text:
\section{Introduction}

The aims of this text are to announce the result in a paper \cite{MT3},
to give proofs of some special cases of it,
and to make comments and remarks for the proof given there.
Because the full proof in \cite{MT3} is much more involved and technical,
we shall give a technical introduction and proofs for weaker statements
in this text (see Theorem \ref{curve} and \ref{reduced}).
This text is basically independent from \cite{MT3}.

\subsection{Result in \cite{MT3}}

Our main concern is the positivity of direct image sheaves
of adjoint bundles $\Rq$, for a K\"ahler morphism $f : X \lra Y$ endowed
with a Nakano semi-positive holomorphic vector bundle $(E, h)$ on $X$.
In our previous paper \cite{MT2},
generalizing a result \cite{B} in case $q = 0$,
we obtained the Nakano semi-positivity of
$\Rq$ with respect to the Hodge metric,
under the assumption that $f : X \lra Y$ is smooth.
However the smoothness assumption on $f$ is rather restrictive,
and it is desirable to remove it.

To state our result precisely, let us fix notations and recall basic facts.
Let $f : X \lra Y$ be a holomorphic map of complex manifolds.
A real $d$-closed $(1,1)$-form $\w$ on $X$ is said to be
{\it a relative K\"ahler form} for $f$, if for every point $y \in Y$,
there exists an open neighbourhood $W$ of $y$ and
a smooth plurisubharmonic function $\psi$ on $W$ such that
$\w + f^*(\ai\rd\rdb \psi)$ is a K\"ahler form on $f^{-1}(W)$.
A morphism $f$ is said to be {\it K\"ahler}, if there exists
a relative K\"ahler form for $f$ (\cite[6.1]{Tk}),
and $f : X \lra Y$ is said to be a {\it K\"ahler fiber space},
if $f$ is proper, K\"ahler, and surjective with connected fibers,

\begin{setup} \label{basicMT}
(1)
Let $X$ and $Y$ be complex manifolds of $\dim X = n + m$ and
$\dim Y = m$, and let $f : X \lra Y$ be a K\"ahler fiber space.
We do not fix a relative K\"ahler form for $f$, unless otherwise stated.
The {\it discriminant locus} of $f : X \lra Y$ is the minimum closed
analytic subset $\Dl \subset Y$ such that $f$ is smooth over $Y \sm \Dl$.

(2)
Let $(E, h)$ be a Nakano semi-positive holomorphic vector bundle on $X$.
Let $q$ be an integer with $0 \le q \le n$.
By Koll\'ar \cite{Ko1} and Takegoshi \cite{Tk},
$\F$ is torsion free on $Y$,
and moreover it is locally free on $Y \sm \Dl$ (\cite[4.9]{MT2}).
In particular we can let $S_q \subset \Dl$ be
the minimum closed analytic subset of $\codim_Y S_q \ge 2$ such that
$\F$ is locally free on $Y \sm S_q$.
Let $\pi : {\mathbb{P}}(\F|_{Y \sm S_q}) \lra Y \sm S_q$
be the projective space bundle, and let $\pi^*(\F|_{Y \sm S_q}) \lra {\mathcal{O}}(1)$
be the universal quotient line bundle.

(3)
Let $\w_f$ be a relative K\"ahler form for $f$.
Then we have the Hodge metric $g$ on the vector bundle
$\F|_{Y \sm \Dl}$ with respect to $\w_f$ and $h$ (\cite[\S 5.1]{MT2}).
By the quotient
$\pi^*(\Rq|_{Y \sm \Dl}) \lra {\mathcal{O}}(1)|_{\pi^{-1}(Y \sm \Dl)}$,
the metric $\pi^*g$ gives the quotient metric $\go$ on
${\mathcal{O}}(1)|_{\pi^{-1}(Y \sm \Dl)}$.
The Nakano, even weaker Griffiths, semi-positivity of $g$
(by \cite[1.2]{B} for $q=0$, and by \cite[1.1]{MT2} for $q$ general)
implies that $\go$ has a semi-positive curvature.
\qed
\end{setup}

In these notations, the main result in \cite{MT3} is as follows.

\begin{thm} \label{MT}
Let $f : X \lra Y$, $(E, h)$ and $0 \le q \le n$ be as in Set up \ref{basicMT}.

(1) Unpolarized case.
Then, for every relatively compact open subset $Y_0 \subset Y$,
the line bundle ${\mathcal{O}}(1)|_{\pi^{-1}(Y_0 \sm S_q)}$ on
${\mathbb{P}}(\Rq|_{Y_0 \sm S_q})$
has a singular Hermitian metric
with semi-positive curvature, and which is smooth on $\pi^{-1}(Y_0 \sm \Dl)$.

(2) Polarized case.
Let $\w_f$ be a relative K\"ahler form for $f$.
Assume that there exists a closed analytic set
$Z \subset \Dl$ of $\codim_Y Z \ge 2$ such that
$f^{-1}(\Dl)|_{X \sm f^{-1}(Z)}$ is a divisor
and has a simple normal crossing support (or empty).
Then the Hermitian metric $\go$ on ${\mathcal{O}}(1)|_{\pi^{-1}(Y \sm \Dl)}$
can be extended as a singular
Hermitian metric $g_{{\mathcal{O}}(1)}$ with semi-positive curvature
of ${\mathcal{O}}(1)$ on ${\mathbb{P}}(\Rq|_{Y \sm S_q})$.
\end{thm}

Theorem \ref{MT}\,(1) is reduced to Theorem \ref{MT}\,(2) for
$f' = f \circ \mu : X' \lra Y$ after a modification $\mu : X' \lra X$.
Then however the induced map $f' : X' \lra Y$ is only locally K\"ahler
in general.
Hence we need to restrict everything on relatively compact subsets
of $Y$ in Theorem \ref{MT}\,(1).

If in particular in Theorem \ref{MT}, $\Rq$ is locally free and
$Y$ is a smooth projective variety, then the vector bundle
$\Rq$ is pseudo-effective in the sense of \cite[\S 6]{DPS}.
This notion \cite[\S 6]{DPS} is a natural generalization of the fact that
on a smooth projective variety, a divisor $D$ is pseudo-effective
(i.e., a limit of effective divisors)
if and only if the associated line bundle ${\mathcal{O}}(D)$ admits a singular
Hermitian metric with semi-positive curvature.
The above curvature property of ${\mathcal{O}}(1)$ leads to the following
algebraic positivity of $\Rq$.

\begin{thm} \label{algebraic}
Let $f : X \lra Y$ be a surjective morphism with connected fibers
between smooth projective varieties, and let $(E, h)$ be
a Nakano semi-positive holomorphic vector bundle on $X$.
Then the torsion free sheaf $\Rq$ is weakly positive over $Y \sm \Dl$
(the smooth locus of $f$), in the sense of Viehweg \cite[2.13]{Vi2}.
\end{thm}

See \cite[\S1]{MT3} for further introduction.

\subsection{Statement in this text} \label{this}

We would like to explain the proofs of the following two theorems
in this text.
Because there is no essential limitations of the number of pages,
we may repeat some arguments and make comments repetitiously.

\begin{setup} \label{basic}
(General set up.) \
Let $f : X \lra Y$ be a holomorphic map of complex manifolds,
which is proper, K\"ahler, surjective with connected fibers,
and $f$ is smooth over the complement $Y \sm \Dl$ of
a closed analytic subset $\Dl \subset Y$.
Let $\w_f$ be a relative K\"ahler form for $f$, and let $(E, h)$
be a Nakano semi-positive holomorphic vector bundle on $X$.
Let $q$ be a non-negative integer.

It is known by Koll\'ar \cite{Ko1} and Takegoshi \cite{Tk} that
$\F$ is torsion free,
and moreover it is locally free where $f$ is smooth (\cite[4.9]{MT2}).
In particular we can let $S_q \subset \Dl$ be
the minimum closed analytic subset of $\codim_Y S_q \ge 2$ such that
$\F$ is locally free on $Y \sm S_q$.
Once we take a relative K\"ahler form $\w_f$ for $f$, we then have
the Hodge metric $g$ on the vector bundle $\F|_{Y \sm \Dl}$
with respect to $\w_f$ and $h$ (\cite[\S 5.1]{MT2} or Remark \ref{defHodge}).
\qed
\end{setup}

\begin{thm} \label{curve}
In Set up \ref{basic}, assume further that $\dim Y = 1$.
Let $L$ be a quotient holomorphic line bundle of $\Rq$.
Then $L$ has a singular Hermitian metric with semi-positive curvature,
whose restriction on $Y \sm \Dl$ is the quotient metric of
the Hodge metric $g$ on $\Rq|_{Y \sm \Dl}$.
\end{thm}

\begin{thm} \label{reduced}
In Set up \ref{basic}, assume further that
$f$ has reduced fibers in codimension 1 on $Y$,
i.e., there exists a closed analytic set $Z \subset \Dl$ of
$\codim_Y Z \ge 2$ such that every fiber of $y \in Y \sm Z$ is reduced.
Let $L$ be a holomorphic line bundle on $Y$ with a surjection
$\Rq|_{Y \sm Z} \lra L|_{Y \sm Z}$.
Then $L$ has a singular Hermitian metric with semi-positive curvature,
whose restriction on $Y \sm \Dl$ is the quotient metric of
the Hodge metric $g$ on $\Rq|_{Y \sm \Dl}$.
\end{thm}

The above assumptions:\
$\dim Y = 1$, and/or with reduced fibers,
or even fibers are semi-stable,
are quite usual in algebraic geometry.
In this sense, the assumptions in Theorem \ref{curve} and \ref{reduced}
are not so artificial.

\subsection{Complement}

Here is a comment on the relation between the statements in \S 1.1
and those in \S 1.2.
Although we will not give proofs, we can pursue the method of proof
of Theorem \ref{curve} and \ref{reduced} to show the following two
statements, as we show Theorem \ref{MT} in \cite{MT3}.

\begin{thm} \label{curve+}
In Set up \ref{basic}, assume further that $\dim Y = 1$.
Then the line bundle ${\mathcal{O}}(1)$ for
$\pi : {\mathbb{P}}(\Rq) \lra Y$ has a singular Hermitian metric $g_{{\mathcal{O}}(1)}$
with semi-positive curvature,
and whose restriction on $\pi^{-1}(Y \sm \Dl)$ is the quotient metric
$\go$ of $\pi^*g$,
where $g$ is the Hodge metric with respect to $\w_f$ and $h$.
\end{thm}

\begin{thm} \label{reduced+}
In Set up \ref{basic}, assume further that
$f$ has reduced fibers in codimension 1 on $Y$.
Then the line bundle ${\mathcal{O}}(1)$ for
$\pi : {\mathbb{P}}(\Rq|_{Y \sm S_q}) \lra Y \sm S_q$ has a singular
Hermitian metric $g_{{\mathcal{O}}(1)}$ with semi-positive curvature,
and whose restriction on $\pi^{-1}(Y \sm \Dl)$ is the quotient metric
$\go$ of $\pi^*g$, where $g$ is the Hodge metric
with respect to $\w_f$ and $h$.
\end{thm}

One clear difference between \S 1.1 and \S 1.2 is
geometric conditions on $f : X \lra Y$.
Another is about line bundles to be considered, namely ${\mathcal{O}}(1)$ or $L$.
For example,
Theorem \ref{curve+} (or \ref{MT}) concerns all rank 1 quotient of $\Rq$,
while Theorem \ref{curve} concerns a rank 1 quotient of $\Rq$,
hence Theorem \ref{curve+} is naturally stronger than Theorem \ref{curve}.
In fact Theorem \ref{curve+} implies Theorem \ref{curve} by a standard
argument (\cite[\S 6.2]{MT3}).
The proof of Theorem \ref{curve+} (as well as Theorem \ref{MT})
requires another uniform estimate which does not depend on
rank 1 quotients $L$ of $\Rq$, other than the uniform estimate
given in Lemma \ref{Ft11} of the proof of Theorem \ref{curve}.

\section{Preliminary Arguments}

\subsection{Localization}

As the next lemma shows, to see our theorems,
we can neglect codimension 2 analytic subsets of $Y$.

\begin{lem} \label{codim2}
Let $Y$ be a complex manifold, and $Z$ a closed analytic subset
of $Y$ with $\codim_Y Z \ge 2$.
Let $L$ be a holomorphic line bundle on $Y$ with a singular Hermitian
metric $h$ on $L|_{Y \sm Z}$ with semi-positive curvature.
Then $h$ extends as a singular Hermitian metric on $L$ with
semi-positive curvature.
\end{lem}

\begin{proof}
Let $W$ be a small open subset of $Y$ with a nowhere vanishing
section $e \in H^0(W, L)$.
Then a function $h(e, e)$ on $W \sm Z$ can be written as
$h(e, e) = e^{-\vph}$ with a plurisubharmonic function
$\vph$ on $W \sm Z$.
By Hartogs type extension for plurisubharmonic functions,
$\vph$ can be extended uniquely as a plurisubhamonic function
$\wtil \vph$ on $W$.
Then $e^{-\wtil \vph}$ gives the desired extension of $h$ on $W$.
\end{proof}

In particular, we can neglect the set $S_q$ (resp.\ $Z$)
in Set up \ref{basic} (resp.\ in Theorem \ref{reduced}),
and only consider codimension 1 part of the discriminant locus $\Dl$.
Once we obtain the Hodge metric $g$ of $\F|_{Y \sm \Dl}$
or the quotient metric $g^\circ_L$ of $L|_{Y \sm \Dl}$,
the extension property of $g^\circ_L$ is a local question.
Hence we can further reduce our situation to the following

\begin{setup} \label{local}
(Generic local set up.) \
Let $Y$ be (a complex manifold which is biholomorphic to)
a unit ball in ${\mathbb{C}}^m$ with coordinates $t = (t_1, \ldots, t_m)$,
$X$ a complex manifold of $\dim X = n+m$ with a K\"ahler form $\w$.
Let $f : X \lra Y$ be a proper surjective holomorphic map
with connected fibers.
Let $(E, h)$ be a Nakano semi-positive holomorphic vector bundle on $X$,
and let $q$ be an integer with $0 \le q \le n$.
Let $K_Y \cong {\mathcal{O}}_Y$ be a trivialization by a nowhere vanishing
section $dt = dt_1 \wed \ldots \wed dt_m \in H^0(Y, K_Y)$.
Let $g$ be the Hodge metric on $\Rq|_{Y \sm \Dl}$ with respect to
$\w$ and $h$.
Let us assume the following:\

(1) $f$ is flat, and the discriminant locus $\Dl \subset Y$ is
$\Dl = \{t_m = 0\}$.

