text:
\section{Introduction}
A generalized flag manifold, namely a simply connected homogeneous space $M=G/K$ where $G$ is a compact Lie group and $K$ is the centralizer of a torus in $G$, can be endowed with invariant complex structures and invariant K\"ahler metrics. Once we fix an invariant complex structure $J$ on $M$, it is well known that there exists precisely one invariant K\"ahler-Einstein metric which is somehow canonically associated to $(M,J)$. A simply connected complex homogeneous space $G/H$ where $G$ is compact and $H$ is a compact subgroup is K\"ahler precisely when $H$ coincides with the centralizer of a torus in $G$, while the problem of finding special (invariant) Hermitian metrics on non-K\"ahler $G/H$ is far from being obvious. \par
Given an Hermitian manifold $(M,J,g)$, there are several connections $D$ which leave both the metric $g$ and the complex structure $J$ parallel. Among them, the Chern connection is the only one with such a property and moreover the torsion $T$ being of type $(2,0)$. The curvature tensor $R$ of the Chern connection can be traced in two different ways yielding two different Ricci tensors $S^{(1)}$ and $S^{(2)}$ which are both Hermitian. While the $(1,1)$ form $\rho$ which can be associated to $S^{(1)}$ is closed and represents the first Chern class $c_1(M)$, the form associated to $S^{(2)}$ is not even closed and there are no obvious relations between these two tensors. When the second Chern-Ricci $S^{(2)}$ is positive definite (or at least nonnegative and positive at least at one point), then the Hodge numbers $h^{p,0}=0$ and therefore the arithmetic genus $\chi(M,\mathcal O)=1$ (see \cite{LY}). More recently (\cite{LY}) the second Chern-Ricci tensor has been involved in defining a Hermitian flow $\frac d{dt}h_t = -S^{(2)}(h_t)$ (called HCF in the sequel), which preserves the Hermitian structure, is strictly parabolic and coincides with the K\"ahler-Ricci flow whenever the initial metric is K\"ahler. This flow is actually a simplified version of the hermitian curvature flow introduced and studied by Streets and Tian (\cite{ST}) and in \cite{U} it has been recently proved that on a compact Hermitian manifold the HCF preserves the Griffiths non-negativeness of the Chern curvature. \par
From this point of view, Hermitian metrics $h$ which are Chern-Einstein, namely whose second Ricci tensor $S^{(2)}$ satisfies $ S^{(2)} = \mu h$ for some $\mu\in C^\infty(M)$, are a distinguished class of metrics which has been first introduced in \cite{G1} and which deserves a special attention. It can be proved that on a generalized flag manifold there might exist several Chern-Einstein invariant metrics beyond the standard K\"ahler-Einstein metric, which on the other hand turns out to be the only one in some particular cases.\par
In this paper we start the investigation of the existence of special metrics on compact simply connected complex homogeneous spaces and in particular we focus on a special subclass $\mathcal C$ of such homogeneous manifolds that includes the Calabi-Eckmann manifolds. A complex manifold in the class $\mathcal C$ is a ${\rm T}^2$-bundle over the product of two compact Hermitian symmetric spaces and can be endowed with a two-parameter family of inequivalent invariant complex structures. We prove that these manifolds, which are non-K\"ahler, do not satisfy the $\partial\bar\partial$-lemma, do not support any balanced nor SKT metrics, while for every invariant complex structure there exists an invariant Chern-Einstein metric with $\mu=1$. We also prove that this special metric is unique whenever the complex structure belongs to a suitable neighborhood of the so called standard complex structure on the manifold.  \par
In Section 2 we recall some basic facts about the compact complex manifolds which are homogeneous under the action of a compact Lie group of biholomophisms. We discuss the $\partial\bar\partial$-lemma and we then focus on special homogeneous manifolds, called {\rm M}-manifolds, and a special subclass $\mathcal C$
which comprises the Calabi-Eckmann manifolds. We then discuss the existence of balanced metric on {\rm M}-manifolds or their suitable products.\par
In Section 3 we review some basic notions about the Chern connection, we introduce the definition of Chern-Einstein metric and give some basic properties. We then state our main result, as Theorem \eqref{main}, where we state that a manifold in the class $\mathcal C$ carries an invariant Hermitian Chern-Einstein metric, but no balanced, nor SKT metric. We then describe the Chern connection of an invariant metric and its curvature algebraically, providing then a proof of our main result. We conclude with a remark on the behaviour of the HCF in a suitable neighborhood of a Chern-Einstein solution on a particular manifold in $\mathcal C$. \par
In Section 4 we discuss the existence of invariant balanced metrics on compact complex homogeneous spaces.\par
 \bigskip
\noindent{\bf Acknowledgment.} We like to thank Andrea Spiro, Daniele Angella and Valentino Tosatti for valuable conversations.

\section{Homogeneous complex manifolds}
Let $M$ be a compact complex manifold with complex structure $J$ and let $G$ be a
compact connected Lie group acting almost effectively, transitively and holomorphically on $(M,J)$. We
will write $M = G/L$ for some compact subgroup $L$. \par
The complexified group $G^c$ acts holomorphically on $G/L$, so that $M=G^c/U$ for
some complex subgroup $U \subset G^c$.
It is well known that the {\em Tits fibration} $\phi$ provides a
holomorphic fibering of the homogeneous space $M$ onto a compact rational homogeneous space $Q:=G^c/P$, where the
parabolic subgroup $P$ is in general defined to be the normalizer
$N_{G^c}(U^o)$ of $U^o$ (see \cite{Ak}). \par
We will now suppose that $G$ is {\it semisimple} and that $M$ is supposed to be simply connected. Then $U$ (and $L$) is connected and the fibres of $\phi$ are complex tori. The flag manifold $G^c/P$ can be written as
$G/H$ endowed with a $G$-invariant complex structure $I$,
where $H$ is the centralizer of some torus in $G$.
Accordingly the Lie algebra $\gg$ can be decomposed as
\begin{equation}\label{dec}
\gg=\mathfrak{l}\oplus\underbrace{\mathfrak{t}\oplus\mathfrak{n}}_{\mathfrak{m}}\,,\end{equation}
where $\mathfrak{m}$ identifies with the tangent space $T_{[eL]}M$, $\mathfrak{t}$ with the tangent space of the fiber $T_{[eL]}F$, $F=\phi^{-1}([eH])$, $\mathfrak{h}=\mathfrak{l}\oplus\mathfrak{t}$ and $\mathfrak{n}$ is an $\Ad(H)$-invariant complement of
$\mathfrak{h}$ in $\gg$. The fiber $F$ being a complex torus implies
$[\mathfrak{t},\mathfrak{t}]=\{0\}$. Moreover the algebra $\mathfrak{h}$ is contained in the normalizer of $\mathfrak{l}$ in $\gg$
by construction, hence $[\mathfrak{l},\mathfrak{t}]\subset\mathfrak{l} \cap \mathfrak{t} = \{0\}$ and $\mathfrak{t}$ is in the center of $\mathfrak{h}$.\par
We can choose a Cartan subalgebra $\mathfrak{s}$ of the form $\mathfrak{s}=\mathfrak{t}_\mathfrak{l}^c\oplus
\mathfrak{t}^c$, where $\mathfrak{t}_\mathfrak{l}$ is a maximal abelian subalgebra of $\mathfrak{l}$.
Denote by $R$ the corresponding root system of $\gg^c$, by $R_\mathfrak{l}$ the subsystem relative to $\mathfrak{l}$ so that
$\mathfrak{l}^c = \mathfrak{t}_\mathfrak{l}^c \oplus \bigoplus_{\alpha\in R_\mathfrak{l}}\gg_\alpha$, and by $R_\mathfrak{n}$ the
symmetric subset of $R$ such that $\mathfrak{n}^c=\bigoplus_{\alpha \in R_\mathfrak{n}}\gg_\alpha$. The $G$-invariant complex structure
$I$ induces an endomorphism of $\mathfrak{n}^c$ that is $\Ad(H)$-invariant and therefore the corresponding subspace
$\mathfrak{n}^{1,0}$ is a sum of root spaces. The integrability of $I$ is equivalent to the condition
$$[\mathfrak{n}^{1,0},\mathfrak{n}^{1,0}]_{\mathfrak{n}^c} \subseteq \mathfrak{n}^{1,0}$$
and one can prove (see e.g. \cite{BFR}) that there is a suitable ordering of $R_\mathfrak{n}=R_\mathfrak{n}^+ \cup R_\mathfrak{n}^-$ such that
$$\mathfrak{n}^{1,0} = \bigoplus_{\alpha\in R_\mathfrak{n}^+} \gg_\alpha, \quad
\mathfrak{n}^{0,1} = \bigoplus_{\alpha\in R_\mathfrak{n}^-} \gg_\alpha.$$
The $G$-invariant complex structure $J$ on $G/L$ induces an $\Ad(L)$-invariant endomorphism, still denoted by $J$,
of $\mathfrak{m}^c$, where $\mathfrak{m} := \mathfrak{t} + \mathfrak{n}$. It leaves both $\mathfrak{t}$ and $\mathfrak{n}$ invariant with $J|_\mathfrak{n} =  I$ and
the integrability of $J$ is equivalent to the vanishing of the Nijenhuis tensor $N_J$, namely  for $X,Y\in \mathfrak{m}$
\begin{equation}\label{NI}
[JX,JY]_\mathfrak{m} - [X,Y]_\mathfrak{m} - J[JX,Y]_\mathfrak{m} - J[X,JY]_\mathfrak{m} = 0.
\end{equation}
Equation \eqref{NI} is trivial for $X,Y\in \mathfrak{t}$ and with $X\in \mathfrak{t}$ and
$Y\in \mathfrak{n}$ it reduces to the $\ad(\mathfrak{t})$-invariance of $I$. When $X,Y\in \mathfrak{n}$, then \eqref{NI} is the integrability of $I$
because $[\mathfrak{n}^{1,0},\mathfrak{n}^{1,0}]\subseteq \mathfrak{n}^{1,0}$. \par
Viceversa, we start with a decomposition as in
\eqref{dec}, where $\mathfrak{l}+\mathfrak{t} = \mathfrak{h}$ and $\mathfrak{h}$ is the centralizer of an abelian
subalgebra.
If we fix an $\ad(\mathfrak{h})$-invariant integrable complex
structure $I$ on $\mathfrak{n}$ and we extend it by choosing an arbitrary complex structure $J_\mathfrak{t}$ on $\mathfrak{t}$,
then $J_\mathfrak{t} + I$ will provide an integrable $L$-invariant complex structure $J$ on the homogeneous space $G/L$.\par
Note that $(G/L,J)$ is K\"ahler if and only if $\mathfrak{t}=\{0\}$, i.e. $L=H$. We are mainly interested in the non-K\"ahler case.
\medskip
\begin{proposition} The compact complex manifold $(G/L,J)$ does not satisfy the $\partial\bar\partial$-Lemma if it is not K\"ahler.\end{proposition}
\begin{proof} We fix a nonzero element $\xi\in \mathfrak{t}$ and consider the $\Ad(L)$-invariant element $\xi^*\in \gg^*$ given by the dual of $\xi$ w.r.t. the
$\Ad(G)$-invariant inner product $B$ on $\gg$. We consider the $2$-form $\omega=d\xi^*$, where we still denote by $\xi^*$ the $G$-invariant $1$-form
on $M$ determined by $\xi^*$. We claim that $\omega$ is a non zero $(1,1)$-form and that it cannot be written as $\o = \partial\bar\partial f$ for
$f\in C^\infty(M)$. The last assertion is clear, because $\o$ is $G$-invariant and the function $f$ can be chosen to be invariant as well, hence a constant. We show that $\o$ is not trivial. Indeed, if $\a\in R_\mathfrak{n}$ we have
$$\o(E_\a,E_{-\a}) = -B(\xi,[E_\a,E_{-\a}]) = B([E_\a,\xi],E_{-\a}) = -\a(\xi)B(E_\a,E_{-\a}).$$
We select a root $\a\in R_\mathfrak{n}$ so that $\a(\xi)\neq 0$ and our claim follows. In order to prove that $\o$ is of type $(1,1)$, we complexify it and
observe that $\o(X,Y)\neq 0$ if and only if $X\in \mathfrak{n}^{1,0}$ and $Y\in \mathfrak{n}^{0,1}$ and therefore $\o(JX,JY) = \o(X,Y)$ holds. \end{proof}
While flag manifolds are K\"ahler and do have a special (invariant) metric which is represented by the unique K\"ahler-Einstein metric, for non-K\"ahler homogeneous spaces the question about the existence of special (invariant) Hermitian metrics is meaningful and deserves a special investigation.\par
Following \cite{W}, an interesting class of complex homogeneous spaces is provided by {\rm M}-manifolds and their products. Given a compact simply connected Lie group $G$, a
{\rm M}-manifold is a $G$-homogeneous space of the form $G/L$ where $L$ is a subgroup of $G$ which coincides with the semisimple part of the centralizer of a torus in $G$. Using the fact that the semisimple $L$ has finite fundamental group we see that $G/L$ is simply connected and has finite second fundamental group. Moreover an even-dimensional {\rm M}-manifold and the product of two odd-dimensional ones carry infinitely many non-equivalent $G$-invariant complex structures (see \cite{W}). Simple examples of this situation is given by the Calabi-Eckmann manifolds, which can be described in group theoretic way as $\SU(n_1)\times\SU(n_2)/(\SU(n_1-1)\times \SU(n_2-1))\cong S^{2n_1-1}\times S^{2n_2-1}$.\par
\begin{proposition}\label{bal0} The even-dimensional {\rm M}-manifolds or the product of two odd-dimensional {\rm M}-manifolds do not admit any balanced metric.\end{proposition}
\begin{proof} Indeed, let $M$ be such a manifold, which is the total space of a holomorphic toric fibration $\pi$ over a flag manifold $Q$. Since $M$
is simply connected and has finite second fundamental group, we see that $H_{2n-2}(M)$ ($n=\dim_{\mathbb C}M$) is trivial. Suppose $M$ admits a balanced metric whose K\"ahler form therefore satisfies $d(\o^{n-1})=0$. If $Z\subset Q$ is a codimension one compact submanifold of $Q$, then $\tilde Z = \pi^{-1}(Z)$ is a codimension one compact submanifold of $M$ that bounds and therefore $\int_{\tilde Z} \o^{n-1} = 0$, a contradiction. \end{proof}
\begin{remark} Note that the $G$-invariance of any balanced metric in the above proposition is not assumed - however, when a balanced metric exists, we can always find an invariant one (see \cite{FG}). Note also that this result has to be contrasted with the K\"ahler case. Indeed, {\it any} invariant Hermitian metric $h$ on a flag manifold $G/K$ {\it is} balanced, since the codifferential $\delta\o$ of the K\"ahler form $\o$ of $h$ is a $G$-invariant $1$-form and a flag manifold supports no non-trivial invariant $1$-forms.\par
In Section 4 we will give a more detailed description of invariant balanced metrics. \end{remark}

