text:
\section{Introduction}

One of the most exciting achievements in string theory is the
remarkable success in counting microscopic counting of black hole
states, starting with the work of \cite{Strominger:1996sh}. A
particularly elegant example of this is provided by considering an
M5-brane wrapped on a complex four-cycle of a Calabi-Yau
\cite{Maldacena:1997de}. This yields a black string in five
dimensions which can be further reduced to four dimensions by
wrapping the string on $S^1$ and including momentum along the $S^1$.
A notable feature of this analysis is that, for a generic
four-cycle, the M5-brane has a smooth worldvolume and hence the only
microscopic information needed is a knowledge of the worldvolume
fields and dynamics of a single M5-brane.

There have been several detailed accounts of the M5-brane wrapped on
cycles of a generic Calabi-Yau manifold, for example see
\cite{Vafa:1997gr,Minasian:1999qn,Lambert:1999ib,Minasian:1999gg,Bonelli},
and also the M5-brane on $K3$ \cite{Cherkis:1997bx}. The case that
we are interested in here concerns an M5-brane whose worldvolume has
non-trivial one-cycles which occurs when the Calabi-Yau degenerates
to $K3\times T^2$ or $T^6$ (see also \cite{Bonelli}). This situation
was discussed in \cite{LopesCardoso} where several puzzles arose. In
particular the number of massless states was not found to be in
accordance with $(0,4)$ supersymmetry and the counting of black hole
microstates failed (albeit at sub-leading order). To resolve these
problems the authors of \cite{LopesCardoso} proposed a novel
mechanism whereby some massless modes are charged with respect to
the worldvolume gauge fields that arise from reduction of the
two-form. The main purpose of this paper is to investigate this
proposal. However we find that the correct resolution comes from
including additional massless modes which are present when the
Calabi-Yau is $K3\times T^2$ or $T^6$.

The rest of this paper is organized as follows. In section two we
present the lowest order equations of motion for an M5-brane which
is wrapped on a smooth cycle $P$ in spacetime. In section three we
consider in detail the case where spacetime is of the form ${\cal M}
= {\bf {R}}^{1,4}\times K3\times T^2$ and ${\cal M} = {\bf
{R}}^{1,4}\times T^6$. We provide a careful counting of the normal
bundle moduli and resolve a puzzle concerning $(0,4)$ supersymmetry
that was observed in \cite{LopesCardoso}. In section four we
consider four-dimensional black hole states that arise by further
compactification on $S^1$. We find that, using our analysis, the
usual counting of left-moving massless modes to determine black hole
entropy does not agree with the supergravity calculations or
arguments using anomalies. To resolve this discrepancy we propose
that  $h_{1,0}(P)$ $(4,4)$ multiplets must become massive and hence
do not appear in the low energy effective action. This provides an
alternative resolution to a second puzzle discussed in
\cite{LopesCardoso}. Finally section five contains a brief
conclusion.

\section{Lowest Order Equations of Motion}

Covariant  equations of motion of the M5-brane were first derived in
\cite{Howe}. We will not need give the full non-linear form of these
equations, however it will be enlightening to give the lowest order
equations (in terms of a derivative expansion). We will work in
static gauge where the six coordinates $x^\mu$,
$\mu,\nu=0,1,2,...,5$, of the M5-brane worldvolume are identified
with the first six coordinates of spacetime. The massless fields
consist of 5 scalars $X^A$, $A,B=6,7,8,...,10$, a two-form
$B_{\mu\nu}$ and a Fermion $\psi$ which satisfies
$\Gamma_{012345}\psi=-\psi$. Here we use a full 32-component spinor
of $SO(1,10)$. It will be sufficient to work at the lowest order in
the fields. $X^A$ represents the coordinates of the M5-brane in the
transverse space and in particular $X^A=0$ corresponds to the
M5-brane wrapped on a calibrated submanifold. We use
$M,N=0,1,2,...,10$ to denote all eleven coordinates. We use an
underline to denote tangent space indices. We will use a hat to
denote eleven-dimensional quantities, the spacetime is denoted by
${\cal M}$ and the M5-brane worldvolume by $\cal W$.

First recall the case where the worldvolume $\cal W$ admits a chiral
Killing spinor $\epsilon$; $D_\mu \epsilon=0$,
$\Gamma_{012345}\epsilon=\epsilon$. For example if ${\cal M} = {\bf
R}^{1,4}\times K3\times T^2$ and ${\cal W}={\bf R}^{1,1}\times K3$.
To lowest order in fluctuations, the equations of motion are just
that of a free theory on a curved background
\begin{eqnarray}\label{oldeqs}
\nonumber
D^2 X^{\underline A} &=&0\\
i\Gamma^\mu D_\mu\psi &=&0\\
\nonumber
H_{\mu\nu\lambda}&=&\frac{1}{3!}\epsilon_{\mu\nu\lambda\rho\sigma\tau}H^{\rho\sigma\tau}
,
\end{eqnarray}
where $H_{\mu\nu\lambda} = 3\partial_{[\mu}B_{\nu\lambda]}$ and
$\epsilon_{\mu\nu\lambda\rho\sigma\tau} $ is totally antisymmetric
with  $\epsilon^{\underline{012345}}=1$. These equations are
invariant under the supersymmetry transformations
\begin{eqnarray}
\nonumber \delta X^{\underline A} &=& i\bar\epsilon\Gamma^{\underline A}\psi\\
\delta B_{\mu\nu} &=& i\bar\epsilon\Gamma_{\mu\nu}\psi\\
\nonumber \delta\psi &=& \partial_\mu X^{\underline
A}\Gamma^\mu\Gamma_{\underline A}\epsilon + \frac{1}{2\cdot
3!}\Gamma^{\mu\nu\lambda}H_{\mu\nu\lambda}\epsilon .
\end{eqnarray}

Next we consider the case where  the spacetime ${\cal M}$ admits a
chiral covariantly constant spinor $\hat \epsilon$, $\hat D_M\hat
\epsilon=0$, $\Gamma_{012345}\hat\epsilon=\hat\epsilon$ but where
this does not descend to a Killing spinor on $\cal W$. For example
we can take ${{\cal M}} ={\bf R}^{1,4}\times K3\times T^2$ but with
${\cal W} = {\bf R}^{1,1}\times \Sigma\times T^2$ where $\Sigma$ is
a 2-cycle in $K3$. We choose a vielbein frame such that, at least
locally,
\begin{equation}
\hat e_M^{\ \underline N} = \left(\matrix{ e_\mu^{\ \ \underline
\nu} &0\cr e_A^{\ \ \underline\nu}&e_A^{\ \ \underline
B}\cr}\right).
\end{equation}
Therefore,  in the static gauge that we are considering, the induced
metric on the M5-brane is simply $g_{\mu\nu} = \hat
g_{\mu\nu}(X^{\underline A}=0)$. We may further choose $\hat
\omega_\mu^{\ \underline {\nu B}}(X^{\underline A}=0)=0$ and $\hat
\omega_\mu^{\ \underline{\nu\lambda}}(X^{\underline A}=0) =
\omega_\mu^{\ \underline{\nu\lambda}}$, where $\omega_\mu^{\
\underline{\nu\lambda}}$ is the spin connection that one would
calculate from the vielbein $e_\mu^{\ \underline \nu}$. Finally we
also see that  $\hat \Gamma_\mu = \hat e_\mu^{\ \underline
\nu}\Gamma_{\nu} = \Gamma_\mu$ is the same $\gamma$-matrix that one
would calculate simply using the worldvolume metric $g_{\mu\nu}$.

This allows us reinterpret the bulk Killing spinor condition on the
worldvolume as
\begin{eqnarray}
\nonumber 0&=&\hat D_\mu \epsilon \\
&=& \partial_\mu\epsilon + \frac{1}{4}\hat\omega_\mu^{\
\underline{\nu\lambda}}\Gamma_{\underline{\nu\lambda}}+
\frac{1}{4}\hat\omega_\mu^{\
\underline{AB}}\Gamma_{\underline{AB}}\\
\nonumber &=& D_\mu\epsilon + A_\mu\epsilon,
\end{eqnarray}
where $\epsilon = \hat \epsilon (X^{\underline A}=0)$,
$\omega_{\mu}^{\ \underline{AB}}=\hat \omega_{\mu}^{\
\underline{AB}}(X^{\underline A}=0)$ and $A_\mu =
\frac{1}{4}\omega_{\mu}^{\ \underline{AB}}\Gamma_{\underline{AB}}$.

We find that, at lowest order in the fields $X^A$, $B_{\mu\nu}$ and
$\psi$, the following symmetries close on-shell into translations,
gauge transformations and local tangent frame rotations
\begin{eqnarray}\label{newsusy}
\nonumber \delta X^{\underline A} &=& i\bar\epsilon\Gamma^{\underline A}\psi\\
\delta B_{\mu\nu} &=& i\bar\epsilon\Gamma_{\mu\nu}\psi\\
\nonumber \delta\psi &=& \nabla_\mu X^{\underline
A}\Gamma^\mu\Gamma_{\underline A}\epsilon + \frac{1}{2\cdot
3!}\Gamma^{\mu\nu\lambda}H_{\mu\nu\lambda}\epsilon,
\end{eqnarray}
where $\nabla_\mu X^{\underline A} = \partial_\mu X^{\underline A} +
\omega_{\mu \underline B}^{\ \ \ \underline A}X^{\underline B}$. The
Fermion equation of motion that is required to close the algebra is
\begin{equation}\label{Feq}
\Gamma^\mu \nabla_\mu\psi=0,
\end{equation}
where $\nabla_\mu\psi = D_\mu\psi + A_\mu\psi$.

