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text:
\section{Introduction}

We begin with the following definition.
  \begin{defn}\label{def-QOBC} A K\"ahler  manifold  $(M, g)$ of complex dimension $n$ is said to have {\it nonnegative  quadratic orthogonal   bisectional  curvature (NQOBC)} at $p\in M$ if $n\geq 2$ and: {\it for any} unitary frame $\{e_1,\dots,e_n\}$ of $T^{(1,0)}_p(M)$ and {\it any} real numbers $\xi_1,...,\xi_n$ we have
\begin{equation}\label{e1}
\sum_{i,j=1}^n R_{i\bar{i}j\bar{j}} (\xi_i -\xi_j)^2 \geq 0.
\end{equation}
We will say that a manifold $(M, g)$ has NQOBC provided it does at every point.
\end{defn}
One can similarly define the notion of nonpositive orthogonal quadratic  bisectional  curvature. Note that if a product of K\"ahler  manifolds $M_1 \times M_2$ has NQOBC then so must each factor $M_1$ and $M_2$. The reverse implication however may be false in general:  $M_1$ and $M_2$ may both have NQOBC while $M_1 \times M_2$ may not.

The condition of NQOBC is weaker than requiring $M$ to have nonnegative orthogonal bisectional curvature: $R(V,\bar V,W,\bar W)\geq 0$ for any orthogonal unitary pair $V, W\in T^{(1,0)}(M)$, while on complex surfaces the two conditions are equivalent.    Li, Wu and Zheng  \cite{LiWuZheng} are able to construct examples of compact K\"ahler
manifolds $M$ with NQOBC which do not admit
any K\"ahler  metrics with nonnegative orthogonal bisectional curvature.

 The structure of compact K\"ahler  manifolds with nonnegative orthogonal
bisectional curvature on the other hand, is now completely understood by
the works of Chen \cite{Chen} and Gu and Zhang \cite{GZ}. Their results
extend those on the generalized Frankel conjecture for compact manifolds
with nonnegative holomorphic bisectional curvature, formulated by Yau \cite[p.677] {Yau}, and
established through the works of Howard-Smyth-Wu \cite{HowardSmythWu1981}, Wu \cite{Wu} and eventually by Bando \cite{Bando} in three-dimension and Mok \cite{Mok} for all dimensions. Frankel's
conjecture for compact manifolds with positive holomorphic bisectional
curvature was established by Mori \cite{M}   and Siu and Yau \cite{SY} independently.

The NQOBC condition arises naturally in the proof of the following fact which appeared implicitly in an earlier work \cite{BG}:
  \begin{lem}\label{l-keyfact}
  If $(M^n, g)$ has NQOBC, then all harmonic $(1,1)$ forms are parallel.
  \end{lem}
The proof uses the Bochner formula for $(1,1)$ forms.  The lemma is implicit from earlier works though was only used in these under the stronger assumptions of positive and nonnegative holomorphic bisectional curvature  (see for example  \cite{BG, GK, HowardSmythWu1981} and references therein).  We will make use of Lemma \ref{l-keyfact} at various points.

 The NQOBC condition was first considered explicitly by Wu-Yau-Zheng in \cite{WuYauZheng} where the authors studied the boundary of the K\"ahler  cone of manifolds with NQOBC.

 In this paper, we try to understand more on  this class of K\"ahler  manifolds.
We first prove that: {\it If a K\"ahler  manifold has NQOBC at a point $p$, then the scalar
curvature is nonnegative at $p$ and is zero if and only if it is flat at $p$, provided
the complex dimension is at least 3}. See Theorem \ref{t-scalarcurvature} for more details.
Using this result we prove: {\it If $(M,g)$ is a compact K\"ahler  manifold with NQOBC which is locally irreducible, then $c_1(M)>0$}. This generalizes \cite[Theorem 2.1]{GZ}.  This naturally leads us to consider the deRham factorization of the universal cover of $M$ for which we prove the following. {\it All compact factors have positive first Chern class.  There is at most one non-compact factor being either: complex Euclidean space (in which case all compact factors will have quasi positive Ricci curvature),  or 1-dimensional (in which case all  compact factors will have positive Ricci curvature)}.  This gives a slight refinement of \cite[Theorem 1.3 (2)]{GZ}.  For more detailed description, see Theorem \ref{t-reduce}.

The organization of the paper is as follows. In Section \ref{s-cone}, we will give a quick proof of a result in \cite{WuYauZheng} on the boundary of the K\"ahler  cone of manifolds with NQOBC which is similar to but independent of the   proof  in \cite{Z} and the idea is similar. In Section \ref{scalar}, we will prove that K\"ahler  manifolds with NQOBC must have nonnegative scalar curvature.  In Section 4 we prove the above mentioned structure results for compact K\"ahler  manifolds with NQOBC.

 The authors would like to thank D. Wu, S.T. Yau and  F. Zheng for helpful comments and interest in this work.

\section{Boundary of the K\"ahler  cone of manifolds with NQOBC}\label{s-cone}

The boundary of the K\"ahler  cone of $(M, g)$ turns out to be rather special: every boundary class $\alpha$ contains a smooth nonnegative representative $\eta \in \alpha$.  This was first proved in \cite{WuYauZheng} by solving a degenerate complex Monge-Amp\`{e}re  equation under the condition of NQOBC. This motivates us to study structure of K\"ahler  manifolds with NQOBC.   In fact, the following holds:

\begin{thm}\label{s2t1}
Suppose $(M^n, g)$ is a compact complex n-dimensional K\"ahler  manifold with NQOBC.  Let $\alpha$ be in the closure of the K\"ahler  cone of $M$ and $\eta$ be the unique harmonic representative in $\alpha$.  Then $\eta$ is nonnegative.  Moreover, $\eta$ is positive if and only if $\alpha^n[M] >0$.
\end{thm}

A short proof of Theorem \ref{s2t1} was provided in \cite{Z} and also independently in \cite{CT}.  We provide details of the proof here for the sake of  completeness.  We follow the presentation in \cite{CT} which we think is more concise.      We refer to \cite{CT} where it was also shown that a rather straight forward observation on the proof in \cite{WuYauZheng} leads to another proof of Theorem \ref{s2t1}.

 Given a complex manifold $M$, recall that a real class $\alpha\in H^{(1,1)}(M)$ is called a K\"ahler  class if $\alpha$ contains a smooth positive definite representative $\eta$.  The space of K\"ahler  classes is a convex cone in $H^{(1,1)}(M)$  referred to as the K\"ahler  cone which we denote by $\mathcal{K}$.  We say that $\alpha$ is in the closure of $\mathcal{K}$ if $[(1-t)\omega+t\eta] \in\mathcal{K}$ for any smooth $\eta\in\alpha$, $\omega\in\mathcal{K}$ and $t\in[0,1)$.  Finally, given any real $\alpha\in H^{(1,1)}(M)$ we use $\alpha^n[M]$ to denote the integral $\int_M  \alpha^n$.

\begin{proof} [Proof of Theorem \ref{s2t1}]
 Let $\eta$ be as in Theorem \ref{s2t1} and let $\omega_0$ be the K\"ahler  form for $(M, g)$.  By the above remarks, $\eta$ is parallel and thus has constant real eigenvalues $a_1,...,a_n$ on $M$ with respect to $\omega_0$.  Also, $[(1-t)\omega_0+t\eta]\in \mathcal{K}$  for every $t\in[0,1)$.

  In other words, for each $t\in[0,1)$ there exists $f_t\in C^{\infty}(M)$ and $\omega_t \in \mathcal{K}$ such that $(1-t)\omega_0+t\eta=\omega_t +dd^c f_t$, giving
$$
\text{Vol}_g (M)\prod_{i=1}^n(1-t+ta_i)= \int_M((1-t)\omega_0+t\eta)^n=\int_M\left(\omega_{t}+dd^c f_{t}\right)^n>0
$$
for all $t\in[0,1)$.  On the other hand, if $a_k<0$ for some $k$ then $1-t+ta_k$ and thus the product on the LHS above would vanish for some $t_0\in(0,1)$ giving a contradiction. Thus $a_i$ must be nonnegative for each $i$, in other words $\eta$ is nonnegative.  In particular, we have $\int_M \eta^n \geq  0$ with strict inequality if and only if $\eta$  is positive.  This completes the proof.
\end{proof}

We end this section with the following description of the boundary of $\mathcal{K}$ from \cite{CT}.  Let $\widetilde M$ be the universal cover of $M$ with projection $\pi:\widetilde M\to M$.  Then by the de Rham decomposition Theorem for K\"ahler  manifolds, we may write $$(\widetilde M, \widetilde \omega_0)=(\widetilde M_0, \widetilde \sigma_0)\times (\widetilde M_1, \widetilde \sigma_1)\times\cdots\times (\widetilde M_k , \widetilde \sigma_k)$$ where $\widetilde \omega_0=\pi^*(\omega_0)$, $\wt M_0$ is flat and  each $\wt M_i$, for $i\ge 1$,  is nonflat irreducible and K\"ahler  and the decomposition is unique up to permutation.

 \begin{cor}\label{c1} Let $\eta$ be a real harmonic (1,1) form on $M$, and let $\widetilde \eta=\pi^*(\eta)$. Then $\widetilde\eta=\widetilde \eta_0\times\prod_{i=1}^k a_i\widetilde \sigma_i$ for some real constants $a_i$ and $\widetilde\eta_0$ is a parallel harmonic form on $\wt M_0$. In particular, the boundary of $\mathcal{K}$ can be identified with the space of harmonic $(1,1)$ forms $\widetilde \eta$ on $\widetilde M$ satisfying: $\widetilde \eta$ is equivariant with respect to $\pi_1(M)$ and $\widetilde\eta=\widetilde\eta_0\times\prod_{i=0}^k a_i\widetilde \sigma_i$, where $\widetilde \eta_0\ge0$ and $a_i\ge0$ for all $i$ with equality holding for some $i$.
\end{cor}
\begin{proof} Since $\eta$ is parallel, then $\widetilde M$ is a product of K\"ahler  manifolds $\widetilde N_i$ such that at each point $T^{(1,0)}(N_i)$ are eigenspace of $\widetilde \eta$. By the uniqueness and irreducibility of the $\wt M_i$'s for $i\ge 1$, the result follows.
\end{proof}

