text:
\section{Introduction}

In 1985, Amari \cite{article.1} introduced the notion of statistical manifolds in the context of information geometry. They may be considered as manifolds consisting of certain probability density functions. Geometrically, they are Riemannian manifolds with a certain affine connection. Beyond the expectations, statistical manifolds are familiar to researchers because they have appeared with alternate names in various research topics in affine geometry \cite{article.09} and in Hessian geometry \cite{article.8}. These manifolds have applications in document classification, face recognition, image analysis, clustering and so on. The oldest examples of statistical structures are the induced structures (consisting of the second fundamental form and the induced connection) on locally strongly convex hypersurfaces in $\mathbb{R}^{n+1}$ endowed with an equiaffine transversal vector field (in other words - with relative normalization). It is natural for researchers to try to build the submanifold theory and the complex manifold theory of statistical manifolds.

In 1978, A. Bejancu \cite{article.2} introduced the notion of a CR-submanifold of a Kaehler manifold with complex structure $\mathcal{J}$. The investigation of CR-submanifolds of Kaehler manifolds has a long history. Many geometers investigated CR-submanifolds in different ambient manifolds such as Hermitian manifolds, Kaehler manifolds, Sasakian manifolds, complex space forms, complex projective space, etc. In this direction, H. Furuhata and I. Hasegawa \cite{article.7} intensively studied their statistical version, that is, CR-statistical submanifolds in holomorphic statistical manifolds. Recently, M.N. Boyom et al. \cite{article.700} studied classification of totally umbilical CR-statistical submanifolds in holomorphic statistical manifolds with constant holomorphic curvature. Also, Milijevi$\acute{c}$ \cite{4mir} and Aliya et al. \cite{article} studied totally real statistical submanifolds in holomorphic statistical manifolds, independently.

The Curvatures invariants are widely used in the field of differential geometry and in physics also. The Ricci curvature is the essential term in the Einstein field equations, which plays a key role in general relativity. It is immensely studied in differential geometry as it gives a way of measuring the degree to which the geometry determined by a given Riemannian metric might differ from that of ordinary Euclidean $n$-space. A Riemannian manifold is said to be an Einstein manifold if the Ricci tensor satisfies the vacuum Einstein equation. The lower bounds on the Ricci tensor on a Riemannian manifold enable one to find global geometric and topological information by comparison with the geometry of a constant curvature space form. Recently, Aydin et al. \cite{a} investigated Chen-Ricci inequality for submanifolds with any codimension of statistical manifolds of constant curvature. Also, A. Mihai et al. established similar inequality with respect to a sectional curvature of the ambient Hessian manifold in \cite{a2}.

In this work, we wish to add some more results to CR-statistical submanifolds in holomorphic statistical manifolds, which are new objects originating from information geometry. We see whether the results in classical Riemannian case hold in our settings or not. Also, we obtain a Chen-Ricci inequality for CR-statistical submanifolds in a holomorphic statistical manifold of constant holomorphic sectional curvature by following an optimization method (see Section 5).

\section{Some Preliminaries}

In this section, we give some basic definitions and fundamental formulae which are related to this paper.

\begin{definition}
\cite{article.7} A statistical manifold is a Riemannian manifold $(\overline{N},g)$ endowed with a pair of torsion-free affine connections $\overline{\nabla}$ and $\overline{\nabla}^{*}$ satisfying
\begin{eqnarray*}
Xg(Y, Z) = g(\overline{\nabla}_{X}Y, Z) + g(Y, \overline{\nabla}_{X}^{*}Z),
\end{eqnarray*}
for any $X, Y, Z \in \Gamma(T\overline{N})$ and $\overline{\nabla}g$ is symmetric. The connections $\overline{\nabla}$ and $\overline{\nabla}^{*}$ are called dual connections and satisfy $\big(\overline{\nabla}^{*}\big)^{*} = \overline{\nabla}$.
\end{definition}

\begin{remark}
\cite{article.7}
\begin{enumerate}
\item[(a)] If $(\overline{\nabla}, g)$ is a statistical structure on $\overline{N}$, then $(\overline{\nabla}^{*}, g)$ is also a statistical structure.
\item[(b)] Any torsion-free affine connection $\overline{\nabla}$ always has a dual connection given by
\begin{eqnarray*}
2\overline{\nabla}^{0} = \overline{\nabla} + \overline{\nabla}^{*},
\end{eqnarray*}
where $\overline{\nabla}^{0}$ is the Levi-Civita connection on $\overline{N}$.
\end{enumerate}
\end{remark}

\begin{definition}
\cite{article.7} Let $\overline{N}$ be a Kaehler manifold with almost complex structure $\mathcal{J} \in \Gamma(T\overline{N}^{(1,1)})$ and metric $g$. A quadruple $(\overline{N}, \overline{\nabla}, g, \mathcal{J})$ is called a holomorphic statistical manifold if
\begin{enumerate}
\item[(a)] $(\overline{\nabla}, g)$ is a statistical structure on $\overline{N}$,
\item[(b)] a $2$-form $\omega$ on $\overline{N}$, given by
\begin{eqnarray*}
\omega(X, Y) = g(X, \mathcal{J}Y),
\end{eqnarray*}
for any $X, Y \in \Gamma(T\overline{N})$, is $\overline{\nabla}-$parallel, that is, $\overline{\nabla}\omega = 0$.
\end{enumerate}
\end{definition}

\begin{remark}
\cite{article.7}
\begin{enumerate}
\item[(a)] For a holomorphic statistical manifold $\big(\overline{N}, \mathcal{J}, g\big)$, we have
\begin{eqnarray}\label{c}
\overline{\nabla}_{X}(\mathcal{J}Y) = \mathcal{J}\overline{\nabla}_{X}^{*}Y,
\end{eqnarray}
for any $X, Y \in \Gamma(T\overline{N})$.
\item[(b)] Since $\omega$ is skew-symmetric, we have $g(X, \mathcal{J}Y) = - g(\mathcal{J}X, Y)$, for any $X, Y \in \Gamma(T\overline{N})$.
\item[(c)] A holomorphic statistical manifold is nothing but a special Kaehler manifold if $\overline{\nabla}$ is flat.
\end{enumerate}
\end{remark}

Let $(\overline{N}, \overline{\nabla}, g)$ be a statistical manifold and $N$ be a submanifold of $\overline{N}$. By $T_{x}^{\perp}N$ we denote
the normal space of $N$, i.e., $T_{x}^{\perp}N = \big\{v \in T_{x}\overline{N} \big | g(u, v) = 0, u \in T_{x}N\big\}$. Define \cite{article.7}
\begin{equation}\label{e}
\left .
\begin{tabular}{ccc}
$\overline{\nabla}_{X}Y = \nabla_{X}Y + \mathcal{B}(X, Y)$,\\
$\overline{\nabla}_{X}^{*}Y = \nabla_{X}^{*}Y + \mathcal{B}^{*}(X, Y)$,\\
$\overline{\nabla}_{X}V = - \mathcal{A}_{V}(X) + \nabla_{X}^{\perp}V$,\\
$\overline{\nabla}_{X}^{*}V = - \mathcal{A}_{V}^{*}(X) + \nabla_{X}^{\perp *}V$,
\end{tabular}
\right \}
\end{equation}
for any $X, Y \in \Gamma(TN)$ and $V \in \Gamma(T^{\perp}N)$. Here $\overline{\nabla}$ and $\overline{\nabla}^{*}$ (respectively, $\nabla$ and $\nabla^{*}$) are the dual connections on $\overline{N}$ (respectively, on $N$), $\mathcal{B}$ and $\mathcal{B}^{*}$ are symmetric and bilinear, called the imbedding curvature tensor of $N$ in $\overline{N}$ for $\overline{\nabla}$ and the imbedding curvature tensor of $N$ in $\overline{N}$ for $\overline{\nabla}^{*}$, respectively. Since $\mathcal{B}$ and $\mathcal{B}^{*}$ are bilinear, the linear transformations $\mathcal{A}_{V}$ and $\mathcal{A}_{V}^{*}$ are related to the imbedding curvature tensors by \cite{article.7}
\begin{eqnarray}\label{f}
\left .
\begin{tabular}{ccc}
$g(\mathcal{B}(X, Y), V) = g(\mathcal{A}_{V}^{*}(X), Y)$,\\
$g(\mathcal{B}^{*}(X, Y), V) = g(\mathcal{A}_{V}(X), Y)$,
\end{tabular}
\right \}
\end{eqnarray}
for any $X, Y \in \Gamma(TN)$ and $V \in \Gamma(T^{\perp}N)$.

Let $\overline{R}$ and $R$ be the curvature tensor fields with respect to $\overline{\nabla}$ and $\nabla$, respectively. Then the Gauss equation is given by \cite{article.7}
\begin{eqnarray}\label{g}
g(\overline{R}(X, Y) Z, W) &=& g(R(X, Y) Z, W) + g(\mathcal{B}(X, Z), \mathcal{B}^{*}(Y, W)) \nonumber \\
&&- g(\mathcal{B}^{*}(X, W), \mathcal{B}(Y, Z)),
\end{eqnarray}
for any $X, Y, Z, W \in \Gamma(TN)$. Similarly, $\overline{R}^{*}$ and $R^{*}$ are respectively the curvature tensor fields with respect to $\overline{\nabla}^{*}$ and $\nabla^{*}$. Then the curvature tensor fields of $\overline{N}$ and $N$ are respectively given by \cite{article.7}
\begin{eqnarray}\label{g1}
\overline{\mathcal{S}} = \frac{1}{2}(\overline{R} + \overline{R}^{*}),  \hspace{0.2 cm} \textrm{and} \hspace{0.2 cm}  \mathcal{S} = \frac{1}{2}(R + R^{*}).
\end{eqnarray}

Suppose that $dim(N)= m$ and $dim(\overline{N}) = 2n$. We consider a local orthonormal tangent frame $\{e_{1}, \dots, e_{m}\}$ of $T_{x}N$ and a local orthonormal normal frame $\{e_{m+1}, \dots, e_{2n}\}$ of $T_{x}^{\perp}N$ in $\overline{N}$, $x \in N$. Then the mean curvature vectors $H$ and $H^{*}$ of $N$ in $\overline{N}$ are
$$H = \frac{1}{m}\sum_{i = 1}^{m}\mathcal{B}(e_{i}, e_{i}),$$
and
$$H^{*} = \frac{1}{m}\sum_{i = 1}^{m}\mathcal{B}^{*}(e_{i}, e_{i}).$$
Also, we set
$$h_{ij}^{r} = g(\mathcal{B}(e_{i}, e_{j}), e_{r}),$$
and
$$h_{ij}^{*r} = g(\mathcal{B}^{*}(e_{i}, e_{j}), e_{r}),$$
for $i, j \in \{1, \dots, m\}$, $r \in \{m+1, \dots, 2n\}$.

Let $X$ be a unit vector such that $||X|| = 1$. We choose an orthonormal frame $\{e_{1}, \dots, e_{m}\}$ of $T_{x}N$, $x \in N$, such that $e_{1} = X$. Then the Ricci curvature at $X$ is given by
\begin{eqnarray*}
\overline{Ric}^{\overline{\nabla}, \overline{\nabla}^{*}}(X) &=& \sum_{i=2}^{m}\overline{\mathcal{K}}^{\overline{\nabla}, \overline{\nabla}^{*}}(X \wedge e_{i}) \\
&=& \frac{1}{2}\bigg\{\sum_{i=2}^{m}\overline{\mathcal{K}}(X \wedge e_{i}) + \sum_{i=2}^{m}\overline{\mathcal{K}}^{*}(X \wedge e_{i})\bigg\},
\end{eqnarray*}
where $\overline{\mathcal{K}}(e_{i} \wedge e_{j})$ denotes the sectional curvature of the 2-plane section spanned by $e_{i}$ and $e_{j}$.

