diagram

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#  CFTP/21-013
Prescription for finite oblique parameters $S$ and $U$
in extensions of the SM with $m_{W}\neq m_{Z}\cos{\theta_{W}}$

Francisco Albergaria and Luís Lavoura
Universidade de Lisboa, Instituto Superior Técnico, CFTP,
Av. Rovisco Pais 1, 1049-001 Lisboa, Portugal E-mail:
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We consider extensions of the Standard Model with neutral scalars in
multiplets of $SU(2)$ larger than doublets. When those scalars acquire vacuum
expectation values, the resulting masses of the gauge bosons $W^{\pm}$ and
$Z^{0}$ are not related by $m_{W}=m_{Z}\cos{\theta_{W}}$. In those extensions
of the Standard Model the oblique parameters $S$ and $U$, when computed at the
one-loop level, turn out to be either gauge-dependent or divergent. We show
that one may eliminate this problem by modifying the Feynman rules of the
Standard Model for some vertices containing the Higgs boson; the modifying
factors are equal to $1$ in the limit $m_{W}=m_{Z}\cos{\theta_{W}}$. We give
the result for $S$ in a model with arbitrary numbers of scalar $SU(2)$
triplets with weak hypercharges either $0$ or $1$.

##  1 Introduction

The present knowledge of particle physics is encapsulated in the Standard
Model (SM) [1, 2, 3]. This is an $SU(3)\times SU(2)\times U(1)$ gauge theory
that describes all the fundamental particles observed until now and the way
they interact with each other. The scalar sector of the SM contains just one
doublet of $SU(2)$; it is responsible for giving masses $m_{W}$ and $m_{Z}$ to
the gauge bosons $W^{\pm}$ and $Z^{0}$, respectively, as well as to the
fermions.

Despite being one of the most accurate theories in Science, there are
phenomena that the SM cannot explain. Therefore particle physicists try to
either extend or complete it. One of the ways to extend the SM is by enlarging
its scalar sector [4]. The most studied extension of that sector is the
addition of another $SU(2)$ doublet, obtaining a two-Higgs-doublet model
(2HDM) [5]. Extensions of the SM with scalar $SU(2)$ singlets are also
frequent in the literature (see for instance Refs. [6, 7, 8]). There are also
some extensions of the SM with scalar $SU(2)$ triplets in the literature (see
for instance Refs. [9, 10, 11, 12, 13, 14, 15, 16, 17]).

A feature of the SM is custodial symmetry. This is a symmetry of the scalar
potential of the SM that is broken both by the Yukawa interactions and by the
gauge coupling of the $U(1)$ group. Custodial symmetry leads at the tree level
to the relation

$m_{W}=m_{Z}c_{W},$ (1)

where $c_{W}$ is the cosine of the Weinberg angle. The relation (1) is in good
agreement with observation: experimental results give [18]

$\rho\equiv\frac{m_{W}^{2}}{m_{Z}^{2}c_{W}^{2}}=1.00038\pm 0.00020.$ (2)

Models in which only scalar $SU(2)$ singlets and doublets have vacuum
expectation values (VEVs) preserve the relation (1) at the tree level. That
relation is broken at the loop level because the custodial symmetry is not
exact, as noted above.

Peskin and Takeuchi have identified three so-called ‘oblique parameters’ that
they named $S$, $T$, and $U$ [19, 20].111At about the same time, Altarelli and
Barbieri defined equivalent parameters $\epsilon_{1}=\alpha T$,
$\epsilon_{2}=-\alpha U\left/\left(4s_{W}^{2}\right)\right.$, and
$\epsilon_{3}=\alpha S\left/\left(4s_{W}^{2}\right)\right.$, where $\alpha$ is
the fine-structure constant and $s_{W}$ is the sine of the Weinberg angle [21,
22]. These three observables parameterize some effects of New Physics (NP),
i.e. physics of extensions of the SM. At the one-loop level, vacuum
polarization produces the tensors $\Pi_{VV^{\prime}}^{\mu\nu}(q)$, where $V$
and $V^{\prime}$ are gauge bosons that may be either $A$ and $A$ ($A$ stands
for the photon), $A$ and $Z$, $Z$ and $Z$, or $W$ and $W$, and $q$ is the
four-momentum of the gauge bosons. We write those tensors as

$\Pi_{VV^{\prime}}^{\mu\nu}(q)=g^{\mu\nu}A_{VV^{\prime}}(q^{2})+q^{\mu}q^{\nu}B_{VV^{\prime}}(q^{2}).$
(3)

We then define $\delta
A_{VV^{\prime}}(q^{2})\equiv\left.A_{VV^{\prime}}(q^{2})\right|_{\mathrm{NP}}-\left.A_{VV^{\prime}}(q^{2})\right|_{\mathrm{SM}}$,
where $\left.A_{VV^{\prime}}(q^{2})\right|_{\mathrm{NP}}$ is
$A_{VV^{\prime}}(q^{2})$ computed in a given extension of the SM and
$\left.A_{VV^{\prime}}(q^{2})\right|_{\mathrm{SM}}$ is
$A_{VV^{\prime}}(q^{2})$ computed in the SM. The oblique parameters are
defined as

$\displaystyle T$ $\displaystyle=$ $\displaystyle\frac{1}{\alpha
m_{Z}^{2}}\left[\frac{\delta A_{WW}(0)}{c_{W}^{2}}-\delta A_{ZZ}(0)\right],$
(4a) $\displaystyle S$ $\displaystyle=$
$\displaystyle\frac{4s_{W}^{2}c_{W}^{2}}{\alpha}\left[\left.\frac{\partial\,\delta
A_{ZZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}-\left.\frac{\partial\,\delta
A_{AA}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}+\frac{c_{W}^{2}-s_{W}^{2}}{c_{W}s_{W}}\left.\frac{\partial\,\delta
A_{AZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}\right],$ (4b) $\displaystyle
U$ $\displaystyle=$
$\displaystyle\frac{4s_{W}^{2}}{\alpha}\left[\left.\frac{\partial\,\delta
A_{WW}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}-c_{W}^{2}\,\left.\frac{\partial\,\delta
A_{ZZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}\right.$ (4c)
$\displaystyle\left.-s_{W}^{2}\,\left.\frac{\partial\,\delta
A_{AA}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}+2c_{W}s_{W}\,\left.\frac{\partial\,\delta
A_{AZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}\right].$

The oblique parameters have already been computed for several extensions of
the SM, in particular for models with arbitrary numbers of scalar doublets and
singlets [23, 24], wherein they are finite. In models where larger scalar
multiplets are added to the SM and have VEVs, $m_{W}$ is in general different
from $m_{Z}c_{W}$. As we will show in this paper, this gives rise to
complications when computing the oblique parameters.222More generally, models
with $m_{W}\neq m_{Z}c_{W}$ lead to various complications in their
renormalization, as emphasized in Ref. [25]. This is because in those models
$\rho$ is an additional parameter that needs to be independently renormalized.
See Refs. [26, 27, 28] too. First of all, the oblique parameter $T$ is
divergent at one-loop level for models that do not have custodial symmetry
[29, 30]. Furthermore, as noted above, when computing the oblique parameters
one needs to subtract the functions $A_{VV^{\prime}}(q^{2})$ computed in the
SM from the same functions computed in the extension of the SM. This
subtraction is non-trivial, since in the SM the masses of the gauge bosons
obey equation (1) while in the extension they do not obey it. This leads to
_divergent_ parameters $S$ and $U$. We have found out that, in order to obtain
_finite_ (and gauge-independent) results for $S$ and $U$ at the one-loop level
in models with $m_{W}\neq m_{Z}c_{W}$, one needs to multiply the usual SM
Feynman rules333The Feyman rules for the SM may be found, for instance, in
Refs. [31] and [32]. for some vertices containing the SM Higgs boson by
factors that become equal to $1$ when $m_{W}=m_{Z}c_{W}$ but are different
from $1$ for $m_{W}\neq m_{Z}c_{W}$.

This paper is organized as follows. In section 2 we consider an extension of
the SM with arbitrary numbers of arbitrarily large $SU(2)\times U(1)$ scalar
multiplets, subject only to the constraint of the preservation of electric
charge. In section 3 we illustrate the problem that arises in the computation
of both $S$ and $U$ in that extension of the SM, and display our proposed
solution to that problem. In section 4 we restrict the model of section 2 to
the case where only scalar $SU(2)$ triplets and singlets with $U(1)$ charges 1
or 0 are added to the SM; we give the explicit expression for $S$ in that
model. Section 5 summarizes our findings. Several appendices contain technical
material that may be avoided by a hurried reader.

##  2 General extension of the SM

We consider an $SU(2)\times U(1)$ electroweak model where the scalar sector
consists of an arbitrary number of multiplets $M_{JY}$ labeled by their weak
isospin $J$ and weak hypercharge $Y$. We restrict ourselves to models where
all the scalars have integer electric charges444In our normalization of $Y$,
the electric charge $Q$ is given by $Q=T_{3}+Y$, where $T_{3}$ is the third
component of weak isospin.; therefore, if $J$ is a (half-)integer then $Y$ is
a (half-)integer too. Furthermore, we consider only complex multiplets; if
some multiplets are real, then our conclusions are still valid but there are
some modifications in the intermediate steps. Finally, for the sake of
simplicity we assume that there is only one $M_{JY}$ for each value of the
pair $\left(J,\,Y\right)$; our conclusions are still valid otherwise, but the
notation would become clumsier.

Each multiplet $M_{JY}$ has a component $M_{JY}^{Q}$ with electric charge $Q$
if and only if $J-\left|Q-Y\right|\in\mathbb{N}_{0}$. Some multiplets $M_{JY}$
may have, for some values of $Q\neq 0$, both a component $M_{JY}^{Q}$ with
electric charge $Q$ and a component $M_{JY}^{-Q}$ with electric charge $-Q$;
in our notation, $M_{JY}^{Q}\neq\left(M_{JY}^{-Q}\right)^{\ast}$.

The covariant derivative for the $SU(2)\times U(1)$ electroweak model is, as
usual,555We use the sign conventions of Ref. [31]. They correspond to setting
$\eta=-1$ and $\eta_{Z}=\eta_{e}=1$ in Ref. [32].

$D_{\mu}=\partial_{\mu}+ieQA_{\mu}-i\,\frac{g}{c_{W}}\left(T_{3}-Qs_{W}^{2}\right)Z_{\mu}-ig\left(W_{\mu}^{+}T_{+}+W_{\mu}^{-}T_{-}\right),$
(5)

where $e=gs_{W}$ is the unit of electric charge, the $T_{i}$ ($i=1,2,3$) are
the three components of weak isospin, and the $T_{\pm}$ are given by
$T_{\pm}=\left(T_{1}\pm iT_{2}\right)\left/\sqrt{2}\right.$. We use the
$SU(2)$ representation for weak isospin $J$:

$\displaystyle\left(T_{3}\right)_{rc}$ $\displaystyle=$
$\displaystyle\delta_{rc}\left(J+1-r\right),$ (6a)
$\displaystyle\left(T_{+}\right)_{rc}$ $\displaystyle=$
$\displaystyle\delta_{r+1,c}\,\sqrt{\frac{r\,\left(2J+1-r\right)}{2}},$ (6b)
$\displaystyle\left(T_{-}\right)_{rc}$ $\displaystyle=$
$\displaystyle\delta_{r-1,c}\,\sqrt{\frac{\left(r-1\right)\left(2J+2-r\right)}{2}},$
(6c)

where $r$ stands for the row of the matrix and $c$ stands for the column of
the matrix, with $1\leq r,c\leq 2J+1$. We write $M_{JY}$ as a column matrix
with $2J+1$ rows upon which the matrices of Eqs. (6) act. Clearly, from Eq.
(6a), $M_{JY}^{Q}$ is in the row $J+1-T_{3}=J+Y+1-Q$ of that column matrix.
Then, using Eqs. (5) and (6),

$\displaystyle D_{\mu}M_{JY}^{Q}$ $\displaystyle=$
$\displaystyle\partial_{\mu}M_{JY}^{Q}+ieQA_{\mu}M_{JY}^{Q}+i\,\frac{g}{c_{w}}\,Z_{\mu}M_{JY}^{Q}\left(Y-Qc_{W}^{2}\right)$
(7c) $\displaystyle-
ig\,W_{\mu}^{+}M_{JY}^{Q-1}\,\sqrt{\frac{\left(J+Y+1-Q\right)\left(J-Y+Q\right)}{2}}$
$\displaystyle-
ig\,W_{\mu}^{-}M_{JY}^{Q+1}\,\sqrt{\frac{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}{2}}.$

We define $\mathcal{A}$ to be the set of multiplets that have a component with
zero electric charge. A multiplet $M_{JY}$ belongs to $\mathcal{A}$ if and
only if $J-\left|Y\right|\in\mathbb{N}_{0}$.

For every $Q\in\mathbb{N}$, we define two sets:

  * •

$\mathcal{R}_{Q}$ is the set of multiplets that have a component with electric
charge $Q$; a multiplet $M_{JY}$ belongs to $\mathcal{R}_{Q}$ if and only if
$J-\left|Q-Y\right|\in\mathbb{N}_{0}$.

  * •

$\mathcal{S}_{Q}$ is the set of multiplets that have a component with electric
charge $-Q$; a multiplet $M_{JY}$ belongs to $\mathcal{S}_{Q}$ if and only if
$J-\left|Q+Y\right|\in\mathbb{N}_{0}$.

We emphasize that both $\mathcal{R}_{Q}$ and $\mathcal{S}_{Q}$ are defined
only for $Q>0$. For those $Q$, let $n_{Q}$ denote the total number of
charge-$Q$ scalars and let $S_{a}^{Q}$ ($a=1,\dots,n_{Q}$) denote the physical
(mass eigenstate) charge-$Q$ scalars. For definiteness, we fix
$S_{1}^{1}\equiv G^{+}$ to be the charged Goldstone boson. We define
$S_{a}^{-Q}\equiv\left(S_{a}^{Q}\right)^{\ast}$.