(2) $\Rq \cong {\mathcal{O}}_Y^{\oplus r}$, i.e.,
globally free and trivialized of rank $r$.

(3)
Let $f^*\Dl = \sum b_iB_i$ be the prime decomposition.
For every $B_i$, the induced morphism $f : \Reg B_i \lra \Dl$
is surjective and smooth.
Here $\Reg B_i$ is the smooth locus of $B_i$.
If $B_i \ne B_j$, the intersection $B_i \cap B_j$ does not contain
any fiber of $f$.

We may replace $Y$ by slightly smaller balls, or
may assume everything is defined over a slightly larger ball.
\qed
\end{setup}

\begin{rem}
(1) For this moment, in Set up \ref{local},
we do not assume that $\dim Y = 1$, nor that $f$ has reduced fibers.

(2) Set up \ref{local}\,(3) is automatically satisfied in case $\dim Y = 1$.

(3) Refer \cite[5.2]{MT2} for the replacement of
a relative K\"ahler form $\w_f$ by a K\"ahler form $\w$.
\qed
\end{rem}

\begin{notation}
(1) For a non-negative integer $d$, we set $c_d = \ai^{d^2}$.

(2)
Let $f : X \lra Y$ be as in Set up \ref{local}.
We set $\Omega_{X/Y}^p = \bigwedge^p \Om_{X/Y}^1$
rather formally, because we will only deal
$\Omega_{X/Y}^p$ on which $f$ is smooth.
For an open subset $U \subset X$ where $f$ is smooth,
and for a differentiable form $\sg \in A^{p,0}(U, E)$,
we say $\sg$ is relatively holomorphic
and write $[\sg] \in H^0(U, \Omega_{X/Y}^p \ot E)$,
if $\sg \wed f^*dt \in H^0(U, \Omega_X^{p+m} \ot E)$.
\qed
\end{notation}

\subsection{Relative hard Lefschetz type theorem}

We discuss in Set up \ref{local}.

One fundamental ingredient, even in the definition of Hodge metrics,
is the following proposition.
In case $q = 0$, this is quite elementary.

\begin{prop} \label{Takegoshi}
\cite[5.2]{Tk}.
There exist $H^0(Y, {\mathcal{O}}_Y)$-module homomorphisms
\begin{equation*}
\begin{aligned}
	* \circ {\mathcal{H}} &: H^0(Y, \F)
		\lra H^0(X, \Om_X^{n+m-q} \ot E), \\
	L^q &: H^0(X, \Om_X^{n+m-q} \ot E)
		\lra H^0(Y, \F)
\end{aligned}
\end{equation*}
such that
(1) $(c_{n+m-q}/q!) L^q \circ (* \circ {\mathcal{H}}) = id$, and
(2)
for every $u \in H^0(Y, \F)$, there exists a relative
holomorphic form
$[\sg_u] \in H^0(X \sm f^{-1}(\Dl), \Om_{X/Y}^{n-q} \ot E)$
such that
$$
	(* \circ {\mathcal{H}} (u))|_{X \sm f^{-1}(\Dl)} = \sg_u \wed f^*dt.
$$
\end{prop}

\begin{proof}
We take a smooth strictly plurisubhamonic exhaustion function $\psi$
on $Y$, for example $\|t\|^2$.
Recalling $\F = K_{Y}^{\ot (-1)} \ot R^qf_*(K_X \ot E)$,
the trivialization $K_{Y} \cong {\mathcal{O}}_Y$ by $dt$ gives an isomorphism
$\F \cong R^qf_*(K_X \ot E)$.
Since $Y$ is Stein, we have also a natural isomorphism
$H^0(Y, R^qf_*(K_X \ot E)) \cong H^q(X, K_X \ot E)$.
We denote by $\ga^q$ the composed isomorphism
$$
 \ga^q : H^0(Y, \F)

	\isom H^q(X, K_X \ot E).
$$

With respect to the K\"ahler form $\w$ on $X$,
we denote by $*$ the Hodge $*$-operator, and by
$$
	L^q : H^0(X, \Om_X^{n+m-q} \ot E)
			\lra H^q(X, K_X \ot E)
$$
the Lefschetz homomorphism induced from ${\w}^q \wed \bullet$.
Also with respect to $\w$ and $h$, we set
${\mathcal{H}}^{n+m,q}(X, E, f^*\psi)
= \{u \in A^{n+m, q}(X, E) ;\ \rdb u = \vth_h u = 0, \
	e(\rdb (f^*\psi))^* u = 0 \}$.
(We do not explain what this space of harmonic forms is,
because the definition is not important in this text.) \
By \cite[5.2.i]{Tk}, ${\mathcal{H}}^{n+m,q}(X, E, f^*\psi)$ represents
$H^q(X, K_X \ot E)$ as an $H^0(Y, {\mathcal{O}}_Y)$-module,
and hence there exists a natural isomorphism
$$
\iota : {\mathcal{H}}^{n+m,q}(X, E, f^*\psi)
		\isom H^q(X, K_X \ot E)
$$
given by taking the Dolbeault cohomology class.
We have an isomorphism
$$
	{\mathcal{H}}  = \iota^{-1} \circ \ga^q :
     H^0(Y, \F) \isom {\mathcal{H}}^{n+m,q}(X, E, f^*\psi).
$$
Also by \cite[5.2.i]{Tk},
the Hodge $*$-operator gives an injective homomorphism
$$
   * : {\mathcal{H}}^{n+m,q}(X, E, f^*\psi)
	\lra H^0(X, \Om_X^{n+m-q} \ot E),
$$
and induces a splitting
$* \circ \iota^{-1} : H^q(X, K_X \ot E)
	\lra H^0(X, \Om_X^{n+m-q} \ot E)$
for the Lefschetz homomorphism $L^q$ such that
$(c_{n+m-q}/q!) L^q \circ * \circ \iota^{-1} = id$.
(The homomorphism $\delta^q$ in \cite[5.2.i]{Tk} with respect to
$\w$ and $h$ is $* \circ \iota^{-1}$ times a universal constant.) \
In particular
$$
	(c_{n+m-q}/q!) ((\ga^q)^{-1} \circ L^q) \circ (* \circ {\mathcal{H}} )
	= id.
$$
All homomorphisms $\ga^q, *, L^q, \iota, {\mathcal{H}}$ are as
$H^0(Y, {\mathcal{O}}_Y)$-modules.

Let $u \in H^0(Y, \F)$.
Then we have $* \circ {\mathcal{H}} (u) \in H^0(X, \Om_X^{n+m-q} \ot E)$,
and then by \cite[5.2.ii]{Tk}
$$
	(* \circ {\mathcal{H}} (u))|_{X \sm f^{-1}(\Dl)}
	= \sg_u \wed f^*dt
$$
for some $[\sg_u] \in H^0(X \sm f^{-1}(\Dl), \Om_{X/Y}^{n-q} \ot E)$.
It is not difficult to see
$[\sg_u] \in H^0(X \sm f^{-1}(\Dl), \Om_{X/Y}^{n-q} \ot E)$
does not depend on the particular choice of a global frame
$dt$ of $K_{Y}$.
\end{proof}

\begin{rem} \label{defHodge}
We recall the definition of the Hodge metric $g$ of $\F|_{Y \sm \Dl}$
with respect to $\w$ and $h$ \cite[5.1]{MT2}.
We only mention it for a global section $u \in H^0(Y, \F)$.
It is given by
$$
 g(u,u)(t)
	= \int_{X_{t}} (c_{n-q}/q!)
	  ({\w}^q \wed \sg_u \wed h \ol{\sg_u})|_{X_{t}}
$$
at $t \in Y \sm \Dl$.
In the relation
$$
	(* \circ {\mathcal{H}} (u))|_{X \sm f^{-1}(\Dl)} = \sg_u \wed f^*dt,
$$
the left hand side is holomorphically extendable across $f^{-1}(\Dl)$,
and is non-vanishing if $u$ is, in an appropriate sense.
In the right hand side, $f^*dt$ may only have zero along $f^{-1}(\Dl)$,
that is ``Jacobian'' of $f$, and hence $\sg_u$ may only have ``pole''
along $f^{-1}(\Dl)$.
This is the main reason why $g(u,u)(t)$ has a positive
lower bound on $Y \sm \Dl$, and which is fundamental
for the extension of positivity (see (5) of the proof of
Proposition \ref{weak} below).
The importance of the role of the Jacobian of $f$ is already
observed by Fujita \cite{Ft}.
\qed
\end{rem}

\subsection{Non-uniform estimate}

Here we state a weak extension property.
This is a basic reason for all extension of positivity
of direct image sheaves of relative canonical bundles,
for example in \cite{Ft}, \cite{Ka1}, \cite{Vi1}, and so on.
However this is not enough to conclude the results in \S 1.

\begin{prop} \label{weak}
In Set up \ref{basic}, let $W \subset Y$ be an open subset, and let
$u \in H^0(W \sm S_q, \Rq)$ which is nowhere vanishing on $W \sm S_q$.
Then the smooth plurisubharmonic function $-\log g(u,u)$ on $W \sm \Dl$
can be extended as a plurisubharmonic function on $W$.
\end{prop}

\begin{proof}
We may assume $W = Y$.
Moreover it is enough to consider in Set up \ref{local} as before.
In particular $S_q = \emptyset$ and $\Dl = \{t_m=0\}$.
We shall discuss the extension property at the origin $t = 0 \in Y$,
and hence we replace $Y$ by a small ball centered at $t=0$.

(1)
By Proposition \ref{Takegoshi},
we have $* \circ {\mathcal{H}} (u) \in H^0(X, \Om_X^{n+m-q} \ot E)$.
This $* \circ {\mathcal{H}} (u)$ does not vanish identically along
$\Dl = \{t_m = 0 \} \subset Y$ as an element of
$H^0(Y, {\mathcal{O}}_Y)$-module $H^0(X, \Om_X^{n+m-q} \ot E)$.
This is saying that there exists at least one component $B_j$
in $f^*\Dl = \sum b_iB_i$ such that $* \circ {\mathcal{H}} (u)$ does not vanish
of order greater than or equal to $b_j$ along $B_j$.
We take one such $B_j$ and denote by
$$
	B = B_j \text{ and } b = b_j.
$$

(2)
We take a general point $x_0 \in B \cap f^{-1}(0)$ so that $x_0$ is a smooth
point on $(f^*\Dl)_{red}$, and take local coordinates
$(U; z = (z_1, \ldots, z_{n+m}))$ centered at $x_0 \in X$.
We may assume $f(U) = Y$ and
$t = f(z) = (z_{n+1}, \ldots, z_{n+m-1}, z_{n+m}^b)$ on $U$.

Over $U$, the bundle $E$ is also trivialized, i.e.,
$E|_U \cong U \times {\mathbb{C}}^{r(E)}$, where $r(E)$ is the rank of $E$.
Using the local trivializations on $U$, we have a constant
$a > 0$ such that
(i) $\w \ge a \w_{eu}$ on $U$,
where $\w_{eu} = \ai/2\sum_{i=1}^{n+m} dz_i \wed d\ol{z_i}$
is the standard complex euclidean K\"ahler form,
and
(ii) $h \ge a \text{Id}$ on $U$ as Hemitian matrixes.
Here we regard $h|_U(x)$ as a positive definite Hermitian matrix
at each $x \in U$ in terms of $E|_U \cong U \times {\mathbb{C}}^{r(E)}$,
and here $\text{Id}$ is the $r(E) \times r(E)$ identity matrix.

