text:
\section{ Introduction }

Recently the study of topologically nontrivial configurations
\cite{3}-\cite{4}-\cite{5} on a noncommutative manifold \cite{1} has
received growing interest. In this framework we have set up a
general method with which all the noncommutative topological
configurations for a $ U(1) $ gauge group could be studied
exhaustively \cite{23}.

Basic elements of such a method are a combination of mathematical
and physical concepts. To identify easily the nontrivial topological
configurations we have made use of the concept of projector
\cite{6}-\cite{7}-\cite{21}, obtaining their classification on a
fuzzy sphere for a $U(1)$ gauge group \cite{2}.

With a subsequent study we have introduced a natural way for
defining noncommutative actions on a fuzzy sphere
\cite{8}-\cite{9}-\cite{10}-\cite{11}-\cite{12}-\cite{13}, by using
the matrix model approach
\cite{14}-\cite{15}-\cite{16}-\cite{17}-\cite{18}-\cite{19}-\cite{20}.
Finally with a simple link between projectors and matrix models we
have associated to the noncommutative projectors the corresponding
noncommutative connections \cite{23}-\cite{22}-\cite{24},
generalizing Dirac monopoles on a fuzzy sphere \cite{23}-\cite{25}.

To demonstrate that our method is quite general, we have studied in
this work the case of a nonabelian gauge group, and , as we will
show next, we are able to obtain the same results known in
literature \cite{25}.

Starting from a noncommutative projector with entries belonging to $
U(2) $, we show that it is possible to reconstruct, along the same
lines of the abelian case, a solution of the matrix model equations
of motion, that coincides, in the classical limit, with the 't
Hooft-Polyakov monopoles for $SU(2)$ gauge group.

In the final part of this work we discuss a possible candidate for
deforming the Chern class and its corresponding topological number.
The action we propose is the $3d$ Chern Simons action, which enjoys
the property of invariance under deformed diffeomorphisms \cite{26}.
While the nonabelian case works fine without any problem, the
abelian one appears more obscure and requires a deeper analysis of
the commutative limit, to reach smoothness with the Chern class.

\section{Review of abelian noncommutative monopoles}

The aim of this work is to describe how abelian and non-abelian
monopoles can be deformed on a non-commutative sphere. Our research
tries to cover both mathematical and physical issues, filling the
gap between the projective module point of view and the matrix model
formalism, which incorporates the natural definition of a
non-commutative gauge theory on a non-commutative manifold.

Since the abelian case has been already worked out in ref.
\cite{23}, we briefly remind the principal features in this section
so that the reader can easily identify common features and
differences with the corresponding non-abelian case.

The starting point is the definition of the fuzzy sphere algebra

\begin{eqnarray} [ \hat{x}_i , \hat{x}_j ] & = & i \alpha \epsilon_{ijk} \hat{x}_k
\ \ \ \ \ \sum_i { ( \hat{x}_i )}^2 = R^2  \nonumber \\
\alpha & = & \frac{2 R}{\sqrt{ N ( N+2 )} } \label{21}
\end{eqnarray}

which produces an useful finite truncation of the
infinite-dimensional function space of the classical sphere.

Such an algebra can be simply obtained by quantizing the Hopf
fibration $S^3 \rightarrow S^2$, which is defined in terms of two
complex coordinates, constrained to the $S^3$ sphere :

\begin{eqnarray}
x_1 & = & z_0 \overline{z}_1 + z_1 \overline{z}_0 \nonumber \\
x_2 & = & i ( z_0 \overline{z}_1 - z_1 \overline{z}_0 ) \nonumber \\
x_3 & = & | z_0  |^2 - | z_1 |^2 \ \ \ \ \ {  | z_0 | }^2 + { | z_1
| }^2 = 1 \label{22} \end{eqnarray}

Substituting into this mapping (\ref{22}) the complex coordinates
with oscillator operators $a_i$ satisfying the algebra

\begin{equation} [ a_i , a_j^{\dagger} ] = \delta_{ij} \label{23} \end{equation}

the corresponding real combinations are representations of the fuzzy
sphere algebra :

\begin{eqnarray}
\hat{x}_1 & = & \frac{\hat{\alpha}}{2} ( a_0 a_1^{\dagger} + a_1
a_0^{\dagger} )  \nonumber \\
\hat{x}_2 & = & i \frac{\hat{\alpha}}{2} ( a_0 a_1^{\dagger} - a_1
a_0^{\dagger} ) \nonumber \\
\hat{x}_3 & = & \frac{\hat{\alpha}}{2} ( a_0 a_0^{\dagger} - a_1
a_1^{\dagger} ) \nonumber \\
\hat{N} & = & a^\dagger_0 a_0 + a^\dagger_1 a_1 \label{24}
\end{eqnarray}

The $\hat{\alpha}$ operator has to be defined yet. Restricting the
action of the oscillators on representations with fixed total number
$ \hat{N} = N $, the $\hat{\alpha}$ operator can be taken as the
$\alpha $ constant

\begin{equation} \hat{\alpha} \rightarrow \alpha = \frac{ 2 R }{ \sqrt{ N ( N+2
)} } \label{25} \end{equation}

This relation is particularly useful to generalize the monopole
projectors at a non-commutative level. Let us recall the classical
$k=1$ case, i.e. the simplest monopole configuration, which is
defined by the following projector $P_1$, function of the complex
coordinates $z_i$:

\begin{eqnarray}
P_1 \ \ & = & | \psi_1 >< \psi_1 | \nonumber \\
| \psi_1 > & = &  \left( \begin{array}{c} z_0 \\ z_1 \end{array}
\right) \label{26} \end{eqnarray}

The normalization condition of this vector

\begin{equation} < \psi_1 | \psi_1 > = {| z_0 |}^2 + {| z_1 |}^2 = 1 \label{27}
\end{equation}

is satisfied, since the complex coordinates belong to $S^3$, as in
formula (\ref{22}).

In our first paper \cite{2} we noticed that the natural
non-commutative extension, based on quantizing the Hopf fibration (
eq. (\ref{23}) )

\begin{eqnarray}
P_1 \ \  & = & | \psi_1 > < \psi_1 | \nonumber \\
| \psi_1 > &  = & N_1 \left( \begin{array}{c} a_0 \\ a_1 \end{array}
\right) \label{28}
 \end{eqnarray}

really works by choosing the normalization factor as

\begin{eqnarray} < \psi_1 | \psi_1 > & =  & 1  \nonumber \\
N_1 =  N_1 ( \hat{N} ) & = & \frac{1}{\sqrt{ \hat{N} + 1 }}
\label{29} \end{eqnarray}

Since the fuzzy sphere algebra is a finite type non-commutative
geometry, the trace of the non-commutative projector is an integer
number :

\begin{equation} Tr P_1 = Tr | \psi_1 > < \psi_1 | =  N+2  < Tr 1_P = 2 ( N+1 )
\label{210} \end{equation}

It is possible to repeat the same procedure for monopoles with
negative charge. For example the projector corresponding to $ k = -1
$ is defined as :

\begin{eqnarray}
P_{-1} \ \  & = & | \psi_{-1} > < \psi_{-1} | \nonumber \\
| \psi_{-1} > &  = & N_{-1} \left( \begin{array}{c} a^{\dagger}_0 \\
a^{\dagger}_1
\end{array} \right) \label{211}
 \end{eqnarray}

where the normalization factor is determined by the condition

\begin{equation} < \psi_{-1} | \psi_{-1} > = 1 \ \ \ \  \Rightarrow N_{-1} =
N_{-1} ( \hat{N} ) = \frac{1}{\sqrt{ \hat{N} +1 }} \label{212} \end{equation}

Again the trace of $P_{-1}$ is an integer number:

\begin{equation} Tr P_{-1} = N < Tr 1_P = 2 ( N+1 ) \label{213} \end{equation}

Proving that these analogues of the topological excitations of
Yang-Mills theory on a sphere satisfy the non-commutative equations
of motion  requires the definition of a Yang-Mills theory on a fuzzy
sphere.

This proof has been firstly achieved in ref. \cite{23} with the
introduction of a matrix model $X_i ( i = 1, .., 3 )$, whose
equations of motion contain the fuzzy sphere algebra as particular
solutions:

\begin{equation} S( \lambda ) = S_0 + \lambda S_1 =
 - \frac{1}{g^2} Tr [
\frac{1}{4} [X_i,X_j][X_i,X_j] - \frac{2}{3}i\lambda \alpha
\epsilon^{ijk} X_i X_j X_k + \alpha^2 ( 1-\lambda ) X_i X_i ]
\label{214} \end{equation}

Both actions $S_0$ and $S_1$ separately contain the fuzzy sphere
algebra as a particular solution, and $\lambda$ is a generic
parameter.

The usual gauge connection is obtained by separating the generic
matrix $X_i$ from the background $L_i$ ( in the following we will
change the variable $X_i $ into $ \alpha X_i $ ) :

\begin{equation} X_i = L_i + A_i \ \ \ \ [ L_i , . ] \rightarrow_{\alpha
\rightarrow 0} - i k_i^a \partial_a \label{215} \end{equation}

The background $ L_i $ is equal to the classical Lie derivative on
the sphere in the $\alpha \rightarrow 0 $ ( classical limit ).
Instead the fluctuation $A_i$ is a linear combination of the
Yang-Mills connection $A_a ( \Omega ) $ and an auxiliary scalar
field ( in the adjoint representation ) $ \phi $:

\begin{equation} A_i ( \Omega ) = k_i^a A_a ( \Omega ) + \frac{x_i}{R} \phi (
\Omega ) \ \ \ \ \Omega = ( \theta, \phi ) \label{216} \end{equation}

A gauge covariant field strength can be defined as

\begin{eqnarray}
F_{ij} & = & [ X_i , X_j ] - i \epsilon_{ijk} X_k = \nonumber \\
& = & [ L_i , A_j ] - [ L_j , A_i ] + [ A_i , A_j ] - i
\epsilon_{ijk} A_k \label{217} \end{eqnarray}

In the classical limit it is useful to calculate it in terms of the
component fields

\begin{equation} F_{ij} ( \Omega ) = k_i^a k_j^b F_{ab} + \frac{i}{R}
\epsilon_{ijk} x_k \phi - \frac{x_i}{R} k_j^a D_a \phi +
\frac{x_j}{R} k_i^a D_a \phi \label{218} \end{equation}

where

\begin{eqnarray}
F_{ab} & = & -i ( \partial_a A_b - \partial_b A_a ) + [ A_a, A_b ]
\nonumber \\
D_a \phi & = & - i \partial_a \phi + [ A_a , \phi ] \label{219}
 \end{eqnarray}

