text:
\section{Introduction} In a recent paper \cite{1} we have described a
formulation of the electroweak theory in terms of
asymptotic fields alone. Since
in our causal approach the asymptotic gauge fields are massive from the very
beginning (except the photon), spontaneous symmetry breaking plays no role.

Let us summarize the strategy. The fundamental problem in gauge theories is
the appearance of {\it unphysical fields} in the gauge potentials. An
additional main difficulty is that already the quantization of the massive
\footnote{This problem appears also for the massless free gauge fields.
The peculiarity of the massive (free) theory is that we need the unphysical
bosonic scalars $\Phi_a$ to restore the nilpotency of $Q$.} free gauge
fields in a $\lambda$-gauge requires an indefinite metric space. To
solve these two problems in the {\it free} theory one has to select the
space of physical states ${\cal H}_{\rm phys}$ which must be a (pre) Hilbert
space, i.e. the inner product must be positive definite. There are two
popular methods at hand: the Gupta Bleuler method (which works well for
abelian gauge theories only) and the BRST-formalism \cite{15}. We take the
latter in the form of Kugo and Ojima \cite{16}: the physical (pre) Hilbert
space is defined as the cohomology of a nilpotent operator $Q$,
i.e. ${\cal H}_{\rm phys}={{\rm Ker}\,Q\over {\rm Ran}\,Q}$. To define such
an operator we introduce unphysical fields: each gauge field $A_a^\mu$ gets
three scalar partners, the fermionic fields $u_a,\,\tilde u_a$
("ghost fields") and the bosonic $\Phi_a$. Then
$$Q\=d \int d^3x\,(\partial}\def\=d{\,{\buildrel\rm def\over =}\,_{\nu}A_a^{\nu}+m_a\Phi_a){\dl}_0u_a,$$
does the job.

To overcome the presence of unphysical fields in the {\it interacting theory}
one has additionally to prove that they {\it decouple}. The latter means
that in the adiabatic limit (if it exists) the $S$-matrix induces a
well-defined unitary operator. For this purpose one needs a notion of
gauge invariance. The most natural formulation which yields the decoupling
is to require that $Q$ commutes with the $S$-matrix in the adiabatic limit
(if it exists). This is essentially the content of our formulation of
gauge invariance
$$d_QT_n\=d [Q,T_n]=i\sum_{l=1}^n{\partial}\def\=d{\,{\buildrel\rm def\over =}\,\over\partial}\def\=d{\,{\buildrel\rm def\over =}\, x_l^\nu}T^\nu_{n/l}
(x_1,\ldots x_l\ldots x_n)\eqno(1.0)$$
(where we use the notations of \cite{1}), which is independent of the
adiabatic limit, i.e. it makes sense also in theories in which this limit
does not exist. This is a pure quantum formulation of gauge invariance.
It has turned out that this symmetry requirement (1.0) cannot be satisfied
with the fields at hand so far. By introducing an additional scalar field
(the "Higgs field"), which is physical, gauge invariance can be saved.
In addition (1.0) is sufficiently strong to determine the couplings of all
fields.

The main steps in this construction are the following. Starting from the pure
Yang - Mills coupling $T_1^A$ at first order, gauge invariance requires
couplings $T_1^u$ of the
ghost fields $u_a\,\tilde u_a$ and fixes them. This step
is the same as in the massless theory \cite{2}. However, the massive gauge
fields need couplings to the bosonic scalar partners $\Phi_a$
to compensate mass terms in the first order
gauge variation. Gauge invariance then determines the couplings $T_1^\Phi$ of
these unphysical scalars to the gauge fields and to the ghosts. But the
resulting theory would not be gauge invariant at second order. This requires
the "Higgs field" $\Phi_0$, its coupling $T_1^{\Phi_0}$
to all other fields is again determined by gauge invariance. The coupling to
leptons is uniquely fixed by first and second order gauge invariance if one
family of fermions is considered \cite{1}. However, it is well known that gauge
invariance is then violated at third order by the triangular anomalies. Hence,
the theory must still be enlarged by including at least one complete generation
of leptons and quarks. In what follows we consider the general situation of
arbitrarily many generations of fermions with arbitrary mixing.
It is our aim to
analyse what gauge invariance has to say about the couplings in this case.

Assuming that the reader is familiar with the main results of \cite{1}, we
present in the following section the general ansatz for the fermionic coupling
$T_1^F$ and determine the constraints imposed by gauge invariance at first
order. At second order (Sect.3) we obtain a long list of interesting conditions
for the coupling matrices. At present we cannot control all solutions of this
system. But inserting a general ansatz which is valid in the vicinity of the
standard model, we get a unique solution that agrees with the standard model. In
Sect.4 we briefly discuss gauge invariance of the third order triangular graphs.
We recover the well-known fact that the cancellation of the axial anomalies
restricts the charges of the fermions.

In Sect.5 we show how gauge invariance of third order tree graphs determines the
last free parameters in the scalar coupling. In this way we {\it derive} the
Goldstone - Higgs potential. Other approaches where this
quartic double-well potential is not postulated but deduced, are based
on methods of non-commutative geometry which have been developed by
Connes \cite{3} and
others \cite{4,5,6,7,12}, or on superconnections \cite{13}.
The scalar potential comes out in the form
$$V(\Phi_0)=\lambda (\Phi_0^4-4a\Phi_0^3+4a^2\Phi_0^2),\eqno(1.1)$$ with
$$a={2\over g}m_W,\eqno(1.2)$$ where $m_W$ is the $W$-mass and $g$ the universal
gauge coupling constant.  The physical 'Higgs' field $\Phi_0$ is realized in an
ordinary Fock representation with unique vacuum and vacuum expectation value
$\langle \Phi_0\rangle=0$.
(In the usual convention $-\Phi_0$ is used instead of
$\Phi_0$.) If we introduce the shifted field
$$\varphi=\Phi_0-a,\eqno(1.3)$$
the potential assumes the usual symmetric form \cite{8}
$$V=\lambda(\varphi^2-a^2)^2.\eqno(1.4)$$
Now the vacuum expectation value
$\langle\varphi\rangle = -a$ is different from 0. If the shift (1.3) is carried
out everywhere in the scalar couplings (see (3.20) of \cite{1}) we get
quadratic
mass terms (with the wrong sign) for the already massive gauge bosons. These
quadratic terms can simply be resummed which changes the W- and Z-masses from
their finite values into zero! That means, we have inverted the
Higgs mechanism. This gives the connection of our approach with the
standard theory. However, this connection does not enter our construction.

\section{Fermionic coupling and first order gauge invariance}

We start from the following generalization of the simple leptonic electroweak
coupling to more than one family
$$T_1^F=ig\sum_{j,k}\Bigl\{W_\mu^+b^1_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu\nu_k+W_\mu^+
b^{\prime 1}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu\gamma^5\nu_k\eqno(2.1.1)$$
$$+W_\mu^-b^2_{jk}\onu_j\gamma^\mu{\rm e}_k+W_\mu^-b^{\prime 2}_{jk}
\onu_j\gamma^\mu\gamma^5{\rm e}_k\eqno(2.1.2)$$ $$+Z_\mu
b^3_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu{\rm e}_k+Z_\mu b^{\prime 3}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j
\gamma^\mu\gamma^5{\rm e}_k\eqno(2.1.3)$$ $$+Z_\mu
b^4_{jk}\onu_j\gamma^\mu\nu_k+Z_\mu b^{\prime 4}_{jk}\onu_j
\gamma^\mu\gamma^5\nu_k\eqno(2.1.4)$$ $$+A_\mu b^5_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu{\rm
e}_k+A_\mu b^{\prime 5}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j \gamma^\mu\gamma^5{\rm e}_k\eqno(2.1.5)$$
$$+A_\mu b^6_{jk}\onu_j\gamma^\mu\nu_k+A_\mu b^{\prime 6}_{jk}\onu_j
\gamma^\mu\gamma^5\nu_k\eqno(2.1.6)$$
$$+\Phi^+c^1_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\nu_k+\Phi^+c^{\prime 1}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5\nu_k
+\Phi^-c^2_{jk}\onu_j{\rm e}_k+\Phi^-c^{\prime 2}_{jk}\onu_j\gamma^5 {\rm
e}_k\eqno(2.1.7)$$ $$+\Phi_3c^3_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j{\rm e}_k+\Phi_3c^{\prime
3}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5 {\rm e}_k+\Phi_3c^4_{jk}\onu_j\nu_k+\Phi_3c^{\prime
4}_{jk}\onu_j \gamma^5\nu_k\eqno(2.1.8)$$ $$+\Phi_0c^0_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j{\rm
e}_k+\Phi_0c^{\prime 0}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5 {\rm
e}_k+\Phi_0c^5_{jk}\onu_j\nu_k+\Phi_0c^{\prime 5}_{jk}\onu_j
\gamma^5\nu_k\Bigl\}.\eqno(2.1.9)$$
Here we have used the same notation as in
\cite{1} (5.1): All products of field operators throughout are normally
ordered (Wick monomials). $W, Z, A$ denote the gauge fields and
$\Phi^\pm=(\Phi_1\pm i\Phi_2)/\sqrt{2}$, $\Phi_3$ the unphysical scalars.
$\Phi_0$ is the physical
scalar, but ${\rm e}_j (x)$ stands for the electron-, muon-, tau-fields, as well
as for the quark fields d, s, b, and $\nu_k(x)$ represents the corresponding
neutrini and the other quark fields u, c, t. In \cite{1} we have only considered
the coupling to leptons. There the terms (2.1.6) are missing because the
neutrini have vanishing electric charge, but here, for the quark couplings, we
must include them. We also assume that the asymptotic Fermi fields
fulfil the Dirac equations $$\hbox{\rlap/$\partial$}} \def\psq{{\overline{\psi}}} \def\tr{{\rm tr}{\rm e_j}=-im_j^e{\rm e_j},\quad
\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu(\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu)=im_j^e\overline{\rm e}} \def\onu{\overline{\nu}_j$$ $$\hbox{\rlap/$\partial$}} \def\psq{{\overline{\psi}}} \def\tr{{\rm tr}\nu_k=-im_k^\nu\nu_k,\quad
\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu(\onu_k\gamma^\mu)=im_k^\nu\onu_k, \eqno(2.2)$$
with arbitrary non-vanishing unequal masses $m^e_j\ne m^e_k\ne 0$,
$m^\nu_j\ne m^\nu_k\ne 0$ for all $j\ne k$.
We do not use further information about the multiplett
structure of the fermions.

According to \cite{1} the gauge structure is introduced as follows. We define a
gauge charge $$Q\=d \int
d^3x\sum_{a=0}^3(\partial}\def\=d{\,{\buildrel\rm def\over =}\,_{\nu}A_a^{\nu}+m_a\Phi_a){\dl}_0u_a,$$ where $A_a^\nu$
stands for $A^\nu, W_1^\nu, W_2^\nu, Z^\nu$ and $u_a$ for the corresponding
ghosts, and a gauge variation $$d_Q F\=d QF-(-1)^{n_F}FQ.$$ Then the gauge
variations of the asymptotic gauge fields are given by $$d_QA^\mu =i\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu
u_0,\> d_Q W_{1,2}^\mu=i\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu u_{1,2},\> d_Q Z^\mu=i\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu u_3,$$ and for the
Higgs and unphysical scalar fields
$$d_Q\Phi_0=0,\>d_Q\Phi_{1,2}=im_Wu_{1,2},\>d_Q\Phi_3=im_Zu_3$$ and finally for
the fermionic ghosts $$d_Qu_a=0,\quad a=0,1,2,3$$ $$d_Q\tilde u_0=-i\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu
A^\mu,\>d_Q\tilde u_{1,2}=-i(\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu W_{1,2} ^\mu+m_W\Phi_{1,2})$$ $$d_Q\tilde
u_3=-i(\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu Z^\mu+m_Z\Phi_3).\eqno(2.3)$$ The gauge variations of the
fermionic matter fields vanish.

