) 2,2 must be made equal to (A min ) 3,1 .
Let A min be the matrix obtained from A min after performing the changes described above.
(2) Study if F( A min ) and H( A min ) admit a Z(c)-rm and a R-rm, respectively.If the answer is yes, β admits a Z(p)-rm.Otherwise by Lemma 4.2, there are two antidiagonals of the Hankel matrix A min , which can by varied so that the matrices F( A min ) and H( A min ) will admit the corresponding measures.Namely, in the notation (4.13), if: (a) a 00 ̸ = 0, then the last two antidiagonals of A min can be changed.(b) a 10 ̸ = 0, then the left-upper and the right-lower corner of A min can be changed.
(c) a 20 ̸ = 0, then the first two antidiagonals of A min can be changed.
To solve the Z(p)-TMP for β one needs to characterize, when it is possible to change these antidiagonals in such a way to obtain a matrix Ȃmin , such that F( Ȃmin ) and H( Ȃmin ) admit a Z(c)-rm and a R-rm, respectively.
In Sections 5 and 6 we solve concretely the TMP on reducible cubic curves in the circular and parabolic type form (see the classification from Proposition 3.1).The parallel lines type form was solved in [Zal22a], while the hyperbolic type forms will be solved in the forthcoming work [YZ+].

CIRCULAR TYPE RELATION
In this section we solve the Z(p)-TMP for the sequence β = {β i,j } i,j∈Z + ,i+j≤2k of degree 2k, k ≥ 3, where p(x, y) = y(ay + x 2 + y 2 ), a ∈ R \ {0}.Assume the notation from Section 4. If β admits a Z(p)-TMP, then M(k; β) must satisfy the relations (5.1) In the presence of all column relations (5.1), the column space C(M(k; β)) is spanned by the columns in the set where Let M(k; β) be as in (4.9).Let (5.3) As described in Remark 4.4, A min might need to be changed to , where η := (A min ) 1,3 − (A min ) 2,2 .
• There exists a (rank M(k; β))-atomic Z(p)-representing measure if and only if any of the following holds: η = 0.
η > 0 and H(A min ) is positive definite.
In particular, a p-pure sequence β with a Z(p)-representing measure admits a (rank M(k; β))atomic Z(p)-representing measure.
Remark 5.2.In this remark we explain the idea of the proof of Theorem 5.1 and the meaning of the conditions in the statement of the theorem.By Lemmas 4.1-4.2, the existence of a Z(p)-rm for β is equivalent to the existence of t, u ∈ R such that F(G(t, u)) admits a Z(ay + x 2 + y 2 )-rm and H(G(t, u)) admits a R-rm.Let We denote by ∂R i and Ri the topological boundary and the interior of the set R i , respectively.By the necessary conditions for the existence of a Z(p)-rm [CF04, Fia95, CF96], M(k; β) must be psd and the relations (5.6) must hold.Using also Theorem 2.6, Theorem 5.1.(1)is equivalent to M(k; β) ⪰ 0, the relations (5.6) hold and In the proof of Theorem 5.1 we show that (5.11) is equivalent to Theorem 5.1.(2): (1) First we establish (see Claims 1 and 2 below) that the form of: • R 1 is one of the following: where the left case occurs if η > 0 and the right if η = 0.The case η < 0 cannot occur.• R 2 is one of the following: where the left case occurs if H 2 /H 22 > 0 and the right if H 2 /H 22 = 0.
The latter statement is further equivalent to Theorem 5.1.(2a).
(3) If η > 0, then by the forms of R 1 and R 2 , I = ∂R 1 ∩ ∂R 2 is one of the following: (i) ∅, (ii) a one-element set, (iii) a two-element set.In the case: • (i), a Z(p)-rm for β clearly cannot exist.
The latter statement is equivalent to Theorem 5.1.(2(b)ii).
• (iii), (5.11) is equivalent to H 2 being positive definite, which is Theorem 5.1.(2(b)i).Moreover, in this case for at least one of the points (t, u) ∈ I, a Z(ay + x 2 + y 2 )-rm and a R-rm exist for F(G(t, u)) and H(G(t, u)), respectively.
First we establish a few claims needed in the proof.Claim 1 (resp.2) describes R 1 (resp.R 2 ) concretely.
Proof of Claim 1.Note that Using (5.14), (5.15) and the definition of R 1 , we have that which proves (5.12).
To prove (5.13) first note that by construction of F(A min ), the columns 1 and X are in the span of the columns indexed by C \ ⃗ X (0,k) .Hence, there are vectors (5.17) of the forms .