What are the remaining equations of motion? The $B$-field has a
self-dual field strength $H= dB$ and hence one finds $d\star H=0$.
This condition is preserved by the supersymmetries (\ref{newsusy}).
Taking a supersymmetry variation of the Fermion equation of motion
(\ref{Feq}) leads the condition
\begin{equation}
\label{Fvar} 0 = \Gamma_{\underline A} \nabla^2 X^{\underline
A}\epsilon +\frac{1}{2}\Gamma_{\underline
A}\Gamma^{\mu\nu}F_{\mu\nu\underline B}^{\ \ \ \ \underline A}
X^{\underline B}\epsilon,
\end{equation}
where
\begin{eqnarray}
\nonumber F_{\mu\nu\underline B}^{\ \ \ \ \underline
A}&=&\partial_{\mu}\omega_{\nu\underline B}^{\ \ \ \underline A}-
\partial_{\nu}\omega_{\mu\underline B}^{\ \ \ \underline A}
+\omega_{\mu\underline B}^{\ \ \ \underline C}\omega_{\mu\underline
C}^{\ \ \ \underline A}-\omega_{\mu\underline B}^{\ \ \ \underline
C}\omega_{\mu\underline C}^{\ \ \ \underline A}\\
&=&\hat R_{\mu\nu\underline B}^{\ \ \ \ \ \underline
A}(X^{\underline A}=0).
\end{eqnarray}
To proceed we assume there is a relation of the form
\begin{equation}\label{Mdef}
\frac{1}{2}\Gamma_{\underline A}\Gamma^{\mu\nu}F_{\mu\nu\underline
B}^{\ \ \ \ \underline A}\epsilon = M^{\underline A}_{\ \underline
B}\Gamma_{\underline A}\epsilon,
\end{equation}
in which case the equation of motion for $X^{\underline A}$, along
with the other fields, is
\begin{eqnarray}\label{Xeq}
\nonumber \nabla^2X^{\underline A} + M^{\underline A}_{\ \underline
{B}}X^{\underline B}&=&0\\
i\Gamma^\mu \nabla_\mu\psi &=&0\\
\nonumber
H_{\mu\nu\lambda}&=&\frac{1}{3!}\epsilon_{\mu\nu\lambda\rho\sigma\tau}H^{\rho\sigma\tau}.
\end{eqnarray}

We need to confirm that these equations are supersymmetric. To this
end we note that the Fermion equation of motion (\ref{Feq}) implies
\begin{equation}
\nabla^2 \psi -\frac{1}{4} R\psi
+\frac{1}{8}\Gamma^{\mu\nu}F_{\mu\nu}^{\ \
\underline{CD}}\Gamma_{\underline{CD}}\psi=0.
\end{equation}
To connect with (\ref{Fvar}) we multiply this on the left by
$\bar\epsilon\Gamma^{\underline A}$ to find
\begin{equation}\label{Feqcon}
\bar\epsilon\Gamma^{\underline A}\nabla^2 \psi -\frac{1}{4}
R\bar\epsilon\Gamma^{\underline A}\psi
+\frac{1}{8}\bar\epsilon\Gamma^{\underline
A}\Gamma^{\mu\nu}F_{\mu\nu}^{\ \
\underline{CD}}\Gamma_{\underline{CD}}\psi=0.
\end{equation}
Next we note that since $\hat \omega_\mu^{\ \underline{\nu A}}=0$ we
have that $\hat R_{\mu\nu}^{\ \ \underline{\lambda A}}=0$ and
therefore the Killing spinor integrability condition $[\hat
D_\mu,\hat D_\nu]\epsilon = \frac{1}{4}\hat R_{\mu\nu}^{\ \
\underline{MN}}\Gamma_{\underline{MN}}\epsilon=0$ implies
\begin{equation}
 R_{\mu\nu}^{\ \
 \underline{\lambda\rho}}\Gamma_{\underline{\lambda\rho}}\epsilon =
 -F_{\mu\nu}^{\ \
 \underline{CD}}\Gamma_{\underline{CD}}\epsilon,
\end{equation}
and hence
\begin{equation}\label{newblah}
\bar\epsilon R
=\frac{1}{2}\bar\epsilon\Gamma_{\underline{CD}}\Gamma^{\mu\nu}F_{\mu\nu}^{\
\ \underline{CD}}.
\end{equation}
Using this we see that (\ref{Feqcon}) implies
\begin{equation}\label{blah}
\bar\epsilon \Gamma^{\underline A}\nabla^2 \psi
+\frac{1}{2}\bar\epsilon\Gamma^{\underline
B}\Gamma^{\mu\nu}F_{\mu\nu\underline B}^{\ \ \ \ \underline A}
X^{\underline B}\psi=0.
\end{equation}
The $X^{\underline A}$ equation can be compared to (\ref{blah}) by
noting that
\begin{equation}
\delta\nabla^2 X^{\underline A} = i\bar\epsilon\Gamma^{\underline
A}\nabla^2\psi,
\end{equation}
and one sees that the equations (\ref{Xeq}) are preserved by
supersymmetry.

Let us consider  for example the case where $\cal W$ is
non-trivially embedded in eight dimensions, so that only
$F_{\mu\nu}^{\ \ \underline{67}}\ne 0$. We see from (\ref{Mdef}) and
(\ref{newblah}) that the only non-vanishing components of
$M^{\underline A}_{\ \underline B}$ are
\begin{equation}
M^{\underline 6}_{\ \underline 6}= M^{\underline 7}_{\ \underline
7}= R.
\end{equation}
Thus we find the scalar equations are
\begin{eqnarray}\label{MXeqs}
\nonumber \nabla^2
X^{\underline 6}+RX^{\underline 6}&=& 0\ , \\
\nabla^2 X^{\underline 7}+RX^{\underline 7}&=&  0\ ,\\
\nonumber \nabla^2 X^{\underline A}&=& 0\ , \qquad A=8,9,10.
\end{eqnarray}

\section{Counting Moduli}

In the previous section we determined the lowest order equation of
motion for an M5-brane wrapped on a general calibrated submanifold
$\cal W$ of $\cal M$. As a result we saw that the Fermions and
scalar fields couple minimally to the gauge field associated to the
structure group of the normal bundle and some scalars develop a mass
term from the curvature. However the three-form remains closed and
self-dual (at the linearized level). In this section we wish to
perform a precise counting of the massless degrees of freedom  for
an M5-brane wrapped on a four-cycle $P\subset {\cal M}$, {\it i.e.}\ ${\cal
W} = {\bf R}^{1,1}\times P$, in a spacetime of the form ${\cal M} =
{\bf R}^{1,4}\times {\cal K}$ where $\cal K$ is some compact
Calabi-Yau space that contains $P$. This has been discussed in great
detail in \cite{Maldacena:1997de,LopesCardoso} and we will largely
follow their discussion.

The simplest field to consider is the dimensional reduction of the
two-form gauge field. As a consequence of the self-duality condition
one finds $b_2^+(P)$ right moving scalars and $b_2^-(P)$ left moving
scalars. For the compact K\"ahler manifolds that we consider here
$b_2^+(P) =2h_{2,0}(P)+1 $ and $b_2^-(P)=h_{1,1}(P)-1$. If
$h_{1,0}(P)$ is non-vanishing then there will be $2h_{1,0}(P)$
Abelian gauge fields in the two-dimensional effective theory.
However these are non-dynamical we will not need them here.

Next we consider reduction of the scalars $X^{\underline A}$. In
total there are five. Three of these, $X^8,X^9,X^{10}$ simply
parameterize the location in the non-compact transverse space. These
always give 3 left and 3 right moving scalars in two dimensions. The
remaining two scalars are in fact sections of the normal bundle of
$P$ inside $\cal K$. As such the number of such zero modes is hard
to calculate. Let us denote the number of normal bundle moduli by
$N(P,{\cal K})$. These are left-right symmetric and we will discuss
them in more detail shortly.

As for the Fermions it is well known (see \cite{Green:1987mn}) that
spinors on a K\"ahler manifold $P$ can be realized as $(0,p)$-forms
on $P$. To see this one first consider complex coordinates for $P$
so that $\{\Gamma^a,\Gamma^b\}=\{\Gamma^{\bar a},\Gamma^{\bar
b}\}=0$ and $\{\Gamma^a,\Gamma^{\bar b}\}=2g^{a\bar b}$ with
$a,b=z,w$. In particular we consider a spinor ground state
$|0\rangle$ which is annihilated by the holomorphic
$\gamma$-matrices; $\Gamma^a|0\rangle=0$. We can then construct a
general spinor by
\begin{equation}
|\psi> = \omega |0\rangle + \Gamma^{\bar a}\omega_{\bar
a}|0\rangle+\frac{1}{2}\Gamma^{\bar a\bar b}\omega_{\bar a\bar
b}|0\rangle.
\end{equation}
By construction $\omega_{\bar a_1...\bar a_p}$ is totally
anti-symmetric and hence represents a $(0,p)$-form on $P$.
Furthermore if we choose the complex $\gamma$-matrices
$\Gamma^z=\Gamma^2+i\Gamma^4$ and $\Gamma^w=\Gamma^3+i\Gamma^5$ then
one sees that $\Gamma_{2345}|0\rangle =|0\rangle$ and more generally
\begin{equation}
\Gamma_{2345}|\omega_p\rangle = (-1)^p|\omega_p\rangle,
\end{equation}
where $|\omega_p\rangle = \frac{1}{p!}\omega_{\bar a_1...\bar
a_p}\Gamma^{\bar a_1..\bar a_p}|0\rangle$. Since  the Fermions on
the M5-brane satisfy $\Gamma_{012345}\psi=-\psi$ we see that
$|\omega_p\rangle$ leads to right and left moving Fermions in two
dimensions if $p$ is even or odd respectively.

To find massless two-dimensional modes we assume that $|0\rangle$ is
Killing with respect to $\nabla$ defined above. In this case one see
that solutions to $(\Gamma^a\nabla_a+\Gamma^{\bar a}\nabla_{\bar
a})\psi=0$ correspond to $\bar \partial_{[\bar b} \omega_{\bar
a_1...\bar a_p]} =0$ and $g^{b\bar a_1}\partial_{[b}\omega_{\bar
a_1...\bar a_p]} =0$, {\it i.e.}\ $\omega_p \in H^{(0,p)}(P)$. Thus one
finds that number of massless left and right moving two-dimensional
Fermions is
\begin{equation}
N_F^L = 4h_{1,0}(P), \qquad N_F^R = 4(h_{0,0}(P)+h_{2,0}(P)).
\end{equation}
Here the factor of 4 comes from the fact  the spinor `groundstate'
$|0\rangle$ can be thought of as having 32 real components but is
subject to the three constraints:
$\Gamma^z|0\rangle=\Gamma^w|0\rangle=0$ and
$\Gamma_{012345}|0\rangle=-|0\rangle$. Thus $|0\rangle$ has four
real independent components.

