text:
\section{Supplemental Material}

\subsection{Deformed wave function and its norm in the loop basis}

In the main text, we have the orthogonal local quantum states $|1\rangle $
and $|\tau \rangle $ on the original lattice and the orthogonal local
quantum states $|\hat{1}\rangle $ and $|\hat{\tau}\rangle $ on the dual
lattice. They are related by the following transformation
\begin{equation}
\left[
\begin{array}{c}
|\hat{1}\rangle \\
|\hat{\tau}\rangle%
\end{array}%
\right] =F\left[
\begin{array}{c}
|1\rangle \\
|\tau \rangle%
\end{array}%
\right] =\frac{1}{\phi }\left[
\begin{array}{cc}
1 & \sqrt{\phi } \\
\sqrt{\phi } & 1%
\end{array}%
\right] \left[
\begin{array}{c}
|1\rangle \\
|\tau \rangle%
\end{array}%
\right] .  \label{transformation}
\end{equation}%
It has been proved that the the quantum net wave function takes the same
form when rewritten on the dual lattice in the representation of $|\hat{1}%
\rangle $ and $|\hat{\tau}\rangle $, exhibiting the quantum self-duality.
This fact can be understood from the completely packed quantum loop model,
from which the Fendley's quantum net originates. The basis for the quantum
loop model is given by the completely packed loop configuration $\mathcal{L}$
consisting of two non-orthogonal states $|\tau \rangle $ and $|\hat{\tau}%
\rangle $, see Fig. \ref{CPL3} (a). Their overlaps are determined as
\begin{equation}
\left[
\begin{array}{cc}
\langle \tau |\tau \rangle & \langle \tau |\hat{\tau}\rangle \\
\langle \hat{\tau}|\tau \rangle & \langle \hat{\tau}|\bar{\tau}\rangle%
\end{array}%
\right] =\left[
\begin{array}{cc}
1 & \lambda \\
\lambda & 1%
\end{array}%
\right] .  \label{overlap}
\end{equation}%
According to Eq.(\ref{transformation}), the value of $\lambda $ is
determined to $-1/\phi $. So the loop configurations $|\mathcal{L}\rangle $ are
non-orthogonal. In the following, we will show that the value of $\lambda $
can be changed by deforming the wave function. In the non-orthogonal loop
basis, the quantum net wave function can be expressed as:
\begin{equation}
|\Psi \rangle =\sum_{\mathcal{L}}\phi ^{n_{\mathcal{L}}}|\mathcal{L}\rangle ,
\label{quantum_loop}
\end{equation}%
where $n_{\mathcal{L}}$ is the number of loops in $\mathcal{L}$ and the
loop fugacity is $\phi $. Under the quantum duality where the $|\tau \rangle $
and $|\hat{\tau}\rangle $ are exchanged with each other, the loop lattice as
the medial lattice of the net lattice and dual net lattice is unchanged, so
it is obvious that the wave function (\ref{quantum_loop}) is invariant.
Therefore, when Eq. (\ref{quantum_loop}) is expressed in the original
lattice on the basis $|1\rangle $ and $|\tau \rangle $ and the dual square
lattice on the basis $|\hat{1}\rangle $ and $|\hat{\tau}\rangle $, the
wavefunction should have the same form.
\begin{figure}[tbp]
\centering
\includegraphics[width=12cm,trim=0 180 10 10,clip]{CPL3.pdf}
\caption{(a) A typical completely packed loop basis $|\mathcal{L}\rangle $ consists
of the local non-orthogonal states $|\protect\tau \rangle $ and $|\hat{%
\protect\tau}\rangle $, the left and right, up and down sides are connected
periodically such that all loops in the configuration are closed. The black
full lines are net lattice, the black dash lines are loop lattice, and the
blue thick lines are completely packed loop configuration. (b) A typical
configuration $(\mathcal{L},\mathcal{L}^{\prime })$ of two-coupled
completely packed loop model consists of the four local configurations. $x$,
$y$ and $z$ are local Boltzmann weights for these local configurations. The
blue full lines stand for the configuration $\mathcal{L}$ form ket, the the
red dash lines stand for the configuration $\mathcal{L}^{\prime }$ form bra.
}
\label{CPL3}
\end{figure}

Actually the norm of the deformed wave function can be expressed in the
non-orthogonal loop basis:
\begin{equation}
\mathcal{Z}=\langle \Psi (h,\hat{h})|\Psi (h,\hat{h})\rangle =\sum_{\mathcal{%
L},\mathcal{L}^{\prime }}\phi ^{n_{\mathcal{L}}+n_{\mathcal{L}^{\prime
}}}x^{n_{\tau \tau }}y^{n_{\hat{\tau}\hat{\tau}}}z^{n_{\tau \hat{\tau}}+n_{%
\hat{\tau}\tau }},  \label{loop_norm}
\end{equation}%
where $\mathcal{L}$ and $\mathcal{L}^{\prime }$ are the loop configurations
coming from ket and bra separately and the loop fugacity is the global part
of Boltzmann weights. $x$, $y$ and $z$ are the local Boltzmann weights for
four local configurations in Fig. \ref{CPL3} (b), while $n_{i,j}$ with $%
i,j=\tau ,\hat{\tau}$ is the total number of local configurations $(i,j)$ in
the global configuration $\left( \mathcal{L},\mathcal{L}^{\prime }\right) $,
see Fig.\ref{CPL3} (b). The values of $x,y$ and $z$ are determined by the
matrix elements of the squared deformation matrix $P^{2}$ in non-orthogonal
basis:
\begin{eqnarray}
x &=&\langle \tau |P^{2}(h,\hat{h})|\tau \rangle =\left( 1-h+\hat{h}\right)
^{2}-\frac{4\hat{h}}{\phi ^{2}}\left( 1-h\right) ,  \notag
\label{Boltzmann_weight} \\
y &=&\langle \hat{\tau}|P^{2}(h,\hat{h})|\hat{\tau}\rangle =\left( 1+h-\hat{h%
}\right) ^{2}-\frac{4h}{\phi ^{2}}\left( 1-\hat{h}\right) ,,  \notag \\
z &=&\langle \tau |P^{2}(h,\hat{h})|\hat{\tau}\rangle =\langle \hat{\tau}%
|P^{2}(h,\hat{h})|\tau \rangle =-\frac{1}{\phi }\left( 1-h-\hat{h}\right)
^{2}+\frac{4h\hat{h}}{\phi ^{2}}.
\end{eqnarray}%
Then the wave function norm can be regarded as the partition function of a
two-coupled loop model. On the self-dual line $h=\hat{h}$, we have
\begin{equation}
x=y=1-\frac{4h}{\phi ^{2}}\left( 1-h\right) ,\text{ }z=-\frac{1}{\phi }%
\left( 1-2h\right) ^{2}+\frac{4h^{2}}{\phi ^{2}},
\end{equation}%
and the partition function (\ref{loop_norm}) is simplified as
\begin{equation}
\mathcal{Z}=\left\langle \Psi \left( h,h\right) |\Psi \left( h,h\right)
\right\rangle \propto \sum_{\mathcal{L},\mathcal{L}^{\prime }}\phi ^{n_{%
\mathcal{L}}+n_{\mathcal{L}^{\prime }}}\lambda ^{n_{\tau \hat{\tau}}+n_{\hat{%
\tau}\tau }}.  \label{norm_loop}
\end{equation}%
with
\begin{equation}
\lambda =\frac{z}{x}=\frac{4h^{2}-\left( 2h-1\right) ^{2}\phi }{\left(
2h-1\right) ^{2}+\phi },  \label{parameter_transformation}
\end{equation}%
which is precisely the overlap between two non-orthogonal local states in
Eq.(\ref{overlap}). This fact can be easily understood from the local
configurations $(\tau ,\hat{\tau})$ and $(\hat{\tau},\tau )$ in Fig. \ref%
{CPL3} (b). So we find that, the effect of tuning the filter strength along
the self-dual line $h=\hat{h}$ is equivalent to changing the overlap $%
\langle \tau |\hat{\tau}\rangle =\langle \hat{\tau}|\tau \rangle $. The
wavefunction norm on the self-dual line in Eq. (\ref{norm_loop}) is nothing
but the self-dual coupled loop model. The
range of parameter on the self-dual line is limited to $h\in \left[ 0,%
\frac{\phi }{2}\right] $, corresponding to the range of $\lambda \in \left[
-1/\phi ,1\right] $.

