# Complete description of invariant Einstein metrics on the generalized flag
manifold $SO(2n)/U(p)\times U(n-p)$

Andreas Arvanitoyeorgos, Ioannis Chrysikos and Yusuke Sakane  University of
Patras, Department of Mathematics, GR-26500 Rion, Greece
<EMAIL_ADDRESS><EMAIL_ADDRESS>Osaka University,
Department of Pure and Applied Mathematics, Graduate School of Information and
Technology, Osaka 560-043, Japan<EMAIL_ADDRESS>
###### Abstract.

We find the precise number of non-Kähler $SO(2n)$-invariant Einstein metrics
on the generalized flag manifold $M=SO(2n)/U(p)\times U(n-p)$ with $n\geq 4$
and $2\leq p\leq n-2$. We use an analysis on parametric systems of polynomial
equations and we give some insight towards the study of such systems. We also
examine the isometric problem for these Einstein metrics.

2000 Mathematics Subject Classification. Primary 53C25; Secondary 53C30,
12D05, 65H10

Keywords: homogeneous manifold, Einstein metric, generalized flag manifold,
algebraic systems of equations, resultant.

The first two authors were partially supported by the C. Carathéodory grant
#C.161 2007-10, University of Patras and the third auther was supported by
Grant-in-Aid for Scientific Research (C) 21540080

## Introduction

A Riemannian metric $g$ is called Einstein if the Ricci tensor
$\operatorname{Ric}_{g}$ satisfies the equation ${\rm Ric}_{g}=e\cdot g$, for
some $e\in\mathbb{R}$. When $M$ is compact, Einstein metrics of volume 1 can
be characterized variationally as the critical points of the scalar curvature
functional $T(g)=\int_{M}S_{g}d{\rm vol}_{g}$ on the space $\mathcal{M}_{1}$
of Riemannian metrics of volume 1. If $M=G/K$ is a compact homogeneous space,
a $G$-invariant Einstein metric is precisely a critical point of $T$
restricted to the set of $G$-invariant metrics of volume 1. As a consequence,
the Einstein equation reduces to a system of non-linear algebraic equations,
which is still very complicated but more manageable, and in some times can be
solved explicity. Thus most known examples of Einstein manifolds are
homogeneous.

In a recent work [AC] the first two authors classified all flag manifolds for
which the isotropy representation decomposes into four pairwise inequivalent
irreducible submodules, and found new invariant Einstein metrics on these
spaces. Recall that a generalized flag manifold is an adjoint orbit of a
compact semisimple Lie group $G$, or equivalently a compact homogeneous space
of the form $M=G/K=G/C(S)$, where $C(S)$ is the centralizer of a torus $S$ in
$G$.

Eventhough the problem of finding all invariant Einstein metrics on $M$ can be
facilitated by use of certain theoretical results (e.g. the work [Grv] on the
total number of $G$-invariant complex Einstein metrics), it still remains a
difficult one, especially when the number of isotropy summands increases. This
difficulty also increases when we pass from exceptional flag manifolds to
classical flag manifolds, because in the later case the Einstein equation
reduces to a parametric system. In particular, eventhough all invariant
Einstein metrics were found for every generalized flag manifold with four
isotropy summands, a partial answer was given for the spaces
$SO(2n)/U(p)\times U(n-p)$ and $Sp(n)/U(p)\times U(n-p)$.

We summarize the results obtained in [AC] about these spaces.

######  Theorem 1.

([AC]) The flag manifold $SO(2n)/U(p)\times U(n-p)$ ($n\geq 4$ and $2\leq
p\leq n-2$) admits at least six $SO(2n)$-invariant Einstein metrics. There are
two non-Kähler Einstein metrics and two pairs of isometric Kähler-Einstein
metrics.

######  Theorem 2.

([AC]) The flag manifold $Sp(n)/U(p)\times U(n-p)$ ($n\geq 2$ and $1\leq p\leq
n-1$) admits at least four $Sp(n)$-invariant Einstein metrics, which are
Kähler.

For the special case $n=2p$ the following results have been obtained:

######  Theorem 3.

([AC]) The flag manifold $SO(4n)/U(p)\times U(p)$ ($p\geq 2$) admits at least
six $SO(4n)$-invariant Einstein metrics. There are two non-isometric non-
Kähler Einstein metrics, and four isometric Kähler-Einstein metrics. In the
special case where $2\leq p\leq 6$ there are two more non-Kähler Einstein
metrics, and the total number of $SO(4n)$-invariant Einstein metrics is
exactly eight.

######  Theorem 4.

([AC]) The flag manifold $Sp(2n)/U(p)\times U(p)$ ($p\geq 1$) admits precisely
six $Sp(n)$-invariant Einstein metrics. There are four isometric Kähler-
Einstein metrics, and two non-Kähler Einstein metrics.

In the present paper we find all $SO(2n)$-invariant Einstein metrics on the
flag manifold $SO(2n)/U(p)\times U(n-p)$, by using a new approach into
manipulating the algebraic systems of equations obtained from the Einstein
equation. The coefficients of the polynomials in such systems involve
parameters, so a major difficulty appears when we try to show existence and
uniqueness of solutions. Therefore, the contribution of the present work is,
besides answering the original problem on Einstein metrics, to give some
insight towards the study of parametric systems of algebraic equations.

Our main result is the following:

Main Theorem. Let $M=SO(2n)/U(p)\times U(n-p)$ with $n\geq 4$ and $2\leq p\leq
n-2$. Then $M$ admits exactly four non-Kähler $SO(2n)$-invariant Einstein
metrics for the pairs $(n,p)=(12,6)$, $(10,5)$, $(8,4)$, $(7,4)$, $(7,3)$,
$(6,4)$, $(6,3)$, $(6,2)$, $(5,3)$, $(5,2)$, $(4,2)$, and two non-Kähler
$SO(2n)$-invariant Einstein metrics for all other cases.

The flag manifold $Sp(n)/U(p)\times U(n-p)$ will be treated in a forthcoming
paper.

##  1\. The Einstein equation on flag manifolds

Let $M=G/K=G/C(S)$ be a generalized flag manifold of a compact simple Lie
group $G$, where $K=C(S)$ is the centralizer of a torus $S$ in $G$. Let $o=eK$
be the identity coset of $G/K$. We denote by $\mathfrak{g}$ and $\mathfrak{k}$
the corresponding Lie algrebras of $G$ and $K$. Let $B$ denote the Killing
form of $\mathfrak{g}$. Since $G$ is compact and simple, $-B$ is a positive
definite inner product on $\mathfrak{g}$. With repsect to $-B$ we consider the
orthogonal decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{m}$. This
is a reductive decomposition of $\mathfrak{g}$, that is
$\operatorname{Ad}(K)\mathfrak{m}\subset\mathfrak{m}$, and as usual we
identify the tangent space $T_{o}M$ with $\mathfrak{m}$. Since $K=C(S)$, the
isotropy group $K$ is connected and the relation
$\operatorname{Ad}(K)\mathfrak{m}\subset\mathfrak{m}$ is equivalent with
$[\mathfrak{k},\mathfrak{m}]\subset\mathfrak{m}$. Thus, for a flag manifold
$M=G/K$ the notion of $\operatorname{Ad}(K)$-invariant and
$\operatorname{ad}(\mathfrak{k})$-invariant is equivalent.

Let $\chi:K\to\operatorname{Aut}(T_{o}M)$ be the isotropy representation of
$K$ on $T_{o}M$. Since $\chi$ is equivalent to the adjoint representation of
$K$ restricted on $\mathfrak{m}$, the set of all $G$-invariant symmetric
covariant 2-tensors on $G/K$ can be identified with the set of all
$\operatorname{Ad}(K)$-invariant symmetric bilinear forms on $\mathfrak{m}$.
In particular, the set of $G$-invariant metrics on $G/K$ is identified with
the set of $\operatorname{Ad}(K)$-invariant inner products on $\mathfrak{m}$.

Let $\mathfrak{m}=\mathfrak{m}_{1}\oplus\cdots\oplus\mathfrak{m}_{s}$ be a
$(-B)$-orthogonal $\operatorname{Ad}(K)$-invariant decomposition of
$\mathfrak{m}$ into pairwise inequivalent irreducible
$\operatorname{Ad}(K)$-modules $\mathfrak{m}_{i}$ $(i=1,\ldots,s)$. Such a
decomposition always exists and can be expressed in terms of
$\mathfrak{t}$-roots (cf. [AP], [AC]). Then, a $G$-invariant Riemannian metric
on $M$ (or equivalently, an $\operatorname{Ad}(K)$-invariant inner product
$\langle\ ,\ \rangle$ on $\mathfrak{m}=T_{o}M$) is given by

$g=\langle\ ,\
\rangle=x_{1}\cdot(-B)|_{\mathfrak{m}_{1}}+\cdots+x_{s}\cdot(-B)|_{\mathfrak{m}_{s}},$
(1)

where $(x_{1},\ldots,x_{s})\in\mathbb{R}^{s}_{+}$. Since
$\mathfrak{m}_{i}\neq\mathfrak{m}_{j}$ as
$\operatorname{Ad}(K)$-representation, any $G$-invariant metric on $M$ has the
above form.

Similarly, the Ricci tensor $\operatorname{Ric}_{g}$ of a $G$-invariant metric
$g$ on $M$, as a symmetric covariant 2-tensor on $G/K$ is given by

$\operatorname{Ric}_{g}=r_{1}\cdot(-B)|_{\mathfrak{m}_{1}}+\cdots+r_{s}\cdot(-B)|_{\mathfrak{m}_{s}},$

where $r_{1},\ldots,r_{s}$ are the components of the Ricci tensor on each
$\mathfrak{m}_{i}$, that is
$\operatorname{Ric}_{g}|_{\mathfrak{m}_{i}}=r_{i}\cdot(-B)|_{\mathfrak{m}_{i}}$.
These components obtain o useful description in terms of the structure
constants $[ijk]$ first introduced in [WZ]. Let $\\{X_{\alpha}\\}$ be a
$(-B)$-orthonormal basis adapted to the decomposition of $\mathfrak{m}$, that
is $X_{\alpha}\in\mathfrak{m}_{i}$ for some $i$, and $\alpha<\beta$ if $i<j$
(with $X_{\alpha}\in\mathfrak{m}_{i}$ and $X_{\beta}\in\mathfrak{m}_{j}$). Set
$A_{\alpha\beta}^{\gamma}=B([X_{\alpha},X_{\beta}],X_{\gamma})$ so that
$[X_{\alpha},X_{\beta}]_{\mathfrak{m}}=\sum_{\gamma}A_{\alpha\beta}^{\gamma}X_{\gamma}$,
and $[ijk]=\sum(A_{\alpha\beta}^{\gamma})^{2}$, where the sum is taken over
all indices $\alpha,\beta,\gamma$ with
$X_{\alpha}\in\mathfrak{m}_{i},X_{\beta}\in\mathfrak{m}_{j},X_{\gamma}\in\mathfrak{m}_{k}$
(where $[\ ,\ ]_{\mathfrak{m}}$ denotes the $\mathfrak{m}$-component). Then
$[ijk]$ is nonnegative, symmetric in all three entries, and independent of the
$(-B)$-orthonormal bases choosen for $\mathfrak{m}_{i},\mathfrak{m}_{j}$ and
$\mathfrak{m}_{k}$ (but it depends on the choise of the decomposition of
$\mathfrak{m}$).

