Non-BPS Supersymmetric 3pt Amplitude for One Massless, Two Equally Massive Particles

In this paper, the non-BPS amplitudes ($Z<2m$) are considered. Utilizing on-shell methods, the three point amplitudes of two equal-mass particles and one massless particle were constructed, where the two massive particles are non-BPS states. We verify the result by matching the $Z = 0$ and BPS limit with $N = 2$ supersymmetry. As an application we derive the non-BPS coupling for $N = 4$ super-Maxwell and supergravity.

Supersymmetry (SUSY) requires on-shell fermionic variables. Spinor helicity formalism then must be formulated in on-shell superspace. The formulation in massless [2] and massive [3,4] on-shell superspace were subsequently constructed. Here we consider extended SUSY with nonvanishing central charge, using superamplitudes in N = 2 SUSY as building blocks. For extended N = 2 SUSY, the algebra takes the form and the central charge of a massive particle has a bound, Z ≤ 2m. BPS states are the states that saturate this bound. The solution to the amplitude of BPS states are studied in [5,6].
M Z (1 (I 1 ···I 2s 1 ) , 2 (J 1 ···J 2s 2 ) , P h ) = 1 (I 1 ···I 2s 1 ) Figure 1: A 3-pt superamplitude of two massive multiplets with equal mass m and a massless multiplet with helicity h. Note that the central charges of the two massive particles are opposite to each other (Z and −Z), due to central charge conservation.
In this paper, we will consider the non-BPS three point superamplitude shown in Figure 1. The amplitude M Z is made up of two massive multiplets, with spin-2s 1 and spin-2s 2 , coupling a spin-h massless multiplet, where the two massive multiplets have equal masses m and opposite central charges (Z and −Z, respectively) 1 . We require M Z satisfy supersymmetry, i.e., where Q A α andQ Ȧ α are the supersymmetric generators, and A = 1, 2 for N = 2 SUSY. By examining the the form of generators, we can show that M Z is a function of δ A , H A , andH A , which are defined in (9). In addition, we also show that we can always factorize the superamplitude with spinning multiplets in N = 2 SUSY into two parts, M Z (1 (I 1 ···I 2s 1 ) , 2 (J 1 ···J 2s 2 ) , P h ) = M(1 (I 1 ···I 2s 1 ) , 2 (J 1 ···J 2s 2 ) , P h ) · A (Z, m) One of them is a bosonic factor M(1 (I 1 ···I 2s 1 ) , 2 (J 1 ···J 2s 2 ) , P h ), which carries the little group (LG) indices, including both the S U(2) indicies for the massive and U(1) for the massless multiplets. The other is a LG neutral and SUSY invariant quantity A (Z, m) 2 . Therefore, to solve M Z , we need to solve A. There are 2 solutions for A, and M Z is a combination of the two solutions, see (57). We verify the solutions of A by matching to known results, i.e., the Z = 0 limit [4] and the BPS limit [6].
As as application, we use our results in N = 4 SUSY. The presence central charge breaks R-symmetry from S U(4) to S U(2) ⊗ S U (2), and the central charge is where A, B = 1, 2, 3, 4. We can then treat A as a "seed" for the superamplitude in N = 4 SUSY.
As an example, where the subscripts 12 or 34 indicate which projected S U(2) group they are describing. As in N = 2 SUSY, the superamplitude for spinning multiplets in N = 4 SUSY can be factorized as well, similar to (5) (see (75) for more details). We then proceed to consider super Maxwell theories and super-gravity (SUGRA) theories by choosing suitable bosonic factors. In the former case, the massless multiplet carries helicity +1, while in the latter case, there are 2 conjugated superamplitudes, and the massless multiplets in the superamplitudes carry helicity 2 and helicity 0.
In section 2, we show that M Z is a function of δ A , H A , andH A by examining the generators. In section 3, we solve for A and M Z by requiring them being SUSY invariant. In section 4, we compare our results in the Z = 0 limit and the BPS limit with previous works. In section 5, we apply the results to N = 4 non-BPS SUSY, and explore super-Maxwell and super-gravity theories.

