text:
\section{Introduction and Results}

The {Barbero--Immirzi connection} is used in Loop Quantum Gravity (LQG) to deal with gravity in lorentzian signature $(3,1)$.

We briefly recall the original definition given by Barbero \cite{Barbero} and Immirzi \cite{Immirzi}. Fix a $4$-dimensional orientable lorentzian manifold $M$ with an Einstein metric $g$ and an embedded $3$-dimensional submanifold $S$ that is spacelike, so that the pull-back metric $h = t^* g$ along the embedding map $t \colon S \hookrightarrow M$ is positive definite.
Choose local coordinates $\left\{ s^A \right\}_{A = 1,2,3}$ on $S$ and a basis $\left\{ \rho_i \right\}_{i=1,2,3}$ for ${\mathfrak{su}}(2)$. An ${\mathfrak{su}}(2)$-valued {\it Barbero--Immirzi (BI) connection} is a $1$-form on $S$ with local coefficients
\begin{equation}
A^k_A(\beta) = \Frac[1/2] \epsilon_{ij}{}^k \, \Gamma^{ij}_A + \beta K^k_A, \quad \beta \in \mathbb{R}
\label{biCoeff}
\end{equation}
where $\epsilon_{ij}{}^k$ is the totally antisymmetric Levi--Civita symbol, $\Gamma^{ij}_A$ are the Christoffel symbols of $h$ and $K^k_A$ are related to the coefficients of the Weingarten operator of $S$ (see \cite{KobaNu}, vol.\ 2, p.\ 14). The real number $\beta$ is the {\it Barbero--Immirzi (BI) parameter} and for each choice of $\beta$ one defines a different BI connection $A(\beta) = A^k_A(\beta) \ ds^A \otimes \rho_k$. In LQG the BI parameter has to be fixed by experimental data, so that one can speak of {\it the} BI connection $A$.

There are several issues with the construction above. First, we want to use the BI connection to reformulate General Relativity, therefore we want to define $A$ without fixing an Einstein metric $g$ a priori. Second, the definition of the BI connection $A$ in terms of $\Gamma$ and $K$ looks arbitrary and there is no apparent explanation for the BI parameter $\beta$. Third, the construction is carried out on a submanifold $S$ of $M$. As Samuel pointed out in \cite{Samuel} one would like to relate the BI connection $A$ to a connection defined on all of $M$.

The aim of this work is to give a geometrically well-defined method for building BI connections, thus solving all the issues outlined above. We will work in the general case of an $(n+1)$-dimensional lorentzian manifold $M$ and discuss the special case of interest $n = 3$.

Ultimately, one is interested in reformulating Einstein's General Relativity as a variational theory for the BI connection $A$ instead of the metric $g$, which is a form more suitable for quantization and the starting point of LQG (see \cite{RovelliBook}). This is one of the main motivations behind the present work.

In Section 2 we review spin structures and spin frames, in this framework we can reformulate the problem without fixing a specific lorentzian metric $g$ on $M$.

In Section 3 we tackle the problem of reducing a principal $\hbox{Spin}(n,1)$-bundle over $M$ to a $\hbox{Spin}(n)$-bundle. The main idea is to define a $\hbox{Spin}(n)$-connection $A$ out of a $\hbox{Spin}(n,1)$-connection $\omega$. In general, reduction of a principal $G$-bundle over $M$ to a closed Lie subgroup $H \subset G$ is subject to topological obstructions. In the case of interest we prove that there are no obstructions, so that a reduction always exists.

In Section 4 we use the concept of reductive pair to explicitly define the BI connection $A$ on the $\hbox{Spin}(n)$-bundle out of a connection $\omega$ on the $\hbox{Spin}(n,1)$-bundle. We prove that $(\hbox{Spin}(n,1), \hbox{Spin}(n))$ is a reductive pair for $n \geq 3$ and that the existence of the BI parameter $\beta$ is related to unique properties of the case $n = 3$. The remarkable result is that there is a {\it unique} way of defining the coefficients of $A$, which is precisely the one given in \eqref{biCoeff}.

\section{Spin Structures and Spin Frames}

Let $M$ be an $m$-dimensional real, smooth manifold and fix a signature $\eta=(r, s)$ on it, i.e.\ with $r+s=m$.
It is well known that all manifolds allow riemannian metrics, i.e.\ in signature $(m,0)$ (see \cite{KobaNu}, vol.\ 1, p.\ 116), whereas a pseudo-riemannian metric of signature $(r,s)$ exists if and only if the tangent bundle $TM$ splits as $TM=T_r\oplus T_s$ where $T_r, T_s$ are two subbundles of rank $r$ and $s$, respectively (see \cite{Steenrod}, p.\ 207).

It is also well known that whenever the manifold $M$ is orientable and admits a global metric of signature $(r,s)$, its frame bundle $L(M)$ admits a reduction to the group $\hbox{SO}(r,s) \subset \hbox{GL}(m)$.
Any such reduction is in fact associated to a metric $g$ of signature $(r,s)$, that is, any $\hbox{SO}(r,s)$-reduction of the frame bundle $L(M)$ is given by a positive orthonormal frame bundle $SO(M, g)$ for some pseudo-riemannian metric $g$ on $M$ (see \cite{KobaNu}, vol.\ 1, p.\ 60).

A further topological obstruction is required for having spin structures on an orientable pseudo-riemannian manifold $(M, g)$.
A {\it spin structure} (see, for instance, \cite{Lawson}, p.\ 80) is a pair $(P, \hat \ell)$, where $P$ is a principal bundle on $M$ with group $\hbox{Spin}(r,s)$ (namely, the connected component of the identity in the spin group of the right signature) and $\hat\ell \colon P \to SO(M, g)$ is a principal bundle morphism with respect to the double cover $\ell \colon \hbox{Spin}(r,s) \to \hbox{SO}(r,s)$ which projects over the identity on $M$, i.e.\
\begin{equation}
\begindc{\commdiag}[10]
\obj(100,30)[M2]{$M$}
\obj(170,30)[M3]{$M$}
\obj(100,80)[P2]{$ P$}
\obj(170,80)[P3]{$SO(M,g)$}
\mor(100,29)(170,29){}[\atleft, \solidline] \mor(100,31)(170,31){}[\atleft, \solidline]
\mor{P2}{M2}{$p$}
\mor{P3}{M3}{$\pi$}
\mor{P2}{P3}{$\hat\ell$}
\enddc
\end{equation}
The obstruction to the existence of a spin structure over $(M, g)$ is the vanishing of the second Stiefel--Whitney class of $M$; see \cite{Antonsen}.

There is also a more general and somehow more geometric framework for spin structures, based on the universal covering of the frame bundle. It is interesting to briefly review it and also to compare to it; see \cite{Dabrowski}.

Whenever the manifold $M$ is orientable we can reduce the structure group of $L(M)$ to $\hbox{GL}_+(m)$, the connected component of the identity in $\hbox{GL}(m)$, to get the bundle of positively oriented frames $L_+(M)$. The double cover of $L_+(M)$ is denoted by $\widetilde L_+(M)$ and is a principal bundle with structure group $\widetilde{\hbox{GL}}_+(m)$, the double cover of $\hbox{GL}_+(m)$, the covering map is $\tilde \pi \colon \widetilde L_+(M) \to L_+(M)$. A spin structure $\Sigma$ is then defined to be a $\hbox{Spin}(r,s)$-subbundle of $\tilde L_+(M)$.

Fixing a metric $g$ of signature $(r,s)$ on $M$ defines the subbundle $SO(M, g)$ of positive $g$-orthonormal frames. Consider the embedding $\iota \colon SO(M, g) \hookrightarrow L_+(M)$ and define the principal $\hbox{Spin}(r,s)$-bundle $\Sigma_g := \iota^* SO(M,g)$. We have the commutative diagram
\begin{equation}
\begindc{\commdiag}[10]
\obj(100, 30)[SO]{$SO(M,g)$}
\obj(100, 80)[S]{$\Sigma_g$}
\obj(170, 30)[L]{$L_+(M)$}
\obj(170, 80)[LL]{$\widetilde L_+(M)$}
\mor{SO}{L}{$\iota$}[\atleft, \injectionarrow]
\mor{S}{SO}{}
\mor{S}{LL}{$\tilde \iota$}[\atleft, \injectionarrow]
\mor{LL}{L}{$\tilde \pi$}
\enddc
\end{equation}
so that $(\Sigma_g, \tilde \pi \circ \tilde\iota)$ is a standard spin structure.

