text:
\section{Setup}

 Let $F$ be a totally real field of degree $d$, and $Y$ the Hilbert modular variety for $F$ of level $U_1(\mathfrak{N}) = \{ g \in \GL_2(\widehat{\mathcal{O}}_F): g = \stbt{\star}{\star}{0}{1} \bmod \mathfrak{N}\}$. Fix a numbering of the embeddings $F \hookrightarrow \mathbf{R}$ as $\sigma_1, \dots, \sigma_d$. Let $f$ be a newform of level $\mathfrak{N}$ and some weight $(\underline{k} + 2, \underline{t})$, where $\underline{k} = (k_1, \dots, k_d) \in \mathbf{Z}_{\ge 0}^d$, and $\underline{t} \in \mathbf{Z}^d$ such that $w = k_i + 2t_i$ is independent of $i$. We choose a prime $p$ which splits completely in $F$ and such that $(p, \mathfrak{N}) = 1$.
 \subsection{The spaces $D_p(f)$ and $V_p(f)$}

  We are interested in the $2^d$-dimensional de Rham cohomology eigenspace
  \[ D_p(f) \coloneqq H^d_{\mathrm{dR}}(Y_{\overline{\mathbf{Q}}_p}, \mathcal{L}_{\mu, \mathrm{dR}})\{f\}, \]
  where $\mathcal{L}_{\mu, \mathrm{dR}}$ is the vector bundle with connection associated to the weight $\mu = (\underline{k}, \underline{t})$, and $\{f\}$ denotes the $f$-generalised eigenspace for the Hecke operators away from $p\mathfrak{N}$. This is equipped with a Hodge filtration $\Fil^\bullet$, and (via comparison with crystalline cohomology) a Frobenius $\varphi$.

  We also have a representation of the Galois group $\Gamma_{\mathbf{Q}} = \Gal(\overline{\mathbf{Q}}/\mathbf{Q})$ given by
  \[ V_p(f) \coloneqq H^{d}_{\text{\textup{\'et}}}(Y_{\overline{\mathbf{Q}}}, \mathcal{L}_{\mu, \text{\textup{\'et}}})\{f\}, \]
  where $\mathcal{L}_{\mu, \text{\textup{\'et}}}$ is the \'etale local system corresponding to $\mu$. The Faltings--Tsuji comparison theorem of $p$-adic Hodge theory gives a canonical isomorphism of filtered $\varphi$-modules
  \begin{equation} \label{eq:faltings} D_p(f) \cong \DD_{\dR}\left(V_p(f) |_{\Gamma_{\QQ_p}}\right). \end{equation}

 \subsection{Plectic structures}

  The space $D_p(f)$ is endowed with a family of $d$ decreasing \emph{partial filtrations} $\Fil_i^\bullet$, for $i = 1, \dots, d$, each with two graded pieces in degrees $(t_i, t_i + k_i + 1)$, for which the associated single filtration is the Hodge filtration. Rougly, the nontrivial subspace in $\Fil_i^\bullet$ corresponds to the part of the cohomology generated by differential forms which are holomorphic at the $i$-th infinite place. The construction is explained in \cite{nekovarscholl16} for $\underline{k} = (0, \dots, 0)$; the extension to general coefficients is routine, but we review it in \cref{sect:BGGcplx} below to fix notations.

  We also have $d$ commuting linear maps $\varphi_1, \dots, \varphi_d$ on $D_p(f)$, the \emph{partial Frobenii}, whose composite is the usual Frobenius $\varphi$. These arise from endomorphisms of the special fibre $Y_0$, sending a Hilbert--Blumenthal abelian variety $A$ to the quotient of $A$ by the $\mathfrak{p}_i$-torsion in the kernel of Frobenius on $A$. We refer to \cite{nekovar-semisimplicity} or \cite{tianxiao16} for detailed accounts of this construction.

 \subsection{Tensor induction}

  Associated to $f$, we also have a 2-dimensional standard Galois representation $V_p^{\mathrm{std}}(f)$ of $\Gamma_F \coloneqq \Gal(\overline{F} / F)$. Via results of \cite{brylinskilabesse84} and \cite{nekovar-semisimplicity}, we may choose an isomorphism of $\Gamma_\mathbf{Q}$-representations
  \begin{equation} \label{eq:defpsi}
   \psi : V_p(f) \cong (\bigotimes-\mathrm{Ind})(V_p^{\mathrm{std}}(f)).
  \end{equation}
  This will often, but not always, be unique up to scalars.

  From $\psi$ we obtain an isomorphism of filtered $\varphi$-modules
  \[ \psi_p: D_p(f) \cong \bigotimes_{i=1}^d D_{\mathfrak{p}_i}(f) \]
  where $D_{\mathfrak{p}_i}(f) = \DD_{\dR}\left(V_p^{\mathrm{std}} |_{\Gamma_{F_{\mathfrak{p}_i}}}\right)$ denotes the filtered $\varphi$-module of the standard representation at the prime $\mathfrak{p}_i$ above $p$.

  The nontrivial graded pieces of $D_{\mathfrak{p}_i}(f)$ are in degrees $t_i$ and $t_i + k_i + 1$; and the ``partial Eichler--Shimura'' congruence relation proved in \cite{nekovar-semisimplicity} shows that we have $(\varphi_i - \alpha_i)(\varphi_i - \beta_i) = 0$ on $D_p(f)$, for each $i$, where $\alpha_i$, $\beta_i$ are the roots of the Hecke polynomial at $\mathfrak{p}_i$. The roots of this polynomial are also the eigenvalues of $\varphi$ on $D_{\mathfrak{p}_i}(f)$. These facts strongly suggest the following conjecture:

  \begin{conjecture}[Plectic comparison conjecture]
   \label{main}
   For some choice of global isomorphism $\psi$ as above, the isomorphism $\psi_p$ intertwines the above partial Frobenii and filtrations on $D_p(f)$ with the obvious partial Frobenii and filtrations on the tensor product.
  \end{conjecture}

 \subsection{Relation to the plectic conjectures}

  \label{rmk:plecticGal}
  It is conjectured in \cite{nekovarscholl16} that $V_p(f)$ has an intrinsically defined action of the \emph{plectic Galois group} $\Gamma_{\mathbf{Q}}^{\mathrm{plec}} \supset \Gamma_\mathbf{Q}$, isomorphic to $S_r \ltimes \Gamma_F$, and that $\psi_p$ can be chosen to intertwine this with the obvious action of $\Gamma_{\mathbf{Q}}^{\mathrm{plec}}$ on the tensor induction.

  If we assume the existence this canonical $\Gamma_\mathbf{Q}^{\mathrm{plec}}$-action on $V_p$, then we can state our conjecture more intrinsically as follows (removing the isomorphism $\psi$ from the picture). By restriction we obtain an action of $\Gamma_{\QQ_p}^{\mathrm{plec}} \cong (\Gamma_{\QQ_p})^d$ on $V_p$, and we can define
  \[ \DD_{\dR}^{\mathrm{plec}}\left(V_p(f) |_{\Gamma_{\QQ_p}^{\mathrm{plec}}}\right) \coloneqq \left[V_p(f) \otimes (\mathbf{B}_{\mathrm{dR}} \otimes \dots \otimes \mathbf{B}_{\mathrm{dR}})\right]^{\Gamma_{\QQ_p}^{\mathrm{plec}}}. \]
  This gives a vector space equipped with $d$ partial Frobenii and partial filtrations, whose underlying (usual) filtered $\varphi$-module is $\DD_{\dR}\left(V_p(f)|_{\Gamma_{\QQ_p}}\right)$. Then the ``correct'' conjecture is that \eqref{eq:faltings} is in fact an isomorphism
  \[ \DD_{\dR}^{\mathrm{plec}}\left(V_p(f) |_{\Gamma_{\QQ_p}^{\mathrm{plec}}}\right) \cong D_p(f) \]
  commuting with the partial Frobenii and filtrations. This explains the name we have given to our conjecture: we are seeking a plectic refinement of the comparison isomorphism of $p$-adic Hodge theory. (In the formulation of \cref{main} above, lacking a canonically defined $\Gamma^{\mathrm{plec}}_{\QQ_p}$-action on $V_p$, we have substituted the $\Gamma^{\mathrm{plec}}_{\QQ_p}$-action transported from the tensor induction via $\psi$.)

  \begin{remark}
   It would be interesting to extend the conjecture to the case when $p$ is not totally split in $F$. In this case we do not know how to define the functor $\DD_{\dR}^{\mathrm{plec}}$ (or even what the target category of this functor should be).
  \end{remark}

 \subsection*{Acknowledgements} It is a pleasure to thank Jan Nekov\'a\v{r} and Tony Scholl for informative conversations in connection with this paper.

\section{Results for general $d$ : statements}

 We shall prove the following theorems in this paper. We note that all of these would be immediate consequences of \cref{main}, but our proofs are unconditional, and our purpose is to derive evidence for the conjecture. For brevity, we shall write $\Fil_i^+ = \Fil_i^{(t_i + 1)}$ for the unique nontrivial step in the $i$-th filtration, and for $S \subset \{1, \dots, d\}$ we put $\Fil^S D_p(f) = \bigcap_{i \in S} \Fil_i^+$.

 \begin{definition}
  For $i \in \{1, \dots, d\}$ and $\alpha_i$ a root of the Hecke polynomial at $\mathfrak{p}_i$, we say $\alpha_i$ has \emph{strictly small slope} if $v_p(\alpha_i) < k_i + t_i$.
 \end{definition}

 This is a special case of the notion of ``strictly small slope'' defined in \cite{boxerpilloni20} for general reductive groups. Note that $v_p(\alpha_i)$ is always in the interval $[t_i, t_i + k_i + 1]$, and the slopes of the two roots are symmetric about the midpoint of the interval; so there is always at least one strictly-small-slope root if $k_i > 1$.

 \begin{theorem}
  \label{thm:main}
  Let $S \subseteq \{1, \dots, d\}$. For each $i \in S$, assume that the Hecke polynomial at $\mathfrak{p}_i$ has distinct roots, and that one of these roots $\alpha_i$ has strictly small slope.