We will now focus on a particular class $\mathcal C$ of homogeneous non-K\"ahler complex spaces given by a product of two {\rm M}-manifolds. This class $\mathcal C$ comprises the Calabi-Eckmann manifolds. We first describe them as a homogeneous space and then we study special hermitian invariant metrics on them. \par
We consider two irreducible compact Hermitian symmetric spaces $G_1/T^1\cdot H_1$, $G_2/T^1\cdot H_2$, where $G_1,G_2$ are two compact simply connected simple Lie groups. The product of the corresponding {\rm M}-manifolds provides a homogeneous manifold
$$M:= (G_1/H_1) \times (G_2/H_2)$$
which can be endowed with a family of invariant complex structures, already considered in \cite{T} (see also the more recent results in \cite{ST1} concerning also non invariant complex structures). Indeed, we consider the Cartan decompositions
$$\gg_i = \underbrace{\mathbb R\cdot Z_i \oplus \mathfrak{h}^s_i }_{\mathfrak{h}_i} \oplus \mathfrak{n}_i,\qquad [\mathfrak{n}_i,\mathfrak{n}_i]\subseteq \mathfrak{h}_i,\qquad i=1,2,$$
where  $\mathfrak{h}^s_i$ denotes the simple part of $\mathfrak{h}_i$ and $Z_i$ in the center of $\mathfrak{h}_i$ determines the complex structure $I_i$ on $\mathfrak{n}_i$ by $I_i = \ad(Z_i)$. Therefore we have $\mathfrak{l} := \mathfrak{h}^s_1\oplus \mathfrak{h}^s_2$ and $\mathfrak{t}$ is spanned by $Z_1,Z_2$. The complex structure $J_\mathfrak{t}\in \mbox{End}(\mathfrak{t})$ can be represented by the matrix $\left(\begin{smallmatrix} a& {\frac{-1-a^2}b}\\ b&-a\end{smallmatrix}\right)$ w.r.t. the basis $\{Z_1,Z_2\}$, where $a,b\in\mathbb R$, $b\neq 0$. The complex structure $J_o$ with $a=0,b=1$ will be called {\it standard}. \par

\section{ The Chern connection and the main Theorem}
\par \medskip
Given a Hermitian manifold ($M,h,J$), the associated Chern connection is the unique hermitian connection, i.e. which leaves $h$ and $J$ parallel, and such that its torsion tensor $T$ is of type $(2,0)$, namely
\begin{equation}\label{tor}T(JX,Y) = JT(X,Y)\end{equation}
for every vector fields $X,Y$ on $M$. Since the torsion of any hermitian connection on a complex manifold has vanishing $(0,2)$-component, \eqref{tor} is equivalent to (see \cite{G})
\begin{equation} \label{tor1} T(JX,JY)=-T(X,Y)\end{equation}
for every $X,Y$. This in turn is equivalent to saying that $T(Z,\overline W)=0$ for sections $Z,W\in \Gamma(T^{10}M)$. The curvature $R$ is a section of
$\Lambda^{1,1}(T^*M)\otimes\mathfrak{u}(TM)$ and in local holomorphic coordinates it has the expression
$$R_{i\bar j k \bar l } = -\frac{\partial h_{k\bar l}}{\partial z_i\partial \overline{z}_j} + h^{p\bar q}\frac{\partial h_{k\bar q}}{\partial z_i}
\frac{\partial h_{p\bar l}}{\partial \bar z_j}.$$
The curvature $R$ can be traced in two different ways. The {\it first} Ricci tensor $S^{(1)}$ is defined by tracing the endomorphism part, namely
$$S^{(1)}_{i\bar j} = h^{p\bar q}R_{i\bar jp\bar q} = -\frac{\partial^2 \log(\det(h))}{\partial z_i\partial \overline{z}_j} $$
and its associated $(1,1)$ form is closed and represents the first Chern class $c_1(M)$. The second Ricci tensor is given by the trace
$$S^{(2)}_{i\bar j} = h^{p\bar q}R_{p\bar qi\bar j}$$
and still there exists an associated $(1,1)$ form, which is not necessarily closed. The two Ricci tensors differ by a term which depends on the covariant derivative of the torsion (see e.g. Lemma 2.4 in \cite{ST}). It is known (see \cite{LY},\cite{KW})  that when $S^{(2)}$ is positive definite (negative definite resp.) the Hodge numbers $h^{p,0}=0$ for $p=1,\ldots,\dim_{\mathbb C}M$ ($M$ has no holomorphic vector fields resp.). We are led to the following definition, which was first considered in \cite{G1}
\begin{definition} A Hermitian metric $h$ is called Chern-Einstein if there exists $\mu \in C^\infty(M)$ so that
$$S^{(2)} = \mu\cdot  h.$$
\end{definition}
\begin{remark} It follows using general formulas (see e.g. \cite{G2}, ~p. 501) that a metric conformal to a K\"ahler-Einstein metric turns out to be Chern-Einstein. For this reason the original definition in \cite{G1} included the hypothesis that the metric is Gauduchon, namely $\partial\bar\partial(\o^{n-1}) =0$, in order to exclude this less significant situation. In case of homogeneous manifold, it is clear that $\mu$ has to be constant and indeed every invariant Hermitian metric is Gauduchon. In \cite{LY} and \cite{G1}  it is shown that the canonical metric on the Hopf manifold $S^{2n+1}\times S^1$ is Chern-Einstein with $\mu >0$. This also shows that, in contrast with the K\"ahler-Einstein case, the existence of a positive Chern-Einstein metric does not imply the simply connectedness of the manifold. Nevertheless, a compact complex manifold $M$ with finite fundamental group is simply connected when it carries a positive Chern-Einstein metric. This indeed follows from the the fact the aritmetic genus  $\chi(M,\mathcal O) =1$ and this invariant is actually multiplicative with finite coverings.\end{remark}
\begin{remark} We also recall that there exists a third Ricci tensor $S^{(3)}$ which is defined as $S^{(3)}_{\a,\bar\b} := h^{p\bar q}R_{\a\bar q p\bar\b}$. The Einstein condition $S^{(1)} = \mu h$ or $S^{(3)} = \mu h$ for some constant $\mu\neq 0$ is easily seen to imply $h$ to be K\"ahler (see \cite{B}).
\end{remark}
We may now state our main result
\begin{theorem}\label{main} Let $M$ be a manifold in the class $\mathcal C$ endowed with an invariant complex structure $J$. Then $M$ is simply connected, non K\"ahler and
\begin{itemize}
\item[a)] $M$ does not admit any balanced or SKT Hermitian metric;
\item [b)] $M$ admits an invariant Hermitian metric $\bar g$ which is Chern-Einstein with $S^{(2)}(\bar g) = \bar g$. Moreover, if $J$ belongs to a suitable neighborhood of the standard complex structure , the metric $\bar g$ is the only invariant Chern-Einstein metric satisfying $S^{(2)}(\bar g) = \bar g$;
\item[c)] $c_1(M)\geq 0$.
\end{itemize}
\end{theorem}
Before starting with the proof of the main Theorem, we describe the Chern connection of an invariant Hermitian metric and prove some basic facts.\par

