# The index of symmetry for a left-invariant metric on a solvable three-
dimensional Lie group

Robert D. May

Abstract: We find the index of symmetry for all solvable three-dimensional Lie
groups with a left-invariant metric. When combined with the work of Reggiani
on unimodular three-dimensional Lie groups, the index of symmetry is then
known for all three-dimensional Lie groups with a left-invariant metric.
Similar to the unimodular case, for every solvable three-dimensional Lie
algebra there is at least one left-invariant metric for which the index of
symmetry is positive. Also, the index is never equal to 2.

Keywords: index of symmetry, solvable Lie group, Killing fields, Ricci matrix

2010 Mathematics Subject Classification: 22E15, 22E25, 53C30

##  1 Introduction

The index of symmetry, $i(G)$, for a Lie group $G$ with identity element $e$
and a left-invariant metric $g$ can be defined as follows: Define a vector
field $X$ on $G$ to be parallel at the identity (or parallel at $e$) if
$\nabla_{Y}(X)$ vanishes at $e$ for every vector field $Y$ (that is,
$\nabla_{Y}(X)\mid_{e}=0$). Then define a subspace $S_{e}\in\mathfrak{g}$
where $\mathfrak{g}$ is the Lie algebra for $G$ considered as the tangent
space to $G$ at $e$: $S_{e}$ consists of all vectors $x\in\mathfrak{g}$ such
that $x=X\mid_{e}$ for a vector field $X$ on $G$ which is a Killing field for
the metric $g$ and which is parallel at $e$. Then the index of symmetry for
$G$ is the dimension of $S_{e}$: $i(G)=i(G,g)=\text{dim}(S_{e})$.

Reggiani in [3] has computed this index of symmetry for all three-dimensional
Lie groups with a left-invariant metric which are unimodular (that is, whose
Lie algebras are unimodular). In this paper we compute the index of symmetry
for all three-dimensional Lie groups with a left-invariant metric which are
solvable (that is, whose Lie algebras are solvable). Since any three-
dimensional Lie group is either unimodular or solvable (or both), the index of
symmetry is determined for any three-dimensional Lie group with a left-
invariant metric.

In section 2 we give a table listing $i(G)$ for all cases where the index of
symmetry is positive. We note that, analogous to the results of Reggiani [3]
for the unimodular Lie algebras, for every solvable three-dimensional Lie
algebra there is at least one metric for which the index of symmetry is
positive. Also $i(G)$ is never equal to $2$.

In section 3 we review the classification of solvable metric Lie algebras as
given by May and Wears in [1] in terms of standard bases and defining triples
$(a,b,c)$. We note which underlying Lie algebras correspond to which triples.

In section 4 we give matrix representations for the solvable metric Lie
algebras and the associated simply-connected Lie groups $G$. We find global
coordinates $\\{x_{1},x_{2},x_{3}\\}$ for each such $G$ with associated global
coordinate frame $\\{\partial/\partial x_{1},\partial/\partial
x_{2},\partial/\partial x_{3}\\}$ for the tangent bundle of $G$. We also find
a global orthonormal frame consisting of left-invariant vector fields $L_{i}$
and a global frame of right-invariant vector fields $R_{i}$ (which are Killing
fields for the given metric on $G$). Finally, we find the relations between
these three frames for the tangent bundle.

In section 5 we review some properties of isometries of $G$ and the related
Killing vector fields. We obtain a condition for a vector field $X$ to be a
Killing field in terms of the left-invariant frame vectors $L_{i}$. We define
a matrix $P(X)$ for any vector field $X$ such that the vanishing of $P(X)$ is
equivalent to the condition that $X$ is parallel at the identity. We then
compute the values of $P(X)$ when $X$ is one of the right invariant vector
fields $R_{i}$. Finally, we obtain a formula for $P(X)$ when $X$ is a Killing
field vanishing at $e=(0,0,0)$.

In section 6 we review the index of symmetry $i(G)$. We then define a subspace
$\bar{S}_{e}\subseteq S_{e}$ and a subindex
$\bar{i}(G)=\text{dim}(\bar{S}_{e})$. $\bar{S}_{e}$ and $\bar{i}_{e}$ involve
only the right-invariant Killing fields. We give a condition under which
$i(G)=\bar{i}(G)$. Then using the formulas for $P(R_{j})$ from section 5 we
calculate the subindex $\bar{i}(G)$ for all cases.

In section 7 we study certain restrictions on the contributions that Killing
fields $X$ vanishing at $e$ can make to the index of symmetry. We define the
Ricci matrix $R$ to be the Ricci tensor for $g$ at $e$ expressed in terms of
the standard basis $\\{e_{i}\\}$. We obtain a formula for $P(X)$ in terms of
the one-parameter group of isometries fixing $e$ corresponding to $X$ and
indicate how the structure of $R$ can influence $P(X)$.

In section 8 we define a Lie algebra (and associated Lie group $G$) to be
generic if the diagonal entries in the diagonalized Ricci matrix are all
distinct. Our main result is that for generic $G$ the subindex and index
agree: $i(G)=\bar{i}(G)$. Then using the results of section 6 we have the
formula for $i(G)$ in all but the nongeneric cases. We catalog all the
nongeneric metric Lie algebras.

In sections 9, 10, and 11 we calculate $i(G)$ for the nongeneric metric Lie
algebras.

In section 12 we sketch the computational proofs of some of our stated
results.

We use the summation convention throughout: unless specified otherwise, any
repeated index implies summation over that index. Also, we use $e$ for the
identity in a Lie group $G$, so $e=(0,0,0)$ for coordinates centered at the
identity. We then write $X\lvert_{e}$ for the value of a vector field $X$ on
$G$ at the identity $e$. So $X\lvert_{e}\in\mathfrak{g}$ when we identify the
Lie algebra $\mathfrak{g}$ with the tangent space to $G$ at $e$.

##  2 Table of results.

Table 1 below summarizes our results for the index of symmetry $i(G)$ for a
simply-connected Lie group $G$ with a left-invariant metric in terms of the
correponding metric Lie algebra $\mathfrak{g}$. We list all cases for which
the index of symmetry is positive. In all other cases the index is $0$. Notice
that for every solvable algebra there is at least one metric for which the
index of symmetry is positive.

$\mathfrak{g}$ | Triple $(a,b,c)$ | $S_{e}$ | $i(G)$
---|---|---|---
$\mathfrak{g}(K)$, $K>1$ |  $(a,b,b)$, $a>0$ | span$(e_{1})$ | 1
$\mathfrak{g}(1)$ | $(a,b,b),\,a>0$ | $\mathfrak{g}$ | 3
$\mathfrak{g}(1)$ | $(a,b,c),\,a>0,\,b\neq c$ | span$(\alpha_{2}e_{2}+\alpha_{3}e_{3})$ | 1
$\mathfrak{g}(K)$, $0<K<1$ |  $(a,b,b)$, $a>0$ | span$(e_{1})$ | 1
$\mathfrak{g}(K)$, $0<K<1$ |  $(a,a+a\sqrt{1-K},a-a\sqrt{1-K})$, $a>0$ | span$(e_{2}-e_{3})$ | 1
$\mathfrak{g}(0)$ |  $(a,2a,0)$, $a>0$ | span$(e_{2}-e_{3})$ | 1
$\mathfrak{g}(K)$, $K<0$ |  $(a,b,-b)$, $a>0$ | $\mathfrak{g}$ | 3
$\mathfrak{g}(K)$, $K<0$ |  $(a,a+a\sqrt{1-K},a-a\sqrt{1-K})$, $a>0$ | span$(e_{2}-e_{3})$ | 1
$\mathfrak{g}_{I}$ |  $(a,0,0)$, $a>0$ | $\mathfrak{g}$ | 3
$\mathfrak{e}(1,1)$ |  $(0,b,b)$, $b>0$ | span$(e_{1})$ | 1
Heisenberg algebra |  $(0,b,0)$, $b>0$ | span$(e_{3})$ | 1
$\mathfrak{e}(2)$ |  $(0,b,-b)$, $b>0$ | $\mathfrak{g}$ | 3
Abelian algebra | (0,0,0) | $\mathfrak{g}$ | 3
Table 1: Index of symmetry $i(G)$

##  3 Classification of solvable three-dimensional Lie algebras

Two metric Lie algebras $(\mathfrak{g}_{1},Q_{1}),(\mathfrak{g}_{2},Q_{2})$
are equivalent if there exists a Lie algebra isomorphism
$f:\mathfrak{g}_{1}\rightarrow\mathfrak{g}_{2}$ which also preserves the
metric, $f^{*}(Q_{2})=Q_{1}$. By the results of May and Wears [1], for any
three-dimensional solvable metric Lie algebra $(\mathfrak{g},Q)$, there exists
an orthonormal basis ${\mathbf{e}}=\\{e_{1},e_{2},e_{3}\\}$ and a matrix
$A=\left[{\begin{array}[]{*{20}c}a&c\\\ b&a\\\ \end{array}}\right]$ with
$a\geq 0,b\geq\mid c\mid$, such that $[e_{1},e_{2}]=ae_{2}+be_{3}$ and
$[e_{1},e_{3}]=ce_{2}+ae_{3}$. We say that the metric Lie algebra
$(\mathfrak{g},Q)$ defines the triple $(a,b,c)$. Equivalence classes of
solvable metric Lie algebras correspond one to one with such matrices $A$ (or
such triples $(a,b,c)$). These solvable metric Lie algebras $(\mathfrak{g},Q)$
correspond one to one with simply-connected three-dimensional solvable Lie
groups $G$ with a left-invariant metric.

The underlying Lie algebra $\mathfrak{g}$ corresponding to a given matrix
$A=\left[{\begin{array}[]{*{20}c}a&c\\\ b&a\\\ \end{array}}\right]$ with
$a\geq 0,b\geq\mid c\mid$ can be described as follows:

First, $\mathfrak{g}$ is unimodular if and only if $a=0$.

If $a=0,\,b\geq c>0$, then $\mathfrak{g}=\mathfrak{e}(1,1)$, the Lie algebra
for the Lie group $E(1,1)$ of rigid motions of Minkowski 2-space. (The simply-
connected Lie group $G$ is the universal cover of the connected component of
the identity of $E(1,1)$.)

If $a=0,\,c<0$, then $\mathfrak{g}=\mathfrak{e}(2)$, the Lie algebra for the
Lie group $E(2)$ of rigid motions of Euclidean 2-space. (The simply-connected
Lie group $G$ is the universal cover of the connected component of the
identity of $E(2)$.)

If $a=0,\,b>0,\,c=0$, then $\mathfrak{g}$ is the Heisenberg algebra and $G$ is
the associated Heisenberg group.

If $a=b=c=0$, then the algebra is abelian.

Next, consider the non-unimodular cases where
$a>0\,,\,b\geqslant\left|c\right|$. If $b=c=0$ , then $\mathfrak{g}$ is the
special case of a (nonunimodular) Lie algebra with a basis
$\\{b_{1},b_{2},b_{3}\\}$ for which
$[b_{1},b_{2}]=b_{2}\,,\,[b_{1},b_{3}]=b_{3}\,,\,[b_{2},b_{3}]=0$. We will
denote this Lie algebra by $\mathfrak{g}_{I}$,

For all other cases when $a>0,\,b>0$, define

$K=K(\mathfrak{g},Q)=\frac{{bc}}{{a^{2}}}.$

The following result was proved by May and Wears in [1] (and was observed by
Milnor in [2]).

######  Lemma 3.1.

For $i=1,2$, let $(\mathfrak{g}_{i},Q_{i})$ be non-unimodular metric Lie
algebras defining triples $(a_{i},b_{i},c_{i})$ with $a_{i}>0,b_{i}>0$. Then
$\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ are isomorphic as Lie algebras if
and only if $K(\mathfrak{g}_{1},Q_{1})=K(\mathfrak{g}_{2},Q_{2})$.

Thus for each real $K$ there is a Lie algebra $\mathfrak{g}(K)$ such that the
metric Lie algebras $(\mathfrak{g},Q)$ with $\mathfrak{g}$ isomorphic to
$\mathfrak{g}(K)$ are exactly those with $K(\mathfrak{g},Q)=K$. Then for
metric Lie algebras defining triples $(a,b,c)$ with $a>0$ and
$\frac{{bc}}{{a^{2}}}=K$ the underlying Lie algebra is the (non-unimodular)
algebra $\mathfrak{g}(K)$ when $K\neq 0$. When $K=0$, the Lie algebra is still
$\mathfrak{g}(0)$ as long as $b>0$ . When the defined triple is $(a,0,0),a>0$,
the underlying Lie algebra is the special Lie algebra $\mathfrak{g}_{I}$.

##  4 Matrix representations and global coordinates

In this section we consider matrix representations for the solvable Lie
algebras and then construct associated simply-connected matrix Lie groups for
these Lie algebras. Let $A=\left[{\begin{array}[]{*{20}c}a&c\\\ b&d\\\
\end{array}}\right]$ be any non-zero two-by-two real matrix and write
$\mathfrak{g}\left(A\right)$ for the subset of $M_{3}(\mathbb{R})$ consisting
of matrices of the form

$L(\mathbf{x})=L(x_{1},x_{2},x_{3})=\left[{\begin{array}[]{*{20}c}{x_{1}A}&{\begin{array}[]{*{20}c}x_{2}\\\
x_{3}\\\ \end{array}}\\\ {\begin{array}[]{*{20}c}0&0\\\ \end{array}}&0\\\
\end{array}}\right].$

Computing brackets in $M_{3}(\mathbb{R})$ gives
$\left[L(\mathbf{x}),L(\mathbf{y})\right]=L(\mathbf{z})$ where $z_{1}=0$ and

$\left[{\begin{array}[]{*{20}c}{z_{2}}\\\ {z_{3}}\\\
\end{array}}\right]=x_{1}A\left[{\begin{array}[]{*{20}c}{y_{2}}\\\ {y_{3}}\\\
\end{array}}\right]-y_{1}A\left[{\begin{array}[]{*{20}c}{x_{2}}\\\ {x_{3}}\\\
\end{array}}\right]=A\left[{\begin{array}[]{*{20}c}{x_{1}y_{2}-y_{1}x_{2}}\\\
{x_{1}y_{3}-y_{1}x_{3}}\\\ \end{array}}\right].$

Then $\mathfrak{g}\left(A\right)$ is a three dimensional Lie subalgebra of
$M_{3}\left(\mathbb{R}\right)$ with a basis

$e_{1}=L(1,0,0),e_{2}=L(0,1,0),e_{3}=L(0,0,1).$

Evidently the derived algebra $\mathfrak{g}(A)^{\prime}$ is contained in the
span of $e_{2}$ and $e_{3}$ and is therefore of dimension at most two. Compute

$\begin{gathered}\left[{e_{1},e_{2}}\right]=L(0,a,b)=ae_{2}+be_{3}\hfill\\\
\left[{e_{1},e_{3}}\right]=L(0,c,d)=ce_{2}+de_{3}\hfill\\\
\left[{e_{2},e_{3}}\right]=L(0,0,0)=0\hfill\\\ \end{gathered}$

If we put a metric on $\mathfrak{g}(A)$ by declaring
$\left\\{{e_{1},e_{2},e_{3}}\right\\}$ to be an orthonormal frame and then
choose $A$ with $a=d\geqslant 0\,,\,b\geqslant\left|c\right|$, then
$\mathfrak{g}\left(A\right)$ is a matrix representation for the solvable non-
abelian metric Lie algebra corresponding to the triple $\left({a,b,c}\right)$.

