$\lambda=\frac{1}{\int_{M}\frac{\Omega^{n}\wedge\bar{\Omega}^{n}}{(n!)^{2}}}\int_{M}\lambda\frac{\Omega^{n}\wedge\bar{\Omega}^{n}}{(n!)^{2}}=\frac{1}{\int_{M}\frac{\Omega^{n}\wedge\bar{\Omega}^{n}}{(n!)^{2}}}\int_{M}\partial_{J}\alpha\wedge\frac{\Omega^{n-1}\wedge\bar{\Omega}^{n}}{(n!)^{2}}=\frac{1}{n}\frac{1}{\int_{M}\frac{\Omega^{n}\wedge\bar{\Omega}^{n}}{(n!)^{2}}}\int_{M}|\alpha|^{2}\frac{\Omega^{n}\wedge\bar{\Omega}^{n}}{(n!)^{2}}$

This is a striking difference with the usual behaviour of the Einstein factor
in other settings. We summarize this as follows.

######  Proposition 8.5.

Let $(M,\mathsf{H})$ be a compact hypercomplex manifold. Then, a Chern-
Einstein hyperhermitian metric with non-identically zero Einstein factor has
constant Chern scalar curvature if and only if it is HKT. When this is the
case, the Chern scalar curvature is necessarily non-negative.

It is fairly easy to come up with examples of compact hyperhermitian Einstein
manifolds with $\lambda=0$ but not admitting compatible HKT metrics. For
instance, any hypercomplex nilmanifold $M$ with a non-abelian hypercomplex
structure $\mathsf{H}$, by [15, Theorem 4.6], does not have any compatible HKT
metric. However, since hypercomplex nilmanifolds are
$\mathrm{SL}(n,\mathbb{H})$, as in the proof of Lemma 5.3, any left-invariant
hyperhermitian metric $\Omega$ satisfies $\alpha=0$. In Subsection 9.8, we
will provide examples of HKT-Einstein metrics with strictly positive Einstein
factor.

Within our framework, the quaternionic Calabi conjecture formulated by Alesker
and Verbitsky in [4, Conjecture 1.5] can be phrased as follows: on a compact
HKT manifold $(M,\mathsf{H},\Omega)$, for any representative $\Psi\in
c_{1}^{\mathrm{qBC}}(M,\mathsf{H})$ it should always be possible to find
another HKT metric $\Omega_{\varphi}:=\Omega+\partial\partial_{J}\varphi$, for
some $\varphi\in C^{\infty}(M,\mathbb{R})$, such that
$\partial_{J}\alpha_{\Omega_{\varphi}}=\Psi$. If this turns out to be true,
when $c_{1}^{\mathrm{qBC}}(M,\mathsf{H})=0$ we would be able to find HKT-
Einstein metrics with vanishing Einstein factor (cf. [78]), i.e. that are
balanced and Chern-Ricci flat.
In a similar fashion, one is led to speculate on the existence of HKT-Einstein
metrics on compact HKT manifolds with positive first quaternionic Bott-Chern
class. More precisely, we wonder:

######  Question 8.6.

Let $(M^{n},\mathsf{H},\Omega)$ be a compact HKT manifold such that
$c_{1}^{\mathrm{qBC}}(M,\mathsf{H})>0$. Does it always exist $\varphi\in
C^{\infty}(M,\mathbb{R})$ such that $\Omega_{\varphi}$ is a HKT-Einstein
metric?

Proceeding similarly to the Kähler case, the question turns out to be
equivalent to the solvability of the following quaternionic Monge-Ampère
equation:

$\left(\Omega+\partial\partial_{J}\varphi\right)^{n}=\mathrm{e}^{f-\varphi}\Omega^{n}\,,\qquad\Omega+\partial\partial_{J}\varphi>0\,,$

where $\varphi\in C^{\infty}(M,\mathbb{R})$ is the unknown and $f\in
C^{\infty}(M,\mathbb{R})$ is the datum. Unfortunately, this case is the
quaternionic analogue of the Fano case in the Kähler setting, hence, when
approaching the problem with the classical method of continuity, the same
difficulties arise. Furthermore, it is natural to expect that certain
obstructions may emerge in our context, similar to those found by Futaki [35]
and Matsushima [61]. It is extremely likely that more sophisticated tools are
necessary to approach this problem.
As a final remark, since in the compact HKT case the Einstein factor is non-
negative, the quaternionic Monge-Ampère equation corresponding to negative
Chern scalar curvature can never be solved.

##  9\. Examples and constructions

In this Section, we collect several examples and two interesting
constructions, one inspired by Arroyo and Nicolini [11, Section 5], the other
due to Barberis and Fino [16]. The purpose of the next 4 Subsections is to
exhibit examples admitting one type of metrics listed in Definition 3.15 but
none of the one immediately stronger. We shall provide several explicit
examples with different dimensions, because these will serve the purpose of
“building blocks” in Subsection 9.6 to provide examples in any quaternionic
dimension for which it is possible. In Subsection 9.8 we study the examples
constructed by Joyce [55] and Spindel, Sevrin, Troost, Van Proeyen [71]
showing that they are HKT-Einstein with positive Einstein constant. Finally,
Subsection 9.9 is devoted to present some non-compact HKT-Einstein manifolds,
one of which has negative Einstein factor.

In the next Subsections, we will work on Lie algebras endowed with a
hypercomplex structure. So, we will always fix a coframe
$\\{e^{1},\ldots,e^{4n}\\}$ for the Lie algebra and, unless otherwise stated,
use the following hypercomplex structure:

$Ie^{4k-3}=-e^{4k-2}\,,\quad Ie^{4k-1}=-e^{4k}\,,\quad
Je^{4k-3}=-e^{4k-1}\,,\quad Je^{4k-2}=e^{4k}\,,\quad\text{for all
}k=1,\ldots,n\,.$

In view of this, the $(1,0)$-coframe $\\{\zeta^{1},\ldots,\zeta^{2n}\\}$ with
respect to $I$ will be $\zeta^{j}=e^{2j-1}+\sqrt{-1}e^{2j}$, for all
$j=1,\ldots,2n$, and $J$ will act on it as $J\zeta^{2i-1}=-\bar{\zeta}^{2i}$,
for all $i=1,\ldots,n$.

###  9.1. Quaternionic balanced manifolds without HKT metrics.

An example of a quaternionic balanced manifold not admitting any HKT metric
can be found, for instance, in [28]. Here we present three other such
examples. Note that in quaternionic dimension $2$ the quaternionic balanced
condition coincides with the HKT condition, hence we need to work in dimension
at least $3$. The examples we provide are, $3,4$ and $5$-dimensional
respectively.

######  Example 9.1.

We consider the nilpotent Lie algebra in [59, Example 3] with structure
constants:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots,8\,,$ $de^{9}=e^{1}\wedge
e^{5}\,,\qquad de^{10}=e^{1}\wedge e^{6}\,,\qquad de^{11}=e^{1}\wedge
e^{7}\,,\qquad de^{12}=e^{1}\wedge e^{8}\,.$

The $(1,0)$-basis $(\zeta^{1},\dots,\zeta^{6})$ with respect to $I$ satisfies:

$d\zeta^{i}=0\,,\quad\mbox{for all }i=1,\ldots,4\,,\qquad
d\zeta^{5}=\frac{1}{2}(\zeta^{1}\wedge\zeta^{3}+\bar{\zeta}^{1}\wedge\zeta^{3})\,,\qquad
d\zeta^{6}=\frac{1}{2}(\zeta^{1}\wedge\zeta^{4}+\bar{\zeta}^{1}\wedge\zeta^{4})\,.$

Since the hypercomplex structure is not abelian, by [15, Theorem 4.6] the
nilmanifold $N$ does not admit any HKT metric. However, we shall show that $N$
admits an invariant quaternionic balanced metric. Indeed, if we consider the
hyperhermitian metric that makes the coframe unitary, i.e.

$\Omega=\zeta^{1}\wedge\zeta^{2}+\zeta^{3}\wedge\zeta^{4}+\zeta^{5}\wedge\zeta^{6}\,;$

such a form is q-real and q-positive, thus it induces an invariant
hyperhermitian metric on $N$. It is easy to show that $\partial\Omega^{2}=0$
meaning that the metric is quaternionic balanced.
Observe that this example does not satisfy the $\partial\partial_{J}$-Lemma
because of [60, Theorem 5] and the observation of Lejmi and Weber [59, Example
3] that it is not $C^{\infty}$-pure nor $C^{\infty}$-full.

