#  Left invariant Riemannian metrics with harmonic curvature are Ricci-
parallel in solvable Lie groups and Lie groups of dimension $\leq 6$

Ilyes Aberaouze, Mohamed Boucetta  Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakech Maroc
e-mail<EMAIL_ADDRESS>Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakech Maroc
e-mail<EMAIL_ADDRESS>
###### Abstract

We show that any left invariant metric with harmonic curvature on a solvable
Lie group is Ricci-parallel. We show the same result for any Lie group of
dimension $\leq$ 6.

###### keywords:

Left invariant metric , Lie groups , Harmonic curvature ,

###### MSC:

22E15 ,

###### MSC:

53C20 ,

###### MSC:

22E25

##  1 Introduction

A Riemannian manifold $(M,\langle\;,\;\rangle)$ is said to have harmonic
curvature if its curvature tensor $R$ has a vanishing divergence, i.e., for
any $X\in\Gamma(TM)$,

$\sum_{i=1}^{n}\nabla_{E_{i}}(R)(E_{i},X)=0,$

where $\nabla$ is the Levi-Civita connection and $(E_{1},\ldots,E_{n})$ is a
local $\langle\;,\;\rangle$-orthonormal frame field. The study of Riemannian
manifolds with harmonic curvature is old and goes to [10] and it is known [3,
4, 5, 8] that a Riemannian manifold $(M,\langle\;,\;\rangle)$ has a harmonic
curvature if and only if its Ricci tensor $\mathrm{Ric}$ is Codazzi, i.e.,

$\nabla_{X}(\mathrm{Ric})(Y,Z)=\nabla_{Y}(\mathrm{Ric})(X,Z)$ (1)

for any vector fields $X,Y,Z$. Obviously, any Einstein Riemannian metric
$(\mathrm{Ric}=\lambda\langle\;,\;\rangle)$ and more generally any Ricci-
parallel metric $(\nabla(\mathrm{Ric})=0)$ satisfy (1). Thus the interesting
case will be where $\mathrm{Ric}$ satisfies (1) and
$\nabla(\mathrm{Ric})\not=0$. Unfortunately, such metrics are difficult to
find and some examples were given in [2, 4, 5, 8]. To our knowledge, there is
no example of a non Ricci-parallel homogeneous Riemannian manifold with
harmonic curvature which supports the following conjecture.

######  Conjecture 1.

Any homogeneous Riemannian manifold $M$ with harmonic curvature is Ricci-
parallel.

This conjecture is true in dimension four (see [7]) and when $M$ is a sphere
or a projective space (see [12]). It was proven in [1] for nilpotent Lie
groups with left invariant metrics. Moreover, any conformally flat Riemannian
manifold with constant scalar curvature satisfies (1) (see [8, Theorem 5.1])
and any homogeneous conformally flat Riemannian manifold has Ricci-parallel
curvature (see [13]). The purpose of this paper is to show that this
conjecture is true when $M$ is a solvable Lie group endowed with a left
invariant metric which implies that the conjecture is true for any homogeneous
Riemannian manifold $G/K$ where $K$ is a maximal compact subgroup of $G$. We
show also that the conjecture is true for any Lie group of dimension $\leq 6$
endowed with a left invariant metric.

The paper is organized as follows. In Section 2, we recall some basic
properties of the Lie algebra associated to a Lie group endowed with a left
invariant metric with harmonic curvature. In Section 3, we prove Conjecture 1
for a solvable Lie group. This is based on a generalization of the main result
of Lauret in [9]. In Section 4, we prove Conjecture 1 for any Lie group of
dimension $\leq 6$. The proof for the dimensions $\leq 5$ is quite easy but,
for dimension 6, it involves a huge computation where the use of Maple is
needed.

##  2 Preliminaries

In this section, we establish the basic tools needed for the proof of our main
results. The results of this section were first proved in [1].

The study of left invariant Riemannian metrics on Lie groups reduces to the
study of Lie algebras endowed with a Euclidean product. Let
$(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a Euclidean Lie algebra, i.e.,
a Lie algebra endowed with a bilinear nondegenerate form which is positive
definite. The _Levi-Civita product_ of ${\mathfrak{g}}$ is the bilinear map
$\mathrm{L}:{\mathfrak{g}}\times{\mathfrak{g}}\longrightarrow{\mathfrak{g}}$
given by Koszul’s formula

$\displaystyle 2\langle\mathrm{L}_{u}v,w\rangle$ $\displaystyle=$
$\displaystyle\langle[u,v],w\rangle+\langle[w,u],v\rangle+\langle[w,v],u\rangle\quad\mbox{for
any}\;u,v,w\in\mathfrak{g}.$ (2)

For any $u,v\in{\mathfrak{g}}$,
$\mathrm{L}_{u}:{\mathfrak{g}}\longrightarrow{\mathfrak{g}}$ is skew-symmetric
and $[u,v]=\mathrm{L}_{u}v-\mathrm{L}_{v}u$. The curvature of ${\mathfrak{g}}$
is given by

$K(u,v)=\mathrm{L}_{[u,v]}-[\mathrm{L}_{u},\mathrm{L}_{v}]\quad\mbox{for
any}\;u,v,\in\mathfrak{g}.$

The Ricci curvature
$\mathrm{ric}:{\mathfrak{g}}\times{\mathfrak{g}}\longrightarrow\hbox{\bb R}$
and its Ricci operator
$\mathrm{Ric}:{\mathfrak{g}}\longrightarrow{\mathfrak{g}}$ are defined by

$\langle\mathrm{Ric}(u),v\rangle=\mathrm{ric}(u,v)=\mathrm{tr}\left(w\longrightarrow
K(u,w)v\right)\quad\mbox{for any}\;u,v,w\in\mathfrak{g}.$

The operator $\mathrm{Ric}$ is symmetric and hence is diagonalizable. Thus
there exists $\lambda_{1}<\ldots<\lambda_{r}$ such that

$\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$

where $\mathfrak{g}_{i}$ is the eigenspace of $\mathrm{Ric}$ associated to the
eigenvalues $\lambda_{i}$.

We call $(\mathfrak{g},\langle\;,\;\rangle)$ a Euclidean Lie algebra with
harmonic curvature if $\mathrm{Ric}$ is a Codazzi operator, i.e., for any
$u,v,w\in\mathfrak{g}$,

$\langle\mathrm{L}_{u}v,\mathrm{Ric}(w)\rangle+\langle\mathrm{L}_{u}w,\mathrm{Ric}(v)\rangle=\langle\mathrm{L}_{v}u,\mathrm{Ric}(w)\rangle+\langle\mathrm{L}_{v}w,\mathrm{Ric}(u)\rangle.$
(3)

If we take $u\in\mathfrak{g}_{i}$, $v\in\mathfrak{g}_{j}$ and
$w\in\mathfrak{g}_{k}$, we get

$(2\lambda_{k}-\lambda_{i}-\lambda_{j})\langle[u,v],w\rangle=(\lambda_{i}-\lambda_{j})(\langle[v,w],u\rangle+\langle[u,w],v\rangle).$
(4)

For $i=j\not=k$, we get that $\mathfrak{g}_{i}$ is a subalgebra and if we take
$k=i\not=j$, we get

$\langle[v,u],w\rangle+\langle[v,w],u\rangle=0\quad
u,w\in\mathfrak{g}_{i},v\in\mathfrak{g}_{j},i\not=j.$

For any $i<j<k$ and
$u_{i}\in\mathfrak{g}_{i},u_{j}\in\mathfrak{g}_{j},u_{k}\in\mathfrak{g}_{k}$,
we have

$\begin{cases}(2\lambda_{i}-\lambda_{j}-\lambda_{k})\langle[u_{j},u_{k}],u_{i}\rangle-(\lambda_{j}-\lambda_{k})(\langle[u_{k},u_{i}],u_{j}\rangle+\langle[u_{j},u_{i}],u_{k}\rangle)=0,\quad(E1)\\\
(2\lambda_{j}-\lambda_{i}-\lambda_{k})\langle[u_{i},u_{k}],u_{j}\rangle-(\lambda_{i}-\lambda_{k})(\langle[u_{k},u_{j}],u_{i}\rangle+\langle[u_{i},u_{j}],u_{k}\rangle)=0,\quad(E2)\\\
(2\lambda_{k}-\lambda_{i}-\lambda_{j})\langle[u_{i},u_{j}],u_{k}\rangle-(\lambda_{i}-\lambda_{j})(\langle[u_{j},u_{k}],u_{i}\rangle+\langle[u_{i},u_{k}],u_{j}\rangle)=0.\quad(E3)\end{cases}$
(5)

We have

$(E1)-(E2)+(E3)=0$

and if we put

$x=\langle[u_{k},u_{j}],u_{i}\rangle,\;y=\langle[u_{k},u_{i}],u_{j}\rangle\quad\mbox{and}\quad
z=\langle[u_{i},u_{j}],u_{k}\rangle,$

then the system (5) becomes

$\begin{cases}(\lambda_{j}+\lambda_{k}-2\lambda_{i})x+(\lambda_{k}-\lambda_{j})y=(\lambda_{k}-\lambda_{j})z,\\\
(\lambda_{k}-\lambda_{i})x+(\lambda_{i}+\lambda_{k}-2\lambda_{j})y=(\lambda_{i}-\lambda_{k})z.\end{cases}$

So

$x=-{\frac{\left(\lambda_{{j}}-\lambda_{{k}}\right)^{2}z}{\left(\lambda_{{i}}-\lambda_{{j}}\right)^{2}}}\quad\mbox{and}\quad
y={\frac{\left(\lambda_{{i}}-\lambda_{{k}}\right)^{2}z}{\left(\lambda_{{i}}-\lambda_{{j}}\right)^{2}}}.$

Thus

$(\lambda_{i}-\lambda_{j})^{2}\langle[u_{j},u_{k}],u_{i}\rangle+\left(\lambda_{{j}}-\lambda_{{k}}\right)^{2}\langle[u_{j},u_{i}],u_{k}\rangle=0\quad\mbox{and}\quad\left(\lambda_{{i}}-\lambda_{{j}}\right)^{2}\langle[u_{i},u_{k}],u_{j}\rangle+\left(\lambda_{{i}}-\lambda_{{k}}\right)^{2}\langle[u_{i},u_{j}],u_{k}\rangle=0.$

So far, we proved the following theorem.

######  Theorem 2.1.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a Euclidean Lie algebra,
$\lambda_{1}<\ldots<\lambda_{r}$ the eigenvalues of $\;\mathrm{Ric}$ and
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$ is the
splitting of $\mathfrak{g}$ on eigenspaces of $\mathrm{Ric}$. Then
$\mathrm{Ric}$ is a Codazzi tensor if and only if:

  1. 1.

for any $i\in\\{1,\ldots,r\\}$, $\mathfrak{g}_{i}$ is a subalgebra of
$\mathfrak{g}$,

  2. 2.

for any $i,j\in\\{1,\ldots,r\\}$ with $i\not=j$ and for any
$u,v\in\mathfrak{g}_{i}$ and $w\in\mathfrak{g}_{j}$ then

$\langle[w,u],v\rangle+\langle[w,v],u\rangle=0,$ (6)

  3. 3.

for any $i<j<k$ for any
$u\in\mathfrak{g}_{i},v\in\mathfrak{g}_{j},w\in\mathfrak{g}_{k}$ then

$\left(\lambda_{i}-\lambda_{j}\right)^{2}\langle[v,w],u\rangle+\left(\lambda_{j}-\lambda_{k}\right)^{2}\langle[v,u],w\rangle=0\quad\mbox{and}\quad\left(\lambda_{i}-\lambda_{j}\right)^{2}\langle[u,w],v\rangle+\left(\lambda_{i}-\lambda_{k}\right)^{2}\langle[u,v],w\rangle=0.$
(7)

In this case, $\mathrm{Ric}$ is non parallel if and only if there exists a
triple $(i,j,k)$ with $i<j<k$ and $u_{0}\in\mathfrak{g}_{i}$,
$v_{0}\in\mathfrak{g}_{j}$ and $w_{0}\in\mathfrak{g}_{k}$ such that
$\langle[u_{0},v_{0}],w_{0}\rangle\not=0$.

Let us give an interpretation of assertion 3. in the theorem. For any
$u\in\mathfrak{g}$, denote by $u_{i}$ its component in $\mathfrak{g}_{i}$. Fix
an index $k$ and define a new product $\langle\;,\;\rangle_{k}$ on
$\mathfrak{g}$ by

$\langle
u,v\rangle_{k}=\sum_{j=1}^{r}(\delta_{jk}+(\lambda_{j}-\lambda_{k})^{2})\langle
u_{j},v_{j}\rangle,\quad u,v\in\mathfrak{g}.$

Denote by $\mathfrak{g}_{k}^{\perp}$ the orthogonal of $\mathfrak{g}_{k}$. It
is easy to check that, for any $u_{k}\in\mathfrak{g}_{k}$ and for any
$u,v\in\mathfrak{g}$,

$\langle[u_{k},v],w\rangle_{k}+\langle[u_{k},w],v\rangle_{k}=0$ (8)

and

$\rho_{k}:\mathfrak{g}_{k}\longrightarrow\mathrm{so}(\mathfrak{g}_{k}^{\perp},\langle\;,\;\rangle_{k})\quad\mbox{where}\quad\rho_{k}(u_{k})(v)=\sum_{l\not=k}[u_{k},v]_{l},$
(9)

is a representation. The proof of the following proposition is given in [1].

######  Proposition 2.1.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a Euclidean Lie algebra
whose Ricci operator is Codazzi. With the notations above, for any
$i=1,\ldots,r$, let $\\{V_{i,\alpha}\\}$ be an orthonormal basis of
$\mathfrak{g}_{i}$. Then, for any $i=1,\ldots,r$, $u_{i}\in\mathfrak{g}_{i}$,

${\mathrm{ric}}(u_{i},u_{i})={\mathrm{ric}}_{i}(u_{i},u_{i})-\sum_{j\not=k}\sum_{\alpha,\beta}\frac{(\lambda_{k}-\lambda_{i})(\lambda_{j}-\lambda_{i})}{(\lambda_{k}-\lambda_{j})^{2}}\langle[V_{j,\alpha},V_{k,\beta}],u_{i}\rangle^{2},$
(10)

where ${\mathrm{ric}}_{i}$ is the Ricci curvature of $\mathfrak{g}_{i}$
endowed with the restriction of $\langle\;,\;\rangle$. Moreover, the scalar
curvature ${\mathfrak{s}}$ of $\mathfrak{g}$ satisfies

${\mathfrak{s}}={\mathfrak{s}}_{1}+\ldots+{\mathfrak{s}}_{r},$ (11)

where ${\mathfrak{s}}_{i}$ is the scalar curvature of $\mathfrak{g}_{i}$.