For any $X \in \Gamma(TN)$, we put \cite{article.9}
\begin{eqnarray}
\mathcal{J}X = \mathcal{P}X + \mathcal{F}X,
\end{eqnarray}
where $\mathcal{P}X$ and $\mathcal{F}X$ are the tangential and normal components of $\mathcal{J}X$, respectively. Then $\mathcal{P}$ is an endomorphism of the tangent bundle $TN$ and $\mathcal{F}$ is a normal-bundle-valued $1-$form on $TN$.

In the similar way, for any $V \in \Gamma(T^{\perp}N)$, we put \cite{article.9}
\begin{eqnarray}
\mathcal{J}V = tV + fV,
\end{eqnarray}
where $tV$ and $fV$ are the tangential and normal components of $\mathcal{J}V$, respectively. Then $f$ is an endomorphism of the normal bundle $T^{\perp}N$ and $t$ is a tangent bundle-valued $1-$form on $T^{\perp}N$.

The statistical version of the definition of a CR-submanifold is as follows:

\begin{definition}
\cite{article.7} A statistical submanifold $N$ is called a CR-statistical submanifold in a holomorphic statistical manifold $\overline{N}$ of dimension $2m \geq 4$ if $N$ is CR-submanifold in $\overline{N}$, i.e., there exists a differentiable distribution $\mathcal{D} : x \rightarrow \mathcal{D}_{x} \subseteq T_{x}N$ on $N$ satisfying the following conditions:
\begin{enumerate}
\item[(a)] $\mathcal{D}$ is holomorphic, i.e., $\mathcal{J}\mathcal{D}_{x} = \mathcal{D}_{x} \subseteq T_{x}N$ for each $x \in N$, and
\item[(b)] the complementary orthogonal distribution $\mathcal{D}^{\perp} : x \rightarrow \mathcal{D}_{x}^{\perp} \subseteq T_{x}N$ is
totally real, i.e., $\mathcal{J}\mathcal{D}_{x}^{\perp} \subset T_{x}^{\perp}N$ for each $x \in N$.
\end{enumerate}
\end{definition}

\begin{remark}\label{q1}
\cite{article.7} CR-statistical submanifolds are characterized by the condition $\mathcal{FP} = 0$.
\end{remark}

\begin{definition}
\cite{article.7} Let $N$ be a CR-statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Then $N$ is said to be
\begin{itemize}
\item[(a)] mixed totally geodesic with respect to $\overline{\nabla}$ if $\mathcal{B}(X, Y) = 0$ for any $X \in \mathcal{D}$ and $Y \in \mathcal{D}^{\perp}$.
\item[(a)$^*$] mixed totally geodesic with respect to $\overline{\nabla}^{*}$ if $\mathcal{B}^{*}(X, Y) = 0$ for any $X \in \mathcal{D}$ and $Y \in \mathcal{D}^{\perp}$.
\end{itemize}
\end{definition}

\begin{definition}
\cite{article.7} A statistical submanifold $N$ of a holomorphic statistical manifold $\overline{N}$ is called a holomorphic submanifold $(\mathcal{F} = 0$ and $t = 0)$ if the almost complex structure $\mathcal{J}$ of $\overline{N}$ carries each tangent space of $N$ into itself, whereas it is said to be a totally real submanifold $(\mathcal{P} = 0)$ if the almost complex structure $\mathcal{J}$ of $\overline{N}$ carries each tangent space of $N$ into its corresponding normal space. If $\mathcal{J}(TN)= T^{\perp}N$, then $N$ is called a Lagrangian submanifold ($\mathcal{P} = 0$ and $f = 0$). A CR-statistical manifold is said to be generic if $\mathcal{J}\mathcal{D}^{\perp} = T^{\perp}N$ and $\mathcal{D} \neq 0$ $(f = 0)$. If $\mathcal{D} \neq 0$ and $\mathcal{D}^{\perp} \neq 0$, then $N$ is said to be proper.
\end{definition}

For a CR-statistical submanifold $N$ we shall denote by $\mu$ the orthogonal complementary subbundle of $\mathcal{J}\mathcal{D}^{\perp}$ in $T^{\perp}N$, we have
\begin{eqnarray}
T^{\perp}N = \mathcal{J}\mathcal{D}^{\perp} \oplus \mu.
\end{eqnarray}

\begin{definition}
\cite{article.7} A holomorphic statistical manifold $(\overline{N}, \overline{\nabla}, g, \mathcal{J})$ is said to be of constant holomorphic curvature $c \in \mathbb{R}$ if the following curvature equation holds
\begin{eqnarray}\label{d}
\overline{\mathcal{S}}(X, Y)Z &=& \frac{c}{4} \big\{g(Y, Z)X - g(X, Z)Y + g(\mathcal{J}Y, Z)\mathcal{J}X \nonumber \\
&&- g(\mathcal{J}X, Z)\mathcal{J}Y + 2g(X, \mathcal{J}Y)\mathcal{J}Z \big\}.
\end{eqnarray}
for any $X, Y, Z \in \Gamma(T\overline{N})$. It is denoted by $\overline{N}(c)$.
\end{definition}

\section{Some Basic Results}

We prove the following results:

\begin{proposition}
Let $N$ be a statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Then
\begin{eqnarray*}
\mathcal{A}_{V}^{*}tU = \mathcal{A}_{U}^{*}tV,
\end{eqnarray*}
for any $U, V \in \Gamma(T^{\perp}N)$ if and only if
\begin{eqnarray*}
\nabla_{X}^{\perp}fV = f\nabla_{X}^{\perp *}V,
\end{eqnarray*}
for any $X \in \Gamma(TN)$ and $V \in \Gamma(T^{\perp}N)$.
\end{proposition}

\begin{proof}
From \cite{article.7}, we have
\begin{eqnarray*}
g(\nabla_{X}^{\perp}fV, U) = g(f\nabla_{X}^{\perp *}V, U) - g(\mathcal{F}\mathcal{A}^{*}_{V}X, U) - g(\mathcal{B}(X, tV), U).
\end{eqnarray*}
Further, we derive that
\begin{eqnarray*}
g(\mathcal{A}^{*}_{V}X, tU) - g(\mathcal{A}_{U}^{*}tV, X) = 0.
\end{eqnarray*}
Our assertion follows from the last relation and self-adjoint property of $\mathcal{A}^{*}$.
\end{proof}

\begin{proposition}\label{1}
Let $N$ be a statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Then
\begin{eqnarray*}
\nabla_{X}^{\perp}(\mathcal{F}Y) = \mathcal{F}\nabla_{X}^{*}Y,
\end{eqnarray*}
for any $X, Y \in \Gamma(TN)$ if and only if
\begin{eqnarray*}
\nabla_{X}(tV) = t\nabla_{X}^{\perp *}V,
\end{eqnarray*}
for any $X \in \Gamma(TN)$ and $V \in \Gamma(T^{\perp}N)$.
\end{proposition}

\begin{proof}
From \cite{article.7}, we have
\begin{eqnarray*}
g(\nabla_{X}^{\perp}(\mathcal{F}Y) - \mathcal{F}\nabla_{X}^{*}Y, V) &=& g(f\mathcal{B}^{*}(X, Y), V) - g(\mathcal{B}(X, \mathcal{P}Y), V)\\
&=& - g(\mathcal{B}^{*}(X, Y), fV) - g(\mathcal{A}^{*}_{V}X, \mathcal{P}Y)\\
&=& - g(\mathcal{A}_{fV}X, Y) + g(\mathcal{P}\mathcal{A}^{*}_{V}X, Y)\\
&=& - g(\nabla_{X}(tV) - t\nabla_{X}^{\perp *}V, Y).
\end{eqnarray*}
This completes the proof of our proposition.
\end{proof}

\begin{proposition}
Let $N$ be a holomorphic statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Then
\begin{enumerate}
\item[(a)] $tr_{g}\mathcal{B} = tr_{g}\mathcal{B}^{*} = 0$,
\item[(b)] $g(\mathcal{S}(X, \mathcal{J}X)\mathcal{J}X, X) = g(\overline{\mathcal{S}}(X, \mathcal{J}X)\mathcal{J}X, X) - 2 g(\mathcal{B}(X, X), \mathcal{B}^{*}(X, X))$, for $X \in \Gamma(TN)$.
\end{enumerate}
\end{proposition}

\begin{proof}
Let $\{e_{1}, \dots, e_{n}, \mathcal{J}e_{1}, \dots, \mathcal{J}e_{n}\}$ be a local orthonormal frame on $N$. Then we have
\begin{eqnarray*}
tr_{g}\mathcal{B} &=& \sum_{i = 1}^{n}(\mathcal{B}(e_{i},e_{i}) + \mathcal{B}(\mathcal{J}e_{i}, \mathcal{J}e_{i})) = 0,
\end{eqnarray*}
and
\begin{eqnarray*}
tr_{g}\mathcal{B}^{*} &=& \sum_{i = 1}^{n}(\mathcal{B}^{*}(e_{i},e_{i}) + \mathcal{B}^{*}(\mathcal{J}e_{i}, \mathcal{J}e_{i}))\\
&=& - \sum_{i = 1}^{n}(\mathcal{J}\mathcal{B}^{*}(e_{i},\mathcal{J}e_{i}) - \mathcal{J}\mathcal{B}^{*}(\mathcal{J}e_{i}, e_{i})) = 0.
\end{eqnarray*}
Following \cite{article.7}, we have
\begin{eqnarray*}
g(\mathcal{S}(X, \mathcal{J}X)\mathcal{J}X, X) &=& g(\overline{\mathcal{S}}(X, \mathcal{J}X)\mathcal{J}X, X) - g(\mathcal{B}(\mathcal{J}X,X), \mathcal{B}^{*}(X, \mathcal{J}X))\\
&&+ \frac{1}{2}( g(\mathcal{B}^{*}(X,X), \mathcal{B}(\mathcal{J}X, \mathcal{J}X)) + g(\mathcal{B}(X,X), \mathcal{B}^{*}(\mathcal{J}X, \mathcal{J}X))) \\
&=& g(\overline{\mathcal{S}}(X, \mathcal{J}X)\mathcal{J}X, X) - g(\mathcal{B}^{*}(X,X), \mathcal{B}(X,X))\\
&&- g(\mathcal{B}(\mathcal{J}X,X), \mathcal{B}^{*}(X, \mathcal{J}X))\\
&=& g(\overline{\mathcal{S}}(X, \mathcal{J}X)\mathcal{J}X, X) - 2g(\mathcal{B}^{*}(X,X), \mathcal{B}(X,X)),
\end{eqnarray*}
for any $X \in \Gamma(TN)$.
\end{proof}

\begin{proposition}\label{p1}
On a CR-statistical submanifold $N$ of a holomorphic statistical manifold $\overline{N}$, the following
\begin{eqnarray}\label{e21}
\mathcal{A}_{fV}X = - \mathcal{A}^{*}_{V}\mathcal{P}X
\end{eqnarray}
holds for any $X \in \mathcal{D}$ and $V \in \Gamma(fT^{\perp}N)$.
\end{proposition}