For every $Q\in\mathbb{N}$ and $M_{JY}\in\mathcal{R}_{Q}$, we write

$M_{JY}^{Q}=R_{JY}^{Q}\left(\begin{array}[]{c}S_{1}^{Q}\\\\[1.42262pt]
S_{2}^{Q}\\\\[1.42262pt] \vdots\\\\[1.42262pt]
S_{n_{Q}}^{Q}\end{array}\right),$ (8)

where $R_{JY}^{Q}$ is a $1\times n_{Q}$ row matrix. For every $Q\in\mathbb{N}$
and $M_{JY}\in\mathcal{S}_{Q}$, we write

$\left(M_{JY}^{-Q}\right)^{\ast}=S_{JY}^{Q}\left(\begin{array}[]{c}S_{1}^{Q}\\\\[1.42262pt]
S_{2}^{Q}\\\\[1.42262pt] \vdots\\\\[1.42262pt]
S_{n_{Q}}^{Q}\end{array}\right),$ (9)

where $S_{JY}^{Q}$ is a $1\times n_{Q}$ row matrix. We form the $n_{Q}\times
n_{Q}$ unitary (mixing) matrix $U^{Q}$ by piling up all the row matrices
$R_{JY}^{Q}$ and $S_{JY}^{Q}$ (for fixed $Q\in\mathbb{N}$) on top of each
other. Since $U^{Q}$ is unitary,

$\displaystyle\sum_{M_{JY}\in\mathcal{R}_{Q}}\left(R^{Q}_{JY}\right)_{1a}\left(R^{Q}_{JY}\right)^{\ast}_{1a^{\prime}}+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left(S^{Q}_{JY}\right)_{1a}\left(S^{Q}_{JY}\right)^{\ast}_{1a^{\prime}}$
$\displaystyle=$ $\displaystyle\delta_{aa^{\prime}},$ (10a)
$\displaystyle\sum_{a=1}^{n_{Q}}\left(R^{Q}_{JY}\right)_{1a}\left(R^{Q}_{J^{\prime}Y^{\prime}}\right)_{1a}^{\ast}$
$\displaystyle=$ $\displaystyle\delta_{JJ^{\prime}}\delta_{YY^{\prime}},$
(10b)
$\displaystyle\sum_{a=1}^{n_{Q}}\left(S^{Q}_{JY}\right)_{1a}\left(S^{Q}_{J^{\prime}Y^{\prime}}\right)_{1a}^{\ast}$
$\displaystyle=$ $\displaystyle\delta_{JJ^{\prime}}\delta_{YY^{\prime}},$
(10c)
$\displaystyle\sum_{a=1}^{n_{Q}}\left(R^{Q}_{JY}\right)_{1a}\left(S^{Q}_{J^{\prime}Y^{\prime}}\right)_{1a}^{\ast}$
$\displaystyle=$ $\displaystyle 0.$ (10d)

Each multiplet $M_{JY}$ that belongs to $\mathcal{A}$ has VEV $v_{JY}$ (which
may in some cases be zero) in its component $M_{JY}^{0}$ with electric charge
zero. We write

$M_{JY}^{0}=v_{JY}+\frac{A_{JY}+iB_{JY}}{\sqrt{2}}\left(\begin{array}[]{c}S_{1}^{0}\\\\[1.42262pt]
S_{2}^{0}\\\\[1.42262pt] \vdots\\\\[1.42262pt]
S_{n_{0}}^{0}\end{array}\right),$ (11)

where $A_{JY}$ and $B_{JY}$ are _real_ $1\times n_{0}$ row matrices and the
_real_ fields $S_{b}^{0}$ ($b=1,\ldots,n_{0}$) are the physical neutral
scalars. Without lack of generality, we fix $S_{1}^{0}\equiv G^{0}$ to be the
neutral Goldstone boson. We form the $n_{0}\times n_{0}$ real orthogonal
matrix $V$ by piling up all the row matrices $A_{JY}$ and $B_{JY}$ on top of
each other. Since $V$ is orthogonal,

$\displaystyle\sum_{b=1}^{n_{0}}\left[\left(A_{JY}\right)_{1b}\right]^{2}=\sum_{b=1}^{n_{0}}\left[\left(B_{JY}\right)_{1b}\right]^{2}$
$\displaystyle=$ $\displaystyle 1,$ (12a)
$\displaystyle\sum_{b=1}^{n_{0}}\left(A_{JY}\right)_{1b}\left(B_{JY}\right)_{1b}$
$\displaystyle=$ $\displaystyle 0.$ (12b)

Hence,

$\displaystyle\sum_{b=1}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left[\left(A_{JY}\right)_{1b}\left(B_{JY}\right)_{1b^{\prime}}-\left(B_{JY}\right)_{1b}\left(A_{JY}\right)_{1b^{\prime}}\right]^{2}$
$\displaystyle=$ $\displaystyle 1,$ (13a)
$\displaystyle\sum_{b=1}^{n_{0}}\left|\left(A_{JY}+iB_{JY}\right)_{1b}\right|^{2}$
$\displaystyle=$ $\displaystyle 2.$ (13b)

The masses of the gauge bosons are given in terms of the VEVs of the scalar
fields by

$\displaystyle\frac{m_{Z}^{2}}{2}$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{c_{W}^{2}}\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y^{2},$
(14a) $\displaystyle m_{W}^{2}$ $\displaystyle=$ $\displaystyle
g^{2}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left|v_{JY}\right|^{2}\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}\right.$
(14b)
$\displaystyle\left.+\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left|v_{JY}\right|^{2}\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}\right]$
$\displaystyle=$ $\displaystyle
g^{2}\left[\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}+\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}\right]\hskip
14.22636pt$ (14c) $\displaystyle=$ $\displaystyle
g^{2}\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\left(J^{2}-Y^{2}+J\right).$
(14d)

In line (14b) we have used firstly line (7c) with $Q=1$ and secondly line (7c)
with $Q=-1$. In passing from line (14b) to line (14c) we have used

$M_{JY}\in\mathcal{A}\setminus\mathcal{R}_{1}\Rightarrow J=-Y,\quad\quad
M_{JY}\in\mathcal{A}\setminus\mathcal{S}_{1}\Rightarrow J=Y.$ (15)

One sees in Eqs. (14) that $m_{Z}^{2}c_{W}^{2}=m_{W}^{2}$ in general requires
$J^{2}+J=3Y^{2}$ for all the nonzero $v_{JY}$.666The relation (1) may of
course hold by accident, even when $J\left(J+1\right)\neq 3Y^{2}$ for some
nonzero $v_{JY}$. This holds for the standard case of doublets with $J=Y=1/2$,
and also for neutral singlets with $J=Y=0$, but it does not hold for most
other choices of $J$ and $Y$.

##  3 Prescription for $S$ and $U$

We consider the vacuum-polarization one-loop diagram with two scalars as
internal particles, see Fig. 1.

(100,100) i1 o1 boson,label.side=lefti1,w1 boson,label.side=leftw2,o1
dashes,right,tension=.3w1,w2 dashes,left,tension=.3w1,w2
lab=$f_{\mu}$,lab.dist=0.05w,lab.angle=0w1
lab=$f^{\prime}_{\nu}$,lab.dist=0.05w,lab.angle=180w2 $V$i1 $V^{\prime}$o1
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30 down); enddef; marrowd,tension=0w1,w2

Figure 1: One-loop vacuum-polarization diagram with two scalars as internal
particles.

Let the Feynman rules for its vertices be
$f_{\mu}=K\left(p-p^{\prime}\right)_{\mu}$ and
$f^{\prime}_{\nu}=K^{\prime}\left(p-p^{\prime}\right)_{\nu}$. Then the
contribution of that diagram to

$\left.\frac{\partial A_{VV^{\prime}}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}$ (16)

is

$\left.\frac{\partial A_{VV^{\prime}}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}=-KK^{\prime}\,\mathrm{div}+\mathrm{finite\ terms},$
(17)

where “div” is a divergent quantity defined in Eq. (68) of appendix A. That
quantity is independent both of $q^{2}$ and of the masses of the scalars in
the diagram of Fig. 1.

We want to compute the divergent contributions to the quantitites (16) for
$VV^{\prime}=AA$, $AZ$, $ZZ$, and $WW$. Those contributions originate solely
in diagrams like the one in Fig. 1. (Other types of diagrams give either
vanishing or finite contributions.) We therefore need the factors $K$ and
$K^{\prime}$ for all the diagrams, i.e. we need the Lagrangian terms that
describe the interactions between one gauge boson and two scalars. Those terms
are derived from Eq. (7) and may be found in appendix B.

Firstly consider $A_{AA}(q^{2})$. This is generated by loops with charged
scalars $S_{a}^{Q}$ and $S_{a}^{-Q}$. According to Eq. (69) of appendix B,
they generate

$\left.\frac{\partial A_{AA}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}=e^{2}\,\mathrm{div}\,\sum_{Q>0}Q^{2}n_{Q}+\mathrm{finite\
terms}.$ (18)

Secondly consider $A_{AZ}(q^{2})$. This is once again generated by loops with
charged scalars $S_{a}^{Q}$ and $S_{a}^{-Q}$. According to Eqs. (69) and (70)
of appendix B, they generate

$\left.\frac{\partial A_{AZ}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}=\frac{eg}{c_{W}}\
\mathrm{div}\,\sum_{Q>0}\left(-c_{W}^{2}Q^{2}n_{Q}+\\!\\!\sum_{M_{JY}\in\mathcal{R}_{Q}}QY-\\!\\!\sum_{M_{JY}\in\mathcal{S}_{Q}}QY\right)+\mathrm{finite\
terms},$ (19)

where we have used Eqs. (10b) and (10c).

Thirdly consider $A_{ZZ}(q^{2})$. This is generated either by loops with
charged scalars $S_{a}^{Q}$ and $S_{a^{\prime}}^{-Q}$, where $a^{\prime}$ may
be different from $a$, or by loops with neutral scalars $S_{b}^{0}$ and
$S_{b^{\prime}}^{0}$, where $b^{\prime}$ may differ from $b$. The relevant
equations are Eqs. (70) and (71) of appendix B, respectively. They generate

$\displaystyle\left.\frac{\partial A_{ZZ}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{c_{W}^{2}}\ \mathrm{div}\,\sum_{Q>0}\
\sum_{a,a^{\prime}=1}^{n_{Q}}\left|-c_{W}^{2}Q\,\delta_{aa^{\prime}}+\sum_{M_{JY}\in\mathcal{R}_{Q}}Y\left(R^{Q}_{JY}\right)_{1a}\left(R^{Q}_{JY}\right)_{1a^{\prime}}^{\ast}\hskip
19.91692pt\right.$ (20b)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{S}_{Q}}Y\left(S^{Q}_{JY}\right)_{1a}\left(S^{Q}_{JY}\right)_{1a^{\prime}}^{\ast}\right|^{2}$
$\displaystyle+\frac{g^{2}}{c_{W}^{2}}\
\mathrm{div}\,\sum_{b=1}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left\\{\sum_{M_{JY}\in\mathcal{A}}Y\left[\left(A_{JY}\right)_{1b}\left(B_{JY}\right)_{1b^{\prime}}\right.\right.$
$\displaystyle\left.\left.-\left(B_{JY}\right)_{1b}\left(A_{JY}\right)_{1b^{\prime}}\right]\vphantom{\sum_{Y=-J}^{J}Y\left(A_{JY}\right)_{1b}\left(B_{JY}\right)_{1b^{\prime}}}\right\\}^{2}+\mathrm{finite\
terms}.$

This leads to

$\displaystyle\left.\frac{\partial A_{ZZ}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{c_{W}^{2}}\
\mathrm{div}\left[\sum_{Q>0}\left(c_{W}^{4}Q^{2}n_{Q}-2c_{W}^{2}Q\sum_{M_{JY}\in\mathcal{R}_{Q}}Y+2c_{W}^{2}Q\sum_{M_{JY}\in\mathcal{S}_{Q}}Y\right.\right.\hskip
19.91692pt$ (21)
$\displaystyle\left.\left.+\sum_{M_{JY}\in\mathcal{R}_{Q}}Y^{2}+\sum_{M_{JY}\in\mathcal{S}_{Q}}Y^{2}\right)+\sum_{M_{JY}\in\mathcal{A}}Y^{2}\right]+\mathrm{finite\
terms},$

where we have used Eqs. (10b)–(10d) and (13a).

Finally consider $A_{WW}(q^{2})$. This is generated either by loops with one
charged scalar $S_{a}^{Q}$ and one neutral scalar $S_{b}^{0}$, or by loops
with two charged scalars $S_{a}^{\pm Q}$ and
$S_{a^{\prime}}^{\mp\left(Q+1\right)}$. The relevant equations are Eqs. (72)
and (73) of appendix B, respectively. They generate

$\displaystyle\left.\frac{\partial A_{WW}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{2}\,\mathrm{div}\left\\{2\sum_{M_{JY}\in\mathcal{A}}\left(J^{2}-Y^{2}+J\right)\right.$
(22b)
$\displaystyle+\sum_{Q>0}\left[\sum_{M_{JY}\in\mathcal{R}_{Q}}\left(J+Y-Q\right)\left(J-Y+Q+1\right)\right.$
$\displaystyle\left.\left.+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left(J+Y+Q+1\right)\left(J-Y-Q\right)\right]\right\\}+\mathrm{finite\
terms},\hskip 11.38109pt$

where we have used Eqs. (10b)–(10d) and (13b), and also

$M_{JY}\in\mathcal{R}_{Q}\setminus\mathcal{R}_{Q+1}\Rightarrow
J+Y-Q=0,\quad\quad
M_{JY}\in\mathcal{S}_{Q}\setminus\mathcal{S}_{Q+1}\Rightarrow J-Y-Q=0.$ (23)

It is now clear that the combinations relevant for the parameters $S$ and $U$
in Eqs. (4b) and (4c), respectively, namely

$\displaystyle\left.\frac{\partial A_{ZZ}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}-\left.\frac{\partial A_{AA}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}$
$\displaystyle+\frac{c_{W}^{2}-s_{W}^{2}}{c_{W}s_{W}}\,\left.\frac{\partial
A_{AZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ $\displaystyle\equiv$
$\displaystyle\frac{g^{2}}{c_{W}^{2}}\,\mathrm{div}\ \psi+\mathrm{finite\
terms},\hskip 14.22636pt$ (24a) $\displaystyle\left.\frac{\partial
A_{WW}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}-c_{W}^{2}\,\left.\frac{\partial
A_{ZZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ $\displaystyle-
s_{W}^{2}\,\left.\frac{\partial A_{AA}(q^{2})}{\partial
q^{2}}\right|_{q^{2}=0}+2c_{W}s_{W}\,\left.\frac{\partial
A_{AZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ $\displaystyle\equiv$
$\displaystyle\frac{g^{2}}{2}\,\mathrm{div}\ \theta+\mathrm{finite\ terms},$
(24b)

are given by

$\displaystyle\psi$ $\displaystyle=$
$\displaystyle\sum_{M_{JY}\in\mathcal{A}}Y^{2}+\sum_{Q>0}\left[\sum_{M_{JY}\in\mathcal{R}_{Q}}\left(Y^{2}-QY\right)+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left(Y^{2}+QY\right)\right],$
(25a) $\displaystyle\theta$ $\displaystyle=$ $\displaystyle
2\sum_{M_{JY}\in\mathcal{A}}\left(J^{2}+J-2Y^{2}\right)+\sum_{Q>0}\left\\{\vphantom{\sum_{M_{JY}\in\mathcal{S}_{Q}}}-2Q^{2}n_{Q}\right.$
(25b)
$\displaystyle+\sum_{M_{JY}\in\mathcal{R}_{Q}}\left[\left(J+Y-Q\right)\left(J-Y+Q+1\right)-2Y^{2}+4QY\right]$
$\displaystyle\left.+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left[\left(J+Y+Q+1\right)\left(J-Y-Q\right)-2Y^{2}-4QY\right]\right\\}.$

It is proven in appendix C that $\psi$ and $\theta$ are both zero, no matter
what the set of $M_{JY}$ (and no matter what the VEVs of their neutral
components are) present in any particular $SU(2)\times U(1)$ electroweak
model, i.e. either in the SM or in any extension thereof.