######  Proposition 1.

([PaS]) Let $M=G/K$ be a generalized flag manifold of a compact simple Lie
group $G$ and let $\mathfrak{m}=\bigoplus_{i=1}^{s}\mathfrak{m}_{i}$ be a
decomposition of $\mathfrak{m}$ into pairwise inequivalent irreducible
$\operatorname{Ad}(K)$-submodules. Then the components $r_{1},\ldots,r_{s}$ of
the Ricci tensor of a $G$-invariant metric (1) on $M$ are given by

$r_{k}=\frac{1}{2x_{k}}+\frac{1}{4d_{k}}\sum_{i,j}\frac{x_{k}}{x_{i}x_{j}}[ijk]-\frac{1}{2d_{k}}\sum_{i,j}\frac{x_{j}}{x_{k}x_{i}}[kij],\qquad(k=1,\ldots,s).$

In wiew of Proposition 1, a $G$-invariant metric
$g=(x_{1},\ldots,x_{s})\in\mathbb{R}^{s}_{+}$ on $M$, is an Einstein metric
with Einstein constant $e$, if and only if it is a positive real solution of
the system

$\frac{1}{2x_{k}}+\frac{1}{4d_{k}}\sum_{i,j}\frac{x_{k}}{x_{i}x_{j}}[ijk]-\frac{1}{2d_{k}}\sum_{i,j}\frac{x_{j}}{x_{k}x_{i}}[kij]=e,\quad
1\leq k\leq s.$

##  2\. The generalized flag manifold $SO(2n)/U(p)\times U(n-p)$

We review some results related to the generalized flag manifold
$M=G/K=SO(2n)/U(p)\times U(n-p)$ ($n\geq 4,\ 2\leq p\leq n-2$) obtained in
[AC]. Its corresponding painted Dynkin diagram is given by

$\alpha_{1}$1$\alpha_{2}$2$\ldots$$(2\leq
p\leq\ell-2)$$\alpha_{p}$2$\ldots$2$\alpha_{\ell-1}$1$\alpha_{\ell}$1

The isotropy representation of $M$ decomposes into a direct sum
$\chi=\chi_{1}\oplus\chi_{2}\oplus\chi_{3}\oplus\chi_{4}$, which gives rise to
a decomposition
$\mathfrak{m}=\mathfrak{m}_{1}\oplus\mathfrak{m}_{2}\oplus\mathfrak{m}_{3}\oplus\mathfrak{m}_{4}$
of $\mathfrak{m}=T_{o}M$ into four irreducible inequivalent
$\operatorname{ad}(\mathfrak{k})$-submodules. The dimensions
$d_{i}=\dim\mathfrak{m}_{i}\ (i=1,2,3,4)$ of these submodules can be obtained
by use of Weyl’s formula [AC, pp. 204-205, p. 210] and are given by

$d_{1}=2p(n-p),\ d_{2}=(n-p)(n-p-1),\ d_{3}=2p(n-p),\ d_{4}=p(p-1).$

According to (1), a $G$-invariant metric on $M=G/K$ is given by

$\left\langle\ ,\
\right\rangle=x_{1}\cdot(-B)|_{\mathfrak{m}_{1}}+x_{2}\cdot(-B)|_{\mathfrak{m}_{2}}+x_{3}\cdot(-B)|_{\mathfrak{m}_{3}}+x_{4}\cdot(-B)|_{\mathfrak{m}_{4}},$
(2)

for positive real numbers $x_{1},x_{2},x_{3},x_{4}$. We will denote such
metrics by $g=(x_{1},x_{2},x_{3},x_{4})$.

It is known ([Nis]) that if $n\neq 2p$ then $M$ admits two non-equivalent
$G$-invariant complex structures $J_{1},J_{2}$, and thus two non-isometric
Kähler-Einstein metrics which are given (up to scale) by (see also [AC,
Theorem 3])

$g_{1}=(n/2,\ n+p-1,\ n/2+p-1,\ p-1)$
---
$g_{2}=(n/2,\ n-p-1,\ 3n/2-p-1,\ 2n-p-1)$.
(3)

If $n=2p$ then $M$ admits a unique $G$-invariant complex structure with
corresponding Kähler-Einstein metric (up to scale) given by $g=(p,\ p-1,\
2p-1,\ 3p-1)$ (cf. also [AC, Theorem 8] where all isometric Kähler-Einstein
metrics are listed).

The Ricci tensor of $M$ has been computed in [AC] and is given as follows:

######  Proposition 2.

The components $r_{i}$ of the Ricci tensor for a $G$-invariant Riemannian
metric on $M$ determined by (2) are given as follows:

$\ \
\left.\begin{tabular}[]{l}$r_{1}=\displaystyle\frac{1}{2x_{1}}+\frac{c_{12}^{3}}{2d_{1}}\Big{(}\frac{x_{1}}{x_{2}x_{3}}-\frac{x_{2}}{x_{1}x_{3}}-\frac{x_{3}}{x_{1}x_{2}}\Big{)}+\frac{c_{13}^{4}}{2d_{1}}\Big{(}\frac{x_{1}}{x_{3}x_{4}}-\frac{x_{4}}{x_{1}x_{3}}-\frac{x_{3}}{x_{1}x_{4}}\Big{)}$\\\
$r_{2}=\displaystyle\frac{1}{2x_{2}}+\frac{c_{12}^{3}}{2d_{2}}\Big{(}\frac{x_{2}}{x_{1}x_{3}}-\frac{x_{1}}{x_{2}x_{3}}-\frac{x_{3}}{x_{1}x_{2}}\Big{)}$\\\
$r_{3}=\displaystyle\frac{1}{2x_{3}}+\frac{c_{12}^{3}}{2d_{3}}\Big{(}\frac{x_{3}}{x_{1}x_{2}}-\frac{x_{2}}{x_{1}x_{3}}-\frac{x_{1}}{x_{2}x_{3}}\Big{)}+\frac{c_{13}^{4}}{2d_{3}}\Big{(}\frac{x_{3}}{x_{1}x_{4}}-\frac{x_{4}}{x_{1}x_{3}}-\frac{x_{1}}{x_{3}x_{4}}\Big{)}$\\\
$r_{4}=\displaystyle\frac{1}{2x_{4}}+\frac{c_{13}^{4}}{2d_{4}}\Big{(}\frac{x_{4}}{x_{1}x_{3}}-\frac{x_{3}}{x_{1}x_{4}}-\frac{x_{1}}{x_{3}x_{4}}\Big{)},$\\\
\end{tabular}\right\\}$ (4)

where $c_{12}^{3}=[123]$ and $c_{13}^{4}=[134]$.

By taking into account the explicit form of the Kähler-Einstein metrics above,
and substituting these in (4), we can find that the values of the unknown
triples $[ijk]$ are given by
$\displaystyle{c_{12}^{3}=\frac{p(n-p)(n-p-1)}{2(n-1)}}$ and
$\displaystyle{c_{13}^{4}=\frac{p(p-1)(n-p)}{2(n-1)}}$.

A $G$-invariant metric $g=(x_{1},x_{2},x_{3},x_{4})$ on $M=G/K$ is Einstein if
and only if, there is a positive constant $e$ such that
$r_{1}=r_{2}=r_{3}=r_{4}=e$, or equivalently

$r_{1}-r_{3}=0,\quad r_{1}-r_{2}=0,\quad r_{3}-r_{4}=0.$ (5)

By substituting the values of $d_{i}\ (i=1,2,3,4)$ and $c_{12}^{3},c_{13}^{4}$
into the components of the Ricci tensor, System (5) is equivalent to the
following equations:

$\left.\begin{tabular}[]{r}$(x_{1}-x_{3})(-x_{1}x_{2}+px_{1}x_{2}-x_{2}x_{3}+px_{2}x_{3}-x_{1}x_{4}+nx_{1}x_{4}$\\\
$-px_{1}x_{4}+2x_{2}x_{4}-2nx_{2}x_{4}-x_{3}x_{4}+nx_{3}x_{4}-px_{3}x_{4})=0$\\\
$4(n-1)x_{3}x_{4}(x_{2}-x_{1})+(n+p-1)x_{4}(x_{1}^{2}-x_{2}^{2})-(n-3p-1)x_{3}^{2}x_{4}$\\\
$+(p-1)x_{2}(x_{1}^{2}-x_{3}^{2}-x_{4}^{2})=0$\\\
$4(n-1)x_{1}x_{2}(x_{4}-x_{3})+(2n-p-1)x_{2}(x_{3}^{2}-x_{4}^{2})+(2n-3p+1)x_{1}^{2}x_{2}$\\\
$+(n-p-1)x_{4}(x_{3}^{2}-x_{1}^{2}-x_{2}^{2})=0$\\\ \end{tabular}\right\\}$
(6)

##  3\. Proof of the Main Theorem

We consider the equation $r_{1}-r_{3}=0$ of System (5). This is equivalent to

$\displaystyle(x_{1}-x_{3})(-x_{1}x_{2}+px_{1}x_{2}-x_{2}x_{3}+px_{2}x_{3}-x_{1}x_{4}+nx_{1}x_{4}$
$\displaystyle-
px_{1}x_{4}+2x_{2}x_{4}-2nx_{2}x_{4}-x_{3}x_{4}+nx_{3}x_{4}-px_{3}x_{4})=0.$

CASE A Let $x_{1}=x_{3}=1$. Then the system of equations $r_{1}-r_{2}=0,\
r_{3}-r_{4}=0$ becomes

$\displaystyle{x_{2}}^{2}(n+p-1)+4(n-p-1)-4(n-1){x_{2}}+(p-1){x_{2}}{x_{4}}$
$\displaystyle=$ $\displaystyle 0$ (7)
$\displaystyle{x_{2}}{x_{4}}(n-p-1)+{x_{4}}^{2}(2n-p-1)-4(n-1){x_{4}}+4(p-1)$
$\displaystyle=$ $\displaystyle 0.$ (8)

From (7) we get that

$\displaystyle
x_{4}=-\frac{({x_{2}}-2)((n+p-1){x_{2}}-2(n-p-1))}{(p-1){x_{2}}}.$ (9)

Note that $x_{4}>0$ if and only if
$\displaystyle\frac{2(n-p-1)}{n+p-1}<x_{2}<2$. By substituting equation (9)
into equation (8), we obtain the following equation:

$\displaystyle\ \
H_{n,p}(x_{2})=(n-1)n(n+p-1){x_{2}}^{4}-4(n-1)\left(2n^{2}-2n-p^{2}+p\right){x_{2}}^{3}$
$\displaystyle+2\left(12n^{3}-11n^{2}p-25n^{2}-2np^{2}+20np+14n+2p^{3}-2p^{2}-6p-2\right){x_{2}}^{2}$
$\displaystyle-8(n-1)(4n-3p-1)(n-p-1){x_{2}}+8(n-p-1)^{2}(2n-p-1)=0.$ (10)

From (8) we get that

$\displaystyle
x_{2}=-\frac{({x_{4}}-2)({x_{4}}(2n-p-1)-2(p-1))}{{x_{4}}(n-p-1)}.$ (11)

Note that $x_{2}>0$ if and only if
$\displaystyle\frac{2(p-1)}{2n-p-1}<x_{4}<2$. By substituting equation (11)
into equation (7), we obtain the following equation:

$\displaystyle\ \
G_{n,p}(x_{4})=(n-1)n(2n-p-1){x_{4}}^{4}-4(n-1)\left(n^{2}+2np-
n-p^{2}-p\right){x_{4}}^{3}$
$\displaystyle+2\left(n^{3}+9n^{2}p-7n^{2}+4np^{2}-16np+8n-2p^{3}-2p^{2}+6p-2\right){x_{4}}^{2}$
$\displaystyle-8(n-1)(p-1){x_{4}}(n+3p-1)+8(p-1)^{2}(n+p-1)=0.$ (12)

Note that the relation between $H_{n,p}(x_{2})$ and $G_{n,p}(x_{4})$ is given
by

$\displaystyle G_{n,p}(x_{4})=H_{n,n-p}(x_{4}).$ (13)

######  Proposition 3.

The equation $H_{n,p}(x_{2})=0$ has at least two solutions between
$\displaystyle x_{2}=\frac{2(n-p-1)}{n+p-1}$ and $x_{2}=2$.

###### Proof.

We consider the value $H_{n,p}(x_{2})$ at $\displaystyle
x_{2}=\frac{2(n-p-1)}{n+p-1}$ and $x_{2}=2$. We see that

$\displaystyle
H_{n,p}\left(\frac{2(n-p-1)}{n+p-1}\right)=\frac{8(p-1)^{3}(n-p-1)^{2}}{(n+p-1)^{2}}>0\quad\mbox{and
}\quad H_{n,p}\left(2\right)=8(p-1)^{3}>0.$

Now, the value $H_{n,p}(x_{2})$ at $\displaystyle x_{2}=\frac{2(n-p-1)}{n}$ is
given by

$\displaystyle
H_{n,p}\left(\frac{2(n-p-1)}{n}\right)=-\frac{16(p-1)^{2}(n-p-1)^{3}}{n^{3}}<0,$

thus the equation $H_{n,p}(x_{2})=0$ has at least two solutions between
$\displaystyle x_{2}=\frac{2(n-p-1)}{n+p-1}$ and $x_{2}=2$. ∎

We need to show that the polynomial $H_{n,p}(x_{2})$ has only one local
minimum (i.e. the two solutions obtained in Proposition 3 are unique), with
some exceptions which will also be studied.

######  Lemma 1.

For $n\geq 2p+5$ and $p\geq 4$ the equation $H_{n,p}(x_{2})=0$ has exactly two
positive solutions.

