is solvable. Recall that for a given $p\in M$ we consider those Killing fields
in $\mathfrak{g}$ which are $\mathsf{N}$-horizontal at $p$:

$\mathfrak{a}_{p}:=\\{U\in\mathfrak{g}:U_{p}\perp\mathsf{N}\cdot p\\}.$

In other words, $\mathfrak{a}_{p}$ is the ${\boldsymbol{h}}_{p}$-orthogonal
complement of $\mathfrak{n}$ in $\mathfrak{g}$.

######  Corollary 11.7.

Suppose that $\mathsf{G}$ is solvable and acts almost freely on $(M,g)$. Then,
under the assumptions of Theorem 8.1, for all $p\in M$ we have
${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}\in\operatorname{Der}(\mathfrak{g})$.

###### Proof.

By definition we have
${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}={\boldsymbol{{q}}}_{p}{\boldsymbol{{\bar{\beta}}}}^{+}{\boldsymbol{{q}}}^{-1}_{p}$.
Since ${\boldsymbol{{q}}}_{p}\in\mathsf{B}_{\boldsymbol{{\bar{\beta}}}}$, we
know that ${\boldsymbol{{q}}}_{p}$ preserves $\mathfrak{n}$ by Proposition
D.7, and of course ${\boldsymbol{{q}}}_{p}|_{\mathfrak{n}}={q}_{p}$. Thus, by
(34),

$\big{(}{{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}\big{)}\big{|}_{\mathfrak{n}}=q_{p}{\bar{\beta}}^{+}q_{p}^{-1}={{}^{\mathfrak{n}}\beta}^{+}_{p},\qquad\big{(}{{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}\big{)}\big{|}_{\mathfrak{a}_{p}}=0,$

with ${{}^{\mathfrak{n}}\beta}^{+}_{p}\in\operatorname{Der}(\mathfrak{n})$ by
Theorem 8.1.

The above and Corollary 8.7 reduce the claim
${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}\in\operatorname{Der}(\mathfrak{g})$
to proving that for all $A\in\mathfrak{a}_{p}$

(35)
$D:=\big{[}(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}},{{}^{\mathfrak{n}}\beta}^{+}_{p}\big{]}=0.$

To show that, recall that by the rigidity from Theorem 8.1 on $M$ we have
$[L_{A_{p}},\beta_{p}^{+}]=0$, which yields
$[L^{\mathfrak{n}}_{A_{p}},{{}^{\mathfrak{n}}\beta}^{+}_{p}]=0$ on
$\operatorname{End}(\mathfrak{n})$. Thus, by Lemma C.2, $D$ is a symmetric
derivation of $\mathfrak{n}$ with

$0\leq\operatorname{tr}D^{2}=\operatorname{tr}\big{[}(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}},{{}^{\mathfrak{n}}\beta}^{+}_{p}\big{]}D=-\operatorname{tr}{{}^{\mathfrak{n}}\beta}^{+}_{p}\big{[}(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}},D\big{]}=0,$

by Proposition D.6, i. This shows the claim. ∎

Note that so far we have not used that $\mathsf{G}$ admits left-invariant
Einstein metrics.

###### Proof of Proposition 11.4.

Let $g^{S}$ be a $\mathsf{G}$-invariant Einstein metric on $S=\mathsf{G}\cdot
p$. At $p$ there is a corresponding inner product
${\boldsymbol{h}}_{S}={\boldsymbol{{q}}}_{S}\cdot{\boldsymbol{\bar{h}}}$ on
$\mathfrak{g}$. By Theorem 8.1 and Corollary 11.7, both applied to the
manifold $(S,g^{S})$ (the orbit space for the $\mathsf{G}$-action is just a
point), we have that

${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{S}:={\boldsymbol{{q}}}_{S}\,{\boldsymbol{{\bar{\beta}}}}^{+}\,{\boldsymbol{{q}}}_{S}^{-1}\in\operatorname{Der}(\mathfrak{g}).$

On the other hand, from those two results, this time applied to $(M^{n},g)$,
we also know that

${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}={\boldsymbol{{q}}}_{p}\,{\boldsymbol{{\bar{\beta}}}}^{+}\,{\boldsymbol{{q}}}_{p}^{-1}\in\operatorname{Der}(\mathfrak{g}).$

Thus, by Lemma D.8, there exists
${\boldsymbol{a}}\in\operatorname{Aut}(\mathfrak{g})$ such that

${{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{S}={\boldsymbol{a}}\,{{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}\,{\boldsymbol{a}}^{-1}.$

The inner product
$({\boldsymbol{a}}\,{\boldsymbol{{q}}}_{S})\cdot{\boldsymbol{\bar{h}}}$ gives
rise to a $\mathsf{G}$-invariant metric $\tilde{g}^{S}$ on $S$ which is
isometric to $g^{S}$, and in particular also Einstein. By construction, it is
clear that the corresponding $\beta$-endomorphism will satisfy
${{}^{\mathfrak{g}}\tilde{\boldsymbol{\beta}}}^{+}_{S}={{}^{\mathfrak{g}}\boldsymbol{\beta}}^{+}_{p}$.
From Proposition D.7, it follows on one hand that both $\tilde{g}^{S}$ and
$g|_{S}$ define the same $\mathsf{N}$-horizontal distribution on $S$ (this is
the kernel of the endomorphism
${{}^{\mathfrak{g}}\tilde{\boldsymbol{\beta}}}^{+}_{S}$). On the other hand we
also have ${\beta^{+}_{S}}=\beta^{+}|_{S}$ as desired, by restricting to the
$\mathsf{N}$-vertical space.

Applying Corollary 8.8 to $(S,g^{S})$ we obtain a $\mathsf{G}$-invariant,
$\mathsf{N}$-horizontal vector field $N_{\beta}$ on $S$ satisfying

$\nabla^{S}N_{\beta}\,|_{\mathcal{V}}=-\beta^{+},$

where $\nabla^{S}$ denotes the Levi-Civita connection of $(S,g^{S})$. To
conclude, we now claim that

$\nabla^{S}N_{\beta}\,|_{\mathcal{V}}=\nabla N_{\beta}\,|_{\mathcal{V}}.$

By $\mathsf{G}$-invariance, it suffices to prove this at $p$, and via
${i}_{p}$ (23) we may work in $\mathfrak{n}$ instead of ${\mathcal{V}}_{p}$.
Let $A\in\mathfrak{a}_{p}$ be a Killing field with $A_{p}=(N_{\beta})_{p}$ and
set $E:=(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}}$. Lemma C.2
reduces the claim to showing that

(36) $-{{}^{\mathfrak{n}}\beta}^{+}_{p}=\tfrac{1}{2}(E+E^{T_{g}}),$

where the tranpose is with respect to
$h_{p}={i}_{p}^{*}(g_{p}|_{{\mathcal{V}}\times{\mathcal{V}}})$, the inner
product on $\mathfrak{n}$ corresponding to $g$ at $p$. Using again Lemma C.2,
now applied to $(S,g^{S})$, we know that

$-{{}^{\mathfrak{n}}\beta}^{+}_{p}=\tfrac{1}{2}(E+E^{T_{S}}),$

transpose with respect to $h^{S}_{p}:={i}_{p}^{*}\,g^{S}_{p}$. By (35), $E$ is
a normal operator with respect to $h^{S}_{p}$. Since $\mathfrak{g}$ is
completely solvable, this implies that $E$ is self-adjoint and
$E=-{{}^{\mathfrak{n}}\beta}^{+}_{p}$. Furthermore, by construction the two
inner products $h_{p}$ and $h^{S}_{p}$ give the same $\beta$-endomorphism,
thus if we write $h_{p}=q\cdot h^{S}_{p}$ for some
$q\in\mathsf{B}_{{\bar{\beta}}}$, then
$[q,{{}^{\mathfrak{n}}\beta}^{+}_{p}]=0$. Therefore, by (19) we have that
$E^{T_{g}}=E^{T_{S}}=E$ and (36) follows. ∎

We now see how applying Proposition 11.4 to each $\mathsf{G}$-orbit in $M$
gives rise to a smooth vector field on $M$, which in addition induces a
Killing field on $P$:

######  Proposition 11.8.

Let $(M^{n},g)$ be as in Theorem G and assume that 11.1 holds. Then, there
exists a unique smooth, $\mathsf{G}$-invariant, $\mathsf{G}$-vertical,
$\mathsf{N}$-horizontal vector field $N_{\beta}$ on $M$ with

$L_{N_{\beta}}=-\beta^{+}.$

Moreover, the corresponding vector field on $P$ is a Killing field.

###### Proof.

By Proposition 11.4, existence and uniqueness of $N_{\beta}$ are clear. It is
also evident that $N_{\beta}$ varies smoothly in $\mathsf{G}$-vertical
directions. To prove that it is smooth as a vector field on $M$, let $p\in M$,
let $\gamma(t)$ be an $\mathsf{N}$-horizontal curve with $\gamma(0)=p$ (in
$P_{p}$), and let $A_{t}\in\mathfrak{g}$ be Killing fields defined by

$-(A_{t})_{\gamma(t)}=(N_{\beta})_{\gamma(t)}.$

By Corollary 8.7 we have $A_{t}\in\mathfrak{a}_{p}$, thus
$A_{t}-A_{0}\in\mathfrak{a}_{p}$ for all $t$. We claim $A_{t}\equiv A_{0}$ is
constant, thus in $\mathsf{N}$-horizontal directions, $N_{\beta}$ behaves like
a Killing field in $\mathfrak{g}$. To see that, notice that by definition of
$N_{\beta}$ and Lemma C.2, the second fundamental form of
$\mathsf{N}\cdot\gamma(t)$ satisfies

(37) $\displaystyle-\langle\beta^{+}V,V\rangle_{\gamma(t)}=\langle
L_{N_{\beta}}V,V\rangle_{\gamma(t)}=-\langle[A_{t},V],V\rangle_{\gamma(t)}\,.$

Since $\mathfrak{g}$ is completely solvable and $\mathfrak{a}_{p}$ is a
subalgebra of $\mathfrak{g}$ by Corollary 8.7, the eigenvalues of
$(\operatorname{ad}_{\mathfrak{g}}A_{t})|_{\mathfrak{n}}$ are all real.
Moreover, by (35) the symmetric endomorphism
${{}^{\mathfrak{n}}\beta}^{+}_{\gamma(t)}$ and
$(\operatorname{ad}_{\mathfrak{g}}A_{t})|_{\mathfrak{n}}$ commute. Thus, by
(37) we get
${{}^{\mathfrak{n}}\beta}^{+}_{\gamma(t)}=(\operatorname{ad}_{\mathfrak{g}}A_{t})|_{\mathfrak{n}}$
for all $t$, which shows that
$(\operatorname{ad}_{\mathfrak{g}}A_{t})|_{\mathfrak{n}}$ depends smoothly on
$t$. Differentiating this indentity, by using Lemma 5.7 and Corollary 8.4, we
conclude that
$(\operatorname{ad}_{\mathfrak{g}}A_{t})|_{\mathfrak{n}}\equiv(\operatorname{ad}_{\mathfrak{g}}A_{0})|_{\mathfrak{n}}$.
Thus, $A_{t}-A_{0}$ centralises $\mathfrak{n}$. Given that the nilradical
$\mathfrak{n}$ is its own centraliser,
$A_{t}-A_{0}\in\mathfrak{n}\cap\mathfrak{a}_{p}=\\{0\\}$. It follows
$A_{t}\equiv A_{0}$, thus $N_{\beta}|_{P_{p}}=-\mathsf{A}_{0}|_{P_{p}}$ and in
particular, $N_{\beta}$ is smooth on $M$. Another consequence is that the
vector field on $P$ corresponding to $N_{\beta}$ is a Killing field, since by
Corollary 8.7 $A_{0}|_{P_{p}}$ is a Killing field on $P_{p}$ (being locally
isometry to $P$). ∎

We now apply the Bochner technique to prove the main result of this section:

###### Proof of Theorem 11.3.

We will show that $N=N_{\beta}$, where $N_{\beta}$ is the vector field from
Proposition 11.8. To that end, consider the horizontal Einstein equation (14)
for the $\mathsf{N}$-submersion evaluated at $N-N_{\beta}$:

$-\|N-N_{\beta}\|^{2}=\operatorname{Ric}_{P}(N-N_{\beta},N-N_{\beta})=\operatorname{Ric}_{P}(N_{\beta},N_{\beta}),$

where we have used that $A=0$ by Theorem G,
$L_{N-N_{\beta}}=-\beta^{+}+\beta^{+}=0$ by Theorem G and Proposition 11.8,
and the fact that $N={N_{0}}$ is parallel on $P$ (Corollaries 8.3 and 9.9).
Since $N_{\beta}$ is a Killing field (Proposition 11.8), we have the well-
known Bochner formula

$\operatorname{Ric}_{P}(N_{\beta},N_{\beta})=\|\nabla
N_{\beta}\|^{2}-\Delta_{P}(\tfrac{1}{2}\|N_{\beta}\|^{2}),$

an immediate consquence of (3), using that $\nabla N_{\beta}$ is skew-
symmetric. Combining both equations yields

$\Delta_{P}(\tfrac{1}{2}\|N_{\beta}\|^{2})=\|\nabla
N_{\beta}\|^{2}+\|N-N_{\beta}\|^{2}\geq 0.$

By Proposition B.2 applied to the $\mathsf{G}/\mathsf{N}$-action on $P$, we
deduce that equality must hold everywhere on $P$ and in particular $N\equiv
N_{\beta}$. ∎

##  12\. New algebraic formulae for the Ricci curvature of a homogeneous space

In order to use the rigidity from Theorem G in the homogeneous setting, in
this section we obtain formulae that relate the Ricci curvature of a
homogeneous space with that of the $\mathsf{N}$-orbits, under the assumption
that the $\mathsf{N}$-action is polar. The main result is Proposition 12.4.
Its main advantage compared to other known formulae is the fact that it does
not require the Killing fields to be in a reductive complement.