In this more general framework, one does not need to fix a metric a priori. On the contrary we can consider all spin structures at once. In particular, one can lift a diffeomorphism of $M$ to $L(M)$ by using its natural bundle structure, and to $\widetilde L_+(M)$ by using the covering properties. As a result, in this framework one can systematically define isomorphisms of spin structures, possibly with different underlying metrics and define natural-like properties. However, all spin structures end up in the same framework, even those with non-isomorphic $\hbox{Spin}(r,s)$-bundles. This generalization is not necessary whenever we have a fixed metric $g$, while it is needed when discussing properties of all spin structures at once (e.g.\ in a variational setting).

According to this standard setting, we give the definition of spin manifold.

\begin{definition}
A real, smooth, orientable manifold $M$ which allows global metrics of signature $(r,s)$ is called {\it spin manifold} if it admits a spin structure $(P, \hat \ell)$.
\end{definition}

\begin{remark}
From now on we shall fix a signature $(r,s)$ and consider a spin manifold $M$. If $M$ admits a metric $g$, we will also call $(M,g)$ a riemannian manifold, regardless of the signature $(r,s)$.
\end{remark}

Although this setting is very well suited to discuss properties on a fixed riemannian structure $(M, g)$, it is considerably less suited to discuss the gravitational theory in which the metric structure is unknown until one solves the Einstein equations (possibly coupled to other equations). In other words, one needs an equivalent formulation of spin structures, though in the category of spin manifolds rather than in the category of riemannian manifolds.

\begin{definition}
Consider the double cover $\ell \colon \hbox{Spin}(r,s) \rightarrow \hbox{SO}(r,s)$ and the canonical embedding $\iota_{r,s} \colon \hbox{SO}(r,s) \hookrightarrow \hbox{GL}(m)$. Given a spin manifold $M$ in a signature $(r,s)$, a {\it spin frame} is principal bundle map $e \colon P \to L(M)$ with respect to the group morphism $\iota_{r,s} \circ \ell$, from some principal $\hbox{Spin}(r,s)$-bundle $P$ on $M$ into the frame bundle $L(M)$. The commutative diagram is
\begin{equation}
\begindc{\commdiag}[10]
\obj(100, 30)[M1]{$M$}
\obj(100, 80)[P]{$P$}
\obj(170, 30)[M2]{$M$}
\obj(170, 80)[L]{$L(M)$}
\mor(100,29)(170,29){}[\atleft, \solidline] \mor(100,31)(170,31){}[\atleft, \solidline]
\mor{P}{M1}{$p$}
\mor{P}{L}{$e$}
\mor{L}{M2}{$\pi$}
\enddc
\end{equation}
\end{definition}

Denote by $\eta$ the canonical bilinear form on $\mathbb{R}^m$ which is represented in the canonical basis by the diagonal matrix of signature $(r,s)$, namely $\hbox{diag}(\eta)= \hbox{diag}(-1, \dots, -1, 1, \dots, 1)$ with $r$ pluses and $s$ minuses. Fix a trivialization on $P$ over an open set $U\subset M$, given by a local section $\sigma$. We have
\begin{equation}
e(\sigma(x)) = E_x \in L_x(M), \quad \forall x \in U
\end{equation}
Notice that, by equivariance, the values $E_x$ completely determine $e$ on $p^{-1}(U) \simeq U \times \hbox{Spin}(r,s)$.

Since $E_x$ is an isomorphism $E_x \colon \mathbb{R}^m \rightarrow T_xM$ then, for any $X,Y \in T_xM$, the metric $g$ associated to $e$ at $x \in M$ is
\begin{equation}
g_x(X,Y) = \eta((E_x)^{-1}(X), (E_x)^{-1}(Y))
\end{equation}
Reasoning as in \cite{KobaNu}, vol.\ 1, p.\ 60, example 5.7, shows that $g$ is a well-defined smooth, symmetric, non-degenerate metric tensor of signature $(r,s)$ on all $M$.

Notice how a spin frame $e \colon P \rightarrow L(M)$ disentangles the metric structure $g$ from the spin structure: by considering only $P$ we can reduce the structure group without needing to fix a metric $g$. In this sense, the spin frame is a richer and more flexible structure than the metric $g$.

\section{Reduction of $\hbox{Spin}(n,1)$-bundles to $\hbox{Spin}(n)$}

From this section onwards we will fix $\dim M = m = n + 1$ and restrict to lorentzian signature, i.e.\ $(r,s) = (n,1)$. In the case of interest $n = 3$ we want to define an $\hbox{SU}(2)$-connection out of a $\hbox{Spin}(3,1)$-connection on $P$. This suggests that we should investigate the $\hbox{SU}(2)$-reductions of a $\hbox{Spin}(3,1)$-bundle $P$, we recall the definition:

\begin{definition}
Consider a principal $G$-bundle $p \colon P \rightarrow M$ and a closed Lie subgroup $H \subset G$. An {\it $H$-reduction of $P$} is a principal $H$-bundle $q \colon Q \rightarrow M$ with a principal bundle embedding
\begin{equation}
\begindc{\commdiag}[10]
\obj(30,30)[M1]{$M$}
\obj(100,30)[M2]{$M$}
\obj(30,80)[Q]{$Q$}
\obj(100,80)[P]{$P$}

\mor{M1}{M2}{}[\atleft, \solidline] \mor(30,33)(100,33){}[\atleft, \solidline]
\mor{Q}{M1}{$q$}
\mor{P}{M2}{$p$}
\mor{Q}{P}{}[\atleft, \injectionarrow]
\enddc
\end{equation}
When $p \colon P \rightarrow M$ admits an $H$-reduction we say that the structure group $G$ of the bundle can be reduced to $H$.
\end{definition}

Since $\hbox{SU}(2) \simeq \hbox{Spin}(3)$, the generalization to arbitrary dimension is to study $\hbox{Spin}(n)$-reductions of a principal $\hbox{Spin}(n,1)$-bundle $P$. We will also restrict to the case $n \geq 3$ since for $n = 1,2$ the spin groups are not the universal coverings of the respective special orthogonal groups.

The main result of this section is that for lorenztian spin manifolds the reduction from $\hbox{Spin}(n,1)$ to $\hbox{Spin}(n)$ is always possible.

\begin{theorem}\label{th1}
For any lorentzian spin manifold $M$ and any principal $\hbox{Spin}(n,1)$-bundle $p \colon P \rightarrow M$, there always exists a $\hbox{Spin}(n)$-reduction.
\end{theorem}

The proof relies on a definition and three lemmas.

\begin{definition}
For a given principal $G$-bundle $p \colon P \rightarrow M$ and any closed Lie subgroup $H \subset G$, one has the (right) coset $G/H$ which is the set of elements $[g]$
\begin{equation}
[g] = \left\{ g' \in G : g' = gh \mbox{ for some } h \in H\right\}
\end{equation}

The group $G$ acts on the left on $G/H$ by
\begin{equation}
\lambda_H \colon G \times G/H \longrightarrow G/H \colon (g, [g']) \longmapsto [gg']
\end{equation}
The bundle associated to $P$ via this left action has $G/H$ as standard fiber, it is called {\it $H$-coset bundle of $P$} and is denoted by $p_H \colon P_H \rightarrow M$.
\end{definition}

\begin{lemma} (see, \cite{KobaNu}, vol.\ 1, p.\ 57)
The structure group of a principal $G$-bundle $p \colon P \rightarrow M$ is reducible to a closed Lie subgroup $H \subset G$ if and only if the associated $H$-coset bundle $p_H \colon P_H \rightarrow M$ admits a global cross section.
\end{lemma}

\begin{lemma} (see, \cite{KobaNu}, vol.\ 1, p.\ 58)
Consider a fiber bundle $\pi \colon B \rightarrow M$ with standard fiber $F$. If the fiber $F$ is diffeomorphic to $\mathbb{R}^k$, for some $k \in \mathbb{N}$, then the bundle admits a global cross section.
\end{lemma}

We want to prove \hyperref[th1]{Theorem 1} by showing that the coset $\hbox{Spin}(n,1)/\hbox{Spin}(n)$ is diffeomorphic to $\mathbb{R}^n$, we do so using the third lemma.

