The Ricci pinching functional on solvmanifolds II

TL;DR

This paper investigates whether solvsolitons are the global maxima of the Ricci pinching functional on solvable Lie groups, extending previous results to more general cases involving nilradicals and metric restrictions.

Contribution

It proves that solvsolitons are the unique global maxima of the Ricci pinching functional in broader classes of solvable Lie groups, including those with nilradicals of codimension one.

Findings

Solvsolitons are global maxima for the Ricci pinching functional in new cases.

The results extend previous work to groups with nilradicals of codimension one.

The study includes cases where the nilradical is abelian and metrics are restricted.

Abstract

It is natural to ask whether solvsolitons are global maxima for the Ricci pinching functional F:=scal^2/|Ric|^2 on the set of all left-invariant metrics on a given solvable Lie group S, as it is to ask whether they are the only global maxima. A positive answer to both questions was given in a recent paper by the same authors when the Lie algebra s of S is either unimodular or has a codimension-one abelian ideal. In the present paper, we prove that this also holds in the following two more general cases: 1) s has a nilradical of codimension-one; 2) the nilradical n of s is abelian and the functional F is restricted to the set of metrics such that a is orthogonal to n, where a is the orthogonal complement of n with respect to the solvsoliton.

Equations

$$\mathrm{Ric}=cI+D,\text{mbox}forsomec\in \mathbb{R},D\in \mathrm{Der}(\mathfrak{s}),$$

$$F:=\frac{{\mathrm{scal}}^2}{|\mathrm{Ric}{|}^2},$$

$$\overline{h}:=\left[\begin{array}[]{c|ccc}c^{-1}&&0&\\ \hline\cr&\\ X&&h&\\ &\end{array}\right],\qquad c\in{\mathbb{R}}^{*},\quad X\in\mathfrak{n},\quad h\in\mathrm{GL}(\mathfrak{n}),$$

$$\operatorname{Ric}=\left[\begin{array}[]{c|c}-\operatorname{tr}{S(A)^{2}}&\ast\\ \hline\cr&\\ \ast&\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}+\frac{1}{2}[A,A^{t}]-(\operatorname{tr}{A})S(A)\\ &\end{array}\right],$$

$$F(A,[\cdot ,\cdot {]}_{\mathfrak{n}})=\frac{{\left(\mathrm{tr}S(A{)}^2+(\mathrm{tr}A{)}^2+1\right)}^2}{\mathrm{tr}S(A{)}^2\left(\mathrm{tr}S(A{)}^2+(\mathrm{tr}A{)}^2\right)+|{\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}{|}^2+\frac{1}{4}|[A,A^t]{|}^2+G(A,[\cdot ,\cdot {]}_{\mathfrak{n}})},$$

$$[A,A^t]=0,\mathrm{tr}S(A{)}^2=|{\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}{|}^2,{\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}+|{\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}{|}^{2}I\in \mathrm{Der}([\cdot ,\cdot {]}_{\mathfrak{n}}).$$

$$F(ch(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1},h\cdot [\cdot ,\cdot {]}_{\mathfrak{n}})\le F(A,[\cdot ,\cdot {]}_{\mathfrak{n}}).$$

$$A=\left[\begin{array}[]{ccc}\begin{array}[]{l|}A_{1}\\ \hline\cr\end{array}&0&0\\ 0&\ddots&0\\ 0&0&\begin{array}[]{|r}\hline\cr A_{k}\end{array}\end{array}\right],\qquad\operatorname{ad}_{\mathfrak{n}}{h^{-1}X}=\left[\begin{array}[]{ccc}\begin{array}[]{l|}0\\ \hline\cr\end{array}&0&0\\ \ast&\ddots&0\\ \ast&\ast&\begin{array}[]{|r}\hline\cr 0\end{array}\end{array}\right],$$

$$|h(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1}|\ge |A|,$$

$$\mathrm{tr}S(h(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1}{)}^2=\frac{1}{2}|h(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1}{|}^2+\frac{1}{2}\mathrm{tr}{A}^2,$$