\section{Scalar curvature of manifolds with NQOBC}\label{scalar}
Let $(M^n,g)$ be a K\"ahler  manifold. If $M$ has nonnegative holomorphic bisectional curvature, then the Ricci curvature of $M$ is nonnegative. If on the other hand $M$ only has NQOBC, then the Ricci curvature may have negative eigenvalues at some point, see \cite[Example 1.2]{GZ}. However, we have the following:

\begin{thm}\label{t-scalarcurvature} Suppose $(M^n,g)$ has nonnegative (resp. nonpositive) QOBC at $p$.  Then the scalar curvature $S(p)$ is nonnegative (resp. nonpositive) and $S(p)=0$ if and only if for all unitary pairs $V,W$ we have $R(V,\bar V,W,\bar W)=0$ and $R(V,\bar V,V,\bar V)+R(W,\bar W,W,\bar W)=0$. If $n\ge 3$, $S(p)=0$ also implies $R(V,\bar V,V,\bar V)=0$ for all $V\in T_p^{(1,0)}(M)$ and hence $M$ is flat at $p$.
\end{thm}
The nonnegativity (nonpositivity) of the scalar curvature $S(p)$ in the theorem follows from the following slightly more general result.
\begin{lem}\label{l-scalar} Let $(M^n,g)$ be a K\"ahler  manifold and $p\in M$.
 \begin{itemize}
   \item [(a)] Suppose there exists an $n\times n$ matrix $a_{ij}$ with the properties: $\sum_{i\neq j}a_{ij}+2\sum_{i}a_{ii}>0$  and {\it for any} unitary frame $\{e_1,\dots,e_n\}$ of $T^{(1,0)}_p(M)$,
\begin{equation}\label{e-weaker1}
\sum_{i,j=1}^n R_{i\bar{i}j\bar{j}} a_{ij} \geq 0.
\end{equation}
Then $S(p)$ is nonnegative. If in addition, there is a unitary frame such that \eqref{e-weaker1} is a strict inequality  then $S(p)>0$.
   \item [(b)] Suppose there exists an $n\times n$ matrix  $a_{ij}$ with the properties: $\sum_{i\neq j}a_{ij}+2\sum_{i}a_{ii}>0$ and {\it for any} unitary frame $\{e_1,\dots,e_n\}$ of $T^{(1,0)}_p(M)$,
\begin{equation}\label{e-weaker2}
\sum_{i,j=1}^n R_{i\bar{i}j\bar{j}} a_{ij} \le 0.
\end{equation}
Then $S(p)$ is nonpositive. If in addition, there is a unitary frame such that \eqref{e-weaker2} is a strict inequality, then $S(p)<0$.
 \end{itemize}
\end{lem}
\begin{proof} To prove (a), let $\mathcal{F}$ be the set of all ordered unitary bases in $T^{(1,0)}_p(M)$.  For any frame ${\alpha}=\{e_1,\dots,e_n\}\in \mathcal{F}$ and $\mathfrak{u}=(u_{ij})\in U(n)$, we define the frame $$\mathfrak{u}{\alpha}:=\left\{\sum_{j}u_{j1}e_j,\dots,\sum_{j}u_{jn}e_j\right\}\in \mathcal{F}.$$   In other words, $\mathfrak{u}$ is just the change of basis matrix from ${\alpha}$ to $\mathfrak{u}{\alpha}$.  Now we choose some frame $\alpha' \in \mathcal{F}$ to be fixed throughout the proof.

 Define functions $F(ij)$ (for any $i,j$) and $G(i)$ (for any $i$) on $U(n)$ by
\begin{equation}
\begin{array}{clcr}
			&F(ij)(\mathfrak{u}):=R(f_i,\bar f_i,f_j,\bar f_j)\\
			&G(i)(\mathfrak{u}):=R(f_i,\bar f_i,f_i,\bar f_i)\\
		\end{array}
\end{equation}
where $\{f_1,\dots,f_n\}=\mathfrak{u}{\alpha}'$.  Now we establish the following

\begin{bf} Claim:\end{bf} Let $\mu$ be the left invariant Haar measure on $U(n)$.  Then we  have $2\int_{U(n)}F(ij) (\mathfrak{u})d\nu(\mathfrak{u})=\int_{U(n)} G(k)(\mathfrak{u})d\nu$
for any $i\neq j$ and for any $k$.  In particular, the LHS is independent of $i,j$ and the RHS is independent of $k$.

For $i\neq j$, let $\mathfrak{u_0}(ij)\in U(n)$ be the unitary matrix satisfying: for any $\{f_1,\dots,f_n\}\in \mathcal{F}$, $\{h_1,\dots,h_n\}=\mathfrak{u_0}(ij)\{f_1,\dots,f_n\}$ satisfies
\begin{equation}
\begin{array}{clcr}
			h_i=\frac{ 1+\sqrt{-1}}2f_i+\frac{ 1-\sqrt{-1}}2f_j\\
			h_j=\frac{ 1-\sqrt{-1}}2f_i+\frac{ 1+\sqrt{-1}}2f_j\\
		\end{array}
\end{equation}
and $h_k=f_k$ if $k\neq i, j$.  Also let $\mathfrak{v_0}(ij)\in U(n)$ be the unitary matrix satisfying: for any $\{f_1,\dots,f_n\}\in \mathcal{F}$, $\{h_1,\dots,h_n\}=\mathfrak{v_0}(ij)\{f_1,\dots,f_n\}$ satisfies
\begin{equation}
\begin{array}{clcr}
			h_i=\frac1{\sqrt 2}(f_i+f_j)\\
			h_j=\frac1{\sqrt 2}(f_i-f_j)\\
		\end{array}
\end{equation}
 and $h_k=f_k$ if $k\neq i, j$.

 A straight forward computation gives the following for any $i\neq j$ and $\mathfrak{u}\in U(n)$
\begin{equation}\label{eq-uv1}
  F(ij)(\mathfrak{u})+F(ij)(\mathfrak{u_0}(ij)\mathfrak{u})=\frac12\left(G(i)(\mathfrak{v_0}(ij)\mathfrak{u})+G(j)(\mathfrak{v_0}(ij)\mathfrak{u})\right).
\end{equation}
  Thus for $i\neq j$ we have
\begin{equation}\label{eq-integral-1}
\begin{split}
  4\int_{U(n)}F(ij)(\mathfrak{u})d\nu(\mathfrak{u})=&2\int_{U(n)}
  \left(F(ij)(\mathfrak{u})+F(ij)(\mathfrak{u_0}\mathfrak{u})\right)d\nu(\mathfrak{u})\\
  =&\int_{U(n)}\left(G(i)(\mathfrak{v_0}(ij)\mathfrak{u})+G(j)(\mathfrak{v_0}(ij)\mathfrak{u})\right)d\nu(\mathfrak{u})\\
=&\int_{U(n)}\left(G(i)(\mathfrak{u})+G(j)(\mathfrak{u})\right)d\nu(\mathfrak{u}).
\end{split}
\end{equation}
On the other hand, for $i\neq j$ if we let $\mathfrak{w_0}\in U(n)$ be such that for any ${\alpha}\in \mathcal{F}$, the frame $\mathfrak{w_0}{\alpha}\in \mathcal{F}$ is simply obtained by switching the $i$ and $j$th elements in ${\alpha}$, then $G(i)(\mathfrak{u})=G(j)(\mathfrak{w_0}\mathfrak{u})$ and hence
$$
\int_{U(n)}G(i)(\mathfrak{u})d\nu(\mathfrak{u})=\int_{U(n)}G(j)(\mathfrak{u})d\nu(\mathfrak{u}),\
$$ and \eqref{eq-integral-1} thus implies:
\begin{equation}\label{eq-integral-2}
   2\int_{U(n)}F(ij) (\mathfrak{u})d\nu(\mathfrak{u})=\int_{U(n)} G(i)(\mathfrak{u})d\nu(\mathfrak{u}) =\int_{U(n)} G(1)(\mathfrak{u})d\nu(\mathfrak{u})=K(p)
\end{equation}
 for some $K(p)$ depending only on $p$.  In particular, the LHS does not depend on $i, j$.  This establishes the Claim.

 For any $\mathfrak{u}\in U(n)$  we may write

\begin{equation}\label{eq-scalarcurvature-1}
 S(p)=\sum_{i\neq j} F(ij)(\mathfrak{u})+\sum_{i}G(i)(\mathfrak{u}).
\end{equation}

Hence
 \begin{equation}\label{eq-scalarcurvature-2}
 \begin{split}
  S(p)=&\int_{U(n)}\left(\sum_{i\neq j}F(ij)(\mathfrak{u}) +\sum_{i}G(i)(\mathfrak{u})\right)d\nu(\mathfrak{u})\\
  =&
   \frac{n(n+1)}2 K(p)
  \end{split}
 \end{equation}
The equality in \eqref{eq-scalarcurvature-2} may be interpreted as saying that on any K\"ahler  manifold
$S(p)$ is
 either an average of holomorphic sectional curvatures at $p$, a
known
fact from \cite{B} (also see \cite[p189]{Zheng}), or an average of orthogonal holomorphic
bisectional curvatures at $p$.  In particular, if either the holomorphic
sectional curvatures or orthogonal bisectional curvatures are positive
(nonnegative) respectively at $p$, then $S(p)$ is automatically positive
(nonnegative).

Now let  $a_{ij}$ be as in part (a) of the lemma.   Then
\begin{equation}\label{e9}
\begin{split}
  \int_{U(n)}\sum_{i, j=1}^n&F(ij)(\mathfrak{u})a_{ij}d\mu(\mathfrak{u})\\
&= \sum_{i\neq j}a_{ij}\int_{U(n)}F(ij)(\mathfrak{u}) d\mu(\mathfrak{u})+\sum_{i}a_{ii}\int_{U(n)}G(i)(\mathfrak{u}) d\mu(\mathfrak{u})\\
&=\frac {K(p)}2 (\sum_{i\neq j}a_{ij}+2\sum_{i}a_{ii}) .\\
\end{split}
\end{equation}
It follows that if \eqref{e-weaker1} is true for all unitary frames, then $K(p)\ge0$ as $\sum_{i\neq j}a_{ij}+2\sum_{i\neq i}a_{ii}>0$. If in addition \eqref{e-weaker1} is a strict inequality for some unitary frame, then
$$
\sum_{i j}F(ij)(\mathfrak{u})a_{ij}>0
$$ for some $\mathfrak{u}\in U(n)$. By continuity and the fact that \eqref{e-weaker1} is true for all unitary frames, we conclude that $K(p)>0$ and thus $S(p)>0$ by \eqref{eq-scalarcurvature-2}.

The proof of (b) is similar.

\end{proof}

\begin{proof}[Proof of Theorem \ref{t-scalarcurvature}] The first assertion of the theorem follows immediately from Lemma \ref{l-scalar}.   We now prove the other assertions in the Theorem in the case of nonnegative QOBC while the proof nonpositive QOBC is similar.