######  Lemma 12.1.

Let $(M,g)$ be a homogeneous manifold and let
$\mathcal{B}:=\\{E_{i}\\}_{i=1}^{n}$ be Killing fields which at $p$ form an
orthonormal basis of $T_{p}M$. Then, at $p$ we have

(38)
$\operatorname{ric}(X,X)=2\,\sum_{i}\langle\nabla_{E_{i}}X,[X,E_{i}]\rangle+\|\nabla
X\|^{2}-\sum_{i}\langle(\operatorname{ad}X)^{2}E_{i},E_{i}\rangle-\sum_{i}\langle(\nabla_{E_{i}}E_{i}),\nabla_{X}X\rangle,$

for any Killing field $X$.

###### Proof.

By Lemma C.3 for Killing fields $X,Y$ we have $R(X,Y)=[\nabla X,\nabla
Y]+\nabla[X,Y]$. Using this, we compute:

$\displaystyle\operatorname{ric}(X,X)=$ $\displaystyle=$
$\displaystyle\,\,\sum_{i}\langle[\nabla X,\nabla
E_{i}]E_{i},X\rangle+\langle\nabla_{E_{i}}[X,E_{i}],X\rangle$ $\displaystyle=$
$\displaystyle\,\,\sum_{i}-\langle\nabla_{E_{i}}E_{i},\nabla_{X}X\rangle+\langle\nabla_{E_{i}}X,\nabla_{X}E_{i}\rangle-\langle
E_{i},\nabla_{X}[X,E_{i}]\rangle$ $\displaystyle=$
$\displaystyle\,\,-\sum_{i}\langle\nabla_{E_{i}}E_{i},\nabla_{X}X\rangle+\sum_{i}\langle\nabla_{E_{i}}X,[X,E_{i}]+\nabla_{E_{i}}X\rangle-\langle
E_{i},[X,[X,E_{i}]]+\nabla_{[X,E_{i}]}X\rangle$ $\displaystyle=$
$\displaystyle\,\,-\sum_{i}\langle\nabla_{E_{i}}E_{i},\nabla_{X}X\rangle+2\,\sum_{i}\langle\nabla_{E_{i}}X,[X,E_{i}]\rangle+\|\nabla
X\|^{2}-\sum_{i}\langle(\operatorname{ad}X)^{2}E_{i},E_{i}\rangle.$

This shows the claim. ∎

######  Remark 12.2.

For a given homogeneous space $\mathsf{F}/\mathsf{H}$, with reductive
decomposition $\mathfrak{f}=\mathfrak{h}\oplus\mathfrak{m}$, if all the
$E_{i}^{\prime}s$ were in $\mathfrak{m}$ then the fourth term would be
$\langle{\operatorname{H}},\nabla_{X}X\rangle=-\langle[{\operatorname{H}},X],X\rangle$.
Here, ${\operatorname{H}}=-\sum_{i}(\nabla_{E_{i}}E_{i})_{\mathfrak{m}}$ is
the mean curvature vector of the homogeneous space $(\mathsf{F}/\mathsf{H},g)$
([Bes87, (7.32)]) satisfying for $Y\in\mathfrak{m}$

(39)
$\displaystyle\langle{\operatorname{H}},Y\rangle=\operatorname{tr}(\operatorname{ad}_{\mathfrak{f}}Y).$

In general this is not the case: consider an irreducible symmetric space of
the non-compact type $M=\mathsf{F}/\mathsf{K}$. Choosing Killing fields which
span the simply transitive $\mathsf{A}\mathsf{N}$-action
($\mathsf{F}=\mathsf{K}\mathsf{A}\mathsf{N}$ is an Iwasawa decomposition), it
can be seen that ${\operatorname{H}}_{\mathsf{G}}\neq 0$, since the group
$\mathsf{G}=\mathsf{A}\mathsf{N}$ is not unimodular. However, $\mathsf{F}$ is
unimodular, thus the mean curvature vector of $\mathsf{F}/\mathsf{K}$
vanishes.

In the Lie group case, that is $\mathfrak{h}=\\{0\\}$, it is easy to see that
(38) gives [Bes87][(7.38)]. The last terms in both formulae agree by the
above. Moreover setting $\operatorname{ad}(X)=A+S$, $A$ skew-symmetric, $S$
symmetric, we have $\nabla X=-A+J$, with $\langle J\cdot
Y,Z\rangle=\tfrac{1}{2}\langle[Y,Z],X\rangle$. A computation shows that the
first three terms in (38) equal to
$-\operatorname{tr}S^{2}-\operatorname{tr}J^{2}$ and it is easy to see that
this equals to the first three terms in [Bes87][(7.38)].

######  Corollary 12.3.

Let $(\mathsf{N},g^{\mathsf{N}})$ be a nilpotent Lie group with left-invariant
metric, and $U\in\mathfrak{n}$ a Killing field. Then, for any set of Killing
fields $\\{E_{i}\\}$ forming an orthonormal basis at $p\in\mathsf{N}$, we have

$\operatorname{ric}^{\mathsf{N}}(U,U)=2\,\sum_{i}\langle\nabla_{E_{i}}U,[U,E_{i}]\rangle+\|\nabla
U\|^{2}.$

###### Proof.

The third term in (38) equals
$-\operatorname{tr}(\operatorname{ad}_{\mathfrak{n}}U)^{2}$, and vanishes
because $\operatorname{ad}U$ is nilpotent. Moreover, letting
$V\in\mathfrak{n}$ with $V_{p}=(\nabla_{X}X)_{p}$ and using that $\\{E_{i}\\}$
spanns $\mathfrak{n}$, the fourth term in (38) equals
$\operatorname{tr}\operatorname{ad}_{\mathfrak{n}}V=0$ by (39). ∎

We say a connected Lie subgroup $\mathsf{N}\leq\mathsf{F}$ is _nilpotently
embedded_ , if for all $U\in\mathfrak{n}$,
$\operatorname{ad}_{\mathfrak{f}}U\in\operatorname{End}(\mathfrak{f})$ is a
nilpotent endomorphism.

######  Proposition 12.4.

Let $(M=\mathsf{F}/\mathsf{H},g)$ be a homogeneous space,
$\mathsf{N}\leq\mathsf{F}$ a nilpotently embedded subgroup acting almost
freely and polarly on $\mathsf{F}/\mathsf{H}$. Given $p\in M$, let
$\\{E_{i}\\}_{i=1}^{n}=\\{U_{r}\\}\cup\\{Y_{k}\\}$ be any set of Killing
fields in $\mathfrak{f}$ which at $p$ form an orthonormal basis of $T_{p}M$,
with $\\{U_{r}\\}$ a basis of $\mathfrak{n}$. Then, we have that

$\operatorname{scal}^{\mathcal{V}}(p)-\sum_{r}\operatorname{ric}_{g}(U_{r},U_{r})_{p}=\sum_{i}\langle\nabla_{E_{i}}E_{i},N\rangle_{p}+\sum_{i,r}\langle[U_{r},[U_{r},E_{i}]\,],E_{i}\rangle_{p}.$

Here, $\operatorname{scal}^{\mathcal{V}}(p)$ denotes the scalar curvature of
the orbit $\mathsf{N}\cdot p$ with the induced metric, and $N$ its mean
curvature vector.

###### Proof.

For any $U_{r}\in\mathfrak{n}$, Lemma 12.1 and Corollary 12.3 (the latter
applied to the orbit $\mathsf{N}\cdot p$, which is locally isometric to a
left-invariant metric on $\mathsf{N}$) give

$\displaystyle\operatorname{ric}_{g}(U_{r},U_{r})+$
$\displaystyle\sum_{i}\langle[U_{r},[U_{r},E_{i}]\,],E_{i}\rangle+\sum_{i}\langle\nabla_{E_{i}}E_{i},\nabla_{U_{r}}U_{r}\rangle=2\,\sum_{i}\langle\nabla_{E_{i}}U_{r},[U_{r},E_{i}]\rangle+\|\nabla
U_{r}\|^{2}$ $\displaystyle=$
$\displaystyle\,\,2\,\sum_{s}\langle\nabla_{U_{s}}U_{r},[U_{r},U_{s}]\rangle+\sum_{s,t}\langle\nabla_{U_{s}}U_{r},U_{t}\rangle^{2}$
$\displaystyle\,\,+2\,\sum_{k}\langle\nabla_{Y_{k}}U_{r},[U_{r},Y_{k}]\rangle+2\,\sum_{s,k}\langle\nabla_{Y_{k}}U_{r},U_{s}\rangle^{2}$
$\displaystyle=$
$\displaystyle\,\,\operatorname{ric}^{\mathcal{V}}(U_{r},U_{r})+2\,\sum_{k,s}\langle\nabla_{Y_{k}}U_{r},U_{s}\rangle\langle[U_{r},Y_{k}],U_{s}\rangle+2\,\sum_{s,k}\langle\nabla_{Y_{k}}U_{r},U_{s}\rangle^{2}$
$\displaystyle=$
$\displaystyle\,\,\operatorname{ric}^{\mathcal{V}}(U_{r},U_{r})+2\,\sum_{k,s}\langle\nabla_{Y_{k}}U_{r},U_{s}\rangle\langle\nabla_{U_{r}}Y_{k},U_{s}\rangle.$

We have omitted any terms of the form
$\langle\nabla_{Y_{l}}U_{r},Y_{k}\rangle$ due to the polar assumption (the
$Y_{k}$’s are not horizontal vector fields, but they are horizontal at $p$,
and the expression under consideration is tensorial in those entries). Summing
over $r$, we notice that the summands of the second term in the right-hand-
side have two factors: one of them,

$\langle\nabla_{Y_{k}}U_{r},U_{s}\rangle=\tfrac{1}{2}\big{(}\langle[Y_{k},U_{r}],U_{s}\rangle+\langle[Y_{k},U_{s}],U_{r}\rangle\big{)},$

symmetric in $r,s$ (recall that $[U_{r},U_{s}]\perp Y_{k}$ at $p$), and the
other one, $\langle\nabla_{U_{r}}Y_{k},U_{s}\rangle$, skew-symmetric in $r,s$
because $Y_{k}$ is a Killing field. Thus, this term vanishes when summing over
all $r,s$, and we obtain

(40)
$\sum_{r}\operatorname{ric}_{g}(U_{r},U_{r})+\sum_{i,r}\langle[U_{r},[U_{r},E_{i}]\,],E_{i}\rangle+\sum_{i,r}\langle\nabla_{E_{i}}E_{i},\nabla_{U_{r}}U_{r}\rangle=\operatorname{scal}^{\mathcal{V}}(p).$

Finally, for any Killing field $U\in\mathfrak{n}$ we have
$\sum_{r}\langle\nabla_{U_{r}}U_{r},U\rangle=-\operatorname{tr}\operatorname{ad}_{\mathfrak{n}}(U)=0$
by (39). Thus, $(\sum_{r}\nabla_{U_{r}}U_{r})_{p}=N_{p}$. ∎

An important consequence of this formula is the following estimate, a
fundamental ingredient in the proof of Theorem A:

######  Proposition 12.5.

Let $(M^{n},g)$ be as in Proposition 12.4, and choose a Killing field
$A\in\mathfrak{f}$ with $A_{p}=-N_{p}$,
$p:=e\mathsf{H}\in\mathsf{F}/\mathsf{H}$. Assume that
$D:=\operatorname{ad}_{\mathfrak{f}}A\in\operatorname{End}(\mathfrak{f})$ is
diagonalisable with real eigenvalues $(\lambda_{i})_{i=1}^{\dim\mathfrak{f}}$
and respective eigenspaces $\mathfrak{f}_{\lambda_{i}}$. Set

$\mathfrak{f}=\mathfrak{f}_{-}\oplus\mathfrak{f}_{0}\oplus\mathfrak{f}_{+},\qquad\mathfrak{f}_{-}:=\bigoplus_{\lambda<0}{\mathfrak{f}_{\lambda}},\qquad\mathfrak{f}_{+}:=\mathfrak{n}=\bigoplus_{\lambda>0}\mathfrak{f}_{\lambda},\qquad\mathfrak{f}_{0}:=\ker
D,\qquad\sigma_{+}:=\sum_{\lambda_{i}>0}\lambda_{i}.$

Then, the Killing fields $\\{E_{i}\\}_{i=1}^{n}$ from Proposition 12.4 may be
chosen so that

$\sum_{i}\langle\nabla_{E_{i}}E_{i},N\rangle_{p}\leq\sigma_{+},\qquad\sum_{i,r}\langle[U_{r},[U_{r},E_{i}]],E_{i}\rangle_{p}=0.$

In particular, if $(M^{n},g)$ is Einstein with $\operatorname{ric}(g)=-g$,
then

$\operatorname{scal}^{\mathcal{V}}(p)+\dim\mathfrak{n}\leq\sigma_{+}$

and equality holds if and only if
$(\mathfrak{f}_{0}\oplus\mathfrak{f}_{+})\cdot p=T_{p}M$.

###### Proof.

Set $\mathfrak{m}:={\rm span}\\{E_{i}:1\leq i\leq n\\}\subset\mathfrak{f}$,
with $E_{i}$’s yet to be determined. Notice that
$\mathfrak{n}\subset\mathfrak{m}$, by assumption on the set $\\{E_{i}\\}$. If
$D\mathfrak{m}\subset\mathfrak{m}$, then $\mathfrak{m}$ would be a sum of
eigenspaces, and by Lemma C.1 we would clearly have

$\sum_{i}\langle\nabla_{E_{i}}E_{i},N\rangle_{p}=\sum_{i}\langle[A,E_{i}],E_{i}\rangle_{p}=\operatorname{tr}D|_{\mathfrak{m}}\leq\sigma_{+},$

with equality if and only if no eigenspaces with negative eigenvalue are
contained in $\mathfrak{m}$. This is of course equivalent to
$\mathfrak{m}\subset\mathfrak{f}_{0}\oplus\mathfrak{f}_{+}$, and also to
$(\mathfrak{f}_{0}\oplus\mathfrak{f}_{+})\cdot p=T_{p}M$, for
$\mathfrak{m}\cdot p=T_{p}M$.