###### Proof.

We have that

$\displaystyle\ \
\frac{dH_{n,p}}{dx_{2}}=4(n-1)n(n+p-1){x_{2}}^{3}-12(n-1)\left(2n^{2}-2n-p^{2}+p\right){x_{2}}^{2}$
$\displaystyle+4\left(12n^{3}-11n^{2}p-25n^{2}-2np^{2}+20np+14n+2p^{3}-2p^{2}-6p-2\right){x_{2}}$
$\displaystyle-8(n-1)(4n-3p-1)(n-p-1),$ $\displaystyle\ \
\frac{d^{2}H_{n,p}}{d{x_{2}}^{2}}=12(n-1)n(n+p-1){x_{2}}^{2}-24(n-1)\left(2n^{2}-2n-p^{2}+p\right){x_{2}}$
$\displaystyle+4\left(12n^{3}-11n^{2}p-25n^{2}-2np^{2}+20np+14n+2p^{3}-2p^{2}-6p-2\right)$

and

$\displaystyle\ \
\frac{d^{3}H_{n,p}}{d{x_{2}}^{3}}=24(n-1)n(n+p-1){x_{2}}-24(n-1)\left(2n^{2}-2n-p^{2}+p\right).$

Note that the quadratic polynomial
$\displaystyle\frac{d^{2}H_{n,p}}{d{x_{2}}^{2}}$ attains its minimum at
$\displaystyle x_{2}=\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}$ and we see that

$\displaystyle\ \
\frac{d^{2}H_{n,p}}{d{x_{2}}^{2}}\left(\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}\right)=\frac{4}{n(n+p-1)}\left(n^{4}p-n^{4}-n^{3}p^{2}-6n^{3}p+3n^{3}\right.$
$\displaystyle\left.-4n^{2}p^{2}+12n^{2}p-4n^{2}-np^{4}+2np^{3}+5np^{2}-8np+2n+3p^{4}-6p^{3}+3p^{2}\right).$

We set

$\displaystyle\ \
M(n,p)=n^{4}p-n^{4}-n^{3}p^{2}-6n^{3}p+3n^{3}-4n^{2}p^{2}+12n^{2}p-4n^{2}-np^{4}+2np^{3}$
$\displaystyle+5np^{2}-8np+2n+3p^{4}-6p^{3}+3p^{2}$

and we investigate the conditions for $n,p$ such that $M(n,p)>0$ for $n\geq
2p$.

We consider the coefficients of $M(n,p)$ as a polynomial of $n-2p-5$. We can
write $M(n,p)$ as

$\displaystyle\quad
M(n,p)=(p-1)(n-2p-5)^{4}+\left(7p^{2}+6p-17\right)(n-2p-5)^{3}$
$\displaystyle+\left(18p^{3}+41p^{2}-30p-109\right)(n-2p-5)^{2}$
$\displaystyle+\left(19p^{4}+62p^{3}-26p^{2}-274p-313\right)(n-2p-5)$
$\displaystyle+6p^{5}+22p^{4}-64p^{3}-309p^{2}-491p-340.$

We put

$\begin{array}[]{lcl}a_{0}=6p^{5}+22p^{4}-64p^{3}-309p^{2}-491p-340,&&a_{1}=19p^{4}+62p^{3}-26p^{2}-274p-313,\\\
a_{2}=18p^{3}+41p^{2}-30p-109,&&a_{3}=7p^{2}+6p-17.\end{array}$

Note that

$\displaystyle
a_{0}=6(p-4)^{5}+142(p-4)^{4}+1248(p-4)^{3}+4875(p-4)^{2}+7277(p-4)+432,$
$\displaystyle a_{1}=19(p-3)^{4}+290(p-3)^{3}+1558(p-3)^{2}+3296(p-3)+1844,$
$\displaystyle a_{2}=18(p-2)^{3}+149(p-2)^{2}+350(p-2)+139,$ $\displaystyle
a_{3}=7(p-2)^{2}+34(p-2)+23.$

Thus we see that $a_{0}>0$, $a_{1}>0,a_{2}>0,a_{3}>0$ for $p\geq 4$. Therefore
we see that $\displaystyle\frac{d^{2}H_{n,p}}{d{x_{2}}^{2}}>0$ for $n\geq
2p+5$ and $p\geq 4$ and hence, $\displaystyle\frac{dH_{n,p}}{d{x_{2}}}(x_{2})$
is monotone increasing and the polynomial $H_{n,p}(x_{2})$ has only one local
minimum for $n\geq 2p+5$ and $p\geq 4$. Thus the equation $H_{n,p}(x_{2})=0$
has exactly two positive solutions. ∎

Now we examine the values $p=2$ and $p=3$.

######  Lemma 2.

(1) Let $p=2$. Then for $n\geq 7$ the equation $H_{n,2}(x_{2})=0$ has exactly
two positive solutions, and for $4\leq n\leq 6$ it has exactly four positive
solutions.

(2) Let $p=3$. Then for $n\geq 8$ the equation $H_{n,3}(x_{2})=0$ has exactly
two positive solutions, and for $6\leq n\leq 7$ it has exactly four positive
solutions.

###### Proof.

(1) For $p=2$ we have that

$\displaystyle M(n,2)=n^{4}-13n^{3}+4n^{2}+6n+12$
$\displaystyle=(n-13)^{4}+39(n-13)^{3}+511(n-13)^{2}+2307(n-13)+766.$

Thus we see that $\displaystyle\frac{d^{2}H_{n,2}}{d{x_{2}}^{2}}>0$ for $n\geq
13$, and hence, $\displaystyle\frac{dH_{n,2}}{d{x_{2}}}(x_{2})$ is monotone
increasing and the polynomial $H_{n,2}(x_{2})$ has only one local minimum for
$n\geq 13$. Thus the equation $H_{n,2}(x_{2})=0$ has exactly two positive
solutions for $n\geq 13$. For $4\leq n\leq 12$, we consider polynomials
$H_{n,2}(x_{2})$ one by one and we see that, for $7\leq n\leq 12$ the equation
$H_{n,2}(x_{2})=0$ has two positive solutions, and for $4\leq n\leq 6$ the
equation $H_{n,2}(x_{2})=0$ has four positive solutions.

$H_{12,2}(x_{2})$

$H_{11,2}(x_{2})$

$H_{10,2}(x_{2})$

$H_{9,2}(x_{2})$

$H_{8,2}(x_{2})$

$H_{7,2}(x_{2})$

$H_{6,2}(x_{2})$

$H_{5,2}(x_{2})$

(2) For $p=3$ we have that

$\displaystyle M(n,3)=2\left(n^{4}-12n^{3}-2n^{2}-2n+54\right)$
$\displaystyle=(n-13)^{4}+40(n-13)^{3}+544(n-13)^{2}+2650(n-13)+1887.$

Thus we see that $\displaystyle\frac{d^{2}H_{n,3}}{d{x_{2}}^{2}}>0$ for $n\geq
13$, and hence, $\displaystyle\frac{dH_{n,3}}{d{x_{2}}}(x_{2})$ is monotone
increasing and the polynomial $H_{n,3}(x_{2})$ has only one local minimum for
$n\geq 13$. Thus the equation $H_{n,3}(x_{2})=0$ has exactly two positive
solutions for $n\geq 13$. For $6\leq n\leq 12$, we consider polynomials
$H_{n,3}(x_{2})$ one by one and we see that, for $8\leq n\leq 12$ the equation
$H_{n,3}(x_{2})=0$ has two positive solutions, and for $6\leq n\leq 7$ the
equation $H_{n,3}(x_{2})=0$ has four positive solutions.

$H_{12,3}(x_{2})$

$H_{11,3}(x_{2})$

$H_{10,3}(x_{2})$

$H_{9,3}(x_{2})$

$H_{8,3}(x_{2})$

$H_{7,3}(x_{2})$

$H_{6,3}(x_{2})$

∎

Next, we consider the case when $2p\leq n\leq 2p+4$. We may assume that $p\geq
4$.

######  Lemma 3.

Let $n=2p$. Then the equation $H_{2p,p}(x_{2})=0$ has exactly two positive
solutions for $p\geq 7$ and four positive solutions for $4\leq p\leq 6$.

###### Proof.

We see that

$\displaystyle\ \
H_{2p,p}(x_{2})=2\left((2p-1){x_{2}}^{2}-2(2p-1){x_{2}}+2(p-1)\right)\times$
$\displaystyle\left(p(3p-1){x_{2}}^{2}-4p(2p-1){x_{2}}+2(p-1)(3p-1)\right)$

Thus, the four solutions of the equation $H_{2p,p}(x_{2})=0$ are given by

$(a)\ x_{2}=\frac{2p\pm\sqrt{2p-1}-1}{2p-1},\quad(b)\
x_{2}=\frac{2p(2p-1)\pm\sqrt{2}\sqrt{-p\left(p^{3}-7p^{2}+5p-1\right)}}{p(3p-1)}.$
(14)

Since $-p\left(p^{3}-7p^{2}+5p-1\right)$ is negative for $p\geq 7$, we see
that the equation $H_{2p,p}(x_{2})=0$ has exactly two positive solutions
$p\geq 7$ and four positive solutions for $4\leq p\leq 6$.

$H_{12,6}(x_{2})$

$H_{10,5}(x_{2})$

$H_{8,4}(x_{2})$

∎

######  Lemma 4.

Let $n=2p+1$. Then for $p\geq 4$ the equation $H_{2p+1,p}(x_{2})=0$ has
exactly two positive solutions.

###### Proof.

We see that

$\displaystyle\ \
H_{2p+1,p}(x_{2})=6p^{2}(2p+1){x_{2}}^{4}-8p^{2}(7p+5){x_{2}}^{3}+2\left(50p^{3}+36p^{2}+3p-1\right){x_{2}}^{2}$
$\displaystyle-16p^{2}(5p+3){x_{2}}+8p^{2}(3p+1)$

and

$\displaystyle\frac{d^{2}H_{2p+1,p}}{d{x_{2}}^{2}}\left(\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}\right)=\frac{4\left(2p^{4}-18p^{3}-8p^{2}+p-1\right)}{2p+1}$
$\displaystyle=\frac{4\left(2(p-9)^{4}+54(p-9)^{3}+478(p-9)^{2}+1315(p-9)-640\right)}{2p+1}.$

Thus we see that $\displaystyle\frac{d^{2}H_{2p+1,p}}{d{x_{2}}^{2}}>0$ for
$p\geq 10$ and hence, $\displaystyle\frac{dH_{2p+1,p}}{d{x_{2}}}(x_{2})$ is
monotone increasing and the polynomial $H_{2p+1,p}(x_{2})$ has only one local
minimum for $p\geq 10$. Thus the equation $H_{2p+1,p}(x_{2})=0$ has exactly
two positive solutions.

For $4\leq p\leq 9$, we see that
$\displaystyle\frac{d^{2}H_{2p+1,p}}{d{x_{2}}^{2}}\left(\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}\right)$
is negative and two real solutions $\alpha,\beta$ of the quadratic equation
$\displaystyle\frac{d^{2}H_{2p+1,p}}{d{x_{2}}^{2}}=0$ are given by

$\displaystyle\alpha=\frac{2p^{2}(7p+5)-\sqrt{2}\sqrt{-p^{2}\left(2p^{4}-18p^{3}-8p^{2}+p-1\right)}}{6\left(2p^{3}+p^{2}\right)},$
$\displaystyle\beta=\frac{2p^{2}(7p+5)+\sqrt{2}\sqrt{-p^{2}\left(2p^{4}-18p^{3}-8p^{2}+p-1\right)}}{6\left(2p^{3}+p^{2}\right)}.$

Since the polynomial $\displaystyle\frac{dH_{2p+1,p}}{d{x_{2}}}(x_{2})$ of
degree 3 takes a local minimum at $\displaystyle x_{2}=\beta$, we consider the
value $\displaystyle\frac{dH_{2p+1,p}}{d{x_{2}}}(\beta)$. We see that

$\displaystyle\ \ \
\frac{dH_{2p+1,p}}{d{x_{2}}}(\beta)=\frac{2}{9p^{4}(2p+1)^{2}}\left(2(p-1)^{2}\left(8p^{3}-14p^{2}-36p-15\right)p^{4}\right.$
$\displaystyle\left.+2\sqrt{2}\left(2p^{4}-18p^{3}-8p^{2}+p-1\right)\sqrt{-p^{2}\left(2p^{4}-18p^{3}-8p^{2}+p-1\right)}p^{2}\right).$

By evaluating the above expression for the integers $4\leq p\leq 9$, we see
that $\displaystyle\frac{dH_{2p+1,p}}{d{x_{2}}}(\beta)>0$ for $6\leq p\leq 9$
and $\displaystyle\frac{dH_{2p+1,p}}{d{x_{2}}}(\beta)<0$ for $4\leq p\leq 5$.
Thus the polynomial $H_{2p+1,p}(x_{2})$ has only one local minimum for $6\leq
p\leq 9$, and $H_{2p+1,p}(x_{2})$ has two local minima and one local maximum
for $4\leq p\leq 5$. However, we see that for $p=4,5$ the equation
$H_{2p+1,p}(x_{2})=0$ has exactly two roots between
$\displaystyle\frac{2(n-p-1)}{(n+p-1)}=\frac{2}{3}$ and $2$, and this
completes the proof.