Given an invariant Hermitian metric $h$ on a complex homogeneous space $G/L$, we will describe its associated Chern connection $\nabla$. \par
We see $h$ as an $\Ad(L)$-invariant inner product $h$ on the $\Ad(L)$-invariant complement $\mathfrak{m}$ with $\gg = \mathfrak{l} \oplus \mathfrak{m}$. Moreover $h$ is supposed to be Hermitian w.r.t. the invariant complex structure $J$ on $\mathfrak{m}$. Being $G$-invariant, the torsion $T$ can be seen as an element of $\Lambda^2\mathfrak{m}\otimes\mathfrak{m}$ and after complexification, the condition \eqref{tor1} is equivalent to
\begin{equation}\label{tor3} T(\mathfrak{m}^{10},\mathfrak{m}^{01})=0.\end{equation}
Since $\nabla$ is an invariant connection on $G/L$, it is well known that it is completely determined by a map $\Lambda\in \mbox{Hom}(\mathfrak{m},\mbox{End}(\mathfrak{m}))$, where the
correspondence can be described as follows. If $X\in \gg$ we denote by $X^*$ the corresponding vector field on $M=G/L$ and observe that the map
$\mathfrak{m}\ni X\mapsto (X^*)|_{[eL]} \in T_{[eL]}M$ is an isomorphism. Then for $X,Y\in \mathfrak{m}$ and $p=[eL]\in M$,
$$(\Lambda(X)Y)^*|_p = (\nabla_{X^*}Y^* - [X^*,Y^*])|_p.$$
\begin{lemma} The condition $\nabla J = 0$ implies $[\Lambda(X),J]=0$.
\end{lemma}
\begin{proof} Using the definition of $J$ as an endomorphism of $\mathfrak{m}$ (namely $(JX)^*|_p = J X^*|_p$) we see that
$$(\nabla_XJY)^*|_p = \nabla_{X^*}(JY)^* = (\nabla_{(JY)^*}X^* + [X^*,(JY)^*] + T(X^*,(JY)^*))|_p = $$
$$= (\nabla_{JY^*}X^* + [X^*,(JY)^*] + T(X^*,JY^*))|_p =$$
$$= (J\nabla_{X^*}Y^* + [J Y^*,X^*] + T(J Y^*,X^*) + [X^*,(JY)^*] + T(X^*,JY^*))|_p =$$
$$= (J\nabla_{X^*}Y^* - J[X^*,Y^*])|_p  + [X^*,(JY)^*]|_p = J(\Lambda(X)Y)^*|_p + [X^*,(JY)^*]|_p$$
and our claim follows.\end{proof}
Therefore when we extend $\Lambda$ to an element $\Lambda\in \mbox{Hom}(\mathfrak{m}^c,\mbox{End}(\mathfrak{m}^c))$, we have that $\Lambda(\cdot)(\mathfrak{m}^{10})\subseteq\mathfrak{m}^{10}$.
The torsion is given by (see \cite{KN},~p.192)
$$T(X,Y) = \Lambda(X)Y - \Lambda(Y)X - [X,Y]_\mathfrak{m}$$
and condition \eqref{tor3} implies for $A,B\in \mathfrak{m}^{10}$
$$0 = (\Lambda(A)\bar B - \Lambda(\bar B)A - [A,\bar B])^{10} = $$
$$ = (- \Lambda(\bar B)A - [A,\bar B])^{10},$$
i.e.
\begin{equation} \Lambda(\bar B)A = (\Lambda(\bar B)A)^{10} =  [\bar B,A]^{10}.\end{equation}
Conjugation yields
\begin{equation}\label{AbarB} \Lambda(A)\bar B = [A,\bar B]^{01}.\end{equation}
\subsection{The proof of the main Theorem}
\medskip
We already know that any manifold $M$ in $\mathcal C$ does not admit any balanced metric. We start here with some generalities in order to prove our main result.\par
Using the same notations as above, we consider the Cartan subalgebra $\mathfrak{a}_i$ in $\mathfrak{h}_i^c$ given by $\mathfrak{t}_i^c\oplus \mathfrak{s}_i$, where $\mathfrak{t}_i = \mathbb R\cdot Z_i$ and $\mathfrak{s}_i$ is a Cartan subalgebra of $(\mathfrak{h}^s_i)^c$ for $i=1,2$. We denote by $R^{(i)}$ the corresponding root systems, which are then endowed with an invariant ordering corresponding to the invariant complex structure $I_i$ on $\mathfrak{n}_i$ ($i=1,2$). Then $R^{(i)}= R_{\mathfrak{h}_i}\cup R_{\mathfrak{n}_i}^+ \cup R_{\mathfrak{n}_i}^-$. In the sequel we will extend each root $\a\in R^{(i)}$ to a functional on $\mathfrak{a}_1\oplus \mathfrak{a}_2$ by putting $\a|_{\mathfrak{a}_j}\equiv 0$ if $i\neq j$.\par
We are interested in studying special Hermitian metrics on these manifolds. We recall that a Hermitian invariant metric $h$ on $M$ is given by an Hermitian
$\Ad(L)$-invariant inner product on $\mathfrak{m} = \mathfrak{t} +\mathfrak{n}$, where $\mathfrak{n} = \mathfrak{n}_1\oplus \mathfrak{n}_2$. Moreover, whenever $\dim\mathfrak{h}_i > 1$, $i=1,2$, the tangent space
$\mathfrak{m}$ splits as the sum of three inequivalent, and therefore $h$-orthogonal, submodules $\mathfrak{t},\mathfrak{n}_1,\mathfrak{n}_2$. We put $n_i:= \dim_\mathbb C\mathfrak{n}_i$ for $i=1,2$.\par
Moreover, by $L$-irreducibility, the metric $h$ on each $\mathfrak{n}_i$ is a negative multiple  of the restriction of the Cartan Killing form $B_i$ of $\gg_i$ on $\mathfrak{n}_i$ ($i=1,2$). We will also denote by $H$ a non zero element of $\mathfrak{t}^{10}$, say $H= Z_1 - iJ_{\mathfrak{t}}Z_1$, and we  put
$$h_o := h(H,\bar H)>0.$$
Note that if $\alpha,\b$ are  positive roots in $R_{\mathfrak{n}_i}^+$ then by the $\Ad(L)$-invariance for every $v\in \mathfrak{s}_i$
$$0=h([v,E_\a],E_{-\b}) + h(E_\a,[v,E_{-\b}]) = (\a-\b)(v)\cdot h(E_\a,E_{-\b}).$$
Since $\a\neq\b$ implies that $\a\not\equiv\b$ on $\mathfrak{s}_i$, we see that $h(E_\a,E_{-\b}) \neq 0$ only when $\a=\b$. In this case $h(E_\a,\bar E_\a) := g_i$ when we use the normalized root vectors $\{E_\a\}$ given by a Chevalley basis (recall also that $\bar E_\a = - E_{-\a}$). \par\medskip
\begin{lemma} $M$ does not admit any SKT metric.\end{lemma}
\begin{proof} Suppose $h$ is a SKT metric, which can be supposed to be invariant using the compactness of the group $G$. The SKT condition amounts to say that $dd^c \o = 0$, where $\o$ is the corresponding K\"ahler form. We will use the Koszul formula for the differential of an invariant $q$-form $\phi$, where for $v_0,\ldots,v_q\in \mathfrak{m}$
$$d\phi(v_o,\ldots,v_q) = \sum_{i<j}(-1)^{i+j}\phi([v_i,v_j]_\mathfrak{m},v_0,\ldots,\tilde v_i,\ldots,\tilde v_j,\ldots, v_q),$$
where $\tilde{\cdot} $ indicates that the corresponding vector does not appear. Select $\a,\b\in R_{\mathfrak{n}_1}$ and we compute
$$d(d^c\o)(E_\a,E_{-\a},E_\b,E_{-\b}) = - d^c\o((H_\a)_\mathfrak{m},E_\b,E_{-\b}) - d^c\o((H_\b)_\mathfrak{m}, E_\a,E_{-\a}) = $$
$$= - d\o(J(H_\a)_\mathfrak{m},E_\b,E_{-\b}) - d\o(J(H_\b)_\mathfrak{m}, E_\a,E_{-\a}).$$
Now
$$d\o(J(H_\a)_\mathfrak{m},E_\b,E_{-\b}) = - \o(\b(J(H_\a)_\mathfrak{m})E_\b,E_{-\b}) - \o(\b(J(H_\a)_\mathfrak{m})E_{-\b},E_\b) - \o((H_\b)_\mathfrak{m},J(H_\a)_\mathfrak{m}) = - h((H_\a)_\mathfrak{m},(H_\b)_\mathfrak{m})$$
is symmetric in $\a,\b$ and therefore we get
$$0=d(d^c\o)(E_\a,E_{-\a},E_\b,E_{-\b}) = 2 g((H_\a)_\mathfrak{m},(H_\b)_\mathfrak{m}).$$
Now $(H_\a)_\mathfrak{m} = \frac{B_1(H_\a,Z_1)}{B_1(Z_1,Z_1)} Z_1 = -\frac{\sqrt{-1}}{2n_1}Z_1$, so that last equation implies $h(Z_1,Z_1) =0$ , a contradiction.
\end{proof}

In order to prove the existence of a Chern-Einstein metric, we prove the following

\begin{lemma}\label{Lambda} Given $\a,\b\in R_{\mathfrak{n}_i}^+$, $\gamma\in R_{\mathfrak{n}_j}^+$ with $i\neq j$, we have
\begin{itemize}
\item[a)] $\Lambda(E_\a)E_\b = 0$,\quad $\Lambda(E_\a)E_{\pm\gamma} = 0$ and $\Lambda(E_\a)\bar H=0 $;
\item[b)] $\Lambda(E_\a)E_{-\b} = 0$ for $\b\neq \a$;\quad $\Lambda(E_\a)E_{-\a} = -\frac {\sqrt{-1}}{2n_i}(Z_i)^{01}$;
\item[c)] $\Lambda(E_\a)H= \frac{h(H,H_\a)}{g_i}E_\a$;
\item[d)] $\Lambda(v) = \ad(v)$ for $v\in\mathfrak{t}^c$.