$$F(ch(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1},h\cdot [\cdot ,\cdot {]}_{\mathfrak{n}})\le$$

$$\le$$

$$x:=\frac{1}{2}|h(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1}{|}^2\ge \frac{1}{2}|A{|}^2=:x_0,b:=\frac{1}{2}\mathrm{tr}{A}^2,a:=b+(\mathrm{tr}A{)}^2.$$

$$F(A,[\cdot ,\cdot {]}_{\mathfrak{n}})=f(x_0,1)=\frac{x_0+a+1}{x_0+b}.$$

$$q(x):=c^4(x+b)(x+a)+x_0+b,$$

$$f(x_0,c)\le f(x_0,1),\forall c\in \mathbb{R},$$

$$c^4(x_0+a{)}^2+1+2c^2(x_0+a)\le (x_0+a+1)(c^4(x_0+a)+1),$$

$$p(x)=r{x}^2+sx+t\ge 0,\forall x\ge x_0,$$

$$t=a{c}^4(x_0(b-a)+b)-2a{c}^2(x_0+b)+(x_0+a)(x_0+b).$$

$$c^4(a-b)(x_0+a)<-c^4(x_0+a)+c^2(2-c^2)(x_0+b),$$

$$s^{2}-4rt=c^{4}(a-b)(x_{0}+a+1)\Big{(}c^{4}(a-b)(x_{0}+a+1)+4(c^{2}-1)(x_{0}+b)\Big{)},$$

$$-c^4(x_0+a)+c^2(2-c^2)(x_0+b)+c^4(a-b)+4(c^2-1)(x_0+b)=-2(x_0+b)(c^4-3c^2+2)<0.$$

$$f(x,c)=\frac{{\left(c^2(x+a)+1\right)}^2}{c^{2}d(x+a)+d}=\frac{c^2(x+a)+1}{d}=\frac{d+\frac{d}{x+b}(a-b)+1}{d}\le f(x_0,1).$$

$$\underset{x\to \infty }{\lim }f(x,c)=1<f(x_0,1),\underset{c\to \infty }{\lim }f(x,c)=\frac{x+a}{x+b}\le 1+\frac{a-b}{d}<f(x_0,1),$$

$$\mathrm{tr}S(ch(A-{\mathrm{ad}}_{\mathfrak{n}}{h}^{-1}X)h^{-1}{)}^2=$$

$$=$$

$${\mathrm{Ric}}_{h\cdot [\cdot ,\cdot {]}_{\mathfrak{n}}}+|{\mathrm{Ric}}_{h\cdot [\cdot ,\cdot {]}_{\mathfrak{n}}}{|}^{2}I=\tilde{h}({\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}+|{\mathrm{Ric}}_{[\cdot ,\cdot {]}_{\mathfrak{n}}}{|}^{2}I){\tilde{h}}^{-1}\in \mathrm{Der}(\tilde{h}\cdot [\cdot ,\cdot {]}_{\mathfrak{n}})=\mathrm{Der}(h\cdot [\cdot ,\cdot {]}_{\mathfrak{n}}).$$

$$\left(h_2(h_1^{-1}{A}_1)h_2^{-1},\dots ,h_2(h_1^{-1}{A}_r)h_2^{-1}\right),h_1\in \mathrm{GL}(\mathfrak{a}),h_2\in \mathrm{GL}(\mathfrak{n}),$$

$$\operatorname{Ric}=\left[\begin{array}[]{c|c}R&0\\ \hline\cr 0&\frac{1}{2}\sum[A_{i},A_{i}^{t}]-\sum(\operatorname{tr}{A_{i}})S(A_{i})\end{array}\right],\qquad R_{ij}=-\operatorname{tr}{S(A_{i})S(A_{j})}.$$