Let us summarize our counting so far. We find
\begin{eqnarray}\label{Ncount}
\nonumber {N}_B^L&=&  2+  h_{1,1}(P)+N(P,{\cal K})\\
\nonumber {N}_B^R &=& 4+2h_{2,0}(P)+  N(P,{\cal K}) \\
  {N}_F^L &=& 4h_{1,0}(P) \\
  \nonumber {N}_F^R &=& 4h_{2,0}(P)+4,
\end{eqnarray}
where we have assumed that $h_{0,0}(P)=1$. Since the wrapped
M5-brane preserves (at least) $(0,4)$ supersymmetry the right-movers
must have Bose-Fermi degeneracy. This immediately allows us to
determine the number of normal moduli to be
\begin{equation}\label{newnormal}
N(P,{\cal K}) = 2h_{2,0}(P),
\end{equation}
and hence the massless spectrum is
\begin{eqnarray}\label{ourcount}
\nonumber N_B^L&=&  2h_{2,0}(P)+h_{1,1}(P)+2\\
\nonumber N_B^R &=&   4h_{2,0}(P)+4\\
  N_F^L &=&  4h_{1,0}(P)\\
  \nonumber N_F^R &=& 4h_{2,0}(P)+4.
\end{eqnarray}
Note that this also ensures that the number of right-moving modes is
a multiple of 4, as also required by $(0,4)$ supersymmetry. We would
like to emphasis that this formula should apply whenever it make
sense to talk of a classical M-brane that is wrapped on a smooth
complex submanifold of any smooth Calabi-Yau (including $K3\times
T^2$ and $T^6$).

This formula should be contrasted with the result
\begin{equation}\label{normal}
N(P,{\cal K}) = 2h_{2,0}(P) - 2h_{1,0}(P),
\end{equation}
first obtained in \cite{Maldacena:1997de} for an ample four-cycle
$P$ in a  generic Calabi-Yau and extended to $K3\times T^3$ and
$T^6$ in \cite{LopesCardoso}. We see that there is agreement for a
generic Calabi-Yau where $h_{1,0}(P)=0$. However, as pointed out in
\cite{LopesCardoso}, the formula (\ref{normal}) contradicts
supersymmetry when $h_{1,0}(P)\ne 0$. In the rest of this section we
will argue that (\ref{newnormal}) is the correct counting and
identify the missing modes that are absent from (\ref{normal}).

We start with a brief review of  the calculation in
\cite{Maldacena:1997de}. This starts from the observation that a
4-cycle $P\subset {\cal K}$ is defined by the zeros of a section of
a line bundle over ${\cal K}$. The Poincar\'e dual two-form to $P$,
which we denote by $[P]$, determines the Chern class of the line
bundle. Thus counting the number of deformations of $P$ corresponds
to counting the (real) dimension of the dimension of the space of
line bundles. However one must take into account the fact that if
$P$ is described by zeros of a section $s$ then the zeros of
$\lambda s$ describe the same $P$ for any $\lambda \in {\bf
C}^\star$. Thus one needs the real dimension of the projective space
of line bundles. In this way one determines $N(P,{\cal K})$ through
\begin{eqnarray}\label{indexcal}
\nonumber N(P,{\cal K}) &=& 2{\rm dim}(H^0({\cal K},{\cal L}))-2\\
\nonumber &=&2\sum _i (-1)^i{\rm dim}(H^i(P,{\cal L}))-2\\
&=& 2\int_{\cal K} e^{[P]}{\rm Td}({\cal K})-2\\
\nonumber &=& \frac{1}{3}\int_{\cal K} [P]^3+\frac{1}{6}\int_{\cal
K}[P]\wedge c_2({\cal K})-2.
\end{eqnarray}
Here the second line follows from the Kodaira vanishing theorem;
$H^i({\cal K},{\cal L})=\emptyset$ for $i>0$, and the third line
from a Riemann-Roch index formula. For an account of these theorems
see \cite{Hirzebruch,GH}. Next one can use the formula (see
\cite{Maldacena:1997de,LopesCardoso})
\begin{equation}\label{h20}
h_{2,0}(P) = \frac{1}{6}\int_{\cal K} [P]^3 + \frac{1}{12}\int_{\cal
K} [P]\wedge c_{2}({\cal K}) + h_{1,0}(P)-1,
\end{equation}
to obtain (\ref{normal}).

So what is missing from this calculation (\ref{indexcal})? A central
assumption of \cite{Maldacena:1997de} is that $P$ is an ample cycle.
Technically this means that the Poincar\'e dual two-form $[P]$ lies
inside the K\"ahler cone, {\it i.e.}\ it defines a positive volume for all
complex 2-,4- and 6-cycles in $\cal K$. More intuitively an ample
cycle $P$ of a manifold $\cal K$ is one that is sufficiently generic
so that the set of all normal vectors to $P$ spans the entire
tangent space of $\cal K$.

A key assumption of the Kodaira vanishing theorem is that the line
bundle $\cal L$ is positive and hence (\ref{indexcal}) counts the
dimension of the space of positive line bundles. While every ample
four-cycle in $\cal K$ defines a positive line bundle there are zero
modes which do not correspond to positive line bundles. In
particular consider translations of $P$ along any of the $S^1$
factors in ${\cal K}$. These $S^1$ factors are trivial and
describing the location of an M5-brane in $S^1$ simply corresponds
to specifying a value of the coordinate for that $S^1$. As such the
location is simply a section of a trivial $U(1)$ line bundle over
$P$ and this extends to a trivial $U(1)$ bundle over $\cal K$. These
deformations  are not counted in (\ref{normal}) since the associated
line bundle is trivial. There are $2h_{1,0}({\cal K})$ such
translations and, using the Lefschetz hyperplane theorem (valid for
ample four-cycles), we have that $h_{1,0}({\cal K})=h_{1,0}(P) $.
Therefore we find an extra $2h_{1,0}(P)$ normal modes that arise
from translations along the $S^1$ factors of $\cal K$. Including
these modes in (\ref{indexcal}) gives (\ref{newnormal}).

An alternative description of these translational modes is to note
that the $S^1$ factors are orbits of a $U(1)$ Killing isometry that
acts on ${\cal K}$. An ample cycle breaks the symmetries
corresponding to translations along the $S^1$ factors and hence
there must be $2h_{1,0}({\cal K}) = 2h_{1,0}(P)$ Goldstone modes.
There are also smooth but non-ample four-cycles for which $
h_{1,0}(P) \ne h_{1,0}({\cal K})$ and the index theorem does not
apply. In these cases one also finds that the cycle breaks fewer
$U(1)$ isometries and as a result has fewer Goldstone modes. We will
explicitly see in the examples below that nevertheless
(\ref{newnormal}) is valid for all smooth four-cycles, as required
by supersymmetry.

\subsection{Three Examples}

To illustrate this discussion let us  consider some explicit
examples for ${\cal K}= K3\times T^2$. We will consider three
choices for $P$: $P=K3$, $P=\Sigma\times T^2$ and $P=K3+
\Sigma\times T^2$, where $\Sigma$ is a two-cycle in $K3$. For a
useful account of various facts about $K3$ see
\cite{Aspinwall:1996mn}. Following this we will also discuss the
case where ${\cal K}= T^6$.

First we consider the case where $P=K3$ which was first studied in
detail in \cite{Cherkis:1997bx}. The Hodge diamond of $K3$ is
\begin{equation}
K3:\qquad\matrix{&&1&&\cr &0&&0&\cr 1&&20&&1\cr&0&&0\cr &&1&&\cr}.
\end{equation}
In this case it is clear that $N(K3,K3\times T^2)=2$ since $\cal K$
is simply a direct product ${\cal K} = K3\times T^2$. Hence the
normal bundle to $P=K3$ is trivial and there is no obstruction to
moving the $K3$ around inside $\cal K$. Since the Killing spinor on
$K3$ is chiral, reduction on $K3\times T^2$ leads to a
two-dimensional theory with $(0,8)$ supersymmetry. Looking at the
field content we find the massless modes
\begin{equation}
N_B^L = 24, \qquad N_B^R = 8\ ,\qquad N_F^L = 0, \qquad N_F^R = 8,
\end{equation}
which is the same as the worldsheet action for the Heterotic string
on $T^3$.

Next we consider the case where $P=\Sigma\times T^2$. Let us suppose
that $\Sigma$ is ample in $K3$. In complex dimension two the
Lefschetz hyperplane theorem does not imply that
$h_{1,0}(\Sigma)=h_{1,0}(K3)=0$ and hence $h_{1,0}(\Sigma)=g$ need
not be zero.  Assuming $\Sigma$ is connected the Hodge diamond of
$\Sigma\times T^2$ is
\begin{equation}\label{Hodge}
\Sigma\times T^2:\qquad\matrix{&&1&&\cr &1+g&&1+g&\cr
g&&2+2g&&g\cr&1+g&&1+g\cr &&1&&\cr}.
\end{equation}
To determine $N(\Sigma\times T^2,K3\times T^2)$ we note that
$N(\Sigma\times T^2,K3\times T^2) = N(\Sigma,K3)$. Since $K3$ does
not have any $S^1$ factors we may use a similar calculation as in
(\ref{indexcal}), suitably adapted to 2 complex dimensions. We find
that
\begin{eqnarray}\label{K3calc}
\nonumber {\rm dim}(H^0(K3,{\cal L}))&=& \sum_{i=0}^2 (-1)^i {\rm dim}(H^i(K3,{\cal L})) \\
&=& \int_{K3} e^{[\Sigma]}{\rm Td}(K3)\\
\nonumber &=& \int_{K3}\frac{1}{12}c_2(K3) + \frac{1}{2}[\Sigma]^2\\
\nonumber &=& 2 + \frac{1}{2}\int_{K3}[\Sigma]^2.
\end{eqnarray}
Thus proceeding as before and taking account of the projective
equivalence we find
\begin{equation}
N(\Sigma,K3) = 2 ({\rm dim}(H^0(K3,{\cal L}))-1) = 2
+\int_{K3}[\Sigma]^2.
\end{equation}
Continuing we observe that $[\Sigma]$ is the Poincare dual to
$\Sigma$ and hence we find
 \begin{eqnarray}\label{NK3}
 \nonumber \int_{K3}[\Sigma]^2  &=& \int_{\Sigma}[\Sigma]\\
&=& -\int_{\Sigma} c_1({\cal L})\\
\nonumber &=& 2g-2,
\end{eqnarray}
where in the second line we have used the adjunction formula to
identify $[\Sigma]=-c_1({\cal L})$ and the last line follows from
the well known formula for the Euler number of a two-dimensional
surface. Note that ample implies that $g\ge 2$. Thus
\begin{equation}
N(\Sigma\times T^2,K3\times T^2)=N(\Sigma,K3) = 2g.
\end{equation}
Putting this all together we see that the field content is
\begin{equation}
N_B^L = 4+4g\qquad N_B^R =4+4g\qquad N_F^L =4+4g\qquad N_F^R = 4+4g.
\end{equation}
Note that the spectrum is non-chiral which is a consequence of the
fact that in this case $(4,4)$ supersymmetry is preserved.