The building blocks for the 3pt amplitude
As mentioned in the introduction, we are dealing with the 3pt amplitude M(1, 2, P) consisting of two equal-mass m, non-BPS particles (leg 1 and 2) and a massless particle (leg P), see Figure  1. In this chapter, we will introduce N = 2, non-BPS generators. Given the generators, we find the general form of M(1, 2, P) is where are the building blocks for the 3pt amplitude ( ζ| and [ξ| are reference spinors, and the dependence on the reference spinors will drop out if H andH is evaluated on the support of A δ A ). In other words, the superamplitude is proportional to a delta function A δ A , and the rest of the amplitude is a function of H andH.

The generators
The generators for massless particles satisfy the anti-commutation rules while for massive particles, the generators satisfy where Z AB = Z · ǫ AB is the central charge. Note that Z has a bound, 0 ≤ |Z| ≤ 2m, and BPS states are states that satisfy |Z| = 2m, where calculations will be lot simpler. However, we are interested in the amplitude for general Z.
In the 3pt amplitude we are considering, the central charges of the two massive particles must carry opposite signs, Z 1 + Z 2 = 0, due to central charge conservation. We then define The bound − π 2 ≤ θ Z ≤ π 2 follows directly from the bound of the central charge 0 ≤ |Z| ≤ 2m, and BPS limit happens at θ Z = π 2 . For convenience, we also define c z = cos(θ Z ) and s z = sin(θ Z ). 4 To be a superamplitude, M(1, 2, P) should satisfy where Q A α andQα A are the sum of the supersymmetric generators of the particles By introducing a set of Grassmann variables {η A iI , η A P } (i = 1, 2) 3 , we can express the generators as where (σ 1 , σ 2 ) = (1, −1). One can verify them by substituting them into (10) and (11). Therefore, the sum of the generators are

The fermionic delta function
From (17), we see that the generators can be separated into a multiplicative part and a differential part. By suitably combining the generators, we are able to subtract the differential part, and the superamplitude will be found to be proportional to the remaining multiplicative part. More explicitly, we can show that M(1, 2, P) is proportional to the product of two fermionic delta functions where δ A ≡ 2 i=1 Pi I η A iI . Let's contract (13) with λ α P andλ Pα to get rid of η P 's, and multiply them with some proper coefficients, we get 3 In this paper, the Grassmann variables' indices are defined as: i: the label of the massive particles, i = 1, 2 I: the LG indices for the massive particles A: the R-charge indices. .
5 By subtracting the two equations, we can see that if c 2 z − s 2 z 0, then This result implies that M(1, 2, P) is proportional to the delta function We then write M(1, 2, P) as We now want to see what constraints SUSY places on M res . Since the calculations of M res later in this section will be on the support of the delta function δ 2 (η), we introduce a notation " δ =", and denote A is equal to B on the support of the delta function as A δ = B. In other words, To see what constraints SUSY places on M res , first observe that when one imposes momentum conservation p 1 + p 2 + P = 0, the following two commutators vanishes Therefore, according to (13), M res should satisfy Let's introduce two reference spinors ζ| and [ξ|, which are not parallel to P| and [P|, 4 To simplify notations, let's further define 4 In the rest of this paper, ζ| and [ξ| stand for reference spinors, and they are not parallel to P| and [P|. 6 and we have The plus and minus signs in the subscripts are indicating that they raise/lower the η i -order 5 .

H andH
On the delta function, we have (simply because they are proportional to δ A ) therefore, Let's define the proportionality coefficient to be H andH H andH are independent to those reference spinors when evaluated on the support of the delta function. For more details about H andH, see Appendix C.
In Appendix D, we show that M res is a function of H andH only. More precisely, all η i (η i = η 1 or η 2 ) in M res must be of the form δ A , H A ,H A play central roles in obtaining the superamplitude. In fact, if a superamplitude satisfies (13), then all its η i 's (massive Grassmann variables) will appear in the form of δ A , H A ,H A (proven in Appendix D). Therefore, δ A , H A ,H A serve as "building blocks" of the superamplitude. To simplify notations, we also define 5 In this paper, the order of η i means the order of the sum of η 1 and η 2 . For instance, η A 1I η B 2J has order 2, η A 1I η B 2J η 1KA has order 3. Note that η A 1I η B 2J η C P has order 2, since η A P doesn't increase the order of η i .

