Massless NC N3LO flavor decomposition is wrong

[giacomomagni]

The N3LO factorization of the NC ($ZZ$ contributions) coupling is wrong. Let me explain the problem.
At N3LO (for the first time) there are 2 different flavor classes of diagrams (or topologies) as reported in Fig1 [1].

• The fl2 diagrams, in which the incoming and outgoing boson is coupled to the same fermionic line (these diagrams are the only one present up to NNLO)
• The fl11 diagrams, in which the incoming and outgoing boson is coupled to different fermionic lines

1. For CC the second class of diagrams vanishes as a single $W$ can't couple to flavor loop, so we are fine.
2. For NC PV we are also fine as this class does not shows up (or better, if it shows up (is allowed by symmetry ?!)
   it has not been computed yet as we are currently getting the F3 NC coefficient from the CC $\nu - \bar{\nu}$. ).

fl11 issue

Now all our references deals only with the photon exchange case, in which only VV couplings are present,
and here we are making our mistake. Following [2], see eq 55 and discussion above, the coefficient functions are factorized as follow:

$$g_{fl2} ( C_{fl2} + g_{fl11} / g_{fl2} C_{fl11} ) \quad \text{Eq (1)}$$

where $g_{fl2}$ is the coupling of the flavor class fl2 i.e. the standard one, while $g_{fl11}$ is the coupling of the new flavor which are then given in table 2 of [2] for photon exchange as a ration. However, in the case of Z exchange both Vector-vector and Axial-vector type coupling are present.

Currently we factorize the partonic coupling as

$$F_{i}^{bb'} = (g^b_A g^{b'}_A + g^b_V g^{b'}_V) C_i \otimes f_i \quad i = g,q$$

so we have two different solutions.

• Either we extend this equation to be:

F_{i}^{bb'} = [ 
               (g^b_{A,fl2} g^{b'}_{A,fl2} + g^b_{V,fl2} g^{b'}_{V,fl2}) C_{i,fl2} +
               (g^b_{A,fl11} g^{b'}_{A,fl11} + g^b_{V,fl11} g^{b'}_{V,fl11})  C_{i,fl11} 
        ] \otimes f_i \quad i = g, q

• Either we recover Eq(1) and we compute all the necessary $g_{fl11} / g_{fl2}$ ratios which can be quite complicated
  especially for the ZZ channel where both AA and VV contribute.

Finally $g,ps$ and $ns$ work in a similar way and we might not have to compute any VA combination.

fl02 issue

I suspect that we have a problem also for the diagram class fl02.
These diagrams are singlet diagrams, which should be taken into account in the PV violating SF when doing the combination $W^+ + W^-$.
Here the misunderstanding is caused by the convention of Vogt and & being different from our decomposition.
https://yadism.readthedocs.io/en/latest/theory/nonsinglet.html

The Fortran files (and papers), adopt the decomposition $C_{ns}$ ans $C_{s}$, clearly table2 of [2] states that fl02 has to be left out in the first case but, kept in the second.
Now since we rewrite

$$ C_s = C_{ns} + C_{ps} $$

This means that $C_{ps}$ doesn't vanish anymore, although here the name ps is highly confusing as the whole thing still couple to $q^-$ (i.e to the full valence)

As written explicitly this in [3] (see Fig3 and text above) these diagrams are precisely the one spoiling the correspondence $g_4 \to F_2$ and $2 x g_1 \to x F_3$, as they contain different traces in the polarized/unpolarized case.

Summary

• fl11 spoils the symmetry CC ($W^+ + W^-$) and NC, both for $ns,ps,g$. Still as F3 is only computed in CC we don't know yet the fl11 contributions. This implies that we need to add some different coupling for the AA and VV combinations for ($\gamma \gamma$, $\gamma Z$, $ZZ$) . Massless coefficients are still the same for VV and AA,
  but we need to distinguish 2 flavor classed.

• fl02 spoils the symmetry unpolarized to polarized as they develop "pure singlet" terms in the PV structure functions, which are coupled to $q^-$ (or the full Valence) with a $ps$ (i.e. averaged) coupling.

cc @felixhekhorn, what do you think?