In the loop representation, the decoupled point $D$ on the self-dual line is
obviously when $\lambda =1$ or $h=\frac{\phi }{2}$, and the partition
function is $\mathcal{Z}\propto \sum_{\mathcal{L},\mathcal{L}^{\prime }}\phi
^{n_{\mathcal{L}}+n_{\mathcal{L}^{\prime }}}$. At this point each loop
model is critical. However, the strongly coupled point $S$ is given by $%
\lambda =0$ corresponding to $h=\frac{\phi ^{2}}{2}\left( 1-\sqrt{\phi }%
\right) $ and $\mathcal{Z}\propto \sum_{\mathcal{L}}\phi ^{2n_{\mathcal{L}%
}} $. This loop model is non-critical because the fugacity $\phi ^{2}>2$. In
addition to these special points, it is pointed out that there is a new
critical point $C$ on the self-dual line. The transfer operator of the
self-dual coupled Potts model can be written in terms of a special $%
SO(4)_{3}$ Birman-Murakami-Wenzl algebra. By the
level-rank duality of the algebra, the position of the new critical point is
determined by $\lambda =-\sqrt{2}\sin \frac{\pi }{20}$ corresponding to
$h\approx 0.196952$.

\subsection{Tensor network representation of the Fendley's quantum net}

So far the quantum net wave function on a square lattice have not been written
in the tensor network representation, because the chromatic polynomial as a
non-local topological part of the ground state weight is difficult to deal with.
Here we would like to construct a triple-line tensor network state (TNS) for
the quantum net on a square lattice.

First of all, we start from the known TNS for the string-net model on a honeycomb
lattice. The physical degrees of freedom located at the edges has two
orthogonal quantum states labelled by $|1\rangle $ and $|\tau \rangle $, as
shown in Fig.\ref{nets} (a). The branching symbol $\delta _{ijk}$ stemming
from the non-Abelian fusion rule imposes the branching rule on the vertices.
It allows the following five configurations shown in Fig.\ref{nets} (a),
corresponding to the non-zero elements of $\delta _{ijk}$:
\begin{equation}
\delta _{111}=\delta _{1\tau \tau }=\delta _{\tau 1\tau }=\delta _{\tau \tau
1}=\delta _{\tau \tau \tau }=1.
\end{equation}%
Therefore, in the ground state wavefunction, $|\tau \rangle $ states form a
closed trivalent net $\mathcal{N}$ consisting of $\tau $ strings, a typical
net is displayed in Fig.\ref{nets} (b).
\begin{figure}[tbp]
\centering
\includegraphics[width=9cm,trim=0 10 0 0,clip]{string-net.pdf}
\caption{(a) Two orthogonal quantum states $|1\rangle$ and $|\protect\tau%
\rangle$ on an edge, and the five local configurations allowed by the branching rule
on a vertex of trivalent lattice. The circles represent the physical degrees
of freedom. (b) A typical net configuration on a honeycomb lattice, the $%
\protect\tau$ strings formed by $|\protect\tau\rangle$ states are closed and
allowed to branch. (c) Up panel: The physical indices $i,j,k$ of the local tensor
locate at the edges and auxiliary indices $t,s,u$ live in the plaquette of the
lattice. Lower panel: The triplet-line local tensor $T^{ijk}_{tsu}$ on the
vertex of the honeycomb lattice for the string-net ground state. The legs of the
tensor connected by lines with only one index for simplicity are locked by
delta function, and the legs pointed out side of the paper are physical indices.}
\label{nets}
\end{figure}

With the chromatic polynomial, the ground state wavefunction for the
Fibonacci string-net can be explicitly expressed as
\begin{equation}
|\Psi _{\text{SN}}\rangle =\sum_{\mathcal{N}}\phi ^{3t_{\mathcal{N}}/4}\chi
_{\hat{\mathcal{N}}}\left( \phi ^{2}\right) |\mathcal{N}\rangle ,
\label{string_net}
\end{equation}%
where $t_{\mathcal{N}}$ is the number of vertices in the net $\mathcal{N}$ and
three $\tau $ strings meet together. So $\phi ^{3/4}$ can be viewed as
the vertex fugacity. The chromatic polynomial is non-local, and it only
relates to the topological properties of the net $\mathcal{N}$. To represent
the non-local wavefunction locally, the auxiliary degrees of freedom have to
be introduced, and the wavefunction can be decomposed into a local
structure, i.e., the triplet-line tensor network states (TNS):
\begin{equation}
|\Psi _{\text{SN}}\rangle =\sum_{\{ijk\}}\text{tTr}\left[ \underset{\text{%
vertex}}{\bigotimes }T_{\alpha \beta \gamma }^{ijk}\right] |\cdots ijk\cdots
\rangle ,  \label{string-net TNS}
\end{equation}%
where $T_{\alpha \beta \gamma }^{ijk}=\left( a_{\alpha }a_{\beta }a_{\gamma
}\right) ^{1/6}\sqrt{v_{i}v_{j}v_{k}}G_{\alpha \beta \gamma }^{ijk}$ is the
local tensor for the string-net model, and "tTr" denotes the contraction
over all virtual indices $\alpha $, $\beta $, $\gamma $. $G$, $v$ and $a$
tensors are uniquely determined by the Fibonacci topological order, and $%
v_{i}=\sqrt{d_{i}}$, $a_{i}=d_{i}/\left( \sum_{i}d_{i}^{2}\right) $ with $%
d_{1}=1$, $d_{\tau }=\phi $ as the quantum dimensions. Moreover, $G_{\alpha
\beta \gamma }^{ijk}=\delta _{ij\gamma }\delta _{\alpha \beta \gamma }\delta
_{i\beta k}\delta _{j\alpha k}\frac{F_{\alpha \beta \gamma }^{ijk}}{%
v_{k}v_{\gamma }}$, and the $F$-symbol is given by:
\begin{equation}
F=\left(
\begin{array}{cc}
F_{\tau \tau 1}^{\tau \tau 1} & F_{\tau \tau \tau }^{\tau \tau 1} \\
F_{\tau \tau 1}^{\tau \tau \tau } & F_{\tau \tau \tau }^{\tau \tau \tau }%
\end{array}%
\right) =\frac{1}{\phi }\left(
\begin{array}{cc}
1 & \sqrt{\phi } \\
\sqrt{\phi } & -1%
\end{array}%
\right) .  \label{fusion_matrix}
\end{equation}%
The graphical representation of the $T$ tensor is shown in Fig.\ref{nets}(c).