Raising and lowering η order.
For an arbitrary quantity X, we can raise its η i -order by acting D A +α andD +αA (see (27) for their definitions) On the other hand, if a quantity meets D A −α orD −αA , the η i -order is lowered,

Solutions to N = 2
In this section, we will consider N = 2 super symmetry, and look for the explicit solution of A introduced in (5) (which is LG neutral and SUSY invariant), using the building blocks introduced in section 2. We will also consider spinning multiplets, and show that they can always be written in the form of (5).
The massless super field with vacuum of helicity h in N = 2 SUSY is On the other hand, the scalar massive multiplet in N = 2 SUSY is [5] Note that in the above equations, I, J stand for LG indices, while A, B stand for R-charge indices. From the above expansion, we can see that there are 5 spin-0 components (1 φ, 1φ, and 3 φ (AB) ), 4 spin-1 2 components (2 ψ I A , and 2ψ I A ), and 1 spin-1 component (W (IJ) ). Since a spin-s particles carries (2s + 1) degree of freedom, our massive multiplet has 8 bosonic d.o.f and 8 fermionic d.o.f.

The SUSY invariant quantity A
A is the quantity that satisfies

invariant under all LG transformations of the external particles 2. invariant under SUSY
Since A is annihilated by the super charges, all the discussions about M(1, 2, P) in section 2 applies to A as well. According to (22), where f, g, h are all functions of H andH. According to (28), f, g, and h should satisfy To get the full superamplitude, let's first solve h, which should satisfyDα A h = 0. We can do this by first writing down all possible combinations of H andH at each η i order, and fix their coefficients by demanding that h satisfiesDα A h = 0. After solving h, we can obtain g A and f by resorting to (40).
Since the R-charge indices should be fully contracted, the η i orders of the terms in h should be even numbers. Otherwise, there will be at least one η i that cannot find a partner to contract with. At each even orders of η (n) i , the possible terms are shown, respectively, and the most general form of h is and resorting to (34) and (35), we arrive at the relations between c i 's: This implies there are two solutions, since according to (44), the general form (42) is decoupled into two linearly independent terms (the result should not be surprising, since there are also two solutions in the massless case), Our final solutions are where a = 1, 2, labeling the two solutions, and C (1) = c 2 z , C (2) = −x, so as to make both solutions LG neutral 6 (and also make S (1) and S (2) look more "symmetrical"). The solutions depend on c z and s z , which are functions of Z.
Up to this point, we have two solutions to A. However, they are written in terms of H A and H A , which are not manifestly Lorentz covariant quantities themselves, see (31). Only when they are combined with the delta functions δ A can they be represented in a Lorentz covariant form. Let's first define 7 Q By combining f (a) , g (a) A , and h (a) with the delta functions, we can finally write down the three point amplitudes in a Lorentz covariant and R-charge symmetric form. (For more details, see Appendix C). 6 One may wonder why adding x-factors is the only way to change the helicity of vacuum state of the massless multiplet. There are after all several quantities that carries helicity, including λ Pα andλ Pα , and of course the x-factor. However, both λ Pα andλ Pα carry an extra S L(2, C) index, i.e., α orα, and should be contracted with λ I iα orλ I iα (i = 1, 2). But all λ I iα orλ I iα dependence are encoded in S (1) and S (2) , and their S L(2, C) indices are already contracted. As a result, x-factors is the only choice remains. 7 Note that Q A α andQα A (normal character Q) stand for generators, while Q A α andQ Ȧ α (curly character Q) stand for the little group covariant supersymmetric components of the solutions.
The first solution is and the second solution is Note that both of the solutions are inhomogeneous in Grassmann degree. The first term in S (1) (θ Z ) has Grassmann degree 4, followed by terms with Grassmann degree 6,8,4,6,4. Similar thing happens to S (2) (θ Z ). The two solutions are related. If we do Grassmann Fourier transform (defined in [4]) to either of the solutions, and change all angle brackets to square brackets (and vice versa), we get the other solution (up to an overall constant). This is in fact a direct consequence of the form of the generators (17). If we change all η's into ∂ ∂η 's (and vice versa) in Q α , and change all λ's intoλ's, we get exactlyQα. We can do the same transformation toQα to get Q α . Since the A is invariant under transformation generated by both Q α andQα, if the generators are symmetric under certain transformation, the solutions would inherit this property, and come in pairs consequently. The symmetry of the generators is the origin of why the solutions must come in pairs.
Observe that the solution (50) can be factorized into a product of two components The solution (51) can be factorized in a similar expression This observation will later play a crucial role in the discussion of BPS limit. To sum up, A is the linear combination of the two solutions Note that all η dependencies are encoded in S (1) and S (2) . There are two unfixed coefficients α 1 and α 2 , which can be fixed by requiring parity symmetry. We will go back to this later.

