# Stout smearing and Wilson flow in
lattice perturbation theory

Maximilian Ammer$\,{}^{a}$, Stephan Dürr$\,{}^{a,b}$

###### Abstract

We present the expansion of stout smearing and the Wilson flow in lattice
perturbation theory to order $g_{0}^{3}$, which is suitable for one-loop
calculations. As the Wilson flow is generated by infinitesimal stout smearing
steps, the results are related to each other by taking the appropriate limits.
We show how to apply perturbative stout smearing or Wilson flow to the Feynman
rules of any lattice fermion action and and illustrate them by calculating the
self-energy of the clover-improved Wilson fermion.

aDepartment of Physics, University of Wuppertal, 42119 Wuppertal, Germany
bJülich Supercomputing Centre, Forschungszentrum Jülich, 52425 Jülich, Germany

##  1 Introduction

Link smearing is a standard recipe for reducing UV fluctuations in lattice
gauge theory calculations, e.g. lattice QCD. It is particularly useful when
applied to the covariant derivative in a lattice fermion action. Long ago it
was noticed that the taste breaking of staggered fermions gets reduced by link
smearing [1, 2, 3], and similarly the additive mass shift (taken as a measure
for the severity of chiral symmetry breaking) of Wilson/clover fermions gets
reduced by link smearing [4, 5, 6, 7, 8].

In recent years the “stout smearing” procedure, proposed in Ref. [9], has
become very popular, since it yields a smeared link $U^{(n)}_{\mu}(x)$ which
depends in a smooth (differentiable) manner on the unsmeared gauge link
$U_{\mu}(x)$ and its neighbors forming the “staple” around it. This allows to
use smeared fermion actions in the HMC/RHMC algorithm [10, 11, 12], which
crucially relies on the differentiability with respect to the gauge field.

A related recipe is known as “gradient flow” or “Wilson flow” [13, 14, 15].
The flow time $t$ (with the dimension of a length squared) parameterises the
smooth deformation from the unsmeared $U_{\mu}(x)$ to the smeared/flowed
$U_{\mu}(x,t)$, and the recipe was originally formulated as a differential
equation (of the diffusion type) in $t$. In fact, a sequence of stout
smearings whose cumulative smearing parameter matches the flow time is the
simplest possible integration scheme (“forward Euler”) to implement the
gradient flow – see the discussion around Eqn. (205) below for details.

The main difference, in contemporary use, between “stout smearing” and
“gradient/Wilson flow” concerns what is kept fixed in the continuum limit
(i.e. if the lattice spacing is sent to zero, $a\to 0$, while the box size $L$
and the renormalized quark masses $m_{q}$, all in physical units, remain
constant). With “stout smearing” it is common practice to keep both the
smearing parameter $\rho$ and the number of smearings $n_{\mathrm{stout}}$
fixed. With “gradient/Wilson flow” it is common practice to keep the flow time
$t$ (in physical units) constant as the continuum limit is taken, whereupon
the flow time in lattice units $t/a^{2}$ (dimensionless) grows in proportion
to the cut-off squared.

The rationale behind the former choice is that it results in an ultra local
modification of the action. Accordingly, a study with, say, seven stout
smearings is guaranteed to yield the same continuum limit as with an unsmeared
action (see e.g. Ref. [16] for an example of this philosophy). The rationale
behind the latter choice is that $t^{-1/2}$ introduces a second regulator,
independent of the cut-off $a^{-1}$, which persists (and stays finite) in the
continuum limit. One ends up, with a flow-scale regulated theory (or a
perfectly legitimate renormalization scheme and scale), which, contrary to the
lattice regulation, shows universal features [17, 18, 19, 20, 21, 22, 23, 24].
Hence, in the future, one might calculate a light quark mass or
$\alpha_{\mathrm{s}}$ with staggered fermions at the fixed flow scale
$t^{-1/2}=3\,\mathrm{GeV}$. Upon taking the continuum limit, all memory of the
gluon/fermion action combination, which has been used in the computation, is
lost, and one ends up with a genuine physics result in the universal flow
scheme. Hence, no conversion to the $\overline{\mathrm{MS}}$ scheme is needed
any more, the advantage being that the renormalization scheme employed is
well-defined beyond perturbation theory. There are encouraging signs that the
continuum community is getting aware of the advantages of the flow scheme and
provides the necessary computational tools [25, 26, 27, 28, 29, 30].

So far, we discussed the smoothing of gauge fields. A second vital ingredient
in the field-theoretic setup of many lattice QCD computations is the concept
of Symanzik improvement [31]. Specifically for Wilson fermions this program
offers the opportunity of relegating the leading cut-off effects from $O(a)$
to $O(\alpha_{\mathrm{s}}^{k}a)$ or $O(a^{2})$, provided the coefficient
$c_{\mathrm{SW}}$ in front of the Sheikholeslami-Wohlert term is properly
chosen [32, 33, 34]. This improvement program may be carried out either
perturbatively or non-perturbatively, but even in the latter case the
perturbative results are vital for cross-checks and to stabilize the non-
perturbative fits [34].

In order to combine the merits of link smoothing and Symanzik improvement, a
valid framework for carrying out perturbative calculations with smoothed
actions is mandatory. Early steps in this direction have been taken in Refs.
[5, 35, 36, 8]. More recently, the mutual benefits which come from combining
link smearing with clover improvement for undoubled fermion actions have been
emphasized in Ref. [37]. What is still missing is a universal framework that
allows for an easy derivation of Feynman rules for fermion actions with
arbitrary smearing recipes, i.e. an arbitrary $(\varrho,n_{\mathrm{stout}})$
combination or an arbitrary flow time $t/a^{2}\in\mathbb{R}$. In the present
paper we try to fill this gap. In particular we try to convince the reader
that a “brute force” approach for deriving the Feynman rules (i.e. inserting
the perturbative expansion of the smeared/flowed variable into the underlying
action) is not recommendable for $n_{\mathrm{stout}}>1$. We advocate a more
elegant method, based on an $SU(N_{c})$ expansion, which keeps intermediate
expressions much shorter.

The remainder of this article is organized as follows. In Sec. 2 we work out
the perturbative expansion of stout smearing, and in Sec. 3 we do the same
thing for the gradient/Wilson flow. As a byproduct the perturbative matching
between these recipes is discussed, see Eqn. (205). In Sec. 4 we discuss the
procedure which yields, for any smoothing recipe, the Feynman rules in a
manageable form. As an application of this procedure we study in Sec. 5 the
self-energy of a stout smeared clover fermion with variable $\varrho$ and
$1\leq n_{\mathrm{stout}}\leq 4$ or under the gradient flow. Finally, Sec. 6
gives a short summary and an outlook. Some preliminary work on this topic is
found in Ref. [38].

##  2 Perturbative expansion of stout smearing

###  2.1 Definition of stout smearing

Stout smearing is a unitary transformation of the link variable $U_{\mu}(x)$,
which depends on the plaquettes containing $U_{\mu}(x)$. Thereby the
configuration of link variables is smoothed. Often multiple smearing steps are
performed to increase the smoothing effect. We write $U^{(n)}_{\mu}(x)$ for
the link variable after $n$ smearing steps. The transformation to obtain
$U^{(n+1)}_{\mu}(x)$ is given by

$\displaystyle U^{(n+1)}_{\mu}(x)$ $\displaystyle=\exp\big{\\{}i\varrho
Q^{(n)}_{\mu}(x)\big{\\}}U^{(n)}_{\mu}(x)$ (1) $\displaystyle
U^{(0)}_{\mu}(x)$ $\displaystyle=U_{\mu}(x),$ (2)

with the smearing parameter $\varrho$ and the hermitian operator
$Q^{(n)}_{\mu}(x)$, which is constructed from link variables of the previous
smearing step according to

$\displaystyle Q^{(n)}_{\mu}(x)$
$\displaystyle=\frac{1}{2i}\left(W^{(n)}_{\mu}(x)-\frac{1}{N_{c}}\mathrm{Tr}\left[W^{(n)}_{\mu}(x)\right]\right)$
(3) $\displaystyle W^{(n)}_{\mu}(x)$
$\displaystyle=S^{(n)}_{\mu}(x)U^{(n)\dagger}_{\mu}(x)-U^{(n)}_{\mu}(x)S^{(n)\dagger}_{\mu}(x)$
(4) $\displaystyle S^{(n)}_{\mu}(x)$
$\displaystyle=\sum\limits_{\nu\neq\mu}\left(U^{(n)}_{\nu}(x)U^{(n)}_{\mu}(x+\hat{\nu})U^{(n)\dagger}_{\nu}(x+\hat{\mu})+U^{(n)\dagger}_{\nu}(x-\hat{\nu})U^{(n)}_{\mu}(x-\hat{\nu})U^{(n)}_{\nu}(x+\hat{\mu}-\hat{\nu})\right).$
(5)

The superscript $(n)$ denotes quantities that are constructed from link
variables after $n$ smearing steps. The quantity $S^{(n)}_{\mu}(x)$ is
referred to as the sum of staples around the link $U^{(n)}_{\mu}(x)$ and
$iQ^{(n)}_{\mu}(x)$ is constructed as the projection of the product
$S^{(n)}_{\mu}(x)U^{(n)\dagger}_{\mu}(x)$ onto the algebra
$\mathfrak{su}(N_{c})$.

###  2.2 Stout smearing in perturbation theory

In lattice perturbation theory the group element $U_{\mu}(x)\in SU(N_{c})$ is
written as the exponential of a linear combination of generators $T^{a}$ and
gluon fields $A_{\mu}^{a}(x)$

$\displaystyle U_{\mu}(x)$ $\displaystyle=e^{ig_{0}T^{a}A^{a}_{\mu}(x)},$ (6)

where the summation over repeated adjoint color indices
$a,b,c,\ldots\in\\{1,\ldots,N_{c}^{2}-1\\}$ is implied. Expanding the action
to the desired order in the bare coupling $g_{0}$ then gives the Feynman
rules. For one-loop calculations we need to expand up to third order in
$g_{0}$

$\displaystyle U_{\mu}(x)$
$\displaystyle=1+ig_{0}T^{a}A^{a}_{\mu}(x)-\frac{g_{0}^{2}}{2}T^{a}T^{b}A_{\mu}^{a}(x)A_{\mu}^{b}(x)-i\frac{g_{0}^{3}}{6}T^{a}T^{b}T^{c}A^{a}_{\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)+\mathcal{O}(g_{0}^{4}).$
(7)

The Fourier transform of the gluon fields is chosen to be

$\displaystyle
A^{a}_{\mu}(x)=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)k}A^{a}_{\mu}(k)\;.$
(8)

The shift $x+\hat{\mu}/2$ is needed to eliminate overall phase factors in our
results. In order for $U_{\mu}(x)$ to be unitary, the exponent
$ig_{0}T^{a}A^{a}_{\mu}(x)$ has to be anti-hermitian. This means the gluon
fields $A^{a}_{\mu}(x)$ are real (because the generators are hermitian). This
implies that the momentum space field $A_{\mu}^{a}(k)$ fulfils

$\displaystyle A^{a}_{\mu}(k)=A^{a}_{\mu}(-k)\,.$ (9)

As we are going to be working in momentum space, it is helpful to keep in mind
that taking the hermitian conjugate in position space involves the
transformation $k\to-k$ in momentum space.

For a link variable after stout smearing things are more complicated. As the
smeared link variable $U_{\mu}^{(n)}(x)$ is again a group element of
$SU(N_{c})$ [9] we expect it to have an exponential representation similar to
the unsmeared one

$\displaystyle U^{(n)}_{\mu}(x)$
$\displaystyle=e^{ig_{0}T^{a}\tilde{A}^{(n)a}_{\mu}(g_{0},x)}$ (10)
$\displaystyle=1+ig_{0}T^{a}\tilde{A}^{(n)a}_{\mu}(g_{0},x)-\frac{g_{0}^{2}}{2}T^{a}T^{b}\tilde{A}_{\mu}^{(n)a}(g_{0},x)\tilde{A}_{\mu}^{(n)b}(g_{0},x)$
$\displaystyle-i\frac{g_{0}^{3}}{6}T^{a}T^{b}T^{c}\tilde{A}^{(n)a}_{\mu}(g_{0},x)\tilde{A}^{(n)b}_{\mu}(g_{0},x)\tilde{A}^{(n)c}_{\mu}(g_{0},x)+\mathcal{O}(T^{4}).$
(11)

This expansion in terms of generators $T$ does, however, no longer coincide
with the expansion in the bare coupling $g_{0}$. The new fields
$\tilde{A}^{(n)a}_{\mu}(g_{0},x)$ are themselves functions of $g_{0}$. By
expanding $\tilde{A}^{(n)a}_{\mu}(g_{0},x)$ in powers of $g_{0}$ we will be
able to determine its coefficients by comparison with the perturbative
expansion of the right hand side of Eq. (1). This also means that, as we will
see below, we will be able to infer the general structure of the coefficients
of $\tilde{A}^{(n)a}_{\mu}(g_{0},x)$ from the perturbative expansion of only
the first smearing step

$\displaystyle U^{(1)}_{\mu}(x)$ $\displaystyle=\exp\big{\\{}i\varrho
Q^{(0)}_{\mu}(x)\big{\\}}U^{(0)}_{\mu}(x),$ (12)

which in turn allows us to use the following strategy: Expand Eq. (12) to the
desired order in $g_{0}$ and replace $A^{(0)}_{\mu}(x)$ by the expansion of
$\tilde{A}^{(n)a}_{\mu}(g_{0},x)$. This will generate new contributions to
higher orders in $g_{0}$ and give the results for a general smearing step
$n+1>1$.

We start by defining expansions of $S_{\mu}(x)$ and $Q_{\mu}(x)$ in $g_{0}$
(we drop the superscript $(0)$ for now):

$\displaystyle S_{\mu}(x)$
$\displaystyle=6+ig_{0}T^{a}S^{a}_{1\mu}(x)-g_{0}^{2}T^{a}T^{b}S^{ab}_{2\mu}(x)-ig_{0}^{3}T^{a}T^{b}T^{c}S^{abc}_{3\mu}(x)+\mathcal{O}(g_{0}^{4})$
(13) $\displaystyle Q_{\mu}(x)$
$\displaystyle=g_{0}T^{a}Q^{a}_{1\mu}(x)+ig_{0}^{2}\Big{(}T^{a}T^{b}-T^{b}T^{a}\Big{)}Q^{ab}_{2\mu}(x)$
$\displaystyle+g_{0}^{3}\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}Q^{abc}_{3\mu}(x)+\mathcal{O}(g_{0}^{4})\;.$
(14)

Using the Baker-Campbell-Hausdorff-formula

$\displaystyle
e^{X}e^{Y}=e^{X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}([X,[X,Y]]+[Y,[Y,X]])+\ldots}\;,$
(15)

we therefore get

$\displaystyle\exp$ $\displaystyle\big{\\{}i\varrho
Q_{\mu}(x)\big{\\}}\exp\big{\\{}ig_{0}T^{a}A^{a}_{\mu}(x)\big{\\}}=\exp\bigg{\\{}ig_{0}T^{a}A^{a}_{\mu}(x)+i\varrho
g_{0}T^{a}Q^{a}_{1\mu}(x)-\varrho g_{0}^{2}[T^{a},T^{b}]Q^{[ab]}_{2\mu}(x)$
$\displaystyle+i\varrho
g_{0}^{3}\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}Q^{abc}_{3\mu}(x)$
$\displaystyle+\frac{1}{2}\Big{(}-\varrho
g_{0}^{2}[T^{a},T^{b}]Q^{a}_{1\mu}(x)A^{b}_{\mu}(x)-i\varrho
g_{0}^{3}[[T^{a},T^{b}],T^{c}]Q^{[ab]}_{2\mu}(x)A^{c}_{\mu}(x)\Big{)}$
$\displaystyle+\frac{1}{12}\Big{(}-i\varrho^{2}g_{0}^{3}[T^{a},[T^{b},T^{c}]]Q^{a}_{1\mu}(x)Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)-i\varrho
g_{0}^{3}[T^{a},[T^{b},T^{c}]]A^{a}_{\mu}(x)A^{b}_{\mu}(x)Q^{c}_{1\mu}(x)\Big{)}$
$\displaystyle+\mathcal{O}(g_{0}^{4})\bigg{\\}}$ (16)
$\displaystyle=1+ig_{0}T^{a}\big{(}A^{a}_{\mu}(x)+\varrho
Q^{a}_{1\mu}(x)\big{)}-\frac{1}{2}g_{0}^{2}T^{a}T^{b}\big{(}A^{a}_{\mu}(x)+\varrho
Q^{a}_{1\mu}(x)\big{)}\big{(}A^{b}_{\mu}(x)+\varrho Q^{b}_{1\mu}(x)\big{)}$
$\displaystyle-\varrho
g_{0}^{2}[T^{a},T^{b}]\Big{(}\frac{1}{2}Q^{a}_{1\mu}(x)A_{\mu}^{b}(x)+Q^{ab}_{2\mu}(x)\Big{)}$
$\displaystyle-i\frac{g_{0}}{6}^{3}T^{a}T^{b}T^{c}\Big{(}A^{a}_{\mu}(x)+\varrho
Q^{a}_{1\mu}(x)\Big{)}\Big{(}A^{b}_{\mu}(x)+\varrho
Q^{b}_{1\mu}(x)\Big{)}\Big{(}A^{c}_{\mu}(x)+\varrho Q^{c}_{1\mu}(x)\Big{)}$
$\displaystyle-i\frac{g_{0}^{3}}{2}\\{T^{a},[T^{b},T^{c}]\\}\Big{(}A^{a}_{\mu}(x)+\varrho
Q^{a}_{1\mu}(x)\Big{)}\bigg{(}\frac{1}{2}\varrho
Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)+\varrho Q^{bc}_{2\mu}(x)\bigg{)}$
$\displaystyle+i\frac{g_{0}^{3}}{2}[T^{a},[T^{b},T^{c}]]A^{a}_{\mu}(x)\varrho
Q^{[bc]}_{2\mu}(x)$
$\displaystyle+\frac{1}{12}\Big{(}-i\varrho^{2}g_{0}^{3}[T^{a},[T^{b},T^{c}]]Q^{a}_{1\mu}(x)Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)-i\varrho
g_{0}^{3}[T^{a},[T^{b},T^{c}]]A^{a}_{\mu}(x)A^{b}_{\mu}(x)Q^{c}_{1\mu}(x)\Big{)}$
$\displaystyle+i\varrho
g_{0}^{3}\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}Q^{abc}_{3\mu}(x)+\mathcal{O}(g_{0}^{4})$
(17)

where we have used

$\displaystyle
e^{ax+bx^{2}+cx^{3}}=1+ax+\big{(}\tfrac{1}{2}a^{2}+b\big{)}x^{2}+\big{(}\tfrac{1}{6}a^{3}+\tfrac{1}{2}(ab+ba)+c\big{)}x^{3}+\mathcal{O}(x^{4})$
(18)

in the second step. Based on this, it is convenient to define fields
$A^{(n)a}$, $A^{(n)ab}$, $\bar{A}^{(n)abc}$, and $\hat{A}^{(n)abc}$ at each
order in $g_{0}$ for a general $n$ such that

$\displaystyle
U^{(n)}_{\mu}(x)=1+ig_{0}T^{a}A^{(n)a}_{\mu}(x)-\frac{g_{0}^{2}}{2}\Big{(}T^{a}T^{b}A^{(n)a}_{\mu}(x)A^{(n)b}_{\mu}(x)+[T^{a},T^{b}]A_{\mu}^{(n)ab}(x)\Big{)}$
$\displaystyle-i\frac{g_{0}^{3}}{6}\bigg{(}T^{a}T^{b}T^{c}A^{(n)a}_{\mu}(x)A^{(n)b}_{\mu}(x)A^{(n)c}_{\mu}(x)+\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}A^{(n)a}_{\mu}(x)A_{\mu}^{(n)bc}(x)$
$\displaystyle+[T^{a},[T^{b},T^{c}]]\bar{A}^{(n)abc}_{\mu}(x)+\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}\hat{A}^{(n)abc}_{\mu}(x)\bigg{)}$
$\displaystyle+\mathcal{O}(g_{0}^{4})\;.$ (19)

We can now also incorporate $\bar{A}$ into $\hat{A}$ because

$\displaystyle[T^{a},[T^{b},T^{c}]]\bar{A}^{(n)abc}_{\mu}(x)$
$\displaystyle=2[T^{a},T^{b}T^{c}]\bar{A}^{(n)a[bc]}_{\mu}(x)=2(T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a})\bar{A}^{(n)a[bc]}_{\mu}(x)$
(20)

and

$\displaystyle\mathrm{Tr}[T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}]\bar{A}^{(n)a[bc]}_{\mu}(x)$
$\displaystyle=\mathrm{Tr}[T^{a}T^{b}T^{c}-T^{b}T^{c}T^{a}]\bar{A}^{(n)a[bc]}_{\mu}(x)=0\;.$
(21)