\end{itemize}

\end{lemma}
\begin{proof} a) Given $A\in \mathfrak{n}^{10}$ we have by \eqref{AbarB} and by $[\mathfrak{n}_i,\mathfrak{n}_i]\subseteq \mathfrak{h}_i$
$$h(\Lambda(E_\a)E_\b,\bar A) = -h(E_\b,\Lambda(E_\a)\bar A) = -h(E_\b,[E_\a,\bar A]) = 0.$$
Moreover,
$$h(\Lambda(E_\a)E_\b,\bar H) = -h(E_\b,\Lambda(E_\a)\bar H) = -h(E_\b,[E_\a,\bar H]) = h(E_\b,\a(\bar H)E_\a)=0.$$
The same kind of arguments shows the second assertion. Finally by \eqref{AbarB}, $\Lambda(E_\a)\bar H = [E_\a,\bar H]^{01} = \a(\bar H)E_\a^{01}=0$.\par
\noindent b) We have $\Lambda(E_\a)E_{-\b} = [E_\a,E_{-\b}]^{01} \neq 0$ only if $\a=\b$ and in this case $\Lambda(E_\a)E_{-\a}=
[E_\a,E_{-\a}]^{01}= [H_\a]^{01}$. Now
$$ [H_\a]^{01} = \frac{B_i(H_\a,Z_i)}{B_i(Z_i,Z_i)} Z_i^{01} = -\frac {\sqrt{-1}}{2n_i}Z_i^{01}.$$\par
\noindent c) We have $h(\Lambda(E_\a)H,E_{-\a}) = - h(H,H_\a^{01}) = - h(H,H_\a)$ and our claim follows.\par
\noindent d) First note that $\Lambda(H)\bar H = [H,\bar H]^{01} = 0$ and $\Lambda(H)E_{-\a}=-\a(H)E_{-\a}$ by \eqref{AbarB}. From this we see that $\Lambda(H)H=0$. Now
$$h(\Lambda(H)E_\a,\bar H) = -h(\Lambda(H)\bar H,E_\a)=0,$$
$$h(\Lambda(H)E_\a,E_{-\b}) = -h(E_\a,-\b(H)E_{-\b}) = \d_{\a,\b}\a(H) h(E_\a,E_{-\a})$$
so that $\Lambda(H)E_\a= \a(H)E_\a$. From this we see that $\Lambda(\bar H)E_\a = -\overline{\a(H)}E_\a$. Since $\a|_{\mathfrak{t}}\in \sqrt{-1}\mathbb R$, we see that $\overline{\a(v)} = -\a(\bar v)$ for $v\in \mathfrak{t}^c$, hence $\Lambda(\bar H)E_\a= \a(\bar H)E_\a$. Similarly for $\Lambda(\bar H)E_{-\a} = -\a(\bar H)E_{-\a}$.\end{proof}
In order to compute the second Ricci tensor $S^{(2)}$ ($S$ for brevity throughout the following), we compute the curvature. We use the general formula for the curvature of an invariant metric (see e.g. \cite{KN}, p. 192) for $v,w\in \mathfrak{m}$
\begin{equation}\label{hcurv}R(v,w) = [\Lambda(v),\Lambda(w)] - \Lambda([v,w]_{\mathfrak{m}}) - \ad([v,w]_{\mathfrak{l}}).\end{equation}
Given $\a,\b\in R_{\mathfrak{n}_i}^+$ we have
\begin{equation}\label{curv}R(E_{\a},\bar E_{\a})E_\b = [\Lambda(E_\a),\Lambda(\bar E_\a)]E_\b - \Lambda((H_\a)_{\mathfrak{m}^c})E_\b - [(H_\a)_{\mathfrak{l}^c},E_\b]=\end{equation}
$$= \Lambda(E_\a)\Lambda(\bar E_\a)E_\b - \Lambda((H_\a)_{\mathfrak{m}^c})E_\b - [(H_\a)_{\mathfrak{l}^c},E_\b]=$$
$$= \Lambda(E_\a)\Lambda(\bar E_\a)E_\b - \b(H_\a)E_\b,$$
where we have used Lemma \ref{Lambda}, (a)(d). Note also that $R(H,\bar H)E_\a=0$. \par

Using Lemma \ref{Lambda} we see that for $\b\in R_{\mathfrak{n}_i}$
$$S(E_\b,\bar E_\b) = \sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i}h(R(E_\a,\bar E_\a)E_\b,\bar E_\b) + \frac 1{h_o}h(R(H,\bar H)E_\b,\bar E_\b) = $$
$$= -\sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i}h(\Lambda(\bar E_\a)E_\b,\Lambda( E_\a)\bar E_\b) -\sum_{\a\in R_{\mathfrak{n}_i}^+} \b(H_\a)= $$
$$ = -\frac 1{g_i}h(\Lambda(\bar E_\b)E_\b,\Lambda( E_\b)\bar E_\b) -\sum_{\a\in R_{\mathfrak{n}_i}^+} \b(H_\a)=$$
$$ = -\frac 1{g_i}h(H_\b^{01},\overline{H_\b^{01}}) -\sum_{\a\in R_{\mathfrak{n}_i}^+} \b(H_\a)=$$
$$ = \frac 1{g_i}h(H_\b^{01},H_\b^{10}) -\sum_{\a\in R_{\mathfrak{n}_i}^+} \b(H_\a)=$$
$$ = -\frac{1}{4g_in_i^2} h(Z_i^{10},Z_i^{01}) -\sum_{\a\in R_{\mathfrak{n}_i}^+} \b(H_\a)= -\frac 1{8g_in_i^2}h(Z_i,Z_i) + \frac 12,$$

where we have used that $\sum _{\a\in R_{\mathfrak{n}_i}^+} H_\a =
-\frac {\sqrt{-1}}{2}Z_i$ and $\b(Z_i)=\sqrt{-1}$. Now it is immediate to see that
$$h(Z_1,Z_1)=\frac 12 h_o,\qquad h(Z_2,Z_2)= \frac{1+a^2}{b^2}h(Z_1,Z_1) = \frac{1+a^2}{2b^2}h_o,$$
so that
\begin{equation}\label{S1} S(E_\b,\bar E_\b) =  -\frac {h_o}{16g_1n_1^2} + \frac 12,\qquad \b\in R_{\mathfrak{n}_1},\end{equation}
\begin{equation}\label{S} S(E_\b,\bar E_\b) =  -\frac {(1+a^2)h_o}{16b^2g_2n_2^2} + \frac 12,\qquad \b\in R_{\mathfrak{n}_2},\end{equation}
\par
We now compute
$$S(H,\bar H) = \sum_{i=1}^2\sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i}h(R(E_\a,\bar E_\a)H,\bar H) =$$
$$= -\sum_{i=1}^2 \sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i}h(\Lambda(\bar E_\a)\Lambda(E_\a)H,\bar H) =
\sum_{i=1}^2 \sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i}h(\Lambda(E_\a)H,\Lambda(\bar E_\a)\bar H) =$$
$$= \sum_{i=1}^2 \sum_{\a\in R_{\mathfrak{n}_i}^+}\frac 1{g_i^2}|h(H,H_\a)|^2.$$
Now it is immediate to see that
$$h(H,H_\a) = -\frac{\sqrt{-1}}{2n_1}h(Z_1,Z_1),\qquad \a\in R_{\mathfrak{n}_1}^+,$$
$$h(H,H_\a) = \frac{\sqrt{-1}}{2n_2}(-\frac ab - \frac{\sqrt{-1}}b)h(Z_1,Z_1),\qquad \a\in R_{\mathfrak{n}_2}^+$$
so that
$$S(H,\bar H) = \frac{h_0^2}{16}\left( \frac{1}{g_1^2n_1} + \frac{1+a^2}{b^2g_2^2n_2}\right).$$