Regarding the second formula, notice that

$\sum_{i,r}\langle[U_{r},[U_{r},E_{i}]],E_{i}\rangle_{p}=\sum_{r}\operatorname{tr}\,\operatorname{pr}_{\mathfrak{m}}\circ(\operatorname{ad}_{\mathfrak{f}}U_{r})^{2}|_{\mathfrak{m}},$

where
$\operatorname{pr}_{\mathfrak{m}}:\mathfrak{f}=\mathfrak{h}\oplus\mathfrak{m}\to\mathfrak{m}$
denotes the projection onto the second factor. Hence, it suffices to choose
$\\{E_{i}\\}$ so that $\mathfrak{m}$ is $D$-invariant and
$\operatorname{pr}_{\mathfrak{m}}\circ(\operatorname{ad}_{\mathfrak{f}}U)^{2}|_{\mathfrak{m}}$
is traceless for each $U\in\mathfrak{n}$.

Let $\\{F_{i}\\}_{i=1}^{\dim\mathfrak{f}}$ denote an eigenbasis for $D$ with
eigenvalues sorted in non-increasing order, so that in particular
$\\{F_{r}\\}_{r=1}^{\dim\mathfrak{n}}$ spans $\mathfrak{n}$. Notice that

(41) $[F_{k+1},\mathfrak{n}]\subset{\rm span}\\{F_{i}:1\leq i\leq k\\}.$

Indeed, for each eigenvector $F_{r}\in\mathfrak{n}$ with eigenvalue
$\lambda_{r}>0$, by the Jacobi identity we have that

$[D,[F_{k+1},F_{r}]]=[[D,F_{k+1}],F_{r}]+[F_{k+1},[D,F_{r}]]=(\lambda_{k+1}+\lambda_{r})[F_{k+1},F_{r}],$

so either $[F_{k+1},F_{r}]=0$, or it is an eigenvector with eigenvalue
$\lambda_{k+1}+\lambda_{r}>\lambda_{k+1}$.

We now construct $\mathfrak{m}$ inductively. Set $\mathfrak{m}_{0}=0$, and for
each $k\geq 0$ define

$\mathfrak{m}_{k+1}:=\begin{cases}\mathfrak{m}_{k},&\hbox{if
}F_{k+1}\in\mathfrak{h}+\mathfrak{m}_{k}\,;\\\
\mathfrak{m}_{k}\oplus{\mathbb{R}}F_{k+1},&\hbox{otherwise.}\end{cases}$

Notice that the dimension of $\mathfrak{m}_{k}$ increases by at most one in
each step. There is a corresponding subset of indices
$1=i_{1}<i_{2}<\cdots<i_{n}$ with

$\dim\mathfrak{m}_{i_{r}}=r,\qquad\hbox{and}\qquad\mathfrak{m}_{i_{r}}=\mathfrak{m}_{i_{r+1}-1}\subsetneq\mathfrak{m}_{i_{r+1}},$

for each $r=1,\ldots,n-1$. Observe that $i_{r}=r$ for all
$r\leq\dim\mathfrak{n}$, since $\mathfrak{n}\cap\mathfrak{h}=0$ and
$\mathfrak{n}$ (which is non-trivial) is spanned by the first
$\dim\mathfrak{n}$ vectors in the eigenbasis $\\{F_{i}\\}$.

We then set $\mathfrak{m}:=\mathfrak{m}_{i_{n}}$. By induction, one can show
that

$\mathfrak{m}_{k}\cap\mathfrak{h}=0,\qquad{\rm span}\\{F_{i}:1\leq i\leq
k\\}\subset\mathfrak{h}\oplus\mathfrak{m}_{k},$

for all $k\geq 1$. In particular,
$\mathfrak{f}=\mathfrak{h}\oplus\mathfrak{m}$. Moreover, using (41) we deduce
that

$[F_{k+1},\mathfrak{n}]\subset\mathfrak{h}\oplus\mathfrak{m}_{k},\qquad\forall\,k\geq
1.$

Refining this slightly, we can write it as

$[F_{i_{r+1}},\mathfrak{n}]\subset\mathfrak{h}\oplus\mathfrak{m}_{i_{r}},\qquad\forall\,r\geq
1.$

Notice that $\mathfrak{m}$ need not be a reductive complement. (In fact, if
$\mathfrak{f}$ is not solvable, then no reductive complement for the
homogeneous space $\mathsf{F}/\mathsf{H}$ will contain $\mathfrak{n}$.)
However, by construction we still have that $\mathfrak{m}$ is $D$-invariant,
as it is a sum of $D$-eigenspaces. In addition,

$(\operatorname{ad}_{\mathfrak{f}}U)^{2}\,\mathfrak{m}_{i_{r+1}}\subset\mathfrak{h}\oplus\mathfrak{m}_{i_{r}},\qquad\forall\,U\in\mathfrak{n},\quad\forall\,r\geq
1.$

Thus,

$\operatorname{pr}_{\mathfrak{m}}\circ(\operatorname{ad}_{\mathfrak{f}}U)^{2}\,\mathfrak{m}_{i_{r+1}}\subset\mathfrak{m}_{i_{r}},\qquad\forall\,r\geq
1.$

from which it follows that
$\operatorname{pr}_{\mathfrak{m}}\circ(\operatorname{ad}_{\mathfrak{f}}U)^{2}|_{\mathfrak{m}}$
is a nilpotent endomorphism of $\mathfrak{m}$, and in particular traceless.
Hence $\mathfrak{m}$ satisfies the required conditions, and it is now enough
to choose a basis $\\{E_{i}\\}$ for it so that it is orthonormal with respect
to the inner product induced by $g$ under the isomorphism
$\mathfrak{m}\simeq\mathfrak{m}\cdot p\simeq T_{p}M$. ∎

##  13\. Semi-direct product of Einstein solvmanifolds

Let $\mathsf{F}/\mathsf{K}$ be a homogeneous space with effective presentation
and global Levi decomposition $\mathsf{F}=\mathsf{L}\ltimes\mathsf{S}$, where
$\mathsf{L}$ is a maximal connected semisimple subgroup with Iwasawa
decomposition $\mathsf{L}=\mathsf{K}\mathsf{A}\mathsf{N}$, $\mathsf{K}$ a
maximal compact subgroup, $\mathsf{S}$ the solvable radical, and
$\mathsf{L}\cap\mathsf{S}=\\{e\\}$. The semi-direct product is defined by a
Lie group homomorphism $\Phi:\mathsf{L}\to\operatorname{Aut}(\mathsf{S})$ with

$\phi:=d\Phi|_{e}:\mathfrak{l}\to\operatorname{Der}(\mathfrak{s})\subset\operatorname{End}(\mathfrak{s}),\qquad\phi(Y):=(\operatorname{ad}_{\mathfrak{f}}Y)|_{\mathfrak{s}},\quad
Y\in\mathfrak{l},$

the corresponding Lie algebra homomorphism defining the semi-direct product
$\mathfrak{f}=\mathfrak{l}\ltimes\mathfrak{s}$.

Since it acts simply-transitively on the symmetric space
$\mathsf{L}/\mathsf{K}$, the solvable Lie group $\mathsf{A}\mathsf{N}$ admits
a left-invariant Einstein metric. Assuming that $\mathsf{S}$ also admits a
left-invariant Einstein metric $g^{\mathsf{S}}$, and that a certain
compatibility condition for $\phi$ and $g^{\mathsf{S}}$ holds, our goal in
this section is to prove that $(\mathsf{A}\mathsf{N})\ltimes\mathsf{S}$ also
admits a left-invariant Einstein metric. Moreover:

######  Theorem 13.1.

Let $g^{\mathsf{S}}$ be a left-invariant Einstein metric on $\mathsf{S}$
defined by an inner product $g^{\mathsf{S}}_{e}$ on $\mathfrak{s}$, and assume
that $\phi(\mathfrak{l})^{T}=\phi(\mathfrak{l})$, transpose with respect to
$g^{\mathsf{S}}_{e}$. Then, $\mathsf{F}/\mathsf{K}$ admits an
$\mathsf{F}$-invariant Einstein metric.

Let $\mathfrak{l}=\mathfrak{k}\oplus\mathfrak{p}$ be a Cartan decomposition
corresponding to a Cartan involution $\theta$. Then,
$\mathfrak{f}=\mathfrak{k}\oplus(\mathfrak{p}\oplus\mathfrak{s})$ is a
reductive decomposition for $\mathsf{F}/\mathsf{K}$. Observe that, since
$(\mathsf{S},g^{\mathsf{S}})$ is an Einstein solvmanifold, $\mathsf{S}$ is
simply-connected by [Jab15a, Thm. 1.1].

######  Lemma 13.2.

After pulling-back $g^{\mathsf{S}}$ by an automorphism of $\mathsf{S}$, we may
assume that

$\phi(K)^{T}=-\phi(K),\quad\forall
K\in\mathfrak{k},\qquad\quad\phi(X)^{T}=\phi(X),\quad\forall
X\in\mathfrak{p}.$

###### Proof.

The kernel of $\phi$ is an ideal in the semisimple Lie algebra $\mathfrak{l}$.
By working on a complementary semisimple ideal, we may assume without loss of
generality that $\phi$ is injective. Thus, we have a Lie algebra isomorphism

$\phi:\mathfrak{l}\to\phi(\mathfrak{l})\subset\operatorname{End}(\mathfrak{s}).$

The assumption $\phi(\mathfrak{l})^{T}=\phi(\mathfrak{l})$ implies that the
map $E\mapsto-E^{T}$ is a Cartan involution on $\phi(\mathfrak{l})$. Hence,
the corresponding map on $\mathfrak{l}$, given by

$\tilde{\theta}:\mathfrak{l}\to\mathfrak{l},\qquad\tilde{\theta}(Y):=-Y^{\prime},\qquad\hbox{where}\quad\phi(Y^{\prime})=\phi(Y)^{T},$

is a Cartan involution of $\mathfrak{l}$. By definition, it satisfies

(42) $\phi(\tilde{\theta}(Y))=-\phi(Y)^{T},\qquad Y\in\mathfrak{l}.$

By uniqueness of Cartan involutions [Kna02, Cor. 6.19], $\theta$ and
$\tilde{\theta}$ are conjugate by an inner automorphism
$\operatorname{Ad}_{\mathsf{L}}(x)=:a\in\operatorname{Aut}(\mathfrak{l})$,
$x\in\mathsf{L}$. The corresponding automorphism
$\operatorname{Ad}_{\mathsf{F}}(x)$ of $\mathfrak{f}$ preserves the Levi
decomposition, and is given by

$\operatorname{Ad}_{\mathsf{F}}(x)|_{\mathfrak{l}}=\operatorname{Ad}_{\mathsf{L}}(x)=a,\qquad\operatorname{Ad}_{\mathsf{F}}(x)|_{\mathfrak{s}}=\Phi(x)=:q.$

Here we view $q=\Phi(x)\in\operatorname{Aut}(\mathfrak{s})$ under the natural
isomorphism
$\operatorname{Aut}(\mathsf{S})\simeq\operatorname{Aut}(\mathfrak{s})$, using
the fact that $\mathsf{S}$ is simply-connected.

We claim that the left-invariant Einstein metric on $\mathsf{S}$ defined by
the inner product $q\cdot g^{\mathsf{S}}_{e}$ on $\mathfrak{s}$ satisfies the
required properties. To see that, we first notice that

(43) $\phi(a^{-1}K)=-\phi(a^{-1}K)^{T},\quad\forall
K\in\mathfrak{k},\qquad\quad\phi(a^{-1}X)=-\phi(a^{-1}X)^{T},\quad\forall
X\in\mathfrak{p}.$

Indeed, if $K\in\mathfrak{k}$ then $K=\theta K=a\tilde{\theta}a^{-1}K$, from
which $a^{-1}K$ is fixed by $\tilde{\theta}$. Thus, by (42),

$\phi(a^{-1}K)=\phi\big{(}\tilde{\theta}a^{-1}K\big{)}=-\phi\left(a^{-1}K\right)^{T},$

and analogously for $X\in\mathfrak{p}$.

Secondly, we observe that

(44) $q^{-1}\phi(Y)q=\phi(a^{-1}Y),\qquad\forall Y\in\mathfrak{l}.$

This follows from the fact that
$\operatorname{Ad}_{\mathsf{F}}(x)^{-1}\in\operatorname{Aut}(\mathfrak{f})$,
which yields

$\operatorname{Ad}_{\mathsf{F}}(x)^{-1}\circ\operatorname{ad}_{\mathfrak{f}}(Y)\circ\operatorname{Ad}_{\mathsf{F}}(x)=\operatorname{ad}_{\mathfrak{f}}(a^{-1}Y),\qquad\forall
Y\in\mathfrak{l},$

and then one simply restricts to $\mathfrak{s}$.

Let $\tilde{T}$ denote transpose in $\operatorname{End}(\mathfrak{s})$ with
respect to $q\cdot g^{\mathsf{S}}_{e}$. Using (19), (43) and (44) we obtain

$\phi(K)^{\tilde{T}}\overset{\eqref{eqn_transposes}}{=}q(q^{-1}\phi(K)q)^{T}q^{-1}\overset{\eqref{eqn_phiKX2}}{=}q\phi(a^{-1}K)^{T}q^{-1}\overset{\eqref{eqn_phiKX1}}{=}-q\phi(a^{-1}K)q^{-1}\overset{\eqref{eqn_phiKX2}}{=}-\phi(K),$

and analogously for $X\in\mathfrak{p}$. ∎

From now on we choose $g^{\mathsf{S}}$ so that the conclusion of Lemma 13.2 is
satisfied. In particular, since $\mathsf{K}$ is connected, the inner product
defined by $g^{\mathsf{S}}$ on $\mathfrak{s}\simeq T_{e}\mathsf{S}$ is
$\operatorname{Ad}(\mathsf{K})$-invariant. We extend it to an
$\operatorname{Ad}(\mathsf{K})$-invariant inner-product $g^{E}$ on
$\mathfrak{p}\oplus\mathfrak{s}$ by setting

(45) $g^{E}|_{\mathfrak{s}\times\mathfrak{s}}:=g^{\mathsf{S}}_{e},\qquad
g^{E}(\mathfrak{p},\mathfrak{s})=0,\qquad
g^{E}|_{\mathfrak{p}\times\mathfrak{p}}:=\big{(}\operatorname{B}_{\mathfrak{f}}-\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}}\big{)}|_{\mathfrak{p}\times\mathfrak{p}},$

where $\operatorname{B}_{\mathfrak{g}}$ denotes the Killing form of
$\mathfrak{g}$. Notice that Lemma 13.2 gives

$\operatorname{B}_{\mathfrak{f}}(X,X)=\operatorname{B}_{\mathfrak{l}}(X,X)+\operatorname{tr}\phi(X)^{2}\geq\operatorname{B}_{\mathfrak{l}}(X,X),\qquad\forall
X\in\mathfrak{p},\quad X\neq 0.$

Thus,
$\operatorname{B}_{\mathfrak{f}}-\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}}\geq\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}}$
is positive-definite on $\mathfrak{p}$, by definition of Cartan decomposition.