 We show $S(p)=0$ iff $R(V,\bar V,W,\bar W)=0$ for all unitary pairs $U,V$:  Suppose $R(V,\bar V,W,\bar W)=0$ for all unitary pair.  Then if $K(p)$ is as in the proof Lemma \ref{l-scalar} we have $K(p)=0$, and thus $S(p)=0$ by \eqref{eq-scalarcurvature-2}.
  Conversely, suppose $S(p)=0$.  By Lemma \ref{l-scalar}(a), \eqref{e1} is then an equality for all
 frames and choices of $\xi$.    Thus for any unitary frame $\{e_1,\dots,e_n\}$, if we let $a_{ij}=R_{i\bar ij\bar j}=a_{ji}$ then the function $f$ of $\xi=(\xi_1,\dots,\xi_n)\in R_\abb^n$ given by
$$
f(\xi)=\sum_{i\neq j}a_{ij}(\xi_i-\xi_j)^2
$$
is identically zero. Hence for all $k\neq l$,
$$
0=\frac{\partial^2f}{\partial \xi_k\partial\xi_l}=a_{kl}+a_{lk}=-2a_{kl}.
$$
Hence $R_{k\bar kl\bar l}=0$. Since $\{e_i\}$ was arbitrary it follows that for any unitary pair $V, W$, $R(V,\bar V,W,\bar W)=0$.
By  \eqref{eq-uv1}, we also have $R(V,\bar V,V,\bar V)+R(W,\bar W,W,\bar W)=0.$

 We now consider the case when $n\geq 3$. If $n\ge3$, then for any unitary pair $V, W$ we can find $U$ such that $U,V,W$ forms a unitary triple.
$$
0=R(U,\bar U,U,\bar U)+  R(W,\bar W,W,\bar W)=-2R(V,\bar V,V,\bar V).
$$
This completes the proof of the Theorem.
\end{proof}

In case $n=2$, the last assertion of the theorem may not be true.  This can be seen by Example 1.2 in \cite{GZ}. Namely, if we let $\Sigma$ be a compact Riemann surface with constant curvature -1 and let $\mathbb{CP}^1$ be the standard sphere. Then $\Sigma\times \mathbb{CP}^1$ has NQOBC. But the scalar curvature is zero everywhere.

\section{Irreducible and reducible manifolds with NQOBC}\label{structure}

We first use Theorem \ref{t-scalarcurvature} to prove that a locally irreducible K\"ahler  manifold with NQOBC must have positive first Chern class. This generalizes Theorem 2.1 in \cite{GZ}. Note that by Lemma \ref{l-keyfact}, $h^{1,1}(M)=1$ for such a manifold.

\begin{thm}\label{t-chern} Let $(M^n,\omega)$, $n\ge 2$ be compact K\"ahler manifold  with K\"ahler form $\omega$  and with quasi nonnegative OBC at all points. Suppose $h^{1,1}(M)=1$.  Then $c_1(M)=\l[\omega] $ for some $\l> 0$.
 \end{thm}
 \begin{proof} Since $h^{1,1}(M)=1$ we have $c_1(M)=\l[\omega]$ for some $\l$. By Theorem \ref{t-scalarcurvature}, the scalar curvature $S$ of $M$ is nonnegative.
 Hence
 $$
 0\le \int_{M}S\omega^m=\int_{M}\text{\rm Ric}\wedge \omega^{n-1}=\l\, \text{Vol}(M)
 $$
 and thus $\l\ge0$. We now show that $l>0$.  Suppose otherwise, and that $\l=0$ and thus $S=0$ everywhere.

  Suppose $n\ge 3$.  Then Theorem  \ref{t-scalarcurvature} implies $M$ is flat which is impossible since $h^{1,1}(M)=1$.
Suppose $n=2$.  At any point we may choose local coordinates $z^i$ such that  $\{\frac{\partial}{\partial z^i}\}$ are unitary and eigenvectors of $\text{\rm Ric}$. Let $a_i:=R_{i\bar i}$.  Then as $S=0$ we have $a_1=-a_2$ and we further calculate
 \begin{equation}
 \begin{split}
  \text{\rm Ric}\wedge \text{\rm Ric}=&\left(\sum_{i=1}^2 a_idz^i\wedge d\bar z^i \right)\wedge\left(\sum_{j=1}^2 a_jdz^j\wedge d\bar z^j \right)\\
  =&2a_1a_2 dz^1\wedge d\bar z^1\wedge dz^2\wedge d\bar z^2 \\
  =&-2a_1^2dz^1\wedge d\bar z^1\wedge dz^2\wedge d\bar z^2.
  \end{split}
  \end{equation}
Since $\l=0$,
$$
\int_{M}\text{\rm Ric}^2=0
$$
which implies $a_1=a_2=0$ everywhere.  On the other hand, Theorem  \ref{t-scalarcurvature} implies $R_{1\bar 12\bar2}=0$ and thus $0=a_i=R_{i\bar i}=R_{i\bar ii\bar i}$ for $i=1, 2$.    Thus $M$ is flat which is impossible is impossible since $h^{1,1}(M)=1.$
This concludes the proof of the Theorem by contradiction.  The above argument for $n=2$ can actually be used for all $n\geq 2$.
 \end{proof}

Next we want to study  the case when $(M, g)$ is possibly reducible.  We will obtain a partial classification of the possible deRham factorizations of the universal cover of $M$ (Theorem \ref{t-reduce}).   Let $\wt M$ be the universal cover of $M$ with covering map $\pi$. Let $$\wt M=\wt M_0\times\wt M_1\times\cdots \times \wt M_k$$ be the deRham decomposition of $\wt M$ where $\wt M_0$ is Euclidean with possible zero dimension  and $\wt M_i$, $i\ge 1$  (with positive dimension) are irreducible non-flat factors. Then the product of any subcollection of $\tilde M_i$'s still has NQOBC provided the product has dimension at least 2.  By Lemma \ref{l-keyfact},  $c_1(M)$ can be represented by a parallel harmonic (1,1) form $\eta$ so that $\text{\rm Ric}=\eta+\sqrt{-1}\partial\bar\partial f$ for some smooth function $f$ on $M$.  Then this pullls back to $\tilde M$ to give
\begin{equation}\label{eq-Ric}
   \widetilde \text{\rm Ric}=\widetilde \eta+\sqrt{-1}\partial\bar\partial \tilde f
\end{equation}
where $\tilde f=\pi^*f$ and $\widetilde \eta=\pi^* \eta$.  Moreover, by Corollary \ref{c1} $\widetilde \eta$ has the form
$$\widetilde \eta =\widetilde\eta_0\times\l_1\tilde \omega_1\times\dots \cdots \times \l_k\tilde\omega_k$$ where $\tilde \omega_i$ is the K\"ahler form of $\tilde M_i$, $\l_i$ are constants and $\widetilde\eta_0$ is parallel in $\wt M_0$.  In the following for a K\"ahler  manifold $(N,h)$ we use the notation $\Delta_Nu=h^{{i\jbar}}u_{{i\jbar}}$.
\begin{lem}\label{l-irr-1} We have $\widetilde\eta_0=0$.  Also, if $i\ge1$ and $\dim \wt M_i\ge2$ then $\l_i>0$.
\end{lem}
\begin{proof} Suppose $\l_1\le 0$ and $\dim \wt M_1\ge2$ say.  Let $$\tilde p=(\tilde p_0,\tilde p_1,\dots,\tilde p_k)\in \wt M_0\times\wt M_1\times\cdots \times \wt M_k=\wt M $$
be such that $f(\pi \tilde p)=\max_M f$. Consider the function $h(\tilde q)=\tilde f (\tilde p_0,\tilde q,\dots,\tilde p_k)$ on $\wt M_1$ for $\tilde q\in \wt M_1$. Then $h(\tilde p_1)=\max_{\tilde M_1} h$. On the other hand, $\widetilde\text{\rm Ric}_{(1)}=\l_1\tilde\omega_1+\sqrt{-1} \partial\bar\partial h$. By Theorem \ref{t-scalarcurvature}, the scalar curvature of $\wt M_1$ is nonnegative because $\dim \wt M_1\ge2$ and we conclude that
$$
\Delta_{\wt M_1} h\ge 0.
$$
By the strong maximum principle, it follows that $h$ is constant and thus $\widetilde\text{\rm Ric}_{(1)}=\l_1\omega_1$. By   Theorem \ref{t-scalarcurvature} again  we have $\l_1\ge0$, and so $\l_1=0$ and $\widetilde\text{\rm Ric}_{(1)}=0$. In particular, the scalar curvature of $\wt M_1$ is zero. By Theorem \ref{t-scalarcurvature} and the fact that $\widetilde\text{\rm Ric}^{(1)}=0$,  we conclude that $\wt M_1$ in fact has zero holomorphic sectional curvature everywhere.   This contradicts the fact that $\wt M_1$ is nonflat.

To prove that $\widetilde \eta_0=0$, first note that for the function $\tilde f_0 (\tilde q)=\tilde f(\tilde q,\tilde p_1,\dots,\tilde p_k)$ on $\wt M_0$ with $\tilde p_i$ being fixed:
$$
\Delta_{\wt M_0}\tilde f_0=C
$$
for some constant because $\wt M_0$ is flat and $\widetilde \eta_0$ is parallel. As before, we may conclude that $\tilde f_0$ is constant. Hence $\widetilde\eta_0=0$.
\end{proof}

\begin{lem}\label{l-irr-2} Suppose $M$ has NQOBC at $p$ and $M=M_1\times M_2$ with $\dim(M_1)=1$. Let $e_1,\dots,e_n$ be a unitary frame at $p$ such that $e_1$ is tangent to $M_1$ and $e_2,\dots,e_n$ are tangent to $M_2$. Then
$$
\sum_{j=2}^nR(e_2,\bar e_2,e_j,\bar e_j)\ge-R(e_1,\bar e_1,e_1,\bar e_1).
$$
\end{lem}
\begin{proof} Let $f_1=\frac{e_1+e_2}{\sqrt 2}$, $f_2=\frac{e_1-e_2}{\sqrt 2}$, $f_j=e_j$ for $j\ge 3$. Then
$$
R(f_1,\bar f_1,f_2,\bar f_2)=\frac14\left(R(e_1,\bar e_1,e_1,\bar e_1)+R(e_2,\bar e_2,e_2,\bar e_2)\right),
$$
and for $j\ge 3$
$$
R(f_1,\bar f_1, f_j,\bar f_j)=\frac12 R(e_2,\bar e_2,e_j,\bar e_j)=R(f_2,\bar f_2, f_j,\bar f_j).
$$
Let $\xi_1=1$, $\xi_2=-1$ and $\xi_j=0$ for $j\ge 3$, then
\begin{equation}
\begin{split}
0\le &\sum_{i<j}R(f_i,\bar f_i,f_j,\bar f_j)(\xi_i-\xi_j)^2\\
=&4R(f_1,\bar f_1,f_2,\bar f_2)+\sum_{j=3}^nR(f_1,\bar f_1, f_j,\bar f_j)+\sum_{j=3}^nR(f_2,\bar f_2, f_j,\bar f_j)\\
=&R(e_1,\bar e_1,e_1,\bar e_1)+R(e_2,\bar e_2,e_2,\bar e_2)+\sum_{j=3}^nR(e_2,\bar e_2,e_j,\bar e_j).
\end{split}
\end{equation}
From this the result follows.
\end{proof}