\begin{proof}
From \cite{article.7}, it follows that
\begin{eqnarray*}
0 &=& g(\nabla_{Y}(tV) - t\nabla_{Y}^{\perp *}V, X)\\
&=& g(\mathcal{A}_{fV}Y - \mathcal{P}\mathcal{A}^{*}_{V}Y, X)\\
&=& g(\mathcal{A}_{fV}X, Y) + g(\mathcal{A}^{*}_{V}\mathcal{P}X, Y),
\end{eqnarray*}
for any $X \in \mathcal{D}$, $Y \in \Gamma(TN)$ and $V \in \Gamma(fT^{\perp}N)$. Thus, we get the desired result.
\end{proof}

\begin{proposition}\label{4}
Let $N$ be a CR-statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Then
\begin{eqnarray*}
\nabla_{X}^{\perp}\mathcal{J}Y - \nabla_{Y}^{\perp}\mathcal{J}X \in \mathcal{J}\mathcal{D}^{\perp},
\end{eqnarray*}
for any $X, Y \in \mathcal{D}^{\perp}$.
\end{proposition}

\begin{proof}
For any $X, Y \in \mathcal{D}^{\perp}$ and $V \in \Gamma(\mu)$, we obtain
\begin{eqnarray*}
g(\nabla_{X}^{\perp}\mathcal{J}Y - \nabla_{Y}^{\perp}\mathcal{J}X, V) &=& g(\overline{\nabla}_{X}\mathcal{J}Y - \overline{\nabla}_{Y}\mathcal{J}X, V) \\
&=& g(\mathcal{J}(\overline{\nabla}_{X}^{*}Y - \overline{\nabla}_{Y}^{*}X), V)\\
&=& - g(\overline{\nabla}_{X}^{*}Y - \overline{\nabla}_{Y}^{*}X, \mathcal{J}V)\\
&=& - g(\mathcal{B}^{*}(X, Y) - \mathcal{B}^{*}(X, Y), \mathcal{J}V)\\
&=& 0.
\end{eqnarray*}
This proves our assertion.
\end{proof}

\begin{definition}
Let $S$ be a $r$-dimensional distribution on a statistical manifold $N$. Then $S$ is said to be
\begin{itemize}
\item[(a)] minimal with respect to $\nabla$ if \hspace{0.01 cm} $\sum_{i = 1}^{r} \nabla_{e_{i}}e_{i} \in S$.
\item[(a)$^*$] minimal with respect to $\nabla^{*}$ if \hspace{0.01 cm} $\sum_{i = 1}^{r} \nabla_{e_{i}}^{*}e_{i} \in S$.
\end{itemize}
\end{definition}

For minimality of the holomorphic distribution $\mathcal{D}$ with respect to $\nabla$ and $\nabla^{*}$, we have the following:

\begin{proposition}\label{12}
Let $N$ be a CR-statistical submanifold of a holomorphic statistical manifold $\overline{N}$, then the holomorphic distribution $\mathcal{D}$ is minimal with respect to $\nabla$ and $\nabla^{*}$.
\end{proposition}

\begin{proof}
For any $X \in \mathcal{D}$ and $Z \in \mathcal{D}^{\perp}$, we have
\begin{eqnarray}\label{9}
g(Z, \nabla_{X}^{*}X) &=& g(Z, \overline{\nabla}_{X}^{*}X) \nonumber \\
&=& g(\mathcal{J}Z, \mathcal{J}\overline{\nabla}_{X}^{*}X) \nonumber \\
&=& g(\mathcal{J}Z, \overline{\nabla}_{X}\mathcal{J}X) \nonumber \\
&=& - g(\overline{\nabla}_{X}^{*}\mathcal{J}Z, \mathcal{J}X) \nonumber \\
&=& g(\mathcal{A}_{\mathcal{J}Z}^{*}X, \mathcal{J}X),
\end{eqnarray}
where we have used
\begin{eqnarray*}
0 = Xg(\mathcal{J}Z, \mathcal{J}X) = g(\mathcal{J}Z, \overline{\nabla}_{X}\mathcal{J}X) + g(\overline{\nabla}_{X}^{*}\mathcal{J}Z, \mathcal{J}X).
\end{eqnarray*}
Replacing $X$ by $\mathcal{J}X$ in equation (\ref{9}), we find that
\begin{eqnarray}\label{10}
g(Z, \nabla_{\mathcal{J}X}^{*}\mathcal{J}X) = - g(\mathcal{A}_{\mathcal{J}Z}^{*}X, \mathcal{J}X).
\end{eqnarray}
By adding (\ref{9}) and (\ref{10}), we get
\begin{eqnarray}\label{1.1}
g(Z, \nabla_{X}^{*}X + \nabla_{\mathcal{J}X}^{*}\mathcal{J}X) = 0.
\end{eqnarray}
From (\ref{1.1}), we conclude that $\mathcal{D}$ is minimal with respect to $\nabla^{*}$. By simple computation, we also prove that $\mathcal{D}$ is minimal with respect to $\nabla$. This completes our proof.
\end{proof}

\begin{definition}
A distribution $S$ is called involutive if $[X, Y] \in S$ for any $X, Y \in S$.
\end{definition}

\begin{lemma}\label{14}
\cite{article.7} Let $N$ be a CR-statistical submanifold of a holomorphic statistical manifold $\overline{N}$. If $\mathcal{D}$ is involutive, then
\begin{eqnarray}\label{22}
\mathcal{B}(\mathcal{J}X, Y) = \mathcal{B}(\mathcal{J}Y, X)
\end{eqnarray}
holds for any $X, Y \in \mathcal{D}$.
\end{lemma}

In the light of Lemma \ref{14}, we have the following proposition:

\begin{proposition}\label{p}
Let $N$ be a CR-statistical submanifold of a holomorphic statistical manifold $\overline{N}$. Suppose that $\mathcal{D}$ is involutive. If $N$ is mixed totally geodesic with respect to $\overline{\nabla}$, then $\mathcal{A}_{V}^{*}\mathcal{J}X = - \mathcal{J}\mathcal{A}_{V}^{*}X$.  If $N$ is mixed totally geodesic with respect to $\overline{\nabla}^{*}$, then $\mathcal{A}_{V}\mathcal{J}X = - \mathcal{J}\mathcal{A}_{V}X$, for any $X \in \mathcal{D}$ and $V \in \Gamma(\mu)$.
\end{proposition}

\begin{proof}
For any $X \in \mathcal{D}$, $Y \in \mathcal{D}$ and $V \in \Gamma(\mu)$, we obtain
\begin{eqnarray*}
g(\mathcal{J}\mathcal{A}_{V}^{*}X, Y) &=& - g(\mathcal{A}_{V}^{*}X, \mathcal{J}Y)\\
&=& - g(\mathcal{B}(X, \mathcal{J}Y), V)\\
&=& - g(\mathcal{B}(\mathcal{J}X, Y), V)\\
&=& - g(\mathcal{A}_{V}^{*}\mathcal{J}X, Y),
\end{eqnarray*}
where we have used Lemma \ref{14}. Hence, our first assertion follows. In the same manner, we prove our second assertion.
\end{proof}

\section{Riemannian Products in Holomorphic Statistical Manifolds}

The statistical version of CR-product defined in \cite{article.50} is as follows:

\begin{definition}
Let $N$ be a CR-statistical submanifold in a holomorphic statistical manifold $\overline{N}$. Then $N$ is called a CR-product if it is a CR-product
as a CR-submanifold in a Kaehler manifold $\overline{N}$, that is, it is locally a Riemannian product of a holomorphic submanifold $N^{T}$ and a totally real submanifold $N^{\perp}$ of $\overline{N}$.
\end{definition}

We recall some results from \cite{article.7}.

\begin{proposition}\label{pp}
Let $(N, \nabla, g)$ be a CR-statistical submanifold in a holomorphic statistical manifold $\overline{N}$. If $N$ is mixed totally geodesic with respect to $\overline{\nabla}$ and $\overline{\nabla}^{*}$, then each leaf of $\mathcal{D}^{\perp}$ is totally geodesic in $N$ with respect to $\nabla$ and $\nabla^{*}$.
\end{proposition}

\begin{proposition}\label{pp1}
Let $(N, \nabla, g)$ be a CR-statistical submanifold in a holomorphic statistical manifold $\overline{N}$. If $N$ is $\mathcal{D}-$totally geodesic with respect to $\overline{\nabla}$ and $\overline{\nabla}^{*}$, then $\mathcal{D}$ is completely integrable and each leaf of $\mathcal{D}$ is totally geodesic in $N$ with respect to $\nabla$ and $\nabla^{*}$.
\end{proposition}

Now, if we assume that $\mathcal{A}_{\mathcal{F}\mathcal{D}^{\perp}} \mathcal{D} = 0$ and $\mathcal{A}_{\mathcal{F}\mathcal{D}^{\perp}}^{*} \mathcal{D} = 0$, then for any $X \in \mathcal{D}, Y \in \Gamma(TN)$ and $Z \in \mathcal{D}^{\perp}$, we have
\begin{eqnarray}\label{eq1}
g(\mathcal{B}(X, Y), \mathcal{F}Z) = g(\mathcal{B}^{*}(X, Y), \mathcal{F}Z) = 0.
\end{eqnarray}
In particular, if $Y \in \mathcal{D}$, then the relations in (\ref{eq1}) give $N$ is $\mathcal{D}-$totally geodesic with respect to $\overline{\nabla}$ and $\overline{\nabla}^{*}$. Thus, from Proposition \ref{pp1}, we know that $\mathcal{D}$ is completely integrable and each leaf of $\mathcal{D}$ is totally geodesic in $N$ with respect to $\nabla$ and $\nabla^{*}$. Now, for $Y \in \mathcal{D}^{\perp}$, again relations in (\ref{eq1}) imply that $N$ is mixed totally geodesic with respect to $\overline{\nabla}$ and $\overline{\nabla}^{*}$. From this and Proposition \ref{pp}, we can easily say that each leaf of $\mathcal{D}^{\perp}$ is totally geodesic in $N$ with respect to $\nabla$ and $\nabla^{*}$. Hence, $N$ is CR-product in $\overline{N}$.