Since the quantities $\psi$ and $\theta$ vanish for the New Physics model and,
identically, also for the Standard Model, one might think that $S$ and $U$
would turn out finite. However, if we subtract the SM result for $\psi$ and
$\theta$, computed by using the usual SM Feynman rules for the Goldstone-boson
vertices, from the New Physics result, which uses a _different_ set of Feynman
rules for those vertices, then we get a gauge-dependent result for both $S$
and $U$. This happens because the gauge-dependent parts of the NP and SM
results are different when Eq. (1) does not hold in the NP model. Indeed, as
is shown in appendix D, some Goldstone-boson vertices _depend on $\rho$_,
hence they are different in the SM and in the NP model. In Eqs. (26) we
display the relevant Goldstone-boson vertices and their $\rho$-dependent
value, as derived in appendix D; we also display, for each vertex and crossed
over, its value in the SM, i.e. when $\rho=1$, as given in Ref. [31].

(75,75) i1,i2 o1 wiggly_arrowi1,w1 photoni2,w1 scalaro1,w1 $W^{\mp}_{\nu}$i1
$Z_{\mu}$i2 $G^{\pm}$o1 $\displaystyle=$ $\displaystyle\
\xcancel{-igs_{W}^{2}m_{Z}g_{\mu\nu}}\quad igm_{W}\left(c_{W}-\frac{1}{\rho
c_{W}}\right)g_{\mu\nu},$ (26a) (75,75) i1,i2 o1 scalari1,w1 scalari2,w1
photonw1,o1 $G^{-}$i1 $G^{+}$i2 $Z_{\mu}$o1 style_def marrowa expr p =
drawarrow subpath (1/4, 4/5) of p shifted 10 left withpen pencircle scaled
0.5; label.lft(btex $p_{-}$ etex, point 0.5 of p shifted 12 left); enddef;
marrowa,tension=0i1,w1 style_def marrowb expr p = drawarrow subpath (1/4, 4/5)
of p shifted 10 left withpen pencircle scaled 0.5; label.lft(btex $p_{+}$
etex, point 0.5 of p shifted 12 left); enddef; marrowb,tension=0i2,w1
$\displaystyle=$ $\displaystyle\ \xcancel{-i\
\frac{g\left(c_{W}^{2}-s_{W}^{2}\right)}{2c_{W}}\,\left(p_{-}-p_{+}\right)_{\mu}}\quad
i\,\frac{g}{2c_{W}}\left(\frac{1}{\rho}-2c_{W}^{2}\right)\left(p_{-}-p_{+}\right)_{\mu},\hskip
25.60747pt$ (26b) (75,75) i1,i2 o1 scalari1,w1 dashesi2,w1 wiggly_arrowo1,w1
$G^{\mp}$i1 $G^{0}$i2 $W_{\mu}^{\pm}$o1 style_def marrowa expr p = drawarrow
subpath (1/4, 4/5) of p shifted 10 left withpen pencircle scaled 0.5;
label.lft(btex $q$ etex, point 0.5 of p shifted 12 left); enddef;
marrowa,tension=0i1,w1 style_def marrowb expr p = drawarrow subpath (1/4, 4/5)
of p shifted 10 left withpen pencircle scaled 0.5; label.lft(btex $p$ etex,
point 0.5 of p shifted 12 left); enddef; marrowb,tension=0i2,w1
$\displaystyle=$ $\displaystyle\
\xcancel{\frac{g}{2}\left(q-p\right)_{\mu}}\quad\frac{g}{2\sqrt{\rho}}\left(q-p\right)_{\mu}.$
(26c)

We thus have a problem. We may obtain $\psi=\theta=0$ in both the SM and its
extension by using, respectively, the Feynman rules adequate for each of them,
with different values of $\rho$ (viz. $\rho=1$ in the SM and $\rho\neq 1$ in
the extension); but, since the Feynman rules for the Goldstone-boson vertices
differ in the two cases, we will then end up with a gauge-dependent result. Or
else we may use in both the SM and its extension the same Feynman rules, as
they are written in Eqs. (26), i.e. with the same value of $\rho$, and then we
will get a gauge-independent result—but $\psi$ and $\theta$ will be non-zero
in the SM and, therefore, $S$ and $U$, respectively, will diverge.

We propose to make $S$ and $U$ finite by multiplying other SM Feynman
rules—not the ones in Eqs. (26)—by factors that are equal to $1$ when
$\rho=1$. This is possible because in the SM there is the Higgs particle $H$,
with mass $m_{h}$, that participates in the computation of
$\left.A_{ZZ}(q^{2})\right|_{\mathrm{SM}}$ through a loop with $G^{0}$ and
$H$, and in the computation of $\left.A_{WW}(q^{2})\right|_{\mathrm{SM}}$
through a loop with $G^{+}$ and $H$. We thus propose to use, in the SM, the
following Feynman rules for the vertices $ZG^{0}H$, $ZZH$, $W^{\pm}G^{\mp}H$,
and $W^{+}W^{-}H$, respectively:

(68,68) i1,i2 o1 dashesi1,w1 dashesi2,w1 photonw1,o1 $H$i1 $G^{0}$i2
$Z^{\mu}$o1 style_def marrowa expr p = drawarrow subpath (1/4, 4/5) of p
shifted 10 left withpen pencircle scaled 0.5; label.lft(btex $p$ etex, point
0.5 of p shifted 12 left); enddef; marrowa,tension=0i1,w1 style_def marrowb
expr p = drawarrow subpath (1/4, 4/5) of p shifted 10 left withpen pencircle
scaled 0.5; label.lft(btex $q$ etex, point 0.5 of p shifted 12 left); enddef;
marrowb,tension=0i2,w1 $\displaystyle=$
$\displaystyle\frac{g}{2c_{W}}\left(q-p\right)^{\mu}\,\sqrt{\aleph},$ (27a)
(68,68) i1,i2 o1 photoni1,w1 photoni2,w1 dashesw1,o1 $Z^{\nu}$i1 $Z^{\mu}$i2
$H$o1 $\displaystyle=$ $\displaystyle i\
\frac{gm_{Z}}{c_{W}}\,g^{\mu\nu}\,\sqrt{\aleph},$ (27b) (68,68) i1,i2 o1
dashesi1,w1 scalari2,w1 wiggly_arrowo1,w1 $H$i1 $G^{\mp}$i2 $W^{\pm\,\mu}$o1
style_def marrowa expr p = drawarrow subpath (1/4, 4/5) of p shifted 10 left
withpen pencircle scaled 0.5; label.lft(btex $p$ etex, point 0.5 of p shifted
12 left); enddef; marrowa,tension=0i1,w1 style_def marrowb expr p = drawarrow
subpath (1/4, 4/5) of p shifted 10 left withpen pencircle scaled 0.5;
label.lft(btex $q$ etex, point 0.5 of p shifted 12 left); enddef;
marrowb,tension=0i2,w1 $\displaystyle=$ $\displaystyle\mp i\
\frac{g}{2}\left(q-p\right)^{\mu}\,\sqrt{\beth},$ (27c) (68,68) i1,i2 o1
wiggly_arrowi1,w1 wiggly_arrowi2,w1 dashesw1,o1 $W^{-\,\nu}$i1 $W^{+\,\mu}$i2
$H$o1 $\displaystyle=$ $\displaystyle igm_{W}g^{\mu\nu}\,\sqrt{\beth}.$ (27d)

The vertices $ZG^{0}H$ and $ZZH$ are multiplied by the same factor
$\sqrt{\aleph}$ due to gauge invariance. The same happens with the vertices
$W^{\pm}G^{\mp}H$ and $W^{+}W^{-}H$, which are both multiplied by
$\sqrt{\beth}$. Obviously, in the true SM, i.e. when relation (1) holds, both
$\aleph$ and $\beth$ are 1.

In the computation of the SM contribution to $S$ and $U$, we must consider six
diagrams like the one in Fig. 1:

  1. 1.

A diagram with inner scalars $G^{+}$ and $H$ contributing to $A_{WW}(q^{2})$.

  2. 2.

A diagram with inner scalars $G^{+}$ and $G^{0}$ contributing to
$A_{WW}(q^{2})$.

  3. 3.

A diagram with inner scalars $G^{0}$ and $H$ contributing to $A_{ZZ}(q^{2})$.

  4. 4.

A diagram with inner scalars $G^{+}$ and $G^{-}$ contributing to
$A_{ZZ}(q^{2})$.

  5. 5.

A diagram with inner scalars $G^{+}$ and $G^{-}$ contributing to
$A_{AA}(q^{2})$.

  6. 6.

A diagram with inner scalars $G^{+}$ and $G^{-}$ contributing to
$A_{AZ}(q^{2})$.

A factor $\aleph$ multiplies the diagram 3 and a factor $\beth$ multiplies the
diagram 1.

Using Eqs. (24) and the results for the quantities (16) computed in the SM, we
get for the parameter $S$

$\displaystyle\frac{\alpha}{4s_{W}^{2}c_{W}^{2}}\ S$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{c_{W}^{2}}\,\mathrm{div}\left[\psi_{\mathrm{NP}}-\frac{\aleph}{4}-c_{W}^{4}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)^{2}+s_{W}^{2}c_{W}^{2}\right.$
(28b)
$\displaystyle\left.+c_{W}^{2}\left(c_{W}^{2}-s_{W}^{2}\right)\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)\right]+\mathrm{finite\
terms},$

where the second to fifth terms inside the square brackets originate in
diagrams 3–6, respectively. We have used Eq. (26b) in the third and fifth
terms inside the square brackets. Since $\psi_{\mathrm{NP}}=0$, we obtain a
finite result for $S$ if

$\aleph=4\left[-c_{W}^{4}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)^{2}+s_{W}^{2}c_{W}^{2}+c_{W}^{2}\left(c_{W}^{2}-s_{W}^{2}\right)\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)\right].$
(29)

Thus, we must choose

$\aleph=\frac{2}{\rho}-\frac{1}{\rho^{2}}.$ (30)

Notice that $\aleph=1$ when Eq. (1) holds, viz. when the extension of the SM
only has scalar doublets and/or singlets.

For the parameter $U$ we get

$\displaystyle\frac{\alpha}{4s_{W}^{2}}\,U$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{2}\,\mathrm{div}\left[\theta_{\mathrm{NP}}-\frac{\beth}{2}-\frac{m_{Z}^{2}c_{W}^{2}}{2m_{W}^{2}}+\frac{\aleph}{2}+2c_{W}^{4}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)^{2}\right.$
(31b)
$\displaystyle\left.+2s_{W}^{4}+4s_{W}^{2}c_{W}^{2}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)\right]+\mathrm{finite\
terms},$

where the second to seventh terms inside the square brackets originate in the
diagrams 1–6, respectively. We have used Eq. (26c) in the third term and Eq.
(26b) in the fifth and seventh terms inside the square brackets. Since
$\theta_{\mathrm{NP}}=0$, we get a finite $U$ if

$\beth=-\frac{m_{Z}^{2}c_{W}^{2}}{m_{W}^{2}}+\aleph+4c_{W}^{4}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right)^{2}+4s_{W}^{4}+8s_{W}^{2}c_{W}^{2}\left(1-\frac{m_{Z}^{2}}{2m_{W}^{2}}\right).$
(32)

Using Eq. (30), we obtain

$\beth=4-\frac{3}{\rho}.$ (33)

Befittingly, $\beth=1$ when Eq. (1) holds, viz. in an extension of the SM with
only scalar doublets and singlets.

##  4 The parameter $S$ in a model with scalar triplets

###  4.1 The model

In this section we consider an $SU(2)\times U(1)$ electroweak model where the
scalar sector comprises arbitrary numbers of

  * •

$SU(2)$ doublets with weak hypercharge $1/2$

$\Phi_{k}=\begin{pmatrix}\phi_{k}^{+}\\\\[2.84526pt]
\phi_{k}^{0}\end{pmatrix}\quad(k=1,\ldots,n_{d}),$ (34)

  * •

$SU(2)$ triplets with weak hypercharge $1$

$\Xi_{p}=\begin{pmatrix}\xi_{p}^{++}\\\\[2.84526pt] \xi_{p}^{+}\\\\[2.84526pt]
\xi_{p}^{0}\end{pmatrix}\quad(p=1,\ldots,n_{t_{1}}),$ (35)

  * •

real $SU(2)$ triplets with weak hypercharge $0$

$\Lambda_{q}=\begin{pmatrix}\lambda_{q}^{+}\\\\[2.84526pt]
\lambda_{q}^{0}={\lambda_{q}^{0}}^{\ast}\\\\[2.84526pt]
-{\lambda_{q}^{+}}^{\ast}\end{pmatrix}\quad(q=1,\ldots,n_{t_{0}}),$ (36)

  * •

$SU(2)$ singlets with weak hypercharge $1$

$\chi_{j}^{+}\quad(j=1,\ldots,n_{s_{1}}),$ (37)

  * •

real $SU(2)$ singlets with weak hypercharge $0$

$\chi_{l}^{0}={\chi_{l}^{0}}^{\ast}\quad(l=1,\ldots,n_{s_{0}}),$ (38)

  * •

and $SU(2)$ singlets with weak hypercharge $2$

$\chi_{r}^{++}\quad(r=1,\ldots,n_{s_{2}}).$ (39)

There is then a total of

$n_{2}=n_{t_{1}}+n_{s_{2}}$ complex scalar fields with electric charge $2$,

$n_{1}=n_{d}+n_{t_{1}}+n_{t_{0}}+n_{s_{1}}$ complex scalar fields with
electric charge $1$,

$n_{0}=2n_{d}+2n_{t_{1}}+n_{t_{0}}+n_{s_{0}}$ real scalar fields with electric
charge $0$.