By inserting these formulas into the action $S(\lambda)$, its value
in the classical limit is the general action:

\begin{eqnarray}
S( \lambda ) & = & - \frac{1}{4 g^2_{YM} } \int d \Omega [ ( F_{ab}
+ ( 4 - 2 \lambda ) i \epsilon_{ab} \phi \sqrt{g} )( F^{ab} + ( 4 -
2 \lambda ) i \epsilon^{ab} \frac{\phi }{\sqrt{g}} ) +
\nonumber \\
& + & 2 g^{ab} D_a \phi D_b \phi +  8 (  \lambda - 2 ) ( \lambda -
\frac{3}{2} )  \phi^2 ] \label{220} . \end{eqnarray}

We note that the auxiliary scalar field $\phi$ can be decoupled from
the pure Yang-Mills theory only in the abelian $U(1)$ case and for
the special case $ \lambda = 2 $:

\begin{eqnarray} S(2) & = & - \frac{1}{4 g^2_{YM} } \int d \Omega ( F_{ab}
F^{ab} - 2 \partial_a \phi \partial^a \phi ) \nonumber \\
d \Omega & = & \sqrt{g} d \theta d \phi = sin \theta d \theta d \phi
\ \ \ \ \ \ F^{ab} = g^{a a'} g^{b b'} F_{a' b'} \label{221} .
\end{eqnarray}

When searching a connection between projectors and the matrix model
variable $X_i$, we don't have many choices since the background
matrix $ L_i $  is the only possible definition of derivative, and
we must play  with the vector $ | \psi > $. The natural guess is

\begin{equation} X_i = < \psi | L_i | \psi > = L_i + < \psi | [ L_i , | \psi > ]
\label{222} \end{equation}

which implies the following representation for the fluctuation field
$A_i$ :

\begin{equation} A_i = < \psi | [ L_i , | \psi > ] \label{223} \end{equation}

Since $ [ L_i , . ] \rightarrow_{\alpha \rightarrow 0} - i k_i^a
\partial_a $, we obtain the well-known classical formula for the
monopole connection:

\begin{equation} A_i \rightarrow - i k_i^a < \psi | \partial_a | \psi >
\label{224} \end{equation}

There is however a problem that may ruin the classical limit. If we
use the vectors $ | \psi_{\pm 1 } > $, depending on the oscillator
algebra, rather than the fuzzy sphere algebra, the action of $ L_i $
on $ | \psi_{\pm 1 } > $ is discontinuous in the classical limit.
The only way out is redefining the vectors $ | \psi_{\pm 1 } > $
with an operator acting on the right such that the new vector can be
restricted to a fixed total oscillator number $ \hat{N} = N $.

It turns out that the correction is only possible for $ | \psi_{-1}
> $ using a quasi-unitary operator :

\begin{equation} | \psi_{-1} > \rightarrow | \psi_{-1}' > = | \psi_{-1} > U \ \
\ \ U U^{\dagger} = 1 \label{225} \end{equation}

The quasi-unitary condition keeps the projector $P_{-1}$ invariant.
A possible choice turns out to be :

\begin{equation} U_1 = \sum_{n_1, n_2 = 0}^{\infty} | n_1 , n_2 >< n_1+1, n_2 |
\label{226} \end{equation}

It is not difficult to show that the combination $X_i$ is
proportional to a reducible representation of the Lie algebra :

\begin{eqnarray}
X_i & = & < \psi'_{-1} | L_i | \psi'_{-1} > = \frac{ N+2 }{ N+1 }
U^{\dagger} L_i U \nonumber \\
F_{ij} & = & \frac{ N+2 }{ {( N+1 )}^2 } i \epsilon_{ijk}
U^{\dagger} L_k U  \label{227} \end{eqnarray}

that is indeed a solution of the non-commutative equations of motion
for :

\begin{equation} \lambda = 2 + \frac{1}{ N+1 } \label{228} \end{equation}

This is a direct deformation of the classical monopole solution.
However one can also choose to redefine $ X_i $ as

\begin{equation} X_i = U^{\dagger} L_i U \ \ \ \ F_{ij} = 0 \ \ \ \ \lambda = 2
\label{229} \end{equation}

remembering that in this case the classical limit contains not only
the monopole field but also a constant scalar field, that due to the
$U(1)$ property, remains totally decoupled. In the following we will
shift from one to the other formulation indifferently.

\section{ $U(2)$ projectors and 't Hooft-Polyakov monopoles }

In the case of $ U(1) $ projectors, our strategy was to extend the
well-known classical case, studied in detail in ref. \cite{21}.

When generalizing to the non-abelian 't Hooft-Polyakov monopoles,
the presence of a non-trivial Higgs field complicates the classical
analysis, and it is more convenient starting directly from the
matrix model formalism, which simplifies the whole picture.

We therefore prefer to postulate some non-commutative $ U(2) $
projectors, whose form is dictated by internal consistency, and then
we connect them to known solutions ( see ref. \cite{25} ), leading
to non-commutative extensions of 't Hooft-Polyakov monopoles. The
form of $ U(2) $ projectors, given in terms of oscillators, can be
guessed from the 4d case, where our guide was the classical case of
$ SU(2) $ BPST instantons,  discussed in ref. \cite{24}. This
analogy suggests us to take the following form for $ U(2) $
projectors ( with the simplest topological charge $ k = 1 $ ):

\begin{eqnarray}
P & = & | \psi > < \psi | \nonumber \\
| \psi > &  = & \left( \begin{array}{c} \left(  \begin{array}{cc}
a_0 & -
a^{\dagger}_1 \\ a_1 & a^{\dagger}_0 \end{array} \right) \\
\left(  \begin{array}{cc} a_1 & - a^{\dagger}_0 \\ a_0 &
a^{\dagger}_1 \end{array} \right)
\end{array} \right) f( \hat{N} ) \label{31}
 \end{eqnarray}

It is easy to recognize that to obtain a projector different from
identity, we need to play with the interference between the first
and the second element of the vector $ | \psi > $, since a single
component would produce no functional dependence on the elements of
the fuzzy sphere. Note the exchange of indices ( $ 0 \leftrightarrow
1 $ ) in the second element of the vectors which is responsible for
a non-trivial interference.

We will notice in the following that adding the second element with
interchanged indices is also important to find a simple result for
the expectation value $ < \psi | L_i | \psi > $, the form which has
enabled us to derive the non-commutative abelian monopoles in the
matrix model formalism.

As in the $U(1)$ case, the function $ f( \hat{N} )$ can be
determined by imposing the normalization condition on $ | \psi > $:

\begin{equation} < \psi | \psi > = f^2 ( \hat{N} ) \left(
\begin{array}{cc} 2 \hat{N} & 0 \\ 0 & 2 ( \hat{N} + 2 ) \end{array}
\right) \label{32} \end{equation}

Therefore the form of $ f ( \hat{N} ) $ is fixed as the following
diagonal form :

\begin{equation} f ( \hat{N} ) = \left(
\begin{array}{cc} \frac{1}{\sqrt{2 \hat{N}}} & 0 \\ 0 & \frac{1}{\sqrt{2 ( \hat{N} + 2
)}} \end{array} \right) \label{33} \end{equation}

Using the commutation rule of the oscillators, the final form of the
vector $ | \psi > $ can be simplified as :

\begin{equation} | \psi >   = \frac{1}{\sqrt{ 2 ( \hat{N} + 1 )}} \left(
\begin{array}{c} \left(
\begin{array}{cc} a_0 & -
a^{\dagger}_1 \\ a_1 & a^{\dagger}_0 \end{array} \right) \\
\left(  \begin{array}{cc} a_1 & - a^{\dagger}_0 \\ a_0 &
a^{\dagger}_1 \end{array} \right)
\end{array} \right)  \label{34}
\end{equation}

The corresponding non-commutative $ U(2) $ projector, that will be
further elaborated for a possible connection with the 't
Hooft-Polyakov monopoles, is given by :

\begin{equation} P = | \psi > < \psi | = \frac{1}{ 2 ( \hat{N} + 1 )} \left(
\begin{array}{cc}
\left(  \begin{array}{cc} \hat{N} + 1  & 0 \\ 0 & \hat{N} + 1
\end{array} \right) &
\left( \begin{array}{cc} 2 a_0 a^{\dagger}_1 &
  a_0 a^{\dagger}_0 - a^{\dagger}_1 a_1  \\
  a_1 a^{\dagger}_1 - a^{\dagger}_0 a_0  &
 2 a^{\dagger}_0 a_1 \end{array}  \right)       \\
\left(   \begin{array}{cc} 2 a^{\dagger}_0 a_1 &   a_1 a^{\dagger}_1
- a^{\dagger}_0 a_0
\\  a_0 a^{\dagger}_0 - a^{\dagger}_1 a_1
 & 2 a_0 a^{\dagger}_1
\end{array} \right) & \left( \begin{array}{cc}  \hat{N} + 1  & 0 \\ 0 &
\hat{N} + 1
\end{array} \right)
\end{array}
\right) \label{35} \end{equation}

At this level, it is safe substituting to the number operator $
\hat{N} $ its eigenvalue $ N $, and considering the entries of $ P $
as elements of the fuzzy sphere function space.

The trace of the projector is obviously :

\begin{equation} Tr P = 2 \ Tr I < Tr 1_P = 4 Tr I \label{36} \end{equation}

Being this projector non-trivial, it can be taken as a natural
candidate for a non-abelian topological configuration. To be sure,
we must study its connection with matrix models.

We already know what can be expected from our previous study of $
U(1) $ projectors. The combined presence of operators of type $ a (
a^{\dagger} ) $ implies that this projector will act on the
background in order to increase ( decrease ) the dimension of the
representation used in the background. This is exactly the
characteristic of a topologically non-trivial configuration that in
ref. \cite{25} has been shown to reproduce in the classical limit
the 't Hooft-Polyakov monopoles. We feel therefore to be on the
right way and the strategy of projectors combined with the use of
matrix models can also, as a byproduct, teach us how to treat the
non-abelian topology on the classical sphere, generalizing the
mathematical work of \cite{21}.