We can state the following:

\begin{theorem}\label{11}
A CR-statistical submanifold $N$ of a holomorphic statistical manifold $\overline{N}$ is a CR-product if
\begin{eqnarray*}
\mathcal{A}_{\mathcal{F}\mathcal{D}^{\perp}} \mathcal{D} = \mathcal{A}_{\mathcal{F}\mathcal{D}^{\perp}}^{*} \mathcal{D} = 0.
\end{eqnarray*}
\end{theorem}

Next, we prove the following:

\begin{theorem}
Let $N$ be an $n$-dimensional generic statistical submanifold of a $2m-$dimensional holomorphic statistical manifold $\overline{N}$. If
\begin{eqnarray}\label{y}
\mathcal{P}\nabla_{X}^{*}Y = \nabla_{X}\mathcal{P}Y
\end{eqnarray}
holds, and $\mathcal{B}^{*}(X, \mathcal{D}) = 0$, for any $X, Y \in \Gamma(TN)$. Then $N$ is locally a Riemannian product of a holomorphic submanifold $N^{T}$ and a totally real submanifold $N^{\perp}$ of $\overline{N}$.
\end{theorem}

\begin{proof}
Following (\ref{y}) and \cite{article.7}, we arrive at
\begin{eqnarray*}
\mathcal{A}_{\mathcal{F}Y}X = - t \mathcal{B}^{*}(X, Y) = - \mathcal{J}\mathcal{B}^{*}(X, Y),
\end{eqnarray*}
for any $X, Y \in \Gamma(TN)$. Replacing $Y$ by $\mathcal{P}Y$, we get
\begin{eqnarray}\label{ya}
\mathcal{B}^{*}(X, \mathcal{P}Y) = 0.
\end{eqnarray}
For any $Y \in \mathcal{D}$ and $X \in \Gamma(TN)$, we obtain
\begin{eqnarray*}\label{e4}
\mathcal{F}\nabla_{X}Y = \mathcal{B}^{*}(X, \mathcal{P}Y) - f\mathcal{B}(X, Y) = 0,
\end{eqnarray*}
and
\begin{eqnarray*}\label{e4}
\mathcal{F}\nabla_{X}^{*}Y = \mathcal{B}(X, \mathcal{P}Y) - f\mathcal{B}^{*}(X, Y) = 0.
\end{eqnarray*}
Thus, the distribution $\mathcal{D}$ is parallel with respect to $\nabla$. Now, we consider $Y \in \mathcal{D}^{\perp}$, then for any $X, Z \in \Gamma(TN)$, we have
\begin{eqnarray*}\label{e5}
g(\mathcal{P}\nabla_{X}^{*}Y, \mathcal{P}Z) &=& - g(\mathcal{B}^{*}(X, \mathcal{P}Z), \mathcal{F}Y) + g(\mathcal{B}^{*}(X, Y), \mathcal{FP}Z) = 0,
\end{eqnarray*}
and
\begin{eqnarray*}\label{e5}
g(\mathcal{P}\nabla_{X}Y, \mathcal{P}Z) &=& - g(\mathcal{B}(X, \mathcal{P}Z), \mathcal{F}Y) + g(\mathcal{B}(X, Y), \mathcal{FP}Z) = 0,
\end{eqnarray*}
Thus, the distribution $\mathcal{D}^{\perp}$ is parallel with respect to $\nabla^{*}$. Consequently, $N$ is locally a Riemannian product of a holomorphic submanifold $N^{T}$ and a totally real submanifold $N^{\perp}$ of $\overline{N}$.
\end{proof}

\section{Chen-Ricci Inequality for Statistical Submanifolds}

In this section, we obtain an optimal inequality for Ricci curvature in terms of squared norm of mean curvature and canonical structure by using an optimization technique. Following \cite{oprea}, we have

\begin{theorem}\label{oth}
Let $(N, g)$ be  a Riemannian submanifold of a Riemannian manifold $(\overline{N}, \overline{g})$ and $f : N \rightarrow \mathbb{R}$ be a differentiable function. If $x_{0} \in N$ is the solution of the optimum problem $\min\limits_{x_{0} \in N} f(x_{0})$, then
\begin{enumerate}
\item[(a)] $(grad \hspace{0.1 cm} f)(x) \in T_{x}^{\perp}N$,
\item[(b)] the bilinear form $A : T_{x}N \times T_{x}N \rightarrow \mathbb{R}$,
$$A(X, Y) = Hess_{f}(X, Y) + \overline{g}(h^{'}(X, Y), (grad \hspace{0.1 cm} f)(x))$$
\end{enumerate}
is positive semi-definite, where $h^{'}$ is the second fundamental form of $N$ in $\overline{N}$ and $grad \hspace{0.1 cm} f$ denotes the gradient of $f$.
\end{theorem}

For nice applications of the above result, follow \cite{toprea}. Here we prove the following optimal inequality in the light of Theorem \ref{oth}:

\begin{theorem}\label{t}
Let $(N, \nabla, g)$ be an $m$-dimensional statistical submanifold in a holomorphic statistical manifold $\overline{N}(c)$ of constant holomorphic sectional curvature $c$. For each unit vector $X \in T_{x}N$, $x \in N$, we have
\begin{eqnarray}\label{r}
Ric^{\nabla, \nabla^{*}}(X) &\geq& 2 Ric^{0}(X) - \frac{c}{4}(m - 1 + 3||\mathcal{P} X||^{2}) \nonumber  \\
&&- \frac{m^{2}}{8} [||\mathcal{H}||^{2} + ||\mathcal{H}^{*}||^{2}],
\end{eqnarray}
where $Ric^{0}$ denotes the Ricci curvature with respect to Levi-Civita connection.
\end{theorem}

\begin{proof}
We choose $\{e_{1}, \dots, e_{m}\}$ as the orthonormal frame of $T_{x}N$ such that $e_{1} = X$ and $||X|| = 1$, and $\{e_{m+1}, \dots, e_{2n}\}$ as the the orthonormal frame of $T_{x}N$ in $\overline{N}$. Then by (\ref{g}) and (\ref{g1}), we have
\begin{eqnarray*}
2\overline{\mathcal{S}}(e_{1}, e_{i}, e_{1}, e_{i}) &=& 2\mathcal{S}(e_{1}, e_{i}, e_{1}, e_{i}) - g(\mathcal{B}(e_{1}, e_{1}), \mathcal{B}^{*}(e_{i}, e_{i}))\\
&& - g(\mathcal{B}^{*}(e_{1}, e_{1}), \mathcal{B}(e_{i}, e_{i}))
+ 2g(\mathcal{B}(e_{1}, e_{i}), \mathcal{B}^{*}(e_{1}, e_{i}))\\
&=& 2\mathcal{S}(e_{1}, e_{i}, e_{1}, e_{i}) - \{4g(\mathcal{B}^{0}(e_{1}, e_{1}), \mathcal{B}^{0}(e_{i}, e_{i}))\\
&&- g(\mathcal{B}(e_{1}, e_{1}), \mathcal{B}(e_{i}, e_{i})) - g(\mathcal{B}^{*}(e_{1}, e_{1}), \mathcal{B}^{*}(e_{i}, e_{i}))\\
&&- 4g(\mathcal{B}^{0}(e_{1}, e_{i}), \mathcal{B}^{0}(e_{1}, e_{i})) + g(\mathcal{B}(e_{1}, e_{i}), \mathcal{B}(e_{1}, e_{i}))\\
&&+ g(\mathcal{B}^{*}(e_{1}, e_{i}), \mathcal{B}^{*}(e_{1}, e_{i}))\}\\
&=& 2\mathcal{S}(e_{1}, e_{i}, e_{1}, e_{i}) - 4\sum_{r=m+1}^{2n}(h^{0r}_{11}h^{0r}_{ii} - (h^{0r}_{1i})^{2})\\
&&+ \sum_{r=m+1}^{2n}(h^{r}_{11}h^{r}_{ii} - (h^{r}_{1i})^{2}) + \sum_{r=m+1}^{2n}(h^{*r}_{11}h^{*r}_{ii} - (h^{*r}_{1i})^{2}),
\end{eqnarray*}
where we have used the notations $\overline{\mathcal{S}}(X,Y,Z,W) = g(\overline{\mathcal{S}}(X,Y)W,Z)$. Summing over $2 \leq i \leq m$, we have
\begin{eqnarray*}
\frac{c}{2}(m-1 + 3||\mathcal{P} X||^{2}) &=& 2Ric^{\nabla, \nabla^{*}}(X) - 4\sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{0r}_{11}h^{0r}_{ii} - (h^{0r}_{1i})^{2}) \\
&&+ \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{r}_{11}h^{r}_{ii} - (h^{r}_{1i})^{2}) + \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{*r}_{11}h^{*r}_{ii} \\
&&- (h^{*r}_{1i})^{2}),
\end{eqnarray*}
where $Ric^{\nabla, \nabla^{*}}(X)$ denotes the Ricci curvature of $N$ with respect to $\nabla$ and $\nabla^{*}$ at $x$. Further, we derive
\begin{eqnarray}\label{g2}
2Ric^{\nabla, \nabla^{*}}(X) - \frac{c}{2}(m-1 + 3||\mathcal{P} X||^{2}) &=& 4\sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{0r}_{11}h^{0r}_{ii} - (h^{0r}_{1i})^{2}) \nonumber \\
&&- \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{r}_{11}h^{r}_{ii} - (h^{r}_{1i})^{2})\nonumber \\
&&- \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{*r}_{11}h^{*r}_{ii} - (h^{*r}_{1i})^{2}).\nonumber \\
\end{eqnarray}
By Gauss equation with respect to Levi-Civita connection, it follows that
\begin{eqnarray*}
Ric^{0}(X) - \frac{c}{4}(m-1 + 3||\mathcal{P} X||^{2}) = \sum_{r=m+1}^{n}\sum_{i=2}^{m}(h^{0r}_{11}h^{0r}_{ii} - (h^{0r}_{1i})^{2}).
\end{eqnarray*}
Substituting into (\ref{g2}), we arrive at
\begin{eqnarray*}
2Ric^{\nabla, \nabla^{*}}(X) - \frac{c}{2}(m-1 + 3||\mathcal{P} X||^{2}) &=& 4[Ric^{0}(X) - \frac{c}{4}(m-1 + 3||\mathcal{P} X||^{2})] \\
&&- \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{r}_{11}h^{r}_{ii} - (h^{r}_{1i})^{2})\nonumber \\
&&- \sum_{r=m+1}^{2n}\sum_{i=2}^{m}(h^{*r}_{11}h^{*r}_{ii} - (h^{*r}_{1i})^{2}).\nonumber \\
\end{eqnarray*}
On simplifying the previous relation, we get
\begin{eqnarray}\label{g3}
&&-2Ric^{\nabla, \nabla^{*}}(X) - \frac{c}{2}(m-1 + 3||\mathcal{P} X||^{2}) + 4Ric^{0}(X)\nonumber \\
&=& \sum_{r=m+1}^{n}\sum_{i=2}^{m}(h^{r}_{11}h^{r}_{ii} - (h^{r}_{1i})^{2})\nonumber \\
&&+ \sum_{r=m+1}^{n}\sum_{i=2}^{m}(h^{*r}_{11}h^{*r}_{ii} - (h^{*r}_{1i})^{2})\nonumber \\
&\leq& \sum_{r=m+1}^{n}\sum_{i=2}^{m}h^{r}_{11}h^{r}_{ii} + \sum_{r=m+1}^{n}\sum_{i=2}^{m}h^{*r}_{11}h^{*r}_{ii}.
\end{eqnarray}
Let us define the quadratic form $f_{r}, f_{r}^{*} : \mathbb{R}^{m} \rightarrow \mathbb{R}$ by
$$f_{r}(h_{11}^{r}, h_{22}^{r}, \dots, h_{mm}^{r}) = \sum_{r=m+1}^{2n}\sum_{i=2}^{m}h^{r}_{11}h^{r}_{ii},$$ and
$$f_{r}^{*}(h_{11}^{*r}, h_{22}^{*r}, \dots, h_{mm}^{*r}) = \sum_{r=m+1}^{2n}\sum_{i=2}^{m}h^{*r}_{11}h^{*r}_{ii}.$$
We consider the constrained extremum problem $\max f_{r}$ subject to $$Q : \sum_{i=1}^{m}h^{r}_{ii} = \alpha^{r},$$ where $\alpha^{r}$ is a real constant. The gradient vector field of the function $f_{r}$ is given by $$grad \hspace{0.1 cm} f_{r} = (\sum_{i=2}^{m}h^{r}_{ii}, h^{r}_{11}, h^{r}_{11}, \dots, h^{r}_{11}).$$ For an optimal solution $p = (h_{11}^{r}, h_{22}^{r}, \dots h_{mm}^{r})$ of the problem in question, the vector $grad \hspace{0.1 cm} f_{r}$ is normal to $Q$ at the point $p$. It follows that $$h^{r}_{11} = \sum_{i=2}^{m}h^{r}_{ii} = \frac{\alpha^{r}}{2}.$$
Now, we fix $x \in Q$. The bilinear form $A : T_{x}Q \times T_{x}Q \rightarrow \mathbb{R}$ has the following expression:
\begin{eqnarray*}
A(X, Y) = Hess_{f_{r}}(X, Y) + <h^{'}(X, Y), (grad \hspace{0.1 cm} f_{r})(x)>,
\end{eqnarray*}
where $h^{'}$ denotes the second fundamental form of $Q$ in $\mathbb{R}^{m}$ and $<\cdot,\cdot>$ denotes the standard inner product on $\mathbb{R}^{m}$. The Hessian matrix of $f_{r}$ is given by
\begin{eqnarray*}
Hess_{f_{r}} = \left(
 \begin{array}{ccccc}
    0 & 1 & \dots & 1 \\
    1 & 0 & \dots& 0 \\
    \vdots & \vdots \ & \ddots & \vdots \\
    1 & 0  & \dots &  0 \\
    1 & 0  & \dots &  0
  \end{array}
\right).
\end{eqnarray*}
We consider a vector $X \in T_{x}Q$, which satisfies a relation $\sum_{i=2}^{m} X_{i} = -X_{1}$. As $h^{'} = 0$ in $\mathbb{R}^{m}$, we get
\begin{eqnarray*}
A(X, X) = Hess_{f_{r}}(X, X) &=& 2 \sum_{i=2}^{m} X_{1}X_{i} \\
&=& (X_{1} + \sum_{i=2}^{m} X_{i})^{2} - (X_{1})^{2} - (\sum_{i=2}^{m} X_{i})^{2}\\
&=& - 2(X_{1})^{2} \leq 0.
\end{eqnarray*}
However, the point $p$ is the only optimal solution, i.e., the global maximum point of problem. Thus, we obtain
\begin{eqnarray}\label{g4}
f_{r} \leq \frac{1}{4}(\sum_{i=1}^{m} h_{ii}^{r})^{2} = \frac{m^{2}}{4}(H^{r})^{2},
\end{eqnarray}
Next, we deal with the constrained extremum problem $\max f_{r}^{*}$ subject to $$Q^{*} : \sum_{i=1}^{m}h^{*r}_{ii} = \alpha^{*r},$$ where $\alpha^{*r}$ is a real constant. By similar arguments as above, we find
\begin{eqnarray}\label{g410}
f_{r}^{*} \leq \frac{1}{4}(\sum_{i=1}^{m} h_{ii}^{*r})^{2} = \frac{m^{2}}{4}(\mathcal{H}^{*r})^{2}.
\end{eqnarray}
On combining (\ref{g3}), (\ref{g4}) and (\ref{g410}), we get our desired inequality (\ref{r}).
\end{proof}