Summing up the Hermitian Einstein equations are
\begin{equation}\label{eq} \left\{\begin{matrix}  \frac{h_0}{16}\left( \frac{1}{g_1^2n_1} + \frac{1+a^2}{b^2g_2^2n_2}\right) &= 1 \\
-\frac{h_o}{16n_1^2g_1} + \frac 12 &= g_1 \\
-\frac {(1+a^2)h_o}{16b^2g_2n_2^2} + \frac 12 &= g_2.    \end{matrix}\right.
\end{equation}
Putting $x:= 1/g_1$, $y:= 1/g_2$ and $z:= 16/h_o$, the system \eqref{eq} can be written as
\begin{equation}\label{eq} \left\{\begin{matrix}  z &=\frac 1{n_1}x^2 + \frac {1+a^2}{b^2n_2} y^2  \\
z(x-2)  &= \frac{2}{n_1^2} x^2 \\
z(y-2) &= \frac{2(1+a^2)}{b^2n_2^2} y^2     \end{matrix}\right.
\end{equation}
which is equivalent to
\begin{equation}\label{eq1} \left\{\begin{matrix}  z &=\frac 1{n_1}x^2 + \frac {1+a^2}{b^2n_2} y^2  \\
z(x-2)  &= \frac{2}{n_1^2} x^2 \\
n_1 x + n_2 y &= 2n_1+2n_2 + 2 .    \end{matrix}\right.
\end{equation}
We have an admissible solution $x,y,z\in \mathbb R^+$ if and only if there is a solution $x$ of the polynomial equation
\begin{equation}\label{eq}\phi(x):= \left[ \frac 1{n_1}x^2 + \frac {1+a^2}{b^2n_2}(\frac{2n_1+2n_2+2-n_1x}{n_2})^2\right]\cdot (x-2)- \frac 2{n_1^2}x^2 = 0\end{equation}
satisfying the conditions
$$x> 0,\qquad x < \frac {2n_1+2n_2+2}{n_1}.$$
This follows immediately from the fact that
$$\phi(0)<0,\qquad \phi(\frac {2n_1+2n_2+2}{n_1}) = \frac{8n_2(n_1+n_2+1)^2}{n_1^4} > 0.$$
We now put $\frac {1+a^2}{b^2}=1$ throughout the following and prove the uniqueness of the Hermitian Einstein metric. \par
\begin{lemma} Any solution of the equation $\phi(x)=0$ satisfies $x\in [2,2+\frac 2{n_1}]$. If there are two distinct solutions $x_1<x_2$, then the equation
$\phi'(x)=0$ has two distinct solutions $y_1<y_2$ in $[2,2+\frac 2{n_1}]$.
\end{lemma}
\begin{proof} It is immediate to see that $\phi(2)<0$ and $\phi(2+\frac 2{n_1}) = \frac 8{n_1n_2} > 0$. Moreover any solution $x$ satisfies
$$0 < \frac 2{n_1^2}x^2 =  \left[ \frac 1{n_1}x^2 + (\frac{2n_1+2n_2+2-n_1x}{n_2})^2\right]\cdot (x-2) \geq \frac 1{n_1}x^2 \cdot (x-2), $$
hence $x\in [2,2+\frac 2{n_1}]$. The second claim follows immediately from the fact that $\phi'$ is a polynomial of degree $2$. \end{proof}
If we suppose that there are two distinct solutions $x_1,x_2$ of \eqref{eq}, then by the previous Lemma we get $2< \frac 12(y_1+y_2)< 2+\frac 2{n_1}$. Using Maple, this last condition is given by
$$\left\{ \begin{matrix} 2 n_1 n_2^3 - 2 n_2n_1^3 - n_2^3 - 2 n_1^3 &\leq 0\\
2 n_2 n_1^3 - 2 n_1n_2^3 - n_1^3 - 2 n_2^3 &\leq 0\end{matrix}\right.$$
that can be rewritten as
\begin{equation}\label{diseq}\left\{ \begin{matrix} \frac{n_1^3}{n_1+1} &\leq 2 \frac{n_2^3}{2n_2-1}\\
\frac{n_2^3}{n_2+1} &\leq 2 \frac{n_1^3}{2n_1-1}\end{matrix}\right.\end{equation}
Now we observe that if, say, $n_1 = 1$, then $n_2^3\leq 2n_2+2$, giving $n_2=1$ and similarly if $n_2=1$ we get $n_1=1$. The cubic equation $\phi(x)=0$ with $n_1=n_2=1$ can be easily checked to have only one solution, so that we can suppose $n_1,n_2\geq 2$. By \eqref{diseq} we see that
\begin{equation}\label{diseq1}\left\{ \begin{matrix} n_1&\leq \sqrt{2} n_2\\
n_2&\leq \sqrt{2} n_1.\end{matrix}\right.\end{equation}
Now the discriminant $d$ of the equation $\phi'(x)=0$ is given by
$$0< d:= n_1^6 + n_2^6 + 2 n_1^6 n_2 + 2 n_2^6 n_1 + n_1^6 n_2^2 + n_1^2 n_2^6 + n_1^3 n_2^3 + $$
$$- (8 n_1^4n_2^4 + 3 n_1^3n_2^5 + 3 n_1^5 n_2^3 + 2 n_1^3n_2^4 + 2 n_1^4 n_2^3)  $$
which is a symmetric expression in $n_1,n_2$. We can suppose $n_2\leq n_1$ and using $n_2\geq \frac1{\sqrt 2} n_1$ by \eqref{diseq1}, we see that
$$0 < d \leq 2 n_1^8 + 4 n_1^7 + 3 n_1^6 - ((2+\frac 9{4\sqrt 2})n_1^8 + (\frac1{\sqrt 2}+\frac 12)n_1^7) =$$
$$= - n_1^6\left( \frac 9{4\sqrt 2} n_1^2 - (\frac 72 - \frac 1{\sqrt 2}) n_1 - 3\right) < 0$$
for $n_1\geq 3$. So we are left with the case $n_1 = n_2 = 2$. In general for $n_1=n_2$ the equation admits only one solution, which is explicitely given by
$$g_1=g_2 = \frac {n_1}{2n_1+1},\ h_o = \frac {8 n_1^3}{(2n_1+1)^2}.$$
In order to prove (c) in Theorem \eqref{main}, we compute the first Chern-Ricci tensor $\rho$ of an invariant metric $h$. Using the general formula \eqref{hcurv} and Lemma \ref{Lambda},(d),  we see that for $\a\in R_{\mathfrak{n}_1}$,
$$\rho(E_\a,\bar E_\a) = -\sum_{\b\in R_{\mathfrak{n}_1}}\b(H_a) = \frac 12.$$
Similarly for $\a\in R_{\mathfrak{n}_2}$, we see that $\rho(E_\a,\bar E_\a) = \frac 12$. Since $R(H,\bar H)=0$, we have $\rho(H,\bar H)=0$ and therefore $\rho\geq 0$. \par\bigskip
\noindent{\bf The Chern-Ricci flow.} As a last remark, we consider the Chern-Ricci flow
\begin{equation}\label{flow}h_t' = h_t - S(h_t),\end{equation}
which has a Chern-Einstein metric as an equilibrium point. It is known that there exists a solution for some interval $t\in [0,T)$ for any initial metric $\bar h$. Moreover it is immediate to observe that the solution $h_t$ still has the full group $G$ acting by isometric biholomorphisms. Using a special case
given by some manifold $M\in \mathcal C$ with $n_1=n_2=2$, we see numerically that the long time existence is not guaranteed and that even when the initial
 metric $\bar h$ has positive Ricci tensor $S$, the flow does not necessarily converge to the Chern-Einstein metric.
\section{Invariant balanced metrics}
We keep the same notations as in the previous sections and we consider a complex homogeneous space $M=G/L$ of complex dimension $n$ as in Section 2. We like to study the existence of invariant balanced metrics. We recall that a Hermitian metric $h$ is called {\it balanced} if $d(\o^{n-1})=0$ where $\o$ denotes the K\"ahler form. This definition is actually equivalent to requiring that $\d\o = 0$, where $\d$ denotes the co-differential w.r.t. the metric $h$. \par
We also recall that if a balanced metric exists, then an invariant balanced metric exists too (see \cite{FG}). We now focus on the possible construction of {\it adapted} balanced metrics on $G/L$,namely metrics which submerge an invariant Hermitian metric on the corresponding flag manifold $G/H$ with $\mathfrak{t}$ and $\mathfrak{n}$ being orthogonal (note that any invariant metric on $G/L$ is of this form whenever $\mathfrak{h}$ coincides with the centralizer in $\gg$ of its semisimple part). The condition of being adapted balanced has been already investigated in \cite{GGP}, Lemma 2; here we give a direct proof using some standard computations on the Levi Civita connection, which might be useful for further research. We start proving the following Lemma, where we denote by $D$ the Levi Civita connection of $h$.
\begin{lemma}\label{lemma1} The metric $h$ is balanced if and only if , given $\{e_i\}_{i=1,\ldots,2n}$ an orthonormal basis of the tangent space $\mathfrak{m} \cong T_{[eL]}M$
we have
$$\sum_i JD_{e_i}e_i - D_{Je_i}e_i = 0.$$
\end{lemma}
\begin{proof} We know that $\d\o(v)=-\sum_i (D_{e_i}\o)(e_i,v)$ for $v\in \mathfrak{m}$. We extend any element of $\mathfrak{m}$ to the corresponding Killing vector field
which will be denoted by the same letter with $^*$. We have
$$-(D_{e_i^*}\o)(e_i^*,v^*) = - e_i^*\o(e_i^*,v^*) + \o(D_{e_i^*}e_i^*,v^*) + \o(e_i^*,D_{e_i^*}v^*) = $$
$$= - \o(e_i^*,[e_i^*,v^*]) + \o(D_{e_i^*}e_i^*,v^*) + \o(e_i^*,D_{e_i^*}v^*) = \o(D_{e_i^*}e_i^*,v^*) + \o(e_i^*,D_{v^*}{e_i^*}) = $$
$$= \o(D_{e_i^*}e_i^*,v^*) + h(Je_i^*,D_{v^*}{e_i^*}) = h(JD_{e_i^*}e_i^*,v^*) - h(v^*,D_{Je_i^*}{e_i^*})$$
and our claim follows.\end{proof}