First order gauge invariance means that the gauge variation of
$$T_1=T_1^A+T_1^u+T_1^\Phi+T_1^{\Phi_0}+T_1^F$$ has divergence form
$$d_QT_1(x)=i\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu T_{1/1}^\mu.\eqno(2.4)$$ To verify this for the fermionic
coupling $T_1^F$ we calculate the gauge variation of (2.1) and take out the
derivatives of the ghost fields. In the additional terms with derivatives on the
matter fields we use the Dirac equations (2.2):
$$d_QT_1^F=-g\sum_{j,k}\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu\Bigl\{u^+(b^1_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu\nu_k+ b^{\prime
1}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu\gamma^5\nu_k) +u^-(b^2_{jk}\onu_j\gamma^\mu{\rm
e}_k+b^{\prime 2}_{jk} \onu_j\gamma^\mu\gamma^5{\rm e}_k)\eqno(a)$$
$$+u_3(b^3_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu{\rm e}_k+b^{\prime 3}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j
\gamma^\mu\gamma^5{\rm e}_k +b^4_{jk}\onu_j\gamma^\mu\nu_k+b^{\prime
4}_{jk}\onu_j \gamma^\mu\gamma^5\nu_k)\eqno(b)$$
$$+u_0(b^5_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^\mu{\rm e}_k+b^{\prime 5}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j
\gamma^\mu\gamma^5{\rm e}_k +b^6_{jk}\onu_j\gamma^\mu\nu_k+b^{\prime
6}_{jk}\onu_j \gamma^\mu\gamma^5\nu_k)\Bigl\}\eqno(c)$$
$$+g\Bigl\{iu^+[b^1_{jk}(m_j^e-m_k^\nu)\overline{\rm e}} \def\onu{\overline{\nu}_j\nu_k+ b^{\prime
1}_{jk}(m_j^e+m_k^\nu)\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5\nu_k]\eqno(d)$$
$$+iu^-[b^2_{jk}(m_j^\nu-m_k^e)\onu_j{\rm e}_k+b^{\prime 2}_{jk}
(m_j^\nu+m_k^e)\onu_j\gamma^5{\rm e}_k]$$ $$+iu_3[b^3_{jk}(m_j^e-m_k^e)\overline{\rm e}} \def\onu{\overline{\nu}_j{\rm
e}_k+b^{\prime 3}_{jk}(m_j^e+m_k^e) \overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5{\rm e}_k$$
$$+b^4_{jk}(m_j^\nu-m_k^\nu)\onu_j\nu_k+b^{\prime 4}_{jk}(m_j^\nu+m_k^\nu)
\onu_j\gamma^5\nu_k]$$ $$+iu_0[b^5_{jk}(m_j^e-m_k^e)\overline{\rm e}} \def\onu{\overline{\nu}_j{\rm e}_k+b^{\prime
5}_{jk}(m_j^e+ m_k^e)\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5{\rm e}_k$$
$$+b^6_{jk}(m_j^\nu-m_k^\nu)\onu_j\nu_k+b^{\prime 6}_{jk}(m_j^\nu+m_k^\nu)
\onu_j\gamma^5\nu_k]\Bigl\}$$ $$+m_W[u^+c^1_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\nu_k+u^+c^{\prime
1}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5\nu_k +u^-c^2_{jk}\onu_j{\rm e}_k+u^-c^{\prime
2}_{jk}\onu_j\gamma^5 {\rm e}_k]$$ $$+m_Zu_3[c^3_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j{\rm e}_k+c^{\prime
3}_{jk}\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5 {\rm e}_k+c^4_{jk}\onu_j\nu_k+c^{\prime 4}_{jk}\onu_j
\gamma^5\nu_k]\Bigl\}.\eqno(2.5)$$

Now, to have first order gauge invariance, the terms ($d$) until the end of
(2.5) which are not of divergence form must cancel. This implies
$$b_{jk}^{\prime 5}=0=b_{jk}^{\prime 6},\>\forall j,k,\quad
b_{jk}^5=0=b_{jk}^6\>{\rm for}\>j\ne k\eqno(2.6)$$ and $$c_{jk}^1={i\over
m_W}(m_j^e-m_k^\nu)b_{jk}^1,\quad c_{jk}^{\prime 1}={i\over
m_W}(m_j^e+m_k^\nu)b_{jk}^{\prime 1}$$ $$c_{jk}^2={i\over
m_W}(m_j^\nu-m_k^e)b_{jk}^2,\quad c_{jk}^{\prime 2}={i\over
m_W}(m_j^\nu+m_k^e)b_{jk}^{\prime 2}$$ $$c_{jk}^3={i\over
m_Z}(m_j^e-m_k^e)b_{jk}^3,\quad c_{jk}^{\prime 3}={i\over
m_Z}(m_j^e+m_k^e)b_{jk}^{\prime 3}$$ $$c_{jk}^4={i\over
m_Z}(m_j^\nu-m_k^\nu)b_{jk}^4,\quad c_{jk}^{\prime 4}={i\over
m_Z}(m_j^\nu+m_k^\nu)b_{jk}^{\prime 4}.\eqno(2.7)$$ The result (2.6) means that
the photon has no axial-vector coupling and no mixing in the vector coupling.
This is due to the fact that it has no scalar partner because it is massless.

\section{Gauge Invariance at Second Order}

As discussed in Sect.4 and 5 of \cite{1}, the essential problem in
second order gauge
invariance is whether the anomalies in the tree graphs cancel out.  These
anomalies are the local terms in $\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu^x T_{2/1}^\nu
\vert^0_{\rm tree}(x,y)+\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu^y T_{2/2}^\nu\vert^0_{\rm tree} (x,y)$
and come from two sources. First, if the terms ($a$)-($c$) in (2.5) are
combined with the terms in (2.1) by a fermionic contraction we get the Feynman
propagator $S^F_m(x-y)$ in $T_{2/1}^\nu
\vert^0_{\rm tree}(x,y)$ and $T_{2/2}^\nu\vert^0_{\rm tree} (x,y)$. Taking
the divergence with respect to the $Q$-vertex a $\delta$-distribution is
generated due to
$$i\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu^x\gamma^\mu S_m^F(x-y)=mS_m^F(x-y)+\delta(x-y).  \eqno(3.1)$$
This $\delta$-term is
the anomaly. Secondly, we can perform a bosonic contraction between the
terms in
(2.1) and the terms (4.5) in \cite{1} which are the anomaly-producing part in
$T_{1/1}^{A\Phi\,\mu}$, coming from the Yang-Mills and scalar couplings:
$$T_{1/1}^{A\Phi\,\mu}\vert_{\rm an}=ig\Bigl\{\sin\Theta(u_1W_2^\nu-
u_2W_1^\nu) \partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu A_\nu \eqno(3.2.1)$$ $$+\sin\Theta
(u_2A^\nu-u_0W_2^\nu)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu W_{1\nu}+\sin\Theta (u_0 W_1^\nu-u_1A^\nu)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu
W_{2\nu} \eqno(3.2.2)$$ $$+\cos\Theta\Bigl[(u_2Z^\nu-u_3W_2^\nu)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu W_{1\nu}
+(u_3W_1^\nu-u_1Z^\nu)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu W_{2\nu}\eqno(3.2.3)$$
$$+(u_1W_2^\nu-u_2W_1^\nu)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu Z_\nu\Bigl]\eqno(3.2.4)$$ $$+\sin\Theta
(u_0u_1\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde u_2+u_2u_0\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde u_1+ u_1u_2\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde
u_0)\eqno(3.2.5)$$ $$+\cos\Theta (u_2u_3\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde u_1+u_3u_1\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde u_2
+u_1u_2\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\tilde u_3)\eqno(3.2.6)$$ $$+\sin\Theta
u_0(\Phi_2\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_1-\Phi_1\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_2)+\Bigl(1-{m_Z^2 \over
2m_W^2}\Bigl)\cos\Theta u_3(\Phi_2\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_1-\Phi_1\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_2)\eqno(3.2.7)$$
$$+{1\over 2}{m_Z\over m_W}\cos\Theta\Bigl[(u_2\Phi_1-u_1\Phi_2)\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_3
+u_1\Phi_3\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_2-u_2\Phi_3\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_1\Bigl] \eqno(3.2.8)$$ $$+{1\over
2}u_1(\Phi_0\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_1-\Phi_1\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0) +{1\over
2}u_2(\Phi_0\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_2-\Phi_2\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0)\eqno(3.2.9)$$ $$+{1\over
2\cos\Theta}u_3(\Phi_0\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_3- \Phi_3\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0)\Bigl\}.\eqno(3.2.10)$$

To give a representative example, we calculate the anomalies with external
field
operators $u_3\Phi_3\overline{\rm e}} \def\onu{\overline{\nu}_j\gamma^5{\rm e}_k$. Combining the first term in (2.5)
($b$) with the second term in (2.1.8) we get an anomaly from the
contraction of ${\rm e}_{k'}(x)$ with $\overline{\rm e}} \def\onu{\overline{\nu}_{j'}(y)$ ($S^F[{\rm e}_{k'}(x),
\overline{\rm e}} \def\onu{\overline{\nu}_{j'}(y)]$ denotes the corresponding Feynman propagator)
$$-iu_3(x)b^3_{jk'}\overline{\rm e}} \def\onu{\overline{\nu}_j(x)\gamma^\mu
S^F[{\rm e}_{k'}(x),\overline{\rm e}} \def\onu{\overline{\nu}_{j'}(y)]\gamma^5{\rm e}_k(y) \Phi_3(y)
c^{\prime 3}_{j'k}.$$
It results the anomaly $\sim -ib^3c^{\prime 3}\delta(x-y)$ involving
the matrix product of $b^3$ with $c^{\prime 3}$. Combining the two terms with
reversed order, we obtain $c^{\prime 3}b^3$ with a different sign so that both
terms together yield the commutator $i[c^{\prime 3},b^3]$.
Similarly, the second
term in (2.5) ($b$) together with the first term in (2.1.8) gives the
anticommutator $\{c^3,b^{\prime 3}\}$, because the $\gamma^5$ is at a different
place. An anomaly of the second source comes from the last term in (3.2.10)
contracted by the two $\Phi_0$-fields with the second term in (2.1.9):
$$-{i\over 2\cos\Theta}u_3(x)\Phi_3(x)D^F[\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0(x),\Phi_0(y)]
\overline{\rm e}} \def\onu{\overline{\nu}_j(y)\gamma^5 {\rm e}_k(y)c_{jk}^{\prime 0}.$$
Altogether we obtain the following matrix equation
$$-{1\over 2\cos\Theta}c^{\prime 0}=i[c^{\prime 3},b^3]+i\{c^3, b^{\prime
3}\}.\eqno(3.3)$$

We now give the complete list of all second order conditions. We specify the
corresponding external legs, then the origin of the terms is pretty clear. Every
combination of external field operators has a corresponding one with an
additional $\gamma^5$. To save space we do not write down the external legs once
more for the $\gamma^5$-term.

$u_0\Phi_0\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>[c^0,b^5]=0,\quad [c^{\prime 0},b^5]=0$

$u_0\Phi_3\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>[c^3,b^5]=0,\quad [c^{\prime 3},b^5]=0$

$u_0Z\overline{\rm e}} \def\onu{\overline{\nu}\gamma{\rm e}:\>[b^3,b^5]=0,\quad [b^{\prime 3},b^5]=0$

$u_0\Phi_0\onu\nu:\>[c^5,b^6]=0,\quad [c^{\prime 5},b^6]=0$

$u_0\Phi_3\onu\nu:\>[c^4,b^6]=0,\quad [c^{\prime 4},b^6]=0$

$u_0Z\onu\gamma\nu:\>[b^4,b^6]=0,\quad [b^{\prime 4},b^6]=0$ \qquad (3.4)

$u^+A\overline{\rm e}} \def\onu{\overline{\nu}\gamma\nu:\>\sin\Theta b^1=b^5b^1-b^1b^6,\quad \sin\Theta b^{\prime
1}=b^5b^{\prime 1}-b^{\prime 1}b^6$

$u^-A\onu\gamma{\rm e}:\>\sin\Theta b^2=b^2b^5-b^6b^2,\quad \sin\Theta
b^{\prime 2}=b^{\prime 2}b^5-b^6b^{\prime 2}$ \qquad (3.5)

$u_0\Phi^+\overline{\rm e}} \def\onu{\overline{\nu}\nu:\>\sin\Theta c^1=b^5c^1-c^1b^6,\quad \sin\Theta c^{\prime
1}=b^5c^{\prime 1}-c^{\prime 1}b^6$

$u_0\Phi^-\onu{\rm e}:\>\sin\Theta c^2=c^2b^5-b^6c^2,\quad \sin\Theta c^{\prime
2}=c^{\prime 2}b^5-b^6c^{\prime 2}$ \qquad (3.6)

$u_3W^+\overline{\rm e}} \def\onu{\overline{\nu}\gamma\nu:\>\cos\Theta b^1=b^3b^1+b^{\prime 3}b^{\prime 1}
-b^1b^4-b^{\prime 1}b^{\prime 4},\quad \cos\Theta b^{\prime 1}=b^{\prime
3}b^1+b^3b^{\prime 1} -b^{\prime 1}b^4-b^1b^{\prime 4}$