We now construct a (simply-connected) matrix Lie group $G(A)$ associated with
$\mathfrak{g}(A)$. Given the matrix $A$ and a real number $t$, define a two by
two matrix $M(t)=M(A,t)=e^{At}$. Note that
$M(t_{1})M\left({t_{2}}\right)=M(t_{1}+t_{2})$ and that
$M\left(t\right)^{-1}=M(-t)$. Define a subset $G(A)$ of $GL(3,\mathbb{R})$ to
be all matrices of the form

$G_{A}(\mathbf{x})=G_{A}(x_{1},x_{2},x_{3})=\left[{\begin{array}[]{*{20}c}{M(x_{1})}&{\begin{array}[]{*{20}c}x_{2}\\\
x_{3}\\\ \end{array}}\\\ {\begin{array}[]{*{20}c}0&0\\\ \end{array}}&1\\\
\end{array}}\right].$

Then

$G_{A}(\mathbf{x})G_{A}(\mathbf{y})=G_{A}(\mathbf{z})$

where $z_{1}=x_{1}+y_{1}$ while

$\left[{\begin{array}[]{*{20}c}{z_{2}}\\\ {z_{3}}\\\
\end{array}}\right]=\left[{\begin{array}[]{*{20}c}{x_{2}}\\\ {x_{3}}\\\
\end{array}}\right]+M(x_{1})\left[{\begin{array}[]{*{20}c}{y_{2}}\\\
{y_{3}}\\\ \end{array}}\right].$

Also,

$G_{A}(x_{1},x_{2},x_{3})^{-1}=G_{A}(\bar{x}_{1},\bar{x}_{2},\bar{x}_{3})$

where $\bar{x}_{1}=-x_{1}$ and

$\left[{\begin{array}[]{*{20}c}{\bar{x}_{2}}\\\ {\bar{x}_{3}}\\\
\end{array}}\right]=-M(x_{1})^{-1}\left[{\begin{array}[]{*{20}c}x_{2}\\\
x_{3}\\\ \end{array}}\right]=-M(-x_{1})\left[{\begin{array}[]{*{20}c}x_{2}\\\
x_{3}\\\ \end{array}}\right].$

So $G(A)$ is a subgroup of $GL(3,\mathbb{R})$ as desired.

For future computations it will be useful to have an explicit formula for
$M(t)=e^{At}=\left[{\begin{array}[]{*{20}c}{m_{11}(t)}&{m_{12}(t)}\\\
{m_{21}(t)}&{m_{22}(t)}\\\ \end{array}}\right].$ From $Ae^{At}=e^{At}A$ we
first find

$m_{22}(t)=m_{11}(t),\,\,m_{12}(t)=\frac{c}{b}m_{21}(t)$ (1)

Then

$M^{\prime}(t)=Ae^{At}=AM(t)=\left[{\begin{array}[]{*{20}c}{am_{11}(t)+cm_{21}(t)}&{\frac{ac}{b}m_{21}(t)+cm_{11}}\\\
{bm_{11}(t)+am_{21}(t)}&{cm_{21}(t)+am_{11}(t)}\\\ \end{array}}\right].$ (2)

This yields a pair of linear first order DE’s which can be solved for
$m_{11}(t)\,,\,m_{21}(t)$. The results are as follows:

  * •

For $c>0$, $M(t)=\left[{\begin{array}[]{*{20}c}{e^{at}\cosh(b\lambda
t)}&{\lambda e^{at}\sinh(b\lambda t)}\\\
{\frac{1}{\lambda}e^{at}\sinh(b\lambda t)}&{e^{at}\cosh(b\lambda t)}\\\
\end{array}}\right]$ where $\lambda=\sqrt{c/b}$.

  * •

For $c<0$, $M(t)=\left[{\begin{array}[]{*{20}c}{e^{at}\cos(b\lambda
t)}&{-\lambda e^{at}\sin(b\lambda t)}\\\ {\frac{1}{\lambda}e^{at}\sin(b\lambda
t)}&{e^{at}\cos(b\lambda t)}\\\ \end{array}}\right]$ where
$\lambda=\sqrt{-c/b}$.

  * •

For $c=0$, $M(t)=\left[{\begin{array}[]{*{20}c}{e^{at}}&{0}\\\
{bte^{at}}&{e^{at}}\\\ \end{array}}\right].$

  * •

Notice that in all cases $0<\lambda\leq 1$ and $\text{det}(M(t))=e^{2at}$.

  * •

Replacing $t$ by $-t$, notice that

$\displaystyle m_{11}(-t)=m_{22}(-t)=e^{-2at}m_{11}(t)$ $\displaystyle
m_{21}(-t)=-e^{-2at}m_{21}(t)$ (3) $\displaystyle
m_{12}(-t)=-\frac{c}{b}e^{-2at}m_{21}(t).$

The $(x_{1},x_{2},x_{3})$ are global coordinates for $G\left(A\right)$ and
$\left({\partial/\partial x_{1},\partial/\partial x_{2},\partial/\partial
x_{3}}\right)$ is a global frame for the tangent bundle. Identifying the Lie
algebra $\mathfrak{g}(A)$ with the tangent space at the identity
$G_{A}(0,0,0)$ gives $e_{i}=\partial/\partial x_{i}|_{(0,0,0)}$.

If we left translate
$\left({e_{1},e_{2},e_{3}}\right)=\left({\partial/\partial
x_{1},\partial/\partial x_{2},\partial/\partial x_{3}}\right)|_{(0,0,0)}$ we
get a global, left-invariant, orthonormal frame
$\left({L_{1},L_{2},L_{3}}\right)$ for the tangent bundle. To express
$\left({L_{1},L_{2},L_{3}}\right)$ in terms of the frame
$\left({\partial/\partial x_{1},\partial/\partial x_{2},\partial/\partial
x_{3}}\right)$, we have

$\left[{\begin{array}[]{*{20}c}{L_{1}}\\\ {L_{2}}\\\ {L_{3}}\\\
\end{array}}\right]_{(x_{1},x_{2},x_{3})}=\psi_{g}^{*}\left[{\begin{array}[]{*{20}c}{\partial/\partial
x_{1}}\\\ {\partial/\partial x_{2}}\\\ {\partial/\partial x_{3}}\\\
\end{array}}\right]_{(0,0,0)}$

where $\psi_{g}:G(A)\to G(A)$ is left translation by
$g=G_{A}(x_{1},x_{2},x_{3})$.

Evaluating gives

$\left[{\begin{array}[]{*{20}c}{L_{1}}\\\ {L_{2}}\\\ {L_{3}}\\\
\end{array}}\right]_{(x_{1},x_{2},x_{3})}=\left[{\begin{array}[]{*{20}c}{\partial/\partial
x_{1}}\\\ {M(x_{1})^{T}\left[{\begin{array}[]{*{20}c}{\partial/\partial
x_{2}}\\\ {\partial/\partial x_{3}}\\\ \end{array}}\right]}\\\
\end{array}}\right]$

or explicitly

$\displaystyle L_{1}$ $\displaystyle=\partial/\partial x_{1}$ $\displaystyle
L_{2}$ $\displaystyle=m_{11}(x_{1})\partial/\partial
x_{2}+m_{21}(x_{1})\partial/\partial x_{3}$ (4) $\displaystyle L_{3}$
$\displaystyle=m_{12}(x_{1})\partial/\partial
x_{2}+m_{22}(x_{1})\partial/\partial x_{3}.$

Solving then yields

$\left[{\begin{array}[]{*{20}c}{\partial/\partial x_{1}}\\\ {\partial/\partial
x_{2}}\\\ {\partial/\partial x_{3}}\\\
\end{array}}\right]=\left[{\begin{array}[]{*{20}c}{L_{1}}\\\
{M(-x_{1})^{T}\left[{\begin{array}[]{*{20}c}{L_{2}}\\\ {L_{3}}\\\
\end{array}}\right]}\\\ \end{array}}\right]$ (5)

or explicitly, using (• ‣ 4),

$\displaystyle\partial/\partial x_{1}$ $\displaystyle=L_{1}$
$\displaystyle\partial/\partial x_{2}$
$\displaystyle=e^{-2ax_{1}}m_{11}(x_{1})L_{2}-e^{-2ax_{1}}m_{21}(x_{1})L_{3}$
(6) $\displaystyle\partial/\partial x_{3}$
$\displaystyle=-e^{-2ax_{1}}\frac{c}{b}m_{21}(x_{1})L_{2}+e^{-2ax_{1}}m_{11}(x_{1})L_{3}.$

If we right translate $(e_{1},e_{2},e_{3})$ we get a global right-invariant
frame $(R_{1},R_{2},R_{3})$ for the tangent bundle. These right-invariant
vector fields $R_{i}$ turn out to be Killing fields for the left-invariant
metric on $G$. To express $\left({R_{1},R_{2},R_{3}}\right)$ in terms of the
frame $\left({\partial/\partial x_{1},\partial/\partial
x_{2},\partial/\partial x_{3}}\right)$, we have

$\left[{\begin{array}[]{*{20}c}{R_{1}}\\\ {R_{2}}\\\ {R_{3}}\\\
\end{array}}\right]_{(x_{1},x_{2},x_{3})}=\phi_{g}^{*}\left[{\begin{array}[]{*{20}c}{\partial/\partial
x_{1}}\\\ {\partial/\partial x_{2}}\\\ {\partial/\partial x_{3}}\\\
\end{array}}\right]_{(0,0,0)}$

where $\phi_{g}:G(A)\to G(A)$ is right translation by
$g=G_{A}(x_{1},x_{2},x_{3})$. A lengthy evaluation gives

$\begin{gathered}\left[{\begin{array}[]{*{20}c}{R_{1}}\\\ {R_{2}}\\\
{R_{3}}\\\
\end{array}}\right]_{(x_{1},x_{2},x_{3})}=\left[{\begin{array}[]{*{20}c}{\partial/\partial
x_{1}+(ax_{2}+cx_{3})\partial/\partial x_{2}+(bx_{2}+ax_{3})\partial/\partial
x_{3}}\\\ {\partial/\partial x_{2}}\\\ {\partial/\partial x_{3}}\\\
\end{array}}\right]\end{gathered}$

Using (4), we can then express the frame $\\{R_{i}\\}$ in terms of the left-
invariant frame $\\{L_{i}\\}$:

$R_{1}=L_{1}+e^{-2ax_{1}}\left[m_{11}(x_{1})(ax_{2}+cx_{3})-\frac{c}{b}m_{21}(x_{1})(bx_{2}+ax_{3})\right]L_{2}\\\
+e^{-2ax_{1}}\left[-m_{21}(x_{1})(ax_{2}+cx_{3})+m_{11}(x_{1})(bx_{2}+ax_{3})\right]L_{3}$
(7) $\displaystyle R_{2}$
$\displaystyle=e^{-2ax_{1}}m_{11}(x_{1})L_{2}-e^{-2ax_{1}}m_{21}(x_{1})L_{3}$
(8) $\displaystyle R_{3}$
$\displaystyle=-e^{2ax_{1}}\frac{c}{b}m_{21}(x_{1})L_{2}+e^{-2ax_{1}}m_{11}(x_{1})L_{3}.$

##  5 Isometry groups and Killing vector fields

The isometries of a Lie group $G$ with a left-invariant metric $g$ themselves
form a Lie group, $\text{Isom}(G,g)$. Given a one-parameter subgroup
$\\{\psi_{t}:G\rightarrow G:t\in\mathbb{R}\\}$ of $\text{Isom}(G,g)$, we get a
Killing vector field $X=X_{i}\partial/\partial x_{i}$ on $G$, where
$X_{i}=d(x_{i}\circ\psi_{t})/dt\arrowvert_{t=0}$.

If $\psi_{t}$ is the product of two one parameter subgroups,
$\psi_{t}=\psi_{1,t}\circ\psi_{2,t}$, then the Killing vector field $X$
corresponding to $\psi_{t}$ is given by $X=X_{1}+X_{2}$, where $X_{i}$ is the
Killing vector field correponding to $\psi_{i,t}$.

For any $x\in G$, left translation by $x$ gives an isometry $L_{x}$ of $G$,
and $G$ can be regarded as a subgroup of $\text{Isom}(G,g)$. Then a one
parameter subgroup of $G$ will give a one parameter subgroup of
$\text{Isom}(G,g)$ and hence a Killing vector field. The one parameter
subgoups of $G$ correspond to elements of the Lie algebra $\mathfrak{g}$. The
Killing vector field corresponding to an element $e\in\mathfrak{g}$ is just
the right invariant vector field obtained by right translating $e$. In
particular, the basis elements $e_{1},e_{2},e_{3}$ give rise to the Killing
fields $R_{1},R_{2},R_{3}$.

Write $\text{Isom}_{e}(G,g)$ for the subgroup of $\text{Isom}(G,g)$ consisting
of isometries leaving the identity $e\in G$ fixed. Then $\text{Isom}(G,g)$ is
a product $\text{Isom}_{e}(G,g)\cdot G$, and any Killing vector field can be
written as a sum of a Killing vector field vanishing at $e$ and a right-
invariant vector field.