  Then the simultaneous eigenspace $\bigcap_{i \in S} D_p(f)^{\varphi_i = \alpha_i}$ has dimension $2^{d - |S|}$, and has zero intersection with the sum $\sum_{i \in S} \Fil_i^+$, so the projection map
  \[ \bigcap_{i \in S} D_p(f)^{\varphi_i = \alpha_i} \xrightarrow{\ \cong\ } \frac{D_p(f)}{\sum_{i \in S} \Fil_i^+}\]
  is an isomorphism. Moreover, this isomorphism is strictly compatible with the partial filtrations, in the following sense: for each $T \subseteq \{1, \dots, d\}$, this map restricts to a bijection
  \[ \left(\Fil^T D_p(f)\right)\cap \left(\bigcap_{i \in S} D_p(f)^{\varphi_i = \alpha_i}\right)
   \xrightarrow{\ \cong\ } \frac{\Fil^T D_p(f)}{\Fil^T D_p(f) \cap\left(\sum_{i \in S} \Fil_i^+ D_p(f)\right)}, \]
  with both sides being zero unless $S \cap T = \varnothing$.
 \end{theorem}

 \begin{remark}
  The condition that the Hecke polynomial have distinct roots at each $i \in S$ is not actually needed for the theorem, but handling the case of a multiple root complicates the proof; and it is known that this case can never occur if the Tate conjecture holds, so we shall not pursue it here.
 \end{remark}

 Using this result, we can prove \cref{main} under some mild assumptions when $[F : \mathbf{Q}] = 2$, see \S\ref{sect:quadratic}. In general we have the following corollaries:

 \begin{corollary}\label{cor:phipreserves}
  Let $j \in \{1, \dots, d\}$ be given, and suppose that for all $i \ne j$, the Hecke polynomial of $f$ at $\mathfrak{p}_i$ has two distinct roots which both have strictly small slope. Then the subspace $\Fil_j^+ D_p$ is stable under the operators $\varphi_i$ for $i \ne j$.
 \end{corollary}

 \begin{proof}
  We apply the theorem with $S = \{1, \dots, d\} - \{j \}$ and each of the $2^{|S|-1}$ choices of roots of the Hecke polynomials, which all satisfy the conditions. This gives $2^{|S|-1}$ linearly independent lines in $\Fil_j^+$, all of which are stable under $\varphi_i$ for $i \ne j$; and the sum of these subspaces must be all of $\Fil_j^+$.
 \end{proof}

 \begin{corollary}
  Suppose that for all $i \in \{1, \dots, d\}$, the Hecke polynomial has distinct roots, and that one of these roots $\alpha_i$ has strictly small slope. For each $S \subseteq \{1, \dots, d\}$, define
  \[ X(S) \coloneqq  (\Fil^S D_p(f)) \cap \left(\bigcap_{i \notin S} D_p(f)^{(\varphi_i = \alpha_i)}\right). \]
  Then $X(S)$ is one-dimensional for all $S$, and we have
  \[ D_p(f) = \bigoplus_{S \subseteq \{1, \dots, d\}} X(S). \]
  Moreover, the subspaces $X(S)$ split both the Hodge filtration, and the filtration by $\varphi_i$-eigenspaces, in the sense that for every $i \in \{1, \dots, d\}$ we have
  \[ \Fil_i^+ D_p(f) = \bigoplus_{\substack{S \subseteq \{1, \dots, d\} \\ i \in S} } X(S),\qquad\text{and}\qquad D_p(f)^{(\varphi_i = \alpha_i)} = \bigoplus_{\substack{S \subseteq \{1, \dots, d\} \\ i \notin S} } X(S). \]
 \end{corollary}

 These last two formulae should be seen as $p$-adic counterparts of the plectic Hodge decomposition on the cohomology over $\mathbf{C}$ described in \cite{nekovarscholl16}.

 \begin{proof}
  We shall prove the following claim: for all $S$ we have $\Fil^S D_p = \bigoplus_{T \supseteq S} X(T)$. By downward induction on $|S|$, we may suppose that the assertion is true for every strict superset of $S$ (which is vacuously satisfied if $S = \{1, \dots, d\}$).

  We firstly claim that
  \[ \sum_{T \supsetneq S} \Fil^T D_p = \bigoplus_{T \supsetneq S} X(T). \]
  The induction hypothesis shows that $\sum_{T \supsetneq S} \Fil^T D_p = \sum_{T \supsetneq S} X(T)$, so what we must prove is that the sum is direct; but this follows by dimension-counting, since we know that $\sum_{T \supsetneq S} \Fil^T D_p$ has dimension $2^{d-|S|} - 1$, and each $X(T)$ has dimension one.

  Having proved the claim, it remains to show that $X(S)$ maps isomorphically onto $\Gr^S D_p$. Since $X(S)$ and $\Gr^S D_p$ are both one-dimensional, this amounts to the assertion that $X(S)$ is not contained in $\sum_{T \supsetneq S} \Fil^T$. But $\sum_{T \supsetneq S} \Fil^T \subset \sum_{i \notin S} \Fil_i^+$, and the theorem shows that this space has zero intersection with $\bigcap_{i \in S} D_p^{(\varphi_i = \alpha_i)}$.

  Having proved the claim, setting $S = \varnothing$ we conclude that $D_p$ is the direct sum of the $X(T)$'s, and the remaining assertions of the theorem are obvious.
 \end{proof}

\section{Filtrations on the BGG complex}\label{sect:BGGcplx}

 Since we do not know a good reference for the existence of partial filtrations on $D_p(f)$ when $\underline{k} \ne (0, \dots, 0)$, we sketch the construction below. We choose a smooth projective toroidal compactification $\imath: Y \hookrightarrow X$.

 \begin{definition} \
  \begin{enumerate}[(i)]
   \item For $U \subseteq \{1, \dots, d\}$, let $\kappa_U$ denote the (possibly non-dominant) weight of the maximal torus of $G$ whose $i$-th entry is $(k_i + 2, t_i)$ if $i \in U$, and $(-k_i, t_i + k_i + 1)$ otherwise. We let $\omega_U$ denote the line bundle on $X$ corresponding to $\kappa_U$, so that $f$ itself is a global section of $\omega_{\{1, \dots, d\}}$.
   \item The \emph{BGG complex} is the complex of sheaves
   \[ \BGG = \left[ \omega_\varnothing \to \bigoplus_{|U| = 1} \omega_U \to \bigoplus_{|U| = 2} \omega_U \to \dots \to \omega_{\{1, \dots, d\}}\right], \]
   with differentials as in \cite{tianxiao16}.
  \end{enumerate}
 \end{definition}

 The BGG complex is known to be quasi-isomorphic to $\imath_* \left(\mathcal{L}_\mu \otimes \Omega^\bullet_Y\right)$; so we have $D_p = \mathbf{H}^d(X, \BGG)\{f\}$.

 \begin{definition}
  For $T \subset \{1, \dots, d\}$, the direct sum of the $\omega_U$ with $T \subseteq U$ is a subcomplex, and we denote this by $\Fil^T \BGG$. We define a subspace $\Fil^T D_p(f) \subseteq D_p(f)$ by
  \[ \Fil^T D_p = \operatorname{image}\left[ \mathbf{H}^d(X, \Fil^T \BGG)\{f\} \to \mathbf{H}^d(X, \BGG)\{f\}\right].\]
 \end{definition}

 \begin{proposition}\label{prop:plecfiltration}
  The collection of subspaces $\Fil^S$ for $S \subseteq \{1, \dots, d\}$ forms a decreasing \textbf{$I$-filtration} in the sense of \cite[Definition 1.2.1]{NS-plectic1}, where $I$ is the poset of subsets of $\{1, \dots, d\}$. Moreover, for each $S \subset \{1, \dots, d\}$, the graded piece
  \[ \Gr^S D_p \coloneqq \Fil^S D_p / \Big(\sum_{T \supsetneq S} \Fil^T D_p\Big) \]
  is isomorphic to $H^{d - |S|}(X, \omega_S)\{f\}$.
 \end{proposition}

\begin{proof}
 In \cite{NS-plectic1} this is proved for $\underline{k} = (0, \dots, 0)$ using Dolbeault cohomology over $\mathbf{C}$, as a consequence of the stronger statement that the complex-analytic cohomology has a plectic Hodge structure. We shall outline a slightly different proof for general coefficients, as this foreshadows the $p$-adic computations we shall make below. The crucial analytic input is the following:

 \begin{quotation}
  For any $U \subseteq \{1, \dots, d\}$, the generalised eigenspace $H^*(X, \kappa_U)\{f\}$ is zero except in degree $d - |U|$.
 \end{quotation}

 By Proposition 1.7.2 of \cite{poliposi05}, the assertion that the $\Fil^T$ define an $I$-filtration is equivalent to the assertion that the lattice of subspaces of $D_p$ generated by the $\Fil_j^+$, under the operations of sum and intersection, is a \emph{distributive} lattice.

 Let $\Lambda$ be the lattice of subcomplexes of $\BGG$ generated by the $\Fil^T$ (under sum and intersection). Then $\Lambda$ is clearly distributive, because it comes from a grading on each individual degree (although the differentials do not respect the grading). Moreover, for each $\mathcal{C} \in \Lambda$, the $f$-part of the cohomology of $\BGG / \mathcal{C}$ vanishes outside degree $d$, so we can identify $H^d(X, \mathcal{C})\{f\}$ with its image in $D_p$.