This inner product extends to an $\mathsf{F}$-invariant Riemannian metric on
$\mathsf{F}/\mathsf{K}$, also denoted by $g^{E}$. Notice that the action of
$\mathsf{S}$ on $\mathsf{F}/\mathsf{K}$ by left-multiplication is free and
isometric, thus it induces a Riemannian submersion cf. [BB78, $\S$9]

$\pi:\mathsf{F}/\mathsf{K}\to\mathsf{S}\backslash\mathsf{F}/\mathsf{K}\,.$

Since $\mathsf{S}$ is normal in $\mathsf{F}$,
$\mathsf{F}/\mathsf{S}\simeq\mathsf{L}$ acts on the base, and this action is
of course isometric and transitive. Thus, the base is isometric to
$\mathsf{L}/\mathsf{K}$, endowed with the $\mathsf{L}$-invariant metric
defined by the $\operatorname{Ad}(\mathsf{K})$-invariant inner product
$\big{(}\operatorname{B}_{\mathfrak{f}}-\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}}\big{)}|_{\mathfrak{p}\times\mathfrak{p}}$
on $\mathfrak{p}$.

The Riemannian submersion $\pi$ has integrable horizontal distribution.
Indeed, since $\mathfrak{p}\perp\mathfrak{s}$, an integral submanifold through
$e\mathsf{K}$ is given by the $\mathsf{L}$-orbit $\mathsf{L}\cdot
e\mathsf{K}\subset\mathsf{F}/\mathsf{K}$. Through different points
$s\mathsf{K}$, $s\in\mathsf{S}$, the integral submanifold will be the orbit of
the Levi subgroup $s\mathsf{L}s^{-1}$.

Furthermore, the $\mathsf{S}$-orbits are minimal submanifolds. It suffices to
show this at the point $p:=e\mathsf{K}$, since by normality of $\mathsf{S}$ in
$\mathsf{F}$, different $\mathsf{S}$-orbits are isometric by an ambient
isometry of $\mathsf{F}/\mathsf{K}$. To that end, recall that $\langle
X,N\rangle=-\operatorname{tr}L_{X}$ for all horizontal $X$, by (16). By Lemmas
13.2 and C.2 we have

(46) $L^{\mathfrak{s}}_{X_{p}}=\phi(X),$

in the notation of Section 5. This implies that
$\operatorname{tr}L_{X}=\operatorname{tr}L^{\mathfrak{s}}_{X_{p}}=\operatorname{tr}\phi(X)=0$,
since $\phi$ is a representation of a semisimple Lie algebra and so its image
consists of traceless endomorphisms.

Using that $\phi(X)$ is self-adjoint for $X\in\mathfrak{p}$, it also follows
that

$\|L_{X_{p}}\|^{2}=\operatorname{tr}\phi(X)^{2}=\operatorname{B}_{\mathfrak{f}}(X,X)-\operatorname{B}_{\mathfrak{l}}(X,X).$

We are now in a position to prove Theorem 13.1:

###### Proof of Theorem 13.1.

We claim that the $\mathsf{F}$-invariant metric $g^{E}$ on
$\mathsf{F}/\mathsf{K}$ defined in (45) is Einstein. To see this, we will use
the Riemannian submersion $\pi:\mathsf{F}/\mathsf{K}\to\mathsf{L}/\mathsf{K}$
determined by the $\mathsf{S}$-action, and compute its Ricci curvature at
$p:=e\mathsf{K}$, which is enough by homogeneity.

By Theorem 2.2 and Proposition 3.6, together with the above observations, we
have that

$\displaystyle\operatorname{ric}_{g^{E}}(U,U)=$
$\displaystyle\,\,-g^{E}(U,U)-\sum\langle(\nabla_{X_{k}}L)_{X_{k}}U,U\rangle,$
$\displaystyle\operatorname{ric}_{g^{E}}(U,X)=$ $\displaystyle\,\,-\langle
L_{X}U,{\operatorname{H}_{L}}\rangle-\langle\nabla^{\mathcal{V}}U,L_{X}\rangle,$
$\displaystyle\operatorname{ric}_{g^{E}}(X,X)=$
$\displaystyle\,\,\operatorname{ric}_{\mathsf{L}/\mathsf{K}}(X,X)-\operatorname{B}_{\mathfrak{f}}(X,X)+\operatorname{B}_{\mathfrak{l}}(X,X),$

at the point $p\in\mathsf{F}/\mathsf{K}$, and for all $U\in\mathfrak{s}$,
$X\in\mathfrak{p}$, (After rescaling, we may assume without loss of generality
that $\operatorname{ric}_{g^{\mathsf{S}}}=-g^{\mathsf{S}}$.) In the off-
diagonal equation, $U$ is not a Killing field but is $\mathsf{S}$-invariant:
see Proposition 3.6.

Regarding the horizontal part, since
$[\mathfrak{p},\mathfrak{p}]\subset\mathfrak{k}$, it is well-known that

$\operatorname{ric}_{\mathsf{L}/\mathsf{K}}=-\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}},$

see e.g. [Bes87, 7.38]. Thus,

$\operatorname{ric}_{g^{E}}(X,X)=-\big{(}\operatorname{B}_{\mathfrak{f}}(X,X)-\tfrac{1}{2}\operatorname{B}_{\mathfrak{l}}(X,X))=-g^{E}(X,X).$

For the off-diagonal terms, we identify –only for this paragraph–

${\mathcal{V}}_{p}=T_{p}(\mathsf{S}\cdot p)\simeq\mathfrak{s}$

via evaluation of $\mathsf{S}$-left-invariant vector fields (as opposed to
Killing fields as in the rest of the paper). The Koszul formula for the Levi-
Civita connection implies that, under this identification,
$L_{X}\in\operatorname{End}({\mathcal{V}}_{p})$ gets identified with
$-\phi(X)\in\operatorname{Der}(\mathfrak{s})$, $X\in\mathfrak{p}$, and
$\nabla^{\mathcal{V}}U$ gets identified with
$-S(\operatorname{ad}_{\mathfrak{s}}U)$, where $S(E)=\tfrac{1}{2}(E+E^{T})$
denotes the symmetric part. Thus, using that
$[\mathfrak{f},\mathfrak{s}]\subset\mathfrak{s}$, we deduce that

$\langle\nabla^{\mathcal{V}}U,L_{X}\rangle=-\operatorname{tr}_{\mathfrak{f}}\operatorname{ad}_{\mathfrak{f}}U\operatorname{ad}_{\mathfrak{f}}X=-\operatorname{B}_{\mathfrak{f}}(U,X)=0,$

since the Levi decomposition is orthogonal with respect to the Killing form
[Bou71, Ch. I, $\S$5, Prop. 5, b)]. For the first term $-\langle
L_{X}U,{\operatorname{H}_{L}}\rangle$, recall that the nilradical
$\mathfrak{n}_{\mathfrak{f}}$ of $\mathfrak{s}$ is unimodular, and from this
it follows that, again under the above identification,
$({\operatorname{H}_{L}})_{p}\in\mathfrak{n}_{\mathfrak{f}}^{\perp}$. Since
$\phi(X)$ is a self-adjoint derivation of $\mathfrak{s}$, it preserves
$\mathfrak{n}_{\mathfrak{f}}$ and thus vanishes on
$\mathfrak{n}^{\mathfrak{f}}$. These observations yield

$\operatorname{ric}_{g^{E}}(U,X)=0,\qquad\forall\,U\in\mathfrak{s},\quad
X\in\mathfrak{p}.$

Finally, for the vertical equation to hold, we must show that

$\sum_{k}\langle(\nabla_{X_{k}}L)_{X_{k}}U,U\rangle_{p}=0,\qquad\forall\,U\in\mathfrak{s}.$

Here $\\{X_{k}\\}$ is a basis of Killing fields in $\mathfrak{p}$, orthonormal
at $p$. Notice that $\sum\nabla_{X_{k}}X_{k}=0$, as this is the mean curvature
vector of the homogeneous space $\mathsf{L}/\mathsf{K}$. Thus,

$\sum_{k}\langle(\nabla_{X_{k}}L)_{X_{k}}U,U\rangle_{p}=\left\langle\big{(}\nabla_{X_{k}}(L_{X_{k}})\big{)}U,U\right\rangle.$

We now claim that $\nabla_{X}(L_{X})=0$ for all Killing fields
$X\in\mathfrak{p}$. Indeed, by Lemmas 5.7 and 13.3 and the formula (46) we
have

$\big{(}\nabla_{X}(L_{X})\big{)}_{p}=D_{X_{p}}(L^{\mathfrak{s}}_{X})=[\phi(X),L^{\mathfrak{s}}_{X_{p}}]=0,\qquad\phi(X)=(\operatorname{ad}_{\mathfrak{f}}X)|_{\mathfrak{s}},$

where in the first equality we are using the identification via Killing fields
evaluation. ∎

######  Lemma 13.3.

Let $X$ be a Killing field with $X_{p}\perp\mathsf{S}\cdot p$, whose flow
normalises $\mathsf{S}$. Then,

$D_{X_{p}}(L^{\mathfrak{s}}_{X})=[\phi(X),L^{\mathfrak{s}}_{X_{p}}],\qquad\phi(X)=(\operatorname{ad}X)|_{\mathfrak{s}}.$

###### Proof.

We use the notation introduced in $\S$5 for the free isometric action of
$\mathsf{S}$ on $M$.

Let $f:M\to M$ be an isometry of $M$ normalising $\mathsf{S}$. On one hand,
$f$ maps the orbit $\mathsf{S}\cdot p$ to $\mathsf{S}\cdot q$, $q:=f(p)$.
Hence, for the second fundamental form we have

$df^{-1}\,L_{X_{q}}\,df=L_{df^{-1}X_{q}}.$

On the other hand, for a Killing field $X$, it is well-known that $x\mapsto
df^{-1}X_{f(x)}$ defines the Killing field
$\tilde{X}=\operatorname{Ad}f^{-1}X$ (see e.g. [BL22, (20)]). This implies
firstly that we may rewrite the above formula as

(47) $df^{-1}\,L_{X_{f(p)}}\,df=L_{(\operatorname{Ad}f^{-1}X)_{p}}.$

And secondly, under the identifications
${i}_{x}:\mathfrak{s}\to{\mathcal{V}}_{x}$, $x=p,f(p)$,
$df|_{p}:{\mathcal{V}}_{p}\to{\mathcal{V}}_{q}$ corresponds to
$\operatorname{Ad}f$. That is,

(48) $df|_{p}={i}_{f(p)}\circ\operatorname{Ad}f\circ{i}_{p}^{-1}.$

Thus, we may use (48) to translate (47) into an equation on
$\operatorname{End}(\mathfrak{s})$ by means of (24):

(49)
$L^{\mathfrak{s}}_{X_{f(p)}}=\operatorname{Ad}f\,\circ\,L^{\mathfrak{s}}_{(\operatorname{Ad}f^{-1}X)_{p}}\,\circ\,\operatorname{Ad}f^{-1}.$

Now if $\varphi_{t}=\exp(tX)$ denotes the flow of $X$, then by definition we
have

$(D_{X}(L^{\mathfrak{s}}_{X}))_{p}=\tfrac{{\rm d}}{{\rm
d}t}\big{|}_{0}L^{\mathfrak{s}}_{X_{\varphi_{t}(p)}}.$

By assumption, $\varphi_{t}$ consists of isometries which normalise
$\mathsf{S}$. Applying (49), we obtain the stated formula after a
straightforward computation. ∎

##  14\. Proof of Theorem A: the Alekseevskii conjecture

Let $(M_{1}^{n},g_{1})$ be a connected homogeneous Einstein space with
Einstein constant $-1$. By Jablonski’s Theorem E.1, for proving Theorem A we
may assume that $M^{n}_{1}$ is simply-connected, provided we end up showing
that it admits a transitive solvable group of isometries.

The proof of Theorem A is organised into the following steps:

  1. (1)

There exists a homogeneous quotient $(M^{n},g)$ of $(M_{1}^{n},g_{1})$
admitting a presentation $M=\mathsf{F}/\mathsf{H}$ such that $g$ is
$\mathsf{F}$-invariant, $\mathsf{F}=\mathsf{K}\mathsf{G}$ with
$\mathsf{H}\leq\mathsf{K}$, $\mathsf{K}$ compact, and $\mathsf{G}$ is
completely solvable and admits an Einstein left-invariant metric.

  2. (2)

The action of the nilradical $\mathsf{N}$ of $\mathsf{G}$ on
$\mathsf{F}/\mathsf{H}$ is polar.

  3. (3)

The mean curvature vector $N$ of the $\mathsf{N}$-orbits is
$\mathsf{G}$-vertical.

  4. (4)

The normaliser $N_{\mathsf{F}}(\mathsf{G})$ of $\mathsf{G}$ in $\mathsf{F}$
acts transitively on $\mathsf{F}/\mathsf{H}$.

  5. (5)

There exists a closed solvable Lie subgroup of $\mathsf{F}$ acting
transitively on $\mathsf{F}/\mathsf{H}$.