We separate three cases according to r.
Proof of Claim 5. Note that H(G(t i , u i )), i = 1, 2, is of the form Assume on the contrary that none of H(G(t 1 , u 1 )) and H(G(t 2 , u 2 )) admits a R-rm.Theorem 2.5 implies that the column X k of H(G(t i , u i )), i = 1, 2, is not in the span of the other columns.Using this fact, the facts that H(G(t i , u i )), i = 1, 2, are not pd (by (t i , u i ) ∈ ∂R 2 , i = 1, 2) and H 2 is pd, it follows that there is a column relation Since the first column of H(G(t i , u i )) ⪰ 0, i = 1, 2, is in the span of the others, (5.22) is equivalent to If v 1 ̸ = 0, this contradicts to (5.24) since u 1 ̸ = u 2 .Hence, v 1 = 0.By the Hankel structure of H(G(t i , u i )), i = 1, 2, we have that Then (5.24) and v 1 = 0 imply that (5.25) , is a principal submatrix of H 2 , (5.25) contradicts to H 2 being pd.This proves Claim 5. ■ Now we prove the implication (5.11) ⇒ Theorem 5.1.(2).Since (t 0 , u 0 ) ∈ R 1 , it follows that R 1 ̸ = ∅.By (5.12), η ≥ 0. We separate two cases according to the value of η.
Case 1: η = 0. We separate two cases according to the invertibility of H 2 .
Case 1.1: H 2 is not pd.Since H 2 is not pd, then by Theorem 2.5, the last column of H(G(t 0 , u 0 )) is in the span of the previous ones.But then by rg, the last column of H 2 is in the span of the previous ones.This is the case Theorem 5.1.(2(a)ii).
Case 1.2: H 2 is pd.We separate two cases according to the invertibility of (H(A min )) ⃗ X (0,k−1) .
Case 1.2.2: rank(H(A min ) ⃗ X (0,k−1) ) < k.We will prove that this case cannot occur.It follows from the assumption in this case that rank H(A min ) = rank H 2 = k.Further on, the last column of H(A min ) cannot be in the span of the previous ones (otherwise rank H(A min ) < k).Hence, by Theorem 2.5, H(A min ) = H(G(0, 0)) does not admit a R-rm.Using this fact and Claim 3, (0, 0) ∈ ∂R 2 .If t 0 = 0, then R 1 ∩ R 2 = {(0, 0)}, which contradicts to the third condition in (5.11).So 0 < t 0 must hold.Since η = 0, Claim 1 implies that R 1 = {(t, 0) : t ≥ 0} is a horizontal half-line.By the form of ∂R 2 , which is the union of the graphs of two square root functions on the interval (−∞, t 0 ], intersecting in the point (t 0 , u 0 ) and such that (t 0 , u 0 ) ∈ ∂R 2 , it follows that R 1 ∩ R 2 = {(0, 0)}.Note that by H 2 ≻ 0, we have H 2 /H 22 > 0 and hence h(t) ̸ ≡ 0 (see (5.8)), which implies that the square root functions are indeed not just a horizontal half-line.As above this contradicts to the third condition in (5.11).Hence, Case 1.2.2 cannot occur.
Case 2: η > 0. By assumptions, (t 0 , u 0 ) ∈ R 1 ∩ R 2 .By Claim 4, I ̸ = ∅ and I has one or two elements.We separate two cases according to the number of elements in I.
Case 2.1: I has two elements.By Claim 4, H 2 /H 22 > 0. If H 2 is not pd, then the fact that H(G(t 0 , u 0 )) has a R-rm, implies that H 2 /H 22 = 0, which is a contradiction.Indeed, if H 2 /H 22 > 0 and H 2 is not pd, then there is a nontrivial column relation among columns X 2 , . . ., X k in H 2 .By Proposition 2.3, the same holds for H(G(t 0 , u 0 )).
Case 2.2: I has one element.Let us denote this element by ( t, ũ).By Claim 4, We separate two cases according to these two possibilities.
• If H(A min ) is pd, then rank H(A min ) = rank H 22 + 2. This and (5.26) imply that β admits a (rank ) being psd.Hence, in this case β has a (rank M(k; β)+1)atomic Z(p)-rm.Moreover, there cannot exist a (rank M(k; β))-atomic Z(p)-rm in this case.Indeed, Using (5.13) and (5.21), in every point from (R 1 ∩ R 2 ) \ I at least rank F(A min ) + rank H 22 + 2 atoms are needed in a Z(p)-rm.