$$F(A_1,\dots ,A_r)=\frac{{\left(\sum \mathrm{tr}S(A_i{)}^2+(\mathrm{tr}{A}_1{)}^2\right)}^2}{\sum {\left(\mathrm{tr}S(A_i)S(A_j)\right)}^2+(\mathrm{tr}{A}_1{)}^2\mathrm{tr}S(A_1{)}^2+\frac{1}{4}|\sum [A_i,A_i^t]{|}^2}.$$

$$F\left(h_2(h_1^{-1}{A}_1)h_2^{-1},\dots ,h_2(h_1^{-1}{A}_r)h_2^{-1}\right)\le F(A_1,\dots ,A_r),$$

Full text

This research was partially supported by grants from FONCYT and SeCyT (Universidad Nacional de Córdoba)

1. Introduction

A left-invariant metric on a simply connected solvable Lie group $S$ is called a solvsoliton when its Ricci operator satisfies

[TABLE]

where $\mathfrak{s}$ denotes the Lie algebra of $S$ (see [L3]). The definition is a neat combination of geometric and algebraic aspects of a Lie group and the following facts explain very well and from different points of view why these metrics are quite distinguished:

• •

Ricci solitons. Solvsolitons are all Ricci solitons, they are precisely the left-invariant Ricci solitons such that the Ricci flow evolves by just scaling and pullback by automorphisms (see [L3] and [J1, Lemma 5.3]). Moreover, if S is of real type, then any scalar-curvature normalized Ricci flow solution converges in Cheeger-Gromov topology to a non-flat solvsoliton on a possibly different solvable Lie group, which does not depend on the initial metric (see [BL, Theorem A]).

• •

Uniqueness. On a given $S$, there is at most one solvsoliton up to scaling and pullback by automorphisms of $S$ (see [L3, Section 5] and [BL, Corollary 4.3]).

• •

Structure. If $\mathfrak{n}$ is the niradical of $\mathfrak{s}$ and $\mathfrak{s}=\mathfrak{a}\oplus\mathfrak{n}$ is the orthogonal decomposition with respect to a solvsoliton, then $[\mathfrak{a},\mathfrak{a}]=0$ (which already imposes an algebraic constraint on $\mathfrak{s}$, as the Lie algebra $[e_{1},e_{3}]=e_{3}$, $[e_{2},e_{4}]=e_{4}$, $[e_{1},e_{2}]=e_{5}$ shows) and any $\operatorname{ad}{Y}|_{\mathfrak{n}}$ must be a semisimple operator, although the strongest and less understood obstruction is that $\mathfrak{n}$ has to admit itself a solvsoliton, called a nilsoliton in the nilpotent case (see [L3, Theorem 4.8] and [L2]). We refer to [W, FC] and the references therein for low-dimensional classification results for solvsolitons.

• •

Maximal symmetry. The dimension of the isometry group of a non-flat solvsoliton on $S$ is maximal among all left-invariant metrics on $S$ (see [BL, Corollary C]). A stronger maximality condition holds in the case of a unimodular $S$: the isometry group of a solvsoliton contains all possible isometry groups of left-invariant metrics on $S$ up to conjugation by a diffeomorphism (see [J2, Corollary 1.3], and see [GJ] and [BL, Corollary D] for the (non-unimodular) Einstein case, where such a diffeomorphism is actually an automorphism of $S$).

• •

Ricci-pinched. On a given unimodular $S$, solvsolitons are the only global maxima for the Ricci pinching functional

[TABLE]

restricted to the set of all left-invariant metrics on $S$ (see [LW1] for nilsolitons and [LW2, Theorem 1.2, (v)] for the general unimodular case). Note that $F$ is measuring in a sense how far is a metric from being Einstein.

We are concerned in this paper with the last property of solvsolitons above. Beyond the unimodular case, solvsolitons were proved in [LW2] to be the only global maxima of $F$ for any almost-abelian $S$ (i.e. $\mathfrak{s}$ has a codimension-one abelian ideal). Our purpose here is to prove that this also holds among the following two much broader classes of solvable Lie groups.