The characterisation of equality cases in Theorem \ref{t}:
\begin{theorem}
Let $(N, \nabla, g)$ be an $m$-dimensional statistical submanifold in a holomorphic statistical manifold $\overline{N}(c)$ of constant holomorphic sectional curvature $c$. The equality holds in the inequality (\ref{r}) if and only if $$h(X, X) = \frac{m}{2} \mathcal{H}(x), \textrm{             } h^{*}(X, X) = \frac{m}{2} \mathcal{H}^{*}(x)$$ and $$h(X, Y) = 0, \textrm{               } h^{*}(X, Y) = 0$$ for all $Y \in T_{x}N$ orthogonal to $X$.
\end{theorem}

\begin{proof}
The vector field $X$ satisfies the equality case if and only if
$$h_{1i}^{r} = 0, \textrm{             } h_{1i}^{*r} = 0, \textrm{             } i \in \{2, \dots, m\},$$
and
$$h_{11}^{r} = \sum_{i=2}^{m}h_{ii}^{r}, \textrm{             } h_{11}^{*r} = \sum_{i=2}^{m}h_{ii}^{*r}, \textrm{             } r \in \{m+1, \dots, 2n\},$$
which can be rewritten as
$$h_{11}^{r} = \frac{m}{2}\mathcal{H},$$
and $$h_{11}^{*r} = \frac{m}{2}\mathcal{H}^{*}.$$
Thus, it proves our assertion.
\end{proof}

An immediate consequence of Theorem \ref{t} is as follows:

\begin{corollary}
Let $(N, \nabla, g)$ be an $m$-dimensional CR-statistical submanifold in a holomorphic statistical manifold $\overline{N}(c)$ of constant holomorphic sectional curvature $c$. Then
\begin{enumerate}
\item[(a)] For each unit vector $X \in \mathcal{D}_{x}$
\begin{eqnarray*}
Ric^{\nabla, \nabla^{*}}(X) \geq 2 Ric^{0}(X) - \frac{c}{4}(m +2) - \frac{m^{2}}{8} [||\mathcal{H}||^{2} + ||\mathcal{H}^{*}||^{2}].
\end{eqnarray*}
\item[(b)] for each unit vector $X \in \mathcal{D}_{x}^{\perp}$
\begin{eqnarray*}
Ric^{\nabla, \nabla^{*}}(X) \geq 2 Ric^{0}(X) -  \frac{c}{4}(m -1) - \frac{m^{2}}{8} [||\mathcal{H}||^{2} + ||\mathcal{H}^{*}||^{2}].
\end{eqnarray*}
\end{enumerate}
\end{corollary}

\section{Some Related Examples}

\begin{example}
Let $(\overline{N}, g)$ be a family of exponential distributions of mean 0:
$$\overline{N}:= \{p(u,\Phi)  |   p(u,\Phi)=\Phi e^{-\Phi u}, u \in [0,\infty), \Phi \in (0, \infty)\},$$
a Riemannian metric is given by $g:= \Phi^{-2}(d \Phi)^{2}$, and an $\alpha-$connection $(\alpha \in \mathbb{R})$ on $\overline{N}$ is defined by
$$\overline{\nabla}_{\frac{\partial}{\partial \Phi}}^{\alpha}\frac{\partial}{\partial \Phi} = (\alpha -1)\Phi^{-1}\frac{\partial}{\partial \Phi}.$$
Then $(\overline{N}, \overline{\nabla}^{\alpha}, g)$ is a $1-$dimensional statistical manifold.

We remark that one can also construct examples for higher dimension by defining Fisher information metric and $\alpha-$connection on a family of statistical distribution (for example \cite{f3,article.7}).
\end{example}

We state the following lemma:

\begin{lemma}\label{L}
\cite{f3} Let $(\overline{N}, g, \mathcal{J})$ be a Kaehler manifold and a connection $\overline{\nabla}$ is defined as $\overline{\nabla}:= \nabla^{g} + K$, where $K$ is a $(1,2)$-tensor field satisfying the following conditions:
\begin{eqnarray}\label{c1}
K(X, Y) &=& K(Y, X),
\end{eqnarray}
\begin{eqnarray}\label{c2}
g(K(X, Y), Z) &=& g(Y, K(X, Z))
\end{eqnarray}
and
\begin{eqnarray}\label{c3}
K(X, \mathcal{J}Y) + \mathcal{J} K(X, Y) &=& 0,
\end{eqnarray}
for any $X, Y, Z \in \Gamma(T\overline{N})$. Then $(\overline{N}, \overline{\nabla}, g, \mathcal{J})$ is a holomorphic statistical manifold.
\end{lemma}

By using the above Lemma \ref{L}, we construct the following examples:

\begin{example}\label{1e}
\cite{article} Let us consider a Kaehler manifold $(\overline{N} = \{(x^{1}, x^{2})^{\prime} \in \mathbb{R}^{2} | x^{1} > 0\}, g, \mathcal{J})$, where a Riemanian metric $g$ and the standard complex structure $\mathcal{J}$ on $\overline{N}$ are defined by
$$g = x^{1}\{(dx^{1})^{2} + (dx^{2})^{2}\}$$
and
$$\mathcal{J} \partial_{1} = \partial_{2}, \hspace{0.2 cm} \mathcal{J} \partial_{2} = - \partial_{1},$$
where $\partial_{i} = \frac{\partial}{\partial x^{i}}$ for $i = 1, 2$. Now, for any $\lambda \in \mathbb{R}$, we define a $(1,2)$-tensor field $K$ on $\mathbb{R}^{2}$ as follows:
$$K = \sum_{i,j,l=1}^{2} k_{ij}^{l}\partial_{l}\otimes dx^{i} \otimes dx^{j},$$
where
$- k_{11}^{1} = k_{12}^{2} = k_{21}^{2} = k_{22}^{1} = \lambda$ and $k_{11}^{2} = k_{12}^{1} = k_{21}^{1} = k_{22}^{2} = 0$. Then $K$ satisfies all three conditions of Lemma \ref{L}, and hence we get a holomorphic statistical manifold $(\overline{N}, \overline{\nabla}:= \nabla^{g} + K, g, \mathcal{J})$, where an affine connection $\overline{\nabla}$ on $\overline{N}$ is given by
\begin{eqnarray*}
\overline{\nabla}_{\partial_{1}}\partial_{1} &=& \bigg(\frac{1}{2}(x^{1})^{-1} - \lambda \bigg)\partial_{1},\\
\overline{\nabla}_{\partial_{1}}\partial_{2} &=& \overline{\nabla}_{\partial_{2}}\partial_{1} = \bigg(\frac{1}{2}(x^{1})^{-1} + \lambda\bigg)\partial_{2},\\
\overline{\nabla}_{\partial_{2}}\partial_{2} &=& -\bigg(\frac{1}{2}(x^{1})^{-1} + \lambda \bigg)\partial_{1}.
\end{eqnarray*}
\end{example}

\begin{example}\label{ex}
\cite{article} Let $(g, \mathcal{J})$ be a Kaehler structure on $\overline{N}$. We take a vector field $\Lambda \in \Gamma(T\overline{N})$ and set a tensor field $K_{1} \in \Gamma(T\overline{N}^{(1,2)})$ as follows:
\begin{eqnarray}\label{e7}
K_{1}(X, Y) &=& \big[g(\mathcal{J}\Lambda, X)g(\mathcal{J}\Lambda, Y) - g(\Lambda, X)g(\Lambda, Y)\big] \Lambda \nonumber \\
&&+ \big[g(\mathcal{J}\Lambda, X)g(\Lambda, Y) + g(\Lambda, X)g(\mathcal{J}\Lambda, Y)\big]\mathcal{J}\Lambda
\end{eqnarray}
for any $X, Y \in \Gamma(T\overline{N})$. Then, by simple computation, we see that $K_{1}$ satisfies three conditions of Lemma \ref{L}, and hence a holomorphic statistical manifold $(\overline{N}, \overline{\nabla}:= \nabla^{g} + K_{1}, g, \mathcal{J})$ is obtained.
\end{example}

\begin{example}\label{ex1}
\cite{article} For a Kaehler manifold $(\overline{N}, g, \mathcal{J})$, we take a vector field $\Lambda \in \Gamma(T\overline{N})$ and set $K_{2}$ as follows:
\begin{eqnarray}\label{e8}
K_{2}(X, Y) &=& \big[g(\Lambda, \mathcal{J}X)g(\Lambda, \mathcal{J}Y) - g(\Lambda, X)g(\Lambda, Y) \nonumber \\
&&- g(\Lambda, \mathcal{J}X)g(\Lambda, Y) - g(\Lambda, X)g(\Lambda, \mathcal{J}Y)\big] \Lambda \nonumber \\
&&+ \big[g(\Lambda, X)g(\Lambda, Y) - g(\Lambda, \mathcal{J}X)g(\Lambda, \mathcal{J}Y) \nonumber \\
&&- g(\Lambda, \mathcal{J}X)g(\Lambda, Y) - g(\Lambda, X)g(\Lambda, \mathcal{J}Y)\big]\mathcal{J}\Lambda
\end{eqnarray}
for any $X, Y \in \Gamma(T\overline{N})$. Then $K_{2} \in \Gamma(T\overline{N}^{(1,2)})$ satisfies three conditions of Lemma \ref{L} as in Example \ref{ex}, and hence $(\overline{N}, \overline{\nabla}:= \nabla^{g} + K_{2}, g, \mathcal{J})$ becomes a holomorphic statistical manifold.
\end{example}

\begin{remark}
We remark that $K_{1} = K_{3}(\mathcal{J} \Lambda)$ ($K_{3} = K_{3}(\Lambda)$ is found in \cite{article.f}, as stated in Example \ref{ex2}) and $K_{2} = (1 - \mathcal{J})K_{1}$.
\end{remark}