###  14.1. Step 1: Choosing the presentation as homogeneous space

Since $(M_{1},g_{1})$ is a simply-connected homogeneous Einstein space with
$\operatorname{Ric}_{g_{1}}=-g_{1}$, by the structure theory for these spaces
developed in [LL14, JP17, AL17] (see [JP17, Thm. 0.2] and [AL17, Thm. 2.1,
Thm. 2.4 and Cor. 2.8] for the precise statements), there exists a
presentation $M_{1}=\mathsf{F}/\mathsf{H}$ with the following properties:

  1. (i)

$\mathsf{F}$ acts effectively on $\mathsf{F}/\mathsf{H}$ and the isotropy
$\mathsf{H}$ is a compact subgroup ;

  2. (ii)

There is a global Levi decomposition $\mathsf{F}=\mathsf{L}\ltimes\mathsf{S}$
(in particular, $\mathsf{L}\cap\mathsf{S}=\\{e\\}$) with $\mathsf{L}$ a
maximal connected semisimple Lie subgroup, $\mathsf{S}$ the solvable radical,
which is simply-connected and completely solvable, $\mathsf{H}\leq\mathsf{L}$,
and $\mathsf{L}$ has no compact simple factors;

  3. (iii)

The orbits of $\mathsf{L}$ and $\mathsf{S}$ are orthogonal at
$p:=e\mathsf{H}$;

  4. (iv)

The induced metric $g^{\mathsf{S}}$ on $\mathsf{S}\cdot p$ is Einstein with
$\operatorname{ric}(g^{\mathsf{S}})=-g^{\mathsf{S}}$ .

Moreover, by [JP17, Prop. 3.9], the Lie algebra representation
$\phi:\mathfrak{l}\to\operatorname{End}(\mathfrak{s})$ defining the semi-
direct product structure of $\mathfrak{f}=\mathfrak{l}\ltimes\mathfrak{s}$
(that is, $\phi(X)=(\operatorname{ad}_{\mathfrak{f}}X)|_{\mathfrak{s}}$ for
$X\in\mathfrak{l}$) satisfies the following important compatibility with the
geometry of $\mathsf{S}\cdot p$:

  1. (v)

We have that $\phi(\mathfrak{l})=\phi(\mathfrak{l})^{T}$, where the transpose
is taken with respect to the inner product on $\mathfrak{s}$ induced by the
Einstein metric $g^{\mathsf{S}}$ at $p$.

Let now $\mathsf{L}=\mathsf{K}\mathsf{A}\mathsf{N}$ be an Iwasawa
decomposition with $\mathsf{H}\leq\mathsf{K}$, and denote the corresponding
Borel subgroup by $\mathsf{B}:=\mathsf{A}\mathsf{N}$. Conditions iv and v
allow us to apply Theorem 13.1 and conclude that the following also holds:

  1. (vi)

The homogeneous space $\mathsf{F}/\mathsf{K}$ admits an $\mathsf{F}$-invariant
Einstein metric.

Thus far we have constructed a presentation $M_{1}=\mathsf{F}/\mathsf{H}$
satisfying i–vi. Consider now $\Lambda:=Z(\mathsf{F})\cap\mathsf{L}$, a
discrete central subgroup of $\mathsf{F}$ and $\mathsf{L}$. The quotient
$M:=\Lambda\backslash M_{1}$ is homogeneous, locally isometric to
$(M_{1},g_{1})$, and has a presentation
$(\mathsf{F}/\Lambda)/(\mathsf{H}/\mathsf{H}\cap\Lambda)$ which clearly also
satisfies i–vi. Moreover, this presentation has the advantage that the Levi
factor $\mathsf{L}/\Lambda$ has finite center. Indeed, by construction
$\operatorname{Ad}_{\mathsf{F}/\Lambda}(\mathsf{L}/\Lambda)\subset\mathsf{GL}(\mathfrak{f})$
is a linear semisimple Lie group isomorphic to $\mathsf{L}/\Lambda$, because
$\ker\operatorname{Ad}_{\mathsf{F}}|_{\mathsf{L}}=\Lambda$.

In what follows we work on $M:=M_{1}/\Lambda$, and by abuse of notation we
drop the $\Lambda$ quotients in the groups notation. In addition, we may
assume that

  1. (vii)

The center of $\mathsf{L}$ is finite, and in particular $\mathsf{K}$ is
compact [Kna02, Thm. 6.31].

To summarise, we re-state what we have proved in this first step:

######  Proposition 14.1.

For any simply-connected homogeneous Einstein manifold $(M^{n}_{1},g_{1})$
with Einstein constant $-1$, there exists a homogeneous Einstein space
$(M^{n}:=\mathsf{F}/\mathsf{H},g)$ satisfying i–vii above, whose universal
cover is $(M^{n}_{1},g_{1})$.

For the remaining of the proof, we focus on proving that there exists a closed
solvable Lie subgroup of $\mathsf{F}$ acting transitively on $M$. This would
imply that $(M^{n}_{1},g_{1})$ itself is a solvmanifold, and by Theorem E.1
this is enough for proving Theorem A.

###  14.2. Step 2: The $\mathsf{N}$-action is polar

Consider the subgroup $\mathsf{G}\leq\mathsf{F}$ given by

$\mathsf{G}=\mathsf{B}\ltimes\mathsf{S}.$

By well-known properties of the Iwasawa decomposition (see for instance
[Kna02, Ch. VI, $\S$4]), $\mathsf{B}$ is a closed, simply-connected,
completely solvable subgroup of $\mathsf{L}$ intersecting $\mathsf{K}$
trivially. Thus, $\mathsf{G}$ is a closed, completely solvable, simply-
connected Lie subgroup of $\mathsf{F}$, which by ii does not intersect
$\mathsf{K}$, and we furthermore have $\mathsf{F}=\mathsf{K}\mathsf{G}$ with
$\mathsf{K}$ compact by vii. This implies that the action of $\mathsf{G}$ on
$\mathsf{F}/\mathsf{H}$ by left-multiplication is free, proper and isometric,
and the quotient space is $\mathsf{G}\backslash\mathsf{F}/\mathsf{H}$,
diffeomorphic to the compact homogeneous space $\mathsf{K}/\mathsf{H}$.

Hence, we are in a position to apply Theorem 8.1 and conclude that the action
of the nilradical $\mathsf{N}$ of $\mathsf{G}$ on $M$ induces a Riemannian
submersion $M\to\mathsf{N}\backslash M$ with integrable horizontal
distribution. Recall also that, thanks to the equivariant modified Helmholtz
decomposition, the mean curvature vector $N$ of the $\mathsf{N}$-orbits can be
decomposed as

$N=-\nabla\log v+{N_{0}},\qquad{N_{0}}\in\mathfrak{X}(M)^{\mathsf{G}},\quad
v\in\mathcal{C}^{\infty}_{+}(M)^{\mathsf{G}}.$

Theorem 8.1 and Corollary 9.9 imply that

(50) $N={N_{0}},\qquad\nabla^{P}N=0,\qquad L_{N}=-\beta^{+}.$

###  14.3. Step 3: $N$ is $\mathsf{G}$-vertical

Condition vi satisfied by the presentation $\mathsf{F}/\mathsf{H}$ implies
that the simply-connected solvable Lie group $\mathsf{G}$ admits a left-
invariant Einstein metric. Thus, Assumption 11.1 holds, and we may apply
Theorem 11.3 to conclude that $N$ is $\mathsf{G}$-vertical.

###  14.4. Step 4: $N_{\mathsf{F}}(\mathsf{G})$ acts transitively on
$\mathsf{F}/\mathsf{H}$

For convenience of the reader we briefly recall our setup on Lie algebra
level: $\mathfrak{f}=\mathfrak{l}\ltimes\mathfrak{s}$ is a Levi decomposition;
$\mathfrak{l}=\mathfrak{k}\oplus\mathfrak{a}_{\mathfrak{l}}\oplus\mathfrak{n}_{\mathfrak{l}}$
is an Iwasawa decomposition with $\mathfrak{k}$ containing the isotropy
subalgebra $\mathfrak{h}$; by setting
$\mathfrak{n}_{\mathfrak{l}}^{-}:=\theta(\mathfrak{n}_{\mathfrak{l}})$ and
$\mathfrak{l}_{0}:=Z_{\mathfrak{l}}(\mathfrak{a}_{\mathfrak{l}})$, where
$\theta\in\operatorname{Aut}(\mathfrak{l})$ is the corresponding Cartan
involution, we get

$\mathfrak{l}=\mathfrak{n}_{\mathfrak{l}}^{-}\oplus\mathfrak{l}_{0}\oplus\mathfrak{n}_{\mathfrak{l}}.$

Regarding the solvable radical, we write
$\mathfrak{s}=\mathfrak{a}_{\mathfrak{s}}\oplus\mathfrak{n}_{\mathfrak{f}}$
where $\mathfrak{n}_{\mathfrak{f}}$ is the nilradical of $\mathfrak{s}$ (also
of $\mathfrak{f}$) and $\mathfrak{a}_{\mathfrak{s}}$ is the orthogonal
complement of $\mathfrak{n}_{\mathfrak{f}}$ in $\mathfrak{s}$ with respect to
the inner product induced by $g^{\mathsf{S}}_{p}$ on $\mathfrak{s}$. The Lie
bracket satisfies

(51)
$[\mathfrak{f},\mathfrak{s}]\subset\mathfrak{n}_{\mathfrak{f}},\qquad[\mathfrak{a}_{\mathfrak{s}},\mathfrak{a}_{\mathfrak{s}}]=0,\qquad[\mathfrak{l},\mathfrak{a}_{\mathfrak{s}}]=0.$

Indeed, the first condition is well-known, see e.g. [Var84, Thm. 3.8.3,(iii)].
The second one follows from the integrability of the $\mathsf{N}$-horizontal
distribution in $\mathsf{S}\cdot p$. The last one follows from the first one,
together with v and the fact that
$[\mathfrak{l},\mathfrak{n}_{\mathfrak{f}}]\subset\mathfrak{n}_{\mathfrak{f}}$.

Finally, we may also write

$\mathfrak{g}:=(\mathfrak{a}_{\mathfrak{l}}\oplus\mathfrak{a}_{\mathfrak{s}})\ltimes(\mathfrak{n}_{\mathfrak{l}}\ltimes\mathfrak{n}_{\mathfrak{f}})=\mathfrak{a}\ltimes\mathfrak{n},\qquad\mathfrak{a}:=\mathfrak{a}_{\mathfrak{l}}\oplus\mathfrak{a}_{\mathfrak{s}},\qquad\mathfrak{n}:=\mathfrak{n}_{\mathfrak{l}}\ltimes\mathfrak{n}_{\mathfrak{f}},$

with $\mathfrak{a}$ abelian by (51).

By the previous steps, $N_{p}$ is tangent to the orbit $\mathsf{G}\cdot p$.
Moreover, by [Heb98, Cor. 2.10], after changing $p$ to another point $x\cdot
p$ for some $x\in\mathsf{N}$, we may assume without loss of generality that
for some Killing field $A\in\mathfrak{a}$ we have $-N_{p}=A_{p}$.

######  Lemma 14.2.

Let $D:=\operatorname{ad}_{\mathfrak{f}}A\in\operatorname{End}(\mathfrak{f})$,
and decompose $\mathfrak{f}$ as

$\mathfrak{f}=\mathfrak{f}_{-}\oplus\mathfrak{f}_{0}\oplus\mathfrak{f}_{+},\qquad\mathfrak{f}_{-}:=\mathfrak{n}_{\mathfrak{l}}^{-},\qquad\mathfrak{f}_{0}:=\mathfrak{l}_{0}\oplus\mathfrak{a}_{\mathfrak{s}},\qquad\mathfrak{f}_{+}:=\mathfrak{n}.$

Then, $D$ is positive-definite on $\mathfrak{f}_{+}$, negative-definite on
$\mathfrak{f}_{-}$, and zero on $\mathfrak{f}_{0}$.

###### Proof.

Lemma C.2 and (50) yield

$\tfrac{1}{2}\left((\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}}+(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}}^{T}\right)=({{}^{\mathfrak{n}}\beta}_{p})^{+}.$

Using (35) and the fact that
$(\operatorname{ad}_{\mathfrak{f}}A)|_{\mathfrak{n}}=(\operatorname{ad}_{\mathfrak{g}}A)|_{\mathfrak{n}}$
has only real eigenvalues (for $\mathfrak{g}$ is completely solvable), we
deduce that

$(\operatorname{ad}_{\mathfrak{f}}A)|_{\mathfrak{n}}=({{}^{\mathfrak{n}}\beta}_{p})^{+}.$

In particular, $D:=\operatorname{ad}_{\mathfrak{f}}A$ is positive-definite on
$\mathfrak{n}$. By (51) and the fact that $\mathfrak{a}$ is abelian,
$\mathfrak{f}_{0}\subset\ker D$. Regarding $\mathfrak{f}_{-}$, let us write
$A=A_{\mathfrak{l}}+A_{\mathfrak{s}}$ with
$A_{\mathfrak{l}}\in\mathfrak{a}_{\mathfrak{l}}$,
$A_{\mathfrak{s}}\in\mathfrak{a}_{\mathfrak{s}}$. Again by (51),
$[A_{\mathfrak{s}},\mathfrak{l}]=0$, thus
$\operatorname{ad}_{\mathfrak{l}}A_{\mathfrak{l}}=D|_{\mathfrak{n}_{\mathfrak{l}}}>0.$
Using that $\theta A_{\mathfrak{l}}=-A_{\mathfrak{l}}$ and the fact that
$\theta=\theta^{-1}\in\operatorname{Aut}(\mathfrak{l})$ it is clear that if
$U\in\mathfrak{n}_{\mathfrak{l}}$ is an eigenvector of
$\operatorname{ad}_{\mathfrak{l}}A_{\mathfrak{l}}$ with eigenvalue
$\lambda>0$, then $\theta(U)\in\mathfrak{n}_{\mathfrak{l}}^{-}$ is an
eigenvector with eigenvalue $-\lambda<0$. It follows that
$D|_{\mathfrak{f}_{-}}<0$. ∎

######  Proposition 14.3.

The group $N_{\mathsf{F}}(\mathsf{G})$ acts transitively on
$M=\mathsf{F}/\mathsf{H}$.