 Now, for any two objects $\mathcal{C}$, $\mathcal{D}$ of $\Lambda$, we have a short exact sequence of complexes
 \[ 0 \to \mathcal{C}\cap \mathcal{D} \to \mathcal{C} \oplus \mathcal{D} \to \mathcal{C} + \mathcal{D} \to 0, \]
 and since the $f$-part of the cohomology vanishes outside degree $d$, we obtain a short exact sequence
 \[ 0 \to H^d(\mathcal{C} \cap \mathcal{D})\{f\} \to H^d(\mathcal{C})\{f\} \oplus H^d(\mathcal{D})\{f\} \to H^d(\mathcal{C} + \mathcal{D})\{f\} \to 0.\]
 The exactness at the right-hand end shows that $H^d(X, -)\{f\}$ commutes with sums; and the exactness at the middle term shows that it commutes with intersections. So $H^d(X, -)\{f\}$ defines a morphism of lattices from $\Lambda$ to the lattice of subspaces of $D_p$ generated by the $\Fil^T$, and since the former lattice is distributive, so is the latter.
\end{proof}

\section{An application of higher Coleman theory}

 \subsection{The $S$-ordinary locus} We now embark on the proof of the theorem.
  \begin{definition}
   Let $X_0$ be the special fibre of $X$, and $X^{S-\ord}_0$ the non-vanishing locus of the partial Hasse invariants for the primes in $S$.
  \end{definition}

  This is the locus where the $\mathfrak{p}_i^\infty$-torsion of the universal semiabelian variety over $X_0$ is ordinary for all $i \in S$. We write $\mathcal{X}^{S-\ord}$ for the tube of $X^{S-\ord}_0$ in the dagger analytification $\mathcal{X} = X^{\mathrm{an}, \dagger}$.

  \begin{proposition}\label{prop:rigcoh-iso}
   The natural pushforward map
   \[ \mathbf{H}^d_{c}\left(\mathcal{X}^{S-\ord}, \BGG \right)\{f\} \to \mathbf{H}^d\left(\mathcal{X}, \BGG \right)\{f\} = D_p \]
   is a bijection.
  \end{proposition}

  \begin{proof}
   The complement of $X_0^{S-\ord}$ in $X_0$ is a normal-crossing divisor $\bigcup_{i \in S} Z_i$; and for each $T \subset S$, the intersection $Z_T = \bigcap_{i \in T} Z_i$ is itself the special fibre of a Shimura variety, associated to the quaternion algebra over $F$ ramified at the finite places $\mathfrak{p}_i$ and the infinite places $\sigma_i$ for $i \in T$. See \cite{tianxiao16} for further details. The rigid cohomology of each $Z_T$ can therefore be computed in terms of quaternionic Hilbert modular forms, whose Jacquet--Langlands transfers to $\GL_2 / F$ are of Steinberg type at the primes $\mathfrak{p}_i$ for $i \in T$. So the $f$-generalised eigenspaces in these cohomology groups are zero. Since the kernel and cokernel of the pushforward map can be computed in terms of the rigid cohomology of the $Z_i$ using the Gysin sequence, it follows that the pushforward is an isomorphism on the $f$-generalised eigenspace.
  \end{proof}

  \begin{remark}
   This is essentially the same argument as the main theorem of \cite{tianxiao16}. More precisely our setting is the ``Poincar\'e dual'' of theirs: in \emph{op.cit.}, the rigid cohomology groups appearing have compact support towards the boundary divisor and non-compact support towards the non-ordinary loci, whereas ours are compactly-supported towards the $Z_i$ but non-compact towards the cuspidal boundary.
  \end{remark}

  From the decomposition of the BGG complex we obtain a new filtration on $D_p$:

  \begin{notation}
   For $T \subset \{1, \dots, d\}$, define $\mathscr{F}_T^+ D_p$ to be the $f$-generalised eigenspace in the space
    \[ \operatorname{image}\Big(\mathbf{H}^d_{c}\left(\mathcal{X}^{S-\ord}, \Fil^T \BGG \right) \to
   \mathbf{H}^d_{c}\left(\mathcal{X}^{S-\ord}, \BGG \right)\Big). \]
  \end{notation}

  Note that \emph{a priori} the subspaces $\mathscr{F}_T^+$ only define a ``$\mathbf{Z}^d$-prefiltration'' in the sense of \cite{NS-plectic1}; it is not clear if it is a $\mathbf{Z}^d$-filtration (or even a weak $\mathbf{Z}^d$-filtration). Moreover, it is clear that there are inclusions
  \[ \mathscr{F}_T^+ D_p \subseteq \Fil^T D_p, \]
  but it is far from obvious if equality holds

 \subsection{Input from higher Coleman theory}

  The first basic input from higher Coleman theory that we shall use is the following two propositions.

  \begin{proposition}
   For each $T \subset \{1, \dots, d\}$, the operator $\varphi_S = \prod_{i \in S} \varphi_i$ on $H^d_c\left(\mathcal{X}^{S-\ord}, \omega_T \right)$ is compact.
  \end{proposition}

  \begin{proof}
   This follows from the dynamics of our Frobenius lifting applied to neighbourhoods of the $S$-ordinary locus: we can find a cofinal family of strict neighbourhoods $\mathcal{N}$ of $\mathcal{X}^{S-\ord}$ with the property that $\varphi_S\left(\mathcal{N}\right)$ is a strict neighbourhood of $\mathcal{N}$, so the map on cohomology induced by Frobenius factors through pushforward along the inclusion $\mathcal{N} \hookrightarrow \varphi_S(\mathcal{N})$ and is therefore compact. (Compare Corollary 5.20 of \cite{boxerpilloni22}.)
  \end{proof}

  So it makes sense to pass to the $\alpha_S$ \emph{generalised} eigenspace for the partial Frobenii, and the resulting space is finite-dimensional, so we can decompose it according to the action of the spherical Hecke operators.

  \begin{proposition} Let $T \subseteq \{1, \dots, d\}$. Then:
   \begin{itemize}
    \item If $S \cap T \ne \varnothing$, then $H^n_c\left(\mathcal{X}^{S-\ord},\omega_T\right)^{\{\alpha_S\}}\{f\}  = 0$ for all $n$.
    \item If $S \cap T = \varnothing$, then $H^n_c\left(\mathcal{X}^{S-\ord},\omega_T\right)^{\{\alpha_S\}}\{f\} = 0$ for $n \ne d - |T|$.
    \item If $S \cap T = \varnothing$ and $n = d - |T|$, then the natural pushforward map is an isomorphism
    \[
    H^n_c\left(\mathcal{X}^{S-\ord}, \omega_T\right)^{\{\alpha_S\}}\{f\}
    \xrightarrow{\ \cong\ }
    H^n_c\left(\mathcal{X}, \omega_T\right)\{f\}.
    \]
   \end{itemize}
  \end{proposition}

  \begin{proof}
   This is a special case of the results of \cite{boxerpilloni20}. The choice of $S$ determines a higher-level Shimura variety $Y_S \to Y$, parametrising choices of cyclic subgroup $C_i \subset A[\mathfrak{p}_i]$ for each $i \in S$. This has a natural semistable model over $\ZZ_p$, with special fibre $Y_{S, 0}$, and the locus $Y_{S, 0}^{\mathrm{mul}}$ where the level groups $C_i$ are multiplicative is an open subvariety (it is one of the $2^{|S|}$ components of the ordinary locus) which maps isomorphically to $Y_0^{S-\ord}$. This decomposition also extends to toroidal compactifications $X_S \supset Y_S$, for a suitable choice of toroidal boundary data.

   The classicity results of \cite[\S 5]{boxerpilloni20} show that if $S \cap T \ne \varnothing$, then the small-slope eigenspace $H^*_c(\mathcal{X}_S^{\mathrm{mul}}, \omega_T)^{\{\alpha_S\}}$ is zero in all degrees; and if $S \cap T = \varnothing$, then the natural pushforward map
   \[ H^*_c(\mathcal{X}_S^{\mathrm{mul}}, \omega_T)^{\{\alpha_S\}} \to H^*(\mathcal{X}_S, \omega_T)^{\{U'_S = \alpha_S\}}\]
   is an isomorphism, where $\{U'_S = \alpha_S\}$ denotes a generalised eigenspace for the transpose Hecke operators $U'_{\mathfrak{p}_i}$ for $i \in S$ (which coincide on the multiplicative locus with our partial Frobenii $\varphi_i$, but are defined on the whole of $X_S$). Since the cohomology on the right-hand side is concentrated in degree $n = d - |T|$, this gives the first and second statements. For the final statement, it suffices to note that since $\pi_{\mathfrak{p}_i}$ is an irreducible principal series representation of $\GL_2(\QQ_p)$, the trace map from the Iwahori-invariants of this representation to the $\GL_2(\ZZ_p)$-invariants restricts to an isomorphism on the $U' = \alpha_i$ eigenspace.
  \end{proof}

 \subsection{Conclusion of the proof}

  \begin{notation}
   For $S \subseteq \{1, \dots, d\}$, write
   \[ \Fil_S^- D_p = \frac{D_p}{\sum_{i \in S} \Fil_i^+ D_p},\]
   whose dimension is $2^{d - |S|}$. For $T \subseteq \{1, \dots, d\}$, we define $\Fil_j^+ \Fil_S^- D_p$ as the image of $\Fil_j^+ D_p$ in $\Fil_S^-$; note that this is zero unless $S \cap T = \varnothing$.
  \end{notation}

  One checks that this defines an $I$-filtration on $\Fil_S^- D_p$, where $I$ is the poset of subsets of $\{1, \dots, d\}$; and the $T$-th graded piece $\Gr^T\Fil_S^- D_p$ is isomorphic to $\Gr^{T} D_p$ if $S \cap T = \varnothing$, and is zero otherwise.

  Since $H^n_c\left(\mathcal{X}^{S-\ord},\omega_T\right)^{\{\alpha_S\}}\{f\}$ vanishes outside degree $d - |T|$, we can argue exactly as in \cref{prop:plecfiltration} to see that the subspaces $\mathscr{F}^T D_p^{\{\alpha_S\}}$ satisfy the distributive property, and the $T$-th graded piece is $0$ if $S \cap T \ne \varnothing$ and maps isomorphically to the corresponding graded piece of $\Fil^\bullet D_p$ otherwise. This shows that the map of $I$-filtered vector spaces
  \[ \left(D_p^{\{ \alpha_S\} }, \mathscr{F}^\bullet\right) \to \left(\Fil_S^- D_p, \Fil^\bullet\right)\]
  induces an isomorphism on every graded piece, and is hence an isomorphism of $I$-filtered vector spaces, completing the proof.