Theorem 1.1**.**

Let $\mathfrak{s}$ be a solvable Lie algebra with nilradical of codimension-one and assume that $S$ admits a solvsoliton $g$. Then $g$ is a global maximum for the functional $F$ restricted to the set of all left-invariant metrics on $S$. Moreover, any other global maximum $g^{\prime}$ is also a solvsoliton (i.e. $g^{\prime}=c\varphi^{*}g$, for some $c>0$ and $\varphi\in\operatorname{Aut}(S)$).

Theorem 1.2**.**

Let $\mathfrak{s}$ be a solvable Lie algebra with abelian nilradical $\mathfrak{n}$ and assume that $S$ admits a solvsoliton $g$. If $\mathfrak{s}=\mathfrak{a}\oplus\mathfrak{n}$ is orthogonal with respect to $g$, then the solvsoliton $g$ is a global maximum for the functional $F$ restricted to the set of all left-invariant metrics on $S$ such that $\mathfrak{a}\perp\mathfrak{n}$. Moreover, any other global maximum $g^{\prime}$ for $F$ on such a set is also a solvsoliton (i.e. $g^{\prime}=c\varphi^{*}g$, for some $c>0$ and $\varphi\in\operatorname{Aut}(S)$).

The proofs of these theorems will be given in Sections 2 and 3, respectively.

2. The functional $F$ on rank-one solvmanifolds

Let $S$ be a solvable Lie group of dimension $n$ such that the nilradical $\mathfrak{n}$ of its Lie algebra $\mathfrak{s}$ is non-abelian and has dimension $n-1$. If we fix a decomposition $\mathfrak{s}={\mathbb{R}}Y\oplus\mathfrak{n}$, then the Lie bracket of $\mathfrak{s}$ is determined by the pair $(A,[\cdot,\cdot]_{\mathfrak{n}})$, where $A:=\operatorname{ad}{Y}|_{\mathfrak{n}}\in\operatorname{Der}(\mathfrak{n})$ and $[\cdot,\cdot]_{\mathfrak{n}}$ is the Lie bracket of $\mathfrak{n}$.

By fixing in addition an inner-product $\langle\cdot,\cdot\rangle$ on $\mathfrak{s}$ such that $Y\perp\mathfrak{n}$ and $|Y|=1$, each pair $(A,[\cdot,\cdot]_{\mathfrak{n}})$ is identified with the corresponding solvable Lie group $S_{(A,[\cdot,\cdot]_{\mathfrak{n}})}$ endowed with the left-invariant metric defined by $\langle\cdot,\cdot\rangle$. The space of such pairs therefore covers, up to isometry, all left-invariant metrics on Lie groups with a codimension-one nilradical. Indeed, more precisely, the left-invariant metric $\langle\overline{h}\cdot,\overline{h}\cdot\rangle$ on the Lie group $S_{(A,[\cdot,\cdot]_{\mathfrak{n}})}$, where, with respect to a fixed orthonormal basis $\{Y,X_{1},\dots,X_{n-1}\}$ of $(\mathfrak{s},\langle\cdot,\cdot\rangle)$,

[TABLE]

is isometric to the pair $(ch(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1},h\cdot[\cdot,\cdot]_{\mathfrak{n}})$, since $\overline{h}$ is an isomorphism between the corresponding Lie algebras.

It follows from [L3, (25)] that the Ricci curvature of $(A,[\cdot,\cdot]_{\mathfrak{n}})$ is given by,

[TABLE]

where $S(A):=\frac{1}{2}(A+A^{t})$ and $\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}$ denotes de Ricci operator of the nilmanifold $([\cdot,\cdot]_{\mathfrak{n}},\langle\cdot,\cdot\rangle|_{\mathfrak{n}\times\mathfrak{n}})$. If $|[\cdot,\cdot]_{\mathfrak{n}}|=2$, which can be assumed up to scaling, then $\operatorname{tr}{\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}}=-1$ and so

[TABLE]

for some expression $G(A,[\cdot,\cdot]_{\mathfrak{n}})\geq 0$ that vanishes if $A$ is normal.