\section{Connection with matrix models}

Tentatively, we can try to connect the projectors with matrix
models, as successfully done in the $ U(1) $ case,

\begin{equation} X_i = < \psi | L_i | \psi > \label{41} \end{equation}

However we easily recognize that, in the classical limit, this
formula is inconsistent with the presence of a non-trivial Higgs
field, since the $ L_i $ action reduces to $ k_i^a \partial_a $ and
therefore it projects in the tangent plane to the sphere, while the
Higgs field fluctuation is in the orthogonal direction, along the
radius. The explicit calculation of the matrix element $ < \psi |
L_i | \psi > $ will suggest us what we need to add to this formula
to complete the connection with matrix models in the non-abelian
case.

The explicit calculation gives, in details,

\begin{equation} < \psi | L_i | \psi > = \frac{1}{2} \left(
\begin{array}{cc} \frac{1}{\hat{N}} ( a^{\dagger}_0 L_i a_0 +
a^{\dagger}_1 L_i a_1  ) & \frac{1}{\hat{N}} ( - a^{\dagger}_0 L_i
a^{\dagger}_1 + a^{\dagger}_1 L_i a^{\dagger}_0 ) \\
\frac{1}{\hat{N}+ 2} ( - a_1 L_i a_0 + a_0 L_i a_1 ) &
\frac{1}{\hat{N}+ 2} ( a_0 L_i a^{\dagger}_0 + a_1 L_i a^{\dagger}_1
)
\end{array} \right) + ( 0 \leftrightarrow 1 ) \label{42}
 \end{equation}

We note from this formula the importance of the contribution ( $ 0
\leftrightarrow 1 $ ) to cancel the off-diagonal terms.

In summary we obtain:

\begin{equation} < \psi | L_i | \psi > =  \left(
\begin{array}{cc} \frac{1}{\hat{N}} ( a^{\dagger}_0 L_i a_0 +
a^{\dagger}_1 L_i a_1  ) & 0  \\
0  & \frac{1}{\hat{N}+ 2} ( a_0 L_i a^{\dagger}_0 + a_1 L_i
a^{\dagger}_1 )
\end{array} \right) \label{43} \end{equation}

It is not difficult to compute the terms inside the parenthesis
using the oscillator algebra:

\begin{equation} < \psi | L_i | \psi > =  \left(
\begin{array}{cc} \frac{\hat{N}-1}{\hat{N}} L_i & 0  \\
0  & \frac{\hat{N} + 3}{\hat{N} + 2}  L_i \end{array} \right)
\label{44} \end{equation}

As we discussed in our previous article \cite{23}, the action of $
L_i $ on $ \psi $ cannot be smoothly connected to the classical Lie
derivative on the sphere unless we project the vector $ | \psi >$ on
the fuzzy sphere algebra. The only possibility left, with the
constraint of keeping invariant the projector $ P $, is dressing $ |
\psi > $ with a quasi-unitary operator, such that $ | \psi' > $
belongs to the fuzzy sphere function space:

\begin{eqnarray}
P & = & | \psi > < \psi | = | \psi' > < \psi' | \nonumber \\
| \psi' > & = & | \psi > U \label{45} \end{eqnarray}

The quasi-unitary operator $ U $, as in the case of non-commutative
extension of Dirac monopoles, plays an essential role to define the
non-abelian topology. The only consistent choice, apart from unitary
gauge transformation, given the structure of the vector $ | \psi >
$, is the following quasi-unitary operator :

\begin{eqnarray}
U & = & \left( \begin{array}{cc} U_1^{\dagger} & U_{12}^{\dagger} \\
0 & U_2 \end{array}\right) \nonumber \\
U_1 & = & \sum_{n_1, n_2 =0}^{\infty} | n_1, n_2 > < n_1 + 1, n_2 |
\nonumber \\
U_2 & = & \sum_{n_1, n_2 =0}^{\infty} | n_1, n_2 > < n_1, n_2 + 1|
\nonumber \\
U_{12} & = & \sum_{n_1=0}^{\infty} | n_1, 0 > < 0 , n_1 + 1|
\label{46} \end{eqnarray}

This operator satisfies to the following properties:

\begin{equation} U U^{\dagger} = \left( \begin{array}{cc} U_1^{\dagger} & U_{12}^{\dagger} \\
0 & U_2 \end{array} \right)
\left( \begin{array}{cc} U_1 & 0 \\
U_{12} & U^{\dagger}_2 \end{array} \right) = \left(
\begin{array}{cc} U_1^{\dagger} U_1 + U_{12}^{\dagger} U_{12} = 1 -
|0><0| & U^{\dagger}_{12} U^{\dagger}_2 = 0 \\ U_2  U_{12} = 0 & U_2
U_2^{\dagger} = 1
\end{array} \right) \label{47} \end{equation}

Since the operator $ | 0 > < 0 | $ is annihilated by the action of $
a_0 $ and $ a_1 $, the combination $ U U^{\dagger} $ behaves as the
identity when acting on the oscillators.

Moreover the following property holds :

\begin{equation} U^{\dagger} U =  \left( \begin{array}{cc} U_1 & 0 \\
U_{12} & U^{\dagger}_2 \end{array} \right)
\left( \begin{array}{cc} U_1^{\dagger} & U_{12}^{\dagger} \\
0 & U_2 \end{array} \right) = \left(
\begin{array}{cc} U_1 U_1^{\dagger} = 1
 & U_1 U_{12}^{\dagger} = 0 \\ U_{12} U_1^{\dagger} = 0 &
U_2^{\dagger} U_2 + U_{12} U_{12}^{\dagger} = 1 \end{array} \right)
 \label{48} \end{equation}

This operator doesn't change the rank of the background but simply
it changes the dimensions of the component representations.
Therefore we must expect that :

\begin{equation} \left( \begin{array}{cc} ( L_i )_{N+1} & 0 \\
0 & ( L_i )_{N+1} \end{array} \right) \rightarrow \left[ U^{\dagger}
\left( \begin{array}{cc} L_i  & 0 \\
0 & L_i  \end{array} \right) U \right]_{N+1} = \left(
\begin{array}{c|c} {( L_i )}_{N+2} & 0 \\ \hline 0 & {( L_i )}_{N}
\end{array} \right) \label{49} \end{equation}

where the basic building blocks in the last formula have different
size from those of the background.

Now let's apply $ | \psi ' > $ to the formula(\ref{41}). In this
case we obtain

\begin{eqnarray}
X_i & = & < \psi' | L_i | \psi' > = U^{\dagger}
\left( \begin{array}{cc} \frac{\hat{N}-1}{\hat{N}} L_i & 0 \\
0 & \frac{\hat{N} + 3}{\hat{N} + 2} L_i  \end{array} \right) U =
\nonumber \\
& = & \frac{\hat{N}}{\hat{N}+1} \left( \begin{array}{cc} U_1 L_i
U_1^{\dagger} & U_1 L_i U_{12}^{\dagger} \\
U_{12} L_i U_1^{\dagger} & U_{12} L_i U_{12}^{\dagger}
\end{array} \right) +
\frac{\hat{N}+2}{\hat{N}+1} \left( \begin{array}{cc} 0  & 0 \\
0 & U^{\dagger}_{2} L_i U_2 \end{array} \right) \label{410}
\end{eqnarray}

It is not difficult to recognize that this formula is nothing but a
sum of $ SU(2) $ representations ( see Appendix ):

\begin{equation} \left( \begin{array}{cc} U_1 L_i
U_1^{\dagger} & U_1 L_i U_{12}^{\dagger} \\
U_{12} L_i U_1^{\dagger} & U_{12} L_i U_{12}^{\dagger}
\end{array} \right)_{N+1} = \left(
\begin{array}{c|c} {( L_i )}_{N+2} & 0 \\ \hline 0 & 0
\end{array} \right) \label{411}
\end{equation}

while

\begin{equation} \left( \begin{array}{cc} 0  & 0 \\
0 & U^{\dagger}_{2} L_i U_2 \end{array} \right)_{N+1} = \left(
\begin{array}{c|c} 0 & 0 \\ \hline 0 & {( L_i )}_N
\end{array} \right) \label{412} \end{equation}

In summary, the presence of the quasi-unitary operator $ U $ can be
completely worked out into this final formula :

\begin{equation} X_i = < \psi' | L_i | \psi' > =  \left(
\begin{array}{c|c} \frac{\hat{N}}{\hat{N}+1} {( L_i )}_{N+2} & 0 \\
\hline 0 & \frac{\hat{N}+2}{\hat{N}+1} {( L_i )}_{N}
\end{array} \right) \label{413} \end{equation}

This block-diagonal form is still not an explicit solution of the
equations of motion. The nearest solution is very simple to obtain,
redefining $ | \psi ' > $ with a diagonal matrix :

\begin{equation} X_i = \left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right) < \psi' | L_i | \psi' >
\left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right) \label{414} \end{equation}

This final form is not of type $ < \psi' | L_i | \psi' > $, as in
the case of Dirac monopoles; however as we discussed in the
beginning, this modification is necessary to obtain in the classical
limit a non-trivial contribution for the Higgs field.

Finally requiring that

\begin{equation} X_i =   \left(
\begin{array}{c|c}  {( L_i )}_{N+2} & 0 \\
\hline 0 &  {( L_i )}_{N}
\end{array} \right) \label{415}\nonumber \end{equation}

we obtain a condition that fixes the unknown constants $ f_{+},
f_{-} $:

\begin{equation} f_{\pm} = \frac{1}{2} \left[ \sqrt{ \frac{ \hat{N} + 1}{
\hat{N} }} \pm \sqrt{ \frac{ \hat{N} + 1}{ \hat{N} + 2 }} \right]
 \label{416} \end{equation}

This solution can also be expressed, with a gauge transformation,
as:

\begin{equation} X_i = L_i \otimes 1 + 1 \otimes S_i \label{417}  \end{equation}

Now it is simpler to extract the contribution of the fluctuation (
non-abelian monopole field ) from the background:

\begin{equation} X_i - ({ \rm background }) = 1 \otimes S_i \label{418} \end{equation}

An alternative way to express formula (\ref{414}), reaching the same
solution, is redefining both $ | \psi' > $ and the projector $ P $
as

\begin{eqnarray}
| \psi'' > & = & \left( \begin{array}{cc} \frac{1}{f_{+} + f_{-}} & 0 \\
0 & \frac{1}{f_{+} - f_{-}} \end{array} \right) | \psi' >
\left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right) \nonumber \\
P' & = & | \psi'' > < \psi'' | = \left( \begin{array}{cc} \frac{1}{f_{+} + f_{-}} & 0 \\
0 & \frac{1}{f_{+} - f_{-}} \end{array} \right) P
\left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right)
 \label{419} \end{eqnarray}

The formula, analogous to (\ref{414}), linking projectors to
connections, can be expressed in this case as

\begin{eqnarray}  X_i & = &
< \psi'' | \left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right) L_i
\left( \begin{array}{cc} f_{+} + f_{-} & 0 \\
0 & f_{+} - f_{-} \end{array} \right) | \psi'' >
 \nonumber \\
& = & < \psi'' | \left( \begin{array}{cc} \frac{ \hat{N} +
1}{\hat{N}} L_i & 0 \\ 0 & \frac{ \hat{N} + 1}{\hat{N}+ 2} L_i
\end{array} \right) | \psi'' > = < \psi'' | X^{0}_i | \psi''
> \label{420} \end{eqnarray}

This version allows us to define a gauge invariant version of matrix
models, generalizing what we have done in the $ U(1) $ case ( ref.
\cite{23} ), built directly on the projectors

\begin{equation} X_i = P' X^{0}_i P' \label{421}\end{equation}

that, by construction, satisfies the same equations of motion.