In the main text, the wavefunction of quantum net is defined by
\begin{equation}
|\Psi \rangle =\sum_{\mathcal{N}}\phi ^{\left. -L_{\mathcal{N}}\right/
2}\chi _{\hat{\mathcal{N}}}\left( \phi ^{2}\right) |\mathcal{N}\rangle ,
\label{Fendley_wavefunction}
\end{equation}%
where $L_{\mathcal{N}}$ is total length of $\tau $ string in net $\mathcal{N}
$. Note that the difference between string-net wavefunctions in Eq. (\ref%
{string_net}) and quantum net in Eq. (\ref{Fendley_wavefunction}) is the local
parts of the ground state weights and the types of the lattice. The local parts
(string tension and vertex fugacity) are very easy to deal with, so the key
point is to represent the chromatic polynomials for nets on the square
lattice in terms of tensor networks. From Eqs. (\ref{string_net}) and (\ref%
{string-net TNS}), we notice the correspondence between the chromatic
polynomials for nets on the honeycomb lattice and the tensor networks is
\begin{equation}
\chi _{\hat{\mathcal{N}}_{\text{hyc}}}\left( \phi ^{2}\right)
\Leftrightarrow \text{tTr}\left( \underset{\text{vertex}}{\bigotimes }\tilde{%
T}_{\alpha \beta \gamma }^{ijk}\right) ,  \label{correspondence}
\end{equation}%
where \textquotedblleft hyc\textquotedblright\ stands for the honeycomb
lattice and $\tilde{T}_{tsu}^{ijk}=\phi ^{-\frac{3}{4}\delta _{i,j}\delta
_{j,k}\delta _{k,\tau }}T_{tsu}^{ijk}$. Since the ground states are
degenerate, two above expressions might not be equal.

Thanks to the equations among the chromatic polynomials shown in Fig. \ref{derive_tensor} (a), we can
obtain the relation between $\chi _{\hat{\mathcal{N}}_{\text{hyc}}}(Q)$ and $%
\chi _{\hat{\mathcal{N}}_{\text{sq}}}(Q)$. More specifically, we can
transform the square lattice into a honeycomb lattice by splitting all the
sites and adding new edges, as shown in Fig. \ref{derive_tensor}. There
exist two different ways of splitting a single site, but the results are
equivalent to each other. A new created edge can be either occupied by $\tau
$ string or unoccupied. For a net on the square lattice, it might include
the bivalent, trivalent and tetra-valent vertices. After the lattice
transformation, there is no ambiguity for both bivalent and trivalent
vertices, because the chromatic polynomials for open nets are zero. For a
net with one tetra-valent vertex, its chromatic polynomial becomes a sum of
two chromatic polynomials on the honeycomb lattice, as shown in Fig. \ref%
{derive_tensor} (b). If a net on the square lattice has $n$ tetra-valent
vertices, its chromatic polynomial is a sum over $2n$ chromatic polynomials
on the honeycomb lattice.
\begin{figure}[tbp]
\centering
\includegraphics[width=8.5 cm,,trim=0 0 0 0,clip]{derive_tensor_sm.pdf}
\caption{(a) The square lattice can be transformed to a honeycomb lattice by
splitting a lattice site into two sites and adding new edges (in the green ovals),
notice that on the new edges there is no physical degree of freedom. There
are two equivalent ways of doing this. The single $\protect\tau $ string on
the vertex of square lattice is transformed into a single $\protect\tau $
string on the vertices of the honeycomb lattice, displayed by the first three
equations. The fourth equation in the box shows that the tetra-valent vertex
on a square lattice where two $\protect\tau $ stings meet can be reduced to
the sum of two graphs on the honeycomb lattice. (b) The chromatic polynomial for
a square lattice net with a tetra-valent vertex (in the box with dash line)
is equal to the sum of two chromatic polynomials for the honeycomb lattice nets.
(c) Contracting two tensors $\tilde{T}$ along one bond direction of the string-net TNS,
we obtain the tensor $T_{sq}$ for Fendley's wavefunction. Because there two equivalent
ways of transforming the vertex of a square lattice into two vertices of the
honeycomb lattice, there are two kinds of $T_{sq}$ tensors. Notice that the
physical index between two $\tilde{T}$ tensors belonging to the new edges is summed.}
\label{derive_tensor}
\end{figure}

Conversely, for the wave function with the weight $\chi _{\hat{\mathcal{N}}_{%
\text{hyc}}}(\phi ^{2})$ on the honeycomb lattice, summing over physical
degrees of freedom on the new edges (NE) of the honeycomb lattice
yields the wave function weighted by $\chi _{\hat{\mathcal{N}}_{\text{sq}%
}}(\phi ^{2})$ on the square lattice:
\begin{equation}
\sum_{\mathcal{N}_{\text{sq}}}\chi _{\hat{\mathcal{N}}_{\text{sq}}}\left(
\phi ^{2}\right) |\mathcal{N}_{\text{sq}}\rangle =\left[ \underset{\text{NE}}%
{\bigotimes }\sum_{i=1,\tau }\langle i|\right] \left[ \sum_{\mathcal{N}_{%
\text{hyc}}}\chi _{\hat{\mathcal{N}}_{\text{hyc}}}\left( \phi ^{2}\right) |%
\mathcal{N}_{\text{hyc}}\rangle \right] .  \label{relation}
\end{equation}%
This is the relation between $\chi _{\hat{\mathcal{N}}_{\text{hyc}}}(Q)$ and
$\chi _{\hat{\mathcal{N}}_{\text{sq}}}(Q)$, which only involves some local
operations. So the connection between the local tensors for a string-net and
local tensors for a quantum net is clear. Although the TNS representation
and chromatic polynomial representation of the ground state might not be
equal, the two expressions can not be distinguished locally. The TNSs for a
string-net on the honeycomb lattice and the corresponding quantum net on the
square lattice obey the same relation:
\begin{equation}
\sum_{\{ijkl\cdots \}}\text{tTr}\left( \underset{\text{vertex}}{\bigotimes
}T_{sq}\right) |\ ijkl\cdots \rangle
=\left[ \underset{\text{ne}}{\bigotimes }\sum_{i=1}^{\tau }\langle i|%
\right] \left[ \sum_{\{ijk\cdots \}}\text{tTr}\left( \underset{\text{vertex}}%
{\bigotimes }\tilde{T}\right) |\cdots ijk\cdots \rangle \right] .
\end{equation}
In this TNS, the sum over physical degrees of freedom on new edges is
equivalent to that there is no physical degrees on these new edges. Hence, the
local tensor $T_{\text{sq}}$ for a square lattice quantum net is given by
combining two tensors $\tilde{T}$ along one bond direction of the honeycomb
lattice, as displayed in Fig. \ref{derive_tensor} (c). Taking the string
tension into account, the TNS for the quantum net on the square lattice is
thus given by
\begin{equation}
|\Psi \rangle =\sum_{\{ijkl\cdots \}}\text{tTr}\left( \underset{\text{vertex}%
}{\bigotimes }\phi ^{-\frac{i+j+k+l}{4}}T_{\text{sq}}\right) |ijkl\cdots
\rangle ,
\end{equation}%
where the string tension on each tensor reduces to $\phi ^{1/4}$.

\subsection{Transfer operator and matrix product operator}

In order to detect the mechanism of the quantum phase transition, we
analysis the fate of anyons with matrix product operator (MPO). The MPO acts
on virtual indices of the tensor networks, satisfying the pulling through
condition, so the invisible MPO can be moved freely. Although our TNS for the quantum net is
different from the string-net TNS, their MPOs are the same. The reason is
that the local tensor $T_{\text{sq}}$ comes from local tensor $T$ of the
string-net, we only do some local operations on the physical degrees of
freedom of the tensor, while the auxiliary degrees of freedom that MPO acts
on remain unchanged. This is also an evidence that the quantum net on the
square lattice and the string-net have the same topological order, because
the MPO algebra is a necessary condition for topologically ordered TNS.

