compat=1.0.0

Chiral fermions, dimensional regularization,
and the trace anomaly

Fiorenzo Bastianelli${}^{\,a,b}$ and Luca Chiese${}^{\,a}$

aDipartimento di Fisica e Astronomia “Augusto Righi”, Università di Bologna,
via Irnerio 46, I-40126 Bologna, Italy

bINFN, Sezione di Bologna, via Irnerio 46, I-40126 Bologna, Italy

###### Abstract

We investigate the trace anomaly of a chiral fermion in dimensional
regularization, considering in detail the simplest case of coupling to an
abelian gauge field. We apply the Breitenlohner-Maison/’t Hooft-Veltman
prescription for dealing with the chiral matrix $\gamma^{5}$ and verify that
no parity-odd term arises in the trace anomaly. The issue treated here is
analogous to that of a chiral fermion in curved spacetime, discussed in recent
literature. The advantage of having a simplified background minimizes the
amount of algebraic calculations and allows pinpointing better the subtle
points carried by dimensional regularization.

##  1 Introduction

The structure of the trace anomaly of a chiral fermion has been the focus of a
recent debate. The main issue concerns the presence or absence of a parity-odd
term in the trace anomaly of a chiral fermion in a curved background. While
such a term was conjectured to be a possibility in certain chiral theories
[1], it is generically unexpected for the case of a Weyl fermion. However, an
explicit calculation performed in dimensional regularization seemed to
indicate its presence [2]. Several groups have been unable to verify that
claim, either directly or indirectly, [3, 4, 5, 6, 7, 8], and so far the only
confirmations [9, 10, 11] are from groups related to Ref. [2].

We believe that the arguments and proofs on the absence of parity-odd terms in
the trace anomaly, discussed in the above papers and references therein, are
sufficiently solid. However, we find it important to clarify as much as
possible the calculational details in support of general arguments, as to
debug them (in one sense or the other) from errors. This is particularly
desirable regarding the use of dimensional regularization. Dimensional
regularization is notoriously subtle in chiral theories. In the case of the
trace anomaly of a Weyl fermion it has been used in [2] and [8], with the
latter presenting a critical assessment of the former, see also [12]. Here we
wish to offer a similar application of dimensional regularization in the
simplified context of a chiral fermion coupled to an abelian gauge field. The
algebraic structure of the Feynman graphs is simpler compared to the one
arising from the coupling to gravity, and all terms entering the trace anomaly
can be computed by hand. The strategy of employing a simplified background has
already been considered in [13] (and [14] for the non-abelian case), where
Pauli-Villars regularization and heat kernel techniques were used to compute
the trace anomaly for a Weyl fermion coupled to a gauge field, thus proving
the absence of parity-odd terms in that case. Here we wish to apply
dimensional regularization to the same problem, and check that result. We
report explicitly the technical details of the calculation in dimensional
regularization, as they may contribute to dispel doubts—present in the
scientific community, see e.g. Ref. [15]—suggesting that the issue is still
unsettled.

One reason why a CP odd contribution to the trace anomaly cannot appear is
related to the CP invariance of the Weyl fermion lagrangian. If the
regularization adopted could manifestly preserve this symmetry, then the
absence of CP odd terms in the trace anomaly would be assured. This is the
case discussed in ref. [13], were a regularization with Pauli-Villars fields
with Majorana mass could manifestly preserve CP invariance. However,
dimensional regularization does not preserve chiral properties, and the result
is not guaranteed a priori. However, as we shall see, the use of the charge
conjugation matrix $C$ will help in proving the cancellation of the CP odd
terms in dimensional regularization.

In the following, we start by briefly reviewing the lagrangian of a Weyl
fermion coupled to an abelian gauge field and present the corresponding
classical conservation laws. Then, we turn to the perturbative calculations of
the anomalies. We first review the well-known case of the chiral anomaly to
set the stage, and then turn to the calculation of the trace anomaly, showing
that no parity-odd term may arise. Finally, we check the Ward identities
related to the conservation of the stress tensor, to make sure that no
gravitational anomaly is induced by dimensional regularization. Our
conventions follow those of Ref. [13] and are recapitulated in appendix A. We
report the main points of the Breitenlohner-Maison-’t Hooft-Veltman
prescription in appendix B, while computational details are collected in
appendices C and D.

##  2 Lagrangian and classical conservations laws

A Weyl fermion $\lambda(x)$ coupled to an abelian gauge field $A_{a}(x)$ is
described by the lagrangian

$\mathcal{L}=-\bar{\lambda}\not{D}\lambda$ (1)

with $\not{D}=\gamma^{a}D_{a}$, where $D_{a}=\partial_{a}-iA_{a}$ is the gauge
covariant derivative acting on $\lambda$. We have taken $\lambda$ to be left-
handed, $\lambda=P_{L}\lambda$, while its Dirac conjugate satisfies
$\bar{\lambda}=\bar{\lambda}P_{R}$, with chiral projectors defined by

$P_{L}=\frac{1}{2}\left(\mathbb{1}+\gamma^{5}\right)\;,\qquad
P_{R}=\frac{1}{2}\left(\mathbb{1}-\gamma^{5}\right)\;.$ (2)

Chirality forbids a Dirac mass, while a Majorana mass is incompatible with the
gauge symmetry.

The gauge current that couples to the abelian gauge field $A_{a}$ is given by

$J^{a}=i\bar{\lambda}\gamma^{a}\lambda$ (3)

and is conserved on-shell because of the gauge symmetry,

$\partial_{a}J^{a}=0\;.$ (4)

It develops a well-known gauge anomaly at the quantum level.

The stress tensor can be obtained by coupling the theory to gravity through a
vielbein $e_{\mu}{}^{a}$, and defined by the functional derivative of the
action $S$ with respect to the vielbein,
$T^{\mu}{}_{a}=\frac{1}{e}\frac{\delta S}{\delta e_{\mu}{}^{a}}$. In the flat
space limit, dropping terms that vanish upon the use of the equations of
motion, it takes the form

$T^{ab}=\frac{1}{4}\bar{\lambda}\big{(}\gamma^{a}\overset{\leftrightarrow}{D^{b}}+\gamma^{b}\overset{\leftrightarrow}{D^{a}}\big{)}\lambda$
(5)

where $\overset{\leftrightarrow}{D_{a}}=D_{a}-\overset{\leftarrow}{D_{a}}$ and
$\overset{\leftarrow}{D_{a}}=\overset{\leftarrow}{\partial_{a}}+iA_{a}$. It is
manifestly symmetric and satisfies the equations

$\displaystyle\partial_{a}T^{ab}$ $\displaystyle=-J_{a}F^{ab}$ (6)
$\displaystyle T^{a}_{\ a}$ $\displaystyle=0$ (7)

which are valid on-shell. The first one is due to translational invariance,
broken classically by the background field $A_{a}$ with field strength
$F_{ab}=\partial_{a}A_{b}-\partial_{b}A_{a}$. It does not develop any quantum
anomaly. The second relation follows from the Weyl invariance in curved space
and is related to conformal invariance in flat space. It develops a trace
anomaly. In the following, we want to make sure that this trace anomaly does
not contain any parity-odd contribution. Of course, eqs. (4), (6), (7) may be
verified directly using the classical equations of motion of $\lambda$ and
$\bar{\lambda}$.

Let us close this section with a few comments on CP invariance, which help in
interpreting our results on the trace anomaly, as mentioned in the
introduction. The Weyl theory breaks parity, as the parity transformation of a
generic Dirac spinor $\psi$

$\psi(x)\quad\stackrel{{\scriptstyle
P}}{{\longrightarrow}}\quad\psi_{{}_{P}}(x_{{}_{P}})=\eta_{{}_{P}}\beta\psi(x)$
(8)

is incompatible with the chiral properties of the Weyl spinor $\lambda$ (with
the constraint $\lambda=\gamma^{5}\lambda$). Here $x_{{}_{P}}=(t,-\vec{x})$,
$\eta_{{}_{P}}$ is a phase that can be chosen to be either $\pm 1$ or $\pm i$,
and $\beta=i\gamma^{0}$ is the matrix that sits in the definition of the Dirac
conjugate spinor ($\bar{\psi}=\psi^{\dagger}\beta$, we follow the notations
spelled out in appendix A of [13]). However, one can combine it with charge
conjugation, that acts on a generic Dirac spinor as

$\psi(x)\quad\stackrel{{\scriptstyle
C}}{{\longrightarrow}}\quad\psi_{{}_{C}}(x)=\eta_{{}_{C}}C\bar{\psi}^{T}(x)$
(9)

and that by itself would again be incompatible with the chirality of the Weyl
spinor $\lambda$. The charge conjugation matrix $C$ satisfies
$C\gamma^{a}C^{-1}=-\gamma^{a\,T}$, while $\eta_{{}_{C}}$ is an arbitrary
phase. The combination of C and P gives rise to the CP transformation, that is
now compatible with the chiral properties of a Weyl spinor and reads

$\lambda(x)\quad\stackrel{{\scriptstyle
CP}}{{\longrightarrow}}\quad\lambda_{{}_{CP}}(x_{{}_{P}})=\eta_{{}_{CP}}C\lambda^{*}(x)\;.$
(10)

The phase $\eta_{{}_{CP}}$ is unobservable for a single fermion and could be
set to one. Then, it is easy to check that the action from the lagrangian (1)
is invariant (integrals of total derivatives are discarded and the jacobian
from changing variables from $x_{{}_{P}}$ to $x$ is unity) if the background
gauge field is also CP transformed as usual.

##  3 The chiral anomaly

As a preparation, we start repeating the exercise of computing the chiral
anomaly. Though standard—it is material that may be found on various QFT
textbooks— it sets the stage for our subsequent calculations. We regularize
the Feynman diagrams with dimensional regularization, extended by the
Breitenlohner-Maison/’t Hooft-Veltman prescription for dealing with the chiral
matrix $\gamma^{5}$, see appendix B.

In order to derive the chiral anomaly, we expand in perturbation theory the
expectation value $\langle J^{c}(x)\rangle_{{}_{\\!A}}$ of the axial current
$J^{c}(x)=i\bar{\lambda}(x)\gamma^{c}\lambda(x)$ in the background of the
abelian gauge field $A_{a}$, and check the Ward identity that follows from eq.
(4). Perturbation theory is generated by setting $\langle
J^{c}(x)\rangle_{{}_{\\!A}}=\langle J^{c}(x)\,e^{iS_{int}}\rangle$, with
$S_{int}=\int d^{4}x\,A_{a}(x)J^{a}(x)$ and using the perturbative propagator

$\langle\lambda(x)\bar{\lambda}(y)\rangle=\int{d^{4}p\over(2\pi)^{4}}\,e^{ip(x-y)}P_{L}\frac{-\not{p}}{p^{2}}P_{R}$
(11)

where we have kept explicitly both projectors matching the chiral properties
of the left-hand side and denoted $px\equiv p_{a}x^{a}$.