$u_3W^-\onu\gamma{\rm e}:\>\cos\Theta b^2=b^2b^3+b^{\prime 2}b^{\prime 3}
-b^4b^2-b^{\prime 4}b^{\prime 2},\quad \cos\Theta b^{\prime 2}=b^{\prime
2}b^3+b^2b^{\prime 3} -b^{\prime 4}b^2-b^4b^{\prime 2}$

\rightline{(3.7)}

$u^+W^-\overline{\rm e}} \def\onu{\overline{\nu}\gamma{\rm e}:\>\cos\Theta b^3=b^1b^2+b^{\prime 1}b^{\prime 2}
-\sin\Theta b^5,\quad \cos\Theta b^{\prime 3}=b^{\prime 1}b^2+b^1b^{\prime 2}$

$u^-W^+\onu\gamma\nu:\>\cos\Theta b^4=-b^2b^1-b^{\prime 2}b^{\prime 1}
-\sin\Theta b^6,\quad \cos\Theta b^{\prime 4}=-b^{\prime 2}b^1-b^2b^{\prime 1}
$\qquad (3.8)

$u^+\Phi_3\overline{\rm e}} \def\onu{\overline{\nu}\nu:\>c^1=2(b^{\prime 1}c^{\prime 4}+c^{\prime 3}b^{\prime 1}
+c^3b^1-b^1c^4),\quad c^{\prime 1}=2(b^{\prime 1}c^4+c^3b^{\prime 1} +c^{\prime
3}b^1-b^1c^{\prime 4})$

$u^-\Phi_3\onu{\rm e}:\>c^2=2(-b^{\prime 2}c^{\prime 3}-c^{\prime 4} b^{\prime
2}-c^4b^2+b^2c^3),\quad c^{\prime 2}=2(-b^{\prime 2}c^3-c^4b^{\prime 2}
-c^{\prime 4}b^2+b^2c^{\prime 3})$

\rightline{(3.9)}

$u^+\Phi_0\overline{\rm e}} \def\onu{\overline{\nu}\nu:\>\delta_1c^1=i(c^0b^1+c^{\prime 0}b^{\prime 1} +b^{\prime
1}c^{\prime 5}-b^1c^5),\quad \delta_1c^{\prime 1}=i(c^{\prime 0}b^1+c^0b^{\prime
1} +b^{\prime 1}c^5-b^1c^{\prime 5})$

$u^-\Phi_0\onu{\rm e}:\>\delta_1c^2=i(c^5b^2+c^{\prime 5} b^{\prime
2}+b^{\prime 2}c^{\prime 0}-b^2c^0),\quad \delta_1c^{\prime 2}=i(c^{\prime
5}b^2+c^5b^{\prime 2} +b^{\prime 2}c^0-b^2c^{\prime 0})$

\rightline{(3.10)}

$u_3\Phi_3\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>\delta_3c^0=i[b^3,c^3]-i\{b^{\prime 3},c^{\prime
3}\},\quad\delta_3c^{\prime 0}=i[b^3,c^{\prime 3}]-i\{b^{\prime 3},c^3\}$

$u_3\Phi_0\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>\delta_3c^3=i[c^0,b^3]+i\{c^{\prime 0},b^{\prime
3}\},\quad\delta_3c^{\prime 3}=i[c^{\prime 0},b^3]+i\{c^0 ,b^{\prime 3}\}$
\qquad (3.11)

$u_3\Phi_3\onu\nu:\>\delta_3c^5=i[b^4,c^4]-i\{b^{\prime 4},c^{\prime
4}\},\quad\delta_3c^{\prime 5}=i[b^4,c^{\prime 4}]-i\{b^{\prime 4},c^4\}$

$u_3\Phi_0\onu\nu:\>\delta_3c^4=i[c^5,b^4]+i\{c^{\prime 5},b^{\prime
4}\},\quad\delta_3c^{\prime 4}=i[c^{\prime 5},b^4]+i\{c^ 5,b^{\prime 4}\}$
\qquad (3.12)

$u^+\Phi^-\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>\delta_1c^0=i(b^1c^2-b^{\prime 1}c^{\prime
2})-ic^3/2,\quad\delta_1c^{\prime 0}=i(b^1c^{\prime 2}-b^{\prime 1}c^2)
-ic^{\prime 3}/2$

$u^-\Phi^+\overline{\rm e}} \def\onu{\overline{\nu}{\rm e}:\>\delta_1c^0=-i(c^1b^2+c^{\prime 1}b^{\prime
2})+ic^3/2,\quad\delta_1c^{\prime 0}=-i(c^{\prime 1}b^2+c^1b^{\prime 2})
+ic^{\prime 3}/2$\qquad (3.13)

$u^+\Phi^-\onu\nu:\>\delta_1c^5=i(b^2c^1-b^{\prime 2}c^{\prime
1})-ic^4/2,\quad\delta_1c^{\prime 5}=i(b^2c^{\prime 1}-b^{\prime 2}c^1)
-ic^{\prime 4}/2$

$u^-\Phi^+\onu\nu:\>\delta_1c^5=-i(c^2b^1+c^{\prime 2}b^{\prime
1})+ic^4/2,\quad\delta_1c^{\prime 5}=-i(c^{\prime 2}b^1+c^2b^{\prime 1})
+ic^{\prime 4}/2$\qquad (3.14)

$u_3\Phi^+\overline{\rm e}} \def\onu{\overline{\nu}\nu:\>\delta_4c^1=b^3c^1-b^{\prime 3}c^{\prime 1}-c^1b^4-c^{\prime
1}b^{\prime 4},\quad\delta_4c^{\prime 1}=-b^{\prime 3}c^1+b^3c^{\prime
1}-c^1b^{\prime 4}-c^{\prime 1}b^4$

$u_3\Phi^-\onu{\rm e}:\>\delta_4c^2=-b^4c^2+b^{\prime 4}c^{\prime
2}+c^2b^3+c^{\prime 2}b^{\prime 3},\quad\delta_4c^{\prime 2}=b^{\prime
4}c^2-b^4c^{\prime 2}+c^2b^{\prime 3}+c^{\prime 2}b^3$,

\rightline{(3.15)}

where $$\delta_1={1\over 2},\quad\delta_3={1\over 2\cos\Theta},
\quad\delta_4=\cos\Theta-{1\over 2\cos\Theta}.$$ The terms with these $\delta$'s
and with the electroweak mixing angle obviously come from (3.2). There are
further combinations of external field operators which have not been written
down, because they give no new condition.

In case of one family, assuming $b^6=0$ and taking pseudounitarity into account
(\cite{1} (5.21)), the corresponding system of scalar equations has a unique
solution, which agrees with the lepton coupling of the standard model. The
solution of the above matrix equations (3.4-15) is not so simple. We start from
the equations (3.5). If we write these equations with matrix elements, using the
fact that $b^5$ and $b^6$ are diagonal (2.6), we easily conclude that $b^5$ and
$b^6$ are actually multiples of the unit matrix $$b^5=\alpha{\bf 1},\quad
b^6=(\alpha-\sin\Theta){\bf 1}.\eqno(3.16)$$
Here $\alpha$ is a free parameter (the electric charge of the upper
quarks or leptons)
and we have assumed that the matrices $b^1, b^{\prime 1}, b^2, b^{\prime 2}$
are nontrivial and non-diagonal. This is not a serious limitation because we shall see that these matrices are essentially the unitary mixing matrices.
For the leptons we assume $b_6=0$ instead. This first consequence of gauge
invariance is
the universality of the electromagnetic coupling: the members $e_k(x)$  and
$\nu_k(x)$ of different generations all couple in the same way to the photon,
with a constant charge difference $q_e-q_\nu=g\sin\Theta$, which is the
electronic charge.

Next we turn to the conditions (3.11). It is convenient to introduce the
diagonal mass matrices $$m^e={\rm diag}(m^e_j),\quad m^\nu={\rm
diag}(m^\nu_j),\> j=1,\ldots n_g,$$ where $n_g$ is the number of generations.
Then (2.7) can be written as follows $$c^1={i\over m_W}(m^eb^1-b^1m^\nu),$$
$$c^{\prime 1}={i\over m_W}(m^eb^{\prime 1}+b^{\prime 1}m^\nu),\eqno(3.17) $$
etc. Then the first two equations in (3.11) read $$c^{\prime
0}={1\over\delta_3m_Z}\Bigl(2b^{\prime 3}m^eb^3-2b^3m^e b^{\prime 3}$$
$$+m^eb^{\prime 3}b^3+m^eb^3b^{\prime 3}-b^3b^{\prime 3}m^e-b^{\prime 3}
b^3m^e\Bigl)$$ $$c^0={1\over\delta_3m_Z}\Bigl(2b^{\prime 3}m^eb^{\prime
3}-2b^3m^eb^3 +(b^3)^2m^e+m^e(b^3)^2+(b^{\prime 3})^2m^e+m^e(b^{\prime
3})^2\Bigl).  \eqno(3.18)$$ Substituting this into the last two equations of
(3.11), we arrive at the following coupled matrix equations for $b^3, b^{\prime
3}$: $$\delta_3^2(m^eb^{\prime 3}+b^{\prime 3}m^e)=3b^{\prime 3}m^e(b^3)^2
+3(b^3)^2m^eb^{\prime 3}+3(b^{\prime 3})^2m^eb^{\prime 3}+3b^{\prime 3}
m^e(b^{\prime 3})^2$$ $$-3b^3m^eb^{\prime 3}b^3-3b^3b^{\prime
3}m^eb^3-3b^{\prime 3}b^3m^eb^3 -3b^3m^eb^3b^{\prime 3}$$ $$+m^eb^{\prime
3}(b^3)^2+m^eb^3b^{\prime 3}b^3+m^e(b^{\prime 3})^3+ m^e(b^3)^2b^{\prime 3}$$
$$+(b^3)^2b^{\prime 3}m^e+b^3b^{\prime 3}b^3m^e+b^{\prime 3}(b^3)^2m^e
+(b^{\prime 3})^3m^e\eqno(3.19)$$ $$\delta_3^2(m^eb^3-b^3m^e)=3(b^{\prime
3})^2m^eb^3+3(b^3)^2m^eb^3+ 3b^{\prime 3}m^eb^{\prime 3}b^3+3b^{\prime
3}m^eb^3b^{\prime 3}$$ $$-3b^3m^e(b^3)^2-3b^3b^{\prime 3}m^eb^{\prime
3}-3b^{\prime 3}b^3m^e b^{\prime 3}-3b^3m^e(b^{\prime 3})^2$$
$$+m^e(b^3)^3-(b^3)^3m^e+m^e(b^{\prime 3})^2b^3-(b^{\prime 3})^2b^3m^e$$
$$+m^eb^3(b^{\prime 3})^2-b^3(b^{\prime 3})^2m^e+m^eb^{\prime 3}b^3 b^{\prime
3}-b^{\prime 3}b^3b^{\prime 3}m^e.\eqno(3.20)$$

The coupled cubic equations (3.19-20) have many solutions in general. To
determine the solutions in the neighbourhood of the standard model, we
substitute $$b^3=\delta_3(\beta{\bf 1}+x),\quad b^{\prime 3}=\delta_3(\beta'
{\bf 1}+y),\quad \beta,\beta'\in {\bf C},\eqno(3.21)$$
and assume the matrices $x, y$ to be small so that only
terms linear in $x$ and $y$ must be taken with in (3.19-20). Then the equations
collapse to the simple form $$2\beta'(1-4\beta^{\prime 2})m+(1-12\beta^{\prime
2})(my+ym)=0 \eqno(3.22)$$ $$(1-12\beta^{\prime 2})(mx-xm)=0.\eqno(3.23)$$ Now,
(3.22) yields a unique solution if $\beta'=O(1)$ is assumed
$$\beta'={\eps_2\over 2},\quad\eps_2=\pm 1,\quad y=0,\eqno(3.24)$$
where the
last result follows by writing the vanishing anticommutator $\{m^e,y\}$ with
matrix elements, using $m^e_j>0, \forall j$. Then (3.23) implies
$$(m^e_i-m^e_k)x_{ik}=0,\quad \hbox{\rm no sum over}\>i,k\eqno(3.25)$$
that
means $x$ and $b^3$ are diagonal, taking into account that the masses
$m^e_j$ are not
degenerate. All matrices in (3.18) commute, thus $$c^{\prime 0}=0\eqno(3.26)$$
$$c^0={4m^e\over\delta_3m_Z}(b^{\prime 3})^2={\delta_3\over m_Z}m^e,
\eqno(3.27)$$ and (3.11) gives $$c_3=0,\quad c^{\prime 3}=i\eps_2{\delta_3\over
m_Z}m^e.\eqno(3.28)$$

The same reasoning can be carried through for (3.12) which leads to $$b^{\prime
4}=-\eps_2{\delta_3\over 2},\quad c^4=0,\eqno(3.29)$$ for the sign see below,
and $$c^{\prime 4}=-i\eps_2{\delta_3\over m_Z}m^\nu,\quad c^{\prime 5}
=0\eqno(3.30)$$ $$c^5={\delta_3\over m_Z}m^\nu.\eqno(3.31)$$ With this knowledge
we turn to (3.9). Substituting $c^1$ in the first equation by (3.17) we arrive
at $$m^e(b^1-\eps_2b^{\prime 1})=(b^1-\eps_2b^{\prime 1})m^\nu,$$ leading to
$$b^{\prime 1}=\eps_2b^1,\eqno(3.32)$$ assuming non-degenerate masses again. In
the same way the second equation in (3.9) yields $$b^{\prime
2}=\eps_2b^2.\eqno(3.33)$$ This is the chiral coupling of all fermion
generations. The sign in (3.29) follows from (3.8).