\begin{thm}\label{t-reduce} We have the following mutually exclusive cases:

\begin{itemize}
  \item [(i)]  $\dim\wt M_0\ge1$ and for each $i\ge 1$: $\wt M_i$ is compact with $\l_i>0$ and quasi positive Ricci curvature, i.e., the Ricci curvature of $\wt M_i$ is nonnegative and is positive at some point. In particular, if $\dim\wt M_i=1$ then $\wt M_i$ is biholomorphic to $ \mathbb{CP}^1$.
  \item [(ii)] $\dim\wt M_0=0$ and for each $i\ge 1$: $\wt M_i$ is compact with $\l_i>0$.  In particular, if $\dim\wt M_i=1$ then $\wt M_i$ is biholomorphic to $ \mathbb{CP}^1$ and all other factors have nonnegative (positive, respectively) Ricci curvature provided $\wt M_i$ has nonpositive (negative, respectively) Gauss curvature somewhere.
  \item [(iii)] $\dim\wt M_0=0$ and for some $i\ge 1$: $\wt M_i$ is biholomorphic either to $\Bbb C$ with $\l_i=0$ or to the unit disk in $\Bbb C$ with $\l_i<0$, and all other factors are compact and have $\l_j>0$ with  Ricci curvature bounded below by a positive constant.
\end{itemize}
\end{thm}
\begin{proof}   If $\dim\wt M_0\ge1$, by Lemma \ref{l-irr-2} and the fact that $\wt M_0$ is flat, we conclude that the Ricci curvature of $\wt M_i$ is nonnegative for all $i\ge 1$. Hence $M$ has nonnegative Ricci curvature. By \cite{CheegerGromoll}, $\wt M_i$ is compact for all $i\ge 1$. If $\dim \wt M_i\ge 2$, then $\l_i>0$ by Lemma \ref{l-irr-1}. If $\dim\wt M_i=1$, then it is biholomorphic to $\mathbb{CP}^1$ because it is compact and has nonnegative Gaussian curvature. Since $\text{\rm Ric}=\eta+\sqrt{-1}\partial\bar \partial f$, the Gauss-Bonnet theorem implies:
$$
4\pi=\int_{\wt M_i}K_i=\l_i\,\text{Vol}(\wt M_i)
$$
where $K_i$ is the Gaussian curvature of $\wt M_i$. Hence $\l_i>0$.  That $\tilde M_i$ has nonnegative Ricci curvature for each $i \geq 1$ follows from  Lemma \ref{l-irr-2}.  Now let $$\tilde p=(\tilde p_0,\tilde p_1,\dots,\tilde p_k)\in \wt M_0\times\wt M_1\times\cdots \times \wt M_k=\wt M $$
be such that $f(\pi \tilde p)=\min_M f$. Then at $\tilde p$,
$$
\widetilde \text{\rm Ric}\ge \widetilde \eta.
$$
from which it is easy to see that $\wt M_i$ has positive Ricci curvature at $\tilde p_i$.  Thus $\tilde M_i$ has quasi positive Ricci curvature for each $i \geq 1$. Hence we are in the situation of case (i) in the Theorem.

  Suppose $\dim\wt M_0=0$ and $\l_i>0$ for all $i\ge 1$.  By Yau's theorem on Calabi conjecture \cite{Yau1978}, there is a real-valued function $u$ on $M$ such that $g_{{i\jbar}}+u_{ijb}>0$ is K\"ahler  metric with Ricci form being $\eta$. Pulling this metric back by $\pi$ gives a K\"ahler  metric on $\wt M$ with Ricci form being $\widetilde \eta$ which is positive and is bounded away from 0, because $\dim\wt M_0=0$ and $\l_i>0$. Hence $\wt M$ is compact by Myer's theorem. Using Lemma \ref{l-irr-2}, we are thus in the situation of case (ii) in the Theorem.

  Suppose $\dim\wt M_0=0$ and $\l_1\le 0$, say. Then $\dim\wt M_1=1$ by Lemma \ref{l-irr-1}. The induced metric on $\wt M_1$ is of the form $e^{2\lambda}|dz|^2$ where $|dz|^2$ is the standard Euclidean metric on $\Bbb C$ or the unit disk in $\Bbb C$. Let $h(\tilde q)=\tilde f(\tilde q,\tilde p_2,\dots,\tilde p_k)$. Then the Gaussian curvature of $\wt M_1$ satisfies:
  $$
  K_1=\l_1+\Delta_{\wt M_1}h=\l_1+\frac12e^{-2\lambda}\Delta_0h
  $$
  by \eqref{eq-Ric} and where $\Delta_0$ is the Euclidean Laplacian. On the other hand, we also have
  $$
  K_1=-\Delta_{\wt M_1}\lambda=-\frac12e^{-2\lambda}\Delta_0\lambda
  $$
   Hence
  $$
 \l_1= -\frac12 e^{-2\lambda}\Delta_0\lambda(\lambda+h)
  $$
 which is just the Gaussian curvature of the metric $e^{2\lambda+2h}|dz|^2$.  Since $h$ is bounded, the metric $e^{2\lambda+2h}|dz|^2$ is complete. Hence $\l_1=0$ if and only if $\wt M_1$ is biholomorphic to $\Bbb C$ and $\l_1<0$ if and only if $\wt M_1$ is biholomorphic to the unit disk in $\Bbb C$.  In case $\l_1=0$, the Gaussian curvature of $\wt M_1$ must be negative somewhere. Otherwise, $\wt M_1$ must be flat by the proof of \cite[Theorem 3]{CheegerGromoll}. This is impossible, because $\dim\wt M_0=0$. In case $\l_1<0$, it is easy to see that the Gaussian curvature of $\wt M_1$ is negative somewhere. Using Lemma \ref{l-irr-2}, we are thus in the situation of case  (iii) in the Theorem.

\end{proof}
 \bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


Since we've been married we've been debating what to do with this site.
We've finally decided 6 months later that we should turn this into a small blog for both of us to use. What that means for you is that our site is currently under construction. There will be lots of changes coming soon, but for now please bare with us!


text:
Bristow Group Inc. Announces Expiration And Final Results Of Cash Tender Offer For Its 8.75% Senior Secured Notes Due 2023
HOUSTON, Sept. 10, 2019 /PRNewswire/ -- Bristow Group Inc. (OTC: BRSWQ) ("Bristow" or the "Company") announced today the expiration and final results of its previously announced cash tender offer (the "Tender Offer") to purchase the notes identified in the table below (the "Notes") up to an aggregate principal amount of the Notes that, together with accrued and unpaid interest to, but not including, the settlement date equals $75,000,000 (the "Aggregate Maximum Tender Amount").
The following table sets forth the approximate aggregate principal amount of the Notes that were tendered and not withdrawn at or prior to midnight, New York City time, at the end of the day on September 9, 2019 (the "Expiration Date") and the principal amount accepted for purchase.
Title of Notes
CUSIP/ISIN Numbers(1)
Tendered Notes
Accepted for
8.75% Senior Secured Notes due 2023
144A: 110394 AG8 / US110394AG86
Reg S: U1104M AB7 / USU1104MAB73
No representation is made as to the correctness or accuracy of the CUSIP/ISIN numbers listed in the table above or printed on the Notes. They are provided solely for the convenience of holders of the Notes.
As previously announced, the Company will only accept for purchase the Notes tendered in the Tender Offer subject to the Aggregate Maximum Tender Amount and proration. As the Tender Offer is oversubscribed, the Company has accepted for purchase tendered Notes on a prorated basis in the manner described in the Offer to Purchase. Notes tendered and not accepted for purchase will be promptly returned or credited to the applicable holders' account.
All conditions of the Tender Offer have either been satisfied or waived, and the settlement date with respect to the Tender Offer is expected to occur on September 11, 2019.
D.F. King & Co., Inc. acted as the tender agent and information agent for the Tender Offer.
This announcement does not constitute an offer to purchase Notes or a solicitation of an offer to sell Notes and shall not be deemed to be an offer to purchase or a solicitation of an offer to sell with respect to any securities of the Company or its subsidiaries.
Bristow Group Inc. is the world's leading industrial aviation service provider offering helicopter transportation, search and rescue (SAR) and aircraft support services to government and civil organizations worldwide. Bristow's strategically located global fleet supports operations in the North Sea, Nigeria and the U.S. Gulf of Mexico; as well as in most of the other major offshore oil and gas producing regions of the world, including Australia, Brazil, Canada, Guyana and Trinidad. Bristow provides SAR services to the private sector worldwide and to the public sector for all of the U.K. on behalf of the Maritime and Coastguard Agency.
This press release contains certain forward-looking information and forward-looking statements as defined in applicable securities laws (collectively referred to as "forward-looking statements"). Forward-looking statements include, but are not limited to, statements regarding the acceptance for purchase of any Notes validly tendered and the expected settlement date thereof.
Forward-looking statements involve known and unknown risks, uncertainties and other factors, which may cause the actual results, performance or achievements of the Company to be materially different from any future results, performance or achievements expressed or implied by the forward-looking statements. Factors that may cause actual results to vary include, but are not limited to, the risk factors as detailed from time to time in the Company's reports filed with the U.S. Securities and Exchange Commission.
Readers are cautioned against unduly relying on forward-looking statements. Forward-looking statements are made as of the date of the relevant document and, except as required by law, the Company undertakes no obligation to update publicly or otherwise revise any forward-looking statements, whether as a result of new information or future events or otherwise.
Global Media Relations
Director, Global Communications
Adam.morgan@bristowgroup.com
Reevemark
Paul Caminiti/Hugh Burns/Delia Cannan
BristowGroup@reevemark.com
Linda McNeill
Linda.mcneill@bristowgroup.com
View original content:http://www.prnewswire.com/news-releases/bristow-group-inc-announces-expiration-and-final-results-of-cash-tender-offer-for-its-8-75-senior-secured-notes-due-2023--300915550.html
SOURCE Bristow Group Inc.

redpajama_set_name: RedPajamaCommonCrawl


1. Alternate Nesting of Whitebellied Sea Eagle in Konkan, Maharashtra, India.
4. A Note on the Breeding of Crimson Sunbird (Aethopyga siparaja) in Ratnagiri Dist., Maharashtra, India.
10. Breeding Biology of Green Bee eater (Merops orientalis) in Chiplun, Ratnagiri district, Maharashtra, India.
11. First record of Blue-eared Kingfisher (Alcida meninting) from Konkan.
13. Breeding biology of Lesser Pied Kingfisher (Ceryle rudis) in Chiplun city, Ratnagiri district, Maharashtra, India.
16. First record of Grey-Headed Lapwing (Vanellus cinereus) from Konkan.
1. Breeding notes of six species of birds from Chiplun, Ratnagiri district, Maharashtra, India.
1. Status of White-bellied Sea Eagle in Ratnagiri district, Maharashtra, India.
Vol. 100 (1) (Apr. 2003), Page No. 148 - 149Vishwas (Bhau) Katdare, Rohan Lovalekar & Ameya Modak3. Status of White-bellied Sea Eagle in Sindhudurga district, Maharashtra, India.
Vol. 101 (2) (Aug. 2004), Page No. 314 - 316Vishwas (Bhau) Katdare, Ram Mone & Pramod Joshi4. Nesting of Terns on Vengurla Rocks, District Sindhudurga, Maharashtra, India.
Turtle Conservation in Konkan, Maharashtra, India.