The chiral anomaly arises at second order in the gauge field

$\displaystyle\langle J^{c}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$
$\displaystyle=-\frac{1}{2}\int d^{4}y\int d^{4}z\,A_{a}(y)A_{b}(z)\,\langle
J^{c}(x)J^{a}(y)J^{b}(z)\rangle$ $\displaystyle=-\frac{1}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\quad\quad\times
e^{-ikx}e^{ipy}e^{iqz}\,(2\pi)^{4}\delta^{(4)}(k-p-q)\mathcal{M}^{cab}(p,q)$
(12)

where the Wick contractions for evaluating the correlator $\langle
J^{c}(x)J^{a}(y)J^{b}(z)\rangle$ in momentum space111 We Fourier transform the
three-point function $\langle J^{c}(x)J^{a}(y)J^{b}(z)\rangle$ by  $\int
d^{4}x\int d^{4}y\int d^{4}z\,e^{ikx}e^{-ipy}e^{-iqz}\langle
J^{c}(x)J^{a}(y)J^{b}(z)\rangle=(2\pi)^{4}\delta^{(4)}(k-p-q)\mathcal{M}^{cab}(p,q)$
with incoming momentum $k$ at vertex $J^{c}$, and outgoing momenta $p$ and $q$
at vertices $J^{a}$ and $J^{b}$, respectively. give rise to the usual triangle
diagrams

$\leavevmode\hbox to0pt{\vbox to0pt{\pgfpicture\makeatletter\hbox{\hskip
0.0pt\lower 0.0pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \vertex(a); \vertex[above
right= of a] (b); \vertex[below right= of a] (c); \vertex[left = of a] (i1)
{$c$}; \vertex[right= of b] (f1) {$a$}; \vertex[right= of c] (f2) {$b$};
\par\diagram* { (i1) -- [boson, momentum=$k$] (a) -- [anti fermion, edge
label= $l-p$] (b) -- [boson, momentum=$p$] (f1) [ particle= $e$ ], \par(a) --
[fermion, edge label'=$l+q$] (c) --[fermion, edge label=$l$] (b), (c) --
[boson, momentum'=$q$] (f2), }; \pgfsys@invoke{\lxSVG@closescope
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to0pt{\vbox to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \vertex(a); \vertex[above
right= of a] (b); \vertex[below right= of a] (c); \vertex[left = of a] (i1)
{$c$}; \vertex[right= of b] (f1) {$b$}; \vertex[right= of c] (f2) {$a$};
\par\diagram* { (i1) -- [boson, momentum=$k$] (a) -- [ anti fermion, edge
label= $l-q$] (b) -- [boson, momentum=$q$] (f1) [ particle= $e$ ], \par(a) --
[fermion, edge label'=$l+p$] (c) --[ fermion, edge label=$l$] (b), (c) --
[boson, momentum'=$p$] (f2), }; \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{
{}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ (13)

and produce the expression

$\begin{split}\mathcal{M}^{cab}(p,q)=\
&-i\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{c}P_{L}(\not{l}-\not{p})P_{R}\gamma^{a}P_{L}\not{l}P_{R}\gamma^{b}P_{L}(\not{l}+\not{q})P_{R}\right\\}}{l^{2}(l-p)^{2}(l+q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\;.\end{split}$
(14)

External momenta are kept four-dimensional, while the loop momentum $l$ has
been extended to $n$ dimensions by dimensional regularization. It splits into
a four-dimensional part (denoted by a bar) and a $(n-4)$-dimensional part
(denoted by a hat), as discussed in appendix B, i.e.
$l=\bar{l}+\hat{l}=\bar{l}+s$ (we find it notationally easier to denote the
$(n-4)$-dimensional part by $\hat{l}=s$). Taking the divergence of eq. (12) is
equivalent to contract the above expression by $-ik_{c}$. Considering
cyclicity of the trace, we write it as

$\begin{split}-ik_{c}\mathcal{M}^{cab}(p,q)=\
&-\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{P_{R}\not{k}P_{L}(\not{l}-\not{p})P_{R}\gamma^{a}P_{L}\not{l}P_{R}\gamma^{b}P_{L}(\not{l}+\not{q})\right\\}}{l^{2}(l-p)^{2}(l+q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\end{split}$
(15)

and using the property $P_{R}\gamma^{a}P_{L}=\bar{\gamma}^{a}P_{L}$, which
further enforces the indices $a$, $b$, $c$ to be four-dimensional (as they
contract with external four-dimensional momenta), leads to

$\begin{split}-ik_{c}\mathcal{M}^{cab}(p,q)=\
&-\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{k}P_{L}(\not{l}-\not{p})\bar{\gamma}^{a}P_{L}\not{l}\bar{\gamma}^{b}P_{L}(\not{l}+\not{q})\right\\}}{l^{2}(l-p)^{2}(l+q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\;.\end{split}$
(16)

The above expression contains sixteen terms, obtained by expanding the
projectors, but most of them cancel pairwise. Indeed, considering first
parity-even terms, we may use222 As explained in appendix B, in parity-even
calculations one can use a fully anticommuting $\gamma^{5}$ without running
into any mathematical inconsistency.
$\left\\{\gamma^{5},\gamma^{a}\right\\}=0$ together with
$(\gamma^{5})^{2}=\mathbb{1}$, so that all parity-even terms become equal to

$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{k}(\not{l}-\not{p})\bar{\gamma}^{a}\not{l}\bar{\gamma}^{b}(\not{l}+\not{q})\right\\}}{l^{2}(l-p)^{2}(l+q)^{2}}+\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{k}(\not{l}-\not{q})\bar{\gamma}^{b}\not{l}\bar{\gamma}^{a}(\not{l}+\not{p})\right\\}}{l^{2}(l-q)^{2}(l+p)^{2}}\
.$ (17)

Then, using the identity
$\not{k}=\not{p}+\not{q}=\not{l}+\not{q}-(\not{l}-\not{p})$ in the first
integral, and $\not{p}+\not{q}=\not{l}+\not{p}-(\not{l}-\not{q})$ in the
second one, allows to remove one of the propagators from the integrands. The
remaining terms cancel out pairwise after changing integration variable
$l\to-l$ and using cyclicity of the trace. Parity-odd terms can be simplified
in a similar way, using the following identities

$\displaystyle\not{k}\gamma^{5}=-\gamma^{5}\not{k}=-\gamma^{5}(\not{p}+\not{q})=\gamma^{5}(\not{l}-\not{p})+(\not{l}+\not{q})\gamma^{5}-2\gamma^{5}\not{s}$
(18)
$\displaystyle\not{k}\gamma^{5}=-\gamma^{5}\not{k}=-\gamma^{5}(\not{p}+\not{q})=\gamma^{5}(\not{l}-\not{q})+(\not{l}+\not{p})\gamma^{5}-2\gamma^{5}\not{s}\
.$ (19)

At the end one is left with the terms

$\displaystyle-ik_{c}\mathcal{M}^{cab}(p,q)=\ $
$\displaystyle\frac{1}{8}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\big{\\{}\gamma^{5}\not{s}(\bar{\not{l}}+\not{s}-\not{p})\bar{\gamma}^{a}(\bar{\not{l}}+\not{s})\bar{\gamma}^{b}(\bar{\not{l}}+\not{s}+\not{q})\big{\\}}}{l^{2}(l-p)^{2}(l+q)^{2}}$
$\displaystyle+\frac{1}{8}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\big{\\{}\gamma^{5}\not{s}(\bar{\not{l}}+\not{s}-\not{p})\bar{\gamma}^{a}(\bar{\not{l}}-\not{s})\bar{\gamma}^{b}(\bar{\not{l}}+\not{s}+\not{q})\big{\\}}}{l^{2}(l-p)^{2}(l+q)^{2}}$
$\displaystyle+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}$ (20)

where in the second line the properties of $\gamma^{5}$ have been used to
bring two $\gamma^{5}$ matrices together, leading effectively to the
replacement of the factor $\bar{l}+\not{s}$ by $\bar{l}-\not{s}$, thus
relating the second line to the first one. Let us focus on these first two
integrals, and rewrite the denominator in symmetric form by using the Feynman
parametric formula

$\frac{1}{l^{2}(l-p)^{2}(l+q)^{2}}=2\int_{0}^{1}dx\int_{0}^{1-x}dy\
\Big{(}\bar{r}^{2}+s^{2}+f\Big{)}^{-3}$ (21)

where $\bar{r}=\bar{l}+xq-yp$ is four dimensional, while $f\equiv
f(x,y,p,q)=xq^{2}+yp^{2}-(xq-yp)^{2}$. Since the denominator is symmetric in
the internal momentum $s$, only terms having an even number of $s$ can
contribute, otherwise the integral vanishes by symmetric integration.
Moreover, a term like ${\rm
tr}\left(\gamma^{5}\not{s}\not{s}\bar{\gamma}^{a}\not{s}\bar{\gamma}^{b}\not{s}\right)=s^{4}{\rm
tr}\left(\gamma^{5}\bar{\gamma}^{a}\bar{\gamma}^{b}\right)=0$, so that only
terms proportional to $s^{2}$ may give a nonvanishing result. These terms
contain traces of $\gamma^{5}$ with four-dimensional gamma matrices and can be
evaluated directly. After shifting333We assume that one may shift the momentum
integration variable in dimensional regularization, and take it as a defining
property of dimensional regularization, as discussed for example in [16]. the
integration variable $\bar{l}\to\bar{r}$, neglecting terms linear in $\bar{r}$
that vanish by symmetric integration and terms that vanish by contraction with
the antisymmetric Levi-Civita symbol, one obtains for the numerator

$-8is^{2}\epsilon^{abcd}p_{c}q_{d}(x+y)\ .$ (22)

Integration over $r$ (with measure $d^{n}r=d^{4}\bar{r}\,d^{n-4}s$, see
appendix C for further details) gives a finite result since

$\int\frac{d^{n}r}{(2\pi)^{n}}\frac{s^{2}}{\left(r^{2}+f\right)^{3}}=-\frac{i}{32\pi^{2}}$
(23)

and integration over $x$ and $y$ yields $\frac{2}{3}$. Similar steps hold also
for the crossed diagram and lead to an identical result. Finally, we obtain
the finite result

$-ik_{c}\mathcal{M}^{cab}(p,q)=-\frac{8}{3}\frac{1}{32\pi^{2}}\epsilon^{abcd}p_{c}q_{d}\
.$ (24)

It has to be inserted in the divergence of (12), that is

$\displaystyle\partial_{c}\langle J^{c}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$
$\displaystyle=\frac{i}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\qquad\qquad\times
e^{-ikx}e^{ipy}e^{iqz}\,(2\pi)^{4}\delta^{(4)}(k-p-q)\
k_{c}\mathcal{M}^{cab}(p,q)$
$\displaystyle=\frac{4}{96\pi^{2}}\epsilon^{abcd}\int d^{4}y\int d^{4}z\
A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\qquad\qquad\times e^{-ikx}e^{ipy}e^{iqz}\ p_{c}q_{d}\
(2\pi)^{4}\delta^{(4)}(k-p-q)$
$\displaystyle=-\frac{4}{96\pi^{2}}\epsilon^{abcd}\int d^{4}y\int
d^{4}z\Big{(}\partial_{c}A_{a}(y)\Big{)}\Big{(}\partial_{d}A_{b}(z)\Big{)}\delta^{(4)}(y-x)\delta^{(4)}(z-x)$
$\displaystyle=\frac{4}{96\pi^{2}}\epsilon^{abcd}\Big{(}\partial_{a}A_{b}(x)\Big{)}\Big{(}\partial_{c}A_{d}(x)\Big{)}$
$\displaystyle=\frac{1}{96\pi^{2}}\epsilon^{abcd}F_{ab}(x)F_{cd}(x)$ (25)

where in the last line the abelian field strength
$F_{ab}=\partial_{a}A_{b}-\partial_{b}A_{a}$ is introduced. This is the
correct chiral anomaly, that signals the breakdown of gauge symmetry. It
prevents a consistent quantization of the gauge field, unless other chiral
fermions are added to cancel it.

##  4 The trace anomaly

We now come to consider the quantum properties of the stress tensor, starting
with the calculation of the trace anomaly in dimensional regularization. The
stress tensor of the Weyl fermion has been defined in equation (5). It is
manifestly symmetric and this property is not modified by dimensional
regularization.