The previous two cases are not generic and in particular the cycle
$P$ is not ample in $\cal K$. In these cases the formula
(\ref{normal}) does not necessarily apply and indeed it doesn't
always agree with our results. However the formula (\ref{newnormal})
is valid and agrees with our discussion. Our final case $P=K3+
\Sigma\times T^2$ is generic in that $P$ is ample. Therefore the
calculation (\ref{indexcal}) is valid. However since $h_{1,0}(P)\ne
0$ we will be able to test whether (\ref{normal}) or
(\ref{newnormal}) reproduces the correct number of normal modes. To
see what these formulae give us we need to compute $h_{2,0}(P)$.
From (\ref{h20}) we find
\begin{equation}
h_{2,0}(P) = \frac{1}{6}\int_{K3\times T^2} [P]^3 +
\frac{1}{12}\int_{K3\times T^2} [P]\wedge c_{2}(K3\times T^2) +
h_{1,0}(P)-1.
\end{equation}
Writing $[P]=dvol_{T^2}+[\Sigma]$, where $dvol_{T^2}$ is the unit
volume form of $T^2$, we find
\begin{eqnarray}\label{ghgh}
\nonumber h_{2,0}(P) &=& \frac{1}{2}\int_{K3}[\Sigma]^2 + \frac{1}{12}\int_{K3}c_2(K3)+h_{1,0}(P)-1 \\
  &=& g+h_{1,0}(P),
\end{eqnarray}
where we have used (\ref{NK3}) and $\chi(K3)=24$ in the second line.
Thus since $P$ is ample $h_{1,0}(P)=1$ and (\ref{normal}) predicts
$2g+2$ normal modes and (\ref{normal}) only $2g$ normal modes.
However one expects that there are always two normal modes which
come from translations along $T^2$ and also that all of the normal
modes which exist in the embedding of $\Sigma$ in $K3$ should also
exist here. This example therefore demonstrates that the formula
(\ref{normal}) fails to include the translational modes whereas
(\ref{newnormal}) correctly accounts for all zero-modes.

For completeness we note that \cite{Maldacena:1997de,LopesCardoso}
\begin{eqnarray}\label{gghh}
\nonumber h_{1,1}(P) &=& \frac{2}{3}\int_{K3\times T^2} [P]^3 + \frac{5}{6}\int_{K3\times T^2} [P]\wedge c_{2}(K3\times T^2)+2h_{1,0}(P)\\
  &=& 4g+16+2h_{1,0}(P).
\end{eqnarray}
Thus  we find, from (\ref{ourcount}),
\begin{equation}
N_B^L = 6g+22, \qquad N_B^R =4g+8 \ ,\qquad N_F^L = 4, \qquad N_F^R
= 4g+8.
\end{equation}

Finally we can follow the above discussion and consider what happens
to these three cases when $K3$ is replaced by $T^4$, {\it i.e.}\ ${\cal
K}=T^6$. In the first case where $P=T^4$ the same arguments give
$N(T^4,T^6)=2$ and hence
\begin{equation}
N_B^L = 8, \qquad N_B^R =8 \ ,\qquad N_F^L = 8, \qquad N_F^R = 8,
\end{equation}
which of course is just a straightforward reduction of the M5-brane
on $T^4$ and has $(8,8)$ supersymmetry.

In the second case all we need to do is replace $c_2(K3)=24$ by
$c_2(T^4)=0$ in (\ref{K3calc}) and we now find that the index
theorem gives $2{\rm dim}(H^0(\Sigma,T^4)-2=2g-4$ normal modes.
However we claim that, in addition to the modes counted by the index
theorem, to obtain the number of normal mode deformations of
$\Sigma$ inside $T^4$ we must also include  4 translational
Goldstone modes and hence $N(\Sigma,T^4)=2g$. Thus we find
$N(\Sigma\times T^2,T^6)=N(\Sigma,T^4)=2g$. From (\ref{Hodge}) we
find the total spectrum is
\begin{equation}
N_B^L = 4g+4, \qquad N_B^R =4g+4 \ ,\qquad N_F^L = 4g+4, \qquad
N_F^R = 4g+4.
\end{equation}
Just as in the $K3\times T^2$ case this spectrum is non-chiral as
result of enhanced $(4,4)$ supersymmetry.

In the third case where $P = T^4 + \Sigma\times T^2$ the cycle is
ample we have that $h_{1,0}(P)=3$. The calculations (\ref{ghgh}) and
(\ref{gghh}) give (replacing $c_2(K3)=24$ by $c_2(T^4)=0$ and
setting $h_{1,0}(P)=3$)
\begin{equation}
h_{2,0}(P) = g+1\qquad h_{1,1}(P)=4g+2.
\end{equation}
Here we see that (\ref{normal}) gives $N(P,T^6)=2g-4$ and
(\ref{newnormal}) gives $N(P,T^6)=2g+2$. The difference is $6$ and
these are clearly the translational modes along $T^6$ which must
exist for a generic cycle which breaks all the translational
symmetries. In total we find
\begin{equation}
N_B^L = 6g+6, \qquad N_B^R =4g+8 \ ,\qquad N_F^L = 12, \qquad N_F^R
= 4g+8,
\end{equation}

In all these cases one finds that $N(P,T^6) = 2h_{2,0}(P)$ as
predicted by (\ref{newnormal}). Let us make a comment on the first
two cases where the cycles are not ample. In these cases the
four-cycles preserve some of the symmetries of the torus and hence
the total number of translational Goldstone modes (equal to $2$ or
$4$ respectively) is less than $2h_{1,0}({\cal K})=6$. Nevertheless
we still find that the total number of normal modes is
$2h_{2,0}(P)$.

\section{Counting Black Holes}

Following \cite{Maldacena:1997de} we can obtain black hole solutions
of four-dimensional extended supergravity by further compactifying
the remaining spatial direction of the wrapped M5-brane on $S^1$. A
static wrapped M5-brane will be a magnetic source for the
four-dimensional Abelian gauge fields that arise from Kaluza-Klein
reduction of the M-theory three-form. One may also consider electric
charges by including M2-branes. According to Beckenstein and Hawking
the entropy of a macroscopic black hole is given by one quarter if
its horizon area. For the solutions at hand one finds
\cite{Maldacena:1997de}
\begin{equation}
S_{BH} = 2\pi\sqrt{\frac{1}{6}|q|\chi(P)},
\end{equation}
where $\chi(P) = 2 - 4h_{1,0}(P)+ 2h_{2,0}(P) + h_{1,1}(P)$ is the
Euler number of $P$. Here $q$ is the momentum carried by the
M5-brane along $S^1$, shifted by a contribution that arises from the
electric charges \cite{Maldacena:1997de}.

Following the  work of \cite{Strominger:1996sh} one obtains the
microscopic black hole degeneracy by counting all the modes of the
low energy M5-brane theory which preserve the supersymmetries of the
vacuum.  For an effective theory with $(0,4)$ supersymmetry this
amounts to counting the number of left moving modes with the right
movers in their vacuum. From Cardy's theorem this is determined by
the left-moving central charge and we find
\begin{equation}\label{macroentropy}
S = {2\pi\sqrt{\frac{1}{6}|q|c_L}} =
2\pi\sqrt{\frac{1}{6}|q|(\chi(P)+6h_{1,0}(P))},
\end{equation}
where we have used (\ref{ourcount}). For a generic Calabi-Yau, where
$h_{1,0}(P) = 0$, we see that the entropy is precisely reproduced by
a microscopic counting of the degrees of freedom of the M5-brane,
including the electric charges which in the microscopic picture
arise from shifts of the vacuum energy \cite{Maldacena:1997de}.
However for ${\cal K}=K3\times T^2$ or ${\cal K} = T^6$ we find
$h_{1,0}(P) \ne 0$ and the two entropy calculations do not agree, as
was pointed out in \cite{LopesCardoso}.

There is a further discrepancy. It is possible to compute the left
and right central charges of the superconformal $(0,4)$ fixed point
of the M5-brane  using gravitational and R-symmetry anomalies
\cite{Harvey:1998bx,Kraus:2005vz}. These arguments give $c_L =
h_{2,0}(P)+h_{1,1}(P)+2-4h_{1,0}(P)$ and
$c_R=6(h_{2,0}(P)-h_{1,0}(P)+1)$. This correctly accounts for the
black hole entropy but also differs from our counting by
$6h_{1,0}(P)$ for both the left and right central charges.

We propose the following resolution. Our field content naturally
splits into that of a `pure' $(0,4)$ supersymmetric sector with
\begin{eqnarray}\label{ournewcount}
\nonumber {N}_B^L&=&  2h_{2,0}(P)+h_{1,1}(P)+2-4h_{1,0}(P)\\
\nonumber {N}_B^R &=&   4h_{2,0}(P)+4-4h_{1,0}(P)\\
  {N}_F^L &=&  0\\
  \nonumber {N}_F^R &=& 4h_{2,0}(P)+4-4h_{1,0}(P),
\end{eqnarray}
and $h_{1,0}(P)$  $(4,4)$ multiplets with
\begin{equation}
{N}_B^L=4\qquad{N}_B^R=4\qquad{N}_F^L=4\qquad{N}_F^R=4.
\end{equation}
Note that we are not assuming that there is a left-moving
supersymmetry which acts on the $(4,4)$ multiplets, we are just
using them as a counting device.  The correct black hole degeneracy
and central charges are readily obtained if we only count the modes
of the $(0,4)$ sector. Furthermore both the black hole entropy and
anomaly arguments only count the degrees of freedom that are
massless at the conformal fixed point. Since the extra states that
we find fall into non-chiral $(4,4)$ multiplets it is reasonable to
conjecture that they become massive and hence do not appear in
spectrum of the conformal fixed.