Thus we can write

$\displaystyle U^{(n)}_{\mu}(x)$
$\displaystyle=1+ig_{0}T^{a}A^{(n)a}_{\mu}(x)-\frac{g_{0}^{2}}{2}\Big{(}T^{a}T^{b}A^{(n)a}_{\mu}(x)A^{(n)b}_{\mu}(x)+[T^{a},T^{b}]A_{\mu}^{(n)ab}(x)\Big{)}$
$\displaystyle-i\frac{g_{0}^{3}}{6}$
$\displaystyle\bigg{(}T^{a}T^{b}T^{c}A^{(n)a}_{\mu}(x)A^{(n)b}_{\mu}(x)A^{(n)c}_{\mu}(x)+\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}A^{(n)a}_{\mu}(x)A_{\mu}^{(n)bc}(x)$
$\displaystyle+\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}A^{(n)abc}_{\mu}(x)\bigg{)}+\mathcal{O}(g_{0}^{4})$
(22)

with

$\displaystyle
A^{(n)abc}_{\mu}(x)=\hat{A}^{(n)abc}_{\mu}(x)+2\bar{A}^{(n)a[bc]}_{\mu}(x)$
(23)

We define their Fourier transforms as

$\displaystyle A^{(n)a}_{\mu}(x)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)k}A^{(n)a}_{\mu}(k)$
(24) $\displaystyle A^{(n)ab}_{\mu}(x)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{1}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{2}}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)(k_{1}+k_{2})}A^{(n)ab}_{\mu}(k_{1},k_{2})$
(25) $\displaystyle A^{(n)abc}_{\mu}(x)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{1}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{2}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{3}}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)(k_{1}+k_{2}+k_{3})}A^{(n)abc}_{\mu}(k_{1},k_{2},k_{3})\;.$
(26)

These fields are ultimately functions of the original gluon fields

$\displaystyle A^{(n)a}_{\mu}(x)$
$\displaystyle=\sum\limits_{\nu}\sum\limits_{y}\tilde{g}^{(n)}_{\mu\nu}(\varrho,y)A^{(0)a}_{\nu}(x+y)$
(27) $\displaystyle A^{(n)ab}_{\mu}(x)$
$\displaystyle=\sum\limits_{\nu\rho}\sum\limits_{yz}\tilde{g}^{(n)}_{\mu\nu\rho}(\varrho,y,z)A^{(0)a}_{\nu}(x+y)A^{(0)b}_{\rho}(x+z)$
(28) $\displaystyle A^{(n)abc}_{\mu}(x)$
$\displaystyle=\sum\limits_{\nu\rho\sigma}\sum\limits_{yzr}\tilde{g}^{(n)}_{\mu\nu\rho\sigma}(\varrho,y,z,r)A^{(0)a}_{\nu}(x+y)A^{(0)b}_{\rho}(x+z)A^{(0)c}_{\sigma}(x+r)$
(29) $\displaystyle A^{(n)a}_{\mu}(k)$
$\displaystyle=\sum\limits_{\nu}\tilde{g}^{(n)}_{\mu\nu}(\varrho,k)A^{(0)a}_{\nu}(k)$
(30) $\displaystyle A^{(n)ab}_{\mu}(k_{1},k_{2})$
$\displaystyle=\sum\limits_{\nu\rho}\tilde{g}^{(n)}_{\mu\nu\rho}(\varrho,k_{1},k_{2})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})$
(31) $\displaystyle A^{(n)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\sum\limits_{\nu\rho\sigma}\tilde{g}^{(n)}_{\mu\nu\rho\sigma}(\varrho,k_{1},k_{2},k_{3})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})A^{(0)c}_{\sigma}(k_{3})$
(32)

with some complicated functions simply denoted by $\tilde{g}$ here. Sections
2.3, 2.4, and 2.5 will be largely concerned with determining the momentum
space $\tilde{g}$ explicitly.

####  2.2.1 Connecting the perturbative and $SU(N_{c})$-expansions

In order to relate the smeared gluon field $\tilde{A}^{(n)a}_{\mu}(g_{0},x)$
to the perturbative functions $A_{\mu}^{(n)a}(x)$, $A_{\mu}^{(n)ab}(x)$, and
$A_{\mu}^{(n)abc}(x)$, we need to expand the products of color matrices such
that only terms with a single $T^{a}$ remain. We use

$\displaystyle[T^{a},T^{b}]$ $\displaystyle=if^{abc}T^{c}\;,$ $\displaystyle
f^{abc}$ $\displaystyle=-2i\mathrm{Tr}\big{(}T^{a}[T^{b},T^{c}]\big{)}$ (33)
$\displaystyle\\{T^{a},T^{b}\\}$
$\displaystyle=\frac{1}{N_{c}}\delta^{ab}+d^{abc}T^{c}\;,$ $\displaystyle
d^{abc}$ $\displaystyle=2\mathrm{Tr}\big{(}T^{a}\\{T^{b},T^{c}\\}\big{)}$ (34)

and

$\displaystyle
T^{a}T^{b}=\frac{1}{2N_{c}}\delta^{ab}+\frac{1}{2}\big{(}d^{abc}+if^{abc}\big{)}T^{c}$
(35)
$\displaystyle\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}\big{]}=\frac{1}{2}\mathrm{Tr}\big{[}T^{a}\\{T^{b},T^{c}\\}+T^{a}[T^{b},T^{c}]\big{]}=\frac{1}{4}\big{(}d^{abc}+if^{abc}\big{)}\,.$
(36)

With the above color identities we can now rewrite the following expressions
in a way that only one generator $T$ remains:

$\displaystyle[T^{a},T^{b}]A^{(n)[ab]}_{\mu}(x)=if^{abc}T^{a}A^{(n)[bc]}_{\mu}(x)$
(37)
$\displaystyle\bigg{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\bigg{)}A^{(n)abc}_{\mu}(x)$
$\displaystyle=\bigg{(}\frac{1}{2}\Big{(}\\{T^{a},\\{T^{b},T^{c}\\}\\}+[T^{a},[T^{b},T^{c}]]\Big{)}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\bigg{)}A^{(n)abc}_{\mu}(x)$
(38)
$\displaystyle=T^{a}\bigg{[}\frac{1}{N_{c}}\delta^{ae}\delta^{bc}+\frac{1}{2}\big{(}d^{bcd}d^{eda}-f^{bcd}f^{eda}\big{)}\bigg{]}A^{(n)ebc}_{\mu}(x)\;.$
(39)

Thus we can rearrange our perturbative expansion (19) to look like

$\displaystyle U^{(n)}_{\mu}(x)=1+ig_{0}T^{a}$
$\displaystyle\times\Bigg{\\{}A^{(n)a}_{\mu}(x)-\frac{g_{0}}{2}f^{abc}A^{(n)[bc]}_{\mu}(x)-\frac{g_{0}^{2}}{6}\bigg{[}\frac{1}{N_{c}}\delta^{ae}\delta^{bc}+\frac{1}{2}\big{(}d^{bcd}d^{eda}-f^{bcd}f^{eda}\big{)}\bigg{]}A^{(n)ebc}_{\mu}(x)+\mathcal{O}(g_{0}^{3})\Bigg{\\}}$
$\displaystyle-\frac{g_{0}^{2}}{2}T^{a}T^{b}\Bigg{\\{}A_{\mu}^{(n)a}(x)A_{\mu}^{(n)a}(x)-\frac{g_{0}}{2}\Big{(}A^{(n)a}_{\mu}(x)f^{bcd}+f^{acd}A^{(n)b}_{\mu}(x)\Big{)}A^{(n)[cd]}_{\mu}(x)+\mathcal{O}(g_{0}^{2})\Bigg{\\}}$
$\displaystyle-i\frac{g_{0}^{3}}{6}T^{a}T^{b}T^{c}\Bigg{\\{}A_{\mu}^{(n)a}(x)A_{\mu}^{(n)b}(x)A_{\mu}^{(n)c}(x)+\mathcal{O}(g_{0})\Bigg{\\}}+\mathcal{O}(T^{4}).$
(40)

We can see that the expressions in the large braces correspond to the first,
second, and third order of an expansion of the exponential function, each cut
off to give an overall highest order of $g_{0}^{3}$. Hence we can give the
first few orders of the smeared gluon field as

$\displaystyle\tilde{A}_{\mu}^{(n)a}(g_{0},x)$
$\displaystyle=A^{(n)a}_{\mu}(x)-\frac{g_{0}}{2}f^{abc}A^{(n)[bc]}_{\mu}(x)-\frac{g_{0}^{2}}{6}\bigg{[}\frac{1}{N_{c}}\delta^{ae}\delta^{bc}+\frac{1}{2}\big{(}d^{bcd}d^{eda}-f^{bcd}f^{eda}\big{)}\bigg{]}A^{(n)ebc}_{\mu}(x)$
$\displaystyle+\mathcal{O}(g_{0}^{3})\,.$ (41)

Thus, we have expressed the new gluon field in terms of the perturbative
fields.

###  2.3 Leading order

As we have seen, at leading order the perturbative and $SU(N_{c})$ expansions
are identical:
$\tilde{A}^{(n)a}_{\mu}(g_{0},x)=A^{(n)a}_{\mu}(x)+\mathcal{O}(g_{0})$. The
sum of staples at leading order is:

$\displaystyle S^{(n)}_{\mu}(x)$
$\displaystyle=6+ig_{0}T^{a}\bigg{(}\sum\limits_{\nu=1}^{4}\Big{[}A_{\nu}^{(n)a}(x)+A_{\mu}^{(n)a}(x+\hat{\nu})-A_{\nu}^{(n)a}(x+\hat{\mu})-A_{\nu}^{(n)a}(x-\hat{\nu})$
$\displaystyle+A_{\mu}^{(n)a}(x-\hat{\nu})+A_{\nu}^{(n)a}(x+\hat{\mu}-\hat{\nu})\Big{]}-2A_{\mu}^{(n)a}(x)\bigg{)}+\mathcal{O}(g_{0}^{2})$
(42)

where the sum now includes the $\nu=\mu$ term, which is compensated by the
subtraction of $2A_{\mu}^{(n)a}(x)$. Therefore

$\displaystyle Q^{(n)a}_{1\mu}(x)=$
$\displaystyle\sum\limits_{\nu=1}^{4}\Big{[}A_{\nu}^{(n)a}(x)+A_{\mu}^{(n)a}(x+\hat{\nu})-A_{\nu}^{(n)a}(x+\hat{\mu})-A_{\nu}^{(n)a}(x-\hat{\nu})$
$\displaystyle+A_{\mu}^{(n)a}(x-\hat{\nu})+A_{\nu}^{(n)a}(x+\hat{\mu}-\hat{\nu})-2A_{\mu}^{(n)a}(x)\Big{]}$
(43) $\displaystyle=-6A^{(n)a}_{\mu}(x)+S^{(n)a}_{1\mu}(x)\,.$ (44)

The smeared gluon field in position space can then be written as a convolution
[5]:

$\displaystyle A_{\mu}^{(n+1)a}(x)$
$\displaystyle=A^{(n)a}_{\mu}(x)+\varrho\,Q^{(n)a}_{1\mu}(x)$ (45)
$\displaystyle=A^{(n)a}_{\mu}(x)+\varrho\sum\limits_{\nu=1}^{4}\sum\limits_{y}g_{\mu\nu}(y)A_{\nu}^{(n)a}(x+y)$
(46)
$\displaystyle=:\sum\limits_{\nu}\sum\limits_{y}\tilde{g}_{\mu\nu}(\varrho,y)A_{\nu}^{(n)a}(x+y)$
(47)

with $\tilde{g}_{\mu\nu}(\varrho,y)=\delta_{\mu\nu}+\varrho g_{\mu\nu}(y)$ and

$\displaystyle g_{\mu\nu}(y)$
$\displaystyle=\delta(y,0)-\delta(y,\hat{\mu})-\delta(y,-\hat{\nu})+\delta(y,\hat{\mu}-\hat{\nu})+\delta_{\mu\nu}\bigg{[}\sum\limits_{\tau=1}^{4}\big{(}\delta(y,\hat{\tau})+\delta(y,-\hat{\tau})\big{)}-8\delta(y,0)\bigg{]}.$
(48)

Performing a Fourier-transformation using
$\delta(x,y)=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}p}{(2\pi)^{4}}e^{i(x-y)p}$,
we get:

$\displaystyle
e^{iyp}\Bigg{(}1-e^{-ip_{\mu}}-e^{ip_{\nu}}+e^{-i(p_{\mu}-p_{\nu})}+\delta_{\mu\nu}\bigg{[}-8+\sum\limits_{\tau=1}^{4}\big{(}e^{-ip_{\tau}}+e^{ip_{\tau}}\big{)}\bigg{]}\Bigg{)}$
$\displaystyle=e^{iyp}e^{-\frac{i}{2}(p_{\mu}-p_{\nu})}\Big{(}-\delta_{\mu\nu}\hat{p}^{2}+\hat{p}_{\mu}\hat{p}_{\nu}\Big{)}$
(49)

with $\hat{p}_{\mu}=2\sin(\tfrac{1}{2}p_{\mu})$ and
$\hat{p}^{2}=\sum_{\mu=1}^{4}\hat{p}_{\mu}^{2}$. Now we can express the
Fourier transform of the first order smeared gluon field as

$\displaystyle A_{\mu}^{(n+1)a}(x)$
$\displaystyle=\sum_{\nu,y}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}p}{(2\pi)^{4}}e^{iyp}e^{-\frac{i}{2}(p_{\mu}-p_{\nu})}\Big{(}\delta_{\mu\nu}+\varrho\Big{(}-\delta_{\mu\nu}\hat{p}^{2}+\hat{p}_{\mu}\hat{p}_{\nu}\Big{)}\Big{)}e^{i(x+y+\hat{\nu}/2)k}A^{(n)a}_{\nu}(k)$
(50)
$\displaystyle=\sum_{\nu}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)k}\Big{(}\delta_{\mu\nu}+\varrho\Big{(}-\delta_{\mu\nu}\hat{k}^{2}+\hat{k}_{\mu}\hat{k}_{\nu}\Big{)}\Big{)}A^{(n)a}_{\nu}(k)\;,$
(51)

where we have used $\sum\limits_{x}e^{-ix(p-q)}=(2\pi)^{4}\delta(p-q)$ and
performed the integration over $p$. Thus at leading order the result in
momentum space is

$\displaystyle
A_{\mu}^{(n+1)a}(k)=A_{\mu}^{(n)a}(k)+\varrho\sum\limits_{\nu}g_{\mu\nu}(k)A_{\nu}^{(n)a}(k)$
(52)

with

$\displaystyle g_{\mu\nu}(k)$
$\displaystyle=-\delta_{\mu\nu}\hat{k}^{2}+\hat{k}_{\mu}\hat{k}_{\nu}$ (53)

Expressing it in terms of the function $f(k)=1-\varrho\hat{k}^{2}$, it becomes
easier to iterate

$\displaystyle
A_{\mu}^{(n+1)a}(k)=\sum\limits_{\nu}\left(f(k)\delta_{\mu\nu}-(f(k)-1)\frac{\hat{k}_{\mu}\hat{k}_{\nu}}{\hat{k}^{2}}\right)A^{(n)a}_{\nu}(k).$
(54)

In this form, the form factor retains the same structure after multiple
iterations, only the powers of $f$ increase [5][37]:

$\displaystyle A_{\mu}^{(n)a}(k)$
$\displaystyle=\sum\limits_{\nu}\bigg{(}f^{n}(k)\delta_{\mu\nu}-(f^{n}(k)-1)\frac{\hat{k}_{\mu}\hat{k}_{\nu}}{\hat{k}^{2}}\bigg{)}A^{(0)a}_{\nu}(k)$
(55)
$\displaystyle=:\sum\limits_{\nu}\tilde{g}^{n}_{\mu\nu}(\varrho,k)A^{(0)a}_{\nu}(k)$
(56)

We note that $\tilde{g}^{n}_{\mu\nu}(\varrho,k)$ is not exactly the Fourier
transform of $\tilde{g}^{n}_{\mu\nu}(\varrho,y)$ but rather fulfils the same
purpose in momentum space of relating the newly smeared field to the one of
the previous smearing step.

###  2.4 Next-to-leading order

At higher orders the expansions deviate, and Eqs. (16) and (17) would need to
be modified for a general smearing step $n\to n+1$ to include the higher order
perturbative fields $A^{(n)[ab]}_{\mu}(x)$, $\bar{A}^{(n)abc}_{\mu}(x)$, and
$\hat{A}^{(n)abc}_{\mu}(x)$. This is why we start again by considering the
first smearing step. At order $g_{0}^{2}$, the sum of staples
$S^{(0)}_{\mu}(x)$ is given by:

$\displaystyle S^{(0)ab}_{2\mu}(x)$
$\displaystyle=\sum\limits_{\nu\neq\mu}\bigg{[}A_{\nu}^{a}(x)A_{\mu}^{b}(x+\hat{\nu})-A_{\nu}^{a}(x)A_{\nu}^{b}(x+\hat{\mu})-A_{\mu}^{a}(x+\hat{\nu})A_{\nu}^{b}(x+\hat{\mu})$
$\displaystyle-
A_{\nu}^{a}(x-\hat{\nu})A_{\mu}^{b}(x-\hat{\nu})-A_{\nu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})$
$\displaystyle+A_{\mu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})+\frac{1}{2}\Big{\\{}A_{\nu}^{a}(x)A_{\nu}^{b}(x)+A_{\mu}^{a}(x+\hat{\nu})A_{\mu}^{b}(x+\hat{\nu})$
$\displaystyle+A_{\nu}^{a}(x+\hat{\mu})A_{\nu}^{b}(x+\hat{\mu})+A_{\nu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x-\hat{\nu})+A_{\mu}^{a}(x-\hat{\nu})A_{\mu}^{b}(x-\hat{\nu})$
$\displaystyle+A_{\nu}^{a}(x+\hat{\mu}-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})\Big{\\}}\bigg{]}.$
(57)

The part in braces $\\{\ldots\\}$ is symmetric in $a\leftrightarrow b$ and
will therefore disappear in $A^{(1)[ab]}_{\mu}(x)$.111We use the following
shorthand notation for symmetrised and anti-symmetrised expressions:
$A^{\\{ab\\}}=\frac{1}{2}(A^{ab}+A^{ba})$ and
$A^{[ab]}=\frac{1}{2}(A^{ab}-A^{ba})$. We get

$\displaystyle Q^{(0)ab}_{2\mu}(x)$
$\displaystyle=\frac{1}{2}\Big{(}-S^{(0)a}_{1\mu}(x)A^{(0)b}_{\mu}(x)+S^{(0)ab}_{2\mu}(x)\Big{)}$
(58)

which leaves

$\displaystyle A_{\mu}^{(1)[ab]}(x)$
$\displaystyle=2\varrho\Big{(}\frac{1}{2}Q^{(0)[a}_{1\mu}(x)A^{(0)b]}_{\mu}(x)+Q^{(0)[ab]}_{2\mu}(x)\Big{)}=\varrho
S^{(0)[ab]}_{2\mu}(x)\;,$ (59)

which we can write as a double convolution:

$\displaystyle S^{(0)[ab]}_{2\mu}(x)$
$\displaystyle=\sum\limits_{y,z}\sum\limits_{\nu,\rho=1}^{4}g_{\mu\nu\rho}(y,z)\cdot
A^{[a}_{\nu}(x+y)A^{b]}_{\rho}(x+z)$ (60)
$\displaystyle=\sum\limits_{y,z}\sum\limits_{\nu,\rho=1}^{4}\frac{1}{2}\big{(}g_{\mu\nu\rho}(y,z)-g_{\mu\rho\nu}(z,y)\big{)}\cdot
A^{a}_{\nu}(x+y)A^{b}_{\rho}(x+z)$ (61)

with

$\displaystyle g_{\mu\nu\rho}(y,z)$
$\displaystyle=\delta_{\mu\nu}\left[\delta(y,-\hat{\rho})\delta(z,\hat{\mu}-\hat{\rho})-\delta(y,\hat{\rho})\delta(z,\hat{\mu})\right]$
$\displaystyle+\delta_{\mu\rho}\left[\delta(y,0)\delta(z,\hat{\nu})-\delta(y,-\hat{\nu})\delta(z,-\hat{\nu})\right]$
$\displaystyle-\delta_{\nu\rho}\left[\delta(y,0)\delta(z,\hat{\mu})+\delta(y,-\hat{\nu})\delta(z,\hat{\mu}-\hat{\nu})\right]\;.$
(62)

Fourier transforming gives us

$\displaystyle
A_{\mu}^{(1)[ab]}(k_{1},k_{2})=\varrho\sum\limits_{\nu,\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})\;,$
(63)

where

$\displaystyle g_{\mu\nu\rho}(k_{1},k_{2})$
$\displaystyle=2i\Big{(}\delta_{\mu\rho}c(k_{1\mu})s(2k_{2\nu}+k_{1\nu})-\delta_{\mu\nu}c(k_{2\mu})s(2k_{1\rho}+k_{2\rho})$
$\displaystyle+\delta_{\nu\rho}s(k_{1\mu}-k_{2\mu})c(k_{1\nu}+k_{2\nu})\Big{)}\;.$
(64)

Here we use the common shorthand for the trigonometric functions

$\displaystyle s(k_{\mu})$
$\displaystyle=\sin\big{(}\tfrac{1}{2}k_{\mu}\big{)}$ $\displaystyle
c(k_{\mu})$ $\displaystyle=\cos\big{(}\tfrac{1}{2}k_{\mu}\big{)}$ (65)
$\displaystyle\bar{s}(k_{\mu})$ $\displaystyle=\sin(k_{\mu})$
$\displaystyle\bar{c}(k_{\mu})$ $\displaystyle=\cos(k_{\mu})$ (66)
$\displaystyle s^{2}(k)$ $\displaystyle=\sum\limits_{\mu}s^{2}(k_{\mu})$ (67)

Although $g_{\mu\nu\rho}(k_{1},k_{2})$ is imaginary (and therefore so is
$A_{\mu}^{(1)[ab]}(k_{1},k_{2})$), it also fulfils
$g_{\mu\nu\rho}(-k_{1},-k_{2})=-g_{\mu\nu\rho}(k_{1},k_{2})$, which means that
$A_{\mu}^{(1)[ab]}(x)$ is real.