######  Example 9.2.

Consider the nilpotent Lie algebra with structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots,12\,,$ $\displaystyle de^{13}$
$\displaystyle=e^{1}\wedge e^{5}+e^{1}\wedge e^{9}\,,$ $\displaystyle de^{15}$
$\displaystyle=e^{1}\wedge e^{7}+e^{1}\wedge e^{11}\,,$ $\displaystyle
de^{14}$ $\displaystyle=e^{1}\wedge e^{6}+e^{1}\wedge e^{10}\,,$
$\displaystyle de^{16}$ $\displaystyle=e^{1}\wedge e^{8}+e^{1}\wedge
e^{12}\,.$

It is easy to show that the hypercomplex structure is non-abelian. The
$(1,0)$-basis $(\zeta^{1},\dots,\zeta^{8})$ with respect to $I$ satisfies:

$d\zeta^{i}=0\,,\quad\mbox{for all }i=1,\ldots,6\,,$ $\displaystyle
d\zeta^{7}$
$\displaystyle=\frac{1}{2}\left(\zeta^{1}\wedge\zeta^{3}+\bar{\zeta}^{1}\wedge\zeta^{3}+\zeta^{1}\wedge\zeta^{5}+\bar{\zeta}^{1}\wedge\zeta^{5}\right)\,,$
$\displaystyle d\zeta^{8}$
$\displaystyle=\frac{1}{2}\left(\zeta^{1}\wedge\zeta^{4}+\bar{\zeta}^{1}\wedge\zeta^{4}+\zeta^{1}\wedge\zeta^{6}+\bar{\zeta}^{1}\wedge\zeta^{6}\right)\,.$

Again, the metric that makes the coframe unitary is quaternionic balanced.

######  Example 9.3.

Consider the nilpotent Lie algebra with structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots,16\,,$ $\displaystyle de^{17}$
$\displaystyle=e^{1}\wedge e^{5}+e^{9}\wedge e^{13}\,,$ $\displaystyle
de^{19}$ $\displaystyle=e^{1}\wedge e^{7}+e^{9}\wedge e^{15}\,,$
$\displaystyle de^{18}$ $\displaystyle=e^{1}\wedge e^{6}+e^{9}\wedge
e^{14}\,,$ $\displaystyle de^{20}$ $\displaystyle=e^{1}\wedge
e^{8}+e^{9}\wedge e^{16}\,.$

Again, the hypercomplex structure is non-abelian. The $(1,0)$-basis
$(\zeta^{1},\dots,\zeta^{10})$ with respect to $I$ satisfies:

$d\zeta^{i}=0\,,\quad\mbox{for all }i=1,\ldots,8\,,$ $\displaystyle
d\zeta^{9}$
$\displaystyle=\frac{1}{2}\left(\zeta^{1}\wedge\zeta^{3}+\bar{\zeta}^{1}\wedge\zeta^{3}+\zeta^{5}\wedge\zeta^{7}+\bar{\zeta}^{5}\wedge\zeta^{7}\right)\,,$
$\displaystyle d\zeta^{10}$
$\displaystyle=\frac{1}{2}\left(\zeta^{1}\wedge\zeta^{4}+\bar{\zeta}^{1}\wedge\zeta^{4}+\zeta^{5}\wedge\zeta^{8}+\bar{\zeta}^{5}\wedge\zeta^{8}\right)\,.$

Also in this case, the hyperhermitian metric with respect to which the given
frame is unitary satisfies the quaternionic balanced condition.

###  9.2. Quaternionic strongly Gauduchon manifolds without quaternionic
balanced metrics.

Here we exhibit examples of compact quaternionic strongly Gauduchon
nilmanifolds on which no quaternionic balanced metric exists. Notice that the
minimum quaternionic dimension for which it is possible to provide such an
example as a nilmanifold is $3$, indeed all hypercomplex nilmanifolds are
$\mathrm{SL}(n,\mathbb{H})$, see [15, Corollary 3.3]. Therefore, in real
dimension $8$, they admit a quaternionic strongly Gauduchon metric if and only
if they admit a HKT metric by [59, Theorem 10.1].

######  Example 9.4.

Consider the nilpotent Lie algebra with structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots,8\,,$ $de^{9}=e^{1}\wedge
e^{3}\,,\qquad de^{10}=e^{1}\wedge e^{4}+e^{7}\wedge e^{8}\,,\qquad
de^{11}=e^{5}\wedge e^{7}\,,\qquad de^{12}=-e^{3}\wedge e^{4}+e^{5}\wedge
e^{8}\,.$

The structure equations can be rewritten in terms of the $(1,0)$-coframe
$(\zeta^{1},\dots,\zeta^{6})$ as follows:

$d\zeta^{i}=0\,,\qquad\text{for all }i=1,\dots,4\,,$
$d\zeta^{5}=\frac{1}{2}(\zeta^{1}\wedge\zeta^{2}+\bar{\zeta}^{1}\wedge\zeta^{2}-\zeta^{4}\wedge\bar{\zeta}^{4})\,,\qquad
d\zeta^{6}=\frac{1}{2}(\zeta^{3}\wedge\zeta^{4}+\bar{\zeta}^{3}\wedge\zeta^{4}+\zeta^{2}\wedge\bar{\zeta}^{2})\,.$

In particular, we have that:

$\displaystyle\partial\zeta^{1}=\partial\zeta^{2}=\partial\zeta^{3}=\partial\zeta^{4}=0\,,$
$\displaystyle\partial\zeta^{5}$
$\displaystyle=\frac{1}{2}\zeta^{1}\wedge\zeta^{2}\,,$
$\displaystyle\partial\zeta^{6}$
$\displaystyle=\frac{1}{2}\zeta^{3}\wedge\zeta^{4}\,,$
$\displaystyle\partial_{J}\zeta^{1}=\partial_{J}\zeta^{2}=\partial_{J}\zeta^{3}=\partial_{J}\zeta^{4}=0\,,$
$\displaystyle\partial_{J}\zeta^{5}$
$\displaystyle=-\frac{1}{2}\zeta^{3}\wedge\zeta^{4}\,,$
$\displaystyle\partial_{J}\zeta^{6}$
$\displaystyle=\frac{1}{2}\zeta^{1}\wedge\zeta^{2}\,.$

The invariant hyperhermitian metric $\Omega$ that renders the coframe unitary
is quaternionic strongly Gauduchon as

$\displaystyle\partial\Omega^{2}$
$\displaystyle=2\partial_{J}(\zeta^{3}\wedge\zeta^{4}\wedge\zeta^{5}\wedge\zeta^{6}-\zeta^{1}\wedge\zeta^{2}\wedge\zeta^{5}\wedge\zeta^{6})\,.$

Since the structure constant are rational, the Lie algebra admits a lattice
and the metric above descends to the compact quotient $N$. On the other hand,
we claim that there are no quaternionic balanced metrics on the nilmanifold
$N$. Indeed, the form $\zeta^{1}\wedge\zeta^{2}=2\partial\zeta^{5}$ induces a
q-positive, $\partial$-exact, $(2,2n)$-current $T$ given by integration

$T(\gamma):=\int_{N}\zeta^{1}\wedge\zeta^{2}\wedge\gamma\wedge\bar{\Theta}\,,\qquad\mbox{for
every }\gamma\in\Lambda_{I}^{2n-2,0}N\,,$

where $\Theta$ is any holomorphic volume form in $\Lambda^{2n,0}_{I}N$. We
conclude applying Proposition 6.3.

######  Example 9.5.

Consider the nilpotent Lie algebra with structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots 12\,,$ $\displaystyle de^{13}$
$\displaystyle=e^{1}\wedge e^{3}\,,$ $\displaystyle de^{15}$
$\displaystyle=e^{5}\wedge e^{7}+e^{9}\wedge e^{11}\,,$ $\displaystyle
de^{14}$ $\displaystyle=e^{1}\wedge e^{4}+e^{7}\wedge e^{8}+e^{11}\wedge
e^{12}\,,$ $\displaystyle de^{16}$ $\displaystyle=-e^{3}\wedge
e^{4}+e^{5}\wedge e^{8}+e^{9}\wedge e^{12}\,.$

The complex structure equations are:

$d\zeta^{i}=0\,,\qquad\text{for all }i=1,\dots,6\,,$
$d\zeta^{7}=\frac{1}{2}(\zeta^{1}\wedge\zeta^{2}+\bar{\zeta}^{1}\wedge\zeta^{2}-\zeta^{4}\wedge\bar{\zeta}^{4}-\zeta^{6}\wedge\bar{\zeta}^{6})\,,\qquad
d\zeta^{8}=\frac{1}{2}(\zeta^{3}\wedge\zeta^{4}+\bar{\zeta}^{3}\wedge\zeta^{4}+\zeta^{5}\wedge\zeta^{6}+\bar{\zeta}^{5}\wedge\zeta^{6}+\zeta^{2}\wedge\bar{\zeta}^{2})\,.$

The invariant hyperhermitian metric that makes such coframe unitary is
quaternionic strongly Gauduchon. On the other hand, as in the previous
example, it is easy to see that there are no quaternionic balanced metrics.