\begin{lemma} (see \cite{Steenrod}, p.\ 30)
Consider a smooth manifold $X$ and a transitive Lie group action $\lambda \colon G \times X \rightarrow X$, fix a point $x_0 \in X$. Let $H \subset G$ be the stabilizer subgroup of $x_0$, which is a closed Lie subgroup. If the map
\begin{equation}
\lambda_{x_0} \colon G \rightarrow X \colon g \mapsto g \cdot x_0
\end{equation}
is an open map, then
\begin{equation}
G/H \simeq X
\end{equation}
is a diffeomorphism.
\end{lemma}

\begin{theorem}
The right coset ${}{\hbox{Spin}(n,1)}/{\hbox{Spin}(n)}$ is diffeomorphic to the euclidean space $\mathbb{R}^n$.
\end{theorem}
\begin{proof}
We prove this fact using the previous lemma. Consider the submanifold of $\mathbb{R}^{n+1}$
\begin{equation}
X = \left\{ (t, {\bf x}) \in \mathbb{R}^{n+1} : -t^2 + |{\bf x}|^2 = -1 , t > 0\right\}
\end{equation}
with the subspace topology. The action of $\hbox{Spin}(n,1)$ is given by
\begin{equation}
S \cdot x = \ell(S) x, \quad \forall S \in \hbox{Spin}(n,1)
\end{equation}
where $\ell \colon \hbox{Spin}(n,1) \rightarrow \hbox{SO}(n,1)$ is the two-to-one covering map and $\hbox{SO}(n,1)$ acts through its fundamental representation.

By definition, an element $\bar S \in \hbox{Spin}(n) \subset \hbox{Spin}(n,1)$ acts only on the ${\bf x}$ part of a vector $v = (t, {\bf x})$. Therefore $\hbox{Spin}(n)$ fixes the point $x_0 = (1, {\bf 0})$. Since $\hbox{Spin}(n,1)$ is a path-connected Lie group and since
\begin{equation}
X = \hbox{Spin}(n,1) \cdot x_0
\end{equation}
any open neighborhood $U$ of the identity $\mathbb{I} \in \hbox{Spin}(n,1)$ is mapped into an open neighborhood of $x_0 \in X$. The hypotheses of the lemma above are all verified and we have
\begin{equation}
\hbox{Spin}(n,1)/{\hbox{Spin}(n)} \simeq X \simeq \mathbb{R}^n
\end{equation}
\end{proof}

\section{Barbero--Immirzi Connections through Reductive Pairs}

The main result of the previous section implies that for any $\hbox{Spin}(n,1)$-bundle {$p \colon P \rightarrow M$} over a lorentzian spin manifold $M$ there exists a reduction to a $\hbox{Spin}(n)$-bundle $q \colon Q \rightarrow M$. The commutative diagram is
\begin{equation}
\begindc{\commdiag}[10]
\obj(30,30)[M1]{$M$}
\obj(100,30)[M2]{$M$}
\obj(30,80)[Q]{$Q$}
\obj(100,80)[P]{$P$}

\mor{M1}{M2}{}[\atleft, \solidline] \mor(30,33)(100,33){}[\atleft, \solidline]
\mor{Q}{M1}{$q$}
\mor{P}{M2}{$p$}
\mor{Q}{P}{}[\atleft, \injectionarrow]
\enddc
\end{equation}

In this section we define the geometric framework for building BI connections on all $M$: given a principal connection $\omega$ on a $\hbox{Spin}(n,1)$-bundle $p \colon P \rightarrow M$ and a $\hbox{Spin}(n)$-reduction $q \colon Q \rightarrow M$, we define a unique principal connection $A$ on $Q$ out of $\omega$.

The main result relies on the definition of reductive pair of Lie groups.

\begin{definition}
(see \cite{KobaNu} vol.\ 1, p.\ 83) A pair of Lie groups $(G, H)$, with Lie algebras ${\mathfrak{g}}$ and ${\mathfrak{h}} \subset {\mathfrak{g}}$ respectively, is a {\it reductive pair} if:

\begin{enumerate}[(i)]
\item $H \subset G$ is a closed Lie subgroup;

\item there exists a {\it reductive splitting}, i.e.~a vector space splitting
\begin{equation}
{{\mathfrak{g}}} = {{\mathfrak{h}}} \oplus {{\mathfrak{m}}} \quad \hbox{and} \quad \hbox{Ad}_G(H) \, {{\mathfrak{m}}} \subseteq {{\mathfrak{m}}}
\end{equation}
\end{enumerate}
where $\hbox{Ad}_G(H)$ is the adjoint representation of $G$ restricted to $H$.
\end{definition}

\begin{remark}
Notice that the vector subspace ${\mathfrak{m}} \subset {\mathfrak{g}}$ in the previous definition need not be a Lie subalgebra.
\end{remark}

The definition of reductive pair is particularly useful due to the following theorem.

\begin{theorem}
(see, \cite{KobaNu}, vol.\ 1, p.\ 83)
Let $(G, H)$ be a reductive pair with Lie algebras ${\mathfrak{g}}, {\mathfrak{h}}$ respectively. Consider a principal $G$-bundle $p \colon P \rightarrow M$ with an $H$-reduction $q \colon Q \rightarrow M$. Choose a connection $1$-form $\omega$ on $P$, call $A$ its ${\mathfrak{h}}$-component and $K$ its ${\mathfrak{m}}$-component.

With this splitting the restriction the $A|_Q$ is a connection $1$-form on $Q$ while the restriction $K|_Q$ is a tensorial $1$-form of type $(\hbox{Ad}_G, {\mathfrak{m}})$.
\end{theorem}

\begin{remark}
For a fixed reductive splitting ${\mathfrak{g}} = {\mathfrak{h}} \oplus {\mathfrak{m}}$ as in the previous theorem, we have a one-to-one correspondence between connections $\omega$ on $P$ and pairs $(A,K)$ on $Q$.
\end{remark}

We can now state the fundamental result of this work.

\begin{theorem}\label{th4}
If $n >  3$,  $(\hbox{Spin}(n,1), \hbox{Spin}(n))$ has a unique reductive splitting
\begin{equation}
\mathfrak{spin}(n,1) = \mathfrak{spin}(n) \oplus {\mathfrak{m}}_0,
\end{equation}

If $n =3 $, we have a $1$-parameter family of reductive splittings
\begin{equation}
\mathfrak{spin}(3,1) = \mathfrak{spin}(3) \oplus {\mathfrak{m}}_\beta
\label{BIReductiveSplitting}
\end{equation}
The parameter $\beta$ is called the {\it Barbero--Immirzi parameter} of the splitting.
\end{theorem}

In the proof the algebras $\mathfrak{spin}(n,1)$ and $\mathfrak{spin}(n)$ are seen as subalgebras of the Clifford algebra, let us recall the basic definitions and fix our notations.

Consider $V = \mathbb{R}^m$ equipped with a non-degenerate, symmetric bilinear form $\eta$ of signature $(r,s)$ and the relative Clifford algebra $\hbox{Cl}(V,\eta)$. We have an injective map $V \hookrightarrow \hbox{Cl}(V,\eta)$, for any vector $v \in V$ we denote with the boldface ${\bf v}$ its image in the Clifford algebra.

The metric $\eta$ induces a non-degenerate bilinear form $Q$ on $\hbox{Cl}(V,\eta)$, if $\eta$ has signature $(r,s)$ we define the spin group $\hbox{Spin}(r,s)$ as the group generated by products of elements $S \in \hbox{Cl}(V,\eta)$ such that $Q(S) = 1$. One can the prove that the adjoint action
\begin{equation}
{\bf v} \longmapsto \hbox{Ad}(S) {\bf v} = S \, {\bf v} \, S^{-1}, \quad \forall v \in V
\end{equation}
is an $\eta$-orthogonal transformation for any $S \in \hbox{Spin}(r,s)$.