We have not been able to provide any additional arguments to support
this proposal. However this claim is essentially a consequence of
our counting along with the results of
\cite{Harvey:1998bx,Kraus:2005vz}. To state this another way we note
that the quantum anomaly arguments determine the central charges at
the conformal fixed point and, combining this with our counting
(which is just a classical counting at lowest order), we deduce that
$h_{1,0}(P)$ $(4,4)$ multiplets become massive at the IR fixed
point.

We note that $(4,4)$ supersymmetry implies that the potential must
arise as the length-squared of a tri-holomorphic Killing vector on
the moduli space \cite{AlvarezGaume:1983ab}. When $h_{1,0}(P)\ne 0$
$\cal K$ has $U(1)$ isometries and these will induce Killing vectors
on the moduli space. We expect that, as a consequence of the
geometrical action of R-symmetry, the moduli space Killing vectors
should be tri-holomorphic. Therefore they can in principle lead to
the required  potential.

Let us make some comments on the mechanism that would provide such a
mass. One could object that the M5-brane moduli cannot become
massive because the equations of motion only involve derivatives of
the fields. In particular, although in section 2 we only gave the
lowest order equations of motion, the full non-linear equations have
been worked out for a general embedding into supergravity  and
indeed these only involve derivatives of the fields. These equations
of motion were derived in \cite{Howe} using the superembedding
formalism applied to the two derivative approximation to M-theory,
{\it i.e.}\ standard eleven-dimensional supergravity of
\cite{Cremmer:1978km}. Another approach to obtaining the M5-brane
equations of motion comes from an analysis of the Goldstone modes of
the supergravity solution \cite{Adawi:1998ta}. The M5-brane
three-form $H$ is identified with zero-modes arising from gauge
transformations of the bulk three-form $C$. Again one would expect
that, as Goldstone modes, the equations of motion of the M5-brane
fields would only involve derivatives, even if higher derivative
terms were added to eleven-dimensional supergravity.

However there is an important caveat. It is well-known that at
next-to-leading order the M-theory effective Lagrangian contains the
anomaly $C\wedge I_8$ term \cite{Duff:1995wd}. This leads to a
source for $C$ and hence the also three-form $H$ on the M5-brane
worldvolume. Furthermore it is precisely the $C\wedge I_8$
term
in the effective action which is needed for cancelation of
the anomalies and which ultimately leads to the correct prediction
of the central charges in \cite{Harvey:1998bx,Kraus:2005vz}. Thus
one might suspect that this term induces a mass for the extra
$(4,4)$ multiplets that we have found.

We finish this section by noting that, in the examples above with
 $P=\Sigma\times T^2$  and $(4,4)$ supersymmetry, this mechanism
removes all the massless modes. Thus the M5-brane will behave as
though it is wrapped on a rigid cycle even though the cycle has
moduli.

\section{Conclusion}

In this paper we have discussed in detail the low energy dynamics of
an M5-brane wrapped on a smooth but otherwise arbitrary, complex
four-cycle in $K3\times T^2$ or $T^6$. In particular we gave the
lowest order equations of motion and determined the spectrum of
massless modes. This required a careful treatment of the zero-modes
that arise from translations along the $S^1$ factors and leads to a
spectrum in a agreement with supersymmetry. Finally we discussed the
counting of black hole microstates obtained by further reduction to
four-dimensions on another $S^1$. The naive counting of massless
modes does not reproduce the correct entropy and is not in agreement
with anomaly cancelation arguments. To resolve this we proposed that
$h_{1,0}(P)$ $(4,4)$ multiplets become massive and are removed from
the low energy spectrum. It would be very interesting to study this
mass mechanism in greater detail and verify that it indeed at work
here. In particular it would be interesting to incorporate the
effect of the $C\wedge I_8$ term on the M5-brane equations of
motion.

\section*{ Acknowledgements}

I am grateful to B. de Wit, T. Mohaupt, G. Moore and S. Trivedi for
useful discussions. This work is supported in part by the PPARC
grant PP/C507145/1 and the EU grant MRTN-CT-2004-512194.

redpajama_set_name: RedPajamaArXiv


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redpajama_set_name: RedPajamaGithub


Bungalow for sale kemer Turkey.
This beautiful bungalow sits in a quiet location in Kemer.
Kemer is approx 25 minutes drive from the stunning town of Fethiye with its beautiful parks, marinas and outstanding bay with the new promenade.
Kemer has now gone from village status and will soon become a town. Already the infrastructure seems to be changing overnight. The town of Kemer has many restaurants, cafes and parks. All needs are catered for with shops, markets and also the local open air market once a week. Kemer is also close to all the mountains and forests in the area and very green. The ancient ruins of Roman cities are sprawled across the area and the beautiful Saklikent Gorge is only 15 minutes drive away.
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2 double bedrooms, lounge, kitchen and shower room.
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We're back with the second part of our four-part series on edible gardening! Did you miss us? We missed you! Last time we introduced you to the basics of water-wise edible gardening, and now that we've given you a week to mulch, plant and tend to your bountiful harvest (maybe), it's time to talk about another favorite topic of ours: edible plants!
You know about tomatoes, artichokes, carrots and other common foods we get from gardens. Did you know that Arizona has some unique alternatives that come from our indigenous plants and are just as tasty? The best part is, our edible plants require very little in the way of watering or maintenance because they are used to the desert heat (unlike those snowbird plants from Ohio).
Here's a quick list of nutritional and delicious desert plants.
So how can you incorporate some of these prickly parts into your diet? Well, those mesquite tree pods that you see everywhere in the summer can be baked and ground into sweet and low-sugar mesquite flour, which can substitute for one-third of the flour content in most recipes. Plus, it's low in fat and a good source of calcium, magnesium, and iron.
You can also grab some tongs and gloves and harvest prickly pear fruits and pads. Popular in Mexican cuisine, prickly pear pads called 'nopales' can be boiled like green beans, dried into jerky or pickled. Prickly pear fruit can be made into jam, added to smoothies or strained for fruit juice – a nice natural sweetener for drinks and desserts.
If you're feeling particularly adventurous, grab a long stick and a bucket and harvest fruit from your Saguaro (typically in early July). Gently knock the fruit loose, being mindful of spines. Fruit pulp can be used for pancake syrup, jam and fruit juice. Dry the seeds in the sun and use as a tasty textural addition to bread, salad or anything else that demands an extra crunch.
If you're thinking of adding a little local flavor to your cooking, just look in your backyard! Each of these edible plants is a delicious and healthy alternative to some of the ingredients you already use.


text:
Genevieve Tregor, M.S.
M.S. in Kinesiology
California Polytechnic State University, San Luis Obispo CA
MBSR Trained Instructor
with Bob Stahl in the UMass Center for Mindfulness in Medicine Oasis approved teacher practicum
Radiant Heart Qi Gong Instructor certified with Teja Fudo Myoo Bell
Advanced Somatic Experiencing (SE) Provider & Trauma-Informed Mindfulness Instruction
completed Peter Levine's 3-year Somatic Experiencing Trauma Institute professional training program
Intensive Mindfulness Meditation Experience
over 13 years of dedicated training and 10,000+ hours of formal practice experience, including a year and a half of intensive long-term (retreat) practice under the guidance of senior Western and Monastic teachers in the Insight Tradition
Faculty for Jack Kornfield & Tara Brach's Mindfulness Meditation Teacher Certification Program
Genevieve offers mindfulness education and consulting from an extensive and in-depth background in mindfulness practice, paired with the expertise of a health and wellness professional.
She has practiced Vipassana meditation in the Buddhist tradition - the rigorous, systematic practice of mindfulness from which the original Mindfulness-Based Stress Reduction (MBSR) program was developed - since 2008, and has practiced in long-term, dedicated retreat periods equaling 1.5 years exploring the depths of mindfulness with renowned Western and monastic teachers. This training enhances her ability to teach mindfulness comprehensively and guide students effectively in the many subtleties of the practice; beyond mere techniques.
Growing up with chronic illness inspired her to pursue a wellness focused education. She received a MS in Kinesiology (Exercise Science and Health Promotion), from California Polytechnic State University in San Luis Obispo; and a BA in Classics, from the University of Arizona. Her graduate level studies focused on incorporating physiology, disease process and pathology, health behavior change and psychology into health education and training from a wellness perspective. She has worked in healthcare, public health and health education including research, planning and delivery.
Specialized training, experience and accomplishments include:
Completion of the MBSR Teacher Training Practicum as required by the University of Massachusetts Medical School's Center for Mindfulness in Medicine advanced teacher training
Hand-selected by Jack Kornfield and Tara Brach as faculty for their Mindfulness Meditation Teacher Certification Program, for which she mentors teachers-in-training internationally
Certified in Radiant Heart Qigong, an ancient energy healing practice bringing mindfulness to movement, with Teja Fudo Myoo Bell
Completion of Peter Levine's 3-year Somatic Experiencing (SE) Trauma Professional Training
Genevieve has studied with and been mentored by senior teachers in the Buddhist mindfulness tradition (AKA Vipassana or Insight Meditation), her primary teachers have been Gil Fronsdal and Howard Cohn. In the MBSR curriculum she trained with Bob Stahl, PhD, senior teacher with the UMass Center for Mindfulness Oasis Institute. Other influential teachers who she's studied and practiced with in the Western Vipassana tradition include Joseph Goldstein, Jack Kornfield, John Travis and Winnie Nazarko. In the Theravada monastic tradition she has practiced with Sayadaw U Vivekananda, Sayadaw U Jagara, and Thanissaro Bhikkhu, among others.
In addition to offering courses through Solution Mindfulness, Genevieve teaches in various capacities: as the founder and facilitating teacher for Insight Meditation Scottsdale, offering instruction and retreats in the Western Vipassana Buddhist tradition, a drop-in group that also meets regularly on Monday nights in Scottsdale; she teaches in the Mindfulness Meditation Teacher Certification Program; offers Insight Meditation Retreats through www.oneworldmindfulness.com; teaches through Cloud Sangha; and additionally teaches a course on Cultural Aspects of Health and Illness as Adjunct Faculty at South Mountain Community College.
"Mindfulness is not a technique or one more tool to put in [a] toolbox; it is a way of relating to life, of experiencing life, all of it. This way of being can be taught only through a deep understanding. Otherwise, mindfulness becomes a caricature of itself, losing its deep and transformational essence."
~ Ruth A. Baer, PhD, from "Mindfulness Based Treatment Approaches"
What people are saying about working with Genevieve...