######  Example 9.6.

Consider the nilpotent Lie algebra with structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots 16\,,$ $\displaystyle de^{17}=$
$\displaystyle\,e^{1}\wedge e^{3}+e^{5}\wedge e^{7}\,,\qquad\quad
de^{18}=e^{1}\wedge e^{4}+e^{5}\wedge e^{8}+e^{11}\wedge e^{12}+e^{15}\wedge
e^{16}\,,$ $\displaystyle de^{19}=$ $\displaystyle\,e^{9}\wedge
e^{11}+e^{13}\wedge e^{15}\,,\quad\,\,de^{20}=-e^{3}\wedge e^{4}-e^{7}\wedge
e^{8}+e^{9}\wedge e^{12}+e^{13}\wedge e^{16}\,.$

The structure equations can be rewritten in terms of a $(1,0)$-coframe with
respect to $I$ as follows:

$d\zeta^{i}=0\,,\qquad\text{for all }i=1,\dots,8\,,$
$\begin{split}d\zeta^{9}&=\frac{1}{2}(\zeta^{1}\wedge\zeta^{2}+\bar{\zeta}^{1}\wedge\zeta^{2}+\zeta^{3}\wedge\zeta^{4}+\bar{\zeta}^{3}\wedge\zeta^{4}-\zeta^{6}\wedge\bar{\zeta}^{6}-\zeta^{8}\wedge\bar{\zeta}^{8})\,,\\\
d\zeta^{10}&=\frac{1}{2}(\zeta^{5}\wedge\zeta^{6}+\bar{\zeta}^{5}\wedge\zeta^{6}+\zeta^{7}\wedge\zeta^{8}+\bar{\zeta}^{7}\wedge\zeta^{8}+\zeta^{2}\wedge\bar{\zeta}^{2}+\zeta^{4}\wedge\bar{\zeta}^{4})\,.\end{split}$

The invariant hyperhermitian metric that makes the $(1,0)$-coframe unitary is
quaternionic strongly Gauduchon and the same argument of Example 9.4 shows
that there exists no quaternionic balanced metric.

######  Remark 9.7.

Both the quaternionic balanced condition and the strongly Gauduchon condition
are not preserved under small deformations of the hypercomplex structure.
Indeed, in [27], the authors provide an example of an $8$-dimensional
nilmanifold on which the existence of HKT metrics is not preserved along a
deformation of the hypercomplex structure. However, HKT metrics in
quaternionic dimension 2 coincide with quaternionic balanced ones while, as
remarked above, on hypercomplex nilmanifolds of real dimension 8 the existence
of HKT metrics is equivalent to the existence of quaternionic strongly
Gauduchon metrics.
It would be interesting to know if the same phenomenon occurs for the
quaternionic Gauduchon condition.

###  9.3. Quaternionic Gauduchon manifolds without quaternionic strongly
Gauduchon metrics.

We already mentioned that any $8$-dimensional hypercomplex nilmanifold with
non-abelian hypercomplex structure admits no quaternionic strongly Gauduchon
metric. On the other hand it admits a quaternionic Gauduchon metric by Lemma
5.2. Notice that, as a consequence of Lemma 5.7, no example in this class can
satisfy the $\partial\partial_{J}$-Lemma.
Here we present an example in each dimension.

######  Example 9.8.

For $n\geq 2$, consider the $4n$-dimensional nilpotent Lie algebra with
structure equations:

$de^{i}=0\,,\qquad\text{for all }i=1,\dots,4n-3\,,$ $\displaystyle
de^{4n-2}=\sum_{k=1}^{n-1}e^{4k-3}\wedge e^{4k-2}\,,\qquad
de^{4n-1}=\sum_{k=1}^{n-1}e^{4k-3}\wedge e^{4k-1}\,,\qquad
de^{4n}=\sum_{k=1}^{n-1}e^{4k-3}\wedge e^{4k}\,.$

The $(1,0)$-basis $(\zeta^{1},\dots,\zeta^{2n})$ with respect to $I$
satisfies:

$d\zeta^{i}=0\,,\quad\mbox{for all }i=1,\ldots,{2n-2}\,,$
$d\zeta^{2n-1}=-\frac{1}{2}\sum_{k=1}^{n-1}\zeta^{2k-1}\wedge\bar{\zeta}^{2k-1}\,,\qquad
d\zeta^{2n}=\frac{1}{2}\sum_{k=1}^{n-1}(\zeta^{2k-1}\wedge\zeta^{2k}+\bar{\zeta}^{2k-1}\wedge\zeta^{2k})\,.$

We shall check that the generic invariant hyperhermitian metric

$\Omega=A_{ij}\zeta^{i}\wedge\zeta^{j}\,,\qquad A_{ij}\in\mathbb{C}\,,$

is not quaternionic strongly Gauduchon. We compute

$\begin{split}\partial\Omega^{n-1}&=\partial\left(A_{ij}\zeta^{i}\wedge\zeta^{j}\right)^{n-1}\\\
&=(n-1)!\,\partial\left(\mathrm{pf}(A(i,j))\zeta^{1}\wedge\dots\wedge\hat{\zeta}^{i}\wedge\dots\wedge\hat{\zeta}^{j}\wedge\dots\wedge\zeta^{2n}\right)\\\
&=-(n-1)!\sum_{1\leq i<j\leq
2n-1}\mathrm{pf}(A(i,j))\zeta^{1}\wedge\dots\wedge\hat{\zeta}^{i}\wedge\dots\wedge\hat{\zeta}^{j}\wedge\dots\wedge\zeta^{2n-1}\wedge\partial\zeta^{2n}\\\
&=-\frac{(n-1)!}{2}\left(\sum_{k=1}^{n-1}\mathrm{pf}(A(2k-1,2k))\right)\zeta^{1}\wedge\dots\wedge\zeta^{2n-1}\,,\end{split}$

where the hat symbols means that the term is missing and $A(i,j)$ denotes the
matrix obtained from the skew-symmetric matrix $A=(A_{rs})$ removing the $i$th
and $j$th rows and columns. Then $\partial\Omega^{n-1}$ is non-zero, because
by q-positivity of $\Omega$ we must have $\mathrm{pf}(A(2k-1,2k))>0$ for all
$k=1,\dots,n$. Moreover, it is not $\partial_{J}$-exact since

$\partial_{J}\zeta^{i}=0\,,\qquad\text{for all }i\neq
2n-1\,,\qquad\partial_{J}\zeta^{2n-1}=-\frac{1}{2}\sum_{k=1}^{n-1}\zeta^{2k-1}\wedge\zeta^{2k}\,.$

On the other hand, these are all nilpotent Lie algebras with rational
structure constants, therefore the corresponding nilmanifolds are
$\mathrm{SL}(n,\mathbb{H})$ and the existence of a quaternionic Gauduchon
metric is guaranteed by Lemma 5.2.

###  9.4. A hypercomplex manifold without quaternionic Gauduchon metrics.

The example we describe appeared in [8, Example 6.3] to provide a negative
answer to a question posed by Verbitsky. It is constructed as a compact
quotient $S=\Gamma\backslash G$ of a solvable Lie group by a lattice $\Gamma$
equipped with an invariant hypercomplex structure $\mathsf{H}$. Andrada and
Tolcachier show that the canonical bundle of $(S,I)$ is holomorphically
trivial but that of $(S,J)$ is not, indeed, they even show that
$c_{1}^{\mathrm{BC}}(S,J)\neq 0$. On the other hand, the holomorphic
triviality of $K(S,I)$ implies $c_{1}^{\mathrm{qBC}}(S,\mathsf{H})=0$ (cf.
Remark 4.5). We then conclude that a compatible quaternionic Gauduchon metric
cannot exists from Corollary 5.5.
For the sake of completeness we also show this in another way. Thanks to the
symmetrization process, it is enough to show that $(S,\mathsf{H})$ does not
admit invariant quaternionic Gauduchon metrics. We may easily conclude by
means of Theorem 5.9. Indeed, $(S,\mathsf{H})$ cannot satisfy condition (24)
in any conformal class (although it does satisfy condition (27)). For any
invariant Gauduchon hyperhermitian metric $\Omega_{\mathrm{G}}$, we thus must
have $s^{\mathrm{Ch}}(\Omega_{\mathrm{G}})=0$ by the triviality of $K(S,I)$
and so if (25) were to hold, it would imply $\alpha_{\Omega_{\mathrm{G}}}=0$
which contradicts the fact that $(S,\mathsf{H})$ is not
$\mathrm{SL}(n,\mathbb{H})$.