###### Proof.

By Lemma 14.2 we may apply the estimate in Proposition 12.5 yielding

$\operatorname{scal}^{\mathcal{V}}(p)+\dim\mathfrak{n}\leq\operatorname{tr}\beta^{+},$

$p=e\mathsf{H}$. But by Proposition 9.7, equality must hold. Hence, the
rigidity in Proposition 12.5 implies that the Killing fields in
$\mathfrak{f}_{0}\oplus\mathfrak{f}_{+}$ span the entire tangent space
$T_{p}M$. Notice that $\mathfrak{f}_{0}\oplus\mathfrak{f}_{+}$ is the Lie
algebra of the normalizer $N_{\mathsf{F}}(\mathsf{G})$ of $\mathsf{G}$ in
$\mathsf{F}$. Since $N_{\mathsf{F}}(\mathsf{G})$ is closed in $\mathsf{F}$,
the orbit $N_{\mathsf{F}}(\mathsf{G})\cdot p$ is an embedded submanifold, in
particular a closed subset. It is also open, since it has full dimension. By
conectedness, $N_{\mathsf{F}}(\mathsf{G})$ must act transitively. ∎

###  14.5. Step 5: $\mathsf{F}/\mathsf{H}$ is a solvmanifold

Proposition 14.3 implies that the Einstein manifold $(M^{n},g)$ admits a
transitive group of the form
$N_{\mathsf{F}}(\mathsf{G})_{0}=\mathsf{M}\ltimes\mathsf{G}$, where
$\mathsf{G}$ is solvable, and $\mathsf{M}\leq\mathsf{L}$, the identity
component of the normalizer of $\mathfrak{a}$ in $\mathsf{K}$, is compact. We
may further decompose $\mathsf{M}$ as $\mathsf{M}_{ss}\mathsf{M}_{z}$, where
$\mathsf{M}_{ss}$ is semisimple and $\mathsf{M}_{z}$ central in $\mathsf{M}$.
Then, the transitive group

$\mathsf{M}\ltimes\mathsf{G}=\mathsf{M}_{ss}\ltimes(\mathsf{M}_{z}\mathsf{G})$

has a compact Levi factor $\mathsf{M}_{ss}$. By [JP17, Thm. 0.1] (see also
[AL15, Cor.A.2]), the solvable group $\mathsf{M}_{z}\ltimes\mathsf{G}$ acts
transitively on $M^{n}$. Thus, $(M^{n},g)$ is an Einstein solvmanifold.
Clearly, the same applies for its universal cover $(M^{n}_{1},g_{1})$, and
Theorem A now follows from Theorem E.1.

##  Appendix A Modified Helmholtz decomposition

In this section we study the following second order linear elliptic PDE

(52) $\mathcal{L}v:=\operatorname{div}(\nabla v+vX)=\Delta v+\langle\nabla
v,X\rangle+v\operatorname{div}X=0,$

on a closed Riemannian manifold $B^{d}$, where $X$ is a given smooth vector
field (corresponding to $N$ in the previous sections) and
$\operatorname{div}X:=\operatorname{tr}\nabla X$. Our main goal is to show

######  Proposition A.1.

Up to scaling, there exists a unique smooth solution to (52), and this
solution does not change sign.

######  Example A.2.

If $X=\nabla u$ is a gradient vector field, then a solution to (52) is given
by $v=e^{-u}$. On the other hand, if $X$ is divergence free, then $v$ must be
constant.

######  Corollary A.3 (Modified Helmholtz decomposition).

Let $X$ be a vector field on a compact Riemannian manifold $B$. Then, there
exists $X_{0}\in\mathfrak{X}(B)$ and a positive
$v\in\mathcal{C}^{\infty}_{+}(B)$ such that

$X=-\nabla\log v+X_{0},\qquad\operatorname{div}(vX_{0})=0.$

###### Proof.

By Proposition A.1, there exists a positive smooth solution $v>0$ to (52). We
then simply set $X_{0}=\nabla\log v+X$. ∎

We are indebted to H.J. Hein for the following proof:

###### Proof of Proposition A.1.

Consider the Sobolev spaces $H_{k}:=W_{k}^{2}(B)$, $k\in{\mathbb{N}}$, with
induced norm

$\|v\|^{2}_{k}=\int_{B}\sum_{i=0}^{k}\|\nabla^{i}v\|^{2}_{g}\,\,d\mu_{g}\,,$

where $\nabla^{0}v:=v$, $\nabla^{1}v=\nabla v$ is the gradient, and
$\nabla^{i}v$ denotes the $(i-1)$-th covariant derivative of $\nabla v$ with
respect to the Levi-Civita connection of $(B,g)$. For each $k\in{\mathbb{N}}$,
$H_{k}$ is a Hilbert space.

Let $k\geq k_{\ast}:=\tfrac{d}{2}+11$. Then by [Tay11, Ch. 4, Prop. 3.3] we
have

(53) $H_{k}\subset\mathcal{C}^{10}(B).$

For $t\in[0,1]$ we consider the family of uniformly elliptic operators

$\mathcal{L}_{t}:H_{k+2}\to H_{k}\,,\qquad v\mapsto\operatorname{div}(\nabla
v+t\cdot v\cdot X)\,.$

The corresponding adjoints $\mathcal{L}^{*}_{t}:H_{k}\to H_{k+2}$ are also
uniformly elliptic and are given by

$\mathcal{L}^{*}_{t}(v)=\Delta v-t\,\langle\nabla v,X\rangle\,.$

This family yields a continuous path connecting $\mathcal{L}_{1}=\mathcal{L}$
with the Laplace-Beltrami operator $\mathcal{L}_{0}=\Delta$. The proof
strategy is to use the maximum principle for $\mathcal{L}_{t}^{*}$ to show
that $\dim\ker\mathcal{L}_{t}=1$, and that there is a continuous curve
$(v_{t})_{t\in[0,1]}$ in $H_{k+2}\subset\mathcal{C}^{10}(B)$ with $v_{0}\equiv
1$ constant on $B$, and $v_{t}\in\ker\mathcal{L}_{t}$ for each $t\in[0,1]$.
Notice that by elliptic regularity, $v_{t}$ is in fact smooth for all
$t\in[0,1]$ [GT01, Cor. 8.11]. The fact that $v_{1}>0$ will then follow from
Harnack’s inequality.

Firstly, we show that $\mathcal{L}_{t},\mathcal{L}_{t}^{*}$ are Fredholm for
all $t\in[0,1]$. By [Bes87, App. I, Thm. 31], $\ker\mathcal{L}_{t}$ is finite-
dimensional, and by [GT01, Thm. 8.10] we have an estimate of the form

$\|v\|_{k+2}\leq C\cdot(\|\mathcal{L}_{t}(v)\|_{k}+\|v\|_{k})\,.$

We write $H_{k+2}=\ker\mathcal{L}_{t}\oplus(\ker\mathcal{L}_{t})^{\perp}$ and
show that the range $\operatorname{ran}\mathcal{L}_{t}$ of $\mathcal{L}_{t}$
is closed. Let $(v_{n})_{n\in\mathsf{N}}$ be a sequence in
$(\ker\mathcal{L}_{t})^{\perp}$ with $\lim_{n\to\infty}w_{n}=w\in H_{k}$,
$w_{n}:=\mathcal{L}_{t}(v_{n})$. It is enough to show, that $(v_{n})$
subconverges to $v_{\infty}\in H_{k+2}$. Firstly, we assume that $(v_{n})$ is
a bounded sequence. Then, since $H_{k+2}$ is compactly embedded into $H_{k}$,
see [Tay11, Ch. 4, Prop. 3.4], we may assume that the sequence $(v_{n})$
converges in $H_{k}$. Thus

$\|v_{n}-v_{m}\|_{k+2}\leq C\cdot(\|w_{n}-w_{m}\|_{k}+\|v_{n}-v_{m}\|_{k})$

and it follows that $(v_{n})$ converges in $H_{k+2}$. It remains to consider
the case $\|v_{n}\|_{k+2}\to\infty$ for $n\to\infty$. We set
$\tilde{v}_{n}:=\tfrac{v_{n}}{\|v_{n}\|_{k+2}}$. Then
$\mathcal{L}_{t}(\tilde{v}_{n})\to 0$ for $n\to\infty$. As above it follows
that $\tilde{v}_{n}\to\tilde{v}\in H_{k+2}$, along a subsequence possibly. But
then $\mathcal{L}_{t}(\tilde{v})=0$, a contradiction. By [Tay11, App. A, Prop.
5.7], we have
$(\operatorname{ran}\mathcal{L}_{t})^{\perp}=\ker\mathcal{L}_{t}^{*}$. Since
$\ker\mathcal{L}_{t}^{*}$ is finite-dimensional it follows that
$\mathcal{L}_{t}$ is Fredholm. Moreover, since
$\operatorname{ran}\mathcal{L}_{t}$ is closed, by [Tay11, App. A, Prop. 5.7]
also $\mathcal{L}_{t}^{*}$ is Fredholm.

Since the index does not change along a continuous path of Fredholm operators
[Tay11, App. A, Prop. 7.4], and $\mathcal{L}_{0}=\mathcal{L}_{0}^{*}$ is the
Laplace-Beltrami operator, for all $t\in[0,1]$ we have

${\rm
ind}\,\mathcal{L}_{t}=\dim\ker\mathcal{L}_{t}-\dim\ker\mathcal{L}^{*}_{t}=0.$

Notice that $\mathcal{L}_{t}^{*}$ satisfies a strong maximum principle: see
[PW84, $\S$3, Thm. 5]. Thus, $\dim\ker\mathcal{L}^{*}_{t}=1$, the kernel
consisting of constant functions, and we deduce that
$\dim\ker\mathcal{L}_{t}=1$ for all $t\in[0,1]$.

We now claim that the constant function $1$ is not in the range of
$\mathcal{L}_{t}^{*}$. This follows by contradiction, since at a maximum of
$v$ we have $\mathcal{L}_{t}^{*}(v)\leq 0$. Hence,
$1\notin(\ker\mathcal{L}_{t})^{\perp}$ by the closed range theorem. By
[Bro65], the orthogonal projections $P_{t}:H_{k+2}\to\ker\mathcal{L}_{t}$
depend continuously on $t$. We thus have a continuous family
$(v_{t}:=P_{t}(1))_{t\in[0,1]}\subset H_{k+2}$, with $\mathcal{L}_{t}v_{t}=0$
and $v_{t}\neq 0$ thanks to the previous claim. Moreover, $v_{0}\equiv 1$ is
constant.

Finally, we claim that $v_{t}>0$ for all $t\in[0,1]$. This holds for
$v_{0}\equiv 1$, and it is an open condition, since $v_{t}$ depends
continuously on $t$ in $C^{0}$-topology. Closedness follows from the Harnack
inequality applied locally in $B$: see [GT01, Thm. 8.20]. ∎

######  Remark A.4.

Another approach for proving Proposition A.1 is as follows. The second order
linear elliptic operator $\mathcal{L}u=-\Delta u+\langle\nabla
u,X\rangle+c\,u$, $c\in\mathcal{C}^{\infty}(B)$, acting on functions on a
closed smooth Riemannian manifold $B$, has a _principal eigenvalue_
$\lambda_{1}$, characterised by the following properties:

  1. (1)

$\lambda_{1}$ admits a positive eigenfunction $v$;

  2. (2)

$\lambda_{1}$ is a simple eigenvalue;

  3. (3)

For any other complex eigenvalue $\lambda\neq\lambda_{1}$ of $\mathcal{L}$ one
has ${\rm Re}(\lambda)>\lambda_{1}$.

(Moreover, in case $c\geq 0$ and $c\not\equiv 0$, we have $\lambda_{1}>0$, but
we will not need this.) This statement is well-known in the case of bounded
domains with smooth boundary in ${\mathbb{R}}^{n}$, see [Eva98, Theorem 3 in
$\S$6.5] or [Ni14] (it also holds in more general domains, see e.g. [BNV94]).
According to some experts, the proof carries over to the closed manifold case
without any issues.

It follows that there exists a smooth $v>0$ and $\lambda_{1}\in{\mathbb{R}}$
with

$\operatorname{div}(\nabla v+vX)=\lambda_{1}v.$

Integrating both sides and using that $v>0$ we deduce that $\lambda_{1}=0$,
thus $v$ solves (52).

##  Appendix B An equivariant divergence theorem

The main aim of this section is to prove equivariant versions of the
divergence theorem and the modified Helmholtz decomposition (Corollary A.3),
for vector fields on a Riemannian manifold $(M^{n},g)$ with a co-compact
isometric action of a unimodular Lie group $\mathsf{G}$ for which the orbit
space $B:=M/\mathsf{G}$ is a closed manifold. We denote by
$\mathfrak{X}(M)^{\mathsf{G}}$, $\mathcal{C}_{+}^{\infty}(M)^{\mathsf{G}}$ the
spaces of $\mathsf{G}$-invariant vector fields and positive
$\mathsf{G}$-invariant smooth functions on $M$, respectively. As usual, we
endow $B$ with the quotient Riemannian metric $g^{B}$, so that $\pi:M\to B$ is
a Riemannian submersion.

######  Proposition B.1 (Equivariant modified Helmholtz decomposition).

Let $(M^{n},g)$, $\mathsf{G}$ be as above, and let
$E\in\mathfrak{X}(M)^{\mathsf{G}}$. Then, there exists
$v\in\mathcal{C}_{+}^{\infty}(M)^{\mathsf{G}}$ and
$E_{0}\in\mathfrak{X}(M)^{\mathsf{G}}$ such that

$E=-\nabla\log v+E_{0},\qquad\operatorname{div}_{M}(vE_{0})=0.$

Moreover, $E_{0}$ is unique with those properties, and $v$ is unique up to
scaling.

######  Proposition B.2.

Let $(M^{n},g)$, $\mathsf{G}$ be as above, and let
$E\in\mathfrak{X}(M)^{\mathsf{G}}$ such that $\operatorname{div}_{M}E\geq 0$.
Then, $\operatorname{div}_{M}E\equiv 0$.

We remark that the vector field $E$ in the above statements is not necessarily
$\mathsf{G}$-horizontal.