To derive the trace anomaly, we express in perturbation theory the expectation
value $\langle T^{cd}(x)\rangle_{{}_{\\!A}}=\langle
T^{cd}(x)e^{iS_{int}}\rangle$ of the stress tensor (5) with $S_{int}=\int
d^{4}x\,A_{a}(x)J^{a}(x)$. We find it convenient to split the stress tensor in
powers of the background field $A_{a}$ as

$T^{cd}=T_{0}^{cd}+T_{1}^{cd}$ (26)

where

$T_{0}^{cd}=\frac{1}{4}\bar{\lambda}\big{(}\gamma^{c}\overset{\leftrightarrow}{\partial^{d}}+\gamma^{d}\overset{\leftrightarrow}{\partial^{c}}\big{)}\lambda\;,\qquad
T_{1}^{cd}=-\frac{1}{2}\left(J^{c}A^{d}+J^{d}A^{c}\right)$ (27)

with the current $J^{a}$ already defined in (3). Then, at second order in the
abelian field $A_{a}$ we find

$\displaystyle\langle T^{cd}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$
$\displaystyle=-\frac{1}{2}\langle T_{0}^{cd}(x)S_{int}S_{int}\rangle+i\langle
T_{1}^{cd}(x)S_{int}\rangle$ $\displaystyle=-\frac{1}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\,\Gamma^{cdab}(x,y,z)$
$\displaystyle=-\frac{1}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\quad\quad\times
e^{-ikx}e^{ipy}e^{iqz}\,(2\pi)^{4}\delta^{(4)}(k-p-q)\mathcal{T}^{cdab}(p,q)$
(28)

where in the second line

$\displaystyle\Gamma^{cdab}(x,y,z)$ $\displaystyle=\langle
T^{cd}_{0}(x)J^{a}(y)J^{b}(z)\rangle$ $\displaystyle\
-\frac{i}{2}\delta^{(4)}(z-x)\left(\eta^{bd}\langle
J^{c}(x)J^{a}(y)\rangle+\eta^{bc}\langle J^{d}(x)J^{a}(y)\rangle\right)$
$\displaystyle\ -\frac{i}{2}\delta^{(4)}(y-x)\left(\eta^{ad}\langle
J^{c}(x)J^{b}(z)\rangle+\eta^{ac}\langle J^{d}(x)J^{b}(z)\rangle\right)\;,$
(29)

while in the last line we have Fourier transformed $\Gamma^{cdab}(x,y,z)$ into
$\mathcal{T}^{cdab}(p,q)$. The matrix element
$\mathcal{T}^{cdab}(p,q)=\mathcal{T}^{cdab}_{(1)}(p,q)+\mathcal{T}^{cdab}_{(2)}(p,q)$
is then given by the following two sets of diagrams

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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \vertex(a); \vertex[above
right= of a] (b); \vertex[below right= of a] (c); \vertex[left = of a] (i1)
{$cd$}; \vertex[right= of b] (f1) {$a$}; \vertex[right= of c] (f2) {$b$};
\par\diagram* { (i1) -- [gluon, momentum=$k$] (a) -- [ fermion, edge label=
$l$] (b) -- [boson, momentum=$p$] (f1) [ particle= $e$ ], \par(a) -- [ anti
fermion, edge label'=$l-p-q$] (c) --[ anti fermion, edge label'=$l-p$] (b),
(c) -- [boson, momentum'=$q$] (f2), }; \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{
{}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\quad+\quad\leavevmode\hbox
to0pt{\vbox to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \vertex(a); \vertex[above
right= of a] (b); \vertex[below right= of a] (c); \vertex[left = of a] (i1)
{$cd$}; \vertex[right= of b] (f1) {$b$}; \vertex[right= of c] (f2) {$a$};
\par\diagram* { (i1) -- [gluon, momentum=$k$] (a) -- [fermion, edge label=
$l$] (b) -- [boson, momentum=$q$] (f1) [ particle= $e$ ], \par(a) -- [ anti
fermion, edge label'=$l-p-q$] (c) --[ anti fermion, edge label'=$l-q$] (b),
(c) -- [boson, momentum'=$p$] (f2), }; \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{
{}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ (30)
$\leavevmode\hbox to0pt{\vbox to0pt{\pgfpicture\makeatletter\hbox{\hskip
0.0pt\lower 0.0pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \par\vertex(a);
\vertex[below=of a] (b); \vertex[left = of a ] (i1) { $cd$}; \vertex[above
right = of a] (f1) { $b$}; \vertex[below right= of b] (f2) {$a$};
\par\diagram* { \par(i1) -- [gluon, momentum = $k$] (a); (a) -- [ boson,
momentum = $q$] (f1); (a) -- [ fermion, half left, edge label=$l$] (b); (b) --
[ fermion, half left, edge label=$l-p$] (a); (b)-- [boson, momentum' = $p$]
(f2); \par}; \par \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{
{}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\quad+\quad\leavevmode\hbox
to0pt{\vbox to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynman \par\vertex(a);
\vertex[below=of a] (b); \par\vertex[left = of a ] (i1) { $cd$}; \vertex[above
right = of a] (f1) { $a$}; \vertex[below right= of b] (f2) {$b$};
\par\diagram* { \par(i1) -- [gluon, momentum = $k$] (a); (a) -- [ boson,
momentum = $p$] (f1); (a) -- [ fermion, half left, edge label=$l$] (b); (b) --
[ fermion, half left, edge label=$l-q$] (a); (b)-- [boson, momentum' = $q$]
(f2); }; \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{
{}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ (31)

leading respectively to the expressions

$\mathcal{T}^{cdab}_{(1)}=-\frac{i}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{\mathcal{N}^{cdab}_{(1)}}{l^{2}(l-p)^{2}(l-p-q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}$
(32)

with numerator

$\begin{split}\mathcal{N}^{cdab}_{(1)}={\rm
tr}&\left\\{P_{R}\left((2l-p-q)^{c}\gamma^{d}+(2l-p-q)^{d}\gamma^{c}\right)P_{L}\not{l}P_{R}\gamma^{a}P_{L}(\not{l}-\not{p})P_{R}\gamma^{b}P_{L}(\not{l}-\not{p}-\not{q})\right\\}\end{split}$
(33)

and

$\mathcal{T}^{cdab}_{(2)}=\frac{i}{2}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{\mathcal{N}^{cdab}_{(2)}}{l^{2}(l-p)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}$
(34)

with

$\mathcal{N}^{cdab}_{(2)}={\rm
tr}\left\\{P_{R}\left(\gamma^{c}\eta^{bd}+\gamma^{d}\eta^{bc}\right)P_{L}\not{l}P_{R}\gamma^{a}P_{L}(\not{l}-\not{p})\right\\}.$
(35)

The loop momentum $l$ is extended by dimensional regularization as before and
the integration measure is $d^{n}l=d^{4}\bar{l}d^{n-4}s$.

Again, using the identities
$P_{R}\gamma^{a}P_{L}=\bar{\gamma}^{a}P_{L}=P_{R}\bar{\gamma}^{a}$ and
$P_{L}\gamma^{a}P_{R}=\bar{\gamma}^{a}P_{R}=P_{L}\bar{\gamma}^{a}$, together
with the idempotence of the projectors, these expressions can be simplified
and the number of projectors reduced to one. Then, the two numerators can be
rewritten as

$\displaystyle\mathcal{N}^{cdab}_{(1)}$ $\displaystyle={\rm
tr}\left\\{P_{R}\left(\
(2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}$
$\displaystyle=\frac{1}{2}{\rm
tr}\left\\{\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}$
$\displaystyle\quad-\frac{1}{2}{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}$
(36)

and

$\displaystyle\mathcal{N}^{cdab}_{(2)}$ $\displaystyle={\rm
tr}\left\\{P_{R}\left(\bar{\gamma}^{c}\eta^{bd}+\bar{\gamma}^{d}\eta^{bc}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\right\\}$
$\displaystyle=\frac{1}{2}{\rm
tr}\left\\{\left(\bar{\gamma}^{c}\eta^{bd}+\bar{\gamma}^{d}\eta^{bc}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\right\\}$
$\displaystyle\quad-\frac{1}{2}{\rm
tr}\left\\{\gamma^{5}\left(\bar{\gamma}^{c}\eta^{bd}+\bar{\gamma}^{d}\eta^{bc}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\right\\}.$
(37)

In the above expressions, there is no need of putting a bar over the momenta
$p$ and $q$ because, being external, they are kept four-dimensional and only
the internal momentum $l$ is extended to $n$ dimensions. We also notice that
all gamma matrices are effectively reduced to the four-dimensional ones.

Now, one can show that terms containing traces of $\gamma^{5}$, which would
produce parity-odd contributions, cancel with those of the crossed diagrams.
This can be done by using the invariance of the trace under transposition, the
known properties of the gamma matrices under transposition, namely
$C\bar{\gamma}^{a}C^{-1}=-\bar{\gamma}^{a\,T}$ and
$C\gamma^{5}C^{-1}=\gamma^{5\,T}$ with $C$ the charge conjugation matrix, a
shift on the integration variable, and the property of the anticommutator
$\left\\{\gamma^{5},\bar{\gamma}^{a}\right\\}=0$. It proves that parity-odd
terms do not appear in the expectation value of the stress tensor, and thus in
the trace anomaly. This is a crucial point of our investigation, so let us
show in more detail the precise steps we have been following.

To start with, it is easy to verify that the parity-odd term from eq. (37)
vanishes

$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left(\bar{\gamma}^{c}\eta^{bd}+\bar{\gamma}^{d}\eta^{bc}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\right\\}}{l^{2}(l-p)^{2}}=0$
(38)

and similarly for the corresponding crossed term. This implies the vanishing
of the parity-odd part of (34). Now, let us turn to the last line of eq. (36)
that also contains $\gamma^{5}$

$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}}{l^{2}(l-p)^{2}(l-p-q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\
\;.$ (39)

These terms cancel among themselves, as we shall see immediately. As already
noticed, the gamma matrices under the trace are now four-dimensional. Then,
using the invariance of the trace under transposition and the relations
involving the charge conjugation matrix $C$

$C\bar{\gamma}^{a}C^{-1}=-\bar{\gamma}^{a\,T}\ ,\qquad
C\gamma^{5}C^{-1}=\gamma^{5\,T}\;,$ (40)

one can write

$\displaystyle(2\bar{l}-p-q)^{c}\ {\rm
tr}\left\\{\gamma^{5}\bar{\gamma}^{d}\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}$
$\displaystyle=(2\bar{l}-p-q)^{c}\ {\rm
tr}\left\\{\gamma^{5}\bar{\gamma}^{d}\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}^{T}$
$\displaystyle=(2\bar{l}-p-q)^{c}\ {\rm
tr}\left\\{(\bar{\not{l}}-\not{p}-\not{q})^{T}(\bar{\gamma}^{b})^{T}(\bar{\not{l}}-\not{p})^{T}(\bar{\gamma}^{a})^{T}(\bar{\not{l}})^{T}(\bar{\gamma}^{d})^{T}\left(\gamma^{5}\right)^{T}\right\\}$
$\displaystyle=(2\bar{l}-p-q)^{c}\ {\rm
tr}\left\\{(\bar{\not{l}}-\not{p}-\not{q})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p})\bar{\gamma}^{a}\bar{\not{l}}\bar{\gamma}^{d}\gamma^{5}\right\\}$
$\displaystyle=-(2\bar{l}-p-q)^{c}\ {\rm
tr}\left\\{\gamma^{5}\bar{\gamma}^{d}(\bar{\not{l}}-\not{p}-\not{q})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p})\bar{\gamma}^{a}\bar{\not{l}}\right\\}$
(41)

where in the last line the anticommutator
$\left\\{\gamma^{5},\bar{\gamma}^{a}\right\\}=0$ has been used. Following
similar steps one finds