Spinning multiplets
A 3-pt amplitude of two equal massive particles with mass m and a massless particle with helicity h can be written in spinor helicity basis, see [1], where .
The amplitude for spinning multiplets can always be factorized into a product of a bosonic part M(1 (I 1 ···I 2s 1 ) , 2 (J 1 ···J 2s 2 ) , P h ) and A, In other words, if a spinor helicity variable that carries LG indices of the massive multiplets (e.g., I 1 · · · I 2s 1 ) have their S L(2, C) indices (e.g., α in λ I α ) contract with a spinor helicity variable whose LG indices contract with a Grassmann variable η A iI (we say the amplitude is "polluted"), then we can always use Schouten identity to reorder the S L(2, C) indices to make the amplitude "unpolluted". More explicitly, we can always write the amplitude without terms such as 1 I 1 1 I η A 1I . The reason is if such terms do exist, then according to the discussions in Appendix D, they must come in combinations λ I , and can be re-written in terms of H orH. For example, if 1 I 1 1 I η A 1I do exist, then it must appear in the form of i 1 I 1 i I η A iI ; but according to (31), i 1 I 1 i I η A iI = 1 I 1 P H A , and thus make the term "unpolluted".

Parity.
Parity symmetry relates each amplitude and its conjugate amplitude, demanding they have the same couplings, and that constraints the coefficients α 1 and α 2 in (57). In N = 2 supersymmetry, the massless multiplets with helicity +h and −h + 1 are We can see the components are related by parity transformation, and parity invariance requires the coefficients of the superamplitudes with (+h)-helicity and (−h + 1)-helicity multiplet should be related. Let's first write down the form of the superamplitudes where Parity invariance requires (β 1 , β 2 ) = (α 2 , α 1 ) and g i =ḡ i .

Limits of the amplitude
In this section, the Z = 0 limit and the BPS limit of A are examined. The former is a product of two N = 1 SUSY invariant delta functions, and the later is a product of two components that are related to each other by Grassmann Fourier transformation.

The Z = 0 limit
When Z 1 = Z 2 = 0, θ Z → 0, and the solution takes the form (see (52)) where we defined This is exactly the square of N = 1 SUSY invariant delta function defined in [4], which is not surprising, since we can see from (17) that Q A α is purely multiplicative in η's, whileQα A is purely differential in η's, and there is no mixing terms between two R-charge indices. Therefore, the solution being multiplicative of two solutions corresponding to two R-charges is an expected result.

13
The other solution is which is related to (61) by Grassmann Fourier transformation.
In the high energy limit, we can take η A i1 → η A i , η A i2 →η A i , and the above amplitudes have the limits 8 (see [1,4]) We can see they are equal to the MHV and anti-MHV amplitudes in [2].

The BPS limit
When the massive particles saturate the BPS limit, i.e., Z 1 = Z 2 = 2m, θ Z → π/4, and (17) becomes Although the anti-commutation relations (11) still hold, (54) is no longer valid. To see this, note that A is proportional to the delta function (21) if and only if c 2 z − s 2 z = 0 (See subsection 2.2), and the BPS limit doesn't meet this criteria. As a result, our solutions (54) do not apply to the BPS limit. However, in the BPS limit, the η i degree of freedom is half the number of that in the non-BPS limit. Therefore, to obtain the amplitude of the BPS limit, we shouldn't have started with (17), but rather one of the following set (either of them is sufficed to serve as a set of generators in the BPS limit), Although (50) and (51) are not the solutions for BPS limit, interesting things happen if we take θ Z = π/4. In this limit, the two factorized components in (52) are related. If we do Grassmann Fourier transformation to one of the components, then complex conjugate spinor helicity variables (λ ↔λ), and transform η P properly, it will become the other component, where By comparing with (B.5), we see that η 2 iI act as the barred Grassmann number of η 1 iI , and serve as another η-basis. In other words, (50) has unnecessary Grassmann variables in the BPS limit. We can use either of the factorized components in (52) to describe a BPS amplitude.
Let's see how the components in (52) are related to BPS amplitudes. If we rename the η iI in the LHS of (66) as η 1 iI , and the η iI in the RHS as η 2 iI , then the SUSY invariant amplitudes are (the first of them is the solution of the LHS set, while the second is that of the RHS set) [6], where the definition of δ A , H A ,H A are same as the previous ones 9 . Note that the index A in the previous non-BPS calculations stands for R-charge, while in (69), A is a label, labeling the two solutions. More concretely, η 1 iI and η 2 iI are more like a conjugate pair in (69). Despite the meaning of A in BPS limit departs radically from non-BPS case, the solutions in (69) can be related to non-BPS solution. The product of the two solutions in (69) is 9 Note that in [6], the amplitude reads which, according to (31), equals to 15 If we replace all η A P replaced by 1 √ 2 η A P , it will be equal to S (1) (θ Z = π 4 ), One may wonder why there should be a η P → 1 √ 2 η P in (73), and we can trace its origin from the generators (17). In the limit θ Z = π 4 , if we sum the two sets of generators Q A α in (66), and modify η P by 1 √ 2 η P , it will be proportional to the generators Q A α in (17). This explains why there should be a 1 √ 2 factor. One may argue that once we make η P → 1 √ 2 η P , the differential part inQα will acquire a √ 2 factor, so that the sums of Q Ȧ α in (66) are not proportional to Q Ȧ α in (17). But since non-BPS solution is the product of BPS solutions, when applying a differential operator, we would need to impose the product law of differentiation, which induces an extra factor of 2, and 1 √ 2 × 2 = √ 2 explains the √ 2 factor.