To prove these we will first establish some basic lemmas. Let $N$ be the mean
curvature vector of the $\mathsf{G}$-orbits, which by Lemma 3.2 can be written
as

$N=-\nabla\log v_{\mathsf{G}},\qquad
v_{\mathsf{G}}\in\mathcal{C}_{+}^{\infty}(M)^{\mathsf{G}}.$

######  Lemma B.3.

Let $X\in\mathfrak{X}(M)^{\mathsf{G}}$ be a basic (horizontal) vector field.
Then,

$\operatorname{div}_{M}X=(\operatorname{div}_{\\!B}X)\circ\pi-\langle
X,N\rangle.$

###### Proof.

Let $\\{X_{i}\\}$, $\\{U_{j}\\}$ be basic and vertical orthonormal frames,
respectively. Then,

$\displaystyle\operatorname{div}_{M}X$
$\displaystyle=\,\,\sum_{i}\langle\nabla^{M}_{X_{i}}X,X_{i}\rangle+\sum_{j}\langle\nabla^{M}_{U_{j}}X,U_{j}\rangle$
$\displaystyle=\,\,\sum_{i}\langle\nabla^{B}_{X_{i}}X,X_{i}\rangle-\sum_{j}\langle
X,\nabla^{M}_{U_{j}},U_{j}\rangle$
$\displaystyle=\,\,\operatorname{div}_{\\!B}X-\langle X,N\rangle.$

∎

######  Lemma B.4.

Let $U\in\mathfrak{X}(M)^{\mathsf{G}}$ be a vertical, $\mathsf{G}$-invariant
vector field. Then, $\operatorname{div}_{M}U=0$.

###### Proof.

We first prove the result for a transitive $\mathsf{G}$-action, that is,
$M^{n}=\mathsf{G}/\mathsf{H}$. Let $d\operatorname{vol}_{g}$ denote the
$\mathsf{G}$-invariant volume form on $(M^{n},g)$, and recall that
$\operatorname{div}_{M}X=\mathcal{L}_{X}d\operatorname{vol}_{g}$. The pull-
back of $d\operatorname{vol}_{g}$ under the $\mathsf{G}$-equivariant
projection $\mathsf{G}\to\mathsf{G}/\mathsf{H}$ is a left-invariant $n$-form
on $\mathsf{G}$. By unimodularity, said pull-back is also right-invariant. It
follows that $d\operatorname{vol}_{g}$ is
right-$\mathsf{N}_{\mathsf{G}}(\mathsf{H})$-invariant. In particular, since
$\mathsf{G}$-invariant vector fields on $\mathsf{G}/\mathsf{H}$ are precisely
those which are tangent to the right
$\mathsf{N}_{\mathsf{G}}(\mathsf{H})$-action on $\mathsf{G}/\mathsf{H}$, we
have $\operatorname{div}_{M}X=0$.

In the general case, let $p\in M$, and let $\\{X_{k}\\}$ and $\\{U_{i}\\}$ be
respectively horizontal and vertical orthonormal frames. Then,

$\displaystyle\operatorname{div}_{M}U=\,\,\sum_{k}\langle\nabla_{X_{k}}U,X_{k}\rangle+\sum_{i}\langle\nabla_{U_{i}}U,U_{i}\rangle=\,\,\operatorname{div}_{\mathsf{G}\cdot
p}U=0.$

where the second equality follows from skew-symmetry of the $A$-tensor in the
horizontal entries, and the last one from the homogeneous case. ∎

######  Lemma B.5.

Let $E\in\mathfrak{X}(M)^{\mathsf{G}}$. Then,

$\operatorname{div}_{M}E=v_{\mathsf{G}}^{-1}\,\operatorname{div}_{\\!B}(v_{\mathsf{G}}\,\mathcal{H}E).$

###### Proof.

By Lemmas B.3 and B.4 we have

$\operatorname{div}_{M}E=\operatorname{div}_{M}\mathcal{H}E=\operatorname{div}_{\\!B}\mathcal{H}E+v_{\mathsf{G}}^{-1}\langle\mathcal{H}E,\nabla
v_{\mathsf{G}}\rangle.$

Thus,

$v_{\mathsf{G}}^{-1}\,\operatorname{div}_{\\!B}(v_{\mathsf{G}}\,\mathcal{H}E)=v_{\mathsf{G}}^{-1}\,\left(v_{\mathsf{G}}\,\operatorname{div}_{\\!B}\mathcal{H}E+\langle\mathcal{H}E,\nabla
v_{\mathsf{G}}\rangle\right)=\operatorname{div}_{M}E.$

∎

We are now in a position to prove both main results of this section:

###### Proof of Proposition B.1.

Since $B$ is compact, by Corollary A.3 there is a generalised Helmholtz
decomposition for the vector field $N+\mathcal{H}E\in\mathfrak{X}(B)$,

$N+\mathcal{H}E=-\nabla\log
v_{1}+X_{0},\qquad\operatorname{div}_{\\!B}(v_{1}\,X_{0})=0.$

We lift $X_{0}$ and $v_{1}$ to $\mathsf{G}$-invariant objects on $M$, and
notice that

$E=-N+(N+\mathcal{H}E)+{\mathcal{V}}E=-\nabla\log(v_{1}/v_{\mathsf{G}})+X_{0}+{\mathcal{V}}E.$

We thus set $v:=v_{1}/v_{G}$, $E_{0}\in X_{0}+{\mathcal{V}}E$, both clearly
$\mathsf{G}$-invariant. Finally, by Lemmas B.4 and B.5 we have

$\operatorname{div}_{M}\left((v_{1}/v_{G})\,(X_{0}+{\mathcal{V}}E)\right)=\operatorname{div}_{M}\left((v_{1}/v_{G})\,X_{0})\right)=v_{\mathsf{G}}^{-1}\,\operatorname{div}_{\\!B}(v_{1}X_{0})=0.$

Finally, since $\mathcal{H}E=-\nabla\log v+\mathcal{H}E_{0}$ is a generalised
Helmholtz decomposition for $\mathcal{H}E\in\mathfrak{X}(B)$, uniqueness
follows from Corollary A.3. ∎

###### Proof of Proposition B.2.

By Lemma B.5 the assumptions imply

$\operatorname{div}_{\\!B}(v_{\mathsf{G}}\mathcal{H}E)=v_{\mathsf{G}}\,\operatorname{div}_{M}E\geq
0.$

Integrating over $B$, the divergence theorem yields equality everywhere. ∎

##  Appendix C Some formulae involving Killing fields

In this section we recall some useful formulae for computing with Killing
fields.

######  Lemma C.1.

[Bes87, Lemma 7.27] Let $X,Y,Z$ be Killing fields on a Riemannian manifold
$(M^{n},g)$. Then, the Levi-Civita connection $\nabla$ satisfies

(54)
$2\,\langle\nabla_{X}Y,Z\rangle=\langle[X,Y],Z\rangle+\langle[X,Z],Y\rangle+\langle
X,[Y,Z]\rangle.$

From this, one can deduce the following:

######  Lemma C.2.

Let $\mathsf{N}$ act isometrically on $(M^{n},g)$, and let $X$ be a Killing
field of $(M,g)$ with $X_{p}\perp(\mathsf{N}\cdot p)$. Then, for all Killing
fields $U,V\in\mathfrak{n}$, the shape operator satisfies

$\langle
L_{X}U,V\rangle_{p}=\tfrac{1}{2}\big{(}\langle[X,U],V\rangle_{p}+\langle
U,[X,V]\rangle_{p}\big{)}.$

In particular, if $[X,\mathfrak{n}]\subset\mathfrak{n}$, then in the notation
of $\S$5 we have

$L^{\mathfrak{n}}_{X_{p}}=S(\operatorname{ad}X|_{\mathfrak{n}}):=\tfrac{1}{2}\big{(}(\operatorname{ad}X|_{\mathfrak{n}})+(\operatorname{ad}X|_{\mathfrak{n}})^{T}\big{)},$

transpose with respect to $h_{p}$.

###### Proof.

By the symmetries of the second fundamental form and (54), we have

$\langle
L_{X}U,V\rangle_{p}=-\langle\nabla_{U}V,X\rangle_{p}=-\tfrac{1}{2}\,\left(\langle[U,V],X\rangle_{p}+\langle[U,X],V\rangle_{p}+\langle
U,[V,X]\rangle_{p}\right),$

and the lemma follows since $[U,V]_{p}\perp X_{p}$. ∎

Even though the next lemma is well known, we provide a proof for convenience
(see e.g. [Pet16, Proposition 8.1.3] or [KN96a, Ch. VI, Prop. 2.6 (2)],
however notice that the latter contains a sign mistake: see [KN96b, p. 469]):

######  Lemma C.3.

Let $X,Y$ be Killing fields. Then, $R(X,Y)=[\nabla X,\nabla Y]+\nabla[X,Y]$.

###### Proof.

Since $X$ is a Killing field, the Lie derivative $\mathcal{L}_{X}$ preserves
the Levi-Civita connection:

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

Using $\nabla X=\nabla_{X}-\mathcal{L}_{X}$ and
$[\mathcal{L}_{X},\mathcal{L}_{Y}]=\mathcal{L}_{[X,Y]}$, we compute:

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

∎

##  Appendix D The endomorphism $\beta$ associated to a Lie algebra

In this appendix we follow the notation from $\S$9. We refer the reader to
[BL20] for further details. Let $\mathfrak{g}$ be a non-abelian Lie algebra
endowed with a background inner product $\bar{h}$ and set
$\mathsf{O}(\mathfrak{g}):=\mathsf{O}(\mathfrak{g},\bar{h})$. This induces an
inner product $\langle\langle\cdot,\cdot\rangle\rangle_{\bar{h}}$ on
$V_{\mathfrak{g}}:=\Lambda^{2}(\mathfrak{g}^{*})\otimes\mathfrak{g}$, see
again $\S$9. By studying the natural ’change of basis’ linear action of
$\mathsf{GL}(\mathfrak{g})$ on $V_{\mathfrak{g}}$, defined in (31), from a
real geometric invariant theory point of view, one obtains a
$\mathsf{GL}(\mathfrak{g})$-invariant stratification

$V_{\mathfrak{g}}\backslash\\{0\\}=\bigcup_{\beta\in\mathcal{B}}\mathcal{S}_{\beta},$

where the union is disjoint. Here, $\mathcal{B}$ is a finite set of
$\bar{h}$-self-adjoint endomorphisms of $\mathfrak{g}$ which are uniquely
determined up to conjugation by $\mathsf{O}(\mathfrak{g})$. This
stratification, first obtained by Kirwan and Ness in the complex setting
[Kir84, Nes84], and in the real setting by [HSS08, Lau10], has a number of
remarkable properties. We describe in Proposition D.1 below the most important
one of them regarding applications in this article.

The Lie bracket $\mu_{\mathfrak{g}}$ of $\mathfrak{g}$, being a non-zero
element in $V_{\mathfrak{g}}$, belongs to a unique stratum

$\mu_{\mathfrak{g}}\in\mathcal{S}_{{\bar{\beta}}}.$

With respect to an ordered $\bar{h}$-orthonormal basis of eigenvectors of
${\bar{\beta}}$ with eigenvalues in non-decreasing order, we consider the
solvable Lie subgroup
${\mathsf{B}_{\bar{\beta}}}\leq\mathsf{GL}^{+}(\mathfrak{g})$ represented in
said basis by lower triangular matrices with positive diagonal entries. Notice
that

$\mathsf{GL}(\mathfrak{g})={\mathsf{B}_{\bar{\beta}}}\mathsf{O}(\mathfrak{g}),\qquad{\mathsf{B}_{\bar{\beta}}}\cap\mathsf{O}(\mathfrak{g})=\\{{\operatorname{Id}}\\}.$

Set
${\bar{\beta}}^{+}:={\bar{\beta}}/\operatorname{tr}({\bar{\beta}}^{2})+{\operatorname{Id}}_{\mathfrak{g}}$.
The endomorphism
$\tau({\bar{\beta}}^{+})\in\operatorname{End}(V_{\mathfrak{g}})$ (see (31)) is
also self-adjoint, and we denote by $V_{{\bar{\beta}}^{+}}^{\geq 0}$ the sum
of eigenspaces of $\tau({\bar{\beta}}^{+})$ with non-negative eigenvalues.

######  Proposition D.1.

[BL20, Lemma 1.7.13] We have that
$\mathcal{S}_{\bar{\beta}}=\mathsf{O}(\mathfrak{g})\cdot
U_{{\bar{\beta}}^{+}}^{\geq 0}$, where $U_{{\bar{\beta}}^{+}}^{\geq 0}\subset
V_{{\bar{\beta}}^{+}}^{\geq 0}$ is a certain open,
${\mathsf{B}_{\bar{\beta}}}$-invariant subset.

Since $U_{\alpha^{+}}^{\geq 0}=k\cdot U_{{\bar{\beta}}^{+}}^{\geq 0}$ for any
$\alpha=k{\bar{\beta}}k^{-1}$, $k\in\mathsf{O}(\mathfrak{g})$, an upshot of
Proposition D.1 is that we may choose the stratum label ${\bar{\beta}}$ so
that $\mu_{\mathfrak{g}}\in U_{{\bar{\beta}}^{+}}^{\geq 0}$. We then say that
$\mu_{\mathfrak{g}}$ is _gauged correctly_ with respect to ${\bar{\beta}}$.
Clearly, this yields

(55)
$\langle\langle\,\tau\left({\bar{\beta}}^{+}\right)\mu,\mu\,\rangle\rangle_{\bar{h}}\geq
0,\qquad\forall\mu\in{\mathsf{B}_{\bar{\beta}}}\cdot\mu_{\mathfrak{g}}.$

######  Lemma D.2.

Given a Lie algebra $\mathfrak{g}$ with Lie bracket $\mu_{\mathfrak{g}}\in
V_{\mathfrak{g}}\backslash\\{0\\}$ and a background inner product $\bar{h}$ on
$\mathfrak{g}$, there exists a unique $\bar{h}$-self-adjoint
${\bar{\beta}}\in\operatorname{End}(\mathfrak{g})$ such that
$\mu_{\mathfrak{g}}\in\mathcal{S}_{\bar{\beta}}$ is gauged correctly with
respect to ${\bar{\beta}}$. In particular, (55) holds.