(3)
By Proposition \ref{Takegoshi}, we can write as
$(* \circ {\mathcal{H}} (u))|_{X \sm f^{-1}(\Dl)} = \sg_u \wed f^*dt$ for some
$\sg_u \in A^{n-q,0}(X \sm f^{-1}(\Dl), E)$.
We write
$\sg_u = \sum_{I \in I_{n-q}} \sg_{I} dz_I + R$ on $U \sm B$.
Here $I_{n-q}$ is the set of all multi-indexes
$1 \le i_1 < \ldots < i_{n-q} \le n$ of length $n-q$
(not including $n+1, \ldots, n+m$),
$\sg_{I} = {}^t(\sg_{I,1}, \ldots, \sg_{I,r(E)})$ is a vector valued
holomorphic function with $\sg_{I,i} \in H^0(U \sm B, {\mathcal{O}}_X)$,
and here $R = \sum_{k=1}^m R_{k} \wed dz_{n+k} \in A^{n-q,0}(U \sm B, E)$.
Now
$$
	\sg_u \wed f^*dt
	= b z_{n+m}^{b-1} \bigg(\sum_{I \in I_{n-q}} \sg_{I} dz_I \bigg)
		\wed dz_{n+1} \wed \ldots \wed dz_{n+m}
$$
on $U \sm B$.
Since $\sg_u \wed f^*dt = (* \circ {\mathcal{H}} (u))|_{X \sm f^{-1}(\Dl)}$ and
$* \circ {\mathcal{H}} (u) \in H^0(X, \Om_X^{n+m-q} \ot E)$,
all $z_{n+m}^{b-1} \sg_{I}$ can be extended holomorphically on $U$.
By the non-vanishing property of $* \circ {\mathcal{H}} (u)$ along $bB$,
we have at least one $\sg_{J_0,i_0} \in H^0(U \sm B, {\mathcal{O}}_X)$
whose divisor is
$$
	\divisor (\sg_{J_0,i_0}) = - pB|_U+D
$$
with some integer $0 \le p \le b-1$, and
an effective divisor $D$ on $U$ not containing $B|_U$.
We take such
$$
	J_0 \in I_{n-q} \text{ and } i_0 \in \{1, \ldots, r(E)\}.
$$
(Now $\divisor (\sg_{J_0,i_0}) = - pB|_U+D$ is fixed.) \
We set
$$
	Z_u = \{ y \in \Dl ; \ D \text{ contains } B|_U \cap f^{-1}(y) \}.
$$
We can see that $Z_u$ is not Zariski dense in $\Dl$, because
otherwise $D$ contains $B|_U$, and also that $Z_u$ is Zariski closed
of $\codim_Y Z_u \ge 2$ (particularly using $f$ is flat).

(4)
We take any point $y_1 \in \Dl \sm Z_u$, and
a point $x_1 \in B|_U \cap f^{-1}(y_1)$ such that $x_1 \not\in D$.
Let $0 < \ep \ll 1$ be a sufficiently small number so that,
on the $\ep$-polydisc neighbourhood
$U(x_1, \ep) = \{z = (z_1, \ldots, z_{n+m}) \in U ; \
|z_i - z_i(x_1)| < \ep \text{ for any } 1 \le i \le n+m \}$,
we have
$$
	A := \inf\{ |\sg_{J_0,i_0}(z)| ; \ z \in U(x_1, \ep) \sm B \} > 0.
$$
We should note that $\sg_{J_0,i_0}$ may have a pole along $B$,
but no zeros on $U(x_1, \ep)$.
We set $Y' := f(U(x_1, \ep))$ which is an open neighbourhood of
$y_1 \in Y$, since $f$ is flat (in particular it is an open mapping).
Then for any $t \in Y' \sm \Dl$, we have
\begin{equation*}
\begin{aligned}
\int_{X_{t}}(c_{n-q}/q!)
(\w^q \wed \sg_u \wed h\ol{\sg_u})|_{X_{t}}
&
\ge
a \int_{X_{t} \cap U}(c_{n-q}/q!)
	(\w^q \wed \sg_u \wed \ol{\sg_u})|_{X_{t} \cap U} \\
& =
a^{q+1} \int_{z \in X_{t} \cap U}
	\sum_{I \in I_{n-q}} \sum_{i=1}^r |\sg_{I,i}(z)|^2 dV_n \\
& \ge
a^{q+1} \int_{z \in X_{t} \cap U(x_1, \ep)} A^2 \ dV_n \\
& = a^{q+1} A^2 (\pi \ep^2)^n.
\end{aligned}
\end{equation*}
Here $dV_n = (\ai/2)^n \bigwedge_{i=1}^n dz_i \wed d\ol{z_i}$
is the standard euclidean volume form in ${\mathbb{C}}^n$.
Namely we have $g(u, u)(t) \ge a^{q+1} A^2 (\pi \ep^2)^n$
for any $t \in Y' \sm \Dl$.

(5)
We proved that $-\log g(u,u)$ is bounded from above
around every point of $\Dl \sm Z_u$.
This means that a plurisubharmonic function $-\log g(u,u)$ on $Y \sm \Dl$
can be extended as a plurisubharmonic function on $Y \sm Z_u$
by Riemann type extension, and hence as
a plurisubharmonic function on $Y$ by Hartogs type extension.
\end{proof}

\begin{rem}
Here are some remarks when we try to generalize the proof above
to obtain Theorem \ref{curve} and \ref{reduced}.
The point is the set $Z_u$ above depends on $u$.
This is the main difficulty when we consider an extension property of
quotient metrics.
In that case, we need to obtain a uniform estimate of $g(u_s, u_s)$
for a family $\{u_s\}$.
If $s$ moves, then $Z_{u_s}$ also may move and cover a larger subset
of $\Dl$, which may not be negligible for the extension of
plurisubharmonic functions.

The intersection $B|_U \cap D$ is a set of indeterminacies.
If (a part of) a fiber $f^{-1}(y)$ is contained in $B|_U \cap D$,
the analysis of the behavior of $g(u, u)$ around such $y$
is quite hard and in fact indeterminate.
This is why we do not want to touch $Z_u$.
In some geometric setting as below, we can avoid such phenomena.
We can delete one of two in the right hand side of
$\divisor (\sg_{I,i}) = -pB|_U + D$.

(i)
In case $\dim Y = 1$, we can take $D = 0$.
This is because, if a prime divisor $\Gamma$ on $U$ contains
$B|_U \cap f^{-1}(y)$, then $\Gamma = B|_U$.
In case when $\dim Y = 1$, $q = 0$ and $E = {\mathcal{O}}_X$,
a uniform estimate is cleared by Fujita \cite[1.11]{Ft}
(as we will see below).
This will lead Theorem \ref{curve}.

(ii)
In case the fibers of $f$ are reduced, we can take $p = 0$
(cf.\ $0 \le p \le b-1$ in (3) of the proof above).
This will lead Theorem \ref{reduced}.

To deal with a general case in \cite{MT3}, we use a semi-stable
reduction for $f$.
A computation of Hodge metrics is a kind of an estimation of integrals,
which usually can be done only after a good choice of local coordinates.
A semi-stable reduction can be seen as a resolution of singularities
of a map $f : X \lra Y$.
Then the crucial point is to compair two Hodge metrics:\
the original one and the one after taking a semi-stable reduction.
\qed
\end{rem}

\section{Proof of Theorems}

\subsection{Quotient metric} \label{quot}

We discuss in Set up \ref{local}.

We denote by $F = \Rq$ which is locally free on $Y$,
and by $r$ the rank of $F$.
We have a smooth Hermitian metric $g$ defined on $Y \sm \Dl$ (not on $Y$).
Let $F \lra L$ be a quotient line bundle with the kernel $M$:\
$0 \lra M \lra F \lra L \lra 0$ (exact).
We first describe the quotient metric on $L|_{Y \sm \Dl}$.
We take a frame $e_1, \ldots, e_r \in H^0(Y, F)$ over $Y$
such that $e_1, \ldots, e_{r-1}$ generate $M$.
Then the image
$$
	\what e_r \in H^0(Y, L)
$$
of $e_r$ under $F \lra L$ generates $L$.
We represent the Hodge metric $g$ on $Y \sm \Dl$ in terms of this frame
as $g_{i\ol j} = g(e_i, e_j) \in A^0(Y \sm \Dl, {\mathbb{C}})$.
At each point $t \in Y \sm \Dl$, $(g_{i\ol j}(t))_{1 \le i, j \le r}$
is a positive definite Hermitian matrix, in particular,
$(g_{i\ol j}(t))_{1 \le i, j \le r-1}$ is also positive definite.
We let $(g^{\ol i j}(t))_{1 \le i, j \le r-1}$ be the inverse matrix.
Then the pointwise orthogonal projection of $e_r$ to
$(M|_{Y \sm \Dl})^\perp$ with respect to $g$ is given by
$$
  P(e_r)
  = e_r - \sum_{i=1}^{r-1}\sum_{j=1}^{r-1}e_ig^{\ol i j}g_{j \ol r}
  \ \in A^0(Y \sm \Dl, F).
$$
We have in fact $P(e_r) - e_r \in A^0(Y \sm \Dl, M)$ and
$g(P(e_r), s) = 0$ for any $s \in A^0(Y \sm \Dl, M)$.
Then the quotient metric on $L|_{Y \sm \Dl}$
is defined by
$$
	g^\circ_L(\what e_r, \what e_r) = g(P(e_r), P(e_r))
$$
on $Y \sm \Dl$.

It is well-known after Griffiths that the curvature
does not decrease by a quotient.
In our setting, the Nakano semi-positivity of $(F|_{Y \sm \Dl}, g)$
\cite[1.1]{MT2}, or even weaker the Griffiths semi-positivity
implies that $(L|_{Y \sm \Dl}, g^\circ_L)$ is semi-positive.
In particular if we write $g^\circ_L(\what{e_r}, \what{e_r}) = e^{-\vph}$
with $\vph \in A^0(Y \sm \Dl, {\mathbb{R}})$, this $\vph$ is plurisubharmonic
on $Y \sm \Dl$.
If we can show $\vph$ is extended as a plurisubharmonic function on $Y$,
then $g^\circ_L$ extends as a singular Hermitian metric on $L$ over $Y$
with semi-positive curvature.
By virtue of Riemann type extension for plurisubharmonic functions,
it is enough to show that $\vph$ is bounded from above
(i.e., $g^\circ_L(\what e_r, \what e_r)$ is bounded from below
by a positive constant) around every point $y \in \Dl$.
In the next two subsections, we shall prove the following

\begin{lem} \label{Ft13+}
In Set up \ref{local} and the notations above, assume further that
$\dim Y = 1$, or that $f$ has reduced fibers.
Let $y \in \Dl$.
Then there exists a neighbourhood $Y'$ of $y \in Y$ and a positive number
$N$ such that $g^\circ_L(\what e_r, \what e_r)(t) \ge N$ for any
$t \in Y' \sm \Dl$.
\end{lem}

\begin{cor}
Theorem \ref{curve} and Theorem \ref{reduced} hold true.
\end{cor}

We introduce the following notations for the following arguments.
For $s = (s_1, \ldots, s_r) \in {\mathbb{C}}^r$, we let
$u_s = \sum_{i=1}^r s_ie_i \in H^0(Y, F)$.
We note that $u_s$ is nowhere vanishing on $Y$ as soon as $s \ne 0$.
We also note that, with respect to the standard topology of ${\mathbb{C}}^r$ and
the topology of uniform convergence on compact sets for
$H^0(X, \Om_X^{n+m-q} \ot E)$,
the map ${\mathbb{C}}^r \lra H^0(X, \Om_X^{n+m-q} \ot E)$ given by
$s \mapsto u_s \mapsto * \circ {\mathcal{H}} (u_s)
= \sum_{i=1}^r s_i (* \circ {\mathcal{H}} (e_i))$,
is continuous.
Let $S^{2r-1} = \{ s \in {\mathbb{C}}^r ; \ |s| = (\sum |s_i|^2)^{1/2} = 1 \}$.

\subsection{Over curves}

We shall prove Theorem \ref{curve} by showing Lemma \ref{Ft13+} in this case.
It is enough to consider in Set up \ref{local} with $\dim Y =1$.
In particular $Y = \{ t \in {\mathbb{C}} ; \ |t| < 1 \}$ a unit disc, and
$\Dl = 0 \in Y$ the origin.
We will use both $\Dl \subset Y$ and $t=0 \in Y$ to compair
our argument here with a general case.
Let $F = \F \lra L$ be a quotient line bundle, and use the same notation
in \S \ref{quot}, in particular we have a frame
$e_1, \ldots, e_r \in H^0(Y, F)$, $\what e_r \in H^0(Y, L)$
generates $L$ and so on.
We use $u_s = \sum_{i=1}^r s_ie_i \in H^0(Y, F)$ for
$s = (s_1, \ldots, s_r) \in {\mathbb{C}}^r$.
The key is to obtain the following uniform bound.

\begin{lem}   \label{Ft11}
(cf.\ \cite[1.11]{Ft}.) \
In Set up \ref{local} with $\dim Y =1$ and the notation above,
let $s_0 \in S^{2r-1}$.
Then there exist a neighbourhood $S(s_0)$ of $s_0$ in $S^{2r-1}$,
a neighbourhood $Y'$ of $0 \in Y$ and a positive number $N$ such that
$g(u_s, u_s)(t) \ge N$ for any $s \in S(s_0)$ and any $t \in Y' \sm \Dl$.
\end{lem}

\begin{proof}
We denote by $f^*\Dl = \sum b_iB_i$.