##  3 Left invariant Riemannian metrics with harmonic curvature on solvable
Lie groups are Ricci-parallel

The proof of our main result is based on the following theorem which is a
generalization of the famous theorem of Lauret [9] which shows that any
solvable Einstein Euclidean Lie algebra is standard in the sense that the
orthogonal of its derived ideal is abelian. Actually, the proof of Lauret’s
theorem is still valid if we suppose that
$\mathrm{Ric}_{|[\mathfrak{g},\mathfrak{g}]}=c\mathrm{Id}_{[\mathfrak{g},\mathfrak{g}]}$
instead of $\mathrm{Ric}=c\mathrm{Id}_{\mathfrak{g}}$. To convince the reader
we copy Lauret’s proof and we give the minor modifications needed to adapt to
our hypothesis.

######  Theorem 3.1.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a solvable Lie algebra
such that there exists a constant $c$ such that
$\mathrm{Ric}_{|[\mathfrak{g},\mathfrak{g}]}=c\mathrm{Id}_{[\mathfrak{g},\mathfrak{g}]}$.
Then $\mathfrak{g}$ is standard, i.e., $[\mathfrak{g},\mathfrak{g}]^{\perp}$
is abelian.

###### Proof.

Put ${\mathfrak{n}}=[\mathfrak{g},\mathfrak{g}]$ and
$\mathfrak{a}=[\mathfrak{g},\mathfrak{g}]^{\perp}$. We have

$\mathrm{Ric}=R-\frac{1}{2}B-S({\mathrm{ad}}_{H})$

where
$S({\mathrm{ad}}_{H})=\frac{1}{2}({\mathrm{ad}}_{H}+{\mathrm{ad}}_{H}^{*})$,
$\langle H,u\rangle={\mathrm{tr}}({\mathrm{ad}}_{u})$, $\langle
Bu,v\rangle={\mathrm{tr}}({\mathrm{ad}}_{u}\circ{\mathrm{ad}}_{v})$ and

$\begin{cases}\displaystyle\langle
Ru,v\rangle=-\frac{1}{2}\sum_{i,j}\langle[u,e_{i}],e_{j}\rangle\langle[v,e_{i}],e_{j}\rangle+\frac{1}{4}\sum_{i,j}\langle[e_{i},e_{j}],u\rangle\langle[e_{i},e_{j}],v\rangle,\\\
\displaystyle{\mathrm{tr}}(\mathrm{Ric}+\frac{1}{2}B+S({\mathrm{ad}}_{H}))E=\frac{1}{4}\sum_{i,j}\langle
E([e_{i},e_{j}])-[E(e_{i}),e_{j}]-[e_{i},E(e_{j})],[e_{i},e_{j}]\rangle,\end{cases}$
(12)

for any endomorphism $E$ of $\mathfrak{g}$ and where $(e_{1},\ldots,e_{n})$ is
an orthonormal basis of $\mathfrak{g}$. Since $\mathrm{Ric}$ leaves
${\mathfrak{n}}$ invariant it leaves $\mathfrak{a}$ invariant and for any,
$s\in\mathfrak{a}$,

$\displaystyle{\mathrm{ric}}(s,s)$
$\displaystyle=-\frac{1}{2}\sum_{i,j}\langle[s,e_{i}],e_{j}\rangle^{2}-\frac{1}{2}{\mathrm{tr}}({\mathrm{ad}}_{s}^{2})=-\frac{1}{2}{\mathrm{tr}}({\mathrm{ad}}_{s}\circ{\mathrm{ad}}_{s}^{*})-\frac{1}{2}{\mathrm{tr}}({\mathrm{ad}}_{s}^{2})=-\frac{1}{4}{\mathrm{tr}}(({\mathrm{ad}}_{s}+{\mathrm{ad}}_{s}^{*})^{2})\leq
0$ (13)

and hence ${\mathrm{ric}}(u,u)\leq 0$ for any $u\in\mathfrak{g}$. If
$\mathfrak{g}$ is unimodular then $\mathfrak{g}$ is standard according [6,
Theorem 1]. We suppose that $\mathfrak{g}$ is not unimodular and we choose
$(u_{1},\ldots,u_{p})$ and orthonormal basis of ${\mathfrak{n}}$ and
$(v_{1},\ldots,v_{q})$ and orthonormal of $\mathfrak{a}$.

We have ${\mathrm{tr}}({\mathrm{ad}}_{H})=|H|^{2}>0$,
${\mathrm{tr}}(S({\mathrm{ad}}_{H}){\mathrm{ad}}_{H})={\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})$
and since ${\mathrm{ad}}_{H}$ is a derivation,
${\mathrm{tr}}(B\circ{\mathrm{ad}}_{H})=0$. Moreover,

${\mathrm{tr}}(\mathrm{Ric}\circ{\mathrm{ad}}_{H})=\sum_{i}\langle[H,u_{i}],\mathrm{Ric}(u_{i})\rangle+\sum_{j}\langle[H,v_{j}],\mathrm{Ric}(v_{j})\rangle=c\sum_{i}\langle[H,u_{i}],u_{i}\rangle=c{\mathrm{tr}}({\mathrm{ad}}_{H}).$

So, by taking $E={\mathrm{ad}}_{H}$ in (12), we get

$c=-\frac{{\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})}{{\mathrm{tr}}(S({\mathrm{ad}}_{H}))}.$
(14)

We identify ${\mathfrak{n}}=[\mathfrak{g},\mathfrak{g}]$ to $\hbox{\bb R}^{p}$
via $(u_{1},\ldots,u_{p})$ and we set
$\mu=[\;,\;]_{|{\mathfrak{n}}\times{\mathfrak{n}}}$. In this way, $\mu$ can be
viewed as an element of $\mathscr{N}\subset V_{p}$. If $\mu\not=0$ then $\mu$
lies in a unique stratum $\mathscr{S}_{\beta}$, $\beta\in\mathscr{B}$, by [9,
Theorem 2.10, $(i)$]. The argument used in the proof of Lauret’s theorem to
show that one can assume $\mu\in Y^{ss}_{\beta}=\mathscr{S}_{\beta}\cap
W_{\beta}$ can be used in our case, so we make the assumption.

We apply (12) to the endomorphism $E$ given by $E_{|\mathfrak{a}}=0$ and
$E_{|{\mathfrak{n}}}=\beta+|\beta|^{2}\mathrm{Id}_{\mathfrak{n}}$. The right
hand of the second equation in $\eqref{eq4}$ becomes

$\displaystyle\frac{1}{4}\sum_{i,j}\langle
E([u_{i},u_{j}])-[E(u_{i}),u_{j}]-[u_{i},E(u_{j})],[u_{i},u_{j}]\rangle+\frac{1}{4}\sum_{r,s}\langle
E([v_{r},v_{s}]),[v_{r},v_{s}]\rangle$
$\displaystyle+\frac{1}{2}\sum_{i,r}\langle
E([u_{i},v_{r}]),[u_{i},v_{r}]\rangle-\frac{1}{2}\sum_{r,i}\langle[v_{r},E(u_{i})],[v_{r},u_{i}]\rangle,$

which in turn equal to

$\displaystyle\frac{1}{2}\langle\pi(\beta+|\beta|^{2}\mathrm{Id}_{{\mathfrak{n}}})\mu,\mu\rangle+\frac{1}{4}\sum_{r,s}\langle(\beta+|\beta|^{2}\mathrm{Id}_{{\mathfrak{n}}})[v_{r},v_{s}],[v_{r},v_{s}]\rangle+\frac{1}{2}\sum_{i,r}\langle(\beta{\mathrm{ad}}_{v_{r}}-{\mathrm{ad}}_{v_{r}}\beta)(u_{i}),{\mathrm{ad}}_{v_{r}}(u_{i})\rangle.$
(15)

The first and second terms in (15) are $\geq 0$ by [9, Lemma 2.16, Lemma 2.17]
and the last one equals
$\frac{1}{2}\langle[\beta,{\mathrm{ad}}_{v_{r}}],{\mathrm{ad}}_{v_{r}}\rangle$
which is $\geq 0$ by [9, (10)].

Therefore, since
$\mathrm{Ric}_{|{\mathfrak{n}}}=c\mathrm{Id}_{{\mathfrak{n}}}$ and
$E_{|\mathfrak{a}}=0$, we obtain from (12) and (14)

$-\frac{{\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})}{{\mathrm{tr}}(S({\mathrm{ad}}_{H}))}{\mathrm{tr}}(E)+{\mathrm{tr}}(S({\mathrm{ad}}_{H}))E\geq
0.$ (16)

But ${\mathrm{tr}}(\beta)=-1$ and so

${\mathrm{tr}}(E^{2})=|\beta|^{2}(-1+p|\beta|^{2})=|\beta|^{2}{\mathrm{tr}}(E).$
(17)

On the other hand, we have

${\mathrm{tr}}(S({\mathrm{ad}}_{H})E)={\mathrm{tr}}(({\mathrm{ad}}_{H})_{|{\mathfrak{n}}}(\beta+|\beta|^{2}\mathrm{Id_{\mathfrak{n}}})=|\beta|^{2}{\mathrm{tr}}(S({\mathrm{ad}}_{H})),$
(18)

by [9, Lemma 2.15]. We use now (16), (17) and (18) to obtain

${\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2}){\mathrm{tr}}(E^{2})\leq({\mathrm{tr}}(S({\mathrm{ad}}_{H})E)^{2},$
(19)

a ”backward” Cauchy-Schwartz inequality. This turns all inequalities mentioned
after (16) into equalities, in particular the second term:

$\frac{1}{4}\sum_{r,s}\langle(\beta+|\beta|^{2}\mathrm{Id}_{{\mathfrak{n}}})[v_{r},v_{s}],[v_{r},v_{s}]\rangle=0.$

We therefore get that $[\mathfrak{a},\mathfrak{a}]=0$ since
$\beta+|\beta|^{2}\mathrm{Id}_{{\mathfrak{n}}}$ is positive definite by [9,
Lemma 2.17].

It only remains to consider the case $\mu=0$. Here we argue in the same way
but with $E$ chosen as $E_{|{\mathfrak{n}}}=\mathrm{Id}_{\mathfrak{n}}$ and
$E_{|\mathfrak{a}}=0$. It then follows from (12) that

$\frac{1}{4}\sum_{r,s}|[v_{r},v_{s}]|^{2}=-\frac{{\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})}{{\mathrm{tr}}(S({\mathrm{ad}}_{H}))}p+{\mathrm{tr}}(S({\mathrm{ad}}_{H}))=\frac{{\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})}{{\mathrm{tr}}(S({\mathrm{ad}}_{H}))}\left(\frac{({\mathrm{tr}}(S({\mathrm{ad}}_{H})))^{2}}{{\mathrm{tr}}(S({\mathrm{ad}}_{H})^{2})}-p\right)\leq
0$

and thus $[{\mathfrak{n}}^{\perp},{\mathfrak{n}}^{\perp}]=0$. This concludes
the proof of the theorem. ∎

Let $(\mathfrak{g},\langle\;,\;\rangle)$ be a Euclidean Lie algebra with
harmonic curvature. According to Theorem 2.1,
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$, for any
$i\in\\{1,\ldots,r\\}$, $\mathfrak{g}_{i}$ is a subalgebra of $\mathfrak{g}$
and the equations (6) and (7) hold. For any $u\in\mathfrak{g}$, we denote by
$u_{i}$ its component in $\mathfrak{g}_{i}$. For any $i=1,\ldots,r$, we
consider

${\mathfrak{p}}_{i}=\sum_{j\not=i,k\not=i,j\not=k}[\mathfrak{g}_{j},\mathfrak{g}_{k}]_{i}\quad\mbox{and}\quad{\mathfrak{h}}_{i}={\mathfrak{p}}_{i}^{\perp}\cap\mathfrak{g}_{i}.$
(20)

######  Proposition 3.1.

${\mathfrak{h}}_{i}$ is a subalgebra which contains
$[\mathfrak{r}_{i},\mathfrak{r}_{i}]$ where $\mathfrak{r}_{i}$ is the radical
of $\mathfrak{g}_{i}$. Moreover,
$[[\mathfrak{r}_{i},\mathfrak{r}_{i}],\mathfrak{g}]\subset\mathfrak{g}_{i}$
and for any $u\in{\mathfrak{h}}_{i}$ and any $v\in\mathfrak{g}_{i}$

${\mathrm{ric}}_{i}(u,v)=\lambda_{i}\langle u,v\rangle.$

###### Proof.

By virtue of (7), $u\in{\mathfrak{h}}_{i}$ if and only if for any $j\not=i$,
$[u,\mathfrak{g}_{j}]\subset\mathfrak{g}_{i}\oplus\mathfrak{g}_{j}$. So if
$u,v\in{\mathfrak{h}}_{i}$ and $w\in\mathfrak{g}_{j}$ with $j\not=i$ then

$\displaystyle[[u,v],w]$ $\displaystyle=[u,[v,w]]-[v,[u,w]]$
$\displaystyle=[u,[v,w]_{i}]+[u,[v,w]_{j}]-[v,[u,w]_{i}]-[v,[u,w]_{j}]\subset\mathfrak{g}_{i}\oplus\mathfrak{g}_{j},$

and hence ${\mathfrak{h}}_{i}$ is a subalgebra.

Let $\mathfrak{r}_{i}$ be the radical of $\mathfrak{g}_{i}$ and $\rho_{i}$ the
representation of $\mathfrak{g}_{i}$ given by (9). Then
$\rho_{i}(\mathfrak{r}_{i})$ is a solvable subalgebra of
$\mathrm{so}(\mathfrak{g}_{i}^{\perp})$ and hence it is abelian. So
$\rho_{i}([\mathfrak{r}_{i},\mathfrak{r}_{i}])=0$ and hence, for any
$X\in[\mathfrak{r}_{i},\mathfrak{r}_{i}]$, and any $j\not=i$,
$[X,\mathfrak{g}_{j}]\subset\mathfrak{g}_{i}$ which shows that
$[\mathfrak{r}_{i},\mathfrak{r}_{i}]\subset{\mathfrak{h}}_{i}$ and
$[[\mathfrak{r}_{i},\mathfrak{r}_{i}],\mathfrak{g}]\subset\mathfrak{g}_{i}$.
The last assertion is a consequence of (10). ∎

We can prove now our main theorem.

######  Theorem 3.2.

Let $\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$ be a
Euclidean Lie algebra with harmonic curvature. If $\mathfrak{g}_{i}$ is
solvable for any $i\in\\{1,\ldots,e\\}$ then $\mathfrak{g}$ is Ricci-parallel.

###### Proof.

We will show that for any $\\{i,j,k\\}$ which are mutually different,
$\langle\mathfrak{g}_{i},[\mathfrak{g}_{j},\mathfrak{g}_{k}]\rangle=0$. This
is equivalent to ${\mathfrak{p}}_{i}=\\{0\\}$ for any $i\in\\{1,\ldots,r\\}$,
where ${\mathfrak{p}}_{i}$ is given by (20).