\section{Quadratic case}
 \label{sect:quadratic}

 Throughout this section we suppose $[F : \mathbf{Q}] = 2$. Then we can prove a stronger result by making use of the self-duality of $D_p$.

 \begin{theorem}\label{thm:nonBC}
  Suppose the following conditions hold:
  \begin{enumerate}
   \item[(a)] The character of $f$ is trivial.
   \item[(b)] For each prime $\mathfrak{p}_i$ above $p$, at least one of the two roots $\alpha_i, \beta_i$ of the Hecke polynomial of $f$ has strictly small slope.
   \item[(c)] The set of pairwise products $\{ \alpha_1 \alpha_2, \alpha_1 \beta_2, \beta_1 \alpha_2, \beta_1 \beta_2\}$ has four distinct elements.
  \end{enumerate}
  Then \cref{main} is true.
 \end{theorem}

 We also have a complementary result for base-change forms (which never satisfy condition (c)).

 \begin{theorem}\label{thm:BC}
  Suppose the following conditions hold:
  \begin{enumerate}
   \item[(a')] $f$ is the base-change of an elliptic modular form $f_0$ with trivial character (so in particular $k_1 = k_2$).
   \item[(b')] The roots $\alpha_0, \beta_0$ of the Hecke polynomial of $f$ at $p$ are distinct, and at least one has strictly small slope.
   \item[(c')] We have $\alpha_0 / \beta_0 \notin \{\pm 1\}$.
  \end{enumerate}
  Then \cref{main} is true.
 \end{theorem}

\subsection{Proof of \cref{thm:nonBC}}

 The condition $[F : \mathbf{Q}] = 2$, together with hypothesis (a), implies that $V_p(f)$ carries a canonical orthogonal form $\lambda$ arising from Poincar\'e duality; and the tensor induction has a canonical-up-to-scalars orthogonal form, arising as the tensor induction of the symplectic form on $V_p^{\mathrm{std}}(f)$. We can, and do, choose the isomorphism $\psi$ of \eqref{eq:defpsi} to be compatible with the orthogonal forms (up to scalars). The orthogonal form $\lambda$ defines an orthogonal form on $D_p$, which we denote by the same letter.

 \begin{proposition} For $i = 1, 2$, we have:
  \begin{enumerate}[(1)]
   \item The partial Frobenius $\varphi_i$ satisfies $\lambda(\varphi_i x, \varphi_i y) = p^{(w + 1)} \lambda(x, y)$.
   \item The space $\Fil_i^+ D_p$ is a maximal isotropic subspace of $D_p$.
   \item The eigenspaces $D_p^{(\varphi_i = \alpha_i)}$ and $D_p^{(\varphi_i = \beta_i)}$ are maximal isotropic.
  \end{enumerate}
 \end{proposition}

 \begin{proof}
  Part (1) is straightforward from the definitions. For (2), it follows easily from the shape of the cup-product on the BGG complex that $\Fil_i^+$ is an isotropic subspace, and since it is 2-dimensional it is maximal isotropic. For part (3), we note that assumption (c) implies $\alpha_i \ne \beta_i$; so we see from part (1) that $\lambda$ must vanish on the $\alpha_i$ and $\beta_i$ eigenspaces and identify each with the dual of the other. So they must each be 2-dimensional and maximal isotropic.
 \end{proof}

 \begin{corollary}
  If $k_1 \ne k_2$, the isomorphism $\psi_p$ respects the partial filtrations. If $k_1 = k_2$, then $\psi_p$ either respects the partial filtrations, or interchanges them (so the image of $\Fil_1^+$ is $D_{\mathfrak{p}_1} \otimes (\Fil^+  D_{\mathfrak{p}_2})$ and vice versa).
 \end{corollary}

 \begin{proof}
  Since $\psi_p$ must respect the Hodge filtration, we know that $\Fil_1^+ D_p + \Fil_2^+ D_p$ and $\Fil_1^+ D_p \cap \Fil_2^+ D_p$ map to their analogues in the tensor product. So $\psi_p$ descends to a bijection
  \[ \frac{\Fil_1^+ D_p + \Fil_2^+ D_p}{\Fil_1^+ \cap \Fil_2^+} \to \frac{(\Fil^+ D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2}) + (D_{\mathfrak{p}_1} \otimes \Fil^+ D_{\mathfrak{p}_2})}{\Fil^+ \otimes \Fil^+}. \]

  The orthogonal form $\lambda$ descends to a nondegenerate orthogonal form on $\frac{\Fil_1^+ D_p + \Fil_2^+ D_p}{\Fil_1^+ \cap \Fil_2^+}$. Since a nondegenerate orthogonal form cannot have more than two isotropic lines, it follows that $\Fil_1^+$ and $\Fil_2^+$ are the only two isotropic subspaces intermediate between $\Fil_1^+ + \Fil_2^+$ and $\Fil_1^+ \cap \Fil_2^+$. Since $\psi$ is compatible with the orthogonal forms, it follows that $\psi_p$ must map these spaces to $(\Fil^+ D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2})$ and $(D_{\mathfrak{p}_1} \otimes \Fil^+ D_{\mathfrak{p}_2})$ \emph{in some order}. If $k_1 \ne k_2$, then only one of these lines is contained in the middle Hodge filtration step, so we conclude that $\psi_p$ sends $\Fil_1^+ D_p$ to $\Fil^+ D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2}$, and similarly for $\Fil_2^+ D_p$.
 \end{proof}

 We now consider the Frobenius action, using hypothesis (c). Then the $(\varphi_1 = \alpha_1, \varphi_2 = \alpha_2)$ simultaneous eigenspace of $D_p$ coincides with the  $\varphi = \alpha_1 \alpha_2$ eigenspace, and similarly for the $(\varphi \otimes 1 = \alpha_1, 1 \otimes \varphi = \alpha_2)$ eigenspace of $D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2}$. Since $\psi_p$ commutes with $\varphi$, it must therefore send the $(\varphi_1 = \alpha_1, \varphi_2 = \alpha_2)$ eigenspace to the $(\varphi \otimes 1 = \alpha_1, 1 \otimes \varphi = \alpha_2)$ eigenspace. Repeating the argument for the other three pairs of roots, we conclude that $\psi_p$ commutes with the partial Frobenii. In particular, if $k_1 \ne k_2$ the proof of \cref{thm:nonBC} is complete.

 In the more delicate $k_1 = k_2$ case, we use \cref{thm:main}. The theorem tells us that $\Fil_2^+ D_p \cap D_p^{(\varphi_1 = \alpha_1)}$ is one-dimensional. Since we have established that $\psi_p$ is compatible with the partial Frobenii, it cannot send $\Fil_2^+ D_p$ to $(\Fil^1 D_{\mathfrak{p}_1}) \otimes D_{\mathfrak{p}_2}$, since the former has nonzero intersection with the $\varphi_1 = \alpha_1$ eigenspace, while the latter does not (because of the weak admissibility of $D_{\mathfrak{p}_1}$). This completes the proof of Theorem \cref{thm:nonBC}.

\subsection{Repeated eigenvalues}

 We now consider, briefly, the contrary situation where (a), (b) of \cref{thm:nonBC} hold, but rather than (c), we suppose that the set of pairwise products has size exactly 3; so without loss of generality we may suppose that $\alpha_1 \beta_2 = \beta_1 \alpha_2$ but there are no other repetitions. Then the same proof as above shows that the $(\alpha_1, \alpha_2)$ and $(\beta_1, \beta_2)$ eigenspaces are sent to their analogues in $D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2}$. Meanwhile, the 2-dimensional $(\varphi = \alpha_1\beta_2 = \beta_1 \alpha_2)$-eigenspace is the sum of the $(\alpha_1, \beta_2)$ and $(\beta_1, \alpha_2)$ simultaneous eigenspaces, which are isotropic lines. So these must be sent to the $(\alpha_1, \beta_2)$ and $(\beta_1, \alpha_2)$ eigenspaces in the tensor product \emph{in some order}. That is, we have two cases:

 \begin{itemize}
  \item Case A: $\psi_p$ commutes with the partial Frobenii.
  \item Case B: $\psi_p$ intertwines $\varphi_1$ and $\varphi_2$ with  $\xi \cdot (1 \otimes \varphi)$ and $\xi^{-1} \cdot (\varphi \otimes 1)$, where $\xi = \tfrac{\alpha_1}{\alpha_2}$ (so $\xi = \pm 1$).
 \end{itemize}

 In case A, we can argue exactly as before to show that $\psi_p$ commutes with the plectic structures. In case B, the same argument shows that we must have $k_1 = k_2$, and the isomorphism $\psi_p$ is an ``anti-plectic isomorphism'' (up to a twist), interchanging the roles of the two embeddings.

 \begin{remark}
  As noted above, the cases when $\alpha_1 = \beta_1$ or $\alpha_2 = \beta_2$ are conjectured never to occur. The only other case we have not considered is when $\alpha_1 / \beta_1 = \alpha_2 / \beta_2 = -1$; this definitely can arise, e.g.~if $f$ has complex multiplication by a totally-imaginary quadratic extension of $F$ in which both $\mathfrak{p}_i$ are inert, but it does not seem to be possible to treat it using the above methods.
 \end{remark}

 The above discussion applies, in particular, in the setting of \cref{thm:BC}: the assumptions (a') and (b') imply (a), (b) of \cref{thm:nonBC}, while (c') implies that the set
 \[ \{\alpha_1 \alpha_2,\, \dots,\, \beta_1 \beta_2\} = \{ \alpha_0^2,\, \alpha_0 \beta_0,\, \beta_0^2\}\]
 has three distinct elements. So $\psi_p$ must be either a plectic isomorphism, or a plectic anti-isomorphism. However, in this case $\psi$ is not itself uniquely determined (even up to scalars): since $f^{\sigma} = f$, the Galois representation $V_p(f)$ has an additional order 2 involution compatible with the orthogonal forms, which corresponds to swapping the factors of the tensor product. Replacing $\psi$ with its composite with this involution if necessary, we conclude that there is some isomorphism $\psi$ fulfilling the conditions of \cref{main}, thus proving \cref{thm:BC}.

 \begin{remark}
  If $f$ has $k_1 = k_2$ and $\alpha_1 \beta_2 = \beta_1 \alpha_2$, but $f$ is \emph{not} globally a twist of a base-change form (and is non-CM), then we are stuck. In this case, an isomorphism between $D_p$ and $D_{\mathfrak{p}_1} \otimes D_{\mathfrak{p}_2}$ which is compatible with the $\varphi$-module structure and the orthogonal forms must be either plectic, or anti-plectic (and both cases can occur). However, since $V_p(f)$ is irreducible in this case, only one of the two possibilities will be compatible with the global Galois action, and we cannot rule out the bizarre possibility that the global isomorphisms are the locally anti-plectic ones!
 \end{remark}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}[1]{%
 MR \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#1}.
}
\providecommand{\href}[2]{#2}
\newcommand{\articlehref}[2]{\href{#1}{#2}}

redpajama_set_name: RedPajamaArXiv


Former 'The Wanted' singer Max George is embarking on a solo career.
The Manchester-born pop star is currently promoting his debut single, 'Barcelona', across the UK.
Max, who's 29, stopped by Sunderland on his tour and had a chat with Dave Sherwood here at Sun FM.