If we were to repeat the previous calculation for a smearing step $n\to n+1$,
we would not only have to replace the $A^{(0)a}_{\mu}(x)$ by
$A^{(n)a}_{\mu}(x)$ but the expansion of $S^{(n)ab}_{2\mu}(x)$ would also
contain terms linear in $A^{(n)[ab]}_{\mu}(x)$. Instead, it is easier to
realize that we can get the same contribution by replacing

$\displaystyle T^{a}A^{(0)a}_{\mu}(x)\to
T^{a}\tilde{A}^{(n)a}_{\mu}(g_{0},x)=T^{a}A^{(n)a}_{\mu}(x)-\frac{g_{0}}{2i}[T^{a},T^{b}]A^{(n)[ab]}_{\mu}(x)+\mathcal{O}(g_{0}^{2})$
(68)

in the _leading order_ calculation. Then we get for a general smearing step

$\displaystyle A_{\mu}^{(n+1)[ab]}(k_{1},k_{2})=$
$\displaystyle\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2})A_{\nu}^{(n)[ab]}(k_{1},k_{2})+\varrho\sum\limits_{\nu,\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{(n)a}_{\nu}(k_{1})A^{(n)b}_{\rho}(k_{2})$
(69)

By iterating once

$\displaystyle A_{\mu}^{(2)[ab]}(k_{1},k_{2})=$
$\displaystyle\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2})\varrho\sum\limits_{\rho\sigma}g_{\nu\rho\sigma}(k_{1},k_{2})A^{(0)a}_{\rho}(k_{1})A^{(0)b}_{\sigma}(k_{2})$
$\displaystyle+\varrho\sum\limits_{\nu,\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{(1)a}_{\nu}(k_{1})A^{(1)b}_{\rho}(k_{2})$
(70)

and twice

$\displaystyle A_{\mu}^{(3)[ab]}(k_{1},k_{2})=$
$\displaystyle\sum\limits_{\nu}\tilde{g}^{2}_{\mu\nu}(\varrho,k_{1}+k_{2})\varrho\sum\limits_{\rho\sigma}g_{\nu\rho\sigma}(k_{1},k_{2})A^{(0)a}_{\rho}(k_{1})A^{(0)b}_{\sigma}(k_{2})$
$\displaystyle+\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2})\varrho\sum\limits_{\rho\sigma}g_{\nu\rho\sigma}(k_{1},k_{2})A^{(1)a}_{\rho}(k_{1})A^{(1)b}_{\sigma}(k_{2})$
$\displaystyle+\varrho\sum\limits_{\nu,\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{(2)a}_{\nu}(k_{1})A^{(2)b}_{\rho}(k_{2})\;,$
(71)

we can infer the general form

$\displaystyle A_{\mu}^{(n)[ab]}(k_{1},k_{2})=$
$\displaystyle\sum\limits_{\nu\rho\sigma}\varrho\,g_{\nu\rho\sigma}(k_{1},k_{2})\sum_{m=0}^{n-1}\tilde{g}^{n-m-1}_{\mu\nu}(\varrho,k_{1}+k_{2})A^{(m)a}_{\rho}(k_{1})A^{(m)b}_{\sigma}(k_{2})\;.$
(72)

Finally, we can also express the first order fields in the sum in terms of the
original unsmeared ones

$\displaystyle A_{\mu}^{(n)[ab]}(k_{1},k_{2})$
$\displaystyle=\sum\limits_{\alpha\beta\gamma\nu\rho}\varrho\,g_{\alpha\beta\gamma}\Bigg{[}\sum_{m=0}^{n-1}\tilde{g}^{n-m-1}_{\mu\alpha}(\varrho,k_{1}+k_{2})\tilde{g}^{m}_{\beta\nu}(\varrho,k_{1})\tilde{g}^{m}_{\gamma\rho}(\varrho,k_{2})\Bigg{]}A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})$
$\displaystyle=:\sum\limits_{\nu\rho}\tilde{g}^{(n)}_{\mu\nu\rho}(\varrho,k_{1},k_{2})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})\;.$
(73)

###  2.5 Next-to-next-to-leading order

As before, we start again by considering the first smearing step and
generalise later to an arbitrary number of smearing steps. At third order we
have defined two perturbative fields $\bar{A}$ and $\hat{A}$ in Eq. (19),
where the former can be read off of Eq. (17) to be

$\displaystyle\bar{A}^{(1)abc}_{\mu}(x)$ $\displaystyle=-3\varrho
A^{a}_{\mu}(x)Q^{[bc]}_{2\mu}(x)+\frac{1}{2}\varrho^{2}Q^{a}_{1\mu}(x)Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)+\frac{1}{2}\varrho
A^{a}_{\mu}(x)A^{b}_{\mu}(x)Q^{c}_{1\mu}(x)$ (74)
$\displaystyle=-\frac{3}{2}A^{a}_{\mu}(x)A^{(1)[bc]}_{\mu}(x)+\frac{3}{2}\varrho
A^{a}_{\mu}(x)Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)+\frac{1}{2}\varrho^{2}Q^{a}_{1\mu}(x)Q^{b}_{1\mu}(x)A^{c}_{\mu}(x)$
$\displaystyle+\frac{1}{2}\varrho A^{a}_{\mu}(x)A^{b}_{\mu}(x)Q^{c}_{1\mu}(x)$
(75)
$\displaystyle=-\frac{3}{2}A^{a}_{\mu}(x)A^{(1)[bc]}_{\mu}(x)+\frac{1}{2}A^{a}_{\mu}(x)A^{(1)b}_{\mu}(x)A^{c}_{\mu}(x)+\frac{1}{2}A^{(1)a}_{\mu}(x)A^{(1)b}_{\mu}(x)A^{c}_{\mu}(x)\,.$
(76)

We have dropped all terms that were symmetric in $b\leftrightarrow c$ in the
last step. Thus $\bar{A}$ is completely given by already known quantities,
which is not surprising as it stems only from the commutator terms of the BCH-
formula.

From the third order in the expansion of $Q_{\mu}(x)$ we get

$\displaystyle\hat{A}^{(1)abc}_{\mu}(x)=-6\varrho Q^{abc}_{3\mu}(x)$ (77)

with

$\displaystyle Q^{abc}_{3\mu}(x)$
$\displaystyle=A^{a}_{\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)-\frac{1}{2}S^{a}_{1\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)+S^{ab}_{2\mu}(x)A^{c}_{\mu}(x)-\frac{1}{2}S^{abc}_{3\mu}(x)$
(78)
$\displaystyle=-2A^{a}_{\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)-\frac{1}{2}Q^{a}_{1\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)+S^{ab}_{2\mu}(x)A^{c}_{\mu}(x)-S^{abc}_{3\mu}(x)\;.$
(79)

We revisit the sum of staples at second order and define

$\displaystyle S^{ab}_{2\mu}(x)$
$\displaystyle=s_{1\mu}^{ab}(x)+\frac{1}{2}s_{2\mu}^{ab}(x)$ (80)

with

$\displaystyle s_{1\mu\nu}^{ab}(x)$
$\displaystyle=\sum\limits_{\nu}\Big{[}A_{\nu}^{a}(x)A_{\mu}^{b}(x+\hat{\nu})-A_{\nu}^{a}(x)A_{\nu}^{b}(x+\hat{\mu})-A_{\mu}^{a}(x+\hat{\nu})A_{\nu}^{b}(x+\hat{\mu})$
$\displaystyle-
A_{\nu}^{a}(x-\hat{\nu})A_{\mu}^{b}(x-\hat{\nu})-A_{\nu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})+A_{\mu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})\Big{]}$
(81) $\displaystyle s_{2\mu}^{ab}(x)$
$\displaystyle=\sum\limits_{\nu}\Big{[}A_{\nu}^{a}(x)A_{\nu}^{b}(x)+A_{\mu}^{a}(x+\hat{\nu})A_{\mu}^{b}(x+\hat{\nu})+A_{\nu}^{a}(x+\hat{\mu})A_{\nu}^{b}(x+\hat{\mu})$
$\displaystyle+A_{\nu}^{a}(x-\hat{\nu})A_{\nu}^{b}(x-\hat{\nu})+A_{\mu}^{a}(x-\hat{\nu})A_{\mu}^{b}(x-\hat{\nu})+A_{\nu}^{a}(x+\hat{\mu}-\hat{\nu})A_{\nu}^{b}(x+\hat{\mu}-\hat{\nu})$
$\displaystyle-2\delta_{\mu\nu}A^{a}_{\mu}(x)A^{b}_{\mu}(x)\Big{]}\;.$ (82)

Their Fourier transforms are

$\displaystyle s_{1\mu}^{ab}(x)$
$\displaystyle=\sum\limits_{yz}\sum\limits_{\nu\rho}a_{1\mu\nu\rho}(y,z)A^{a}_{\nu}(x+y)A^{b}_{\rho}(x+z)$
$\displaystyle=\sum\limits_{\nu\rho}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}}{(2\pi)^{8}}e^{-i(x+\frac{1}{2}\hat{\mu})(k_{1}+k_{2})}\cdot
a_{1\mu\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})$ (83)
$\displaystyle s_{2\mu}^{ab}(x)$
$\displaystyle=\sum\limits_{y}\sum\limits_{\nu}a_{2\mu\nu}(y)A^{a}_{\nu}(x+y)A^{b}_{\nu}(x+y)$
$\displaystyle=\sum\limits_{\nu}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}}{(2\pi)^{8}}e^{-i(x+\frac{1}{2}\hat{\mu})(k_{1}+k_{2})}\cdot
a_{2\mu\nu}(k_{1},k_{2})A^{a}_{\nu}(k_{1})A^{b}_{\nu}(k_{2})$ (84)

with

$\displaystyle a_{1\mu\nu\rho}(y,z)$
$\displaystyle=\delta_{\mu\nu}\left[\delta(y,-\hat{\rho})\delta(z,\hat{\mu}-\hat{\rho})-\delta(y,\hat{\rho})\delta(z,\hat{\mu})\right]$
$\displaystyle+\delta_{\mu\rho}\left[\delta(y,0)\delta(z,\hat{\nu})-\delta(y,-\hat{\nu})\delta(z,-\hat{\nu})\right]$
$\displaystyle-\delta_{\nu\rho}\left[\delta(y,0)\delta(z,\hat{\mu})+\delta(y,-\hat{\nu})\delta(z,\hat{\mu}-\hat{\nu})\right]$
(85) $\displaystyle a_{2\mu\nu}(y)$
$\displaystyle=\delta_{\mu\nu}\sum\limits_{\alpha=1}^{4}\Big{[}\delta(y,\hat{\alpha})+\delta(y,-\hat{\alpha})-2\delta(y,0)\Big{]}+6\delta_{\mu\nu}\delta(y,0)$
$\displaystyle+\delta(y,0)+\delta(y,\hat{\mu})+\delta(y,-\hat{\nu})+\delta(y,\hat{\mu}-\hat{\nu})$
(86)

and

$\displaystyle a_{1\mu\nu\rho}(k_{1},k_{2})$
$\displaystyle=-\delta_{\mu\nu}\left[2i\sin\left(k_{1\rho}+\tfrac{1}{2}k_{2\rho}\right)e^{\frac{i}{2}k_{2\mu}}\right]+\delta_{\mu\rho}\left[2i\sin\left(k_{2\nu}+\tfrac{1}{2}k_{1\nu}\right)e^{-\frac{i}{2}k_{1\mu}}\right]$
$\displaystyle-\delta_{\nu\rho}\left[2\cos\left(\tfrac{1}{2}k_{1\nu}+\tfrac{1}{2}k_{2\nu}\right)e^{-\frac{i}{2}(k_{1\mu}-k_{2\mu})}\right]$
(87)
$\displaystyle=-h_{\mu\nu\rho}(k_{1},k_{2})+g_{\mu\nu\rho}(k_{1},k_{2})\;.$
(88)

Here $h_{\mu\nu\rho}$ and $g_{\mu\nu\rho}$ are essentially the real and
imaginary parts of $a_{1\mu\nu\rho}$, i.e.

$\displaystyle g_{\mu\nu\rho}(k_{1},k_{2})$
$\displaystyle=2i\Big{(}-\delta_{\mu\nu}c(k_{2\mu})s(2k_{1\rho}+k_{2\rho})+\delta_{\mu\rho}c(k_{1\mu})s(2k_{2\nu}+k_{1\nu})$
$\displaystyle+\delta_{\nu\rho}s(k_{1\mu}-k_{2\mu})c(k_{1\nu}+k_{2\nu})\Big{)}$
(89) $\displaystyle h_{\mu\nu\rho}(k_{1},k_{2})$
$\displaystyle=2\Big{(}-\delta_{\mu\nu}s(k_{2\mu})s(2k_{1\rho}+k_{2\rho})-\delta_{\mu\rho}s(k_{1\mu})s(2k_{2\nu}+k_{1\nu})$
$\displaystyle+\delta_{\nu\rho}c(k_{1\mu}-k_{2\mu})c(k_{1\nu}+k_{2\nu})\Big{)}\;.$
(90)

We know the anti-symmetric $g_{\mu\nu\rho}$ already from the second order.

Furthermore,

$\displaystyle a_{2\mu\nu}(k_{1},k_{2})$
$\displaystyle=6\delta_{\mu\nu}+\delta_{\mu\nu}\sum\limits_{\alpha=1}^{4}\bigg{[}2\cos\big{(}k_{1\alpha}+k_{2\alpha}\big{)}-2\bigg{]}+4\cos\big{(}\tfrac{1}{2}(k_{1\mu}+k_{2\mu})\big{)}\cos\big{(}\tfrac{1}{2}(k_{1\nu}+k_{2\nu})\big{)}$
$\displaystyle=6\delta_{\mu\nu}-\delta_{\mu\nu}4s^{2}(k_{1}+k_{2})+4c(k_{1\mu}+k_{2\mu})c(k_{1\nu}+k_{2\nu})$
(91) $\displaystyle=:6\delta_{\mu\nu}-g^{(i)}_{\mu\nu}(k_{1}+k_{2})$ (92)

where we have defined $g^{(i)}_{\mu\nu}(k)$ which is a variation of
$g_{\mu\nu}(k)$ with cosines in the second term instead of sines.

For the sum of staples at third order we define

$\displaystyle S^{abc}_{3\mu}(x)=$ $\displaystyle
s_{1\mu}^{abc}(x)+\frac{1}{2}s_{2\mu}^{abc}(x)+\frac{1}{6}s_{3\mu}^{abc}(x)$
(93)

with

$\displaystyle s_{1\mu}^{abc}$
$\displaystyle(x)=\sum\limits_{\nu}\Big{[}-A_{\nu}^{a}(x)A_{\mu}^{b}(x+\hat{\nu})A_{\nu}^{c}(x+\hat{\mu})-A_{\nu}^{a}(x-\hat{\nu})A_{\mu}^{b}(x-\hat{\nu})A_{\nu}^{c}(x+\hat{\mu}-\hat{\nu})\Big{]}$
(94) $\displaystyle s_{2\mu}^{abc}$
$\displaystyle(x)=\sum\limits_{\nu}\Big{[}\big{\\{}A^{a}_{\nu}(x)A^{b}_{\mu}(x+\hat{\nu})A^{c}_{\mu}(x+\hat{\nu})+A^{a}_{\nu}(x)A^{b}_{\nu}(x+\hat{\mu})A^{c}_{\nu}(x+\hat{\mu})$
$\displaystyle+A^{a}_{\mu}(x+\hat{\nu})A^{b}_{\nu}(x+\hat{\mu})A^{c}_{\nu}(x+\hat{\mu})-A^{a}_{\nu}(x-\hat{\nu})A^{b}_{\mu}(x-\hat{\nu})A^{c}_{\mu}(x-\hat{\nu})$
$\displaystyle-A^{a}_{\nu}(x-\hat{\nu})A^{b}_{\nu}(x+\hat{\mu}-\hat{\nu})A^{c}_{\nu}(x+\hat{\mu}-\hat{\nu})+A^{a}_{\mu}(x-\hat{\nu})A^{b}_{\nu}(x+\hat{\mu}-\hat{\nu})A^{c}_{\nu}(x+\hat{\mu}-\hat{\nu})\big{\\}}$
$\displaystyle+\big{\\{}-A^{a}_{\mu}(x+\hat{\nu})A^{b}_{\mu}(x+\hat{\nu})A^{c}_{\nu}(x+\hat{\mu})+A^{a}_{\nu}(x)A^{b}_{\nu}(x)A^{c}_{\mu}(x+\hat{\nu})-A^{a}_{\nu}(x)A^{b}_{\nu}(x)A^{c}_{\nu}(x+\hat{\mu})$
$\displaystyle+A^{a}_{\mu}(x-\hat{\nu})A^{b}_{\mu}(x-\hat{\nu})A^{c}_{\nu}(x+\hat{\mu}-\hat{\nu})+A^{a}_{\nu}(x-\hat{\nu})A^{b}_{\nu}(x-\hat{\nu})A^{c}_{\mu}(x-\hat{\nu})$
$\displaystyle+A^{a}_{\nu}(x-\hat{\nu})A^{b}_{\nu}(x-\hat{\nu})A^{c}_{\nu}(x+\hat{\mu}-\hat{\nu})\big{\\}}\Big{]}$
(95) $\displaystyle s_{3\mu}^{abc}$
$\displaystyle(x)=\sum\limits_{\nu}\Big{[}-2\delta_{\mu\nu}A^{a}_{\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)+A^{a}_{\nu}(x)A^{b}_{\nu}(x)A^{c}_{\nu}(x)+A^{a}_{\mu}(x+\hat{\nu})A^{b}_{\mu}(x+\hat{\nu})A^{c}_{\mu}(x+\hat{\nu})$
$\displaystyle-A^{a}_{\nu}(x+\hat{\mu})A^{b}_{\nu}(x+\hat{\mu})A^{c}_{\nu}(x+\hat{\mu})-A^{a}_{\nu}(x-\hat{\nu})A^{b}_{\nu}(x-\hat{\nu})A^{c}_{\nu}(x-\hat{\nu})$
$\displaystyle+A^{a}_{\mu}(x-\hat{\nu})A^{b}_{\mu}(x-\hat{\nu})A^{c}_{\mu}(x-\hat{\nu})+A^{a}_{\nu}(x+\hat{\mu}-\hat{\nu})A^{b}_{\nu}(x+\hat{\mu}-\hat{\nu})A^{c}_{\nu}(x+\hat{\mu}-\hat{\nu})\Big{]}\;.$
(96)

They can be written as

$\displaystyle s_{1\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{y,z,r}\sum\limits_{\nu\rho\sigma}b_{1\mu\nu\rho\sigma}(y,z,r)A^{a}_{\nu}(x+y)A^{b}_{\rho}(x+z)A^{c}_{\sigma}(x+r)$
(97) $\displaystyle s_{2\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{y,z}\sum\limits_{\nu\rho}b_{21\mu\nu\rho}(y,z)A^{a}_{\nu}(x+y)A^{b}_{\rho}(x+z)A^{c}_{\rho}(x+z)$
$\displaystyle+\sum\limits_{y,z}\sum\limits_{\nu\rho}b_{22\mu\nu\rho}(y,z)A^{a}_{\nu}(x+y)A^{b}_{\nu}(x+y)A^{c}_{\rho}(x+z)$
(98) $\displaystyle s_{3\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{y}\sum\limits_{\nu}b_{3\mu\nu}(y)A^{a}_{\nu}(x+y)A^{b}_{\nu}(x+y)A^{c}_{\nu}(x+y)$
(99)

with

$\displaystyle b_{1\mu\nu\rho\sigma}(y,z,r)$
$\displaystyle=-\delta_{\mu\rho}\delta_{\nu\sigma}\Big{(}\delta(y,0)\delta(z,\hat{\nu})\delta(r,\hat{\mu})+\delta(y,-\hat{\nu})\delta(z,-\hat{\nu})\delta(r,\hat{\mu}-\hat{\nu})\Big{)}$
(100) $\displaystyle b_{21\mu\nu\rho}(y,z)$
$\displaystyle=\delta_{\mu\nu}\Big{(}\delta(y,\hat{\rho})\delta(z,\hat{\mu})+\delta(y,-\hat{\rho})\delta(z,\hat{\mu}-\hat{\rho})\Big{)}$
$\displaystyle+\delta_{\mu\rho}\Big{(}\delta(y,0)\delta(z,\hat{\nu})-\delta(y,-\hat{\nu})\delta(z,-\hat{\nu})\Big{)}$
$\displaystyle+\delta_{\nu\rho}\Big{(}\delta(y,0)\delta(z,\hat{\mu})-\delta(y,-\hat{\nu})\delta(z,\hat{\mu}-\hat{\nu})\Big{)}$
(101) $\displaystyle b_{22\mu\nu\rho}(y,z)$
$\displaystyle=\delta_{\mu\nu}\Big{(}-\delta(y,\hat{\rho})\delta(z,\hat{\mu})+\delta(y,-\hat{\rho})\delta(z,\hat{\mu}-\hat{\rho})\Big{)}$
$\displaystyle+\delta_{\mu\rho}\Big{(}\delta(y,0)\delta(z,\hat{\nu})+\delta(y,-\hat{\nu})\delta(z,-\hat{\nu})\Big{)}$
$\displaystyle+\delta_{\nu\rho}\Big{(}-\delta(y,0)\delta(z,\hat{\mu})+\delta(y,-\hat{\nu})\delta(z,\hat{\mu}-\hat{\nu})\Big{)}$
(102) $\displaystyle b_{3\mu\nu}(y)$
$\displaystyle=6\delta_{\mu\nu}\delta(y,0)+g_{\mu\nu}(y)\;.$ (103)