The local tensor of MPO is also determined by $G$ and $v$ tensors. There are
two local tensors
\begin{equation}
B_{1,(jl),(im)}^{(jki),(lkm)}=G_{lm1}^{ijk}\sqrt{v_{i}v_{j}v_{l}v_{m}}\text{%
, \ }B_{\tau ,(jl),(im)}^{(jki),(lkm)}=G_{lm\tau }^{ijk}\sqrt{%
v_{i}v_{j}v_{l}v_{m}},
\end{equation}%
where the superscribes are the physical indices and the subscribes are
virtual indices, as displayed in the right panel of Fig. \ref%
{Transfer_Operator} (a). The MPOs generated by $B_{1}$ and $B_{\tau }$ are
given by
\begin{eqnarray}
O_{1} &=&\sum_{\{(jki),(lkm)\}}\text{tr}\left( \prod_{l}B_{1}^{\left(
j_{l}k_{l}i_{l}\right) ,\left( l_{1}k_{1}m_{1}\right) }\right)
|\{jki\}\rangle \langle \{lkm\}|,  \notag \\
O_{\tau } &=&\sum_{\{(jki),(lkm)\}}\text{tr}\left( \prod_{l}B_{\tau
}^{\left( j_{l}k_{l}i_{l}\right) ,\left( l_{1}k_{1}m_{1}\right) }\right)
|\{jki\}\rangle \langle \{lkm\}|.
\end{eqnarray}%
$O_{1}$ is trivial in the sense that it is equivalent to an identity
operator, while $O_{\tau }$ is non-trivial, because it has two eigenvalues $%
\phi $ and $-\phi ^{-1}$. Both $O_{1}$ and $O_{\tau }$ satisfy the Fibonacci
fusion rule. Notice that these two MPOs are unchanged for different values
of $h$ and $\hat{h}$. To investigate the mechanism of the phase transition,
we need to systematically classify the topological sectors of the tensor
networks with MPOs.
\begin{figure}[tbp]
\centering
\includegraphics[width=8.5cm, trim=0 0 0 0,clip]{Transfer_Operator_sm.pdf}
\caption{(a) A simplified graphic representation for the local tensor $T_{sq}
$ of the deformed wavefunction $|\Psi (h,\hat{h})\rangle $. (b) The local
tensor $B_{n}$ generating the MPO $O_{n}$, where $n=1,\protect\tau $, and
its simplified graphic representation. (c) The tensor networks for
wavefunction $|\Psi (h,\hat{h})\rangle $ and $|\Psi _{\protect\tau }(h,\hat{h})\rangle $ with MPO $O_{%
\protect\tau }$ (blue line) insertion in $x$ direction. (d) Three
inequivalent transfer operators: $\mathbb{T}$ without MPO tensor insertion; $%
\mathbb{T}_{\protect\tau }$ with MPO tensor $B_{\protect\tau }$ insertion in
ket layer and $\mathbb{T}_{\protect\tau }^{\protect\tau }$ with MPO tensor $%
B_{\protect\tau }$ insertion in both layers. }
\label{Transfer_Operator}
\end{figure}

On a torus, the ground state is degenerate. The other
wavefunction denoted by $|\Psi _{\tau }(h,\hat{h})\rangle $ can be obtained
by inserting MPO $O_{\tau }$ into the tensor network, as shown in Fig .\ref%
{Transfer_Operator} (b). The corresponding topological order can be
identified from the transfer operator $\mathbb{T}$ of the wavefunction norm
\begin{equation}
\langle \Psi (h,\hat{h})|\Psi (h,\hat{h})\rangle =\text{Tr}\left( \mathbb{T}%
^{L_{x}}\right) ,
\end{equation}%
where $L_{x}$ is the length of the lattice in the $x$-direction and the
circumference of the transfer operator $\mathbb{T}$ is denoted by $L_{y}$. Since
the wavefunction norm is regarded as two coupled $\phi ^{2}$ Potts models,
the effective dimension of the transfer operator $\mathbb{T}$ is thus given
by $\phi ^{4L_{y}}\times \phi ^{4L_{y}}$ for a large $L_{y}$. The
wavefunction norm thus has the two layers, one is the bra layer and the
other is the ket layer. Inserting the MPO $O_{\tau }$ in either bra layer or
ket layer, or both layers, results in the norms
\begin{eqnarray}
\langle \Psi (h,\hat{h})|\Psi _{\tau }(h,\hat{h})\rangle  &=&\text{Tr}\left[
(\mathbb{T}_{\tau })^{L_{x}}\right] ,  \notag \\
\langle \Psi _{\tau }(h,\hat{h})|\Psi _{\tau }(h,\hat{h})\rangle  &=&\text{Tr%
}\left[ (\mathbb{T}_{\tau }^{\tau })^{L_{x}}\right] .
\end{eqnarray}%
So there are three inequivalent transfer operators $\mathbb{T}$, $\mathbb{T}%
_{\tau }^{\tau }$, and $\mathbb{T}_{\tau }\mathbb{\ }$or $\mathbb{T}^{\tau }$%
, as shown in Fig. \ref{Transfer_Operator} (d), where the subscribe
(superscribe) stands for the MPO tensor $B_{\tau }$ insertion in the ket
(bra) layers.

The anyon sectors of the transfer operators can be classified for the bra
and ket layers individually, and they are denoted by $(\alpha |\beta )$,
where $\alpha ,\beta =1,\tau ,\bar{\tau},b$ are anyon sectors of the bra and
ket, separately. There are in total 16 sectors, in which the topological
sectors $(b|b)$, $(\tau |\tau )$ and $(\bar{\tau}|\bar{\tau})$ of the
transfer operator measuring the confinement of $b$, $\tau $ and $\bar{\tau}$
anyons and sectors $(1|b)$, $(1|\tau )$ and $(1|\bar{\tau})$ measuring the
condensation of $b$, $\tau $ and $\bar{\tau}$ anyons. The dominant
eigenvalues of these sectors determine the property of anyons in the
thermodynamic limit.

Next, we illustrate how to find dominant eigenvalues and the properties of anyons.
By analysis the central idempotent, the layer of a transfer operator
without MPO tensor $B_{\tau }$ insertion contains the anyon sectors $1$ and $b$,
while the layer with MPO tensor $B_{\tau }$ insertion contains the anyon sector $b$,
$\tau $ and $\bar{\tau}$. Therefore, the transfer operator $\mathbb{T}$ includes
three anyon sectors:
\begin{equation}
(1|1)\text{; }(1|b)\text{ \& }(b|1)\text{;\ }(b|b).  \label{sector}
\end{equation}%
$\mathbb{T}_{\tau }^{\tau }$ has the following anyon sectors:
\begin{eqnarray}
&&(b|b)\text{; }(b|\tau )\text{ \& }(b|\bar{\tau})\text{ \& }(\tau |b)\text{
\& }(\left. \bar{\tau}\right\vert b)\text{; }  \notag \\
&&(\tau |\bar{\tau})\text{ \& }(\left. \bar{\tau}\right\vert \tau )\text{, }%
(\tau |\tau )\text{ \& }(\bar{\tau}|\bar{\tau}).
\end{eqnarray}%
And $\mathbb{T}_{\tau }$ contains sectors:
\begin{equation}
(1|b)\text{, }(1|\tau )\text{ \& }(1\left\vert \bar{\tau}\right. )\text{, }%
(b|\tau )\text{ \& }(b|\bar{\tau})\text{, }(b|b).
\end{equation}%
Because of the exchanging $\tau $ and $\bar{\tau}$ under the time reversal
symmetry or the symmetry between bra and ket, some sectors are always
exactly degenerate, and they are denoted by the symbol \textquotedblleft
\&\textquotedblright. Moreover, $\mathbb{T}$ commutes with $O_{\tau
}\otimes O_{\tau }$, the eigenstates of $\mathbb{T}$ are also eigenstates of
$O_{\tau }\otimes O_{\tau }$ with eigenvalue $\phi ^{2}$, $-1$, $-1$, $%
1/\phi ^{2}$ for sectors $(1|1)$, $(1|b)$ \& $(b|1)$, $(b|b)$, separately.
According to the degeneracy of a given level and the associated transfer
operator, we can determine the sectors to which the level belongs.