The neutral fields are allowed to have non-zero VEVs

$\left\langle
0\left|\phi_{k}^{0}\right|0\right\rangle=\frac{v_{k}}{\sqrt{2}},\quad\quad\left\langle
0\left|\xi_{p}^{0}\right|0\right\rangle=\frac{w_{p}}{\sqrt{2}},\quad\quad\left\langle
0\left|\lambda_{q}^{0}\right|0\right\rangle=x_{q},\quad\quad\left\langle
0\left|\chi_{l}^{0}\right|0\right\rangle=u_{l},$ (40)

where the $v_{k}$ and the $w_{p}$ are in general complex and the $x_{q}$ and
the $u_{l}$ are real. The masses of the gauge bosons $W^{\pm}$ and $Z^{0}$ are
given in terms of the VEVs of the scalar fields as

$m_{Z}^{2}=\frac{g^{2}}{4c_{W}^{2}}\left(v^{2}+4w^{2}\right),\qquad
m_{W}^{2}=\frac{g^{2}}{4}\left(v^{2}+2w^{2}+4x^{2}\right),$ (41)

where we have defined

$v=\sqrt{\sum_{k=1}^{n_{d}}\left|v_{k}\right|^{2}},\qquad
w=\sqrt{\sum_{p=1}^{n_{t_{1}}}\left|w_{p}\right|^{2}},\qquad
x=\sqrt{\sum_{q=1}^{n_{t_{0}}}x_{q}^{2}}.$ (42)

Comparing the second Eq. (41) with Eq. (14d), note that there is an extra
factor $1/2$ multiplying $x$. This is because the field $\lambda_{q}^{0}$ is
real, hence it has a factor $1/2$ in its gauge-kinetic term.

We expand the neutral fields around their VEVs as

$\phi_{k}^{0}=\frac{v_{k}+{\phi_{k}^{0}}^{\prime}}{\sqrt{2}},\quad\quad\xi_{p}^{0}=\frac{w_{p}+{\xi_{p}^{0}}^{\prime}}{\sqrt{2}},\quad\quad\lambda_{q}^{0}=x_{q}+{\lambda_{q}^{0}}^{\prime},\quad\quad\chi_{l}^{0}=u_{l}+{\chi_{l}^{0}}^{\prime}.$
(43)

We define the mixing matrices of the scalar fields as

$\xi_{p}^{++}=\sum_{c=1}^{n_{2}}\left(T_{2}\right)_{pc}S_{c}^{++},\quad\chi_{r}^{++}=\sum_{c=1}^{n_{2}}\left(T_{4}\right)_{rc}S_{c}^{++},$
(44)
$\phi_{k}^{+}=\sum_{a=1}^{n_{1}}\left(U_{1}\right)_{ka}S_{a}^{+},\quad\xi_{p}^{+}=\sum_{a=1}^{n_{1}}\left(U_{2}\right)_{pa}S_{a}^{+},\quad\lambda_{q}^{+}=\sum_{a=1}^{n_{1}}\left(U_{3}\right)_{qa}S_{a}^{+},\quad\chi_{j}^{+}=\sum_{a=1}^{n_{1}}\left(U_{4}\right)_{ja}S_{a}^{+},$
(45)
${\phi_{k}^{0}}^{\prime}=\sum_{b=1}^{n_{0}}\left(V_{1}\right)_{kb}S_{b}^{0},\quad{\xi_{p}^{0}}^{\prime}=\sum_{b=1}^{n_{0}}\left(V_{2}\right)_{pb}S_{b}^{0},\quad{\lambda_{q}^{0}}^{\prime}=\sum_{b=1}^{n_{0}}\left(V_{3}\right)_{qb}S_{b}^{0},\quad{\chi_{l}^{0}}^{\prime}=\sum_{b=1}^{n_{0}}\left(V_{4}\right)_{lb}S_{b}^{0},$
(46)

where the dimensions of the matrices are

$\displaystyle T_{2}:\ n_{t_{1}}\times n_{2},\quad\quad T_{4}:\
n_{s_{2}}\times n_{2},$ (47a) $\displaystyle U_{1}:\ n_{d}\times n_{1},\qquad
U_{2}:\ n_{t_{1}}\times n_{1},\qquad U_{3}:\ n_{t_{0}}\times n_{1},\qquad
U_{4}:\ n_{s_{1}}\times n_{1},$ (47b) $\displaystyle V_{1}:\ n_{d}\times
n_{0},\qquad V_{2}:\ n_{t_{1}}\times n_{0},\qquad V_{3}:\ n_{t_{0}}\times
n_{0},\qquad V_{4}:\ n_{s_{0}}\times n_{0}.$ (47c)

The matrices $V_{3}$ and $V_{4}$ are real while the other ones are complex.
The matrix

$\tilde{T}=\begin{pmatrix}T_{2}\\\ T_{4}\end{pmatrix}$ (48)

is $n_{2}\times n_{2}$ unitary; it diagonalizes the Hermitian mass matrix of
the scalars with charge $2$. The matrix

$\tilde{U}=\begin{pmatrix}U_{1}\\\ U_{2}\\\ U_{3}\\\ U_{4}\end{pmatrix}$ (49)

is $n_{1}\times n_{1}$ unitary; it diagonalizes the Hermitian mass matrix of
the scalars with charge $1$. The matrix

$\tilde{V}=\begin{pmatrix}\real V_{1}\\\ \imaginary V_{1}\\\ \real V_{2}\\\
\imaginary V_{2}\\\ V_{3}\\\ V_{4}\end{pmatrix}$ (50)

is $n_{0}\times n_{0}$ real orthogonal; it diagonalizes the symmetric mass
matrix of the real components of the neutral scalar fields.

One may obtain the form of the Goldstone bosons by applying to the vacuum
state the generators of the gauge group that are spontaneously broken. Using
this method, we obtain

$\left(U_{1}\right)_{k1}=\frac{v_{k}}{\sqrt{v^{2}+2w^{2}+4x^{2}}},\
\quad\left(U_{2}\right)_{p1}=\frac{\sqrt{2}w_{p}}{\sqrt{v^{2}+2w^{2}+4x^{2}}},\quad\left(U_{3}\right)_{q1}=\frac{2x_{q}}{\sqrt{v^{2}+2w^{2}+4x^{2}}},$
(51)
$\left(V_{1}\right)_{k1}=\frac{iv_{k}}{\sqrt{v^{2}+4w^{2}}},\qquad\left(V_{2}\right)_{p1}=\frac{2iw_{p}}{\sqrt{v^{2}+4w^{2}}},$
(52)

while the first columns of $U_{4}$, $V_{3}$, and $V_{4}$ are identically zero.

###  4.2 The formula for $S$

There are two kinds of diagrams that produce the parameter $S$: the ones like
in Fig. 1 and those like in Fig. 2.

(100,100) i1 o1 boson,label.side=lefti1,w1 boson,label.side=leftw2,o1
boson,right,tension=.3w1,w2 dashes,left,tension=.3w1,w2 $V_{\mu}$i1
$V^{\prime}_{\nu}$o1 style_def marrowa expr p = drawarrow subpath (1/4, 2/3)
of p shifted 10 down withpen pencircle scaled 0.5; label.bot(btex $q$ etex,
point 0.5 of p shifted 12 down); enddef; marrowa,tension=0i1,w1 style_def
marrowb expr p = drawarrow subpath (1/3, 3/4) of p shifted 10 down withpen
pencircle scaled 0.5; label.bot(btex $q$ etex, point 0.5 of p shifted 12
down); enddef; marrowb,tension=0w2,o1 style_def marrowc expr p = drawarrow
subpath (1/4, 2/3) of p shifted 28 up withpen pencircle scaled 0.5;
label.top(btex $p$ etex, point 0.5 of p shifted 30 up); enddef;
marrowc,tension=0w1,w2 style_def marrowd expr p = drawarrow subpath (1/4, 2/3)
of p shifted 28 down withpen pencircle scaled 0.5; label.bot(btex $p^{\prime}$
etex, point 0.5 of p shifted 30 down); enddef; marrowd,tension=0w1,w2

Figure 2: One-loop vacuum-polarization diagram with one scalar and one gauge
boson as internal particles.

Diagrams of the type of Fig. 1 produce a function $K\left(I,\,J\right)$, where
$I$ and $J$ are the squared-masses of the two scalars in the loop. Diagrams of
the type of Fig. 2 produce a function
$K\left(I,\,J\right)-6J\,\tilde{K}\left(I,\,J\right)$, where $I$ is the
squared-mass of the scalar and $J$ is the squared-mass of the gauge boson in
the loop. The functions $K$ and $\tilde{K}$ are defined in Eqs. (67) of
appendix A.

Let $M_{c}$ denote the mass of the scalar $S_{c}^{++}$, $m_{a}$ denote the
mass of the scalar $S_{a}^{+}$, and $\mu_{b}$ denote the mass of $S_{b}^{0}$.
Using the Feynman diagrams from appendix E and the prescription described in
section 2, we have obtained for the oblique parameter $S$ the following
result:

$\displaystyle S$ $\displaystyle=$
$\displaystyle\frac{1}{12\pi}\left\\{4\,\sum_{c,c^{\prime}=1}^{n_{2}}\left|2s_{W}^{2}\delta_{cc^{\prime}}-\left(T_{2}^{\dagger}T_{2}\right)_{cc^{\prime}}\right|^{2}\left[K\left(M_{c}^{2},\,M_{c^{\prime}}^{2}\right)+\frac{\ln{M_{c}^{2}}+\ln{M_{c^{\prime}}^{2}}}{2}\right]\right.$
(53)
$\displaystyle+8\,\sum_{c=1}^{n_{2}}\left\\{\left(c_{W}^{2}-s_{W}^{2}\right)\left[2s_{W}^{2}-\left(T_{2}^{\dagger}T_{2}\right)_{cc}\right]-2c_{W}^{2}s_{W}^{2}\right\\}\ln{M_{c}^{2}}$
$\displaystyle+\sum_{a,a^{\prime}=2}^{n_{1}}\left|2s_{W}^{2}\delta_{aa^{\prime}}-\left(U_{1}^{\dagger}U_{1}\right)_{aa^{\prime}}-2\left(U_{3}^{\dagger}U_{3}\right)_{aa^{\prime}}\right|^{2}\left[K\left(m_{a}^{2},\,m_{a^{\prime}}^{2}\right)+\frac{\ln{m_{a}^{2}}+\ln{m_{a^{\prime}}^{2}}}{2}\right]$
$\displaystyle+2\,\sum_{a=2}^{n_{1}}\left\\{\left(c_{W}^{2}-s_{W}^{2}\right)\left[2s_{W}^{2}-\left(U_{1}^{\dagger}U_{1}\right)_{aa}-2\left(U_{3}^{\dagger}U_{3}\right)_{aa}\right]-2c_{W}^{2}s_{W}^{2}\right\\}\ln{m_{a}^{2}}$
$\displaystyle+\sum_{b=2}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{bb^{\prime}}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{bb^{\prime}}\right]^{2}\left[K\left(\mu_{b}^{2},\,\mu_{b^{\prime}}^{2}\right)+\frac{\ln{\mu_{b}^{2}}+\ln{\mu_{b^{\prime}}^{2}}}{2}\right]$
$\displaystyle+2\,\sum_{a=2}^{n_{1}}\left|\left(U_{2}^{\dagger}U_{2}\right)_{a1}-\left(U_{3}^{\dagger}U_{3}\right)_{a1}\right|^{2}\left[K\left(m_{a}^{2},\,m_{W}^{2}\right)+\frac{\ln{m_{a}^{2}}+\ln{m_{W}^{2}}}{2}\right.$
$\displaystyle\left.\hskip
170.71652pt-6m_{W}^{2}\,\tilde{K}\left(m_{a}^{2},\,m_{W}^{2}\right)\right]$
$\displaystyle+\sum_{b=2}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{1b}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{1b}\right]^{2}\left[K\left(\mu_{b}^{2},\,m_{Z}^{2}\right)+\frac{\ln{\mu_{b}^{2}}+\ln{m_{Z}^{2}}}{2}\right.$
$\displaystyle\left.\hskip
204.85983pt-6m_{Z}^{2}\,\tilde{K}\left(\mu_{b}^{2},\,m_{Z}^{2}\right)\right]$
$\displaystyle\left.-\left(\frac{2}{\rho}-\frac{1}{\rho^{2}}\right)\left[K\left(m_{h}^{2},\,m_{Z}^{2}\right)+\frac{\ln{m_{h}^{2}}+\ln{m_{Z}^{2}}}{2}-6m_{Z}^{2}\,\tilde{K}\left(m_{h}^{2},\,m_{Z}^{2}\right)\right]\right\\}.$

One may use the unitarity of the matrices $\tilde{T}$ and $\tilde{U}$ and the
orthogonality of $\tilde{V}$ to write the following relations:

$\displaystyle\sum_{a,a^{\prime}=2}^{n_{1}}\left|2s_{W}^{2}\delta_{aa^{\prime}}-\left(U_{1}^{\dagger}U_{1}\right)_{aa^{\prime}}-2\left(U_{3}^{\dagger}U_{3}\right)_{aa^{\prime}}\right|^{2}\,\ln
m_{a}^{2}$ (54) $\displaystyle=$
$\displaystyle\sum_{a=2}^{n_{1}}\left[4s_{W}^{4}+\left(1-4s_{W}^{2}\right)\left(U_{1}^{\dagger}U_{1}\right)_{aa}+4\left(1-2s_{W}^{2}\right)\left(U_{3}^{\dagger}U_{3}\right)_{aa}\right]\ln
m_{a}^{2}$
$\displaystyle-\sum_{a=2}^{n_{1}}\left|\left(U_{2}^{\dagger}U_{2}\right)_{a1}-\left(U_{3}^{\dagger}U_{3}\right)_{a1}\right|^{2}\ln
m_{a}^{2},$
$\displaystyle\frac{1}{2}\,\sum_{b=2}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{bb^{\prime}}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{bb^{\prime}}\right]^{2}\left(\ln\mu_{b}^{2}+\ln\mu_{b^{\prime}}^{2}\right)$
(55) $\displaystyle=$
$\displaystyle\frac{1}{2}\,\sum_{b=2}^{n_{0}}\left[\left(V_{1}^{\dagger}V_{1}\right)_{bb}+4\left(V_{2}^{\dagger}V_{2}\right)_{bb}\right]\ln\mu_{b}^{2}$
$\displaystyle-\frac{1}{2}\,\sum_{b=2}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{1b}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{1b}\right]^{2}\ln\mu_{b}^{2},$
$\sum_{b=2}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{1b}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{1b}\right]^{2}=\left(V_{1}^{\dagger}V_{1}\right)_{11}+4\left(V_{2}^{\dagger}V_{2}\right)_{11},$
(56)
$\sum_{a=2}^{n_{1}}\left|\left(U_{2}^{\dagger}U_{2}\right)_{a1}-\left(U_{3}^{\dagger}U_{3}\right)_{a1}\right|^{2}=\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-\left(1-\frac{1}{\rho}\right)^{2},$
(57)
$\sum_{c,c^{\prime}=1}^{n_{2}}\left|2s_{W}^{2}\delta_{cc^{\prime}}-\left(T_{2}^{\dagger}T_{2}\right)_{cc^{\prime}}\right|^{2}\,\ln
M_{c}^{2}=\sum_{c=1}^{n_{2}}\left[4s_{W}^{4}+\left(1-4s_{W}^{2}\right)\left(T_{2}^{\dagger}T_{2}\right)_{cc}\right]\ln
M_{c}^{2}.$ (58)

We use Eqs. (54)–(58) to simplify Eq. (53), obtaining

$\displaystyle S$ $\displaystyle=$
$\displaystyle\frac{1}{12\pi}\left\\{4\,\sum_{c,c^{\prime}=1}^{n_{2}}\left|2s_{W}^{2}\delta_{cc^{\prime}}-\left(T_{2}^{\dagger}T_{2}\right)_{cc^{\prime}}\right|^{2}K\left(M_{c}^{2},\,M_{c^{\prime}}^{2}\right)\right.$
(59k)
$\displaystyle+\sum_{a,a^{\prime}=2}^{n_{1}}\left|2s_{W}^{2}\delta_{aa^{\prime}}-\left(U_{1}^{\dagger}U_{1}\right)_{aa^{\prime}}-2\left(U_{3}^{\dagger}U_{3}\right)_{aa^{\prime}}\right|^{2}K\left(m_{a}^{2},\,m_{a^{\prime}}^{2}\right)$
$\displaystyle+\sum_{b=2}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{bb^{\prime}}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{bb^{\prime}}\right]^{2}K\left(\mu_{b}^{2},\,\mu_{b^{\prime}}^{2}\right)$
$\displaystyle+2\,\sum_{a=2}^{n_{1}}\left|\left(U_{2}^{\dagger}U_{2}\right)_{a1}-\left(U_{3}^{\dagger}U_{3}\right)_{a1}\right|^{2}\left[K\left(m_{a}^{2},\,m_{W}^{2}\right)-6m_{W}^{2}\,\tilde{K}\left(m_{a}^{2},\,m_{W}^{2}\right)\right]$
$\displaystyle+\sum_{b=2}^{n_{0}}\left[\imaginary\left(V_{1}^{\dagger}V_{1}\right)_{1b}+2\,\imaginary\left(V_{2}^{\dagger}V_{2}\right)_{1b}\right]^{2}\left[K\left(\mu_{b}^{2},\,m_{Z}^{2}\right)-6m_{Z}^{2}\,\tilde{K}\left(\mu_{b}^{2},\,m_{Z}^{2}\right)\right]$
$\displaystyle-\left(\frac{2}{\rho}-\frac{1}{\rho^{2}}\right)\left[K\left(m_{h}^{2},\,m_{Z}^{2}\right)-6m_{Z}^{2}\,\tilde{K}\left(m_{h}^{2},\,m_{Z}^{2}\right)\right]$
$\displaystyle-4\,\sum_{c=1}^{n_{2}}\left(T_{2}^{\dagger}T_{2}\right)_{cc}\ln{M_{c}^{2}}-\sum_{a=2}^{n_{1}}\left(U_{1}^{\dagger}U_{1}\right)_{aa}\ln{m_{a}^{2}}$
$\displaystyle+\sum_{b=2}^{n_{0}}\left[\left(V_{1}^{\dagger}V_{1}\right)_{bb}+4\left(V_{2}^{\dagger}V_{2}\right)_{bb}\right]\frac{\ln{\mu_{b}^{2}}}{2}$
$\displaystyle+\left[\left(V_{1}^{\dagger}V_{1}\right)_{11}+4\left(V_{2}^{\dagger}V_{2}\right)_{11}\right]\frac{\ln{m_{Z}^{2}}}{2}$
$\displaystyle+\left[\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-\left(1-\frac{1}{\rho}\right)^{2}\right]\ln
m_{W}^{2}$
$\displaystyle\left.-\left(\frac{2}{\rho}-\frac{1}{\rho^{2}}\right)\frac{\ln{m_{h}^{2}}+\ln{m_{Z}^{2}}}{2}\right\\}.$

The coefficients of the logarithms in lines (59k)–(59k) add up tp zero as they
should:

$\displaystyle-4\,\sum_{c=1}^{n_{2}}\left(T_{2}^{\dagger}T_{2}\right)_{cc}-\sum_{a=2}^{n_{1}}\left(U_{1}^{\dagger}U_{1}\right)_{aa}+\frac{1}{2}\
\sum_{b=1}^{n_{0}}\left[\left(V_{1}^{\dagger}V_{1}\right)_{bb}+4\left(V_{2}^{\dagger}V_{2}\right)_{bb}\right]$
(60c)
$\displaystyle+\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-\left(1-\frac{1}{\rho}\right)^{2}-\left(\frac{2}{\rho}-\frac{1}{\rho^{2}}\right)$
$\displaystyle=$
$\displaystyle-4\,\mathrm{tr}\left(T_{2}T_{2}^{\dagger}\right)-\mathrm{tr}\left(U_{1}U_{1}^{\dagger}\right)+\left(U_{1}^{\dagger}U_{1}\right)_{11}+\frac{1}{2}\,\mathrm{tr}\left(V_{1}V_{1}^{\dagger}\right)+2\,\mathrm{tr}\left(V_{2}V_{2}^{\dagger}\right)$
$\displaystyle+\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-1$
$\displaystyle=$
$\displaystyle-4n_{t_{1}}-n_{d}+\left(U_{1}^{\dagger}U_{1}\right)_{11}+n_{d}+4n_{t_{1}}+\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-1$
(60d) $\displaystyle=$
$\displaystyle\left(U_{1}^{\dagger}U_{1}\right)_{11}+\left(U_{2}^{\dagger}U_{2}\right)_{11}+\left(U_{3}^{\dagger}U_{3}\right)_{11}-1$
(60e) $\displaystyle=$ $\displaystyle 0,$ (60f)

where in the last step we have used Eqs. (51).

Equation (59) for $S$ generalizes the result given in ref. [24] for the case
without any scalar $SU(2)$ triplets and without charge-$2$ singlets. In that
case the matrices $T_{2}$, $T_{4}$, $U_{2}$, $U_{3}$, $V_{2}$, and $V_{3}$ and
zero. Ref. [24] then gave

$\displaystyle S$ $\displaystyle=$
$\displaystyle\frac{1}{24\pi}\left\\{\sum_{a=2}^{n_{1}}\left[2s_{W}^{2}-\left(U_{1}^{\dagger}U_{1}\right)_{aa}\right]^{2}G\left(m_{a}^{2},\,m_{a}^{2},\,m_{Z}^{2}\right)\right.$
(61e)
$\displaystyle+2\,\sum_{a=2}^{n_{1}-1}\sum_{a^{\prime}=a+1}^{n_{1}}\left|\left(U_{1}^{\dagger}U_{1}\right)_{aa^{\prime}}\right|^{2}G\left(m_{a}^{2},\,m_{a^{\prime}}^{2},\,m_{Z}^{2}\right)$
$\displaystyle+\sum_{b=2}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left[\mathrm{Im}\left(V_{1}^{\dagger}V_{1}\right)_{bb^{\prime}}\right]^{2}G\left(\mu_{b}^{2},\,\mu_{b^{\prime}}^{2},\,m_{Z}^{2}\right)$
$\displaystyle+\sum_{b=2}^{n_{0}}\left[\mathrm{Im}\left(V_{1}^{\dagger}V_{1}\right)_{1b}\right]^{2}\hat{G}\left(\mu_{b}^{2},\,m_{Z}^{2}\right)-\hat{G}\left(m_{h}^{2},\,m_{Z}^{2}\right)$
$\displaystyle\left.-\ln{m_{h}^{2}}-2\,\sum_{a=2}^{n_{1}}\left(U_{1}^{\dagger}U_{1}\right)_{aa}\ln{m_{a}^{2}}+\sum_{b=2}^{n_{0}}\left(V_{1}^{\dagger}V_{1}\right)_{bb}\ln{\mu_{b}^{2}}\right\\}.$

It must be stressed that ref. [24] used a different definition of $S$ [33],
wherein

$\left.\frac{\partial\,\delta A_{ZZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$
(62)

in Eq. (4b) was substituted by

$\frac{\delta A_{ZZ}\left(m_{Z}^{2}\right)-\delta
A_{ZZ}\left(0\right)}{m_{Z}^{2}}.$ (63)

This caused the appearance of the functions $G\left(I,\,J,\,m_{Z}^{2}\right)$
and $\hat{G}\left(J,\,m_{Z}^{2}\right)$ in Eq. (61). In order to make the
connection with the present work, one must use

$\displaystyle G\left(I,\,J,\,m_{Z}^{2}\right)$ $\displaystyle=$
$\displaystyle 2\,K\left(I,\,J\right)+O\left(m_{Z}^{2}\right),$ (64a)
$\displaystyle\hat{G}\left(J,\,m_{Z}^{2}\right)$ $\displaystyle=$
$\displaystyle
2\left[K\left(J,\,m_{Z}^{2}\right)-6m_{Z}^{2}\,\tilde{K}\left(J,\,m_{Z}^{2}\right)\right]+O\left(m_{Z}^{2}\right).$
(64b)

##  5 Conclusions

In this paper we have discovered a problem that arises in the computation of
the oblique parameters $S$ and $U$ whenever the New-Physics model at hand does
not obey relation (1), because of scalars in multiplets of $SU(2)$ larger than
doublets having VEVs. Namely, the Feynman rules for some Goldstone-boson
vertices depend on $\rho$—see Eqs. (26)—and this causes a mismatch between the
New-Physics model, wherein $\rho\neq 1$, and the Standard Model, wherein
$\rho=1$. Depending on how one (mis)handles this mismatch, $S$ and $U$ may
turn out either gauge-dependent or divergent. We have proceeded by suggesting
a solution for this problem through a redefinition of the Standard-Model
Feynman rules for some vertices containing the Higgs boson $H$. Namely, we
have inserted by hand into those Feynman rules factors $\sqrt{\aleph}$ and
$\sqrt{\beth}$—see Eqs. (27)—that are equal to $1$ when relation (1) holds.
One easily finds out that, if $\aleph$ and $\beth$ take the values in Eqs.
(30) and (33), respectively, then the problem is solved.

We have utilized our insight to explicitly compute the value of $S$ in a New-
Physics model containing arbitrary numbers of triplets of $SU(2)$ (and also
extra doublets and singlets) with weak hypercharges either $0$ or $1$.

In the future, it may be interesting to study whether the changes to the
Feynman rules that we have suggested may somehow be justified, whether they
cure the problems in the computation of $S$ and $U$ also at the two-loop
level, and whether they may cure other possible problems elsewhere. We leave
these questions for consideration by other authors.

#### Acknowledgements:

This work was supported by the Portuguese Foundation for Science and
Technology through projects CERN/FIS-PAR/0004/2019, CERN/FIS-PAR/0008/2019,
PTDC/FIS-PAR/29436/2017, UIDB/00777/2020, and UIDP/00777/2020. F.A.
acknowledges a fellowship from Project CERN/FIS-PAR/0008/2019.