\section{ Classical limit }

The lagrangian of the matrix model ( eq. (\ref{214}) ) for $ \lambda
= 2 $ in the non-abelian case leads to the following classical
action on the sphere:

\begin{equation} S = - \frac{1}{ 4 g^2_{YM} }\int d \Omega ( F_{ab} F_{a'b'}
g^{aa'} g^{bb'} + 2 g^{aa'} D_a \phi D_{a'} \phi ) \label{51} \end{equation}

where we define

\begin{eqnarray}
D_a \phi & = & - i \partial_a \phi + [ A_a, \phi ] \nonumber \\
F_{ab} & = & - i ( \partial_a A_b - \partial_b A_a ) + [ A_a , A_b ]
 \label{52} \end{eqnarray}

The variation with respect to $ A_a $ and $ \phi $ is vanishing by
simply requiring that

\begin{equation} D_a ( \sqrt{ g } g^{aa'} g^{bb'} F_{a' b'} ) = D_a \phi = 0
\label{53}\end{equation}

On the other hand, we know that at non-commutative level

\begin{equation} X_i = L_i \otimes 1 + 1 \otimes S_i \label{54}\end{equation}

satisfies the constraint

\begin{equation} F_{ij} = [ X_i, X_j ] - i \epsilon_{ijk} X_k = 0 \label{55}
\end{equation}

since $ X_i $ satisfies the commutation rules of the angular
momentum. In the classical limit the fluctuation field $ A_i $

\begin{equation} A_i = 1  \otimes S_i = k_i^a A_a + n_i \phi \label{56} \end{equation}

can be projected over the tangent plane of the sphere and on the
orthogonal direction

\begin{eqnarray}
\phi & = & n_i \otimes S_i \nonumber \\
A_a & = & g_{ab} \ k_i^b \otimes S_i
 \label{57} \end{eqnarray}

We must obtain, as a check, that

\begin{eqnarray}
F_{ij} & = & \frac{1}{R} \epsilon_{ijk} x_k \left( i \phi +
\frac{\epsilon^{ab}}{ 2 \sqrt{g}} F_{ab} \right) + \frac{1}{R} x_j
k_i^a D_a \phi - \frac{1}{R} x_i k_j^a D_a \phi = 0 \nonumber \\
D_a \phi & = & 0 \nonumber \\
F_{ab} & = & - i \epsilon_{ab} \sqrt{g} \phi \label{58}
\end{eqnarray}

We note that these constraints are enough to solve the classical
equations of motion ( eq. (\ref{53}) ) for $ \lambda = 2 $. This
check permits us to have an explicit formula for $ \phi $ and $ A_a
$:

\begin{eqnarray}
\phi & = & n_i \otimes S_i = \frac{1}{2} \left(
\begin{array}{cc} cos \theta & sin \theta e^{- i \phi } \\
sin \theta e^{i \phi} & - cos \theta \end{array}
\right)\nonumber \\
A_\theta & = & k_i^\theta \otimes S_i = \frac{i}{2} \left(
\begin{array}{cc} 0 & - e^{- i \phi } \\
e^{i \phi} & 0 \end{array}
\right)\nonumber \\
A_\phi & = & sin^2 \theta \ k_i^\phi \otimes S_i = \frac{sin
\theta}{2} \left(
\begin{array}{cc} sin \theta & - cos \theta e^{- i \phi } \\
- cos \theta e^{i \phi} & - sin \theta \end{array} \right)
 \label{59}\end{eqnarray}

It is easy to deduce that

\begin{equation} F_{\theta\phi} = - \frac{i}{2} ( \partial_\theta A_\phi -
\partial_\phi A_\theta ) = - [ A_\theta , A_\phi ] = - i \sqrt{g}
\phi \label{510} \end{equation}

that is equivalent to eq. (\ref{58}), with the notation $
\epsilon_{\theta\phi} = 1 $.

Moreover

\begin{eqnarray}
- i \partial_\theta \phi & = & - [ A_\theta, \phi ] = - \frac{i}{2}
\left(
\begin{array}{cc} - sin \theta & cos \theta e^{- i \phi } \\
cos \theta e^{i \phi} & sin \theta \end{array} \right) \nonumber \\
- i \partial_\phi \phi & = & - [ A_\phi, \phi ] =
\frac{sin\theta}{2} \left(
\begin{array}{cc} 0 & -  e^{- i \phi } \\
 e^{i \phi} & 0 \end{array} \right) \label{511}
\end{eqnarray}

The classical limit induced by our formula (\ref{414}) produces a
slightly different solution. Starting from the classical limit of
the vector $ | \psi' > $:

\begin{equation} | \psi' > \rightarrow_{N \rightarrow \infty} | \psi_{cl} >  =
\frac{1}{\sqrt{2}} \left(
\begin{array}{c} \left(
\begin{array}{cc} a_0 & -
a^{*}_1 \\ a_1 & a^{*}_0 \end{array} \right) \\
\left(  \begin{array}{cc} a_1 & - a^{*}_0 \\ a_0 & a^{*}_1
\end{array} \right)
\end{array} \right) \ \ \ \ \ < \psi_{cl} | \psi_{cl} > = 1
\label{512}\end{equation}

we obtain that the variable $ X_i $ of the classical matrix model is
of the type:

\begin{equation} X_i \simeq ( f_{+}^2 + f_{-}^2 ) < \psi_{cl} | L_i | \psi_{cl}
> + 2 f_{+} f_{-} \left(
\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) < \psi_{cl} | L_i
| \psi_{cl} > \label{513} \end{equation}

Since

\begin{equation} f_{+} \rightarrow 1 \ \ \ \ f_{-} \rightarrow \frac{1}{2 N}
\label{514}\end{equation}

we can deduce that

\begin{equation} X_i \simeq < \psi_{cl} | L_i | \psi_{cl} > +  \frac{1}{2}
\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) <
\psi_{cl} | \frac{\hat{x}_i}{R} | \psi_{cl} > \label{515}\end{equation}

where $ \hat{x}_i = \alpha L_i $ is the operator deforming the
coordinate $x_i$. The Higgs field contribution comes exactly from
this last term:

\begin{eqnarray}
\phi & = & \frac{1}{2}  \left(
\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \nonumber \\
A_{\theta} & = & - i < \psi_{cl} | \partial_{\theta} | \psi_{cl} > =
0
\nonumber \\
A_{\phi} & = & - i < \psi_{cl} | \partial_{\phi} | \psi_{cl} > = -
\frac{ cos\theta}{2} \left(
\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)
\label{516} \end{eqnarray}

with the choice

\begin{eqnarray} a_0 & = & cos \frac{\theta}{2}
e^{-i \frac{\phi}{2}} \nonumber \\
a_1 & = & sin \frac{\theta}{2} e^{i \frac{\phi}{2}} \label{518}
\end{eqnarray}

Naturally there exists a classical gauge transformation that
connects the solution (\ref{59}) to the solution (\ref{516})

\begin{eqnarray}
\phi_g & = & n_i \otimes  S_i \ \ \ \ \ \  \ \ \ \ \ \ \ \phi  = S_3 \nonumber \\
A_{ag} & = & g_{ab} k_i^b \otimes S_i \ \ \ \ \ \ \ \ \ \ A_\theta
= 0 \ \ A_\phi = - cos \theta S_3 \label{519}\end{eqnarray}

In both cases the constraint $ F_{ij} = 0 $ is satisfied. Starting
for example from the Higgs field :

\begin{equation} \phi_g = g^{-1} \phi g = \frac{1}{2} \left( \begin{array}{cc}
\overline{a} & - b \\ \overline{b} & a \end{array} \right) \left(
\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}
\right) \left( \begin{array}{cc} a & b \\ - \overline{b} &
\overline{a}
\end{array} \right)
\ \ \ \ \ \ a \overline{a} + b \overline{b} = 1
 \label{520}\end{equation}

the gauge transformation is defined by :

\begin{eqnarray}
a \overline{a} - b \overline{b} &  =& cos \theta \nonumber \\
2 \overline{a} b & = & sin \theta e^{ - i \phi } \nonumber \\
a \overline{a} + b \overline{b} & = & 1 \label{521} \end{eqnarray}

which is solved by

\begin{eqnarray}
a & = & cos \frac{\theta}{2} e^{ i \frac{\phi}{2}} \nonumber \\
b & = &  sin \frac{\theta}{2} e^{ - i \frac{\phi}{2}} \label{522}
\end{eqnarray}

We have checked that $ A_{ag} $ and $ A_a $ are related by the same
gauge transformation.

\section{ Deforming the Chern Class}

In literature a candidate for an eventual non-commutative
topological index has been proposed, by using an action taking
values only in the integer numbers \cite{25}. In this paper we want
to suggest an alternative, more traditional, definition, by taking
an action taking values not necessarily in the integers.

Our candidate for the non-commutative Chern class is the
Chern-Simons term

\begin{equation} S_{CS} = \frac{1}{N+1} Tr \left[ \frac{2i}{3} \epsilon_{ijk}
X_i X_j X_k + X_i X_i \right] \label{61} \end{equation}

This action, as we have found in a previous paper ( ref. \cite{26}
), is invariant under deformed diffeomorphisms, a property which
doesn't hold for the standard Yang-Mills action.