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text:
Noble Research Institute President Sets Retirement Date
Sept. 19, 2018 01:45 PM
Bill Buckner announces his plans to retire as president and CEO of the Noble Research Institute during an employee meeting in Ardmore, Oklahoma, on Sept. 18, 2018. Buckner will retire at the end of 2018 after leading the organization for seven years.
Bill Buckner announced today that he will retire as president and chief executive officer of the Noble Research Institute at the end of 2018.
Buckner was selected as Noble's eighth president in 2011 and has led the organization for seven years. "There are no words to properly express the thankfulness I have for my time at Noble," Buckner said. "Every day at Noble, I have the opportunity to positively impact agriculture and contribute to something greater than myself. I am honored to be a part of the Noble legacy, and I'm excited to see the next chapter unfold for myself and the organization."
Russell "Rusty" Noble, a member of the Board of Director's executive committee and founder Lloyd Noble's grandson, hailed Buckner's tireless pursuit to advance agriculture. "When you meet Bill Buckner, you know one thing for sure: he loves agriculture. He is a passionate advocate for the sector," Noble said. "During his time at Noble, Bill worked tirelessly to promote soil health and create innovative new solutions that could open up new opportunities for farmers and ranchers. The Noble Foundation Board of Directors and all the employees offer a whole-hearted thanks to Bill for all of his work."
The Board of Directors has retained Spencer Stuart, a global executive search firm, to assist in identifying Buckner's successor. If a successor has not been identified by the end of 2018, Buckner will continue in his current role until his replacement can be found. "We are confident in the continuity of the Noble Research Institute's business activities," Noble said. "When a successor is identified, the transition will be seamless."
A Missouri native, Buckner earned a Bachelor of Science in agricultural economics from the University of Missouri-Columbia in 1979. He spent the next four decades working in the agriculture industry, including 18 years with Bayer. Buckner retired as North American regional head and president and CEO of Bayer CropScience, LP in 2011. Later that year, he agreed to lead Noble.
During Buckner's tenure as president and CEO, Noble experienced one of its most dramatic organizational shifts in its 73-year history. Originally known as The Samuel Roberts Noble Foundation, Noble separated its research and education operations from its philanthropic activities in 2017.
The organization's research, education and consultation activities continued forward under a new name, the Noble Research Institute, LLC, and became one of the country's first agricultural research organizations, a new type of 501(c)(3). The philanthropic activities, including grant-making and scholarship programs, of the original organization were placed in a new, private foundation, which carried the name traditionally associated with the organization's community giving, The Samuel Roberts Noble Foundation.
The Buckner era saw Noble expand its educational and community service efforts, dive deeper into cover crop research, establish the Land Stewardship Program, increase its national collaborations, and re-establish soil health as a core competency.
Noble brought together farmers, ranchers, soil scientists, economists, environmental interests, agribusinesses, NGOs and government agencies together to examine the role of soil health. Their work identified the need for a national organization to serve as a hub for measurement standards, economic data and coordinated research. Noble then launched the Soil Health Institute in 2015. The soil-focused nonprofit aims to safeguard and enhance the vitality and productivity of the soil through scientific research and advancement.
"Soil health is very much at the heart of the Noble Research Institute's mission," Buckner said. "Our founder, Lloyd Noble, lived through the Dust Bowl, and he established our organization to help combat the agricultural challenges our region faced. Launching the Soil Health Institute continues to bring Mr. Noble's vision to the national stage."
In recent years at the Noble Research Institute, Buckner has been spearheading the creation of the Ecosystem Services Market, an ambitious national effort to incentivize farmers and ranchers to improve soil health systems through a large-scale program that would finance, generate and sell ecosystem service credits from agricultural working lands.
"I am so proud of the tremendous accomplishments of the past seven years," Buckner said. "As our founder once said, 'No individual accomplishes anything worthwhile by his effort alone." I know that to be true. Any success I have experienced is a direct result of the tremendous men and women I've worked with every day here at Noble and our collaborators across the country."
Looking forward to 2019, Buckner said he anticipates spending more time with his family. While he is retiring from his organizational leadership role, Buckner will continue to serve the agriculture sector.
"Time with my children and grandchildren is first on the agenda," Buckner said. "However, I'm not done yet. The encore of my career will be spent helping shepherd the Ecosystem Services Market project on behalf of Noble as well as the Soil Health Institute, where I serve as chairman of the board. While I will no longer have the daily responsibilities of leading an organization, I will continue to be involved in advancing agriculture. That's something I will never stop doing."
AGCO Launches AGCO Agriculture Foundation Next
Corteva Agriscience™ Announces New Leadership Structure Previous


Lighthouse Leaders from the Academy of Leadership and Technology at Mound represented their school and showed off leadership schools at the Leader in Me Symposium Feb. 1 and 2.
Students showcased the seven habits of leadership to representatives from schools all over the world.
Students were chosen to greet and give guided tours to more than 60 visitors from parts of Texas, Oklahoma, Arkansas, Louisiana, Hawaii and Guatemala.
There was a question and answer session where students discussed how to use the seven habits, and some students presented a song about them.
"The students were showing some amazing communication skills," Penigian said. "They were really, really good ambassadors for our school."
The teachers benefitted from the experience by getting to visit College Street Elementary in Lewisville ISD.
Fifth grader Hannah Thompson was one of the students who gave student-led tours.
Thompson gave speeches about her school and what she does at Mound. During the tours, the data charts were one of the most popular attractions.
Thompson said it was an honor to be chosen to go to the Leader in Me Symposium.
"It told me that people saw what we were doing, and they thought we were leaders," she said.


text:
402 Reduction for Parolees Too
September 20, 2021 By Stephen Howard
Exciting news for anyone who has served prison time and then successfully completed parole! Utah's legislature in 2021 enacted amendments that now allow a 402 reduction after successfully completing either probation OR parole.
Utah Code 76-3-402 – Probation or Parole
Utah Code 76-3-402 now allows a person who has served their prison time and successfully completed parole to request a reduction in the level of the conviction. This can have HUGE benefits by reducing certain felony convictions down to the misdemeanor level.
The law previously distinguished between parole and probation, and allowed reductions only after successfully completing probation. This meant that someone who was never given a chance at probation or who violated probation and went to prison could not qualify for a reduction. Under the new law, if you serve your prison sentence and then successfully complete parole, you still qualify to request the reduction.
Instead of making technical distinctions between "probation" or "parole" the law now focuses on whether the person has made changes and turned their life around. Under the new law, when you successfully complete either probation OR parole, you can get that reduction if the court finds that it is "in the interest of justice."
Jurisdiction for the District Court
As a general rule, the district court loses jurisdiction over a criminal case if the defendant is sentenced to prison (not a "suspended" prison term). Under Utah's indeterminate sentencing system, the Board of Pardons and Parole then has jurisdiction to determine how much of a prison sentence will actually be served, whether the person will be given a chance on parole, what conditions will be imposed on parole, when parole will begin, and how long parole will last. But under the 2021 amendments, the legislature has explicitly given the original sentencing court jurisdiction over a request under Utah Code 76-3-402 to reduce the level of a conviction, even after serving prison time and on successful completion of parole.
Filed Under: 402 Reductions, Criminal Defense Attorneys Utah
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text:
Good Broadband Needs Good Maps
July 2019 •
By Trevor Jones |
OTELCO
Insufficient government maps, plant records and marketing are a few of the challenges to accessing accurate data.
If you're involved in advocating for better broadband access, chances are you've found yourself in an exasperating conversation about maps. It doesn't matter whether you're an ISP, a policymaker or a consumer – all three have been frustrated at some point by bad mapping, specifically the deficiency of accurate, geolocated data.
Maps play multiple roles in the business of building broadband networks. Those who build broadband infrastructure use maps to keep location records of their facilities and to identify the target market for a service in which location matters even more than it does in the retail world. More important, maps help determine where service is available commercially and where government funding may be necessary to stimulate deployment.
Map and Broadband Policy Problems
Government agencies spend lots of money mapping broadband to help guide their broadband policy decisions, but they get surprisingly poor results for the expense. The reasons are complex, rooted in issues such as granularity, availability of records, and the competitive motivations of private industry.
Most government data maps – and certainly the FCC's current broadband map – dissect the country at the census-block level. The census block seems like a reasonably small unit of measure when assessing the entire country, except that in most cases a block is considered "served" for funding decisions if a single address in the block is served – sometimes with as little as 10 Mbps / 1 Mbps. In such cases, hundreds of homes may lose out because one person on their block is at the very end of a cable TV line.
In addition, until now government maps have relied on service providers to self-report data. Often, service provider maps are only marginally more granular than government maps. This produces grossly inaccurate results.
Recently, when OTELCO looked at funding solutions for a broadband project in the unorganized township of Argyle, Maine, we found that the state broadband map indicated that Argyle already was covered by 25 Mbps / 3 Mbps broadband from the cable company. Yet, to our knowledge, the cable company did not serve Argyle.
Upon investigation, we found that because Argyle shares a ZIP code with nearby Old Town and the cable company's broadband availability data was at the ZIP-code level, users of both the state and cable company's websites would be told their address qualified, only to find that service was unavailable when they attempted to order it. Fortunately, the state program provides a process to prove an address is unserved and funds projects at a much more granular level than census blocks or ZIP codes. Unfortunately, however, this adds another step to the application process.
Simply building fiber down a road doesn't guarantee that every address can be served, creating a unique set of challenges for both service providers and municipalities as they work to sell and service the network.
The first challenge is what industry folk call "plant records." This is an inventory of available fibers, ONTs and other network components; details about whether components are aerial or buried; and a record of what components are assigned to specific customers. Understanding the location of available facilities in relation to customers is essential to fulfill orders, resolve problems and forecast the need for new facilities when populations grow and networks near capacity. Maintaining such records is a big challenge and generally requires specialized software, which may be costly.
Another challenge that comes from the street-by-street nature of fiber networks is marketing. Most digital marketing tools work at the ZIP-code level, and most broadcast media are even less granular. This requires service providers to use geotargeted marketing vehicles such as direct mail and door-to-door canvassing. To minimize aggravation and waste in the marketing process, providers will want to accurately plot street addresses in relation to fiber facilities to identify the addresses they wish to serve. Again, this is challenging even in the era of Google Maps because matching an address to a latitude and longitude, or "geocoding," is as much an art as a science.
Whether the broadband industry is funding, building, operating or marketing a broadband network, mapping will be an important tool to reach goals – and poorly done maps will be among the chief obstacles.
Trevor Jones is vice president of marketing, sales and customer service for OTELCO, which owns independent telephone companies in seven states and partners with several community networks in Massachusetts. Contact Trevor at trevor.jones@otelco.com.
Read what others have to say, and share your own thoughts with the community.
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Grounds for celebration? Anniversary of City and United's historic stadium-share deal
Bomb damage at Old Trafford led to a unique period in the history of Manchester's rival football teams
Chris Slater
Updated 17:04, 4 JUN 2014
A bomb-damaged Old Trafford in 1945. The disaster led to an historic ground-share between City and United
The thought of City and United sharing a ground is one that would send a shiver down the spine of most fans.
However 73 years ago today, it became a reality.
The now arch-rivals agreed a historic deal for United to move into City's Maine Road home after Old Trafford was badly damaged in a Second World War bombing raid
And it sparked the start of a unique period in Manchester footballing history.
Competitive fixtures were suspended during the conflict but a 'War league' was set up and ran until 1945. Remarkably, the rental contract for the ground in Moss Side was agreed in June 1941 by the club's two managers Wilf Wild and Walter Crickmer, with payments starting at just £10 per match, increasing on a sliding scale according to the gate.
And United notched up their record attendance whilst playing at the ground, with 83,260 watching their match with Arsenal in 1948, which remains a football league record.
City had earlier bettered that with 84,569 seeing an FA Cup tie with Stoke City in 1934.
Maine Road memories: 1940 to 1955
To this day that remains the record attendance at an English football match outside London.
The clubs had previously discussed sharing a ground in the 1920s but it never materialised.
And author Gary James, whose book Manchester: A Football History covers the period extensively, puts the huge crowds down to both sets of fans watching each other.
He said: "It is forgotten to some extent.
"United's support definitely increased during that period.
"You had a situation where fans would go to Maine Road every week to watch both teams.
"City fans and players would always want United to win if they were in a cup final for example and vice versa.
"And the attitude from the City was 'let's help United any way we can.'
Football researcher Gary James has written a book about United and City's post-war ground-share deal
"There were other options but they chose to share with City, which is indicative of the relationship between the clubs, which was very close."
The only time the share was said to cause problems, was on derby day when City players resorted to getting to the ground early to take charge of the home dressing room, even if United were officially the home team.
United remained in Moss Side until repairs to Old Trafford were finished in 1949.
Moss Side
Manchester United Transfer NewsManchester United transfer news RECAP Harry Maguire to Man Utd and Romelu Lukaku exit latestWe'll bring you all the latest Man United news and transfer gossip after their pre-season victory over Leeds United in Australia.