######  Remark 9.9.

This example, together with Proposition 4.3 also shows that $\kappa(S,I)=0$
while $\kappa(S,J)=-\infty$, so, in general, it may happen that different
complex structures in the same hypercomplex structure yield different Kodaira
dimensions.
From this, it also follows that the complex deformation

$I_{t}=\cos(\pi t)I+\sin(\pi t)J\,,\qquad t\in[0,1/2]\,,$

does not preserve the Kodaira dimension, a fact that is in contrast with the
projective case as shown by Siu [70]. We are grateful to G. Grantcharov for
this observation.

###  9.5. Dependence of the quaternionic strongly Gauduchon condition from the
complex structure

Example 9.4 can be used to show that the quaternionic strongly Gauduchon
condition depends on the choice of the pair of anticommuting complex
structures in the same hypercomplex structure. Indeed, let us show that with
respect to the pair $(J,I)$ there are no quaternionic strongly Gauduchon
metrics. Consider the $(1,0)$-coframe with respect to $J$ given by
$w^{2k-1}=e^{4k-3}+\sqrt{-1}e^{4k-1}$ and $w^{2k}=e^{4k-2}-\sqrt{-1}e^{4k}$
for $k=1,2,3$, and let $d=\partial+\bar{\partial}$ be the splitting of the
exterior differential with respect to $J$. Then

$\displaystyle\partial w^{i}$ $\displaystyle=0\,,\qquad\text{for all
}i=1,\dots,5\,,$ $\displaystyle\partial w^{6}$
$\displaystyle=\frac{1}{2}(\sqrt{-1}w^{1}\wedge w^{2}+w^{3}\wedge w^{4})\,,$
$\displaystyle I^{-1}\bar{\partial}Iw^{i}$ $\displaystyle=0\,,\qquad\text{for
all }i\neq 5\,,$ $\displaystyle I^{-1}\bar{\partial}Iw^{5}$
$\displaystyle=\frac{1}{2}(\sqrt{-1}w^{1}\wedge w^{2}-w^{3}\wedge w^{4})\,.$

It is now easy to check that the generic invariant hyperhermitian metric
$\Omega$ cannot be quaternionic strongly Gauduchon, since $\partial\Omega^{2}$
is proportional to $w^{1}\wedge w^{2}\wedge w^{3}\wedge w^{4}\wedge w^{5}$
which is never $I^{-1}\bar{\partial}I$-exact.

###  9.6. Arroyo-Nicolini’s construction

In [11, Section 5], Arroyo and Nicolini give a procedure to construct $2$-step
nilpotent SKT Lie algebras starting from previous ones. Here we adapt their
argument to build new hyperhermitian nilpotent Lie algebras.

Let $(\mathfrak{g}_{1},[\cdot,\cdot]_{1},\mathsf{H}_{1})$ and
$(\mathfrak{g}_{2},[\cdot,\cdot]_{2},\mathsf{H}_{2})$ be two nilpotent Lie
algebras equipped with a hypercomplex structure generated by $(I_{1},J_{1})$
and $(I_{2},J_{2})$ respectively. Then, provided
$\dim[\mathfrak{g}_{i},\mathfrak{g}_{i}]<\dim Z(\mathfrak{g}_{i})$, we can
choose $e_{i}\in
Z(\mathfrak{g}_{i})\setminus[\mathfrak{g}_{i},\mathfrak{g}_{i}]$, where
$Z(\mathfrak{g}_{i})$ is the center of $\mathfrak{g}_{i}$. Then, we can define
the new Lie algebra
$\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\oplus\langle
X,Y,Z,W\rangle$ with non-zero Lie brackets:

$[\cdot,\cdot]|_{\mathfrak{g}_{i}\times\mathfrak{g}_{i}}:=[\cdot,\cdot]_{i}\,,\qquad[X,Y]=-[Z,W]:=e_{1}+e_{2}$

and hypercomplex structure $\mathsf{H}$ determined by:

$\displaystyle I|_{\mathfrak{g}_{i}}=I_{i}\,,$ $\displaystyle IX=Y\,,$
$\displaystyle IZ=W\,,$ $\displaystyle J|_{\mathfrak{g}_{i}}=J_{i}\,,$
$\displaystyle JX=Z\,,$ $\displaystyle JY=-W\,.$

Observe that the nilpotency step of $\mathfrak{g}$ is the maximum of the
nilpotency steps of $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$.

######  Theorem 9.10.

The hypercomplex Lie algebra $(\mathfrak{g},\mathsf{H})$ admits HKT (resp.
quaternionic balanced, quaternionic strongly Gauduchon) if and only if
$(\mathfrak{g}_{1},\mathsf{H}_{1})$ and $(\mathfrak{g}_{2},\mathsf{H}_{2})$
do.

###### Proof.

Clearly, $\mathsf{H}$ is abelian if and only if both $\mathsf{H}_{1}$ and
$\mathsf{H}_{2}$ are, therefore, thanks to [15, Theorem 4.6], the Theorem is
true for HKT metrics. We can be more precise: $\Omega_{i}$ is a HKT metric on
$\mathfrak{g}_{i}$ for $i=1,2$ if and only if

(38) $\Omega:=\Omega_{1}+\Omega_{2}+\zeta^{1}\wedge\zeta^{2}$

is HKT, where $(\zeta^{1},\zeta^{2})$ is the $(1,0)$ coframe with respect to
$I$ dual to $(X-\sqrt{-1}Y,Z-\sqrt{-1}W)$. This is because
$\partial(\zeta^{1}\wedge\zeta^{2})=0$ and the only modification that occurs
in the structure constants of $\mathfrak{g}_{i}$ is that now
$de^{i}=-\sqrt{-1}\zeta^{1}\wedge\bar{\zeta}^{1}+\sqrt{-1}\zeta^{2}\wedge\bar{\zeta}^{2}$,
which before was zero. In particular, $\partial e^{i}=0$ is unaltered.

The same phenomenon occurs for the other kinds of special metrics. Let
$\Omega_{i}$ be a hyperhermitian metric on $\mathfrak{g}_{i}$ and define
$\Omega$ on $\mathfrak{g}$ as in (38). Let $n_{i}$ be the quaternionic
dimension of $\mathfrak{g}_{i}$ and let $n=n_{1}+n_{2}+1$ be that of
$\mathfrak{g}$. Since

$\Omega^{n-1}=\binom{n_{1}}{n}\Omega_{1}^{n_{1}}\wedge\Omega_{2}^{n_{2}}+n_{2}\binom{n_{1}}{n}\Omega_{1}^{n_{1}}\wedge\Omega_{2}^{n_{2}-1}\wedge\zeta^{1}\wedge\zeta^{2}+n_{1}\binom{n_{2}}{n}\Omega_{1}^{n_{1}-1}\wedge\Omega_{2}^{n_{2}}\wedge\zeta^{1}\wedge\zeta^{2}\,,$

we see that

$\partial\Omega^{n-1}=n_{2}\binom{n_{1}}{n}\Omega_{1}^{n_{1}}\wedge\partial\Omega_{2}^{n_{2}-1}\wedge\zeta^{1}\wedge\zeta^{2}+n_{1}\binom{n_{2}}{n}\partial\Omega_{1}^{n_{1}-1}\wedge\Omega_{2}^{n_{2}}\wedge\zeta^{1}\wedge\zeta^{2}\,.$

Therefore $\Omega$ is quaternionic balanced or quaternionic strongly Gauduchon
if and only if $\Omega_{1}$ and $\Omega_{2}$ are. ∎

Assume that $(\mathfrak{g}_{i},\mathsf{H}_{i})$ are equipped with metrics
$g_{i}$ that makes them _indecomposable_ , that amounts to say that they
cannot be written as a $g_{i}$-orthogonal sum of $\mathsf{H}_{i}$-invariant
ideals for $i=1,2$. If we choose $e_{i}\in
Z(\mathfrak{g}_{i})\cap[\mathfrak{g}_{i},\mathfrak{g}_{i}]^{\perp_{g_{i}}}$,
the resulting hypercomplex Lie algebra $(\mathfrak{g},\mathsf{H})$ is then
also indecomposable, a fact that can be seen with the same argument of Arroyo
and Nicolini [11, Subsection 5.1]. Furthermore, it is not hard to check that
$\dim[\mathfrak{g},\mathfrak{g}]<\dim Z(\mathfrak{g})$ so this Lie algebra can
be used again in the process to obtain higher-dimensional indecomposable
examples.