Denote by $\ell \colon \hbox{Spin}(r,s) \rightarrow \hbox{SO}(r,s)$ the double covering of the spin group into the (connected component of the identity of the) special orthogonal group. The action of $\hbox{SO}(r,s)$ on $V$ is assumed to be the standard representation. Since $V$ is mapped injectively into $\hbox{Cl}(V,\eta)$ and since $\hbox{Ad}(S) = \hbox{Ad}(-S)$, we have that
\begin{equation}
\hbox{Ad}(S) {\bf v} = \ell(S) v, \quad \forall v \in V
\end{equation}

Now choose an $\eta$-orthogonal basis for $V$ of vectors $\left\{ e_1, \dots, e_m \right\}$, the spin algebra $\mathfrak{spin}(r,s)$ is the real vector subspace of $\hbox{Cl}(V, \eta)$ generated by products of two elements, that is
\begin{equation}
\mathfrak{spin}(r,s) = \langle {\bf e}_a {\bf e}_b : a,b = 1, \dots, m \mbox{ and } a \neq b \rangle
\end{equation}
Recall that ${\bf e}_a {\bf e}_b = -{\bf e}_b {\bf e}_a$ so that only elements with $a < b$ form a basis. The adjoint representation of $\hbox{Spin}(r,s)$ on its algebra $\mathfrak{spin}(r,s)$ has the simple form
\begin{equation}
\hbox{Ad}(S) \colon {\bf e}_a {\bf e}_b \longmapsto S \, {\bf e}_a {\bf e}_b \, S^{-1}, \quad \forall S \in \hbox{Spin}(r,s)
\end{equation}
where the products are in $\hbox{Cl}(V,\eta)$.

Let us now return to the case of lorentzian signature $(n,1)$. In $V = \mathbb{R}^{n+1}$ it is customary to denote the $\eta$-orthogonal basis elements by $e_0, e_1, \dots, e_n$, that is
\begin{equation}
\begin{aligned}
&\eta(e_0, e_0) = -1
	\\
&\eta(e_i, e_j) = \delta_{ij}, \quad i,j = 1, \dots, n
	\\
&\eta(e_0, e_i) = 0, \quad i = 1,\dots, n
\end{aligned}
\end{equation}
Define $W = \mathbb{R}^n$ as the span of the last $n$-vectors $e_1, \dots, e_n$ and denote by $\delta$ the restricted metric, which is euclidean. We have an isometric embedding $\iota \colon (W,\delta) \rightarrow (V,\eta)$. From here onwards, latin indices from the beginning of the alphabet (e.g.\ $a,b,c$) will range from $0$ to $n$ while latin indices from the middle of the alphabeth (e.g.\ $i,j,k$) will range from $1$ to $n$.

By functoriality of Clifford algebras (see \cite{Lawson}) we have an injection $\hbox{Spin}(n) \hookrightarrow \hbox{Spin}(n,1)$ which is characterised by
\begin{equation}
\hbox{Ad}(S) {\bf e}_0 = {\bf e}_0, \quad \forall S \in \hbox{Spin}(n)
\end{equation}

Similarly, we have an injection $\mathfrak{spin}(n) \hookrightarrow \mathfrak{spin}(n,1)$ through which the generators ${\bf e}_i {\bf e}_j$ are mapped to themselves. Therefore the basis for $\mathfrak{spin}(n)$ can be completed to a basis for $\mathfrak{spin}(n,1)$ by adding the vectors $\left\{ {\bf e}_0 {\bf e}_k \right\}_{k = 1, \dots, n}$.

We can now prove \hyperref[th4]{Theorem 4}.

\begin{proof}
Let $U:={\mathfrak{spin}(n,1)} /{\mathfrak{spin}(n)}$. There is a short exact sequence
\begin{equation}
\begindc{\commdiag}[10]
\obj(-30,30)[o1]{$0$}
\obj(20,30)[h]{$\mathfrak{spin}(n)$}
\obj(80,30)[g]{$\mathfrak{spin}(n,1)$}
\obj(140,30)[m]{$U$}
\obj(180,30)[o2]{$0$}

\mor{o1}{h}{}
\mor{h}{g}{}
\mor{g}{m}{}
\mor{m}{o2}{}
\enddc
\end{equation}
Identifying $U$ with $W$ via the isomorphism $ [{\bf e} _0 {\bf e}_k] :={\bf e} _0 {\bf e}_k + \mathfrak{spin} (n) \mapsto e_k$, a splitting   corresponds to a linear injection $\phi \colon W \hookrightarrow \mathfrak{spin}( n,1 ) $ such that $\phi (W)  \oplus  \mathfrak{spin} (n) =\mathfrak{spin}(n+1) $.
The most general choice of $\phi$  is
\begin{equation}
   \phi({e}_k) = {\bf e}_0 {\bf e}_k + \psi (e_k )
\end{equation}
for some linear map $\psi \colon W \rightarrow \mathfrak{spin}(n)$.  Setting ${\mathfrak{m}} := \phi(W)$, we obtain the splitting
\begin{equation}
\mathfrak{spin}(n,1) = \mathfrak{spin}(n) \oplus {\mathfrak{m}}.
\end{equation}
We now determine under which conditions the vector space ${\mathfrak{m}}$ is $\hbox{Ad}(\hbox{Spin}(n))$-invariant. The space ${\mathfrak{m}}$ is spanned by vectors of the form ${\bf e} _0 {\bf e} _k  + \psi (e_k ) $. For any $S \in \hbox{Spin}(n)$ the element
\begin{equation}
\begin{aligned}
\hbox{Ad}(S)({\bf e} _0 {\bf e} _k  + \psi ({e} _k ))
&=S {\bf e } _0 S^{-1} \, S {\bf e} _k S^{-1} + (\hbox{Ad} (S) \circ \psi) ({e} _k )
	\\
&={\bf e } _0 \, \hbox{Ad} (S) ({\bf e} _k ) + (\hbox{Ad} (S) \circ \psi) ({e} _k )
\end{aligned}
\end{equation}
belongs to ${\mathfrak{m}} $ if and only if
\begin{equation}
(\hbox{Ad} (S) \circ \psi )({e} _k ) = (\psi \circ \hbox{Ad} (S) )({\bf e } _k ).
\end{equation}
Since $\hbox{Ad} (S) ({\bf e } _k )  = \ell (S) ({ e } _k ) $,
the splitting is reductive if and only if
$\hbox{Ad}(S) \circ \psi = \psi \circ \ell(S)$,
that is if $\psi$ is an intertwiner between the $\hbox{Ad}(S) \in \hbox{End}(\mathfrak{spin}(n))$ and $\ell(S) \in \hbox{End}(W)$ representations of $\hbox{Spin}(n)$.

If $n \neq 4$ the group $\hbox{SO}(n) $ acts irreducibly on $W$. The adjoint representation of $\mathfrak{spin}(n) $ on itself is irreducible since ${\mathfrak{so}}(n) =\mathfrak{spin} (n) $ is a simple Lie algebra, hence the adjoint representation of $\hbox{Spin} (n) $ on $\mathfrak{spin} (n) $ is also irreducible. Given that both $\hbox{Ad}$ and $\ell$ are irreducible representations of $\hbox{Spin}(n)$, by Schur's Lemma $\psi$ is either the null map or an isomorphism. In the latter case we must have $n = \dim W = \dim( \mathfrak{spin}(n) )= n(n-1)/2$
which is possible only if  $n = 0$ or $n = 3$.

For $n =3 $ we  identify $W $ with $  \mathfrak{spin} (3) \simeq {\mathfrak{su}}(2)$ via the map  $ {e} _k  \mapsto \tau_k$, where $\left\{\tau_k\right\}_{k = 1,2,3}$ are the Pauli matrices multiplied by the immaginary unit $i$. Then $\psi$ becomes an equivariant map with respect to the adjoint representation of $\hbox{Spin} (3) $ on $\mathfrak{spin} (3) $. Since any endomorphism of an odd dimensional real vector space has a real eigenvalue, it follows by Schur's lemma that $\psi$ is a constant multiple of the identity, i.e.\
\begin{equation}
\psi({e}_k) = \beta \, \tau_k, \quad \beta \in \mathbb{R}
\end{equation}

For $n = 4$ we use the fact that $\hbox{Spin}(4) \simeq \hbox{SU}(2) \times \hbox{SU}(2)$ and $\mathfrak{spin}(4) \simeq {\mathfrak{su}}(2) \oplus {\mathfrak{su}}(2)$. Each of the two copies of ${\mathfrak{su}}(2)$ is a $3$-dimensional invariant subspace for the adjoint action of $\hbox{Spin}(4)$ on $\mathfrak{spin}(4)$. On the other hand, as shown in \cite{Lawson}, the fundamental representation $\ell$ has two invariant subspaces which are both $2$-dimensional.