N = 4 Super-Maxwell and Supergravity
We have been working on N = 2 SUSY amplitudes in the previous sections, and now we consider N = 4 SUSY, where the central charge matrix can be put in the standard block-diagonal form, In central charge free N = 4 SUSY, the R-symmetry group is S U(4). However, central charge extension breaks the full S U(4) group into S U(2) ⊗ S U(2) subgroup. Since the N = 1, 2 part and the N = 3, 4 part do not mix with each other, the supersymmetric part of the amplitude is the square of A, with θ Z 's properly modified (c.f. (57)), The subscripts that S (1) AB and S (2) AB carries, i.e., 12 or 34, indicate which projected S U(2) group they are describing.

N = 4 Super-Maxwell
If the massless multiplet carries helicity +1, then the helicities of the component fields span from +1 to −1, which describes Super-Maxwell theory. On the other hand, if the vacuum states of the massive particles are scalars, then the component amplitudes will have massive particles' spins up to spin 2. Demanding that parity symmetry should be preserved, see (79), we have where e is the electrical charge and S (1) AB ≡ S (1) (θ Z AB ). Note that θ Z AB , which are functions of Z AB , determines all the coefficients of the component amplitudes, including minimal and non-minimal couplings. Let's explicitly write down the form of the component amplitude M(1 s=2 , 2 s=2 , P +1 ) as an example, where both massive external states are spin 2, and the massless external state has helicity +1, with all the LG indices symmetrized. We can see that the component amplitude consists of nonminimal couplings. In addition, the non-minimal couplings of M(1 s=2 , 2 s=2 , P +1 ) vanish if and only if Z AB = 0.
The g 1 problem of gravitational interaction.
The most general form of the 3-pt amplitude of two massive spinning particles with equal masses m interacting with a massless particle, where the two massive particles are spin S 1 and S 2 , and the massless particle has helicity h, is of the form (see [1]) It is shown in [7] that, if h = 2 in the above equation, i.e., 3-pt amplitude includes a graviton, then g 1 -term must vanish. Otherwise, we can't write down a Lagrangian with local operators that give raise to the amplitude.
Let's consider the component amplitude M(1 s=2 , 2 s=2 , P +2 ) (one of the component amplitudes of M(Φ, Φ, P +2 )), where both massive external states are spin 2, and the massless external state has helicity +2 (i.e. the graviton), and the coupling constants are We can see the component amplitudes contains g 1 term, which is forbidden. In order to get rid of the g 1 -term, we are forced to choose α 2 = α 4 = 0 10 . This choice not only excludes the g 1 term in M(1 s=2 , 2 s=2 , P +2 ), but g 1 term in all component amplitudes that includes graviton, e.g., M(1 s=1 , 2 s=1 , P +2 ). Substitute α 2 = α 4 = 0 into (82), we get In fact, not only g 1 -terms are excluded in all component amplitudes of this superamplitude, but also g 2 terms, leaving with only minimal coupling terms. This indicates that supersymmetry implies all components in massive scalar multiplets interact with graviton only through minimal couplings.