Suppose that $\lambda_{1}<\ldots<\lambda_{r}$. From (10), we deduce that for
any $u\in\mathfrak{g}_{r}$,

$\lambda_{r}|u|^{2}\leq{\mathrm{ric}}_{r}(u,u)$

and hence $\lambda_{r}\dim\mathfrak{g}_{r}\leq{\mathfrak{s}}_{r}$, where
${\mathfrak{s}}_{r}$ is the scalar curvature of $\mathfrak{g}_{r}$. But
$\mathfrak{g}_{r}$ is solvable and, according to [11, Theorem 3.1],
${\mathfrak{s}}_{r}<0$ or $\mathfrak{g}_{r}$ is flat, thus $\lambda_{r}\leq
0$.

Denote by $S$ the set of $j$ such that $\mathfrak{g}_{j}$ is not flat and let
$i\in S$. Then, according to Proposition 3.1,
$[[\mathfrak{g}_{i},\mathfrak{g}_{i}],\mathfrak{g}]\subset\mathfrak{g}_{i}$,
$[\mathfrak{g}_{i},\mathfrak{g}_{i}]\subset{\mathfrak{h}}_{i}$ and hence for
any $u\in\mathfrak{g}_{i}$ and $v\in[\mathfrak{g}_{i},\mathfrak{g}_{i}]$

${\mathrm{ric}}_{i}(u,v)=\lambda_{i}\langle u,v\rangle.$

By virtue of Theorem 3.1,
$[\mathfrak{g}_{i},\mathfrak{g}_{i}]^{\perp}\cap\mathfrak{g}_{i}$ is abelian.
Write

$\mathfrak{g}_{i}=[\mathfrak{g}_{i},\mathfrak{g}_{i}]\oplus\mathfrak{u}_{i}\oplus{\mathfrak{k}}_{i}\quad\mbox{and}\quad[\mathfrak{g}_{i},\mathfrak{g}_{i}]^{\perp}\cap\mathfrak{g}_{i}=\mathfrak{u}_{i}\stackrel{{\scriptstyle\perp}}{{\oplus}}{\mathfrak{k}}_{i},$

where

${\mathfrak{k}}_{i}=\\{u\in[\mathfrak{g}_{i},\mathfrak{g}_{i}]^{\perp}\cap\mathfrak{g}_{i},({\mathrm{ad}}_{u})_{|[\mathfrak{g}_{i},\mathfrak{g}_{i}]}\;\mbox{is
skew-symmetric}\\}.$

Let us show that ${\mathfrak{p}}_{i}\subset{\mathfrak{k}}_{i}$. Indeed, for
any $u_{j}\in\mathfrak{g}_{j}$, $u_{k}\in\mathfrak{g}_{k}$, and $j\not=k$ both
different from $i$, and for any $d\in[\mathfrak{g}_{i},\mathfrak{g}_{i}]$, by
using the fact that
$[[\mathfrak{g}_{i},\mathfrak{g}_{i}],\mathfrak{g}]\subset\mathfrak{g}_{i}$
and (6), we get

$0=\langle[[u_{j},u_{k}],d],d\rangle+\langle[[u_{k},d],u_{j}],d\rangle+\langle[[d,u_{j}],u_{k}],d\rangle=\langle[[u_{j},u_{k}]_{i},d],d\rangle.$

On the other hand, by virtue of (10) and since
${\mathfrak{p}}_{i}\subset{\mathfrak{k}}_{i}$,
$\mathfrak{u}_{i}\subset{\mathfrak{h}}_{i}$ and for any
$u\in\mathfrak{u}_{i}$,

${\mathrm{ric}}_{i}(u,u)=\lambda_{i}\langle u,u\rangle.$

Moreover, for any $u\in{\mathfrak{k}}_{i}$, ${\mathrm{ad}}_{u}$ is skew-
symmetric in restriction to $[\mathfrak{g}_{i},\mathfrak{g}_{i}]$. But
$\mathfrak{u}_{i}\oplus{\mathfrak{k}}_{i}$ is abelian so ${\mathrm{ad}}_{u}$
is skew-symmetric in restriction to $\mathfrak{g}_{i}$ and, by virtue of (13),
${\mathrm{ric}}_{i}(u,u)=0$. So far we have shown that

${\mathrm{ric}}_{i}(u,u)=\begin{cases}\lambda_{i}|u|^{2}\;\mbox{if}\;u\in[\mathfrak{g}_{i},\mathfrak{g}_{i}]\oplus\mathfrak{u}_{i},\\\
0\;\mbox{if}\;u\in{\mathfrak{k}}_{i}\end{cases}$

which implies that
${\mathfrak{s}}_{i}=\lambda_{i}(\dim\mathfrak{u}_{i}+\dim[\mathfrak{g}_{i},\mathfrak{g}_{i}])$.
Then the formula (11) can be written

${\mathfrak{s}}=\lambda_{1}\dim\mathfrak{g}_{1}+\ldots+\lambda_{r}\dim\mathfrak{g}_{r}=\sum_{j\in
S}\lambda_{j}(\dim[\mathfrak{g}_{j},\mathfrak{g}_{j}]+\dim\mathfrak{u}_{j}).$

So $\displaystyle\sum_{j\in S}\lambda_{j}\dim{\mathfrak{k}}_{j}+\sum_{j\notin
S}\lambda_{j}\dim\mathfrak{g}_{j}=0$ and, since
$\lambda_{1}<\ldots<\lambda_{r}\leq 0$,

$\sum_{j\in S}\lambda_{j}\dim{\mathfrak{k}}_{j}=\sum_{j\notin
S}\lambda_{j}\dim\mathfrak{g}_{j}=0.$

If $\lambda_{r}<0$ then for any $j$, $\mathfrak{g}_{j}$ is not flat and
${\mathfrak{k}}_{j}=\\{0\\}$ which implies ${\mathfrak{p}}_{j}=\\{0\\}$ and
the result follows. If $\lambda_{r}=0$ then for any $j\leq r-1$,
$\mathfrak{g}_{j}$ is not flat and
${\mathfrak{k}}_{j}={\mathfrak{p}}_{j}=\\{0\\}$. We will have also
${\mathfrak{p}}_{r}=\\{0\\}$ by virtue of (7) which completes the proof. ∎

######  Corollary 3.1.

Let $\mathfrak{g}$ be a solvable Euclidean Lie algebra with harmonic
curvature. Then $\mathfrak{g}$ is Ricci-parallel.

######  Corollary 3.2.

Let $\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$ be a
Euclidean Lie algebra with harmonic curvature and not Ricci-parallel. Then
there exists $i$ such that $\dim\mathfrak{g}_{i}\geq 3$.

######  Corollary 3.3.

Let $G/K$ be a homogeneous Riemannian manifold with harmonic curvature and $K$
is a maximal compact subgroup of $G$. Then $G/K$ is Ricci-parallel.

###### Proof.

Since $K$ is a maximal compact subgroup of $G$, there exists a solvable
subgroup of $G$ which acts simply transitively on $G/K$ and hence $G/K$ is
isometric to a solvable Lie group with a left invariant metric. ∎

##  4 Left invariant Riemannian metrics with harmonic curvature on Lie groups
of dimension $\leq 6$ are Ricci-parallel

In this section, we prove that any left invariant Riemannian metric with
harmonic curvature on a Lie group of dimension $\leq 6$ is Ricci-parallel.
Through-out this section
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\ldots\oplus\mathfrak{g}_{r}$ is a
Euclidean Lie algebra with harmonic curvature and we will not make any
assumption on the order of the eigenvalues unless it is mentioned.

######  Theorem 4.1.

Let
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}$ be
a Euclidean Lie algebra with harmonic curvature and
$\dim\mathfrak{g}_{1}=\dim\mathfrak{g}_{2}=1$. Then $\mathfrak{g}$ is Ricci-
parallel.

###### Proof.

Put $\mathfrak{g}_{1}=\hbox{\bb R}u_{1}$, $\mathfrak{g}_{2}=\hbox{\bb R}u_{2}$
with $|u_{1}|=|u_{2}|=1$. By virtue of (6),
$\omega=[u_{1},u_{2}]\in\mathfrak{g}_{3}$ and $\mathfrak{g}$ is Ricci-parallel
if and only if $\omega=0$. Suppose that $\omega\not=0$. As consequence of
Proposition 3.1, we get $\mathfrak{g}_{3}={\mathfrak{h}}_{3}\oplus\hbox{\bb
R}\omega$ and for any $u\in{\mathfrak{h}}_{3}$,
${\mathrm{ric}}_{3}(u,u)=\lambda_{3}\langle u,u\rangle$ and
${\mathrm{ric}}_{3}(u,\omega)=0$.

Denote by $A_{i}:\mathfrak{g}_{3}\longrightarrow\mathfrak{g}_{3}$,
$v\mapsto[u_{i},v]_{3}$. Both $A_{1}$ and $A_{2}$ are skew-symmetric. By
virtue of (6), for any $v\in\mathfrak{g}_{3}$, $[u_{2},v]_{2}=[u_{1},v]_{1}=0$
and hence

$\displaystyle\;[\omega,v]$
$\displaystyle=[u_{1},[u_{2},v]]-[u_{2},[u_{1},v]]$
$\displaystyle=[u_{1},A_{2}(v)]-[u_{2},A_{1}(v)]$
$\displaystyle=[A_{1},A_{2}](v)+\langle[u_{1},A_{2}(v)],u_{2}\rangle
u_{2}-\langle[u_{2},A_{1}(v)],u_{1}\rangle u_{1}.$

Thus

$({\mathrm{ad}}_{\omega})_{|\mathfrak{g}_{3}}=[A_{1},A_{2}]\quad\mbox{and}\quad
A_{1}(\omega)=A_{2}(\omega)=0.$

On the other hand, for any $v,w\in\mathfrak{g}_{3}$,

$\displaystyle 0$
$\displaystyle=\langle[u_{1},[v,w]],u_{2}\rangle+\langle[v,[w,u_{1}]],u_{2}\rangle+\langle[w,[u_{1},v]],u_{2}\rangle$
$\displaystyle=\langle[u_{1},[v,w]],u_{2}\rangle+\langle[v,[w,u_{1}]_{2}],u_{2}\rangle+\langle[w,[u_{1},v]_{2}],u_{2}\rangle$
$\displaystyle\stackrel{{\scriptstyle\eqref{sk},\eqref{3}}}{{=}}-\frac{(\lambda_{1}-\lambda_{3})^{2}}{(\lambda_{1}-\lambda_{2})^{2}}\langle[v,w],[u_{1},u_{2}]\rangle-\langle[w,u_{1}]_{2},[v,u_{2}]\rangle-\langle[u_{1},v]_{2},[w,u_{2}]\rangle$
$\displaystyle=-\frac{(\lambda_{1}-\lambda_{3})^{2}}{(\lambda_{1}-\lambda_{2})^{2}}\langle[v,w],[u_{1},u_{2}]\rangle$

since $[w,u_{2}]_{2}=[v,u_{2}]_{2}=0$. Thus
$\omega\in[\mathfrak{g}_{3},\mathfrak{g}_{3}]^{\perp}\cap\mathfrak{g}_{3}$. In
conclusion, $({\mathrm{ad}}_{\omega})_{|\mathfrak{g}_{3}}$ is skew-symmetric
and $\omega\in[\mathfrak{g}_{3},\mathfrak{g}_{3}]^{\perp}\cap\mathfrak{g}_{3}$
and hence, by virtue of (13), ${\mathrm{ric}}_{3}(\omega,\omega)=0$ and hence
${\mathfrak{s}}_{3}=\lambda_{3}(\dim\mathfrak{g}_{3}-1)$. By virtue of (11),

${\mathfrak{s}}=\lambda_{1}+\lambda_{2}+\lambda_{3}\dim\mathfrak{g}_{3}={\mathfrak{s}}_{1}+{\mathfrak{s}}_{2}+{\mathfrak{s}}_{3}=\lambda_{3}(\dim\mathfrak{g}_{3}-1)$

and hence

$\lambda_{1}+\lambda_{2}+\lambda_{3}=0.$ (21)

Let $(V_{1},\ldots,V_{p})$ and orthonormal basis of $\mathfrak{g}_{3}$. By
using (10), we get

$\displaystyle\lambda_{1}$
$\displaystyle={\mathrm{ric}}(u_{1},u_{1})={\mathrm{ric}}_{1}(u_{1},u_{1})-2\frac{(\lambda_{3}-\lambda_{1})(\lambda_{2}-\lambda_{1})}{(\lambda_{3}-\lambda_{2})^{2}}\sum_{i}\langle[u_{2},V_{i}],u_{1}\rangle^{2}$
$\displaystyle\stackrel{{\scriptstyle\eqref{3}}}{{=}}-2\frac{(\lambda_{3}-\lambda_{1})(\lambda_{2}-\lambda_{1})(\lambda_{2}-\lambda_{3})^{4}}{(\lambda_{3}-\lambda_{2})^{2}(\lambda_{1}-\lambda_{2})^{4}}\sum_{i}\langle
V_{i},\omega\rangle^{2}$
$\displaystyle=2\frac{(\lambda_{3}-\lambda_{1})(\lambda_{2}-\lambda_{3})^{2}}{(\lambda_{1}-\lambda_{2})^{3}}|\omega|^{2},$
$\displaystyle\lambda_{2}$
$\displaystyle={\mathrm{ric}}(u_{2},u_{2})={\mathrm{ric}}_{2}(u_{2},u_{2})-2\frac{(\lambda_{3}-\lambda_{2})(\lambda_{1}-\lambda_{2})}{(\lambda_{3}-\lambda_{1})^{2}}\sum_{i}\langle[u_{1},V_{i}],u_{2}\rangle^{2}$
$\displaystyle\stackrel{{\scriptstyle\eqref{3}}}{{=}}-2\frac{(\lambda_{3}-\lambda_{2})(\lambda_{1}-\lambda_{2})(\lambda_{1}-\lambda_{3})^{4}}{(\lambda_{3}-\lambda_{1})^{2}(\lambda_{1}-\lambda_{2})^{4}}\sum_{i}\langle
V_{i},\omega\rangle^{2}$
$\displaystyle=-2\frac{(\lambda_{3}-\lambda_{2})(\lambda_{1}-\lambda_{3})^{2}}{(\lambda_{1}-\lambda_{2})^{3}}|\omega|^{2},$
$\displaystyle\lambda_{3}|\omega|^{2}$
$\displaystyle={\mathrm{ric}}_{3}(\omega,\omega)-2\frac{(\lambda_{1}-\lambda_{3})(\lambda_{2}-\lambda_{3})}{(\lambda_{1}-\lambda_{2})^{2}}\langle[u_{1},u_{2}],\omega\rangle^{2}$
$\displaystyle=-2\frac{(\lambda_{1}-\lambda_{3})(\lambda_{2}-\lambda_{3})}{(\lambda_{1}-\lambda_{2})^{2}}|\omega|^{4}.$

Put $\lambda_{ij}=\lambda_{i}-\lambda_{j}$. So we must have

$\lambda_{1}+\lambda_{2}+\lambda_{3}=0,\lambda_{1}=-2\frac{\lambda_{23}^{2}\lambda_{13}}{\lambda_{12}^{3}}|\omega|^{2},\;\lambda_{2}=2\frac{\lambda_{13}^{2}\lambda_{23}}{\lambda_{12}^{3}}|\omega|^{2}\quad\mbox{and}\quad\lambda_{3}=-2\frac{\lambda_{13}\lambda_{23}}{\lambda_{12}^{2}}|\omega|^{2}.$

This is equivalent to

$\lambda_{1}\lambda_{12}-\lambda_{23}\lambda_{3}=\lambda_{2}\lambda_{12}+\lambda_{3}\lambda_{13}=0\quad\mbox{and}\quad\lambda_{3}=-2\frac{\lambda_{13}\lambda_{23}}{\lambda_{12}^{2}}|\omega|^{2}.$

This implies

$\lambda_{1}^{2}-\lambda_{1}\lambda_{2}-\lambda_{1}(\lambda_{1}+\lambda_{2})=0\quad\mbox{and}\quad-\lambda_{2}^{2}+\lambda_{1}\lambda_{2}+(\lambda_{1}+\lambda_{2})\lambda_{2}=0.$

Thus $\lambda_{1}\lambda_{2}=0$ which is impossible. So $\omega=0$ which
completes the proof. ∎

######  Corollary 4.1.

Any left invariant Riemannian metric with harmonic curvature on a Lie group
$G$ of dimension $\leq 5$ is Ricci-parallel.