[1] https://arxiv.org/pdf/hep-ph/0504242.pdf
[2] https://arxiv.org/pdf/hep-ph/9605317.pdf
[3] https://arxiv.org/pdf/2210.12014.pdf

[giacomomagni]

So to wrap up my understanding and to proceed to correct the implementation:

fl02

• contributes both to NC and CC in F3 with an averaged coupling, and proportional to valence contribution.
• contributes to PV and PC [3, figure 3] and [5], but differently [4]
• contributes to AA and VV in the same manner [5].
• does not contribute to CC \nu - \bar{\nu} ✅.
• it's directly related to the splitting function $P_{ns}^{s}$

fl11

• does not contribute to CC ✅.
• contributes to VV and AA in the same manner (?? guess) [5].
• does not contribute to g1 and F3 from [4] you can see it does not show up in eq 918 of g1 (??, guess)

Finally:

• N3LO polarized and unpolarized are different because the fl02 and fl11 contribute differently to PV and PC structure functions thus spoiling the relation between non singlet coeffs [4] ( cf eq 1074 and eq 918).

@felixhekhorn please correct me.

[4] https://arxiv.org/pdf/2208.14325.pdf
[5] https://arxiv.org/pdf/1506.04517.pdf

[giacomomagni]

Okay I might have a solution also for the fl11 problem.

I tries an explicit calculation with mathematica

(*Define SUN rappresentation *)
SU[n_] := Module[{a, b, c, basis},
  	a = 1/Sqrt[2] Flatten[Table[SparseArray[{{i, j} -> I, {j, i} -> I}, {n, n}], {i, 1, n}, {j, i + 1, n}], 1];
  	    b = 1/Sqrt[2] Flatten[Table[SparseArray[{{i, j} -> -1, {j, i} -> 1}, {n, n}], {i, 1, n}, {j, i + 1, n}], 1];
  	    c = DiagonalMatrix@*SparseArray /@ Orthogonalize[Table[SparseArray[{{i} -> I, {i + 1} -> -I}, {n}], {i, 1, n - 1}]];
  	    basis = Join[a, b, c];
          basis
  ];
(* u d s c b t *)
phV = {2/3, -1/3, -1/3, 2/3, -1/3, 2/3};
ZA = {1/2, -1/2, -1/2, 1/2, -1/2, 1/2};
ZV = ZA - 2 phV * sinW;

nf = 4;
coupling1 = ZA;
coupling2 = ZV;

(* Compute k eq 86, k is a constant *)
Q1 = DiagonalMatrix[coupling1[[;; nf]]];
Q2 = DiagonalMatrix[coupling2[[;; nf]]];
num = Tr[Dot[Q1, # ]] & /@  SU[nf]
den = Tr[Dot[Q1, Q2, # ]] & /@  SU[nf]
k = Total[num] / Total[den] // Simplify

(* Compute fl11 for non singlet (w2) *)
w2 = k Total[Total[Q2]] / nf
(* Compute fl11 for singlet and gluon (w3) *)
w3 = (Total[Total[Q2]] Total[Total[Q1]]) / (nf Total[Total[Q1.Q2]])

And this show explicitly that eq 54 of [2] or 86 of [4] is:

• $k_{\gamma \gamma} = 3$

• $k_{\gamma, Z_V} = \frac{6}{1 - 4 \sin{\theta_W}}$

• $k_{Z_V, \gamma} = \frac{6 - 12 \sin{\theta_W}}{1 - 4 \sin{\theta_W}} $

• $k_{Z_V,Z_V} = \frac{-3}{2\sin{\theta_W}}$

• $k_{Z_A, Z_A} = 0$ (or actually here you get inf as the denominator vanish...) for the axial Z coupling

And eventually for the VA couplings

• $k_{Z_A,\gamma} = k_{\gamma, Z_A} = 6 $
• $k_{Z_V, Z_A} = 6 - \frac{3}{\sin{\theta_W}}$
• $k_{Z_A, Z_V} = \frac{-3}{2\sin{\theta_W}}$

So I'd say that diagrams fl11 diagrams do not contribute to AA,
but are only present in VV.

Now it remains to understand if, they can contribute to the PS
Moreover in the yadism convention the couplings of the fl11 should be:

• c_g = <g_{g,b}> <g_{g,b'}>
• c_{q,ns} = K_bb' <g_{q,b'}> g_{q,b} g_{q,b'}
• c_{q,ps} = <g_{q,b}> <g_{q,b'}> - < g_{q,b'} g_{q,b} > K_bb' <g_{q,b'}>

with b,b' = \gamma, Z. And we can factor out the standard fl2 couplings obtaining:

• w_g = w_3 = <g_{g,b}> <g_{g,b'}> / < g_{g,b'} g_{g,b} >
• w_{q,ns} = w_2 = K_bb' <g_{q,b'}>
• w_{q,ps} = w_4 = w_3 - w_2

which are easier to implement.

[giacomomagni]

Actually I'm confused again, the calculation before shows that the non-singlet projection of the standard fl2 diagrams is 0 for AA case, but non 0 for the fl11 diagrams. While the singlet and gluon projections are both present.