(1)
By Proposition \ref{Takegoshi}, we have
$* \circ {\mathcal{H}} (u_{s_0}) \in H^0(X, \Om_X^{n+1-q} \ot E)$.
This $* \circ {\mathcal{H}} (u_{s_0})$ does not vanish at $t = 0$ as an element
of $H^0(Y, {\mathcal{O}}_Y)$-module $H^0(X, \Om_X^{n+1-q} \ot E)$.
Then there exists a component
$B_j$ in $f^*\Dl = \sum b_iB_i$ such that
$(* \circ {\mathcal{H}})(u_{s_0})$ does not vanish
of order greater than or equal to $b_j$ along $B_j$.
We take one such $B_j$ and denote by $B = B_j \text{ and } b = b_j$.

(2)
We take a general point $x_0 \in B$ so that
$x_0$ is a smooth point on $(f^*\Dl)_{red} = f^{-1}(0)$,
and take local coordinates $(U; z = (z_1, \ldots, z_{n+1}))$
centered at $x_0 \in X$ such that $t = f(z) = z_{n+1}^b$ on $U$.
Over $U$, the bundle $E$ is also trivialized.
Using the local trivializations on $U$, we have a constant
$a > 0$ such that
(i) $\w \ge a \w_{eu}$ on $U$,
where $\w_{eu} = \ai/2\sum_{i=1}^{n+1} dz_i \wed d\ol{z_i}$,
and
(ii) $h \ge a \text{Id}$ on $U$ as Hemitian matrixes,
as in the proof of Proposition \ref{weak}.

(3)
Let $s \in S^{2r-1}$.
By Proposition \ref{Takegoshi}, we can write as
$(* \circ {\mathcal{H}} (u_s))|_{X \sm f^{-1}(\Dl)} = \sg_s \wed f^*dt$ for some
$\sg_s \in A^{n-q,0}(X \sm f^{-1}(\Dl), E)$.
We write
$\sg_s = \sum_{I \in I_{n-q}} \sg_{sI} dz_I + R_s \wed dz_{n+1}$
on $U \sm B$.
Here $I_{n-q}$ is the set of all multi-indexes
$1 \le i_1 < \ldots < i_{n-q} \le n$ of length $n-q$,
$\sg_{sI} = {}^t(\sg_{sI,1}, \ldots, \sg_{sI,r(E)})$ with
$\sg_{sI,i} \in H^0(U \sm B, {\mathcal{O}}_X)$,
and here $R_s \wed dz_{n+1} \in A^{n-q,0}(U \sm B, E)$.
Now
$$
	\sg_s \wed f^*dt
	= b z_{n+1}^{b-1} \bigg(\sum_{I \in I_{n-q}} \sg_{sI} dz_I \bigg)
		 \wed dz_{n+1}
$$
on $U \sm B$.
Since $\sg_s \wed f^*dt = (* \circ {\mathcal{H}}(u_s))|_{X \sm f^{-1}(\Dl)}$ and
$* \circ {\mathcal{H}}(u_s) \in H^0(X, \Om_X^{n+1-q} \ot E)$,
all $z_{n+1}^{b-1} \sg_{sI}$ can be extended holomorphically on $U$.

At the point $s_0 \in S^{2r-1}$,
by the non-vanishing property of $* \circ {\mathcal{H}}(u_{s_0})$ along $bB$,
we have at least one $\sg_{s_0J_0,i_0} \in H^0(U \sm B, {\mathcal{O}}_X)$
whose divisor is $\divisor (\sg_{s_0J_0,i_0}) = - p_0B|_U$
with some integer $0 \le p_0 \le b-1$ (being $x_0 \in B|_U$
general, and $U$ sufficiently small).
Here we used $\dim Y = 1$.
We take such $J_0 \in I_{n-q}$ and $i_0 \in \{1, \ldots, r(E)\}$.
By the continuity of $s \mapsto u_s \mapsto * \circ {\mathcal{H}}(u_s)$,
we can take the same $J_0$ and $i_0$ for any $s \in S^{2r-1}$ near $s_0$,
so that $\divisor (\sg_{sJ_0,i_0}) = - p(s)B|_U$ with
the order $p(s)$ satisfies $0 \le p(s) \le p_0 = p(s_0)$
for any $s \in S^{2r-1}$ near $s_0$.

(4)
By the continuity of $s \mapsto u_s \mapsto * \circ {\mathcal{H}}(u_s)$,
we can take an $\ep$-polydisc neighbourhood
$U(x_0, \ep) = \{z = (z_1, \ldots, z_{n+1}) \in U ; \
|z_i - z_i(x_0)| < \ep \text{ for any } 1 \le i \le n+1 \}$ for some
$\ep > 0$, and a neighbourhood $S(s_0)$ of $s_0$ in $S^{2r-1}$ such that
$A := \inf\{ |\sg_{sJ_0,i_0}(z)| ; \
	 s \in S(s_0), \ z \in U(x_0, \ep) \sm B \} > 0$.
We should note that $\sg_{s J_0, i_0}$ may have a pole along $B$,
but no zeros on $U(x_0, \ep)$.
We set $Y' := f(U(x_0, \ep))$ which is an open neighbourhood of
$0 \in Y$, since $f$ is flat.
Then for any $s \in S(s_0)$ and any $t \in Y' \sm \Dl$, we have
$g(u_s, u_s)(t) \ge a^{q+1} A^2 (\pi \ep^2)^n$ as in Proposition \ref{weak}.
\end{proof}

\begin{lem}  \label{Ft12}
(cf.\ \cite[1.12]{Ft}.) \
There exist a neighbourhood $Y'$ of $0 \in Y$ and a positive number
$N$ such that $g(u_s, u_s)(t) \ge N$ for any $s \in S^{2r-1}$
and any $t \in Y' \sm \Dl$.
\end{lem}

\begin{proof}
Since $S^{2r-1}$ is compact, this is clear from Lemma \ref{Ft11}.
\end{proof}

\begin{lem} \label{Ft13}
(cf.\ \cite[1.13]{Ft}.) \
There exists a neighbourhood $Y'$ of $0 \in Y$ and a positive number
$N$ such that $g^\circ_L(\what e_r, \what e_r)(t) \ge N$ for any $t \in Y' \sm \Dl$.
\end{lem}

\begin{proof}
We take a neighbourhood $Y'$ of $0 \in Y$ and a positive number $N$
in Lemma \ref{Ft12}.
We may assume $Y'$ is relatively compact in $Y$.
We put $s_i = - \sum_{j=1}^{r-1} g^{\ol i j}g_{j \ol r}
\in A^0(Y \sm \Dl, {\mathbb{C}})$ for $1 \le i \le r-1$, and $s_r = 1$.
Then $P(e_r) = \sum_{i=1}^r s_i e_i$ on $Y \sm \Dl$.
For every $t \in Y' \sm \Dl$, we have
$s = (s_1, s_2, \ldots, s_r) \in {\mathbb{C}}^r \sm \{0\}$, and
$s(t)/|s(t)| \in S^{2r-1}$.
Then for any $t \in Y' \sm \Dl$, we have
$g^\circ_L(\what e_r, \what e_r)(t) = g(u_{s(t)}, u_{s(t)})(t)
= |s(t)|^2 g(u_{s(t)/|s(t)|}, u_{s(t)/|s(t)|})(t) \ge N$,
since $s/|s| \in S^{2r-1}$.
\end{proof}

\subsection{Fiber reduced}

We shall prove Theorem \ref{reduced} by the same strategy
in the previous subsection.
By Lemma \ref{codim2} we may assume the set $Z$
in Theorem \ref{reduced} is empty.
It is enough to consider in Set up \ref{local} with $f^*\Dl = \sum B_i$.
Let $F = \F \lra L$ be a quotient line bundle, and use the same notation
in \S \ref{quot}, in particular we have a frame
$e_1, \ldots, e_r \in H^0(Y, F)$, $\what e_r \in H^0(Y, L)$
generates $L$ and so on.
We use $u_s = \sum_{i=1}^r s_ie_i \in H^0(Y, F)$ for
$s = (s_1, \ldots, s_r) \in {\mathbb{C}}^r$.
As we observed in the previous subsection, it is enough to show
the following

\begin{lem}
(cf.\ \cite[1.11]{Ft}.) \
In Set up \ref{local} and the notation above, let $s_0 \in S^{2r-1}$.
Then there exist a neighbourhood $S(s_0)$ of $s_0$ in $S^{2r-1}$,
a neighbourhood $Y'$ of $0 \in Y$ and a positive number $N$ such that
$g(u_s, u_s)(t) \ge N$ for any $s \in S(s_0)$ and any $t \in Y' \sm \Dl$.
\end{lem}

\begin{proof}
(1)
By Proposition \ref{Takegoshi}, we have
$* \circ {\mathcal{H}}(u_{s_0}) \in H^0(X, \Om_X^{n+m-q} \ot E)$.
This $* \circ {\mathcal{H}}(u_{s_0})$ does not vanish at $t = 0$ as an element
of $H^0(Y, {\mathcal{O}}_Y)$-module $H^0(X, \Om_X^{n+m-q} \ot E)$.
There exists a component $B_j$ in $f^*\Dl = \sum B_i$
such that $* \circ {\mathcal{H}}(u_{s_0})$ does not vanish identically along
$B_j \cap f^{-1}(0)$.
Here we used our assumption in Theorem \ref{reduced} that
$f$ has reduced fibers.
In fact, if $* \circ {\mathcal{H}}(u_{s_0})$ does vanish identically along
all $B_i \cap f^{-1}(0)$ in $f^*\Dl = \sum B_i$, then
$* \circ {\mathcal{H}}(u_{s_0})$ vanishes at $t = 0$ as an element
of $H^0(Y, {\mathcal{O}}_Y)$-module $H^0(X, \Om_X^{n+m-q} \ot E)$,
and leads a contradiction.
We take one such $B_j$ and denote by $B = B_j$ (with $b = b_j = 1$).

(2)
We take a point $x_0 \in B \cap f^{-1}(0)$ such that
$* \circ {\mathcal{H}}(u_{s_0})$ does not vanish at $x_0$,
and that $f^*\Dl$ is smooth at $x_0$.
We then take local coordinates
$(U; z = (z_1, \ldots, z_{n+m}))$ centered at $x_0 \in X$
such that $t = f(z) = (z_{n+1}, \ldots, z_{n+m-1}, z_{n+m})$ on $U$.
Over $U$, the bundle $E$ is also trivialized.
Using the local trivializations on $U$, we have a constant
$a > 0$ such that
(i) $\w \ge a \w_{eu}$ on $U$,
where $\w_{eu} = \ai/2\sum_{i=1}^{n+m} dz_i \wed d\ol{z_i}$,
and
(ii) $h \ge a \text{Id}$ on $U$ as Hemitian matrixes,
as in the proof of Proposition \ref{weak}.

(3)
Let $s \in S^{2r-1}$.
By Proposition \ref{Takegoshi}, we can write as
$(* \circ {\mathcal{H}}(u_s))|_{X \sm f^{-1}(\Dl)} = \sg_s \wed f^*dt$ for some
$\sg_s \in A^{n-q,0}(X \sm f^{-1}(\Dl), E)$.
We write
$\sg_s = \sum_{I \in I_{n-q}} \sg_{sI} dz_I + R_s$ on $U \sm B$.
Here $I_{n-q}$ is the set of all multi-indexes
$1 \le i_1 < \ldots < i_{n-q} \le n$ of length $n-q$,
$\sg_{sI} = {}^t(\sg_{sI,1}, \ldots, \sg_{sI,r(E)})$ with
$\sg_{sI,i} \in H^0(U \sm B, {\mathcal{O}}_X)$,
and here $R_s = \sum_{k=1}^m R_{sk} \wed dz_{n+k} \in A^{n-q,0}(U \sm B, E)$.
Now
$$
	\sg_s \wed f^*dt
	= \bigg(\sum_{I \in I_{n-q}} \sg_{sI} dz_I \bigg)
		 \wed dz_{n+1} \wed \ldots \wed dz_{n+m}
$$
on $U \sm B$.
Since $\sg_s \wed f^*dt = (* \circ {\mathcal{H}}(u_s))|_{X \sm f^{-1}(\Dl)}$ and
$* \circ {\mathcal{H}}(u_s) \in H^0(X, \Om_X^{n+m-q} \ot E)$,
all $\sg_{sI}$ can be extended holomorphically on $U$.

At the point $s_0 \in S^{2r-1}$,
by the non-vanishing property of $* \circ {\mathcal{H}}(u_{s_0})$ at $x_0$,
we have at least one $\sg_{s_0J_0,i_0} \in H^0(U \sm B, {\mathcal{O}}_X)$
whose divisor is $\divisor (\sg_{s_0J_0,i_0}) = D_0$ with some
effective divisor $D_0$ on $U$ not containing $x_0$.
This is because, if all $\sg_{s_0I,i}$ vanish at $x_0$,
we see $* \circ {\mathcal{H}}(u_{s_0}) = \sg_{s_0} \wed f^*dt$ (now on $U$)
vanishes at $x_0$, and we have a contradiction.
We take such $J_0 \in I_{n-q}$ and $i_0 \in \{1, \ldots, r(E)\}$.
By the continuity of $s \mapsto u_s \mapsto * \circ {\mathcal{H}}(u_s)$,
we can take the same $J_0$ and $i_0$ for any $s \in S^{2r-1}$ near $s_0$.
By the same token, the divisor $D(s)$ may depend on $s \in S^{2r-1}$, but
we can keep the condition that $D(s)$ does not contain
$B|_U \cap f^{-1}(0)$ if $s \in S^{2r-1}$ is close to $s_0$.