$(2\bar{l}-p-q)^{d}\ {\rm
tr}\left\\{\gamma^{5}\bar{\gamma}^{c}\bar{\not{l}}\bar{\gamma}^{a}(\bar{\not{l}}-\not{p})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p}-\not{q})\right\\}=-(2\bar{l}-p-q)^{d}\
{\rm
tr}\left\\{\gamma^{5}\bar{\gamma}^{c}(\bar{\not{l}}-\not{p}-\not{q})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p})\bar{\gamma}^{a}\bar{\not{l}}\right\\}$
(42)

and (39) becomes

$\displaystyle-\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)(\bar{\not{l}}-\not{p}-\not{q})\bar{\gamma}^{b}(\bar{\not{l}}-\not{p})\bar{\gamma}^{a}\bar{\not{l}}\right\\}}{l^{2}(l-p)^{2}(l-p-q)^{2}}$
$\displaystyle+\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{b}(\bar{\not{l}}-\not{q})\bar{\gamma}^{a}(\bar{\not{l}}-\not{p}-\not{q})\right\\}}{l^{2}(l-q)^{2}(l-p-q)^{2}}$
(43)

where in the second line the crossed term has been made explicit. After
shifting the integration variable of the first integral as444The change of
variable can also be written as $\bar{l}\to-\bar{l}+p+q,\ s\to-s$.
$l\to-l+p+q$, one finally obtains

$\displaystyle-\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{b}(\bar{\not{l}}-\not{q})\bar{\gamma}^{a}(\bar{\not{l}}-\not{p}-\not{q})\right\\}}{l^{2}(l-q)^{2}(l-p-q)^{2}}$
$\displaystyle+\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{5}\left((2\bar{l}-p-q)^{c}\bar{\gamma}^{d}+(2\bar{l}-p-q)^{d}\bar{\gamma}^{c}\right)\bar{\not{l}}\bar{\gamma}^{b}(\bar{\not{l}}-\not{q})\bar{\gamma}^{a}(\bar{\not{l}}-\not{p}-\not{q})\right\\}}{l^{2}(l-q)^{2}(l-p-q)^{2}}=0\
.$ (44)

This proves the vanishing of the parity-odd part of (32). Thus, parity-odd
terms do not appear in the expectation value of the stress tensor and in the
corresponding trace anomaly. Similar manipulations with the charge conjugation
matrix have been used in [17]. One could interpret physically these algebraic
results by recalling the CP invariance of the unregulated Weyl theory, which
describes a spin 1/2 particle composed of two states, interpretable as a left
handed particle plus its right handed antiparticle, as well-known for the
original Weyl theory of the massless neutrino. These two states of the Weyl
theory have chiral effects that balance each other out in the trace of the
stress tensor.

Now, as explained in appendix B, in parity-even calculations one can use a
fully anticommuting $\gamma^{5}$ without running into any mathematical
inconsistency. This implies that in the above expressions of the numerators
$\mathcal{N}^{cdab}_{(1)}$ and $\mathcal{N}^{cdab}_{(2)}$ the four-dimensional
gamma matrices can be replaced by $n$-dimensional ones, that is

$\mathcal{N}^{cdab}_{(1)}=\frac{1}{2}{\rm
tr}\left\\{\left((2l-p-q)^{c}\gamma^{d}+(2l-p-q)^{d}\gamma^{c}\right)\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{p}-\not{q})\right\\}$
(45)

and

$\mathcal{N}^{cdab}_{(2)}=\frac{1}{2}{\rm
tr}\left\\{\left(\gamma^{c}\eta^{bd}+\gamma^{d}\eta^{bc}\right)\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}\
.$ (46)

To obtain the trace anomaly, we contract the expectation value (28) with the
four-dimensional metric tensor $\bar{\eta}^{ab}$

$\displaystyle\langle T^{a}_{\
a}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}\equiv\bar{\eta}_{cd}\langle
T^{cd}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$ $\displaystyle=-\frac{1}{2}\int
d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\quad\quad\times
e^{-ikx}e^{ipy}e^{iqz}\,(2\pi)^{4}\delta^{(4)}(k-p-q)\mathcal{T}^{ab}(p,q)$
(47)

where

$\displaystyle\mathcal{T}^{ab}(p,q)\equiv\bar{\eta}_{cd}\mathcal{T}^{cdab}(p,q)=$
$\displaystyle\ -\frac{i}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{(2\not{\bar{l}}-\not{p}-\not{q})\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{p}-\not{q})\right\\}}{l^{2}(l-p)^{2}(l-p-q)^{2}}$
$\displaystyle\quad+\frac{i}{2}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{b}\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
$\displaystyle\quad+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\ .$ (48)

Of course, $\gamma^{a}$ and $\gamma^{b}$ remain essentially four-dimensional
as they are contracted with external gauge fields which are kept four-
dimensional. This expression is further simplified by rewriting
$\not{\bar{l}}=\not{l}-\not{s}$ and using the identity

$\frac{(\not{l}-\not{p}-\not{q})(2\not{l}-\not{p}-\not{q})}{l^{2}(l-p)^{2}(l-p-q)^{2}}=\frac{2}{l^{2}(l-p)^{2}}+\frac{(\not{l}-\not{p}-\not{q})(\not{p}+\not{q})}{l^{2}(l-p)^{2}(l-p-q)^{2}}\
.$ (49)

From the first term of the right-hand side we get an expression which cancels
the integral in the second line of (4). A similar identity can be written for
the crossed terms. Moreover, the second term of the above identity gives a
term that cancels with that one of the crossed diagram. At the end, one is
left with the expression

$\mathcal{T}^{ab}(p,q)=\frac{i}{2}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{s}\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{p}-\not{q})\right\\}}{l^{2}(l-p)^{2}(l-p-q)^{2}}+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}\
.$ (50)

Using the Feynman parametric formula to rewrite the denominator in symmetric
form, see appendix C, the integral becomes

$i\int_{0}^{1}dx\int_{0}^{1-x}dy\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\
\frac{N^{ab}}{\left(\bar{l}^{2}+s^{2}+f\right)^{3}}$ (51)

where the integration variable has been shifted by
$\bar{l}\to\bar{l}+p(x+y)+qx$, $f\equiv
f(x,y,p,q)=p^{2}[x(1-x)+y(1-y)-2xy]+x(1-x)q^{2}+2pqx(1-x-y)$, and the
numerator becomes

$N^{ab}={\rm
tr}\left\\{\not{s}(\not{\bar{l}}+\not{s}+\not{p}(x+y)+\not{q}x)\gamma^{a}(\not{\bar{l}}+\not{s}+\not{p}(x+y-1)+\not{q}x)\gamma^{b}(\not{\bar{l}}+\not{s}+\not{p}(x+y-1)+\not{q}(x-1))\right\\}\
.$ (52)

By symmetry only terms proportional to even powers of $s$ give a nonvanishing
contribution. There are three terms proportional to $s^{2}$ and one
proportional to $s^{4}$. After working the traces out, neglecting linear term
in $\bar{l}$, replacing
$\bar{l}^{a}\bar{l}^{b}=\frac{1}{4}\bar{\eta}^{ab}{\bar{l}}^{2}$, and
evaluating the loop integrals using

$\displaystyle\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{2}}{(\bar{l}^{2}+s^{2}+f)^{3}}=-\frac{i}{32\pi^{2}}$
(53)
$\displaystyle\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{4}}{(\bar{l}^{2}+s^{2}+f)^{3}}=\frac{i}{32\pi^{2}}f\;,$
(54)

we obtain

$\frac{1}{8\pi^{2}}\int_{0}^{1}dx\int_{0}^{1-x}dy\ \bar{N}^{ab}$ (55)

where

$\begin{split}\bar{N}^{ab}&=2p^{a}p^{b}(2x^{2}+2y^{2}+4xy-3x-3y+1)+p^{a}q^{b}(4x^{2}+4xy-4x-2y+1)\\\
&\quad+q^{a}p^{b}(4x^{2}+4xy-4x+1)+q^{a}q^{b}(4x^{2}-2x)\\\
&\quad-\eta^{ab}\Big{(}p^{2}(2x^{2}+4xy-3x+2y^{2}-3y+1)+q^{2}(2x^{2}-x)+pq(4x^{2}+4xy-4x+1)\Big{)}\
.\end{split}$ (56)

Integrating over $x$ and $y$, the only contribution is from
$\int_{0}^{1}dx\int_{0}^{1-x}dy\ (4x^{2}+4xy-4x+1)=\frac{1}{3}$. Similar steps
hold also for the crossed terms and lead to an identical result, so that at
the end we find

$\mathcal{T}^{ab}(p,q)=\frac{1}{12\pi^{2}}\left(q^{a}p^{b}-\eta^{ab}pq\right)\
.$ (57)

This result has to be inserted in (4) and gives the final expression of the
trace anomaly

$\displaystyle\langle T^{a}_{\ a}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$
$\displaystyle=-\frac{1}{2}\int d^{4}y\int d^{4}z\
A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\qquad\qquad\times
e^{-ikx}e^{ipy}e^{iqz}(2\pi)^{4}\delta^{(4)}(k-p-q)\mathcal{T}^{ab}(p,q)$
$\displaystyle=-\frac{1}{24\pi^{2}}\int d^{4}y\int d^{4}z\
A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\qquad\qquad\times
e^{-ikx}e^{ipy}e^{iqz}(2\pi)^{4}\delta^{(4)}(k-p-q)\Big{(}q^{a}p^{b}-\eta^{ab}pq\Big{)}$
$\displaystyle=-\frac{1}{24\pi^{2}}\int d^{4}y\int
d^{4}z\left(\eta^{ab}\partial_{c}^{y}\partial^{c}_{z}-\partial^{a}_{z}\partial^{b}_{y}\right)A_{a}(y)A_{b}(z)\times$
$\displaystyle\qquad\qquad\times\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}e^{-ikx}e^{ipy}e^{iqz}(2\pi)^{4}\delta^{(4)}(k-p-q)$
$\displaystyle=-\frac{1}{24\pi^{2}}\int d^{4}y\int
d^{4}z\Big{[}\Big{(}\partial_{a}A_{b}(y)\Big{)}\Big{(}\partial^{a}A^{b}(z)\Big{)}-\Big{(}\partial_{a}A_{b}(z)\Big{)}\Big{(}\partial^{b}A^{a}(y)\Big{)}\Big{]}\delta^{(4)}(y-x)\delta^{(4)}(z-x)$
$\displaystyle=-\frac{1}{24\pi^{2}}\Big{[}\Big{(}\partial_{a}A_{b}(x)\Big{)}\Big{(}\partial^{a}A^{b}(x)\Big{)}-\Big{(}\partial_{a}A_{b}(x)\Big{)}\Big{(}\partial^{b}A^{a}(x)\Big{)}\Big{]}$
$\displaystyle=-\frac{1}{48\pi^{2}}F^{ab}(x)F_{ab}(x)\ .$ (58)

This is the trace anomaly of a chiral fermion [13]. As one may check, it
corresponds precisely to half the trace anomaly of a Dirac fermion. However,
we cannot yet assert with certainty that this is the exact expression of the
trace anomaly of a Weyl fermion without first checking the conservation of the
stress tensor. That is because the Breitenlohner-Maison/’t Hooft-Veltman
scheme breaks $n$-dimensional Lorentz covariance of chiral theories and the
conservation of the stress tensor is not guaranteed at the quantum level. As a
consequence, it may happen that to preserve the conservation one needs to
introduce counterterms555Conservation is guaranteed, as there are no genuine
gravitational anomalies in four dimensions [18]., which in turn may modify the
expression of the trace anomaly. As we shall see, this is not the case.