redpajama_set_name: RedPajamaCommonCrawl


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By Nick Rocco Scalia	| June 12, 2018
Efficient pacing and a couple of genuinely surprising developments elevate the eminently watchable if flawed Remote, a Canadian-produced short film that plays something like a much darker and (thankfully) much shorter variation on the little-loved 2006 Sandra Bullock/Keanu Reeves romantic melodrama The Lake House.
That largely forgotten Hollywood effort, as sappy as it could be (and, good God, could it be sappy), at least contained an intriguing, Rod Serling-lite sort of premise: two people who live years apart from one another in the same house find a way to communicate even though they're separated by time. That's not a terrible idea, and it's definitely ripe for something other than the woefully mushy big-studio treatment it got.
Remote, from writer/director Marc Roussel, isn't mushy at all, though it does hint at a potential romance between its lead characters and cleverly plays its genre intentions fairly close to the vest for a good amount of running time. If the film doesn't quite hold together overall, it at least has the good sense to keep viewers guessing and not outstay its welcome.
"…their Facetime-through-time session is eventually interrupted when the film takes a turn into horror/suspense…"
The protagonist is Matt (Ron Basch), an average Joe who lives by himself as he glumly goes through divorce proceedings with his unseen wife. Holed up alone in his suburban home while a snowstorm rages outside, he's startled awake by a blast of decidedly old-fashioned-looking static from his modern flat-screen TV. When he's able to tune to a stable picture (hence the film's title), he sees a room that strangely seems to be a mirror image of the one he's sitting in – except that, in the TV-world, there also happens to be a young woman lounging in her underwear, who quickly discovers (to her understandable terror) that she can see and talk to Matt, too.
Her name is Justine (Sarah Silverthorne), and it's revealed before too long that she lives in Matt's house, 30 years in the past – and, thus, the person she's improbably interacting with through her own television set is somehow coming to her from three decades in the future. You'd think this situation might be of the brain-breaking type, but for some reason, the idea of their respective TVs acting as a sort of weird time portal is gotten over almost immediately by both Justine and Matt; they accept the situation so implausibly quickly and calmly that it's almost comical. Luckily for the audience, though, their Facetime-through-time session is eventually interrupted when the film takes a turn into horror/suspense territory – it's not fair to spoil just how and why, but the shift in tone happens precisely when it needs to.
"Misdirection and smart foreshadowing are among Remote's strongest assets…"
Misdirection and smart foreshadowing are among Remote's strongest assets, and they keep things interesting even when the staging, acting, and dialogue fall somewhat short; moments intended to be creepy, such as a protracted villain monologue, often come off a little too silly and over-the-top to fully connect. Similarly, the camerawork is slick and professional enough to be easy on the eyes throughout, but the visuals aren't particularly atmospheric or expressive – they serve the story, but only just.
Remote does mostly work, though, and its 20 minutes breeze by so briskly that it's definitely worth seeing through to the end. Roussel's storytelling is clean, his characters are fairly sympathetic, and his concept is solid. Given Hollywood's unimpressive track record with similar material, he's made a decent case for small-scale filmmakers' abilities to do a whole lot more with a whole lot less.
Remote (2011) Written and directed by Marc Roussel. Starring Ron Basch, Sarah Silverthorne, George Komorowski, Peter Racanelli, Julie McCarthy.
6 out of 10 stars
Sandow
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TWEEKED
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SEQUIN RAZE
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Oh Firefox, why aren't thou working?
Once in a while, you stop responding.
But you're ignoring our bonding.
The fear, the weakness and the horror.
I can't go on with this last strike.
Be there, when I need her most.
I am sorry. Our love is lost.


Bronze Radio Return would like to thank: Chad, Mark, John, Andrew, Jarod, and Nathan for working so hard on this record; our managers, Chris Maltese and Keith Mahler; and Alan at Telefunken.
Thank you to our loving and supportive families, our devoted and faithful friends, our caring and helpful mentors, and our true and loyal fans. We hope we tell you often enough how much we appreciate you and your contributions to our lives.


The Cuisinart 12.25" BBQ Melting Dome is a multi-functional cooking dome that will inspire your inner chef. The structure of the Melting Dome allows heat to concentrate under the dome and cook your food more evenly. Perfect for quickly melting cheese on burgers. Add woodchips under the dome on your griddle to infuse a smoky flavor in your dish. Looking for a healthier dinner option? Steam any protein or vegetable with the innovated vent on the dome. Fast melting cheese, better tasting meals, what more could you ask for?
LWH: 12.5" x 12.5" x 6"


Joerg is the newest member and joined the party in 2008, replacing Cliff on drums, who had to turn down the sticks, lacking time. Our samples were recorded shortly before he showed up, so he hadn't been taped. What a pity, he's a fond of timing, hey, you'll feel it, I'm sure about that!!


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All is well that Ends well
It's over, peacefully, bloodless and as far as I can tell, without any change. The rebels were sent back to their barracks. I can only hope that someone was disciplined but here it seems it's one extreme or the other and in this case I'm guessing nothing happened. They stood down, and that was the primary goal.
American citizens had been encouraged to evacuate the neighborhoods directly around the Ayala Center. The Embassy provided hotel rooms to all those interested. We were not part of that system, thankfully, as we have a major road between our neighborhood and the business district.
What did we do? We stayed home, made chocolate chip cookies, built a pillow/couch/table fort in the living room and figured God would forgive us for missing Mass. We didn't explain what was going on to the kids. Too close to home, and it did not interfere with our lives directly. If the standoff had continued, we had been in jeopardy, the building exploded, people died or the worst, we couldn't go to summer camp on Monday (perhaps something else that directly affected them), we would have given an abbreviated version of the news. But as it was, it ended without them needing to know anything.
Ian was called in to work on the task force at the Embassy and write a cable to Washington, then was sent to Makati to be an official American set of eyes on the situation. The roles switched between people every few hours so he wouldn't be working all night, just until 6 or 7.
While sitting in his bullet proof van, he was at the right spot to inform some new Japanese arrivals heading to check-in at the hotel that it really wasn't a good time. Once he filled them in on what was happening, Ian escorted them to find another cab and payed the driver a day's wages to take the Japanese business to a hotel on the other side of the city. All in a day's work around here.
Yesterday when our housekeeper arrived, one of the first things she said was "This was your first time, huh?" Doesn't that speak volumes?
In other news about things ending well, my dad is out of surgery. It only lasted a couple hours and he was in the recovery room and actually got up on his own. He had a disk removed in his back and replaced with a portion of his hip. Who came up with -that- idea? Sounds horrendous, but if all is well, he could be home tomorrow, off his feet the rest of the week, in a brace for 3 weeks and then almost all normal. In fact better than normal since this was to relieve a lot pain he's been having the past few years.
It's a serious drawback to our lifestyle when too much happens all at once. My parents really had a hard time, especially my mom. On one side she was hoping and praying my dad didn't die in the hospital (irrational? Possibly, but that's completely irrelevant). On the other, she was hoping and praying we didn't die 8000 miles away, or at all for that matter (irrational? Possibly, but that's completely irrelevant). I wanted to be home so that she wouldn't have to sit in the hospital all alone, and to be there for my dad. The kids are worried about grandpa but they made him cards which they hope will make him feel better. We wanted to make it very clear how much we love him.
My dad is someone who most people would call passionate. I don't feel that term adequately describes him. Passion suggests a single focus, something that draws you in and when the project is done it is a work of art, something to behold, something you realize and hold dear. Above all, I feel that passion suggests an inward acknowledgment of talent and pride and a power over your chosen medium.
My father is all that, but in everything he does. Everything he touches becomes his own. He does nothing half way. He reminds me of Midas where a touch of his finger transforms a potentially plain basic subject into a treasure. Only with my dad, he gives his soul to his music, his work, his family, his life. He touches so many people and transforms so much into gold. And most of the time, doesn't even know it.
Camp Critters Final Field Trip (and other stuff)
To coup or not to coup
So this is the life of an expat
Yesterday was busier than normal
Random thoughts...
It's not every day you get old.. I mean, turn 30!