Since $\ker \psi$ is an invariant subspace for $\ell$, its dimension must be $0, 2$ or $4$. By the rank-nullity theorem $\dim W = \dim (\ker \psi) + \dim (\im \psi)$, so that $\dim (\im \psi)$ is $4, 2$ or $0$. However $\im \psi$ is an invariant subspace for $\hbox{Ad}$, so that its dimension is $0, 3$ or $6$. The only possibiity therefore is $\dim(\ker \psi) = 4$ which implies that $\psi = 0$.

The proof is thus complete.
\end{proof}

\begin{corollary}\label{cor1}
For $n > 3$ we have the unique reductive splitting $\mathfrak{spin}(n,1) = \mathfrak{spin}(n) \oplus {\mathfrak{m}}_0$ with
\begin{equation}
{\mathfrak{m}}_0 = \langle {\bf e}_0 {\bf e}_k : k = 1, \dots, n \rangle
\end{equation}

For $n = 3$ we have a one-parameter family of reductive splittings $\mathfrak{spin}(3,1) = {\mathfrak{su}}(2) \oplus {\mathfrak{m}}_\beta$ with
\begin{equation}
{\mathfrak{m}}_\beta = \langle \sigma_k + \beta \tau_k : k = 1,2,3\rangle, \quad \beta \in \mathbb{R}
\end{equation}
where $\left\{\sigma_k\right\}_{k=1,2,3}$ are the Pauli matrices and $\tau_k = i \sigma_k$.
\end{corollary}

\

We close this section by recovering the expression for the coefficient of a BI connection $A$ in the case $n = 3$. Choose a basis $\left\{ e_a \right\}_{a = 0,\dots 3}$ in $\mathbb{R}^4$ and define
\begin{equation}
T_{ab} = \Frac[1/2] e_a \wedge e_b
\end{equation}
Since $\Lambda^2 \mathbb{R}^4 \simeq {\mathfrak{so}}(3,1) \simeq \mathfrak{spin}(3,1)$ (see \cite{Lawson} for details), the bivectors $\left\{ T_{ab} \right\}_{a < b}$ form a basis for the spin algebra $\mathfrak{spin}(3,1)$. For any vector $v \in \mathbb{R}^4$ we have that $T_{ab}$ acts as
\begin{equation}
T_{ab}(v) = \Frac[1/2] (\eta(e_a, v) e_b - \eta(e_b, v) e_a)
\end{equation}
By looking at all commutators $[T_{ab}, T_{cd}]$ one finds that
\begin{equation}
T_{0k} = -\Frac[1/4] \sigma_k \quad \mbox{and} \quad T_{ij} = \Frac[1/4] \epsilon_{ij}{}^k \tau_k
\end{equation}

Let us now return to the $\hbox{Spin}(3,1)$-bundle $p \colon P \rightarrow M$ with a $\hbox{SU}(2)$-reduction $q \colon Q \rightarrow M$.
\begin{equation}
\begindc{\commdiag}[10]
\obj(30,30)[M1]{$M$}
\obj(100,30)[M2]{$M$}
\obj(30,80)[Q]{$Q$}
\obj(100,80)[P]{$P$}

\mor{M1}{M2}{}[\atleft, \solidline] \mor(30,33)(100,33){}[\atleft, \solidline]
\mor{Q}{M1}{$q$}
\mor{P}{M2}{$p$}
\mor{Q}{P}{}[\atleft, \injectionarrow]
\enddc
\end{equation}
Let us also fix a BI parameter $\beta$ so that we have a reductive splitting $\mathfrak{spin}(3,1) = {\mathfrak{su}}(2) \oplus {\mathfrak{m}}_\beta$. A connection $1$-form $\omega$ on $P$ can be written as
\begin{equation}
\omega = \omega^{ab} \otimes T_{ab}, \quad \mbox{with } \omega^{ab} \in \Omega^1(P)
\end{equation}
Following \hyperref[cor1]{Corollary 1}, the vectors $\left\{ -\Frac[1/2](\sigma_k + \beta \tau_k) \right\}_{k=1,2,3}$ form a basis for the space ${\mathfrak{m}}_\beta$. Therefore we can split $\omega$ into ${\mathfrak{su}}(2)$ and ${\mathfrak{m}}_\beta$ components
\begin{equation}
\begin{aligned}
\omega &= 2 \omega^{0k} \otimes T_{0k} + \omega^{ij} \otimes T_{ij} =
	\\
&= 2\omega^{0k} \otimes \left(- \Frac[1/4] \sigma_k \right) + \omega^{ij} \otimes \left( \Frac[1/4] \epsilon_{ij}{}^k \tau_k \right) =
	\\
&= \omega^{0k} \otimes \left( -\Frac[1/2](\sigma_k + \beta \tau_k) \right) + \left( \Frac[1/2] \epsilon_{ij}{}^k \omega^{ij} + \beta \omega^{0k} \right) \otimes \left( \Frac[1/2] \tau_k \right)
\end{aligned}
\end{equation}

By choosing $\left\{ \Frac[1/2] \tau_k \right\}_{k = 1,2,3}$ as a basis for ${\mathfrak{su}}(2)$, we get that the coefficients of $A$ and $K$ with BI parameter $\beta$ are
\begin{equation}
\begin{cases}
A^k = \Frac[1/2] \, \epsilon_{ij}{}^k \, \omega^{ij} + \beta \, \omega^{0k}
	\\
K^k = \omega^{0k}
\end{cases}
\end{equation}
We thus find that our construction generalizes the definition of BI connections to a generic $(n+1)$-dimensional lorentzian manifold $M$, for $n \geq 3$. The BI parameter $\beta$ comes out naturally in the case $n = 3$ whereas for $n > 3$ it is absent.

\section{Conclusions and Perspectives}

In this work we gave a well-defined geometric framework for building BI connections on any given $(n+1)$-dimensional lorentzian spin manifold $M$. The construction generalises what is found in the physics and mathematical physics literature and improves it by not requiring a fixed metric $g$ on $M$ and producing BI connections on the whole spacetime manifold $M$. The much discussed BI parameter $\beta$ comes out naturally and we proved that it is a feature {\it unique} to the $4$-dimensional case.

As stated in the introduction, the main motivation behind this work was to reformulate General Relativity: instead of using a metric $g$ and a spin connection $\omega$ as field variables, we rewrite the lagrangian in terms of a spin frame $e$ and the pair $(A,K)$ for a fixed BI parameter $\beta$. Since this formulation is the classical starting point of LQG, it is worthwhile to study the variational Euler--Lagrange equations for $(e,A,K)$ and their relation to the Einstein field equations. The equations dependence on the BI parameter $\beta$ is also of interest, especially when we couple General Relativity with other fields (e.g.\ Klein--Gordon, Yang--Mills, Dirac).

Another application of this work is the investigation of the holonomy group of a BI connection $A$, which plays a central role in LQG. Samuel's argument in \cite{Samuel} can be solved by looking at the relation between the holonomy group of $A$ and that of the original spin connection $\omega$, both on $M$ and on a spacelike submanifold $S$. One should also refer to the classification of metric lorentzian holonomies by Berger \cite{HolClass} and Leistner \cite{Leistner} to better characterize the BI connections, this was suggested to one of the authors in private form.

\section*{Acknowledgements}

This article is based upon work from COST Action (CA15117 CANTATA), supported by COST (European Cooperation in Science and Technology).

We also acknowledge the contribution of INFN (Iniziativa Specifica QGSKY), the local research project {\it  Metodi Geometrici in Fisica Matematica e Applicazioni (2019)} of Dipartimento di Matematica of University of Torino (Italy). This paper is also supported by INdAM-GNFM.