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text:
Sanders, Clinton, and Trump: the Good, the Bad, and the Ugly
With the insurgencies of Sanders in the Democratic Party and Trump on the Republican side, the grassroots have spoken out, expressing their smoldering dissatisfaction and alienation with politics as usual. This popular upwelling points to opportunities for independent politics and even political realignments.
Polling results reported in The Washington Post show that insurgency and even socialism has legs in the world beyond the Beltway:
* 43% of likely participants in the Iowa caucus identify as "socialists" compared to only 38% as "capitalists."
* 47% of all American voters are willing to vote socialist; 59% among Democrats.
* 56% of Democratic voters had a positive view of socialism; 49% considered themselves "anti-Wall Street."
Although an independent and self-professed democratic-socialist, Bernie Sanders may well be the "good Democrat" in this race. Sanders has the potential to legitimize the Democratic Party in the eyes of the party's traditional liberal base of workers, students, and environmentalists who feel abandoned and only vote Democrat as the perceived lesser of two evils.
Sanders swept the 2016 Democrat New Hampshire presidential primary with a landslide 60% of the vote. He won without accepting corporate PAC money, with practically all the newspapers endorsing his opponent, and with almost every Democratic Party official in opposition.
The Democrat establishment has yet to warm up to Sanders. To date Sanders has 44 presidential primary delegates to Clinton's 394, the vast bulk of whom are unelected superdelegates.
Nor have many of the liberal Democratic pundits felt the Bern. Noble prize winning economist Paul Krugman lauds Clinton's "fundamental toughness" while warning that there are "bigger problems with his (Sanders') candidacy than lack of political realism."
Interestingly, Sanders has been attacked mostly on his political positions and not on some supposed gaffe. There has been almost no ad hominem demonization campaign against Sanders by the Democrat establishment and its media sycophants as there was when another presidential hopeful from Vermont, Howard Dean, ran on a somewhat insurgency platform – though not nearly as progressive as Sanders.
Dean was pilloried by the establishment media for his supposed "scream" indiscretion, which played 633 times on TV in the four days after he enthusiastically rallied his supporters. This completely manufactured smear campaign effectively eliminated Dean from the race.
So why has the Democrat establishment not stooped to character assassination with their near unanimous opposition to Sanders politically? It could be that these gatekeepers have integrity.
A more likely explanation is that they have a stake in preserving the credibility of Sanders in the hopes that, when Clinton takes the nomination, Sanders will be the sheep dog who herds the dissident masses back into the Democratic Party. Sanders has already chosen not to run as an independent and agreed to support the Democrat nominee.
What if Sanders goes on to win the White House? Would his presidency be any different from Hillary Clinton's? Sanders holds out the hope for a kinder, gentler imperialism with a less rapacious capitalist face. He wants to reform the beast, but not challenge it.
A historical analogy may be pertinent to assess whether Sanders' intentions could match his ultimate deeds in terms of affecting a political revolution in the US. Recall that New Deal liberal from Texas who had little stomach for strident Cold War warriors and who wanted to concentrate on domestic reform while ignoring foreign policy. But when LBJ inherited the Vietnam War from his predecessor JFK, the logic of Johnson's ideological assumptions drove him to become the war monger for which history remembers him.
A similar logic would be operating on a Sanders presidency given the imperatives of imperialism today. Sanders, a supporter of NSA's domestic surveillance at home and "the war on terrorism" abroad, could end up sacrificing his grand domestic reforms to fund the insatiable appetite of the US military.
Is Hillary Clinton feeling the Bern and being pushed to the left? After all she came out against the Trans-Pacific Partnership (TPP). As Glen Ford of the Black Agenda Report comments, Sanders may only have forced Clinton to lie more.
The North American Free Trade Agreement (NAFTA) was Bill Clinton's greatest achievement in his eight years in the White House. NAFTA sold out the unions and the environment. While many Republicans supported NAFTA, only a Democrat was able to force such a betrayal down the throats of the party's own constituency to the cheers of Wall Street.
Recall Obama-the-candidate promising to repeal NAFTA, close Gitmo, forbid companies in bankruptcy to give bonuses to their executives, sign the Employee Free Choice Act (making it easier for workers to unionize), etc. Recall too that Obama was the candidate favored with the most financial contributions from Wall Street, who promptly paid back his sponsors when in office.
There is little reason to expect better from Mrs. Clinton, especially on the TPP, which is like NAFTA on steroids.
Clinton promotes herself as some sort of progressive, but she is demonstratively closer to the Republican candidates on key issues than she is to Sanders:
* Reinforce military alliances against China.
* Strengthen US military alliances generally.
* Maintain Israel's military superiority.
* Establish a no-fly zone, conduct more US airstrikes, arm foreign mercenaries, and allow US Special Forces to engage in combat in Syria.
* Increase sanctions and bolster US weaponry against North Korea.
* Ratchet up US sanctions against Russia and expand US military in Eastern Europe.
Given campaign promises like these, Clinton may well be the one who is out of touch with Democrat base. She may be the "bad Democrat," but the "good Republican," in this race.
As a Republican acquaintance of mine presciently crowed the day that Clinton-the-husband won the presidency in 1992, "the best Republican won," meaning the nominal Democrat would better implement a pro-business/military agenda traditionally associated with the Republican Party. In practice, the Democrats prove to be not the "lesser evil" but the "more effective evil."
Despite leading in the polls, Donald Trump still has a tough slog to go to get the Republican nomination. He won the New Hampshire Republican presidential primary with 35.3% of the votes. However, what passes for moderates in the Republican Party – Kasich, Rubio, and Bush – got a combined vote of 37.4%. As the horse race continues and the anointed "moderate" eventually emerges, Trump may well lose his lead.
Meanwhile Senator Ted Cruz is on Trump's right flank. Cruz, a more fundamentalist Christian and a more consistent right-winger, wants to keep the US embargo of Cuba.
In contrast Trump finds himself in the same camp as Sanders on ending the embargo of Cuba as well as abiding by the Iran nuclear deal, opposing the Iraq war, rejecting the TPP, and denying cuts to Medicare and Social Security. Nor does either Trump or Sanders oppose same-sex marriage. Once both candidates were pro-choice, with Trump now hedging on this key social issue.
Of course Trump, with his Islamophobic and immigrant bashing persona, is UGLY. That could be a fatal flaw to his candidacy in a party that must preserve its share of the heterogeneous but usually pro-immigrant Latino vote to win the presidency.
Revolt of the Insurgents in Both Corporate Parties
As the disaster that is neo-liberal capitalism lurches forward, increasing numbers of disaffected and disposed are left in its aftermath. These are the people who are responding most to the insurgency campaigns of Sanders and Trump.
In the 1850s, one of the two major US political parties experienced an anti-slavery insurgency, which split off to become a third party and is today the Republican Party. Meanwhile the parent party, the Whigs, faded into oblivion.
The Sanders and Trump insurgencies may be harbingers of fractures to come within the Democratic and Republican parties as populist peripheries splinter from the parties' corporatist centers.
Battle for the Working Class
The demographics of both the Sanders and the Trump supporters have some notable similarities…many are working class. The great challenge for the left is to win over these disaffected and disposed or surely they will become the shock troops of the right wing.
This pressing task of winning the working class is not an easy one, because at least part of Trump's appeal is bigotry and the data suggest a disturbing correlation between his supporters and racism. A similar divide relates to the immigration issue.
It is easy enough for the left to dismiss Trump as a racist and nativist. But it will be a hard nut to crack for the left to figure out how to woo working class people with whom we have unity on some but not all fundamentals in a principled manner and not simply write them off. We need to pull these folks to the left, not push them to the right.

redpajama_set_name: RedPajamaCommonCrawl


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WOW! What a busy year. The AfriOceans Warriors Programme 2012, a 3 yr environmental education programme sponsored by the National Lotteries Board, started in earnest in January as we approached schools in the South Peninsula of Cape Town to join the programme. We started with 12 Primary schools as AfriOceans Blue Schools.
In February our AfriOceans Warriors Environmental Programme was in full swing, celebrating World Wetlands day and our wonderful team, led by Terry Corr, would successfully reach 27 000 children by the end of December 2012. Each AfriOceans Warrior went through a right of passage which included hand painting a kelp vuvuzela. In March we ran a very successful shark education programme for the Blue Flag Beaches in partnership with the City of Cape Town reaching 720 children, teaching them about the importance of sharks and shark safety.In April, we rolled out a road show up the South African coast, presenting to 6 schools as part of our HANDS UP FOR SHARKS campaign, and collected over 1000 children's handprints for our Deep Freedive for Sharks campaign.
In June we hosted our first AfriOceans Warriors Challenge involving our 12 Tribes from various schools and communities, and our AfriOceans Warriors from Lavender Hill took part in the Peninsula Paddle at Princess Vlei. We launched a promo video of the AfriOceans Warriors, watch it here….. . We conducted Zibi Wastewise performances teaching over 10 000 township children about litter & our Oceans, and took our young Warriors on excursions to the Two Oceans Aquarium and the Imhoff Snake & Reptile Rehabilitation centre.
In September, we ran International Coastal Cleanup events for a week in partnership with the City of Cape Town, taught our special needs kids to SCUBA dive, developed the Shark Warrior beaded bracelet made by our AfriOceans Warriors and did cleanups at all our Blue Schools. In October our AfriOceans Warriors kids went on a white shark cage diving excursion with African Shark Eco Charters. We participated in the Simonstown Penguin Festival and we ran the Drains to the Oceans Programme with the City of Cape Town in Kraaifontein, where kids picked up thousands of crisp packets up on their school grounds.
In December we held our final AfriOceans Warriors Event, involving our 17 Blue School Tribes – they displayed their eco journals and school and community Blue projects of work done during the year, held a recycled-waste fashion show, showcasing song and dance routines.They competed in an extreme AfriOceans Warriors obstacle course for the coveted AfriOceans Warriors Challenge floating trophy, won convincingly by St. James Primary in Kalk Bay. Inspired by Maddie, we selected our 6 National AfriOceans Warriors Ambassadors.
Wow! What a year! We reached 27 000 school going learners from 100 schools, reached 120 teachers and touched over 45 000 adults with our programme during 2012. 2013 promises to be even busier!
Wishing all our AfriOceans Warriors, supporters, partners and our incredible sponsors an Awesome 2013 and wherever you are, please take ACTION for our Oceans! From our AfriOceans Warriors Project team Lesley Rochat, Dr. Leonard Compagno, Terry Corr, Charmaine Rochat, Ishmael Sabodien, Monica Sokudela, Naomi Mdayi and our three CPUT interns, Ovayo, Zinthatu and Sibulele.