Finally, from the four conditions (3.8) it is easy to conclude $$b^2={1\over
8}(b^1)^{-1}\eqno(3.34)$$ $$b^3={1\over\cos\Theta}\Bigl({1\over
4}-\alpha\sin\Theta\Bigl) \eqno(3.35)$$ $$b^4={1\over\cos\Theta}\Bigl(-{1\over
4}-\alpha\sin\Theta+\sin^2 \Theta\Bigl).\eqno(3.36)$$ This means that $x$ in
(3.21) is actually zero, so that {\it there is no other solution in the
neighbourhood of the standard model}. But solutions "far away" are not excluded.
All values of the $b$'s and $c$'s agree with the standard model \cite{8} for an
arbitrary number $n_g$ of generations. It is not hard to check that with the
results so determined all other second order conditions of gauge invariance are
satisfied. To finish this discussion we notice that pseudo-unitarity implies
$$b^{1+}=b^2={1\over 8}(b^1)^{-1},$$ hence $$b^1={1\over 2\sqrt{2}}V,\quad
b^2={1\over 2\sqrt{2}}V^+,\eqno(3.37)$$ where $V$ is an arbitrary unitary
matrix. This is the CKM mixing matrix \cite{8,9} for the quark coupling. A
similar mixing is possible for the charged leptonic currents. The recently
observed signals of neutrino oscillations show that this mixing probably
occurs.

\section{Gauge invariance at third order: axial anomalies}

Adler, Bell and Jackiw \cite{10} discovered that there exists a possibility to
violate gauge invariance at third order in the triangular graphs. This holds also true in the causal approach to gauge theory (\cite{14} sect.5.3).
The anomalous
graphs contain one axial-vector and two vector couplings (VVA) or three
axial-vector couplings (AAA) of the fermions and three external gauge fields. To
have a more compact notation, we collect all fermionic matter fields into a big
vector $\psi=({\rm e},\mu,\tau,\nu_e,\nu_\mu,\nu_\tau)$ or $=(d,s,b,u,c,t),$
respectively. The gauge fields $A,W^+,W^-,Z$ are denoted by $A^\mu_a$ with
$a=0,+,-,3$. Then the coupling between fermions and gauge fields in (2.1) can be
written as
$$T_1^{FA}=ig\Bigl\{\psq\gamma_\mu M_a\psi
A_a^\mu+\psq\gamma_\mu\gamma^5 M'_a\psi A_a^\mu\Bigl\},\eqno(4.1)$$
where $M_a$
stands for the following matrices of matrices: $$M_+=\pmatrix{0&b^1\cr
0&0\cr},\quad M_-=\pmatrix{0&0\cr b^2&0\cr}\eqno (4.2)$$ $$M_3=\pmatrix{b^3&0\cr
0&b^4\cr},\quad M_0=\pmatrix{b^5&0\cr 0&b^6\cr}, \eqno(4.3)$$ and similarly for
the axial-vector couplings, denoted by a prime.

Each triangular graph gives rise to two diagrams which differ by a permutation
of two vertices. Therefore, we have to compute the traces
$$\tr(M_aM_bM_c)+\tr(M_aM_cM_b)=\tr(M_a\{M_b,M_c\}),$$ where one or three $M$'s
must be axial-vector couplings with a prime.  It is well-known that the
cancellation of the axial anomalies relies on the compensation of these traces
in the sum of the leptonic and hadronic contributions. In this way third order
gauge invariance gives a further restriction of the quark coupling. To work this
out in detail, we consider the following cases.

{\it A. Case $(a=0,b=+,c=-)_{VVA}$:}

$$\tr (M_0\{M'_+,M_-\})+\tr (M_0\{M_+,M'_-\})=$$ $$=\tr[b^5(b^1b^{\prime
2}+b^{\prime 1}b^2)+b^6(b^2b^{\prime 1}+ b^{\prime 2}b^1)]$$ Using the results
of the last section, this is equal to $$=n_g{\eps_2\over
4}(2\alpha-\sin\Theta),\eqno(4.4)$$ where $n_g$ is the number of generations,
i.e. the dimension of the matrices $b^k$, and $\alpha$ is the charge of the
fermions in (3.16).  For leptons we have $b^6=0$, because the neutrini have no
electric charge, so that $$\alpha_L=\sin\Theta.\eqno(4.5)$$ Consequently, to
compensate the triangular anomaly proportional to (4.4), one needs the
compensation between leptons and quarks.  We assume equal number of families in
the lepton and quark sectors. Then for three colors of quarks one must have
$$2\alpha_L-\sin\Theta+3(2\alpha_Q-\sin\Theta)=0,\eqno(4.6)$$ which implies
$$\alpha_Q={1\over 3}\sin\Theta\eqno(4.7)$$ by (4.5).

In most textbooks the electric charge $\alpha_Q$ (4.7) of the d-, s-, b-quarks is put in and then, by requiring cancellation of the anomalies, one concludes
that the number of families in the lepton and quark sectors must be equal. We have simply reversed the argument.

{\it B. Case $(0,3,3)_{VVA}$:}

In this case the trace is simply given by
$$\tr(M_0\{M'_3,M_3\})=\tr[b^5(b^{\prime 3}b^3+b^3b^{\prime 3})+ b^6(b^{\prime
4}b^4+b^4b^{\prime 4})]$$ $$=n_g\eps_2{\delta_3\over\cos\Theta}\Bigl({1\over
4}-\sin^2\Theta \Bigl)\Bigl(2\alpha-\sin\Theta\Bigl).\eqno(4.8)$$ Due to the
same factor $(2\alpha-\sin\Theta)$ as in (4.4) the mechanism of compensation is
the same.

{\it C. Case $(3,+,-)_{VVA}$:}

Here the trace is equal to
$$\tr(M'_3\{M_+,M_-\}+M_3\{M'_+,M_-\}+M_3\{M_+,M'_-\}=$$
$$=\tr[b'_3b_1b_2+b'_4b_2b_1+b^3(b^1b^{\prime 2}+b^{\prime 1}b^2)
+b^4(b^2b^{\prime 1}+b^{\prime 2}b^1)]$$ $$=n_g{\eps_2\sin\Theta\over
4\cos\Theta}(\sin\Theta-2\alpha),\eqno(4.9)$$ with the same consequences as
before.

{\it D. Case $(3,+,-)_{AAA}$:}

Now we have to compute the trace $$\tr(M'_3\{M'_+,M'_-\}=\tr[b^{\prime
3}b^{\prime 1}b^{\prime 2}+ b^{\prime 4}b^{\prime 2}b^{\prime 1}]=0.$$

{\it E. Case $(0,0,3)_{VVA}$:}

Here the relevant trace is equal to $$\tr(M_0^2M'_3)=\tr[(b^5)^2b^{\prime
3}+(b^6)^2b^{\prime 4}$$ $$=n_g\eps_2{\delta_3\over
2}\sin\Theta(2\alpha-\sin\Theta).$$ Due to the same factor as in case A, the
compensation  between leptons and quarks is the same. This is also true in the
next case:

{\it F. Case $(3,3,3)_{VVA}$:}

$$\tr(M_3^2M'_3)=\tr[(b^3)^2b^{\prime 3}+(b^4)^2b^{\prime 4}]$$
$$=n_g\eps_2{\delta_3\over \cos\Theta}\Bigl(\sin^3\Theta-{\sin\Theta \over
2}\Bigl)(2\alpha-\sin\Theta).$$

{\it G. Case $(3,3,3)_{AAA}$:}

This final case is trivial: $$\tr(M'_3)^3=\tr[(b^{\prime 3})^3+(b^{\prime
4})^3]=0.$$ Summing up the axial anomalies completely cancel in each generation,
if and only if $\alpha_Q$ has the value (4.7).

\section{Gauge invariance at third order: tree graphs}

Third order tree graphs are not covered by the general inductive proof of gauge
invariance in massless Yang-Mills theories \cite{11}. They are part of the
beginning of the induction and need an explicit verification of gauge
invariance. The latter can easily be done in the massless Yang-Mills theory, by
using the fact that all couplings are of Yang-Mills type, that means
proportional to $f_{abc}$. But this is no longer true for the scalar couplings
in massive Yang-Mills theories \cite{1}. Therefore, it is not surprising that
gauge invariance of third order tree diagrams fixes the last free parameters in
the scalar coupling of the electroweak theory, namely the Higgs self-coupling
(Higgs potential).

Let us first discuss a simple special case of the standard model, the $U(1)$
Higgs model, which contains all essential features of the scalar self-coupling.
We consider one massive gauge field $W^\mu$ only, in interaction with one
unphysical ($\Phi$) and one physical scalar field $\Phi_0$. Therefore, we let
$W_1=W, u_1=u, \tilde u_1=\tilde u, \Phi_1=\Phi, m_W=m$ in (3.20) of \cite{1}
and omit the other fields. From (3.20.9, 10, 12) we then get $$T_1=i{g\over
2}\Bigl\{W^\nu(\Phi_0\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu\Phi-\Phi\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu\Phi_0) -mW_\nu W^\nu\Phi_0$$
$$+{m_H^2\over 2m}\Phi_0\Phi^2+m\tilde uu\Phi_0+2b\Phi_0^3\Bigl\}, \eqno(5.1)$$
and from (4.5.10) of \cite{1} or (3.2.9) $$T^\mu_{1/1}\vert_{\rm an}=-i{g\over
2}u\{\Phi\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0-\Phi_0 \partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi\}\=d D_1+D_2.\eqno(5.2)$$

The calculation and compensation of the anomalies at second order can now be
directly taken over from the appendix in \cite{1}. Only the following two
sectors 1) and 7) appear:

\noindent {\it 1) Sector $(\Phi_0,\Phi_0,1,1)$}

Here we have found the two normalization terms (A.1, 2) $$N_1={i\over
4}g^2W_\nu W^\nu\Phi_0^2\delta(x-y)\eqno(5.3)$$ $$N_2=ig^2\Bigl({m_H^2\over
4m^2}-{3b\over 2m}\Bigl)\Phi_0^2 \Phi^2\delta(x-y).\eqno(5.4)$$

\noindent {\it 7) Sector $(1,1,1,1)$}

$$N_{10}={i\over 4}g^2W_\nu W^\nu\Phi^2\delta(x-y)\eqno(5.5)$$
$$N_{11}=-ig^2{m_H^2\over 16m^2}\Phi^4\delta(x-y)\eqno(5.6)$$ These results
(5.3-6) are valid for the $U(1)$ Higgs model, too, because all additional
couplings in the standard model do not contribute to these two sectors.
Furthermore, the final normalization term (A.20)
$$N_{20}=ig^2\lambda'\Phi_0^4\delta(x-y)\eqno(5.7)$$
becomes now important.
Until now the coupling parameters $b$ (5.1) and $\lambda'$ (5.7) are arbitrary.