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Wynder participates in Day on the Hill
June 19, 2016 / NPJ
Ramona Wynder, instructor and coordinator of alternative certification for Northwestern State University's Department of Teaching, Leadership and Counseling, participated in the Day on the Hill, the signature component of Washington Week for members of the American Association of Colleges for Teacher Education (AACTE). The event provides educators an opportunity to meet with their representatives and learn the latest information on education policy from Capitol Hill. The event includes training and resources for members to conduct meetings with congressional offices, where they can showcase their successful programs and advocate for relevant legislation or funding.
"I met with Congressman Ralph Abraham and his Legislative Assistant Blake Lasuzzo, Pamela Davidson, senior policy advisor for Senator Bill Cassidy, and Katie Doherty, legislative director for Congressman John Fleming to discuss legislation that will affect teacher preparation program providers and our candidates," Wynder said. "I also took the opportunity to highlight the successes of Northwestern State's teacher preparation programs, including our partnerships with local school districts."
Wynder was one of two delegates from Louisiana. The other was Dr. Don Shillinger, dean of education at Louisiana Tech University.
"Participating in AACTE Day on the Hill was personally and professionally rewarding," Wynder said."It was empowering to converge on Capitol Hill with teacher educators and leaders from across the nation to advocate for our programs. How blessed we are to live in a democratic nation where, regardless of our background, we can voice our thoughts to lawmakers about issues that impact us"
Wynder said that although not every citizen has the opportunity to travel to Washington, D.C., to meet with members of congress, every citizen can write a latter, send an email or make a phone call to their representative.
"Exercising our civic obligation should not stop in the voting booth," she said. "As citizens, we have the responsibility to speak up about laws and policies that affect our communities. We cannot afford to remain silent about things that matter; our collective voices can effectuate change."
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New Year's Eve dining in the San Gabriel…
ReviewReview An assessment or critique of a service, product, or creative endeavor such as art, literature or a performance.
New Year's Eve dining in the San Gabriel Valley
Chinese seafood, Chinese dumplings and Thai food are delicious ways to welcome 2022
Boiled dumplings, foreground, with roasted beef pancakes, left, and pan-fried dumplings, are great options for New Year's Eve dining, says restaurant critic Merrill Shindler. (SCNG file photo)
By Merrill Shindler | MrEats@aol.com |
PUBLISHED: December 18, 2021 at 3:03 p.m. | UPDATED: December 29, 2021 at 7:32 p.m.
My history of New Year's Eve celebrations has run pretty much from sad to sorrowful.
Growing up in New York City, I actually went to Times Square many years ago — long before they started corralling the celebrants for hours on end, forcing them to wear adult diapers. My memory of it was that it was cold … bone-chilling cold.
After the ball descended, the assembled masses all ran for the subway entrances. The New York subway system had, thoughtfully, gone on a post-midnight schedule. Meaning that several hundred thousand of us were clustered on the edge of the platforms, praying for a train — and that no one would push us off.
Then, there was the time I was invited to a gala dinner at a friend's house — only to discover the gala dinner consisted of bags of chips and cans of onion dip, with Lawrence Welk playing on the TV, and a lot of pitiful characters sitting on couches, trying to stay awake.
There was the meal at a favorite Italian restaurant, where I discovered, too late, that what was being served was a meal consisting of two pasta entrées, a green salad, and some sort of store-bought chocolate cake for dessert. I think the meal included a split of champagne. That's half a bottle, for two persons. It was bad champagne. And they charged too much for it.
After moving to the West Coast, to simplify New Year's Eve, I decided to stay on East Coast time the night of Dec. 31. This allowed me to celebrate early, and then go to sleep whenever I wanted. And to get up early enough for the Rose Parade, if I wanted to.
After all these years, there's been only one New Year's Eve that stands out as an unmitigated triumph — and it involved going into a world where Dec. 31 is not New Year's Eve. I organized a crowd to go to one of the sprawling Chinese seafood restaurants of Monterey Park for an early evening banquet. It was quite the experience.
Getting off at the Atlantic Avenue exit, just after 6 p.m., I found myself stuck in a traffic jam, caused by police running a sobriety check. At 6 o'clock? The Chinese New Year wasn't coming up for another month or more, so this was just another night. Except that the larger hall of the restaurant had been taken over by what I was told was a Chinese high school reunion. There were hundreds of celebrants. There were speeches and songs — all in Chinese. There were acrobatic acts. Prizes were awarded. And I sat in an adjacent room, eating shrimp and lobster, and having a very good time.
I notably remember bringing my own wine, and asking if there was a corkage fee. I was told, yes, $10. Per bottle, I asked? No, I was told — for the table, for the evening. What's not to love? I had a great meal — and was home early enough to catch the ball coming down in Times Square.
I've repeated it many times in the years since. A Chinese restaurant New Year's Eve on Dec. 31 has always been a joy. And reasonably priced, too. (I've also varied the feeds with meals at Thai restaurants, Filipino spots, Vietnamese places; Persian, Lebanese and Indian cafes; Korean barbecues, even the occasional Cambodian eatery, of which we have precious few! The prices are always great!
We bring our own fun with us, which does not include funny hats and noisemakers! These days, if you're at Times Square, you're locked in till after midnight. How is that fun? But meals at the following restaurants are always a joy — how could it be otherwise?
Chinese seafood (part one)
Newport Seafood (518 W. Las Tunas Drive, San Gabriel, 626-289-5998; 18441 Colima Road, Rowland Heights, 626-839-1239; www.newportseafood.com) — or, as it's also called, Newport Tan Cang — has long been a cult destination for those who yearn for what's essentially a culinary parlor game. For this is either a Vietnamese restaurant that serves Chinese seafood, or a Chinese restaurant that serves Vietnamese noodles and rice. Or, perhaps, it's a Pan-Asian restaurant that offers Chinese, Vietnamese, Thai and Cambodian dishes.
Whatever it is (and it's probably all of the above, and more), it's amazingly good — and amazingly packed, night in and night out, and especially on weekends.
The original Newport Seafood opened in Orange County in 1989. It moved to a chaotic space in a funky shopping mall about a mile west of its current location. And then, it relocated to what used to be a Marie Callender's — a clear statement about the ongoing tectonic culinary shift of ethnicities in the San Gabriel Valley. (In between, a branch opened in Rowland Heights as well.)
The current Newport Seafood still looks, a bit, like a Marie Callender's. Or at least, it does from the outside, with its peaked-roof and ersatz bricks walls. But inside, the place is deep Asian, with red lacquer wall dividers, numerous areas for private parties, oversized tanks in which some true monsters of the deep roam, and a level of cacophony little known in American pie houses.
The wait for a table at Newport Seafood is never as long as you think it will be. And anyway, while you wait, you can while away the time admiring the outlandishly large lobsters in the tanks. Newport is famed for its outlandishly large lobsters (they claim to serve 10-pounders, though five-pounders seem closer to the average, which is still a lot of lobster!), cooked variously with ginger and green onions, salt and peppers, black bean sauce, steamed and Newport Special style — which involves hacking up the carapace, and frying the critters with ginger, garlic, chiles (just about everything in the kitchen).
Lobster is the signature dish at Newport Seafood, with a seasonal price that can make it a real deal — or a real indulgence. Whatever the price, it's going to be less than you'd pay at any number of local steakhouses. A lot less.
They apparently buy their lobsters by the shipload at Newport. And they go through every one of them. As is often the case at our Chinese seafood palaces, the food emerges from the kitchen with what seems to be impossible speed; it's as if all the dishes are sitting there, ready to go. And, in a way, they are. For the essence of Chinese cooking is preparation. It only takes a couple of minutes (if that) to wok some crab with tamarind sauce, or beef with mushrooms and bamboo shoots. It's getting all the parts ready that takes most of the day.
Anyway, with a menu of 137 items, and a couple of hungry demanding customers, you've got to be ready to roll. Which at Newport, they definitely are.
The menu at Newport Seafood is in English, Chinese and Vietnamese. Though those who read menus as both a diversion and a study of culinary anthropology will fairly come to the conclusion that the lemon tilapia soup is a Cambodian dish, the sort of thing not often found outside of the Cambodian neighborhood of Long Beach.
The rice noodle with beef certainly wouldn't be out of place in a Vietnamese pho shop, and the beef loc lac would fit just right into any of our local French-Vietnamese cafes. But for the most part, this is a Cantonese seafood house that appeals to a wide variety of Asian customers.
After the house special lobster, the dishes I always order at Newport are many. Expect a lot of noise, and a lot of big tables filled with big families. Expect some terrific food, including some of the only baked bean curd with spicy salt around. Expect a journey into a cuisine-within-a-cuisine, at the only restaurant I know of where even the fried rice options are offered in three languages.
Chinese seafood (part two)
Boston Lobster (727 E. Valley Blvd., San Gabriel; 626-288-4388, www.restaurantji.com/ca/san-gabriel/boston-lobster-seafood-restaurant-) isn't in Boston, and it doesn't serve the seafood of Boston. There's no chowdah on the menu, no fried clam bellies.
Naming a Chinese seafood house Boston Lobster was a true oddity. Except that not far away from Boston Lobster, there's Newport Seafood, which is not in Newport; it's in San Gabriel, as is Boston Lobster. Boston Lobster more than satisfies.
The menu, with 127 dishes, and seven pre-set meals, is shorter than some, but certainly long enough. Dishes are listed in Chinese, English and Vietnamese, a further reminder of our multi-ethnicity. And though this isn't a Vietnamese restaurant, there are Vietnamese dishes — the Vietnamese style shrimp and catfish with hot and sour soup. But mostly, there's the Hong Kong style seafood we've come to know and virtually worship over the years, the dishes we ate before Szechuan became all the rage.
Since the restaurant is Boston Lobster, ordering a lobster seems essential, even if it can easily cost as much as the rest of your meal combined. And really, it is, available in a variety of sizes, served steamed, with black bean sauce, with ginger and green onion, with salt and pepper, and "special style," which seems to be spicier than the others.
For $6 extra, it's served over garlic noodles, which absorb the juices and flavors of the lobster — they're worth every penny. But then, although you could easily get nothing but the lobster at Boston Lobster, it would mean passing on so many other really wonderful dishes.
There are 11 shrimp dishes. But as has long been true, none make me as happy as the fried whole shrimp with salt and pepper. The shrimp are crispy, perfectly so — so crispy that biting off the head and chewing on it, and devouring the crunchy tail, is akin to devouring a shrimp-flavored potato chip, salt and pepper flavored, of course.
I didn't expect the sole to be deep-fried, like the world's best fish and chips, but without the chips; I didn't notice the absence. And really, since one must have their vegetables, for the goodness that's in them, the Chinese broccoli with oyster sauce, and the sautéed pea sprouts were a fine addition. Ditto the fried noodles with yet more shrimp. And the crabmeat fried rice.
For dessert, there's Fosselman's, because that's what you do after too much Chinese seafood. Chocolate ice cream tastes so good after a sodium overload. It's part of the ritual. That…and scrubbing your finger when you get home, to make them smell less like lobster. It rarely works.
Chinese seafood (part three)
Sea Harbour Seafood (3939 N. Rosemead Blvd., Rosemead; 626-288-3939, www.seaharbourrosemead.com) is the Southern California outlet of a restaurant based not in Kowloon or Tsim Sha Tsui, but in Richmond Canada, just outside of Vancouver.
The back of one of the restaurant's business cards featured a totem pole juxtaposed with Sleeping Beauty's castle at Disneyland. The menu — one part of which sits inside a binder worthy of the Encyclopedia Britannica, another part of which is in a carved wooden frame suitable for nice pix of mom, pop and the kids — was filled with dishes not often encountered at the several dozen Chinese seafood palaces found in the wide expanse of the San Gabriel Valley. Clearly, this wasn't the same old same old.