##  5 Stress tensor conservation

The Ward identity associated with the conservation of the stress tensor,
corresponding to equation (6), is

$\partial_{c}\langle T^{cd}\rangle_{{}_{\\!A}}=-F^{cd}\langle
J_{c}\rangle_{{}_{\\!A}}\ .$ (59)

To check that there are no anomalies associated with this equation we expand
both sides of this equation at second order in the background gauge field, and
verify their equality666In [19] and [20] this strategy was used to determine
the structure of the Ward identity for the conservation of the stress tensor
of a Dirac fermion. We do the same thing, but for a Weyl fermion.. For the
left-hand side we get

$\displaystyle\partial_{c}\langle T^{cd}(x)\rangle_{{}_{\\!A}}^{{}^{\\!(2)}}$
$\displaystyle=-\frac{1}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\,\partial_{c}^{x}\ \Gamma^{cdab}(x,y,z)$
$\displaystyle=-\frac{1}{2}\int d^{4}y\int
d^{4}z\,A_{a}(y)A_{b}(z)\int\frac{d^{4}k}{(2\pi)^{4}}\int\frac{d^{4}p}{(2\pi)^{4}}\int\frac{d^{4}q}{(2\pi)^{4}}$
$\displaystyle\quad\quad\times
e^{-ikx}e^{ipy}e^{iqz}\,(2\pi)^{4}\delta^{(4)}(k-p-q)\left(-ik_{c}\right)\mathcal{T}^{cdab}(p,q)\
.$ (60)

By expanding perturbatively the right-hand side of equation (59) one finds

$-F^{cd}\langle J_{c}\rangle_{{}_{\\!A}}=-F^{cd}\langle J_{c}\rangle-
iF^{cd}\langle J_{c}S_{int}\rangle+...$ (61)

where $S_{int}=\int d^{4}xA_{a}(x)J^{a}(x)$. Only the second term of the above
expansion contributes, since the first one vanishes, and at second order in
the background gauge field it gives

$\displaystyle-F^{cd}(x)\langle J_{c}(x)\rangle_{{}_{\\!A}}^{(2)}$
$\displaystyle=\frac{i}{2}\int d^{4}y\int d^{4}z\ A_{a}(y)A_{b}(z)\ \times$
$\displaystyle\quad\times\Big{[}\left(\eta^{bd}\delta^{e}_{c}-\delta^{b}_{c}\eta^{de}\right)\langle
J^{c}(x)J^{a}(y)\rangle\ \partial_{e}^{z}\ \delta^{(4)}(z-x)$
$\displaystyle\qquad+\left(\eta^{ad}\delta^{e}_{c}-\delta^{a}_{c}\eta^{de}\right)\langle
J^{c}(x)J^{b}(z)\rangle\ \partial_{e}^{y}\ \delta^{(4)}(y-z)\Big{]}\;.$ (62)

By comparing both sides of equation (59) we obtain the identity

$\displaystyle\partial_{c}^{x}\Gamma^{cdab}(x,y,z)$
$\displaystyle=-i\left(\eta^{bd}\delta^{e}_{c}-\delta^{b}_{c}\eta^{de}\right)\langle
J^{c}(x)J^{a}(y)\rangle\ \partial_{e}^{z}\ \delta^{(4)}(z-x)$
$\displaystyle\quad-i\left(\eta^{ad}\delta^{e}_{c}-\delta^{a}_{c}\eta^{de}\right)\langle
J^{c}(x)J^{b}(z)\rangle\ \partial_{e}^{y}\ \delta^{(4)}(y-z)$ (63)

which in momentum space becomes

$-ik_{c}\mathcal{T}^{cdab}(p,q)=\left(\eta^{bd}q_{c}-\delta^{b}_{c}q^{d}\right)\Pi^{ca}(p)+\left(\eta^{ad}p_{c}-\delta^{a}_{c}p^{d}\right)\Pi^{cb}(q)$
(64)

where $\Pi^{ab}(p)$ is the Fourier transform of the two-point function
$\langle J^{a}(x)J^{b}(y)\rangle$, whose expression is

$\Pi^{ab}(p)=\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{a}P_{L}\not{l}P_{R}\gamma^{b}P_{L}(\not{l}-\not{p})P_{R}\right\\}}{l^{2}(l-p)^{2}}$
(65)

where the integration variable $l$ has been extended by dimensional
regularization. It is easy to see that this integral reduces to half the one
of the vacuum polarization (photon self-energy), i.e.

$\displaystyle\Pi^{ab}(p)$
$\displaystyle=\frac{1}{2}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{a}\not{l}\gamma^{b}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
$\displaystyle\overset{n\to
4}{=}-\frac{i}{24\pi^{2}}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)\left(\frac{2}{4-n}+\frac{5}{3}-\log
p^{2}-\gamma+\log 4\pi\right)$ (66)

where the limit $n\to 4$ has been taken and $\gamma$ is the Euler-Mascheroni
constant. Thus, the right-hand side of equation (64) yields

$\displaystyle\frac{i}{24\pi^{2}}\Big{(}q^{d}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)+\eta^{bd}\left(q^{a}p^{2}-p^{a}pq\right)\Big{)}\left(\frac{2}{4-n}+\frac{5}{3}-\log
p^{2}-\gamma+\log 4\pi+O(4-n)\right)+$ $\displaystyle\
+\frac{i}{24\pi^{2}}\Big{(}p^{d}\left(q^{a}q^{b}-\eta^{ab}q^{2}\right)+\eta^{ad}\left(p^{b}q^{2}-q^{b}pq\right)\Big{)}\left(\frac{2}{4-n}+\frac{5}{3}-\log
q^{2}-\gamma+\log 4\pi+O(4-n)\right)\ .$ (67)

Let us now evaluate the left-hand side of equation (64)

$\displaystyle-ik_{c}\mathcal{T}^{cdab}(p,q)$
$\displaystyle=-\frac{1}{8}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\left[k_{c}(2l-k)^{c}\gamma^{d}+(2l-k)^{d}\not{k}\right]\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}}{l^{2}(l-p)^{2}(l-k)^{2}}$
$\displaystyle\quad+\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\left(\not{k}\eta^{bd}+\gamma^{d}k^{b}\right)\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
$\displaystyle\quad+\Big{(}(p,a)\leftrightarrow(q,b)\Big{)}$ (68)

where $k=p+q$ because of momentum conservation. The following two identities
can be used to simplify the calculation

$\displaystyle k(2l-k)=2lk-k^{2}=l^{2}-(l-k)^{2}$ (69)
$\displaystyle\not{k}=\not{l}-(\not{l}-\not{k})$ (70)

and one has

$\displaystyle-ik_{c}\mathcal{T}^{cdab}(p,q)$
$\displaystyle=-\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}\left(\frac{1}{(l-p)^{2}(l-k)^{2}}-\frac{1}{l^{2}(l-p)^{2}}\right)$
$\displaystyle\quad-\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\left(2l-k\right)^{d}\left(\frac{{\rm
tr}\left\\{\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}}{(l-p)^{2}(l-k)^{2}}-\frac{{\rm
tr}\left\\{\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}\right\\}}{l^{2}(l-p)^{2}}\right)$
$\displaystyle\quad+\frac{1}{4}\ k^{b}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}+\frac{1}{4}\
\eta^{bd}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{q}\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
$\displaystyle\quad+\frac{1}{4}\ k^{a}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{b}(\not{l}-\not{q})\right\\}}{l^{2}(l-q)^{2}}+\frac{1}{4}\
\eta^{ad}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{p}\not{l}\gamma^{b}(\not{l}-\not{q})\right\\}}{l^{2}(l-q)^{2}}$
(71)

where the first two integrals have been multiplied by two since those of the
crossed diagrams are equal after using the invariance of the trace under
transposition and shifting the integration variable. Tadpole integrals have
been neglected because they vanish. The calculation is quite long and we refer
to appendix D for details. Here we give the final result

$\displaystyle-ik_{c}\mathcal{T}^{cdab}(p,q)$
$\displaystyle=\frac{i}{24\pi^{2}}\Bigg{(}q^{d}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)+\eta^{bd}\left(q^{a}p^{2}-p^{a}pq\right)\Bigg{)}\times$
$\displaystyle\qquad\times\left(\frac{2}{4-n}+\frac{5}{3}-\log
p^{2}-\gamma+\log 4\pi+O(4-n)\right)$ $\displaystyle\
+\frac{i}{24\pi^{2}}\Bigg{(}p^{d}\left(q^{a}q^{b}-\eta^{ab}q^{2}\right)+\eta^{ad}\left(p^{b}q^{2}-q^{b}pq\right)\Bigg{)}\times$
$\displaystyle\qquad\times\left(\frac{2}{4-n}+\frac{5}{3}-\log
q^{2}-\gamma+\log 4\pi+O(4-n)\right)$ (72)

that matches precisely eq. (5), thus verifying the Ward identity (59) at
quadratic order in the background field $A_{a}$.

##  6 Conclusions

We have used dimensional regularization to study the trace anomaly of a Weyl
fermion coupled to an abelian gauge field, confirming that no parity-odd term
arises in its expression, as found in [13] by Pauli-Villars regularization and
heat kernel methods. The resulting expression is gauge invariant, even though
the gauge symmetry is anomalous. It equals half the trace anomaly of a Dirac
fermion. This result matches the analogous case of a chiral fermion in curved
spacetime, which has been much debated in the recent literature, as reviewed
in the introduction. The coupling to the abelian gauge field— while
interesting in itself—has allowed to expose in a simpler context the subtle
points of dimensional regularization of chiral theories, which become much
more tedious when the coupling to gravity is turned on. We have given a
detailed description of the strategy adopted and the steps needed to calculate
the trace anomaly. Our exposition might be useful for comparing with
alternative calculations that one may wish to adopt in verifying that no
parity-odd terms arise in the trace anomaly of a Weyl fermion.

##  Appendix A Conventions

We use a mostly plus Minkowski metric $\eta_{ab}$ and gamma matrices
satisfying

$\\{\gamma^{a},\gamma^{b}\\}=2\eta^{ab}\;.$ (73)

The hermitian and traceless chiral matrix $\gamma^{5}$ is defined as

$\gamma^{5}=\frac{i}{4!}\epsilon_{abcd}\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}=-i\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}$
(74)

with the symbol $\epsilon_{abcd}$ normalized as $\epsilon_{0123}=-1$ and
$\epsilon^{0123}=1$. In particular, one has

${\rm
tr}\,(\gamma^{5}\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d})=4i\epsilon^{abcd}.$
(75)

The conjugate Dirac spinor $\bar{\lambda}$ is defined using
$\beta=i\gamma^{0}$ by

$\bar{\lambda}=\lambda^{\dagger}\beta\;.$ (76)

The charge conjugation matrix $C$ is defined to satisfy

$C\gamma^{a}C^{-1}=-{\gamma^{a}}^{T}\;.$ (77)

##  Appendix B The Breitenlohner-Maison-’t Hooft-Veltman prescription

Dimensional regularization is subtle in chiral theories. The main problem
concerns the extension of purely four-dimensional quantities to $n$
dimensions, such as the chiral matrix $\gamma^{5}$ and the Levi-Civita symbol
$\epsilon_{abcd}$. In four dimensions we define $\gamma^{5}$ by eq. (74), so
that it squares to the identity and anticommutes with the other gamma matrices

$\\{\gamma^{5},\gamma^{a}\\}=0\;.$ (78)

In $n$ dimensions there are $n$ gamma matrices satisfying the Clifford algebra

$\\{\gamma^{a},\gamma^{b}\\}=2\eta^{ab},\qquad a,b=0,1,...,n\quad(a,b\neq 5)$
(79)

and the simplest thing one could try is to extend (78) to $n$ dimensions, but
a fully anticommuting $\gamma^{5}$ would lead to inconsistency for parity-odd
calculations, i.e. calculations involving an odd number of $\gamma^{5}$
matrices [21, 22, 23, 24].