###### Proof.

Let $\mathfrak{g}$ be the Lie algebra of $G$. If the metric is non Ricci-
parallel then $\mathrm{Ric}$ must have at least three distinct eigenvalues and
at least one eigenvalue with multiplicity greater than 3 (See Corollary 3.2).
This is not possible in dimensions 3 and 4. In dimension 5, the only
possibility is
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}$
where $\dim\mathfrak{g}_{3}=3$ and
$\dim\mathfrak{g}_{1}=\dim\mathfrak{g}_{2}=1$. This is not possible by virtue
of Theorem 4.1. ∎

Let $\mathfrak{g}$ be a 6-dimensional Euclidean Lie algebra with harmonic
curvature and suppose that $\mathfrak{g}$ is not Ricci-parallel. By virtue of
Corollary 3.2 and Theorem 4.1, we have two possibilities:

  1. 1.

$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}$
with $\dim\mathfrak{g}_{1}=1$, $\dim\mathfrak{g}_{2}=2$ and $\mathfrak{g}_{3}$
is isomorphic either to $\mathrm{su}(2)$ or $\mathrm{sl}(2,\hbox{\bb R})$.

  2. 2.

$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}\oplus\mathfrak{g}_{4}$
with $\dim\mathfrak{g}_{1}=\dim\mathfrak{g}_{2}=\dim\mathfrak{g}_{3}=1$ and
$\mathfrak{g}_{4}$ is isomorphic either to $\mathrm{su}(2)$ or
$\mathrm{sl}(2,\hbox{\bb R})$.

We will see that both these cases are impossible.

######  Theorem 4.2.

Let
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}$ be
a Euclidean Lie algebra with harmonic curvature such that
$\dim\mathfrak{g}_{1}=1$, $\dim\mathfrak{g}_{2}=2$ and
$\dim\mathfrak{g}_{3}=3$. Then $\mathfrak{g}$ is Ricci-parallel.

###### Proof.

We choose an orthonormal family $(X_{1},X_{2},X_{3})$ such that
$\mathfrak{g}_{1}=\hbox{\bb R}X_{1}$,
$\mathfrak{g}_{2}=\mathrm{span}\\{X_{2},X_{3}\\}$ and $[X_{2},X_{3}]=aX_{2}$.
The relation (6) has many consequences. There exists $p\in\mathfrak{g}_{3}$
and $r\in\hbox{\bb R}$ such that, for any $Y\in\mathfrak{g}_{3}$,

$[X_{1},X_{2}]_{2}=rX_{3}\quad\mbox{and}\quad[X_{1},X_{3}]_{2}=-rX_{2},\;[Y,X_{2}]_{2}=\langle
p,Y\rangle X_{3}\quad\mbox{and}\quad[Y,X_{3}]_{2}=-\langle p,Y\rangle X_{2}.$

Moreover, if we put $\omega_{1j}=[X_{1},X_{j}]_{3}\in\mathfrak{g}_{3}$ for
$j=2,3$ and for $i=1,2,3$,
$\rho_{i}:\mathfrak{g}_{3}\longrightarrow\mathfrak{g}_{3}$,
$Y\mapsto[X_{i},Y]_{3}$, the $\rho_{i}$ are skew-symmetric. Finally, for any
$Y\in\mathfrak{g}$, $[X_{1},Y]_{1}=0$.

Now, for any $Y\in\mathfrak{g}_{3}$ and for $j=2,3$,

$\displaystyle\;[[X_{1},X_{j}],Y]$
$\displaystyle=[\omega_{1j},Y]+[[X_{1},X_{j}]_{2},Y]$
$\displaystyle=[X_{1},[X_{j},Y]]-[X_{j},[X_{1},Y]]$
$\displaystyle=[X_{1},\rho_{j}(Y)]+[X_{1},[X_{j},Y]_{2}]-[X_{j},\rho_{1}(Y)]-[X_{j},[X_{1},Y]_{2}]$
$\displaystyle=\rho_{1}\circ\rho_{j}(Y)+[X_{1},\rho_{j}(Y)]_{2}+[X_{1},[X_{j},Y]_{2}]-\rho_{j}\circ\rho_{1}(Y)-[X_{j},\rho_{1}(Y)]_{1}-[X_{j},\rho_{1}(Y)]_{2}-[X_{j},[X_{1},Y]_{2}].$

So we get

$({\mathrm{ad}}_{\omega_{12}})_{|\mathfrak{g}_{3}}=[\rho_{1},\rho_{2}]-r\rho_{3}-\langle
p,\bullet\rangle\omega_{13},\;({\mathrm{ad}}_{\omega_{13}})_{|\mathfrak{g}_{3}}=[\rho_{1},\rho_{3}]+r\rho_{2}+\langle
p,\bullet\rangle\omega_{12},$

and, for any $Y\in\mathfrak{g}_{3}$,

$\displaystyle[[X_{1},X_{j}]_{2},Y]_{1}+[[X_{1},X_{j}]_{2},Y]_{2}$
$\displaystyle=[X_{1},\rho_{j}(Y)]_{2}+[X_{1},[X_{j},Y]_{2}]_{2}-[X_{j},\rho_{i}(Y)]_{2}-[X_{j},\rho_{1}(Y)]_{1}-[X_{j},[X_{1},Y]_{2}].$

So

$\begin{cases}r\langle[X_{3},Y],X_{1}\rangle+\langle[X_{2},\rho_{1}(Y)],X_{1}\rangle=0,\\\
r\langle p,Y\rangle X_{2}=[X_{1},\rho_{2}(Y)]_{2}+r\langle p,Y\rangle
X_{2}+\langle p,\rho_{1}(Y)\rangle
X_{3}-\langle[X_{1},Y]_{2},X_{3}\rangle[X_{2},X_{3}],\\\
-r\langle[X_{2},Y],X_{1}\rangle+\langle[X_{3},\rho_{1}(Y)],X_{1}\rangle=0,\\\
r\langle p,Y\rangle X_{3}=[X_{1},\rho_{3}(Y)]_{2}+r\langle p,Y\rangle
X_{3}-\langle p,\rho_{1}(Y)\rangle
X_{2}-\langle[X_{1},Y]_{2},X_{2}\rangle[X_{3},X_{2}].\end{cases}$

Thus

$\begin{cases}\rho_{1}(\omega_{12})=r\omega_{13}\quad\mbox{and}\quad\rho_{1}(\omega_{13})=-r\omega_{12},\\\
0=[X_{1},\rho_{2}(Y)]_{2}+\langle p,\rho_{1}(Y)\rangle
X_{3}-\langle[X_{1},Y]_{2},X_{3}\rangle[X_{2},X_{3}],\\\
0=[X_{1},\rho_{3}(Y)]_{2}-\langle p,\rho_{1}(Y)\rangle
X_{2}-\langle[X_{1},Y]_{2},X_{2}\rangle[X_{3},X_{2}],\end{cases}$

and finally,

$\begin{cases}({\mathrm{ad}}_{\omega_{12}})_{|\mathfrak{g}_{3}}=[\rho_{1},\rho_{2}]-r\rho_{3}-\langle
p,\bullet\rangle\omega_{13},\\\
({\mathrm{ad}}_{\omega_{13}})_{|\mathfrak{g}_{3}}=[\rho_{1},\rho_{3}]+r\rho_{2}+\langle
p,\bullet\rangle\omega_{12},\\\
\rho_{1}(\omega_{12})=r\omega_{13}\quad\mbox{and}\quad\rho_{1}(\omega_{13})=-r\omega_{12},\\\
0=[X_{1},\rho_{2}(Y)]_{2}+\langle p,\rho_{1}(Y)\rangle
X_{3}-\langle[X_{1},Y]_{2},X_{3}\rangle[X_{2},X_{3}],\\\
0=[X_{1},\rho_{3}(Y)]_{2}-\langle p,\rho_{1}(Y)\rangle
X_{2}-\langle[X_{1},Y]_{2},X_{2}\rangle[X_{3},X_{2}].\end{cases}$ (22)

On the other hand,

$\displaystyle[[X_{1},X_{2}],X_{3}]$
$\displaystyle=[\omega_{12},X_{3}]=-\rho_{3}(\omega_{12})-\langle
p,\omega_{12}\rangle X_{2}+\langle[\omega_{12},X_{3}],X_{1}\rangle X_{1},$
$\displaystyle=[X_{1},[X_{2},X_{3}]]-[X_{2},[X_{1},X_{3}]]=a[X_{1},X_{2}]-[X_{2},\omega_{13}],$
$\displaystyle[[X_{1},X_{3}],X_{2}]$
$\displaystyle=[\omega_{13},X_{2}]=-\rho_{2}(\omega_{13})+\langle
p,\omega_{13}\rangle X_{3}+\langle[\omega_{13},X_{2}],X_{1}\rangle X_{1},$
$\displaystyle=[X_{1},[X_{3},X_{2}]]-[X_{3},[X_{1},X_{2}]]=-a[X_{1},X_{2}]-[X_{3},\omega_{12}].$

So

$\begin{cases}-\rho_{3}(\omega_{12})-\langle p,\omega_{12}\rangle
X_{2}+\langle[\omega_{12},X_{3}],X_{1}\rangle
X_{1}=a\omega_{12}+arX_{3}-\rho_{2}(\omega_{13})+\langle p,\omega_{13}\rangle
X_{3}-\langle[X_{2},\omega_{13}],X_{1}\rangle X_{1},\\\
-\rho_{2}(\omega_{13})+\langle p,\omega_{13}\rangle
X_{3}+\langle[\omega_{13},X_{2}],X_{1}\rangle
X_{1}=-a\omega_{12}-arX_{3}-\rho_{3}(\omega_{12})-\langle p,\omega_{12}\rangle
X_{2}-\langle[X_{3},\omega_{12}],X_{1}\rangle X_{1},\end{cases}$

and hence

$\langle p,\omega_{12}\rangle=\langle p,\omega_{12}\rangle=0,\quad ar=-\langle
p,\omega_{13}\rangle=-\langle
p,\omega_{12}\rangle,\quad-\rho_{3}(\omega_{12})=a\omega_{12}-\rho_{2}(\omega_{13})\quad\mbox{and}\quad-\rho_{2}(\omega_{13})=-a\omega_{12}-\rho_{3}(\omega_{12}).$

Suppose that $\mathfrak{g}$ is not Ricci-parallel then
$\dim\mathrm{span}\\{\omega_{12},\omega_{13}\\}\geq 1$ and, according to
Theorem 3.2, $\mathfrak{g}_{3}$ is isomorphic either to $\mathrm{su}(2)$ or
$\mathrm{sl}(2,\hbox{\bb R})$.

Suppose that $\dim\mathrm{span}\\{\omega_{12},\omega_{13}\\}=1$. Then there
exists $(\mu_{1},\mu_{2})\not=(0,0)$ such that
$\mu_{1}\omega_{12}+\mu_{2}\omega_{13}=0$. The vector
$X=\mu_{2}\omega_{12}-\mu_{1}\omega_{13}\not=0$ and, by virtue of (22),
${\mathrm{ad}}_{X}$ is skew-symmetric. The orthogonal of $X$ is the subalgebra
${\mathfrak{h}}_{3}$ defined in (20) and since ${\mathrm{ad}}_{X}$ is skew-
symmetric it leaves invariant ${\mathfrak{h}}_{3}$ and hence it is an ideal
which is not possible since $\mathfrak{g}_{3}$ is simple.

Suppose now that $\dim\mathrm{span}\\{\omega_{12},\omega_{13}\\}=2$. We have

$ar=\langle p,\omega_{12}\rangle=\langle p,\omega_{13}\rangle=0.$

If $r=0$ then we get from (22) that
$\rho_{1}(\omega_{12})=\rho_{1}(\omega_{13})=0$ and hence $\rho_{1}=0$ and
then $[\omega_{12},\omega_{13}]=0$ which is impossible since
$\mathfrak{g}_{3}$ does not contains an abelian subalgebra.
If $a=0$ and $r\not=0$. From (22), we have for any $Y\in\mathfrak{g}_{3}$,

$[X_{1},\rho_{2}(Y)]_{2}+\langle p,\rho_{1}(Y)\rangle
X_{3}=[X_{1},\rho_{3}(Y)]_{2}-\langle p,\rho_{1}(Y)\rangle X_{2}=0.$

But $p\in\\{\omega_{12},\omega_{13}\\}^{\perp}$ and hence $\rho_{1}(p)=0$. So

$\langle[X_{1},\rho_{2}(Y)]_{2},X_{2}\rangle=\langle[X_{1},\rho_{3}(Y)]_{2},X_{3}\rangle=\langle[X_{1},\rho_{2}(Y)]_{2},X_{3}\rangle=\langle[X_{1},\rho_{3}(Y)]_{2},X_{2}\rangle=0.$

By using (7), we get that $\rho_{2}=\rho_{3}=0$. This implies that
$[\omega_{12},\omega_{13}]=0$ which is impossible. ∎

######  Proposition 4.1.

Let
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}\oplus\mathfrak{g}_{4}$
a Euclidean Lie algebra with harmonic curvature such that
$\dim\mathfrak{g}_{1}=\dim\mathfrak{g}_{2}=\dim\mathfrak{g}_{3}=1$,
$\dim\mathfrak{g}_{4}=3$ and $\lambda_{1}<\lambda_{2}<\lambda_{3}$. If
$\mathfrak{g}$ is not Ricci-parallel then there exists an orthonormal basis
$(e_{1},\ldots,e_{6})$ of $\mathfrak{g}$ with $e_{i}\in\mathfrak{g}_{i}$ for
$i=1,2,3$ and $(e_{4},e_{5},e_{6})$ is a basis of $\mathfrak{g}_{4}$ such that
the non vanishing Lie brackets are

$\begin{cases}\;[e_{4},e_{5}]=ae_{6},\;[e_{4},e_{6}]=be_{5}\quad\mbox{and}\quad[e_{5},e_{6}]=ce_{4},\\\
\;[e_{1},e_{2}]=\mu_{1}e_{4}+\frac{\lambda_{12}^{2}}{\lambda_{23}^{2}}re_{3},[e_{1},e_{3}]=\mu_{2}e_{5}-\frac{\lambda_{13}^{2}}{\lambda_{23}^{2}}re_{2},\;[e_{2},e_{3}]=\mu_{3}e_{6}+re_{1},\\\
[e_{1},e_{4}]=\alpha_{1}e_{5}-\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\mu_{1}e_{2},\;[e_{1},e_{5}]=-\alpha_{1}e_{4}-\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\mu_{2}e_{3},\;\\\
[e_{2},e_{4}]=\alpha_{2}e_{6}+\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\mu_{1}e_{1},\;[e_{2},e_{6}]=-\alpha_{2}e_{4}-\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\mu_{3}e_{3},\\\
[e_{3},e_{5}]=\alpha_{3}e_{6}+\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\mu_{2}e_{1},\;[e_{3},e_{6}]=-\alpha_{3}e_{5}+\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\mu_{3}e_{2},\end{cases}$
(23)

where $a\not=0,b\not=0,c\not=0$, $\mu_{i}>0$ for $i=1,2,3$,
$r,\alpha_{1},\alpha_{2},\alpha_{3}\in\hbox{\bb R}$ and
$\lambda_{ij}=\lambda_{i}-\lambda_{j}$.