We want to show that this action in the classical limit, once
evaluated on the 't Hooft-Polyakov monopole, corresponds to the
classical topological number

\begin{equation} Q = - \frac{1}{8\pi} \int_S d \Omega \epsilon_{ijk} n_i
\epsilon^{abc} n^a (
\partial_j n^b ) ( \partial_k n^c ) = - \frac{1}{4\pi} \int_S d
\Omega = - 1 \label{62}\end{equation}

Comparing with ref. \cite{25}, the integral $ Q $ is equivalent to
the following action

\begin{equation} Q = \frac{1}{8\pi} Tr \int_S d \Omega \epsilon_{ijk} n_i
F_{jk}^{ \perp } \phi \label{63} \end{equation}

where

\begin{eqnarray}
\phi & = & n_i \otimes S_i \nonumber \\
 F_{ij}^{ \perp } &  =  & \partial_i
a_j^{ \perp } -
\partial_j a_i^{ \perp } - i [ a_i^{ \perp } , a_j^{ \perp } ]
\label{64}
 \end{eqnarray}

and $ a_i^{\perp} $ is the orthogonal part of the fluctuation $ A_i
$, i.e. in our notations

\begin{equation} a_i^{\perp} = \epsilon_{ijk} n_j  A_k \label{65}\end{equation}

Extracting the contribution of the transverse part and substituting
the explicit relation between $ F_{ab} $ and $ \phi $ given by
(\ref{510})

\begin{equation} \epsilon_{ijk} n_i F_{ij}^{\perp} = - 2 \phi = -i  \left(
\frac{\epsilon^{ab}}{\sqrt{g}} F_{ab} \right)\label{66}\end{equation}

we can prove that

\begin{equation} Q = \frac{1}{8\pi}Tr \int_S d \Omega \epsilon_{ijk} n_i \
F_{jk}^{\perp} \phi  = -  \frac{1}{8\pi} Tr \int_S d \Omega \left[ \
i \phi \frac{\epsilon^{ab}}{\sqrt{g}} F_{ab} \ \right]\label{67}
\end{equation}

This expression is very similar to the classical limit of the
Chern-Simons topological term (\ref{61}). In fact, from references
\cite{10} and \cite{12} we obtain that

\begin{equation} S_{CS} \rightarrow_{\alpha \rightarrow 0} \frac{1}{8\pi} Tr
\int d \Omega \left( i \phi \frac{\epsilon^{ab}}{\sqrt{g}} F_{ab} -
\phi^2 \right) \label{68} \end{equation}

Applying again the classical limit of the constraint $ F_{ij} = 0 $
i.e.

\begin{eqnarray}
F_{ab} & = & - i \epsilon_{ab} \sqrt{g} \phi \nonumber \\
S & \rightarrow_{\alpha \rightarrow 0}&  \frac{1}{8\pi} Tr \int d
\Omega \left( i \phi \frac{\epsilon^{ab}}{2 \sqrt{g}} F_{ab} \right)
= - \frac{Q}{2} = \frac{1}{2} \label{69}
\end{eqnarray}

and substituting the explicit value for $ \phi $ given by eq.
(\ref{59}) we obtain perfect agreement.

We have proved that, at least in the classical limit, the
Chern-Simons action, evaluated on the classical solution (\ref{56}),
produces the topological number characteristic of non-abelian
monopoles.

Now we want to show that the same topological number can be obtained
evaluating the Chern-Simons action at pure non-commutative level:

\begin{equation} S_{CS}^{\rm monopole} - S_{CS}^{\rm bg} = -\frac{Q}{2} =
\frac{1}{2} \label{610}\end{equation}

and that in the non-abelian case there is a continuity with the
non-commutative extension. If we evaluate $ S_{CS} $ on the
background we obtain :

\begin{equation} S_{CS}^{\rm bg} = \frac{1}{3 ( N+1 )} Tr \left(
\begin{array}{cc} ( L_i L_i )_{N+1} & 0 \\ 0 & ( L_i L_i )_{N+1}
\end{array} \right)= \frac{N(N+2)}{6} \label{611} \end{equation}

while on the monopole configuration

\begin{equation} S_{CS}^{\rm monopole} = \frac{1}{3 ( N+1 )} Tr \left(
\begin{array}{c|c} ( L_i L_i )_{N+2} & 0 \\ \hline 0 & ( L_i L_i )_{N}
\end{array} \right) = \frac{ N^2 + 2 N + 3}{6} \label{612} \end{equation}

Therefore

\begin{equation} S_{CS}^{\rm monopole} - S_{CS}^{\rm bg} = \frac{1}{8\pi} \int_S
Tr \phi^2 = \frac{1}{2} \label{613} \end{equation}

the classical topological number is maintained by the
non-commutative deformation in the non-abelian case.

To complete the picture, we want to analyze what happens for  Dirac
monopoles. At first sight, the classical limit of the Chern-Simons
term doesn't seem to be in agreement with the Chern class

\begin{equation} S_{Ch} = \int_S d \Omega \left( i \frac{ \epsilon^{ab} }{
\sqrt{g} } F_{ab} \right) = Q \label{614} \end{equation}

because of the absence of the Higgs field. This problem can be
easily circumvented by admitting that the classical monopole
configurations have also a constant Higgs field. Since the
lagrangian of the Higgs field is decoupled, this configuration is
still solution of the equations of motion.

Therefore the evaluation of $S_{CS}$ is equivalent to the evaluation
of $S_{Ch}$, by allowing the presence of a constant Higgs field.
This different classical limit can be realized starting from

\begin{equation} X_i^{NC} = U^{\dagger} L_i U \label{615} \end{equation}

instead of

\begin{equation} X_i^{NC} = \frac{ N+2 }{ N+1 } \ U^{\dagger} L_i U \label{616}
\end{equation}

This new solution satifies the same commutation rules of the angular
momentum and therefore to the constraint

\begin{equation} F_{\theta\phi} = - i \partial_{\theta} A_{\phi} = i
\epsilon_{\theta\phi} \sqrt{g} \phi \leftarrow_{\alpha \rightarrow
0} \ F_{ij} = 0 \label{617} \end{equation}

and the explicit solution is

\begin{equation} X_i^{NC} \rightarrow_{\alpha \rightarrow 0} \left\{
\begin{array}{c} A_\phi = \frac{ cos \theta - 1 }{2} \\ \phi = - \frac{1}{2}
\end{array} \right. \label{618} \end{equation}

In summary the classical fluctuation that must be compared with the
solution $X_i^{NC}$ is of the type:

\begin{eqnarray}
a_i & = & k_i^a A_a + \frac{x^i}{R} \phi \nonumber \\
k_i^a A_a & = & k_i^\phi A_\phi = - \frac{1}{2} \left[
\frac{cos\theta}{sin\theta} ( cos \theta - 1 ) cos \phi ,
\frac{cos\theta}{sin\theta} ( cos \theta - 1 ) sin \phi, 1 - cos
\theta \right] \nonumber
\\
n^i \phi & = & - \frac{1}{2} [ sin \theta cos \phi, sin \theta sin
\phi, cos
\theta ] \nonumber \\
a_i & \sim & - \frac{1}{2} \left[ \frac{1 - cos \theta}{ sin \theta
} cos \phi, \frac{1 - cos \theta}{ sin \theta } sin \phi, 1 \right]
\simeq ( L_i )_N - ( L_i )_{N+1} \label{619} \end{eqnarray}

The direct non-commutative evaluation in this case produces a
singularity:

\begin{eqnarray}
S_{CS}^{\rm bg} & = & \frac{1}{3 ( N+1 )} Tr ( L_i L_i )_{N+1} =
\frac{N ( N+2 )}{12} \nonumber \\
S_{CS}^{\rm monopole} & = & \frac{1}{3 ( N+1 )} Tr \left(
\begin{array}{cc}
( L_i L_i )_{N} & 0 \\ 0 & 0 \end{array} \right) = \frac{N(N-1)}{12}
\nonumber \\
S_{CS}^{\rm monopole} &  - & S_{CS}^{\rm bg} = \frac{N(N-1)}{12} -
\frac{N ( N+2 )}{12} = - \frac{N}{4} \label{620} \end{eqnarray}

This negative result requires to analyze more carefully the
classical limit of the Dirac monopoles. Starting from the components
$ a_{\pm}$

\begin{eqnarray} a_{+} & = & ( U^{\dagger} L_{+} U )_{N+1} - ( L_{+} )_{N+1} =
\sum ( \sqrt{ n_1 } - \sqrt{ n_1 + 1 } ) \sqrt{ n_2 + 1 } | n_1 + 1,
n_2 > < n_1 , n_2 + 1 | \nonumber \\
& \simeq & - \frac{1}{2} \sum \sqrt{ \frac{ n_2+1 }{ n_1+1 } } |
n_1+1, n_2 >< n_1, n_2+1 | = - \frac{1}{2} \frac{1}{\hat{z}_1}
\hat{z}_2
 \label{621} \end{eqnarray}

where we have defined

\begin{eqnarray} \frac{1}{\hat{z}_1} & = & \sum_{n_1}
\frac{1}{\sqrt{n_1+1}} | n_1+1 ><n_1 | \ \ \ \ \ \ \ \ z_1 =
\sqrt{2} \ cos
\frac{\theta}{2} e^{-i \frac{\phi}{2}} \nonumber \\
\hat{z}_2 & = & \sum_{n_2} \sqrt{ n_2 + 1 } | n_2 >< n_2 + 1 | \ \ \
\ \ \ \ \ \ \ z_2 = \sqrt{2} \ sin \frac{\theta}{2} e^{i
\frac{\phi}{2}} \nonumber \\
x_{+} & = & \overline{z}_1 z_2 = sin \theta e^{ i \phi } \ \ \ \ x_3
= cos \theta \nonumber \\
a_{+} & \rightarrow & - \frac{1}{2 z_1} z_2 = - \frac{1}{2} \frac{
sin \frac{\theta}{2} }{ cos \frac{\theta}{2} } e^{ i \phi } =
\frac{1}{2} \frac{ cos \theta - 1 }{ sin \theta } e^{ i \phi }
\label{622}
\end{eqnarray}

we find that $ a_{+} $ and $ a_{-} $ are really continuous
deformations of the classical configurations. Different is the case
of $ a_3 $:

\begin{eqnarray}
a_3 & = & ( U^{\dagger} L_3 U )_{N+1} - ( L_3 )_{N+1} = \frac{1}{2}
\left( - \sum_{n_1 \neq 0} | n_1, n_2 >< n_1, n_2 | + \sum_{n_2} n_2
| 0, n_2
><0, n_2 |
\right)_{N+1} = \nonumber \\
& = & - \left( \frac{1 - P_0}{2} \right) + \frac{ N+1 }{2} P_0 \ \ \
\ \ P_0 = | 0, N+1
>< 0, N+1 | \label{623} \end{eqnarray}