Some examples of CR-statistical submanifolds in holomorphic statistical manifolds are given by:

\begin{example}\label{ex2}
Recall the examples of \cite{article.f}. For a Kaehler manifold $(\overline{N}, \overline{g}, \mathcal{J})$ and $\Lambda \in \Gamma(T\overline{N})$, set
\begin{eqnarray}\label{e1}
K_{3}(X, Y) &=& \big[\overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, \mathcal{J}Y) - \overline{g}(\Lambda, X)\overline{g}(\Lambda, Y)\big] \mathcal{J}\Lambda \nonumber \\
&&+ \big[\overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, Y) + \overline{g}(\Lambda, X)\overline{g}(\Lambda, \mathcal{J}Y)\big]\Lambda
\end{eqnarray}
for any $X, Y \in \Gamma(T\overline{N})$.

On the other hand, let $\mathbb{C}^{n + 1}$ be the complex Euclidean space with coordinates $z^{i} = x^{i} + \sqrt{-1}y^{i}$. For a Kaehler manifold $\mathbb{C}^{n + 1}$, we can construct $K_{3}$ as (\ref{e1}) and can get a holomorphic statistical manifold $(\mathbb{C}^{n + 1}, \overline{\nabla}:= \nabla^{\overline{g}} + K_{3}, \overline{g}, \mathcal{J})$, where $\nabla^{\overline{g}}$ is the standard statistical flat connection on $\mathbb{C}^{n + 1}$ and $\Lambda$ is the radial vector field, i.e., $\Lambda = \sum x^{i}\frac{\partial}{\partial x^{i}} + \sum y^{i} \frac{\partial}{\partial y^{i}}$. Now, one can define a statistical immersion as follows:
$$f: \mathbb{C}^{n} \times \mathbb{R} \mapsto \mathbb{C}^{n + 1}$$
and take a standard CR-submanifold of the form $N = \mathbb{C}^{n} \times \mathbb{R}$ in $\mathbb{C}^{n+1}$. Also one can find the induced statistical structure $(\nabla, g)$ on $N$ from $(\overline{\nabla}, \overline{g})$ by an immersion $f$. Then $(N, \nabla, g)$ becomes a CR-product in a holomorphic statistical submanifold $\mathbb{C}^{n+1}$.
\end{example}

\begin{example}\label{ex3}
Let $\mathbb{R}^{2n}$ be a holomorphic statistical manifold, which is topologically the same as $(\mathbb{R}^{+})^{2n}$. Let $\mathbb{R}^{2n}$ be considered as $\mathbb{C}^{n}$ and then we define a statistical submanifold by
{\small \begin{eqnarray*}
\big(-\frac{\pi}{2}, \frac{\pi}{2}\big)^{n} \ni {(x^{1}, \dots, x^{n})}\mapsto {(r_{1}\cos x^{1}, \dots, r_{n}\cos x^{n}, r_{1}\sin x^{1}, \dots, r_{n}\sin x^{n})}\in \mathbb{C}^{n},
\end{eqnarray*}}
where $r_{1}, \dots, r_{n}$ are positive constants. Then $\big(-\frac{\pi}{2}, \frac{\pi}{2}\big)^{n}$ is a Lagrangian submanifold of $\mathbb{C}^{n}$ (see Example $6$ in \cite{article.7}).

Furthermore, we consider a holomorphic statistical manifold $(\mathbb{C}^{m+1}, \overline{\nabla}:= \nabla^{\overline{g}} + K_{4}, \overline{g}, \mathcal{J})$, where $K_{4}$ is as follows (see \cite{article.f}):
\begin{eqnarray}\label{e2}
K_{4}(X, Y) &=& \big[\overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, \mathcal{J}Y) - \overline{g}(\Lambda, X)g(\Lambda, Y)\nonumber \\
&&+ \overline{g}(\Lambda, X)\overline{g}(\Lambda, \mathcal{J}Y) + \overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, Y)\big] \mathcal{J}\Lambda\nonumber \\
&&+ \big[\overline{g}(\Lambda, X)\overline{g}(\Lambda, Y) - \overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, \mathcal{J}Y)\nonumber \\
&&+ \overline{g}(\Lambda, X)\overline{g}(\Lambda, \mathcal{J}Y) + \overline{g}(\Lambda, \mathcal{J}X)\overline{g}(\Lambda, Y)\big]\Lambda,
\end{eqnarray}
for $X, Y \in \Gamma(T\mathbb{C}^{m+1})$. Now, one can consider a statistical immersion $i$, given by
\begin{eqnarray*}
i : \mathbb{C}^{n+1} \times \big(-\frac{\pi}{2}, \frac{\pi}{2}\big)^{n} \mapsto \mathbb{C}^{2n+1},
\end{eqnarray*}
where $\mathbb{C}^{n+1} \times \big(-\frac{\pi}{2}, \frac{\pi}{2}\big)^{n}$ is a product Riemannian manifold, denoted by $N$. Also one can show that $(N = \mathbb{C}^{n+1} \times \big(-\frac{\pi}{2}, \frac{\pi}{2}\big)^{n}, \nabla, g)$ is generic submanifold of $\mathbb{C}^{2n+1}$, where $(\nabla, g)$ is the statistical structure on $N$ induced by $i$ from $(\overline{\nabla}, \overline{g})$. Moreover, $N$ is a CR-statistical submanifold of $\mathbb{C}^{m+1}$, $m > 2n$. Thus, $N$ induces a natural CR-product in a holomorphic statistical manifold $\mathbb{C}^{m+1}$.
\end{example}

redpajama_set_name: RedPajamaArXiv


text:
Ernesto Reyes Milian (born 10 September 1992) is a Cuban badminton player. He competed at the 2015 Pan American Games. Reyes was the bronze medalists at the 2014 Central American and Caribbean Games in the men's doubles and team event. He won the Giraldilla International tournament in the men's doubles event partnered with Ronald Toledo in 2012, and with Leodannis Martínez in 2016. He competed at the 2018 Central American and Caribbean Games in Barranquilla, Colombia, and helped the team win the silver medal.

Achievements

Central American and Caribbean Games
Men's doubles

BWF International Challenge/Series
Men's doubles

  BWF International Challenge tournament
  BWF International Series tournament
  BWF Future Series tournament

References

External links

1992 births
Living people
Cuban male badminton players
Pan American Games competitors for Cuba
Badminton players at the 2015 Pan American Games
Central American and Caribbean Games silver medalists for Cuba
Central American and Caribbean Games bronze medalists for Cuba
Competitors at the 2014 Central American and Caribbean Games
Competitors at the 2018 Central American and Caribbean Games
Central American and Caribbean Games medalists in badminton
20th-century Cuban people
21st-century Cuban people

redpajama_set_name: RedPajamaWikipedia


text:
Ziua internațională a prieteniei (, , , , , , , , , , , , , ) este proclamată, din 2011, ca ziua în care se sărbătorește la nivel mondial prietenia între oameni, țări și culturi. Are loc, anual, în prima duminică din luna august.

Inițial a fost creată de cei din industria producției de felicitări, , în anul 1930, aceștia sperând ca la un moment dat, să ajungă o sărbătoare mondială, iar vânzarile lor să crească. Această sărbătoare presupune cumpararea de felicitări cu diverse mesaje de prietenie și oferirea lor către persoanele la care țin cel mai mult.

Note

 [UN Resolution A/65/L.72]

Legături externe
International Day of Friendship 30 July

Prietenie
Organizația Națiunilor Unite
Relații interpersonale
August
Prietenie
20 iulie


If you love high street favourite New Look, you'll have the time of your life at the enormous branch here at Liverpool ONE.
Spread across three floors in a triple sized unit, New Look Liverpool ONE is a fashion fan's paradise. Coincidentally, the store itself is on Paradise Street right in the heart of Liverpool city centre.
New Look Liverpool offers trend-led, affordable and simply gorgeous fashion for men and women, including teens. This vibrant store stocks a full range of collections, including Curves, Maternity, Petite, label Lounge, Teens and Menswear. There really is something for everyone, and it's all in line with the season's biggest trends and catwalk looks.
Shop everything from statement dresses and hot party looks to beachwear, lingerie and accessories, as well as loungewear and wardrobe staples. If you're in need of a classic tee or a perfect-fitting pair of jeans, you'll find it right here at New Look.
There's a whole department dedicated to shoes, from heels, wedges and boots in all styles to sandals, beach footwear and sneakers. And make sure you top off your look with a glam accessory – New Look Liverpool ONE has bags, belts, jewellery, purses, sunglasses and much more in stock.
If you prefer to shop online, you can place an order from the comfort of your home and use the free Click and Collect service to pick it up from New Look Liverpool ONE as little as three days later. It's that easy, and if you're a student you can even get a 10% discount on top.
Visiting the New Look store at Liverpool ONE by car? Secure, convenient parking is available at Q-Park Hanover Street just a short three-minute stroll away. You can also reach this huge fashion retailer easily by public transport, as Liverpool Central rail station is only five minutes away on foot.


Bookmarklets | MOVABLE TYPE Vow ! Via silicon.com : "The destruction of a mobile phone mast in the West Midlands has led to concern that campaigners against mobile phone base stations are increasingly taking matters into their own hands, and could hamper the rollout of third-generation (3G) mobile services. The base station, at Wishaw near Sutton Coldfield, was pulled down in the early hours of 6 November. It has been sited there for a decade, and people living nearby blame it on a number of cases of serious illnesses in the area."


text:
Q: Is Payone Supports Tokenization with payment method Paypal,Credit and debit card? I want to know that is Payone payment method supports tokenization with above methods ? So I can capture amount using token id in future.