Fourier transforming the $s_{i\mu}^{abc}(x)$ leads to

$\displaystyle s_{1\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{\nu\rho\sigma}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}\,\mathrm{d}^{4}k_{3}}{(2\pi)^{12}}e^{-i\left(x+\frac{1}{2}\hat{\mu}\right)(k_{1}+k_{2}+k_{3})}b_{1\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{c}_{\sigma}(k_{3})$
(104) $\displaystyle s_{2\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{\nu\rho}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}\,\mathrm{d}^{4}k_{3}}{(2\pi)^{12}}e^{-i\left(x+\frac{1}{2}\hat{\mu}\right)(k_{1}+k_{2}+k_{3})}b_{21\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{c}_{\rho}(k_{3})$
$\displaystyle+\sum\limits_{\nu\rho}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}\,\mathrm{d}^{4}k_{3}}{(2\pi)^{12}}e^{-i\left(x+\frac{1}{2}\hat{\mu}\right)(k_{1}+k_{2}+k_{3})}b_{22\mu\nu\rho}(k_{1}+k_{2},k_{3})A^{a}_{\nu}(k_{1})A^{b}_{\nu}(k_{2})A^{c}_{\rho}(k_{3})$
(105) $\displaystyle s_{3\mu}^{abc}(x)$
$\displaystyle=\sum\limits_{\nu}\int\frac{\mathrm{d}^{4}k_{1}\,\mathrm{d}^{4}k_{2}\,\mathrm{d}^{4}k_{3}}{(2\pi)^{12}}e^{-i\left(x+\frac{1}{2}\hat{\mu}\right)(k_{1}+k_{2}+k_{3})}g_{\mu\nu}(k_{1}+k_{2}+k_{3})A^{a}_{\nu}(k_{1})A^{b}_{\nu}(k_{2})A^{c}_{\nu}(k_{3})$
(106)

with

$\displaystyle b_{1\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})$
$\displaystyle=-2\delta_{\mu\rho}\delta_{\nu\sigma}\cos\big{(}\tfrac{1}{2}k_{1\nu}+k_{2\nu}+\tfrac{1}{2}k_{3\nu}\big{)}e^{\frac{i}{2}(k_{1\mu}-k_{3\mu})}$
(107) $\displaystyle b_{21\mu\nu\rho}(k_{1},k_{2}+k_{3})$
$\displaystyle=\delta_{\mu\nu}2\cos\big{(}k_{1\rho}+\tfrac{1}{2}(k_{2\rho}+k_{3\rho})\big{)}e^{-\frac{i}{2}(k_{2\mu}+k_{3\mu})}$
$\displaystyle+\delta_{\mu\rho}(-2i)\sin\big{(}\tfrac{1}{2}k_{1\nu}+k_{2\nu}+k_{3\nu}\big{)}e^{\frac{i}{2}k_{1\mu}}$
$\displaystyle+\delta_{\nu\rho}(-2i)\sin\big{(}\tfrac{1}{2}(k_{1\nu}+k_{2\nu}+k_{3\nu})\big{)}e^{\frac{i}{2}(k_{1\mu}-k_{2\mu}-k_{3\mu})}$
(108) $\displaystyle b_{22\mu\nu\rho}(k_{1}+k_{2},k_{3})$
$\displaystyle=\delta_{\mu\nu}2i\sin\big{(}k_{1\rho}+k_{2\rho}+\tfrac{1}{2}k_{3\rho}\big{)}e^{-\frac{i}{2}k_{3\mu}}$
$\displaystyle+\delta_{\mu\rho}2\cos\big{(}\tfrac{1}{2}(k_{1\nu}+k_{2\nu})+k_{3\nu}\big{)}e^{\frac{i}{2}(k_{1\mu}+k_{2\mu})}$
$\displaystyle+\delta_{\nu\rho}2i\sin\big{(}\tfrac{1}{2}(k_{1\nu}+k_{2\nu}+k_{3\nu})\big{)}e^{\frac{i}{2}(k_{1\mu}+k_{2\mu}-k_{3\mu})}$
(109) $\displaystyle b_{3\mu\nu}(k_{1}+k_{2}+k_{3})$
$\displaystyle=6\delta_{\mu\nu}+g_{\mu\nu}(k_{1}+k_{2}+k_{3})\;.$ (110)

We can symmetrise these expressions according to

$\displaystyle\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}v_{i\mu}^{abc}=\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\frac{1}{2}\big{(}v_{i\mu}^{abc}+v_{i\mu}^{cba}\big{)}$
(111)

and observe that under the integrals

$\displaystyle
b_{1\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})A^{c}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{a}_{\sigma}(k_{3})$
$\displaystyle=b_{1\mu\sigma\rho\nu}(k_{3},k_{2},k_{1})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{c}_{\sigma}(k_{3})$
(112) $\displaystyle
b_{21\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{c}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{a}_{\rho}(k_{3})$
$\displaystyle=b_{21\mu\rho\nu}(k_{3},k_{1}+k_{2})A^{a}_{\nu}(k_{1})A^{b}_{\nu}(k_{2})A^{c}_{\rho}(k_{3})$
(113) $\displaystyle
b_{22\mu\nu\rho}(k_{1}+k_{2},k_{3})A^{c}_{\nu}(k_{1})A^{b}_{\nu}(k_{2})A^{a}_{\rho}(k_{3})$
$\displaystyle=b_{22\mu\rho\nu}(k_{2}+k_{3},k_{1})A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{c}_{\rho}(k_{3})\;.$
(114)

Thus we need

$\displaystyle
b_{1\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})+b_{1\mu\sigma\rho\nu}(k_{3},k_{2},k_{1})$
$\displaystyle=-4\delta_{\mu\rho}\delta_{\nu\sigma}c(k_{1\nu}+2k_{2\nu}+k_{3\nu})c(k_{1\mu}-k_{3\mu})$
(115) $\displaystyle=:g^{(iii)}_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})$ (116)
$\displaystyle
b_{21\mu\nu\rho}(k_{1},k_{2}+k_{3})+b_{22\mu\rho\nu}(k_{2}+k_{3},k_{1})$
$\displaystyle=4\delta_{\mu\nu}c(2k_{1\rho}+k_{2\rho}+k_{3\rho})c(k_{2\mu}+k_{3\mu})$
$\displaystyle+4\delta_{\mu\rho}s(k_{1\mu})s(k_{1\nu}+2(k_{2\nu}+k_{3\nu}))$
$\displaystyle+4\delta_{\nu\rho}s(k_{1\mu}-k_{2\mu}-k_{3\mu})s(k_{1\nu}+k_{2\nu}+k_{3\nu})$
(117) $\displaystyle=:g^{(ii)}_{\mu\nu\rho}(k_{1},k_{2}+k_{3})$ (118)
$\displaystyle
b_{22\mu\nu\rho}(k_{1}+k_{2},k_{3})+b_{21\mu\rho\nu}(k_{3},k_{1}+k_{2})$
$\displaystyle=g^{(ii)}_{\mu\rho\nu}(k_{3},k_{1}+k_{2})\;.$ (119)

Now we can write

$\displaystyle\hat{A}^{(1)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\rho\sum\limits_{\nu\rho\sigma}\bigg{(}3\big{(}h_{\mu\nu\rho}(k_{1},k_{2})-g_{\mu\nu\rho}(k_{1},k_{2})\big{)}\delta_{\mu\sigma}+\frac{3}{2}g_{\mu\nu}(k_{1})\delta_{\mu\rho}\delta_{\mu\sigma}$
$\displaystyle+\frac{3}{2}\Big{(}g^{(i)}_{\mu\nu}(k_{1},k_{2})\delta_{\nu\rho}\delta_{\mu\sigma}+g^{(ii)}_{\mu\nu\rho}(k_{1},k_{2}+k_{3})\delta_{\rho\sigma}+g^{(iii)}_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})\Big{)}$
$\displaystyle+\frac{1}{2}g_{\mu\nu}(k_{1}+k_{2}+k_{3})\delta_{\nu\rho}\delta_{\nu\sigma}\bigg{)}A^{a}_{\nu}(k_{1})A^{b}_{\rho}(k_{2})A^{c}_{\sigma}(k_{3})\;.$
(120)

For a general smearing step $n>1$ we have to take into account that in the
previous step $A^{(n-1)[ab]}_{\mu}\neq 0$, $\bar{A}^{(n-1)abc}_{\mu}\neq 0$,
and $\hat{A}^{(n-1)abc}_{\mu}\neq 0$. Again we replace the unsmeared gluon
field $A^{(0)}$ in the previous orders by the smeared one $\tilde{A}^{(n)}$
and collect all terms of order $g_{0}^{3}$:

$\displaystyle
U^{(n+1)}_{\mu}(k)=1+ig_{0}\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k)\bigg{(}T^{a}A^{(n)a}_{\nu}(k)-\frac{g_{0}}{2i}[T^{a},T^{b}]A^{(n)[ab]}_{\nu}(k_{1},k_{2})$
$\displaystyle-\frac{g_{0}^{2}}{6}\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}[T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}]\Big{)}A^{(n)abc}_{\nu}(k_{3},k_{4},k_{5})\bigg{)}$
$\displaystyle-\frac{g_{0}^{2}}{2}\sum\limits_{\nu\rho}\Big{(}\tilde{g}_{\mu\nu}(\varrho,k_{1})\tilde{g}_{\mu\rho}(\varrho,k_{2})+2g_{\mu\nu\rho}(k_{1},k_{2})\Big{)}\bigg{(}T^{a}A^{(n)a}_{\nu}(k_{1})-\frac{g_{0}}{2i}[T^{a},T^{b}]A^{(n)[ab]}_{\nu}(k_{11},k_{12})\bigg{)}$
$\displaystyle\times\bigg{(}T^{a}A^{(n)a}_{\rho}(k_{2})-\frac{g_{0}}{2i}[T^{a},T^{b}]A^{(n)[ab]}_{\rho}(k_{21},k_{22})\bigg{)}$
$\displaystyle-i\frac{g_{0}^{3}}{6}\sum\limits_{\nu\rho\sigma}\bigg{(}T^{a}T^{b}T^{c}\tilde{g}_{\mu\nu}(\varrho,k_{1})\tilde{g}_{\mu\rho}(\varrho,k_{2})\tilde{g}_{\mu\sigma}(\varrho,k_{3})+\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}\tilde{g}_{\mu\nu}(\varrho,k_{1})\rho
g_{\mu\rho\sigma}(k_{2},k_{3})$
$\displaystyle+[T^{a},[T^{b},T^{c}]]\Big{(}-\frac{3}{2}\delta_{\mu\nu}\varrho
g_{\mu\rho\sigma}(k_{2},k_{3})+\frac{1}{2}\delta_{\mu\nu}\tilde{g}_{\mu\rho}(\varrho,k_{2})\delta_{\mu\sigma}+\frac{1}{2}\tilde{g}_{\mu\nu}(\varrho,k_{1})\tilde{g}_{\mu\rho}(\varrho,k_{2})\delta_{\mu\sigma}\Big{)}$
$\displaystyle+\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}[T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}]\Big{)}\hat{G}_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})\bigg{)}A^{(n)a}_{\nu}(k_{1})A^{(n)b}_{\rho}(k_{2})A^{(n)c}_{\sigma}(k_{3})$
(121)

with $k=k_{1}+k_{2}=k_{3}+k_{4}+k_{5}$, $k_{1}=k_{11}+k_{12}$, and
$k_{2}=k_{21}+k_{22}$. From the third term involving the product
$\tilde{A}_{\nu}\tilde{A}_{\rho}$ the following terms appear at order
$g_{0}^{3}$:

$\displaystyle-i\frac{g_{0}^{3}}{4}\sum\limits_{\nu\rho}\Big{(}\big{(}\tilde{g}_{\mu\nu}(\varrho,k_{1})\tilde{g}_{\mu\rho}(\varrho,k_{2}+k_{3})+2g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})\big{)}T^{a}[T^{b},T^{c}]A^{(n)a}_{\nu}(k_{1})A^{(n)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+\big{(}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2})\tilde{g}_{\mu\rho}(\varrho,k_{3})+2g_{\mu\nu\rho}(k_{1}+k_{2},k_{3})\big{)}[T^{a},T^{b}]T^{c}A^{(n)[ab]}_{\nu}(k_{1},k_{2})A^{(n)c}_{\rho}(k_{3})\Big{)}$
$\displaystyle=-i\frac{g_{0}^{3}}{6}\bigg{(}\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}\sum\limits_{\nu\rho}\tilde{g}_{\mu\nu}(\varrho,k_{1})\tilde{g}_{\mu\rho}(\varrho,k_{2}+k_{3})A^{(n)a}_{\nu}(k_{1})A^{(n)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+3[T^{a},[T^{b},T^{c}]]\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{(n)a}_{\nu}(k_{1})A^{(n)[bc]}_{\rho}(k_{2},k_{3})\bigg{)}\;.$
(122)

Then we have for our two third-order fields $\bar{A}$ and $\hat{A}$:

$\displaystyle\bar{A}^{(n+1)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2}+k_{3})\bar{A}^{(n)abc}_{\nu}(k_{1},k_{2},k_{3})$
$\displaystyle+3\varrho\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{(n)a}_{\nu}(k_{1})A^{(n)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle-\frac{3}{2}\varrho\sum\limits_{\nu\rho\sigma}\delta_{\mu\nu}g_{\mu\rho\sigma}(k_{2},k_{3})A^{(n)a}_{\nu}(k_{1})A^{(n)b}_{\rho}(k_{2})A^{(n)c}_{\sigma}(k_{3})$
$\displaystyle+\frac{1}{2}A^{(n)a}_{\mu}(k_{1})A^{(n+1)b}_{\mu}(k_{2})A^{(n)c}_{\mu}(k_{3})+\frac{1}{2}A^{(n+1)a}_{\mu}(k_{1})A^{(n+1)b}_{\mu}(k_{2})A^{(n)c}_{\mu}(k_{3})$
(123) $\displaystyle\hat{A}^{(n+1)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2}+k_{3})\hat{A}^{(n)abc}_{\nu}(k_{1},k_{2},k_{3})$
$\displaystyle\varrho\sum\limits_{\nu\rho\sigma}\bigg{(}3\big{(}h_{\mu\nu\rho}(k_{1},k_{2})+g_{\mu\nu\rho}(k_{1},k_{2})\big{)}\delta_{\mu\sigma}+\frac{3}{2}g_{\mu\nu}(k_{1})\delta_{\mu\rho}\delta_{\mu\sigma}$
$\displaystyle+\frac{3}{2}\Big{(}g^{(i)}_{\mu\nu}(k_{1},k_{2})\delta_{\nu\rho}\delta_{\mu\sigma}+g^{(ii)}_{\mu\nu\rho}(k_{1},k_{2}+k_{3})\delta_{\rho\sigma}+g^{(iii)}_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})\Big{)}$
$\displaystyle+\frac{1}{2}g_{\mu\nu}(k_{1}+k_{2}+k_{3})\delta_{\nu\rho}\delta_{\nu\sigma}\bigg{)}A^{(n)a}_{\nu}(k_{1})A^{(n)b}_{\rho}(k_{2})A^{(n)c}_{\sigma}(k_{3})\;,$
(124)

and finally

$\displaystyle A^{(n+1)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\hat{A}^{(n+1)abc}_{\mu}(k_{1},k_{2},k_{3})+2\bar{A}^{(n+1)a[bc]}_{\mu}(k_{1},k_{2},k_{3})$
(125)
$\displaystyle=\sum\limits_{\nu}\tilde{g}_{\mu\nu}(\varrho,k_{1}+k_{2}+k_{3})A^{(n)abc}_{\nu}(k_{1},k_{2},k_{3})$
$\displaystyle+6\varrho\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{(n)a}_{\nu}(k_{1})A^{(n)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+\sum\limits_{\nu\rho\sigma}\Bigg{(}\frac{1}{2}\varrho^{2}\Big{(}g_{\mu\nu}(k_{1})g_{\mu\rho}(k_{2})\delta_{\mu\sigma}-g_{\mu\nu}(k_{1})\delta_{\mu\rho}g_{\mu\sigma}(k_{3})\Big{)}+\varrho
G_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})\Bigg{)}$ $\displaystyle\times
A^{(n)a}_{\nu}(k_{1})A^{(n)b}_{\rho}(k_{2})A^{(n)c}_{\sigma}(k_{3})$ (126)

with

$\displaystyle G_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})$
$\displaystyle=\frac{1}{2}g_{\mu\nu}(k_{1})\delta_{\mu\rho}\delta_{\mu\sigma}+\delta_{\mu\nu}g_{\mu\rho}(k_{2})\delta_{\mu\sigma}+3h_{\mu\nu\rho}(k_{1},k_{2})\delta_{\mu\sigma}$
$\displaystyle+\frac{3}{2}\Big{(}g^{(i)}_{\mu\nu}(k_{1},k_{2})\delta_{\nu\rho}\delta_{\mu\sigma}+g^{(ii)}_{\mu\nu\rho}(k_{1},k_{2}+k_{3})\delta_{\rho\sigma}+g^{(iii)}_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})\Big{)}$
$\displaystyle+\frac{1}{2}g_{\mu\nu}(k_{1}+k_{2}+k_{3})\delta_{\nu\rho}\delta_{\nu\sigma}\;.$
(127)

After two smearing steps:

$\displaystyle A^{(2)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\sum\limits_{\alpha}\tilde{g}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\sum\limits_{\nu\rho\sigma}G_{\alpha\nu\rho\sigma}(\varrho,k_{1},k_{2},k_{3})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})A^{(0)c}_{\sigma}(k_{3})$
$\displaystyle+6\varrho\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{(1)a}_{\nu}(k_{1})A^{(1)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+\sum\limits_{\nu\rho\sigma}\tilde{G}_{\mu\nu\rho}(\varrho,k_{1},k_{2},k_{3})A^{(1)a}_{\nu}(k_{1})A^{(1)b}_{\rho}(k_{2})A^{(1)c}_{\sigma}(k_{3})\;.$
(128)

After three smearing steps:

$\displaystyle A^{(3)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=\sum\limits_{\alpha}\tilde{g}^{2}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\sum\limits_{\nu\rho\sigma}\tilde{G}_{\alpha\nu\rho\sigma}(\varrho,k_{1},k_{2},k_{3})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})A^{(0)c}_{\sigma}(k_{3})$
$\displaystyle+\sum\limits_{\alpha}\tilde{g}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\sum\limits_{\nu\rho\sigma}\tilde{G}_{\alpha\nu\rho\sigma}(\varrho,k_{1},k_{2},k_{3})A^{(1)a}_{\nu}(k_{1})A^{(1)b}_{\rho}(k_{2})A^{(1)c}_{\sigma}(k_{3})$
$\displaystyle+\tilde{G}_{\mu\nu\rho}(\varrho,k_{1},k_{2},k_{3})A^{(2)a}_{\nu}(k_{1})A^{(2)b}_{\rho}(k_{2})A^{(2)c}_{\sigma}(k_{3})$
$\displaystyle+6\varrho\sum\limits_{\alpha}\tilde{g}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\sum\limits_{\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2}+k_{3})A^{(1)a}_{\nu}(k_{1})A^{(1)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+6\varrho\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{(2)a}_{\nu}(k_{1})A^{(2)[bc]}_{\rho}(k_{2},k_{3})\;.$
(129)

After $n$ smearing steps:

$\displaystyle A^{(n)abc}_{\mu}(k_{1},k_{2},k_{3})$
$\displaystyle=6\varrho\sum\limits_{\alpha\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2}+k_{3})\sum\limits_{m=1}^{n-1}\tilde{g}^{n-m-1}_{\mu\alpha}(\rho,k_{1}+k_{2}+k_{3})A^{(m)a}_{\nu}(k_{1})A^{(m)[bc]}_{\rho}(k_{2},k_{3})$
$\displaystyle+\sum\limits_{\alpha\nu\rho\sigma}\Bigg{(}\frac{1}{2}\varrho^{2}\Big{(}g_{\alpha\nu}(k_{1})g_{\mu\rho}(k_{2})\delta_{\mu\sigma}-g_{\mu\nu}(k_{1})\delta_{\mu\rho}g_{\mu\sigma}(k_{3})\Big{)}+\varrho
G_{\alpha\nu\rho\sigma}(k_{1},k_{2},k_{3})\Bigg{)}$
$\displaystyle\times\sum\limits_{m=0}^{n-1}\tilde{g}^{n-m-1}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})A^{(m)a}_{\nu}(k_{1})A^{(m)b}_{\rho}(k_{2})A^{(m)c}_{\sigma}(k_{3})\;.$
(130)

Plugging in the relations from leading order and next-to-leading order, we
express the end result as a function of the unsmeared fields:

$\displaystyle A^{(n)abc}_{\mu}$ $\displaystyle(k_{1},k_{2},k_{3})=$
$\displaystyle\sum\limits_{\nu\rho\sigma}\Bigg{(}\sum\limits_{\alpha\beta\gamma}6\varrho
g_{\alpha\beta\gamma}(k_{1},k_{2}+k_{3})\sum\limits_{m=1}^{n-1}\tilde{g}^{n-m-1}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\tilde{g}^{m}_{\beta\nu}(\varrho,k_{1})\tilde{g}^{(m)}_{\gamma\rho\sigma}(\varrho,k_{2},k_{3})$
$\displaystyle+\sum\limits_{\alpha\beta\gamma\delta}\bigg{(}\frac{1}{2}\varrho^{2}\Big{(}g_{\alpha\beta}(k_{1})g_{\alpha\gamma}(k_{2})\delta_{\alpha\delta}-g_{\alpha\beta}(k_{1})\delta_{\alpha\gamma}g_{\alpha\delta}(k_{3})\Big{)}+\varrho
G_{\alpha\beta\gamma\delta}(k_{1},k_{2},k_{3})\bigg{)}$
$\displaystyle\times\sum\limits_{m=0}^{n-1}\tilde{g}^{n-m-1}_{\mu\alpha}(\varrho,k_{1}+k_{2}+k_{3})\tilde{g}^{m}_{\beta\nu}(\varrho,k_{1})\tilde{g}^{m}_{\gamma\rho}(\varrho,k_{2})\tilde{g}^{m}_{\delta\sigma}(\varrho,k_{3})\Bigg{)}A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})A^{(0)c}_{\sigma}(k_{3})$
$\displaystyle=:\sum\limits_{\nu\rho\sigma}\tilde{g}^{(n)}_{\mu\nu\rho\sigma}(\varrho,k_{1},k_{2},k_{3})A^{(0)a}_{\nu}(k_{1})A^{(0)b}_{\rho}(k_{2})A^{(0)c}_{\sigma}(k_{3})\;,$
(131)

and we have found the last of the three $\tilde{g}$-functions from Eq. (32).