We would like to deeply thank Giovanni Russo, for reading the initial draft and giving invaluable advice on how to improve the presentation, and Anna Fino, for useful comments and discussions.

redpajama_set_name: RedPajamaArXiv


BulkGate integration comes with following list of actions and triggers.
Since it's an action, transactional SMS can be placed anywhere in scenario chain. Placing BulkGate module in the chain you will be presented with above mentioned list. Choose transactional SMS action and following modal window will show up. First you'll need to create a connection with BulkGate, which unlike most apps sends authentication information (Application ID, Application Token) with every request. If you do not know how to acquire such information visit BulkGate API setup. For additional information on what to input as other parameters visit BulkGate API specification.
Unlike transactional SMS which is used to send single SMS to single contact, Bulk SMS is used either to send same text SMS to multiple contacts, or different text SMS to multiple contacts. For further information visit BulkGate API specification.
This action is used to receive status of credit. As an action it can be placed in anywhere in scenario chain. All parameters it needs are (Application ID, Application Token).
This action just as others can be placed anywhere in the integromat chain, but unlike others it requires parameter "Message ID". Which can be acquired only by sending SMS by either Transaction SMS, or Promotional SMS and receiving their return data. Ideally this action should be placed behind previsously mentioned actions.
This trigger shares the same name with one of the actions, because it function similarly, but is used differently. Triggers are placed at the beginning of the integromat chain. To properly setup this trigger visit Integromat webhook setup.


Well, I was hoping they'd reveal the last god today, but maybe they're holding off for now, or else it will be revealed, Wizards just decided they wouldn't be the ones to do it (or it will be a Reddit or Twitter thing).
3 mana for a 2/2 which gives everything first strike and opponents lose first strike. Definitely a boost to Boros decks that employ Battalion, and it is great for match-ups.
The way I see it with this cycle is that in match-up games, it becomes a "who can play the Archetypes first" to come out on top as opposed to luck of the draw which sets in halfway through the game. This gives you the edge to win because, if I interpret this correctly, if you cast this first, an opponent casting their copy just gives them a 2/2 for 3 mana because the ability of yours triggers denying their creatures from obtaining first strike by them playing this card. These will be crucial to have for sure.
Here is the second card revealed of the rare "Fated" cycle. These look to be useful for Commander/EDH, but still playable in Standard. Here, 3 mana (all blue) to copy a creature and place a token copy (Copy anything you want...) and you scry for 2 if you play this instant on your turn. Already that's amazing, and both together lead me to call it EDH material, especially since you want to find your combos in EDH and copying creatures (using cards like Clone) is a big thing for blue players.
You may have already met this one, and I like it. Bit difficult to cast, but it will be great in Limited, and on top of it all, you get to scry on your turn. Instant speed removal for Planeswalkers in red.
This has been a great week 1 so far, and we're only a few weeks away from prerelease.
Keep checking the spoiler page for any more cards, and if you haven't figured it out already, the spoiler pages on this site (whenever they get posted), the cards show up in collector numerical order/the same way you'd see them in the Fat Pack books.


text:
Cette page concerne des événements d'actualité qui se sont produits durant l'année 1988 dans la province canadienne de la Saskatchewan.

Politique
 Premier ministre : Grant Devine
 Chef de l'Opposition :
 Lieutenant-gouverneur : Frederick W. Johnson
 Législature :

Événements

Naissances

 14 avril : Eric Gryba (né à Saskatoon) est un joueur professionnel de hockey sur glace canadien.

 18 mai : Krista Phillips, née à Saskatoon, est une joueuse canadienne de basket-ball. Elle évolue au poste de pivot.

 28 mai : Logan Pyett (né à Balgonie) est un joueur professionnel de hockey sur glace canadien.

 18 août : Riley Holzapfel (né à Regina) est un joueur de hockey sur glace.

 6 octobre : Kaylyn Kyle (née à Saskatoon) est une joueuse de football (de soccer) canadienne évoluant au poste de milieu de terrain offensif. Elle joue pour le Seattle Reign FC et est membre de l'Équipe du Canada de soccer féminin (65 sélections en date du ).

 27 novembre : Garrett Klotz (né à Régina) est un joueur professionnel de hockey sur glace canadien.

 18 décembre : Brianne Theisen (née à Saskatoon) est une athlète canadienne, spécialiste des épreuves combinées.

Décès

Voir aussi

Notes et références

Chronologie de la Saskatchewan
1988 au Canada

redpajama_set_name: RedPajamaWikipedia


Indeed. It's a means of weaponising ambiguity into guilt. Everyone knows it's not unlimited, because you can't take 3 months off. But because there's no boundary, the real limit has to be found by stressfully, riskily probing it. Then there's plenty of room for favouritism in who's actually allowed to take the time off.
See "Tyranny of Structurelessness", etc, etc.


Ratatouille, the great vegetable dish of Provence, is the pure expression of those sunny fields and gardens. This is a twist on the traditional. The fresh summer vegetables that go into a classic ratatouille are given a smoky upgrade: marinated in a garlicky vinaigrette, and then cooked over hot coals instead of in a pan.
1. Cut all the vegetables in large pieces for the grill. Slice large eggplants about ½-inch thick and thin ones in half lengthwise. Slice the zucchini lengthwise a little thinner than the eggplant. Cut the bell peppers lengthwise in thirds and trim out the cores and seeds. Thickly slice the onions crosswise. Cut the tomatoes in half.
2. Whisk together all the marinade ingredients, or pulse them briefly in a blender.
3. Layer the vegetables in a large glass or ceramic dish, brushing them generously on both sides with the marinade as you put them in, leaving the tomatoes for last. Drizzle the remaining marinade over the vegetables in the dish and place the tomatoes cut side down on top of them. Leave the vegetables to marinate for at least an hour, longer if you have plenty of time.
4. Fire up the coals! When they are ready, grill the vegetables until they are charred and blistered just the way you like them, and tender enough to be flexible but not mushy. The exact time is impossible to call, as it varies with the heat of the coals and the distance between the grate and the coals, but probably 3 to 4 minutes on a side if the coals are nice and hot I start with the peppers, and then amuse myself by pulling off their charred skins as I grill the eggplants, zucchini and onions. Grill the tomatoes last, giving them a minute or two on the cut side, then turning them and leaving them on the coals, on their skins, until they are completely soft, up to 7 minutes.
5. Let the vegetables cool until you can handle them, then scoop the tomatoes out of their skins into a bowl and discard the skins. Drizzle the marinade left in the dish over the tomatoes and crush them with your bands, breaking them up. That's the sauce.
6. Cut all the other vegetables into generous bite-sized strips and mix them with the tomatoes. Taste, and add more salt, pepper and balsamic vinegar if needed to gel the proper level of zinginess. Serve hot, with a pilaf or rice or orzo and drizzle a little olive oil on top of each serving.


This was a major renovation project that created a new combined servery for Butler and Wilson Colleges. Dining room improvements and a new kitchen were completed. A new entry addition on Goheen Walk gives Wilcox hall a new look while providing a new stair and passenger elevator. This new elevator, as well as new ramps, provides barrier-free access to all floors. Barrier-free toilet rooms are also provided on the second floor. Life safety improvements include upgraded fire alarm systems and extended fire suppression systems. Much of the aging building infrastructure was replaced or improved as part of this project with new energy-efficient systems.


text:
Florida Uses 3 Goals In Second To Claim Series Finale
Cherniwchan Scores Twice For Rays in Losing Effort
(Source: South Carolina Stingrays)
December 1, 2018 at 10:38 PM EST - Updated December 1 at 10:38 PM
ESTERO, Fla. - The Florida Everblades (11-5-4-0) scored three times in the second period and held off the South Carolina Stingrays (11-10-0-0) in the third to take the rubber game of a 3-game set between the two South Division opponents by a score of 4-2 at Hertz Arena on Saturday night.
Rays forward Andrew Cherniwchan scored twice for South Carolina in the contest, while four different skaters added assists. Goaltender Parker Milner started the game and made 24 saves before being replaced by Gordon Defiel in the third period, who stopped all six shots that came his way.
The opening period saw just one goal and it came from Florida's Steven Lorentz at 10:34 to give the Everblades a 1-0 advantage after the first 20 minutes.
Nathan Perkovich pushed the Florida lead to 2-0 just 33 seconds into the second period and John McCarron made it a 3-0 game in favor of the Everblades at 12:00.
Cherniwchan got the offense started for South Carolina late in the frame, scoring his seventh tally of the season at 17:33 with a sweeping wrist shot from the front of the net with assists by forwards Jonathan Charbonneau and Tad Kozun.
But Lorentz notched his second of the game in the final minute of the period during a delayed penalty that was coming to South Carolina. In what became a 5-on-4 advantage for Florida after goaltender Jamie Phillips skated to the bench in favor of an extra attacker, Lorentz found the back of the net with just 34 seconds remaining in the second period and made it 4-1 Everblades.
With the Stingrays on a power play to start the third period after an unsportsmanlike conduct penalty to Florida's Ben Masella, Cherniwchan fired a slap shot from the top of the right circle that found its way to the top left corner of the net for his eighth of the year 18 seconds into the final stanza. Assists on the goal were given to defender Tim Davison as well as forward Shane Eiserman and made the score 4-2 in favor of Florida.
South Carolina was unable to muster any more offense on the scoreboard for the remainder in the third, but out-shot Florida 9-6 in the last 20 minutes. The Everblades had the edge in shots on goal for the game, 34-31.
Cherniwchan's second goal was the only power play strike on either side, with South Carolina finishing 1-for-2 on the man-advantage and Florida ending at 0-for-1. Phillips earned the win for the Blades, stopping 29 shots in the victory.
South Carolina returns home this coming week with some big games including Social Media Night on Dec. 7 and Teddy Bear Toss on Dec. 8. The Rays are back in action at the North Charleston Coliseum to face the Greenville Swamp Rabbits on Tuesday at 7:05 p.m.