This concludes the proof of the moreover part.

PARABOLIC TYPE RELATION
In this section we solve the Z(p)-TMP for the sequence β = {β i } i,j∈Z + ,i+j≤2k of degree 2k, k ≥ 3, where p(x, y) = y(x − y 2 ).Assume the notation from Section 4. If β admits a Z(p)-TMP, then M(k; β) must satisfy the relations (6.1) In the presence of all column relations (6.1), the column space C(M(k; β)) is spanned by the columns in the set (6.2) where Let M(k; β) be as in (4.8).Let (6.3) As described in Remark 4.4, A min might need to be changed to Let F(A) and H(A) be as in (4.10).Define also the matrix function Let us define the matrix and P be a permutation matrix such that moment matrix M(k; β) := P M(k; β)( P ) T has rows and columns indexed in the order T , C \ T .
The solution to the cubic parabolic type relation TMP is the following.
• There exists a (rank M(k; β))-atomic Z(p)-representing measure if and only if any of the following holds: η = 0.
In particular, a p-pure sequence β with a Z(p)-representing measure admits a (rank M(k; β))atomic Z(p)-representing measure.
Remark 6.2.In this remark we explain the idea of the proof of Theorem 6.1 and the meaning of conditions in the statement of the theorem.By Lemmas 4.1-4.2, the existence of a Z(p)-rm for β is equivalent to the existence of t, u ∈ R such that F(G(t, u)) admits a Z(x − y 2 )-rm and H(G(t, u)) admits a R-rm.Let We denote by ∂R i and Ri the topological boundary and the interior of the set R i , respectively.By the necessary conditions for the existence of a Z(p)-rm [CF04, Fia95, CF96], M(k; β) must be psd and the relations (6.8) must hold.Then Theorem 6.1.(1) is equivalent to M(k; β) ⪰ 0, the relations (6.8) hold and ) and H(G(t 0 , u 0 )) admit a Z(x − y 2 )-rm and a R-rm, respectively. (6.12) In the proof of Theorem 6.1 we show that (6.12) is equivalent to Theorem 6.1.(2): (1) First we establish (see Claims 1 and 2 below) that the form of: • R 1 is one of the following: where the left case occurs if η ̸ = 0 and the right if η = 0. • R 2 is one of the following: where the left case occurs if k 12 ̸ = 0 and the right if k 12 = 0.
(3) Assume that F 22 is positive definite and H 22 is only positive semidefinite but not definite. If: • u 1 = 0, then we show that (6.12) is equivalent to (6.13).The latter statement is further equivalent to Theorem 6.1.(2(b)i).
First we establish a few claims needed in the proof.Claim 1 (resp.2) describes R 1 (resp.R 2 ) concretely.
To prove (6.15) first note that by construction of F(A min ), the columns 1 and X k are in the span of the columns indexed by C \ ⃗ X (0,k) .Hence, there are vectors of the forms (6.20) .
We separate three cases according to r.
Proof of Claim 2. Permuting rows and columns of H(G(t, u)) we define where we use the definition (6.23) of K(t, u) in the last equivalence.Moreover, rank H(G(t, u)) = rank H 22 + rank K(t, u).This proves (6.24) and (6.25).
The equality is achieved if: The moreover part follows by noticing that f (0) = 0 and hence on the interval [0, t ± ], f attains all values between 0 and p max .■ In the proof of Theorem 6.1 we will need a few further observations: • Observe that (6.29) where in the first equality we used (6.29) and in the second the definition of t 1 (see (6.9)).
• We have where in the first equality we used (6.29) and in the second the definition of u 1 (see (6.9)).
Case 1: F 22 is not pd.Let β (c) be a sequence corresponding to the moment matrix F(G(t 0 , u 0 )).Let γ = (γ 0 , . . ., γ 4k ) be a sequence defined by where γ = (γ 2 , . . ., γ 4k−2 ).Since F 22 is not pd, it follows that there is a non-trivial column relation in F 22 , which is also a column relation in A γ by Proposition 2.3.By Theorem 2.7, γ has a R-rm, which implies by Theorem 2.5, that A γ is rg.Hence, the last column of A γ = F(G(t 0 , u 0 )) is in the span of the columns in T \ {1, X k }.It follows that (6.33) On the other hand, by construction of F , the column X k is also in the span of the columns in T \ {1, X k }.Hence, (6.34) By (6.33) and (6.34), it follows that (A min ) 2,k = (A min ) 1,k+1 or equivalently η = 0, and u 0 = 0. Note that Further on, F(A min ) has a Z(x − y 2 )-rm by Theorem 2.7 and H(A min ) by Theorem 2.5.Indeed, the column X k of F(A min ) is in the span of the others and since H(G(t 0 , 0)) satisfies the conditions in Theorem 2.5, the same holds for H(A min ).But then the property (6.10) holds (note that η = 0).This is the case Theorem 6.1.(2a).