$X$ is a Killing field for $g$ if the Lie derivative of $g$ with respect to
$X$ vanishes, that is if $(\mathcal{L}_{X}g)(Y,Z)=0$ for all $Y,Z$. We express
this condition in terms of the orthonormal frame $L_{1},L_{2},L_{3}$ of left-
invariant fields.

######  Proposition 5.1.

A vector field $X=X^{i}L_{i}$ is a Killling vector field if and only if

$L_{j}(X^{k})+L_{k}(X^{j})+X^{i}(c^{j}_{ki}+c^{k}_{ji})=0$ (9)

for all $j,k$.

###### Proof.

We have

$\mathcal{L}_{X}(g(Y,Z))=(\mathcal{L}_{X}g)(Y,Z)+g(\mathcal{L}_{X}(Y),Z)+g(Y,\mathcal{L}_{X}(Z)).$

For the standard torsion free, inner product preserving connection on $G$ we
have

$(\mathcal{L}_{X})(g(Y,Z))=X(g(Y,Z))=g(\nabla_{X}Y,Z)+g(Y,\nabla_{X}Z)$

and

$\mathcal{L}_{X}(Y)=[X,Y]=\nabla_{X}Y-\nabla_{Y}X.$

Then the condition $(\mathcal{L}_{X}g)(Y,Z)=0$ reduces to

$g(\nabla_{Y}X,Z)+g(Y,\nabla_{Z}X)=0.$

Take $Y=L_{j},Z=L_{k}$, write $X=X^{i}L_{i}$, and recall that the frame
$\\{L_{i}\\}$ is orthonormal. Then

$g(\nabla_{L_{j}}X,L_{k})=g(L_{j}(X^{i})L_{i}+\Gamma^{l}_{ji}L_{l},L_{k})=L_{j}(X^{k})+X^{i}\Gamma^{k}_{ji}$

and similarly

$g(L_{j},\nabla_{L_{k}}X)=L_{k}(X^{j})+X^{i}\Gamma^{j}_{ki}.$

Here the symbols $\Gamma^{j}_{ik}$ are defined by
$\nabla_{L_{k}}L_{i}=\Gamma^{j}_{ki}L_{j}$ and for this orthnormal frame are
given by $\Gamma^{j}_{ki}=\frac{1}{2}(c^{j}_{ki}-c^{k}_{ij}+c^{i}_{jk})$.
(Recall that the Lie algebra of left-invariant vector fields is isomorphic to
$\mathfrak{g}$, so $[L_{i},L_{j}]=c^{k}_{ij}L_{k}$.) Notice that
$\Gamma^{k}_{ji}+\Gamma^{j}_{ki}=c^{j}_{ki}+c^{k}_{ji}$. Then the condition
for a Killing field becomes

$L_{j}(X^{k})+L_{k}(X^{j})+X^{i}(c^{j}_{ki}+c^{k}_{ji})=0$

for all $j,k$ as claimed. ∎

If we define matrices $D(X)$ and $C(X)$ for any vector field $X=X^{i}L_{i}$ by

$D(X)_{jk}=L_{j}X^{k}\,,\,\,\,\,C(X)_{jk}=X^{i}(c^{j}_{ki}+c^{k}_{ji})$ (10)

then the condition for $X$ to be a Killing field is just

$D(X)+D(X)^{T}+C(X)=0.$ (11)

Recall that the Lie algebra structure constants (for $\mathfrak{g}$ or for the
left-invariant vector fields on $G$ ) are given by

$\displaystyle c^{2}_{12}$ $\displaystyle=-c^{2}_{21}=a$ $\displaystyle
c^{2}_{13}$ $\displaystyle=-c^{2}_{31}=c$ $\displaystyle c^{3}_{12}$
$\displaystyle=-c^{3}_{21}=b$ $\displaystyle c^{3}_{13}$
$\displaystyle=-c^{3}_{31}=a.$

For future use we compute the matrix

$C(X)=\left[{\begin{array}[]{*{20}c}{0}&{aX^{2}+cX^{3}}&{bX^{2}+aX^{3}}\\\
{aX^{2}+cX^{3}}&{-2aX^{1}}&{-(b+c)X^{1}}\\\
{bX^{2}+aX^{3}}&{-(b+c)X^{1}}&{-2aX^{1}}\end{array}}\right]$ (12)

As a tedious exercise, one can check that each right invariant field $R_{i}$
satisfies the condition (9).

A Killing vector field $X$ is “parallel at the identity” if $\nabla_{Y}(X)$
vanishes at $e$ for every vector field $Y$. Evidently it is enough to check
that $\nabla_{L_{i}}(X)\arrowvert_{e}=0$ for each $L_{i}$. Writing
$X=X^{j}L_{j}$, we find

$\nabla_{L_{i}}(X)=L_{i}(X^{j})L_{j}+X^{j}\nabla_{L_{i}}(L_{j})=L_{i}(X^{k})L_{k}+X^{j}\Gamma^{k}_{ij}L_{k}.$

Define a matrix $P(X)$ by

$P(X)_{ki}=\\{L_{i}(X^{k})+X^{j}\Gamma^{k}_{ij}\\}\arrowvert_{e}$ (13)

(where $\arrowvert_{e}$ means evaluated at the identity $e=(0,0,0)$). We then
have

######  Proposition 5.2.

A Killing vector field $X=X^{i}L_{i}$ is parallel at the identity if and only
if $P(X)=0$.

Notice that $P(X)$ is linear in $X$, that is, if $X=\alpha_{r}X_{r}$, then
$P(X)=\alpha_{r}P(X_{r})$.

If we define a matrix $\Gamma(X)$ by $\Gamma(X)_{ki}=X^{j}\Gamma^{k}_{ij}\
\arrowvert_{e}$, then we have

$P(X)=D(X)^{T}+\Gamma(X).$

Using $\Gamma^{j}_{ki}=\frac{1}{2}(c^{j}_{ki}-c^{k}_{ij}+c^{i}_{jk})$ we
compute for future use the matrix

$\Gamma(X)=\left[{\begin{array}[]{*{20}c}{0}&{\left(aX^{2}+\frac{1}{2}(b+c)X^{3}\right)\arrowvert_{e}}&{\left(\frac{1}{2}(b+c)X^{2}+aX^{3}\right)\arrowvert_{e}}\\\
{-\frac{1}{2}(b-c)X^{3}\arrowvert_{e}}&{-aX^{1}\arrowvert_{e}}&{-\frac{1}{2}(b+c)X^{1}\arrowvert_{e}}\\\
{\frac{1}{2}(b-c)X^{2}\arrowvert_{e}}&{-\frac{1}{2}(b+c)X^{1}\arrowvert_{e}}&{-aX^{1}\arrowvert_{e}}\end{array}}\right]$
(14)

Using the explicit values found above in (8) for the right invariant vector
fields $R_{j}$, we can calculate (see section 12) the $P(R_{j})$ matrices for
the $R_{j}$ (which span the space of all right invariant vector fields). The
results are as follows:

$\displaystyle P(R_{1})=\left[{\begin{array}[]{*{20}c}{0}&{0}&{0}\\\
{0}&{0}&{-\frac{b-c}{2}}\\\ {0}&{\frac{b-c}{2}}&{0}\end{array}}\right]$ (18)
$\displaystyle
P(R_{2})=\left[{\begin{array}[]{*{20}c}{0}&{a}&{\frac{b+c}{2}}\\\
{-a}&{0}&{0}\\\ {-\frac{b+c}{2}}&{0}&{0}\end{array}}\right]$ (22)
$\displaystyle
P(R_{3})=\left[{\begin{array}[]{*{20}c}{0}&{\frac{b+c}{2}}&{a}\\\
{-\frac{b+c}{2}}&{0}&{0}\\\ {-a}&{0}&{0}\end{array}}\right]$ (26)

For a Killing vector field $X$ vanishing at $e$ which corresponds to a one
parameter subgroup $\psi_{t}$ of $\text{Isom}_{e}(G,g)$, we can compute $P(X)$
as follows. Since each $x_{j}\circ\psi_{t}$ vanishes at $e=(0,0,0)$, we can
write

$x_{j}\circ\psi_{t}=a(t)_{ji}x_{i}+r_{t}^{j}$

where each $r_{t}^{j}$ and all derivatives $\partial(r_{t}^{j})/\partial
x_{i}$ vanish at $e$. The matrix $a(t)$ is just the Jacobean of $\psi_{t}$
evaluated at $e$, that is,

$a(t)_{ji}=\partial(x_{j}\circ\psi_{t})/\partial(x_{i})\arrowvert_{e}.$ (27)

Each matrix $a(t)$ gives a linear map of the tangent space at $e$ to itself,
and, since each $\psi_{t}$ is an isometry, $a(t)$ will be an orthogonal
transformation on $\mathfrak{g}$ taking the standard basis $e_{1},e_{2},e_{3}$
to a new orthonormal basis $b(t)_{1},b(t)_{2},b(t)_{3}:$

$b(t)_{i}=(\psi_{t})_{*}(e_{i})=a(t)_{ji}e_{j}.$ (28)

This one parameter subgroup $a(t)$ of the orthogonal group $O(3)$ determines
$P(X)$ as follows. Writing $X=F^{j}\partial/\partial x_{j}$ where
$F^{j}=(x_{j}\circ\psi_{t})^{\prime}\arrowvert_{t=0}\,$, we find

$F^{j}=[(a(t)_{ji})^{\prime}x_{i}+(r_{t}^{j})^{\prime}]\arrowvert_{t=0}=\bar{a}_{ji}x_{i}+\bar{r}^{j}$

where $\bar{a}=(a(t))^{\prime}\arrowvert_{t=0}$ and each $\bar{r}^{j}$ and all
its first order partials vanish at $e$. Then, using (5),
$X=F^{j}\partial/\partial x_{j}=X^{j}L_{j}$ where

$X^{1}=F^{1}=\bar{a}_{1i}x_{i}+\bar{r}^{j}$

and

$\left[\begin{array}[]{*{20}c}X^{2}\\\
X^{3}\end{array}\right]=M(-x_{1})\left[\begin{array}[]{*{20}c}{F^{2}}\\\
{F^{3}}\end{array}\right]=M(-x_{1})\left[\begin{array}[]{*{20}c}{\bar{a}_{2i}x_{i}+\bar{r}^{2}}\\\
{\bar{a}_{3i}x_{i}+\bar{r}^{3}}\end{array}\right].$

Recalling that each $X^{i}$ vanishes at $e$ and that $L_{i}=\partial/\partial
x_{i}$ at $e$, the formula for $P(X)_{ki}$ becomes

$P(X)_{ki}=\\{L_{i}(X^{k})+X^{j}\Gamma^{k}_{ij}\\}\arrowvert_{e}=\\{\partial(X^{k})/\partial
x_{i}\\}\arrowvert_{e}.$ (29)

But then since $M(0)=I$ and all $\bar{r}^{j}$ and their first order partial
derivatives vanish at $e=(0,0,0)$, we have the following result:

######  Proposition 5.3.

Let $\psi_{t}$ be a one parameter subgroup of $\text{Isom}_{e}(G,g)$ with
corresponding Killing vector field $X=X^{i}L_{i}$ vanishing at $e$. Then

$P(X)_{ki}=\bar{a}_{ki}=(a(t)_{ki})^{\prime}\lvert_{t=0}=\frac{d}{dt}\left[\partial(x_{j}\circ\psi_{t})/\partial(x_{i})\arrowvert_{e}\right]\lvert_{t=0}.$
(30)

As a corollary of propositions (5.2) and (5.3) we have

######  Corollary 5.1.

If $a(t)$ is independent of $t$, then $P(X)=0$ and $X$ is parallel at the
identity.

##  6 The index of symmetry $i(G)$ and subindex $\bar{i}(G)$.

Write $S_{e}$ for the subspace of $\mathfrak{g}$ (considered as the tangent
space to $G$ at $e$) consisting of all tangent vectors $x=X\arrowvert_{e}$
where $X$ is a Killing vector field parallel at $e$. Then the index of
symmetry for $G$ is $i(G)=i(G,g)=\text{dim}(S_{e})$.

Let $\bar{S}_{e}$ be the subspace of $S_{e}$ consisting of all vectors
$x=R\arrowvert_{e}$ where $R$ is a right invariant vector field parallel at
$e$. Then define the subindex $\bar{i}(G)=\text{dim}(\bar{S}_{e})$. Given the
explicit formulas (22) for the $R_{i}$, which span the right invariant vector
fields, we can compute $\bar{i}(G)$ with relative ease. The following
proposition suggests the usefulness of $\bar{i}(G)$.

######  Proposition 6.1.

If every Killing field which vanishes at $e$ is actually parallel at $e$, then
$\bar{i}(G)=i(G)$.

###### Proof.

Take $x=Y\arrowvert_{e}\in S_{e}$ where $Y$ is a Killing vector field parallel
at $e$. We must show that $x\in\bar{S}_{e}$. Write $Y=R+X$ where $R$ is a
right invariant field and $X$ is a Killing field vanishing at $e$. Then
$x=Y\arrowvert_{e}=R\arrowvert_{e}$, so if $R$ is parallel at $e$, then
$x\in\bar{i}(G)$ and we are done. But $Y$ is parallel at $e$ and $X$ is
parallel at $e$ by hypothesis, so $R=Y-X$ is in fact parallel at $e$. ∎

Then using corollary 5.1 we get

######  Corollary 6.1.

If for every one parameter subgroup $\psi_{t}$ of $\text{Isom}_{e}(G,g)$ the
matrices $a(t)$ are independent of $t$, then $\bar{i}(G)=i(G)$.

To calculate $\bar{i}(G)$, note that a Killing field $X=\alpha_{j}R_{j}$ is
parallel at $e$ if and only if $\alpha_{j}P(R_{j})=0$. $X$ then contributes
$\alpha_{j}e_{j}$ to $\bar{S}_{e}$. Using the formulas (22) for $P(R_{j})$ we
get three independent equations for the $\alpha_{j}$:

$\displaystyle\alpha_{1}\left(\dfrac{b-c}{2}\right)=0$
$\displaystyle\alpha_{2}\,a+\alpha_{3}\left(\dfrac{b+c}{2}\right)=0$ (31)
$\displaystyle\alpha_{2}\left(\dfrac{b+c}{2}\right)+\alpha_{3}\,a=0.$

So when $b=c$, we can take $\alpha_{1}=1$ and $\alpha_{2}=\alpha_{3}=0$ and
get the vector $e_{1}\in\bar{S}_{e}$. This occurs for the algebras
$\mathfrak{g}(K)$ with $K>0$ for the triples $(a,b,b)$, for the algebra
$\mathfrak{g}_{I}$ for the triples $(a,0,0)$, for the unimodular algebra
$\mathfrak{e}(1,1)$ for the triples $(0,b,b)$, and for the abelian algebra
with triple $(0,0,0)$.