##  3 Perturbative expansion of the Wilson flow

The Wilson flow is defined through

$\displaystyle\partial_{t}U_{\mu}(x,t)$
$\displaystyle=iQ_{\mu}(x,t)U_{\mu}(x,t)$ (132) $\displaystyle U_{\mu}(x,0)$
$\displaystyle=U_{\mu}(x)$ (133)

with the same $Q_{\mu}$ as in the definition of stout smearing but made up
from the “flowed” fields $U_{\mu}(x,t)$ instead of the smeared fields. We
already know the perturbative expansion of $Q_{\mu}$

$\displaystyle Q_{\mu}(x,t)$
$\displaystyle=g_{0}T^{a}Q^{a}_{1\mu}(x,t)+ig_{0}^{2}[T^{a},T^{b}]Q^{ab}_{2\mu}(x,t)$
$\displaystyle+g_{0}^{3}\Big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\Big{)}Q^{abc}_{3\mu}(x,t)$
(134)

We define the expansion of the flowed link variable in terms of fields
$A^{a}_{\mu}(x,t)$, $A^{[ab]}_{\mu}(x,t)$, and $A^{abc}_{\mu}(x,t)$ such that
similarly to the smeared case

$\displaystyle U_{\mu}(x,t)$
$\displaystyle=1+ig_{0}T^{a}A^{a}_{\mu}(x,t)-\frac{g_{0}^{2}}{2}\Big{(}T^{a}T^{b}A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)+[T^{a},T^{b}]A^{[ab]}_{\mu}(x,t)\Big{)}$
$\displaystyle-i\frac{g_{0}^{3}}{6}\Bigg{(}T^{a}T^{b}T^{c}A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)+\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}A^{a}_{\mu}(x,t)A^{[ab]}_{\mu}(x,t)$
$\displaystyle+\bigg{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{[}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{]}\bigg{)}A^{abc}_{\mu}(x,t)\Bigg{)}+\mathcal{O}(g_{0}^{4})\;.$
(135)

They fulfil the initial conditions

$\displaystyle A^{a}_{\mu}(x,0)$ $\displaystyle=A^{a}_{\mu}(x)\;,$
$\displaystyle A^{[ab]}_{\mu}(x,0)$ $\displaystyle=0\;,$ $\displaystyle
A^{abc}_{\mu}(x,0)$ $\displaystyle=0\;,$ (136)

and we define their Fourier transforms consistently:

$\displaystyle A^{a}_{\mu}(x,t)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)k}A^{a}_{\mu}(k,t)$
(137) $\displaystyle A^{ab}_{\mu}(x,t)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{1}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{2}}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)(k_{1}+k_{2})}A^{ab}_{\mu}(k_{1},k_{2},t)$
(138) $\displaystyle A^{abc}_{\mu}(x,t)$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{1}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{2}}{(2\pi)^{4}}\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k_{3}}{(2\pi)^{4}}e^{i(x+\hat{\mu}/2)(k_{1}+k_{2}+k_{3})}A^{abc}_{\mu}(k_{1},k_{2},k_{3},t)\;.$
(139)

As in the stout smearing case, we can connect the perturbative fields to the
flowed gluon field $\tilde{A}^{a}_{\mu}(g_{0},x,t)$ in the $SU(N_{c})$
expansion such that

$\displaystyle U_{\mu}(x,t)=e^{ig_{0}T^{a}\tilde{A}^{a}_{\mu}(g_{0},x,t)}$
(140)

and

$\displaystyle\tilde{A}_{\mu}^{a}(g_{0},x,t)$
$\displaystyle=A^{a}_{\mu}(x,t)-\frac{g_{0}}{2}f^{abc}A^{[bc]}_{\mu}(x,t)-\frac{g_{0}^{2}}{6}\bigg{[}\frac{1}{N_{c}}\delta^{ae}\delta^{bc}+\frac{1}{2}\big{(}d^{bcd}d^{eda}-f^{bcd}f^{eda}\big{)}\bigg{]}A^{ebc}_{\mu}(x,t)$
$\displaystyle+\mathcal{O}(g_{0}^{3})\,.$ (141)

###  3.1 Leading order

At leading order the differential equation is

$\displaystyle\partial_{t}A^{a}_{\mu}(x,t)=Q_{1\mu}^{a}(x,t)=\sum\limits_{\nu,y}g_{\mu\nu}(y)A^{a}_{\nu}(x+y,t)$
(142)

which in momentum space becomes

$\displaystyle\partial_{t}A^{a}_{\mu}(k,t)=\sum\limits_{\nu}g_{\mu\nu}(k)A^{a}_{\nu}(k,t)$
(143)

with the same $g_{\mu\nu}$ as defined at leading order of the stout smearing.
With

$\displaystyle
g^{2}_{\mu\nu}(k)=\sum\limits_{\alpha}g_{\mu\alpha}(k)g_{\alpha\nu}(k)=-\hat{k}^{2}g_{\mu\nu}(k)\;,$
(144)

the matrix exponential of $g$ is

$\displaystyle\exp(g(k)t)_{\mu\nu}$
$\displaystyle=\delta_{\mu\nu}+g_{\mu\nu}t+\frac{1}{2}g^{2}_{\mu\nu}t^{2}+\ldots$
$\displaystyle=\delta_{\mu\nu}-\frac{1}{\hat{k}^{2}}\left(e^{-\hat{k}^{2}t}-1\right)g_{\mu\nu}(k)\;.$
(145)

With the initial condition for $t=0$, we arrive at:

$\displaystyle A^{a}_{\mu}(k,t)$
$\displaystyle=\sum\limits_{\nu}\left(e^{-\hat{k}^{2}t}\delta_{\mu\nu}-\left(e^{-\hat{k}^{2}t}-1\right)\frac{\hat{k}_{\mu}\hat{k}_{\nu}}{\hat{k}^{2}}\right)A^{a}_{\nu}(k,0)$
(146) $\displaystyle=:\sum\limits_{\nu}B_{\mu\nu}(k,t)A^{a}_{\nu}(k,0)\;.$
(147)

It is easy to see that for an infinitesimal flow-time $t=\epsilon$, the
exponentials become
$e^{-\hat{k}^{2}\epsilon}=1-\epsilon\hat{k}^{2}+\mathcal{O}(\epsilon^{2})$ and
the flow transformation becomes the same as a stout smearing with smearing
parameter $\varrho=\epsilon$, reiterating the fact that the Wilson flow is
generated by infinitesimal stout smearing.
The same result is obtained by starting from $n$ stout smearings and taking
the limit $n\to\infty$ while keeping the product $n\varrho=:t$ constant. Then
$f(k)^{n}=(1-\tfrac{t}{n}\hat{k}^{2})^{n}\to e^{-\hat{k}^{2}t}$ [38].
Moreover, this means for a small flow time $t<1$:

$\displaystyle B_{\mu\nu}(k,t)$
$\displaystyle=\tilde{g}_{\mu\nu}(t,k)+\frac{t^{2}}{2}(\hat{k}^{2})^{2}g_{\mu\nu}(k)+\mathcal{O}(t^{3})$
(148)
$\displaystyle=\tilde{g}^{2}_{\mu\nu}(t/2,k)+\frac{t^{2}}{4}(\hat{k}^{2})^{2}g_{\mu\nu}(k)+\mathcal{O}(t^{3})$
(149)
$\displaystyle=\tilde{g}^{n}_{\mu\nu}(t/n,k)+\frac{t^{2}}{2n}(\hat{k}^{2})^{2}g_{\mu\nu}(k)+\mathcal{O}(t^{3})$
(150)

Thus the error we get by approximating the flowed field by the field which has
been smeared $n$ times with smearing parameter $\varrho=t/n$ is always of the
order $t^{2}$ but decreases with $1/n$.

###  3.2 Next-to-leading order

At order $g_{0}^{2}$, the differential equation is

$\displaystyle
T^{a}T^{b}\partial_{t}\big{(}A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)\big{)}+[T^{a},T^{b}]\partial_{t}A^{[ab]}_{\mu}(x)$
$\displaystyle=2T^{a}T^{b}Q^{a}_{1\mu}(x,t)A^{b}_{\mu}(x,t)+2[T^{a},T^{b}]Q^{ab}_{2\mu}(x,t)\;.$
(151)

With

$\displaystyle Q^{ab}_{2\mu}(x,t)$
$\displaystyle=\frac{1}{2}\Big{(}-S^{a}_{1\mu}(x,t)A^{b}_{\mu}(x,t)+S^{ab}_{2\mu}(x,t)-6A^{[ab]}_{\mu}(x,t)\Big{)}$
(152)

and the result from the leading order, we arrive at

$\displaystyle\partial_{t}A^{[ab]}_{\mu}(x)$
$\displaystyle=S^{[ab]}_{2\mu}(x,t)-6A^{[ab]}_{\mu}(x,t)$ (153)
$\displaystyle=\sum\limits_{\nu,y}g_{\mu\nu}(y)A^{[ab]}_{\nu}(x+y,t)+\sum\limits_{\nu\rho,yz}g_{\mu\nu\rho}(y,z)A^{a}_{\nu}(x+y,t)A^{b}_{\rho}(x+z,t)\;.$
(154)

In momentum space, we have

$\displaystyle\partial_{t}A^{ab}_{\mu}(k_{1},k_{2},t)$
$\displaystyle=\sum\limits_{\nu}g_{\mu\nu}(k_{1}+k_{2})A^{[ab]}_{\nu}(k_{1},k_{2},t)+\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)$
(155)

with the same $g_{\mu\nu\rho}$ we know from stout smearing. Now we have an
inhomogenous differential equation and make the ansatz

$\displaystyle
A^{[ab]}_{\mu}(k_{1},k_{2},t)=\sum\limits_{\alpha}B_{\mu\alpha}(k_{1}+k_{2},t)C^{[ab]}_{\alpha}(k_{1},k_{2},t)$
(156)

as we know that $B_{\mu\nu}$ solves the homogeneous part. For the
inhomogeneous part we need to solve

$\displaystyle\sum\limits_{\alpha}B_{\mu\alpha}(k_{1}+k_{2},t)\partial_{t}C^{[ab]}_{\alpha}(k_{1},k_{2},t)=\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)\;,$
(157)

or equivalently

$\displaystyle\partial_{t}C^{[ab]}_{\mu}(k_{1},k_{2},t)=\sum\limits_{\alpha}B_{\mu\alpha}(k_{1}+k_{2},-t)\sum\limits_{\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)\;,$
(158)

because $\big{(}B_{\mu\nu}(k,t)\big{)}^{-1}=B_{\mu\nu}(k,-t)$. Thus

$\displaystyle
C^{[ab]}_{\mu}(k_{1},k_{2},t)=\sum\limits_{\alpha\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2},-t^{\prime})A^{a}_{\nu}(k_{1},t^{\prime})A^{b}_{\rho}(k_{2},t^{\prime})dt^{\prime}$
(159)

and

$\displaystyle A^{[ab]}_{\mu}(k_{1},k_{2},t)$
$\displaystyle=\sum\limits_{\alpha\beta\nu\rho}B_{\mu\alpha}(k_{1}+k_{2},t)\int\limits_{0}^{t}B_{\alpha\beta}(k_{1}+k_{2},-t^{\prime})g_{\beta\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1},t^{\prime})A^{b}_{\rho}(k_{2},t^{\prime})dt^{\prime}$
$\displaystyle=\sum\limits_{\alpha\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2},t-t^{\prime})A^{a}_{\nu}(k_{1},t^{\prime})A^{b}_{\rho}(k_{2},t^{\prime})dt^{\prime}\;,$
(160)

where we have used
$\sum_{\alpha}B_{\mu\alpha}(k,t_{1})B_{\alpha\nu}(k,t_{2})=B_{\mu\nu}(k,t_{1}+t_{2})$.
We can also go further and express $A^{[ab]}_{\mu}$ in terms of the original
gluon fields

$\displaystyle A^{[ab]}_{\mu}(k_{1},k_{2},t)$
$\displaystyle=\sum\limits_{\alpha\beta\gamma\nu\rho}g_{\alpha\beta\gamma}(k_{1},k_{2})\bigg{[}\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2},t-t^{\prime})B_{\beta\nu}(k_{1},t^{\prime})B_{\gamma\rho}(k_{2},t^{\prime})dt^{\prime}\bigg{]}$
$\displaystyle\times A^{a}_{\nu}(k_{1},0)A^{b}_{\rho}(k_{2},0)$
$\displaystyle=:\sum\limits_{\nu\rho}B_{\mu\nu\rho}(k_{1},k_{2},t)A^{a}_{\nu}(k_{1},0)A^{b}_{\rho}(k_{2},0)\;.$
(161)

Again, we can check that we recover $A^{(1)[ab]}_{\mu}(k_{1},k_{2})$ for an
infinitesimal flow time $t=\epsilon$

$\displaystyle A^{[ab]}_{\mu}(k_{1},k_{2},\epsilon)$
$\displaystyle=\sum\limits_{\alpha\beta\gamma\nu\rho}g_{\alpha\beta\gamma\delta}(k_{1},k_{2})\bigg{[}\epsilon
B_{\mu\alpha}(k_{1}+k_{2},0)B_{\beta\nu}(k_{1},0)B_{\gamma\rho}(k_{2},0)\bigg{]}A^{a}_{\nu}(k_{1},0)A^{b}_{\rho}(k_{2},0)$
(162) $\displaystyle=\sum\limits_{\nu\rho}\epsilon
g_{\mu\nu\rho}(k_{1},k_{2})A^{a}_{\nu}(k_{1},0)A^{b}_{\rho}(k_{2},0)=A^{(1)[ab]}_{\mu}(k_{1},k_{2})\Big{|}_{\rho=\epsilon}$
(163)

If we expand further (for small $t<1$), we get

$\displaystyle B_{\mu\nu\rho}(k_{1},k_{2},t)$
$\displaystyle=\sum\limits_{\alpha\beta\gamma}g_{\alpha\beta\gamma}(k_{1},k_{2})\bigg{(}\delta_{\mu\alpha}\delta_{\beta\nu}\delta_{\gamma\rho}t+\frac{1}{2}\big{(}g_{\mu\alpha}(k_{1}+k_{2})\delta_{\beta\nu}\delta_{\gamma\rho}+\delta_{\mu\alpha}g_{\beta\nu}(k_{1})\delta_{\gamma\rho}$
$\displaystyle+\delta_{\mu\alpha}\delta_{\beta\nu}g_{\gamma\rho}(k_{2})\big{)}t^{2}+\mathcal{O}(t^{3})\bigg{)}$
(164)
$\displaystyle=\tilde{g}^{(1)}_{\mu\nu\rho}(t,k_{1},k_{2})+\frac{t^{2}}{2}(\ldots)+\mathcal{O}(t^{3})$
(165)
$\displaystyle=\tilde{g}^{(n)}_{\mu\nu\rho}(t/n,k_{1},k_{2})+\frac{t^{2}}{2n}(\ldots)+\mathcal{O}(t^{3})$
(166)

which shows the same behavior as the first order fields.

###  3.3 Next-to-next-to leading order

The differential equation at third order is

$\displaystyle-i\frac{g_{0}^{3}}{6}\Bigg{[}T^{a}T^{b}T^{c}\partial_{t}\big{(}A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)\big{)}+\frac{3}{2}\\{T^{a},[T^{b},T^{c}]\\}\partial_{t}\big{(}A^{a}_{\mu}(x,t)A^{[bc]}_{\mu}(x,t)\big{)}$
$\displaystyle+\bigg{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\bigg{)}\partial_{t}A^{abc}_{\mu}(x,t)\Bigg{]}$
$\displaystyle=-ig_{0}^{3}\Bigg{[}\frac{1}{2}T^{a}T^{b}T^{c}Q^{a}_{1\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)+\frac{1}{2}T^{a}[T^{b},T^{c}]Q^{a}_{1\mu}(x,t)A^{[bc]}_{\mu}(x,t)$
$\displaystyle+[T^{a},T^{b}]T^{c}Q^{ab}_{2\mu}(x,t)A^{c}_{\mu}(x,t)-\bigg{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\bigg{)}Q^{abc}_{3\mu}(x,t)\Bigg{]}\;.$
(167)

The right hand side can be rewritten to look like

$\displaystyle-i\frac{g_{0}^{3}}{6}\bigg{(}T^{a}T^{b}T^{c}\partial_{t}(A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t))+\frac{3}{2}\\{T^{a}[T^{b},T^{c}]\\}\partial_{t}(A^{a}_{\mu}(x,t)A^{[bc]}_{\mu}(x,t))$
$\displaystyle-\bigg{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\bigg{)}\Big{(}-\frac{1}{2}Q_{1\mu}^{a}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)$
$\displaystyle-A^{a}Q_{1\mu}^{b}A^{c}_{\mu}(x,t)+3V^{\\{ab\\}}_{2\mu}(x,t)A^{c}_{\mu}(x,t)-3V^{abc}_{3\mu}(x,t)-6A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)$
$\displaystyle+6A^{abc}_{\mu}(x,t)\Big{)}\bigg{)}\;.$ (168)

Thus

$\displaystyle\partial_{t}A^{abc}_{\mu}(x,t)$
$\displaystyle=\frac{1}{2}Q_{1\mu}^{a}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)+A^{a}Q_{1\mu}^{b}A^{c}_{\mu}(x,t)-3V^{\\{ab\\}}_{2\mu}(x,t)A^{c}_{\mu}(x,t)$
$\displaystyle+3V^{abc}_{3\mu}(x,t)+6A^{a}_{\mu}(x,t)A^{b}_{\mu}(x,t)A^{c}_{\mu}(x,t)-6A^{abc}_{\mu}(x,t)\;.$
(169)

In momentum space, we get

$\displaystyle\partial_{t}A^{abc}_{\mu}(k_{1},k_{2},k_{3},t)$
$\displaystyle=\sum\limits_{\nu}g_{\mu\nu}(k_{1}+k_{2}+k_{3})A^{abc}_{\nu}(k_{1},k_{2},k_{3},t)$
$\displaystyle+6\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{a}_{\nu}(k_{1},t)A^{[bc]}_{\rho}(k_{2},k_{3},t)$
$\displaystyle+\sum\limits_{\nu\rho\sigma}G_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)A^{c}_{\sigma}(k_{3},t)\;.$
(170)

We make again the following ansatz

$\displaystyle
A^{abc}_{\mu}(k_{1},k_{2},k_{3},t)=\sum\limits_{\alpha}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t)C_{\alpha}^{abc}(k_{1},k_{2},k_{3},t)$
(171)

as we know from the leading order that $B_{\mu\nu}(k_{1}+k_{2}+k_{3},t)$
satisfies the homogeneous part of the differential equation. Then we are left
with the inhomogeneous part