$H_{11,5}(x_{2})$

$H_{9,4}(x_{2})$

∎

######  Lemma 5.

Let $n=2p+2$. Then for $p\geq 4$ the equation $H_{2p+2,p}(x_{2})=0$ has
exactly two positive solutions.

###### Proof.

We see that

$\displaystyle\ \
H_{2p+2,p}(x_{2})=(2p+1)(2p+2)(3p+1){x_{2}}^{4}-4(2p+1)\left(7p^{2}+13p+4\right){x_{2}}^{3}$
$\displaystyle+4(p+1)\left(25p^{2}+42p+11\right){x_{2}}^{2}-8(p+1)(2p+1)(5p+7){x_{2}}+24(p+1)^{3}$

and

$\displaystyle\frac{d^{2}H_{2p+2,p}}{d{x_{2}}^{2}}\left(\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}\right)=\frac{2(6p^{5}-35p^{4}-88p^{3}-51p^{2}-20p-4)}{(3p+1)(p+1)}$
$\displaystyle=\frac{2\left(6(p-8)^{5}+205(p-8)^{4}+2632(p-8)^{3}+15117(p-8)^{2}+33468(p-8)+4764\right)}{(3p+1)(p+1)}.$

Thus we see that $\displaystyle\frac{d^{2}H_{2p+2,p}}{d{x_{2}}^{2}}>0$ for
$p\geq 8$ and hence, $\displaystyle\frac{dH_{2p+2,p}}{d{x_{2}}}(x_{2})$ is
monotone increasing and the polynomial $H_{2p+2,p}(x_{2})$ has only one local
minimum for $p\geq 8$. Thus the equation $H_{2p+2,p}(x_{2})=0$ has exactly two
positive solutions.

For $4\leq p\leq 7$, we see that
$\displaystyle\frac{d^{2}H_{2p+2,p}}{d{x_{2}}^{2}}\left(\frac{2n^{2}-2n-p^{2}+p}{n(n+p-1)}\right)$
is negative and the two real solutions $\alpha,\beta$ of the quadratic
equation $\displaystyle\frac{d^{2}H_{2p+2,p}}{d{x_{2}}^{2}}=0$ are given by

$\displaystyle\alpha=\frac{3(2p+1)\left(7p^{2}+13p+4\right)-\sqrt{3}\sqrt{(-2p-1)\left(6p^{5}-35p^{4}-88p^{3}-51p^{2}-20p-4\right)}}{6(p+1)(2p+1)(3p+1)},$
$\displaystyle\beta=\frac{3(2p+1)\left(7p^{2}+13p+4\right)+\sqrt{3}\sqrt{(-2p-1)\left(6p^{5}-35p^{4}-88p^{3}-51p^{2}-20p-4\right)}}{6(p+1)(2p+1)(3p+1)}.$

Since the polynomial $\displaystyle\frac{dH_{2p+2,p}}{d{x_{2}}}(x_{2})$ of
degree 3 takes local minimum at $\displaystyle x_{2}=\beta$, we consider the
value $\displaystyle\frac{dH_{2p+2,p}}{d{x_{2}}}(\beta)$. We see that

$\displaystyle\ \ \
\frac{dH_{2p+2,p}}{d{x_{2}}}(\beta)=\frac{1}{9(p+1)^{2}(2p+1)(3p+1)^{2}}\left(18(2p+1)\left(4p^{2}+7p+2\right)\right.\times$
$\displaystyle\left(p^{3}-p^{2}-6p-2\right)(p-1)^{2}+2\sqrt{3}\left(6p^{5}-35p^{4}-88p^{3}-51p^{2}-20p-4\right)\times$
$\displaystyle\left.\sqrt{(-2p-1)\left(6p^{5}-35p^{4}-88p^{3}-51p^{2}-20p-4\right)}\right).$

By substituting integer $4\leq p\leq 7$, we see that
$\displaystyle\frac{dH_{2p+2,p}}{d{x_{2}}}(\beta)>0$ for $5\leq p\leq 7$ and
$\displaystyle\frac{dH_{2p+2,p}}{d{x_{2}}}(\beta)<0$ for $p=4$. Thus the
polynomial $H_{2p+2,p}(x_{2})$ has only one local minimum for $5\leq p\leq 7$,
and $H_{2p+2,p}(x_{2})$ has two local minima and one local maximum for $p=4$.
However, we see that for $p=4$ the equation $H_{2p+2,p}(x_{2})=0$ has exactly
two roots between $\displaystyle\frac{2(n-p-1)}{(n+p-1)}=\frac{2(p+1)}{3p+1}$
and $2$.

$H_{10,4}(x_{2})$

∎

By a similar method we can prove the next two lemmas.

######  Lemma 6.

Let $n=2p+3$. Then for $p\geq 4$ the equation $H_{2p+3,p}(x_{2})=0$ has
exactly two positive solutions.

######  Lemma 7.

Let $n=2p+4$. Then for $p\geq 4$ the equation $H_{2p+4,p}(x_{2})=0$ has
exactly two positive solutions.

Therefore we have obtained the following:

######  Proposition 4.

(1) If $x_{1}=x_{3}$ and $n\geq 2p$, then $M$ admits exactly four
$SO(2n)$-invariant Einstein metrics for the pairs $(n,p)=(12,6)$, $(10,5)$,
$(8,4)$, $(7,3)$, $(6,3)$, $(6,2)$, $(5,2)$, $(4,2)$ and two
$SO(2n)$-invariant Einstein metrics for all other cases.

(2) If $x_{1}=x_{3}$ and $n\leq 2p$, then $M$ admits exactly four
$SO(2n)$-invariant Einstein metrics for the pairs $(n,p)=(12,6)$, $(10,5)$,
$(8,4)$, $(7,4)$, $(6,4)$, $(6,3)$, $(5,3)$, $(4,2)$ and two
$SO(2n)$-invariant Einstein metrics for all other cases.

###### Proof.

Part (1) is a consequence of Proposition 3 and Lemmas 1 – 7. For (2), we
consider the equation $G_{n,p}(x_{4})=0$, and the result follows from the
relation (13). ∎

CASE B Let

$\displaystyle\ \
-\,x_{1}x_{2}+px_{1}x_{2}-x_{2}x_{3}+px_{2}x_{3}-x_{1}x_{4}+nx_{1}x_{4}-px_{1}x_{4}$
$\displaystyle+\,2x_{2}x_{4}-2nx_{2}x_{4}-x_{3}x_{4}+nx_{3}x_{4}-px_{3}x_{4}=0,$
(15)

and set $x_{1}=1$. From equation (15) we obtain that

$\displaystyle
x_{3}=\frac{2(n-1){x_{2}}{x_{4}}-(n-p-1){x_{4}}-(p-1){x_{2}}}{(n-p-1){x_{4}}+(p-1){x_{2}}}.$
(16)

We need to show the following:

######  Proposition 5.

The system of equations $r_{1}-r_{2}=0$ and $r_{3}-r_{4}=0$ has no positive
solutions, except Kähler-Einstein metrics.

###### Proof.

We substitute equation (16) and $x_{1}=1$ into the equations $r_{1}-r_{2}=0$
and $r_{3}-r_{4}=0$, and we obtain the following equations :

$\displaystyle
F(x_{2},x_{4})=-(p-1){x_{2}}^{3}{x_{4}}\left(2n^{2}-4n+p^{2}+2p+1\right)-{x_{2}}^{2}{x_{4}}^{2}(3n^{3}+5n^{2}p-9n^{2}$
$\displaystyle-np^{2}-6np+7n+p^{3}-3p^{2}+3p-1)+(p-1)^{2}{x_{2}}^{4}(n+p-1)$
$\displaystyle+8(n-1)(p-1){x_{2}}^{2}{x_{4}}(n+p-1)-4(p-1)^{2}{x_{2}}^{2}(n+p-1)$
$\displaystyle+(p-1){x_{2}}{x_{4}}^{3}(n-p-1)^{2}+8(n-1){x_{2}}{x_{4}}^{2}(n-p-1)(n+p-1)$
$\displaystyle-8(p-1){x_{2}}{x_{4}}(n-p-1)(n+p-1)-4{x_{4}}^{2}(n-p-1)^{2}(n+p-1)=0,$
(17) $\displaystyle
G(x_{2},x_{4})={x_{2}}{x_{4}}^{3}(n-p-1)\left(3n^{2}-2np-2n+p^{2}-2p+1\right)$
$\displaystyle+{x_{2}}^{2}{x_{4}}^{2}(8n^{3}-6n^{2}p-18n^{2}+2np^{2}+12np+10n-p^{3}-3p^{2}-3p-1)$
$\displaystyle-(p-1)^{2}{x_{2}}^{3}{x_{4}}(n-p-1)-8(n-1)(p-1){x_{2}}^{2}{x_{4}}(2n-p-1)$
$\displaystyle+4(p-1)^{2}{x_{2}}^{2}(2n-p-1)-8(n-1){x_{2}}{x_{4}}^{2}(n-p-1)(2n-p-1)$
$\displaystyle+8(p-1){x_{2}}{x_{4}}(n-p-1)(2n-p-1)-{x_{4}}^{4}(n-p-1)^{2}(2n-p-1)$
$\displaystyle+4{x_{4}}^{2}(n-p-1)^{2}(2n-p-1)=0.$ (18)

We consider the resultant of the polynomials $F(x_{2},x_{4})$ and
$G(x_{2},x_{4})$ with respect to $x_{2}$, which is a polynomial of $x_{4}$,
say $Q(x_{4})$. We factor $Q(x_{4})$ as

$\displaystyle
Q(x_{4})=128(n-1)^{6}(p-1)^{2}{x_{4}}^{8}(n-p-1)^{4}(n{x_{4}}-2p+2)(n{x_{4}}-4n+2p+2)\times$
$\displaystyle(3n{x_{4}}-4n-2p{x_{4}}+2p-2{x_{4}}+2)(n{x_{4}}+2p{x_{4}}-2p-2{x_{4}}+2)\times$
$\displaystyle(6n^{5}{x_{4}}^{4}+8n^{5}{x_{4}}^{3}+2n^{5}{x_{4}}^{2}-3n^{4}p{x_{4}}^{4}-36n^{4}p{x_{4}}^{3}-38n^{4}p{x_{4}}^{2}-8n^{4}p{x_{4}}-17n^{4}{x_{4}}^{4}$
$\displaystyle-12n^{4}{x_{4}}^{3}+22n^{4}{x_{4}}^{2}+8n^{4}{x_{4}}+72n^{3}p^{2}{x_{4}}^{2}+56n^{3}p^{2}{x_{4}}+8n^{3}p^{2}+7n^{3}p{x_{4}}^{4}+116n^{3}p{x_{4}}^{3}$
$\displaystyle+36n^{3}p{x_{4}}^{2}-64n^{3}p{x_{4}}-16n^{3}p+15n^{3}{x_{4}}^{4}-12n^{3}{x_{4}}^{3}-60n^{3}{x_{4}}^{2}+8n^{3}{x_{4}}+8n^{3}$
$\displaystyle+8n^{2}p^{3}{x_{4}}^{3}+44n^{2}p^{3}{x_{4}}^{2}-48n^{2}p^{3}{x_{4}}-24n^{2}p^{3}-24n^{2}p^{2}{x_{4}}^{3}-260n^{2}p^{2}{x_{4}}^{2}-32n^{2}p^{2}{x_{4}}$
$\displaystyle+40n^{2}p^{2}-4n^{2}p{x_{4}}^{4}-104n^{2}p{x_{4}}^{3}+108n^{2}p{x_{4}}^{2}+112n^{2}p{x_{4}}-8n^{2}p-4n^{2}{x_{4}}^{4}+24n^{2}{x_{4}}^{3}$
$\displaystyle+44n^{2}{x_{4}}^{2}-32n^{2}{x_{4}}-8n^{2}-32np^{4}{x_{4}}^{2}-80np^{4}{x_{4}}-8np^{3}{x_{4}}^{3}-8np^{3}{x_{4}}^{2}+256np^{3}{x_{4}}$
$\displaystyle+32np^{3}+24np^{2}{x_{4}}^{3}+216np^{2}{x_{4}}^{2}-192np^{2}{x_{4}}-64np^{2}+24np{x_{4}}^{3}-136np{x_{4}}^{2}+32np$
$\displaystyle-8n{x_{4}}^{3}-8n{x_{4}}^{2}+16n{x_{4}}+32p^{5}{x_{4}}+32p^{5}+32p^{4}{x_{4}}^{2}-96p^{4}-32p^{3}{x_{4}}^{2}-128p^{3}{x_{4}}$
$\displaystyle+96p^{3}-32p^{2}{x_{4}}^{2}+128p^{2}{x_{4}}-32p^{2}+32p{x_{4}}^{2}-32p{x_{4}}).$

We first consider the cases when

$\displaystyle(n{x_{4}}-2(p-1))(n{x_{4}}-2(2n-p-1)\times$
$\displaystyle\left((3n-2(p+1)){x_{4}}-2(2n-p-1)\right)\left((n+2(p-1)){x_{4}}-2(p-1)\right)=0,$

and we claim that we only get Kähler-Einstein metrics on $SO(2n)/U(p)\times
U(n-p)$.