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Q: HTTP API deprected warnings UTF_8, Scheme, ThreadSafeClientConnManager I'm new java. I created a java project to call my webservice(Servlet,Tomcat6.0).
To use http i had added the following jar files
1. httpclient-4.2.1
2. httpcore-4.2.1
For methods(In Bold) below i'm getting type deprecated warnings and not able to run the project.
HTTP.**UTF_8**
aSchemeRegistryL.register(new **Scheme**("http", PlainSocketFactory.getSocketFactory(), 80));
ClientConnectionManager ccm = new **ThreadSafeClientConnManager**(aHttpParamsL, aSchemeRegistryL);

Help me to rectify it.

A: Most of the time - check your java docs - they'll suggest an alternative class.  See:
ThreadSafeClientManager
And it will suggest instead of ThreadSafeClientManager use: PoolingClientConnectionManager
PoolingClientConnectionManager
As for the Scheme class, if you look at the javadocs, the class itself isn't deprecated, but the constructor you are using is:
Scheme

A: For HTTP.UTF_8 the alternative is Consts.UTF_8. It's weird they don't mention that in the docs...

A: ThreadSafeClientConnManager class is deprecated, use PoolingClientConnectionManager and the constructor for Scheme you are using is deprecated too, use the other constructor found in that class.
As for the deprecated warnings, check that in Eclipse (I'm assuming you're using Eclipse, otherwise check for similar options in your IDE) that under Preferences->Java->Compiler->Errors/Warnings you don't have "Deprecated API" set to "Error" status.

redpajama_set_name: RedPajamaStackExchange


Inputting and organizing new submission materials (loss runs, ACORDS, underlying quotes, supplemental applications), into the Agency Information Manager system and Image Right filing system.
Where can I find a Sentinel Bookkeeping Founder resume example in Ashland, Oregon ?
This is an actual resume example of a Founder. LiveCareer's Resume Directory contains real resumes created by subscribers using LiveCareer's Resume Builder.


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Reviewing the state of the Tigers system with Fangraphs' Marc Hulet
We at Tigers Prospect Report have given you our impressions of the Tigers system. It's time to get the opinion of other prospect experts.
By hillmanchad Jul 19, 2013, 8:12am EDT
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Melvin Mercedes
I decided to try something different for this article. We at Tigers Prospect Report have given our views on prospects within the organization, but as Tiger fans, we can obviously be biased. With the recent completion of Jordan's top 30 list, I decided to ask a prospect expert from outside of our inner circle for his thoughts on the organization and the name that popped into my head immediately was Marc Hulet.
If you spend any time following prospects or visit Fangraphs.com, you most likely are familiar with Marc's work. For those of you that aren't familiar with Marc's work, he has been the lead prospect writer at Fangraphs since 2008. In addition to his work at Fangraphs, he also writes a weekly article that appears on Thursday's for MLB Trade Rumors and also contributes at Prospect Insider.
I value Marc's opinion and input and have for a long time. Prior to his work at Fangraphs, I contributed to a few projects that Marc worked on and he has been one of the people that I have relied on learning how to cover prospects. I have written about prospects since 2005 and Marc has truly been a mentor to me.
Marc was kind enough to field my questions. Below is what we discussed and his views on the state of the Tigers.
TPR: Could you give your opinion on the state of the Tigers system? Where would you rank them currently?
MH: Tigers system is in the bottom third in baseball. The club has leveraged its depth well in deals for players like Miguel Cabrera so you definitely have to give them credit for that. I'm a fan of Danry Vasquez, who probably doesn't get enough love because of the weak system. With that said, I really didn't like the Tigers' recent amateur draft.
TPR: You mentioned that you didn't like the Tigers draft. What did you see that struck you as odd? Was it the fact that the Tigers were pitcher heavy in the early rounds? Was that a mistake?
MH: It's never a mistake to focus on pitching. You have to be able to develop your own pitching if you're going to survive with the big boys. I didn't like the Tigers' draft because I don't see a ton of upside there. You've got some hard-throwing arms and some guys that could move quickly but the consensus seems to be that there were a lot of relievers chosen.
TPR: For the Tigers, there obviously is a lot of interest in the fan base in Nick Castellanos, especially with the MLB struggles this season of Andy Dirks. I think you may have seen him in the AFL. What are your thoughts on his bat and his defense?
MH: It's obviously a pretty new position for Castellanos so I'm giving him the benefit of the doubt. He's a long-term third or first baseman. Miguel Cabrera isn't going to be able to hold down the hot corner for that much longer so he'll eventually have a spot to move to. He lacks the prototypical power you look for from a third baseman or corner outfielder but he has a sweet swing and enough pop to make things interesting.
TPR: Relief pitching is a major concern for the Tigers right now. What arms do you look at in their system as possible future late inning options? Obviously Rondon is the guy that everyone looks at, but his secondary offerings have been inconsistent.
MH: Rondon is probably the best future option right now, but I agree that he's no sure thing. Melvin Mercedes is another guy to keep an eye on.
TPR: Austin Schotts is a kid that started this season in Low-A West Michigan after having a solid GCL season last year. He was a bit overmatched in the Midwest League and really has continued to struggle since being transferred to Connecticut. Were expectations maybe a bit to high on him coming into this season?
MH: I ranked Schotts fifth in the system prior to 2013 and I think his raw talent warrants that spot. However, the outfielder has been inexplicably rushed and it's Daniel Fields all over again. I respect the front office and their developmental staff but I don't like his handling whatsoever.
TPR: What would you say is the strength of the Tigers system? Are there any positions of excess?
MH: I think there is some outfield depth with the likes of Avisail Garcia, Danry Vasquez, Steve Moya and Nick Castellanos.
TPR: We have become huge fans of Devon Travis. He put up great numbers in West Michigan and plays a very consistent and solid second base. I know he is more of a high floor, low ceiling guy, but do you see MLB potential in him?
MH: He doesn't get a lot of love outside the organization and he's undersized but he does enough well that I could see him playing a role on a big league club, but probably off the bench.
TPR: How would you rate the development of Eugenio Suarez? Do you feel he is a possible option to replace Peralta at SS next season?
MH: His bat is still on the light side, so if the club is OK with a glove-only guy at shortstop then he could perhaps take over the job by mid-2014.
TPR: Outside of Rondon and Castellanos, who are some other prospects that maybe we overlook and should keep our eye on?
MH: I've already mentioned Danry Vasquez, Steve Moya and would also toss in Hernan Perez and Jake Thompson.
TPR: Jordan Gorosh, who is the lead prospect guru at Tigers Prospect Report recently completed his top 30 mid-season rankings. I was curious if you could take a look and see what your thoughts on it are?
MH: This was a very well done list… I couldn't find much fault with it… outside perhaps Schotts being a little too low. I wouldn't fault him for Detroit's mistake just yet.
More Roars
•Trade deadline 2013 coverage
•Tigers GIFS | On Twitter: @TigersGIFS
•Tigers minor league scores and coverage


Preparation of drawings for approval construction.
Producing standard and component drawings.
Registers RFI for each project logging onto drawings.
The successful candidate will have worked on multiple projects at once, and will be able to work unsupervised after project briefings.
Ability to visualise 2D objects in 3D.
General understanding of mechanical systems/engineering based products and components.
Accuracy and efficiency, and the ability to work on a number of projects at one time.
Previous experience with drawings essential.


LumoPro has added Phottix Odin 2.4GHz wireless TTL control to the LP180 flash, in the new LP180R.
LumoPro has unveiled an updated flash unit with built-in wireless TTL. The LumoPro LP180R builds on the features of the all-manual LP180 flash by integrating a Phottix Odin 2.4GHz radio receiver, allowing you to use it as a manual flash on-camera and a manual or TTL flash off-camera.
The Phottix Odin system, with the respective transmitter control unit (TCU), provides wireless remote control in manual or TTL modes, as well as high-speed sync and second-curtain sync, all over a radio connection that is longer-ranged and more reliable than optical triggering. The system also includes its own Odin receivers, letting you add other TTL flashes to your setup, the Mitros+ TTL flash and the Indra360 and Indra500 TTL studio flashes. If you just want built-in triggering and can spare the remote adjustment and TTL functions, the LP180R's Odin receiver is also compatible with the Phottix Strato-series manual flash triggers.
"We think people are going to be really fired up over the new features. The LP180R's built-in receiver adds a whole new level of fine-tuning to an already great flash," said LumoPro brand manager Janae Miller in a press release. "We're not going to ask you to train your dragon, because we're giving you complete control straight out of its lair." Whatever that means.
Apart from the addition of the radio receiver, the LumoPro LP180R is similar to the non-radio LP180. The hotshoe is still a single-pin so you cannot get on-camera TTL or other features beside basic triggering with any other camera or radio system. And it still has the great-on-paper-but-useless-in-practice gel holder for colour correction filters. See our full review of the original LumoPro LP180 here.
The LumoPro LP180R is available to order now for $229.99 from Midwest Photo Exchange. For more information, visit the official LumoPro product page.