text:
'use strict';

// MODULES //

var tape = require( 'tape' );
var sqrt = require( '@stdlib/math/base/special/sqrt' );
var SQRT_HALF = require( './../lib' );

// TESTS //

tape( 'main export is a number', function test( t ) {
	t.equal( typeof SQRT_HALF, 'number', 'main export is a number' );
	t.end();
});

tape( 'the exported value is a double-precision floating-point number which equals the square root of 1/2', function test( t ) {
	var expected = sqrt( 1/2 );
	t.equal( SQRT_HALF, expected, 'equals sqrt(1/2)' );
	t.end();
});

redpajama_set_name: RedPajamaGithub


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text:
Erin Lucas Is Leaving The City to Be a Gypsy
Published at 6:59 pm on July 9, 2009
Whitney Port'sCity
sidekick, Erin Lucas, announced last week she won't be part of the second season of the reality series. "I decided to do something different," she told us this week at the launch party for Alizé's online reality series,Concrete and Cashmere
(on which Lucas is a guest judge). "I have a project in the mix, but I want to get into more hosting. That's really what I love." She also admitted she hasn't spoken to Port inmonths
, though she claims there's no hard feelings and she'd give Port a big hug if she ever saw her again. "We were [close], but things change," she said. "There was a falling out" — reality-TVgold
. However, we spottedThe City
crew filming at a Gen-Art benefit two weeks ago, which Erin attended along with Whitney. And you know what? There was no hugging. "I wasn't filming [that night], I was just there," Lucas tried to explain. She said running into the old camera crews and co-stars was awkward. "But Sammy's still one of my best friends." So why'd she give it up? "It was just hard," she said. "It was a chapter in my life and it was nothing but positive. I'm glad I did it, but on to the next thing. I'm kind of a gypsy. I released a statement a couple of days ago, but this is the first interview with people asking me about it, so I don't know what to say." Life is hard when no one gives you a script.
Previously on The Cut...
Agyness Deyn: 'Even Though I'm in Fashion, I Don't, Like, Do Fashion'
Pharrell Williams Makes a Booty Call
Dolce & Gabbana Publicist Arrested for Hacking into Socialite's Voicemail
Copyright FASHN - NY Mag


text:
Q: Unable to get provider - rarely crash on kitkat Few times in the week i receive crash report in GP console:
java.lang.RuntimeException: Unable to get provider mypackage.MyProvider: java.lang.ClassNotFoundException: Didn't find class "mypackage.MyProvider" on path: DexPathList[[zip file "/mnt/asec/mypackage-1/pkg.apk"],nativeLibraryDirectories=[/mnt/asec/mypackage-1/lib, /vendor/lib, /system/lib]]
    at android.app.ActivityThread.installProvider(ActivityThread.java:5018)
    at android.app.ActivityThread.installContentProviders(ActivityThread.java:4589)
    at android.app.ActivityThread.handleBindApplication(ActivityThread.java:4522)
    at android.app.ActivityThread.access$1500(ActivityThread.java:151)
    at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1381)
    at android.os.Handler.dispatchMessage(Handler.java:110)
    at android.os.Looper.loop(Looper.java:193)
    at android.app.ActivityThread.main(ActivityThread.java:5299)
    at java.lang.reflect.Method.invokeNative(Method.java)
    at java.lang.reflect.Method.invoke(Method.java:515)
    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:829)
    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:645)
    at dalvik.system.NativeStart.main(NativeStart.java)
Caused by: java.lang.ClassNotFoundException: Didn't find class "mypackage.MyProvider" on path: DexPathList[[zip file "/mnt/asec/mypackage-1/pkg.apk"],nativeLibraryDirectories=[/mnt/asec/mypackage-1/lib, /vendor/lib, /system/lib]]
    at dalvik.system.BaseDexClassLoader.findClass(BaseDexClassLoader.java:56)
    at java.lang.ClassLoader.loadClass(ClassLoader.java:497)
    at java.lang.ClassLoader.loadClass(ClassLoader.java:457)
    at android.app.ActivityThread.installProvider(ActivityThread.java:5003)
    ... 12 more
    Suppressed: java.io.IOException: unable to open DEX file
        at dalvik.system.DexFile.openDexFileNative(DexFile.java)
        at dalvik.system.DexFile.openDexFile(DexFile.java:296)
        at dalvik.system.DexFile.<init>(DexFile.java:80)
        at dalvik.system.DexFile.<init>(DexFile.java:59)
        at dalvik.system.DexPathList.loadDexFile(DexPathList.java:263)
        at dalvik.system.DexPathList.makeDexElements(DexPathList.java:230)
        at dalvik.system.DexPathList.<init>(DexPathList.java:112)
        at dalvik.system.BaseDexClassLoader.<init>(BaseDexClassLoader.java:48)
        at dalvik.system.PathClassLoader.<init>(PathClassLoader.java:65)
        at android.app.ApplicationLoaders.getClassLoader(ApplicationLoaders.java:57)
        at android.app.LoadedApk.getClassLoader(LoadedApk.java:326)
        at android.app.LoadedApk.makeApplication(LoadedApk.java:508)
        at android.app.ActivityThread.handleBindApplication(ActivityThread.java:4514)
        ... 10 more

Also, after adding Firebase features to my project i start to receive same crash reports about com.google.firebase.provider.FirebaseInitProvider.
According to reports, this errors happens only on:
Android 4.4 98,7 %
Android 4.2 1,3 %
I've try all this ClassNotFoundException for a ContentProvider, but no result.
Is this really a system bug and every developer deals with it, or it's my error? I really want to stop annoying users. Thanks.

A: Your case really looks like bug described there in Android's Issue Tracker.
There you got issue when user starting app immediately after apk update.
There is workaround from comments which could help you to prevent this bug:
public class DevToolsApplication extends Application {
private static final String TAG = "DevToolsApplication";

    @Override
    public void onCreate() {
        super.onCreate();
        AppLogger.i(TAG, "app start...");
        checkAppReplacingState();
    }

    private void checkAppReplacingState() {
        if (getResources() == null) {
            AppLogger.w(TAG, "app is replacing...kill");
            Process.killProcess(Process.myPid());
        }
    }
}