There is a nontrivial solution for $\alpha_{2}$ and $\alpha_{3}$ only if the
determinant of the coefficents, $a^{2}-\left((b+c)/2\right)^{2}$, vanishes,
that is, if $a=\left(\dfrac{b+c}{2}\right)$. In this case, when $2a=b+c$, we
can take $\alpha_{2}=-\alpha_{3}=1$ and get the vector
$e_{2}-e_{3}\in\bar{S}_{e}$. This occurs for the algebras $\mathfrak{g}(K)$
with $K\leq 1$ and the triples $(a,a+a\sqrt{1-K},a-a\sqrt{1-K})$ and for the
unimodular algebra $\mathfrak{e}(2)$ with the triples $(0,b,-b)$ (and for the
abelian algebra with the triple $(0,0,0)$).

The vectors found above give a basis for $\bar{S}_{e}$, so we obtain the
following results for the subindex $\bar{i}(G)$:

  * •

For $\mathfrak{g}(K)$ with $K>0\,,\,K\neq 1$, triple $(a,b,b)$:
$\bar{S}_{e}=\text{span}(e_{1})$ and $\bar{i}(G)=1$.

  * •

For $\mathfrak{g}(1)$ with triple $(a,b,b)$:
$\bar{S}_{e}=\text{span}(e_{1},e_{2}-e_{3})$ and $\bar{i}(G)=2$.

  * •

For $\mathfrak{g}(K)$ with $K<1$, triple $(a,a+a\sqrt{1-K},a-a\sqrt{1-K})$:
$\bar{S}_{e}=\text{span}(e_{2}-e_{3})$ and $\bar{i}(G)=1$.

  * •

For $\mathfrak{g}_{I}$ with triple $(a,0,0)$: $\bar{S}_{e}=\text{span}(e_{1})$
and $\bar{i}(G)=1$.

  * •

For the unimodular algebra $\mathfrak{e}(1,1)$ with triple $(0,b,b)$:
$\bar{S}_{e}=\text{span}(e_{1})$ and $\bar{i}(G)=1$.

  * •

For the unimodular algebra $\mathfrak{e}(2)$ with triple $(0,b,-b)$:
$\bar{S}_{e}=\text{span}(e_{2}-e_{3})$ and $\bar{i}(G)=1$.

  * •

For the abelian algebra: $\bar{S}_{e}=\mathfrak{g}$ and $\bar{i}(G)=3$.

  * •

For all other cases we have $\bar{S}_{e}=\\{0\\}$ and $\bar{i}(G)=0$.

##  7 Killing fields vanishing at $e$

In this section we show how the Ricci tensor, Ric, of $G$ evaluated at
$e=(0,0,0)$ can influence the contribution of a Killing vector field vanishing
at $e$ to the index of symmetry $i(G)$.

First note that for any one parameter group of isometries
$\psi_{t}:G\rightarrow G$ taking $e$ to $e$, each $\psi_{t}$ will preserve the
Ricci tensor, $\psi_{t}^{*}\text{Ric}=\text{Ric}$, and, in particular,
$\psi_{t}^{*}\text{Ric}(e)=\text{Ric}(e)$. Define matrices for $\text{Ric}(e)$
and $\psi_{t}^{*}\text{Ric}(e)$ for the standard basis
$\mathbf{e}=(e_{1},e_{2},e_{3})$: $R_{ij}=\text{Ric}(e)(e_{i},e_{j})$ and
$(\psi_{t}^{*}R)_{ij}=\psi_{t}^{*}\text{Ric}(e)(e_{i},e_{j})$. (So we must
have $\psi_{t}^{*}R=R$.) Then since $\psi_{t}(e)=e$, we find using equation
(28)

$\displaystyle(\psi_{t}^{*}R)_{ij}$
$\displaystyle=\psi_{t}^{*}\text{Ric}(e)(e_{i},e_{j})=\text{Ric}(e)((\psi_{t})_{*}e_{i},(\psi_{t})_{*}e_{j})$
$\displaystyle=\text{Ric}(e)(a(t)_{ki}e_{k},a(t)_{lj}e_{l})$
$\displaystyle=a(t)_{ki}a(t)_{lj}\text{Ric}(e)(e_{k},e_{l})$
$\displaystyle=a(t)_{ki}a(t)_{lj}R_{kl}$

where $a(t)_{ki}=\partial(x_{k}\circ\psi_{t})/\partial x_{i}$. In matrix terms
we have

$R=\psi_{t}^{*}R=a(t)^{T}R\,\,a(t).$

Recall that the matrices $a(t)$ are orthogonal, so $a(t)^{T}=a(t)^{-1}$. Then

$R=a(t)^{-1}R\,\,a(t)=a(t)R\,\,a(t)^{-1}.$

For the standard basis $e=\\{e_{i}\\}$ for a metric Lie algebra $\mathfrak{g}$
corresponding to the triple $(a,b,c)$, the Ricci matrix can be calculated as

$R=\left[{\begin{array}[]{*{20}c}{-2a^{2}-\frac{{(b+c)^{2}}}{2}}&0&0\\\
0&{-2a^{2}-\frac{{b^{2}-c^{2}}}{2}}&{-a(b+c)}\\\
0&{-a(b+c)}&{-2a^{2}-\frac{{c^{2}-b^{2}}}{2}}\\\ \end{array}}\right].$ (32)

One can diagonalize this matrix by a “rotation” of the $e_{2}e_{3}$-plane in
$\mathfrak{g}$ through an angle $\theta$ where $\tan(2\theta)=2a/(b-c)$:
Define a rotation matrix $r(\theta)=\left[\begin{array}[]{*{20}c}1&0&0\\\
0&\cos(\theta)&-\sin(\theta)\\\
0&\sin(\theta)&\cos(\theta)\end{array}\right]$. Notice that
$r(\theta)^{-1}=r(-\theta)=r(\theta)^{T}$, so $r(\theta)$ is orthogonal.
Define a smooth change of coordinates with $y_{j}=r(\theta)_{i,j}x_{i}$. Then
the new orthonormal basis $d=\\{d_{j}\\}$ for $\mathfrak{g}$ where
$d_{j}=\partial/\partial y_{j}\lvert_{e}$ is obtained by rotating the
$e_{2}e_{3}$ plane through the angle $\theta$: $d_{j}=r(\theta)_{i,j}e_{i}$.
Notice that $e_{k}=r(-\theta)_{j,k}d_{j}$. Then when $\tan(2\theta)=2a/(b-c)$
one finds that the Ricci matrix relative to this new basis $d$ has the
diagonal form

$R_{d}=\left[{\begin{array}[]{*{20}c}R_{11}&0&0\\\ 0&R_{22}&0\\\ 0&0&R_{33}\\\
\end{array}}\right].$

where

$\displaystyle R_{11}=-2a^{2}-\frac{(b+c)^{2}}{2}$ $\displaystyle
R_{22}=-2a^{2}-\frac{(b+c)}{2}\sqrt{4a^{2}+(b-c)^{2}}$ (33) $\displaystyle
R_{33}=-2a^{2}+\frac{(b+c)}{2}\sqrt{4a^{2}+(b-c)^{2}}\,\,.$

The matrices $R$ and $R_{d}$ are related by

$R_{d}=r(\theta)^{T}R\,\,r(\theta)=r(\theta)^{-1}R\,\,r(\theta)$

where $r(\theta)$ is the orthogonal rotation matrix relating the standard
basis $e_{i}$ to the “diagonalizing” basis $d_{i}$. Then a calculation shows
that $R=a(t)R\,\,a(t)^{-1}$ implies $R_{d}=a(d,t)^{-1}R_{d}\,\,a(d,t)$ where
$a(d,t)=r(\theta)^{-1}a(t)^{-1}\,r(\theta)$. Then using proposition (5.3) we
have

######  Proposition 7.1.

Let $\psi_{t}$ be a one parameter subgroup of $\text{Isom}_{e}(G,g)$ with
corresponding Killing vector field $X=X^{i}L_{i}$ vanishing at $e$. Then there
exists a smoothly varying family of orthogonal matrices $a(d,t)$ with
$a(d,0)=I$, the identity, such that

(1) Each $a(d,t)$ commutes with the diagonalized Ricci matrix $R_{d}$, that
is, $R_{d}=a(d,t)^{-1}R_{d}\,\,a(d,t)$.

(2) $a(t)=r(\theta)a(d,t)^{-1}\,r(\theta)^{-1}$.

(3)
$P(X)=(a(t)^{\prime}\lvert_{t=0})=\left(r(\theta)(a(d,t)^{-1})^{\prime}\lvert_{t=0}\,\,r(\theta)^{-1}\right)$.

As we will see, the condition that $a(d,t)$ commutes with the diagonal matrix
$R_{d}$ places restrictions on $a(d,t)$ and therefor on $a(t)$ and $P(X)$.

##  8 Index of symmetry for generic $G$

Define a metric Lie algebra $\mathfrak{g}$ (and the corresponding simply
connected Lie group $G$) to be generic if the diagonal entries
$R_{11},R_{22},R_{33}$ for the diagonalized Ricci operator $R_{d}$ are all
distinct. Our first main result is the following:

######  Proposition 8.1.

For generic $G$, $S_{e}=\bar{S}_{e}$ and $i(G)=\bar{i}(G)$.

###### Proof.

For a generic $G$, take any one parameter group $\psi_{t}$ of isometries
fixing $e$ with corresponding Killing vector field $X_{e}$ vanishing at $e$.
As in proposition (7.1) we have a smoothly varying family $a(d,t)$ of
orthogonal matrices which commute with the diagonal matrix $R_{d}$. Since $G$
is generic, the diagonal entries of $R_{d}$ are all distinct. Then one finds
that commuting with $R_{d}$ requires each $a(d,t)$ to be diagonal. But then
since $a(d,t)$ is orthogonal, each diagonal entry must be $\pm 1$. Finally,
since $a(d,0)$ is the identity and $a(d,t)$ varies continuously with $t$, all
diagonal entries must be $1$, that is, $a(d,t)=I=a(d,t)^{-1}$ for all $t$. But
then $a(t)=r(\theta)a(d,t)^{-1}\,r(\theta)^{-1}=I$ is independent of $t$. Then
$\bar{i}(G)=i(G)$ by corollary (6.1). ∎

Combined with the results of section 6, the proposition gives $i(G)$ for all
but the nongeneric $G$. We now find the triples $(a,b,c)$ which give
nongeneric $G$, that is, for which the Ricci matrix $R_{d}$ does not have
distinct diagonal entries.

First notice that if $b+c=0$, then $R_{11}=R_{22}=R_{33}=-2a^{2}$.

If $b+c\neq 0$, then $R_{11}=R_{22}$ if and only if $a^{2}=bc$.

If $b+c\neq 0$, then $R_{11}\neq R_{33}$.

If $b+c\neq 0$, then $R_{22}=R_{33}$ if and only if $a=0$ and $c=b$.

We can then catalog the nongeneric (and nonabelian) metric Lie algebras as
follows:

  * •

$\mathfrak{g}(1)$ for any triple $(a,b,c)$ (with $a>0,c=a^{2}/b$).

  * •

$\mathfrak{g}(K)$ for $K<0$ and any triple $(a,b,-b)$ with $a>0,b=a\sqrt{-K}$.

  * •

$\mathfrak{g}_{I}$ for any triple $(a,0,0)$ (with $a>0$).

  * •

The unimodular algebra $\mathfrak{e}$(2) with any triple $(0,b,-b)$ ($b>0$).

  * •

The unimodular algebra $\mathfrak{e}(1,1)$ with any triple $(0,b,b)$ ($b>0$).

  * •

The Heisenberg algebra with any triple $(0,b,0)$ ($b>0$).

We will compute $i(G)$ for these non-generic $\mathfrak{g}$ case by case.

##  9 The case $\mathfrak{g}(1)$.

Consider the metric Lie algebras $\mathfrak{g}(1)$ corresponding to a matrix
$A=\left[{\begin{array}[]{*{20}c}a&c\\\ b&a\end{array}}\right]$ where $a>0$
and $a^{2}=bc$. In this section we will verify the following result:

######  Proposition 9.1.

Let $G$ be a simply-connected Lie group with left-invariant metric and let the
corresponding metric Lie algebra be $\mathfrak{g}(1)$ with triple $(a,b,c)$,
$a>0$ and $a^{2}=bc$. Then: (1) If $b=c$, $i(G)=3$. (2) If $b\neq c$,
$i(G)=1$.