In [25] ’t Hooft and Veltman proposed a generalization of $\gamma^{5}$ to $n$
dimensions such that $\gamma^{5}$ anticommutes with the first four gamma
matrices and commutes with the remaining $n-4$ matrices, and derived the
standard chiral anomaly within this scheme. This proposal was further
developed by Breitenlohner and Maison [21], who proved its consistency to all
orders in perturbation theory.

According to this scheme, the $n$-dimensional Minkowski spacetime splits into
the product of a four-dimensional subspace and a $(n-4)$-dimensional subspace.
Any $n$-dimensional object, such as metric tensor, gamma matrices, momenta,
etc., decomposes into a four-dimensional part (denoted by a bar) and a
$(n-4)$-dimensional part (denoted by a hat), for instance

$\eta^{ab}=\bar{\eta}^{ab}+\hat{\eta}^{ab},\quad\gamma^{a}=\bar{\gamma}^{a}+\hat{\gamma}^{a},\quad
p^{a}=\bar{p}^{a}+\hat{p}^{a},\quad...$ (80)

Contractions of indices belonging to different subspaces vanish. The chiral
matrix $\gamma^{5}$ is defined as in four dimensions by (74)

$\gamma^{5}=\frac{i}{4!}\epsilon_{abcd}\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}=-i\bar{\gamma}^{0}\bar{\gamma}^{1}\bar{\gamma}^{2}\bar{\gamma}^{3}$
(81)

where $\epsilon_{abcd}$ is purely four-dimensional, and anticommutes with the
gamma matrices of the four-dimensional subspace, while it commutes with those
belonging to the $(n-4)$-dimensional one

$\displaystyle\left\\{\gamma^{5},\bar{\gamma}^{a}\right\\}=0\quad\text{for}\
a=0,1,2,3$ (82)
$\displaystyle\left[\gamma^{5},\hat{\gamma}^{a}\right]=0\quad\text{for}\ a\geq
4.$

This prescription is such to preserve the square of $\gamma^{5}$ to unity and
the cyclicity of the trace. It breaks $n$-dimensional Lorentz covariance for
chiral objects and, as a consequence, spurious noncovariant terms may appear
in the calculations, which however are expected to be removable by finite
noncovariant counterterms.

The Breitenlohner-Maison prescription is not the unique one for dealing with
$\gamma^{5}$ in dimensional regularization, and a comparison between different
proposals can be found in [22]. However, among these, only the Breitenlohner-
Maison scheme has been shown to give mathematically consistent results at
arbitrary loop orders [21, 26].

At this stage, we wish to stress an important remark. As explained above, a
prescription for $\gamma^{5}$ in $n$ dimensions is necessary to overcome
inconsistencies in parity-odd calculations. However, whenever parity-even
calculations are concerned, in which traces containing only an even number of
$\gamma^{5}$ matrices appear, no such inconsistencies arise, and one can
safely extend (78) to $n$ dimensions and use the square
$(\gamma^{5})^{2}=\mathbb{1}$ to completely eliminate $\gamma^{5}$ from the
traces [22, 23, 24].

We conclude this appendix collecting a list of useful relations. The metric
tensor is split as $\eta^{ab}=\bar{\eta}^{ab}+\hat{\eta}^{ab}$ with

$\eta_{ab}\eta^{ab}=n,\qquad\bar{\eta}_{ab}\bar{\eta}^{ab}=4,\qquad\hat{\eta}_{ab}\hat{\eta}^{ab}=n-4,\qquad\bar{\eta}_{ab}\hat{\eta}^{ab}=0.$
(83)

The last shows that contractions between indices belonging to different
subspaces vanish.

Any vector decomposes as

$k^{a}=\bar{k}^{a}+\hat{k}^{a}$ (84)

and metric tensors act as projectors onto different subspaces

$\displaystyle k^{a}=\eta^{ab}k_{b},\qquad
k_{a}=\eta_{ab}k^{b},\qquad\bar{k}_{a}=\bar{\eta}_{ab}k^{b},\qquad\hat{k}_{a}=\hat{\eta}_{ab}k^{b},\qquad
k^{2}=\bar{k}^{2}+\hat{k}^{2},$ $\displaystyle
k^{2}=k^{a}k_{a}=\eta^{ab}k_{a}k_{b}=\eta_{ab}k^{a}k^{b},\qquad\bar{k}^{2}=\bar{k}_{a}\bar{k}^{a}=\bar{\eta}^{ab}k_{a}k_{b}=\bar{\eta}_{ab}k^{a}k^{b},$
$\displaystyle\hat{k}^{2}=\hat{k}_{a}\hat{k}^{a}=\hat{\eta}^{ab}k_{a}k_{b}=\hat{\eta}_{ab}k^{a}k^{b},\qquad\bar{\eta}_{ab}\hat{k}^{b}=0,\qquad\hat{\eta}_{ab}\bar{k}^{b}=0.$
(85)

Gamma matrices decompose as

$\gamma^{a}=\bar{\gamma}^{a}+\hat{\gamma}^{a}$ (86)

and satisfy

$\displaystyle\\{\gamma^{a},\gamma^{b}\\}=2\eta^{ab},\qquad\gamma^{a}\gamma_{a}=n,\qquad{\rm
tr}\,\gamma^{a}=0,$
$\displaystyle\\{\gamma^{a},\bar{\gamma}^{b}\\}=\\{\bar{\gamma}^{a},\bar{\gamma}^{b}\\}=2\bar{\eta}^{ab},\qquad\gamma^{a}\bar{\gamma}_{a}=\bar{\gamma}^{a}\bar{\gamma}_{a}=4,\qquad{\rm
tr}\,\bar{\gamma}^{a}=0,$
$\displaystyle\\{\gamma^{a},\hat{\gamma}^{b}\\}=\\{\hat{\gamma}^{a},\hat{\gamma}^{b}\\}=2\hat{\eta}^{ab},\qquad\gamma^{a}\hat{\gamma}_{a}=\hat{\gamma}^{a}\hat{\gamma}_{a}=n-4,\qquad{\rm
tr}\,\hat{\gamma}^{a}=0,$
$\displaystyle\\{\bar{\gamma}^{a},\hat{\gamma}^{b}\\}=0,\qquad\bar{\gamma}^{a}\hat{\gamma}_{a}=0.$
(87)

The matrix $\gamma^{5}$ is defined as in four dimensions, see eq. (81). It
anticommutes with the gamma matrices of the four-dimensional subspace and
commutes with those of the $(n-4)$-dimensional subspace

$\left\\{\gamma^{5},\bar{\gamma}^{a}\right\\}=0,\qquad\left[\gamma^{5},\hat{\gamma}^{a}\right]=0$
(88)

which implies

$\left\\{\gamma^{5},\gamma^{a}\right\\}=\left\\{\gamma^{5},\hat{\gamma}^{a}\right\\}=2\gamma^{5}\hat{\gamma}^{a},\qquad\left[\gamma^{5},\gamma^{a}\right]=\left[\gamma^{5},\bar{\gamma}^{a}\right]=2\gamma^{5}\bar{\gamma}^{a}.$
(89)

From the definition of $\gamma^{5}$ in (81), its square
$(\gamma^{5})^{2}=\mathbb{1}$, and the definition of the chiral projectors
(2), one can derive the following identities

$P_{R}\gamma^{a}P_{L}=\bar{\gamma}^{a}P_{L}=P_{R}\bar{\gamma}^{a},\qquad
P_{L}\gamma^{a}P_{R}=\bar{\gamma}^{a}P_{R}=P_{L}\bar{\gamma}^{a}.$ (90)

At last, we list the explicit expression of traces involving two, four, and
six gamma matrices in $n$ even dimensions

${\rm tr}\,(\gamma^{a}\gamma^{b})=2^{\frac{n}{2}}\eta^{ab}$ (91)

${\rm
tr}\,(\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d})=2^{\frac{n}{2}}(\eta^{ab}\eta^{cd}-\eta^{ac}\eta^{bd}+\eta^{ad}\eta^{bc})$
(92) $\displaystyle{\rm
tr}\,(\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g})$
$\displaystyle=2^{\frac{n}{2}}(\eta^{de}\eta^{af}\eta^{bg}-\eta^{de}\eta^{ab}\eta^{fg}+\eta^{de}\eta^{ag}\eta^{fb}$
$\displaystyle\quad\qquad-\eta^{da}\eta^{ef}\eta^{bg}+\eta^{da}\eta^{eb}\eta^{fg}-\eta^{da}\eta^{eg}\eta^{fb}$
$\displaystyle\quad\qquad+\eta^{ea}\eta^{df}\eta^{bg}-\eta^{ea}\eta^{db}\eta^{fg}+\eta^{ea}\eta^{dg}\eta^{fb}$
$\displaystyle\quad\qquad-\eta^{df}\eta^{eb}\eta^{ag}+\eta^{df}\eta^{ab}\eta^{eg}-\eta^{ef}\eta^{ab}\eta^{dg}$
$\displaystyle\quad\qquad+\eta^{ef}\eta^{db}\eta^{ag}-\eta^{af}\eta^{db}\eta^{eg}+\eta^{af}\eta^{eb}\eta^{dg})\
.$ (93)

##  Appendix C Loop integrals and dimensional regularization

In order to combine propagator denominators in loop integrals, we have used
Feynman parametric formulae

$\displaystyle\frac{1}{AB}=\int_{0}^{1}dx\,\frac{1}{\left[xA+(1-x)B\right]^{2}}$
(94)
$\displaystyle\frac{1}{ABC}=2\int_{0}^{1}dx\int_{0}^{1-x}dy\,\frac{1}{\left[xA+yB+(1-x-y)C\right]^{3}}$
(95)

which make the denominators quadratic functions of the loop integration
variable $l$ used in the main text. Then, one completes the square and shifts
the integration variable to absorb linear terms in $l$. The denominator takes
the form $(l^{2}+f)^{m}$, where $m=2,3$ and $f$ is a function of the Feynman
parameters and external momenta. Performing integration over the loop momentum
$l$, terms with odd powers of $l$ in the numerator vanish by symmetry.
Symmetry allows also to replace

$\displaystyle l^{a}l^{b}\to\frac{1}{n}\eta^{ab}l^{2}$ (96) $\displaystyle
l^{a}l^{b}l^{c}l^{d}\to\frac{1}{n(n+2)}\
l^{4}\left(\eta^{ab}\eta^{cd}+\eta^{ac}\eta^{bd}+\eta^{ad}\eta^{bc}\right)$
(97)

where $n$ is the spacetime dimension. It is most convenient to evaluate the
integrals by Wick-rotating the integration variable to Euclidean space, i.e.
by replacing $l^{0}\to il^{0}$.

In the following table we collect $n$-dimensional integrals in Minkowski space

$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{1}{\left(l^{2}+f\right)^{m}}=\frac{i}{(4\pi)^{\frac{n}{2}}}\frac{\Gamma\left(m-\frac{n}{2}\right)}{\Gamma(m)}\left(\frac{1}{f}\right)^{m-\frac{n}{2}}$
(98)
$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{l^{2}}{(l^{2}+f)^{m}}=\frac{i}{(4\pi)^{2}}\frac{n}{2}\frac{\Gamma(m-\frac{n}{2}-1)}{\Gamma(m)}\left(\frac{1}{f}\right)^{m-\frac{n}{2}-1}$
(99)
$\int\frac{d^{n}l}{(2\pi)^{n}}\frac{l^{4}}{(l^{2}+f)^{m}}=\frac{i}{(4\pi)^{\frac{n}{2}}}\frac{n(n+2)}{4}\frac{\Gamma(m-\frac{n}{2}-2)}{\Gamma(m)}\left(\frac{1}{f}\right)^{m-\frac{n}{2}-2}$
(100)

where the overall factor $i$ comes from the Wick rotation of the integration
variable. We need also the following expansion

$\frac{\Gamma(2-\frac{n}{2})}{f^{2-\frac{n}{2}}}\overset{n\to
4}{=}\frac{2}{4-n}-\log f-\gamma+\log 4\pi+O(4-n)$ (101)

where $\gamma$ is the Euler-Mascheroni constant. Thanks to this table we can
easily evaluate integrals appearing in the calculations.