Gauge invariance (1.0) at third order can only be violated by local terms
$\sim D\delta (x_1-x_3,x_2-x_3)$ (where $D$ denotes a differential operator).
This can easily be seen by inserting the causal factorization of the
time-ordered products and using gauge invariance at lower orders (cf. (4.3)
in \cite{1}). We only consider the tree diagrams
$$d_QT_3\vert_{\rm tree}=i\sum_{k=1,2,3}\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu^k T_{3/k}^\nu \vert_{\rm tree}
\eqno(5.8)$$
and adopt the notations and terminology from the corresponding calculation
at second order (sect. 4 of \cite{1}).\footnote{Note that e.g. a term
$\sim\delta(x_1-x_2)\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu D_F(x_2-x_3)$ ($\mu$ and $\nu$ not
contracted) is non-local.} There are no
local terms in $T_3\vert_{\rm tree}$ and $T_{3/k}\vert_{\rm tree}$, because
a term $\sim\delta (x_1-x_3,x_2-x_3):B_1B_2B_3B_4B_5:$ would violate
renormalizability by power counting (terms with derivatives on $\delta
(x_1-x_3,x_2-x_3)$ are even worse); especially
there is no freedom of normalization (cf. (4.2) in \cite{1}). Hence the only
local terms in (5.8) are the anomalies which are the terms $\sim\delta
(x_1-x_3,x_2-x_3)$ in $\sum_{k=1,2,3}\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\nu^k T_{3/k}^\nu \vert_{\rm tree}
(x_1,x_2,x_3)$. (Derivatives of the $\delta$-distribution do not appear
as can be seen by power counting.) Up to permutation of the vertices there
is only one possibility to generate an anomaly: due to causal factorization,
e.g. $T_{3/1}^\mu (x_1,x_2,x_3)=T_{1/1}^\mu (x_1) T_2(x_2,x_3)$ for
$x_1\not\in(\{x_2,x_3\}+\bar V^-)$, the normalization term $N_{(2)}(x_2,x_3)
\sim\delta (x_2-x_3)$ of $T_2\vert_{\rm tree}(x_2,x_3)$ is contained in
$T_{3/1}^\mu (x_1,x_2,x_3)$. Hence, if $\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi_0$ in $D_1$, or
$\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu\Phi$ in $D_2$ (5.2), is contracted with a field operator in
$N_{(2)}(x_2,x_3)$ we obtain a term
$$T_{3/1}\vert_{\rm tree}^\mu (x_1,x_2,x_3)=\sim\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu D_F(x_1-x_2)\delta
(x_2-x_3):B_1(x_1)B_2(x_1)B_3(x_2)B_4(x_3)B_5(x_3):+...\>.$$
Taking now the divergence with respect to the $Q$-vertex $x_1$, an anomaly
appears due to \break $\partial}\def\=d{\,{\buildrel\rm def\over =}\,_\mu\partial}\def\=d{\,{\buildrel\rm def\over =}\,^\mu D_F(x_1-x_2)=-m^2D_F(x_1-x_2)+\delta
(x_1-x_2)$. Considering the external field operators $u\Phi\Phi_0^3$
there come anomalies from the combinations of $D_1$ with $N_{20}$ and of
$D_2$ with $N_2$. Gauge invariance requires the cancellation of these
anomalies
$$(D_1,N_{20})_{\rm loc}+(D_2,N_2)_{\rm loc}=0,$$
which gives the condition
$$-i{g\over 2}\>4ig^2\lambda'+i{g\over
2}\>2ig^2\Bigl( {m_H^2\over 4m^2}-{3b\over 2m}\Bigl)=0.\eqno(5.9)$$
The factors 4 and 2 are due to the number of possible contractions between
the two members $D$ and $N$. Because $N_{20}$ and $N_2$ depend on the
unknown parameters $\lambda^\prime$ and $b$, this is a first equation to
determine these parameters.

The anomalies with field operators $u\Phi\Phi_0W_\nu W^\nu$ come from the
terms
$$(D_1,N_1)_{\rm loc}+(D_2,N_{10})_{\rm loc}=0.$$
This gives the condition
$$-i{g\over 2}\>2i{g^2\over 4}+i{g\over 2}\>2{i\over 4}g^2=0,$$ which is
automatically satisfied. Finally, in the sector $u\Phi_0 \Phi^3$ gauge
invariance requires
$$(D_1,N_2)_{\rm loc}+(D_2,N_{11})_{\rm loc}=0$$
and this leads to the constraint
$$-i{g\over 2}\>2ig^2\Bigl({m_H^2\over 4m^2}-{3b\over
2m}\Bigl) +i{g\over 2}\>4(-ig^2){m_H^2\over 16m^2}=0.\eqno(5.10)$$
>From (5.9) and (5.10) we find the values of the two coupling parameters
$$b={m_H^2\over 4m},\quad\lambda'=-{m_H^2\over 16m^2}.\eqno(5.11)$$

The same argument applies to the whole electroweak theory: In the sector
$u_1\Phi_1\Phi_0^3$ we get the equation (5.9) (with $m=m_W$), and in the
sector $u_1\Phi_0\Phi_1^3$ we find the condition (5.10) again. It is easy to
see that there are no additional terms contributing in the bigger theory. The
resulting values $$b={m_H^2\over 4m_W},\quad\lambda'=-{m_H^2\over
16m_W^2}\eqno(5.12)$$ agree with those obtained from the Higgs potential in the
standard theory as shown below. We have verified that, with these parameters,
all other anomalies from third order tree graphs cancel out, without giving
further information.

For comparison with the standard model we collect all self-couplings of scalar
fields. The first order terms are contained in (3.20.10-12) of \cite{1}:
$$V_1(\Phi)=ig{m_H^2\over 4m_W}\Phi_0(\Phi_0^2+\Phi_1^2+\Phi_2^2
+\Phi_3^2).\eqno(5.13)$$ The second order terms are given by the normalization
terms in the appendix of \cite{1} $$V_2(\Phi)=ig^2{m_H^2\over
16m_W^2}\Bigl[-(\Phi_0^2+\Phi_1^2+\Phi_2^2 +\Phi_3^2)^2\Bigl].\eqno(5.14)$$
Remembering that the second order must be multiplied by 1/2, we obtain the
following total scalar potential $$V(\Phi)=V_1+{1\over 2}V_2=-ig^2{m_H^2\over
32m_W^2}\Bigl[(\Phi_0^2+ \Phi_1^2+\Phi_2^2+\Phi_3^2)^2-8{m_W\over
g}\Phi_0(\Phi_0^2+\Phi_1^2 +\Phi_2^2+\Phi_3^2) \Bigl].\eqno(5.15)$$ To compare
this with the Goldstone-Higgs potential of the standard theory we set
$\Phi_1=\Phi_2=\Phi_3=0$ and add a mass term ${1\over 2}m_H^2\Phi_0^2$.
 Omitting the factor $-i$, the potential is then
equal to $$V(\Phi_0)=\lambda(\Phi_0^4-4a\Phi_0^3+4a^2\Phi_0^2),\eqno(5.16)$$
where $$\lambda={1\over 2}\biggl({gm_H\over 4m_W}\biggl)^2,\quad a={2\over
g}m_W.\eqno(5.17)$$ This is the shifted double-well potential discussed in the
introduction (1.1). Our point is that this structure is not obtained from a
clever choice of a Lagrangean and subsequent symmetry breaking, but comes out as
the necessary consequence of gauge invariance in the massive situation.
\vfill\eject

\vskip 0.5cm

redpajama_set_name: RedPajamaArXiv


text:
Forums > Fierce Board > Cheerleading > College Cheerleading, NCATA, & STUNT >
How Much Does Business Play Into Scoring?
Discussion in 'College Cheerleading, NCATA, & STUNT' started by Sterling von Shimmer, Jan 10, 2019.
Sterling von Shimmer When all else fails.... I shimmy
This will be a college UDA question, but it still applies to all levels of cheer/dance:
I'm sure we've all heard variations of this rumor before, but I personally have always dismissed it. But then UDA happened last year, and now I'm wondering if there isn't some credence to it (which, if true, would break my heart a little).
To be clear: I'm not a "conspiracy theorist" about judging. I'm don't scapegoat low scores back to some wild card factor — usually business-related — to explain our failures. But I know teams that firmly believe they've received lower scores or lost titles because they didn't attend the right camp, or didn't get uniforms from the right vendor, or whatever. In short, they didn't grease the wheels with the right organisation when they should've, and then they paid for it at Nationals. Whether their suspicions are correct or not isn't for me to say. I'm just saying I've never personally experienced such a thing, and therefore don't think much about those allegations.
But then UDA happened last year and the University of Minnesota officially lost their 7-year winning streak (that they'd been maintaining since their inaugural year) to Ohio State. Not only in one, but TWO divisions (pom and jazz).
A lot of people — including myself — were surprised at this. To be fair, U of M had their problems, but so did Ohio. And yet Ohio unseated U of M in two categories (Minnesota got second to Ohio in both). So it has people wondering if there wasn't something else happening.
The theory is: UDA finally gave the titles to Ohio (in a year where Ohio performed competently enough that UDA could successfully defend their decision). Why? To show it could be done. Because to keep solidifying U of M as the team that could not be beat in two of their most impacted dance divisions would mean fewer entrants — and therefore, fewer entrance fees — down the line. So basically they broke a team's 7-year streak and handed the title to another team that didn't earn it for the money. That's the theory, anyway.
Of course, there's the possibility that Ohio was just better that day. But most people believe that while they were good, they were NOT Minnesota good. And certainly not twice over.
The romantic in me doesn't want to believe this is true. But also, I kind of get it from a "cheer-as-business" standpoint. I don't like it or condone it, but I get it.
For reference, here are the teams in question (I'm only posting the pom division because I'm more familiar with it). If you're not sure what you're looking for, it's the teams' mega-hard full-squad turn sections that usually go a long way in determining who wins (Minnesota has traditionally been famous for the synchronisation, difficulty, and speed changes in theirs):
Minnesota, 7-time champ:
Ohio:
Does anyone else wrestle with these suspicions? Can anyone with more experience into the cheer industry provide some insight? Am I paranoid to suspect something happened or am I naive to think nothing happened?
Thanks, all.
Sterling von Shimmer, Jan 10, 2019
alpaca Best Flyer.. on a parent team
I absolutely believe that politics can come into play during sports that are judged based on a rehearsed performance as opposed to a reactionary ball sport. Politics comes into play even in ball sports with instant replay.
That said, in this case I thought Ohio had a better performance. Keep in mind that I have zero experience in pom so I don't know the scoresheet. This is the first time I have ever watched a pom routine. But I will say that Ohio was more enjoyable to watch and had more performance value. They started off strong with one of their their turn sections and kept the energy. It felt more modern. I liked it better.
alpaca, Jan 10, 2019
oncecoolcoachnowmom Bestest Newbie '14
Disclaimer: My dance knowledge is limited so I'm going to relate this to cheer.
By this logic, UCA should have politically colluded with some other University to have Kentucky come in 2nd like 15 titles ago.
They've only taken second 3 times in 20 years but when it does happen, I think the other team does have the better technique and execution (ex: the year UCF beat them.) It's not a political ploy.
oncecoolcoachnowmom, Jan 10, 2019
catlady Slow your roll, Sparkle.
After watching both, I noticed Ohio provided additional skills (aerials, front walkovers, and some type of handstand split kick something or other) that Minnesota did not that could have boosted their difficulty or creativity. Synchronization and speed favored Minn IMO, but I didn't feel it was enough to exceed Ohio's additional content. Strictly an opinion, I know nothing about pom.
In the eight years we were in AS, I felt good about the placements my kids teams received.
catlady, Jan 10, 2019
ScottyB I have my own cheer message board
i don't necessarily agree that last year's UDA result was due to any conspiracy. mostly because UMN still is a major camp, apparel, and competition customer, with their 5 teams (all girl, coed, mascot, dance(s), and hockey cheer).
that said, the question of having the EP have all the control with scoring and placing for a national championship is questionable when they are in business in other areas. i've been in the room when top teams' data was checked regarding their last orders, when they went to camp last, and other business related info. Varsity didn't become a 2 billion dollar company by being dumb..
ScottyB, Jan 10, 2019
ScottyB said: ↑
i've been in the room when top teams' data was checked regarding their last orders, when they went to camp last, and other business related info. Varsity didn't become a 2 billion dollar company by being dumb..
We differ on what we think is "dumb" then. Questions about "influencers" and what they wear and where they are seen at are truly basic everyday business questions in fashion. I guarantee they are more concerned about getting "inflencers" fashion business, versus pissing them off. The fact is many top teams don't wear Varsity and they are watched intently for favoritism and bias not only by the gym owners and coaches, but fans as well.
IMO, I tend to think individual judge bias and politics is probably more of an issue. Cheer, dance, and skating are all very incestuous businesses. For those that think releasing scores would prevent biases and incestuous politics, lol, hello Olympic ice skating.
catlady said: ↑
I said I was in the room. Idk how more specific I can get in saying that it happens.
Many of the top teams do all wear varsity, especially on the school/college side of spirit.
When you are a figurative monopoly, you can piss off whoever you want, you will still get everyone else's business.
oncecoolcoachnowmom said: ↑
That's true. But to your Kentucky comparison I will say this: Kentucky at least has been beat before. Minnesota never had — not once — since their debut year. Which begs the question: maybe UDA thought it was just time for them to lose lest people think there was a conspiracy afoot in that direction? I could see it.
The aerials Ohio had were a nice flourish — especially out of the turns — but they're not valued all that much. I think it's because they're considered more to be acrobatic skills than dance. A few years ago U of M had only half their team execute aerials and they still beat a team with full aerials. They're nice, but there are better ways to spend your time. So I'm not totally convinced that's what pushed them over the edge. But who knows. They did look REALLY pretty.
Informative x 1
CPop They call me Susie
Coach / Athlete
Sterling von Shimmer said: ↑
What? Aerials are HUGE in pom now. Dance evolves just like cheer... Difficulty isn't everything, but Ohio was more difficult, had higher energy, while still maintaining cleanliness. U of M was slightly sharper and cleaner especially in their transitions but imo was one of their more "meh" routines in terms of choreography they've ever brought to nationals. Ohio had a high energy opening and had good pacing throughout the routine. U of M switched choreographers last year. Choreography is EVERYTHING when you're at that level.
CPop, Jan 10, 2019
I didn't imply you weren't in the room, I said it was foolish for the very reason of you coming on here and saying "it happens." Do I believe everyone in a 2 billion corp is that foolish, "no". #2, "yes" some top teams do and don't wear Varsity, and as I said above, gym owners, coaches and fans are going to be watching carefully for that bias. #3 We will have to agree to disagree on the term "monopoly", there are a plethora of cheer uniform companies out there. As far as EP's, it's either an empire or it isn't lucrative. Based on all the mergers and lack of people trying to break into the cheer EP business, I'm going with the latter. It wouldn't be the first time a corp has to stay in a lower profiting business to support their highly profitable business(es). And lastly, as a once CS exec for a division of an $11 billion corp.... that no longer exists because of a merge..... "no" you can't keep ticking people off when your business isn't a necessity in life. Everyone likes to ponder the "conspiracy theories," but I don't know of anyone that will voluntarily keep throwing thousands of dollars at a business they don't have any trust in or value at some level.
lol 'figurative monopoly' =/= a monopoly that the courts could pass judgement/penalties onto. That is why I said it like that. The Summit was an abomination last year, yet people are still going en masse this year. Why? The only other end of year options for lower level teams are few and far between.
but I don't know of anyone that will voluntarily keep throwing thousands of dollars at a business they don't have any trust in or value at some level.
that's how Varsity's business model has worked for years:
competitions were marketing for uniforms and camp. that's why only now are we seeing all other costs go up, Varsity is owned by two new companies in the past 7 years that don't want to take a dive on these massive competitions. it will only get worse. Varsity didn't create their competitions and fund all star events to grow the sport, they just want to grow their bottom line (which for most of their history is based around apparel).
Which is why Flocheer/VarsityTV is now a thing (they finally wanna profit off the media side). For years all the video and promotion work came as a marketing cost. For years we lost thousands of dollars doing video and live streaming so that the sport would get exposure and more teams would get involved, boosting up the company.
CPop said: ↑
Aerials as a means of picking up difficulty points are a popular choice, but there is no "aerial" section in the score sheet. It only mentions "tricks," of which U/M had (the backward rolls, the walkover thing, and the lift) that were faster and more creative than Ohio's, and that's in addition to the handful of aerials that U/M also had.
I'll admit, this wasn't U/M's best routine ever. If I'm being totally honest, I don't feel the girls enjoyed performing the routine that much. Compare this one to their 2016 routine which you should watch if you're a fan of outright perfection:
But I still feel like the "meat" of the routine should've put them on top. Ohio was flashier and more energetic, but Minn was cleaner and their fouetté turn sequences were clearly superior.
All in all, I understand the judges' decision, but nonetheless was not expecting it.
With UDA right around the corner, I'm practically salivating to see what both teams bring in a week. Cannot wait.
Why was Summit bad last year?
Lisa Seye I'm an announcer on CBS for Worlds (or should be)
Waaaaay overcrowded. Parents fighting because they cant get in the buildings to see their kids, rude staff, ran out of medals, wrong placements/wrong winners announced......
Lisa Seye, Jan 11, 2019