$ a_3 $ has a problem, i.e. a discontinuous term that has support on
a single state:

\begin{eqnarray}
a_3^m & = & - \left(  \frac{ 1 - P_0 }{2} \right) \nonumber \\
a_3^s & = & \frac{N+1}{2} P_0
 \label{624} \end{eqnarray}

The effect of $ a^s_3 $ seems at first sight negligible, if we
compare a single state with respect to an infinite number. However
on the non-commutative action it makes a difference and produces a
nasty discontinuity with the classical topological number. Let us
try to exclude the contribution of this state, redefining the
solution as

\begin{eqnarray}
X_i & = & ( L_i + a_i^s ) + a_i^m \nonumber \\
X'_i & = & X_i - a_i^s = L_i + a_i^m \nonumber \\
S_{CS}^{\rm monopole} - S_{CS}^{\rm bg} & = & S_{X'} - S_{L_i} =
\frac{ N^2+2N+3 }{12} - \frac{N(N+2)}{12}  = \frac{1}{4} = \nonumber \\
& = & \frac{1}{4\pi} \int_S d \Omega \  \left( i \phi \frac{
\epsilon^{ab}}{2 \sqrt{g}} F_{ab} \right) = - \frac{Q}{4} =
\frac{1}{4}
 \label{625} \end{eqnarray}

The agreement now is perfect.

\section{Conclusions}

In this work we have shown how to characterize easily the
topologically nontrivial configurations leading to 't Hooft-Polyakov
monopoles with a noncommutative $ U(2) $ projector. The
non-triviality of the $U(2)$ projector is assured in the classical
limit by the use of the Hopf fibration $ \pi : S^3 \rightarrow S^2
$, since the projector cannot be decomposed in terms of vectors
belonging to the sphere $S^2$ but only to $S^3$.

At noncommutative level we can surely state that there exist
solutions to the matrix model equations of motion extending in a
smooth way classical topology.

It remains an open question how to characterize the topological
meaning of these noncommutative configurations, whether they remain
stable and so on.

As a first step in this direction we have suggested a candidate for
deforming the Chern class, maintaining invariant the classical
topological number. However, this subject still requires a deeper
investigation in the future.

redpajama_set_name: RedPajamaArXiv


text:
Q: android - Latency hiccup with setting imageView content I already searched and found a lot on this topic, but none of the answers helped me.
My program is a drumcomputer, actually it's working fine, execpt for this latency bug.
The visual architecture is the following:
A HorizontalScrollView contains a LinearLayout (LayoutRhythm, vertical).
This Layout contains one Linearlayout for each Track.
The TrackLayout (horizontal) contains multiple ImageViews for each sound.
Just for the Imagination:
HorizontalScrollView
rhythmLayout
trackLayout ImageView ImageView ImageView ... ImageView ImageView trackLayout
trackLayout ImageView ImageView ImageView ... ImageView ImageView trackLayout
trackLayout ImageView ImageView ImageView ... ImageView ImageView trackLayout
/rhythmLayout
/HorizontalScrollView
The first location is the buildRhythm function, where I have to check the rhythm, and the set the corresponging png-image (which is just around 665 bytes!) to all imageViews.
The second Location is the OnClick Listener of the ImageView, which changes the image of itself.
In my buildRhythm method I call the follwoing method to get an ImageView:
private ImageView getRhythmImage(int playerTrackindex, int playerSoundIndex, int soundIndexViews, int zoomLevel){

    final ImageView sound = new ImageView(this);
    final int trackIndexViews=playerTrackindex+1;
    final int trackIndexPlayer = playerTrackindex;
    final int soundIndexPlayer=playerSoundIndex;

    if (soundIndexViews % zoomLevel == 0) {
    //sound.setBackgroundResource(R.drawable.strich);
    sound.setBackgroundDrawable(drawableStrich);
    }
    else {
    //sound.setBackgroundResource(0);
    sound.setBackgroundDrawable(null);
    }

    if(this.awesomePlayer.actualRhythm.tracks.get(playerTrackindex).sounds.get(playerSoundIndex).equals("1")){
         //sound.setImageResource(R.drawable.bob);
         sound.setImageBitmap(bitmapBob);
    }
    else{
        sound.setImageBitmap(bitmapnobob);
    }
    //sound.setImageDrawable(getRhythmDrawable(playerTrackindex, playerSoundIndex));
    //sound.setImageResource(getRhythmImageId(playerTrackindex, playerSoundIndex));
    sound.setId(playerSoundIndex);

    sound.setOnClickListener(new OnClickListener() {
        public void onClick(View v) {
            awesomePlayer.changeRhythmElement(trackIndexPlayer, soundIndexPlayer);
            if(awesomePlayer.actualRhythm.tracks.get(trackIndexViews-1).sounds.get(soundIndexPlayer).equals("1")){

                sound.setImageBitmap(bitmapBob);
                //sound.setImageDrawable(drawableBob);
                //sound.setImageResource(R.drawable.bop);
                sound.invalidate();
            }
            else {
                sound.setImageBitmap(bitmapnobob);
                //sound.setImageDrawable(drawablenobob);
                //sound.setImageResource(R.drawable.nobop);
                sound.invalidate();
            }
        }
    });
    System.gc();
    return sound;
}

You see, at first I used sound.setImageResource(R.drawable.bop) for setting the ImageView Content.
In the android doc (http://developer.android.com/reference/android/widget/ImageView.html) I found out, that this method decodes the resource in the UI thread and that it can cause latency
"Awesome" I thought :)
"I got the solution"
So I changed everything to drawables and most important:
I set the drawables in my OnCreate method, so that they already exist and just have to be connected to the imageViews. But nothing changed.
Then I changed everything into bitmap using
this.bitmapBob = BitmapFactory.decodeResource(getResources(), R.drawable.bop);
this.bitmapnobob = BitmapFactory.decodeResource(getResources(), R.drawable.nobop);

in my OnCreate method. But nothing changed.
I have multiple track with around 100-200 ImageViews in each, but I'm too new in android to know why it's lagging.
The latency to change an image of an ImageView is about 2-3 seconds.
Am I using Bitmap the wrong way?
Are the tons of ImageViews the problem?
Can anyone help me?

A: Your basic problem is that Android doesn't really like putting multiple image views on the screen at the same (more or less) time. Especially when you're trying to make it work for 200 of them.
To solve your problem, please tell me why did you choose to use multiple imageViews over drawing just one?
I know this is kinda late answer, hope you've already solved it. If so, please share your solution.

redpajama_set_name: RedPajamaStackExchange


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For the unsuspecting shopper, the go-to area upon touchdown in Edinburgh will often be Princes Street: the capital's mile-long stretch of identikit high street emporiums. But as any fashion savvy local knows, there are plenty of incredible independent shops and one-of-a-kind boutiques to choose from, several of which are stocked with, and run by, Edinburgh's most talented designers, craft-makers and fashionistas – fresh from the illustrious Edinburgh College of Art. (Alumni include the likes of Christopher Kane, Jonathan Saunders and Holly Fulton.) So, where to look"¦?
Edinburgh's indies are in rude health, and for those looking for something a little special, Broughton Street is a great place to start the search. Life Story, run by Fee and Adam Storey, is home to all manner of hidden treasures, including Edinburgh College of Art graduate Fee's own handiwork. Set in an uber-contemporary minimalist space, Life Story sells everything from printed textiles to simple but effective hand-dyed tops. Ceramics, prints, bags and shoes from home and abroad also feature, as well as some striking artwork on sale by local artist Paul Muzni.
Nearby, the longer established Concrete Wardrobe has been a bit of a local institution since it opened in 2000, and sells some fantastic homeware pieces alongside a great array of purses and one-off dresses from Scottish based and trained designers.
Just a couple of doors away, local designer Joey D uses his Edinburgh studio to imagine some pretty natty recycled, bespoke fashion pieces. We're partial to his tweedy, military-inspired handbags, as well as the crazy shoe lounge out back. Clued-up fashion fiends might recognise his pieces from their outings on the Dressed to Kilt catwalk in New York.
Across town, sartorial savvy abounds around the West Port and Grassmarket area, playing host to some of the most dynamic young designers in the city. Tartan tourist tat gets short shrift here in a sea of brightly coloured boutiques and hip designer-led collections. Working down a cobbled Victoria Street and into the Grassmarket make time for Totty Rocks, set up by two Edinburgh College of Art fashion course lecturers, namely Lynsey Blackburn and Holly Mitchell. The pair can often be found behind the counter putting the last stitch or button onto one of their handmade creations. Just next door, The Red Door Gallery offers up works by local artists and illustrators, as well as crafters and jewellers (the handmade brooches, in particular, putting any ordinary high street knock-off to shame).
Head over to West Port and you'll find our favourite: Godiva, where a cornucopia of skirts and frocks by local designers co-exist with a backroom of vintage goodies. It's all about one-offs here and local in-house designer Rowan McIntosh's playful, feminine style can be made to measure. Last but certainly not least, make sure to stop off at Herman Brown on your fashion pilgrimage, also on West Port, whose vintage and retro fashions are a must.
Virgin Atlantic Little Red is our new UK domestic service, offering great value, full service multiple daily flights between London and Manchester, London and Aberdeen and London and Edinburgh. Little Red flights can be booked now, bringing you closer to all our destinations.
Know of any other must-visit Edinburgh boutiques? Share your favourites in the comments below.
Anna is a Glasgow born and Edinburgh based freelance writer and editor specialising in arts and travel. When she's not exploring the Highlands and Islands or reviewing Scotland's festival scene, she's likely to be found propping up the bar at one of New York's finest watering holes or exploring Europe's untapped corners.