redpajama_set_name: RedPajamaStackExchange


text:
Success! Kisembo from Uganda raised $249 to fund hernia repair surgery.
Kisembo's treatment was fully funded on January 8, 2018.
Kisembo's story
Kisembo's update
Kisembo underwent hernia repair surgery.
Kisembo's hernia repair was a success. He is no longer at risk of getting hernia obstruction and strangulation. After recovery he will be able to continue with his studies.
His grandmother says, "Kisembo is now much better. He is able to walk very well around the ward and he can move outside. After recovery he will be able to resume his studies when the holiday ends. I am very excited by Kisembo's recovery because now he will concentrate in class. Thank you very much Watsi for paying for my grandson's surgery. I cannot thank you enough. Keep up the spirit of helping. God bless the works of your hands.''
Kisembo's hernia repair was a success. He is no longer at risk of getting hernia obstruction and strangulation. After recovery he will be ab...
Kisembo is a five-year-old boy from Uganda. He is the eldest of two children, and he has a younger brother. Kisembo currently attends nursery school. His father is a builder, and his mother is a farmer.
About three months ago, Kisembo developed a swelling around his umbilical region. A nurse at his school gave him treatment, but the swelling did not subside. Kisembo could not sleep because of the pain, and the painkillers were not working. At the end of the school term, Kisembo's father took him to the hospital, where Kisembo was diagnosed with an epigastric hernia. If not treated, Kisembo may experience intestinal tissue twisting and blocking.
On September 6, Kisembo will undergo hernia repair surgery. Our medical partner, African Mission Healthcare Foundation, has requested $249 to fund Kisembo's procedure. This will cover the full cost of treatment, including a three-night hospital stay, supplies, labs, medication and physician time.
Kisembo is a five-year-old boy from Uganda. He is the eldest of two children, and he has a younger brother. Kisembo currently attends nurser...
Kisembo's case was submitted by Joan Kadagaya at African Mission Healthcare.
Kisembo's Timeline
Kisembo was submitted by Joan Kadagaya, Curative Medical Support Program-Partner Representative at African Mission Healthcare.
Kisembo received treatment at Holy Family Virika Hospital in Uganda. Medical partners often provide care to patients accepted by Watsi before those patients are fully funded, operating under the guarantee that the cost of care will be paid for by donors.
Kisembo's profile was published to start raising funds.
Kisembo's treatment was successful. Read the update.
Kisembo's treatment was fully funded.
Share Kisembo's story
Funded by 3 donors
On average, it costs $249 for Kisembo's treatment
A hernia occurs when a portion of the intestine protrudes through the lower abdominal wall, usually for one of two reasons. The first is a congenital abnormality in which the tissues did not close. The second is excessive stress in an adult, often due to heavy physical labor or pregnancy. Patients experience a bulge or lump in the affected area. The hernia may cause the patient to feel pain, discomfort, weakness, pressure, and sensations of heaviness or aching. These symptoms are often exacerbated when the patient coughs, bends over, or lifts heavy objects. In some cases, hernias have no symptoms and are only detected during routine medical exams.
Patients with uncomplicated hernias may experience only annoyance or discomfort. As the hernia opening expands, the discomfort will increase. Small openings are more likely to trap the intestine, potentially leading to intestinal damage or death.
Hernias are common in Africa. People often do very hard physical labor and lift heavy objects. Women tend to have more children than those in the United States. It is possible that some hernias have infectious or genetic causes.
Surgery lasts for three to eight hours, depending on the age of the patient and the severity of the hernia. The patient will stay in the hospital anywhere from two days to eight weeks, again depending on the age of the patient and the severity of the hernia. The patient is continually monitored.
Treatment is curative. The chance of intestinal strangulation or bowel obstruction reduces significantly.
Hernia repair is not a risky procedure, and it comes with few side effects.
Many patients will ignore a hernia until it becomes uncomfortable and seek care at that time. Some people will wear tighter pants or a tight band around the waist to prevent the intestine from protruding.
If the hernia is not "stuck," patients tend to ignore it and adapt to living with it. However, this could lead to future complications.
Meng Heng is a three-month-old boy and the youngest child in his family. His mother works as a tailor, while his father is a soldier. He also has a six-year-old brother in first grade. When Meng Heng was born, his parents noticed that he had a small growth at the front of his left ear. The growth has been identified as a skin pit, or a sinus tract traveling under the skin that doesn't belong there. It is marked by a tiny opening to the tract, in front of the ear and above his ear canal. His parents noticed that the tag has grown, and also that Meng Heng experiences drainage from his ear. Fortunately, our medical partner, Children's Surgical Centre, is helping Meng Heng receive treatment. On January 10th, he will undergo a preauricular skin pit excision and drainage. Now, he and his family need help raising $296 to fund his procedure and care so he can grow up healthy. His mother shared, "I want his ear to be fixed so he can grow up strong and healthy and not have problems hearing."
0%funded
$0raised
Tabitha is a mother of four children and shared that her husband passed away few years ago. She works hard to find jobs, mostly doing laundry for people, to provide for her family. Over the years, she has managed to take care of her four children with the limited income she gets from her work. Last November, Tabitha noticed a worrisome bump and consulted with her daughters who advised her to visit a hospital. She gathered some funds and visited a hospital nearby, but was not able to receive any help. She went to another facility that referred her to our Medical Partner's care center Kijabe Hospital. There doctors have told Tabitha that she needs an urgent surgery. Tabitha has been diagnosed with breast cancer. Without treatment, the cancer may spread to other organs. A mastectomy, a surgery to remove breast tissue, has been advised to rid her body of breast cancer and to prevent the cancer from metastasizing. Our medical partner, African Mission Healthcare Foundation, is requesting $1,110 to cover the cost of a mastectomy for Tabitha. The procedure is scheduled to take place on January 14th. After treatment, Tabitha will hopefully return to a cancer-free life. Tabitha says, "I was very shocked when I learned about my condition and up to now, I do not know what to do as I cannot afford the surgery."
Ananiya is a 5-year-old boy from Ethiopia. He loves playing football and playing with toys. Ananiya is the firstborn in a family of two children. Ananiya's father passed away five years ago so his uncle helps to take care of them. His mom is worried and concerned about her son's wellbeing and is seeking financial assistance to help Ananiya get the second-stage surgery that he needs to be healthy. Ananiya was born with hypospadias, a congenital abnormality that causes urinary dysfunction. Without treatment, he will continue to experience uncomfortable symptoms and will be at risk of infertility. Fortunately, Ananiya is scheduled to undergo corrective surgery on November 16th. Our medical partner, African Mission Healthcare Foundation, is requesting $1,293 to cover the total cost of his procedure and care. His mom is feeling a renewed sense of hope for the future. She said, "I hope my child goes to school after the treatment. If it's God's will, he may even be a doctor in the future."


text:
Bright Health to Expand into California through Planned Acquisition of Brand New Day Health Plan
- This transaction furthers the company's commitment to helping all consumers, including seniors and vulnerable Americans, build long-term provider relationships and receive higher quality, more affordable healthcare
- This transaction reaffirms Bright Health's commitment to the growing Medicare Advantage market
- This expansion will bring Bright Health Plan's national footprint to a total of 34 markets in 13 states, serving approximately 200,000 members
Bright Health
MINNEAPOLIS, Jan. 8, 2020 /PRNewswire/ -- Today, Bright Health, a diversified consumer-focused healthcare company, announced it has signed an agreement to acquire California-based Universal Care (d.b.a. Brand New Day) health plan. The transaction will combine Brand New Day's local expertise and specialized clinical programs with Bright Health's nationally recognized leadership, next-generation technology and intelligence platform, and innovative personalized approach to healthcare. The transaction is subject to regulatory approval and expected to close in 2020.
"For 37 years, the Davis family and Brand New Day have been committed to improving their members' health in local communities throughout California," said Mike Mikan, President and Vice Chairman of Bright Health. "Their integrated approach to serving vulnerable populations with complex conditions aligns closely with our Health Plan Care Partner model. By combining our core strengths, we will be able to deliver an even better experience for our members across the country."
Brand New Day was founded in 1983 by the Davis family to improve health outcomes through an emphasis on complex care management, population health and strong relationships between patients and their primary care physicians. Current leadership, including Jeff and Jay Davis, will join the Bright Health Plan leadership team and will continue to lead local market operations, ensuring continuity for Brand New Day members and alignment across Bright Health Plan's business.
"We are impressed by Bright Health's seasoned leadership, national scale and depth of expertise. And the speed with which they've grown demonstrates the strength of their consumer-focused healthcare model," said Jeff Davis, CEO of Brand New Day. "We look forward to working alongside Bright Health leadership to deliver robust healthcare options to more consumers in California and beyond."
"This transaction gives Bright Health Plan a strong presence in California with an established and philosophically aligned partner," said Mikan. "But beyond that, it positions us to leverage this partnership to bring key elements of Brand New Day's patient-first, integrated clinical model of care to all our members across product lines and geographies."
About Bright Health
Bright Health is a diversified consumer-focused healthcare and technology company, providing a broad range of innovative healthcare products and services for consumers and care providers in local markets throughout the U.S. Working in close alignment with Care Partners and doctors to create a virtually integrated healthcare system, Bright Health connects their members to Personalized Care Teams, combining care and coverage to deliver high quality, simpler and more affordable everyday healthcare. Their health plans, which include individual, family and Medicare Advantage plans, are available in 22 markets across 12 states. Learn more at www.brighthealthinc.com.
About Brand New Day / Universal Care
Universal Care (d.b.a. Brand New Day) has been a leader in providing healthcare services in California for over 35 years. Brand New Day serves Medicare eligible seniors and special needs populations in 12 counties through their extensive provider network of over 7,300 PCPs and 22,300 specialists. They combine deep analytics and evidence-based clinical programs with aligned provider relationships to provide high quality, affordable care for complex and vulnerable populations. They have a long-standing commitment to providing local, community-based care to individuals in a personalized manner to improve health and quality of life. For more information, visit www.BNDHMO.com or call (866) 255-4795.
SOURCE Bright Health
https://brighthealthplan.com
Bright Health Appoints New Chief Financial Officer...
Bright Health Announces $635M Series D Funding Led by NEA...


Established in 2004, Geo Fun is a partnership founded by two friends who have been exploring the vast biological and geological patrimony of the Azores since 1992.
Geo Fun is a company which focuses on active tourism with activities that range from hiking, canoeing and jeep tours, among many others.
With experienced and specialist guides to explain all the details of the surrounding landscape, Geo Fun caters for the increasing demands of the today´s explorer.


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text:
cambridge map
cambridge accommodation
Cambridge Hotels
Cambridge Campsites
Cambridge Restaurants
Cambridge Videos
Cambridge Area
Cambridge News
Cambridge Golf Courses
Cambridge History
Cambridge Airports
Cambridge Airports (1 of 5) More Cambridge Airports >
Cambridge Airports Here you can find Local Airports in Cambridge and near Cambridge .
Duxford Aerodrome
Map | More Details
Situated as the crow flyes, 8.6 miles away from Cambridge's center, Duxford Aerodrome is located 8 nautical miles (14.8 km) south of Cambridge in the village of Duxford, Cambridgeshire, England. The airfield is owned by the Imperial War Museum and is the site of the Imperial War Museum Duxford and the American Air Mu .. read more
Bourn Airport
Situated as the crow flyes, 9.8 miles away from Cambridge's center, Bourn Airport is located 7 nautical miles (13.0 km) west of Cambridge, Cambridgeshire, England. (...) .. read more
Little Gransden Airfield
Situated as the crow flyes, 15.1 miles away from Cambridge's center, Little Gransden Airfield is a licensed airfield located 5 nautical miles (9.3 km) southeast of St Neots, Cambridgeshire, England. Little Gransden is the home of Mark Jefferies, 2005 and 2006 UK Aerobatics Champion, who flies an Extra 300. It is also .. read more
Wyton Royal Air Force Base
This airport is situated as the crow flyes, 17.0 miles away from Cambridge's center. .. read more
RAF Wyton
Situated as the crow flyes, 17.0 miles away from Cambridge's center, RAF Wyton is a Royal Air Force station near St. Ives, Cambridgeshire, England. In terms of organisation RAF Wyton is now part of the combined station RAF Brampton Wyton Henlow, a merger of Wyton with two previously separate bases, RAF Brampton and RA .. read more
Mildenhall Royal Air Force Base
RAF Mildenhall
Situated as the crow flyes, 18.2 miles away from Cambridge's center, station located at Mildenhall in Suffolk, England. It is currently the home of the United States Air Force 100th Air Refueling Wing. The 100th RW supports some 16,000 personnel, three geographically separated units, and 15 associated units. RAF Milde .. read more
Alconbury Royal Air Force Base
(1 of 5) More Cambridge Airports >