1) Let $\displaystyle x_{4}=\frac{2(p-1)}{n}$. Then equations (17) and (18)
reduce to

$\displaystyle\frac{(p-1)^{2}(n\,{x_{2}}-2(n+p-1))}{n^{3}}\left(n^{2}(n+p-1){x_{2}}^{3}-2(n-2)n(n-2p){x_{2}}^{2}\right.$
$\displaystyle\left.-4(n-p-1)\left(n^{2}+2np-4n-p^{2}+1\right){x_{2}}+8n(n-p-1)^{2}\right)=0,$
$\displaystyle\frac{2(p-1)^{2}(n-p-1)}{n^{4}}(n\,{x_{2}}-2(n+p-1))(n\,{x_{2}}-2(n-p+1))\times$
$\displaystyle\left(2(n-p-1)(2n-p-1)-n(p-1){x_{2}}\right)=0.$

If $n\,{x_{2}}-2(n-p+1)\neq 0$, we have

$\displaystyle\ \
\left(n^{2}(n+p-1){x_{2}}^{3}-2(n-2)n(n-2p){x_{2}}^{2}\right.$
$\displaystyle\left.-4(n-p-1)\left(n^{2}+2np-4n-p^{2}+1\right){x_{2}}+8n(n-p-1)^{2}\right)=0,$
$\displaystyle\ \
(n\,{x_{2}}-2(n-p+1))\left(2(n-p-1)(2n-p-1)-n(p-1){x_{2}}\right)=0.$

By taking the resultant of these polynomials with respect to $x_{2}$, we get

$-2048(n-1)^{2}n^{6}\left((n-p)^{2}+n-1\right)(n-p-1)^{3}(n-p),$

and we see that the resultant is non-zero for $2\leq p\leq n-2$. Thus we get
only $\displaystyle{x_{2}}=\frac{2(n+p-1)}{n}$ for a solution of equations
(17) and (18). From (16), we see $\displaystyle{x_{3}}=\frac{n+2p-2}{n}$. Thus
we obtain a Kähler-Einstein metric in this case.

Notice that this metric corresponds (up to scale) to the Kähler-Einstein
metric $g_{1}$ of (3)

2) Let $\displaystyle x_{4}=\frac{2(2n-p-1)}{n}$. Then equations (17) and (18)
reduce to

$\displaystyle\ \
-\frac{(n\,{x_{2}}-2(n-p-1))}{n^{3}}\left(-n^{2}(p-1)^{2}(n+p-1){x_{2}}^{3}\right.$
$\displaystyle+2n(p-1)\left(4n^{3}-3n^{2}p-9n^{2}+2np^{2}+10np+4n-4p^{2}-4p\right){x_{2}}^{2}$
$\displaystyle+4(2n-p-1)\left(6n^{4}+5n^{3}p-19n^{3}-13n^{2}p^{2}+21n^{2}+5np^{3}+9np^{2}\right.$
$\displaystyle\left.\left.-5np-9n-p^{4}-2p^{3}+2p+1\right){x_{2}}-8n(n-p-1)(2n-p-1)^{2}(n+p-1)\right)=0,$
$\displaystyle\ \
\frac{2(2n-p-1)(n\,{x_{2}}-2(n-p-1))}{n^{4}}\left(-n^{2}(p-1)^{2}(n-p-1){x_{2}}^{2}\right.$
$\displaystyle+4n\left(4n^{3}-5n^{2}p-7n^{2}+np^{2}+8np+3n-2p^{2}-2p\right)(2n-p-1){x_{2}}$
$\displaystyle\left.+4(n-p-1)^{2}(3n-p-1)(2n-p-1)^{2}\right)=0.$

If $n\,{x_{2}}-2(n-p-1)\neq 0$, we have

$\displaystyle\ \
\left(-n^{2}(p-1)^{2}(n+p-1){x_{2}}^{3}+2n(p-1)\left(4n^{3}-3n^{2}p-9n^{2}+2np^{2}+10np+4n\right.\right.$
$\displaystyle\left.-4p^{2}-4p\right){x_{2}}^{2}+4(2n-p-1)\left(6n^{4}+5n^{3}p-19n^{3}-13n^{2}p^{2}+21n^{2}+5np^{3}+9np^{2}\right.$
$\displaystyle\left.\left.-5np-9n-p^{4}-2p^{3}+2p+1\right){x_{2}}-8n(n-p-1)(2n-p-1)^{2}(n+p-1)\right)=0,$
$\displaystyle\ \
\left(-n^{2}(p-1)^{2}(n-p-1){x_{2}}^{2}+4n\left(4n^{3}-5n^{2}p-7n^{2}+np^{2}+8np+3n-2p^{2}-2p\right)\times\right.$
$\displaystyle\left.(2n-p-1){x_{2}}+4(n-p-1)^{2}(3n-p-1)(2n-p-1)^{2}\right)=0.$

By taking the resultant of these polynomials with respect to $x_{2}$, we get

$\displaystyle-2048(n-1)^{2}n^{6}(p-1)^{2}(n-p-1)(n-p)(2n-p-1)^{6}\left(p(n-p)+(n-1)^{2}\right)\times$
$\displaystyle\left(26n^{5}-48n^{4}p-92n^{4}+14n^{3}p^{2}+160n^{3}p+124n^{3}+12n^{2}p^{3}-64n^{2}p^{2}-180n^{2}p\right.$
$\displaystyle\left.-80n^{2}-4np^{4}-7np^{3}+63np^{2}+83np+25n+3p^{4}-2p^{3}-16p^{2}-14p-3\right).$

Now we have

$\displaystyle\ \
26n^{5}-48n^{4}p-92n^{4}+14n^{3}p^{2}+160n^{3}p+124n^{3}+12n^{2}p^{3}-64n^{2}p^{2}-180n^{2}p$
$\displaystyle-80n^{2}-4np^{4}-7np^{3}+63np^{2}+83np+25n+3p^{4}-2p^{3}-16p^{2}-14p-3$
$\displaystyle=26(n-p-1)^{5}+2(41p+19)(n-p-1)^{4}+2\left(41p^{2}+60p+8\right)(n-p-1)^{3}$
$\displaystyle+2p\left(13p^{2}+55p+30\right)(n-p-1)^{2}+\left(29p^{3}+49p^{2}+11p-1\right)(n-p-1)+8p^{2}(p+1)$

which is positive for $2\leq p\leq n-2$. Thus we see that the resultant is
non-zero and we only get $\displaystyle{x_{2}}=\frac{2(n-p-1)}{n}$ for a
solution of equations (17) and (18). From (16), we see
$\displaystyle{x_{3}}=\frac{3n-2p-2}{n}$. Thus we obtain a Kähler-Einstein
metric in this case.

Notice that this metric corresponds (up to scale) to the Kähler-Einstein
metric $g_{2}$ of (3)

3) Let $\displaystyle x_{4}=\frac{2(2n-p-1)}{3n-2(p+1)}$. By a similar method
we obtain that for $2\leq p\leq n-2$,
$\displaystyle{x_{2}}=\frac{2(n-p-1)}{3n-2p-2}$ is the only solution of
equations (17) and (18), and from (16) we see that
$\displaystyle{x_{3}}=\frac{n}{3n-2p-2}$. Thus we obtain a Kähler-Einstein
metric in this case.

4) Let $\displaystyle x_{4}=\frac{2(p-1)}{n+2(p-1)}$. By a similar method we
obtain that for $2\leq p\leq n-2$,
$\displaystyle{x_{2}}=\frac{2(n+p-1)}{n+2p-2}$ is the only positive solution
of the equations (17) and (18) for $\displaystyle\frac{n}{2}\leq p\leq n-2$,
and from (16) we see that $\displaystyle{x_{3}}=\frac{n}{n+2p-2}$.

Therefore, we obtain a Kähler-Einstein metric in all four cases.

We now denote by $T(x_{4})$ the factor of degree $4$ in the factorization of
$Q(x_{4})$. Then we can write

$\displaystyle T(x_{4})=(n-1)n^{2}(3n-4)(2n-p-1){x_{4}}^{4}$
$\displaystyle+4(n-1)n(2n-p-1)\left(n^{2}-4np-2p^{2}+8p-2\right){x_{4}}^{3}$
$\displaystyle+2(n^{5}-19n^{4}p+11n^{4}+36n^{3}p^{2}+18n^{3}p-30n^{3}+22n^{2}p^{3}-130n^{2}p^{2}+54n^{2}p$
$\displaystyle+22n^{2}-16np^{4}-4np^{3}+108np^{2}-68np-4n+16p^{4}-16p^{3}-16p^{2}+16p){x_{4}}^{2}$
$\displaystyle-8(p-1)(n-2p)(n+p-1)\left(n^{2}-6np+2n+2p^{2}+4p-2\right){x_{4}}$
$\displaystyle+8(p-1)^{2}(n-2p)^{2}(n+p-1).$

The case $n=2p$ has been studied in [AC].

We now proceed in two steps.

STEP 1. We will show that for $n\geq 4$ and $2\leq p<n/2$ the equation
$T(x_{4})=0$ has no positive solutions.

Note that $T(0)=8(p-1)^{2}(n-2p)^{2}(n+p-1)>0$ for $2\leq p<n/2$.

We have that

$\displaystyle\frac{dT}{dx_{4}}(x_{4})=4(n-1)n^{2}(3n-4)(2n-p-1){x_{4}}^{3}$
$\displaystyle+12(n-1)n(2n-p-1)\left(n^{2}-4np-2p^{2}+8p-2\right){x_{4}}^{2}$
$\displaystyle+4\left(n^{5}-19n^{4}p+11n^{4}+36n^{3}p^{2}+18n^{3}p-30n^{3}+22n^{2}p^{3}-130n^{2}p^{2}+54n^{2}p\right.$
$\displaystyle\left.+22n^{2}-16np^{4}-4np^{3}+108np^{2}-68np-4n+16p^{4}-16p^{3}-16p^{2}+16p\right){x_{4}}$
$\displaystyle-8(p-1)(n-2p)(n+p-1)\left(n^{2}-6np+2n+2p^{2}+4p-2\right).$

Note that the coefficient of ${x_{4}}^{3}$ is $4(n-1)n^{2}(3n-4)(2n-p-1)>0$.

The polynomial $T(x_{4})$ of degree $4$ attains a local minimum at
$x_{4}=u_{1}$, a local maximum at $x_{4}=u_{2}$, and a local minimum at
$x_{4}=u_{3}$.