##  Appendix A Feynman integrals

We define the function

$D\left(Q,I,J,x\right)=Qx\left(x-1\right)+I\left(1-x\right)+Jx.$ (65)

We have used the following results:

$\displaystyle\int_{0}^{1}\mathrm{d}x\left.\frac{\partial}{\partial
Q}\left\\{\frac{4}{d}\
\mu^{4-d}\int\frac{\mathrm{d}^{d}k}{\left(2\pi\right)^{d}}\,\frac{k^{2}}{\left[k^{2}-D\left(Q,I,J,x\right)+i\epsilon\right]^{2}}\right\\}\right|_{Q=0}$
$\displaystyle=$ $\displaystyle
i\left[\text{div}+\frac{K\left(I,J\right)}{48\pi^{2}}\right.$ (66a)
$\displaystyle\left.+\frac{\ln I+\ln J}{96\pi^{2}}\right],$
$\displaystyle\int_{0}^{1}\mathrm{d}x\left.\frac{\partial}{\partial
Q}\left\\{\mu^{4-d}\int\frac{\mathrm{d}^{d}k}{\left(2\pi\right)^{d}}\
\frac{1}{\left[k^{2}-D\left(Q,I,J,x\right)+i\epsilon\right]^{2}}\right\\}\right|_{Q=0}$
$\displaystyle=$ $\displaystyle\frac{i}{32\pi^{2}}\
\tilde{K}\left(I,J\right),$ (66b)

where

$\displaystyle K\left(I,J\right)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{lcl}{\displaystyle-\frac{5}{6}+\frac{2IJ}{\left(I-J\right)^{2}}+\frac{I^{3}+J^{3}-3IJ\left(I+J\right)}{2\left(I-J\right)^{3}}\,\ln{\frac{I}{J}}}&\Leftarrow&I\neq
J,\\\ 0&\Leftarrow&I=J,\end{array}\right.$ (67c)
$\displaystyle\tilde{K}\left(I,J\right)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{lcl}{\displaystyle\frac{I+J}{\left(I-J\right)^{2}}-\frac{2IJ}{\left(I-J\right)^{3}}\,\ln{\frac{I}{J}}}&\Leftarrow&I\neq
J,\\\\[14.22636pt]
{\displaystyle\frac{1}{3I}}&\Leftarrow&I=J,\end{array}\right.$ (67f)

and

$\mathrm{div}=\frac{-1}{48\pi^{2}}\left[\frac{2}{4-d}-\gamma+\ln{\left(4\pi\mu^{2}\right)}\right].$
(68)

In Eqs. (66) and (68), $\mu$ is an arbitrary mass scale. In Eq. (68), $\gamma$
is Euler’s constant. The quantity named “$\mathrm{div}$” diverges in the limit
$d\to 4$.

##  Appendix B Interaction terms of one gauge boson with two scalars

In this appendix we derive some consequences of Eq. (7).

The interaction terms of the photon with the charge-$Q$ scalars ($Q>0$) are
given by

$\displaystyle\mathcal{L}_{AS^{Q}S^{-Q}}$ $\displaystyle=$ $\displaystyle
ieQA^{\mu}\left\\{\sum_{M_{JY}\in\mathcal{R}_{Q}}\left[M_{JY}^{Q}\,\partial_{\mu}\left(M_{JY}^{Q}\right)^{\ast}-\left(M_{JY}^{Q}\right)^{\ast}\partial_{\mu}M_{JY}^{Q}\right]\right.$
(69b)
$\displaystyle\left.+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left[\left(M_{JY}^{-Q}\right)^{\ast}\partial_{\mu}M_{JY}^{-Q}-M_{JY}^{-Q}\,\partial_{\mu}\left(M_{JY}^{-Q}\right)^{\ast}\right]\right\\}$
$\displaystyle=$ $\displaystyle
ieQA^{\mu}\sum_{a,a^{\prime}=1}^{n_{Q}}\left(S_{a}^{Q}\,\partial_{\mu}S_{a^{\prime}}^{-Q}-S_{a^{\prime}}^{-Q}\,\partial_{\mu}S_{a}^{Q}\right)$
(69d)
$\displaystyle\times\left[\sum_{M_{JY}\in\mathcal{R}_{Q}}\left(R_{JY}^{Q}\right)_{1a}\left(R_{JY}^{Q}\right)^{\ast}_{1a^{\prime}}+\sum_{M_{JY}\in\mathcal{S}_{Q}}\left(S_{JY}^{Q}\right)_{1a}\left(S_{JY}^{Q}\right)^{\ast}_{1a^{\prime}}\right]$
$\displaystyle=$ $\displaystyle
ieQA^{\mu}\sum_{a=1}^{n_{Q}}\left(S_{a}^{Q}\,\partial_{\mu}S_{a}^{-Q}-S_{a}^{-Q}\,\partial_{\mu}S_{a}^{Q}\right),$
(69e)

where we have used Eq. (10a).

The interaction terms of the $Z$ with two charged scalars are given by

$\displaystyle\mathcal{L}_{ZS^{Q}S^{-Q}}$ $\displaystyle=$ $\displaystyle
i\,\frac{g}{c_{W}}\,Z^{\mu}\left\\{\sum_{M_{JY}\in\mathcal{R}_{Q}}\left(Y-Qc_{W}^{2}\right)\left[M_{JY}^{Q}\,\partial_{\mu}\left(M_{JY}^{Q}\right)^{\ast}-\left(M_{JY}^{Q}\right)^{\ast}\partial_{\mu}M_{JY}^{Q}\right]\right.\hskip
19.91692pt$ (70b)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{S}_{Q}}\left(Y+Qc_{W}^{2}\right)\left[\left(M_{JY}^{-Q}\right)^{\ast}\partial_{\mu}M_{JY}^{-Q}-M_{JY}^{-Q}\,\partial_{\mu}\left(M_{JY}^{-Q}\right)^{\ast}\right]\right\\}$
$\displaystyle=$ $\displaystyle-
igc_{W}QZ^{\mu}\sum_{a=1}^{n_{Q}}\left(S_{a}^{Q}\,\partial_{\mu}S_{a}^{-Q}-S_{a}^{-Q}\,\partial_{\mu}S_{a}^{Q}\right)$
(70e)
$\displaystyle+i\,\frac{g}{c_{W}}\,Z^{\mu}\sum_{a,a^{\prime}=1}^{n_{Q}}\left(S_{a}^{Q}\,\partial_{\mu}S_{a^{\prime}}^{-Q}-S_{a^{\prime}}^{-Q}\,\partial_{\mu}S_{a}^{Q}\right)$
$\displaystyle\times\left[\sum_{M_{JY}\in\mathcal{R}_{Q}}Y\left(R_{JY}^{Q}\right)_{1a}\left(R_{JY}^{Q}\right)^{\ast}_{1a^{\prime}}-\sum_{M_{JY}\in\mathcal{S}_{Q}}Y\left(S_{JY}^{Q}\right)_{1a}\left(S_{JY}^{Q}\right)^{\ast}_{1a^{\prime}}\right].$

The interaction terms of the $Z$ with two neutral scalars are given by

$\displaystyle\mathcal{L}_{ZS^{0}S^{0}}$ $\displaystyle=$ $\displaystyle
i\,\frac{g}{c_{W}}\,Z^{\mu}\\!\\!\sum_{M_{JY}\in\mathcal{A}}Y\left[M_{JY}^{0}\,\partial_{\mu}\left(M_{JY}^{0}\right)^{\ast}-\left(M_{JY}^{0}\right)^{\ast}\partial_{\mu}M_{JY}^{0}\right]$
(71a) $\displaystyle=$ $\displaystyle
i\,\frac{g}{2c_{W}}\,Z^{\mu}\sum_{b,b^{\prime}=1}^{n_{0}}\left(S_{b}^{0}\,\partial_{\mu}S_{b^{\prime}}^{0}-S_{b^{\prime}}^{0}\,\partial_{\mu}S_{b}^{0}\right)$
(71c)
$\displaystyle\times\sum_{M_{JY}\in\mathcal{A}}Y\left[\left(A_{JY}\right)_{1b}+i\left(B_{JY}\right)_{1b}\right]\left[\left(A_{JY}\right)_{1b^{\prime}}-i\left(B_{JY}\right)_{1b^{\prime}}\right]$
$\displaystyle=$
$\displaystyle\frac{g}{c_{W}}\,Z^{\mu}\sum_{b=1}^{n_{0}-1}\sum_{b^{\prime}=b+1}^{n_{0}}\left(S_{b}^{0}\,\partial_{\mu}S_{b^{\prime}}^{0}-S_{b^{\prime}}^{0}\,\partial_{\mu}S_{b}^{0}\right)$
(71e)
$\displaystyle\times\sum_{M_{JY}\in\mathcal{A}}Y\left[\left(A_{JY}\right)_{1b}\left(B_{JY}\right)_{1b^{\prime}}-\left(A_{JY}\right)_{1b^{\prime}}\left(B_{JY}\right)_{1b}\right].$

The interaction terms of the $W$ with one singly-charged scalar and one
neutral scalar are given by

$\displaystyle\mathcal{L}_{W^{\pm}S^{\mp}S^{0}}$ $\displaystyle=$
$\displaystyle
ig\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}$
(72f)
$\displaystyle\times\left\\{W^{\mu+}\left[\left(M_{JY}^{0}\right)^{\ast}\,\partial_{\mu}M_{JY}^{-1}-M_{JY}^{-1}\,\partial_{\mu}\left(M_{JY}^{0}\right)^{\ast}\right]\right.$
$\displaystyle\left.+W^{\mu-}\left[\left(M_{JY}^{-1}\right)^{\ast}\,\partial_{\mu}M_{JY}^{0}-M_{JY}^{0}\,\partial_{\mu}\left(M_{JY}^{-1}\right)^{\ast}\right]\right\\}$
$\displaystyle+ig\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}$
$\displaystyle\times\left\\{W^{\mu+}\left[\left(M_{JY}^{1}\right)^{\ast}\partial_{\mu}M_{JY}^{0}-M_{JY}^{0}\,\partial_{\mu}\left(M_{JY}^{1}\right)^{\ast}\right]\right.$
$\displaystyle\left.+W^{\mu-}\left[\left(M_{JY}^{0}\right)^{\ast}\partial_{\mu}M_{JY}^{1}-M_{JY}^{1}\,\partial_{\mu}\left(M_{JY}^{0}\right)^{\ast}\right]\right\\}$
$\displaystyle=$ $\displaystyle
i\,\frac{g}{2}\,\sum_{M_{JY}\in\mathcal{A}}\,\sum_{a=1}^{n_{1}}\,\sum_{b=1}^{n_{0}}\,\sqrt{\left(J+Y+1\right)\left(J-Y\right)}$
(72l)
$\displaystyle\times\left[W^{\mu+}\left(A_{JY}-iB_{JY}\right)_{1b}\left(S_{JY}^{1}\right)^{\ast}_{1a}\left(S_{b}^{0}\,\partial_{\mu}S_{a}^{-}-S_{a}^{-}\,\partial_{\mu}S_{b}^{0}\right)\right.$
$\displaystyle\left.+W^{\mu-}\left(A_{JY}+iB_{JY}\right)_{1b}\left(S_{JY}^{1}\right)_{1a}\left(S_{a}^{+}\,\partial_{\mu}S_{b}^{0}-S_{b}^{0}\,\partial_{\mu}S_{a}^{+}\right)\right]$
$\displaystyle+i\,\frac{g}{2}\,\sum_{M_{JY}\in\mathcal{A}}\,\sum_{a=1}^{n_{1}}\,\sum_{b=1}^{n_{0}}\,\sqrt{\left(J+Y\right)\left(J-Y+1\right)}$
$\displaystyle\times\left[W^{\mu+}\left(A_{JY}+iB_{JY}\right)_{1b}\left(R_{JY}^{1}\right)^{\ast}_{1a}\left(S_{a}^{-}\,\partial_{\mu}S_{b}^{0}-S_{b}^{0}\,\partial_{\mu}S_{a}^{-}\right)\right.$
$\displaystyle\left.+W^{\mu-}\left(A_{JY}-iB_{JY}\right)_{1b}\left(R_{JY}^{1}\right)_{1a}\left(S_{b}^{0}\,\partial_{\mu}S_{a}^{+}-S_{a}^{+}\,\partial_{\mu}S_{b}^{0}\right)\right],$

where we have used Eqs. (15).

The interaction terms of the $W$ with two charged scalars are given by

$\displaystyle\mathcal{L}$ $\displaystyle=$
$\displaystyle\cdots+i\,\frac{g}{\sqrt{2}}\,\sum_{a=1}^{n_{Q}}\,\sum_{a^{\prime}=1}^{n_{Q+1}}\,\left\\{W_{\mu}^{+}\,\left(S_{a}^{Q}\,\partial^{\mu}S_{a^{\prime}}^{-Q-1}-S_{a^{\prime}}^{-Q-1}\,\partial^{\mu}S_{a}^{Q}\right)\right.$
(73f)
$\displaystyle\times\left[\sum_{M_{JY}\in\mathcal{S}_{Q}\cap\mathcal{S}_{Q+1}}\\!\sqrt{\left(J+Y+Q+1\right)\left(J-Y-Q\right)}\,\left(S_{JY}^{Q}\right)_{1a}\left(S_{JY}^{Q+1}\right)^{\ast}_{1a^{\prime}}\right.$
$\displaystyle\left.-\\!\\!\sum_{M_{JY}\in\mathcal{R}_{Q}\cap\mathcal{R}_{Q+1}}\\!\sqrt{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}\,\left(R_{JY}^{Q}\right)_{1a}\left(R_{JY}^{Q+1}\right)_{1a^{\prime}}^{\ast}\right]$
$\displaystyle+W_{\mu}^{-}\left(S_{a^{\prime}}^{Q+1}\,\partial_{\mu}S_{a}^{-Q}-S_{a}^{-Q}\,\partial_{\mu}S_{a^{\prime}}^{Q+1}\right)$
$\displaystyle\times\left[\sum_{M_{JY}\in\mathcal{S}_{Q}\cap\mathcal{S}_{Q+1}}\\!\sqrt{\left(J+Y+Q+1\right)\left(J-Y-Q\right)}\,\left(S_{JY}^{Q}\right)^{\ast}_{1a}\left(S_{JY}^{Q+1}\right)_{1a^{\prime}}\right.$
$\displaystyle\left.\left.-\\!\\!\sum_{M_{JY}\in\mathcal{R}_{Q}\cap\mathcal{R}_{Q+1}}\\!\sqrt{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}\,\left(R_{JY}^{Q}\right)_{1a}^{\ast}\left(R_{JY}^{Q+1}\right)_{1a^{\prime}}\right]\right\\}.$

##  Appendix C Proof of $\psi=\theta=0$

We show in this appendix that the quantities $\psi$ and $\theta$ in Eqs. (25)
are equal to $0$. We shall use the sums

$\displaystyle 1+2+3+\cdots+n$ $\displaystyle=$
$\displaystyle\frac{n\left(n+1\right)}{2},$ (74a) $\displaystyle
1+4+9+\cdots+n^{2}$ $\displaystyle=$
$\displaystyle\frac{n\left(n+1\right)\left(2n+1\right)}{6},$ (74b)

which are valid for any positive integer $n$.

We shall separately consider the following types of multiplets $M_{JY}$:

  1. 1.

$Y<-J$. A multiplet of this type belongs neither to $\mathcal{A}$ nor to
$\mathcal{R}_{Q}$ for any $Q>0$. It belongs to $\mathcal{S}_{Q}$ for
$Q=-Y-J,\ldots,-Y+J$.