redpajama_set_name: RedPajamaCommonCrawl


We understand that looking for a job can be a tiresome necessity so we have tried to keep it as simple as possible for you.
Once you've applied for a position via our online application process, our Recruitment team will use the information you have provided to determine your suitability for the position.
If your application has been successful in initial screening, a member of our Recruitment team will contact you to arrange an interview.
Unfortunately we are unable to respond to all of the applications that we receive. Therefore, if you do not hear back from us after 21 days from the closing date, your application has been unsuccessful. However, we may try to find you employment with other possible matching vacancies; therefore we will use your submitted details in connection with the recruitment process.


The new Honda HR-V is now available at Yonkers Honda in Yonkers, NY. When you need an all-wheel drive vehicle that keeps you connected to the road and to your compatible smart device you need the new Honda HR-V. Honda's latest HR-V offers convenient cargo space and versatility through three cargo modes: Utility, Tall, and Long. Drivers can transport a surfboard, kayak, or ficus plant in the nearly sixty cubic feet of cargo space. Enjoy multiple seating configurations through the second-row Magic Seat to create the extra room you need. Contact Yonkers Honda to set up a test drive. Our sales associates can review the available features, safety systems, and warranty packages available on the entire line of new Honda HR-V models.
The multi-angle, rear-view camera tells you what is behind you from three different vantage points to help you park and see rear cross traffic in time. Protect you, your passengers and those around you with the new Honda HR-V. Braking assist activates if the HR-V detects understeer or oversteer. LED Daytime running lights to increase your visibility and alert other drivers to your presence on the road. Stability assist will help keep you upright in rough driving conditions.
Make the quick trip from the Bronx or Westchester to discover your Honda HR-V financing options. Visit Yonkers Honda in Yonkers, NY, for a competitive price on the new Honda HR-V!


Small churches can and do have validity. One of the great benefits and blessings that small churches have is the relational factor that can and should go on within a small church. Join Karl as he talks about how you can lean into this to maximize discipleship and mission.


When I try to watch any video fr0m YouTube or any other video site, it's taking a long time to come out. But still I have enough internet speed. Also good uploading speed is there.
How can I avoid this problem? The magic is, after all the video will open without any buffering. I feel like it's downloaded first and spreading.
Could I just mention one point of interest. You-Tube offers 360 and 480 which are flash, in addition the HD 720p and 1080p are essentially .mp4's.
Playing the HD's requires significantly more bandwidth as well as cpu resources. I have experienced the issues described in the question quite often when viewing 1080p on a dual core 2.6ghz 6gig optimized intel based system with 10mbps internet.
So try a 360, if that works and 1080 does not you will have an idea where the issue is, if it is not one already mentioned.
Make sure that your browser is set to clean cache, temporary files at exit. Also you could download crapcleaner to clean the history, cache, temporary files, cookies, etc. Install crapcleaner select the stuff you want deleted and once finished restart your computer. now open browser, go to youtube or any other site where buffering takes a long time and then open the video. Do you see any improvement or difference?
>update your browser to the latest version .
>update the Flash to the latest version .
>clear out Cache and History . use CCleaner .
>use an AdBlocker . like ABP for firefox or chrome .
>use a faster DNS server . like Google DNS or Ultra DNS .
first check if you have the latest flash player.
install and then try again.
if problem is still there, consdier this points.
does your computer runs slow over all ? specially the browsers when you open 4 or 5 tabs?
sometimes it freezes and then plays, meanwhile it has completed some buffering.
so it will play directly.
do you have 1 gb or more RAM ?
which are the broswers you use?
use chrome as it is genreally update it self too latest version.
if you have only 512 RAM, that may be the reason.