$\displaystyle\sum\limits_{\alpha}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t)\partial_{t}C^{abc}_{\alpha}(k_{1},k_{2},k_{3},t)$
$\displaystyle=6\sum\limits_{\nu\rho}g_{\mu\nu\rho}(k_{1},k_{2}+k_{3})A^{a}_{\nu}(k_{1},t)A^{[bc]}_{\rho}(k_{2},k_{3},t)$
$\displaystyle+\sum\limits_{\nu\rho\sigma}G_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)A^{c}_{\sigma}(k_{3},t)\;,$
(172)

or equivalently

$\displaystyle\partial_{t}C^{abc}_{\alpha}(k_{1},k_{2},k_{3},t)$
$\displaystyle=6\sum\limits_{\alpha\nu\rho}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},-t)g_{\alpha\nu\rho}(k_{1},k_{2}+k_{3})A^{a}_{\nu}(k_{1},t)A^{[bc]}_{\rho}(k_{2},k_{3},t)$
$\displaystyle+\sum\limits_{\alpha\nu\rho\sigma}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},-t)G_{\alpha\nu\rho\sigma}(k_{1},k_{2},k_{3})A^{a}_{\nu}(k_{1},t)A^{b}_{\rho}(k_{2},t)A^{c}_{\sigma}(k_{3},t)\;.$
(173)

Integrating this and putting everything together, we get

$\displaystyle A_{\mu}^{abc}$ $\displaystyle(k_{1},k_{2},k_{3},t)=$
$\displaystyle\sum\limits_{\alpha\nu\rho}g_{\alpha\nu\rho}(k_{1},k_{2}+k_{3})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t-t^{\prime})A_{\nu}^{a}(k_{1},t^{\prime})A_{\rho}^{[bc]}(k_{2},k_{3},t^{\prime})\,dt^{\prime}$
$\displaystyle+\sum\limits_{\alpha\nu\rho\sigma}G_{\alpha\nu\rho\sigma}(k_{1},k_{2},k_{3})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t-t^{\prime})A_{\nu}^{a}(k_{1},t^{\prime})A_{\rho}^{b}(k_{2},t^{\prime})A_{\sigma}^{c}(k_{3},t^{\prime})\,dt^{\prime}$
(174)

and we can define a function $B_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3},t)$ for
the third order of the Wilson flow:

$\displaystyle A_{\mu}^{abc}$ $\displaystyle(k_{1},k_{2},k_{3},t)=$
$\displaystyle\sum\limits_{\nu\rho\sigma}\Bigg{(}\sum\limits_{\alpha\beta\gamma}g_{\alpha\beta\gamma}(k_{1},k_{2}+k_{3})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t-t^{\prime})B_{\beta\nu}(k_{1},t^{\prime})B_{\gamma\rho\sigma}(k_{2},k_{3},t^{\prime})\,dt^{\prime}$
$\displaystyle+\sum\limits_{\alpha\beta\gamma\delta}G_{\alpha\beta\gamma\delta}(k_{1},k_{2},k_{3})\int\limits_{0}^{t}B_{\mu\alpha}(k_{1}+k_{2}+k_{3},t-t^{\prime})B_{\beta\nu}(k_{1},t^{\prime})B_{\gamma\rho}(k_{2},t^{\prime})B_{\delta_{\sigma}}(k_{3},t^{\prime})\,dt^{\prime}\Bigg{)}$
$\displaystyle\times
A_{\nu}^{a}(k_{1},0)A_{\rho}^{b}(k_{2},0)A_{\sigma}^{c}(k_{3},0)$
$\displaystyle=:\sum\limits_{\nu\rho\sigma}B_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3},t)A_{\nu}^{a}(k_{1},0)A_{\rho}^{b}(k_{2},0)A_{\sigma}^{c}(k_{3},0)\;.$
(175)

Again, we can check that the error one makes by approximating the Wilson flow
by stout smearing is of the order $t^{2}/(2n)$:

$\displaystyle
B_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3},t)=\tilde{g}^{(n)}_{\mu\nu\rho\sigma}(t/n,k_{1},k_{2},k_{3})+\frac{t^{2}}{2n}(\ldots)\;.$
(176)

Thus we have found three $B$-functions that are the Wilson flow equivalent to
the $\tilde{g}$-functions from the expansion of stout smearing.

###  3.4 Continuum limit

In the continuum limit $a\to 0$, we replace $\cos(k_{\mu})\to 1$ and
$\sin(k_{\mu})\to k_{\mu}$, and expand to first order in the momenta. We get

$\displaystyle B_{\mu\nu}(k,t)$
$\displaystyle\to\sum\limits_{\nu}\bigg{(}e^{-k^{2}t}\delta_{\mu\nu}-\left(e^{-k^{2}t}-1\bigg{)}\frac{k_{\mu}k_{\nu}}{k^{2}}\right)$
(177) $\displaystyle g_{\mu\nu\rho}(k_{1},k_{2})$ $\displaystyle\to
i\Big{(}-\delta_{\beta\nu}\big{(}2k_{1\rho}+k_{2\rho}\big{)}+\delta_{\beta\rho}\big{(}2k_{2\nu}+k_{1\nu}\big{)}+\delta_{\nu\rho}\big{(}k_{1\mu}-k_{2\mu}\big{)}\Big{)}$
(178) $\displaystyle G_{\mu\nu\rho\sigma}(k_{1},k_{2},k_{3})$
$\displaystyle\to
6\big{(}\delta_{\mu\nu}\delta_{\rho\sigma}-\delta_{\mu\rho}\delta_{\nu\sigma}\big{)}\;,$
(179)

and the flowed field $\tilde{A}_{\mu}$ from Eq. (141) in this limit fulfils
the differential equation of the continuum gradient flow

$\displaystyle\partial_{t}\tilde{A}_{\mu}(x,t)$
$\displaystyle=\sum\limits_{\nu}\tilde{D}_{\nu}\tilde{F}_{\nu\mu},$
$\displaystyle\tilde{A}_{\mu}(x,0)$ $\displaystyle=A_{\mu}(x)$ (180)

with

$\displaystyle\tilde{D}_{\mu}$
$\displaystyle=\partial_{\mu}+\big{[}\tilde{A}_{\mu}(x,t),\;\cdot\;\big{]},$
$\displaystyle\tilde{F}_{\mu\nu}$
$\displaystyle=\partial_{\mu}\tilde{A}_{\nu}(x,t)-\partial_{\nu}\tilde{A}_{\mu}(x,t)+\big{[}\tilde{A}_{\mu}(x,t),\tilde{A}_{\nu}(x,t)\big{]}\;.$
(181)

These results agree with the ones given in Refs. [13] and [14].

##  4 Feynman rules of fermion actions with stout smearing or Wilson flow

The Feynman rules are obtained by inserting the expansion of the link
variables (182) into the Fermion action. This results in expressions for anti-
quark-quark-$n$-gluon vertices at each order $g_{0}^{n}$. These vertices are
proportional to $a^{n-1}$ such that only the $\bar{q}qg$ vertex survives in
the continuum limit $a\to 0$, as is expected from continuum perturbation
theory. For the purpose of one-loop calculations, we will need the
$\bar{q}qg$-, $\bar{q}qgg$\- and $\bar{q}qggg$-vertex, i.e. anything up to
order $g_{0}^{3}$.

For deriving the Feynman rules of the fermion action including stout smearing
or Wilson flow, there are two possible strategies. One is to simply insert the
perturbative expansion of the smeared or flowed link variable into the action
and expand to the needed order in $g_{0}$. This “brute force” approach leads
to very large intermediate expressions and involves many steps where indices
have to be renamed etc. This is not feasible beyond one step of stout
smearing. The second much more effective method is based on the
$SU(N_{c})$-expansion of the link variable. One simply derives the Feynman
rules for generic link variables of the form

$\displaystyle
U_{\mu}(x)=1+ig_{0}T^{a}A^{a}_{\mu}(x)-\frac{g_{0}^{2}}{2}T^{a}T^{b}A^{a}_{\mu}(x)A^{b}_{\mu}(x)-i\frac{g_{0}^{3}}{6}T^{a}T^{b}T^{c}A^{a}_{\mu}(x)A^{b}_{\mu}(x)A^{c}_{\mu}(x)$
(182)

which results in fairly simple expressions. One then couples the stout or flow
“form factors” to these generic Feynman rules to obtain the complete results.
This also practically allows us to easily split lengthy expressions into
smaller parts for Mathematica to work on. The following sections show the
derivations of the generic Feynman rules and how to couple them to stout
smearing or the Wilson flow.

We start with the Wilson quark action

$\displaystyle S_{W}=\sum\limits_{x,y}\sum\limits_{\mu=\pm 1}^{\pm
4}\bar{\psi}(x)\frac{1}{2}\bigg{[}U_{\mu}(x)\delta_{x+\hat{\mu},y}\gamma_{\mu}-r\Big{(}U_{\mu}(x)\delta_{x+\hat{\mu},y}-\delta_{x,y}\Big{)}\bigg{]}\psi(y)\;.$
(183)

Links with negative indices are given by
$U_{-\mu}(x)=U^{\dagger}_{\mu}(x-\hat{\mu})$ and
$\gamma_{-\mu}=-\gamma_{\mu}$. Up to order $g_{0}^{3}$ we get the well known
three vertices of a quark anti-quark pair coupling to one, two, and three
gluons respectively.

$\displaystyle V^{a}_{1W\mu}(p,q)$
$\displaystyle=-g_{0}T^{a}\Big{(}i\gamma_{\mu}c(p_{\mu}+q_{\mu})+rs(p_{\mu}+q_{\mu})\Big{)}$
(184) $\displaystyle V^{ab}_{2W\mu\nu}(p,q)$
$\displaystyle=\frac{g_{0}^{2}}{2}T^{a}T^{b}\delta_{\mu\nu}\Big{(}i\gamma_{\mu}s(p_{\mu}+q_{\mu})-rc(p_{\mu}+q_{\mu})\Big{)}$
(185) $\displaystyle V^{abc}_{3W\mu\nu\rho}(p,q)$
$\displaystyle=\frac{g_{0}^{3}}{6}T^{a}T^{b}T^{c}\delta_{\mu\nu}\delta_{\mu\rho}\Big{(}i\gamma_{\mu}c(p_{\mu}+q_{\mu})+rs(p_{\mu}+q_{\mu})\Big{)}$
(186)

These appear in the expansion of the action contracted with the (original)
gluon fields

$\displaystyle S_{W}$
$\displaystyle\sim\sum\limits_{\mu}\bar{\psi}(q)T^{a}V_{1\,\mu}(p,q)A^{a}_{\mu}(q-p)\psi(p)$
$\displaystyle+\sum\limits_{\mu\nu}\bar{\psi}(q)T^{a}T^{b}V_{2\,\mu\nu}(p,q)A^{a}_{\mu}(k_{1})A^{b}_{\nu}(k_{2})\delta(q-p-
k_{1}-k_{2})\psi(p)$
$\displaystyle+\sum\limits_{\mu\nu\rho}\bar{\psi}(q)T^{a}T^{b}T^{c}V_{3\,\mu\nu\rho}(p,q)A^{a}_{\mu}(k_{1})A^{b}_{\nu}(k_{2})A^{c}_{\rho}(k_{3})\delta(q-p-
k_{1}-k_{2}-k_{3})\psi(p)$ (187)

In order to derive the Feynman rules with smearing or gradient flow, we
replace the gluon fields $A^{a}_{\mu}$ by $\tilde{A}^{(n)a}_{\mu}$ or
$\tilde{A}^{a}_{\mu}(t)$ respectively. Thereby new terms are added at higher
orders of $g_{0}$. For example, replacing the $A^{a}_{\mu}$ which couples to
$V^{a}_{1\,\mu}$ by a smeared or flowed field, provides the new $\bar{q}qg$
vertex with a form factor and adds terms to the $\bar{q}qgg$ and $\bar{q}qggg$
vertices proportional to $A^{[ab]}_{\mu}$ and $A^{abc}_{\mu}$ respectively.
Specifically, the new stout smeared vertices look like

$\displaystyle V^{(n)a}_{1\mu}(p,q)$
$\displaystyle=T^{a}\sum\limits_{\nu}\tilde{g}^{n}_{\mu\nu}(\varrho,q-p)V_{1\nu}(p,q)$
(188) $\displaystyle V^{(n)ab}_{2\mu\nu}(p,q,k_{1},k_{2})$
$\displaystyle=T^{a}T^{b}\sum\limits_{\rho\sigma}\tilde{g}^{n}_{\mu\rho}(\varrho,k_{1})\tilde{g}^{n}_{\nu\sigma}(\varrho,k_{2})V_{2\rho\sigma}(p,q,k_{1},k_{2})$
$\displaystyle-\frac{g_{0}}{2i}[T^{a},T^{b}]\sum\limits_{\rho}\tilde{g}^{(n)}_{\rho\mu\nu}(k_{1},k_{2})V_{1\rho}(p,q)$
(189) $\displaystyle V^{(n)abc}_{3\mu\nu\rho}(p,q,k_{1},k_{2},k_{3})$
$\displaystyle=T^{a}T^{b}T^{c}\sum\limits_{\alpha\beta\gamma}\tilde{g}^{n}_{\mu\alpha}(\varrho,k_{1})\tilde{g}^{n}_{\nu\beta}(\varrho,k_{2})\tilde{g}^{n}_{\rho\gamma}(\varrho,k_{3})V_{3\alpha\beta\gamma}(p,q,k_{1},k_{2},k_{3})$
$\displaystyle-\frac{g_{0}}{2i}\\{T^{a},[T^{b},T^{c}]\\}$
$\displaystyle\sum\limits_{\alpha\beta}\tilde{g}^{n}_{\mu\alpha}(\varrho,k_{1})\tilde{g}^{(n)}_{\beta\nu\rho}(\varrho,k_{2},k_{3})V_{2\alpha\beta}(p,q,k_{1},k_{2}+k_{3})$
$\displaystyle-\frac{g_{0}^{2}}{6}\bigg{[}T^{a}T^{b}T^{c}+T^{c}$
$\displaystyle
T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\bigg{]}\sum\limits_{\alpha}\tilde{g}^{(n)}_{\alpha\mu\nu\rho}(\varrho,k_{1},k_{2},k_{3})V_{1\alpha}(p,q)$
(190)

with the $\tilde{g}$ given in Eqs. (56),(73), and (131). The vertices with
Wilson flow are analogously

$\displaystyle V^{a}_{1\mu}(p,q,t)$
$\displaystyle=T^{a}\sum\limits_{\nu}B_{\mu\nu}(q-p,t)V_{1\nu}(p,q)$ (191)
$\displaystyle V^{ab}_{2\mu\nu}(p,q,k_{1},k_{2},t)$
$\displaystyle=T^{a}T^{b}\sum\limits_{\rho\sigma}B_{\mu\rho}(k_{1},t)B_{\nu\sigma}(k_{2},t)V_{2\rho\sigma}(p,q,k_{1},k_{2})$
$\displaystyle-\frac{g_{0}}{2i}[T^{a},T^{b}]\sum\limits_{\rho}B_{\rho\mu\nu}(k_{1},k_{2},t)V_{1\rho}(p,q)$
(192) $\displaystyle V^{abc}_{3\mu\nu\rho}(p,q,k_{1},k_{2},k_{3},t)$
$\displaystyle=T^{a}T^{b}T^{c}\sum\limits_{\alpha\beta\gamma}B_{\mu\alpha}(k_{1},t)B_{\nu\beta}(k_{2},t)B_{\rho\gamma}(k_{3},t)V_{3\alpha\beta\gamma}(p,q,k_{1},k_{2},k_{3})$
$\displaystyle-\frac{g_{0}}{2i}\\{T^{a},[T^{b},T^{c}]\\}$
$\displaystyle\sum\limits_{\alpha\beta}B_{\mu\alpha}(k_{1},t)B_{\beta\nu\rho}(k_{2},k_{3},t)V_{2\alpha\beta}(p,q,k_{1},k_{2}+k_{3})$
$\displaystyle-\frac{g_{0}^{2}}{6}\bigg{[}T^{a}T^{b}T^{c}+T^{c}$
$\displaystyle
T^{b}T^{a}-\frac{1}{N_{c}}\mathrm{Tr}\big{(}T^{a}T^{b}T^{c}+T^{c}T^{b}T^{a}\big{)}\bigg{]}\sum\limits_{\alpha}B_{\alpha\mu\nu\rho}(k_{1},k_{2},k_{3},t)V_{1\alpha}(p,q)$
(193)

with the $B$ given in Eqs. (147), (161), and (175).

For the calculation of the self-energy below we need the $\bar{q}qg$\- and
$\bar{q}qgg$-vertices and we use the Feynman rules of the Wilson clover action

$\displaystyle V_{1\mu}(p,q)$ $\displaystyle=V_{1W\mu}(p,q)+V_{1C\mu}(p,q)$
(194) $\displaystyle V_{2\mu\nu}(p,q,k_{1},k_{2})$
$\displaystyle=V_{2W\mu\nu}(p,q)+V_{2C\mu\nu}(p,q,k_{1},k_{2})$ (195)

with the clover vertices

$\displaystyle V_{1C\mu}(p,q)$
$\displaystyle=i\frac{g_{0}}{2}c_{\mathrm{SW}}\sum\limits_{\nu}\sigma_{\mu\nu}c(p_{\mu}-q_{\mu})\bar{s}(p_{\nu}-q_{\nu})$
(196) $\displaystyle V_{2C\mu\nu}(p,q,k_{1},k_{2})$
$\displaystyle=ig_{0}^{2}c_{\mathrm{SW}}\Bigg{(}\frac{1}{4}\delta_{\mu\nu}\sum\limits_{\rho}\sigma_{\mu\rho}s(q_{\mu}-p_{\mu})\big{(}\bar{s}(k_{2\rho})-\bar{s}(k_{1\rho})\big{)}$
$\displaystyle+\sigma_{\mu\nu}\bigg{(}c(k_{1\nu})c(k_{2\mu})c(q_{\mu}-p_{\mu})c(p_{\nu}-p_{\nu})-\frac{1}{2}c(k_{1\mu})c(k_{2\nu})\bigg{)}\Bigg{)}$
(197)

where $\sigma_{\mu\nu}=\frac{i}{2}[\gamma_{\mu},\gamma_{\nu}]$ and the tree-
level value $c_{\mathrm{SW}}=r$.

##  5 Fermion self-energy with stout smearing or Wilson flow

Figure 1: Tadpole (left) and sunset (right) diagrams of the quark self-energy.

To illustrate the use of the smeared Feynman rules derived in the previous
section, we give in this section the result of the self-energy of a clover
fermion coupled to a gluon background through stout-smeared or gradient-flowed
vertices. The smoothing thus affects both the covariant derivative (in the
unimproved part of the fermion action) and the field strength tensor (in the
part proportional to $c_{\mathrm{SW}}$).

At the one-loop order the fermion self-energy $\Sigma$ receives contributions
from the “tadpole” and the “sunset” diagrams shown in Fig. 1. We define
$\Sigma_{0}$ through the linearly divergent part and $\Sigma_{1}$ as the
finite piece in an expansion in the lattice spacing [39, 40, 41]

$\displaystyle\Sigma=\frac{g_{0}^{2}C_{F}}{16\pi^{2}}\bigg{(}\frac{\Sigma_{0}}{a}+\Sigma_{1}i\not{p}+\mathcal{O}(a)\bigg{)}\;.$
(198)

where $p_{\mu}$ is the momentum of the external fermion (see Fig. 1), and the
additive mass shift follows as

$\displaystyle
am_{\mathrm{crit}}=\frac{g_{0}^{2}C_{F}}{16\pi^{2}}\Sigma_{0}\;.$ (199)

The contributions to $\Sigma$ from the two diagrams read

$\displaystyle\Sigma^{\mathrm{(tadpole)}}$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}\sum\limits_{\mu,\nu,a}G_{\mu\nu}(k)V_{2\,\mu\nu}^{aa}(p,p,k,-k)$
(200) $\displaystyle\Sigma^{\mathrm{(sunset)}}$
$\displaystyle=\int\limits_{-\pi}^{\pi}\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}\sum\limits_{\mu,\nu,a}V_{1\,\mu}^{a}(p,k)G_{\mu\nu}(p-k)S(k)V_{1\,\nu}^{a}(k,p)$
(201)

respectively, and from the sum we obtain $\Sigma_{0}$ and $\Sigma_{1}$ by

$\displaystyle\frac{g_{0}^{2}C_{F}}{16\pi^{2}}\frac{\Sigma_{0}}{a}$
$\displaystyle=\Sigma\big{|}_{p=0}$ (202)
$\displaystyle\frac{g_{0}^{2}C_{F}}{16\pi^{2}}\Sigma_{1}$
$\displaystyle=\frac{1}{4}\mathrm{Tr}\bigg{[}\frac{\partial}{\partial
p_{\mu}}\Sigma\cdot\gamma_{\mu}\bigg{]}_{p=0}$ (203)

where $\mu$ is kept fixed, i.e. not summed over. While $\Sigma_{0}$ is gauge
independent, $\Sigma_{1}$ is part of the fermion field renormalization and
does depend on the gauge. We use Feynman gauge and all results for
$\Sigma_{1}$ are given in this gauge. We refer the reader to Ref. [42] for the
explicit form of the gluon and fermion propagators in this gauge. Furthermore,
for the gluon propagator we employ both the Wilson “plaquette” gauge action
and the tree-level “Symanzik improved” gluon action, since these are the most
popular choices in contemporary non-perturbative studies. These results will
be labeled as “plaq” and “sym”, respectively, in the figures and tables below.
A somewhat technical point is that in Feynman gauge $\Sigma_{1}$ depends
logarithmically on an external momentum $p\to 0$

$\displaystyle\Sigma_{1}=\log(p^{2})+\Sigma_{10}$ (204)

and this is why we shall plot or quote the constant part $\Sigma_{10}$.