Splitting the loop momentum $l=\bar{l}+s$ we can evaluate the integrals

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{2}}{(\bar{l}^{2}+s^{2}+f)^{3}}$
(102)

and

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{4}}{(\bar{l}^{2}+s^{2}+f)^{3}}\
.$ (103)

Let us first perform integration over $s$ and define $t=\bar{l}^{2}+f$

$\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{2}}{(s^{2}+t)^{3}}=\frac{n-4}{2(4\pi)^{\frac{n-4}{2}}}\frac{\Gamma(4-\frac{n}{2})}{\Gamma(3)}\left(\frac{1}{t}\right)^{4-\frac{n}{2}}$
(104)

then, integrating over $\bar{l}$

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\frac{1}{(\bar{l}^{2}+f)^{4-\frac{n}{2}}}=\frac{i}{(4\pi)^{2}}\frac{\Gamma(2-\frac{n}{2})}{\Gamma(4-\frac{n}{2})}\left(\frac{1}{f}\right)^{2-\frac{n}{2}}\
.$ (105)

Putting everything together and using the expansion (101) we get the finite
result

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{2}}{(\bar{l}^{2}+s^{2}+f)^{3}}=-\frac{i}{32\pi^{2}}\;.$
(106)

Following similar steps one has

$\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{4}}{(s^{2}+t)^{3}}=\frac{(n-4)(n-2)}{(4\pi)^{\frac{n-4}{2}}4}\frac{\Gamma(3-\frac{n}{2})}{\Gamma(3)}\left(\frac{1}{t}\right)^{3-\frac{n}{2}}$
(107)

from the integration over $s$, and

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\frac{1}{(\bar{l}^{2}+f)^{3-\frac{n}{2}}}=\frac{i}{(4\pi)^{2}}\frac{\Gamma(1-\frac{n}{2})}{\Gamma(3-\frac{n}{2})}\left(\frac{1}{f}\right)^{1-\frac{n}{2}}$
(108)

from integrating over $\bar{l}$. Putting everything together we obtain the
finite result

$\int\frac{d^{4}\bar{l}}{(2\pi)^{4}}\int\frac{d^{n-4}s}{(2\pi)^{n-4}}\frac{s^{4}}{(\bar{l}^{2}+s^{2}+f)^{3}}=\frac{i}{32\pi^{2}}f\
.$ (109)

We notice that the factor $(n-4)$ arising from the integration over $s$ kills
the singularity in the expansion (101) and cancels all other terms, once the
limit $n\to 4$ is taken, leading to a finite result.

For completeness, we collect here the relevant integrals employed during the
calculations derived from the above table

$\displaystyle\left(\frac{2}{n}-1\right)\int\frac{d^{n}l}{(2\pi)^{n}}\frac{l^{2}}{(l^{2}+f)^{2}}$
$\displaystyle=\left(\frac{2}{n}-1\right)\frac{i}{(4\pi)^{\frac{n}{2}}}\frac{n}{2}\frac{\Gamma(1-\frac{n}{2})}{\Gamma(2)}\left(\frac{1}{f}\right)^{1-\frac{n}{2}}$
$\displaystyle=\frac{i}{(4\pi)^{\frac{n}{2}}}\left(1-\frac{n}{2}\right)\frac{\Gamma(1-\frac{n}{2})}{f^{1-\frac{n}{2}}}$
$\displaystyle=\frac{i}{(4\pi)^{\frac{n}{2}}}\frac{\Gamma(2-\frac{n}{2})}{f^{2-\frac{n}{2}}}f$
$\displaystyle\overset{n\to 4}{=}\frac{i}{16\pi^{2}}f\left(\frac{2}{4-n}-\log
f-\gamma+\log 4\pi+O(4-n)\right)$ (110)

$\displaystyle\int\frac{d^{n}l}{(2\pi)^{n}}\frac{1}{(l^{2}+f)^{2}}$
$\displaystyle=\frac{i}{(4\pi)^{2}}\frac{\Gamma(2-\frac{n}{2})}{\Gamma(2)}\left(\frac{1}{f}\right)^{2-\frac{n}{2}}$
$\displaystyle\overset{n\to 4}{=}\frac{i}{16\pi^{2}}\left(\frac{2}{4-n}-\log
f-\gamma+\log 4\pi+O(4-n)\right)$ (111)

$\displaystyle\int\frac{d^{n}l}{(2\pi)^{n}}\frac{l^{2}}{(l^{2}+f)^{3}}$
$\displaystyle=\frac{i}{(4\pi)^{2}}\frac{n}{2}\frac{\Gamma(2-\frac{n}{2})}{\Gamma(3)}\left(\frac{1}{f}\right)^{2-\frac{n}{2}}$
$\displaystyle\overset{n\to 4}{=}\frac{i}{16\pi^{2}}\left(\frac{2}{4-n}-\log
f-\gamma+\log 4\pi+O(4-n)\right)\ .$ (112)

##  Appendix D Stress tensor conservation

Let us compute

$\displaystyle-ik_{c}\mathcal{T}^{cdab}=$
$\displaystyle-\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}{\rm
tr}\Big{(}\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\Big{)}\left(\frac{1}{(l-p)^{2}(l-k)^{2}}-\frac{1}{l^{2}(l-p)^{2}}\right)$
(113a)
$\displaystyle-\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\left(2l-k\right)^{d}\left(\frac{{\rm
tr}\left\\{\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}}{(l-p)^{2}(l-k)^{2}}-\frac{{\rm
tr}\left\\{\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}\right\\}}{l^{2}(l-p)^{2}}\right)$
(113b) $\displaystyle+\frac{1}{4}\
k^{b}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
(113c) $\displaystyle+\frac{1}{4}\
\eta^{bd}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{q}\not{l}\gamma^{a}(\not{l}-\not{p})\right\\}}{l^{2}(l-p)^{2}}$
(113d) $\displaystyle+\frac{1}{4}\
k^{a}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{b}(\not{l}-\not{q})\right\\}}{l^{2}(l-q)^{2}}$
(113e) $\displaystyle+\frac{1}{4}\
\eta^{ad}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{p}\not{l}\gamma^{b}(\not{l}-\not{q})\right\\}}{l^{2}(l-q)^{2}}$
(113f)

The terms (113c) and (113e) have the same structure of the integral appearing
in the vacuum polarization (photon self-energy) and yield

$\displaystyle\frac{i}{8\pi^{2}}(p+q)^{b}\left(p^{a}p^{d}-\eta^{ad}p^{2}\right)\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log f-\gamma+\log 4\pi+O(4-n)\right)$ (114a)
$\displaystyle+\frac{i}{8\pi^{2}}(p+q)^{a}\left(q^{b}q^{d}-\eta^{bd}q^{2}\right)\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log g-\gamma+\log 4\pi+O(4-n)\right)$ (114b)

where $f\equiv f(x,p)=p^{2}x(1-x)$, $g\equiv g(q,x)=q^{2}x(1-x)$, $\gamma$ is
the Euler-Mascheroni constant and we have rewritten $k=p+q$ due to momentum
conservation.

Let us consider (113d). Using Feynman parametric formula (94) it becomes

$\frac{1}{4}\ \eta^{bd}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\not{q}(\not{l}+\not{p}x)\gamma^{a}(\not{l}+\not{p}(x-1))\right\\}}{(l^{2}+f)^{2}}$
(115)

where $f\equiv f(x,p)=p^{2}x(1-x)$ and integration variable shifted by $l\to
l+px$. Then, we evaluate the trace

${\rm
tr}\left\\{\gamma^{e}\gamma^{f}\gamma^{a}\gamma^{g}\right\\}q_{e}(l+px)_{f}(l+p(x-1))_{g}={\rm
tr}\left\\{\gamma^{e}\gamma^{f}\gamma^{a}\gamma^{g}\right\\}\Big{(}q_{e}l_{f}l_{g}+q_{e}p_{f}p_{g}x(x-1)\Big{)}$
(116)

where linear terms in $l$ have been neglected because they vanish by symmetric
integration. After replacing $l_{f}l_{g}\to\frac{1}{n}\eta_{fg}l^{2}$, using
(92), (C) and (C) we obtain

$\begin{split}&-\frac{i}{16\pi^{2}}\eta^{bd}\Big{(}2q^{a}p^{2}-2pqp^{a}\Big{)}\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log f-\gamma+\log 4\pi+O(4-n)\right).\end{split}$
(117)

By an analogous reasoning, we can compute (113f) which yields

$\begin{split}&-\frac{i}{16\pi^{2}}\eta^{ad}\Big{(}2p^{b}q^{2}-2pqq^{b}\Big{)}\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log g-\gamma+\log 4\pi+O(4-n)\right)\end{split}$
(118)

where $g\equiv g(q,x)=q^{2}x(1-x)$.

Let us now consider the first integral of (113b), after using the Feynman
parametric formula and shifting the integration variable $l\to l+p+qx$, this
becomes

$-\frac{1}{4}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}\Big{(}2l+p+q(2x-1)\Big{)}^{d}\
\frac{{\rm
tr}\left\\{\gamma^{a}(\not{l}+\not{q}x)\gamma^{b}(\not{l}+\not{q}(x-1))\right\\}}{(l^{2}+g)^{2}}$
(119)

where $g\equiv g(q,x)=q^{2}x(1-x)$. Terms proportional to $(2x-1)$ vanish by
integrating over $x$. The integral proportional to $p^{d}$ has the same
structure of the photon self-energy and yields

$-\frac{i}{8\pi^{2}}p^{d}\left(q^{a}q^{b}-\eta^{ab}q^{2}\right)\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log g-\gamma+\log 4\pi+O(4-n)\right).$ (120)

Now, we evaluate

$-\frac{1}{2}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}l^{d}\frac{{\rm
tr}\left\\{\gamma^{a}(\not{l}+\not{q}x)\gamma^{b}(\not{l}+\not{q}(x-1))\right\\}}{(l^{2}+g)^{2}}\
.$ (121)

Let us compute the trace using (92)

$\begin{split}&{\rm
tr}\left\\{\gamma^{a}\gamma^{c}\gamma^{b}\gamma^{e}\right\\}\left(l+qx\right)_{c}\left(l+q(x-1)\right)_{e}=\\\
&=2^{\frac{n}{2}}\left(\left(l+qx\right)^{a}\left(l+q(x-1)\right)^{b}+\left(l+qx\right)^{b}\left(l+q(x-1)\right)^{a}-\eta^{ab}\left(l+qx\right)\left(l+q(x-1)\right)\right)\
.\end{split}$ (122)

Since the integral is non zero only if even powers of $l$ appear in the
numerator, we keep only terms of this trace with one $l$, and obtain

$-\int_{0}^{1}dx(2x-1)\int\frac{d^{n}l}{(2\pi)^{n}}\frac{l^{d}l^{a}q^{b}+l^{d}l^{b}q^{a}-\eta^{ab}l^{d}lq}{(l^{2}+g)^{2}}=0$
(123)

which vanishes by integration over $x$.