###### Proof.

For $i=1,2,3$, put $\mathfrak{g}_{i}=\hbox{\bb R}X_{i}$ with $|X_{i}|=1$ and
$\omega_{ij}=[X_{i},X_{j}]_{4}$ for $1\leq i<j\leq 3$. By virtue of (6), we
have

$[X_{1},X_{2}]=r_{3}X_{3}+\omega_{12},\;[X_{1},X_{3}]=r_{2}X_{2}+\omega_{13}\quad\mbox{and}\quad[X_{2},X_{3}]=r_{1}X_{1}+\omega_{23}.$

For any $X\in\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}$,
denote by $\rho(X):\mathfrak{g}_{4}\longrightarrow\mathfrak{g}_{4}$,
$Y\mapsto[X,Y]_{4}$ and for $i=1,2,3$ $\rho_{i}=\rho(X_{i})$. For
$Y\in\mathfrak{g}_{4}$ and $1\leq i<j\leq 3$,

$\displaystyle\;[[X_{i},X_{j}],Y]$
$\displaystyle=[\omega_{ij},Y]+\sum_{l\not=4}[[X_{i},X_{j}]_{l},Y],$
$\displaystyle=[X_{i},[X_{j},Y]]-[X_{j},[X_{i},Y]]$
$\displaystyle=[X_{i},\rho_{j}(Y)]+\sum_{l\not=4}[X_{i},[X_{j},Y]_{l}]-[X_{j},\rho_{i}(Y)]-\sum_{l\not=4}[X_{j},[X_{i},Y]_{l}].$

So

$[\omega_{ij},Y]+\sum_{l\not=4}[[X_{i},X_{j}]_{l},Y]=[X_{i},\rho_{j}(Y)]-[X_{j},\rho_{i}(Y)]+\sum_{l\not=4}\left([X_{i},[X_{j},Y]_{l}]-[X_{j},[X_{i},Y]_{l}]\right).$
(24)

By taking the $\mathfrak{g}_{4}$-component, we get

$({\mathrm{ad}}_{\omega_{ij}})_{|\mathfrak{g}_{4}}(Y)=[\rho_{i},\rho_{j}](Y)-\sum_{l\not=4}\rho([X_{i},X_{j}]_{l})(Y)+\sum_{l\not=4}\left([X_{i},[X_{j},Y]_{l}]_{4}-[X_{j},[X_{i},Y]_{l}]_{4}\right).$

Define $F_{ij}:\mathfrak{g}_{4}\longrightarrow\mathfrak{g}_{4}$ by

$F_{ij}(Y)=[X_{i},[X_{j},Y]_{j}]_{4}-[X_{j},[X_{i},Y]_{i}]_{4}+\sum_{l\not=4,l\not=i,l\not=j}\left([X_{i},[X_{j},Y]_{l}]_{4}-[X_{j},[X_{i},Y]_{l}]_{4}\right).$

So we get

$({\mathrm{ad}}_{\omega_{ij}})_{|\mathfrak{g}_{4}}=[\rho_{i},\rho_{j}]-\sum_{l\not=4}\rho([X_{i},X_{j}]_{l})+F_{ij}.$

Having (6) and (7) in mind, we have

$\displaystyle F_{12}(Y)$
$\displaystyle=[X_{1},[X_{2},Y]_{3}]_{4}-[X_{2},[X_{1},Y]_{3}]_{4}=\langle[X_{2},Y],X_{3}\rangle\omega_{13}-\langle[X_{1},Y],X_{3}\rangle\omega_{23}$
$\displaystyle=-\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle
Y,\omega_{23}\rangle\omega_{13}+\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle\omega_{23},$ $\displaystyle F_{13}(Y)$
$\displaystyle=[X_{1},[X_{3},Y]_{2}]_{4}-[X_{3},[X_{1},Y]_{2}]_{4}=\langle[X_{3},Y],X_{2}\rangle\omega_{12}+\langle[X_{1},Y],X_{2}\rangle\omega_{23}$
$\displaystyle=\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle
Y,\omega_{23}\rangle\omega_{12}-\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle
Y,\omega_{12}\rangle\omega_{23},$ $\displaystyle F_{23}(Y)$
$\displaystyle=[X_{2},[X_{3},Y]_{1}]_{4}-[X_{3},[X_{2},Y]_{1}]_{4}=-\langle[X_{3},Y],X_{1}\rangle\omega_{12}+\langle[X_{2},Y],X_{1}\rangle\omega_{13}$
$\displaystyle=-\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle\omega_{12}+\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle
Y,\omega_{12}\rangle\omega_{13}.$

Thus

$\begin{cases}({\mathrm{ad}}_{\omega_{12}})_{|\mathfrak{g}_{4}}=[\rho_{1},\rho_{2}]-r_{3}\rho_{3}-\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle\bullet,\omega_{23}\rangle\omega_{13}+\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle\bullet,\omega_{13}\rangle\omega_{23},\\\
({\mathrm{ad}}_{\omega_{13}})_{|\mathfrak{g}_{4}}=[\rho_{1},\rho_{3}]-r_{2}\rho_{2}+\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle\bullet,\omega_{23}\rangle\omega_{12}-\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle\bullet,\omega_{12}\rangle\omega_{23},\\\
({\mathrm{ad}}_{\omega_{23}})_{|\mathfrak{g}_{4}}=[\rho_{2},\rho_{3}]-r_{1}\rho_{1}-\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle\bullet,\omega_{13}\rangle\omega_{12}+\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle\bullet,\omega_{12}\rangle\omega_{13}.\end{cases}$
(25)

Now, for $(i,j)=(1,2)$ we identify the $\mathfrak{g}_{k}$-component of (24)
for $k=1,2,3$,

$\begin{cases}r_{3}\langle[X_{3},Y],X_{1}\rangle=-\langle[X_{2},\rho_{1}(Y)],X_{1}\rangle-\langle[X_{2},[X_{1},Y]_{3}],X_{1}\rangle,\\\
r_{3}\langle[X_{3},Y],X_{2}\rangle=\langle[X_{1},\rho_{2}(Y)],X_{2}\rangle+\langle[X_{1},[X_{2},Y]_{3}],X_{2}\rangle,\\\
0=\langle[X_{1},\rho_{2}(Y)],X_{3}\rangle-\langle[X_{2},\rho_{1}(Y)],X_{3}\rangle.\end{cases}$
(26)

We do the same for $(i,j)=(1,3)$

$\begin{cases}r_{2}\langle[X_{2},Y],X_{1}\rangle=-\langle[X_{3},\rho_{1}(Y)],X_{1}\rangle-\langle[X_{3},[X_{1},Y]_{2}],X_{1}\rangle,\\\
0=\langle[X_{1},\rho_{3}(Y)],X_{2}\rangle-\langle[X_{3},\rho_{1}(Y)],X_{2}\rangle,\\\
r_{2}\langle[X_{2},Y],X_{3}\rangle=\langle[X_{1},\rho_{3}(Y)],X_{3}\rangle+\langle[X_{1},[X_{3},Y]_{2}],X_{3}\rangle.\end{cases}$
(27)

Finally, we take $(i,j)=(2,3)$,

$\begin{cases}0=\langle[X_{2},\rho_{3}(Y)],X_{1}\rangle-\langle[X_{3},\rho_{2}(Y)],X_{1}\rangle,\\\
r_{1}\langle[X_{1},Y],X_{2}\rangle=-\langle[X_{3},\rho_{2}(Y)],X_{2}\rangle-\langle[X_{3},[X_{2},Y]_{1}],X_{2}\rangle,\\\
r_{1}\langle[X_{1},Y],X_{3}\rangle=\langle[X_{2},\rho_{3}(Y)],X_{3}\rangle+\langle[X_{2},[X_{3},Y]_{1}],X_{3}\rangle.\end{cases}$
(28)

By using (7), we get that (26)-(28) are equivalent to

$\begin{cases}r_{3}\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle=\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle
Y,\rho_{1}(\omega_{12})\rangle+r_{1}\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle,\\\ r_{3}\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle
Y,\omega_{23}\rangle=\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle
Y,\rho_{2}(\omega_{12})\rangle-
r_{2}\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle Y,\omega_{23}\rangle,\\\
\langle\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\rho_{2}(\omega_{13})-\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\rho_{1}(\omega_{23}),Y\rangle=0,\end{cases}$
$\begin{cases}r_{2}\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle
Y,\omega_{12}\rangle=\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle
Y,\rho_{1}(\omega_{13})\rangle-
r_{1}\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle Y,\omega_{12}\rangle,\\\
\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle
Y,\rho_{3}(\omega_{12})\rangle+\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle
Y,\rho_{1}(\omega_{23})\rangle=0,\\\
-r_{2}\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle
Y,\omega_{23}\rangle=\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle
Y,\rho_{3}(\omega_{13})\rangle+r_{3}\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle
Y,\omega_{23}\rangle,\end{cases}$
$\begin{cases}-\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle
Y,\rho_{3}(\omega_{12})\rangle+\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle
Y,\rho_{2}(\omega_{13})\rangle=0,\\\
-r_{1}\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle
Y,\omega_{12}\rangle=\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle
Y,\rho_{2}(\omega_{23})\rangle+r_{2}\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\langle
Y,\omega_{12}\rangle,\\\
-r_{1}\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle=\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle
Y,\rho_{3}(\omega_{23})\rangle-
r_{3}\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}\langle
Y,\omega_{13}\rangle.\end{cases}$

These relations hold for any $Y\in\mathfrak{g}_{4}$ and hence

$\begin{cases}\rho_{1}(\omega_{12})=\frac{\lambda_{12}^{2}}{\lambda_{24}^{2}}\left(r_{3}\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}-r_{1}\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\right)\omega_{13},\\\
\rho_{1}(\omega_{13})=\frac{\lambda_{13}^{2}}{\lambda_{34}^{2}}\left(r_{2}\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}+r_{1}\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\right)\omega_{12},\end{cases},\begin{cases}\rho_{2}(\omega_{12})=\frac{\lambda_{12}^{2}}{\lambda_{14}^{2}}\left(r_{3}\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}+r_{2}\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\right)\omega_{23},\\\
\rho_{2}(\omega_{23})=-\frac{\lambda_{23}^{2}}{\lambda_{34}^{2}}\left(r_{1}\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}+r_{2}\frac{\lambda_{24}^{2}}{\lambda_{12}^{2}}\right)\omega_{12},\end{cases},\begin{cases}\rho_{3}(\omega_{13})=-\frac{\lambda_{13}^{2}}{\lambda_{14}^{2}}\left(r_{2}\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}+r_{3}\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\right)\omega_{23},\\\
\rho_{3}(\omega_{23})=\frac{\lambda_{23}^{2}}{\lambda_{24}^{2}}\left(r_{3}\frac{\lambda_{34}^{2}}{\lambda_{13}^{2}}-r_{1}\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\right)\omega_{13},\\\
\end{cases}$ (29)

and

$\begin{cases}{\lambda_{23}^{2}}{\lambda_{14}^{2}}\rho_{2}(\omega_{13})-{\lambda_{13}^{2}}{\lambda_{24}^{2}}\rho_{1}(\omega_{23})=0,\\\
{\lambda_{23}^{2}}{\lambda_{14}^{2}}\rho_{3}(\omega_{12})+{\lambda_{12}^{2}}{\lambda_{34}^{2}}\rho_{1}(\omega_{23})=0,\\\
-{\lambda_{13}^{2}}{\lambda_{24}^{2}}\rho_{3}(\omega_{12})+{\lambda_{12}^{2}}{\lambda_{34}^{2}}\rho_{2}(\omega_{13})=0.\end{cases}$

The determinant of this system in
$(\rho_{1}(\omega_{23}),\rho_{2}(\omega_{13}),\rho_{3}(\omega_{12}))$ in non
zero and hence

$\rho_{1}(\omega_{23})=\rho_{2}(\omega_{13})=\rho_{3}(\omega_{12})=0.$ (30)

From (25), (29) and (30),
$\mathrm{span}\\{\omega_{12},\omega_{13},\omega_{23}\\}={\mathfrak{p}}_{4}$ is
a subalgebra invariant by $\rho_{i}$ for $i=1,2,3$. Moreover, by virtue of
Proposition 20, its orthogonal ${\mathfrak{h}}_{4}$ is a subalgebra. On the
other hand, the endomorphisms $[\rho_{1},\rho_{2}]-r_{3}\rho_{3}$,
$[\rho_{1},\rho_{3}]-r_{2}\rho_{2}$ and $[\rho_{2},\rho_{3}]-r_{1}\rho_{1}$
are skew-symmetric and leave invariant ${\mathfrak{p}}_{4}$ so they leave
invariant ${\mathfrak{h}}_{4}$. This implies, by virtue of (25), that
${\mathfrak{h}}_{4}$ is an ideal of $\mathfrak{g}_{4}$. But, since
$\mathfrak{g}$ is not Ricci-parallel, $\mathfrak{g}_{4}$ is not solvable by
virtue of Theorem 3.2 and hence it is simple. But
${\mathfrak{p}}_{4}\not=\\{0\\}$ otherwise $\mathfrak{g}_{4}$ is Ricci-
parallel. In conclusion ${\mathfrak{h}}_{4}=0$ and
$\mathfrak{g}_{4}={\mathfrak{p}}_{4}$.