Let us suppose that $(A,[\cdot,\cdot]_{\mathfrak{n}})$ is a solvsoliton with $|[\cdot,\cdot]_{\mathfrak{n}}|=2$, i.e. $\operatorname{Ric}+|\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}|^{2}I\in\operatorname{Der}(\mathfrak{s})$, which is equivalent by [L3, Theorem 4.8] to

[TABLE]

Consider $(ch(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1},h\cdot[\cdot,\cdot]_{\mathfrak{n}})$, where $c\neq 0$, $X\in\mathfrak{n}$, $h\in\mathrm{GL}(\mathfrak{n})$, and assume (up to scaling) that $|h\cdot[\cdot,\cdot]_{\mathfrak{n}}|=2$. In order to prove Theorem 1.1, it is therefore enough to show that

[TABLE]

Let $\mathfrak{n}=\mathfrak{n}_{1}\oplus\dots\oplus\mathfrak{n}_{k}$ be the orthogonal decomposition such that $[\mathfrak{n},\mathfrak{n}]_{\mathfrak{n}}=\mathfrak{n}_{2}\oplus\dots\oplus\mathfrak{n}_{k}$ and so on with the rest of the descending central series. Since $A$ is normal, $A^{t}$ is also a derivation of $\mathfrak{n}$ and thus relative to this decomposition,

[TABLE]

where the blocks correspond to each $\mathfrak{n}_{i}$. This implies that $A$ belongs to the closure of the conjugation class of $A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X}$ (by conjugating with matrices which are multiples of the identity on each block), and so

[TABLE]

where equality holds if and only if $h(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1}$ is normal. Indeed, recall that $A$ is normal and so it is a minimal vector since the moment map for the conjugation $\mathrm{GL}(\mathfrak{n})$-action on $\mathfrak{gl}(\mathfrak{n})$ is given by $m(A)=[A,A^{t}]/|A|^{2}$ (see [RS, HSS]). On the other hand,

[TABLE]

and we also know that $|\operatorname{Ric}_{h\cdot[\cdot,\cdot]_{\mathfrak{n}}}|\geq|\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}|$, by using that $[\cdot,\cdot]_{\mathfrak{n}}$ is a nilsoliton (see [LW2, Section 3.1]). It therefore follows from (2) that

[TABLE]

where $T:=\operatorname{tr}{S(h(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1})^{2}}$ and

[TABLE]

Note that $x_{0}+b=|\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}|^{2}>0$ and that the value of $F$ at the solvsoliton is given by

[TABLE]

Lemma 2.1**.**

$f(x,c)\leq f(x_{0},1)$* for any $x\geq x_{0}$ and $c\in{\mathbb{R}}$, where equality holds if and only if $x=x_{0}$ and $c=\pm 1$.*

Proof.

We first note that if we consider the denominator of $f(x,c)$, given by the parabola

[TABLE]

then $q(x_{0})>0$ and $q^{\prime}(x_{0})>0$, from which follows that $q(x)>0$ for any $x\geq x_{0}$. It follows that

[TABLE]

where equality holds if and only if $c=\pm 1$, as this is equivalent to

[TABLE]

which simplifies to $0\leq(x_{0}+a)(c^{2}-1)^{2}$.

On the other hand, it is straightforward to show that inequality $f(x,c)\leq f(x_{0},1)$ can be written as

[TABLE]

where $r=c^{4}(a-b+1)>0$, s=c^{4}\Big{(}a(a-b+1)-x_{0}(a-b)+b\Big{)}-2c^{2}(x_{0}+b) and

[TABLE]

It follows from (6) that $p(x_{0})\geq 0$, where equality holds if and only if $c=\pm 1$, if $p^{\prime}(x_{0})\geq 0$, then the lemma follows. One can therefore assume that $p^{\prime}(x_{0})=2rx_{0}+s<0$, that is,

[TABLE]

which is easily seen to imply that $c^{2}<1$ by using that $a>b$. A straightforward computation gives that the discriminant of $p$ equals

[TABLE]

and so it is negative since by (7) and the fact that $c^{2}<1$, the factor on the right is smaller than

[TABLE]

Thus $p$ is always positive, concluding the proof.