(4)
Then by the continuity of $s \mapsto u_s \mapsto * \circ {\mathcal{H}}(u_s)$,
we can take an $\ep$-polydisc neighbourhood
$U(x_0, \ep) = \{z = (z_1, \ldots, z_{n+m}) \in U ; \
|z_i - z_i(x_0)| < \ep \text{ for any } 1 \le i \le n+m \}$ for some
$\ep > 0$, and a neighbourhood $S(s_0)$ of $s_0$ in $S^{2r-1}$ such that
$A := \inf\{ |\sg_{sJ_0,i_0}(z)| ; \
	 s \in S(s_0), \ z \in U(x_0, \ep) \sm B \} > 0$.
We set $Y' := f(U(x_0, \ep))$ which is an open neighbourhood of
$0 \in Y$, since $f$ is flat.
Then for any $s \in S(s_0)$ and any $t \in Y' \sm \Dl$, we have
$g(u_s, u_s)(t) \ge a^{q+1} A^2 (\pi \ep^2)^n$ as in
Proposition \ref{weak}.
\end{proof}

\section{Examples}

\newcommand{\mathbb{P}}\newcommand{\cb}{\mathbb{C}}{\mathbb{P}}\newcommand{\cb}{\mathbb{C}}
\newcommand{\mathbb{Z}}\newcommand{\rb}{\mathbb{R}}{\mathbb{Z}}\newcommand{\rb}{\mathbb{R}}
\newcommand{\mathcal{O}}{\mathcal{O}}

Here are some related examples and counter-examples of the positivity of
direct image sheaves, including cases the total space
$X$ can be singular.
These are due to Wi\'sniewski and H\"oring, and taken from \cite{Ho}.

Our general set up is as follows.
We take a vector bundle $E$ of rank $n+2$ over a smooth projective
variety $Y$.
Denote by $p : \mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)\lra Y$ the natural (smooth) projection.
We take a hypersurface $X$ in $\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)$ cut out by a section of
$N:=\mathcal{O}_E(d)\ot p^* \lambda$ for $d>0$ and some line bundle
$\lambda$ on $Y$.
Denote by $f : X \lra Y$ the induced (non necessary smooth) map
of relative dimension $n$.
Because $X$ is a divisor, the sheaf
$\w_{P(E)/Y} \ot N \ot \mathcal{O}_X$ equals $\w_{X/Y}$.
We choose a line bundle $L:=\mathcal{O}_E(k)\ot p^* \mu$ with $k>0$ and with
a line bundle $\mu$ on $Y$, and set $L_X := L|_X$ the resrtiction on $X$.
We then consider the exact sequence
$$
	0 \lra \w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y} \ot L \lra \w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot N\ot L
	  \lra \w_{X/Y}\ot L_X \lra 0.
$$
Note that since $L$ is $p$-ample, we have $R^1 p_* (\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot L)=0$.
We push the sequence forward by $p$ to get the following exact sequence
of sheaves on $Y$:
$$
	0\lra p_*(\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot L)
	\lra p_*(\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot N\ot L)
	\lra f_* (\w_{X/Y}\ot L_X) \lra 0.
$$
Remember that $\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C} (E)/Y}=\mathcal{O}_E(-n-2) \ot p^*\det E$, so that
$p_* (\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot \mathcal{O}_E(k))=0$ for $k< n+2$, and that
$p_* (\w_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E)/Y}\ot \mathcal{O}_E(k))=S^{k-n-2}E\ot\det E$
for $k\geq n+2$.

\begin{exmp} (\cite[2.C]{Ho}.) \
Choose $Y=\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1$, $E = \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(-1)^{\oplus 2}\oplus \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}$,
$N = \mathcal{O}_E(2)$ that is effective and defines $X$,
and $L = \mathcal{O}_E(1)\ot p^* \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(1)$ that is semi-positive.
The push-forward sequence reads
$$
	0\lra 0\lra \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(-1)\lra f_* (\w_{X/Y}\ot L_X)
	\lra 0.
$$
Hence $f_* (\w_{X/Y}\ot L_X)$ is negative.
The point is that here, $X$ is not reduced.
\qed
\end{exmp}

\begin{exmp} (\cite[2.D]{Ho}.) \
Choose $Y=\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1$ and $E = \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(-1)\oplus \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}^{\oplus 3}$.
Take $N = \mathcal{O}_E(4)$ whose generic section defines $X$.
This scheme is a $3$-fold smooth outside the $1$-dimensional base locus
$\mathbb{P}}\newcommand{\cb}{\mathbb{C} (\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(-1))\subset\mathbb{P}}\newcommand{\cb}{\mathbb{C} (E)$, Gorenstein as a divisor,
and normal since smooth in codimension 1.
Choose $L=\mathcal{O}_E(k)\ot p^* \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(k)$ that is semi-positive.
The push-forward sequence shows that for $1\leq k< 4$,
$S^k E\ot \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(k-1) = f_* (\w_{X/Y}\ot L_X)$
is not nef.
For $k \geq 4$, the push-forward sequence reads
$$
	0\lra S^{k-4}E\ot \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(k-1)
	\overset{\sg}{\lra}
	S^k E\ot \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(k-1)\lra f_* (\w_{X/Y}\ot L_X)
	\lra 0.
$$
Here the map $\sigma$ is given by the contraction with the section
$s\in H^0(\mathbb{P}}\newcommand{\cb}{\mathbb{C}(E), N)$ $=$ $H^0(Y,S^4 E)$ $=$
$H^0(Y,S^4\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}^{\oplus 3})$,
whereas the quotient
$S^k E/\text{Im} (S^4\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}^{\oplus 3}\ot S^{k-4}E)$
contains the factor $\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^1}(-1)^{\oplus k}$.
Hence $f_* (\w_{X/Y}\ot L_X)$ is not weakly positive.
The point here is that the locus of non-rational singularities of
$X$ projects onto $Y$ by $f : X \lra Y$.
\qed
\end{exmp}

\begin{exmp} (\cite[2.A]{Ho}.) \
Choose $Y$ to be $\pi : Y = \mathbb{P}}\newcommand{\cb}{\mathbb{C} (F) \lra {\mathbb{P}}^3$, where
$F:=\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^3}(2)^{\oplus 2}\oplus \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^3}$ is semi-ample but not ample.
Choose $E$ to be
$\mathcal{O}_F(1)^{\oplus 2} \oplus (\mathcal{O}_F(1) \ot \pi^* \mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^3}(1))$.
Wi\'sniewski showed that the linear system
$|N|:=|\mathcal{O}_E(2) \ot p^*\pi^*\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^3}(-2)|$
has a smooth member, that we denote by $X$.
Remark that $L:=\mathcal{O}_E(1)$ is semi-positive, but
the push-forward sequence shows that
$$
	\mathcal{O}_F(3) \ot \pi^*\mathcal{O}_{\mathbb{P}}\newcommand{\cb}{\mathbb{C}^3}(-1) = f_* (\w_{X/Y}\ot L_X)
$$
is not nef.
The point here is that the conic bundle $f : X \lra Y$
has some non-reduced fibers,
that make the direct image only weakly positive.
\qed
\end{exmp}

redpajama_set_name: RedPajamaArXiv


text:
Поверхности постоянной средней кривизны — класс поверхностей моделирующий поверхности мыльных плёнок разделяющие области с фиксированной разницей давлений. В частном случае, если давление с обеих сторон равно, модель определяет минимальные поверхности.

Определяются как гладкие поверхности с постоянной средней кривизной.

История исследования
 В 1841 году Шарль-Эжен Делоне доказал, что единственными поверхностями вращения с постоянной средней кривизной были поверхности, полученные вращением кривых получаемых качением коник. Таковыми являются плоскость, цилиндр, сфера, катеноид, ундулоид и нодоид.

 В 1853 году Дж. Джелле показал, что если  компактная звёздчатая поверхность в  с постоянной средней кривизной, то это стандартная сфера. Впоследствии Александр Данилович Александров доказал, что компактная вложенная поверхность в  с постоянной средней кривизной  должна быть сфера.
 На основании этого Хайнц Хопф в 1956 году предположил, что любая погружённая компактная ориентируемая гиперповерхность постоянной средней кривизны в  должна быть круглой сферой.
 Это предположение было опровергнуто в 1982 году Ву-И Сяном с использованием контрпримера в .
 В 1984 году Генри К. Венте построил так называемый тор Венте — погружение в  тора постоянной средней кривизны.

 Существуют методы построения множества примеров. В частности, методы склеивания позволяют произвольно комбинировать поверхности постоянной средней кривизны.

 Микс показал, что не существует вложенных поверхностей постоянной средней кривизны с одним концом в  . Кореваар, Куснер и Соломон доказали, что концы полной вложенной поверхности асимптотические ундулоиды.

Приложения
Помимо мыльных плёнок, поверхности постоянной средней кривизны появляются как границы раздела газ-жидкость на супергидрофобной поверхности.

В архитектуре поверхности постоянной средней кривизны используются в конструкциях с воздушной опорой, таких, как надувные купола и ограждения, а также в качестве источника плавных органических форм.

Примечания

Дифференциальная геометрия поверхностей

redpajama_set_name: RedPajamaWikipedia


text:
5ive Minutes with Hai-Ting Chinn
by Donna | previous seasons, Uncategorized
In advance of our Season 6IX opening concerts featuring mezzo-soprano Hai-Ting Chinn, 5BMF sat down with this talented performer for a chat about Arnold Schoenberg's iconic chamber work Pierrot Lunaire, the focal point of The Pierrot Project.
5BMF: Hai-Ting, thank you so much for taking a few minutes out of your Pierrot prep to chat with us. It's so great to have you open our 6th Season; you've performed Wolf with us in Queens, Bach in the Bronx, and Brahms in Bryant Park, and now Debussy in Staten Island and Schoenberg in SoHo. You're quite the versatile artist!
Hai-Ting Chinn: You have quite the versatile series! What a great thing for the city. And it's an honor to be part of it.
5BMF: What first drew you to Pierrot Lunaire, and when did you first perform it?
HTC: I first learned about Pierrot Lunaire while I was at Eastman School of Music as an undergrad. Jan DeGaetani, who taught at ESM, had recently passed away, and everyone in the voice department was still talking sadly and admiringly about her. She was a champion of new and challenging music, even for young singers, and her recording of Pierrot was the first one I heard. Some combination of her passion and the somehow-mysterious nature of the piece made it fascinating to my teen-age self, and it hasn't gotten any less fascinating over the years. I first performed it myself five or six years ago, at the Greenwich House Music School, with the same ensemble who is joining me this time.
5BMF: You're teaming up with the Proteus Ensemble for these two programs. Besides being a Pierrot Ensemble, what makes them perfect for this assignment?
HTC: Well, first off, they're just great, and they are masters of the music from this period. Also, we have now a mutual history with this piece – this will be the fourth time we've performed it together – plus, I've known several of them since school, and it's amazing to watch and listen to these incredible musicians grow over the course of their lives and careers. I hope they enjoy working with me half as much as I do with them.
5BMF: What do you think it is about Pierrot that still sounds new and revolutionary, 100 years after its premiere?
HTC: I suppose if you could put your finger on exactly what it is that makes it sound revolutionary, it wouldn't sound so revolutionary…but I'll take a crack at it. At the time, one of the revolutionary things about it was its atonality. That was new to the original audiences, and although we are used to some even more radical atonality built on Schoenberg's beginning, I think his use of it at this period is magical; to me, it feels like listening to a language that you knew when you were a infant, that sounds intensely familiar but that you can't translate or attach any meaning to – it feels like the meaning is there but you can't grasp it. It's like the tonality of dreams. Which is absolutely perfect for this piece and these poems.
"It feels like listening to a language that you knew when you were a infant, that sounds intensely familiar but that you can't translate or attach any meaning to – it feels like the meaning is there but you can't grasp it. It's like the tonality of dreams."
Something even more obvious that makes Pierrot sound mysterious to us today is the use of "Sprechstimme," the highly-inflected, half-sung-half-spoken vocal technique. Those who know the history will know that a sort of sing-song performance of poetry was fairly common around this period, especially in German. In fact, the actress who commissioned the piece, Albertine Zehme, was touring Europe declaiming these same poems to another piece of music, by Otto Vrieslander. Apparently she wasn't satisfied with that other piece, and she asked Schoenberg to compose something new. I wonder if she had any inkling what she was getting into! The previous piece was basically just background music to which she gave dramatic readings; but Schoenberg specified exact (and incredibly complicated) rhythmic delivery and pitches that are supposed to be approximated but never dwelt on steadily. Except for a small number of pitches (nine?) that are actually to be sung, Schoenberg is essentially composing the exact inflection he wants the spoken text to have. And it's extreme, and wide ranging. I guess what I'm saying is that this old German technique of declaiming poetry, with a little extra specificity by Schoenberg, sounds revolutionary now because it's not something anyone would think of doing any more. In fact, in some ways, it may sound more revolutionary to us today that in did when it was premiered. Just think of this: at its premiere, it not only got a standing ovation, but the audience demanded a repeat on the spot! I somehow can't imagine a contemporary audience wanting to hear it twice in one night.
5BMF: You've performed everything from cantatas to song cycles to operatic and theatrical works. Where does Pierrot fit in, in terms of the challenges it presents to the performers?
HTC: To me personally, the strict rhythmic challenge is the biggest hurdle, and though that's kind of a boring technical answer, the instrumentalists might agree. Although they have a whole lot of notes to play. Schoenberg also wrote in his performance notes that the speaker should not add any of their own interpretation to his: that it was all there in the music, and anything that the speaker/singer might add based on their own feeling for the words would detract rather than add to the performance. How to add none of your own emphasis or inflection, and yet give a completely dedicated, dramatic, and heart-felt performances of these poems? It's a bit of a gauntlet he threw down a century ago, and still a fascinating challenge today.
5BMF: What would you say to a potential concertgoer who gets a bit queasy seeing the word 'Schoenberg' on a concert announcement?
HTC: Take some Dramamine and suck it up. If you get on board, it's worth the ride.
Hai-Ting Chinn and the Proteus Ensemble will perform Pierrot Lunaire on October 16th, in the second part of 5BMF's The Pierrot Project. Part I, on October 14th, will feature works by Debussy, Ravel and Schoenberg. Buy your tickets today and join us for these exciting concerts!