Besides the choice of the gauge action also the numerical values of the Wilson
parameter $r$ and the clover coefficient $c_{\mathrm{SW}}$ matter and, of
course, the parameter $(\rho,n_{\mathrm{stout}})$ or the gradient flow time
$t/a^{2}$. To enhance verbosity, in this section $n_{\mathrm{stout}}$ is used
to denote the number of stout steps. It goes without saying that we cannot
represent all information in a compact form. This is why we restrict ourselves
to a quick overview on how $\Sigma$ varies as a function of $\rho$ at fixed
$c_{\mathrm{SW}}=r$, and an overview on how it depends on
($c_{\mathrm{SW}}=r$) $c_{\mathrm{SW}}$ and $r$ for a few choices of
$(\rho,n_{\mathrm{stout}})$ or the flow time $t/a^{2}$. Some additional
information is shifted to the appendix, in the hope that the selection made is
broad enough to allow the user to make an informed decision when it comes to
choosing a lattice fermion action for future non-perturbative studies.

In Ref. [37] results for $\Sigma_{0}$ with up to three steps of stout smearing
at the specific value $\varrho=0.1$ (labeled there as $\alpha=0.6$, due to the
perturbative matching between stout and APE smearing) were given, which agree
with our results (see our Appendix for details). In addition, results for
$\Sigma_{0}$ and $\Sigma_{1}$ with one step of SLINC smearing for plaquette
and Lüscher-Weisz glue were given in Ref. [43]. In the SLINC action only the
links entering the Wilson part of the fermion action are subject to one step
of stout smearing (with variable $\varrho$), while the clover part stays
unsmeared. Reproducing these results222There is a typo in Eqn. (53) of Ref.
[43], the number 2235.407087 should read 2335.407087. served as a cross-check
for our code.

Figure 2:  Contributions to the divergent piece $\Sigma_{0}$ of the clover
($c_{\mathrm{SW}}=r=1$) fermion self-energy from the tadpole (top) and sunset
(middle) diagrams and their sum (bottom). Results are shown as a function of
the smearing parameter $\varrho$, with plaquette gluon background (left) and
tree-level Symanzik-improved gluon background (right). In the top row the
results for one step of overall stout smearing ($n_{\mathrm{stout}}=1$)
coincide with those for SLINC-fermions [43].

Figure 3:  Same as Fig. 2, but for the finite piece $\Sigma_{10}$; see Eq.
(204).

Calculating $\Sigma$ and extracting $\Sigma_{0}$ and $\Sigma_{10}$ from it
becomes more challenging as the number of stout steps (or the overall flow
time) increases. We managed to get reasonably accurate results for up to
$n_{\mathrm{stout}}=4$ smearing steps.

| $n=1$ | $n=2$ | $n=3$ | $n=4$
---|---|---|---|---
$\varrho_{\mathrm{max}}$ of $\Sigma_{0}$ (plaq) | 0.12096871(3) | 0.11973422(9) | 0.1189366(3) | 0.1187(4)
$\varrho_{\mathrm{max}}$ of $\Sigma_{0}$ (sym) | 0.12357712(3) | 0.1220287(3) | 0.120999(5) | 0.1204(9)
$\varrho_{\mathrm{min}}$ of $\Sigma_{10}$ (plaq) | 0.13078397(5) | 0.1285409(2) | 0.1271189(9) | 0.1263(6)
$\varrho_{\mathrm{min}}$ of $\Sigma_{10}$ (sym) | 0.13417728(6) | 0.1315944(2) | 0.129866(6) | 0.1287(6)
Table 1: Position of the extreme points in the bottom panels of Figs. 2 and 3.

Fig. 2 displays how the piece $\Sigma_{0}$ of the self-energy of a clover
fermion ($c_{\mathrm{SW}}=r$) depends on $\varrho$ for up to four stout steps.
Regardless of the gauge background adopted (left versus right column),
choosing a reasonable $\varrho$ is found to reduce the additive mass shift
(bottom row) quite drastically, while (in Feynman gauge) the “tadpole” and
“sunset” contributions, inspected individually, do not show any noteworthy
feature (top and middle panels). In the physical result (bottom row) the
initial slope is proportional to $n_{\mathrm{stout}}$ (see below). Compared to
the full curve the SLINC curve (wide-spaced dots) seems to be an improvement
for small values of $\varrho$, but for a value $\varrho\sim 0.08$ the critical
mass of SLINC fermions changes sign. On the other hand, the “overall smearing”
advocated in Ref. [37] seems to be rather benevolent; the more smearing steps
(with $0<\varrho<0.12$) are taken, the more $\Sigma_{0}$ tends to zero.
Moreover, regardless of $n_{\mathrm{stout}}$ and the chosen gluon background,
the maximum of $\Sigma_{0}$ is always near $\varrho\sim 0.12$, see Tab. 1 for
details. This is consistent with the observation made in Ref. [37] that
$\varrho$ in the stout recipe should be kept below 0.125 in order to have a
(first order) smearing form factor smaller than one throughout the Brillouin
zone.

Fig. 3 displays how $\Sigma_{10}$ depends on $\varrho$ for up to four stout
steps (again for $c_{\mathrm{SW}}=r$). Iterated stout smearings with a
parameter in the vicinity of $\varrho\simeq 0.1$ seem to bring this quantity
to a value near $-6$. Again the extremal points are found to be near
$\varrho\sim 0.12$, see Tab. 1 for details.

Figure 4: Same information as in Fig. 2, but plotted versus the effective flow
time $t=\varrho n_{\mathrm{stout}}$. The vertical dashed lines indicate
integer multiples of t=0.125 (in lattice units). The curve of the Wilson flow
(to which these results converge under $n_{\mathrm{stout}}\to\infty$) is
indicated with black crosses.

Figure 5: Same as Fig. 4, but for the finite piece $\Sigma_{10}$; see Eq.
(204).

It is interesting to plot the results for iterated stout smearing versus
$n_{\mathrm{stout}}\varrho$ rather than versus $\varrho$; this is done in
Figs. 4 and 5. With a bit of intuition one can see that for any fixed value of
$n_{\mathrm{stout}}\rho$ these results converge to a universal curve, which we
indicate through black crosses. Given our results in Sec. 3, this should not
come as a surprise. The black curves in Figs. 4 and 5 indicate the results for
the Wilson variety of the gradient flow. Evidently, what we see is a graphical
illustration of the correspondence 333Corrections to (205) start at order
$t^{2}/(a^{4}n_{\mathrm{stout}})$, see Eqns. (150), (166), and (176). Hence,
the left-hand side of (205) is to be dressed with a factor
$(1+\mathcal{O}(t/(a^{2}n_{\mathrm{stout}})))$.

$t/a^{2}=n_{\mathrm{stout}}\cdot\varrho$ (205)

that links the dimensionless flow-time to the cumulative parameter of the
iterated stout recipe. Note that this identification becomes exact under
$n_{\mathrm{stout}}\to\infty$ and $\varrho\to 0$ with the product
$n_{\mathrm{stout}}\cdot\varrho$ held constant.

The thin vertical lines in Figs. 4 and 5 indicate integer multiples of 0.125.
They are meant to separate each curve into a part where it gives a reasonable
approximation to the gradient flow (to the left) and a part where it does not
(to the right). For instance the mark at $3\cdot 0.125=0.375$ separates the
dash-dotted line ($n_{\mathrm{stout}}=3$) into an ascending part (where it
approximates the crosses quite well), and a descending part (where it does
not). In the terminology of numerical mathematics one would say that multiple
stout smearings implement a simple integration scheme (with an integration
error in the flow time $t/a^{2}$ which scales like $1/n_{\mathrm{stout}}$),
and the caveat of Ref. [37] is meant to limit the integration error.

Figure 6: The divergent piece $\Sigma_{0}$ of the clover fermion
($c_{\mathrm{SW}}=r$) self-energy as a function of $r$, with 1 to 4 stout
steps (top to bottom). The background uses plaquette glue (left) or tree-level
Symanzik improved glue (right). The legends give details of linear least-
squares fits.

Figure 7: Same as Fig. 6, but for the finite piece $\Sigma_{10}$; see Eq.
(204).

Let us consider the behavior of the self-energy as a function of the Wilson
parameter $r$ (with $c_{\mathrm{SW}}=r$) as shown in Figs. 6 and 7. Even
though the self-energy at one-loop order is not a linear function in $r$
analytically (since $r$ appears in the denominator of the fermion propagator),
$\Sigma_{0}$ and $\Sigma_{10}$ seem almost linear over the plotted range. The
former quantity decreases and the latter one increases with increasing $r$.
For $\rho$ approaching 0.125 and/or more smearing steps the behavior becomes
gradually flatter. In practical terms this means that the Wilson parameter $r$
is not a useful knob to engineer the self-energy of tree-level clover
fermions.

Figure 8: Contributions to the divergent piece $\Sigma_{0}$ of the clover
($c_{\mathrm{SW}}=r=1$) fermion self-energy from the tadpole (top) and sunset
(middle) diagrams and their sum (bottom). Results for several values of $r$
are given as a function of the gradient-flow time $t$ in lattice units.

Figure 9: Same as Fig. 8, but for the finite piece $\Sigma_{10}$; see Eq.
(204).

In Figs. 8 and 9 we finally shift to results with the gradient flow recipe.
For $r=1$ (green dots) the “tadpole” and the “sunset” contribution seem to
tend to unsuspicious finite values, as $t/a^{2}$ is taken large, and the
physical sum tends to zero. Varying $r$ (and synchronously
$c_{\mathrm{SW}}=r$) seems to affect only the splitting between the “tadpole”
and the “sunset” contributions, but barely the final (physical) result. And
the former observation is only pronounced for $\Sigma_{0}$; for $\Sigma_{10}$
it is rather mild, here the “tadpole” contribution does not depend on $r$ at
all.

Figure 10: The dependence on the Wilson paramter $r$ of $\Sigma_{0}$ (top) and
$\Sigma_{10}$ (bottom) at different flow times $t$ of the Wilson flow. Linear
least-square fits are given in brackets.

For completeness we show in Fig. 10 how $\Sigma_{0}$ and $\Sigma_{10}$ depend
on $r$ for a few Wilson flow-times $t/a^{2}$. Any dependence that is still
present at tiny flow times ($t/a^{2}\simeq 0.1$) is found to quickly disappear
for moderate flow times ($t/a^{2}\simeq 1$).

In our view, the take-away message from this section is that there is a
perturbative backing of the well known benefit that comes from multiple stout
smearing. One step reduces the absolute value of $\Sigma_{0}$ by about 85%,
two steps by 95%, three by 98% and four steps by 99%.

##  6 Conclusions

In this paper we have worked out a systematic approach for deriving Feynman
rules for lattice fermion actions with smeared gauge links (to be used
interchangeably in the covariant derivative, in the species-separating
covariant Laplacian and/or a possible clover term).

The basis for such calculations is provided by a perturbative expansion of the
stout smearing recipe of Ref. [9] and the Wilson variety of the gradient flow
introduced in Refs. [13, 14, 15]; these expansions have been presented in Sec.
2 and Sec. 3, respectively.

With these results in hand, one may derive the Feynman rules for a fermion
action with an arbitrary parameter combination $(\varrho,n_{\mathrm{stout}})$
in case of stout smearing or an arbitrary flow time $t/a^{2}\in\mathbb{R}$ in
case of the gradient flow. In Sec. 4 we argue that the “brute force” approach,
i.e. inserting the perturbative expansion of the smeared/flowed variable into
the underlying action, is not recommendable, since it leads to excessively
long intermediate expressions. We present an approach based on the $SU(N_{c})$
expansion of the original (“unsmeared”) links $U_{\mu}(x)$ to the required
order in $g_{0}$ (typically $g_{0}^{3}$ for one-loop calculations), which
keeps intermediate expressions much shorter. This holds true for both stout
smearing and the gradient flow, and confirms, as a byproduct, the perturbative
matching (205) between these recipes.

To give an immediate application of our recipe, we derive in Sec. 4 the
Feynman rules of a Wilson fermion with clover improvement for arbitrary Wilson
parameter $r$ and Sheikholeslami-Wohlert parameter $c_{\mathrm{SW}}$. With
these rules in hand, we calculate in Sec. 5 the self-energy of the clover
fermion for $c_{\mathrm{SW}}=r$, for variable $\varrho$ and $1\leq
n_{\mathrm{stout}}\leq 4$ in case of stout smearing and the range $0\leq
t/a^{2}\leq 1$ in case of gradient-flowing.

The cross-check with Ref. [37], where numerical results obtained in the “brute
force” approach were presented, proves particularly enlightening. Of course,
reproducing these results felt like a relief, but the difference is in the
details. In Ref. [37] results for $\Sigma_{0}$ were given for a single stout
parameter ($\varrho=0.1$ in our terminology), restricted to
$n_{\mathrm{stout}}\leq 3$ due to the numerical burden. In our approach we
managed to go up to $n_{\mathrm{stout}}=4$ and to evaluate the generic result
for arbitrary $\varrho$ (using rather modest computational resources). This
illustrates that the more elegant approach implies not just aesthetic
pleasures but actual savings compared to the “brute force” approach. Still,
the bottom line is an affirmation of the finding of Ref. [37] that combining
tree-level clover improvement with stout smearing or gradient-flowing with a
cumulative flow time of the order $t/a^{2}\simeq 1$ yields an undoubled
fermion action with surprisingly good chiral properties (see the discussion at
the end of our Sec. 5).

A logical next step would be to work out the 1-loop values of
$c_{\mathrm{SW}}$ for Wilson and/or Brillouin fermions with a limited number
of stout steps (hopefully again up to four iterations) and/or a few values of
the flow time $t/a^{2}$, for instance $t/a^{2}\in\\{0.1,0.2,0.4,0.7,1.0\\}$.
This amounts to combining the work we did in Ref. [42] with the framework for
smeared/flowed Feynman rules presented in this paper, and we hope to present
such results in the not so distant future. In a further perspective, one might
also work out the improvement factors $Z_{S,P,V,A}$ to 1-loop (and perhaps
even 2-loop) order in stouted/flowed perturbation theory.

The alert reader might interject that a fully non-perturbative improvement
strategy is more desirable than a perturbative one, and the link smearing may
imply only mild technical complications in the non-perturbative approach.
However, as mentioned in the introduction and detailed in Ref. [34], the non-
perturbative approach builds, in practical terms, on existing perturbative
results to 1-loop (or even better: 2-loop) order, as such results provide
essential constraints in the final step of summarizing the results through a
rational function in $g_{0}^{2}$. To us this indicates that perturbative work
continues to be useful in lattice gauge theory.

##  Appendix A Clover fermion self-energy details

In this appendix we give more details for the self-energy of a Wilson fermion
(with and without a clover term) in the presence of stout smearing or gradient
flow.

We start with the tree-level improved case ($c_{\mathrm{SW}}=r=1$). The pieces
$\Sigma_{0}$ and $\Sigma_{10}$, see Eqns. (198) and (204), for up to four
stout steps are listed in Tab. 2 for a selection of $\varrho$ values. We refer
to Ref. [37] for an argument why choosing $\varrho$ larger than 0.125 should
be avoided. We believe that a spline interpolation will suffice for practical
purposes. In the event the reader requests higher precision, Tabs. 3 and 4
hold the necessary information to reconstruct $\Sigma_{0}$ and $\Sigma_{10}$
for arbitrary $\varrho$. The first respective lines (i.e. those where the
entry does not depend on $n$) hold the unsmeared values for comparison.

The results with Wilson flow are more cumbersome to communicate. Fig. 11 shows
$\Sigma_{0}$ and $\Sigma_{10}$ for plaquette and Lüscher-Weisz glue as a
function of the flow time in lattice units. It turns out that a two-
exponential fit represents the data with reasonable accuracy over the range
shown. The plots contain this information in the legends. Note that these
results reflect the tree-level choice $c_{\mathrm{SW}}=r=1$.

| $\varrho$ | $n=1$ | $n=2$ | $n=3$ | $n=4$
---|---|---|---|---|---
$\Sigma_{0}$ (plaq) | 0.05 | -13.67830(4) | -6.81841(4) | -3.79814(4) | -2.30(1)
0.09 | -5.89958(4) | -2.12071(5) | -1.00136(6) | -0.56(2)
0.11 | -4.29949(4) | -1.37888(5) | -0.62450(6) | -0.34(2)
0.12 | -4.07175(4) | -1.27800(6) | -0.57513(6) | -0.32(3)
0.125 | -4.10097(4) | -1.31229(6) | -0.60582(7) | -0.34(3)
0.13 | -4.22557(4) | -1.41647(6) | -0.69334(7) | -0.42(3)
$\Sigma_{0}$ (sym) | $0.05$ | -10.50283(3) | -5.36991(3) | -3.05827(4) | -1.89(4)
$0.09$ | -4.70711(3) | -1.76141(5 | -0.85802(8) | -0.49(6)
$0.11$ | -3.43187(3) | -1.15565(5) | -0.5431(2) | -0.31(7)
$0.12$ | -3.19992(3) | -1.04722(6) | -0.4871(2) | -0.28(8)
$0.125$ | -3.18535(3) | -1.05160(6) | -0.4966(2) | -0.29(8)
$0.13$ | -3.23840(3) | -1.10355(6) | -0.5435(3) | -0.33(9)
$\Sigma_{10}$ (plaq) | $0.05$ | 0.75710(4) | -2.32322(4) | -3.80405(4) | -4.599(7)
$0.09$ | -2.66716(4) | -4.65582(5) | -5.34360(5) | -5.66(2)
$0.11$ | -3.53426(4) | -5.11323(5) | -5.60877(5) | -5.83(2)
$0.12$ | -3.75656(4) | -5.23165(5) | -5.67941(6) | -5.88(2)
$0.125$ | -3.81490(4) | -5.26045(5) | -5.69546(6) | -5.89(2)
$0.13$ | -3.83802(4) | -5.26563(5) | ’-5.69366(7) | -5.88(3)
$\Sigma_{10}$ (sym) | $0.05$ | -0.625159(8) | -3.00765(1) | -4.18067(2) | -4.824(9)
$0.09$ | -3.25805(1) | -4.86118(2) | -5.43657(5) | -5.71(2)
$0.11$ | -3.95912(2) | -5.24318(2) | -5.66454(9) | -5.86(2)
$0.12$ | -4.15581(2) | -5.35110(2) | -5.7305(2) | -5.90(3)
$0.125$ | -4.21569(2) | -5.38307(2) | -5.7497(2) | -5.92(3)
$0.13$ | -4.24993(2) | -5.39832(2) | -5.7566(2) | -5.92(3)
Table 2: Values of $\Sigma_{0}$ and $\Sigma_{10}$ for some values of the
smearing parameter $\varrho$ and smearing steps $n\in[1,2,3,4]$ with plaquette
gauge action (plaq) and Lüscher-Weisz gauge action (sym). All information
refers to the choice $c_{\mathrm{SW}}=r=1$.

Sometimes the reader may choose another value of $c_{\mathrm{SW}}$, e.g. no
improvement at all ($c_{\mathrm{SW}}=0$). Since (to one loop order)
$\Sigma_{0}$ and $\Sigma_{10}$ are quadratic in $c_{\mathrm{SW}}$, we split
them according to

$\displaystyle\Sigma_{0}$
$\displaystyle=\Sigma_{0}^{(0)}+c_{\mathrm{SW}}\Sigma_{0}^{(1)}+c_{\mathrm{SW}}^{2}\Sigma_{0}^{(2)}$
(206) $\displaystyle\Sigma_{10}$
$\displaystyle=\Sigma_{10}^{(0)}+c_{\mathrm{SW}}\Sigma_{10}^{(1)}+c_{\mathrm{SW}}^{2}\Sigma_{10}^{(2)}\;.$
(207)

and give $\Sigma_{0}^{(0)}$, $\Sigma_{0}^{(1)}$ and $\Sigma_{0}^{(2)}$ in Tab.
5 and $\Sigma_{10}^{(0)}$, $\Sigma_{10}^{(1)}$ and $\Sigma_{10}^{(2)}$ in Tab.
6 for the same selection of $\varrho$ values as before (the parameter in the
Wilson term is always $r=1$). Again, we expect that a simple spline
interpolation will suffice to obtain useful numbers for a different choice of
$\varrho$. In the event this expectation is wrong, Tabs. 7 through 12 hold the
necessary information to reconstruct $\Sigma_{0}$ and $\Sigma_{10}$ for
arbitrary $\varrho$.

The results with Wilson flow are represented in Figs. 12 and 13 in the same
style as before. Again a two-exponential fit is found to reproduce the data
well over the range shown.

Physics wise, the most impressive lesson to be learned from this appendix
comes from comparing the top panels of Fig. 11 to Fig. 12. With tree-level
improvement (Fig. 11) the additive mass shift (which is proportional to
$\Sigma_{0}$) approaches zero quite quickly with increasing flow time. For
unimproved clover fermions this is not the case (top row in Fig. 12); in fact
the nice feature seen in Fig. 11 stems from a cancellation between
$\Sigma_{0}^{(0)}$, $\Sigma_{0}^{(1)}$ and $\Sigma_{0}^{(2)}$ of Fig. 12.