Let us now focus on the second integral of (113b) which can be rewritten as

$\frac{1}{4}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}\left(2l+p(2x-1)-q\right)^{d}\frac{{\rm
tr}\left\\{(\not{l}+\not{p}x)\gamma^{a}(\not{l}+\not{p}(x-1))\gamma^{b}\right\\}}{(l^{2}+f)^{2}}$
(124)

with $f\equiv f(x,p)=p^{2}x(1-x)$. By following a similar reasoning as before,
the unique non zero term is

$-\frac{i}{8\pi^{2}}q^{d}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)\int_{0}^{1}dx\
x(x-1)\left(\frac{2}{4-n}-\log f-\gamma+\log 4\pi+O(4-n)\right).$ (125)

Let us now evaluate (113a) and start from the first term

$-\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}}{(l-p)^{2}(l-k)^{2}}\
.$ (126)

Introducing Feynman parameter and shifting $l\to l+p+qx$, this becomes

$-\frac{1}{4}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}(\not{l}+\not{p}+\not{q}x)\gamma^{a}(\not{l}+\not{q}x)\gamma^{b}(\not{l}+\not{q}(x-1))\right\\}}{(l^{2}+g)^{2}}$
(127)

where $g\equiv g(q,x)=q^{2}x(1-x)$. In evaluating the trace we neglect terms
containing an odd number of $l$ because they vanish by symmetric integration.
Thus, one has

${\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\Big{(}l_{e}l_{f}q_{g}(x-1)+l_{e}l_{g}q_{f}x+l_{f}l_{g}(p+qx)_{e}+(p+qx)_{e}q_{f}xq_{g}(x-1)\Big{)}\
.$ (128)

By symmetric integration we can replace
$l_{a}l_{b}\to\frac{1}{n}\eta_{ab}l^{2}$, use (B) and compute

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{e}l_{f}q_{g}(x-1)$
$\displaystyle=\frac{1}{n}l^{2}{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\eta_{ef}q_{g}(x-1)$
$\displaystyle=\frac{1}{n}l^{2}2^{\frac{n}{2}}\left(\eta^{ab}(2-n)q^{d}(x-1)+\eta^{db}(n-2)q^{a}(x-1)+\eta^{da}(2-n)q^{b}(x-1)\right)$
$\displaystyle=\left(\frac{2}{n}-1\right)2^{\frac{n}{2}}l^{2}\left(\eta^{ab}q^{d}(x-1)-\eta^{db}q^{a}(x-1)+\eta^{da}q^{b}(x-1)\right)\
,$ (129)

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{f}l_{g}(p+qx)_{e}=\frac{1}{n}l^{2}{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\eta_{fg}(p+qx)_{e}$
$\displaystyle=\left(\frac{2}{n}-1\right)2^{\frac{n}{2}}l^{2}\left(\eta^{ab}(p+qx)^{d}-\eta^{da}(p+qx)^{b}+\eta^{db}(p+qx)^{a}\right)\
,$ (130)

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{e}l_{g}q_{f}x=\frac{1}{n}l^{2}{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\eta_{eg}q_{f}x$
$\displaystyle=\left(\frac{2}{n}-1\right)2^{\frac{n}{2}}l^{2}\left(\eta^{db}q^{a}x+\eta^{da}q^{b}x-\eta^{ab}q^{d}x\right)\
.$ (131)

Putting everything together we find

$\left(\frac{2}{n}-1\right)2^{\frac{n}{2}}l^{2}\left(\eta^{ab}\left(q^{d}(x-1)+p^{d}\right)+\eta^{da}\left(-p^{b}+q^{b}(x-1)\right)+\eta^{db}\left(q^{a}(x+1)+p^{a}\right)\right)\
.$ (132)

Integrating over $l$ using (C), one obtains

$\displaystyle\frac{i}{16\pi^{2}}\int_{0}^{1}dx\Big{(}\eta^{ab}\left(q^{d}(x-1)+p^{d}\right)+\eta^{da}\left(-p^{b}+q^{b}(x-1)\right)+$
$\displaystyle+\eta^{db}\left(q^{a}(x+1)+p^{a}\right)\Big{)}g\left(\frac{2}{4-n}-\log
g-\gamma+\log 4\pi+O(4-n)\right)\ .$ (133)

From the term with no $l$ we obtain

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\left(p+qx\right)_{e}q_{f}q_{g}x(x-1)=$
(134)
$\displaystyle=2^{\frac{n}{2}}x(x-1)\Big{(}\left(p+qx\right)^{d}\left(2q^{a}q^{b}-\eta^{ab}q^{2}\right)+$
$\displaystyle\quad+\left(p+qx\right)^{a}\left(2q^{d}q^{b}-\eta^{db}q^{2}\right)+\eta^{ad}\left(q^{2}(p+qx)^{b}-2q^{b}(p+qx)q\right)\Big{)}$
(135)

and after integrating over $l$ using (C)

$\displaystyle-\frac{i}{16\pi^{2}}\int_{0}^{1}dx\ x(x-1)\times$
$\displaystyle\quad\times\Big{(}\left(p+qx\right)^{d}\left(2q^{a}q^{b}-\eta^{ab}q^{2}\right)+\left(p+qx\right)^{a}\left(2q^{d}q^{b}-\eta^{db}q^{2}\right)+$
$\displaystyle\qquad+\eta^{ad}\left(q^{2}(p+qx)^{b}-2q^{b}(p+qx)q\right)\Big{)}\times$
$\displaystyle\quad\qquad\times\left(\frac{2}{4-n}-\log g-\gamma+\log
4\pi+O(4-n)\right)\ .$ (136)

By adding (114b), (LABEL:g_5), (120), (D) and (D) we obtain

$\displaystyle-\frac{i}{16\pi^{2}}\int_{0}^{1}dx\ x(x-1)\times$
$\displaystyle\times\Big{(}4p^{d}\left(q^{a}q^{b}-\eta^{ab}q^{2}\right)+4\eta^{ad}\left(q^{2}p^{b}-q^{b}pq\right)+$
$\displaystyle\quad(2x-1)\left(-\eta^{ad}q^{b}q^{2}-\eta^{ab}q^{d}q^{2}+2q^{a}q^{b}q^{d}+\eta^{bd}q^{a}q^{2}\right)\Big{)}\times$
$\displaystyle\qquad\times\left(\frac{2}{4-n}-\log g-\gamma+\log
4\pi+O(4-n)\right)$ (137)

and after integrating over $x$

$\frac{i}{24\pi^{2}}\Big{(}p^{d}\left(q^{a}q^{b}-\eta^{ab}q^{2}\right)+\eta^{ad}\left(q^{2}p^{b}-q^{b}pq\right)\Big{)}\left(\frac{2}{4-n}+\frac{5}{3}-\log
q^{2}-\gamma+\log 4\pi+O(4-n)\right)\ .$ (138)

Let us now consider the second term of (113a)

$\frac{1}{4}\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}\not{l}\gamma^{a}(\not{l}-\not{p})\gamma^{b}(\not{l}-\not{k})\right\\}}{l^{2}(l-p)^{2}}\
.$ (139)

Making use of Feynman parametric formula for rewriting the denominator and
shifting the integration variable $l\to l+px$, this becomes

$\frac{1}{4}\int_{0}^{1}dx\int\frac{d^{n}l}{(2\pi)^{n}}\frac{{\rm
tr}\left\\{\gamma^{d}(\not{l}+\not{p}x)\gamma^{a}(\not{l}+\not{p}(x-1))\gamma^{b}(\not{l}-\not{q}+\not{p}(x-1))\right\\}}{(l^{2}+f)^{2}}\
.$ (140)

As before, we keep only terms containing an even number of $l$, and they are

${\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}\Big{(}l_{e}l_{f}(-q+p(x-1))_{g}+l_{e}l_{g}p_{f}(x-1)+l_{f}l_{g}p_{e}x+p_{e}xp_{f}(x-1)(-q+p(x-1))_{g}\Big{)}\
.$ (141)

We compute

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{e}l_{f}(-q+p(x-1))_{g}=$
$\displaystyle=\Big{(}\frac{2}{n}-1\Big{)}2^{\frac{n}{2}}l^{2}\left(\eta^{ab}(p(x-1)-q)^{d}-\eta^{db}(p(x-1)-q)^{a}+\eta^{da}(p(x-1)-q)^{b}\right)\
,$ (142)

${\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{e}l_{g}p_{f}(x-1)=\Big{(}\frac{2}{n}-1\Big{)}2^{\frac{n}{2}}l^{2}\left(\eta^{db}p^{a}(x-1)+\eta^{da}p^{b}(x-1)-\eta^{ab}p^{d}(x-1)\right)\
,$ (143)

${\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}l_{f}l_{g}p_{e}x=\Big{(}\frac{2}{n}-1\Big{)}2^{\frac{n}{2}}l^{2}\left(\eta^{ab}p^{d}x-\eta^{da}p^{b}x+\eta^{db}p^{a}x\right)\
.$ (144)

After putting everything together and integrating over $l$, one has

$\displaystyle-\frac{i}{16\pi^{2}}\int_{0}^{1}dx\Big{(}\eta^{ab}\left(p^{d}x-q^{d}\right)+\eta^{da}\left(p^{b}(x-2)-q^{b}\right)+\eta^{db}\left(p^{a}x+q^{a}\right)\Big{)}f\
\times$ $\displaystyle\qquad\qquad\qquad\times\left(\frac{2}{4-n}-\log
f-\gamma+\log 4\pi+O(4-n)\right)\ .$ (145)

From the term with no $l$ one obtains

$\displaystyle{\rm
tr}\left\\{\gamma^{d}\gamma^{e}\gamma^{a}\gamma^{f}\gamma^{b}\gamma^{g}\right\\}p_{e}xp_{f}(x-1)(-q+p(x-1))_{g}=$
$\displaystyle=2^{\frac{n}{2}}x(x-1)\Big{(}\left(-q+p(x-1)\right)^{b}\left(2p^{a}p^{d}-\eta^{ad}p^{2}\right)+$
$\displaystyle+\left(-q+p(x-1)\right)^{d}\left(2p^{a}p^{b}-\eta^{ab}p^{2}\right)-\eta^{bd}p^{a}p^{2}(x-1)+\eta^{bd}\left(2p^{a}pq-q^{a}p^{2}\right)\Big{)}$
(146)

and integrating over $l$

$\displaystyle\frac{i}{16\pi^{2}}\int_{0}^{1}dx\ x(x-1)\times$
$\displaystyle\quad\times\Big{(}\left(-q+p(x-1)\right)^{b}\left(2p^{a}p^{d}-\eta^{ad}p^{2}\right)+\left(-q+p(x-1)\right)^{d}\left(2p^{a}p^{b}-\eta^{ab}p^{2}\right)+$
$\displaystyle\quad\qquad-\eta^{bd}p^{a}p^{2}(x-1)+\eta^{bd}\left(2p^{a}pq-q^{a}p^{2}\right)\Big{)}\times$
$\displaystyle\qquad\qquad\times\left(\frac{2}{4-n}-\log f-\gamma+\log
4\pi+O(4-n)\right).$ (147)

Adding (114a), (117), (125), (D) and (D), we obtain

$\displaystyle\frac{i}{16\pi^{2}}\int_{0}^{1}dx\ x(x-1)\times$
$\displaystyle\times\Big{(}-4q^{d}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)-4\eta^{bd}\left(q^{a}p^{2}-p^{a}pq\right)+$
$\displaystyle\quad+(2x-1)\left(2p^{a}p^{b}p^{d}-\eta^{ab}p^{2}p^{d}-\eta^{bd}p^{a}p^{2}-p^{b}\eta^{ad}p^{2}\right)\Big{)}\times$
$\displaystyle\qquad\qquad\times\left(\frac{2}{4-n}-\log f-\gamma+\log
4\pi+O(4-n)\right)$ (148)

and integrating over $x$

$\frac{i}{24\pi^{2}}\Big{(}q^{d}\left(p^{a}p^{b}-\eta^{ab}p^{2}\right)+\eta^{bd}\left(q^{a}p^{2}-p^{a}pq\right)\Big{)}\left(\frac{2}{4-n}+\frac{5}{3}-\log
p^{2}-\gamma+\log 4\pi+O(4-n)\right)\ .$ (149)

The final result in momentum space is given by the sum of (138) and (149)
which leads to (5).

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