By evaluating $\displaystyle\frac{dT}{dx_{4}}(x_{4})$ at the point
$\displaystyle\alpha=-\frac{n-2p}{2n}<0$, we have that

$\displaystyle\frac{dT}{dx_{4}}\left(-\frac{n-2p}{2n}\right)=\frac{(n-2p)}{2n}\left(2n^{5}+9n^{4}p-29n^{4}+8n^{3}p^{2}-33n^{3}p+103n^{3}-24n^{2}p^{2}\right.$
$\displaystyle\left.-16n^{2}p-112n^{2}+8np^{4}+48np^{3}-32np^{2}+92np+36n-40p^{4}+8p^{3}+8p^{2}-40p\right).$

Since we can write

$\displaystyle
2n^{5}+9n^{4}p-29n^{4}+8n^{3}p^{2}-33n^{3}p+103n^{3}-24n^{2}p^{2}-16n^{2}p-112n^{2}+8np^{4}$
$\displaystyle+48np^{3}-32np^{2}+92np+36n-40p^{4}+8p^{3}+8p^{2}-40p$
$\displaystyle=2(n-2p)^{5}+(29p-29)(n-2p)^{4}+\left(160p^{2}-265p+103\right)(n-2p)^{3}$
$\displaystyle+\left(424p^{3}-918p^{2}+602p-112\right)(n-2p)^{2}+\left(552p^{4}-1372p^{3}+1140p^{2}-356p+36\right)\times$
$\displaystyle(n-2p)+288p^{5}-768p^{4}+704p^{3}-256p^{2}+32p$

we see that $\displaystyle\frac{dT}{dx_{4}}\left(\alpha\right)>0$, thus
$u_{1}<\alpha$

Also, by evaluating $\displaystyle\frac{dT}{dx_{4}}(x_{4})$ at the point
$\displaystyle x_{4}=\beta=\frac{2(p-1)}{n}>0$, we have that

$\displaystyle\frac{dT}{dx_{4}}\left(\frac{2(p-1)}{n}\right)=-\frac{16(p-1)(n-p-1)}{n}\left(n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n\right.$
$\displaystyle\left.+2p^{3}+6p^{2}-6p-2\right).$

Since we can write

$\displaystyle n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n+2p^{3}+6p^{2}-6p-2$
$\displaystyle=$
$\displaystyle(n-2p)^{3}+(9p-7)(n-2p)^{2}+8(p-1)(2p-1)(n-2p)+2(p-1)^{2}(3p-1))>0,$

we see that $\displaystyle\frac{dT}{dx_{4}}\left(\beta\right)<0$, thus
$\beta<u_{3}$

Therefore, the three real solutions $u_{1}$, $u_{2}$, $u_{3}$ of the
polynomial $\displaystyle\frac{dT}{dx_{4}}(x_{4})$ of degree 3 satisfy

$\displaystyle u_{1}<\alpha<u_{2}<\beta<u_{3}.$

$T(x_{4})$ $\displaystyle\frac{dT}{dx_{4}}(x_{4})$

$\displaystyle\frac{dT}{dx_{4}}(x_{4})$ $T(x_{4})$

Since $T(0)>0$, in order to show that $T(x_{4})>0$ for $x_{4}>0$, we need to
prove the following:

Claim. The local minimum $T(u_{3})$ is positive.

We show our claim by dividing into two cases, namely $p=2$ and $p\geq 3$.

Case 1. $p=2$

The polynomial $T(x_{4})$ is given by

$\displaystyle
T(x_{4})=(n-1)n^{2}(2n-3)(3n-4){x_{4}}^{4}+4(n-1)n(2n-3)\left(n^{2}-8n+6\right){x_{4}}^{3}$
$\displaystyle+2\left(n^{5}-27n^{4}+150n^{3}-214n^{2}+4n+96\right){x_{4}}^{2}-8(n-4)(n+1)\left(n^{2}-10n+14\right){x_{4}}$
$\displaystyle+8(n-4)^{2}(n+1).$

Then the local minimum of $T(x_{4})$ at $x_{4}=u_{3}$ satisfies
$2/n<u_{3}<2/n+(2/n)^{2}$.

Indeed, it is

$\displaystyle\frac{dT}{dx_{4}}(x_{4})=4(n-1)n^{2}(2n-3)(3n-4){x_{4}}^{3}+12(n-1)n(2n-3)\left(n^{2}-8n+6\right){x_{4}}^{2}$
$\displaystyle+4\left(n^{5}-27n^{4}+150n^{3}-214n^{2}+4n+96\right){x_{4}}-8(n-4)(n+1)\left(n^{2}-10n+14\right).$

Then

$\displaystyle\frac{dT}{dx_{4}}(2/n)=-\frac{16(n-3)\left(n^{3}-n^{2}-16n+26\right)}{n}$
$\displaystyle=-\frac{16(n-3)\left((n-4)^{3}+11(n-4)^{2}+24(n-4)+10\right)}{n}<0$

and

$\displaystyle\frac{dT}{dx_{4}}(2/n+(2/n)^{2})=\frac{16\left(n^{6}+7n^{5}-12n^{4}-14n^{3}-152n^{2}+392n-192\right)}{n^{4}}$
$\displaystyle=\frac{16}{n^{4}}\left((n-4)^{6}+31(n-4)^{5}+368(n-4)^{4}+2194(n-4)^{3}+6848(n-4)^{2}\right.$
$\displaystyle\left.+10536(n-4)+6240\right)>0.$

Also, we have that

$\displaystyle\frac{d^{2}T}{d{x_{4}}^{2}}(x_{4})=12(n-1)n^{2}(2n-3)(3n-4){x_{4}}^{2}+24(n-1)n(2n-3)\left(n^{2}-8n+6\right){x_{4}}$
$\displaystyle+4\left(n^{5}-27n^{4}+150n^{3}-214n^{2}+4n+96\right)$
$\displaystyle=12(n-1)n^{2}(2n-3)(3n-4)\left(x_{4}+\frac{n^{2}-8n+6}{n(3n-4)}\right)^{2}+4(n^{5}-27n^{4}+150n^{3}$
$\displaystyle-214n^{2}+4n+96)-\frac{12(n-1)(2n-3)(n^{2}-8n+6)^{2}}{(3n-4)}.$

Note that

$\frac{2}{n}-(-\frac{n^{2}-8n+6}{n(3n-4)})=\frac{n^{2}-2n-2}{n(3n-4)}=\frac{(n-3)^{2}+4(n-3)+1}{n(3n-4)}>0$

and

$\displaystyle\frac{d^{2}T}{d{x_{4}}^{2}}(2/n)=4(n-3)(n-2)\left(n^{3}+2n^{2}-26n+28\right)$
$\displaystyle=4(n-3)(n-2)\left((n-4)^{3}+14(n-4)^{2}+38(n-4)+20\right)>0.$

Hence, the function $T(x_{4})$ is concave up for $x_{4}\geq 2/n$, so the local
minimum $x_{4}=u_{3}$ satisfies $2/n<u_{3}<2/n+(2/n)^{2}$.

We consider the tangent lines of the curve $T(x_{4})$ at $x_{4}=2/n$ and
$x_{4}=2/n+(2/n)^{2}$, given by the equations

$\displaystyle
z_{1}(t)=\frac{16(n-3)^{2}(3n+8)}{n^{2}}-\frac{16(n-3)\left(n^{3}-n^{2}-16n+26\right)}{n}(t-2/n)$
$\displaystyle=-\frac{16(n-3)\left(\left(n^{3}-n^{2}-16n+26\right)n\
t-2n^{3}-n^{2}+33n-28\right)}{n^{2}}$

and

$\displaystyle
z_{2}(t)=\frac{16\left(n^{6}+7n^{5}-12n^{4}-14n^{3}-152n^{2}+392n-192\right)}{n^{4}}\left(t-\frac{4}{n^{2}}-\frac{2}{n}\right)$
$\displaystyle+\frac{16\left(n^{7}+3n^{5}-28n^{4}-40n^{3}+32n^{2}+368n-192\right)}{n^{6}}$
$\displaystyle=\frac{16}{n^{6}}\left(\left(n^{6}+7n^{5}-12n^{4}-14n^{3}-152n^{2}+392n-192\right)n^{2}\
t-n^{7}-18n^{6}\right.\ \ \quad\quad\quad$
$\displaystyle\left.-n^{5}+48n^{4}+320n^{3}-144n^{2}-816n+576\right)$

respectively. These are shown in the figure.

$x_{4}=2/n$ $x_{4}=2/n+(2/n)^{2}$

Let $(x_{0},y_{0})$ be their point of intersection given by

$\displaystyle
x_{0}=\frac{2\left(n^{8}-2n^{7}-9n^{6}+64n^{5}-66n^{4}-160n^{3}+72n^{2}+408n-288\right)}{n^{2}\left(n^{7}-3n^{6}-6n^{5}+62n^{4}-92n^{3}-152n^{2}+392n-192\right)},$
$\displaystyle
y_{0}=\frac{16(n-3)}{n^{3}\left(n^{7}-3n^{6}-6n^{5}+62n^{4}-92n^{3}-152n^{2}+392n-192\right)}\times$
$\displaystyle\left(n^{10}-2n^{9}-17n^{8}+68n^{7}+94n^{6}-500n^{5}+88n^{4}-5368n^{3}+26048n^{2}-35808n+14976\right).$

Note that

$\displaystyle
n^{10}-2n^{9}-17n^{8}+68n^{7}+94n^{6}-500n^{5}+88n^{4}-5368n^{3}+26048n^{2}-35808n+14976$
$\displaystyle=(n-3)^{10}+28(n-3)^{9}+334(n-3)^{8}+2252(n-3)^{7}+9712(n-3)^{6}+29164(n-3)^{5}$
$\displaystyle+65002(n-3)^{4}+102860(n-3)^{3}+99479(n-3)^{2}+47904(n-3)+8064>0$

for $n\geq 3$. Therefore, the local minimal $T(u_{3})$ is greater than
$y_{0}$, and the claim has been proved.

Case 2. $3\leq p<n/2$.

Note that $n-p\geq p$ and

$\displaystyle
T(2(p-1)/n)=\frac{16(p-1)^{2}(n-p-1)^{2}\left(np+n+4(p-1)p\right)}{n^{2}}>0.$

Now we have

$\displaystyle\frac{dT}{dx_{4}}(x_{4})=4(n-1)n^{2}(3n-4)(2n-p-1){x_{4}}^{3}$
$\displaystyle+12(n-1)n(2n-p-1)\left(n^{2}-4np-2p^{2}+8p-2\right){x_{4}}^{2}$
$\displaystyle+4\left(n^{5}-19n^{4}p+11n^{4}+36n^{3}p^{2}+18n^{3}p-30n^{3}+22n^{2}p^{3}-130n^{2}p^{2}+54n^{2}p\right.$
$\displaystyle\left.+22n^{2}-16np^{4}-4np^{3}+108np^{2}-68np-4n+16p^{4}-16p^{3}-16p^{2}+16p\right){x_{4}}$
$\displaystyle-8(p-1)(n-2p)(n+p-1)\left(n^{2}-6np+2n+2p^{2}+4p-2\right)$

and

$\displaystyle\frac{dT}{dx_{4}}(2(p-1)/n)=-\frac{16(p-1)(n-p-1)}{n}\left(n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n\right.$
$\displaystyle\left.+2p^{3}+6p^{2}-6p-2\right).$

Note that

$\displaystyle n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n+2p^{3}+6p^{2}-6p-2$
$\displaystyle=$
$\displaystyle(n-2p)^{3}+(9p-7)(n-2p)^{2}+8(p-1)(2p-1)(n-2p)+2(p-1)^{2}(3p-1))>0,$

thus we see that $\displaystyle\frac{dT}{dx_{4}}(\beta)<0$.

Let $z_{1}(t)$ be the tangent line of the curve $T(x_{4})$ at $x_{4}=\beta$.
This is given by

$\displaystyle
z_{1}(t)=-\frac{16(p-1)(n-p-1)}{n}\left(n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n+2p^{3}+6p^{2}\right.$
$\displaystyle\left.-6p-2\right)\left(t-\frac{2(p-1)}{n}\right)+\frac{16(p-1)^{2}(n-p-1)^{2}\left(np+n+4(p-1)p\right)}{n^{2}}.$

We consider the point $t_{0}$ such that $z_{1}(t_{0})=0$. Then we see that

$\displaystyle
t_{0}=\frac{(p-1)\left(2n^{3}+7n^{2}p-13n^{2}-13np^{2}+2np+15n+12p^{2}-8p-4\right)}{n\left(n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n+2p^{3}+6p^{2}-6p-2\right)}.$

We will show that $\displaystyle\frac{dT}{dx_{4}}(t_{0})>0$ for $3\leq p\leq
n/2$. Indeed, we have