With the dominant eigenvalues of anyon sectors we concern, we can analysis
the mechanism for quantum phase transition. For
convenience, a quantity is defined $E_{(\alpha |\beta )}=-\log (\frac{%
e_{(\alpha |\beta )}}{e_{(1|1)}})$, where $e_{(\alpha |\beta )}$ is the dominant
eigenvalue of the transfer operator in the sector ${(\alpha |\beta )}$.
So $E_{(1|\beta )}$ measures the condensation of $\beta $ anyon. As the
circumference of the transfer operator $L_{y}\rightarrow \infty $, the
anyons $\beta $ are condensed if $E_{(1|\beta )}\rightarrow 0$ \emph{%
exponentially} with $L_{y}$. On the other hand, $E_{(\alpha |\alpha )}$
measures the confinement of $\alpha $ anyons. As the circumference of the
transfer operator $L_{y}\rightarrow \infty $, the anyons are confined if $%
E_{(\alpha |\alpha )}>0$.

From the viewpoint of anyon condensation, the only
possibility of the Fibonacci topological phase transition is the
condensation of $b$ anyons with the confinement of $\tau $ and $\bar{\tau}$
anyons. We are curious about whether the phase transition along the
self-dual line is beyond the anyon condensation. In analysis below, we make
a reasonable assumption that in the gapped phases the splitting between
degenerate eigenvalues $E_{(\alpha |\beta )}$ below the gap of the transfer
operator spectrum is exponential small with the system size.
\begin{figure}[tbp]
\centering
\includegraphics[width=10cm,trim=0 80 30 50,clip]{subplot_anyon_condensation2_sm.pdf}
\caption{The values of $E_{(\alpha |\beta ) }$
for each anyon sectors $( \protect\alpha |\protect\beta ) $
along the self-dual line, obtained from the transfer operator with
circumference $L_{y}=6$. Inset: The spectrum of transfer operator $\mathbb{T}
$ with circumference $L_{y}=6$ along the $\protect\phi ^{4}$-state Potts
lines. The horizontal axis is labelled by angle of polar coordinate so that
the spectrum is symmetric about the self-dual point ($\protect\theta =0.25%
\protect\pi $). }
\label{TO_spectrum}
\end{figure}

In the left panel of Fig. \ref{TO_spectrum} we show the variation of $%
E_{(\alpha |\beta )}$ along the self-dual line. It implies that in the
topological phase, $E_{(b|b)}$, $E_{(\tau |\tau )}$ and $E_{(\bar{\tau}|\bar{%
\tau})}$ approach to zero exponentially, so $b$, $\tau $ and $\bar{\tau}$
anyons are not confined. $E_{(1|b)}$, $E_{(1|\tau )}$ and $E_{(1|\bar{\tau})}
$ are above the spectrum gap, so that the $b$, $\tau $ and $\bar{\tau}$ are
not condensed. On the first-order phase transition line, there is a much
longer correlation length and an approximate conformal invariance.
Thus it is not easy to distinguish the gap and
the degeneracy below the gap on the first-order transition line with such a
small system size. To resolve the problem, we plot the $\phi ^{4}$-state
Potts line in the insert of Fig. \ref{TO_spectrum}, where $\theta =0.25\pi $
corresponds to $h\simeq 0.279$ on the self-dual line. From the inset, the
dominant eigenvalues $E_{(1|b)}$, $E_{(b|1)}$ and $E_{(b|b)}$ and the
subdominant eigenvalues for the three topological sectors are degenerate
with $E_{(1|1)}$ in the thermodynamical limit due to the level crossings at
the first order transition line. The splitting between them is exponentially
small for a finite size system. We can conclude that $E_{(1|b)}\rightarrow 0$
exponentially with $L_{y}\rightarrow \infty $ because it is below the gap.
So along the self-dual line, it still exhibits the $b$ anyon condensation.
In addition, $E_{(b|b)}\rightarrow 0$ exponentially with $L_{y}\rightarrow
\infty $, indicating that $b$ anyons are not confined. Meanwhile,
$E_{(\tau|\tau)}, E_{(\bar{\tau}|\bar{\tau})}>0$ imply the $\tau$ and $\bar{\tau}$
anyons are confined.

\subsection{Finite-size spectrum of conformal field theory}

In the view point of statistical model, the point $D$ in the phase diagram
corresponds to the critical point of two decoupled $\phi ^{2}$-state Potts
models, and we can still calculate the transfer operator spectrum at the
point $D$ to check the validity of our TNS. The single $\phi ^{2}$-state
Potts model at the critical point is described by the second unitary minimal
conformal field theory (CFT) with the central charge $c=7/10$. This CFT is
also the first unitary minimal superconformal model, and it is the only
theory with both Virasoro and super-Virasoro minimal. The scaling dimensions
of primary fields of the CFT are listed in Tab. \ref{7/10 CFT}, and our
numerical results match with such a prediction. The total central charge of
two decoupled models is $7/5$.
\begin{table}[tbp]
\caption{Scaling dimensions of the primary fields in the 2nd unitary minimal
model with the central charge $c=\frac{7}{10}$. The third column is the
results extracted from the transfer operator with circumference $L_{y}=12$
numerically, and the spectrum of the transfer operator is rescaled such that the
scaling dimensions of the first descendents of identity field are 1. The
second column is the Neveu-Schwarz (NS) sector and Ramond (R) sector of the
theory.}%
\begin{tabular}{C{2cm}C{1cm}C{3cm}C{1.2cm}}
\toprule
  scaling dimension& sector & Numerical result&Error \\
\colrule
$0$ & NS &0 &0\\

$3/40$ &R&0.07139148&0.0036\\

$1/5$ & NS &  0.18874674&0.0113\\

 $7/8$ &R &0.84037430&0.0346\\
\botrule
\end{tabular}
\label{7/10 CFT}
\end{table}

At the critical point $A$ on the $\hat{h}$ axis, the critical theory is
described by the seventh unitary minimal CFT with the central charge $c=14/15
$, which belongs to another family of CFT. The corresponding central charge $%
c$ and scaling dimensions $\mathfrak{h}_{r,s}$ of primary fields are given by
\begin{eqnarray}
c &=&1-\frac{6}{m(m+1)},\text{ }m\geqslant 3,  \notag \\
\mathfrak{h}_{r,s} &=&\frac{[(m+1)r-ms]^{2}}{4m(m+1)},\text{ }1\leqslant r<m,%
\text{ }1\leqslant s\leqslant r.
\end{eqnarray}%
Our case corresponds to $m=9$ and the scaling dimensions of the primary
fields are listed in Tab. \ref{14/15CFT}. Notice that only primary fields
with $r=$ odd in the minimal model appear in our microscopic model. The critical theory of the point $B$ has the
same CFT due to the quantum duality.
\begin{table}[tbp]
\caption{Scaling dimensions $\mathfrak{h}_{r,s}+\bar{\mathfrak{h}}_{r,s}$ of
the primary fields in the seventh unitary minimal model with the central charge
$c=\frac{14}{15}$. The third and forth columns are the scaling dimensions
numerically extracted from the transfer operator of the quantum net with
circumference $L_{y}=10$ and their errors. The fifth and sixth columns are
the scaling dimensions numerically extracted from the transfer operator of the
string-net with a circumference $L_{y}=10$ and their errors. The spectrum of
transfer operator is rescaled such that the scaling dimension of the first
descendent of $\mathfrak{h}_{r,s}+\bar{\mathfrak{h}}_{r,s}=0$ is 1. The
absolute errors of data with a star are larger than 10\% compared with exact
scaling dimension. }
\label{14/15CFT}%
\begin{tabular}{C{1cm}C{2cm}C{3cm}C{1.5cm}C{2cm}C{2cm}}
\toprule
 $(r,s)$ &  $\mathfrak{h}_{r,s}+\bar{\mathfrak{h}}_{r,s}$ & Quantum net&Error&String-net&Error  \\
\colrule
 $(1,1)$ & $0$  &$ 0$&$0$&$0$&$0$ \\