What almost every table seems to be chewing is, variously, baked barbecue pork buns (unexpected at night, and equally unexpected in a seafood house), spareribs deep-fried in honey, panfried diced barbecue pork, deep-fried pigeon, Dungeness crab with egg white and steamed live prawns with garlic.
To make sure I don't starve, I add a tureen of the hot and sour seafood stew, some Chinese greens (they looked like collard greens) tossed with yet more pork and a salt-baked chicken. I order a whole lobster in black bean sauce. I order some whole prawns in spicy salt because I always order whole prawns in spicy salt because I love the dish — especially since you can eat the whole prawn, head and tail and all. And a whole rock cod cooked so that the skin and bones were crisp enough to eat.
The prawns are perfect, big and sweet and filled with orangy-coral roe, with a like sprinkling of almost raw garlic. The whole lobster is filled with tender chunks of meat in every nook and cranny — if you were willing to do the work, there was much to be found. The Dungeness crab is fine.
Unexpectedly, it's the egg white dressing that I find the most flavorful — it seems to have absorbed the very essence of the crab, turning into Dungeness-flavored scrambled egg whites. The greens wouldn't be out of line at a Southern eatery like Aunt Kizzy's. The prawns with spicy salt are at the red line when it came to crunchiness; they're prawn cracklin's.
For dessert, they bring some sort of red bean soup, which doesn't really make it as dessert unless you grew up eating the stuff.
Chinese dumplings
Hui Tou Xiang (704 W. Las Tunas Drive, San Gabriel; 626-281-9888, www.huitouxiang.com) is a restaurant named for a dish, which is apparently unique to this one location. Hui tou potstickers are longish, blintz-like creations, filled with either pork or beef, crispy on the outside, juicy on the inside.
The name has a double meaning, for "hui tou" always translates as "to return to." And indeed, you will — for this dish, and three dozen more. It's the sort of mini-mall café where I want to keep going back until I've gone through the entire menu.
You order from a simple checklist, with sections for panfried, steamed, boiled, wontons, soup noodles, dry noodles and appetizers. Wherever you opt to begin, your table will be covered by dishes with manic speed. This is not a restaurant that wastes time with fancy preps. It's all ready to go. And glad you'll be for that. For like a shark in a feeding frenzy, once you begin chewing, it's hard to stop.
This is what I think of as Obsessive Cuisine. With the exception of the hui tou, I've eaten the dumplings and noodles served here all over the Valley. I feel safe in saying this is as good as they get. Din Tai Fung may be glitzier — it's a culinary thrill ride at an amusement park, very high tech, very modern, and making you want to go on the ride again. But then, the wait to get into Din Tai Fung can be daunting. For me, getting into Hui Tou Xiang was easy. Might not be easy all the time. But at least, sometimes.
So…what to order? Certainly it matters. But it also doesn't matter; I'm not sure there's a miss on the menu. The hui tou is, of course, essential because, for goodness sake, the place is named for them, and they were created here. Attention must be paid. But along with the panfried hui tou, there are also panfried pancakes — both pork and leek, and vegetable and leek — and old school potstickers, smaller, but no less wondrous. What the heck, order them all; they taste just great the next day.
As you may know, xiao long bao is the dish of the moment in many dumpling houses. To Anglos, it's better known as soup dumplings, which is what these tasty little flying saucers of dough, chopped meat and vegetables, and hot broth really are — dumplings filled with soup, which can singe your mouth if you don't approach them with care.
Aficionados have many methods, ranging from letting them cool to biting a small hole in one edge and cautiously sipping out the broth. It all works, though once you get into the frenzy, you may burn yourself just out of impatience. The temptations are many.
The steamed pork shu mai is much like the soup dumplings, but without the soup. The boiled dumplings are akin to the steamed ones, only one comes filled with fish, another with pork and bell peppers, yet another with pork, Napa cabbage and celery.
The soup noodles don't get nearly enough credit, for they come in a marvel of a broth, umami heavy, deeply flavorful, not a broth to waste. It's the Chinese version of Jewish chicken noodle soup, advancing my contention that the Chinese are one of the Lost Tribes of Israel.
The soup noodles come flavored with pork ribs and braised meatballs and pork hock as well. The sesame noodles — served "dry" rather than in soup — are presented deconstructed. You get the noodles and the flavorings, and you do it yourself. A fun bit of arts and crafts in between bits of dumplings.
The menu for President Thai (498 S. Rosemead Blvd., Pasadena; 626-578-9814, www.president-thai.com) declares it the "First traditional Thai architecture restaurant in Los Angeles." Though, frankly, I'm not at all sure what the second might be: President Thai may be not only be the first — it may indeed be the only.
President is a reminder that Saladang isn't the only great Thai restaurant in Pasadena. The difference is that Saladang is just south of Old Pasadena, while President is perhaps a tad off the well-trod path. Still, it's a path many have no trouble trodding.
For the record, there's another President Thai on Fullerton Road in Rowland Heights, in the somewhat chaotic Diamond Plaza, where the menu is similar, but not identical. But it's the Pasadena President that was there first, opening in 1994, while I'm told the sibling restaurant was opened some years later by the owner's daughter. Which is a nice bit of genealogy. But it matters little as you dissect the insanely good barbecue chicken at the original.
Exactly how the restaurant manages to get the marinade to ooze into every crevasse and corner of the chicken is a bit of a mystery. But however it's done, it makes for an amazing chicken; the temptation to eat the bones is strong.
President Thai is one of the most efficiently run restaurants in town. You walk through the door, and if a table is available, you get it — even if you're a solitary diner. (Some restaurant treat solitary diners like plague carriers. At President, you're respected as much as a tables of four.)
The menus — and there are several — arrive fast. Drink orders are taken before you start considering what to eat. Drinks arrive while you're still considering what to eat.
After ordering, dishes arrive with almost disconcerting speed — as if they were already made, and in the kitchen waiting for you. And servers don't disappear; this is a big problem these days. And trust me, ordering here can take some study, unless you arrive having already set your mouth for a particular dish.
There are 115 dishes listed on the menu. But that number is actually far higher, since in numerous cases, you've got an option of pork, beef or chicken, and in others of tofu, veggies, shrimp and seafood as well. That's a lot of decisions to make — a good reason to pause over a nice cold Singha beer.
The menu is, no surprise, a greatest hits selection of Thai dishes, with plenty of house specialties tossed in for good measure. The satays are a thing of beauty — large, moist, perfectly cooked strips of chicken, beef or pork, with a first-rate peanut sauce. The mee krob, which in too many Thai restaurants is so sweet it's more a dessert than a savory dish, is actually subtle and under-sweetened, with lots of chicken and shrimp.
And what's not to like about the President's Shrimp — shrimp on a stick, wrapped in bacon and egg noodles and deep-fried; the stick makes it that much better, like a protein lollipop.
If this President is running in 2024, it has my vote.
Merrill Shindler is a Los Angeles-based freelance dining critic. Email mreats@aol.com.
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Merrill Shindler | Restaurant reviewer
MrEats@aol.com
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Schnuppy.ch: Find Suitable Trial Apprenticeships in Just a Few Clicks
Daniela Capaul-Zoppi
The idea for schnuppy.ch was born in 2015, triggered by often frustrating experiences in the search for trial apprenticeships. On the initiative of the Sek Regensdorf and with the support of the Rotary Club, the first version of the online job placement system was designed and developed in many workshops and launched in 2017 in the district of Dielsdorf.
The platform was quickly very widely used and requests followed from other regions. Therefore, schnuppy. ch was completely redesigned and developed last year and launched in May 2021.
Convenient job search & application
schnuppy.ch shows young people the nearest available taster apprenticeships and also enables a targeted search for professions. The application takes just a few clicks, as the necessary documents such as certificates or letters of motivation are stored in the personal dashboard.
Everything at a glance in the personal dashboard
The personal dashboard contains all open applications and accepted trial apprenticeships, as well as a timeline that shows the young person on which dates he or she has already accepted trial apprenticeships or still has open applications. Young people and companies can communicate directly via schnuppy.ch using the chat function. Important events such as the acceptance of a position are also communicated automatically via e-mail.
Advertisements & application management for companies
Registered companies can advertise as many trial apprenticeship positions as they wish in schnuppy.ch. They can advertise different types of apprenticeships, define the time of availability and in which radius of the company location the jobs should be displayed. Received applications can be accepted and rejected via the platform. In order to prevent too many applications from being received for a position, it is possible to define after how many applications or acceptances the availability of a position is automatically set to red. Particularly attractive positions can be advertised in selection mode, i.e. the young people can only apply for an exact date and are pre-informed that the chance of acceptance is rather small.
Close involvement of teachers and parents
A key success factor of schnuppy is that schools, teachers and parents are closely involved in the process. They are kept up to date on applications, acceptances and rejections of young people. Teachers also have a dashboard that shows them the activities of all the students in their class. This provides optimal support for the still very young students to apply correctly and find the right trial apprenticeships.
UX/Design: Focus on an optimal mobile experience
Since the young schnuppy.ch target group is mostly mobile, there was a strong focus on a simple and intuitive mobile application. All functions – from the search, to the application and access to all dashboard functions – are also easily accessible via smartphone or tablet.
On the backend schnuppy.ch is based on the Symfony Framework/PHP7 and a PostgreSQL database. The webfrontend was developed with HTML/Javascript. The UI was developed via Pattern Lab according to the Atomic Design approach.
Who is behind schnuppy.ch?
The overall responsible organization the association schnuppy.ch. It is responsible for the financing and the proper operation of the platform. The association is also responsible for the admission and introduction of new districts. The local networking is a recipe for success of schnuppy.ch and is guaranteed by the various project leaders in the districts. In addition, the respective area management is available to the schools and businesses as a direct contact person. A schoolhouse delegate is appointed for each school, who is available as a responsible person on site for questions or concerns of teachers, school administrators and young people. schnuppy.ch is supported by:
Daniela has many years of Marketing and Leadership experience in various industries. She has built a large Mobile Business expertise since the raise of Smart Mobile Devices.
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Retired Attorney General Eric Holder pushed back on statements made last week by FBI Director James Comey in which Comey suggested the so-called Ferguson effect might be responsible for a recent rise in crime in cities around the country.
The Ferguson effect is a name given to a recent spike in crime in some, but not all, American cities this year. The idea is that police have taken note of the public mood and decided to hold back on more aggressive policing.
The comments by Eric Holder are part of an ongoing war of words within the Obama administration, which escalated last Friday when FBI Director James Comey seemed to endorse the Ferguson effect as an explanation during an address at the University of Chicago Law School.
"Nobody says this on the record. Nobody says this in public," Comey told his audience last Friday, adding that it was, "the one explanation that, to my mind, explains the calendar and the map." He was referring to both the widespread increase in crime in cities across the country and the timing of that increase this year.