redpajama_set_name: RedPajamaCommonCrawl


text:
STILL HAIR ... St Thomas man still rocking natural hat after 36 years
Darain Housen came up with the idea of his hair hat after some of his friends decided to wear hats to a party, but he could not find one.
Darain Housen uses a lot of Aloe Vera in his hair.
Darain Housen walks through his community of New Monkland in St Thomas.
It has been 36 years since Darain Housen took off his hat, and from all indication, he has no plans to put one back on anytime soon.
The 52-year-old St Thomas resident has been wearing a natural headgear since 1982 and he expects that he will be buried in it.
"This is my trademark and I never grow tired of it or feel like I would cut it. It's actually quite light and not as heavy as some persons may think," he said.
The STAR first highlighted Housen's hairstyle in 2005. So eye-catching was his 'hair hat' that a television station in Japan invited him on a two-week trip to Tokyo.
"They took me to a lot of places there and showed off my hairstyle. I also demonstrated some of my barbering styles to them but I never got any money. They promised they would send me money when I get out but they never did. Persons from the United States also contacted me after the article and they took me to Hollywood. Them pay and look bout my visa and everything," he said.
It has been 13 years since Housen got the opportunity to fly as a result of his hair.
"If me get a chance fi guh back a farin me would a guh, but me waah come back with some money inna me pocket," the father of one said.
Housen is convinced that he can offer value to marketers who are seeking to show off their products. But he is not prepared to leave his New Monkland community in St Thomas unless it is worth his while.
"People call me fi come work at a barbershop in MoBay but me never go because me have whole heap a customers up yah," said the barber, who claims to be very creative when carving images in persons' hair.
Meanwhile, for those who are wondering how Housen manages to keep his hair in shape after so many years, the 'hair hat' man said that grooming is quite easy and inexpensive. A weekly 'line up' is a must and sometimes a touch of hair dye adds the right magic.
"I wash it all three or two times a week and use a lot of Aloe Vera in it. This makes my hair healthy and easy to handle. There are persons who would want to try it but it won't look the same. A Japanese man wanted me to cut his hair in a style like mine but his hair texture couldn't work," he said.
Housen told THE WEEKEND STAR that he came up with the idea of his hair hat after some of his friends decided to wear hats to a party, but he could not find one. He told THE WEEKEND STAR that he was the highlight of the party, and his hairstyle created such an impact that he got a crate of beer to share with his friends.
"Persons often stop me to take picture with me and I never tell them no. Whenever I wake in the morning, it is always set in the right position. I never have to tie it down or anything," he said.
Government approves additional funds to truck water


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On the fifth day of the festival we've got plentyof not only theatre treats for our audiences. At noon at the Festival Center another Something's Cooking Backstage meeting. This time our guests, Jacek Poniedziałek and Magdalena Popławska, not only shares some stories about working on Kabaret Warszawski, but also prepare for the audience an oatmeal and a fruit salads – secrets to their great shape. The discussion is facilitated by Łukasz Maciejewski. There's three Polish Contest shows playing today: Women as Lovers, directed by Ewelina Marciniak from Wybrzeże Theatre at 4 PM and 7 PM at Łaźnia Nowa Theatre, Radek Rychcik's The Devil from Współczesny Theatre in Szczecin presented at 6.15 PM at National Stary Theatre and Marcin Liber's For God's Sake from Drama Theatre in Wałbrzych – at 8.15 at National Stary Theatre. The first one, based on Elfriede Jelinek's novel, takes down patriarchal system that reduces women to love objects. The second, refers to pop-cultural images of evil, like Hollywood movies and analyze our most hidden fears. The third show, based on Jarosław Murawski's text, investigates the phenomenon of Polish faith. There's a Dantesque Scenes meeting with the artistic teams held after every show and led by, as follows: Łukasz Maciejewski, Joanna Derkaczew and Łukasz Drewniak. Join us!


A voyage of discovery is on offer in Tysoe when villagers take to the stage to bring to life some precious memories of village life.
A new piece of work called Won't You Come Along With Me will be performed in Tysoe Village Hall this Friday and Saturday, June 29 and 30.
The idea was conceived by consulting engineer and amateur historian, Andrew Baxter after Tysoe Amateur Dramatic Association (TADA!) helped celebrate the official re-opening of Tysoe Village Hall last June.
His wife Sue Baxter said: "Inspired by the vibrant local talent in Tysoe and particularly by long-time resident Doreen Smith's beautiful voice, Andrew has pulled together memories of Tysoe drawn from living there in the late 60s and from close association ever since.
"The audience is promised an uplifting, entertaining and enlightening evening with narration, songs and sketches.
The audience will be taken on a journey interspersing memories from around the parish, with some much-loved songs.
Some of the tales were familiar but Mr Baxter also unearthed fresh facts as he did the research.
Tickets cost £7.50 (£5 concessions) and are available from Tysoe Village Stores or from Sue on 01295 670605 or baxtersue@btconnect.com. Proceeds to the village hall.


Funeral potatoes are essentially a hash brown casserole in the Mormon community. They got their somber name from the fact that the dish is brought at many group functions and gatherings, including funerals.
Wonder which is going to kill him first: Old age, or the fast food.


text:
// Copyright (c) 2012 Ecma International.  All rights reserved.
// This code is governed by the BSD license found in the LICENSE file.

/*---
es5id: 15.4.4.16-2-18
description: >
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    its own property get method
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        function callbackfn1(val, idx, obj) {
            return parseInt(val, 10) > 1;
        }

        function callbackfn2(val, idx, obj) {
            return parseInt(val, 10) > 2;
        }

        var str = new String("432");

            String.prototype[3] = "1";

assert(Array.prototype.every.call(str, callbackfn1), 'Array.prototype.every.call(str, callbackfn1) !== true');
assert.sameValue(Array.prototype.every.call(str, callbackfn2), false, 'Array.prototype.every.call(str, callbackfn2)');

redpajama_set_name: RedPajamaGithub


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This box came earlier this month, and i havent had a chance to write it up, so apologies for that!
This used to be called "Beautyco Box" and as its only been around for a short amount of time, its odd that a name change was done so quickly! But never mind, it still has the same options as it always has.
If you haven't had one of these boxes, they are slightly different to other beauty boxes, in the fact that you have a choice of 3 boxes, and you know what you're getting. In fact, the contents of these boxes are pretty good, great sizes, some full size ones and this is the 2nd one i have got, and pretty impressed with it!
Its such a refreshing change to see good sized items in a beauty box, rather than sachets or teeny vials.
"ideal for partied out peepers, Puffy The Eyebag Slayer is a cooling eye mask that will put a twinkle back into our eyes in no time and leave them looking and feeling less puffy"
"Both to prevent and to cure various signs of ageing. It fills and reduces wrinkles since first applications and removes expression lines in just 30 days"
"This refreshing mist replenishes and moisturises to plump and firm skin. The Duel-phase moisturising system infuses skin with the essential nutrients that increase cellular hydration, repair the skin's barriers and soothe the skin. suitable for all skin types"
"This natural facial wash is a mild, antiseptic rinse off cleanser which leaves the skin scrupulously yet naturally clean, smooth and refreshed. It is particularly suited to normal to oily or acne prone skin with its blend of organic essential oils including antibacterial tea tree for a fresh unisex fragrance"
"This cooling, soothing balm is packed with ingredients to cushion, cool and protect fatigued feet. A power packed formula this ultra reviving feet treat will refresh weary legs with uber cooling peppermint, moisturise and revive with glycerin and instantly soothe with allantoin"
The monthly boxes cost £10 plus £2.95 postage. Looking at the items i got in this box and the RRP's, i think it was more than worth the cost, and im happy with the items ive got and will use them (which with so many other beauty boxes, you just chuck them to the side and leave them!).
What do you think? Personally, i think this is one of the better boxes on the market, and i do always take a look at their site each month but dont always get them. As for some of the other well known, i dont even really look! I would say except for the special edition, like Glossybox released an Espa box (which i got and will review) and was worth the pennies.
Did you get one of these boxes? Whaddya think? worth it?


text:
El Festival Internacional de la Canción de Sopot, o Sopot Festival, es un certamen musical que se celebra en Sopot (Polonia). Junto con el Festival Nacional de la Canción de Opole, es el festival de música más importante de Polonia.

Historia

El creador del festival fue Władysław Szpilman, quien lo puso en marcha junto con la radio polaca. Entre 1977 y 1980, este festival fue reemplazado por el Festival de la Canción de Intervisión, que mantuvo su sede en Sopot, y más tarde volvió a la fórmula original. A diferencia del Festival de la Canción de Eurovisión, Sopot ha realizado varios cambios a lo largo de su historia y ha estado abierto a otras naciones que no formaban parte del Bloque del Este o de Europa. Además, ha incluido las actuaciones de artistas invitados famosos.

El certamen comenzó a perder popularidad en los años 1980 y en la década de 1990 sufrió una grave crisis de audiencia que le llevó incluso al borde de la desaparición, cuando la TVP (televisión pública polaca) dejó de organizar el concurso y mantuvo solo el festival musical con cantantes famosos a partir de 1999. La organización de Sopot, poco convencida con las últimas retransmisiones de la TVP, decidió ceder los derechos del festival en 2005 a la emisora privada TVN, que mostró su interés en recuperar el concurso y la organización del festival.

El Festival de Sopot es considerado como uno de los festivales musicales más importantes de Europa junto con los de Eurovisión y San Remo.

Ganadores

1961-1973: Premio a la mejor canción

 1961 : Jo Rolland (Nous Deux)
 1962 : Jeanne Yovanna (Ti Krima)
 1963 : Tamara Miansarova (Pust vsiegda budiet solnce) y : Simone Langlois (Toi et ton sourire)
 1964 : Nadia Constantinoupoulou (Je te remercie mon cœur)
 1965 : Monique Leyrac (Mon pays)
 1966 : Lana Cantrell (I'm all smiles)
 1967 : Dana Lerska (Po prostu jestem)
 1968 : Urszula Sipińska (Po ten kwiat czerwony)
 1969 : Henri Dès (Maria Consuella). Este año España ganó el tercer premio de interpretación y el premio especial del público por la actuación de Conchita Bautista y el Gran Premio del Disco por la actuación de Michel.
 1970 : Robert Charlebois (Ordinaire)
 1971 : Samantha Jones (He Moves Me)
 1972 : Andrzej Dąbrowski (Do zakochania jeden krok) y : Lev Leshenko - (Ja ne byl z nim znakom)
 1973 : Tony Craig (Can You Feel It y I Think Of You baby)

1974-1976: Gran premio del disco

 1974 : Marion Rung (entre varias canciones)
 1975 : Glen Weston (entre varias canciones)
 1976 : Irina Ponarovskaya (entre varias canciones)

1977-1980: Festival de la Canción de Intervisión

(entre 1981 y 1983, no se celebró el Festival)

1984-1987: Festival musical de Sopot

 1984 : Krystyna Giżowska (Blue Box)
 1985 : Herreys (Summer Party)
 1986 : Mary Getz (Hero Of My Heart)
 1987 : Double Take (Rockola)

1988-1990: Disco de Oro

 1988 : Kenny James (The Magic In You)
 1989 : Dance With The Strangers (entre varias canciones)
 1990 : Lora Szafran (Zły chłopak y Trust Me At Once)

1991-¿?: Festival musical de Sopot

 1991 : New Moon (entre varias canciones)
 1992 : Mark Andrews (entre varias canciones)
 1993 : Arina (Rain Is Coming Down)
 1994 : Varius Manx (Zanim zrozumiesz)
 1995 : Kasia Kowalska (Jak rzecz y A to co mam)
 1996 Solo se celebró el festival, no hubo concurso
 1997 : Total Touch (Somebody Else's Lover)
 1998 : Alex Baroni (Male che fa male)

(entre 1999 y 2004, no se celebró el concurso, solamente el Festival)

-contando solo los ganadores por voto del público-
 2005 : Andrzej Piaseczny (Z głębi duszy) -solo hubo cantantes polacos en ese Festival-
 2006 : Mattafix (Big City Life)
 2007 : Feel (A gdy już jest ciemno)
 2008 : Oh Laura (Release me)
 2009  Australia: Gabriella Cilmi (Sweet about me)

(en 2010 y 2011, no se celebró el Festival)