###### Proof.

In section 12 we sketch the proof that $X=X^{i}L_{i}$ is a Killing vector
field when

$X^{1}=bx_{2}+ax_{3}$
$X^{2}=\frac{1}{8b}\left[e^{-2ax_{1}}\left\\{4a(bx_{2}+ax_{3})^{2}+(a^{2}+b^{2})/a\right\\}+e^{2ax_{1}}(a^{2}-3b^{2})/a+2(b^{2}-a^{2})/a\right]$
$X^{3}=\frac{1}{8a}\left[e^{-2ax_{1}}\left\\{4a(bx_{2}+ax_{3})^{2}+(a^{2}+b^{2})/a\right\\}+e^{2ax_{1}}(b^{2}-3a^{2})/a+2(a^{2}-b^{2})/a\right]\,.$

One easily checks that this vector field $X$ vanishes at $e=(0,0,0)$. Then one
finds (see section 12)

$P(X)=D(X)^{T}\mid_{e}=\left[\begin{array}[]{*{20}c}0&b&a\\\ -b&0&0\\\
-a&0&0\end{array}\right].$

Using the values found above in (22) for the right invariant fields $R_{j}$,
we find for a linear combination $Y=\alpha_{j}R_{j}+\beta X$:

$P(Y)=\alpha_{j}P(R_{j})+\beta P(X)=$
$\left[\begin{array}[]{*{20}c}0&{\alpha_{2}a+\alpha_{3}\left(\dfrac{a^{2}+b^{2}}{2b}\right)+\beta
b}&{\alpha_{2}\left(\dfrac{a^{2}+b^{2}}{2b}\right)+\alpha_{3}a+\beta a}\\\
{-\alpha_{2}a-\alpha_{3}\left(\dfrac{a^{2}+b^{2}}{2b}\right)-\beta
b}&0&{-\alpha_{1}\left(\dfrac{b^{2}-a^{2}}{2b}\right)}\\\
{-\alpha_{2}\left(\dfrac{a^{2}+b^{2}}{2b}\right)-\alpha_{3}a-\beta
a}&\alpha_{1}\left(\dfrac{b^{2}-a^{2}}{2b}\right)&0\\\ \end{array}\right].$

Then for $Y$ to be parallel at $e$ (that is, for $P(Y)=0$) we need

$\alpha_{1}\left(\dfrac{b^{2}-a^{2}}{2b}\right)=0$
$\alpha_{2}a+\alpha_{3}\left(\dfrac{a^{2}+b^{2}}{2b}\right)=-\beta b$
$\alpha_{2}\left(\dfrac{a^{2}+b^{2}}{2b}\right)+\alpha_{3}a=-\beta a\,\,.$

When $a=b$ there is a solution with $\alpha_{1}=1$ and all other coefficients
$0$. Then $Y=R_{1}$ and $Y\lvert_{e}=e_{1}\in S_{e}$. But also if $a=b$ the
last two equations both become $\alpha_{2}b+\alpha_{3}b=-\beta b$. This gives
two independent solutions: $\beta=0,\,\,\alpha_{2}=1,\,\,\alpha_{3}=-1$, for
which $Y=R_{2}-R_{3}$ and $Y\lvert_{e}=e_{2}-e_{3}\in S_{e}$ and
$\beta=1,\,\,\alpha_{2}=0,\,\,\alpha_{3}=-1$ for which $Y=-R_{3}+X$ and
$Y\arrowvert_{e}=-e_{3}\in S_{e}$. Then evidently $S_{e}$ is all of
$\mathfrak{g}$ and $i(G)=3$.

Now consider the case $a\neq b$. We must then have $\alpha_{1}=0$. The two by
two coefficient determinant for the last pair of equations is nonzero, so
there will be a unique solution for $\alpha_{2},\alpha_{3}$ corresponding to
any choice of $\beta$. (Notice that for $\beta=0$ the solution is
$\alpha_{2}=\alpha_{3}=0$, so there is no solution where $Y$ is right
invariant. That is, $\bar{S}_{e}=0$, $\bar{i}(G)=0$.) However, for any nonzero
$\beta$, say $\beta=1$, there will be a unique solution for
$\alpha_{2},\alpha_{3}$. (In fact, for $\beta=1$, the solution is
$\alpha_{2}=\left[-ab+a\left(\frac{a^{2}+b^{2}}{2b}\right)\right]/D$ and
$\alpha_{3}=\left[-a^{2}+b\left(\frac{a^{2}+b^{2}}{2b}\right)\right]/D$ where
$D$ is the determinant $D=a^{2}-\left(\dfrac{a^{2}+b^{2}}{2b}\right)^{2}$ .)
So there is a solution (unique up to nonzero scalar multiple)
$Y=\alpha_{2}R_{2}+\alpha_{3}R_{3}+X$ with
$Y\lvert_{e}=\alpha_{2}e_{2}+\alpha_{3}e_{3}\in S_{e}$.

We claim that up to scalar multiple this $Y\arrowvert_{e}$ is the only element
of $S_{e}$, so that $i(G)=1$ for this case. In fact, take any Killing field
$\bar{X}$ vanishing at $e$ and consider the corresponding one parameter family
of isometries $\psi_{t}$. For the appropriate coordinates, the Ricci operator
is diagonalized as

$R_{d}=\left[{\begin{array}[]{*{20}c}R_{11}&0&0\\\ 0&R_{22}&0\\\ 0&0&R_{33}\\\
\end{array}}\right].$

where

$R_{11}=R_{22}=-4a^{2}-\frac{(b-c)^{2}}{2}$ $R_{33}=\frac{(b-c)^{2}}{2}.$

As seen in section 8, the group $\psi_{t}$ gives rise to a one parameter group
of orthogonal matrices $a(d,t)$ which commute with $R_{d}$. To commute with
$R_{d}$, evidently each orthogonal $a(d,t)$ must leave $d_{3}$ fixed and
perform a rotation in the $d_{1}-d_{2}$ plane. That is,
$a(d,t)=\left[\begin{array}[]{*{20}c}\cos(\lambda t)&-\sin(\lambda t)&0\\\
\sin(\lambda t)&\cos(\lambda t)&0\\\ 0&0&1\end{array}\right]$. Then
$a(d,t)^{-1}=\left[\begin{array}[]{*{20}c}\cos(\lambda t)&\sin(\lambda t)&0\\\
-\sin(\lambda t)&\cos(\lambda t)&0\\\ 0&0&1\end{array}\right]$ and
$\left(a(d,t)^{-1}\right)^{\prime}\mid_{t=0}=\left[\begin{array}[]{*{20}c}0&\lambda&0\\\
-\lambda&0&0\\\ 0&0&0\end{array}\right].$ Then by proposition (7.1),
$P(\bar{X})$ is completely determined (up to a scalar factor $\lambda$)
independent of the choice of the field $\bar{X}$. In particular, $P(\bar{X})$
must equal $P(X)$ up to a scalar factor. Now take any
$x=\bar{Y}\arrowvert_{e}\in S_{e}$ where $\bar{Y}=R+\bar{X}$ is a Killing
field parallel at $e$ with $R$ right invariant and $\bar{X}$ vanishing at $e$.
Then $0=P(\bar{Y})=P(R)+P(\bar{X})=P(R)+\kappa P(X)$ for some scalar $\kappa$.
Then $Z=R+\kappa X$ is parallel at $e$, so $Z$ must be a scalar multiple of
the field $Y$ found above, say $Z=\gamma Y$. Then
$Z\arrowvert_{e}=R\arrowvert_{e}=\bar{Y}\arrowvert_{e}=x$, but also
$Z\arrowvert_{e}=\gamma
Y\arrowvert_{e}=\gamma(\alpha_{2}e_{2}+\alpha_{3}e_{3})$. So every element
$x\in S_{e}$ is a scalar multiple of $\alpha_{2}e_{2}+\alpha_{3}e_{3}$. Then
$i(G)=1$ for the cases $a\neq b$ as claimed. ∎

##  10 The cases $a>0$, $b+c=0$.

In this section we consider the nongeneric metric Lie algebras
$\mathfrak{g}(K)$, $K<1$, with $b+c=0$ and the special case
$\mathfrak{g}_{I}$. We will derive the following result:

######  Proposition 10.1.

Let $G$ be a simply-connected Lie group with left-invariant metric and let the
corresponding metric Lie algebra be either $\mathfrak{g}(K)$ with $K<1$ and
$b+c=0$ or $\mathfrak{g}_{I}$. Then $i(G)=3$.

The Ricci matrix for this case is already diagonalized for the standard
coordinates with $d_{i}=e_{i}$:

$R=R_{d}=\left[{\begin{array}[]{*{20}c}R_{11}&0&0\\\ 0&R_{22}&0\\\
0&0&R_{33}\\\ \end{array}}\right].$

where

$R_{11}=R_{22}=R_{33}=-2a^{2}.$

All sectional curvatures are negative and equal to $-a^{2}$, so (by the
Killing-Hopf theorem, see e.g. Wolf [5]) $G$ is isometric to a hyperbolic
three-space $H^{3}=H^{3}(a^{2})$ (with a metric having constant negative
curvature $-a^{2}$). So it is not surprising that $G$ has a maximal amount of
symmetry, $i(G)=3$.

A standard Poincare model (see e.g. Wielenberg [4]) identifies $H^{3}(a^{2})$
with the subspace $\\{(y_{1},y_{2},y_{3}):y_{1}>0\\}$ of $\mathbb{R}^{3}$ with
the metric $(dy_{1}^{2}+dy_{2}^{2}+dy_{3}^{2})/(a^{2}y_{1}^{2})$. Meanwhile
the metric for $G$ with the standard coordinates is
$dx_{1}^{2}+(dx_{2}^{2}+dx_{3}^{2})/e^{2ax_{1}}$. Then one can check that the
map $G\rightarrow H^{3}(a^{2})$ given by

$y_{1}=e^{ax_{1}}/a,\,\,y_{2}=x_{2},\,\,y_{3}=x_{3}$

is an isometry.

It is known (see Wielenberg [4]) that the isometry group of $H^{3}$ (and
therefor also of $G$) can be identified with the Lie group $PSL(2,\mathbb{C})$
of all two by two complex matrices with determinant 1. So any one parameter
subgroup of $PSL(2,\mathbb{C})$ should give a Killing vector field for $G$.

For the one parameter subgroup $\left[\begin{array}[]{*{20}c}1&0\\\
t&1\end{array}\right]$ one can show that the corresponding Killing field on
$G$ is $X_{1}=X_{1}^{i}L_{i}$ where

$X_{1}^{1}=\dfrac{-2x_{2}}{a}$
$X_{1}^{2}=e^{-ax_{1}}\left[\cos(bx_{1})(x_{3}^{2}-x_{2}^{2})-\sin(bx_{1})(2x_{2}x_{3})\right]+\dfrac{e^{ax_{1}}\cos(bx_{1})}{a^{2}}$
$X_{1}^{3}=e^{-ax_{1}}\left[-\sin(bx_{1})(x_{3}^{2}-x_{2}^{2})-\cos(bx_{1})(2x_{2}x_{3})\right]-\dfrac{e^{ax_{1}}\sin(bx_{1})}{a^{2}}$

In section 12 we sketch a proof that $X_{1}$ is in fact a Killing field.

One can easily see that $X_{1}\arrowvert_{e}=e_{2}/a^{2}$. In section 12 we
compute

$P(X_{1})=\left[\begin{array}[]{*{20}c}0&-1/a&0\\\ 1/a&0&0\\\
0&0&0\end{array}\right].$

Then let $Y_{1}=X_{1}+R_{2}/a^{2}$. Using (22) and $b+c=0$, we find
$P(Y_{1})=0$. So $Y_{1}$ is parallel at $e$ and $Y_{1}|_{e}=2e_{2}/a^{2}\in
S_{e}$.

Similarly, for the one parameter subgroup $\left[\begin{array}[]{*{20}c}1&0\\\
it&1\end{array}\right]$ the corresponding Killing field on $G$ is
$X_{2}=X_{2}^{i}L_{i}$ where

$X_{2}^{1}=\dfrac{2x_{3}}{a}$
$X_{2}^{2}=e^{-ax_{1}}\left[\sin(bx_{1})(x_{3}^{2}-x_{2}^{2})+\cos(bx_{1})(2x_{2}x_{3})\right]-\dfrac{e^{ax_{1}}\sin(bx_{1})}{a^{2}}$
$X_{2}^{3}=e^{-ax_{1}}\left[\cos(bx_{1})(x_{3}^{2}-x_{2}^{2})-\sin(bx_{1})(2x_{2}x_{3})\right]-\dfrac{e^{ax_{1}}\cos(bx_{1})}{a^{2}}$

One finds $X_{2}\arrowvert_{e}=-e_{3}/a^{2}$ and that

$P(X_{2})=\left[\begin{array}[]{*{20}c}0&0&1/a\\\ 0&0&0\\\
-1/a&0&0\end{array}\right].$

Then if $Y_{2}=X_{2}-R_{3}/a^{2}$ we find $P(Y_{2})=0$. So $Y_{2}$ is parallel
at $e$ and $Y_{2}|_{e}=-2e_{3}/a^{2}\in S_{e}$.

Finally, for the one parameter subgroup
$\left[\begin{array}[]{*{20}c}e^{it}&0\\\ 0&e^{-it}\end{array}\right]$ the
corresponding Killing field on $G$ is $X_{3}=X_{3}^{i}L_{i}$ where

$X_{3}^{1}=0$
$X_{3}^{2}=e^{-ax_{1}}\left[-2x_{3}\cos(bx_{1})+2x_{2}\sin(bx_{1})\right]$
$X_{3}^{3}=e^{-ax_{1}}\left[2x_{3}\sin(bx_{1})+2x_{2}\cos(bx_{1})\right]$

One finds $X_{3}\arrowvert_{e}=0$ and that

$P(X_{3})=\left[\begin{array}[]{*{20}c}0&0&0\\\ 0&0&-2\\\
0&2&0\end{array}\right].$

Then if $Y_{3}=X_{3}-2R_{1}$ we find $P(Y_{3})=0$. So $Y_{3}$ is parallel at
$e$ and $Y_{3}|_{e}=-2e_{1}\in S_{e}$.

Then since the $Y_{i}|e$ for $i=1,2,3$ are in $S_{e}$ and span all of
$\mathfrak{g}$, we must have $S_{e}=\mathfrak{g}$ and $i(G)=3$ as claimed.

##  11 The unimodular nongeneric cases.

In this section we determine $i(G)$ for the three unimodular nongeneric cases.
The results are

######  Proposition 11.1.

Let $G$ be a simply-connected Lie group with left-invariant metric and
corresponding metric Lie algebra $\mathfrak{g}$. Then

(1) if $\mathfrak{g}=\mathfrak{e}(1,1)$ with triple $a=0$ and $b=c$, then
$i(G)=1$.

(2) if $\mathfrak{g}$ is the Heisenberg algebra with triple $a=c=0$ and $b>0$,
then $i(G)=1$.

(3) if $\mathfrak{g}=\mathfrak{e}(2)$ with triple $a=0$ and $b+c=0$, then
$i(G)=3$.