$\Sigma_{0}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq -31.98644(3) -31.98644(3)
-31.98644(3) -31.98644(3) sym -23.83234(3) -23.83234(3) -23.83234(3)
-23.83234(3) $\varrho^{1}$ plaq 461.5488(1) 923.0975(2) 1384.645(4) 1846.19(6)
sym 334.1999(1) 668.3997(1) 1002.5996(2) 1336.8(6) $\varrho^{2}$ plaq
-1907.719658(3) -11446.31795(2) -28615.795(2) -53416.150(3) sym -1352.19157(4)
-8113.149(2) -20282.874(1) -37861.370(2) $\varrho^{3}$ plaq 69587.70312(7)
347938.5(8) 974228(2) sym 48476.67898(2) 242383.40(2) 678673.4(2)
$\varrho^{4}$ plaq -171121.8788(3) -2566828(2) -11978531(7) sym
-117482.6334(5) -1762239.50(1) -8223785(5) $\varrho^{5}$ plaq 10728080(5)
100128744(41) sym 7273681(5) 67887689(29) $\varrho^{6}$ plaq -19627720(22)
-549575266(290) sym -13163619.12(3) -368581513(347) $\varrho^{7}$ plaq
1795933838(5106) sym 1193026157(2578) $\varrho^{8}$ plaq -2658101497(19216)
sym -1750940862(13091)

Table 3: Coefficients of the polynomials in the smearing parameter $\varrho$
for $\Sigma_{0}$ for smearing steps $n\in[1,2,3,4]$ with plaquette gauge
action (plaq) and Lüscher-Weisz gauge action (sym).

$\Sigma_{10}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq 8.20627(3) 8.20627(3) 8.20627(3)
8.20627(3) sym 4.973633(5) 4.973633(5) 4.973633(5) 4.973633(5) $\varrho^{1}$
plaq -184.19270(6) -368.3854(2) -552.58(5) -736.77(7)) sym -137.61668(5)
-275.2334(1) -412.85002(9) -550.5(2) $\varrho^{2}$ plaq 704.18682(5)
4225.1209(3) 10562.8(2) 19717.2(3) sym 512.81660(2) 3076.8996(2) 7692.2490(3)
14358.9(2) $\varrho^{3}$ plaq -24227.54502(9) -121137.7(5) -339186(1) sym
-17289.6263(2) -86448.13(4) -242054.9(4) $\varrho^{4}$ plaq 56866.2480(2)
852993.70(3) 3980637(6) sym 39893.8762(1) 598408.143(2) 2792571.48(4)
$\varrho^{5}$ plaq -3431931(1) -32031358(9) sym -2372221(1) -22140728(2)
$\varrho^{6}$ plaq 6083759(7) 170345240(173) sym 4150857.326(4) 116224020(336)
$\varrho^{7}$ plaq -542124208(1774) sym -365651055(67) $\varrho^{8}$ plaq
784643279(12269) sym 523847362(2021)

Table 4: Coefficients of the polynomials in the smearing parameter $\varrho$
for $\Sigma_{10}$ for smearing steps $n\in[1,2,3,4]$ with plaquette gauge
action (plaq) and Lüscher-Weisz gauge action (sym).

Figure 11: Double exponential fits of the form $c_{0}+c_{1}e^{-c_{2}t}+c_{3}e^{-c_{4}t}$ of $\Sigma_{0}$ (top) and $\Sigma_{10}$ (bottom) with Wilson flow for different values of $r$ (with $c_{\mathrm{SW}}=r$) plotted on a logarithmic $y$-axis. The fit coeffcients $[c_{0},c_{1},c_{2},c_{3},c_{4}]$ are given in brackets.  | $\varrho$ | $n=1$ | $n=2$ | $n=3$ | $n=4$
---|---|---|---|---|---
$\Sigma_{0}^{(0)}$ (plaq) | 0.05 | -26.65808(5) | -16.28605(5) | -11.14760(7) | -8.268(9)
0.09 | -15.24425(5) | -8.17389(7) | -5.4550(2) | -4.08(2)
0.11 | -12.33982(5) | -6.44793(8) | -4.3213(2) | -3.26(2)
0.12 | -11.58822(6) | -5.96517(9) | -3.9821(2) | -3.00(2)
0.125 | -11.38758(6) | -5.83167(9) | -3.8853(3) | -2.92(2)
0.13 | -11.30371(6) | -5.7827(1) | -3.8560(3) | -2.90(2)
$\Sigma_{0}^{(0)}$ (sym) | 0.05 | -21.88153(4) | -13.84082(5) | -9.73971(9) | -7.38(4)
0.09 | -13.10032(4) | -7.34394(7) | -5.0335(2) | -3.83(6)
0.11 | -10.73243(5) | -5.88197(8) | -4.0428(3) | -3.09(7)
0.12 | -10.05417(5) | -5.43642(9) | -3.7247(4) | -2.85(8)
0.125 | -9.84145(5) | -5.29131(9) | -3.6179(4) | -2.76(8)
0.13 | -9.71302(5) | -5.2058(1) | -3.5577(5) | -2.72(8)
$\Sigma_{0}^{(1)}$ (plaq) | 0.05 | 9.7541(2) | 7.4558(4) | 5.9965(6) | 5.001(4)
0.09 | 7.4250(4) | 5.0954(7) | 3.882(1) | 3.138(6)
0.11 | 6.5452(4) | 4.3673(8) | 3.288(2) | 2.641(8)
0.12 | 6.1764(5) | 4.0752(9) | 3.054(2) | 2.45(1)
0.125 | 6.0098(5) | 3.9442(9) | 2.950(2) | 2.36(2)
0.13 | 5.8551(5) | 3.8227(9) | 2.853(2) | 2.28(2)
$\Sigma_{0}^{(1)}$ (sym) | 0.05 | 8.6852(2) | 6.7573(3) | 5.5094(4) | 4.644(2)
0.09 | 6.7474(3) | 4.7368(5) | 3.6600(7) | 2.987(3)
0.11 | 6.0027(3) | 4.0976(6) | 3.1260(9) | 2.532(4)
0.12 | 5.6865(4) | 3.8381(7) | 2.913(1) | 2.353(5)
0.125 | 5.5423(4) | 3.7209(7) | 2.818(1) | 2.273(5)
0.13 | 5.4075(4) | 3.6115(7) | 2.729(1) | 2.198(6)
$\Sigma_{0}^{(2)}$ (plaq) | 0.05 | 3.2256(3) | 2.0118(3) | 1.3529(3) | 0.9637(9)
0.09 | 1.9196(3) | 0.9578(3) | 0.5719(3) | 0.380(2)
0.11 | 1.4952(3) | 0.7018(3) | 0.4088(3) | 0.267(3)
0.12 | 1.3400(3) | 0.6120(3) | 0.3527(3) | 0.229(4)
0.125 | 1.2768(3) | 0.5751(3) | 0.3296(3) | 0.213(5)
0.13 | 1.2230(3) | 0.5436(3) | 0.3097(3) | 0.199(5)
$\Sigma_{0}^{(2)}$ (sym) | 0.05 | 2.6935(2) | 1.7136(2) | 1.1721(2) | 0.8470(3)
0.09 | 1.6458(2) | 0.8457(2) | 0.5155(2) | 0.348(2)
0.11 | 1.2978(2) | 0.6287(2) | 0.3736(2) | 0.248(3)
0.12 | 1.1678(2) | 0.5511(2) | 0.3240(2) | 0.214(5)
0.125 | 1.1138(2) | 0.5188(2) | 0.3034(2) | 0.200(6)
0.13 | 1.0671(2) | 0.4907(2) | 0.2853(2) | 0.187(7)
Table 5: Same as Tab. 2 but for $\Sigma_{0}^{(0)}$, $\Sigma_{0}^{(1)}$ and $\Sigma_{0}^{(2)}$ from Equation (206).  | $\varrho$ | $n=1$ | $n=2$ | $n=3$ | $n=4$
---|---|---|---|---|---
$\Sigma_{10}^{(0)}$ (plaq) | 0.05 | 3.60121(2) | -0.01039(3) | -1.86437(3) | -2.934(4)
0.09 | -0.33688(3) | -2.95019(5) | -4.00186(5) | -4.551(6)
0.11 | -1.41830(3) | -3.61776(5) | -4.45331(7) | -4.886(7)
0.12 | -1.73711(3) | -3.82460(6) | -4.59974(9) | -4.999(9)
0.125 | -1.84104(3) | -3.89407(6) | -4.65020(9) | -5.039(9)
0.13 | -1.90798(3) | -3.93775(7) | -4.6807(1) | -5.06(1)
$\Sigma_{10}^{(0)}$ (sym) | 0.05 | 1.956497(6) | -0.882122(8) | -2.38041(2) | -3.266(8)
0.09 | -1.114339(7) | -3.26513(2) | -4.16661(3) | -4.65(2)
0.11 | -2.000100(8) | -3.83374(2) | -4.56328(5) | -4.95(2)
0.12 | -2.280566(8) | -4.02084(2) | -4.69847(6) | -5.06(2)
0.125 | -2.380196(9) | -4.08938(2) | -4.74919(7) | -5.10(2)
0.13 | -2.452757(9) | -4.13934(2) | -4.78580(8) | -5.13(2)
$\Sigma_{10}^{(0)}$ (plaq) | 0.05 | 3.60121(2) | -0.01039(3) | -1.86437(3) | -2.934(4)
0.09 | -0.33688(3) | -2.95019(5) | -4.00186(5) | -4.551(6)
0.11 | -1.41830(3) | -3.61776(5) | -4.45331(7) | -4.886(7)
0.12 | -1.73711(3) | -3.82460(6) | -4.59974(9) | -4.999(9)
0.125 | -1.84104(3) | -3.89407(6) | -4.65020(9) | -5.039(9)
0.13 | -1.90798(3) | -3.93775(7) | -4.6807(1) | -5.06(1)
$\Sigma_{10}^{(0)}$ (sym) | 0.05 | 1.956497(6) | -0.882122(8) | -2.38041(2) | -3.266(8)
0.09 | -1.114339(7) | -3.26513(2) | -4.16661(3) | -4.65(2)
0.11 | -2.000100(8) | -3.83374(2) | -4.56328(5) | -4.95(2)
0.12 | -2.280566(8) | -4.02084(2) | -4.69847(6) | -5.06(2)
0.125 | -2.380196(9) | -4.08938(2) | -4.74919(7) | -5.10(2)
0.13 | -2.452757(9) | -4.13934(2) | -4.78580(8) | -5.13(2)
$\Sigma_{10}^{(2)}$ (plaq) | 0.05 | -1.09264(2) | -0.88380(2) | -0.73530(2) | -0.6262(7)
0.09 | -0.89206(2) | -0.64168(2) | -0.49704(2) | -0.404(2)
0.11 | -0.80614(2) | -0.55755(2) | -0.42364(2) | -0.341(2)
0.12 | -0.76678(2) | -0.52223(2) | -0.39401(2) | -0.316(3)
0.125 | -0.74800(2) | -0.50598(2) | -0.38060(3) | -0.305(3)
0.13 | -0.72981(2) | -0.49059(2) | -0.36803(3) | -0.295(3)
$\Sigma_{10}^{(2)}$ (sym) | 0.05 | -0.986186(4) | -0.807676(5) | -0.678888(6) | -0.5831(4)
0.09 | -0.815953(5) | -0.597649(7) | -0.468699(8) | -0.3850(8)
0.11 | -0.742363(6) | -0.523351(7) | -0.40253(1) | -0.327(2)
0.12 | -0.708450(6) | -0.491894(8) | -0.37558(1) | -0.304(2)
0.125 | -0.692213(6) | -0.477374(8) | -0.36334(1) | -0.294(2)
0.13 | -0.676458(6) | -0.463584(8) | -0.35183(2) | -0.284(2)
Table 6: Same as Tab. 2 but for $\Sigma_{10}^{(0)}$, $\Sigma_{10}^{(1)}$ and
$\Sigma_{10}^{(2)}$ from Equation (207).

$\Sigma^{(0)}_{0}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq -51.43471(4) -51.43471(4)
-51.43471(4) -51.435(8) sym -40.44323(3) -40.44323(3) -40.44323(3) -40.443(7)
$\varrho^{1}$ plaq 612.3029(1) 1224.6058(2) 1836.9087(3) 2449.21(3) sym
455.5139(1) 911.0278(2) 1366.5417(9) 1822.1(6) $\varrho^{2}$ plaq
-2335.4071(3) -14012.442(2) -35031.106(4) -65391.6(2) sym -1685.5973(3)
-10113.584(3) -25283.960(4) -47196.889(4) $\varrho^{3}$ plaq 81277.32696(7)
406386.6344(6) 1137883.43(5) sym 57399.24192(3) 286996.21(2) 803589.90(8)
$\varrho^{4}$ plaq -193630.40(3) -2904456.0(4) -13554132(2) sym -134388.50(2)
-2015827.5(3) -9407196.9(4) $\varrho^{5}$ plaq 11863189.778(7) 110723094(4)
sym 8114869(5) 75738794(14) $\varrho^{6}$ plaq -21329904.75(2) -597236493(336)
sym -14410242.41(3) -403487007(319) $\varrho^{7}$ plaq 1925422531(5277) sym
1286849516(2631) $\varrho^{8}$ plaq -2819282113(19240) sym -1866581334(13091)

Table 7: Same as Tab. 3 but for $\Sigma^{(0)}_{0}$ from Eq. (206).

$\Sigma^{(1)}_{0}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq 13.73313(4) 13.73313(4) 13.73313(4)
13.732(2) sym 11.94822(3) 11.94822(3) 11.94822(3) 11.947(2) $\varrho^{1}$ plaq
-91.442(4) -182.884(7) -274.33(1) -365.76(3) sym -74.603(3) -149.206(5)
-223.808(8) -298.403(4) $\varrho^{2}$ plaq 237.247540(3) 1423.48523(3)
3558.71310(5) 6642.94(9) sym 186.845029(2) 1121.070182(3) 2802.6754(3)
5231.66(3) $\varrho^{3}$ plaq -6094.07640(6) -30470.3820(4) -85317(2) sym
-4689.94425(4) -23449.721432(8) -65659.2(8) $\varrho^{4}$ plaq 11191.09479(3)
167866.4219(4) 783374(2) sym 8465.23735(4) 126978.5602(6) 592567(2)
$\varrho^{5}$ plaq -543236.523(8) -5070208(26) sym -405150.278(4)
-3781420.66(5) $\varrho^{6}$ plaq 789161.8462(9) 22096532(104) sym
581383.242(7) 16278817(190) $\varrho^{7}$ plaq -58433357(1223) sym
-42575391.8(3) $\varrho^{8}$ plaq 71058490(163) sym 51248188(2163)

Table 8: Same as Tab. 3 but for $\Sigma^{(1)}_{0}$ from Eq. (206).

$\Sigma^{(2)}_{0}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq 5.7151(3) 5.7151(3) 5.7151(3)
5.7150(5) sym 4.6627(2) 4.6627(2) 4.6627(2) 4.66255(5) $\varrho^{1}$ plaq
-59.3118850(8) -118.623770(2) -177.935654(4) -237.248(4) sym -46.7112574(3)
-93.4225147(6) -140.1337727(3) -186.845(1) $\varrho^{2}$ plaq 190.439888(2)
1142.639327(9) 2856.59832(3) 5332.32(7) sym 146.560758(2) 879.36454(1)
2198.41137(2) 4103.70(5) $\varrho^{3}$ plaq -5595.54740(2) -27977.73695(9)
-78337.66(7) sym -4232.61867(2) -21163.09330(3) -59256.7(2) $\varrho^{4}$ plaq
11317.4275(3) 169761.412(4) 792220(5) sym 8440.6307(2) 126609.461(2)
590846.979(8) $\varrho^{5}$ plaq -591871.37(2) -5524133(26) sym -436037.438(2)
-4069704(48) $\varrho^{6}$ plaq 913021.209(7) 25564594(2) sym 665240.039(6)
18626734(383) $\varrho^{7}$ plaq -71058490(163) sym -51248188(2163)
$\varrho^{8}$ plaq 90125153(5113) sym 64387254(3655)

Table 9: Same as Tab. 3 but for $\Sigma^{(2)}_{0}$ from Eq. (206).

$\Sigma^{(0)}_{10}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq 11.852396(8) 11.852396(8)
11.852396(8) 11.852(2) sym 8.231254(4) 8.231254(4) 8.231254(4) 8.231(2)
$\varrho^{1}$ plaq -202.0079(2) -404.0158(3) -606.0237(2) -808.03(4) sym
-152.56416(3) -305.12832(6) -457.6925(2) -610.3(2) $\varrho^{2}$ plaq
739.6836(2) 4438.102(1) 11095.255(3) 20711.14(3) sym 541.38051(4) 3248.2831(3)
8120.7077(3) 15158.66(4) $\varrho^{3}$ plaq -24956.6947(4) -124783.4729(5)
-349393.7(5) sym -17858.09191(3) -89290.4593(2) -250013.3(2) $\varrho^{4}$
plaq 57973.619(4) 869604.29(5) 4058153.0(5) sym 40735.013(3) 611025.19(5)
2851450.82(9) $\varrho^{5}$ plaq -3477653.965(2) -32458091(8) sym -2406174(1)
-22457620(2) $\varrho^{6}$ plaq 6141663(3) 171966439(99) sym 4192996.284(4)
117403911(55) $\varrho^{7}$ plaq -545947087(1804) sym -368383101(142)
$\varrho^{8}$ plaq 788877857(5503) sym 526824035(1817)

Table 10: Same as Tab. 4 but for $\Sigma^{(0)}_{10}$ from Eq. (207).

$\Sigma^{(1)}_{10}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq -2.248869(6) -2.248869(6)
-2.248869(6) -2.249(2) sym -2.015426(5) -2.015426(5) -2.015426(5) -2.01543(5)
$\varrho^{1}$ plaq 11.123836(8) 22.24767(2) 33.37151(3) 44.497(3) sym
9.346879(5) 18.693758(9) 28.040642(8) 37.388(3) $\varrho^{2}$ plaq
-23.518665(6) -141.11198(3) -352.77996(7) -658.52(8) sym -18.958350(2)
-113.75010(2) -284.37525(2) -530.83(5) $\varrho^{3}$ plaq 525.716776(9)
2628.58386(7) 7360.035(6) sym 411.66550(4) 2058.3275(2) 5763.32(2)
$\varrho^{4}$ plaq -878.20974(7) -13173.1458(7) -61475(3) sym -672.98584(5)
-10094.788(2) -47107.302(7) $\varrho^{5}$ plaq 39917.11(2) 372559.6(2) sym
30072.058(6) 280672.57(9) $\varrho^{6}$ plaq -55331.94(5) -1549295(216) sym
-41099.241(6) -1150778.8(7) $\varrho^{7}$ plaq 3959099.6(6) sym 2904907(3)
$\varrho^{8}$ plaq -4693170(2) sym -3406008(4)

Table 11: Same as Tab. 4 but for $\Sigma^{(1)}_{10}$ from Eq. (207).

$\Sigma^{(2)}_{10}$
$n=1$ $n=2$ $n=3$ $n=4$ $\varrho^{0}$ plaq -1.397261(9) -1.397261(9)
-1.397261(9) -1.3973(3) sym -1.242203(4) -1.242203(4) -1.242203(4) -1.24223(9)
$\varrho^{1}$ plaq 6.69138(4) 13.38277(7) 20.07415(9) 26.762(8) sym 5.60061(2)
11.20122(4) 16.80183(6) 22.399(5) $\varrho^{2}$ plaq -11.978147(2)
-71.868869(6) -179.67220(2) -335.388(8) sym -9.6055598(7) -57.633356(1)
-144.083397(9) -268.980(2) $\varrho^{3}$ plaq 203.43279(2) 1017.16394(6)
2848.1(5) sym 156.799846(8) 783.99926(3) 2195.2(4) $\varrho^{4}$ plaq
-229.16130(5) -3437.4195(4) -16041.293(2) sym -168.15048(2) -2522.2572(4)
-11770.532(2) $\varrho^{5}$ plaq 5804.3207(6) 54173.62(4) sym 3880.9479(9)
36222.19(2) $\varrho^{6}$ plaq -2569.25(2) -71939.0(4) sym -1039.729(6)
-29112.3(3) $\varrho^{7}$ plaq -136222(83) sym -172862(90) $\varrho^{8}$ plaq
458914(149) sym 428921(211)

Table 12: Same as Tab. 4 but for $\Sigma^{(2)}_{10}$ from Eq. (207).

Figure 12: Double exponential fits of the form
$c_{0}+c_{1}e^{-c_{2}t}+c_{3}e^{-c_{4}t}$ of $\Sigma^{(0)}_{0}$,
$\Sigma^{(1)}_{0}$, and $\Sigma^{(2)}_{0}$ (top to bottom) with Wilson flow
for different values of $r$ plotted on a logarithmic $y$-axis. The fit
coeffcients $[c_{0},c_{1},c_{2},c_{3},c_{4}]$ are given in brackets.

Figure 13: Same as Fig. 12 but for $\Sigma^{(0)}_{10}$, $\Sigma^{(1)}_{10}$,
and $\Sigma^{(2)}_{10}$ (top to bottom).

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