Note that, if $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ have rational
structure constants the same is true for $\mathfrak{g}$, therefore we can find
lattices. This discussion, together with the observation that the examples
produced in Subsections 9.1, 9.2, 9.3 are indecomposable nilpotent Lie
algebras with rational structure constants, shows that we can use them as
building blocks to construct compact nilmanifolds carrying a certain type of
special metric but no metrics of the class immediately stronger among those we
have studied. This can be achieved in any quaternionic dimension except for
those low dimensions for which we explained that this cannot happen.

###  9.7. Barberis-Fino’s construction.

In this Subsection, we study the behaviour of the metric conditions under a
construction due to Barberis and Fino [16]. The idea is to take a hypercomplex
Lie algebra and build a new one via a quaternionic representation.

Let $\mathfrak{g}$ be a $4n$-dimensional Lie algebra and
$\rho\colon\mathfrak{g}\to\mathfrak{gl}(k,\mathbb{H})$ a Lie algebra
homomorphism. Define on
$T_{\rho}\,\mathfrak{g}:=\mathfrak{g}\ltimes_{\rho}\mathbb{H}^{k}$ the Lie
bracket

$[(X,U),(Y,V)]:=([X,Y],\rho_{X}(V)-\rho_{Y}(U))$

for every $X,Y\in\mathfrak{g}$ and $U,V\in\mathbb{H}^{k}$. If $\mathfrak{g}$
is endowed with a hypercomplex structure $\mathsf{H}$

$\tilde{L}(X,U):=(LX,lU)\,,$

defines a hypercomplex structure $\tilde{\mathsf{H}}$ on
$T_{\rho}\,\mathfrak{g}$, where if $L=aI+bJ+cK$ then $l=ai+bj+ck$, being
$i,j,k$ the quaternion units in $\mathbb{H}$. Finally, if $g$ is a
hyperhermitian metric on $(\mathfrak{g},\mathsf{H})$, the metric $\tilde{g}$
induced by $g$ and the natural metric on $\mathbb{H}^{k}$ in such a way that
$\mathfrak{g}$ is orthogonal to $\mathbb{H}^{k}$ is hyperhermitian on
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}})$. Denote by $\Omega$ and
$\tilde{\Omega}$ the respective $(2,0)$-forms with respect to $I$ and
$\tilde{I}$.

######  Theorem 9.11 ([16]).

Let $(\mathfrak{g},\mathsf{H},g)$ and
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}},\tilde{g})$ be as above and denote
$p\colon T_{\rho}\,\mathfrak{g}\to\mathfrak{g}$ the orthogonal projection.
Then,

  1. (1)

$\tilde{\nabla}^{{\rm Ob}}\tilde{g}=0$ if and only if $\nabla^{{\rm Ob}}g=0$
and $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$, where $\tilde{\nabla}^{{\rm
Ob}}$ and $\nabla^{{\rm Ob}}$ are the Obata connections of
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}},\tilde{g})$ and
$(\mathfrak{g},\mathsf{H},g)$ respectively;

  2. (2)

$\tilde{\Omega}$ is HKT if and only if $\Omega$ is HKT and
$\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$;

  3. (3)

if $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$ then $T^{\tilde{L}}=T^{L}\circ
p$ and $dT^{\tilde{L}}=dT^{L}\circ p$, where $T^{\tilde{L}}$ and $T^{L}$ are
the torsions of the Bismut connections of $(\tilde{g},\tilde{L})$ and $(g,L)$
respectively, for $L\in\mathsf{H}$;

  4. (4)

if $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$ then
$\theta_{\tilde{\Omega}}=\theta_{\Omega}\circ p$.

In particular, when $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$,
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}},\tilde{\Omega})$ is strong HKT
(resp. weak HKT, hyperkähler, balanced) if and only if
$(\mathfrak{g},\mathsf{H},\Omega)$ is.

We can refine this result by proving the following result.

######  Theorem 9.12.

Let $(\mathfrak{g},\mathsf{H},\Omega)$ and
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}},\tilde{\Omega})$ be as above,
denote $p\colon T_{\rho}\,\mathfrak{g}\to\mathfrak{g}$ the orthogonal
projection and assume $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$. Then,
$\alpha_{\tilde{\Omega}}=\alpha_{\Omega}\circ p$,
$\beta_{\tilde{\Omega}}=\beta_{\Omega}\circ p$,
$\mathrm{Ric}^{\mathrm{Ch}}_{\omega_{\tilde{L}}}=\mathrm{Ric}^{\mathrm{Ch}}_{\omega_{L}}\circ
p$ and
$\mathrm{Ric}^{\mathrm{Bis}}_{\omega_{\tilde{L}}}=\mathrm{Ric}^{\mathrm{Bis}}_{\omega_{L}}\circ
p$. In particular,
$(T_{\rho}\,\mathfrak{g},\tilde{\mathsf{H}},\tilde{\Omega})$ is quaternionic
balanced (resp. quaternionic Gauduchon) if and only if
$(\mathfrak{g},\mathsf{H},\Omega)$ is.

###### Proof.

Let $\Omega^{\prime}$ be the standard metric on $\mathbb{H}^{k}$. Then

$\tilde{\Omega}((X,U),(Y,V))=\Omega(X,Y)+\Omega^{\prime}(U,V)\,.$

It follows that

$\begin{split}d\tilde{\Omega}((X,U),(Y,V),(Z,W))=&\,d\Omega(X,Y,Z)-\Omega^{\prime}((\rho_{X}+\rho_{X}^{*})V,W)-\Omega^{\prime}((\rho_{Y}+\rho_{Y}^{*})W,U)\\\
\,\,&-\Omega^{\prime}((\rho_{Z}+\rho_{Z}^{*})U,V)\,.\end{split}$

Therefore, since we assumed $\rho\colon\mathfrak{g}\to\mathfrak{sp}(k)$ we get
$d\tilde{\Omega}=d\Omega\circ p$. Therefore, taking $(3,0)$-parts and applying
$\Lambda$ yields

(39)
$\beta_{\tilde{\Omega}}(X,U)=(\Lambda_{\tilde{\Omega}}\partial\tilde{\Omega})(X,U)=(\Lambda_{\Omega}\partial\Omega)(X)=\beta_{\Omega}(X)\,.$

Now, using Proposition 3.4 (a), Theorem 9.11 (4) and (39), we deduce the claim
for $\alpha_{\tilde{\Omega}}$. This, in turn, implies
$\mathrm{Ric}^{\mathrm{Ch}}_{\omega_{\tilde{L}}}=\mathrm{Ric}^{\mathrm{Ch}}_{\omega_{L}}\circ
p$. Similarly, we also have
$\mathrm{Ric}^{\mathrm{Bis}}_{\omega_{\tilde{L}}}=\mathrm{Ric}^{\mathrm{Bis}}_{\omega_{L}}\circ
p$. ∎

As a consequence of the above, we notice that

$\partial_{\tilde{J}}\alpha_{\tilde{\Omega}}=\partial_{J}\alpha_{\Omega}\circ
p\,.$

Hence, if $\tilde{\Omega}$ is Einstein

$\partial_{J}\alpha_{\Omega}(X,Y)=\partial_{\tilde{J}}\alpha_{\tilde{\Omega}}((X,U),(Y,V))=\lambda\tilde{\Omega}((X,U),(Y,V))=\lambda\Omega(X,Y)+\lambda\Omega^{\prime}(U,V)\,,$

for all $X,Y\in\mathfrak{g}$ and $U,V\in\mathbb{H}^{k}$, which is possible if
and only if $\lambda=0$.

###  9.8. Joyce’s examples.

In this Subsection, we shall prove Theorem 1.5. To do so, it will be useful to
briefly overview the construction of Joyce [55]. Let $G$ be a compact
semisimple Lie group of rank $r$ and fix a maximal torus $H$ in $G$. Within
this framework, structure theory can be performed, which allows to obtain a
decomposition of $\mathfrak{g}={\rm Lie}(G)$ of the following form:

$\mathfrak{g}=\mathfrak{b}\oplus\bigoplus_{j=1}^{m}\mathfrak{d}_{j}\oplus\bigoplus_{j=1}^{m}\mathfrak{f}_{j}\,,$

where $\mathfrak{b}$ is abelian of dimension $r-m$,
$\mathfrak{d}_{j}\subseteq\mathfrak{g}$ are subalgebras isomorphic to
$\mathfrak{su}(2)$, and $\mathfrak{f}_{j}\subseteq\mathfrak{g}$ are subspaces
satisfying the following properties:

  1. (1)

$[\mathfrak{d}_{j},\mathfrak{b}]=0$, for any $j=1,\dots,m$, and
$\mathfrak{h}:={\rm
Lie}(H)\subseteq\mathfrak{b}\oplus\bigoplus_{j=1}^{m}\mathfrak{d}_{j}$;

  2. (2)

$[\mathfrak{d}_{j},\mathfrak{d}_{i}]=0$, for $j\neq i$;

  3. (3)

$[\mathfrak{d}_{j},\mathfrak{f}_{i}]=0$, for $j<i;$

  4. (4)

$[\mathfrak{d}_{j},\mathfrak{f}_{j}]\subseteq\mathfrak{f}_{j}$, for any
$j=1,\dots,m$, and this Lie bracket action is isomorphic to the direct sum of
a finite amount of copies of the $\mathfrak{su}(2)$-action on $\mathbb{C}^{2}$
by matrix multiplication from the left.