Let us show now that $\\{\omega_{12},\omega_{13},\omega_{23}\\}$ is
orthonormal. This is true if $\rho_{i}\not=0$ for some $i$ by virtue of (29)
and (30). Suppose now that $\rho_{1}=\rho_{2}=\rho_{3}=0$. From (25), we get

$[\omega_{12},\omega_{13}]=-\frac{\lambda_{24}^{2}}{\lambda_{23}^{2}}\langle\omega_{13},\omega_{23}\rangle\omega_{13}+\frac{\lambda_{14}^{2}}{\lambda_{13}^{2}}\langle\omega_{13},\omega_{13}\rangle\omega_{23}=-\frac{\lambda_{34}^{2}}{\lambda_{23}^{2}}\langle\omega_{12},\omega_{23}\rangle\omega_{12}+\frac{\lambda_{14}^{2}}{\lambda_{12}^{2}}\langle\omega_{12},\omega_{12}\rangle\omega_{23}$

This shows that
$\langle\omega_{13},\omega_{23}\rangle=\langle\omega_{12},\omega_{23}\rangle=0$
and a similar argument gives $\langle\omega_{13},\omega_{12}\rangle=0$.

In conclusion, we take $e_{i}=X_{i}$ for $i=1,2,3$ and
$(\omega_{12},\omega_{13},\omega_{23})=(\mu_{1}e_{4},\mu_{2}e_{5},\mu_{3}e_{6})$
to get the desired expression of the Lie brackets. ∎

######  Theorem 4.3.

Let
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{3}\oplus\mathfrak{g}_{4}$
a Euclidean Lie algebra with harmonic curvature such that
$\dim\mathfrak{g}_{1}=\dim\mathfrak{g}_{2}=\dim\mathfrak{g}_{3}=1$,
$\dim\mathfrak{g}_{4}=3$ and $\lambda_{1}<\lambda_{2}<\lambda_{3}$. Then
$\mathfrak{g}$ is Ricci-parallel.

###### Proof.

Suppose that $\mathfrak{g}$ is not Ricci-parallel. Then we can apply
Proposition 4.1 and there exists an orthonormal basis $\hbox{\bb
B}=(e_{1},\ldots,e_{6})$ of $\mathfrak{g}$ such that (23) holds. For any
$u,v,w\in\mathfrak{g}$, put

$J(u,v,w)=[u,[v,w]]+[v,[w,u]]+[w,[u,v]]$

and denote by $u_{i}$ the $i$-component of $u$ in the basis 𝔹. A direct
computation gives

$\begin{cases}J(e_{1},e_{4},e_{6})_{4}=-b\alpha_{{1}}-{\frac{\alpha_{{2}}{\lambda_{{14}}}^{2}\mu_{{1}}}{{\lambda_{{12}}}^{2}}}-\alpha_{{1}}c=0,\\\
J(e_{1},e_{5},e_{6})_{5}=\alpha_{{1}}c-{\frac{\alpha_{{3}}{\lambda_{{14}}}^{2}\mu_{{2}}}{{\lambda_{{13}}}^{2}}}+b\alpha_{{1}}=0,\\\
J(e_{2},e_{4},e_{5})_{4}=-a\alpha_{{2}}+{\frac{\alpha_{{1}}{\lambda_{{24}}}^{2}\mu_{{1}}}{{\lambda_{{12}}}^{2}}}+\alpha_{{2}}c=0,\\\
J(e_{2},e_{5},e_{6})_{6}=\alpha_{{2}}c-{\frac{\alpha_{{3}}{\lambda_{{24}}}^{2}\mu_{{3}}}{{\lambda_{{23}}}^{2}}}-a\alpha_{{2}}=0,\\\
J(e_{3},e_{4},e_{5})_{5}=-a\alpha_{{3}}+{\frac{\alpha_{{1}}{\lambda_{{34}}}^{2}\mu_{{2}}}{{\lambda_{{13}}}^{2}}}-\alpha_{{3}}b=0,\\\
J(e_{3},e_{4},e_{6})_{6}=\alpha_{{3}}b+{\frac{\alpha_{{2}}{\lambda_{{34}}}^{2}\mu_{{3}}}{{\lambda_{{23}}}^{2}}}+a\alpha_{{3}}=0.\\\
\end{cases}$

If $E_{i}$ designs the $i$-equation in this system then $E_{1}+E_{2}=0$,
$E_{3}-E_{4}=0$ and $E_{5}+E_{6}=0$ are equivalent to

$\begin{cases}\displaystyle-{\frac{\alpha_{{2}}{\lambda_{{14}}}^{2}\mu_{{1}}}{{\lambda_{{12}}}^{2}}}-{\frac{\alpha_{{3}}{\lambda_{{14}}}^{2}\mu_{{2}}}{{\lambda_{{13}}}^{2}}}=0,\\\
\displaystyle{\frac{\alpha_{{1}}{\lambda_{{24}}}^{2}\mu_{{1}}}{{\lambda_{{12}}}^{2}}}+{\frac{\alpha_{{3}}{\lambda_{{24}}}^{2}\mu_{{3}}}{{\lambda_{{23}}}^{2}}}=0,\\\
\displaystyle{\frac{\alpha_{{1}}{\lambda_{{34}}}^{2}\mu_{{2}}}{{\lambda_{{13}}}^{2}}}+{\frac{\alpha_{{2}}{\lambda_{{34}}}^{2}\mu_{{3}}}{{\lambda_{{23}}}^{2}}}=0.\end{cases}$

This is equivalent to

$M\left(\begin{array}[]{c}\mu_{1}\\\ \mu_{2}\\\
\mu_{3}\end{array}\right)=\left(\begin{array}[]{c}0\\\ 0\\\
0\end{array}\right)\quad\mbox{and}\quad
M=\left(\begin{array}[]{ccc}-{\frac{\alpha_{{2}}{\lambda_{{14}}}^{2}}{{\lambda_{{12}}}^{2}}}&-{\frac{\alpha_{{3}}{\lambda_{{14}}}^{2}}{{\lambda_{{13}}}^{2}}}&0\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr{\frac{\alpha_{{1}}{\lambda_{{24}}}^{2}}{{\lambda_{{12}}}^{2}}}&0&{\frac{\alpha_{{3}}{\lambda_{{24}}}^{2}}{{\lambda_{{23}}}^{2}}}\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&{\frac{\alpha_{{1}}{\lambda_{{34}}}^{2}}{{\lambda_{{13}}}^{2}}}&{\frac{\alpha_{{2}}{\lambda_{{34}}}^{2}}{{\lambda_{{23}}}^{2}}}\end{array}\right).$

Since $(\mu_{1},\mu_{2},\mu_{3})\not=(0,0,0)$ we get

$\det
M=2\,{\frac{\alpha_{{2}}{\lambda_{{14}}}^{2}\alpha_{{3}}{\lambda_{{24}}}^{2}\alpha_{{1}}{\lambda_{{34}}}^{2}}{{\lambda_{{12}}}^{2}{\lambda_{{23}}}^{2}{\lambda_{{13}}}^{2}}}=0.$

We conclude that $\alpha_{1}=\alpha_{2}=\alpha_{3}=0$.

On the other hand,

$\begin{cases}\displaystyle
J(e_{1},e_{2},e_{5})_{1}=-{\frac{\mu_{{2}}r\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{14}}\lambda_{{23}}\right)\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{14}}\lambda_{{23}}\right)}{{\lambda_{{23}}}^{2}{\lambda_{{13}}}^{2}}}=0\\\
\displaystyle
J(e_{1},e_{2},e_{6})_{2}=-{\frac{\mu_{{3}}r\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{13}}\lambda_{{24}}\right)\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{13}}\lambda_{{24}}\right)}{{\lambda_{{23}}}^{4}}}=0\\\
\displaystyle
J(e_{1},e_{3},e_{4})_{1}={\frac{\mu_{{1}}r\left(\lambda_{{13}}\lambda_{{24}}-\lambda_{{14}}\lambda_{{23}}\right)\left(\lambda_{{13}}\lambda_{{24}}+\lambda_{{14}}\lambda_{{23}}\right)}{{\lambda_{{12}}}^{2}{\lambda_{{23}}}^{2}}}=0.\end{cases}$

If $r\not=0$ this is equivalent to

$\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{14}}\lambda_{{23}}\right)\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{14}}\lambda_{{23}}\right)=\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{13}}\lambda_{{24}}\right)\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{13}}\lambda_{{24}}\right)=0.$

But

$\begin{cases}\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{13}}\lambda_{{24}}\right)=(\lambda_{2}-\lambda_{3})(\lambda_{4}-\lambda_{1}),\\\
\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{14}}\lambda_{{23}}\right)=(\lambda_{1}-\lambda_{3})(\lambda_{2}-\lambda_{4}),\\\
\left(\lambda_{{12}}\lambda_{{34}}-\lambda_{{14}}\lambda_{{23}}\right)+\left(\lambda_{{12}}\lambda_{{34}}+\lambda_{{13}}\lambda_{{24}}\right)=3(\lambda_{1}-\lambda_{2})(\lambda_{3}-\lambda_{4}),\end{cases}$

which shows that this situation is impossible and hence $r=0$.

So far we have shown that $r=\alpha_{1}=\alpha_{2}=\alpha_{3}=0$. Now,

$J(e_{1},e_{2},e_{5})_{6}={\frac{{\lambda_{{14}}}^{2}\mu_{{2}}\mu_{{3}}}{{\lambda_{{13}}}^{2}}}-\mu_{{1}}a,\;J(e_{1},e_{2},e_{6})_{5}=-{\frac{{\lambda_{{24}}}^{2}\mu_{{3}}\mu_{{2}}}{{\lambda_{{23}}}^{2}}}-\mu_{{1}}b,\;J(e_{1},e_{3},e_{6})_{4}={\frac{{\lambda_{{34}}}^{2}\mu_{{3}}\mu_{{1}}}{{\lambda_{{23}}}^{2}}}-\mu_{{2}}c$

and hence

$a={\frac{{\lambda_{{14}}}^{2}\mu_{{2}}\mu_{{3}}}{{\lambda_{{13}}}^{2}\mu_{{1}}}},\;b=-{\frac{{\lambda_{{24}}}^{2}\mu_{{3}}\mu_{{2}}}{{\lambda_{{23}}}^{2}\mu_{{1}}}}\quad\mbox{and}\quad
c={\frac{{\lambda_{{34}}}^{2}\mu_{{3}}\mu_{{1}}}{{\lambda_{{23}}}^{2}\mu_{{2}}}}.$

With all the relations established above, the vanishing of the Jacobi identity
is equivalent to

${\lambda_{{12}}}^{2}{\mu_{{2}}}^{2}-{\lambda_{{13}}}^{2}{\mu_{{1}}}^{2}={\lambda_{{12}}}^{2}{\mu_{{3}}}^{2}-{\lambda_{{23}}}^{2}{\mu_{{1}}}^{2}={\lambda_{{13}}}^{2}{\mu_{{3}}}^{2}-{\lambda_{{23}}}^{2}{\mu_{{2}}}^{2}=0,$

which is equivalent to

$\mu_{2}=\frac{\lambda_{13}\mu_{1}}{\lambda_{12}}\quad\mbox{and}\quad\mu_{3}=\frac{\lambda_{23}\mu_{1}}{\lambda_{12}}.$

We replace in $a,b,c$ and we get

$a=\frac{\lambda_{14}^{2}\lambda_{23}\mu_{1}}{\lambda_{12}^{2}\lambda_{13}},\;b=-\frac{\lambda_{24}^{2}\lambda_{13}\mu_{1}}{\lambda_{12}^{2}\lambda_{23}}\quad\mbox{and}\quad
c=\frac{\lambda_{34}^{2}\mu_{1}}{\lambda_{23}\lambda_{13}}.$

Now, by using Maple we get

$\begin{cases}\displaystyle\nabla_{e_{1}}(\mathrm{Ric})(e_{2},e_{4})-\nabla_{e_{2}}(\mathrm{Ric})(e_{1},e_{4})=4\,{\frac{{\mu_{{1}}}^{3}\left(\lambda_{{1}}-\lambda_{{4}}\right)\left(\lambda_{{2}}-\lambda_{{4}}\right)\left({\lambda_{{1}}}^{2}+\left(-\lambda_{{2}}-\lambda_{{4}}\right)\lambda_{{1}}+{\lambda_{{2}}}^{2}-\lambda_{{2}}\lambda_{{4}}+{\lambda_{{4}}}^{2}\right)}{\left(\lambda_{{1}}-\lambda_{{2}}\right)^{4}}},\\\
\displaystyle\nabla_{e_{1}}(\mathrm{Ric})(e_{3},e_{5})-\nabla_{e_{3}}(\mathrm{Ric})(e_{1},e_{5})=4\,{\frac{{\mu_{{1}}}^{3}\left(\lambda_{{1}}-\lambda_{{4}}\right)\left(\lambda_{{3}}-\lambda_{{4}}\right)\left({\lambda_{{1}}}^{2}+\left(-\lambda_{{3}}-\lambda_{{4}}\right)\lambda_{{1}}+{\lambda_{{3}}}^{2}-\lambda_{{3}}\lambda_{{4}}+{\lambda_{{4}}}^{2}\right)}{\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)^{3}}},\\\
\displaystyle\nabla_{e_{2}}(\mathrm{Ric})(e_{3},e_{6})-\nabla_{e_{3}}(\mathrm{Ric})(e_{2},e_{6})=4\,{\frac{\left(\lambda_{{2}}-\lambda_{{4}}\right){\mu_{{1}}}^{3}\left({\lambda_{{2}}}^{2}+\left(-\lambda_{{3}}-\lambda_{{4}}\right)\lambda_{{2}}+{\lambda_{{3}}}^{2}-\lambda_{{3}}\lambda_{{4}}+{\lambda_{{4}}}^{2}\right)\left(\lambda_{{3}}-\lambda_{{4}}\right)}{\left(\lambda_{{2}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)^{3}}}.\end{cases}$

So we must have

$\begin{cases}q_{1}:={\lambda_{{1}}}^{2}+\left(-\lambda_{{2}}-\lambda_{{4}}\right)\lambda_{{1}}+{\lambda_{{2}}}^{2}-\lambda_{{2}}\lambda_{{4}}+{\lambda_{{4}}}^{2}=0,\\\
q_{2}:={\lambda_{{1}}}^{2}+\left(-\lambda_{{3}}-\lambda_{{4}}\right)\lambda_{{1}}+{\lambda_{{3}}}^{2}-\lambda_{{3}}\lambda_{{4}}+{\lambda_{{4}}}^{2}=0,\\\
q_{3}:={\lambda_{{2}}}^{2}+\left(-\lambda_{{3}}-\lambda_{{4}}\right)\lambda_{{2}}+{\lambda_{{3}}}^{2}-\lambda_{{3}}\lambda_{{4}}+{\lambda_{{4}}}^{2}=0.\end{cases}$