 $(3,3)$ & $2/45$ &$0.04973837$& $0.0053$&$0.04369855$&$0.0007$\\

 $(5,5)$ & $2/15$  &$0.14089491$&$0.0076$ &$0.12985540$&$0.0035$ \\

  $(5,5)$ & $2/15$  &$0.14089491$& $0.0076$&$0.12985540$&$0.0035$ \\

 $(7,7)$  & $4/15$ & $0.23521085 $&$0.0315$&$0.25426793$&$0.0124$  \\

 $(2,1)$ & $2/3$   & $0.94338622^*$& $0.2767$&$0.72615165$&$0.0595$  \\

 $(4,3)$ & $14/15$  & $0.94457473$ &$0.0112$&$0.93859015$&$0.0053$  \\

 $(6,5)$  & $56/45$ &$0.95672807^*$&$0.2877$&$1.19624200$&$0.0482$\\

 $(6,5)$  & $56/45$ &$0.95672807^* $ &  $0.2877$&$1.19624200$&$0.0482$ \\

 $(8,7)$ & $8/5$ &$0.98182889^* $ & $0.6182$&$1.44124861$&$0.1588$ \\
\botrule
\end{tabular}
\end{table}

Along the $h$ axis, the effective dimension of the transfer operator $%
\mathbb{T}$ can be reduced from $\phi ^{4L_{y}}\times \phi ^{4L_{y}}$ to $%
(\phi +2)^{L_{y}}\times (\phi +2)^{L_{y}}$, and the number of degrees of
freedom coincides with the $(\phi +2)$-state Potts model. We also calculate
the transfer operator spectrum of the string-net model, which can be
exactly mapped to the $(\phi +2)$-state Potts model. From the comparison
between the transfer operator spectra of the string-net and quantum net, we
suspect that the larger errors for the scaling dimensions of some primary
fields seen Tab. \ref{14/15CFT} may be caused by the fact that the quantum net
model does not exactly correspond to the $(\phi+2)$-Potts model.

Finally, the tri-critical point $C$ in the phase diagram is characterized
by the coset CFT of $\frac{SU(2)_{3}\times
SU(2)_{3}}{SU(2)_{6}}$, which belongs to the series of models with
fractional superconformal symmetry. The central charge $c$ and conformal
dimensions $\mathfrak{h}_{p,q}$ of primary fields for the $\frac{SU(2)_{k}\times
SU(2)_{m}}{SU(2)_{k+m}}$ coset CFT are given by
\begin{eqnarray}
c &=&1-\frac{6m}{(k+2)(k+m+2)}+\frac{2(m-1)}{m+2}, \notag \\
\mathfrak{h}_{p,q} &=&\frac{[(k+m+2)p-(k+2)q]^{2}-k^{2}}{4m(k+2)(k+m+2)}+%
\frac{t (m-t)}{2m(m+2)},
\end{eqnarray}
where $t=(p-q)\bmod m$, $1\leqslant p\leqslant k+1$ and $1\leqslant q\leqslant k+1+m$. The
tricritical point $C$ corresponds to $m=k=3$, and the corresponding
finite-size spectrum has been included in the main text.

\end{widetext}

\end{document}

redpajama_set_name: RedPajamaArXiv


The Health and Social Care Board has reaffirmed that residents of statutory care homes will be able to remain in their existing home for as long as they wish and as long as their care needs can continue to be met. The board has now discussed and approved a report on the consultation exercise that it carried out between November and March.
This report sets out four criteria which trusts will use when making decisions about the role and function of the homes: quality of care; care trends; availability and accessibility of alternative services; and best use of public money.
Local commissioning groups (which are committees of the board) have been asked to assess the level of need for services for older people in each locality. This will cover the needs of older people, the availability and range of services to respond to those needs, any planned developments, and the capacity to meet current and emerging needs.


text:
Ghost in the Shell - Stand Alone Complex -Volume 05
$4.99 Was $9.98 You Save 50%!
is currently unavailable.
Director: Kenji Kamiyama
Actors: Akio , Atsuko Tanaka , K, Osamu Saka
The subplot about the Tachikoma robots developing an awareness of their existence falls by the wayside in two episodes focusing on Chief Aramaki's life outside Section 9. He visits an old flame in London and is drawn into a bizarre money-laundering scheme, then outwits a former co-worker's son who's become an assassin. In episode 20, the filmmakers finally return to the central storyline of the Laughing Man, which may be tied to J.D. Salinger's 1949 story of the same name. The cyber-criminal may also be linked to the Sunflower Society, a group that files class action lawsuits against large corporations, including one involving a cure for "cyberbrain sclerosis. The tangled multiple storylines detract from an otherwise engaging series with strong characters. The extras include interviews with the mecha designers, and the directors of photography and 3-D computer graphics. (Rated 13 and older: violence, grotesque imagery, alcohol and tobacco use).
Format size: Widescreen
Languages: English, Japanese

redpajama_set_name: RedPajamaCommonCrawl


Imagine being in a room with more than 1500 original pieces of art – all created and donated to support a fantastic organization. That's exactly what happened on an unseasonably warm February night at NYC's Bortolami Gallery, The Visual AIDS Benefit "Postcards From The Edge" held its fundraiser for the 21st year – and my second as a participating artist! The large gallery space was packed with artists scanning the walls in anticipation of finding their art. I loved watching that process of discovery and the child-like joy everyone had when they found it. "Here it is! I found it" echoes around the room, followed by calls out to the friends that were also there to "Come see!" Every artist looks for their work in the gallery, which does take a bit a patience – since there are nearly as many artists present as cards and everyone is vying for selfie space. Here's mine!
After you find your own work, the next step is wandering through the room – more slowly – taking time to look at all of the amazing pieces that were produced for this event: prints, paintings, sculptures, photographs, collages, carvings, calligraphy, and drawings all line the walls of the gallery.
Although I'm able to draw on my background as a visual artist and photographer in my work at Eventive, it's always great to find additional opportunities to challenge myself. It's also incredibly inspiring to see all of the new work. This year's event raised more than $107,000 and I'm thrilled to say that mine was one of the cards that sold. The call for entries usually goes out in early November, I'd encourage all who are interested to submit a card.
See you at the artist's preview party next year!
Postcards From the Edge offers a rare opportunity to acquire original, postcard-sized artwork from internationally renowned and emerging artists for only $85 each. Offered on a first-come, first-served basis, over 1500 works are exhibited anonymously, and the identity of the artist is revealed only after the work is purchased. With the playing field leveled, all participants can take home a piece by a famous artist, or one who's just making their debut in the art world. Visual AIDS utilizes art to fight AIDS by provoking dialogue, supporting HIV+ artists, and preserving a legacy, because AIDS is not over. Visual AIDS is committed to preserving and honoring the work of artists with HIV/AIDS and the artistic contributions of the AIDS movement. For additional information on Visual AIDS, please visit www.VisualAIDS.org.