Summary and outlook
In this paper, we calculated the supersymmetric part in (5) Two special cases are discussed, the Z = 0 case and the BPS case (Z = 2m). In the Z = 0 case, the solutions can be factorized into products of N = 1 amplitudes, see (61) and (63). In the BPS limit, we know that half of the Grassmannian degrees drop out. It is shown in (73) that in BPS limit, the amplitude can be factorized into two components that are related by Grassmann Fourier transform, each of which is a BPS amplitude. In other words, non-BPS amplitude is the product of two BPS amplitudes, explaining the drop out of half of the Grassmannian degrees in the BPS limit.
The N = 4 supersymmetry is also considered, and especially super-Maxwell amplitude and SUGRA amplitude. The super-Maxwell amplitude indicates that there is non-minimal coupling of photons, see (80). The SUGRA amplitude showed that, if we require g 1 = 0, then all non-minimal couplings in graviton exchange vanish as well, see (87).
Any massive spinning body can be describes as a spinning particle at large distances, and black holes are no exception. For instance, a Kerr black hole can be describe as a massive particle with large spin, interacting with graviton fields through minimal couplings [8]. It would be interesting to see what spinning particles are able to describe a supersymmetric black hole at large distance. Recently, relative entanglement entropy of binary Kerr black holes is found to be nearly zero for minimal coupling in the Eikonal limit, and increases when spin multipole moments are turned on [9]. We are interested in the relative entanglement entropy of supersymmetric black holes, and examine whether BPS limit results the lowest entropy, relative to non-BPS amplitudes [10].
The Massless and Massive Momenta. The momentum of the massless particle P can be written as a product of two two-component spinors: and a massive momentum p i can be written as a product of two 2-by-2 matrices: where the I is the SU(2) index.

Explicit kinematics.
For massless particles with momentum we have For massive particles with momentum we have where c ≡ cos θ 2 e iφ , s ≡ sin θ 2 .

Contractions of Massive Spinors.
The on-shell condition for massive particle is given by where we choose which will be repeatedly used throughout massive amplitude calculations With these conventions, we find that momentum contracting with the massive spinors are: For massive spinors associated with the same particle whose LG indices contracted:

(A.13)
Definition and convention of the x-factor. The external momenta satisfy p 1 + p 2 + P = 0 . (A.14) The momentum conservation condition (A.14) and the on-shell condition yields: so that λ Pα is proportional to p 1ααλα P . This allow us to define the x-factor:

Appendix B. Grassmann variables
Convention of the Grassmann variables η.
When contract with Levi-Civita tensors, the massive Grassmann variables η's transform as To simplify equations, let's denote the contraction rules for massive Grassmann variables η i and for massless Grassmann variables η P Grassmann Fourier transformation. The Grassmann Fourier transformation of a function f (η) in η basis toη basis is For Grassmann variables of massless particles, the Grassmann Fourier transformation is while for Grassmann variables of massive particles Appendix C. More on the building blocks H A andH A play crucial rules in this paper, they serve as basic building blocks of the superamplitude. This appendix is devoted to introducing several important properties of them, which will be useful if one wants to reproduce the calculations we have done in this paper. To simplify equations, let's denote which is just a special case of (B.2).

Proof.
Let's just proof the first equation in the lemma, since the proof of the second is similar to that of the first.
where the last equality follows from Lemma Appendix C.1. Proof.
where we used Schouten identity. The other equations can be proven in similar ways.

(C.8)
Proof. Let m = 1, The second equation can be proven in similar fashion.

Appendix D. More on M res
The amplitude is proportional to δ 1 δ 2 follows straightforwardly given the generators, see (22). The main goal in this section is to study the rest of the amplitude, i.e., M res , in more detail, and proof that it is a function of H andH, in other words, all η i (η i = η 1 or η 2 ) in M res must be of the form Observe the form of the generators (17), they have a multiplicative part in η i and a differential part in η i . Therefore, since M res must satisfy (28), we have (see (27)  This implies whenever λ I iα η A iI appears in M (n) res , it must be either ζ1 I η A 1I + ζ2 I η A 2I or P1 I η A (D.13) The last identity follows from Lemma Appendix C.1. As a result, M (n) res is a function of H and H.
Given Lemma Appendix D.2, and given M (−1) res = 0, we can conclude that M (1) res is function of H andH. By iteration, we can conclude all M (n) res , and thus M res , are functions of H andH.