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Home » Universal Attracts The Walking Dead this Summer!
Universal Attracts The Walking Dead this Summer!
Nicola Odeku March 14, 2016 No Comments 68 views 0 likes
The Walking Dead has been a part of Universal's seasonal Halloween Horror Nights for the past four years, featuring creepy mazes based on the hugely popular drama. The Golden-Globe nominated, Emmy Award-winning AMC zombie apocalypse television shows inclusion to the theme parks Halloween Horror Nights, has been credited for increasing attendance to the popular themed nights.
Fans will soon be able to walk among the undead all-year-round with a permanent The Walking Dead attraction coming to the Los Angeles theme park this summer! The announcement of The Walking Dead Attraction came Sunday on Talking Dead, the chat show that follows weekly episodes of The Walking Dead.
The stomach-churning walk-through experience will boast authentic walker make-up effects, sophisticated animatronic walkers, substantially more detailed set design and costuming, and highly recognizable props replicated from the series.
The Walking Dead is an American horror, drama television series developed by Frank Darabont, based on the graphic novel written by Robert Kirkman and published by Image Comics. The series follows a group of survivors, months and years after a zombie apocalypse, led by sheriff Rick Grimes, who travel in search of a safe and secure home. AMC has renewed the series each year because of its consistently increasing, Nielsen ratings, which have been unprecedentedly high for a cable series, including averaging the most 18- to 49-year-old viewers of any cable or broadcast television series during its third through (current) sixth seasons.
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Check out the promo for The Walking Dead Attraction, below.
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Tags: Horror Attractions, horror fan, the walking dead, Universal Studios, Universal Studios Hollywood, walking dead, Walking Dead Ride
Written by Nicola Odeku
Nicola Odeku is the co-founder of Wicked Horror. After managing a fashion and lifestyle magazine, in addition to writing her own published relationship articles, Nicola set up team with Arturo and followed her genuine love for horror. Also known as Miss Horror, Nicola is a big fan of hardcore gore, anything demonic, and disturbing cinema. Also cats. She looooves cats. Nicola enjoys traveling, sitting on a beach and eating cake.
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The ticker I inserted in a row goes beyond the right marging limit! have a look at the right size of the image, there is only a ticker in this image. It ends greatly at the left margin but it exceeds to the right! Suggestions please!
Well, my question was rather idiot! I fixed it playing with PADDING, MARGIN and BORDER in VisuaL Composer Row Settings! In the beginning it seemed to never work, probably was me not making the cache get empty!


During his three-year tenure at LSU, Dr. Bond lectured to medical students, residents, and fellows, and was the Medical Program Director for the LSU Sports Medicine Fellowship Program. In 2007, Dr. Bond left the world of medical academia behind to pursue the inception of TotalCare Health & Wellness Medical Center, thus enabling him to set his own professional mission and goals at the head of his career. Today, Dr. Bond continues to lecture to physician colleagues, as well as, medical students and residents. Dr. Bond has served as a Team Physician for all levels of competition, amoungst a multitude of sporting activities – from football and baseball, to Triathlons and Marathons. Most recently, in prior years, he served as a Team Physician for both the University of Louisiana at Lafayette and also the Louisiana IceGators. At this time, Dr.Bond continues to serve as a Team Physician the U.S. Ski & Snowboarding Team, and is the exclusive Physician provider of the Regenexx suite of Regenerative Orthopedic stem-cell procedures.
Dr. Bond was the first physician to perform US-guided Regenerative Injection Procedures in the state of Louisiana. He has since performed thousands of these injections, making him far and away the most experienced practitioner of Regenerative Injections in Louisiana and the southern region. Dr. Bond is excited and proud to be a part of this approach to Orthopedic/Musculoskeletal pain, in which his experience in Wellness Medicine is laid at the forefront of each procedure.


The Notary Signing Agent Course is given completely online, which means you can take it at your leisure and even spend extra time on those more important sections.
You'll have up to 90 days to complete the course, and when you're done you'll receive your Notary Signing Agent certificate by mail.
Note: You must have an active Notary commission to act as a Notary Signing Agent.
The Notary Signing Agent course is not a requirement to become a Notary. But if you want to bring out the true earning potential of your Notary stamp, it's the best online tool you'll ever use.
Register now and get started today.
Note: This course is written for general applicability. Notaries should be aware that some states have specific regulations concerning loan signings. It is your responsibility as a Notary to understand the laws and requirements of your state.


The final rough drawing for a spread in an forthcoming picture book, mentioned earlier.
I've been told that the layout for an upcoming book of mine, authored by an Ontario writer, should be in my inbox in the next month. I'm actually quite excited to see it; the artwork was shipped off in the summer and I've forgotten what most of it looks like.


Destin, Fla., Jan. 23 – On Thursday, March 28, the Destin and Greater Fort Walton Beach chambers of commerce will present a new event to the community – the Destin-Fort Walton Beach Taste & Business Showcase. This event will highlight the best of the area's businesses, from restaurants and breweries to retail, home, and personal and professional services. Guests will enjoy samples of delicious foods and drinks, promotional items, discounts, and goodies from local businesses.
The event takes place at the Emerald Coast Convention Center from 4 – 7 p.m., and the public is invited to attend at no charge. Local businesses that would like to participate may contact Bryana Bahmer at the Destin Chamber at (850) 837-6241 or bryana@DestinChamber.com.


This universal controller functions as a transmitter, receiver and/or repeater. The number of connected sensors does not depend volume as typical in industry standard applications, but is determined by the topological conditional of a project location.
With its compact form factor, the ParQSense Base Station is designed to require minimal installation efforts. Power requirements are low-voltage with communication through various possible communication methods such as GSM/3G, Ethernet, RS-485, etc. depending on availability.
The ParQSense Base Station can be installed on street lighting poles, building roof-tops/sides etc. Thousands of ParQSense Smart Sensors can communicate wirelessly to a single ParQSense Base Station. The cloud based Q-Free HUB is the central software engine for ParQSense monitoring projects. Access to the Q-Free HUB can be achieved through ParQ, a web-based user portal, or through the ParQSense API.


MINING an unprotected gap between Federal and state insurance regulations, unscrupulous operators have sold virtually worthless health-care policies to dozens of small-business groups desperate for low-cost coverage. Government officials say hundreds of thousands of workers and their families have been saddled with tens of millions of dollars in unpaid medical bills.
After the Federal indictments last week of a group in North Carolina charged with selling worthless health insurance, Federal and state officials urged prospective buyers to be wary of bargain-priced offers. Buyers should determine what group is ultimately responsible for paying medical bills and ask state insurance officials if that group is an authorized insurer.
At least 22 million working people have been left without insurance since the late 1980's, as health costs have soared and insurers have narrowed coverage. Some employers rejected by insurers bought health coverage through trade associations. Others joined outside groups known as multiple-employer welfare arrangements, or MEWA's.
The multiple-employer groups are so loosely regulated that Federal and state officials do not even know how many there are. Most are believed to be legitimate; Michigan, Florida and Virginia have adopted rules to see that medical bills are paid. But in at least 70 cases Federal officials say the organizers appear to have illegally diverted members' health contributions.
When Elizabeth Hanford Dole took office as Secretary of Labor last year, she promised to improve Federal procedures for dealing with the muliple-employer groups. But state insurance officials saythe Labor Department has been slow to make good on Mrs. Dole's pledge. At hearings last spring before the Investigations subcommittee of the Senate Governmental Affairs Committee. Senator Sam Nunn, Democrat of Georgia, the subcommittee chairman, asked Labor officials about reports of months-long delays in replying to state requests for information about dubious health plans. The department says response times are shorter now.
The multiple-employer plans typically claim exemption from state rules under Federal pension laws, which exempt single employers that are self-insured. The states say few, if any, multiple-employer plans are exempt, so the confusion persists.
When Secretary Dole resigned in October, Bob Dole, her husband and the Senate Republican leader, introduced a Labor Department bill to require Federal registration of multiple-employer health plans. Department officials say it would be up to Lynn Martin, the Illinois Congresswoman nominated to succeed Mrs. Dole, to pursue the issue.
Jim Long, the North Carolina insurance commissioner, said in a telephone interview that the stateswere still "not getting the support we need" from Washington. But Mr. Long, who is president of the National Association of Insurance Commissioners, said there had been "great cooperation" by state and Federal authorities in preparing the Federal fraud indictments filed last week against a group based in Charlotte, N.C., which sold health policies to small employer groups.
Like many insurance buyers, these employers apparently had trouble with the fine print. "Many employers haven't got a clue what they bought," said Patricia Petersen, Washington's deputy state insurance commissioner. "They just know something is wrong when they are not getting their claims paid."
Ms. Petersen said one group tried to escape state rules by selling $20-a-month "associate memberships" in a small union, the International Union of Petroleum and Industrial Workers. Union health plans are federally regulated. Ms. Peterson said her state outlawed multiple employer groups as "unauthorized insurers" unless they could qualify as fully documented insurance companies.
But in Michigan, which regulates the groups, Jay Matthew, president of Benefit Actuaries in Grand Rapids, said his company administered six health plans for 25,000 employees. He said the plans were required to keep reserves of 35 percent of the previous year's health claims or 25 percent of member contributions, whichever was greater. Small employer groups like the Michigan Auto Parts Association typically disburse 19 percent to 22 percent of contributions each year for health bills, he said.
"Multiple-employer arrangements are a real tough issue for us, but we wouldn't want to close that option off," said Karen Brigham, health policy manager with the United States Chamber of Commerce. "Obviously one way to reduce the cost of health insurance for small businesses is pooling arrangements."


MIDDLETOWN — An overturned work van on the Garden State Parkway slowed the end of Tuesday morning's commute.
The blue Mercedes-Benz Sprinter van with "Tony Boloney's" painted on the side flipped on the northbound local lanes near Exit 114 for Red Hill Road in Middletown, according to New Jersey State Police spokesman Lawrence Peele.
Tony Boloney's owner Mike Hauke told New Jersey 101.5 the van was taking the ingredients from their famous taco pizza, which sells for $80, from Atlantic City to Hoboken. Hauke said the vehicle flipped three times, but the driver was OK. About $10,000 worth of toppings, however, was lost, including taco shells, brisket and cheese.
The food made quite a mess inside the van, according to Hauke.
Taco pizza is only served on Tuesday at the Hoboken location and Hauke said despite the crash it will be available as usual. The pizza has been featured on "Live with Kelly" and Conan O'Brien's show.