text:
Michelangelo Paintings And Where To Find Them In Europe
August 22, 2018 10:14 am August 22, 2018
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As you may know, Michelangelo Buonarroti is one of humanity's greatest sculptors and painters. Everyone who knows a bit about art has heard of David or the Sistine Chapel's ceiling….
The Italian's immense influence on other artists and the world in general is something yet to be seen again. Even though he spent his life in his homeland, Italy, Michelangelo's works are now part of exhibitions around the world.
Thousands of tourists and art enthusiasts form queues in front of museums just to take a peek at his creations. His legacy is so powerful that witnessing his art is looking into the eyes of history and revolution. Just knowing that you're looking at something he once stared at is exciting to say the least.
Although he may be more famous for his sculptures, Michelangelo's paintings are something extraordinary as well.
Here's a list of some of them you can (and should!) find in museums across Europe!
The Sistine Chapel Ceiling, Vatican
When it comes to Michelangelo paintings, this group of frescoes is by far the most grandiose one. They cover the entire ceiling of Vatican's most important chapel and have a legendary status in the art world.
Michelangelo was the type of person who wouldn't settle for less when it came to his work. The Sistine Chapel is the perfect example. When the pope asked him to depict the 12 Apostles on the ceiling, he ended up painting over 300 figures. What was supposed to be a simple frescoe became one of the most complex paintings in history.
The centerpiece is The Creation of Adam, an iconic scene from the Book of Genesis. Wherever you're from, you must have seen it at least once in your lifetime. God and Adam almost touching fingers is not only familiar to art historians, but important for pop culture as well!
You will need a moment just to take it all in. All the faces, the shadows and details. It's hard to imagine someone standing in the same position for years until they've created perfection in their own eyes. But that's what truly makes a genius stick out from everyone else!
The Vatican is famous for having one of the most stunning art collections in the entire world. It is the only state whose cultural property is completely protected in case of war. Among many world class artists, Michelangelo's paintings and sculptures have a prestigious position.
They say all roads lead to Rome. If you find yourself there, make sure the road leads you to the Vatican as well. The Sistine Chapel and Vatican Museums are must see!
Doni Tondo, Uffizi Gallery, Florence
This Michelangelo painting is a perfect representation of his signature techniques and colors. Even though the painting hasn't been signed, there is no doubt that this is his work of art.
It is unique since this is the only saved panel painting by Michelangelo that has been completely finished. The theme is once again Biblical. It portrays the Holy Family – Mary, Joseph and Jesus.
This Michelangelo painting is so authentic and obviously his that experts have noted it even has many characteristics of sculptures. Buonarroti, primarily a sculptor, had successfully transferred his famous skills and style from marble to wood.
The name Doni comes from its owner, Agnolo Doni, who had it made to celebrate his marriage. Tondo refers to the circular form of the panel, which was very popular during the Renaissance.
The Doni Tondo can be found in the Gallerie degli Uffizi in Florence, Italy. The Uffizi has a rich collection of paintings and sculptures, especially works of famous Italian artists. Have in mind that there's an entire room on the 2nd floor dedicated to Michelangelo only. Grab a slice of pizza and stop by the Gallery later to enjoy some food for the soul! Get your timed entrance ticket here.
The Manchester Madonna, National Gallery, London
This unfinished painting was attributed to Michelangelo long after his death. This is a panel, an art form so rare with Buonarotti, that there are only a couple left.
Originally called The Virgin and Child with Saint John and Angels, it shows Virgin Mary in a heartbreaking scene. She's holding an open book in her hand and trying to keep it away from Jesus. Many have interpreted this as her trying to save him from reading about his pain and suffering in the future.
The name Manchester Madonna comes from the fact it caused a real frenzy when it first came to Manchester in 1857. The people were beyond excited about it and over a million of them saw it in only a few months.
Today, it can be seen in the UK capital's National Gallery. This magnificent art museum has an impressive collection of some of the most famous art ever. You can book a walking tour of the National Gallery and the British Museum here. Van Gogh's Sunflowers and Venus and Mars by Botticelli are only a couple you must see with your own eyes. Michelangelo is high up on that list as well!
The Last Judgement, Sistine Chapel, Vatican
We've come full circle and returned to the Vatican. Il Giudizio Universale covers the entire wall behind the altar and it took Buonarroti 9 years to finish!
Many believe that this Michelangelo painting is his greatest masterpiece ever. It was heavily inspired by Dante's Divine Comedy and it was just as revolutionary.
Traditionally, the Earth, heaven and hell were always shown as horizontal layers, completely different in characteristics. However, Michelangelo distanced himself from the usual depiction. In his painting, all three form a single large space and neither section is entirely spared from chaos.
The figure of Jesus Christ takes the very central position. He is looking down and, just by waving his hand, deciding the fate of the whole human race. Compared to him, everyone looks powerless.
Many believe Michelangelo incorporated his self-portrait in the frescoe. A morbid detail of St. Bartholomew holding his own skin in his hands might distract you from it. But, if you look closely, the face he's holding is very similar to Michelangelo's.
If you only had to see one Michelangelo painting, it might as well be this one!
Cities & Culture Culture Explore Europe Poland Travel Tips
Budget Travel Tips for Exploring Warsaw in 24 hours
October 9, 2015 6:51 pm October 9, 2015
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This is a guest post by Agness of eTramping. eTramping is famous for its travel guides 'How to travel for less than 25$ a day' and it was created by…
This is a guest post by Agness of eTramping. eTramping is famous for its travel guides 'How to travel for less than 25$ a day' and it was created by two Poles- Agness and Cez. Quickly it became one of the most popular travel blogs and the number one source for budget travel tips. You can also follow eTramping on Facebook and Twitter.
If you have never been to Warsaw, you don't know what you've missed. It is the largest city of Poland, located in the eastern part of the country, in Mazowsze, on the Vistula River. What is so special about Warsaw? – You may ask. First of all it is the scientific, cultural, political and economic center of Europe commonly known as the "phoenix city" as it has survived so many wars throughout its history. Secondly, you will be astonished here by historical museums, cathedrals, churches, its rich history and traditions as well as delicious Polish cuisine. In fact, nobody can resist the taste of pierogi (traditional Polish dumplings) or rosol (chicken soup). By the way, did you know that Warsaw is considered to be the most sophisticated city in Poland where you can experience the real local hospitality? Yes, it surely is, trust us! Here are some tips on exploring Warsaw in 24 hours;
Must-see spots in Warsaw in 24h
Warsaw is divided into two parts: the Old Town and the New Town and it can be simply explored in just 1 day for less than $21. Atmospheric alleys, squares and cafés create a very unique atmosphere, and these two historic squares become a stage for musical and theatrical performances and open-air galleries in the summer.
When in Warsaw, you should start exploring the city from the Old Town. Here is what you can see there:
The Royal Castle
It was built in the 15th century as the residence of the Dukes of Mazovia. With the transfer of the capital from Krakow, it became the seat of the king and the government. It was rebuilt several times when destroyed during the 2nd World War. Today, the segment of the clock tower (from which daily at. 11.15 bugle call is played) opens the way to the Old Town. Museum attractions include two original paintings by Rembrandt and works by Bernard Bellotto.
Column of King Zygmunt III Waza
That's the oldest and tallest secular monument in Warsaw, built in 1644. The monument is 22 meters high, and the figure of the king measures 275 centimeters. Held in the right hand sword symbolizes his bravery, and the cross on the left – a constant readiness to fight evil. According to legend, the King's sword downwards will herald the imminent defeat of the city.
Old Town Square
It was founded between 13th and 14th century and it is currently one of the most picturesque corners of the city and the main square of Warsaw. Visiting the Old Town Square will certainly be a highlight of the day. You will be surrounded by affordable cafés and restaurants as well as charming alleys.
The Monument of Warsaw Mermaid
This monument of mermaid has been a symbol of Warsaw from the beginning of the 20th century. The statue stands in the center of the Old Town and it is surrounded by a fountain, where guests can cool off on hot days.