$\displaystyle\frac{dT}{dx_{4}}(t_{0})=\frac{4(p-1)(n-p-1)A(n,p)}{n\left(n^{3}+3n^{2}p-7n^{2}-8np^{2}+4np+8n+2p^{3}+6p^{2}-6p-2\right)^{3}},$

where

$\displaystyle
A(n,p)=n^{12}p-3n^{12}+15n^{11}p^{2}-68n^{11}p+85n^{11}+73n^{10}p^{3}-447n^{10}p^{2}$
$\displaystyle+1135n^{10}p-985n^{10}+68n^{9}p^{4}-730n^{9}p^{3}+3590n^{9}p^{2}-8134n^{9}p+6102n^{9}$
$\displaystyle-388n^{8}p^{5}+1743n^{8}p^{4}-3118n^{8}p^{3}-6724n^{8}p^{2}+28594n^{8}p-22347n^{8}-590n^{7}p^{6}$
$\displaystyle+3284n^{7}p^{5}-13140n^{7}p^{4}+38772n^{7}p^{3}-26930n^{7}p^{2}-48968n^{7}p+51156n^{7}$
$\displaystyle+1180n^{6}p^{7}-3852n^{6}p^{6}+17728n^{6}p^{5}-22616n^{6}p^{4}-72692n^{6}p^{3}+123252n^{6}p^{2}$
$\displaystyle+29464n^{6}p-76048n^{6}+961n^{5}p^{8}+148n^{5}p^{7}-20352n^{5}p^{6}-18940n^{5}p^{5}$
$\displaystyle+140330n^{5}p^{4}-14084n^{5}p^{3}-190872n^{5}p^{2}+29804n^{5}p+75053n^{5}-2356n^{4}p^{9}$
$\displaystyle-6225n^{4}p^{8}+24308n^{4}p^{7}+60596n^{4}p^{6}-94876n^{4}p^{5}-165630n^{4}p^{4}+164044n^{4}p^{3}$
$\displaystyle+136388n^{4}p^{2}-67312n^{4}p-49449n^{4}+1068n^{3}p^{10}+11644n^{3}p^{9}-9136n^{3}p^{8}$
$\displaystyle-59672n^{3}p^{7}-27216n^{3}p^{6}+194496n^{3}p^{5}+28056n^{3}p^{4}-177768n^{3}p^{3}-35740n^{3}p^{2}$
$\displaystyle+52804n^{3}p+21464n^{3}+32n^{2}p^{11}-6532n^{2}p^{10}-5992n^{2}p^{9}+24252n^{2}p^{8}+53120n^{2}p^{7}$
$\displaystyle-41016n^{2}p^{6}-130288n^{2}p^{5}+63304n^{2}p^{4}+77920n^{2}p^{3}-7492n^{2}p^{2}-21416n^{2}p$
$\displaystyle-5892n^{2}-64np^{12}+1024np^{11}+5664np^{10}-5120np^{9}-11616np^{8}-23168np^{7}$
$\displaystyle+44864np^{6}+27264np^{5}-37376np^{4}-12672np^{3}+5792np^{2}+4480np+928n$
$\displaystyle-768p^{11}-832p^{10}+2688p^{9}+960p^{8}+4608p^{7}-12672p^{6}+1792p^{5}+5248p^{4}$
$\displaystyle+256p^{3}-832p^{2}-384p-64.$

We shall show that $A(n,p)>0$ for $3\leq p\leq n/2$. We can write $A(n,p)$ as
a polynomial of $y=n-2p$ of the form

$A(n,p)=(p-3)y^{12}+a_{11}y^{11}+a_{10}y^{10}+a_{9}y^{9}+a_{8}y^{8}+a_{7}y^{7}+a_{6}y^{6}+a_{5}y^{5}+a_{4}y^{4}+a_{3}y^{3}+a_{2}y^{2}+a_{1}y+a_{0},$

where $a_{j}\ (j=0,\dots,11)$ can be written as follows:

$\displaystyle a_{11}=39(p-3)^{2}+94(p-3)+16$ $\displaystyle
a_{10}=667(p-3)^{3}+3268(p-3)^{2}+4604(p-3)+1424$ $\displaystyle
a_{9}=6588(p-3)^{4}+49146(p-3)^{3}+131552(p-3)^{2}+146040(p-3)+53568$
$\displaystyle
a_{8}=41696(p-3)^{5}+420063(p-3)^{4}+1666118(p-3)^{3}+3239144(p-3)^{2}$
$\displaystyle+3068032(p-3)+1121664$ $\displaystyle
a_{7}=177618(p-3)^{6}+2258984(p-3)^{5}+11898662(p-3)^{4}+33203396(p-3)^{3}$
$\displaystyle+51731888(p-3)^{2}+42627136(p-3)+14495232$ $\displaystyle
a_{6}=521336(p-3)^{7}+8010196(p-3)^{6}+52618296(p-3)^{5}+191551956(p-3)^{4}$
$\displaystyle+417348472(p-3)^{3}+544194848(p-3)^{2}+393195520(p-3)+121432064$
$\displaystyle
a_{5}=1062393(p-3)^{8}+19117036(p-3)^{7}+150329840(p-3)^{6}+674768512(p-3)^{5}$
$\displaystyle+1890947640(p-3)^{4}+3387906256(p-3)^{3}+3789854976(p-3)^{2}$
$\displaystyle+2420175872(p-3)+675510272$ $\displaystyle
a_{4}=1493910(p-3)^{9}+30767865(p-3)^{8}+281434708(p-3)^{7}+1500619596(p-3)^{6}$
$\displaystyle+5140194384(p-3)^{5}+11730160160(p-3)^{4}+17833993024(p-3)^{3}$
$\displaystyle+17419011328(p-3)^{2}+9918241792(p-3)+2508337152$ $\displaystyle
a_{3}=1416852(p-3)^{10}+32818860(p-3)^{9}+341869872(p-3)^{8}+2109020632(p-3)^{7}$
$\displaystyle+8532907744(p-3)^{6}+23658308832(p-3)^{5}+45523459968(p-3)^{4}$
$\displaystyle+60028498688(p-3)^{3}+51913028096(p-3)^{2}+26587561984(p-3)+6123782144$
$\displaystyle
a_{2}=862488(p-3)^{11}+22162788(p-3)^{10}+258695208(p-3)^{9}+1810579704(p-3)^{8}$
$\displaystyle+8442449008(p-3)^{7}+27537781712(p-3)^{6}+64116833984(p-3)^{5}$
$\displaystyle+106560650432(p-3)^{4}+123887801600(p-3)^{3}+95957073920(p-3)^{2}$
$\displaystyle+44563972096(p-3)+9401008128$ $\displaystyle
a_{1}=303264(p-3)^{12}+8551008(p-3)^{11}+110430432(p-3)^{10}+863710128(p-3)^{9}$
$\displaystyle+4556601456(p-3)^{8}+17082048928(p-3)^{7}+46660844352(p-3)^{6}$
$\displaystyle+93574409856(p-3)^{5}+136732708864(p-3)^{4}+141973649408(p-3)^{3}$
$\displaystyle+99432382464(p-3)^{2}+42173857792(p-3)+8192524288$
$\displaystyle
a_{0}=46656(p-3)^{13}+1430784(p-3)^{12}+20235744(p-3)^{11}+174764304(p-3)^{10}$
$\displaystyle+1028302272(p-3)^{9}+4352962512(p-3)^{8}+13638809216(p-3)^{7}$
$\displaystyle+32024909952(p-3)^{6}+56352955904(p-3)^{5}+73394750720(p-3)^{4}$
$\displaystyle+68769538048(p-3)^{3}+43897815040(p-3)^{2}+17110138880(p-3)+3075473408.$

We see that the coefficients $a_{j}$ ($j=0,\dots,11$) are positive for $p\geq
3$, which means that $A(n,p)>0$ for $3\leq p<n/2$. Therefore,
$\displaystyle\frac{dT}{dx_{4}}(t_{0})>0$ for $3\leq p\leq n/2$.

Now we compute $\displaystyle\frac{d^{2}T}{d{x_{4}}^{2}}(x_{4})$. We see that

$\displaystyle\frac{d^{2}T}{d{x_{4}}^{2}}(x_{4})=12(n-1)n^{2}(3n-4)(2n-p-1){x_{4}}^{2}$
$\displaystyle+24(n-1)n(2n-p-1)\left(n^{2}-4np-2p^{2}+8p-2\right){x_{4}}$
$\displaystyle+4\left(n^{5}-19n^{4}p+11n^{4}+36n^{3}p^{2}+18n^{3}p-30n^{3}+22n^{2}p^{3}-130n^{2}p^{2}+54n^{2}p+22n^{2}\right.$
$\displaystyle\left.-16np^{4}-4np^{3}+108np^{2}-68np-4n+16p^{4}-16p^{3}-16p^{2}+16p\right)$
$\displaystyle=12(n-1)n^{2}(3n-4)(2n-p-1)\left(x_{4}+\frac{n^{2}-4np-2p^{2}+8p-2}{n(3n-4)}\right)^{2}$
$\displaystyle+4\left(n^{5}-19n^{4}p+11n^{4}+36n^{3}p^{2}+18n^{3}p-30n^{3}+22n^{2}p^{3}-130n^{2}p^{2}+54n^{2}p+22n^{2}\right.$
$\displaystyle\left.-16np^{4}-4np^{3}+108np^{2}-68np-4n+16p^{4}-16p^{3}-16p^{2}+16p\right)$
$\displaystyle-12(n-1)(2n-p-1)\frac{(n^{2}-4np-2p^{2}+8p-2)^{2}}{(3n-4)}.$

Note that

$\displaystyle\beta-(-\frac{n^{2}-4np-2p^{2}+8p-2}{n(3n-4)})=\frac{n^{2}+2np-6n-2p^{2}+6}{n(3n-4)}$
$\displaystyle=\frac{(n-2p)^{2}+6(p-1)(n-2p)+6(p-1)^{2}}{n(3n-4)}>0,$

and

$\displaystyle\frac{d^{2}T}{d{x_{4}}^{2}}\left(\beta\right)=4(n-p-1)\left(n^{4}+6n^{3}p-12n^{3}+6n^{2}p^{2}-60n^{2}p+66n^{2}-8np^{3}\right.$
$\displaystyle\left.+32np^{2}+48np-80n+8p^{3}-40p^{2}+8p+24\right)$
$\displaystyle=4(n-p-1)\left((n-2p)^{4}+2(7p-6)(n-2p)^{3}+66(p-1)^{2}(n-2p)^{2}\right.$
$\displaystyle\left.+8(p-1)\left(15p^{2}-29p+10\right)(n-2p)+8(p-1)(3p-1)\left(3p^{2}-7p+3\right)\right)>0.$

Therefore, the function $T(x_{4})$ is concave up for $x_{4}\geq 2(p-1)/n$.

$x_{4}=\beta$ $x_{4}=t_{0}$

Consider the tangent line $l_{1}$ of the curve $T(x_{4})$ at $x_{4}=\beta$,
which intersects $x$-axis at a point $t_{0}$, and the tangent line $l_{2}$ of
the curve $T(x_{4})$ at $x_{4}=t_{0}$. Since
$\displaystyle\frac{dT}{dx_{4}}(\beta)<0$ and
$\displaystyle\frac{dT}{dx_{4}}(t_{0})>0$, the tangent lines $l_{1},l_{2}$
intersect at a point $(x_{0},y_{0})$ with $y_{0}>0$. Since $T(x_{4})$ is
concave up, we see that the curve $(x_{4},T(x_{4}))$ $(\beta\leq x_{4}\leq
t_{0})$ lies inside the triangle given by the three points $(\beta,T(\beta)$,
$(x_{0},y_{0})$ and $(t_{0},T(t_{0}))$.

Since the point $(u_{3},T(u_{3}))$ is inside of this triangle, it follows that
the local minimum $T(u_{3})$ is greater than $y_{0}>0$, and the claim has also
been shown in this case.

STEP 2. We consider the case that $n\geq 4$ and $n-2\geq p>n/2$.

This reduces to case STEP 1 as follows.