 2012 : Eric Saade (Hotter than fire)
 2013 : Imany (You Will Never Know)
 2014 : Ewa Farna (Cicho)

(entre 2015 y 2017, no se celebró el Festival)

2018 : Michail (Who You Are)

Ganadores por país

Estrellas del Festival de Sopot
El Festival de Sopot cuenta con la presencia de otros artistas famosos fuera de concurso, que son el reclamo principal para la audiencia por delante del concurso. Estos son:

 1964: Nadia Constantinoupoulou, Karel Gott
 1966: Angela Zilia
 1968: Zsuzsa Koncz
 1969: Karel Gott, Muslim Magomajew
 1977: Helena Vondráčková
 1978: Drupi, Pussycat, The Temptations
 1979: Demis Roussos, Boney M
 1984: Charles Aznavour
 1986: Bonnie Tyler
 1988: Kim Wilde, Sabrina
 1989: Savage, Blue System, C. C. Catch
 1990: Ronnie Hawkins, Duo Datz, Erasure, Victor Lazlo, Curiosity Killed The Cat
 1991: Dannii Minogue, Johnny Hates Jazz, Technotronic, Deacon Blue, Alison Moyet, Aztec Camera, OMD, Bros, Jimmy Somerville
 1992: Kim Wilde, Bobby Kimbell, Sonia, Marillion, Simone Angel
 1993: Boney M, La Toya Jackson, Marc Almond, Helena Vondráčková, Jiří Korn
 1995: Chuck Berry, Vanessa Mae, Annie Lennox
 1996: The Kelly Family, Vaya Con Dios, Deep Forest, La Bouche
 1997: Khadja Nin, Alexia, Secret Garden, Boyz II Men
 1998: The Corrs, Chris Rea, Tanita Tikaram, Ace of Base, Era
 1999: Lionel Richie, Whitney Houston
 2000: Bryan Adams
 2001: Goran Bregović, Lou Bega, UB40
 2002: Zucchero, Garou
 2003: Björk, Ricky Martin
 2004: In-Grid, Kate Ryan, Patricia Kaas
 2005: Patrizio Buanne, Lemar, Gordon Haskell, Scorpions, Simply Red, Bonnie Tyler, Blondie
 2006: Katie Melua, Elton John, Karel Gott, Helena Vondráčková, Demis Roussos, Drupi
 2007: Norah Jones, Village People, Gloria Gaynor, Hot Chocolate, Kool & The Gang, Sister Sledge, Mattafix
 2008: Sabrina, Samantha Fox, Kim Wilde, Limahl, Thomas Anders.

Véase también
 Festival de la Canción de Eurovisión
Festival de la Canción de AfriMusic
 Festival de la Canción de Intervisión
 Festival de la Canción Mediterránea
 Festival de la Canción de San Remo
 Festival Internacional da Canção
 Festival Internacional de la Canción de Benidorm
 Festival Internacional de la Canción de Viña del Mar
 Festival Mundial de la Canción Latina
 Festival OTI de la Canción
 Festival Yamaha Music
 Olimpíada de la Canción de Atenas

Referencias

Enlaces externos
 Web del Festival de Sopot
 Festival de Festivales - España en Sopot

Festival de la Cancion de Sopot
Festivales de música de Polonia
Sopot


text:
Les strangozzi ou stringozzi sont une variété de pâtes de farine de blé essentiellement consommée dans la région de l'Ombrie et les zones limitrophes. Ce sont des pâtes longues de section rectangulaire. Elles sont généralement servies accompagnées de truffe noire de production locale, de ragù ou encore de sauce à base de tomate.

Les strangozzi bénéficient de l'appellation Produit agroalimentaire traditionnel.

Le nom de cette variété de pâtes est issu de sa ressemblance avec les lacets de chaussure appelés stringhe en italien.

On les trouve aussi sous les noms de strengozzi, strengozze, strozzapreti ou strangozzo et ciriole dans la province de Terni.

Il s'agit d'un produit dit pauvre qui, selon la recette issue de la fiche agroalimentaire traditionnelle, doit être réalisé à la main en abaisse (sfoglia) d'épaisseur comprise entre deux et quatre millimètres à partir de farine de blé tendre, d'eau et d'une pincée de sel. La sfoglia, après une période de repos au frais, est coupée en bandelettes de quatre millimètres de largeur et trente centimètres de long.

Le produit peut aussi être produit industriellement à partir de farine de blé dur avec un procédé d'extrusion et de laminage et amassé dans sa typique forme de « 8 » ou plus artisanalement à l'aide de la chitarra.

Notes et références

Annexes

Articles connexes
 Liste de pâtes

Pâte alimentaire
Cuisine ombrienne
Produit agroalimentaire traditionnel en Ombrie


text:
David O'Doherty: Big Time
Starring: David O'Doherty
David O'Doherty – the Ryanair Enya, the Aldi Bublé – returns to the Fringe with last year's hit show Big Time, an hour of talking and songs in a haunted hall on a hill filled with portraits of men who look like cocker spaniels.
Sorry. We have no upcoming listings for David O'Doherty: Big Time.
David O'Doherty: Grown Up
David O'Doherty: Whoa Is Me
David O'Doherty Is My Name
David O'Doherty in World Champion of Some Things
David O'Doherty Creates Something New Under The Su
David O'Doherty: Small Things
Brighton Comedy Festival: Best Of The Fest
Carlsberg Comedy Carnival 2009
Pecha Kucka KLK with David O'Doherty
A Film With Me In It
David O'Doherty: It's David O'Doherty Time
David O'Doherty: Let's Comedy
Honourable Men Of Art 2008
I Can't Sleep by David O'Doherty
O'Comic Gala
The Alternative Show
Jimmy Fallon Gala
O'Comics 2008
David O'Doherty: David O' Doh-party
I Can't Sleep
David O'Doherty: Somewhere Over The David O'Doherty
David O'Doherty Is Looking Up
David O'Doherty Presents: Rory Sheridan's Tales Of The Antarctica
David O'Doherty Presents 403 Second Masterworks
David O'Doherty: Seize The David O'Doherty
Mark Watson's Edinborolympics
David O'Doherty Will Try to Fix Everything
David O'Doherty Has Checked Everything
David O'Doherty: We Are All in the Gutter, But Some of Us Are Looking at David O'Doherty
David O'Doherty: You Have to Laugh
David O'Doherty: Ultrasound
"The only honest art form is laughter, comedy. You can't fake it...try to fake three laughs in an hour -- ha ha ha ha ha -- they'll take you away, man."
– Lenny Bruce


This kolam should be drawn below to the pooja mandab. Cleand and Wipe the place with wet cloth, then draw this kolam.
After drawing the kolam, place the wooden stool / manapalagai / halage / peta and cover with cloth. Above to it place the deity and do the pooja.


text:
Marion George Romney (September 19, 1897 – May 20, 1988) was an apostle and a member of the First Presidency of the Church of Jesus Christ of Latter-day Saints (LDS Church).

Early life
Romney was born in Colonia Juárez, Chihuahua, Mexico, to expatriated parents, George S. Romney and Terressa Artemesia Romney ( Redd).

His father, George Romney, was the uncle of Michigan governor George W. Romney, who was born in nearby Colonia Dublan. Marion Romney is the first cousin, once removed of Mitt Romney. Terressa Redd was the daughter of Lemuel Hardison Redd and Sariah Louisa Chamberlain. Terressa and George married in 1894 in Colonia Juárez.

Romney was the second of ten children. His younger sister, Lurlene Romney Cheney, later converted to Catholicism and entered a religious order as Sister Mary Catherine, a Carmelite nun at the Carmel of the Immaculate Heart of Mary in Holladay, Utah.

Romney studied at Academia Juárez until his family left Mexico in 1912, when violence from the ongoing Mexican revolution spread to their region. He lived with them for the remainder of his youth in California and Idaho. In 1917, the Romneys moved to Rexburg, Idaho, where his father became principal of Ricks Academy. Romney graduated valedictorian of Ricks high school in 1918.

From 1920 to 1923, Romney served as an LDS Church missionary in Australia. After returning, he worked in construction in Salt Lake City for his uncle, Gaskell Romney (father of George W. Romney).

Higher education and family
Romney studied at Brigham Young University (BYU) for a year. While there, he renewed his acquaintance with Ida Jensen, a former teacher at Ricks and then a post-graduate candidate at BYU. Romney and Jensen married September 12, 1924, in the Salt Lake Temple, officiated by Joseph Fielding Smith.

Romney next studied at the University of Utah, receiving a bachelor's degree in political science and history in 1926. Romney studied law at the University of Utah, but did not complete course work there. He passed the Utah bar exam in 1929.

Romney and his wife had three children together; two died in infancy. Their son, George Jensen Romney, survived to adulthood. They also adopted a child, Richard Jensen Romney. George served an LDS Church mission. In April 1983, he delivered a General Conference sermon written by his father. Ida Romney died in 1979 at age 88.

Church service and politics
Romney first worked for the post office, and later became an assistant prosecuting attorney in Salt Lake City. In 1934, he successfully campaigned for the Utah House of Representatives as a Democrat. While campaigning, he was called as an LDS Church bishop by stake president Bryant S. Hinckley. After his election, his bishop ordination was delayed until after the legislature's term in April 1935, with him seeking and serving only one term. While in the legislature, Romney helped author Utah's liquor control law.

General authority

Romney's 47 years as a general authority of the LDS Church began as one of the first five Assistants to the Quorum of the Twelve Apostles called to the new position in 1941. In 1951, he was called to the Quorum of the Twelve. In 1960, Romney helped develop the home teaching program of the LDS Church.

In 1961, Romney was appointed area supervisor for the LDS Church in Mexico. Although he had lived his first fifteen years in Mexico, it was in the mainly American Mormon colonies, and he knew very little Spanish. Assisted by Eduardo Balderas in learning Spanish, Romney supervised the church's growth in Mexico for the next eleven years. He oversaw construction of many meetinghouses in Mexico, along with organization of the first Spanish-language stakes in Mexico.

Romney became Second Counselor to church president Harold B. Lee in 1972.

Following Lee's death the following year, Spencer W. Kimball became the church president and he retained Romney as the Second Counselor. As the First Presidency, Kimball, Tanner, and Romney announced the 1978 Revelation on Priesthood, canonized as "Official Declaration 2" in the Doctrine and Covenants.

When Kimball, Tanner, and Romney all aged and developed health problems at similar rates, Gordon B. Hinckley was added as an additional counselor in 1981. Upon Tanner's death in 1982, Romney became First Counselor and Hinckley the Second Counselor, though Romney was relatively inactive in his position due to poor health. When Kimball died in 1985, press reports indicated that Romney had not been seen in public for many months.

Ezra Taft Benson, who had been President of the Quorum of the Twelve Apostles, then became the church president and named Hinckley as First Counselor, with Thomas S. Monson as Second Counselor. Romney, as the apostle with the second-longest seniority in the church, became the quorum president. However, "because President Romney's health [kept] him from taking an active part in Church administration," Howard W. Hunter, the next in seniority, served as acting president.

Romney died from natural causes at his home in Salt Lake City at age 90.  He served 47 years as a church general authority. Funeral services were held in the Salt Lake Tabernacle on May 23, 1988, presided over by Benson. Romney was buried at Wasatch Lawn Memorial Park in Salt Lake City, beside his wife, approximately 10 years after her death. The Deseret News Church Almanac remembered him as a "renowned Church Welfare pioneer and Book of Mormon scholar".

See also
 Thomas E. McKay
 Nicholas G. Smith

References

General references

1897 births
1988 deaths
20th-century Mormon missionaries
American Mormon missionaries in Australia
American general authorities (LDS Church)
Apostles (LDS Church)
Assistants to the Quorum of the Twelve Apostles
Brigham Young University alumni
Brigham Young University–Idaho alumni
Citizens of the United States through descent
Counselors in the First Presidency (LDS Church)
Doctrine and Covenants people
Latter Day Saints from California
Latter Day Saints from Idaho
Latter Day Saints from Utah
Democratic Party members of the Utah House of Representatives
Mexican emigrants to the United States
Mexican general authorities (LDS Church)
Mexican people of American descent
People from Colonia Juárez, Chihuahua
Presidents of the Quorum of the Twelve Apostles (LDS Church)
Romney family
University of Utah alumni
Utah lawyers


text:
Q: Random() happening only once I have this for() loop where I am randomizing the selection of slices of a picture, to display 16 slices of an image in a random order.
I'm picking those slices from an array and I have a variable that picks up what slice is going to be selected in the array.
The problem being that I'd think that the random function would be triggered for every frame, but it's triggered only once.
Here's the code :
void setup() {

    size(720,720);

    slices = new PImage[16];

        slices[0] = loadImage("1.png");
        slices[1] = loadImage("2.png");
        slices[2] = loadImage("3.png");
        slices[3] = loadImage("4.png");
        slices[4] = loadImage("5.png");
        slices[5] = loadImage("6.png");
        slices[6] = loadImage("7.png");
        slices[7] = loadImage("8.png");
        slices[8] = loadImage("9.png");
        slices[9] = loadImage("10.png");
        slices[10] = loadImage("11.png");
        slices[11] = loadImage("12.png");
        slices[12] = loadImage("13.png");
        slices[13] = loadImage("14.png");
        slices[14] = loadImage("15.png");
        slices[15] = loadImage("16.png");

        frameRate(1);

}

void draw() {

        for (int a = 0; a < 16; a++){

        int rand = int(random(slices.length));

        image(slices[rand],x,y,size,size);
         x += size;

         if (a % 4 == 3){
            y += size;
            x = 0;
            }

        }

It's dispalying the randomized slices only once and then I end up with a fix image. What I'd like to have is random slices appearing at every frame.
Thanks for your help !