Such a decomposition will be called a _Joyce decomposition_ of the Lie algebra
$\mathfrak{g}.$
Now, denote $T^{2m-r}\cong\mathrm{U}(1)^{2m-r}$ the $(2m-r)$-dimensional
torus, so that the Lie algebra of $T^{2m-r}\times G$ decomposes as

$(2m-r)\mathfrak{u}(1)\oplus\mathfrak{g}\cong\mathbb{R}^{m}\oplus\bigoplus_{j=1}^{m}\mathfrak{d}_{j}\oplus\bigoplus_{j=1}^{m}\mathfrak{f}_{j}\,.$

We define a hypercomplex structure
$I,J,K\in\mathrm{End}((2m-r)\mathfrak{u}(1)\oplus\mathfrak{g})$ in the
following manner. For every $j=1,\dots,m$, denote
$(e_{1}^{j},e_{2}^{j},e_{3}^{j},e_{4}^{j})$ a basis of
$\mathbb{R}\oplus\mathfrak{d}_{j}$ such that
$(e_{1}^{1},e_{1}^{2},\dots,e_{1}^{m})$ is the standard basis of
$\mathbb{R}^{m}$ and $e_{2}^{j},e_{3}^{j},e_{4}^{j}$ satisfy the commutation
relations:

(40) $\displaystyle[e_{2}^{j},e_{3}^{j}]=2e^{j}_{4}\,,$
$\displaystyle[e_{4}^{j},e^{j}_{2}]=2e^{j}_{3}\,,$
$\displaystyle[e^{j}_{3},e^{j}_{4}]=2e^{j}_{2}\,.$

We can regard $\mathbb{R}\oplus\mathfrak{d}_{j}$ as a copy of the space of
quaternions.

  1. (a)

For any $j=1,\dots,m$, let $I,J,K$ act on $\mathbb{R}\oplus\mathfrak{d}_{j}$
as:

$\displaystyle\qquad\,\,\,Ie^{j}_{1}=e^{j}_{2}\,,\qquad
Ie^{j}_{3}=e^{j}_{4}\,,\qquad Je^{j}_{1}=e^{j}_{3}\,,\qquad
Je^{j}_{2}=-e^{j}_{4}\,,\qquad Ke^{j}_{1}=e^{j}_{4}\,,\qquad
Ke^{j}_{2}=e^{j}_{3}\,.$

We obviously further require $I^{2}=J^{2}=-\mathrm{Id}$.

  2. (b)

For any $j=1,\dots,m$, let $I,J,K$ act on $\mathfrak{f}_{j}$ as:

$If=[e^{j}_{2},f]\,,\qquad Jf=[e^{j}_{3},f]\,,\qquad Kf=[e^{j}_{4},f]\,,$

for each $f\in\mathfrak{f}_{j}$.

It is clear that $I,J,K$ induce a hypercomplex structure $\mathsf{H}$ on
$\mathbb{R}^{m}\oplus\bigoplus_{j=1}^{m}\mathfrak{d}_{j}$, the fact that it is
also a hypercomplex structure on $\bigoplus_{j=1}^{m}\mathfrak{f}_{j}$ follows
from (3). At this point Joyce uses an argument due to Samelson [69] to prove
that $I$ and $J$ must be integrable.

Since the group $G$ is semisimple, the opposite of the Cartan-Killing form
$-B$ is a positive-definite inner product on the Lie algebra $\mathfrak{g}$.
We now follow the argumentation of Grantcharov and Poon [44] in order to show
that $-B$ can be extended to a HKT metric on $G$. This observation is
originally due to Opfermann and Papadopoulos [66], who also generalized the
construction to certain homogeneous spaces.
It can be seen that the Joyce decomposition is orthogonal with respect to $B$,
see [44, Lemma 2]. Let then $H_{j},X_{j},Y_{j}$ be an orthogonal basis for
$\mathfrak{d}_{j}$ such that $H_{j}\in\mathfrak{h}$ and

$B(H_{j},H_{j})=B(X_{j},X_{j})=B(Y_{j},Y_{j})=-\lambda_{j}^{2},$

for some $\lambda_{j}\in\mathbb{R}$. Since
$(2m-r)\mathfrak{u}(1)\oplus\mathfrak{b}\cong\mathbb{R}^{m}$, we can extend
$B$ to $\mathbb{R}^{m}$ simply by setting
$B(e_{i},e_{j})=-\delta_{ij}\lambda^{2}$, where $(e_{1},\dots,e_{m})$ is the
canonical basis of $\mathbb{R}^{m}.$ An easy inspection shows that this
extension is a hyperhermitian metric on
$(2m-r)\mathfrak{u}(1)\oplus\mathfrak{g}$. Let $g^{\prime}$ be the metric
induced on $T^{2m-r}\times G$ by such hyperhermitian inner product. Consider
the left-invariant connection $\nabla$ on $T^{2m-r}\times G$ such that all
left-invariant vector fields are parallel. Since the metric $g^{\prime}$ and
the hypercomplex structure $\mathsf{H}$ on $T^{2m-r}\times G$ are left-
invariant, they are preserved by $\nabla$. The torsion tensor is
$T^{\nabla}(X,Y)=-[X,Y]$, therefore

(41)
$T^{\nabla}(X,Y,Z):=g^{\prime}(T^{\nabla}(X,Y),Z)=B([X,Y],Z)\in\Lambda^{3}(T^{2m-r}\times
G)\,.$

It follows that the Bismut connections of $I$ and $J$ coincide with $\nabla$.
Therefore, the metric $g^{\prime}$ is HKT and it is easy to see that it is
strong.

We shall now show that such a metric can be appropriately rescaled at each
level of the Joyce decomposition in order to obtain an HKT-Einstein metric.

######  Theorem 9.13.

Let $G$ be a compact semisimple Lie group of rank $r$ and $\mathsf{H}$ an
invariant hypercomplex structure defined on $T^{2m-r}\times G$ as above. Then,
$(T^{2m-r}\times G,\mathsf{H})$ admits an invariant strong HKT-Einstein
metric.

###### Proof.

Here we can work at the algebra level. Let

$\mathfrak{g}=\mathfrak{b}\oplus\bigoplus_{j=1}^{m}\mathfrak{d}_{j}\oplus\bigoplus_{j=1}^{m}\mathfrak{f}_{j}$

be a Joyce decomposition of the Lie algebra $\mathfrak{g}$ and let $B$ be the
Cartan-Killing form on $\mathfrak{g}$. Denote
$d_{j}=\dim_{\mathbb{H}}(\mathfrak{f}_{j})$ and set
$\mu_{j}=\frac{1}{\sqrt{2(1+d_{j})}}$, for any $j=1,\dots,m$. Then, we claim
that

$g=-\sum_{j=1}^{m}\mu_{j}^{2}B|_{\mathbb{R}\oplus\mathfrak{d}_{j}\oplus\mathfrak{f}_{j}}$

is an HKT-Einstein metric on $(2m-r)\mathfrak{u}(1)\oplus\mathfrak{g}$
satisfying

(42) $\partial_{J}\alpha=\Omega\,,$

where $\Omega$ denote the corresponding $(2,0)$ form with respect to $I$.

First of all, it is clear that $g$ is a hyperhermitian metric with respect to
$\mathsf{H}$. Second, the same discussion as above shows that the metric
induced by $g$ on $T^{2m-r}\times G$ is strong HKT. Since on both sides of
(42) we have q-real $(2,0)$-forms, it is enough to show the identity along the
diagonal, i.e.