redpajama_set_name: RedPajamaCommonCrawl


I own a small Revlon branded makeup mirror with a lightbulb and reflector inside. I find it fairly convenient as I can apply makeup while sitting at my desk and listening to music or watching videos online. Unfortunately, it has issues. It's not very bright, and so doesn't really give you an accurate image of your finished look. It also generates a fair bit of heat, which makes it uncomfortable to use in the summer. Recently, I found out that there are 12V LED light strips of many colors available on eBay for cheap (usually less than $10 for a 5 meter roll, that's about 15 feet). I figured that with a cool-white LED strip, a little dimmer switch, some wire, some glue, and a source of 12V DC power, I could turn the vertical wall mirror I own into a giant lit makeup mirror.
I acquired a 5 meter cool-white LED strip with 60 LEDs per meter. The LEDs are of the "5050" variety, which are a bit more expensive but much brighter than the alternative "3528" kind. The total cost of the strip after shipping was about $9. The dimmer I bought is a digital part, that is, it controls brightness through pulse-width modulation (PWM), and is specifically made for 12V LED systems. I found it on ebay for less than $2 shipped. The total cost of all the parts for this project, including the glue and wire and clamps, is about $30. It's important to note though that I still have leftover LED strip, glue, wire and of course the clamps to use on other projects.
Once the strip segments were glues down, I soldered wires in the corners of the mirror, connecting the strips to each other, and also to the dimmer. I applied epoxy glue over the connections to solidify and insulate them. I also added a long wire at the back of the mirror, which goes from the dimmer's input to a female DC power jack, the kind one would connect to a typical wall-wart power adapter.
Assembly was more time-consuming than I expected, it took me about 4 total hours of work, and was done over 3 days, since I needed to leave time for the glue to solidify. I started by drilling a hole on the side so I could make the power input wires of the dimmer go to the back of the mirror, I then glued the dimmer over this hole. The second step was cutting the LED strips to length. These can only be cut every 3 LEDs, and so it wasn't possible to make them go exactly all around the mirror, covering the whole perimeter, I needed to leave gaps, and connect the multiple pieces of strips with wire.
The LED strips found on eBay have an "auto-adhesive" backing, you're supposed to be able to just peel off the protective tape, and stick them to things. In my experience, this adhesive backing is very much worthless, the strips will peel off immediately, and so some other kind of glue is needed. I decided to go with epoxy, which is sold at the dollar stores here. I also purchased two adjustable clamps to hold the strips in place while the epoxy solidifies. Epoxy glues have several advantages: they solidify quickly, the resin they form is extremely solid, and they are not electrically conductive. In my experience, epoxy can solidify in as little as 15-20 minutes.
The total power consumption for the mirror, when on maximum brightness, is less than two amps. I hooked it to the 12V 15A DC power distribution box I already built for other LED lighting shenanigans. Your bedroom probably doesn't come with one of these, but fortunately you can get by with a good old 12V 2A DC wall-wart, which you can find on eBay for about $5 shipped. Just make sure to get one that fits your country's electrical outlets.
The finished product, once wired up, can put out quite a bit of light: about 3000 lumens, which is more than a 100W incandescent lightbulb. Fortunately, the dimmer allows tuning down the brightness to the point where you can look at the LEDs directly without visual discomfort. I usually just leave it on maximum brightness, however. I've found the lights are not blinding if you get up close to the mirror to look at your face. The cool-white color also seems to have a pleasant daylight-like glow. I'm actually considering buying more of this strip just to make my bedroom brighter.
That's awesome! I love projects like this. Purchasing a lit mirror like that is hundreds of dollars and you just proved you can make your own for less than $50. Keep it up!


* Please Note: orders from outside the EU may be subject to local taxes, import duties, or customs charges. All import duties & customs taxes are the responsiblity of the buyer, as customs policies vary widely from country to country. If you would like more detailed information, please contact your local customs office. Also, please be aware that cross-border shipments may be subject to inspection by customs authorities. Thank you.


The colour, music and energy of Bollywood is to be brought to life in Luxembourg at the next annual Diwali celebration.
Tickets are now available for The Indian Business Chamber of Luxembourg's 2012 Diwali celebration, which promises to be the biggest yet.
Hosted by the Alvisse Parc Hotel, the event attracts up to 600 people each year with a mouth-watering traditional Indian meal followed by live music and dance.
This year, guests can expect to witness one of the best Bollywood dance and musical acts in the world with top-class performers from KSPARK entertainments.
The event takes place on October 27 from 6.30pm onwards.
Tickets cost 59 euros for non-member adults and 30 euros for children aged 7 to 12 years old. Members pay 39 euros for adults or 25 euros per child. Entry includes food, mineral water, entertainment and a free tombola.
Anyone wishing to attend is urged to register early to avoid disappointment.


How to install: Leon runs on Windows. After you downloaded the program run install.exe and follow the instructions.
Leon doesn't run or crashes: Please contact us. Send, if possible, a screenshot attached to your email and your system's version. Information on doing this is provided in the section "About us".
Does Leon cheat?: Definitely no!. Humans often claim that backgammon software cheats. If you don't believe me, you can prove it by yourself. The dice values are selected by a so called "pseudo" number generator. Main property of "pseudo" number generators is the use of a seed. For the same seed value the random numbers sequence that is generated is always the same. Start a game and write down the seed that is used for this game. You will find it in the bottom/right corner of Leon's main window. Play some moves and write down the dice used. Now go to the Options menu of Leon. Write the seed used by your game. Start a new game and play different moves. You will see that the sequence of the dice, for the same seed, will always be the same, independent of the game played or state of the game.If Leon cheats then it would thrown different dice sequences, depending on the game state. But as you will see, this doesn't happen.
If you don't believe even this mathematical proof, Leon provides a way to input your own dice. If you still find Leon to be very lucky then... you should better reduce the playing strength level of the program.
Why not the doubling cube?: There are already some programs around that play the Backgammon variation in a very decent level of play. With Leon i didn't want to write just another strong Backgammon program. Leon wants to be a very strong Tavli player. You can play Portes with Leon, whice is almost the same with Backgammon except that there is no doubling cube and no tripple games. Of course, depending on the resonance of Leon to the community, a Backgammon engine could be made available in the future.