A: Every transaction regarding PAYONE methods can be captured or refunded with one id. This one is called 'txid' in our API.
This unique id will be created with the first payment request to PAYONE as response parameter. But you can handle captures and refunds in between the backend of magento. You can use the normal finish order or refunding flow. If you need more details, please contact the developer.

redpajama_set_name: RedPajamaStackExchange


Our September Job of the Month was close to home - literally! This gorgeous apartment was sold by members of the Blue Door Painting team, but not before a modern interior refresh.
The Avondale apartment already had a lot going for it - a great location, tons of space and natural light, as well as owners who had cared for it - but the key to fetching a good price from a buyer was presenting all the positives of the space itself.
Aside from choosing clean and attractive colors like Grey Cloud and Revere Pewter by Benjamin Moore to update the rooms, the Blue Door Painting team was careful to stage the home to highlight all the wonderful attributes that a new buyer might enjoy. Especially for those transitioning from renter to owner, the pride of ownership extends far beyond being able to choose paint colors, so it's important to showcase all the elements that make your house a home.
When staging a home or apartment, don't forget the basics - make sure the space is clean, organized and inviting before hosting an open house. Our interior design contacts would be the first to tell you to keep the details in mind; pillows, plants, curtains, even candles or fragrance can create a cozy, attractive atmosphere for potential buyers.
Sticking with bright neutrals wasn't just about staying on trend - this also allows for buyers to better visualize their own tastes when viewing the rooms. Overwhelming reds or dark blues or any color that's striking and bold might work for you, but your buyer might be turned off by the daring choice.
The formal dining room started off as a playroom, but you would never know! It was reverted back to its original purpose after clearing away clutter and setting up a more natural flow of traffic through the home.
Buyers are looking for a great bedroom, so it's savvy to focus a lot of energy on staging an impressive master suite. In this space, the bedrooms were tricky because of the attic ceilings, but the team was still able to set up a demonstrative room with appealing aesthetics.
Don't be surprised if you have to spend a little money on staging - pictures, posters, rugs and other decorative pieces might cost a few bucks, but they will add measurable value to your home. Because this space was attractive and move-in ready, it took only one day in the property listings to pull in numerous offers. Better staging = more offers, more offers = more demand and value.
Aside from paint and decor updates, it can't hurt to refinish floors, touch up trim and update kitchen cabinets and other hardware like bathroom faucets. Lighting fixtures need a facelift? Invest in some updated options that cast the room in the best light.
Keep in mind the kitchens and bathrooms can make or break a sale. Swap that old bathroom vanity for a modern sink and add a beautiful backsplash to your kitchen for an instant revitalization.
All set on the interior? Don't forget curb appeal. Freshly painted front doors and finished decks can present a tidy image of a home that will look great for years to come.
Selling soon? Contact us today for a FREE estimate so we can help you present the best version of your home to potential buyers. Follow us on Instagram for daily inspirations and read our 5 STAR reviews on Yelp, Angie's List or Houzz.


Spark Conference 2019 (5th ed) - Create.Elevate.Celebrate!
Join us for the 5th anniversary edition of Spark's annual conference - Create, Elevate, Celebrate!
We have a full day program filled with open plenaries, masterclasses, and networking all with the purpose of making sure you leave the conference well connected, inspiried, and ready to take the next steps to grow your business. Business owners and founders are welcome for knowledge-sharing and skillset building with the help of local experts in the field including academics with the latest research. Join us for what will be a memorable day in your career as a business owner.


text:
Alleppy Coir refers to the Coir products made in Ambalappuzha and Cherthala Taluks of Alappuzha district, in the state of Kerala, India. Coir production of this region dates back to as early as 1859. The World Trade Organization (WTO) granted Geographical Indication status to "Alleppey Coir" in 2007.

Main Products

 Coir Bed
 Coir Mat
 Coir Carpet

History

The production of Coir in Alappuzha dates back to centuries. There are mentions about coir in travelogues of Arabs and in Mythology like Ramayana.

The production of coir in Alappuzha were non mechanized and unorganized until the mid 19th century. The first coir factory in India at Alleppey Darragh Smail & Company was established in 1859 by an Irish American entrepreneur James Darragh. Coir was called Cocoa Mats in American English and Darragh Smail & Co. Ltd. had their office and salesrooms at 177, Water Street, New York City.
In the following one century many prominent European firms manufactured and exported coir from Alappuzha. The prominent ones were Aspinwall &Co., Pierce Lesley &Co. Ltd., William Goodacre & Sons India Pvt Ltd., Volkart Brothers etc.

In 1940's following the Indian Independence movement and Labour movement in Travancore the European companies exited and decentralization of coir production happened. Presently Coir Board of India is the prime regulatory and promotion body that encourage coir production.

See also

 List of geographical indications in India

References

Geographical indications in Kerala
Agriculture in Kerala
Culture of Alappuzha district


text:
A Syrian refugee child sits on the window of his family's trailer home painted by refugee artists in a camp near Mafraq, Jordan. AP/Raad Adayleh
5 ways the Syrian revolution continues
March 15, 2019 6.43am EDT • Updated March 15, 2019 11.08am EDT
Wendy Pearlman, Northwestern University
Wendy Pearlman
Associate Professor of Political Science, Northwestern University
Wendy Pearlman does not work for, consult, own shares in or receive funding from any company or organization that would benefit from this article, and has disclosed no relevant affiliations beyond their academic appointment.
Bashar al-Assad has "won" the war in Syria – or so many analysts tell us.
His regime has reconquered swaths of territory from rebel forces with starvation-and-surrender sieges, barrel bombs, chemical weapons and what one human rights investigator called "industrial scale" torture and killing of detainees.
Still, the regime might have fallen had Russia not stepped in and begun bombarding opposition strongholds in 2015.
In areas now under Assad's grip, Syrians speak of exhaustion, forced complicity with government rule and the return of the very walls of fear and silence that they sacrificed so much to tear down.
No one should underestimate the crushing toll of this violence or overestimate the capacity of any people to endure more than Syrians already have.
"We're tired and we can't bear any more blood. We're afraid for Syria," an activist told me. And that was in 2012.
But that does not mean that the struggle for freedom, dignity and justice that Syrians launched eight years ago is over.
Since 2012, I have interviewed hundreds of displaced Syrians who championed the uprising. I have seen that their revolution persists wherever Syrians continue to believe in their capacity to make change.
Five realms of activism and resilience stand out.
1. Nurturing civil society
Forty years of dictatorships stifled independent civil society in Syria. But the years since 2011 have witnessed a flourishing of citizen-led activities in fields from education to media and the creative arts.
In Syrian towns where rebels forced the state to withdraw, communities labored to create local councils through which people governed themselves. The regime subjected these experiments in civic participation to siege and bombardment, and then collapsed them in areas retaken by its troops.
In this 2012 photo, Syrian shopkeeper Yasser Mamaar speaks about working as a member of his town's revolutionary council to distribute fuel, broker a truce with the army and win the release of dozens of locals kidnapped by armed groups. AP/ Khalil Hamra
Still, civic work continues inside and outside Syria. Citizens for Syria, itself a Syrian initiative, identifies hundreds of organizations and projects spearheaded by and for Syrians.
Among them are the Molham Volunteering Team which, beginning as a group of students, is now an essential nonprofit provider of humanitarian relief. Women Now for Development offers vocational training, literacy courses and small grants that have empowered hundreds of Syrian women.
Read more: How the Syrian uprising began and why it matters
In these and other efforts, Syrians are demonstrating their will to build a democratic society from the bottom up, against enormous odds. This is a realm in which individual and state contributions can help.
For example, the Trump administration froze stabilization funds for Syria in 2018. That forced to a halt at least 150 civil society organizations. Some laid-off staff had no choice but to seek salaries with the only entities that offered them: armed factions, including those that the United States identified as terrorist groups.
2. Mobilizing for detainees
Syrians continue to take action on behalf of the more than 100,000 citizens who have become forcibly disappeared. Most of them were nonviolent civilians arrested arbitrarily by the government.
Copious documentation details the torture, death, rape and sexual violence synonymous with imprisonment in Syria, and even the suspected use of crematoriums to dispose of prisoners executed en masse.
The members of Families for Freedom advocate for 'freedom and justice for their loved ones who have been detained and forcibly disappeared by the Syrian regime and armed groups.' Families for Freedom
Defying despair, groups such as the Caesar Families Association and the women-led Families for Freedom demand release of information about detainees and humanitarian access to detention facilities. Their mobilization, from the streets of London to the halls of U.N. and EU conferences, is keeping attention to this issue that has left few Syrian families untouched.
3. Ending impunity
Syrians are working for justice and accountability. Teams are amassing exhaustive evidence of atrocities, from photographs of prisoners' emaciated corpses to documents incriminating leaders by name.
Several European states are using the principle of universal jurisdiction to investigate, prosecute and convict Syrian officials implicated in crimes against humanity.
In recent weeks, authorities arrested three Syrian officers in Germany and France for torture constituting crimes against humanity. And Syrian refugees submitted dossiers of evidence against Assad to the International Criminal Court in The Hague.
While state and international institutions have an important role to play, the real engine of the fight to end impunity in Syria is coming from Syrians themselves.
4. Demanding oversight for reconstruction
Assad will need hundreds of billions of dollars to rebuild the country that he played a large role in shattering.
Reports suggest that the Syrian government is already using reconstruction as "war by other means."
For example, the Law No. 10 of 2018 proposed to give citizens limited time to register proof of ownership with the Syrian state before it would confiscate properties for "redevelopment." Other legislation enables the regime to transfer these and other expropriated assets to private companies run by regime cronies.
Some Syrians are working to make sure that the Assad regime's plans for reconstructing devastated towns do not enrich loyalists and rob the properties of citizens forced to flee as refugees. AP/Hassan Ammar
Critics decry such "lawfare" as a strategy to to control society and the economy, further dispossess those forced to flee the country and enrich loyalists and foreign allies. It might also serve as demographic engineering that cleanses areas of some ethnic or religious groups and replaces them with others.
Syrian activists drawing attention to the dangerous politics of reconstruction call for transparency and accountability that were missing in the realm of humanitarian relief.
Since 2011, for example, United Nations aid programs in Syria awarded tens of millions of dollars in contracts to Assad family members and associates, thereby propping up the regime. In working against repetition of such oversight failures, Syrians continue the revolution's fight against corruption and unfairness.
5. Building dignified lives
The approximately 5.6 million refugees in countries on Syria's borders continue to work to achieve their aspirations despite severe legal and economic precariousness.
Many lack minimal access to health care and decent housing. Some 40 percent of children are not in school.
As war dies down in Syria, host states and international agencies increasingly discuss refugees' return. The small numbers who have gone back to Syria report that they did so because they could no longer endure the harshness of life in neighboring states. Yet return is not a safe solution; already some returnees have been disappeared or killed upon arrival.
Displaced persons are eager to continue that most revolutionary act of building dignified lives, but they cannot do so unless states offer them a modicum of safety and opportunity.
Past and continuing atrocities in Syria will haunt history for generations to come. That Syrians retain any hope in a better future is not only a testament to their sheer will. It is also a sign, I believe, that the revolution they began eight years ago continues today.
Syria's uprising did not happen in a vacuum. Read the first part of this series, "How the Syrian uprising began and why it matters," to hear the voices of Syrians describing the oppression that characterized their lives for decades, and how they mobilized against it.
Syrian war
Syrian refugee crisis