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ACT Test Centers and Dates in Serbia
Your search found 1 match. The following is the full list of ACT testing locations in Serbia among which you can pick one to take the exam. Please know that on the test day, test takers can use any 4-function, scientific, or graphing calculator. On the table below, you can also find all test dates through 2019.
2019-2020 ACT Test Dates in Serbia
ACT Test Centers in Serbia
Belgrade International Academic Center 872280
More about Serbia
IAMHIGHER: Latest statistics of population in the country of Serbia, including languages spoken, urban population, birth rate, fertility rate and life expectancy for both men and women.
After the declaration of independence of Montenegro (June 3, 2006), Serbia declared itself the legal successor of Serbia and Montenegro in accordance with the constitutional charter of the State Union by a parliamentary resolution of June 5, 2006. The referendum of 28./29. 10. New constitution adopted in 2006, which characterizes Serbia as a parliamentary republic and in which Kosovo is designated as an integral part of the country. The head of state of Serbia is the president, who is directly elected for a period of 5 years; his term of office is limited to 2 terms. He is the commander in chief of the armed forces and can declare a state of emergency at the suggestion of the government. The highest legislative body is the People's Assembly, whose 250 deputies are elected for 4 years. The executive rests with the government under the chairmanship of the Prime Minister, which is responsible to Parliament. As an autonomous province within Serbia, Vojvodina has self-government rights and a provincial assembly with 120 members and its own government. The Constitutional Court of Serbia has 15 members with a nine-year mandate, 5 members are elected by the parliament, and 5 more are appointed by the President and the Supreme Court of Cassation. Check 3rjewelry to see Serbia Travel Guide.
Serbia has a widely diversified party system, which is characterized by frequent splits and mergers. For the elections, the various parties usually form joint electoral lists or alliances. Important parties are the national-conservative Serbian Progressive Party (SNS; formed in 2008 as a split from the Serbian Radical Party), the Socialist Party of Serbia (SPS; successor organization to the Union of Communists of Serbia [SKS] formed in 1990), and the right-wing nationalist Serbian Radical Party (SRS; founded 1991), the social democratic Democratic Party (DS; founded in 1990; merged with the Democratic Center [DC, founded in 1996] in 2005), the protest movement »Enough« (DJB; founded in 2014), the right-wing conservative Democratic Party of Serbia (DSS; founded in 1992),
Around a third of the workers are unionized. The largest umbrella organizations are the Confederation of Independent Trade Unions of Serbia (SSSS), whose history dates back to 1893, and the United Industry Union Confederation (UGS), founded in 1991. The largest single union is the Self-Employed Metalworkers Union of Serbia (SSMS).
After the end of the international community of Serbia and Montenegro, the Army of Serbia was formed in 2006. The total strength of the armed forces (2011: suspension of compulsory military service) is around 28,000 men. The army is a. divided into 4 mechanized brigades, 1 mixed artillery brigade and a special brigade, the air force and air defense in 1 fighter and 1 fighter-bomber squadron.
The Republic of Serbia is divided into central Serbia and the autonomous province of Vojvodina (since February 17, 2007), on the lower administrative level in 29 districts (okruži), 22 of them in central Serbia and 7 in Vojvodina, as well as on the municipal level in 23 cities (gradovi) and 153 municipalities (opštine).
The 2006 constitution guarantees the independence of the judiciary. The Supreme Judicial Council, the supreme body of the judiciary, has 11 members. The highest instances of jurisdiction are the Supreme Court of Cassation, the Administrative Court and the High Criminal Court. Further judicial instances are appellate courts, district and municipal courts. The former Yugoslav federal laws (adjusted) continue to apply to the law of obligations and property law; Added to this are the Serbian Inheritance Act (1995) and the Family Act (2005). There are new regulations in particular in economically relevant areas, e.g. B. The Law on Business Companies (2004) and Laws on Register Liens and Finance Leases (2003). Criminal procedural law (2001) and civil procedural law (2004) have been fundamentally renewed.
There is general compulsory schooling for children between the ages of 7 and 15, which is covered by attending eight-year elementary school (four-year lower and upper grades). A structured system of secondary schools follows the primary level. Vocational and technical secondary schools are attended more often than general secondary schools (grammar schools), which lead to higher education entrance qualifications in four years. The universities in Belgrade (founded in 1863), Kragujevac (founded in 1976), Niš (founded in 1965) and Novi Sad (founded in 1960), the Belgrade University of the Arts (founded in 1973) and over 100 higher universities form the basis of higher education educational institutions comparable to German universities of applied sciences.
SAT Test Centers and Dates in Serbia
ACT Test Centers and Dates in Montserrat
ACT Test Centers and Dates in Aruba
Category: ACT Test Centers Tags: Study in Serbia
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Kuit, princesse, Loender, more kuit… lately slowly but surely Dutch beers surface that are based on a rediscovered beer type from the past. It's not yet a complete renaissance, but all in all there's more and more Dutch beer history on the market, and then I'm not even mentioning the home brews. Time for a recap.
One of the advantages of writing about lost beers, is that you don't have to brew them all yourself. More and more other people do it for you. That's why last Wednesday on a terrace in Utrecht I could taste a Breda Ersatz beer made by an enthusiastic home brewer, Laurens Potman, who had used the original 1934 recipe. And was it any good? The result, based on things like skimmed milk, starch and cane sugar, was at least better than could be expected from the ingredients. It didn't actually taste a lot like beer, but more like… Rivella. Maybe that's not such a strange outcome, after all this Swiss soft drink is made from whey, a leftover product from cheese making.
Dutch home brewers, including myself, are doing nice experiments, but there's more: historical beer that's actually available commercially. Some based on recipes found by me, but by others too. Of course since 2014 there already has been the medieval kuit beer, redeveloped by the Campaign Dutch Beer Styles, followed by a Luiks beer in 2015. The latter is currently available as Luikse Vechter at Van Moll (Eindhoven) and Tulpomania by Jopen (Haarlem). Of the kuit I was able to sample another variant last weekend at the Utrecht Beer brewer's Festival, made by De Dikke from Wijk bij Duurstede. It has been based on a 1495 ordonnance and was already elected as the festival's most appreciated beer last year.
A nice upcoming trend when it comes to historic Dutch beers is the princesse beer. Last year the spicy princesse by Oranjeboom (United Dutch Breweries) was released in two colour variants, and various home brewers gave it a go. Even the renowned Anchor Brewery from San Francisco made a one-off Wild Princess for the Amsterdam In de Wildeman pub, even though for some unknown reason they decided to add a substantial amount of anise.
This year a princesse came on draught at Poesiat & Kater in Amsterdam, also known for their wonderful Van Vollenhoven Stout. And there was a very well-made one-time princesse beer by Jopen in collaboration with the Stadsbrouwhuis ('City brew house') in Leiden, called 'Behouden Vaart', and I wouldn't mind tasting that one more often. The Behouden Vaart ('Safe journey') contains rye and at Oranjeboom and Poesiat & Kater they don't mind using wheat. Not every beer history purist agrees with that, but all in all a very interesting palette of new 'old' spicy beers with a nineteenth-century flavour is emerging.
Another temporary historical beer by Jopen is the koninginne ('queen's') beer they brewed lately under the name of 'Enjoy the ride'. The reason was the similarly called charity bike ride in Nijmegen, that uses the poppy as its symbol. I like that: in Nijmegen they wanted a poppy beer, and instead of just making one up, the Jopen people thought of the historical koninginne beer with poppies. This beer is a one-off too, just like the Jopen Dark Gerste no. 1 from 2015, brewed after a barley beer recipe from 1911 found by Amsterdam-based English beer historian Ron Pattinson.
Finally, Groningen is in the game with their kluin beer, which has been reconstructed several times since the eighties. At this moment it's available in a version by Toon van den Broek and by brewery Vechter. In my book there's a reconstruction of my own. And a golden oldie classic is the Mestreechs Aajt by Gulpener, which is becoming more available.


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Q: Unboundedness of Quadratic Function I was reading about Trust Region Methods for solving Nonlinear Optimization problems are came across this statement in my notes:
If the quadratic approximation to the function $f(x)$ which is given by $m(x+p)=f(x)+\nabla f(x)^Tp + 0.5p^T\nabla^2 f(x) p$ has $\nabla^2f(x)$ which is not positive definite, then the quadratic model is unbounded below.
I can't seem to appreciate why this is true.
I tried looking at it from the following perspective but that leads me to a dead end.
\begin{align}
m(x+p)&=f(x)+\nabla f(x)^Tp + 0.5p^T\nabla^2 f(x) p\\\nabla^2f(x)&\not \succ 0\\
\exists y\;s.t.\;\\
y^T\nabla^2f(x)\;y&<0 \\
Let \;p&=y\\
m(x+y)&=\underbrace{f(x)}_{Known}+\underbrace{\nabla f(x)^Ty}_{?????} + \underbrace{0.5y^T\nabla^2 f(x) y}_{<0}\\
\end{align}

A: I disagree; it's the case where $Hf$ has a negative eigenvalue that is trivial.
Let $v$ be an eigenvector with negative eigenvalue. Then
$$m(x+vt) = f(x) + t\left[\nabla f(x) \cdot v\right] + t^2 \lambda/2$$
is a polynomial with negative leading coefficient, and so is unbounded below as $t\to\infty$.
When $Hf$ is positive semi-definite, it's trickier, and I don't think your statement is universally true. In particular, I claim that $m$ is bounded below if and only if all eigenvectors of $Hf$ with zero eigenvalue are orthogonal to $\nabla f(x)$.
Only if direction: let $v$ be an eigenvector with eigenvalue 0, and with $v\cdot \nabla f(x) \neq 0$. Since the sign of $v$ is arbitrary we can assume WLOG that $v\cdot \nabla f(x) < 0$, and then
$$m(x+vt) = f(x) + t \left[\nabla f(x)\cdot v\right] + 0 \to -\infty$$
as $t\to\infty$.
If direction: Since $m$ is a polynomial, it suffices to show that $\lim_{t\to\infty} m(x+vt)$ is bounded below for all $v$. Expand $v$ in the eigenbasis of $Hf(x)$, with $v_0$ the component in the span of the eigenvectors with zero eigenvalue, and $v_+$ the component in the span of the eigenvectors with positive eigenvalue. Then
$$m(x+vt) = f(x) + t\left[\nabla f(x)\cdot v_+\right] + t^2v_+^THf(x)v_+/2.$$
If $v_+=0$, $m(x+tv)$ is constant; otherwise, $m(x+vt)$ has positive leading coefficient, and $m(x+vt)\to\infty$ as $t\to\infty$.