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redpajama_set_name: RedPajamaCommonCrawl


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TOTTENHAM take on Manchester City in the Premier League on Monday - but who will start for Spurs?
Mauricio Pochettino's men are on a four-game winning streak in the Premier League.
Their latest victory came away at London rivals West Ham, with Erik Lamela's 44th-minute strike enough to see off the Hammers.
They suffered another setback in the Champions League on Wednesday, however, by drawing 2-2 away at PSV Eindhoven.
But it is back to league duty on Monday night when the champions come to town.
City are still unbeaten this season, taking 23 points from their first nine outings.
Spurs currently trail them by two points in the table after winning seven and losing two of their league games so far.
And they will be without Vincent Janssen and Jan Vertonghen on Monday, while Dele Alli and Danny Rose are both doubts to feature.
Hugo Lloris should start in goal.
In defence; Kieran Trippier, Toby Alderweireld, Davinson Sanchez and Ben Davies are likely to get the nod.
Eric Dier will be expected to partner Mousa Dembele in a two-man central midfield.
Pochettino should then start Erik Lamela, Christian Eriksen and Lucas Moura in the three attacking midfield slots in front of the pair.
And Harry Kane is bound to be spearheading the Spurs frontline.


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Professor of World Religions at ACT, Greek ambassador of Globethics, head of the environmental campaign at the Center for Ecumenical, Missiological and Environmental Studies (CEMES), a frontman in the indie-rock band "The Gang". Nikos Dimitriadis is a genuine performer - a "rock star" as the Huffington Post wrote.
He gives concerts for thousands of people, promotes environmental justice among young people worldwide through his music project called "Smile", and at the same time keeps his energy and spirit up for his students at ACT.
Globethics, Religion, Environmental Consciousness. And rock music. What is the common ground between these fields?
You can find common ground in themes that, at first sight, look different and then start building a relationship on them. This is how I live my life. For me it's so natural to live with all those mentioned and even more in my everyday practice. First, they all have to do with ethics, conducts and performances.
Take for example, rock music. Rock is not a genre; is an attitude. Playing my music is an exhilarating experience. I have exactly the same feeling when I am in class teaching ethics. Singing and teaching are both performances. You give your best in front of an audience.
What are the updates and the results of your world tour? How do people respond to this a music project that carries a message?
To start with, it was a great experience. This world is so colorful and diverse and that's its beauty. There are so many identities! People think in different ways, act differently but there is common ground that unite us all. I think this is a common, yet unwritten code of ethics. In order to understand someone's perspective you should first understand their context. A world tour helps you realize this. The Smile World Tour does not convey a common message to all. It's more of a suggestion to look things from a different perspective. I would say that "SMILE" it's not a theory. It's more of an action. It enhances you to perform, which brings us to performance again.
What has this experience offered you on a personal level? Is it useful for your teaching?
A World Tour can definitely offer you a lot. As a Historian of Religions, I experienced and materialized many things I was reading all those years during my personal research. Through World Religions one can examine the religious beliefs of persons as they develop through time and space and as we comprehend them in a framework of cultural and psychological data.
A tour offers a unique approach to those data. Interacting with people of different cultural, religious or, I'd better say, spiritual beliefs and experiences, can bring you closer to the comprehension of humanity in general through the study of sentiments and attitudes. I think this experience added value also to my teaching. The students' comments in their evaluations of my teaching make a special reference to the fact that I use real life examples in class during my course on World Religions and this greatly facilitates their understanding of the religious traditions of several spiritual backgrounds we examine.
Can you share with us some of your methods and teaching approaches?
"Imagination is more important than knowledge" said Albert Einstein. I'm trying to light up students' imagination.
Books provide knowledge and the role of the educator is to open this window of knowledge and provide students with the tools to build their own worldview.
Something else I constantly keep in mind is that when a student fails it's not only a student's fault. Educators should be always alert to the fact that education and pedagogies are changing and they should try to find ways to make their courses more attractive to their students.
If I would use only one word to describe my method, I would use the word "interactive". In class we share experiences and real life examples and we try to understand ourselves, each other and the universality and multiplicity of the various traditions.
What do you enjoy most about ACT? How would you describe it?
Being in class! When I say "class" at ACT I don't only mean the curricular but also the co-curricular and extracurricular experiences. ACT not only provides but also encourage this way of "learning in action".
Thessaloniki's ACT is a crossroad of civilizations and more than any other institution in the region it offers this intercultural opportunity to our students.
If I would like to describe this teaching experience at ACT I would say that those classes are a journey of discovery both personal and of this world.
Through that journey we become more appreciative of the "experience of being human in the universe".


They say it takes more muscles to frown than to smile. This group of Mets fans is getting quite the workout.
Belle of the Ball: I don't know…Jesus? I mean he died for our sins, and there was nothing more sinful than this game. Shows real dedication to the team to die for an early season loss.
Smell of the Ball: Where do I begin? Oh, Ian Desmond. That's where. Yeah, this was a near full 25-man all around sucking effort, but Desmond's error on the first Mets batter made me realize I was about to sit through one of the worst experiences of my life. And I've seen Norbit.
…That was a lie. Nobody saw Norbit.
I love attending Nationals games. I hate attending extended sessions of grown men shitting all over the place and ruining my night. Tonight, thinking I was attending the former, I sadly attended the latter. It is an easy mistake to make though. If only they made the announcements on the NYC Subway sound clearer, maybe I would've correctly gotten off at Mets – Willets Point instead of Awful – Poop Central.


Since time began, people throughout the world have passed stories down from one generation to the next. Adirondack Storytellers, a new WMHT production, spotlights men and women who are not only keeping old regional stories alive, but adding new stories to the repertoire.
From Jim Bruchac �and his father Joseph, tradition bearers of the Abenaki, to Fran Yardley, who has presented her Adirondack-inspired stories at the National Storytelling Conference, numerous storytellers reveal the intricacies of their craft, weaving fairy tales, legends, anecdotes and tall tales from the experiences of women and men among the Abenaki, French-Canadians, lumberjacks and mountaineers.
First broadcast on Tuesday, Nov. 27, 2001, 8:45 pm - 10:30 pm.


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Four Thoughts on Nick Caserio's First Meeting With Houston Media
New COVID Vaccine Strategies for Texas as We're "Right in the Middle of the Storm"
Penelope Cruz, Daniel Day-Lewis and Amy Adams: New On DVD Today
Olivia Flores Alvarez
Olivia Flores Alvarez | May 4, 2010 | 4:31pm
Yep, that's Penelope Cruz looking like a high-class stripper. It's a scene from the just-released DVD Nine.
Cruz is just one of seven women that surround Nine's central character, Daniel Day-Lewis as frustrated film director Guido Contini. Just days away from starting to shoot his next film, he's in the middle of a personal and professional crisis. As he tries to pull himself together, he reflects on all the women in his life.
There's bombshell Penelope Cruz as his mistress, Marion Cotillard as his wife and Fergie as a whore from his youth. There's also Dame Judi Dench as the costumer for his yet-to-be-shot film, Kate Hudson as an American journalist writing about him and Nicole Kidman as his star and muse. Finally, there's Sophia Loren as his mother.
All of them have played a major part of Guido's past, which he must sort out in order to move into his future.
A few of the women are surprisingly well-suited to the musical format. Cruz and Hudson pull off especially sexy dance numbers and surprise viewers with their singing voices. Kidman excels as the blonde princess and Loren is, well, brilliantly Loren. (There's a reason she's a legend. Besides her considerable acting chops, she's 75 years old and still one of the most beautiful women in the world. One look from those famous almond-shaped eyes and the younger women pale in comparison -- no offense, Penelope.)
From the splashy, big-budget film packed with stars, we turn our attention to Paper Covers Rock.
Made for just $6,000 and shot in ten days, Paper stars newcomer Jeannine Kaspar as Sam, a troubled young mom.
Okay, we've done a splashy musical, a serious low-budget indie, now it's time for the romantic comedy Leap Year.
Amy Adams, the perky redhead seen in Wedding Crashers, Doubt and Julie & Julia, is Anna, an American businesswoman who follows her boyfriend to Ireland to take advantage of the old world custom that allows young women to propose to their beloved on February 29 during Leap Year (rather than wait around for their boyfriends to take the hint and spring for an engagement ring, we guess).
Each of Leap Year's characters is right out of the standard romantic comedy arsenal. Amy is uptight and determined to get her man. She, of course, gets waysided by Irish charmer Declan (handsomely played by Matthew Goode) who is ready and willing to step in for her jughead boyfriend. Circumstances keep throwing the two together and eventually into each other's arms.
There are no real surprises in Leap Year and fans of the cinema (you snobs, you) will find this a bit of innocuous fluff but Amy Adams fans will want to tuck this one away for a rainy afternoon. (We suggest you add lots of chocolate and keep some tissues handy.)
Rockets Will Have a Long, Slow Road to Improvement
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Betting Odds on Next Houston Texans Head Coach
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Kohn to Receive 2019 Henry W. Bloch Human Relations Award
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Herbert M. Kohn
hmkohn@bclplaw.com hmkohn@bclplaw.com
BCLP Senior Counsel Herb Kohn has been selected to receive the 2019 Henry W. Bloch Human Relations Award in Kansas City. Kohn will be presented with the award Nov. 24, at the 43rd Annual Human Relations Dinner sponsored by the Jewish Community Relations Bureau/American Jewish Committee (JCRB/AJC).
For 75 years, the Jewish Community Relations Bureau/AJC has worked to fight anti-Semitism, racism and bigotry, and to build bridges of understanding throughout the Greater Kansas City community. This event has a long tradition of bringing the community together to celebrate the honoree and recognize the importance of seeking justice, avoiding discrimination and promoting interfaith relations.
After a harrowing early childhood on the run from Nazis across Europe, Herb arrived in Kansas City as a third grader with his parents. Today, he's well-known in the KC business community, where he practices corporate, business and banking law at BCLP, concentrating in the areas of acquisitions, financial transactions, banking and public finance. Kohn was chosen "Best Rainmaker in Missouri" in 2009 and 2011 by Missouri Lawyers Weekly.
His civic roles have included serving as chairman of Kansas City, Mo., Mayor Kay Barnes' transition team in 1999 and chairman of the mayor's Charter Review Commission in 2000. He serves on the Executive Committee of the Board of Trustees for the Truman Presidential Museum & Library and is the past chairman of the Board of Trustees for the Kansas City Art Institute. Kohn served as director of Missouri Boys State, as chairman of the Missouri Gaming Commission and chairman of the Missouri Cancer Commission.
"JCRB|AJC is proud to present Herb Kohn with the Henry W. Bloch Human Relations Award, and to add his name to the list of previous award honorees, to whom we owe our great respect and appreciation, and who have set an example for our agency's current and future endeavors," JCRB|AJC Executive Director Gavriela Geller noted in The Kansas City Jewish Chronicle.
This document provides a general summary and is for information/educational purposes only. It is not intended to be comprehensive, nor does it constitute legal advice. Specific legal advice should always be sought before taking or refraining from taking any action.