text:
Life for My Friend in Afghanistan
September 19, 2021 FriendsTaliban, Taliban atrocities, Taliban atrocities against womenTeri
Much of this blog post about my Afghan friend will be intentionally vague. I will refer to her as Fatima, although that is not her real name.
I've changed her name and the circumstances under which we met to protect her identity and safety and ensure the organization's anonymity that helped facilitate our friendship.
Soon after 9/11, in the capacity of the publisher and chief operating officer of World Press Review Magazine, I was invited to a two-day conference with students and educators from war-torn areas of the world—primarily the Middle East, South Asia, and the Balkans.
I was looking forward to the conference because there would be a delegation from Afghanistan, and I wanted to meet them first hand, hoping for an exclusive interview with one or more of them.
I brought along a spiral notebook hoping to fill it with enough material about my interaction with the Afghan contingent to write a compelling, knock-your-socks-off article for Worldpress.org.
The notebook had the following words on the cover: NOT ALL WHO WANDER ARE LOST.
As I entered the reception room, I saw a handful of young women in pale blue full-body cloaks, otherwise known as burkas. Their faces were completely covered except for a small area around the eyes, camouflaged by heavy netting. As they huddled closely together, I walked over and introduced myself.
When I stuck my hand out in greeting, I realized that their cloaks had no armholes. I awkwardly apologized while they all silently bowed their heads up and down.
The conference organizer informed me that I would be sitting with Fatima, an Afghan teacher, and her students.
As we went around the table offering our names, Fatima quietly prompted her students to introduce themselves. To be honest, if she hadn't spoken to her students, I wouldn't have known the difference between student and teacher. With all that material covering Fatima's face and body, I wouldn't have known if she was sixteen or sixty.
I made a few quick observations: At first glance, I was unnerved by this hooded creature. The woman looked like a blue ghost, and the semi-transparent mesh fabric covering her eyes made it impossible to garner any sort of emotion from them.
Fatima was eyeless and faceless, and I wasn't sure where to focus my own eyes. I can usually tell a lot about someone through their eyes. Do they make eye contact? Do their eyes reveal sadness or gladness? Are they happy to see me?
Because the semi-transparent mesh obscured her eyes, the fabric made it impossible for me to size her up. I felt self-conscious as I tried to focus on where her eyes should be, but I forced myself to do so anyway.
Her burka was nylon—you know, the kind of fabric that doesn't breathe. And as I tried to make small talk, I couldn't help but imagine how uncomfortably warm she must have felt.
Surprisingly to me, we hit it off right away. I suppose my knowledge of Afghanistan and the cruelty of the Taliban helped to promote easy conversation. Her English was excellent, and we were equally interested in each other's stories.
The burka served as a roadblock between us, though. Here I was in my designer dress—hair and makeup accentuating my persona while she sat there visibly invisible. I unfairly imagined what she looked like: Dark eyes, swarthy skin, with teeth in need of an orthodontist.
She asked me about my family background, and I told her about recently finding my paternal family.
I explained to Fatima that until June of 2001, I knew nothing about my father or that side of my family. She listened in fascination as I told her about discovering that my father was a Syrian Christian. And that his mother, my paternal grandmother, was in all likelihood a Syrian Jew and that I had five half brothers and sisters I never knew existed.
I divulged something to her that I had never uttered out loud before: That I had finally found peace and relief, and although I never knew I needed it, I felt almost whole and more complete than ever before. I shared a photo of my daughter and half-sister, and she enthusiastically agreed that they looked eerily alike, although not surprising.
She delighted in the story and the discovery of my heritage and newly found siblings. The woman without a face listened in amazement and peppered me with question after question. I will remember forever what she said to me after I finished telling her my story: "You can't see it, but I'm smiling."
And then it was Fatima's turn to tell me about herself. She was in her early 30's, and coming to America was a life-long dream.
She, too, had found peace in 2001, when in October, the United States invaded Afghanistan. "Until then, the Taliban treated girls and women worse than animals," Fatima whispered in her thick accent while furtively looking to the right and left, as though there were spies among us.
Her head hung low as Fatima explained that when the Taliban came to prominence in the fall of 1994, life as she knew it changed for her. "The Taliban imposed strict and oppressive rules and orders based on their misinterpretation of Islamic law. Women had effectively committed the crime of being born a girl."
Fatima pointed out a student of hers that was old enough to be in the seventh grade, but she had just finished second grade because, under the Taliban, she had not attended school for almost seven years.
In 1994, the Taliban's assault on women began immediately. They barred women from attending classes or working at Kabul University. The Taliban forced nearly all women to quit their jobs, which had a devastating impact on household incomes, especially widow-headed households, which, according to Fatima, was common in Afghanistan.
They drastically restricted women's access to medical care, brutally enforced a burka dress code, and made the ability of women to move about Kabul impossible. The Taliban forced them to quit their jobs as teachers, doctors, nurses, journalists, government officials, and clerical workers.
She further explained that domestic violence had become rampant in Afghanistan—the physical evidence conveniently hidden under the burka.
"Under the Taliban regime, there was a complete ban on women working outside of the home, which made it impossible for me to teach," Fatima explained. "And there were no schools for girls anyway, so my profession was useless," she continued, barely audible between her whispering and all that material covering her mouth. When I told her it was hard to hear her, she apologized, saying that as a veiled woman, it was also hard for her to hear and that the burka was claustrophobic and unbearably warm.
Her depiction of the Taliban was haunting.
The Taliban banned movies, music, dancing, clapping during sports events, beard trimming, shaving, card and board games, cameras, children's toys including stuffed animals and dolls, television, and paper bags. They outlawed hanging pictures in homes, pet parakeets, satellite dishes, chess, cigarettes, alcohol, magazines, newspapers, most books, anything made from human hair, nail polish, statues, pictures, paintings, or photos of any living thing. Children were forbidden to fly kites or sing songs.
"They even forbid applause, although the ban was a moot point since, as a woman, there was absolutely nothing left to applaud. We used to describe ourselves as the living dead."
The only public transportation permitted for women were special buses, which were rarely available, and all of their windows, except the driver's, was covered with thick, filthy blankets.
The Taliban indiscriminately beat Afghans with heavy clubs and long sticks daily. They publicly stoned adulterers to death and amputated the hands of thieves. They banned films with women, and images of females in newspapers, books, shops, or the home. Every visual depiction of a woman was forbidden.
When paying any merchant, a woman's hand could never be exposed when handing over money or receiving their purchase. Makeup and nail polish were illegal, as well as white socks and white shoes. The Taliban frequently cut off fingers with nail polish.
While the burka existed before the Taliban, its wearing was not a requirement. It was only when the Taliban came into power that the burka became mandatory. Even girls as young as eight or nine years old had to wear a burka. They enforced the wearing of the burka with threats, fines, and severe punishments. And even the accidental showing of a foot or ankle resulted in brutal on-the-spot beatings or, in some unfortunate situations, amputation.
Fatima explained that a burka is expensive and can cost the equivalent of five month's Afghan salary. And any woman unable to afford a burka faced house arrest. In some neighborhoods, women would share a single garment, many of them waiting days and weeks for their turn to go out, despite their lack of food and medical needs. Fatima described women and girls as wingless birds.
She quoted me an often-used Taliban phrase: "There are only two places for Afghan women. In her husband's house and the graveyard."
When I asked her why she was still wearing a burka, she answered that even after the fall of the Taliban regime, many women felt that there was still no safe alternative. "The majority of women who don't wear a burka face the possibility of being single for the rest of their lives," said Fatima. She emphasized that it was still a struggle for a woman to gain employment, so they had no choice but to continue relying on men for money. "Men don't want to marry women who do not abide by hijab."
She looked around to see if anyone was listening before she continued. "We had to paint our windows black so that no one could see inside, and I could not see outside. So, you see, for seven years, my world was dark." She paused then, and I imagined that perhaps she was holding back tears. I tried especially hard to see her eyes but to no avail. And yet, I didn't need to see her face to feel her pain as she continued.
"Yes, my world was dark for seven years because there was a complete ban on women's activities outside the home. Unless a close male relative could accompany me, the Taliban forced me to spend most of my life in my house. So, when the Americans arrived, I was silently hoping that the worst was over for us. And that my seven years of misery were over."
We spent hours talking about the horrific life of an Afghan woman. From eight years old, girls were not allowed direct contact with males other than a close blood relative, husband, or in-law. Women and girls were not allowed to be treated by male doctors unless accompanied by a male chaperone, which caused many illnesses to go untreated. Women faced public flogging and execution if they violated Taliban laws.
The Taliban perpetrated egregious and unending violence against women, including rape, abduction, and forced marriages.
Women were not allowed to speak, laugh, or make any sound in public because it was deemed improper for a stranger to hear their voices. Women were also barred from being involved in politics or speaking publicly and could not appear in the streets without wearing a burka. And they could not wear high-heeled shoes because if a man would hear a woman's footsteps, it might excite him.
Fatima further explained that the light blue burka was commonly worn in Kabul and was native to Afghanistan. The cutwork by her eyes pricked her skin, leaving bloody marks and very little room to breathe and rendered her unable to eat. The small mesh panel allowed such limited vision that even safely crossing the street was difficult. And wearing the burka regularly often led to headaches, poor eyesight, hearing loss, asthma, and other severe disorders.
But worse than all of it was that Fatima longed to feel the sun on her skin.
I shocked her when I suggested that she take off the burka. I tried to assure her that no one in Afghanistan would ever know. After all, we were safe and sound in New York. She silently shook her head no.
We said goodnight, and I gently hugged her. She couldn't hug me back because the burka constricted her arms. As I awkwardly patted her back, she leaned her head on my shoulder, and we stayed that way for a good while.
I awoke very early the next day, having had a fitful and sleepless night. I walked to the dining room, where I sat in quiet solitude at one of the many long tables. I ordered a coffee and mentally played back all I had learned and heard from my Afghan friend.
As I feverishly wrote in my notebook, a beautiful light-haired brunette woman sat across from me.
Let's just say she had me at "good morning" because it was my burka-less friend!
Through my tears, I gazed into her piercing hazel eyes and attempted to speak, but I had to pause for fear of crying. She had the whitest of skin, probably because it had barely seen the light of day. I was finding it difficult to breathe. But she was calm.
Her smile was radiant. She had an endearing space between her two front teeth. It was a tiny gap but adorable and unforgettable.
And then, out of the corner of my eye, I saw her burka-less students. They were visibly unsure of themselves and wary as their eyes darted around the room. The girls clustered together and hunched over each other. Fatima looked over at the girls and gave them a head bow which they all respectfully returned.
When Fatima looked back at me, she said, "The girls are quiet because they're used to being voiceless. Most of them have been kept inside and unable to go to school. Some of us women ran underground schools in our homes for girls and women under the guise of sewing and knitting classes. Many of my student's parents were arrested and lost their jobs. They have seen teachers shot and executed for secretly schooling girls like them. My students have witnessed atrocities that children should never know or see. And I fear that even with the American presence in my country, their voices will never be heard."
During one of our conference breaks, we walked outside, and Fatima tilted her pale face up toward the sky and basked in the sun, her hazel eyes closed.
As she turned away, with her back to me, the sun revealed the shiny red highlights in her hair. Her head no longer hung low; she was walking tall and strong. Her mighty shadow towered larger than life alongside her.
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2 thoughts on "Life for My Friend in Afghanistan"
Joanne Romano-Csonka says:
Most will never get to know the atrocities they only see on tv and in newspapers. I hope there's more to this story.
Joanne, You are so right. Unfortunately, there is no happy ending to this story. Afghan women and girls will now go back to be being brutalized by the Taliban.


Most advanced market democracies are pervaded by laws and rules that structure social and economic relationships. Navigating these rules requires significant amount of legal information and guidance. The vast majority of ordinary Americans, however, lack significant access to legal resources. In this paper we explore the limited amounts of empirical data available to document the legal resource landscape for ordinary Americans. We demonstrate that Americans experience legal problems with at least the same frequency as do people in other countries but by and large they are left to manage these problems with little legal help. A higher fraction of Americans "lump" their problems—doing nothing to resolve significant difficulties with housing, employment, family relations etc.—or represent themselves in court than is the case in places such as the U.K. and the Netherlands. We highlight what we believe are two important reasons for the limited options available to ordinary Americans dealing with legal problems. First, the U.S. devotes lower levels of public resources to the legal system when we adjust measures of public expenditures on the court system, public and charitable legal aid, and numbers of lawyers and judges to take account of the much higher numbers of cases in U.S. court systems. Second, the U.S. legal profession exercises a much tighter grip on legal markets than is the case in places such as the U.K. and the Netherlands, where there is much more robust set of options, beyond attorneys, to provide ordinary people with legal assistance.


We are happy to announce that our Lovey- Spark babies arrived May 23, 2006. We have 3 boys , all sable and white. Two boys appear to be tri factored and 1 boy appears to be pure for sable.
These little boys are doing great and growing well.


text:
Police: Appraiser stole medications from two homes in Palmyra
David Levinsky
dlevinsky@thebct.com
PALMYRA — An employee of a company contracted to perform a revaluation of all borough properties has been charged with stealing prescription medications from two homes on Highland Avenue and East Henry Street.
Dylan Croft, 25, employed by Professional Property Appraisers in Delran, was charged with two counts of theft of a prescription medication after his arrest Wednesday, police said.
Police did not specify the type and amount of medications stolen or when the suspected thefts occurred, but they said Croft stole the drugs while completing evaluations at the homes.
Croft, who lives on Rockcress Place in Delran, was released Wednesday pending a court hearing, police said.
Mandated by the Burlington County Board of Taxation, revaluations are conducted to bring assessed property values back in line with true market values for tax purposes.
Professional Property Appraisers was awarded a $224,807 contract to conduct the revaluation in April after submitting the low bid. The firm was also the low bidder on two previous occasions, but the Borough Council rejected those bids over concerns about potential conflicts after officials learned that the tax assessor at the time was the spouse of the company's president.
Former Assessor Karen Davis resigned from her part-time position in October 2010 after notifying the borough that her husband's business was the only bidder for the project. When the borough rebid the work, Professional Property Appraisers again submitted the lowest bid, but was rejected because Davis had been involved in preparing the request for proposals for the revaluation.
Officials said the firm was finally awarded the contract this year after submitting the low bid from new specifications drawn up by a different tax assessor.
Officials at Professional Property Appraisers could not be reached for comment Wednesday.
Borough police urged any residents who have noticed missing medications after their homes were appraised to call detectives at 856-829-0191.
David Levinsky: 609-871-8154;
email: dlevinsky@phillyBurbs.com;
Twitter: @davidlevinsky


The U.S. Department of Agriculture has released a scathing animal welfare inspection report against Lagoon, citing the park with a "critical" violation, the most serious level of offense that can be assessed by an inspector. The violation stems from an incident when an elk escaped through a hole in his enclosure and was gored by a bison in an adjacent enclosure. The injuries were so severe that he was euthanized. Utah Animal Rights Coalition (UARC) has submitted FOIA and GRAMA requests with state and federal authorities to learn more about the incident.
UARC has also scheduled an emergency protest outside Lagoon on Sunday, July 22nd at 11 AM, which will be the 5-month anniversary of the elk's death. Following the demonstration, the protestors will walk to the site of the death and lay flowers in memoriam.