We separate two cases according to the value of u 1 .
Case 2.2: H 22 is pd.We separate two cases according to the value of η.
Assume that Theorem 6.1.(2(b)ii)holds.We separate two cases according to the value of η: • η = 0. We separate two cases according to the existence of a R-rm of H(A min ): -The last column of H(A min ) is in the span of the previous ones.Then as in the previous paragraph, F(A min ) and H(A min ) admit a (rank F(A min ))-atomic Z(x − y 2 )-rm and a (rank H(A min ))-atomic R-rm, respectively.Hence, β has a (rank M(k; β))-atomic Z(p)-rm.-The last column of H(A min ) is not in the span of the previous ones.Since also t 1 > 0, it follows that rank H(A min ) = rank H 22 + 2. But then rank H(G(t 1 , u 1 )) = rank H 22 and rank F(G(t 1 , u 1 )) = rank F(A min ) + 2 (see (6.15)).This implies that M(β; k) admits a (rank M(k); β)-atomic Z(p)-rm.• η ̸ = 0. We separate two cases according to rank H(A min ), which can be either rank H 22 +2 or rank H 22 + 1 (since t 1 > 0).
Since for a p-pure sequence with M(k; β)) ⪰ 0, (6.42) implies that H(A min ) is pd, it follows by the moreover part that the existence of a Z(p)-rm implies the existence of a (rank M(k; β))atomic Z(p)-rm.□ The following example demonstrates the use of Theorem 6.1 to show that there exists a bivariate y(x − y 2 )-pure sequence β of degree 6 with a positive semidefinite M(3) and without a Z(y(x − y 2 ))-rm.By Theorem 6.1, β does not have a Z(y(x − y 2 ))-rm, since by (2(c)ii) of Theorem 6.1, (6.43) should be positive.
be a real symmetric matrix where A ∈ S n , B ∈ R n×m and C ∈ S m .Then: (1) The following conditions are equivalent: (a) M ⪰ 0. (b) C ⪰ 0, C(B T ) ⊆ C(C) and M/C ⪰ 0. (c) A ⪰ 0, C(B) ⊆ C(A) and M/A ⪰ 0.
) T A 33 ⪰ 0, it follows by Theorem 2.2, used for (M, A) = (B, A 22 ), that rank B = rank A 22 .Using this and the Claim, (4.23) implies the statement (4).Since M(k; β) satisfies the relations (4.5), it follows that the restrictionF( A min ) C\ ⃗ X (0,k) ,C satisfies the column relations X i Y j c(X, Y ) = 0 for i, j ∈ Z + such that i + j ≤ k − 2.By Proposition 2.3, these relations extend to F( A min ), which proves (3).□ Remark 4.4.By Lemmas 4.1-4.3,solving the Z(p)-TMP for the sequence β = {β i } i∈Z 2 + ,|i|≤2k , where k ≥ 3, with p being any but the mixed type relation from Proposition 3.1, the natural procedure is the following:
[Zal21,Lemma 2.11].The proof of [Zal21, Lemma 2.11] is separated into two cases:A 2 /A 1 > 0 and A 2 /A 1 = 0.The case A 2 /A 1 = 0 does not use (2.7).Assume now that A 2 /A 1 > 0 or equivalently rank A 2 > rank A 1 .Invertibility of A 1 (and by A 2 /A 1 > 0 also A 2 is invertible) is used in the proof of[Zal21, Lemma 2.11]for the application of the quotient formula ([CH69]) H(A min ) ⪰ 0. Permuting rows and columns, this implies that By Theorem 2.2, used for (M, C) = ( H(A min ), H 22 ), it follows that H 22 ⪰ 0 and h 12 , h 23 ∈ C(H 22 ).Let L : S 2 → S k+1 , L(A) = We are searching for the maximum of f (t) on the interval [0, k 11 ].The stationary points of f are t ± and Claim 3 is clear.Assume that k 12 ̸ = 0. Then clearly tu is maximized in some point satisfying (k 11 − t)(k 22 − u) = Example 6.3.Let β be a bivariate degree 6 sequence given by