(43) $\partial_{J}\alpha(X,JX)=\lambda\Omega(X,JX)\,,\qquad\text{for every
}X\in(2m-r)\mathfrak{u}(1)\oplus\mathfrak{g}\,.$

Let $\theta=\theta_{\Omega}$ be the Lee form of $\Omega$. Then, we have

$\partial_{J}\alpha=\partial_{J}\theta^{1,0}=\frac{1}{4}J^{-1}(d-\sqrt{-1}d^{c}_{I})(J\theta-\sqrt{-1}K\theta)\,,$

whence

$4\partial_{J}\alpha(X,JX)=J\theta([JX,X])-J\theta([KX,IX])-\sqrt{-1}K\theta([JX,X])+\sqrt{-1}K\theta([KX,IX])\,.$

On a HKT manifold the Lee form satisfies

$\theta(X)=-\frac{1}{2}\sum_{i=1}^{4n}T^{\nabla}(IX,e_{i},Ie_{i})=-\frac{1}{2}\sum_{i=1}^{4n}T^{\nabla}(JX,e_{i},Je_{i})=-\frac{1}{2}\sum_{i=1}^{4n}T^{\nabla}(KX,e_{i},Ke_{i})\,,$

where $(e_{1},\dots,e_{4n})$ is an orthonormal basis of the tangent bundle
(see [53]). In this setting, the torsion $3$-form is defined as in (41) with
respect to $g$, therefore, we obtain

$\begin{split}8\partial_{J}\alpha(X,JX)=\sum_{i=1}^{4n}\bigl{(}&T^{\nabla}([JX,X],e_{i},Je_{i})-T^{\nabla}([KX,IX],e_{i},Je_{i})\\\
&-\sqrt{-1}T^{\nabla}([JX,X],e_{i},Ke_{i})+\sqrt{-1}T^{\nabla}([KX,IX],e_{i},Ke_{i})\bigr{)}\,.\end{split}$

But, since

$T^{\nabla}([X,Y],Z,W)=T^{\nabla}([Z,W],X,Y)\,,$

we get

(44)
$\begin{split}8\partial_{J}\alpha(X,JX)=\sum_{i=1}^{4n}\bigl{(}&-g([[e_{i},Je_{i}],JX],X)+g([[e_{i},Je_{i}],KX],IX)\\\
&+\sqrt{-1}g([[e_{i},Ke_{i}],JX],X)-\sqrt{-1}g([[e_{i},Ke_{i}],KX],IX)\bigr{)}\,.\end{split}$

At this point, we choose the orthonormal basis $(e_{1},\dots,e_{4n})$ to be
the following: for any $j=1,\dots,m$, let $(e_{1}^{j},\dots,e^{j}_{4})$ be an
orthonormal basis of $\mathbb{R}\oplus\mathfrak{d}_{j}$ with only non-zero
commutators:

$[e_{2}^{j},e_{3}^{j}]=2\mu_{j}e^{j}_{4}\,,\qquad[e_{4}^{j},e^{j}_{2}]=2\mu_{j}e^{j}_{3}\,,\qquad[e^{j}_{3},e^{j}_{4}]=2\mu_{j}e^{j}_{2}\,.$

On $\mathfrak{f}_{j}$, we fix any orthonormal basis
$\\{f^{j}_{1},f^{j}_{2},\dots,f^{j}_{4d_{j}}\\}$ where
$f^{j}_{4k-2}=If^{j}_{4k-3}$, $f^{j}_{4k-1}=Jf^{j}_{4k-3}$,
$f^{j}_{4k}=Kf^{j}_{4k-3}$ for $k=1,\dots,d_{j}$. Notice that
$(e_{1}^{j},\dots,e^{j}_{4})$ is, in general, different from the basis chosen
in (40) and, with this new orthonormal basis, the definition of $\mathsf{H}$
on $\mathfrak{f}_{j}$ is:

$If=\frac{1}{\mu_{j}}[e_{2}^{j},f]\,,\qquad
Jf=\frac{1}{\mu_{j}}[e_{3}^{j},f]\,,\qquad
Kf=\frac{1}{\mu_{j}}[e_{4}^{j},f]\,.$

Thus, we have

$[e_{1}^{j},Je_{1}^{j}]+[e_{2}^{j},Je_{2}^{j}]+[e_{3}^{j},Je_{3}^{j}]+[e_{4}^{j},Je_{4}^{j}]=4\mu_{j}e_{3}^{j}\,,$
$[e_{1}^{j},Ke_{1}^{j}]+[e_{2}^{j},Ke_{2}^{j}]+[e_{3}^{j},Ke_{3}^{j}]+[e_{4}^{j},Ke_{4}^{j}]=4\mu_{j}e_{4}^{j}\,.$

On the other hand, we claim that, for any $j=1,\dots,m$,

(45) $\sum_{i=1}^{4d_{j}}[f_{i}^{j},Jf_{i}^{j}]=4d_{j}\mu_{j}e^{j}_{3}\,.$

Indeed, we observe that

$\begin{split}[f^{j}_{4k-3},Jf^{j}_{4k-3}]+[f^{j}_{4k-2},Jf^{j}_{4k-2}]+[f^{j}_{4k-1},Jf^{j}_{4k-1}]+[f^{j}_{4k},Jf^{j}_{4k}]&=2\left([f^{j}_{4k-3},f^{j}_{4k-1}]-[f^{j}_{4k-2},f^{j}_{4k}]\right).\end{split}$

By definition of $(I,J,K)$ on $\mathfrak{f}_{j}$ and using Jacobi’s identity,
we have

$\begin{split}[f^{j}_{4k-2},f^{j}_{4k}]&=[f^{j}_{4k-2},Kf^{j}_{4k-3}]=\frac{1}{\mu_{j}}[[f^{j}_{4k-2},e^{j}_{4}],f^{j}_{4k-3}]+\frac{1}{\mu_{j}}[e^{j}_{4},[f^{j}_{4k-2},f^{j}_{4k-3}]]\\\
&=[f^{j}_{4k-3},f^{j}_{4k-1}]+\frac{1}{\mu_{j}}[e^{j}_{4},[f^{j}_{4k-2},f^{j}_{4k-3}]]\,,\end{split}$

hence

$\sum_{i=1}^{4d_{j}}[f_{i}^{j},Jf_{i}^{j}]=\frac{2}{\mu_{j}}\sum_{k=1}^{d_{j}}[e^{j}_{4},[f^{j}_{4k-3},f^{j}_{4k-2}]]\,.$

To prove the claim, we fix $k\in\\{1,\ldots,d_{j}\\}$ and consider the
components of $[f^{j}_{4k-3},f^{j}_{4k-2}]$ with respect to the Joyce
decomposition:

$[f^{j}_{4k-3},f^{j}_{4k-2}]=\sum_{p=1}^{m}\left(D_{1}^{p}e^{p}_{1}+D^{p}_{2}e^{p}_{2}+D^{p}_{3}e^{p}_{3}+D^{p}_{4}e^{p}_{4}+\sum_{q=1}^{4d_{p}}F^{p}_{q}f^{p}_{q}\
\right).$

Then, we have that $[e^{j}_{4},D_{1}^{p}e^{p}_{1}]=0$, for $p=1,\dots,m$ and
$[e^{j}_{4},D^{p}_{q}e^{p}_{q}]=0$, for $p\neq j$ and $q\in\\{2,3,4\\}$, by
property (1) and (2). Moreover, since the chosen basis is orthonormal, we have

$\begin{split}D^{j}_{2}&=g(f^{j}_{4k-3},[f^{j}_{4k-2},e^{j}_{2}])=-\mu_{j}g(f^{j}_{4k-3},If^{j}_{4k-2})=\mu_{j}\,,\end{split}$

while, proceeding similarly,

$D^{j}_{3}=\mu_{j}g(f^{j}_{4k-3},f^{j}_{4k})=0\,\mbox{ and
}\,D^{j}_{4}=-\mu_{j}g(f^{j}_{4k-3},f^{j}_{4k-1})=0\,.$

Furthermore, for $p<j$ and $q\in\\{1,\dots,4d_{p}\\}$, we have

$\begin{split}F^{p}_{q}&=-g([f^{j}_{4k-3},f^{j}_{4k-2}],I^{2}f^{p}_{q})=-\frac{1}{\mu_{j}}g([f^{j}_{4k-3},f^{j}_{4k-2}],[e^{p}_{2},If^{p}_{q}])\\\
&=-\frac{1}{\mu_{j}}g([[f^{j}_{4k-3},f^{j}_{4k-2}],e^{p}_{2}],If^{p}_{q})=-\frac{1}{\mu_{j}}g([[f^{j}_{4k-3},e^{p}_{2}],f^{j}_{4k-2}]+[f^{j}_{4k-3},[f^{j}_{4k-2},e^{p}_{2}]],If^{p}_{q})=0\,,\end{split}$

where in the last equality we used (3). For $p=j$, as above, we have that

$\begin{split}F^{j}_{q}&=-\frac{1}{\mu_{j}}g([[f^{j}_{4k-3},e^{j}_{2}],f^{j}_{4k-2}]+[f^{j}_{4k-3},[f^{j}_{4k-2},e^{j}_{2}]],If^{j}_{q})\\\
&=g([If^{j}_{4k-3},f^{j}_{4k-2}]+[f^{j}_{4k-3},If^{j}_{4k-2}],If^{j}_{q})=g([f^{j}_{4k-2},f^{j}_{4k-2}]-[f^{j}_{4k-3},f^{j}_{4k-3}],If^{j}_{q})=0\end{split}$