This is equivalent

$\begin{cases}q_{1}-q_{2}=-\left(\lambda_{{2}}-\lambda_{{3}}\right)\left(-\lambda_{{2}}+\lambda_{{1}}+\lambda_{{4}}-\lambda_{{3}}\right)=0,\\\
q_{1}-q_{3}=\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}-\lambda_{{4}}+\lambda_{{3}}\right)=0,\\\
q_{3}=0\end{cases}$

which is equivalent to

$\begin{cases}p_{1}=\left(-\lambda_{{2}}+\lambda_{{1}}+\lambda_{{4}}-\lambda_{{3}}\right)=0,\\\
p_{2}=\left(\lambda_{{1}}-\lambda_{{2}}-\lambda_{{4}}+\lambda_{{3}}\right)=0,\\\
q_{3}=0.\end{cases}$

This implies that $p_{1}+p_{2}=2(\lambda_{1}-\lambda_{2})=0$ which is
impossible. This completes the proof. ∎

Recall that a left-invariant symmetric tensor field $A$ on $\mathfrak{g}$ is
called essential Codazzi if it satisfies the Codazzi equation, it is non
parallel and none of its eigenspace subalgebras $\mathfrak{g}_{k}$ is an
ideal. The proof of the precedent result gives the following:

######  Proposition 4.2.

We consider the 6-dimensional Euclidean Lie algebra with an orthonormal basis
$(e_{1},\ldots,e_{6})$ in which the non-vanishing Lie brackets are given by

$\begin{cases}[e_{1},e_{2}]=\mu_{1}e_{4},[e_{1},e_{3}]={\frac{\left(\lambda_{{1}}-\lambda_{{3}}\right)\mu_{{1}}}{\lambda_{{1}}-\lambda_{{2}}}}e_{5},[e_{1},e_{4}]=-{\frac{\left(\lambda_{{1}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{1}}-\lambda_{{2}}\right)^{2}}}e_{2},[e_{1},e_{5}]=-{\frac{\left(\lambda_{{1}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)}}e_{3},\\\
[e_{2},e_{3}]={\frac{\left(\lambda_{{2}}-\lambda_{{3}}\right)\mu_{{1}}}{\lambda_{{1}}-\lambda_{{2}}}}e_{6},[e_{2},e_{4}]={\frac{\left(\lambda_{{2}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{1}}-\lambda_{{2}}\right)^{2}}}e_{1},[e_{2},e_{6}]=-{\frac{\left(\lambda_{{2}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{2}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)}}e_{3},[e_{3},e_{5}]={\frac{\left(\lambda_{{3}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)}}e_{1},\\\
[e_{3},e_{6}]={\frac{\left(\lambda_{{3}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{2}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)}}e_{2},[e_{4},e_{5}]={\frac{\left(\lambda_{{1}}-\lambda_{{4}}\right)^{2}\mu_{{1}}\left(\lambda_{{2}}-\lambda_{{3}}\right)}{\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{1}}-\lambda_{{2}}\right)^{2}}}e_{6},[e_{4},e_{6}]=-\frac{\lambda_{24}^{2}\lambda_{13}\mu_{1}}{\lambda_{12}^{2}\lambda_{23}}e_{5},\;[e_{5},e_{6}]={\frac{\left(\lambda_{{3}}-\lambda_{{4}}\right)^{2}\mu_{{1}}}{\left(\lambda_{{1}}-\lambda_{{3}}\right)\left(\lambda_{{2}}-\lambda_{{3}}\right)}}e_{4}.\end{cases}$

Then the operator
$A=\mathrm{Diag}(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{4},\lambda_{4})$
is a symmetric essential Codazzi tensor. This Lie algebra is compact (its
Killing form is negative definite) and hence it is isomorphic to
$\mathrm{su}(2)\oplus\mathrm{su}(2)$.

## References

  * [1] J. E. d’Atri, Codazzi tensors and harmonic curvature for left invariant metrics, Geometriae Dedicata 19 (1985) 229-236.
  * [2] Arthur L. Besse, Einstein manifolds, Springer-Verlag Berlin Heidelberg 1987.
  * [3] Bourguignon, J. P., Les variétés de dimension 4 à courbure harmonique et à signature non-nulle sont d’Einstein, Invent. Math. 63 (1981), 263-286.
  * [4] Derdzinski, A., Classification of Certain Compact Riemannian Manifolds with Harmonic Curvature and Non-parallel Ricci Tensor, Math. Z. 172 (1980), 273-280.
  * [5] Derdzinski, A., On Compact Riemannian Manifolds with Harmonic Curvature, Mathematische Annalen volume 259, 145-152 (1982).
  * [6] Dotti I. M., Ricci Curvature of Left Invariant Metrics on Solvable Unimodular Lie Groups, Math. Z. 180 257-263 (1982).
  * [7] Fabio Podesta and Andrea Spiro, Four-Dimensional Einstein-Like Manifolds and Curvature Homogeneity, Geometriae Dedicata 54 225-243 (1995).
  * [8] Gray, A., Einstein-like manifolds which are not Einstein, Geom. Dedicata 7 (1978), 259-280.
  * [9] J. Lauret, Einstein solvmanifolds are standard, Annals of Mathematics, 172 (2010), 1859-1877.
  * [10] Lichnerowicz, A.: Ondes et radiations électromagnètiques et gravitationnelles en relativité gńérale, Ann. di Mat. Pura ed Appl. 50, 1-95 (1960).
  * [11] Milnor J., Curvature of left invariant metrics on Lie groups, Adv. in Math. 21 (1976), 283-329.
  * [12] ChiaKuei Peng and Chao Qian, Homogeneous Einstein-like metrics on spheres and projective spaces, Differential Geometry and its Applications Volume 44, February 2016, Pages 63-76.
  * [13] Hitoshi Takagi, Conformally flat Riemannian manifolds admitting a transitive group of isometries, Tôhoku Math. Journal, 27 (1975) 103-110.


# Going in Style: Audio Backdoors through Stylistic Transformations

###### Abstract

This work explores stylistic triggers for backdoor attacks in the audio
domain: dynamic transformations of malicious samples through guitar effects.
We first formalize stylistic triggers – currently missing in the literature.
Second, we explore how to develop stylistic triggers in the audio domain by
proposing JingleBack. Our experiments confirm the effectiveness of the attack,
achieving a 96% attack success rate. Our code is available in
https://github.com/skoffas/going-in-style.

##  1 Introduction

††∗Equal Contribution

Backdoor attacks are a class of machine learning threats where the attacker
embeds a secret functionality into the victim’s model, which can be triggered
at the testing time from malicious inputs [1, 2]. Backdoor triggers can be
grouped into two major families: static, when the trigger is a fixed pattern
attached to the poisoned sample [1], and dynamic when the trigger’s properties
vary for each poisoned sample [3, 4]. Dynamic triggers are generally stronger
as they can be effective under different conditions and potentially bypass
state-of-the-art countermeasures [3]. Recently, a new way of creating dynamic
triggers has emerged: stylistic triggers. For instance, paintings and writing
styles can be used as triggers in computer vision (CV) [5] and text domain
[6].

Motivation. Stylistic backdoors are powerful but not yet studied in the audio
domain. Such backdoors highly depend on the targeted application resulting in
domain-specific challenges. In audio, the most important challenge is to
create a style that alters the original signal in a way that is
distinguishable by the trained models but also keeps the audio quality at an
acceptable level. To our best knowledge, no work has explored stylistic
backdoors in audio, and we aim to fill this gap. We investigate the effect of
six stylistic triggers in a speech classification task in both clean and
dirty-label settings. Triggers are generated through electric guitar effects
that do not require training of complex generative models [5, 6], making our
attack easy to deploy. Our contributions are:

  * •

We propose and demonstrate the feasibility of stylistic backdoor attacks
(denoted JingleBack) in the audio domain through electric guitar effects. For
our experiments, we trained 444 models, and JingleBack reached up to 96%
attack success rate.

  * •

We are the first to formally describe stylistic backdoor attacks and establish
a domain-agnostic framework that can be used for stylistic backdoors in any
application.

##  2 Background

Evasion Attack. The attacker aims to find a small perturbation $\epsilon$ for
a sample $x$ to produce a misclassification in a target classifier
$\mathbf{C}$. The definition of $\epsilon$ is task-dependent [7]. For images,
$\epsilon$ is usually an additive noise at a pixel level computed via a
gradient-based procedure [8]. For audio, $\epsilon$ is a small waveform
computed through an optimization process [9].

Poisoning Attack. Attackers able to manipulate victims’ training data
$\mathcal{D}^{tr}$ can inject adversarial samples to increase $\mathbf{C}$’s
testing error and decrease the model’s performance [10]. Data manipulation is
a concrete threat because building datasets often involves untrusted sources,
and complex model training is outsourced to third parties.

Backdoor Attack. Data poisoning leads to backdooor attacks. The attackers
insert backdoors in the model resulting in targeted misclassifications [1].
Most backdoors are source-agnostic, as the trigger can be applied to any
dataset’s class. There are two different approaches to creating such a
backdoor: dirty-label [1], where the adversary adds the trigger to some
training samples but also alters their labels, and clean-label [11], where the
attacker only poisons samples from the target class. The dirty-label attack
produces stronger backdoors and requires less poisoned data, but the clean-
label attack is a more realistic threat as poisoned samples cannot be easily
identified by manual inspection [11]. For example, crowdsourced datasets like
Mozilla’s Common Voice [12] use volunteers to validate each audio file and its
transcription. When the attack is dirty-label, the volunteer can easily spot
inconsistencies between the provided audio and its transcription. However, in
a clean-label attack, the trigger will remain unnoticed because the audio and
its transcription will be correct. This work explores the effects of both
approaches on a style-based backdoor attack in speech classification.

##  3 Style Backdoor Attack

###  3.1 Threat Model

Attacker’s Capabilities. We assume a gray-box threat model where the adversary
has access to a small portion of the training data, which can be altered
without restriction, but has no knowledge of the training algorithm and the
model’s architecture. Such capabilities are realistic as modern datasets are
based on crowdsourcing [12], and malicious data may evade security checks
[13].

Attacker’s Aim. The attacker’s purpose is to insert a secret functionality
into the deployed model, which is activated when the trigger is present in the
model’s input. Usually, this functionality causes targeted misclassifications
with a very high probability. Furthermore, the model should behave normally
for any other input to avoid raising suspicions.

###  3.2 Attack Formulation

Prior studies in backdoor attacks mainly focus on static triggers, where the
attack trivially inserts a trigger $\epsilon$ in victim samples. Such addition
is sample-independent. Let $x$ be a sample, $\epsilon$ a trigger, the backdoor
$\mathcal{F}$ can be described as:

$\mathcal{F}(x,\epsilon)=x+\epsilon.$ (1)

In BadNets [1] (CV), the trigger is a patch at a pixel level that replaces a
part of the original sample. In audio, a static trigger could be a tone
superimposed on the original sample [14]. Static triggers can thus be
considered as a constant parameter of the attack, as a batch of benign samples
$\\{x_{0},\ldots,x_{n}\\}$ is transformed in nothing more than
$\\{x_{0}+\epsilon,\ldots,x_{n}+\epsilon\\}$. Thus, the victim’s model
$\mathbf{C}_{back}$ learns to associate the static pattern $\epsilon$ to a
target label $y*$.

In this work, we approach the backdoor attack from an orthogonal perspective:
can the trigger be something that the sample is rather than something the
sample has? Static backdoors answer the has proposition: the poisoned sample
contains a specific (and constant) pattern that $\mathbf{C}_{back}$ associates
with the target class. Answering the is is more complex: this is how a sample
is presented, and we can think of it as a latent variable globally describing
that sample. In our study, global descriptors are defined by stylistic
properties. Examples are image exposure (CV), writing complexity (text), and
the signal’s pitch (audio). Once the stylistic trigger $\epsilon$ is
identified, we need a function $\mathcal{S}_{\epsilon}$ that embeds such a
style to a given sample $x$. Formally, Eq. (1) changes to:

$\mathcal{F}(x,\mathcal{S}_{\epsilon})=\mathcal{S}_{\epsilon}(x).$ (2)

We explore different $\mathcal{S}_{\epsilon}$ for speech recognition in the
following sections. With the formulation given in Eq. (2), the batch of
adversarial samples$\\{\mathcal{S}_{\epsilon}(x_{0}),\ldots,$
$\mathcal{S}_{\epsilon}(x_{n})\\}$ now contains dynamic triggers since the
outcome of the backdoor generation varies based on the inputted sample. Now,
suppose that $\mathcal{S}_{\epsilon}$ exists. The stylistic backdoor is
effective if the model $\mathbf{C}_{back}$ learns an association between
$\mathcal{S}_{\epsilon}$ and the target class $y*$. This problem can be
formulated with two sub-questions:

  1. 1.

Can $\mathbf{C}$ learn to recognize the presence of $\mathcal{S}_{\epsilon}$?
We need proof or at least an indication that stylistic properties can be
learned by the victim’s model.

  2. 2.

Can we let $\mathbf{C}$ learn the association between $\mathcal{S}_{\epsilon}$
and $y*$? If the previous question is positively answered, we need to
understand further how to create the backdoor and what are the possible
obstacles at this stage.

Geirhos et al. [15] answered the first question, showing that CNNs focus more
on textures rather than object shapes. For example, an image with ‘cat’ shapes
and ‘elephant’ texture is classified as an elephant. Similarly, styles can be
learned in the speech domain [16, 17].

Finally, we need to understand what are the conditions to create the stylistic
backdoor in $\mathbf{C}$. In source-agnostic backdoors, the trigger is present
only in one of the classes of the training data. This condition is easily
satisfied by the classic backdoor attacks injecting artifacts as triggers.
Conversely, with stylistic backdoors, such conditions might not always be met.
For example, high exposure (image domain) and low-frequency tunes (audio
domain) might be present in many classes of clean data. We conclude that the
choice of $\mathcal{S}_{\epsilon}$ impacts attack success. In our experiments,
we use stylistic functions that are unlikely to be present in the training
data.

##  4 JingleBack Design

###  4.1 Stylistic Generation

This work focuses on backdoors in audio aiming into an easy-to-deploy attack.
Thus, we investigate if such an attack can be implemented through simple
digital music effects like chorus or gain. We used Spotify’s
pedalboard111https://github.com/spotify/pedalboard to implement six styles by
combining effects like PitchShift, Distortion, Chorus, Reverb, Gain,
Ladderfilter, and Phaser. We explain the adopted effects briefly and show
their analytical mathematical expressions in Table 1.