(1) For $\mathfrak{e}(1,1)$ with triple $a=0,\,b=c$, the Ricci matrix is
already diagonalized for the standard coordinates with $d_{i}=e_{i}$:

$R=R_{d}=\left[{\begin{array}[]{*{20}c}R_{11}&0&0\\\ 0&R_{22}&0\\\
0&0&R_{33}\\\ \end{array}}\right].$

where

$R_{11}=-2b^{2}$ $R_{22}=R_{33}=0.$

As in section 5, any one parameter group of isometries preserving $e$ gives a
family of orthogonal transformations $a(t)$ which commute with $R=R_{d}$. Then
$a(t)$ must leave $e_{1}$ invariant while rotating the $e_{2}-e_{3}$ plane. So
$a(t)^{\prime}\mid_{t=0}=\left[\begin{array}[]{*{20}c}0&0&0\\\ 0&0&\lambda\\\
0&-\lambda&0\end{array}\right]$ for some constant $\lambda$. Then by
proposition (5.3),

$P(X)=a(t)^{\prime}\mid_{t=0}=\left[\begin{array}[]{*{20}c}0&0&0\\\
0&0&\lambda\\\ 0&-\lambda&0\end{array}\right]$

for the Killing field $X$ associated with the one parameter group. Then for
any Killing field $Y=\alpha_{i}R_{i}+\beta X$ we find (using the formulas (22)
and recalling that $a=0,\,b=c>0$) that

$P(Y)=\left[\begin{array}[]{*{20}c}0&\alpha_{3}b&\alpha_{2}b\\\
-\alpha_{3}b&0&\beta\lambda\\\
-\alpha_{2}b&-\beta\lambda&0\end{array}\right].$

So $Y$ can be parallel at $e$ only if $\alpha_{2}=\alpha_{3}=\beta=0$. But
then $Y=\alpha_{1}R_{1}$ and $Y\lvert_{e}=\alpha_{1}e_{1}$. Then
$S_{e}=\bar{S}_{e}$ and $i(G)=\bar{i}(G)=1$ as claimed.

(2) For the Heisenberg algebra with triple $a=c=0$, $b>0$, the Ricci matrix is
already diagonalized for the standard coordinates with $d_{i}=e_{i}$:

$R=R_{d}=\left[{\begin{array}[]{*{20}c}R_{11}&0&0\\\ 0&R_{22}&0\\\
0&0&R_{33}\\\ \end{array}}\right].$

where

$R_{11}=R_{22}=-b^{2}/2$ $R_{33}=b^{2}/2.$

As in section 5, any one parameter group of isometries preserving $e$ gives a
family of orthogonal transformations $a(t)$ which commute with $R=R_{d}$. Then
$a(t)$ must leave $e_{3}$ invariant while rotating the $e_{1}-e_{2}$ plane. So
$a(t)^{\prime}\mid_{t=0}=\left[\begin{array}[]{*{20}c}0&\lambda&0\\\
-\lambda&0&0\\\ 0&0&0\end{array}\right]$ for some constant $\lambda$. Then by
proposition (5.3),

$P(X)=a(t)^{\prime}\mid_{t=0}=\left[\begin{array}[]{*{20}c}0&\lambda&0\\\
-\lambda&0&0\\\ 0&0&0\end{array}\right]$

for the Killing field $X$ associated with the one parameter group. Then for
any Killing field $Y=\alpha_{i}R_{i}+\beta X$ we find (using the formulas (22)
and recalling that $a=c=0,\,b>0$) that

$P(Y)=\left[\begin{array}[]{*{20}c}0&\alpha_{3}b/2+\beta\lambda&\alpha_{2}b/2\\\
-\alpha_{3}b/2-\beta\lambda&0&-\alpha_{1}b/2\\\
-\alpha_{2}b/2&\alpha_{1}b/2&0\end{array}\right].$

So $Y$ can be parallel at $e$ only if $\alpha_{1}=\alpha_{2}=0$ and
$\alpha_{3}b/2+\beta\lambda=0$. Then $Y=\alpha_{3}R_{3}+\beta X$ and
$Y\lvert_{e}=\alpha_{3}e_{3}$. So $S_{e}=\text{span}(e_{3})$ is one
dimensional if there exists any such Killing field $X$ vanishing at
$e=(0,0,0)$. In section 12 we show that the vector field
$X=x_{2}L_{1}-x_{1}L_{2}+\frac{b}{2}(x_{1}^{2}+x_{2}^{2})L_{3}$ is in fact a
Killing field which vanishes at $e$. So $S_{e}=\text{span}(e_{3})$ and
$i(G)=1$ as claimed.

(3) Finally, for $\mathfrak{g}=\mathfrak{e}(2)$ with triple $a=0$ and $b+c=0$,
the Ricci operator $R$ vanishes completely for the standard coordinates. In
fact all sectional curvatures vanish and $G$ is isometric to Euclidean space.
(By the Killing-Hopf theorem, see [5]. The metric for the standard coordinates
is just $dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}$.)

The right invariant fields $R_{2}$ and $R_{3}$ are already parallel at $e$ in
this case (see equations (22)), so $e_{2},e_{3}\in S_{e}$. For $R_{1}$ we have
$P(R_{1})=\left[\begin{array}[]{*{20}c}0&0&0\\\ 0&0&-b\\\
0&b&0\end{array}\right]$.

The vector field

$X=\left[x_{3}\cos(bx_{1})-x_{2}\sin(bx_{1})\right]L_{2}+\left[-x_{3}\sin(bx_{1})-x_{2}\cos(bx_{1})\right]L_{3}$

clearly vanishes at $e$ and in section 12 we check that $X$ is in fact a
Killing field. We also check in section 12 that
$P(X)=\left[\begin{array}[]{*{20}c}0&0&0\\\ 0&0&1\\\
0&-1&0\end{array}\right]$. Then $R_{1}+bX$ is a Killing field parallel at $e$
and $(R_{1}+bX)\lvert_{e}=e_{1}\in S_{e}$. So $S_{e}$ contains al1 $e_{i}$,
$i=1,2,3$, and $i(G)=3$ as claimed.

##  12 Selected proofs.

In this section we sketch the proofs of some of the results given above.

We begin by checking the formulas (22) for the matrices $P(R_{s})$
corresponding to the right invariant fields $R_{s}$ of equations (8).

###### Proof.

We have

$R_{1}=L_{1}+e^{-2ax_{1}}\left[m_{11}(x_{1})(ax_{2}+cx_{3})-\frac{c}{b}m_{21}(x_{1})(bx_{2}+ax_{3})\right]L_{2}\\\
+e^{-2ax_{1}}\left[-m_{21}(x_{1})(ax_{2}+cx_{3})+m_{11}(x_{1})(bx_{2}+ax_{3})\right]L_{3}$
(34) $\displaystyle R_{2}$
$\displaystyle=e^{-2ax_{1}}m_{11}(x_{1})L_{2}-e^{-2ax_{1}}m_{21}(x_{1})L_{3}$
$\displaystyle R_{3}$
$\displaystyle=-e^{2ax_{1}}\frac{c}{b}m_{21}(x_{1})L_{2}+e^{-2ax_{1}}m_{11}(x_{1})L_{3}.$

We need to compute

$P(R_{s})=D(R_{s})^{T}\arrowvert_{e}+\Gamma(R_{s}).$

For $D(R_{s})^{T}\arrowvert_{e}$, recall that we are evaluating derivatives at
$e=(0,0,0)$ and that by (1) we have $m_{11}(0)=1$ and $m_{21}(0)=0$. Also
recall that $M^{\prime}(0)=A$. We then find

$\displaystyle
D(R_{1})^{T}\arrowvert_{e}=\left[{\begin{array}[]{*{20}c}{0}&{0}&{0}\\\
{0}&{a}&{c}\\\ {0}&{b}&{a}\end{array}}\right]$ (38) $\displaystyle
D(R_{2})^{T}\arrowvert_{e}=\left[{\begin{array}[]{*{20}c}{0}&{0}&{0}\\\
{-a}&{0}&{0}\\\ {-b}&{0}&{0}\end{array}}\right]$ (42) $\displaystyle
D(R_{3})^{T}\arrowvert_{e}=\left[{\begin{array}[]{*{20}c}{0}&{0}&{0}\\\
{-c}&{0}&{0}\\\ {-a}&{0}&{0}\end{array}}\right]$ (46)

Next notice that $R_{s}^{j}\arrowvert_{e}=R_{s}^{j}(e)=1$ if $j=s$ and
vanishes if $j\neq s$. Then the formula (14) gives

$\displaystyle\Gamma(R_{1})=\left[{\begin{array}[]{*{20}c}{0}&{0}&{0}\\\
{0}&{-a}&{-(b+c)/2}\\\ {0}&{-(b+c)/2}&{-a}\end{array}}\right]$ (50)
$\displaystyle\Gamma(R_{2})=\left[{\begin{array}[]{*{20}c}{0}&{a}&{(b+c)/2}\\\
{0}&{0}&{0}\\\ {(b-c)/2}&{0}&{0}\end{array}}\right]$ (54)
$\displaystyle\Gamma(R_{3})=\left[{\begin{array}[]{*{20}c}{0}&{(b+c)/2}&{a}\\\
{-(b-c)/2}&{0}&{0}\\\ {0}&{0}&{0}\end{array}}\right]$ (58)

Adding the pairs of matrices we get the formulas (22) for $R_{s}(k,i)$. ∎

We now sketch proofs that certain vector fields are Killing fields as claimed
above. We begin with a field from section 9.

###### Proof.

Recall that $a^{2}=bc>0$ for this case. Then using the formula for $M(t)$ from
section 3 when $c=a^{2}/b>0$ and $\lambda=a/b$ we find

$m_{11}(t)=m_{22}(t)=e^{at}\cosh(at)=1/2+e^{2at}/2$

and

$m_{21}(t)=\frac{b}{c}m_{12}(t)=\frac{b}{a}e^{at}\sinh(at)=\frac{b}{a}(-1/2+e^{2at}/2).$

Then (4) gives

$L_{1}=\partial/\partial x_{1}$
$L_{2}=(1/2+e^{2ax_{1}}/2)\,\,\partial/\partial
x_{2}+\frac{b}{a}(-1/2+e^{2ax_{1}}/2)\,\,\partial/\partial x_{3}$
$L_{3}=\frac{a}{b}(-1/2+e^{2ax_{1}}/2)\,\,\partial/\partial
x_{2}+(1/2+e^{2ax_{1}}/2)\,\,\partial/\partial x_{3}\,\,.$

Consider the vector field $X=X^{i}L_{i}$ where

$X^{1}=bx_{2}+ax_{3}$
$X^{2}=\frac{1}{8b}\left[e^{-2ax_{1}}Y+e^{2ax_{1}}(a^{2}-3b^{2})/a+2(b^{2}-a^{2})/a\right]$
$X^{3}=\frac{1}{8a}\left[e^{-2ax_{1}}Y+e^{2ax_{1}}(b^{2}-3a^{2})/a+2(a^{2}-b^{2})/a\right]\,$

and $Y=4a(bx_{2}+ax_{3})^{2}+(a^{2}+b^{2})/a$.

The condition for $X$ to be a Killing field is

$L_{j}(X^{k})+L_{k}(X^{j})+X^{i}(c^{j}_{ki}+c^{k}_{ji})=0$

for all $j,k$. or

$D(X)+D(X)^{T}+C(X)=0.$

Using $D(X)_{jk}=L_{j}X^{k}$, a tedious but straightforward computation gives
the matrix

$D(X)=\left[{\begin{array}[]{*{20}c}{0}&{-aX^{2}-cX^{3}-be^{2ax_{1}}}&{-bX^{2}-aX^{3}-ae^{2ax_{1}}}\\\
{be^{2ax_{1}}}&{aX^{1}}&{bX^{1}}\\\
{ae^{2ax_{1}}}&{cX^{1}}&{aX^{1}}\end{array}}\right].$

By (12) we also have

$C(X)=\left[{\begin{array}[]{*{20}c}{0}&{aX^{2}+cX^{3}}&{bX^{2}+aX^{3}}\\\
{aX^{2}+cX^{3}}&{-2aX^{1}}&{-(b+c)X^{1}}\\\
{bX^{2}+aX^{3}}&{-(b+c)X^{1}}&{-2aX^{1}}\end{array}}\right]$

It is then easy to check that ∎

We next sketch a proof that the field $X_{1}$ from section 10 is a Killing
field and that $P(X_{1})=\left[\begin{array}[]{*{20}c}0&-1/a&0\\\ 1/a&0&0\\\
0&0&0\end{array}\right].$

###### Proof.

For this case the matrix is $\left[\begin{array}[]{*{20}c}a&-b\\\
b&a\end{array}\right]$ and we have (see section 3)

$M(t)=e^{At}=e^{at}\left[{\begin{array}[]{*{20}c}\cos(bt)&-\sin(bt)\\\
\sin(bt)&\cos(bt)\end{array}}\right]$

Then

$L_{1}=\partial/\partial x_{1}$
$L_{2}=e^{ax_{1}}\cos(bx_{1})\,\,\partial/\partial
x_{2}+e^{ax_{1}}\sin(bx_{1})\,\,\partial/\partial x_{3}$
$L_{3}=-e^{ax_{1}}\sin(bx_{1})\,\,\partial/\partial
x_{2}+e^{ax_{1}}\cos(bx_{1})\,\,\partial/\partial x_{3}\,\,.$

We must show that the field $X_{1}=X_{1}^{i}L_{i}$ satisfies

$D(X)+D(X)^{T}+C(X)=0.$

when

$X_{1}^{1}=\dfrac{-2x_{2}}{a}$
$X_{1}^{2}=e^{-ax_{1}}\left[\cos(bx_{1})(x_{3}^{2}-x_{2}^{2})-\sin(bx_{1})(2x_{2}x_{3})\right]+\dfrac{e^{ax_{1}}\cos(bx_{1})}{a^{2}}$
$X_{1}^{3}=e^{-ax_{1}}\left[-\sin(bx_{1})(x_{3}^{2}-x_{2}^{2})-\cos(bx_{1})(2x_{2}x_{3})\right]-\dfrac{e^{ax_{1}}\sin(bx_{1})}{a^{2}}$

Another tedious but straightforward computation gives the matrix $D(X_{1})$:

$\left[{\begin{array}[]{*{20}c}{0}&{-aX_{1}^{2}+bX_{1}^{3}+2e^{ax_{1}}\cos(bx_{1})/a}&{-bX_{1}^{2}-aX_{1}^{3}-2e^{ax_{1}}\sin(bx_{1})/a}\\\
{-2e^{ax_{1}}\cos(bx_{1})/a}&{-2x_{2}}&{-2x_{3}}\\\
{2e^{ax_{1}}\sin(bx_{1})/a}&{2x_{3}}&{-2x_{2}}\end{array}}\right]$

Then using (12) with $c=-b$ and $X_{1}^{1}=(-2x_{2})/a$, we also have

$C(X)=\left[{\begin{array}[]{*{20}c}{0}&{aX_{1}^{2}-bX_{1}^{3}}&{bX_{1}^{2}+aX_{1}^{3}}\\\
{aX_{1}^{2}-bX_{1}^{3}}&{4x_{2}}&{0}\\\
{bX_{1}^{2}+aX_{1}^{3}}&{0}&{4x_{2}}\end{array}}\right].$

It is then easy to check that $D(X)+D(X)^{T}+C(X)=0$, so $X_{1}$ is in fact a
Killing field.