Finally, $[e^{j}_{4},F^{p}_{q}f^{p}_{q}]=0$, for $p>j$ and
$q\in\\{1,\dots,4d_{p}\\}$, by (3). Then, putting everything together we infer

$[e^{j}_{4},[f^{j}_{4k-3},f^{j}_{4k-2}]]=\mu_{j}[e^{j}_{4},e^{j}_{2}]=2\mu_{j}^{2}e^{j}_{3}$

and summing up over $k$ we obtain (45), as claimed. In the same way, one can
verify that

$\sum_{i=1}^{4d_{j}}[f_{i}^{j},Kf_{i}^{j}]=4d_{j}\mu_{j}e^{j}_{4}\,\,.$

Therefore

$\sum_{i=1}^{4n}[e_{i},Je_{i}]=4\sum_{j=1}^{m}(1+d_{j})\mu_{j}e^{j}_{3}\,,\qquad\sum_{i=1}^{4n}[e_{i},Ke_{i}]=4\sum_{j=1}^{m}(1+d_{j})\mu_{j}e^{j}_{4}$

and (44) becomes

$\begin{split}\partial_{J}\alpha(X,JX)=\frac{1}{2}\sum_{j=1}^{m}(1+d_{j})\mu_{j}\bigl{(}&-g([e_{3}^{j},JX],X)+g([e_{3}^{j},KX],IX)\\\
&+\sqrt{-1}g([e_{4}^{j},JX],X)-\sqrt{-1}g([e_{4}^{j},KX],IX)\bigr{)}\,.\end{split}$

We have that, for any $X$ in the chosen basis, $\Omega(X,JX)=\frac{1}{2}$. So,
we aim to prove $\partial_{J}\alpha(X,JX)=\frac{1}{2}$ which will conclude the
proof. We will prove it case by case. Suppose, firstly, that $X=e_{1}^{k}$,
for some $k=1,\dots,m$. Then

$\partial_{J}\alpha(X,JX)=\frac{1}{2}(1+d_{k})\mu_{k}g([e_{3}^{k},e_{4}^{k}],e_{2}^{k})=(1+d_{k})\mu_{k}^{2}=\frac{1}{2}\,.$

Moreover, if $X=e_{2}^{k}$, for some $k=1,\dots,m$, then

$\partial_{J}\alpha(X,JX)=\frac{1}{2}(1+d_{k})\mu_{k}g([e_{3}^{k},e^{k}_{4}],e_{2}^{k})=\frac{1}{2}\,.$

Furthermore, if $X=e_{3}^{k}$, for some $k=1,\dots,m$, then

$\partial_{J}\alpha(X,JX)=-\frac{1}{2}(1+d_{k})\mu_{k}g([e_{3}^{j},e_{2}^{k}],e_{4}^{k})=\frac{1}{2}\,.$

If $X=e_{4}^{k}$, for some $k=1,\dots,m$, then

$\partial_{J}\alpha(X,JX)=-\frac{1}{2}(1+d_{k})\mu_{k}g([e_{3}^{j},e_{2}^{k}],e_{4}^{k})=\frac{1}{2}\,.$

Finally, if $X\in\mathfrak{f}_{k}$, for some $k=1,\dots,m$, the terms in the
sum for $j\neq k$ vanish because

$\begin{split}g([e^{j}_{3},JX],X)&=-\frac{1}{\mu_{j}}g([e^{j}_{3},[e^{k}_{2},KX]],X)=-\frac{1}{\mu_{j}}g([[e^{j}_{3},e^{k}_{2}],KX],X)-\frac{1}{\mu_{j}}g([e^{k}_{2},[e^{j}_{3},KX]],X)\\\
&=\frac{1}{\mu_{j}}g([e^{j}_{3},KX],[e^{k}_{2},X])=g([e^{j}_{3},KX],IX)\,,\end{split}$

where in the second to last equality we used (2). In the same way,
$g([e^{j}_{4},JX],X)=g([e^{j}_{4},KX],IX)$. Therefore we only need to consider
the terms for $k=j$, which yields

$\begin{split}\partial_{J}\alpha(X,JX)=\,&\frac{1}{2}(1+d_{k})\mu_{k}^{2}\bigl{(}-g(J^{2}X,X)+g(JKX,IX)\\\
&\qquad\qquad\qquad\quad+\sqrt{-1}g(KJX,X)-\sqrt{-1}g(K^{2}X,IX)\bigr{)}=\frac{1}{2}\end{split}$

and the proof is concluded. ∎

###  9.9. Some non-compact HKT-Einstein manifolds

We conclude this Section providing other examples of invariant HKT-Einstein
manifolds. This time the examples are non-compact and they are simply-
connected solvable Lie groups. We shall consider all $4$-dimensional solvable
Lie algebras, which are classified in [14]. By dimensional reasons they all
are HKT and we show here that they actually are HKT-Einstein. Except for the
abelian Lie algebra we have three cases to consider. Again, we adopt the
convention explained at the beginning of the Section regarding the coframe and
the hypercomplex structure.

######  Example 9.14.

Here we take into account the Lie algebra $\mathfrak{aff}(\mathbb{C})$ of the
affine motion group of $\mathbb{C}$. The structure equations are

$de^{1}=-e^{1}\wedge e^{4}+e^{2}\wedge e^{3}\,,\qquad de^{2}=0\,,\qquad
de^{3}=e^{1}\wedge e^{2}-e^{3}\wedge e^{4}\,,\qquad de^{4}=0\,,$

which translate to the complex coframe to

$d\zeta^{1}=\frac{\sqrt{-1}}{2}(\bar{\zeta}^{1}\wedge\zeta^{2}-\zeta^{1}\wedge\bar{\zeta}^{2})\,,\qquad
d\zeta^{2}=\frac{\sqrt{-1}}{2}(\zeta^{1}\wedge\bar{\zeta}^{1}-\zeta^{2}\wedge\bar{\zeta}^{2})\,.$

The diagonal metric $\Omega=\zeta^{1}\wedge\zeta^{2}$ satisfies
$\alpha=\sqrt{-1}\zeta^{2}$ and thus $\partial_{J}\alpha=0$, so that the
metric $\Omega$ is HKT-Einstein with vanishing Einstein constant.

######  Example 9.15.

Consider the $4$-dimensional solvable Lie algebra with structure equations

$de^{1}=0\,,\qquad de^{2}=e^{1}\wedge e^{2}\,,\qquad de^{3}=e^{1}\wedge
e^{3}\,,\qquad de^{4}=e^{1}\wedge e^{4}\,.$

In terms of the complex $(1,0)$-coframe we have

$d\zeta^{1}=-\frac{1}{2}\zeta^{1}\wedge\bar{\zeta}^{1}\,,\qquad
d\zeta^{2}=\frac{1}{2}(\zeta^{1}\wedge\zeta^{2}+\bar{\zeta}^{1}\wedge\zeta^{2})\,.$

The diagonal metric $\Omega=\zeta^{1}\wedge\zeta^{2}$ satisfies
$\alpha=\zeta^{1}$ and so

$\partial_{J}\alpha=-\frac{1}{2}\Omega\,,$

showing that $\Omega$ induces on the corresponding simply connected solvable
Lie group a HKT-Einstein metric with negative Einstein constant.

######  Example 9.16.

We conclude with the solvable Lie algebra with structure equations

$de^{1}=0\,,\qquad de^{2}=-e^{1}\wedge e^{2}+2e^{3}\wedge e^{4}\,,\qquad
de^{3}=-\frac{1}{2}e^{1}\wedge e^{3}\,,\qquad de^{4}=-\frac{1}{2}e^{1}\wedge
e^{4}\,,$

equivalently,

$d\zeta^{1}=\frac{1}{2}\zeta^{1}\wedge\bar{\zeta}^{1}-\zeta^{2}\wedge\bar{\zeta}^{2}\,,\qquad
d\zeta^{2}=-\frac{1}{2}(\zeta^{1}\wedge\zeta^{2}+\bar{\zeta}^{1}\wedge\zeta^{2})\,.$

The diagonal metric $\Omega=\zeta^{1}\wedge\zeta^{2}$ satisfies
$\alpha=-\zeta^{1}$ and so $\partial_{J}\alpha=\frac{1}{2}\Omega$.

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