text:
VoIP Service Rebtel Trounces Skype On ARPU
By Desire Athow 10 March 2011
Rebtel which claims to be the world's second largest VoIP company in the world, after Skype, announced today that it had enlisted its 10 millionth user worldwide.
While this is desperately low compared to Skype, which had more than 663 million user accounts towards the end of 2010, Rebtel's doing pretty well when it comes to revenue with the average revenue per user reaching $4.00 based on last year's revenue figures (at the current run-rate, it reaches $5.50).
That compares very favourably with Skype whose ARPU for 2010 stood at just over $1.20. Five-year old Rebtel is a relative newcomer in the world of VoIP but has managed to double its user base in the last 12 months and one billionth international minute milestone reached (Skype in comparison did 3.1 billion SkypeOUT minutes in Q3 2009).
Rebtel approach differs from Skype in that it focuses mainly, like many VoIP player, on the lucrative market of international phone calls rather than free calls. Indeed, Rebtel piggybacks on Skype for PC to Phone calls.
In addition, Rebtel offers free overseas calls between two Rebtel countries, a list that currently includes 53 countries including like Algeria and Jordan. The announcement by the firm comes as analyst company, Frost & Sullivan said that it nominated Rebtel for its "emerging company of the year award" because it was impressed by its revenue figures and high growth rates.
The VoIP market though is an increasingly crowded place as Google and others plan to make a splash by trying to bundle other services. With Skype's in the process of launching an IPO in 2011, now is a good time for Rebtel to get a share of voice and get anyone to focus on what's probably the most important factors in the business, growing ARPU and profitabilty with minimum investment.
With revenues expected to double to $75 million by the end of the year, Rebtel could prove to be an attractive target for an aggressive acquirer (Facebook or Apple maybe), given that O2/Telefonica paid $200m for Jajah back in 2009.
Mobile & Telco
Covid-19 phishing attacks have evolved yet again


text:
Heart of Europe
Brain City
Suchen Sie hier:
Israelians in Berlin: Guy Sternberg
Capturing Moments of Berlin Sound
Name: Guy Sternberg
City of origin: Tel Aviv
Position: Music Producer
Guy Sternberg, music producer from Israel, talks about his decision to go all analog with his label, LowSwing records, and how it encourages musicians and listeners to be more in the moment.
Why did you want to open your own studio and what got you interested in Berlin?
I moved from Tel Aviv to Berlin in 1998 to join Babelsberg Filmstudio. In 2002, I opened LowSwing studios with the idea to create something exotic and unusual with vintage equipment. The production and engineering LowSwing does is all human without processing or programming. The sound is vintage and old school. We capture performances in the moment.
Guy Sternberg in his studio
How have things changed in Berlin over the past 20 years?
It used to be very raw and alternative in the 90s, after the wall fell. Hackesche Höfe had a lot of tiny ateliers for artists to rent for a small fee. To make their rent, they would turn the ateliers into illegal bars every Wednesday, selling beer from the crate. Just after the 2000s, Berlin drew more and more attention. Everybody wanted to be here. Things have changed a lot now, but what's cool about Berlin is that it's still hanging on to this free spirit. People are proud to be different.
"What's cool about Berlin is that it's still hanging on to this free spirit. People are proud to be different."
Berlin is also the capital of music of Germany. Why do you think that is?
Nick Cave, David Bowie, U2.... Berlin is magnetic! To start with: its cheap. So it's very good for young musicians and artists. You can just share a flat with somebody, work two shifts in a cafe and hang out in the right places to make your way. Also, Berlin is very international. You can work with bands from three different continents in the same week. There are not many cities where you get such a variety of people and sounds.
What would you say is the key to creating good music?
The main thing is the human factor. You have to have a person with the right personality and something to say. A little bit of life conflict won't hurt either! In a place like Berlin you can dedicate yourself to art and try something extreme. Berlin is low risk. Even if you fail, it's not the end of the world. Just try. This kind of atmosphere really helps people let go and trust their process.
"Berlin is very international. You can work with bands from three different continents in the same week."
LowSwing has evolved into a label in the last year. What has driven you to transform the studio into something more?
LowSwing Records helps me say what I want. I went completely analog with my label. Everything I'm doing is recorded and mixed just with tape. It isn't digitized in the process and is not available anywhere on the Internet. If you want to listen to the music, you have to buy the vinyl.
Why did you refuse to joing the music streaming industry and instead decided to go completely analog?
There's a different sound. With digital recording, you can sing out of time, out of tune, and you can fix it later. With tapes, what you do is what's going to be on it. It's so direct. You're capturing a moment. In some great classical jazz albums, the first note in the trumpet is slightly out of tune and today nobody would leave it in. For me, it's not a mistake, but in the right context, it's actually a feature. You get this fragility, you hear that somebody is playing and his lips are not really there yet.
Are people liking your new venture? Have you received criticism for not being more commercial?
There's a depth to music like this and the reception to it has been more than positive. There's a really wide vinyl community, from audiophiles who always listened to vinyl to hipsters who just discovered it in the last two years. Financial pressures are much less in Berlin, you can actually afford to try things out and follow your dream. Here, I've never had to apologize for not doing commercial stuff. You can just feel good about things.
Images & Text: blogfabrik - Paul Rikeit & Jacob Schickler
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Fair rental prices, cheaper living costs than in many other major European cities, that's what Berlin is known for.
Brits in Berlin: Jon Sykes
Name: Jon Sykes, Age: 33, I'm based in the original Berlin coworking space – betahaus. It's the ideal first stop for anyone looking to establish...
Brits in Berlin: Temi Adeniyi
Temi felt restless in London and found a new, inspiring workplace in Berlin. She talks about her phtography, rush hours and language barriers.
Brits in Berlin: Andrew
I had some friends from university that moved here a year before I did. When one of them came back over for a visit six months later all he could talk...
Where Berlin feels British...
...or Scottish or Irish. With all these places, there are plenty of reasons why Berlin could easily become your new home away from home.
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When most of us look in the mirror we don't see all that is right with ourselves - we see all that we think is wrong. This best-selling message presents a very simple yet profound truth: It is God's will for you to get free from other people's opinions and to learn to be happy in your own skin. When God made you, He made no mistakes. Learn to see yourself as God does and celebrate the masterpiece that you were created to be.


text:
package com.kirak.mock;

import com.kirak.matchers.ModelMatcher;
import com.kirak.model.Vote;

import java.time.Month;

import static com.kirak.mock.HotelTestData.HOTEL1;
import static com.kirak.mock.HotelTestData.HOTEL2;
import static com.kirak.mock.HotelTestData.HOTEL3;
import static com.kirak.mock.UserTestData.USER1;
import static com.kirak.mock.UserTestData.USER2;
import static com.kirak.mock.UserTestData.USER3;
import static com.kirak.model.abstraction.BaseIntEntity.START_SEQ;
import static java.time.LocalDateTime.of;

/**
 * Created by Kir on 20.06.2017.
 */
public class VoteTestData {

    public static final ModelMatcher<Vote> VOTE_MATCHER = ModelMatcher.of(Vote.class);

    public static final Integer VOTE1_ID = START_SEQ;
    public static final Integer VOTE2_ID = START_SEQ + 1;
    public static final Integer VOTE3_ID = START_SEQ + 2;

    public static final Vote VOTE1 = new Vote(VOTE1_ID, 9.5, "User Vote comment  1",
            of(2017, Month.JUNE, 16, 19, 32), USER1, HOTEL1);
    public static final Vote VOTE2 = new Vote(VOTE2_ID, 6.5, "User Vote comment  2",
            of(2017, Month.MAY, 4, 9, 45), USER2, HOTEL1);
    public static final Vote VOTE3 = new Vote(VOTE3_ID, 5.5, "User Vote comment  3",
            of(2017, Month.JUNE, 10, 13, 45), USER3, HOTEL2);

    public static Vote getCreatedVote() {
        return new Vote(null, 9.0, "User Vote comment  NEW",
                of(2017, Month.JUNE, 11, 14, 32), USER2, HOTEL3);
    }

    public static Vote getUpdatedVote() {
        return new Vote(VOTE1_ID, 8.5, "User Vote comment  EDITED",
                of(2017, Month.JUNE, 17, 12, 35), USER1, HOTEL1);
    }

}


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text:
Ten Thoughts on the NFL: Chicago Bears and beyond
It's time for that Monday afternoon staple here at Windy City Gridiron, our Ten Thoughts on all things NFL. We'll have a slight Chicago Bears slant of course, and be sure to use the comment thread to chime in on any Thoughts you may have.
By Lester A. Wiltfong Jr.@wiltfongjr Mar 4, 2013, 3:00pm CST
Share All sharing options for: Ten Thoughts on the NFL: Chicago Bears and beyond
1) Jake Long is not worth $11 million a year. That's just silly talk, especially for a player whose 2012 season resembled that of J'Marcus Webb.
2) With a couple more combine participants stating that teams are asking about sexual orientation, the NFL is skating on some very thin ice. They need to investigate thoroughly, and come down hard with a hefty fine.
3) I'm not sure what to make of the female kicker that had a try out with the NFL. On one hand, she deserves kudos for following her dream, but she was terrible. Like really terrible. She apparently injured herself on her first kick, but there was noting in her approach that told me she could have hit an NFL quality kickoff.
4) Reports have Darrelle Revis, Pro Bowl corner back of the New York Jets, on the trading block, but he expects top dollar, the Jets will no doubt want a ton in return, and he's coming off a knee injury. Do you think Revis is worth it at this point in his career?
5) Now that the Joe Flacco deal is official (six-years, $120.6 million, with $52 million guaranteed) I think we've all learned one thing. Timing is everything.
6) The buzz surrounding the NFL Draft big boards now has Central Michigan left tackle Eric Fisher sitting on top, and a lot of the mocks are reflecting the move by having the Kansas City Chiefs picking him at #1. I see a handful of can't miss offensive linemen in this draft, but who's your number one overall prospect?
MORE: Who is the one guy you would trade up for in the 2013 NFL Draft?
7) And speaking of Mock Drafts, the latest from the Mother-ship still has the Bears taking North Carolina offensive guard Jonathan Cooper at #20.
8) The Atlanta Falcons released running back Michael Turner, but now have interest in free agent running back Steven Jackson. It's evident that Jackson is an upgrade, but I'd interested to see how much the soon to be 30 year old tail back will set the Falcons back. There's a lot of mileage on the former Rams workhorse.
9) Don't worry Notre Dame / Manti Te'o fans, the controversial linebacker claims that he'll get his 40 time in the 4.6 range for his March 26th Pro Day. If Te'o does improve, will that be enough to quell the doubters?
10) I just don't see any scenario playing out that doesn't end with Brian Urlacher signing a short deal to stay in Chicago.
MORE: Brian Urlacher can't wait for next season to start