Alternatively, a simple analytic argument using the partial derivatives of $f$ gives that $(x_{0},1)$ is a local maximum of the function $f$ on the half plane $\{(x,c):x\geq x_{0}\}$ and that the only critical points of $f$ in this region are $\{(x,c):c^{2}(x+b)=d\}$, with critical values,

[TABLE]

Since

[TABLE]

there are positive numbers $M_{1},M_{2}$ such that the value of $f$ out of the compact region $\{(x,c):x_{0}\leq x\leq M_{1},\;|c|\leq M_{2}\}$ is always strictly less than $f(x_{0},1)$. All this implies that $(x_{0},\pm 1)$ are actually the only global maxima of $f$ on the half plane $\{(x,c):x\geq x_{0}\}$, as desired. ∎

Inequality (4) therefore follows from (5) and the inequality given in Lemma 2.1. On the other hand, if equality holds in (5) and Lemma 2.1, then $h(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1}$ is normal, $|h(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1}|=|A|$, $c=\pm 1$ and $|\operatorname{Ric}_{h\cdot[\cdot,\cdot]_{\mathfrak{n}}}|^{2}=|\operatorname{Ric}_{[\cdot,\cdot]_{\mathfrak{n}}}|^{2}$. This implies that

[TABLE]

and it follows from [LW2, Section 3.1] that $h\cdot[\cdot,\cdot]_{\mathfrak{n}}$ is a nilsoliton. Thus $h\cdot[\cdot,\cdot]_{\mathfrak{n}}=\tilde{h}\cdot[\cdot,\cdot]_{\mathfrak{n}}$ for some $\tilde{h}\in\mathrm{O}(\mathfrak{n},\langle\cdot,\cdot\rangle)$ by the uniqueness of nilsolitons (see [L1, Theorem 3.5]) and hence

[TABLE]

All this implies that $(ch(A-\operatorname{ad}_{\mathfrak{n}}{h^{-1}X})h^{-1},h\cdot[\cdot,\cdot]_{\mathfrak{n}})$ is a solvoliton (see (3)), concluding the proof of Theorem 1.1.

3. The functional $F$ on solvmanifolds with an abelian nilradical

In this section, we consider a solvable Lie group $S$ of dimension $n$ such that the nilradical $\mathfrak{n}$ of its Lie algebra $\mathfrak{s}$ is abelian, say with $\dim{\mathfrak{n}}=n-r$. After fixing a decomposition $\mathfrak{s}=\mathfrak{a}\oplus\mathfrak{n}$ and a basis $\{Y_{i}\}$ of $\mathfrak{a}$, the Lie bracket of $\mathfrak{s}$ is determined by an $r$-tuple $(A_{1},\dots,A_{r})$ of linearly independent linear operators of $\mathfrak{n}$ such that $[A_{i},A_{j}]=0$ for all $i,j$, where $A_{i}:=\operatorname{ad}{Y_{i}}|_{\mathfrak{n}}\in\mathfrak{gl}(\mathfrak{n})$, and a bilinear map $\lambda:\mathfrak{a}\times\mathfrak{a}\rightarrow\mathfrak{n}$. We assume that $S$ admits a solvsoliton, hence $\lambda=0$ for the corresponding orthogonal decomposition $\mathfrak{s}=\mathfrak{a}\oplus\mathfrak{n}$ (see [L3, Theorem 4.8]).