  2. 2.

$Y=-J$. A multiplet of this type belongs to $\mathcal{A}$. It also belongs to
$\mathcal{S}_{Q}$ for $Q=1,2,\ldots,-2Y$. It does not belong to
$\mathcal{R}_{Q}$ for any value of $Q>0$.

  3. 3.

$-J<Y<J$. A multiplet of this type belongs to $\mathcal{A}$, to
$\mathcal{R}_{Q}$ for $Q=1,\ldots,J+Y$, and to $\mathcal{S}_{Q}$ for
$Q=1,\ldots,J-Y$.

  4. 4.

$Y=J$. A multiplet of this type belongs to $\mathcal{A}$ and also to
$\mathcal{R}_{Q}$ for $Q=1,2,\ldots,2Y$. It does not belong to
$\mathcal{S}_{Q}$ for any value of $Q>0$.

  5. 5.

$Y>J$. A multiplet of this type belongs neither to $\mathcal{A}$ nor to
$\mathcal{S}_{Q}$ for any $Q>0$. It belongs to $\mathcal{R}_{Q}$ for
$Q=Y-J,\ldots,Y+J$.

Firstly consider

$\psi=\sum_{M_{JY}\in\mathcal{A}}Y^{2}+\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{R}_{Q}}Y^{2}-\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{R}_{Q}}QY+\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{S}_{Q}}Y^{2}+\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{S}_{Q}}QY.$
(75)

Take a multiplet of type 1. It contributes to $\psi$ as

$\displaystyle\left(0\right)+\left(0\right)-\left(0\right)+\left[\left(-Y+J\right)-\left(-Y-J\right)+1\right]Y^{2}$
(76a) $\displaystyle+\left[\left(-Y-J\right)+\cdots+\left(-Y+J\right)\right]Y$
$\displaystyle=$
$\displaystyle\left(2J+1\right)Y^{2}+\frac{\left(-Y+J\right)\left(-Y+J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)}{2}\
Y$ (76b) $\displaystyle=$ $\displaystyle 0.$ (76c)

Take a multiplet of type 2. It gives the following contribution to $\psi$:

$\displaystyle\left(Y^{2}\right)+\left(0\right)-\left(0\right)+\left(-2Y\right)Y^{2}+\left[1+2+\cdots+\left(-2Y\right)\right]Y$
(77a) $\displaystyle=$
$\displaystyle\left(-2Y+1\right)Y^{2}+\frac{\left(-2Y\right)\left(-2Y+1\right)}{2}\
Y$ (77b) $\displaystyle=$ $\displaystyle 0.$ (77c)

Take a multiplet of type 3. Its contribution to $\psi$ is

$\displaystyle
Y^{2}+\left(J+Y\right)Y^{2}-\left[1+\cdots+\left(J+Y\right)\right]Y$ (78b)
$\displaystyle+\left(J-Y\right)Y^{2}+\left[1+\cdots+\left(J-Y\right)\right]Y$
$\displaystyle=$
$\displaystyle\left(2J+1\right)Y^{2}+\frac{\left(J-Y\right)\left(J-Y+1\right)-\left(J+Y\right)\left(J+Y+1\right)}{2}\
Y$ (78c) $\displaystyle=$ $\displaystyle 0.$ (78d)

Take a multiplet of type 4. It contributes to $\psi$ as

$\displaystyle\left(Y^{2}\right)+\left(2Y\right)Y^{2}-\left(1+2+\cdots+2Y\right)Y+\left(0\right)+\left(0\right)$
(79a) $\displaystyle=$
$\displaystyle\left(2Y+1\right)Y^{2}-\frac{\left(2Y\right)\left(2Y+1\right)}{2}\
Y$ (79b) $\displaystyle=$ $\displaystyle 0.$ (79c)

Take a multiplet of type 5. Its contribution to $\psi$ is

$\displaystyle\left(0\right)+\left[\left(Y+J\right)-\left(Y-J\right)+1\right]Y^{2}$
(80a)
$\displaystyle-\left[\left(Y-J\right)+\cdots+\left(Y+J\right)\right]Y+\left(0\right)+\left(0\right)$
$\displaystyle=$
$\displaystyle\left(2J+1\right)Y^{2}-\frac{\left(Y+J\right)\left(Y+J+1\right)-\left(Y-J-1\right)\left(Y-J\right)}{2}\
Y$ (80b) $\displaystyle=$ $\displaystyle 0.$ (80c)

We conclude that no multiplet produces a nonzero contribution to $\psi$, hence
$\psi$ is zero.

We next consider

$\displaystyle\theta$ $\displaystyle=$ $\displaystyle
2\sum_{M_{JY}\in\mathcal{A}}\left(J^{2}+J-2Y^{2}\right)$ (83)
$\displaystyle+\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{R}_{Q}}\left[\left(J^{2}+J-3Y^{2}+Y\right)+\left(6Y-1\right)Q-3Q^{2}\right]$
$\displaystyle+\sum_{Q>0}\,\sum_{M_{JY}\in\mathcal{S}_{Q}}\left[\left(J^{2}+J-3Y^{2}-Y\right)-\left(6Y+1\right)Q-3Q^{2}\right].$

Take a multiplet of type 1. It belongs to neither $\mathcal{A}$ nor
$\mathcal{R}_{Q}$; it belongs to $\mathcal{S}_{Q}$ from $Q=-Y-J$ to $Q=-Y+J$.
Its contribution to $\theta$ reads

$\displaystyle\left(J^{2}+J-3Y^{2}-Y\right)\left[\left(-Y+J\right)-\left(-Y-J\right)+1\right]$
(84b)
$\displaystyle-\left(6Y+1\right)\left[\left(-Y-J\right)+\cdots+\left(-Y+J\right)\right]-3\left[\left(-Y-J\right)^{2}+\cdots+\left(-Y+J\right)^{2}\right]$
$\displaystyle=$ $\displaystyle\left(J^{2}+J-3Y^{2}-Y\right)\left(2J+1\right)$
$\displaystyle-\left(6Y+1\right)\frac{\left(-Y+J\right)\left(-Y+J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)}{2}$
$\displaystyle-\frac{\left(-Y+J\right)\left(-Y+J+1\right)\left(-2Y+2J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)\left(-2Y-2J-1\right)}{2}$
$\displaystyle=$ $\displaystyle\left(J^{2}+J-3Y^{2}-Y\right)\left(2J+1\right)$
(84h)
$\displaystyle-\left(6Y+1\right)\frac{\left(-Y+J\right)^{2}-Y+J-\left(-Y-J\right)^{2}-Y-J}{2}$
$\displaystyle-\frac{2\left(-Y+J\right)^{3}+3\left(-Y+J\right)^{2}-Y+J-2\left(-Y-J\right)^{3}+3\left(-Y-J\right)^{2}+Y+J}{2}$
$\displaystyle=$ $\displaystyle 0.$ (84i)

Take a multiplet of type 2, which has $J=-Y$. Its contribution to $\theta$ is

$\displaystyle
2\left(Y^{2}-Y-2Y^{2}\right)+\left(Y^{2}-Y-3Y^{2}-Y\right)\left(-2Y\right)$
(85b)
$\displaystyle-\left(6Y+1\right)\left[1+2+\cdots+\left(-2Y\right)\right]-3\left[1+4+\cdots+\left(-2Y\right)^{2}\right]$
$\displaystyle=$
$\displaystyle-2Y-2Y^{2}+4Y^{2}+4Y^{3}+\left(6Y+1\right)Y\left(-2Y+1\right)+Y\left(-2Y+1\right)\left(-4Y+1\right)\hskip
19.91692pt$ (85c) $\displaystyle=$ $\displaystyle 0.$ (85d)

Now consider a multiplet of type 3. It produces the following contribution to
$\theta$:

$\displaystyle
2\left(J^{2}+J-2Y^{2}\right)+\left(J^{2}+J-3Y^{2}+Y\right)\left(J+Y\right)+\left(6Y-1\right)\frac{\left(J+Y\right)\left(J+Y+1\right)}{2}\hskip
34.1433pt$ (86c)
$\displaystyle-\frac{\left(J+Y\right)\left(J+Y+1\right)\left(2J+2Y+1\right)}{2}+\left(J^{2}+J-3Y^{2}-Y\right)\left(J-Y\right)$
$\displaystyle-\left(6Y+1\right)\frac{\left(J-Y\right)\left(J-Y+1\right)}{2}-\frac{\left(J-Y\right)\left(J-Y+1\right)\left(2J-2Y+1\right)}{2}$
$\displaystyle=$ $\displaystyle 0.$ (86d)

Take a multiplet of type 4, i.e. with $J=Y$. It contributes to $\theta$ as

$\displaystyle
2\left(Y^{2}+Y-2Y^{2}\right)+\left(Y^{2}+Y-3Y^{2}+Y\right)\left(2Y\right)$
(87b)
$\displaystyle+\left(6Y-1\right)\left[1+2+\cdots+\left(2Y\right)\right]-3\left[1+4+\cdots+\left(2Y\right)^{2}\right]$
$\displaystyle=$ $\displaystyle
2Y-2Y^{2}+4Y^{2}-4Y^{3}+\left(6Y-1\right)Y\left(2Y+1\right)-Y\left(2Y+1\right)\left(4Y+1\right)$
(87c) $\displaystyle=$ $\displaystyle 0.$ (87d)

Take a multiplet of type 5, that has only $\mathcal{R}_{Q}$ from $Q=Y-J$ to
$Q=Y+J$. It contributes to $\theta$ as

$\displaystyle\left(J^{2}+J-3Y^{2}+Y\right)\left[\left(Y+J\right)-\left(Y-J\right)+1\right]$
(88b)
$\displaystyle+\left(6Y-1\right)\left[\left(Y-J\right)+\cdots+\left(Y+J\right)\right]-3\left[\left(Y-J\right)^{2}+\cdots+\left(Y+J\right)^{2}\right]$
$\displaystyle=$ $\displaystyle\left(J^{2}+J-3Y^{2}+Y\right)\left(2J+1\right)$
(88e)
$\displaystyle+\left(6Y-1\right)\frac{\left(Y+J\right)\left(Y+J+1\right)-\left(Y-J-1\right)\left(Y-J\right)}{2}$
$\displaystyle-\frac{\left(Y+J\right)\left(Y+J+1\right)\left(2Y+2J+1\right)-\left(Y-J-1\right)\left(Y-J\right)\left(2Y-2J-1\right)}{2}\hskip
14.22636pt$ $\displaystyle=$
$\displaystyle\left(J^{2}+J-3Y^{2}+Y\right)\left(2J+1\right)$ (88h)
$\displaystyle+\left(6Y-1\right)\frac{\left(Y+J\right)^{2}+Y+J-\left(Y-J\right)^{2}+Y-J}{2}$
$\displaystyle-\frac{2\left(Y+J\right)^{3}+3\left(Y+J\right)^{2}+Y+J-2\left(Y-J\right)^{3}+3\left(Y-J\right)^{2}-Y+J}{2}$
$\displaystyle=$ $\displaystyle 0.$ (88i)

Thus, all the multiplets produce a null contribution to $\theta$, which means
that $\theta=0$.

##  Appendix D The Goldstone bosons

The terms that describe the mixing of the gauge bosons with the scalars allow
one to obtain the form of the Goldstone bosons. For the mixing of the $W$
boson with the charged scalars we get

$\displaystyle\mathcal{L}_{W^{\pm}G^{\mp}}$ $\displaystyle=$ $\displaystyle
ig\,W_{\mu}^{-}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}v_{JY}^{\ast}\,\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\
\partial^{\mu}M_{JY}^{1}\right.$ (89b)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}v_{JY}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\
\partial^{\mu}\left(M_{JY}^{-1}\right)^{\ast}\right]+\mathrm{H.c.}$

We identify the quantity in Eq. (89) as being equal to
$im_{W}W_{\mu}^{-}\partial^{\mu}G^{+}+\mathrm{H.c.}$ [31]. This implicitly
defines the phase of

$\displaystyle G^{+}$ $\displaystyle=$
$\displaystyle\frac{g}{m_{W}}\left[\sum_{M_{JY}\in\mathcal{R}_{1}}v_{JY}^{\ast}\,\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\
M_{JY}^{1}\right.$ (90b)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{S}_{1}}v_{JY}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\
\left(M_{JY}^{-1}\right)^{\ast}\right],$

where we have used

$M_{JY}\in\mathcal{R}_{1}\setminus\mathcal{A}\Rightarrow J-Y+1=0,\quad\quad
M_{JY}\in\mathcal{S}_{1}\setminus\mathcal{A}\Rightarrow J+Y+1=0.$ (91)

It follows from Eq. (90) that the row matrices defined in Eqs. (8) and (9)
have their first matrix elements given by

$\left(R^{1}_{JY}\right)_{11}=\frac{gv_{JY}}{m_{W}}\,\sqrt{\frac{J^{2}-Y^{2}+J+Y}{2}},\qquad\left(S^{1}_{JY}\right)_{11}=-\frac{gv_{JY}^{\ast}}{m_{W}}\,\sqrt{\frac{J^{2}-Y^{2}+J-Y}{2}}.$
(92)

For the mixing of the $Z$ boson with the neutral scalars we get

$\mathcal{L}_{ZG^{0}}=i\,\frac{g}{c_{W}}\,Z_{\mu}\sum_{M_{JY}\in\mathcal{A}}Y\,v_{JY}\,\partial^{\mu}\left(M_{JY}^{0}\right)^{\ast}+\mathrm{H.c.}$
(93)

We identify this as being equal to
$m_{Z}Z_{\mu}\partial^{\mu}G^{0}+\mathrm{H.c.}$ [31]. This implicitly defines
the sign of

$G^{0}=\frac{ig}{c_{W}m_{Z}}\sum_{M_{JY}\in\mathcal{A}}Y\left[v_{JY}\left(M^{0}_{JY}\right)^{\ast}-v_{JY}^{\ast}\,M^{0}_{JY}\right].$
(94)

Therefore, the row matrices defined in Eq. (11) have their first matrix
elements given by