Which comics from the past year are just begging for a big screen adaptation?
2014 has been a landmark year for comic book movies. Not only did we get two of Marvel's best-ever films (Guardians of the Galaxy and Captain America: The Winter Soldier), but the major studios announced enough future superhero adaptations to take us all the way up to 2019.
It's been a pretty landmark year for comic books too. It feels as though more genuinely exciting new titles launched in 2014 than in the past ten years combined, with DC, Marvel and especially Image taking big risks and creating some seriously compelling comics in the process.
The Comic: On paper, Spread doesn't sound like the most original book on the market. Like The Walking Dead, it's set in a post-apocalyptic world where survivors fight off gruesome monsters and each other. As with The Thing, those monsters are giant shape-shifting creatures made mostly of flesh and teeth. Just like Lone Wolf & Cub, our hero is a ronin warrior protecting an infant with bladed weapons. And, as with Saga, the comic is narrated by that infant.
But Spread combines those disparate elements to create a vividly original and intensely emotional experience, packed with set pieces we can't wait to see played out on IMAX.
The Movie: Cast Hugh Jackman as the Wolverine-esque protagonist named 'No' (so called because that's pretty much all he says) and Leonardo DiCaprio as the mysterious blonde tracking him. Hire Mad Max: Fury Road's George Miller to direct. Get KNB to team with WETA digital to do the effects. Watch box offices shatter.
The Comic: Earl Tubb turns up in Craw County, Alabama after a 40-year absence to put some family affairs in order. His old home is darkened by a huge tree growing from his father's grave, and flashbacks reveal the tough old man casts his own long shadow over Earl's life.
Tubb's aim is to get in and out of town as quickly as possible. But when an old friend gets in trouble with some local heavies, Tubb finds himself forced to extend his stay.
The Movie: The very definition of 'speak softly and carry a big stick you've crafted from the tree growing from your dead dad's grave'. Earl Tubb could be cinema's next great tough guy.
He's not a young man, so we'd be looking at Clint Eastwood, or JK Simmons in age make-up for the lead. But all we have to do is remind Quentin Tarantino how much he loves movies like Macon County Line (1974), Walking Tall (1973) and White Lightning (1973) and we reckon he would be up for directing it.
The Comic: Launched in February, then tragically cancelled after 12 issues, She-Hulk is the best 2014 comic that no-one really bought. But it has so much potential.
A strong, smart, cool, career-focused, funny protagonist (can you tell we've got a bit of a crush on her?) She-Hulk's Jennifer Walters balances being a superhero with practicing law - her first case is against her old mate Tony Stark - while finding time to be a good friend to fellow female heroes, and drinking pints of beer with Thor.
Beautifully drawn and critically acclaimed, let's hope that She-Hulk finds a new life on a big screen soon.
Movie Version: Let's chuck some green paint onto Gwendoline Christie and do a very different kind of superhero movie, one that's more influenced by 1959's Anatomy of a Murder than The Avengers. Imagine the best courtroom thriller you've ever seen, but starring superheroes. Sounds great, right?
The Comic: Robert Kirkman's first horror series since The Walking Dead is dark, rich and complex. It follows Kyle Barnes, whose experience of being possessed by demonic forces has turned him into one of the most unlikeable leads on offer in your local comic book store.
Well, initially - until the chance at redemption pulls Kyle out of his self-imposed exile from society, transforming him into one of the most engaging characters of the year.
Outcast is a book that respects and requires readers' patience - but stick with it and the rewards are frequent and intense.
Movie Version: It's actually more likely that we'll see Outcast on the small screen first - it's at the pilot stage at Cinemax with Patrick Fugit playing Barnes. But should that fail, we'd like Blumhouse Productions to pick it up, then convince Benedict Cumberbatch to star.
The Comic: Every now and then, an idea enters pop culture so fully-formed and instantly brilliant that you're flabbergasted it hasn't been done before.
The Kitchen takes place in the late seventies, in Hell's Kitchen, New York. It features Irish American gangsters violently breaking the law to maintain dominance of their territory. So far, so Scorsese.
Only, here's the twist - these gangsters are women, struggling to maintain the status quo after their husbands are locked up.
It's early days for the comic, but on the strength of the first issue end-twist, The Kitchen could end up being a female-led masterpiece.
Movie Version: We've mentioned him once already, so it would be rude not to offer it to Martin Scorsese. Or, actually, Michelle MacLaren - the concept of innocent people exploring their inner darkness to support their family is as much Breaking Bad as it is The Departed.
The Comic: Scott Snyder is one of mainstream comics' most successful writers. He's in charge of DC's New 52's main Bat-book, Batman - a responsibility which this year involved over-seeing approximately a billion different spin-offs - Batman Eternal, Arkham Manor, Gotham Academy to name but a few.
Because of his current Bat-association, it's easy to forget Snyder had his roots in horror comics, building his reputation on titles like American Vampire, Severed and The Wake.
Somehow, Snyder found time to return to those roots in 2014 with Wytches, a terrifying modern take on Wiccan mythology which gets so nasty you'll want to throw your digital reading device across the room. You know, in a good way.
Movie Version: He's one of the most hated men in movies, but we've got a soft spot for Rob Zombie. We'd love to see him return to some of the themes he explored in Lords of Salem, combining some of his Devil's Rejects craziness with the intensity of his Halloween II. Honestly, we're being completely serious about this. Please don't get too angry in the comments.
The Comic: Every 90 years, twelve gods are reincarnated in the bodies of young people. They have near-limitless powers, and the adoration of millions. The only catch? In two years, they'll be dead.
The year is 2014, and this generation's gods are pop stars - rock, dance, rap, more - but some of them aren't happy with their time limit, and set out to rewrite the rules.
Essentially a combination of Garth Ennis's Preacher, Neil Gaiman's Sandman series and smart-phone social media culture, The Wicked + The Divine is one of the most exciting - and thought-provoking - comics to launch this year.
Movie Version: Plenty of lead characters to chose from, but we'd centre our film around androgynous David Bowie analogue Luci - this generation's Lucifer - sticking Cate Blanchett into an iconic white suit then sitting back to watch the literal fireworks (and occasional head-explosions).
The Comic: Oh, Starlight. This comic's giving us goosebumps just thinking about it - it's easily one of the most fun and emotional experiences we had this year, and the thought of it becoming a movie is unspeakably exciting.
If you haven't had the joy of Mark Millar's return to form yet, the concept is simple. A Flash Gordon/Buck Rogers style hero saves the universe, then returns home to earth - to discover no-one believes his adventures. He grows old, unappreciated. That is until a spaceship lands on his front lawn, and he has to remember how to be a hero.
Movie Version: Starlight wears its influences on its sleeve, right next to its heart. We mentioned a couple above, but you can add Star Wars (key character Tilda Starr is basically a female Han Solo) to the list.
Oh, and it's also a bit like The Incredibles - Duke has a very strong Mr Incredible vibe. But forget the homages, Starlight is very much its own beast, as sincere and kind-hearted as anything Millar's ever done.
If studios are scrambling to green-light sci-fi flicks to follow The Force Awakens, they should look no further than Starlight. Fox have the rights, so let's hope they fast-track it soon.
The Comic: New Sheriff in town and single mother Clara Bronson attempts to get to the bottom of a complex murder mystery, whilst protecting her young son and dealing with the grumpiest alien deputy this side of Chewbacca.
Movie Version: As you might have guessed from the high-concept pitch above, Copperhead nods so heavily to Star Wars we sometimes worry its head is going to fall off.
But whereas that franchise took its lead from old Japanese movies, Copperhead owes more to American westerns, feeling more like The Searchers than Seven Samurai.
And with a strong female lead - we'd love to see Jessica Chastain playing Clara - dealing with a brilliant cast of weird alien creatures, we reckon this one has James Cameron's name all over it.
The Comic: When Marvel announced that they were rebooting Ms Marvel with a teenage Muslim girl as the new lead, fan sites practically exploded with excitement.
For evidence, buy the first issue to see a letters section packed with praise; letters sent long before a single page had been read.
Thankfully, the comic lives up to the hype. Kamala Khan is arguably Marvel's most relatable lead since Peter Parker, as she negotiates friendships, unexpected power and her family's expectations via social media, Avengers fan fiction and homemade costumes.
Movie Version: Cast Sajal Ali as Kamala Khan, and give her a full costume cameo in Captain Marvel (2018).
Ali will be 24 by that point, so a tiny bit old for the role - but with the likes of Andrew Garfield (age 31) playing a teenager in Sony's Amazing Spider-Man movies, that shouldn't be too much of a problem.
Spider-Man should be a key influence on a Ms Marvel movie, given that the comic explores the responsibility of power in a very similar way. Instead of Marvel Studios buying the rights to Spidey back, they should consider putting their money behind Ms Marvel instead.
The Comic: In January, kings of graphic novel noir Ed Brubaker and Sean Phillips signed a landmark five-year deal at Image Comics, a contract that kept them exclusive to the company, whilst allowing them to create pretty much whatever they want.
The Fade Out was their first title as part of the deal, and if everything else they do is as good as this, then it'll be money well spent.
A dark murder mystery set in the studio system of 1940s Hollywood, The Fade Out's mixture of attention to detail and realistic historical representation with pure pulp fiction entertainment makes it an essential monthly purchase.
Movie Version: Brubaker (who also had a cameo in The Winter Soldier this year) describes the title as David Lynch directing Mad Men, so let's hope the director wraps up Twin Peaks season three quickly in 2015, so he can get to work on this.
The Comic: Despite the continued success of Game of Thrones on telly this year, there aren't a lot of successful fantasy comics being picked up by mainstream readers.
But Joshua Williamson's stunning series seems to be changing that, with the first three issues selling out almost instantly.
We don't want to go too much into the plot, because so many wonderful reveals are packed into issue one, it'd spoil the fun. But let's just say that it involves a boy named Mikey, who goes missing - before turning up in a very different situation.
Movie Version: Again, without spoiling too much, we'd definitely need to get The Jim Henson Company involved. And possibly half the cast of the aforementioned Game of Thrones.
The Comic: Stunning art and compelling storytelling combined to produce the most impressive creator-owned comic of the year when Deadly Class launched way back in January.
Set in the '80s - complete with an era-appropriate soundtrack - it centres around homeless Nicaraguan teen Lopez Arguello, and follows a bunch of outcast teenagers who have been enrolled into a school for assassins (think Harry Potter meets Wanted but, you know, more punk rock than either of those things), examining the trials, tribulations, and occasional death-traps they encounter as a result.
Movie Version: It would have to be rated 18 and contain the most kinetic action this side of Shaw Brothers movies.
We've already got Tarantino busily making Southern Bastards, so we'll have to hire his mate Edgar Wright instead. We hope Ant-Man hasn't put him off comic-book movies, as he'd be so perfect for this one.
The Comic: Where to begin with The Multiversity? Split into a series of one-shots set in DC's classic vast universe of parallel realities that's so packed with meta-references there are entire websites devoted to picking it apart panel by panel, Grant Morrison's masterpiece has consistently been the best book on shelves whatever week it's been released.
It has a high concept - a cursed comic travels through DC's Multiverse - but the giddy joy we get reading it isn't gleaned from the plot, so much as it is the way it's delivered.
With one issue - Pax Americana - daring to update Watchmen (we're sure Morrison's 'old mate' Alan Moore is busily creating a 23-page fan letter as we speak), and another putting a black Superman on the cover, The Multiversity has been progressive, challenging and - most of all - exciting.
Movie Version: Once Warner Brothers and DC have finished their first wave of superhero movies, we'd love for them to hire Morrison to create their very own Phase 2 - a series of spin-offs that deconstruct and reassemble their most iconic stars in strange and unexpected ways.
And if The Just - Morrison's interpretation of the Justice League as a bunch of entitled celebrity teenagers - isn't made into a six-film franchise, there truly is no justice.