We consider the resultant of $F(x_{2},x_{4})$ and $G(x_{2},x_{4})$ with
respect to $x_{4}$, which is a polynomial of $x_{2}$ (instead of $x_{4}$), and
we denote this resultant by $R(x_{2})$. By factorizing $R(x_{2})$ we have that

$\displaystyle
R(x_{2})=128(n-1)^{6}(p-1)^{4}(n-p-1)^{2}{x_{2}}^{8}(n{x_{2}}-2n-2p+2)(n{x_{2}}-2n+2p+2)\times$
$\displaystyle(3n{x_{2}}-2n-2p{x_{2}}+2p-2{x_{2}}+2)(n{x_{2}}-2n+2p{x_{2}}-2p-2{x_{2}}+2)$
$\displaystyle\left(3n^{5}{x_{2}}^{4}-20n^{5}{x_{2}}^{3}+48n^{5}{x_{2}}^{2}-48n^{5}{x_{2}}+16n^{5}+3n^{4}p{x_{2}}^{4}+12n^{4}p{x_{2}}^{3}-110n^{4}p{x_{2}}^{2}\right.$
$\displaystyle+200n^{4}p{x_{2}}-104n^{4}p-10n^{4}{x_{2}}^{4}+72n^{4}{x_{2}}^{3}-178n^{4}{x_{2}}^{2}+168n^{4}{x_{2}}-40n^{4}+24n^{3}p^{2}{x_{2}}^{3}$
$\displaystyle+12n^{3}p^{2}{x_{2}}^{2}-248n^{3}p^{2}{x_{2}}+256n^{3}p^{2}-7n^{3}p{x_{2}}^{4}-44n^{3}p{x_{2}}^{3}+380n^{3}p{x_{2}}^{2}-640n^{3}p{x_{2}}$
$\displaystyle+224n^{3}p+11n^{3}{x_{2}}^{4}-92n^{3}{x_{2}}^{3}+232n^{3}{x_{2}}^{2}-200n^{3}{x_{2}}+32n^{3}-8n^{2}p^{3}{x_{2}}^{3}+84n^{2}p^{3}{x_{2}}^{2}$
$\displaystyle+48n^{2}p^{3}{x_{2}}-296n^{2}p^{3}-48n^{2}p^{2}{x_{2}}^{3}-92n^{2}p^{2}{x_{2}}^{2}+736n^{2}p^{2}{x_{2}}-440n^{2}p^{2}+4n^{2}p{x_{2}}^{4}$
$\displaystyle+56n^{2}p{x_{2}}^{3}-444n^{2}p{x_{2}}^{2}+656n^{2}p{x_{2}}-152n^{2}p-4n^{2}{x_{2}}^{4}+48n^{2}{x_{2}}^{3}-124n^{2}{x_{2}}^{2}$
$\displaystyle+96n^{2}{x_{2}}-8n^{2}-32np^{4}{x_{2}}^{2}+80np^{4}{x_{2}}+160np^{4}+8np^{3}{x_{2}}^{3}-120np^{3}{x_{2}}^{2}-256np^{3}{x_{2}}$
$\displaystyle+352np^{3}+24np^{2}{x_{2}}^{3}+120np^{2}{x_{2}}^{2}-576np^{2}{x_{2}}+224np^{2}-24np{x_{2}}^{3}+200np{x_{2}}^{2}$
$\displaystyle-256np{x_{2}}+32np-8n{x_{2}}^{3}+24n{x_{2}}^{2}-16n{x_{2}}-32p^{5}{x_{2}}-32p^{5}+32p^{4}{x_{2}}^{2}-96p^{4}$
$\displaystyle\left.+32p^{3}{x_{2}}^{2}+128p^{3}{x_{2}}-96p^{3}-32p^{2}{x_{2}}^{2}+128p^{2}{x_{2}}-32p^{2}-32p{x_{2}}^{2}+32p{x_{2}}\right).$

We denote by $S(x_{2})$ the factor of degree $4$ in the above factorization.
Then we can write

$\displaystyle S(x_{2})=(n-1)n^{2}(3n-4)(n+p-1){x_{2}}^{4}$
$\displaystyle-4(n-1)n(n+p-1)\left(5n^{2}-8np-8n+2p^{2}+8p+2\right){x_{2}}^{3}$
$\displaystyle+2\left(24n^{5}-55n^{4}p-89n^{4}+6n^{3}p^{2}+190n^{3}p+116n^{3}+42n^{2}p^{3}-46n^{2}p^{2}-222n^{2}p\right.$
$\displaystyle\left.-62n^{2}-16np^{4}-60np^{3}+60np^{2}+100np+12n+16p^{4}+16p^{3}-16p^{2}-16p\right){x_{2}}^{2}$
$\displaystyle-8(n-2p)(n-p-1)(2n-p-1)\left(3n^{2}-2np-6n-2p^{2}+4p+2\right){x_{2}}$
$\displaystyle+8(n-2p)^{2}(n-p-1)^{2}(2n-p-1).$

If we replace $p$ with $n-p$ in the polynomial $S(x_{2})$, we get exactly the
same polynomial as $T(x_{2})$, and thus we see that the equation $S(x_{2})=0$
has no positive solutions for $n-2\geq p>n/2$.

∎

The Main Theorem now follows from Propositions 4 and 5.

##  4\. The isometry problem

In this section we study the isometry problem for the new homogeneous Einstein
metrics of $M=SO(2n)/U(p)\times U(n-p)$, corresponding to the pairs $(n,p)$
which are presented in the Main Theorem. Recall that when $n=2p$, it was
proved in [AC] that the non-Kähler homogeneous Einstein metrics of the form
$g=(1,x_{2},1,x_{2})$, where $x_{2}$ is given by part $(a)$ of (14), are not
isometric. However for the special case of $2\leq p\leq 6$, the isometry
problem for the remaining two new Einstein metrics $g=(1,x_{2},1,x_{4})$,
where $x_{2}$ and $x_{4}$ are determined by part $(b)$ of (14), and
$(\ref{3})$ respectively, has not been studied yet.111Note that the first two
non-Kähler Einstein metrics on $M=SO(4p)/U(p)\times U(p)$, were obtained in
[AC, Theorem. 8] with respect to the normalization $g=(x_{1},1,x_{1},1)$. For
the special case $2\leq p\leq 6$ the new Einstein metrics are given with
respect to the normalization $g=(x_{1},1,x_{1},x_{4})$.

Let us recall the method used in [AC]. For any $G$-invariant Einstein metric
$g=(x_{1},x_{2},x_{3},x_{4})$ on $M=SO(2n)/U(p)\times U(n-p)$ (with $2\leq
p\leq n-2$) we determine a (normalized) scale invariant given by
$H_{g}=V_{g}^{1/d}S_{g}$, where $S_{g}$ is the scalar curvature of the given
metric $g$, $V_{g}=\prod_{i=1}^{4}x_{i}^{d_{i}}$ is the volume of $g$ and
$d=\sum_{i=1}^{4}d_{i}=\dim M$. In particular, the scalar curvature of $g$ is
given by

$S_{g}=\frac{1}{2}\sum_{i=1}^{4}\frac{d_{i}}{x_{i}}-\frac{[123]}{2}(\frac{x_{1}}{x_{2}x_{3}}+\frac{x_{2}}{x_{1}x_{3}}+\frac{x_{3}}{x_{1}x_{2}})-\frac{[134]}{2}(\frac{x_{1}}{x_{3}x_{4}}+\frac{x_{3}}{x_{1}x_{4}}+\frac{x_{4}}{x_{1}x_{3}})$

where $d_{i}$ and $[123]$, $[234]$ are given in Section 2. Note that $S_{g}$
is a homogeneous polynomial of degree $-1$ on the variables $x_{i}$, and the
volume $V_{g}$ is a monomial of degree $d$. Thus $H_{g}=V_{g}^{1/d}S_{g}$ is a
homogeneous polynomial of degree 0, and it is invariant under a common scaling
of the variables $x_{i}$. If two metrics are isometric then they have the same
scale invariant, so if the scale invariants $H_{g}$ and $H_{g^{\prime}}$ are
different, then the metrics $g$ and $g^{\prime}$ can not be isometric. If
$H_{g}=H^{\prime}_{g}$ we can not draw an immediate decision and conclude if
the metrics $g$ and $g^{\prime}$ are isometric or not. Finally, Kähler-
Einstein metrics which correspond to equivalent invariant complex structures
on $M$ are isometric (cf. [AC]).

In order to detect which pairs of Einstein metrics in the Main Theorem are
isometric or not, first we need to give their approximate values. Note that
the non-Kähler Einstein metrics are of the form $g=(1,x_{2},1,x_{4})$, where
$x_{2}$ is obtained by solving equation $H_{n,p}(x_{2})=0$ (see $(\ref{4})$),
if we first substitute the corresponding values of $n$ and $p$. Next, $x_{4}$
is easily obtained from (9). In the following table we present the case of
$n\neq 2p$.

Table 1 Approximate values of Einstein metrics on $M$ for pairs $(n,p)$ with
$n\neq 2p$

Pair | Einstein metrics |  |  |
---|---|---|---|---
$(n,p)$ | $g_{1}=(1,x_{2},1,x_{4})$ | $g_{2}=(1,x_{2},1,x_{4})$ | $g_{3}=(1,x_{2},1,x_{4})$ | $g_{4}=(1,x_{2},1,x_{4})$
$(7,4)$ | $(1,0.4661,1,0.7256)$ | $(1,0.6614,1,1.7636)$ | $(1,1.4144,1,1.3999)$ | $(1,1.5722,1,1.0631)$
$(7,3)$ | $(1,0.7256,1,0.4661)$ | $(1,1.7636,1,0.6614)$ | $(1,1.3999,1,1.4144)$ | $(1,1.0631,1,1.5722)$
$(6,4)$ | $(1,0.2680,1,0.8876)$ | $(1,0.3631,1,1.9057)$ | $(1,1.3782,1,1.5645)$ | $(1,1.5461,1,1.1658)$
$(6,2)$ | $(1,0.8876,1,0.2680)$ | $(1,1.9057,1,0.3631)$ | $(1,1.5645,1,1.3782)$ | $(1,1.1658,1,1.5461)$
$(5,3)$ | $(1,0.3241,1,0.6954)$ | $(1,0.4361,1,1.8876)$ | $(1,1.4331,1,1.5883)$ | $(1,1.6922,1,0.8952)$
$(5,2)$ | $(1,0.6954,1,0.3241)$ | $(1,1.8876,1,0.4361)$ | $(1,1.5883,1,1.4331)$ | $(1,0.8952,1,1.6922)$

Note that the values $(7,4)$ and $(7,3)$, $(6,4)$ and $(6,2)$, $(5,3)$ and
$(5,2)$, determine the quotients

$M_{1}=SO(14)/U(4)\times U(3)$, |  $M^{1}=SO(14)/U(3)\times U(4)$,
---|---
$M_{2}=SO(12)/U(4)\times U(2)$, |  $M^{2}=SO(12)/U(2)\times U(4)$,
$M_{3}=SO(10)/U(3)\times U(2)$, |  $M^{3}=SO(10)/U(2)\times U(3)$,

respectively. In particular, as we can see from Table 1, the Einstein metrics
on $M^{i}$ are obtained from the Einstein metrics on $M_{i}$, by a permutation
of the components $x_{2},x_{4}$, for any $i=1,2,3$, and conversely.222In
general, the flag manifolds $SO(2n)/U(n-p)\times U(p)$ and $SO(2n)/U(p)\times
U(n-p)$ are isometric via an element of the Weyl group of $G$. Thus we obtain
the isometries $M_{1}\cong M^{1}$, $M_{2}\cong M^{2}$ and $M_{3}\cong M^{3}$.
This result is also obtained from Table 2, where we give the values of the
corresponding scale invariants for the Einstein metrics $g_{1},g_{2},g_{3}$
and $g_{4}$.

Also, from Table 2 we easily conclude that all non Kähler invariant Einstein
metrics on $M_{1}\cong M^{1}$, $M_{2}\cong M^{2}$ and $M_{3}\cong M^{3}$ are
not isometric, since for any case it is $H_{g_{1}}\neq H_{g_{2}}\neq
H_{g_{3}}\neq H_{g_{4}}$. This completes the examination of the case $n\neq
2p$.

Table 2 The values of the corresponding scale invariants

Scale invariants | $(7,4)$ | $(7,3)$ | $(6,4)$ | $(6,2)$ | $(5,3)$ | $(5,2)$
---|---|---|---|---|---|---
$H_{g_{1}}$ | $25.2814$ | $25.2814$ | $17.9698$ | $17.9698$ | $12.4373$ | $12.4373$
$H_{g_{2}}$ | $25.5264$ | $25.5264$ | $18.1243$ | $18.1243$ | $12.6088$ | $12.6088$
$H_{g_{3}}$ | $25.6020$ | $25.6020$ | $18.2540$ | $18.2540$ | $12.7050$ | $12.7050$
$H_{g_{4}}$ | $25.5943$ | $25.5943$ | $18.2446$ | $18.2446$ | $12.6700$ | $12.6700$

For the special case $n=2p$ with $2\leq p\leq 6$, the scale invariants
corresponding to the new non-Kähler Einstein metrics on $M=SO(4p)/U(p)\times
U(p)$ given by $g=(1,x_{2},1,x_{4})$, where $x_{2}$ and $x_{4}$ are determined
by part $(b)$ of (14), and $(\ref{3})$, respectively, are equal. However, for

$x_{2}=\frac{2p(2p-1)-\sqrt{2}\sqrt{-p\left(p^{3}-7p^{2}+5p-1\right)}}{p(3p-1)},$

$x_{4}$ is given by

$x_{4}=\frac{2p(2p-1)+\sqrt{2}\sqrt{-p\left(p^{3}-7p^{2}+5p-1\right)}}{p(3p-1)},$

and for

$x_{2}=\frac{2p(2p-1)+\sqrt{2}\sqrt{-p\left(p^{3}-7p^{2}+5p-1\right)}}{p(3p-1)},$

$x_{4}$ is given by

$x_{4}=\frac{2p(2p-1)-\sqrt{2}\sqrt{-p\left(p^{3}-7p^{2}+5p-1\right)}}{p(3p-1)}.$

Thus these two Einstein metrics on $M$ are isometric.

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