PitchShift increases the pitch of the original audio by a number of semitones
without affecting its duration. The pitch is the lowest frequency $f_{0}$ of a
signal $S$ [18, p. xv]. A semitone $sem$ is the smallest musical interval used
in music denoting a different tone. Distortion applies a distortion with a
$tanh()$ waveshaper. $\beta$ controls the signal’s amplitude increase. Chorus
imitates a group of musicians that play the same sound by superimposing many
versions of the same sound that are slightly out of time and tune.
Pedalboard’s chorus uses one unsynchronized version with the provided delay
$d$ where $\alpha$ controls the chorus’s amplitude. Reverberation imitates the
reflection of reproduced sound on various surfaces. Spotify’s pedalboard is
based on FreeVerb222https://ccrma.stanford.edu/~jos/pasp/Freeverb.html, which
uses eight parallel Schroeder-Moorer filtered-feedback comb-filters that
create eight delayed versions of the input signal that are added and fed into
four Schroeder all-pass filters in series. In Table 1, $AP_{1-4}$ is the
combined effect of the four cascaded all-pass filters, and $CF_{i}$ is the
$i^{th}$ comb-filter. Gain changes the signal amplitude by a factor $G$.
Spotify pedalboard implements a Moog Ladder filter $L(\cdot)$ [19]. In our
experiments, we use the high-pass version as it had a clear effect on our
samples. Phaser is based on special filters that can change the frequencies
they block over time through a low-frequency oscillator $P$. The phaser
superimposes the original signal with its altered version. This results in a
soft-moving sound. $\alpha$ controls the effect’s intensity.

Table 1: Equations of the chosen effects.

Effect Equation
---
$\mathrm{PitchShift}(S(f_{0}),sem)=S(f_{0}\cdot e^{sem/12})$
$\mathrm{Distortion}(S(t),\beta)=\beta\cdot tanh(S(t))$
$\mathrm{Chorus}(S(t),d,\alpha)=S(t)+\alpha\cdot S(t-d)$
$\mathrm{Reverb}(S(t))=AP_{1-4}[\sum_{i=1}^{8}CF_{i}(S(t))]$
$\mathrm{Gain}(S(t),G)=G\cdot S(t)$
$\mathrm{Ladder}(S(t))=\alpha\cdot L(S(t))$
$\mathrm{Phaser}(S(t))=S(t)+\alpha\cdot P(t,S(t))$

###  4.2 Experimental Settings

Dataset and Features. We used Google’s Speech Commands (GSC) dataset [20],
containing 30 classes of audio keywords like “yes”, “no”, “up”, and “down”. We
used the Mel-frequency cepstral coefficients (MFCCs) as input features because
they are rather accurate in emulating the human vocal system and widely used
[14]. We use common settings described in the literature [21], i.e., 40-mel
bands, a step of 10ms, and a window length of 25ms.

Backdoor. We split our data into training, validation, and test sets in a
64/16/20 way. We poisoned up to 1% of the training data and used two backdoor
settings: clean-label and dirty-label attacks. We chose the class “yes” as the
target class without loss of generality since we expect similar behavior
regardless of the target class. Triggers are generated with the six styles
described in Table 2. Parameters are selected to limit sample distortion and
preserve their quality.

Table 2: Stylistic triggers deployed in our experiments.

Style | Effect
---|---
0 | $\mathrm{PitchShift}(S,10)$
1 | $\mathrm{Distortion}(S,30dB)$
2 | $\mathrm{Chorus}(S,10ms,5)$
3 | $\mathrm{Chorus}(\mathrm{Distortion}(\mathrm{PitchShift}(S,10),20dB),8ms,5)$
4 | $\mathrm{Reverb(Distortion(Chorus}(S,15ms,0.25),20dB))$
5 | $\mathrm{Phaser(Ladder(Gain}(S,12dB)))$

User study. We conducted a user study (250 samples, 30 participants) to verify
that the adversarial samples preserve the original audio. Each sample is
reviewed by three participants. We analyze the perturbation effect in terms of
users’ accuracy (random guessing at 3%). 95% of legitimate samples are
correctly classified, implying that the dataset contains audio that is not
always comprehensible by target users. Stylistic transformation, instead,
slightly degrades samples quality; in order, from style 0 to style 5, we found
the following scores: 71%, 80%, 89%, 45%, 86%, and 86%.

Models. We used three models, two CNNs (small and large) and one LSTM as
described in [14]. Experiments are repeated four times to limit the randomness
in results. Each model is trained for a maximum of 300 epochs, with an early
stopping (patience of 20 epochs) based on the validation loss. In total, by
considering stylistic triggers (6), backdoor settings (2), models (3),
poisoning rates (3), and repetitions (4), 432 poisoned models are trained. We
also trained 12 clean models (4 repetitions for each model) to use as a
reference when investigating the backdoor’s effect on the original task.

##  5 Experimental Results

###  5.1 Effect on Clean Accuracy

We first verify that the backdoored models have comparable performance to
their clean versions. Clean models show, on average, high performance
(expressed as F1-score): large CNN $93.8\pm 0.2$, small CNN $87.2\pm 0.3$, and
LSTM $90.8\pm 1.1$. We notice that our attack is stealthy since we observed
only a small drop in the performance of the backdoored models. On average,
models drop $0.24\pm 0.9$ pp (points percentage) in the F1-score, while the
worst model drops 4.87 pp. Among the 432 poisoned models, the performance drop
is $>2$ pp in 23 cases and decreases to 10 cases when $>3$ pp.

###  5.2 Backdoor

We analyze the performance of our proposed stylistic backdoor under clean and
dirty-label settings. Figure 1 shows the results. The y-axis shows the attack
success rate (ASR), which represents the percentage of successfully triggered
backdoors over a number of tries, and the x-axis shows the poisoning rate,
which is the percentage of the poisoned training samples used for our attack.

Fig. 1: Backdoor attack success rate.

Our results are in line with the literature [14, 11]. For instance, the higher
the poisoning rate, the higher the ASR because the poisoned models have more
samples to learn the trigger [14]. Additionally, the dirty-label attack is
more effective as it almost always results in higher ASR than the clean-label
attack, as shown in [11]. In particular, dirty-label attacks include cases in
which even a small poisoning rate ($0.1\%$) is effective (ASR $>50\%$):
examples are style 2 in Large-CNN, style 3 in Large-CNN, and style 5 in both
Large-CNN and Small-CNN. Furthermore, Large-CNN – the best-performing model on
average – is the most vulnerable, showing that backdoor effectiveness is
connected to the model’s ability to learn.

Generally, we can notice that styles have different effects (e.g., style 3 is
always more effective than style 1), leading to the following observations.
(i) The clean-label attack is effective only with styles 3 and 5. (ii) Dirty-
label attack shows better performance in all cases. (iii) The addition of
effects does not always result in a performance boost (e.g., style 0
outperforms style 4 in clean-label settings).

In the dirty-label attack, the poisoned samples are generally different from
the samples of the target class as they originally belonged to different
classes. As a result, the distance between these samples in the feature space
is large and easy to learn by our models, even by applying simple effects to
them. However, in the clean-label attack, the poisoned samples belong to the
target class, and there is a higher probability that their features are not
very different from the clean samples. For that reason, the dirty-label attack
is more effective than the clean-label attack, which explains (i) and (ii). We
believe that adding more effects is not the best way to create distinguishable
triggers because what is most important is to change the signal’s
representation in the feature space. For that reason, a simpler transformation
that alters the frequency representation of the original signal, like
$PitchShift$ may result in a larger difference in the MFCCs as they use
Fourier transform internally. This explains (iii).

###  5.3 Ablation Study

To understand the effect of each style’s parameter on ASR, we performed an
ablation study for our most effective styles (styles 2 and 5) on GSC. In this
study, we varied the chorus’s amplitude for style 2 ($[2,4,6,8,10]$) and the
gain level for style 5 ($[4,8,12,16,20]$). Figure 2 shows that larger
distortion increases ASR. The effect is more visible when the attack performs
poorly, i.e., for the clean-label attack. Future investigations should analyze
the trade-off between ASR and audio quality (or stealthiness).

Fig. 2: ASR for different distortion levels for styles 2 and 5.

###  5.4 Evasion

Cao et al. [22] showed that – in CV – stylistic transformations can lead to
evasion attacks. Following such intuition, we tested our malicious samples on
unpoisoned models and analyzed the evasion rate (misclassification). Style 0
and style 3 are the most effective attacks (70% evasion, on average); however,
the user study showed that these styles degrade the samples’ quality. The
other triggers show better evasion performance, e.g., style 5 with an evasion
rate equal to 42.8% (Large-CNN), 70.2% (Small-CNN), and 50.7% (LSTM).

##  6 Generalization on Different Datasets

We now demonstrate that the attack generalizes in different datasets. We use
the large-CNN for styles 2 and 5 (the most effective ones), with a poisoning
rate equal to 1% for the dirty-label scenario. Other settings are the same as
in Section 4.2.

TIMIT [23]. We use the dataset’s core test set, consisting of 240 audio clips
from 24 speakers (speaker classification). Each sample is clipped to 1 second
to maintain consistency with the experimental settings adopted in this paper.
This task is challenging (F1-score 0.57), but the results show that ASR is, on
average, 41.4% (with 9.4 std) and 77.4% (with 4.5 std) for styles 2 and 5,
respectively.

Keras Speaker
Recognition333https://keras.io/examples/audio/speaker_recognition_using_cnn is
a speaker classification task of 5 users. The task is easier than the previous
one (F1-score 0.99). The results show that ASR is, on average, 89.9% (with 5.3
std) and 99.5% (with 0.6 std) for style 2 and style 5, respectively.

##  7 Conclusions and Future Work

This work contribution is twofold: a formal description of stylistic backdoors
and JingleBack, the first stylistic backdoor in the audio domain. We
demonstrated the potential of our attack through an extensive evaluation
considering 444 models, reaching up to 96% of the attack success rate. Future
investigations are required to better understand the stylistic backdoors in
the audio domain, for example, by considering the impact of the target class
or, more in general, how to find an optimal style effective in both clean and
dirty-label settings. Finally, further investigation is needed regarding the
stylistic backdoor’s behavior against state-of-the-art countermeasures.

## References

  * [1] T. Gu, B. Dolan-Gavitt, and S. Garg, “Badnets: Identifying vulnerabilities in the machine learning model supply chain,” arXiv preprint arXiv:1708.06733, 2017.
  * [2] T. Zhai, Y. Li, Z. Zhang, B. Wu, Y. Jiang, and S.-T. Xia, “Backdoor attack against speaker verification,” in ICASSP 2021-2021 IEEE International Conference on Acoustics, Speech and Signal Processing (ICASSP), pp. 2560–2564, IEEE, 2021.
  * [3] A. Salem, R. Wen, M. Backes, S. Ma, and Y. Zhang, “Dynamic backdoor attacks against machine learning models,” in 2022 IEEE 7th European Symposium on Security and Privacy, pp. 703–718, 2022.
  * [4] S. Koffas, S. Picek, and M. Conti, “Dynamic backdoors with global average pooling,” in 2022 IEEE 4th International Conference on Artificial Intelligence Circuits and Systems, pp. 320–323, 2022.
  * [5] S. Cheng, Y. Liu, S. Ma, and X. Zhang, “Deep feature space trojan attack of neural networks by controlled detoxification,” Proceedings of the AAAI Conference on Artificial Intelligence, vol. 35, pp. 1148–1156, May 2021.
  * [6] F. Qi, Y. Chen, X. Zhang, M. Li, Z. Liu, and M. Sun, “Mind the style of text! adversarial and backdoor attacks based on text style transfer,” in Proceedings of the 2021 Conference on Empirical Methods in Natural Language Processing, pp. 4569–4580, ACM, Nov. 2021.
  * [7] N. Dalvi, P. Domingos, Mausam, S. Sanghai, and D. Verma, “Adversarial classification,” in Proceedings of the Tenth ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, p. 99–108, Association for Computing Machinery, 2004.
  * [8] S. Huang, N. Papernot, I. Goodfellow, Y. Duan, and P. Abbeel, “Adversarial attacks on neural network policies,” 2017.
  * [9] N. Carlini and D. Wagner, “Audio adversarial examples: Targeted attacks on speech-to-text,” in 2018 IEEE security and privacy workshops (SPW), pp. 1–7, IEEE, 2018.
  * [10] B. Biggio, B. Nelson, and P. Laskov, “Poisoning attacks against support vector machines,” in Proceedings of the 29th International Coference on International Conference on Machine Learning, p. 1467–1474, 2012.
  * [11] A. Turner, D. Tsipras, and A. Madry, “Label-consistent backdoor attacks,” arXiv preprint arXiv:1912.02771, 2019.
  * [12] R. Ardila, M. Branson, K. Davis, M. Henretty, M. Kohler, J. Meyer, R. Morais, L. Saunders, F. M. Tyers, and G. Weber, “Common voice: A massively-multilingual speech corpus,” arXiv preprint arXiv:1912.06670, 2019.
  * [13] V. U. Prabhu and A. Birhane, “Large image datasets: A pyrrhic win for computer vision?,” CoRR, vol. abs/2006.16923, 2020.
  * [14] S. Koffas, J. Xu, M. Conti, and S. Picek, “Can you hear it? backdoor attacks via ultrasonic triggers,” in Proceedings of the 2022 ACM Workshop on Wireless Security and Machine Learning, p. 57–62, Association for Computing Machinery, 2022.
  * [15] R. Geirhos, P. Rubisch, C. Michaelis, M. Bethge, F. A. Wichmann, and W. Brendel, “Imagenet-trained CNNs are biased towards texture; increasing shape bias improves accuracy and robustness.,” in International Conference on Learning Representations, 2019.
  * [16] E. Grinstein, N. Q. Duong, A. Ozerov, and P. Pérez, “Audio style transfer,” in 2018 IEEE international conference on acoustics, speech and signal processing, pp. 586–590, IEEE, 2018.
  * [17] E. A. AlBadawy and S. Lyu, “Voice Conversion Using Speech-to-Speech Neuro-Style Transfer,” in Proc. Interspeech 2020, pp. 4726–4730, 2020\.
  * [18] B. Benward, Music in Theory and Practice Volume 1.  McGraw-Hill Higher Education, 2014.
  * [19] R. A. Moog, “A voltage-controlled low-pass high-pass filter for audio signal processing,” in Audio Engineering Society Convention 17, Audio Engineering Society, 1965.
  * [20] P. Warden, “Speech commands: A dataset for limited-vocabulary speech recognition,” CoRR, vol. abs/1804.03209, 2018.
  * [21] L. Wan, Q. Wang, A. Papir, and I. L. Moreno, “Generalized end-to-end loss for speaker verification,” in 2018 IEEE International Conference on Acoustics, Speech and Signal Processing, pp. 4879–4883, IEEE, 2018.
  * [22] Y. Cao, X. Xiao, R. Sun, D. Wang, M. Xue, and S. Wen, “Stylefool: Fooling video classification systems via style transfer,” arXiv preprint arXiv:2203.16000, 2022.
  * [23] V. Zue, S. Seneff, and J. Glass, “Speech database development at mit: Timit and beyond,” Speech communication, vol. 9, no. 4, pp. 351–356, 1990.