A: Could it be because you've forgot to clear the screen (e.g. calling background()) (meaning once you've drawn an image it will stay rendered) ?
You could also make use of the for loop in setup to avoid repeating yourself:
int numSlices = 16;
PImage[] slices = new PImage[numSlices];
float x, y;
float size = 180;

void setup() {

  size(720, 720);

  for(int i = 0 ; i < numSlices; i++){
    slices[i] = loadImage((i+1) + ".png");
  }

  frameRate(1);
}

void draw() {
  background(255);

  for (int a = 0; a < numSlices; a++) {

    int rand = int(random(numSlices));

    image(slices[rand], x, y, size, size);
    x += size;

    if (a % 4 == 3) {
      y += size;
      x = 0;
    }
  }
  y = 0;
}

Additionally you could easily format your code (via CMD+T on OSX or Ctrl+T on Windows/Linux)
Update Kamakura (+1) correctly pointing out y not being reset to 0.
As a distraction I though't point you to IntList's shuffle() method:
int numSlices = 16;
PImage[] slices = new PImage[numSlices];
float x, y;
float size = 180;

IntList indices = new IntList();

void setup() {

  size(720, 720);

  for(int i = 0 ; i < numSlices; i++){
    slices[i] = loadImage((i+1) + ".png");
    indices.append(i);
  }

  frameRate(1);
}

void draw() {
  background(255);
  // shuffle list
  indices.shuffle();
  // reset y
  y = 0;

  for (int a = 0; a < numSlices; a++) {

    int rand = indices.get(a);

    image(slices[rand], x, y, size, size);
    x += size;

    if (a % 4 == 3) {
      y += size;
      x = 0;
    }
  }

}

Extra reason to play with it, other than a learning experience is that fact that it will be unlikely to get the same random index repeated.
Regarding splicing/shuffling, here's a modified version of the Load and Display example:
/**
 * Load and Display
 *
 * Images can be loaded and displayed to the screen at their actual size
 * or any other size.
 */

PImage img;  // Declare variable "a" of type PImage
// shuffled image
PImage imgShuffled;
// list of indices to shuffle
IntList shuffleIndices = new IntList();
// configure image slicing rows/columns
int rows = 4;
int cols = 4;
// total sections
int numSections = rows * cols;
// image section dimensions
int sectionWidth;
int sectionHeight;

void setup() {
  size(640, 360);
  frameRate(1);
  // The image file must be in the data folder of the current sketch
  // to load successfully
  img = loadImage("https://processing.org/examples/moonwalk.jpg");  // Load the image into the program
  // calculate section dimensions
  sectionWidth  = img.width / cols;
  sectionHeight = img.height / rows;
  // allocate a separate image to copy shuffled pixels into
  imgShuffled = createImage(img.width, img.height, RGB);
  // populate image section indices
  for(int i = 0 ; i < numSections; i++){
    shuffleIndices.append(i);
  }
}

void shuffleImage(){
  // shuffle the list
  shuffleIndices.shuffle();
  // Ta-da!
  println(shuffleIndices);
  // loop through each section
  for(int i = 0 ; i < numSections; i++){
    // index to row, col conversion
    int srcCol = i % cols;
    int srcRow = i / cols;
    // convert to pixel coordinates to copy from
    int srcX = srcCol * sectionWidth;
    int srcY = srcRow * sectionHeight;
    // get random / shuffled index
    int index = shuffleIndices.get(i);
    // same row, col, to pixel conversion to copy to
    int dstCol = index % cols;
    int dstRow = index / cols;
    int dstX = dstCol * sectionWidth;
    int dstY = dstRow * sectionHeight;
    // copy from original image to shuffled pixel coordinates
    imgShuffled.copy(img,srcX,srcY,sectionWidth,sectionHeight,dstX,dstY,sectionWidth,sectionHeight);
  }
}

void draw() {
  shuffleImage();
  // Displays the image at its actual size at point (0,0)
  image(imgShuffled, 0, 0);
}

A: You have 2 problems in your code.
First, you may not want to choose a random index.
This is because the same image could be chosen twice.
Instead, you could shuffle the array before drawing the images, like this:
for (int i = slices.length; i > 1; i--) {
  //choose a random index for the i-th element to be swapped with
  int j = (int)random(i);

  //swap them
  PImage temp = slices[j];
  slices[j] = slices[i-1];
  slices[i-1] = temp;
}

Second, the index is chosen on every frame, and the images are drawn, too, but you can't see it, because your code never resets y back to 0, meaning that they are below the screen.
You can fix this by adding
y = 0;

to the top or bottom of your draw().

redpajama_set_name: RedPajamaStackExchange


Welcome to our summer house. The air conditioning is non-existent and the weather humid and sticky but at least we don't have any mosquitoes to bother us. No it's so hot even they have decided to stay in with a nice ice cream and take it easy.
But not us! Oh no May was a very busy month and June looks to be just as busy again. Such is the life of a games developer – cursed to stay indoors all year round and sit down for a living.
First up we have a new videogram showing off all of our lovely new bits and things.


NJ Rev Stat § 39:4-37 (2016) What's This?
Local authorities in their respective jurisdictions may regulate by ordinance, which shall first be approved by the director, the crossing of pedestrians at intersections of roadways where traffic on the roadways is controlled by traffic control signals.
Amended by L.1951, c. 23, p. 75, s. 27.


Munda 6, Pakistan 1998- Munda 6 was a Pakistani show camp for refugees. We obtained government permission to visit the camp with a Pakistani "liaison officer" or "LO". As we drove to the camp, we passed through miles of lush farms and villages. As we approached the camp, it was clearly visible from a distance as the only patch in the valley that had no vegetation. The Pakistanis apparently did not allow refugees farm lest they become too comfortable on what is tenuously Pakistani territory.
Our taxi drove right up to an administrative hut and our LO asked us to wait in the car while he went inside. As soon as he was inside, an old man shoved his way through a crowd of refugees who had closed in around our car and thrust his arm through the back window at me. His leathery hand laid upon me a piece of paper. I glanced at it quickly and stuffed it into my shirt pocket. What I saw appeared to be written in Pashto or Urdu. But I was too slow. The LO had come out and demanded that I give him the slip of paper. I never found out what was on it but no one looked too happy to be there.
We went inside the Administration hut. They were fingerprinting the refugees inside. Some show camp!


The liver performs over 500 different functions in the human body (1).
Every day, your liver supports your everyday life by providing you with energy, fighting off infection, clearing out toxins, helping clot the blood, regulating hormones and so much more (2).
You liver does a lot for you, so when it isn't working optimally, you'll be sure to notice.
The liver, like the kidneys, filters out toxins and harmful substances from the body.
The organ also produces albumin, a protein that helps keep fluid in the bloodstream. This prevent swelling and edema, an accumulation of fluid in bodily tissues (3).
When the liver is overloaded with fat-soluble toxins, it begins to store these substances in your fat cells.
In fact, one of the first steps in getting rid of toxins is the liver converting fat-soluble toxins into water-soluble ones so that may be expelled from the body through bile.
This can leave you feeling tired, weak, nauseous, confused and can cause digestive troubles and weight gain.


At a minimum, this means a shift in American policy toward nonbelligerence toward Iran and Syria, a commitment to a clear timetable for a Palestinian state and a commitment to a true no-weapons-of-mass-destruction zone in the Middle East, which means a commitment to confront Israel over its possession of nuclear weapons.
In interviews last week, he did not bother with campaign rhetoric about bringing peace and pledged to aim for security and nonbelligerence.
A year after official Zionist policy achieved its aim of Jewish statehood in 1949, Buber expressed his fears that after the war peace, when it comes, will not be peace, a real peace which is constructive, creative (but) a stunted peace, no more than nonbelligerence, which at any moment, when any new constellation of forces arises, is liable to turn into war.


I know you can buy them in the States, ones that dont require welding but drill and tap into the chassis rails and have U bolts around the axle tubes, that would be ideal.
Anyone fitted one locally ?
peter , im pretty sure pedders have a universal tyrpe rear you can get...I think i done it years ago on my 71 i use to own. just take your chassis rail measurements. also dave at pontiworld has them for around $300.
I just ordered the camaro one through pedders, they get them made and have the specs already anyway.
When Roger Penske was racing these, they used a standard 11/16" front sway bar and turned up-side-down for the rear.
It did involve making some custom brackets, so if you have a welder and angle grinder, it wont be hard to do.
The details and parts images are in Wayne D. Guinn's "Camaro. Untold Secrets" Pg 20-26.
Thanks for the replies, I will call in and see Pedders. I guess it makes no difference if its a 9 inch diff instead of a chev 12 bolt ??
I run a 9"' as well, had to make up a couple of mounts but no big deal...by the way make sure the u bolts they give you are for the 9'' not a 12 bolt as the 9'' axle tubes are bigger the the 12 bolt ones.
If you need some photos of how I mounted mine let me know.
Last edited by chevguy; 26-04-2010 at 10:05 AM..
Fitted the axle tube mounts and u bolts then tacked the mount onto the axle tube after tightening u bolts.
$50 on ebay for a hk type 3/4 inch..and get the front one for nothing..claims never been fitted.
By the way Pete.. Have you actually felt the need to fit a rear one? Do you suffer from lot's of understeer?
Not sure about you, but my 69 with it's stock 11/16" front only was very flat in the corners. Had some understeer if I really pushed it, but on Bias F60 Firestones, I wasn't about to blame the suspension.


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Irish astronomer says he has successfully tracked UFOs
Roscommon-based astronomer attracted international attention with claims of evidence of extraterrestrial activity.
Jane Walsh
Roscommon-based astronomer attracted international attention with claims of evidence of extraterrestrial activity. Fotomedia
Eamonn Ansbro, a Roscommon-based astronomer, attracted international media attention as he claims his research has unearthed evidence of extraterrestrial activity. Now the Boyle local wants to put the town on the map establishing a centre for all things UFO.
For the past 20 years Ansbro, from the town of Boyle, has been carrying out research on extraterrestrial intelligence. Ansbro gives presentations across the globe, including one to the French government.
Although his work might seem controversial and unusual he insists it is carried out "scientifically".
Ansbro's work has been published by the European Space Agency and he works as a part of a group of scientists called OSETI (Optical Search for Extra-Terrestrial Intelligence).
His recent work shows that alien space-crafts follow defined paths around the earth. In his hometown of Boyle Ansbro has set up an "all sky camera" to monitor the whole sky hemisphere around the clock. The camera which comprises of 11 cameras record anything that moves or is unusual. Ansbro, along with other scientists, has collated the data which suggests that space-crafts are circling the earth.
He explained "When you take all that data and apply it to this theory we have found that it fits with the computerized time graphs and that we can predict when future sightings will occur and we have found that it has happened in Boyle."
He refers to thema as "Identified Flying Objects" rather than UFOs.
Making comparisons to Roswell, New Mexico, he said "the tourism potential is huge." he said, drawing comparisons with theUFO centre in Roswell New Mexico, which attracts huge numbers of visitors every year and also another UFO centre in Japan.
He told his local paper the Roscommon Herald, "We haven't proven our discovery to the scientific community yet and don't have the resources but we can show that extraterrestrial space craft are on specific orbital tracks around the earth."
*Originally published in 2011.


In this video tutorial, we go over all the steps on how to connect and pair the GoPro Remo remote to the GoPro Fusion. The Remo remote is nice in the fact that it is a small water proof remote that is compatible with the Hero 5/6 Black, Hero 5 Session and the GoPro Fusion. It has a simple one-button design and can also assist with your GoPro to hear voice commands in less than ideal conditions. One downfall, however, is that the Remo Remote can only be paired to one GoPro at a time.
Power on the GoPro Fusion and let it fully load.
Press the mode button on the side of the GoPro Fusion to cycle through the modes unit you come to the settings. It will have an icon of a wrench.
Press the record button on the GoPro Fusion to select and enter the settings menu.
There are several sub-menus within the settings mode. press the record button again to cycle through the submenus until you reach the wireless settings. It will have an icon of an antenna.
Now press the mode button to scroll down to the connect option and press the record button to select it.
From the connect menu, press the mode button to scroll down to the Remo option. Press the record button to select it.
At this point, a message will come on the LCD screen of the Fusion telling you to put the Remo Remote into pairing mode. You can set down the Fusion and get your Remo ready.
To put the Remo Remote into pairing mode, long press the power button for 8 seconds.
Once complete your Remo will vibrate to signify a successful pairing, as well as a notification, will appear on the Fusion LCD screen.
At this point, it is always a good idea to test the connection to make sure it is working as it should.


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O Poni é uma província de Burkina Faso localizada na região Sul-Oeste. Sua capital é a cidade de Gaoua.

Departamentos

A província do Poni está dividida em dez departamentos:
 Bouroum-Bouroum
 Bousséra
 Djigoué
 Gaoua
 Gbomblora
 Kampti
 Loropéni
 Malba
 Nako
 Périgban

Províncias de Burquina Fasso