We compute the Levi Civita connection using the standard formula (see e.g. \cite{KN}) for $v,w,z\in \mathfrak{m}$
\begin{equation}\label{LC} 2h(D_vw,z) = h([v,w]_\mathfrak{m},z) + h([z,v]_\mathfrak{m},w) + h([z,w]_\mathfrak{m},v).\end{equation}
We immediately see that for every $v,w\in \mathfrak{t}$ we have $D_vw=0$ because $\ad(v)(\mathfrak{n})\subseteq \mathfrak{n}$ and $h(\mathfrak{t},\mathfrak{n})=0$. For every $\a\in R_\mathfrak{n}^+$ we consider the vectors $e_\a:= \frac{E_\a-E_{-\a}}{\sqrt {2 g_\a}}$ so that $\{e_\a,Je_\a\}_{\a\in R_\mathfrak{n}^+}$ gives an orthonormal basis of $\mathfrak{n}$. A simple computation shows that
\begin{equation}\label{D1} D_{e_\a}e_\a + D_{Je_\a}Je_\a = -\frac 1{g_\a}\left[ D_{E_\a}E_{-\a}+ D_{E_{-\a}}E_\a\right],\qquad  D_{Je_\a}e_\a - D_{e_\a}Je_\a = \frac{i}{g_\a}\left[D_{E_{-\a}}E_{\a} - D_{E_\a}E_{-\a} \right].\end{equation}
Now, using \eqref{LC} we see that for every root $\b\in R_\mathfrak{n}$, $v\in \mathfrak{t}^c$
$$h(D_{E_\a}E_{-\a},E_\b) = 0,\qquad h(D_{E_\a}E_{-\a},v) = \frac 12 h(H_\a,v),$$
so that
$$D_{E_\a}E_{-\a} = \frac 12(H_\a)_{\mathfrak{t}^c},\qquad D_{E_{-\a}}E_{\a} = \overline{D_{E_{\a}}E_{-\a}} = - \frac 12 (H_\a)_{\mathfrak{t}^c}.$$
Therefore $\sum_{i=1}^{2n} D_{e_i}e_i = 0$ by \eqref{D1} and the condition in Lemma~\ref{lemma1} becomes $\sum_{i=1}^{2n} D_{Je_i}e_i = 0$. Therefore by \eqref{D1}, $h$ is balanced if and only if
\begin{equation}\label{bal} \sum_{\a\in R_\mathfrak{n}^+} \frac 1{g_\a} H_\a|_{\mathfrak{t}^c} = 0\qquad\mbox{or\ equivalently}\quad \sum_{\a\in R_\mathfrak{n}^+} \frac 1{g_\a} H_\a \in \sqrt{-1}\mathfrak{l}\end{equation}
We define the vector
\begin{equation}\label{def} \d_h:= \sum_{\a\in R_\mathfrak{n}^+} \frac 1{g_\a} H_\a\end{equation}
and note that $\d_h$ is a slight modification of the standard Koszul element $\d_{\kappa}:= \frac 12 \sum_{\a\in R_\mathfrak{n}^+} H_a$ which lies in $\sqrt{-1}\mathfrak{z}$, where $\mathfrak{z}$ is the center of $\mathfrak{h}$. Indeed, we can prove that $\d_h$ is a non zero vector in  $ \sqrt{-1}\mathfrak{z}$ by the following arguments. First of all we decompose $\mathfrak{n}^{1,0}=\oplus_{j=1}^s \mathfrak{q}_i$ as a sum of irreducible $\mathfrak{h}$-modules $\mathfrak{q}_j$, $j=1,\ldots,s$. Note that there exist $R_j\subset R_\mathfrak{n}^+$ with $\mathfrak{q}_j = \bigoplus_{\a\in R_j}\gg_\a$, $j=1,\ldots,s$. We prove the following
\begin{lemma} Given  $\zeta_j:= \sum_{\a\in R_j}H_\a$, then $\sqrt{-1}\zeta_j\in\mathfrak{z}$.\end{lemma}
\begin{proof} We fix $\gamma\in R_\mathfrak{k}$ and for every $\b\in R_j$ we consider the maximal $\gamma$-string $\{\b+k\gamma,\ p\leq k\leq q\}$. Note that $(R_j\pm\gamma)\cap R\subseteq R_j$ by the $\Ad(H)$-invariance of $\mathfrak{n}_1$. This means that the whole $\gamma$-string belongs to $R_j$. Moreover
$$\sum_p^q (\b+k\gamma,\gamma) = (q-p+1)(\b,\gamma) + \frac 12(q(q+1)+p(1-p)) ||\gamma||^2 = \frac {||\gamma||^2}2[-(p+q)(q-p+1) + q(q+1)+p(1-p)] = 0.$$
Since the whole $R_j$ splits up as the disjoint union of $\gamma$-strings, we can sum up all the scalar products with $\gamma$ and we get that $(\zeta_j,H_\gamma)=0$. Since $\sqrt{-1}\zeta_j$ belongs to the Cartan subalgebra of $\mathfrak{h}$ and is orthogonal to every $H_\gamma$, $\gamma\in R_\mathfrak{h}$, it lies in the center of $\mathfrak{h}$.\end{proof}
Now it is clear that for every $j=1,\ldots,s$ and for every $\a,\b\in R_j$ we have $g_\a=g_\b$ and this common value will be called $g_j$. We can
write $\d_h = \sum_{j=1}^s\frac 1{g_j}\zeta_j$ and therefore it lies in  $\sqrt{-1}\mathfrak{z}$. Moreover, since $(\d_\kappa,H_\a)>0$ for every $\a\in R_\mathfrak{n}^+$ (see e.g. \cite{BFR}), we see that $(\d_\kappa,\d_h)>0$ and therefore $\d_h\neq 0$.
Our result is the following
\begin{theorem} Let $G$ be a compact connected semisimple Lie group.
\begin{itemize}
\item[i)] Let $M=G/L$ be a compact simply connected complex homogeneous space. Then an adapted $G$-invariant Hermitian metric $h$ on $M$ is balanced if and only if  $\sqrt{-1}\d_h$ lies in the center of $\mathfrak{l}$.
\item[ii)] Let $Q:=G/H$ be a flag manifold with $b_2(Q)\geq 3$. Then there exists a complex homogeneous space $G/L$ with Tits fibration $G/L\to Q$, which admits a balanced metric.
\end{itemize}\end{theorem}
In order to prove (ii), we recall some standard facts about flag manifolds and $T$-roots(see e.g. \cite{Al}). It is known that there exists a system of simple roots $\{\a_1,\ldots,\a_p,\b_1,\ldots,\b_t\}$ for $\gg$ such that $\{\a_1,\ldots,\a_k\}$ is a system of simple roots for $\mathfrak{h}_{ss}$ and $\b_1,\ldots,\b_t\in R_\mathfrak{n}^+$. Moreover we can reorder the modules $\mathfrak{n}_j$ ($j=1,\ldots,s$) so that $\b_j\in R_j$ for $j=1,\ldots,t$. This implies that
$\{\zeta_1,\ldots,\zeta_t\}$ is a basis of $\sqrt{-1}\mathfrak{z}$ and there exist non negative integers $n_{ij}$ with
$\zeta_i = \sum_{j=1}^t n_{ij}\zeta_j$ for $i=t+1,\ldots,s$. Now let $\Lambda$ be the integral lattice in $\mathfrak{z}$ given by the kernel of the exponential map. We can find
$v\in \Lambda$ so that $\sqrt{-1}v = \sum_{j=1}^t c_j\zeta_j$ with $c_j >0$ for $j=1,\ldots,t$ and, up to a suitable scaling by a positive real number, we can suppose that $c_j>\sum_{k=t+1}^sn_{kj}$ for $j=1,\ldots,s$. We now put $g_i = 1$ for $i=t+1,\ldots,s$ and $\frac1{g_j}= c_j - \sum_{k=t+1}^sn_{kj} >0$, defining an invariant metric $h$ on $G/H$. The corresponding $\sqrt{-1}\d_h$ will therefore generate a one-dimensional line in $\mathfrak{z}$ which integrates to a closed one-dimensional torus $T$ by construction. Since $b_2(Q) = \dim \mathfrak{z}\geq 3$, we can find a torus $\tilde T$ (of dimension $1$ or $2$) with $T\subseteq \tilde T\subset Z(H)$ so that the codimension of $\tilde T$ in $Z(H)$ is even and this gives the isotropy $L = \tilde T\cdot H_{ss}$.\par
As a final remark, we note that the case when $b_2(Q)=2$ has been already treated in full generality in Proposition \ref{bal0}.

\iffalse

\fi

\vfil\eject
\bigskip\bigskip

redpajama_set_name: RedPajamaArXiv


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How critical is cyber security to the success of enterprise initiatives? Think about it like this — there's no facet of your organization that isn't impacted by IT services. There are obvious links between accounting and technology services and the need for IT oversight when it comes to managing mobile devices. However, that's just the beginning. Now, human resource departments are using cloud-based solutions to manage personnel and recruit new staff; marketers are mining social data to improve product campaigns; and executive staff rely on data analytics tools for long-term decision-making strategies.
The common denominator? Everything depends on an existing IT product, a custom-built tech solution or emerging technology that could take your business to the next level. Nonetheless, increasing integration demands effective IT security management. If cloud apps aren't properly secured, remote access points don't properly authenticate, or databases are compromised by hackers, the results are disastrous for your organization.
Upfront security costs — The average cost of a data breach is just over $3.6 million per organization, while stolen records cost $141. Even one network breach per year could mean the difference between profit and loss. Once hackers know your system is vulnerable, they're more likely to try again.
Public relations problems — Customers are more tech-savvy than ever. While they're willing to provide personal information to access e-commerce and website services, they won't let data breaches slide. Both human error and technological failure lead to the same outcome: Consumers taking their business elsewhere.
Long-term disruptions — Paying to clean systems and restore records is a substantial cost, but it doesn't stop there. IT teams must ensure networks aren't prone to another breach while also ramping up efforts to detect and defeat attacks before they happen.
The result? Enterprises are now working with IT security companies to bring on cyber security consultants with the experience and expertise to manage current defenses, suggest new strategies and respond in real time. For many organizations, the push for outsourced IT network security stems from a need to scale on demand and adapt to changing market conditions. At MNS Group, we call it the "Human Cloud" — the right people at the right time to manage your security needs right now. As a leading cyber security service provider, our living contract and process-based service provide the level of IT excellence you need, combined with the flexibility you want to maximize impact and manage IT costs.
Enterprise cyber security is changing. Make sure you're ready to protect your assets and defend your data with expert IT outsourcing.


Adele has been recognised by the royal family for her services to music, after being named an MBE on the Queen's birthday honours list.
The super-songstress, who bagged an Oscar for her performance in Skyfall back in January, joined a host of other famous names on the rollcall, which was announced over the weekend.
Topping off what has been an incredible year for the singer, who not only won her first Academy Award, but also welcomed her first child, Adele takes up the title at just 25 years old.
It's thought her musical achievements, including having the best-selling album in the world for two years running were influential in her being given the award.
Adele is joined on the honours list by sports presenter Clare Balding, who was appointed OBE for services to broadcasting and journalism, writer Jackie Collins, golfer Paul Lawrie, GQ editor Dylan Jones and author Kate Mosse.
Meanwhile, comedian Rob Brydon, singer Aled Jones, and songwriter PJ Harvey received MBEs, and artist Grayson Perry and Dragon's Den star Hilary Devey all received CBEs.


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Rutgers launches nation's largest study of health care workers exposed to COVID-19
Apr 27, 2020, 2:18pmUpdated on Oct 04, 2020
Rutgers University announced that it has launched the nation's largest prospective study of health care workers exposed to COVID-19.
PHOTOS: COVID-19 Impacts the World
More than 800 employees from Rutgers, Robert Wood Johnson University Hospital and University Hospital are participating in the study.
The study includes a series of clinical trials that will explore antibody testing, new drug treatments, and long-term health tracking in the hope of providing insight into how to treat the disease and prevent its spread.
PHOTOS: Your Hometown Heroes
Those who participate in the study will be tested for COVID-19.
The university has launched two clinical trials as part of the study. The first is for patients who test positive for COVID-19 and are symptomatic. The second clinical trial will focus on those who test positive for COVID-19, but show no symptoms.
PHOTOS: Your Coronavirus Pandemic Experience

redpajama_set_name: RedPajamaCommonCrawl


Write a Java program to sort an array of given integers using the CombSort Algorithm.
The Comb Sort is a variant of the Bubble Sort. Like the Shell sort, the Comb Sort increases the gap used in comparisons and exchanges. Some implementations use the insertion sort once the gap is less than a certain amount. The basic idea is to eliminate turtles, or small values near the end of the list, since in a bubble sort these slow the sorting down tremendously. Rabbits, large values around the beginning of the list do not pose a problem in bubble sort.
In bubble sort, when any two elements are compared, they always have a gap of 1. The basic idea of comb sort is that the gap can be much more than 1.
Previous: Write a Java program to sort an array of given integers using the Cocktail sort Algorithm.
Next: Write a Java program to sort an array of given integers using CountingSort Algorithm.


Beany Teen binding boasts of invaluable properties – it is very light and it provides sensitive support during the ride. Special composition of its polycarbonate mixture ensures optimum firmness and flexibility, as if tailored to the low weights of young snowboarders.
Transition of the force from the foot to the snowboard is a piece of cake with Teen. The riders therefore safely control their movement in any terrain.


For rent at 20,000 THB/month, this modern and fully furnished 1 bed, 1 bath condo unit offers 54.16 sqm of living area.
Located in a modern 28 story building built by Sansiri, you are minutes away from BTS station "Wongwian Yai".
This corner unit features many built-in components. The overall effect is very tasteful and contemporary. 3M protective films have been added on the balcony window to reduce heat and glare.
Sitting on the 17th floor, the unit is next to a Sky Garden only accessible by residents of that floor.


What Licence Do You Have?
To undertake any training course you will need to have a valid UK driving licence with the correct motorcycle entitlement.
You must complete CBT before you ride either mopeds or motorcycles. CBT validates your licence and allows you to ride for two years during which time you must display L-plates. Note: you are not allowed to carry passengers on any machine if you are also required to display L-plates.
You will have full moped (up to 49cc) and provisional motorcycle entitlement. You do not need a CBTcertifcate or L-Plates to ride a moped and can carry passengers on 49cc bikes (but we would recommend it) A CBT is required to ride motorcycles over 49cc.
You must complete CBT before using your full moped entitlement. Once completed you may ride up to 49cc mopeds without L-plates and carry passengers until your car licence expires. You may ride motorcycles up to 125cc on the same certificate, but for two years only with L-plates.
1) CBT Training - This is Compulsory Basic Training and is mandatory to allow you to ride on a road as a learner rider.
5) Refresher Training - Ideal course if you haven't riden for a while or even if you just want some additional training to enhance your riding skills and techniques.
6) Scooter and Moped Training - Courses tailored to suit all age groups from 16year olds and upwards, from CBT to full licence.