By fixing an inner-product $\langle\cdot,\cdot\rangle$ on $\mathfrak{s}$ such that $\mathfrak{a}\perp\mathfrak{n}$ and $\{Y_{i}\}$ is orthonormal, each $(A_{1},\dots,A_{r})$ is identified with the corresponding solvable Lie group $S_{(A_{1},\dots,A_{r})}$ endowed with the left-invariant metric defined by $\langle\cdot,\cdot\rangle$. It is easy to see that any left-invariant metric on the Lie group $S_{(A_{1},\dots,A_{r})}$ for which $\mathfrak{a}\perp\mathfrak{n}$ is isometric to some

[TABLE]

where $h_{1}^{-1}A_{i}:=\operatorname{ad}{h_{1}^{-1}Y_{i}}=\sum c_{ji}A_{j}$ if the matrix of $h_{1}^{-1}$ relative to $\{Y_{i}\}$ is $[c_{ij}]$.

It follows from [L3, (25)] that the Ricci curvature of $(A_{1},\dots,A_{r})$ is given by,

[TABLE]

Up to isometry, it can always be assumed that $\operatorname{tr}{A_{2}}=\dots=\operatorname{tr}{A_{r}}=0$ (i.e. $H=(\operatorname{tr}{A_{1}})Y_{1}$) by considering in (8) $h_{2}=I$ and a suitable $h_{1}\in\mathrm{O}(\mathfrak{a},\langle\cdot,\cdot\rangle)$. In that case,

[TABLE]

Let us suppose that $(A_{1},\dots,A_{r})$ is a solvsoliton, that is, $\operatorname{tr}{S(A_{i})S(A_{j})}=\delta_{ij}$ (up to scaling) and $A_{i}$ is normal for all $i$ (see [L3, Theorem 4.8]). We consider an $r$-uple as in (8) and assume (up to isometry and scaling) that $\operatorname{tr}{h_{1}^{-1}A_{1}}=\operatorname{tr}{A_{1}}$ and $\operatorname{tr}{h_{1}^{-1}A_{2}}=\dots=\operatorname{tr}{h_{1}^{-1}A_{r}}=0$ (i.e. $c_{11}=1$ and $c_{12}=\dots=c_{1r}=0$). Thus what we must show to prove Theorem 1.2 is that

[TABLE]

for any $h_{1}\in\mathrm{GL}(\mathfrak{a})$ and $h_{2}\in\mathrm{GL}(\mathfrak{n})$.

By (10), we have that

[TABLE]

where

[TABLE]

Since $h_{1}^{-1}A_{1}=A_{1}+c_{21}A_{2}+\dots+c_{r1}A_{r}$ is normal, it follows from [LW2, (17)] that

[TABLE]

Note that the value of $F$ at the solvsoliton is given by

[TABLE]

Lemma 3.1**.**

$f(x_{1},\dots,x_{r})\leq f(1,\dots,1)$* for any $x_{1}\geq 1$, $x_{2},\dots,x_{r}>0$, where equality holds if and only if $x_{1}=\dots=x_{r}=1$.*

Proof.

An elementary algebraic manipulation gives that the inequality is equivalent to

[TABLE]

Since the first term is $\geq 0$ by the Cauchy-Schwartz inequality $(\sum x_{i})^{2}\leq r\sum x_{i}^{2}$ and the third one is $\geq a\sum(x_{i}-1)^{2}$, one obtains that both the above inequality and the equality condition in the lemma follow. ∎

Since $F$ is invariant under all the assumptions made above up to isometry and scaling, inequality (11) follows from (12) and Lemma 3.1. Moreover, if equality holds, then $h_{2}(h_{1}^{-1}A_{i})h_{2}^{-1}$ is normal for all $i$ and

[TABLE]

which implies that $\left(h_{2}(h_{1}^{-1}A_{1})h_{2}^{-1},\dots,h_{2}(h_{1}^{-1}A_{r})h_{2}^{-1}\right)$ is a solvsoliton, concluding the proof of Theorem 1.2.