redpajama_set_name: RedPajamaCommonCrawl


text:
Deathkeep Download (1996 Role playing Game)
Role playing Games:
Download full Deathkeep:
Deathkeep screenshots:
For many years, S.S.I. had the sole rights to the AD&D series on the desk-top computer. In the late eighties, they released some fairly popular titles, that were built on campaigns from the AD&D role playing universe. They employed a slightly modified perspective of what was available back then. The movement was based on a small 3D window in the top left corner of the screen (a la Wizardry I), and when combat was engaged, the perspective shifted over to an angular bird's eye view (a la Ultima). Back then role playing offerings were pretty thin, so the AD&D games fared well. Since then much has changed and S.S.I. no longer holds the exclusive rights to the AD&D universe. A multitude of companies have arisen offering many different types of role playing games. S.S.I. has tried to stay in the game by offering up DeathKeep.
DeathKeep seemed to offer a great deal. It was one of the first games that was released exclusively for Windows 95, offered a real time 3D perspective and was once again based in the world of AD&D.
The introduction of the game was nice. The story is narrated by the use of "stain glass" style pictures, and some small amounts of 3D rendered animation. According to the story line, the Oracle of the Keep has summoned one of three heroes (one of the three player classes you choose: a Dwarven Fighter, an Elven Mage or a Half-Elf Fighter/Mage) to battle the evil Necromancer who has escaped from his long imprisonment. It seems that the Necromancer is out to get his long lost power in the ruins of an old dwarven sanctuary called the DeathKeep. Within the Keep are three towers, each of which holds a magical orb key. Naturally, its up to you to get each one of them. kill the evil one and save the day!
This game has ground breaking graphics. Unfortunately they would only be ground breaking if the game had been released in 1980. By today's standards, DeathKeep's graphics couldn't even be considered poor. When it first loaded up, I was sure that the game was only using four colors (that would be CGA graphics for all of you ankle-biters). The graphics are just really, really bad and I'm still in shock. The walls have poorly applied textures to them. The textures are badly designed and colored. The player's face looks as if it was hand drawn by an amateur. The monsters are also poorly drawn and are almost pathetic.
On the other hand, the music and sound effects were good. Naturally, music and sound effects are never enough to save a low-quality game, and in this case, the player feels as though it was a shame that they didn't use the sounds for something a little better.
The controller interface was badly designed. The mouse cannot be configured for different actions, and there are many keys to learn. There was nothing intuitive about the interface, and some functions seemed outright useless. For example, there was a jump button, but you could barely tell that you jumped.
Another disappointing aspect of the game was that the spell system hadn't changed from the late eighties. The AD&D series have suffered in that there is no originality in any of their products. The spells are the same ones in every single game that they release. They never change!!! That is a big problem. AD&D titles seem to be stagnant, and they can't compete with games like Heretic, Hexen, and even the now old Ultima Underworld series. In conclusion, DeathKeep plays like an in-house pre-beta beta. This is a product that should never have been released, and S.S.I. will suffer for it.
People who downloaded Deathkeep have also downloaded:
Descent to Undermountain, Dark Queen of Krynn, The, Dark Sun 2: Wake of the Ravager, Dark Sun: Shattered Lands, Diablo, Dink Smallwood, Death Knights of Krynn, Demise: Rise of the Ku'tan


text:
const uint64_t kExpected_NEON_cmge_2D_2OPIMM[] = {
  0x0000000000000000, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0xffffffffffffffff,
  0xffffffffffffffff, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
  0x0000000000000000, 0x0000000000000000,
};
const unsigned kExpectedCount_NEON_cmge_2D_2OPIMM = 40;

#endif  // VIXL_SIM_CMGE_2D_2OPIMM_TRACE_A64_H_

redpajama_set_name: RedPajamaGithub


text:
Pericoma bipunctata és una espècie d'insecte dípter pertanyent a la família dels psicòdids que es troba a Nord-amèrica: Alaska (els Estats Units) i la Colúmbia Britànica (el Canadà).

Referències

Enllaços externs
 Arctos
 The Animal Diversity Web

bipunctata


Recent times have seen a very positive surge in popularity of silver jewellery. It seems that in times of economic recession such as these, more subtle attitudes are called for in jewellery, and so silver has taken the place formerly occupied by more ostentatious and outwardly showy styles of jewellery, such as gold.
Men's silver jewellery has seen a particularly sharp increase in popularity. The reason for this is that men's jewellery in general has traditionally been grounded in more masculine themes such as simplicity, functionality and endurance, rather than the more feminine themes of visual beauty and complexity in appearance. Silver suits these strong masculine attributes quite well, as evidenced by the classic Italian silver chains and bracelets, which feature thick links and ruggedness to the point of being nearly indestructible. If jewellery is indeed a reflection of the wearer, then it makes sense that men gravitate towards silver.
Today's fashions are also quite rapidly repopularizing the idea of antique silver jewellery. These are usually heirloom pieces that are quite valuable, and can sometimes date back several decades – of course, their value is enhanced not only by their age, but by the quality of craftsmanship as well. Of course, it usually takes superior craftsmanship for a piece to survive decades without any visible blemishes or damage; the condition of the piece adds to its value as well.
The main problem with vintage silver pieces is that they are hard to find – they can surface in the oddest places at the oddest times – such as flea markets, pawnshops, or occasionally as undervalued bargains in jewellery stores – so finding them is oftentimes a matter of luck, and being in the right place at the right time. However, once found, a truly beautiful vintage piece will almost certainly become the centerpiece of anyone's jewellery collection. Antique silver amulets with Latin inscriptions, bracelets with hand-carved filigree scrollwork, even signet rings – items such as these are the absolute pinnacle of jewel-making, and are reminders of the fabulous quality that was once commonplace, a lost era of jewellery.
That is not to say that today's silver lacks anything in terms of quality or craftsmanship. Although few pieces are hand-made today (not only due to the prohibitive cost of creating a piece by hand, but also due to the fact that we have machines that simply do it better), the fact is that we have reached a technological level that has never been seen before. We can now create shapes and objects that have never before been possible – and it is mostly due to the wonderful characteristics of that famous alloy, sterling silver. Sterling silver jewellery offers more creation possibilities than most metals, due to its flexibility, resilience and strength. "Sterling silver" refers to sterling 925 silver, which is so named because it is 92.5% silver, with the other 7.5% being a different metal, usually copper or nickel.
The reason for mixing the metals like this is because silver on its own is quite soft (similar to gold jewellery), making it impractical for everyday wear. The weaving of other metals into pure silver, however, creates an alloy that is powerful and yet light – the perfect material for those rugged Italian silver chains. A keen observer will note that most fine silver jewellery today is made from sterling silver, because jewellers prefer it to any other form of silver.


text:
Q: Using momentjs with recurring dates I have a date which flagged as recurring date (occurs every year), this date is 1st April, 2014, and starting form 2014 this event will occur yearly.
I have a job which run weekly, in this job I insert in the DB events that will occur in the coming 180 days only starting from today, other events will be discarded.
now when in my check, 1st April 2016 is past....so it's discarded (while 1st April 2015 should be added as it's in the range).
I am using momentjs but am not able to find the correct logic for it.
below is my code ....
//desiredDate = today + 180 days

if (isYearlyRecur) {
    var year = eventeDate.getFullYear();
    if (year < currentYear) {
        if (moment(eventeDate).unix() < moment(todayDate).unix()) {
            eventeDate = moment(eventeDate).set('year', currentYear + 1);
        }
        else if (moment(eventeDate).unix() > moment(desiredDate).unix()) {
            eventeDate = moment(eventeDate).set('year', currentYear);
        }
    }
}

what is the correct way to do it and cover if the date comes in next year ...etc

A: You could use rSchedule for this (a javascript recurrence library which I maintain).
Example:

I have a date which flagged as recurring date (occurs every year), this date is 1st April, 2014, and starting form 2014 this event will occur yearly.

import { Schedule } from '@rschedule/rschedule';
import { MomentDateAdapter } from '@rschedule/moment-date-adapter';
import * as moment from 'moment';

const schedule = new Schedule({
  rrules: [{
    frequency: 'YEARLY',
    start: moment([2014, 3, 1]),
  }],
  dateAdapter: MomentDateAdapter,
});

I have a job which run weekly, in this job I insert in the DB events that will occur in the coming 180 days only starting from today, other events will be discarded.

const next180Days = moment().add(180, 'days').endOf('day');

if (schedule.occursBetween(moment(), next180Days)) {
  // do stuff...
}


text:
<?php
/**
 * Extension to {@link ContentController} which handles
 * redirection from main site to mobile.
 *
 * @package mobile
 */
class MobileSiteControllerExtension extends Extension {

	/**
	 * The expiration time of a cookie set for full site requests
	 * from the mobile site. Default is 30 minutes (1800 seconds)
	 * @var int
	 */
	public static $cookie_expire_time = 1800;

	/**
	 * Override the default behavior to ensure that if this is a mobile device
	 * or if they are on the configured mobile domain then they receive the mobile site.
	 */
	public function onBeforeInit() {
		$config = SiteConfig::current_site_config();

		// Redirect users to the full site if requested (cookie expires in 30 minutes)
		if(isset($_GET['fullSite'])) {
			$_COOKIE['fullSite'] = 1;
			setcookie('fullSite', 1, time() + self::$cookie_expire_time);
		}

		// Redirect to the full site if user requested
		if($this->onMobileDomain() && !empty($_COOKIE['fullSite']) && $config->MobileSiteType == 'RedirectToDomain') {
			return $this->owner->redirect($config->FullSiteDomain);
		} elseif(!empty($_COOKIE['fullSite'])) {
			return; // nothing more to be done
		}

		// If the user requested the mobile domain, set the right theme
		if($this->onMobileDomain()) {
			SSViewer::set_theme($config->MobileTheme);
		}

		// User just wants to see a theme, but no redirect occurs
		if(MobileBrowserDetector::is_mobile() && $config->MobileSiteType == 'MobileThemeOnly') {
			SSViewer::set_theme($config->MobileTheme);
		}

		//I added 14/1/2011 to detect ipad and send to theme

		if(MobileBrowserDetector::is_mobile() && $config->MobileSiteType == 'MobileThemeOnly') {
			if(MobileBrowserDetector::is_ipad())
				SSViewer::set_theme($config->Ipadtheme);
		}

		// If on a mobile device, but not on the mobile domain and has been setup for redirection
		if(!$this->onMobileDomain() && MobileBrowserDetector::is_mobile() && $config->MobileSiteType == 'RedirectToDomain') {
			return $this->owner->redirect($config->MobileDomain);
		}
	}

	/**
	 * Return whether the user is on the mobile version of the website
	 * @return boolean
	 */
	public function onMobileDomain() {
		$config = SiteConfig::current_site_config();
		$parts = parse_url($config->MobileDomain);
		if(isset($parts['host']) && $parts['host'] == $_SERVER['HTTP_HOST']) {
			return true;
		} else {
			return false;
		}
	}

	/**
	 * Return a link to the full site.
	 * @return string
	 */
	public function FullSiteLink() {
		return $this->owner->Link() . '?fullSite=1';
	}

	//i added 14/1/2011
	public function IsiPad() {
		return MobileBrowserDetector::is_ipad();
	}
	/**
	 * Is the current HTTP_USER_AGENT a known iPhone or iPod Touch
	 * mobile agent string?
	 *
	 * @return boolean
	 */
	public function IsiPhone() {
		return MobileBrowserDetector::is_iphone();
	}

	/**
	 * Is the current HTTP_USER_AGENT a known Android mobile
	 * agent string?
	 *
	 * @return boolean
	 */
	public function IsAndroid() {
		return MobileBrowserDetector::is_android();
	}

	/**
	 * Is the current HTTP_USER_AGENT a known Opera Mini
	 * agent string?
	 *
	 * @return boolean
	 */
	public function IsOperaMini() {
		return MobileBrowserDetector::is_opera_mini();
	}

	/**
	 * Is the current HTTP_USER_AGENT a known Blackberry
	 * mobile agent string?
	 *
	 * @return boolean
	 */
	public function IsBlackBerry() {
		return MobileBrowserDetector::is_blackberry();
	}

}