Let $\mathsf{G}$ be the simply-connected Lie group with Lie algebra
$\mathfrak{g}$. Given $h\in\operatorname{Sym}^{2}_{+}(\mathfrak{g}^{*})$, we
denote by $\operatorname{Ric}(h)\in\operatorname{End}(\mathfrak{g})$ the Ricci
endomorphism of the corresponding left-invariant metric on $\mathsf{G}$, at
the point $e\in\mathsf{G}$, identifying $T_{e}\mathsf{G}\simeq\mathfrak{g}$.
The following is the key Ricci curvature estimate:

######  Proposition D.3.

[BL18, Lemma 6.2] If $(\mathfrak{g},\mu_{\mathfrak{g}})$ is a nilpotent Lie
algebra with $\mu_{\mathfrak{g}}\in\mathcal{S}_{\bar{\beta}}$ gauged
correctly, then for any
$h=q\cdot\bar{h}\in\operatorname{Sym}^{2}_{+}(\mathfrak{g}^{*})$,
$q\in{\mathsf{B}_{\bar{\beta}}}$,

$\operatorname{tr}\operatorname{Ric}(h)q{\bar{\beta}}^{+}q^{-1}\geq 0,$

for all $h=q\cdot\bar{h}\in\operatorname{Sym}^{2}_{+}(\mathfrak{g}^{*})$,
$q\in{\mathsf{B}_{\bar{\beta}}}$. Equality holds if and only if
$q{\bar{\beta}}^{+}q^{-1}\in\operatorname{Der}(\mathfrak{g})$.

###### Proof.

By Proposition 9.4 applied to $E=q{\bar{\beta}}^{+}q^{-1}$, Lemma 9.2 and (55)
we have

$\displaystyle\operatorname{tr}\operatorname{Ric}(h)E=$
$\displaystyle\,\,\tfrac{1}{4}\,\langle\langle\tau(E)\mu_{\mathfrak{g}},\mu_{\mathfrak{g}}\rangle\rangle_{h}=\tfrac{1}{4}\,\langle\langle
q^{-1}\cdot(\tau(E)\mu_{\mathfrak{g}}),q^{-1}\cdot\mu_{\mathfrak{g}}\rangle\rangle$
$\displaystyle=$
$\displaystyle\,\,\tfrac{1}{4}\,\langle\langle\tau({\bar{\beta}}^{+})(q^{-1}\cdot\mu_{\mathfrak{g}}),(q^{-1}\cdot\mu_{\mathfrak{g}})\rangle\rangle\geq
0.$

In the third eqality we also used the equivariance of $\tau$. Equality holds
if and only if we have equality in (55), which happens if and only if
$q^{-1}\cdot\mu_{\mathfrak{g}}$ is in the kernel of $\tau({\bar{\beta}}^{+})$.
By equivariance of $\tau$ this is equivalent to
$\tau(q{\bar{\beta}}^{+}q^{-1})\mu_{\mathfrak{g}}=0$, which by definition
means $q{\bar{\beta}}^{+}q^{-1}\in\operatorname{Der}(\mathfrak{g})$. ∎

######  Remark D.4.

The estimate in Proposition D.3 holds more generally for arbitrary homogeneous
manifolds (replacing $\operatorname{Ric}(h)$ by the so-called _modified Ricci
curvature_ in the non-unimodular case). This follows essentially from [BL18,
Lemma 6.2].

######  Remark D.5.

Let $\mathsf{G}$ be the simply-connected Lie group with Lie algebra
$(\mathfrak{g},\mu_{\mathfrak{g}})$ and identify $\mathfrak{g}$ with left-
invariant vector fields on $\mathsf{G}$. Each
$q\cdot\bar{h}\in\operatorname{Sym}^{2}_{+}(\mathfrak{g}^{*})$ corresponds to
a left-invariant metric $g_{q}$ on $\mathsf{G}$, and we set
$\bar{g}:=g_{{\operatorname{Id}}}$. If $\\{U_{i}\\}$ is a left-invariant,
$\bar{g}$-orthonormal frame of eigenvectors of ${\bar{\beta}}^{+}$ with
eigenvalues $\beta_{i}$, then $\\{qU_{i}\\}$ is a $g_{q}$-orthonormal, left-
invariant frame, and we have

$\operatorname{tr}\operatorname{Ric}(h)q{\bar{\beta}}^{+}q^{-1}=\sum_{i}\beta_{i}\operatorname{ric}_{g_{q}}^{ii},\qquad\operatorname{ric}_{g_{q}}^{ii}:=\operatorname{ric}_{g_{q}}(qE_{i},qE_{i}).$

In this sense, the estimate in Proposition D.3 asserts the non-negativity of
the _$\beta$ -weighted scalar curvature_. The reason one has to be careful
with the choice of gauge is that, in general, ${\bar{\beta}}^{+}$ and
$\operatorname{Ric}(h)$ do not diagonalise simultaneously.

Besides Proposition D.3, the endomorphism ${\bar{\beta}}^{+}$ satisfies a
number of algebraic properties which are key for our applications:

######  Proposition D.6.

Let $(\mathfrak{g},\mu_{\mathfrak{g}})$ be a non-abelian Lie algebra with
$\mu_{\mathfrak{g}}\in\mathcal{S}_{\bar{\beta}}$ gauged correctly. Then, the
following hold:

  1. (i)

$\operatorname{tr}\left(D\,q{\bar{\beta}}q^{-1}\right)=0$, for all
$q\in\mathsf{B}_{\bar{\beta}}$, $D\in\operatorname{Der}(\mathfrak{g})$;

  2. (ii)

$\operatorname{tr}{\bar{\beta}}^{+}=\operatorname{tr}\big{(}({\bar{\beta}}^{+})^{2}\big{)}$;

  3. (iii)

$\operatorname{tr}[E,E^{T}]{\bar{\beta}}\geq 0$ for all
$E\in\mathfrak{b}_{\bar{\beta}}$, with equality if and only if
$[E,{\bar{\beta}}]=0$. Here $E^{T}$ denotes transpose with respect to
$\bar{h}$. In particular, ${\bar{\beta}}$ commutes with $\bar{h}$-self-adjoint
derivations.

###### Proof.

Given $D\in\operatorname{Der}(\mu_{\mathfrak{g}})$, we have that
$q^{-1}Dq\in\operatorname{Der}(q^{-1}\cdot\mu_{\mathfrak{g}})$. Since
$q\in{\mathsf{B}_{\bar{\beta}}}$, $q^{-1}\cdot\mu_{\mathfrak{g}}\in
U_{{\bar{\beta}}^{+}}^{\geq 0}$ is also gauged correctly. Thus, (i) follows
from [BL20, Corollary 1.9.2]. Regarding (ii), by definition of
${\bar{\beta}}^{+}$ and the fact that $\operatorname{tr}{\bar{\beta}}=-1$, we
have that $\operatorname{tr}{\bar{\beta}}{\bar{\beta}}^{+}=0$, from which

$\operatorname{tr}({\bar{\beta}}^{+})^{2}=\frac{1}{\operatorname{tr}{\bar{\beta}}^{2}}\,\operatorname{tr}{\bar{\beta}}{\bar{\beta}}^{+}+\operatorname{tr}{\bar{\beta}}^{+}=\operatorname{tr}{\bar{\beta}}^{+}.$

Finally, (iii) follows from [BL22, Lemma B.3]. ∎

######  Proposition D.7.

[Lau10], [BL18, Corollary C.2] If $(\mathfrak{n},\mu_{\mathfrak{n}})$ denotes
the nilradical of $\mathfrak{g}$, then

${\bar{\beta}}^{+}|_{\mathfrak{n}^{\perp}}=0,\qquad{\bar{\beta}}^{+}|_{\mathfrak{n}}={\bar{\beta}}^{+}_{\mathfrak{n}}>0.$

In particular, $\mathsf{B}_{\bar{\beta}}$ preserves the subspace
$\mathfrak{n}$. Here, ${\bar{\beta}}_{\mathfrak{n}}$ denotes the stratum label
for the Lie algbera $(\mathfrak{n},\mu_{\mathfrak{n}})$ (with background inner
product $\bar{h}|_{\mathfrak{n}\times\mathfrak{n}}$), chosen so that
$\mu_{\mathfrak{n}}$ is gauged correctly.

Finally, the following GIT technical lemma was needed in the proof of
Proposition 11.4. For its proof we adopt the notation from [BL20].

######  Lemma D.8.

Let $b_{1},b_{2}\in{\mathsf{B}_{\bar{\beta}}}$ be such that
$b_{i}{\bar{\beta}}^{+}b_{i}^{-1}\in\operatorname{Der}(\mathfrak{g})$ for
$i=1,2$. Then, there exists $a\in\operatorname{Aut}(\mathfrak{g})$,
$z\in\mathsf{G}_{\bar{\beta}}:=\\{h\in\mathsf{GL}(\mathfrak{g}):h{\bar{\beta}}h^{-1}={\bar{\beta}}\\}$,
such that

$b_{1}=a\,b_{2}\,z.$

###### Proof.

Set ${\bar{\beta}}^{+}_{i}:=b_{i}{\bar{\beta}}^{+}b_{i}^{-1}$,
$\mu_{i}:=b_{i}^{-1}\cdot\mu^{\mathfrak{g}}$, $i=1,2$. The assumption implies
that

$\tau\left({\bar{\beta}}^{+}_{i}\right)\mu^{\mathfrak{g}}=0,\qquad i=1,2.$

Acting with $b_{i}^{-1}$ and using the equivariance of $\tau$ we obtain

$0=b_{i}^{-1}\cdot\tau\left({\bar{\beta}}^{+}_{i}\right)\mu^{\mathfrak{g}}=\tau\left(b_{i}^{-1}{\bar{\beta}}^{+}_{i}b_{i}\right)(b_{i}^{-1}\cdot\mu^{\mathfrak{g}})=\tau({\bar{\beta}}^{+})\mu_{i},\qquad
i=1,2.$

Thus, $\mu_{i}\in V^{0}_{{\bar{\beta}}^{+}}$, $i=1,2$. Notice that
$\mu_{1}=(b_{1}^{-1}b_{2})\cdot\mu_{2}$. Write
$b_{1}^{-1}b_{2}\in{\mathsf{B}_{\bar{\beta}}}$ as

$b_{1}^{-1}b_{2}=zu,\qquad
z\in\mathsf{G}_{\bar{\beta}},\,\,u\in\mathsf{U}_{\bar{\beta}}.$

Using that $\mathsf{G}_{\bar{\beta}}$ preserves the subspace
$V^{0}_{{\bar{\beta}}^{+}}$, we have that

$u\cdot\mu_{2}=z^{-1}\cdot(b_{1}^{-1}b_{2})\cdot\mu_{2}=z^{-1}\cdot\mu_{1}\in
V^{0}_{{\bar{\beta}}^{+}}.$

Applying the $\mathsf{U}_{\bar{\beta}}$-invariant projection
$p_{\bar{\beta}}:V^{\geq 0}_{{\bar{\beta}}^{+}}\to V^{0}_{{\bar{\beta}}^{+}}$
[BL20, Lemma 7.5] we obtain

$u\cdot\mu_{2}=p_{\bar{\beta}}(u\cdot\mu_{2})=p_{\bar{\beta}}(\mu_{2})=\mu_{2},$

from which $u\in\operatorname{Aut}(\mu_{2})$. Since
$\mu_{2}=b_{2}^{-1}\cdot\mu^{\mathfrak{g}}$, we can write $u=b_{2}^{-1}ab_{2}$
with $a\in\operatorname{Aut}(\mathfrak{g})$. Hence,
$b_{1}^{-1}b_{2}=zb_{2}^{-1}ab_{2}$, from which $b_{1}^{-1}=zb_{2}^{-1}a$.
Inverting both sides yields the desired formula. ∎

##  Appendix E Homogeneous quotients of Einstein solvmanifolds are trivial

Recall that a Riemannian solvmanifold is a Riemannian manifold admitting a
transitive solvable group of isometries. The following key result is due to M.
Jablonski. It holds even more generally for Ricci solitons. However, as
already noted in [Jab15b], the proof simplifies significantly in the Einstein
case. The argument below follows a sketch of proof indicated in [Jab15b].

######  Theorem E.1.

[Jab15b] Let $(M^{n},g)$ be an Einstein solvmanifold and
$\Gamma<\operatorname{Iso}(M^{n},g)$ a subgroup acting properly
discontinuously such that $M/\Gamma$ is homogeneous. Then, $\Gamma$ is
trivial.

###### Proof.

Since the universal cover of a solvmanifold is again a solvmanifold, acted
transitively by the universal cover of the corresponding solvable Lie group,
it is enough to prove the theorem when $M$ is simply-connected. Choosing the
solvable group to act simply transitively [GW88], it follows that $M$ is
diffeomorphic to ${\mathbb{R}}^{n}$.

Recall that $\operatorname{Iso}(M^{n},g)$ is linear by [AC99] (see also the
discussion after Theorem B in [Heb98]). This implies that any connected
transitive group of isometries $\mathsf{F}\leq\operatorname{Iso}(M,g)$ has a
solvable subgroup which is still transitive. Indeed, it suffices to show this
for an effective presentation $M\simeq\mathsf{F}/\mathsf{F}_{p}$, where
$\mathsf{F}_{p}$ is the compact isotropy at $p\in M$. Since
$M\simeq{\mathbb{R}}^{n}$, $\mathsf{F}_{p}$ must be a maximal compact subgroup
of $\mathsf{F}$. On the other hand, recall that from the Iwasawa and Levi
decompositions, any linear Lie group decomposes as
$\mathsf{F}=\mathsf{K}\mathsf{S}$ with $\mathsf{K}$ maximal compact and
$\mathsf{S}$ solvable. By the conjugacy of maximal compact subgroups of a Lie
group, after changing $p$ we may assume that $\mathsf{F}_{p}=\mathsf{K}$, and
it follows that $\mathsf{S}$ acts transitively and isometrically on
$(M^{n},g)$.

Finally, let $M_{2}:=M/\Gamma$ be as in the statement, and let
$\mathsf{F}_{2}$ be the identity component of its full isometry group. The
universal cover $\tilde{\mathsf{F}}_{2}$ acts transitively and isometrically
on $M$. By the argument in the previous paragraph, there is a transitive
solvable Lie subgroup $\mathsf{S}\leq\tilde{\mathsf{F}}_{2}$, and it is clear
that $\mathsf{S}_{2}:=\mathsf{S}/(\Gamma\cap\mathsf{S})$ acts transitively on
$M_{2}$. Hence, $M_{2}$ is also an Einstein solvmanifold. Since these are
simply-connected by [Jab15a], it follows that $\Gamma$ is trivial. ∎

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