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An Embarrassment of Riches: There's Too Many Things to Do in Los Angeles!
I am a product of Generation X, and like the Dave Matthew's song that bemoans the modern state of having too many choices, I scan the pavement and palm tree laden horizon of LA and ponder the endless list of things to do in Los Angeles. There's the California coast, filled with nostalgia and good for the soul, beach boy vibrations. Then there's the endless bar scene from Santa Monica to the swills of Silverlake, where khaki wearing corporate types or wanna-be screenwriting hipsters abound. Lest I forget the hippest daytime spots in LA, which currently includes strolling Abbott Kinney in Venice Beach to window shop and sip on large lattes, or climbing the stairs just north of San Vicente with others equally determined to attain and tighten an array of toned physiques. There's plenty of offbeat, crazy digs to catch in LA. Nothing's too weird in this city!
Choosing Things to Do in Los Angeles
Recently, I've been drawn to less crowds and more views, like the one from the top of the Getty (that's free on Saturdays) or better yet, the Griffith Park Observatory (which may be the only spot to appropriately appreciate that Los Angeles is really four cities in one). Of all the amazing things to do in Los Angeles, sometimes getting outside for the best view is the wisest way to spend your day, especially if you find yourself at home again, unable to make a game time decision.
Los Angeles: An Exciting Melting Pot
Did you know that Olvera Street is the oldest street in Los Angeles? Plaza Olvera, as it's written in Spanish, is a hot spot for authentic Mexican goods, food, drinks and live performances. The cobblestone streets are reminiscent to its time when LA was a smaller city known for older industries that have since then dissipated. The dominant industry in modern day LA is that of film and music. Given the fact that LA is one of the most culturally diverse cities in the world, you can never run out of things do in Los Angeles. Live performances in LA can be costly but there are plenty of free events around the city all year long. Concerts are a hot commodity in the city of angels, so plan ahead of time and purchase tickets to see your favorite artist… your paths may magically cross in tinsel town! Our go-to concert venues in LA are the Hollywood Bowl, the Hollywood Palladium, Bardot, the Echo and Echoplex. If you'd rather do things in LA during the day, we recommend taking the Expo Line from Fig and 7th Street in downtown LA to the end of the line: the Santa Monica Pier. What we love about the Expo Line in LA is the elevated design, so you can see the city, take photos or sit back, relax and enjoy the ride.
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Psychologists Declare Traditional Masculinity 'Harmful', by Susan Berry.
And one small group of men commit just over 50% of the homicides in the US — black men. What does the APA say about them? Are they ever more "harmful", or are they just the victims of "harsher punishments at school"? Hmmm.
Everyone should be normal, like women. Like at school and in sexual mores, "good" is what girls do. Men are increasingly demonized as "toxic". Hmmm.
If women are so superior, why we do we need affirmative action? Hmmm.


0:02Skip to 0 minutes and 2 secondsThis week, we've looked at the figure of the translator in history and in the contemporary world. We've learned that translators come in all sorts of shapes and sizes from courtroom interpreters to multilingual speakers, from monarchs to migrants. We've also explored the differences between translators and interpreters. And we've found that they require different sets of skills, and they follow different codes of conduct. Yet both translators and interpreters encounter linguistic difficulties and ethical dilemmas. And they're both vital to enable communication at different levels of society. Next week, we will be looking at where translation takes place. And we'll be exploring the different environments where translators work-- hospitals, court rooms, museums. And we'll also look at the multilingual landscapes that they inhabit.
In this video, Dr Cristina Marinetti summarises what we have learnt this week about who translators are and the role they play in society.


What is Going on with Oil Prices?!
As you might expect, it's complicated. Oil is a global commodity, and the price of it affects nearly everything we purchase and consume. And you have probably heard that it is dropping like a sky diver whose shoot won't open. Is this true, how bad is it, and most importantly, why?
First, I should state I am not a commodities analyst, but today I will pretend to play one. Let's first consider where the price of oil has been, and where it is now. Most oil in the US is tracked by the index for West Texas Intermediate (WTI) oil. In the last twelve months, WTI peaked at nearly $100/barrel and has since fallen into the mid $50/barrel. While swings like this are not unheard of, the last time WTI was around $50/barrel was back during 2009 in the depths of the recession. Also, this really hasn't been so much a price swing as it has been a gradual 6-month decline.
The decline in price, as far as I can tell, can be tied to three things: Supply, Demand, and Currency. The supply of oil right now is ample, and this has much to do with fracking revolution. Now that many new oil fields in North Dakota and Texas have been developed due to fracking, there is more oil production in the world. This comes at a time when demand is pulling back, because globally, most of the world's countries are not growing as much as previously anticipated. China, Japan, and the European Union are all experiencing sub-par growth, so their energy needs are not growing as rapidly as forecasted.
That leaves us with the last issue, currency. The one major piece of the global economy that is growing is us! The United States is finally realizing nominal growth, not seen since 2009. This is impacting our currency, in particularly, driving up the price of the dollar. The dollar is currently crushing the other major world currencies, and the strength of the dollar is returning to a pre-2009 level when compared against the Euro or the Yen. Why does this impact oil prices? Oil is priced in dollars, and so when the dollar is strong, it buys more oil!
Fear mongers are abound, screaming the price of oil may hit $40, no $30, no $25 dollars a barrel! While they have no evidence to support their claims, they sure get plenty of attention, like someone shouting "fire!" in a crowded movie theater. The facts are, the price of oil seemed to be quite expensive in the past 5 years as it hovered around $100/barrel. Fracking technology supplied the world with additional oil, and yes, there was great incentive to do so at $100/barrel. And as we know in economics, more supply will push down prices.
But, logic would also seem to indicate that fracking isn't going to oversupply the market until fracking isn't feasible. Rather than seeing fracking disappear, the price will balance out fracking production so it is done efficiently and deliver oil at a price the market will tolerate. This isn't a bust in the traditional sense, because the oil isn't drying up; there is just too much of it all of a sudden.
This is bad news to weak companies who were only in the game because of the high prices and could only live off their cash flow and not their own equity. Other companies, which have made more wise investments, will still be able to produce at some price between $50 to $100 a barrel, and they have the financial strength to sit back and wait for the market to figure it out.
It appears the "rush" may be coming to a close, but the fracking and oil production is here to stay. Now the maturing Bakken oil fields will come to settle into their niche in the global market, and this market correction may give wary North Dakota and Montana communities a little time to catch up with their growth.


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"Emerging Writers, Serious Writers"
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Home / News / SRI LANKA: The recent judgement of the Supreme Court on the Singarasa case is an attack on the sovereignty of the people
SRI LANKA: The recent judgement of the Supreme Court on the Singarasa case is an attack on the sovereignty of the people
Statement |Sri Lanka | September 19, 2006
Earlier the Asian Human Rights Commission (AHRC) circulated the complete text of this judgement (please see AS-217-2006). Among the many completely baseless legal positions on which this judgement is premised there are some outstanding issues. The AHRC intends to explore a comprehensive critique on this judgement soon. In the meantime we would like to mention some of the more obvious issues. These relate to some basic fallacies on which the Supreme Court judgement stands:
That the sovereignty of the country as enshrined in the Constitution is compromised by Sri Lanka's accession to the Covenant on Civil and Political Rights in 1980 followed by it being a signatory to the Optional Protocol in 1997;
that when the president of Sri Lanka, as the head of state, signed the Optional Protocol she did so merely as the head of the executive; and,
that Sri Lanka has violated Article 2 of the ICCPR and on that basis the obligations to the Convention by the Sri Lankan state can be abolished.
On the issue of sovereignty:
In Sri Lanka the people are sovereign. The simple issue is as to whether the people's sovereignty is enhanced by their entitlement to a greater degree of freedom and human rights or by having a lesser degree of freedom and human rights. The ICCPR is a universally accepted standard of human rights and gives wider scope for rights than anything contained in the Sri Lankan Constitution. By the accession to the ICCPR the scope of rights available to the people of Sri Lanka has been enhanced; about this there is no dispute. The Supreme Court judgement, in essence, states that if the Constitution has provided for a certain degree of rights further expansion of such rights by the state violates . This is an absurd proposition.
The expansion of rights gained by the people of Sri Lanka was further enhanced by the state becoming a party to the Optional Protocol thus giving the sovereign people of Sri Lanka, the right to complain to an international body when their rights are violated even by any agency of the state. Is the sovereignty of people lessened when they can complain about violations of human rights by anyone including a state agency? Alternatively, if the people's rights are violated and their right to complain is restricted (and thereby a right to a remedy is limited) is that an enhancement of sovereignty? If the state agency that violates rights happens to be a court should not a sovereign people have the right to complain and find a remedy against such a court? By stating that people are sovereign the Constitution has provided an infinite space for the expansion of the rights and freedoms of the people, even against their state and all its branches. In this way Sri Lanka's Constitution has brought in the principles that were developed in the world's great revolutions, in all of which the adoption of a comprehensive bill of rights on the basis of was an integral part. Today the ICCPR is considered the most universally accepted bill of rights. Thus, the Supreme Court's proposition that accession to these international treatises violates is a basic fallacy. In fact the contrary proposition that the ICCPR enhances people's sovereignty is both legally and politically more profound and valid.
The signature of a head of state
The second basic fallacy on which the Supreme Court judgement is based is that when the president of the country signed the Optional Protocol of the ICCPR in 1997 the president did so only as the head of the executive.
When a head of the state signs a treaty he or she does so on behalf of the entire state, meaning the legislature, the executive and the judiciary. This failure to recognise the legislature and the judiciary also as a part of the state is the basic fallacy on which the entire judgement of the Supreme Court is based. The judiciary is a component of the state. Judicial independence in the context of the separation of powers is not about three components with three different heads of state functioning separately. The separation of powers relates to the functioning of the state and is not about who is the head of the state.
The head of the executive or a chief justice as the head of the judiciary is a functionary within the apparatus of the state. As opposed to this the president acting on behalf of the state acts for the whole of the state meaning the legislature, the executive and the judiciary. Internal challenges to the power of the executive, including that of the president acting on behalf of the executive, is regulated by the constitutional principles of the separation of powers. Unfortunately when it comes to the internal matters of the country the Supreme Court of Sri Lanka has gone by the rule that under Article 31 (1) acts or omissions of the president done personally or officially cannot be challenged by the court. Even the incumbent president's open violation of the Constitution of Sri Lanka by his failure to comply with the 17th Amendment to the Constitution was upheld as legal both by the Court of Appeal and the Supreme Court.
On the issue of the Optional Protocol the president has acted as the head of the state. The Supreme Court has tried to treat this as an act of the head of the executive only, thereby claiming a right, as a separate branch of the government to challenge the executive action. While the president is the chief of the executive, the Chief Justice is the chief of the judiciary. However, that is not an equation to the head of the state. Sri Lanka does not have two or three heads of states, one for the executive, one for the legislature and one for the judiciary. The whole misconception on this issue is what has led the judiciary to remove the obligations created by the head of the state by being a party to the said treatises. When the head of the state acts on behalf of the state and expands the scope of human rights by becoming a party to international treatises this is a political act that brings about a new order in terms of the human rights of the entire people. What this implies is that the Supreme Court of Sri Lanka led by the Chief Justice has entered into an area that does not belong to it; the area of politics. Under the pretext of a legal judgement the Supreme Court has tried to act on matters that belong only to a legitimate head of state.
A relevant example on this matter is when the Bush administration attempted to 'unsign' the International Criminal Court Treaty signed by the former president, Bill Clinton on his last day of office. This move was rejected by the United Nations when the then chief of the treaty section for the United Nations, Palitha Kohona, said "it was unheard of for a nation that signed a treaty to withdraw that signature". A Supreme Court trying to undo treaty obligations entered into by the state with the United Nations is also unheard of. However, this is not the first time that the incumbent Chief Justice, Sarath N. Silva has done things unheard of in the entire jurisprudence of the world like for example sentencing a man to one year of rigorous imprisonment for allegedly raising his voice in court as he did in the case of Tony Fernando.
On the issue of Article 2 of the ICCPR (creating confusion on monist and dualist theories)
The Supreme Court judgement throughout distorts the meaning of Article 2 of the ICCPR. Article 2 of the ICCPR obligates the governments who become party to this Covenant to take adequate measures to implement the rights by legislative, judicial, administrative and other measures. Simply put, this means that new laws must be brought in to enhance the rights guaranteed by the Covenant. New judicial remedies must be provided and the judicial interpretation of rights itself must be enriched by the legal incorporation of the ICCPR. It also means that the government must provide adequate funds and undertake other measures to implement the provisions of the ICCPR. The Supreme Court in its judgment put forward a strange logic that since such measures envisaged by Article 2 have not been taken adequately by the state, the state entering into such treatises is in itself invalid. This simply does not stand up to any logic. If the Supreme Court intervened as a branch of the state to take whatever steps might be within its power to improve such laws, judicial remedies and interpretation and made further attempts to obligate the executive to invest funds, and other measures to ensure the rights guaranteed by the ICCPR, that would have been the logical process to be expected. This is exactly what the Indian Supreme Court has done. The distinction between monism and dualism is about following the obligations taken internationally by adequate measures as mentioned above by all branches of the state. The way monist and dualist argument is used in this judgement does not make any sense at all.
The AHRC in earlier statements has pointed out that the Chief Justice, Sarath N. Silva is trying to transform the Supreme Court of Sri Lanka into a political institution. Such a move is against the Constitution of Sri Lanka as well as the principles of constitutionalism in general. Further the attempt to limit the rights of the people by removing the implication of being a state party to human rights treatises is an attack on .
Document Type : Statement
Document ID : AS-220-2006
Countries : Sri Lanka,
Issues : Administration of justice, International human rights mechanisms, Judicial system,
« THAILAND: MILITARY COUP – Restore civilian government immediately
THAILAND: MILITARY COUP – Junta poses grave dangers to lives and liberties »