Top panel: 3–24 keV image of the field around the detected serendipitous source. The green cross marks the position of the blazar LBQS 1319+0039 from Petrov (2011). Bottom panel: NuSTAR X-ray spectrum of LBQS 1319+0039. Black and red crosses denote the measurements from the two NuSTAR Focal Plane Modules A and B, respectively. The corresponding black and red lines show the best-fit models to the X-ray spectrum, which includes a cross-calibration constant and a power-law component. Credit: Privon et al., 2018.
An international team of astronomers reports the detection of the blazar LBQS 1319+0039 in hard X-rays using NASA's Nuclear Spectroscopic Telescope Array (NuSTAR) space telescope. The finding, updating knowledge about this object, is available in a paper published September 26 on the arXiv pre-print repository.
In general, blazars are perceived by astronomers as high-energy engines serving as natural laboratories to study particle acceleration, relativistic plasma processes, magnetic field dynamics and black hole physics. Therefore, observations of blazars in different wavelengths could be essential for improving the understanding of these phenomena.
Based on their optical emission properties, astronomers divide blazars into two classes: flat-spectrum radio quasars (FSRQs) which feature prominent and broad optical emission lines, and BL Lacertae objects (BL Lacs), which do not. LBQS 1319+0039, at a redshift of 1.62, was initially classified as an FSRQ. However, new observations show that this object also exhibits emission in hard X-rays.
The finding was made by a group of researchers led by George Privon of the University of Florida in Gainesville, Florida. The scientists were observing the spiral galaxy NGC 5104 with NuSTAR, and during these observations, they unexpectedly observed LBQS 1319+0039 in X-rays.
"We report a serendipitous hard X-ray (3–24 keV) detection of the blazar LBQS 1319+0039. This detection was obtained during a 20.7 ks Nuclear Spectroscopic Telescope Array (NuSTAR; Harrison et al. 2013) observation of NGC 5104," the astronomers wrote in the paper.
The researchers noted that the hard X-ray emission was identified at a position consistent with that of the blazar LBQS 1319+0039, during the analysis of NuSTAR observations. The analysis of the collected data was conducted with the use of NuSTAR Data Analysis Software (NUSTARDAS) v1.8 within Heasoft v6.2.
"We modeled the X-ray spectrum with a simple power-law model, considering also a constant to account for possible cross-calibration between the two focal plane modules on board NuSTAR," the researchers wrote.
As a result of the analysis, the astronomers found that in the case of LBQS 1319+0039, the X-ray spectrum is consistent with powerlaw emission with a photon index of 1.72, while the cross-calibration constant is 0.93. The observed 2–10 keV luminosity was calculated to be approximately 3.4 quattuordecillion erg/s.
The authors of the paper concluded that their results suggest that X-ray emission in LBQS 1319+0039 is typical for hard X-ray blazars. They noted that the observed photon index is consistent with the median value (1.68) found for the sample of hard X-ray blazars studied by the Burst Alert Telescope (BAT) onboard NASA's Swift space observatory.


Today, 4-year-old Eli stopped by with his mom, grandparents and baby sister to drop off food that he purchased with his own allowance. With his mom matching his allowance, Eli was able to help pick out $12 worth of food to donate.
While touring the Second Harvest Heartland warehouse in Maplewood, Eli was amazed by the large amounts of food and told us, "I wish for a refrigerator full of macaroni and cheese!" This hunger-fighter even said he wanted to work at Second Harvest Heartland when he grows up.
We are thankful for support that comes in all sizes. Eli and his family are a shining example of how you can make an impact on hunger at any age. Thank you Eli for the generous support!


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Q: Passing a value from a selectfield in a view to a store How do i correctly pass the value of my selectfield into the ajax proxy in my store?
Ext.define('FirstApp.view.Home',{
extend:'Ext.Panel',
xtype:'home',
config:{
    title:'Home',
    iconCls:'home',
    html:'
Home Page

Welcome to Sencha Touch 2 Training
',
    layout:'fit',
    scrollable:true,
    styleHtmlContent:true,
    styleHtmlCls:'home',

    items: [
            {
                xtype: 'selectfield',
                id: 'visit',
                label: 'Choose one',
                value:'restaurant',
                options: [
                    {text: 'Museum',  value: 'museum'},
                    {text: 'Pubs', value: 'pub'},
                    {text: 'Attractions',  value: 'attraction'}
                ]
            }
        ]
                }

    })

I am trying to place the value here: '+ REFERENCE HERE +' in the code below. I have tried '+#visit+' and '+#value+' with no success
Ext.define('FirstApp.store.Places',{
extend:'Ext.data.Store',

config:{

    autoLoad:true,
    model:'FirstApp.model.Place',
    proxy:{
        type:'ajax',
        url:'https://maps.googleapis.com/maps/api/place/search/json?  location=52.247983,-7.141113&radius=10000&types=food&name='+ REFERENCE HERE +'&sensor=false&key=KEY',
        reader:{
            type:'json',
            rootProperty:'results'
        }
    }
}
})

A: To get a component by ID in Sencha frameworks you want to use Ext.getCmp, and then use the getValue method. So where you are trying to put the reference, put this:
Ext.getCmp('visit').getValue()

Ideally you want to try to stay away from hardcoded component IDs, and instead use refs/selectors based on the component path, type, etc.


Hi, I was wondering if any of the above treatments would help with hyperpigmentation left by acne and also skin which has darkened due to the sun?
Reply Hi, The best advice that i can give to you is that you can come in for a consultation and allow a therapist to analyse your skin, as they would be able to determine which treatment would be most suitable for you.
We have recently launched a new range, called the Oxypeel. It consists of a mask being applied onto the skin for 45 minutes. It targets acne, scarring and hyperpigmentation. With this product we have seen evidence of very good and faster results.
The best thing would be to come in for a consultation, where we can answer any further questions.


From spring to autumn is "Castellers" season and you can join every Friday night. In the courtyard of a mansion in Vilafranca training takes place from 22.00 – 0,00h. This Tradition is so fascinating, you shouldn´t miss it! It's just 15 minutes from your B&B.
Some interesting links to approach that ancient tradition of building human towers. Arianella B&B will inform you on any special events like competitions or fiesta mayores (Main festival of the towns, where you can be part of this fascinating Catalonia tradition.


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Cushman & Wakefield Forms Strategic Alliance with Service-Disabled-Veteran-Owned Real Estate Services Provider
NEW YORK--(BUSINESS WIRE)-- Cushman & Wakefield (NYSE: CWK) recently announced the formation of a strategic alliance with DHC Real Estate Services, LLC, a service-disabled-veteran-owned small business (SDVOSB) founded by Jonathan Luttwak, a former U.S. Army captain and New York-based commercial real estate professional.
The alliance will enable public- and private-sector enterprises to increase their commitment to supplier diversity, and to meet state and federal mandates, while also maintaining access to the global, multi-disciplinary platform of services offered by Cushman & Wakefield.
"This alliance reaffirms our commitment to the military and veteran community, which is a core component of our corporate culture," said Frank Stanley, Director of Veterans Programs at Cushman & Wakefield. "It also allows us to provide an outstanding solution for our clients who are looking to bolster their commitment to supplier diversity — particularly those looking to partner with a veteran-owned firm."
DHC Real Estate Services was founded in 2018 by Luttwak, a combat veteran who started his commercial real estate career in brokerage services at Cushman & Wakefield. A graduate of West Point, Luttwak served as a U.S. Army officer, reaching the grade of captain. He was deployed to Iraq as part of Operation Iraqi Freedom in 2003 and also supported operation Enduring Freedom in Kuwait following 9/11. He subsequently earned his MBA from Columbia University.
"I founded my business with a belief that real estate is at the heart of everything we do as a nation," Luttwak said. "It is where we innovate, collaborate, live, work, govern and educate. This alliance enables me to provide value to clients while also empowering them with the full depth and breadth of Cushman & Wakefield's global platform of services."
DHC Real Estate Services has been certified as a SDVOSB by the state of New York and the U.S. Department of Veterans Affairs Center for Verification and Evaluation. Services provided through the firm's alliance with Cushman & Wakefield include real estate leasing, acquisitions/dispositions, strategic consulting, property/facility management, valuation and project management.
Additionally, the Military Times recently announced that Cushman & Wakefield has been included on the publication's annual "Best for Vets" list of 132 for profit, non-profit and governmental employers across the U.S. The firm was also selected as one of 198 companies across all industries to be named as a Top Veteran-Friendly Company by U.S. Veterans Magazine.
Grace Wilk
Global Marketing & Communications Associate
grace.wilk@cushwake.com


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Wijnaldum Talks Netflix As Match Prep
Today in "Celebs: they're just like us..."
By rbhasker@mostlyinane Mar 31, 2017, 11:30am BST
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Photo by Clive Mason/Getty Images
Without too much fanfare, Gini Wijnaldum has grown into one of the best midfielders on the squad, someone Klopp has referred to as being a "good combination of an aggressive, defensive midfielder plus a creative, offensive midfield player."
With skills like that, an absolutely perfect smile, and everything he's doing for younger players, it's no wonder Gini has become a firm fan favourite. The fact that he enjoys a Netflix marathon just as much as the next person is really just the icing on a pretty much perfect cake.
"Personally I watch TV shows on my laptop, or movies," he told the club's official website when asked about what he does the night before a game. "I've been watching Narcos a lot in the last few months. I also watch Breaking Bad, and I like Entourage and have watched it all. There's also Power, Empire… there are a lot of things I watch.
"Empire is the one I've been watching the most recently. What I love about watching TV series is that you can watch it for a long time if there are a lot of seasons. 'Binge watching' – I've learned that is what it's called! It's easier when you just want to chill out."
But is he really completely tuned in to whatever he's watching, or is he focused on the game tomorrow? Luckily, Gini answers that burning question for us.
"I never think about what might happen in the game the next day," he said. "I can't control what might happen, I can control what is going to happen in the moment, so if I can control it I will try to. I am not really a person who thinks, 'Oh, this might happen… that might happen'. I try not to look too far forward and stay in the moment. I hope to score, I hope to play well, but I don't think 'I am going to score'.
Seems like a wise approach to me.