Canceling a Backup Assistant subscription or removing it deletes the application from the handset completely. Airtime/megabyte charges apply while removing a subscribed application.
Backup Assistant cannot be removed/canceled if it came preloaded on the device. Reset the application by performing a master reset on the device.
From the main screen, tap Menu.
Tap Options next to the app.
Select Media Center (Get It Now).
Select Browse & Download (Get Going).
Select Backup Assistant then press the right soft key to choose Options.
Select Remove or Cancel Subscription.
Select Get It Now from the handset menu.


text:
Ontario priest charged with child molestation
By Melissa Pinion-Whitt | Melissa.Pinion-Whitt@inlandnewspapers.com |
PUBLISHED: October 26, 2010 at 12:00 a.m. | UPDATED: August 30, 2017 at 10:07 a.m.
Prosecutors on Monday charged an Ontario Catholic priest with molesting a 12-year-old boy, and police say he had at least four other victims.
The Rev. Jose Alejandro Castillo, also known as Father Alex, was arrested at his home in the 600 block of W. Zenia Court. He was booked into West Valley Detention Center in Rancho Cucamonga in lieu of $1 million bail.
Ontario police had been investigating the case since July 21. They confirmed 57-year-old Castillo committed eight sex acts with the boy.
Castillo worked as a priest the past seven years at Our Lady of Guadalupe Church in Ontario.
News of the investigation prompted parishioners to write letters to the Diocese of San Bernardino in support of Castillo. Castillo has denied the allegations.
The San Bernardino County District Attorney's Office issued an arrest warrant for Castillo on Monday, charging him with seven felony counts of lewd and lascivious acts with a child younger than 14-years-old. He was charged with an additional count of committing lewd and lascivious acts with a child, police said.
Police say he could face as much as 22 years in prison.
Investigators are also investigating allegations that Castillo committed sex acts with a 14-year-old boy while serving at Our Lady of Guadalupe Church.
Officers say a 16-year-old boy was identified as a victim, but charges won't be filed due to the statute of limitations on the crime.
Two other people claimed to be victimized by Castillo, but they were older than 18-years-old at the time and Castillo won't be charged in those cases either.
Police are concerned there are more victims. Anyone with information about the case or may have been victimized by Castillo may call Detective Mark Guski, 909-395-2726.
Melissa Pinion-Whitt
Homeless man found shot to death in Pomona


text:
ISA Thwarts Attack Against Security Forces
Stabbing attack planned against security forces in Jerusalem's Kfar Hashiloach has been thwarted by arrests of slew of terrorists.
Uzi Baruch
May 14, 2015, 12:05 PM (GMT+3)
JerusalemterrorismCity Of DavidShin BetShiloach (Silwan)
Arab riots in eastern JerusalemFlash 90
It was cleared for publication on Thursday that the Israeli Security Agency (ISA or Shin Bet) had thwarted a stabbing attack being planned against security forces stationed at "Bet Ovadia" in Shiloach, near the City of David in Jerusalem.
Through the course of the ISA's investigation into the matter, a number of terrorists were arrested in the Shiloach (Silwan) neighborhood of the capital, including one youth.
The main detainees are: Muhammad Nasser Ahmad Abbasi, 21, a Popular Front for the Liberation of Palestine (PFLP) terrorist, and Murad Muhammad Odeh Kostiro, 20.
During his interrogation, Abbasi admitted to the ISA that a month ago he had recruited a minor to reconnoiter security arrangements at Bet Ovadia in Shiloach, as part of preparations to carry out a shooting attack on guards in the building.
Later, after receiving the youth's information, Abbasi decided to carry out the attack with a knife and an axe, hiding the weapons in both his own home and the home of his grandfather.
Additionally, Abbasi admitted to manufacturing pipe bombs and firebombs, which he then sent to PFLP terrorists, guiding them to throw the bombs into Jewish homes in Shiloach.
The detainees admitted to various other acts of terrorism and gave information on dozens of activists in Shiloach, involved last year in riots where firebombs, fireworks, and stones were thrown at Border Patrol forces, police, and civilians.
One of the detainees also admitted to throwing stones at police officers and visitors to the Temple Mount during 2014, and implicated several other Shiloach activists, involved in throwing stones at visitors to the Mount of Olives Jewish Cemetery.
Indictments containing serious security-related charges against the detainees are expected to be filed in coming days


NOTICE: Not currently compatible with Intel HD graphics controllers! Or at least claimed so by the game's Steam page.
This is a DRM-free GOG release. Courtesy of Jasminfever.
This is the February, 7 build but at the time of upload there's a newer 1.0.6.5 build available. This game is still in the works and updates are frequent.


text:
La manacá (Calyptronoma plumeriana) es una especie perteneciente a la familia de las palmeras (Arecaceae).

Distribución y hábitat
Es originaria de Cuba y La Española donde crece en zonas húmedas cerca de las orillas de los arroyos en alturas superiores a 450 .

Descripción
Tiene un estípite solitario que alcanza los  4-10 m de altura, con  10-20 cm de diámetro.

Los pétalos de las flores masculinas, son comestibles.

Taxonomía
Calyptronoma plumeriana fue descrita por (Mart.) Lourteig y publicado en Phytologia 65(6): 484. 1989.
Etimología
Calyptronoma: nombre genérico que deriva de kalyptra = "tapa" y nomos = "que está en uso habitual", en referencia a la parte superior de la corola de la flor pistilada que es empujada fuera como una tapa o tapón.

plumeriana: epíteto latino que significa "con penacho".
Sinonimia
 Geonoma plumeriana Mart.
 Calyptrogyne plumeriana (Mart.) Roncal
 Geonoma dulcis C.Wright ex Griseb.
 Calyptronoma dulcis (C.Wright ex Griseb.) H.Wendl.
 Calyptronoma intermedia H.Wendl.
 Geonoma intermedia (H.Wendl.) B.S.Williams
 Calyptrogyne dulcis (C.Wright ex Griseb.) M.Gómez
 Calyptrogyne clementis León
 Calyptrogyne intermedia M.Gómez ex Léon
 Calyptrogyne microcarpa León
 Calyptronoma clementis (León) A.D.Hawkes
 Calyptronoma microcarpa (León) A.D.Hawkes
 Calyptronoma clementis subsp. orientensis O.Muñiz & Borhidi

Referencias

Enlaces externos
 Imágenes en Google

plumeriana
Flora de Cuba
Flora de La Española
Plantas descritas en 1989
Plantas descritas por Martius
Plantas descritas por Lourteig

redpajama_set_name: RedPajamaWikipedia


text:
Colocleora bilobata is een vlinder uit de familie spanners (Geometridae). De wetenschappelijke naam van deze soort is voor het eerst geldig gepubliceerd in 1975 door Herbulot.

De soort komt voor in tropisch Afrika.

bilobata
Dier uit het Afrotropisch gebied


text:
Q: SQL Server 2008 - Delete data exclusively from stored procedure I would like to stop power users from deleting data using SQL Server Management Studio. I need to archive data and add some info to the audit trail when data gets deleted.

*

*Is there a way to stop them when they attempt to delete the data from SSMS?

*Is there a way to know which process caused the deletion? such as from SSMS, application, stored proc?

*Is there a way to allow only deletes from Stored Procedure?

Thanks

A: Well they are power users aren't they. You could set deny Delete permissions for them.
DENY DELETE TO [Your_User]
GO

A: Create a new login and database user for this login. Then grant delete permission to this user, and revoke it from all others. Write procedure[s] that removes data, add WITH EXECUTE AS [previously created user that can delete data]. Grant other users with execute permissions to the procedure[s].

A: From your second sentence I get the impression that deleting is not the real issue but that you need to archive data and create an audit trail when data is deleted. Why not use a delete trigger?

redpajama_set_name: RedPajamaStackExchange


The shoe, which is the first result of a collaboration with research organisation Fraunhofer Institute, will be presented at the "Materials Village" exhibition held by Material ConneXion Italia at Milano Design Week 2019.
It features a biologically active layer of the shoe-upper that acts like a dense nanosensor net and at a microscopic level maps the areas of the foot which produce heat and sweat. This mapping then creates a matching ventilation pattern, or in other words, living micro-organisms selectively remove material to create a unique fit.
"The shoe is not unique by itself - it comes as a blank canvas. It evolves with you, when you start to wear it, and it becomes more and more an expression of your body," said Charles Johnson, PUMA's Global Director of Innovation.
The project has been conducted under the creative curation of Innovation by Design - a global, multidisciplinary design & strategy studio founded in 2014, as a MIT Design Lab spin-off.


text:
Dephi isn't big enough to have a bus station, so instead, tickets are bought from a cafe on the edge of town, near where the bus stops. As Steph and I had a long day of travel, we caught the earliest bus, around 10AM. We bought tickets for Lamia, knowing we would have to make two more switches.
When the bus arrived, right on time, the bus driver informed us that we would have to change buses at Amfissa, a step we hadn't planned. So we boarded the half-full bus and set off on our merry way. It was another windy journey, down one mountain, up another, and down into a valley. The scenery was spectacular, and thirty minutes later, we reached Amfissa. We didn't see much of the town, but from the bus it seemed a rather work-a-day kind of place. We got out at the bus station, which more resembled an abandoned dry-cleaner than any kind of terminal. We waited on the side-walk for half an hour, before another bus pulled up. This one bound for Larissa.
The bus ride to Larissa continued the beautiful mountain scenery, and, despite Greek buses lacking toilets, it was a comfortable enough journey. I had expected to get a lot of reading done on the trip, but I found myself more inclined to simply stare out of the window as the Greek countryside passed by. Two hours later, we pulled into the bus station at Lamia.
Lamia is a big city and serves as a transportation hub to points both north and south. The bus station we pulled into was large and clean with nice facilities. It is only one of the several bus stations in the city, but thankfully the one we needed for the next leg of our trip. I asked for two tickets to Trikala, which the man behind the counter produced with a slightly pained expression. It seemed we had just missed a bus, and the next was not for another two hours. So, for the first time that day, I did get a lot of reading done, while eating a ham and cheese sandwich. The Greeks don't seem that big into sandwiches, certainly not like the Brits, but passable ones could be found in most places.
Like every bus before it, the one to Trikala pulled up right on time. Before passengers got on, however, the bus driver got off, and gave a package to a waiting woman. We noticed throughout our trip, that the buses in Greece not only carry people, but also the mail and bits of cargo. At one point we stopped and picked up some auto parts which we delivered along our way.
After a short drive through the mountains, our bus descended onto the plain of Thessaly. In stark contrast to the rest of Greece we'd seen, this was wide, flat country, devoted to farming. It reminded me a bit of the rural American South, with its small towns filled with rusting cars and scattered farm machinery in every state of repair. Although I continued my quiet survey of the Greek countryside, I did so with less interest, and the hours sitting stiffly on a bus started to take their toll. It was near to five o'clock when we finally reached the bus station, just outside of Trikala. It appeared that the station was attached to a huge mall, but we didn't have time to explore as our next (and thankfully last) bus was due in fifteen minutes.
This bus proved to be a much more local route, and we stopped frequently to pick up and drop of passengers. This did give us time to get a good look at Trikala, which appeared to be a slightly more attractive city than most. It at least had a nice river running through it. As we closed in on Kalambaka, our destination, we began to realize that we didn't really know where to get off. However, when we saw a big fountain, we assumed that was the centre of town, and the guess proved correct.
We knew our guest house was on the northern edge of town, nestled below the rocks of Meteora. So, in the fading light, we started upwards. Although the great rocks loomed above us on two sides, we were more focused on where we were going. In the dim evening, the upper parts of Kalambaka take on a slightly sinister air. Several houses are half-built or abandoned, junk is piled on the sidewalk, and numerous stray cats lounge on every corner. In truth, it was mainly a product of our tired minds, and after a day in Kalambaka it lost all of its sinister feel, even if much of it could use a fresh coat of paint.
We eventually reached our destination, Alsos house, a beautifully-built and up-kept guest house. After dropping our bags, we went off and had a lovely dinner outside, back down by the fountain, a great place for people watching as the sun disappears. It had been a long day of sitting on buses, but what little we knew of Meteora, promised us several days of great walking.