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text:
Dream Weaver Creations, LLC was started by father, husband, and Baltimore native Devin Hickson. Even though Devin is from a middle-class Baltimore two-parent household, he saw his fair share of challenges throughout his own formative years, but also through the experiences of some of his immediate peers and close friends.
Post graduating from college and obtaining his master's he searched for a way to give back to those whose dreams had been derailed or delayed. It started with the untimely passing of one of his close friends, Peyton . He and Peyton would often chat about making things better for not only themselves and their families, but talked about creating some kind of platform that would help others. After becoming a new father and husband, Devin settled on Children's book Daddy Hero to be the first creative endeavor from Dream Weaver, and has proven himself a one of Baltimore's best black authors of children's books. Since that time he has become involved in several other creative spaces.
"Where Dreams Become Creations" Dream Weaver was started as a simple outlet for Devin's creative content ideas. Dream Weaver is committed to producing high-quality creative content in all of its various forms. D.W.C. is essentially the vehicle for others like Devin, to find an outlet for their creative passions. The company absolutely loves creatives and through various collaborative efforts creates a medium where new ideas and talent translate to distributed content against multiple platforms.
We try to strike a balance between allowing our creators to work within their creative process in order to maximize their unique energy. We believe that when creativity is fostered the clients are the ones who truly benefit from something amazing. People are at the heart of what we do and we prioritize their stories, and their gifts over profit, which sets us apart from other players in the creative media agency space. Collaboration, integrity, and respect are of the utmost importance of what we seek to do and invite others to partner with us and embark on a fantastic journey to push the boundaries of content creation.
Dream Weaver Creations
We are a Baltimore-based entertainment production company producing various forms of content.
Privacy Policy Site By Kiu Solutions


Hip dysplasia (HD) is well known among breeders and owners of larger dogs but surprisingly is quite common in smaller breeds. For many years the Kennel Club and British Veterinary Association (BVA) have run a scheme to improve the conformation of dogs' hips and elbows, by examining x-rays and having them scored by experts for quality. Ideally, all dogs intended for breeding should have a hip score well below the breed average and 0 or 1 for elbows (hips and elbows are scored differently).
All dogs will need sedation or general anaesthesia for the x-rays to be taken safely and the dog positioned appropriately. X-rays are labelled with the dogs' unique identification and the x-ray films sent away to the BVA. The numerical score is then sent back normally 2-4 weeks later.
All dogs must be at least 1 year old and micro-chipped; some unscrupulous people have been known to "borrow" hip scores from other dogs. Thankfully micro-chips have made this form of fraud much more difficult, in addition to proving your ownership and reuniting you with a lost animal.
Another check for potential dog parents is an eye exam to screen for inherited problems. This must be performed every year (unlike the x-rays which last a lifetime) and by a Veterinary Ophthalmologist.
Last, but by no means least are the multitude of blood tests now available to genetically screen your dog for specific diseases. Please contact us at the surgery or your dogs breed health advisor for an up to date list.


text:
Q: EasyNetQ not calling subscriber I have an ASP.NET MVC action that verifies that a connection string properly works to connect to a server using RabbitMQ. How I do this is by simply creating a queue, subscribing to it and then right away publish a message to it. I expect the message to show up on my subscriber at most a few seconds later, but it doesn't. The subscriber only gets notified the first time I call the action (right after I delete the queue from RabbitMQ using its web based manager), but then fails to call the subscriber once I publish any following messages with the queue already created. Please, take a look at my code and let me know if you see something I'm not seeing. Thanks in advance.
    //This is just the action called on a POST request.
    //Here is where the test is done.
    [HttpPost]
    public void DoConnectionVerificationAsync()
    {
        const string paramName = "queueSuccessful";
        try
        {
            //This is my routing key
            const string routingKey = "management.verify";

            //Here I declare the key and the exchange and bind them.
            var queue = Queue.DeclareDurable(routingKey);
            var exchange = Exchange.DeclareDirect("Orchard");
            queue.BindTo(exchange, routingKey);

            //Here I just generate a random int to send in the test message to the queue
            Random random = new Random();
            int randomInt = random.Next();
            //Instantiate the actual message with the random integer.
            var message = new Message<VerifyMessage>(new VerifyMessage
            {
                Content = randomInt.ToString(CultureInfo.InvariantCulture)
            });
            message.Properties.AppId = "CRM";

            //Because this is an asynchronous action, here I hint the beginning of the async operation.
            AsyncManager.OutstandingOperations.Increment();

            //Here I have my subscriber. The subscriber gets called only the first time, when the queue hasn't been created yet. That's a problem, it should be called every time I publish, which is in the following lines of code.
            _bus.Subscribe<VerifyMessage>(queue, (response, messageReceivedInfo) => Task.Factory.StartNew(() =>
            {
                VerifyMessage receivedMessage = response.Body;
                string content = receivedMessage.Content;
                int integer = int.Parse(content);
                //I expect the int received from the queue is the same as the one I sent
                bool success = integer == randomInt;
                AsyncManager.Parameters[paramName] = success;
                AsyncManager.OutstandingOperations.Decrement();
            }));

            //And here I publish the message. This always works, I can see the message stored in the queue using the web based RabbitMQ Manager
            //The problem is that the message gets stuck in the queue and never gets sent to the subscriber defined above
            using (var publishChannel = _bus.OpenPublishChannel(x => x.WithPublisherConfirms()))
            {
                publishChannel.Publish(exchange, routingKey, message, t =>
                    t.OnSuccess(() =>
                        {
                            // If we successfully publish, then there's nothing we really need to do. So this function stays empty.
                        })
                        .OnFailure(() =>
                            {
                                AsyncManager.Parameters[paramName] = false; AsyncManager.OutstandingOperations.Decrement();
                            }));
            }
        }
        catch (EasyNetQException)
        {
            AsyncManager.Parameters[paramName] = false;
            AsyncManager.OutstandingOperations.Decrement();
        }
    }

    //These functions down here don't really matter, but I'm including them just in case so that you can see it all.
    public ActionResult DoConnectionVerificationCompleted(bool queueSuccessful)
    {
        return RedirectToAction("VerifyQueueConnectionResult", new { queueSuccessful });
    }

    public ActionResult VerifyQueueConnectionResult(bool queueSuccessful)
    {
        VerifyQueueConnectionResultModel model = new VerifyQueueConnectionResultModel();
        model.Succeded = queueSuccessful;
        return View(model);
    }

A: Each call to bus.Subscribe creates a new consumer. So the first time it works, you have just one consumer on your queue and when you publish a message to it it's routed to that consumer.
The second time you call bus.Subscribe a second consumer is bound to the same queue. When RabbitMQ has multiple consumers on a single queue it round-robins messages to the consumers. It's a feature. That's how it does work-sharing out-of-the-box. Your message is routed to the first consumer, not the consumer that you just declared. Same for the 3rd, 4th, etc consumers, so it appears that message has not arrived.

A: Well, I found a simple solution to the problem. I ended up searching through the available methods in the EasyNetQ API's and found Queue.SetAsSingleUse(). I simply called the method after binding the queue to the exchange and that pretty much did it. Not sure what queue.SetAsSingleUse() does, but by its name it sounds like it disposes the queue after the first message is published and delivered to a subscriber. In my case, it makes sense to use the queue only once for each test, but someone who might want to keep the queue around for whatever reason might run into trouble.

A: You could use different queues for every different subscriber, like this:
bus.Subscribe<MyMessage>("my_subscription_1"...
bus.Subscribe<MyMessage>("my_subscription_2"...

This way whatever message you publish of type MyMessage will arrive in all these queues.

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