Next note that evaluating the matrix above at $e=(0,0,0)$ gives

$D(X_{1})\arrowvert_{e}=\left[{\begin{array}[]{*{20}c}{0}&{1/a}&{-b/a^{2}}\\\
{-2/a}&{0}&{0}\\\ {0}&{0}&{0}\end{array}}\right].$

Then using $X_{1}^{1}(e)=X_{1}^{3}(e)=0$, $X_{1}^{2}(e)=1/a^{2}$, and $b+c=0$,
the matrix (14) gives

$\Gamma(X_{1})=\left[{\begin{array}[]{*{20}c}{0}&{1/a}&{0}\\\ {0}&{0}&{0}\\\
{b/a^{2}}&{0}&{0}\end{array}}\right]$

Taking a transpose matrix and adding we find

$P(X_{1})=D(X_{1})^{T}\arrowvert_{e}+\Gamma(X_{1})=\left[\begin{array}[]{*{20}c}0&-1/a&0\\\
1/a&0&0\\\ 0&0&0\end{array}\right]$

as claimed in section 9. ∎

Finally we verify that the vector fields in section 11 are in fact Killing
fields.

###### Proof.

For the Heisenberg algebra with matrix $\left[\begin{array}[]{*{20}c}0&0\\\
b&0\end{array}\right]$ we have (see section 4)

$M(t)=e^{At}=\left[{\begin{array}[]{*{20}c}1&0\\\ bt&1\end{array}}\right]$

and then

$L_{1}=\partial/\partial x_{1}$ $L_{2}=\partial/\partial
x_{2}+bx_{1}\partial/\partial x_{3}$ $L_{3}=\partial/\partial x_{3}\,\,.$

The vector field
$X=x_{2}L_{1}-x_{1}L_{2}+\frac{b}{2}(x_{1}^{2}+x_{2}^{2})L_{3}$ clearly
vanishes at $e$ and we claim $X$ is in fact a Killing field. Recalling that
$D(X)_{jk}=L_{j}X^{k}$, we compute the matrix

$D(X)=\left[{\begin{array}[]{*{20}c}{0}&{-1}&{bx_{1}}\\\ {1}&{0}&{bx_{2}}\\\
{0}&{0}&{0}\end{array}}\right].$

Then using (12) with $a=c=0$, $X^{1}=x_{2}$, and $X^{2}=-x_{1}$ we also have

$C(X)=\left[{\begin{array}[]{*{20}c}{0}&{0}&{-bx^{1}}\\\ {0}&{0}&{-bx_{2}}\\\
{-bx_{1}}&{-bx_{2}}&{0}\end{array}}\right].$

It is then easy to check that $D(X)+D(X)^{T}+C(X)=0$, so $X$ is in fact a
Killing field.

For $\mathfrak{g}=\mathfrak{e}(2)$ and the matrix
$\left[\begin{array}[]{*{20}c}0&-b\\\ b&0\end{array}\right]$ we have (see
section 4)

$M(t)=e^{At}=\left[{\begin{array}[]{*{20}c}\cos(bx_{1})&-\sin(bx_{1})\\\
\sin(bx_{1})&\cos(bx_{1})\end{array}}\right]$

and then

$L_{1}=\partial/\partial x_{1}$ $L_{2}=\cos(bx_{1})\partial/\partial
x_{2}+\sin(bx_{1})\partial/\partial x_{3}$
$L_{3}=-\sin(bx_{1})\partial/\partial x_{2}+\cos(bx_{1})\partial/\partial
x_{3}\,\,.$

The field

$X=\left[x_{3}\cos(bx_{1})-x_{2}\sin(bx_{1})\right]L_{2}+\left[-x_{3}\sin(bx_{1})-x_{2}\cos(bx_{1})\right]L_{3}$

clearly vanishes at $e$ and we claim it is a Killing field. We compute the
matrix

$D(X)=\left[{\begin{array}[]{*{20}c}{0}&{bX^{3}}&{-bX^{2}}\\\ {0}&{0}&{-1}\\\
{0}&{1}&{0}\end{array}}\right].$

Then using (12) with $a=0$ and $b+c=0$ we also have

$C(X)=\left[{\begin{array}[]{*{20}c}{0}&{-bX^{3}}&{bX^{2}}\\\
{-bX^{3}}&{0}&{0}\\\ {bX^{2}}&{0}&{0}\end{array}}\right].$

It is then easy to check that $D(X)+D(X)^{T}+C(X=0=0$, so $X$ is in fact a
Killing field.

Since $X$ vanishes at $e$, we have $\Gamma(X)=0$. Evaluating partial
derivatives at $e$, we find
$D(X)\arrowvert_{e}=\left[\begin{array}[]{*{20}c}0&0&0\\\ 0&0&-1\\\
0&1&0\end{array}\right]$. Then

$P(X)=D(X)^{T}\arrowvert_{e}+\Gamma(X)=\left[\begin{array}[]{*{20}c}0&0&0\\\
0&0&1\\\ 0&-1&0\end{array}\right]$

as claimed in section 11.

∎

## References

  * [1] May,R.D. and Wears, T.H., Three-dimensional metric Lie algebras and Ricci flow, J. Lie Theory, 30 (2020).
  * [2] Milnor, J., Curvatures of left invariant metrics on Lie groups, Advances in Math. 21(1976) 293-329.
  * [3] Reggiani, S., The index of symmetry of three-dimensional Lie groups with a left-invariant metric, Adv. Geom., 18 4 (2018) 395-404.
  * [4] Wielenberg, N., Three dimensional hyperbolic spaces, Rocky Mtn. J. Math., 11 4 (1981) 659-671.
  * [5] Wolf, J.A., Spaces of constant curvature, 6th ed., AMS Chelsea Publishing, Amer. Math. Soc., Providence (2011).


# OSSEM: one-shot speaker adaptive speech enhancement
using meta learning

###### Abstract

Although deep learning (DL) has achieved notable progress in speech
enhancement (SE), further research is still required for a DL-based SE system
to adapt effectively and efficiently to particular speakers. In this study, we
propose a novel meta-learning-based speaker-adaptive SE approach (called
OSSEM) that aims to achieve SE model adaptation in a one-shot manner. OSSEM
consists of a modified transformer SE network and a speaker-specific masking
(SSM) network. In practice, the SSM network takes an enrolled speaker
embedding extracted using ECAPA-TDNN to adjust the input noisy feature through
masking. To evaluate OSSEM, we designed a modified Voice Bank-DEMAND dataset,
in which one utterance from the testing set was used for model adaptation, and
the remaining utterances were used for testing the performance. Moreover, we
set restrictions allowing the enhancement process to be conducted in real
time, and thus designed OSSEM to be a causal SE system. Experimental results
first show that OSSEM can effectively adapt a pretrained SE model to a
particular speaker with only one utterance, thus yielding improved SE results.
Meanwhile, OSSEM exhibits a competitive performance compared to state-of-the-
art causal SE systems.

Index Terms—  Speech Enhancement, Speaker Embedding, Meta-learning, Deep
Learning

##  1 Introduction

The goal of speech enhancement (SE) is to improve the quality and
intelligibility of distorted speech. In various speech-related applications,
such as automatic speech recognition [1], speaker recognition [2, 3], and
assistive hearing devices [4], SE serves as an indispensable front-end unit.
Traditional SE methods [5, 6] are derived based on statistical models and
assumed properties of speech and noise signals. Under scenarios in which the
assumptions are unsatisfactory, the traditional SE approaches may yield a
suboptimal performance.

Deep learning (DL) models have recently been widely applied for SE tasks [7,
8, 9, 10]. Although several studies have shown that DL-based SE systems can
outperform traditional methods, their limited generalization ability is still
an issue. One simple solution to improve the generalization of the model
involves collecting as much training data as possible. However, it is almost
impossible to cover all types of conditions (e.g., different SNRs, noise
types, and speakers). Another solution is to adapt the SE model before/during
the inference using auxiliary information. For example, in [11, 12, 13], the
authors first prepared multiple SE models, and then trained them under
different conditions. In the testing stage, a most-match model was selected
based on certain criteria.

Another class utilizes additional noise or speaker information to guide the SE
adaptation to certain types of noise [14, 15, 16, 17] or speaker [18]
conditions. In this study, we go one step further to simultaneously adapt both
the model inputs and model weights through embedding vector and meta-learning,
respectively. Specifically, we propose a novel one-shot speaker-adaptive SE
approach using meta-learning (OSSEM), in which the SE model can be effectively
and efficiently adapted to a particular speaker with only one adaptation
sample. OSSEM consists of two networks: a modified transformer SE network that
aims to achieve an SE and a speaker-specific masking (SSM) network that
generates speaker-specific masks for adapting the SE model. For the SSM
network, we adopted ECAPA-TDNN [19] through the SpeechBrain toolkit [20] to
extract speaker embeddings as the input features. The two networks were
trained in a meta-learning manner, which has been shown to be effective for
one/few-shot learning on several tasks [21, 22]. In contrast to previous
studies that adapt the overall SE model [23], OSSEM can reach a fast
adaptation because only the parameters in the SSM network are adjusted instead
of the entire OSSEM system. In addition, we propose several training
techniques for OSSEM to further improve its stability and performance.

To evaluate the proposed OSSEM system, we slightly modified the Voice Bank-
DEMAND [24] dataset. One utterance from each speaker was selected from the
testing set for model adaptation, and the remaining utterances from the same
speaker were used for testing the performance. Experiment results confirmed
the fast model adaptation of OSSEM and showed that it can yield a competitive
performance comparable to that of state-of-art casual SE systems. The
remainder of this paper is organized as follows. Section 2 reviews related
studies and previous meta-learning. Section 3 then elaborates on the proposed
OSSEM approach. Section 4 reports the experiment setup and results. Finally,
Section 5 provides some concluding remarks regarding the findings and
contributions of this study.

Fig. 1: Flow chart of the OSSEM system, (a) the training stage (offline), and
(b) the one-shot learning and testing stages (online).

##  2 Meta-learning SE system

The aim of OSSEM is to effectively adapt a pretrained SE model to a particular
speaker in a one-shot learning manner. To this end, as the main criterion for
training OSSEM, we adopt the meta-learning algorithm, which has been popularly
used to build one/few-shot machine learning systems [25]. To apply meta-
learning to SE systems, the training set data needs to be reorganized into
task-specific classes, such as noise conditions [23] or speaker identities
(considered in this study). Each class is further divided into support and
query set. The former is used for adaptation by one (or few) shot learning,
whereas the latter is used for model training. To the best of our knowledge,
the proposed OSSEM is the first one-shot speaker-adaptation SE using meta-
learning. To clarify the implementation details, we elaborate on the training,
one-shot learning, and testing stages as follows.

###  2.1 Training stage (offline)

Fig. 1 (a) demonstrates the training stage (offline) of OSSEM. For clarity, we
denote the DL model by $g_{\theta_{n}}$, with weights $\theta_{n}$ of the
$n$-th iteration. We elaborate three steps in this stage as follows: In Step
1, the weights $\theta_{n}$ are copied as $\theta^{{}^{\prime}}_{n}$. In Step
2, the weights $\theta^{{}^{\prime}}_{n}$ are adapted using gradients from the
training on the support set $s_{n}$, where
$\hat{X}^{s_{n}}_{t,f}=g_{\theta^{{}^{\prime}}_{n}}(X^{s_{n}}_{t,f})$ denotes
the prediction using the input feature $X^{s_{n}}_{t,f}$, and
$Y^{s_{n}}_{t,f}$ denotes the clean features. Finally, in Step 3, the adapted
weights $\theta^{{}^{\prime\prime}}_{n}$ are then trained using the
corresponding query set $q_{n}$, where
$\hat{X}^{q_{n}}_{t,f}=g_{\theta^{{}^{\prime\prime}}_{n}}(X^{q_{n}}_{t,f})$
denotes the prediction using the input feature $X^{q_{n}}_{t,f}$, and
$Y^{q_{n}}_{t,f}$ denotes the clean features. The gradients computed from this
step are used to update the original weights $\theta_{n}$ to yield
$\theta_{n+1}$. The two gradients in Step 2 and Step 3 are computed using the
losses as follows:

$\displaystyle
L_{1}(\hat{X}^{s_{n}}_{t,f},Y^{s_{n}}_{t,f})=\frac{1}{N(s_{n})}\sum^{N(s_{n})}_{n=1}\|\hat{X}^{s_{n}}_{t,f}-Y^{s_{n}}_{t,f}\|_{1}$
(1) $\displaystyle
L_{1}(\hat{X}^{q_{n}}_{t,f},Y^{q_{n}}_{t,f})=\frac{1}{N(q_{n})}\sum^{N(q_{n})}_{n=1}\|\hat{X}^{q_{n}}_{t,f}-Y^{q_{n}}_{t,f}\|_{1}$
(2)

where $N(s_{n})$ and $N(q_{n})$ denote the numbers of data in the support and
query sets, respectively. The value $N(s_{n})$ is usually extremely small
(e.g., in this paper, $N(s_{n})$ = 1 for one-shot learning and $N(q_{n})$ = 20
for Step 3). The inner and outer loops represent supervised training cycles on
the support set $s_{n}$ and query set $q_{n}$, respectively.