$\left(A_{JY}\right)_{11}=\frac{-\sqrt{2}g}{c_{W}m_{Z}}\
Y\,\mathrm{Im}\,v_{JY},\qquad\left(B_{JY}\right)_{11}=\frac{\sqrt{2}g}{c_{W}m_{Z}}\
Y\,\mathrm{Re}\,v_{JY}.$ (95)

We shall next demonstrate the Feynman rules in Eqs. (26). Using the result of
Eq. (70), the $ZG^{+}G^{-}$ interaction terms are

$\displaystyle\mathcal{L}_{ZG^{+}G^{-}}$ $\displaystyle=$ $\displaystyle
Z^{\mu}\left(G^{+}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{+}\right)\left\\{-igc_{W}+i\,\frac{g}{c_{W}}\left[\sum_{M_{JY}\in\mathcal{R}_{1}}Y\left|\left(R_{JY}^{1}\right)_{11}\right|^{2}\right.\right.$
(96c)
$\displaystyle\left.\left.-\sum_{M_{JY}\in\mathcal{S}_{1}}Y\left|\left(S_{JY}^{1}\right)_{11}\right|^{2}\right]\right\\}$
$\displaystyle=$ $\displaystyle
Z^{\mu}\left(G^{+}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{+}\right)\left\\{-igc_{W}\right.$
$\displaystyle+i\,\frac{g^{3}}{2m_{W}^{2}c_{W}}\left[\sum_{M_{JY}\in\mathcal{R}_{1}}\left|v_{JY}\right|^{2}Y\left(J^{2}-Y^{2}+J+Y\right)\right.$
$\displaystyle\left.\left.-\sum_{M_{JY}\in\mathcal{S}_{1}}\left|v_{JY}\right|^{2}Y\left(J^{2}-Y^{2}+J-Y\right)\right]\right\\}$
$\displaystyle=$ $\displaystyle
Z^{\mu}\left(G^{+}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{+}\right)\left\\{-igc_{W}\right.$
$\displaystyle+i\,\frac{g^{3}}{2m_{W}^{2}c_{W}}\left[\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y\left(J^{2}-Y^{2}+J+Y\right)\right.$
$\displaystyle\left.\left.-\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y\left(J^{2}-Y^{2}+J-Y\right)\right]\right\\}$
$\displaystyle=$ $\displaystyle
Z^{\mu}\left(G^{+}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{+}\right)\left(-igc_{W}+i\,\frac{g^{3}}{m_{W}^{2}c_{W}}\,\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y^{2}\right)$
(96d) $\displaystyle=$ $\displaystyle
igc_{W}\left(\frac{m_{Z}^{2}}{2m_{W}^{2}}-1\right)Z^{\mu}\left(G^{+}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{+}\right),$
(96e)

where we have used firstly Eq. (92), secondly Eqs. (15) and (91), and thirdly
Eq. (14a). Equation (96e) proves that the Feynman rule in Eq. (26b) is correct
for a model with an arbitrary scalar content.

Using the result of Eq. (72), the $W^{\pm}G^{\mp}G^{0}$ interaction terms are

$\displaystyle\mathcal{L}_{W^{\pm}G^{\mp}G^{0}}$ $\displaystyle=$
$\displaystyle
i\,\frac{g}{2}\,\sum_{M_{JY}\in\mathcal{A}}\,\sqrt{\left(J+Y+1\right)\left(J-Y\right)}$
(97f)
$\displaystyle\times\left[W^{\mu+}\left(A_{JY}-iB_{JY}\right)_{11}\left(S_{JY}^{1}\right)^{\ast}_{11}\left(G^{0}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{0}\right)\right.$
$\displaystyle\left.+W^{\mu-}\left(A_{JY}+iB_{JY}\right)_{11}\left(S_{JY}^{1}\right)_{11}\left(G^{+}\,\partial_{\mu}G^{0}-G^{0}\,\partial_{\mu}G^{+}\right)\right]$
$\displaystyle+i\,\frac{g}{2}\,\sum_{M_{JY}\in\mathcal{A}}\,\sqrt{\left(J+Y\right)\left(J-Y+1\right)}$
$\displaystyle\times\left[W^{\mu+}\left(A_{JY}+iB_{JY}\right)_{11}\left(R_{JY}^{1}\right)^{\ast}_{11}\left(G^{-}\,\partial_{\mu}G^{0}-G^{0}\,\partial_{\mu}G^{-}\right)\right.$
$\displaystyle\left.+W^{\mu-}\left(A_{JY}-iB_{JY}\right)_{11}\left(R_{JY}^{1}\right)_{11}\left(G^{0}\,\partial_{\mu}G^{+}-G^{+}\,\partial_{\mu}G^{0}\right)\right]$
$\displaystyle=$
$\displaystyle-\frac{g^{3}}{2m_{Z}m_{W}c_{W}}\,\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y\left(J+Y+1\right)\left(J-Y\right)$
(97j)
$\displaystyle\times\left[W^{\mu+}\left(G^{0}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{0}\right)-W^{\mu-}\left(G^{+}\,\partial_{\mu}G^{0}-G^{0}\,\partial_{\mu}G^{+}\right)\right]$
$\displaystyle+\frac{g^{3}}{2m_{Z}m_{W}c_{W}}\,\sum_{M_{JY}\in\mathcal{A}}\,\left|v_{JY}\right|^{2}Y\left(J+Y\right)\left(J-Y+1\right)$
$\displaystyle\times\left[W^{\mu+}\left(G^{0}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{0}\right)-W^{\mu-}\left(G^{+}\,\partial_{\mu}G^{0}-G^{0}\,\partial_{\mu}G^{+}\right)\right]$
$\displaystyle=$
$\displaystyle\frac{g^{3}}{m_{Z}m_{W}c_{W}}\,\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}Y^{2}$
(97l)
$\displaystyle\times\left[W^{\mu+}\left(G^{0}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{0}\right)+W^{\mu-}\left(G^{0}\,\partial_{\mu}G^{+}-G^{+}\,\partial_{\mu}G^{0}\right)\right]$
$\displaystyle=$
$\displaystyle\frac{gm_{Z}c_{W}}{2m_{W}}\left[W^{\mu+}\left(G^{0}\,\partial_{\mu}G^{-}-G^{-}\,\partial_{\mu}G^{0}\right)+W^{\mu-}\left(G^{0}\,\partial_{\mu}G^{+}-G^{+}\,\partial_{\mu}G^{0}\right)\right].\hskip
19.91692pt$ (97m)

This proves that the Feynman rule in Eq. (26c) is correct for a model with an
arbitrary scalar content.

From Eq. (7),

$\displaystyle D_{\mu}M_{JY}^{-1}$ $\displaystyle=$
$\displaystyle\partial_{\mu}M_{JY}^{-1}-ieA_{\mu}M_{JY}^{-1}+i\,\frac{g}{c_{W}}\,Z_{\mu}M_{JY}^{-1}\left(Y+c_{W}^{2}\right)$
(98c) $\displaystyle-
ig\,W_{\mu}^{+}M_{JY}^{-2}\,\sqrt{\frac{\left(J+Y+2\right)\left(J-Y-1\right)}{2}}$
$\displaystyle-
ig\,W_{\mu}^{-}M_{JY}^{0}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}},$
$\displaystyle D_{\mu}M_{JY}^{0}$ $\displaystyle=$
$\displaystyle\partial_{\mu}M_{JY}^{0}+i\,\frac{g}{c_{W}}\,Z_{\mu}M_{JY}^{0}Y$
(98f) $\displaystyle-
ig\,W_{\mu}^{+}M_{JY}^{-1}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}$
$\displaystyle-
ig\,W_{\mu}^{-}M_{JY}^{1}\,\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}},$
$\displaystyle D_{\mu}M_{JY}^{1}$ $\displaystyle=$
$\displaystyle\partial_{\mu}M_{JY}^{1}+ieA_{\mu}M_{JY}^{1}+i\,\frac{g}{c_{W}}\,Z_{\mu}M_{JY}^{1}\left(Y-c_{W}^{2}\right)$
(98i) $\displaystyle-
ig\,W_{\mu}^{+}M_{JY}^{0}\,\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}$
$\displaystyle-
ig\,W_{\mu}^{-}M_{JY}^{2}\,\sqrt{\frac{\left(J+Y-1\right)\left(J-Y+2\right)}{2}}.$

Therefore, the interaction terms of the $Z^{0}$ with $W^{\pm}$ and a charged
scalar are given by

$\displaystyle\mathcal{L}_{ZW^{\pm}S^{\mp}}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{c_{W}}\,Z^{\mu}W^{+}_{\mu}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left(Y+c_{W}^{2}\right)\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{-1}\right.$ (99d)
$\displaystyle\left.+\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}Y\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{-1}\right]$
$\displaystyle-\frac{g^{2}}{c_{W}}\,Z_{\mu}W^{-}_{\mu}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}Y\,\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{1}\right.$
$\displaystyle\left.+\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left(Y-c_{W}^{2}\right)\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{1}\right]+\mathrm{H.c.}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{c_{W}}\,Z^{\mu}W^{+}_{\mu}\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left(2Y+c_{W}^{2}\right)\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{-1}$ (99g)
$\displaystyle-\frac{g^{2}}{c_{W}}\,Z_{\mu}W^{-}_{\mu}\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left(2Y-c_{W}^{2}\right)\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\
v_{JY}^{\ast}M_{JY}^{1}$ $\displaystyle+\mathrm{H.c.}$ $\displaystyle=$
$\displaystyle\frac{gm_{W}}{c_{W}}\,Z^{\mu}W^{+}_{\mu}\\!\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left(2Y+c_{W}^{2}\right)\left(S_{JY}\right)_{11}\,\sum_{a=1}^{n_{1}}\left(S_{JY}^{1}\right)^{\ast}_{1a}S_{a}^{-}$
(99j)
$\displaystyle-\frac{gm_{W}}{c_{W}}\,Z_{\mu}W^{-}_{\mu}\\!\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left(2Y-c_{W}^{2}\right)\left(R_{JY}\right)_{11}^{\ast}\,\sum_{a=1}^{n_{1}}\left(R_{JY}^{1}\right)_{1a}S_{a}^{+}$
$\displaystyle+\mathrm{H.c.}$ $\displaystyle=$
$\displaystyle\frac{gm_{W}}{c_{W}}\,Z_{\mu}W^{\mu-}\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left(2Y+c_{W}^{2}\right)\left(S_{JY}^{1}\right)^{\ast}_{11}\,\sum_{a=1}^{n_{1}}\left(S_{JY}^{1}\right)_{1a}S_{a}^{+}$
(99m)
$\displaystyle-\frac{gm_{W}}{c_{W}}\,Z_{\mu}W^{\mu-}\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left(2Y-c_{W}^{2}\right)\left(R_{JY}^{1}\right)^{\ast}_{11}\,\sum_{a=1}^{n_{1}}\left(R_{JY}^{1}\right)_{1a}S_{a}^{+}$
$\displaystyle+\mathrm{H.c.}$

Therefore, the $ZW^{\pm}G^{\mp}$ interaction terms are

$\displaystyle\mathcal{L}_{ZW^{\pm}G^{\mp}}\\!$ $\displaystyle=$
$\displaystyle\\!\frac{gm_{W}}{c_{W}}\,Z_{\mu}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left|\left(S_{JY}^{1}\right)_{11}\right|^{2}\left(2Y+c_{W}^{2}\right)\right.$
(100b)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left|\left(R_{JY}^{1}\right)_{11}\right|^{2}\left(2Y-c_{W}^{2}\right)\right]\left(W^{\mu-}G^{+}+W^{\mu+}G^{-}\right)$
$\displaystyle=$
$\displaystyle\frac{g^{3}}{2m_{W}c_{W}}\,Z_{\mu}W^{\mu-}G^{+}\left[\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{S}_{1}}\left|v_{JY}\right|^{2}\left(J^{2}-Y^{2}+J-Y\right)\left(2Y+c_{W}^{2}\right)\right.$
(100d)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{A}\cap\mathcal{R}_{1}}\left|v_{JY}\right|^{2}\left(J^{2}-Y^{2}+J+Y\right)\left(2Y-c_{W}^{2}\right)\right]+\mathrm{H.c.}$
$\displaystyle=$
$\displaystyle\frac{g^{3}}{2m_{W}c_{W}}\,Z_{\mu}W^{\mu-}G^{+}\left[\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\left(J^{2}-Y^{2}+J-Y\right)\left(2Y+c_{W}^{2}\right)\right.$
(100f)
$\displaystyle\left.-\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\left(J^{2}-Y^{2}+J+Y\right)\left(2Y-c_{W}^{2}\right)\right]+\mathrm{H.c.}$
$\displaystyle=$
$\displaystyle\frac{g^{3}}{m_{W}c_{W}}\,Z_{\mu}W^{\mu-}G^{+}\sum_{M_{JY}\in\mathcal{A}}\left|v_{JY}\right|^{2}\left[-2Y^{2}+c_{W}^{2}\left(J^{2}-Y^{2}+J\right)\right]+\mathrm{H.c.}\hskip
19.91692pt$ (100g) $\displaystyle=$
$\displaystyle\frac{g}{m_{W}c_{W}}\,Z_{\mu}W^{\mu-}G^{+}\left(-c_{W}^{2}m_{Z}^{2}+c_{W}^{2}m_{W}^{2}\right)+\mathrm{H.c.}$
(100h)

This proves the Feynman rule in Eq. (26a) for a model with an arbitrary scalar
content.

##  Appendix E The Feynman diagrams that produce the parameter $S$ in a model
with triplets

In order to compute the oblique parameter $S$ we need to compute the vacuum-
polarization tensors $\Pi_{VV^{\prime}}^{\mu\nu}$, where $V$ and $V^{\prime}$
may be either $A$ and $A$ ($A$ is the photon), $A$ and $Z$, or $Z$ and $Z$.

The diagrams that contribute to $\left.\frac{\partial\,\delta
A_{ZZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ at the one-loop level
are777Here and below we do not include the diagrams that correspond to the
same amplitudes in the model with triplets as in the Standard Model, since
those diagrams do not contribute to $S$.

(101a) (101b) (101c)

The diagrams that contribute to $\left.\frac{\partial\,\delta
A_{AA}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ at one-loop level are

(102a)

The diagrams that contribute to $\left.\frac{\partial\,\delta
A_{AZ}(q^{2})}{\partial q^{2}}\right|_{q^{2}=0}$ at one-loop level are

(103a)

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