###  2.2 One-shot learning and testing stage (online)

We demonstrate the testing stage (online) flowchart of the OSSEM in Fig. 1
(b). We elaborate on two steps in this stage as follows: In Step 1, the
weights $\theta_{meta}$ are copied as $\theta^{{}^{\prime}}_{meta}$. This copy
step allows users to maintain well-trained weights $\theta_{meta}$, whereas
adaptation is only applied on the copied weights. In Step 2, the weights
$\theta^{{}^{\prime}}_{meta}$ are adapted using gradients from the training on
the enrollment set $e_{n}$, where
$\hat{X}^{e_{n}}_{t,f}=g_{\theta^{{}^{\prime}}_{meta}}(X^{e_{n}}_{t,f})$ and
$Y^{e_{n}}_{t,f}$ denotes the clean features. Finally, the adapted weights
$\theta^{spk}_{meta}$ are ready for SE to enhance the noisy features $X_{t,f}$
to $\hat{X}_{t,f}$. The gradients in Step 2 are computed using the loss as
follows:

$\displaystyle
L_{1}(\hat{X}^{e_{n}}_{t,f},Y^{e_{n}}_{t,f})=\frac{1}{N(e_{n})}\sum^{N(e_{n})}_{n=1}\|\hat{X}^{e_{n}}_{t,f}-Y^{e_{n}}_{t,f}\|_{1}$
(3)

where $N(e_{n})$ denotes the number of data in the enrollment set, which is
usually an extremely small number. For example, for a one-shot adaptation,
$N(e_{n})$ = 1. In the proposed OSSEM system, the gradients from Eq. (3)
achieve an efficient adaptation because
$\nabla_{\theta^{{}^{\prime}}_{meta}}L_{1}(\hat{X}^{e_{n}}_{t,f},Y^{e_{n}}_{t,f})$
only adapts to the SSM network of OSSEM.

Fig. 2: Model architecture of the OSSEM system that consists of a modified
Transformer (encoder, multi-head attention, and decoder) SE network and an SSM
network.

##  3 Proposed OSSEM system

###  3.1 Model architecture

####  3.1.1 Modified Transformer-based SE network

The Transformer model [26] has recently been used in SE systems and has shown
a promising performance [27, 28]. In the proposed OSSEM system, we adopted the
modified transformer model proposed in [28]; the model consists of a
convolutional encoder (for replacing the original positional encoder), several
multi-head attention blocks, and a fully connected layer. The convolutional
encoder includes four 1-D convolutional layers, and each attention block
consists of eight heads, with 64 dimensions per head.

####  3.1.2 The SSM Network

As shown in Fig. 2, SSM takes the speaker embedding (from ECAPA) as input and
generates speaker-specific masks. The SSM network consists of a three-layered
dense network with the leaky-Relu activation functions for the first two
layers and the sigmoid function for the last layer. The generated mask is then
multiplied with the noisy spectrogram to yield speaker-adaptive features,
which are the inputs to the Transformer-based SE network.

###  3.2 Training techniques

In our preliminary experiments, directly applying meta-learning to SE tasks
did not achieve good results, probably because a model trained using MAML
becomes easily unstable [29].). To optimally exert the capability of the
proposed OSSEM system, we propose several training techniques and describe
them as follows:

  1. 1.

Speaker-Inner-Loop:
MAML was previously reported to suffer from a trade-off [29] between
burdensome inner loops and better performance. To solve this problem, Raghu et
al. proposed an almost-no-inner-loop (ANIL) [30] that only updates task-
specific parameters in the inner loop. However, it may be difficult to
determine the task-specificity of the parameters in a model without a
predefined task-specific network. In this study, the OSSEM system consists of
two task-specific building blocks. The modified transformer SE network is
responsible for general-purpose SE, whereas the SSM network adapts the input
features based on a specific speaker. We propose a speaker-inner-loop method
to apply adaptation only on the SSM network in the inner loop. As shown in
Fig. 1 (a), the gradients
$\nabla_{\theta^{{}^{\prime}}_{n}}L_{1}(\hat{X}^{s_{n}}_{t,f},Y^{s_{n}}_{t,f})$
(computed using support set data $s_{n}$ of a specific speaker) are only used
to adapt the parameters of the SSM network and keep the parameters in the SE
network unchanged. The speaker-inner-loop is beneficial for two reasons:
First, the updates are completely focused on the adaptation of the task-
specific SSM network parameters. Second, the adaptation efficiency can be
further improved by focusing only on the compact SSM network instead of the
entire OSSEM system.

  2. 2.

Learning Rate Scaling Rule:
As shown in Fig. 1 (a), the training stage of OSSEM has an inner-loop and an
outer-loop. Each loop has a specific learning rate (i.e., inner-loop learning
rate (ILR) and outer-loop learning rate (OLR)). For OLR, we fixed its value.
For ILR, we adopted a modified learning rate scaling rule [31] to stabilize
the training procedure. Note that when ILR is $0$ (removing Step 2 in Fig. 1
(a)), the training stage of OSSEM becomes a normal supervised training. For
each outer loop, the training data in the query set are from the same speaker,
which may make model training difficult. Thus, as in [15], the weights of a
normally supervised trained model are applied as the weight initialization for
OSSEM. To smoothly transform from supervised training to meta-learning, we set
ILR to $0$ in the first five epochs. From the sixth epoch, ILR was linearly
scaled up (warmup) and then fixed during the last 20 epochs to stabilize the
training.

  3. 3.

Feature Re-scaling Inner-Loop:
Feature re-scaling has been confirmed to be effective [32] at stabilizing the
gradients when training DL-based tasks. Because OSSEM relies on the
effectiveness of speaker adaptation, it is crucial to stabilize the training
of the inner loop. Specifically, speaker adaptation functions through an SSM
network, which is highly dependent on the training of the inner loop. To
further stabilize the training of the inner loop, we designed a feature re-
scaling ratio $\alpha$ that is computed between clean and noisy features (the
average energy power ratio between features):

$\displaystyle\alpha=\frac{\|Y^{s}_{t,f}\|_{2}^{2}}{\|X^{s}_{t,f}\|_{2}^{2}}$
(4) $\displaystyle Loss=L_{1}({\rm OSSEM}(\hat{X^{s}}\cdot\alpha),Y^{s})$ (5)

where $X^{s}_{t,f}$ and $Y^{s}_{t,f}$ denote the noisy speech and the
corresponding clean speech in the support set $s$. The ratio $\alpha$ is
multiplied by the model output for only the loss computation. By adopting the
feature re-scaling technique, the inner-loop gradients can be stabilized, and
thus the SSM network can be tuned more accurately, finally enabling OSSEM to
attain a better speaker adaptation performance.

##  4 Experimental Results

In this section, we investigate the effectiveness of OSSEM using the Voice
Bank-DEMAND dataset. We evaluated the proposed OSSEM system using five
standard metrics, PESQ, STOI, CSIG, CBAK, and COVL, which are commonly
reported when using this dataset. We prepared 192-dimensional speaker
embedding of one enrollment speech from each speaker using the ECAPA-TDNN. For
fair comparisons, we trained all of our proposed systems using a total of 100
epochs (including pretrained epochs), with 1000 iterations in each epoch. In
addition, we adopted the same set of hyperparameters of training. Note that we
removed the enrolled speech from the two speakers in the testing set for a
fair evaluation. OSSEM is a causal SE system that can perform enhancement in
real time.

###  4.1 Speaker masks in the SE process

As shown in Fig. 2, OSSEM uses the SSM network to directly modify the noisy
input. To justify this early signal modification, we implemented four systems
with the same SE and SSM networks, whereas the masks generated by the SSM
network were applied in different places during the SE process: before the
encoder, before the multi-head attention, before the decoder, and after the
decoder. These systems are labeled Pre, Mid1, Mid2, and Last, respectively,
and their PESQ scores on the test set are reported in Table 1. An addition
system, called Non, represents a system having only an SE network and serves
as a baseline for comparison. From Table 1, Pre performs better than the other
systems, confirming that early signal modification is a suitable choice for
building an adaptive SE system.

Table 1: Results of different places that the speaker masks (generated by the SSM network) are applied in the SE process. | Pre | Mid1 | Mid2 | Last | Non
---|---|---|---|---|---
PESQ | 2.80 | 2.78 | 2.76 | 2.64 | 2.69
Table 2: Results of OSSEM using different training techniques. System | PESQ | STOI
---|---|---
TransformerSpk | 2.809 | 0.93
OSSEM with technique 1 | 2.818 | 0.93
OSSEM with techniques 1&2 | 2.847 | 0.93
OSSEM with techniques 1&2&3 | 2.896 | 0.93

###  4.2 Ablation study

Next, we investigate the effectiveness of the proposed training techniques
introduced in Section 3.2. Training techniques 1, 2, and 3 denote the speaker
inner-loop, learning rate scaling, and feature re-scaling inner-loop,
respectively. The transformer Spk in Table 2 denotes the SE network, which is
first trained in a supervised manner and using Fig. 1 (b) for online
adaptation. From Table 2, we can observe that each technique is helpful in
improving the PESQ score. Among all of the proposed techniques, the feature
re-scaling inner-loop (technique 3) is the most important.

Table 3: Evaluation results of OSSEM and other causal SE systems on the
VoiceBank-DEMAND dataset.

SE approach |  PESQ |  CSIG |  CBAK |  COVL |  STOI |  causal
---|---|---|---|---|---|---
Noisy[33] |  1.97 |  3.35 |  2.44 |  2.63 |  0.92 |  –
Wiener [33] |  2.22 |  3.23 |  2.68 |  2.67 |  – |  yes
Conv-TasNet [34] |  2.53 |  3.95 |  3.08 |  3.23 |  – |  yes*
Transformer [28] |  2.69 |  4.07 |  3.03 |  3.38 |  0.93 |  yes
STFT-TCN [34] |  2.73 |  4.11 |  3.25 |  3.42 |  – |  yes*
CRN-MSE [35] |  2.74 |  3.86 |  3.14 |  3.30 |  0.93 |  yes
TFSNN [36] |  2.79 |  4.17 |  3.27 |  3.49 |  - |  yes
TransformerSpk |  2.81 |  3.95 |  3.19 |  3.36 |  0.93 |  yes
OSSEM |  2.90 |  4.11 |  3.15 |  3.50 |  0.93 |  yes

yes* denotes the use of look-ahead mechanism, where Conv-TasNet uses 1ms look-
ahead, and STFT-TCN uses 4ms look-ahead.

###  4.3 Comparison of OSSEM with other causal SE systems

Table 3 lists the results of OSSEM and several related SE systems, including
the modified Transformer SE network [28], TransformerSpk (as discussed in
Section 4.2), and some well-known causal SE systems. Note that our testing set
is a slightly modified version of the original testing set of Voice Bank-
DEMAND. In our preliminary experiments, we have confirmed that the results of
Noisy, Wiener, and Transformer tested on the modified testing sets are almost
identical to those tested on the original Voice Bank-DEMAND. Therefore, we
directly copied the results of Noisy, Wiener, and other SE systems from
existing studies listed in Table 3 for comparison.

From Table 3, we first note that OSSEM outperforms both Transformer and
TransformerSpk, confirming the effectiveness of one-shot adaptation through
meta-learning. Next, it was observed that OSSEM outperforms other systems in
most of the five metric scores, confirming the effectiveness of OSSEM as
compared to other systems. We also noticed that, among the five metrics, OSSEM
yields a slightly lower yet comparable performance to DEMUCS [37]. We argue
that the model size of DEMUCS is almost twice that of the OSSEM system (OSSEM:
38MB; DEMUCS: 73MB [38]). Thus, OSSEM achieves a better applicability to real-
world systems, particularly when storage is limited. We also note that DEMUCS
uses a 3-ms look-ahead mechanism, whereas our OSSEM system is completely
causal. To further investigate the capabilities of OSSEM, we implemented a
non-causal version of OSSEM. The results of the non-causal OSSEM do not
notably outperform the causal version, probably owning to the fact that
speaker embedding can mitigate the information gap between causal and non-
causal systems.

###  4.4 Analysis of the SSM network

In this section, we further investigate the function of the SSM network in
OSSEM. Figs. 3 (a) and (b) show the mask plots (output of the SSM network) of
female and male test speakers, where the x- and y- axes of the plots indicate
the frequency and amplitude, respectively. The blue and black lines indicate
the detailed mask values for every frequency bin and the smoothed mask value
over the frequency (showing the trend of the mask). From Fig. 3, we first note
that the frequency range of the first peak in Fig. 3 (a) is higher than that
of Fig. 3 (b), confirming that the mask indeed captures the gender
information. Next, we observed that Figs. 3 (a) and (b) present extremely
different patterns, showing that for each speaker, a specified mask is
generated and assigned, to modify the noisy input.

Fig. 3: The masks generated by the SSM network of speech samples from (a) a
female speaker, and (b) a male speaker.

Fig. 4 (a) shows the results of a t-SNE analysis [39] on the embedding vectors
extracted using ECAPA. The speech utterances were taken from the training set
of Voice Bank-DEMAND, which comprises utterances from 28 speakers. From Fig. 4
(a), we can note that the utterances from each speaker are grouped together,
and there are a total of 28 groups. Moreover, we can observe a clear distance
between each pair of groups. The results confirm that the speaker
representations extracted by ECAPA are informative and discriminative. Fig. 4
(b) shows the results of a t-SNE analysis that was performed on the masks
generated by the SSM network with the same utterances used in Fig. 4 (a). Note
that the masks are also similarly informative and discriminative but are
divided into two groups of two genders. The transition from Fig. 4 (a) to Fig.
4 (b) suggests that the SSM network further emphasized the characteristics of
gender when converting the embedding into masks. The results conform to a
previous study [40] showing that the gender identity of the speaker plays a
major factor in SE performance.

Fig. 4: The t-SNE analysis of (a) ECAPA-TDNN speaker embedding of all
speakers, and (b) the speaker masks of all speakers.

##  5 Conclusion

In this study, we proposed an OSSEM system that can adapt a pretrained SE
model to a new speaker in a one-shot manner. Experiment results confirm the
effectiveness of speaker-specific masking and metal learning for fast SE model
adaptation. The results also show that OSSEM outperforms several well-known
causal SE systems in terms of the standard evaluation metrics. In summary, the
contributions of this study are threefold: (1) The proposed OSSEM is the first
one-shot speaker adaptive SE system based on meta-learning, (2) a novel SSM
network that directly modifies the noisy input is proposed and was proven to
be effective, and (3) several techniques to improve the stability of meta-
learning have been adopted and modified to conform to the SE task. In the
future, we will explore the application of the proposed approach to other
speech generation tasks, such as speech separation and target speaker
extraction.

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