Historical Museum of the City of Warsaw
The museum is housed in several buildings rebuilt after the war and includes three courtyards. The exhibition presents the history of the capital since the dawn of modern times. From Tuesday to Saturday at 12:00 you can watch a documentary film about Warsaw between 1939 and 1945. The video is available in Polish, English, Spanish, French and German language.
The New Town also has a lot to offer. This part of Warsaw was founded in the late 14th century and it functioned as a separate city to the 18th century with its own administration, the town hall and the church. Most of the baroque and neoclassical houses around the Market are a post-war reconstruction as they were destroyed during the Warsaw Uprising (1944).
Here are the highlights of the New Town:
Church of Holy Spirit
The first wooden church was built in the 14th century. It's one of the biggest and oldest churches in the city, definitely worth paying a visit if you are Catholic.
The New Town Square
It was built in the 15th century. There are plenty of historical churches and monasteries here so if you are a big fan of history and architecture, don't miss it out.
Top Warsaw's activities on the cheap
Walking across Royal Lazienki
Royal Lazienki is a place visited by many local and tourists daily. It's a Palace-Garden complex offering one of the most picturesque views in Warsaw. It's placed at Agrykoli Street, nearby the city center. Here you can go for a relaxing walk, have a picnic, jog or even meditate. Many locals go there to breathe some fresh air and have a morning coffee. Moreover, here are some peacocks, pigeons and squirrels you can take a few pictures of.
Get to the top of Palace of Culture and Science
The Palace of Culture and Science is one of the best places to see Warsaw from above. It's definitely the highest building not only in the city itself, but also in the whole country. Built between 1952 and 1955, it quickly became one of the main Warsaw's attractions. There is a free entrance but you should pay 15 PLN (only 10 PLN if you have a student ID card) for getting to the 30th floor of the Palace to see the scenery of Warsaw.
Where and What to Eat and Drink
Warsaw has a lot of cozy restaurants where not only traditional Polish dishes are served, but also Vietnamese, Chinese, Portugal or Italian. If your budget is really tight, you should eat from street vendors or visit milk bars.
Here is the list of cheapest and yummiest restaurants in Warsaw you may want to try:
Street Restaurant & Club
Location: al. Jana Pawła II 19, Wola – the place offers a great variety of dinner options such as gyros kitchen with potatoes and salad (30PLN), Polish dumplings stuffed with meat (19.50 PLN) or cherry pancakes (14 PLN).
Milk Bar "Zlota Kurka"
Location: Marszalkowska Street 55/73 – That's the cheapest milk bar in Warsaw. You can get a homemade soup (chicken, mushroom, vegetable) for only 1.30 PLN, a typical Polish dinner which are potatoes, salad and pork chop for 5 PLN.
Zapiexy Luxusowe
Location: Francuska Street – That's definitely the best place for having a typical Polish zapiekanka which is simply a halved baguette or bread topped mainly with mushrooms and cheese, also ham or other types of meat, and vegetables. Here you can get a massive zapiekanka for only 3-4 PLN.
Ways to Get In and Around on a budget
There are many different ways to get to Warsaw. If you're flying, you will probably arrive at the Frederic Chopin International Airport. It is located 10 km south west from Warsaw city centre. There are two buses no. 175 and 188 which go from the airport to the city centre. They leave every ten minutes. If you need a ride late at night catch the bus N32 and you'll get right to Warsaw Central Station (Dworzec Centralny) from where you can take another local bus going to the city center. Tickets for the bus cost a mere 2.80 PLN.
If you are somewhere nearby like Germany, Belgium, Czech Republic or the Netherlands, you can get to Warsaw by bus or a van. Eurolines and Polski Bus have the cheapest tickets when it comes to bus services (the prices start from $10) and there are plenty of van companies charging around 50/60 euro for a ride (Amsterdam – Warsaw costs 60 euro and Berlin – Warsaw costs 35 euro).
You can also reach Warsaw by train. There is the Berlin-Warsaw Express (commonly knows as the Berlin-Warszawa Express). It is operated by Deutsche Bahn (DB), the German national railway company and PKP, the Polish State Railways. It unites the revitalized city of Berlin in Germany with the city of Warsaw in Poland. The trains are modern with air-conditioned carriages and restaurant cars. One way ticket costs between 29 and 46 euro.
You can't really get around the area easily with a car as there is a heavy traffic in Warsaw. However, you can catch a tram or a metro at any time. The tickets vary from 2 zloty to 5 zloty per ticket (60 cents to$1.50).
If you are short of cash, you can go Couchsurfing, which is extremely common nowadays. Many Polish people want to meet foreigners this way and show them what Poland has to offer to its visitors. There are couchsurfing meetings in Warsaw every Wednesday, so you can meet other people and hang out with them if you want to. The meetings are held in Warsaw at 7PM local time at Lorelei, Widok Street no. 8.
Best budget accommodation options:
1) Hostel: Aga Hostel. Address: Topiel street no. 27. Contact: 694 372 321.
2) Camping: Camping 123. Address: Bitwa Warszawska street no.15/17. Contact: 228229121.
3) 5-Star Hotel: Warsaw Marriott Hotel Address: 79, Aleje Jerozolimskie 65. Contact: 22 630 63 06.
Do you have any tips to share on how to explore Warsaw in 24 hours? Feel free to comment below.
Cities & Culture Culture Explore Europe Slovenia
Underground Culture in Rog Factory Ljubljana
June 22, 2014 11:54 am June 22, 2014
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Soon after World War Two, Rog Factory Ljubljana was a place for manufacturing bicycles and typewriting machines until the early 90's when the production was closed. Since then the abandoned…
Soon after World War Two, Rog Factory Ljubljana was a place for manufacturing bicycles and typewriting machines until the early 90's when the production was closed. Since then the abandoned factory represents an alternative open space for users who developed cultural and artistic scene on site.
The first thing you see when entering the territory of ROG is a panel labeled "Staying at your own risk on the ROG factory area". After 15 years of falling apart in the center of Ljubljana this abandoned factory was occupied by alternative artists.
The city of Ljubljana plans to renovate this area on the basis of public-private partnership which would transform it into Contemporary Arts Centre with exhibition spaces, studios, artist in residence programmes, educational facilities but also with commercial spaces such as apartments, shops, restaurants, hotel and more. Of course, the main problem is that idea tends to be in favour of private commercial space which would make 80% of the total area.
ROG is a place where different people gather and where friendships are formed, mainly among people who work on joint projects. All that lead to creation of a new community which is known as "Rogovci".
Besides these joint projects, they even formed their own media. Through its newsletter ROG notifies the local community with the activities that happen there and through their radio they allow young voices to be heard.
The main motto of the community is based on "temporarity". "We do not plan things that can not be realized in less than three days" they say.
And while the future of ROG is now uncertain, this will still be a place where people of different professions gather and create in a space free of racism, hierarchy and discrimination.
Cities & Culture Croatia Culture Explore Europe Travel Tips
Zagreb's Museum of Broken Relationships
June 9, 2014 7:26 pm June 9, 2014
3 Comments on Zagreb's Museum of Broken Relationships
"The Museum of Broken Relationships encourages discussion and reflection not only on the fragility of human relationships but also on the political, social, and cultural circumstances surrounding the stories being…
"The Museum of Broken Relationships encourages discussion and reflection not only on the fragility of human relationships but also on the political, social, and cultural circumstances surrounding the stories being told. The museum respects the audience's capacity for understanding wider historical, social issues inherent to different cultures and identities and provides a catharsis for donors on a more personal level" the EMF's (European Museum Forum) judging panel noted this when awarding it as Europe's most innovative museum in 2011.
It all started as a traveling exhibition based on the idea of failed love relationships. Anyone can donate an object which represents their broken relationship along with a story and objects could be anything from a fluffy toy and music CD to something more extreme as an axe. You can also become a donor and send your exhibit. Donors come from all over the world, from neighboring Serbia to far Japan.
Stories are usually packed with humor, hatred and hurt. The exhibition was shown in cities like Amsterdam, Paris, Istanbul, Singapore, Belgrade, Berlin and Ljubljana.
"I was determined to marry her. After two months I called her. She told me that she was in love with another guy who lived near her home. No words came from my mouth, tears ran like a new river from my eyes, and I hung up. I never called her again. My heart was broken very badly, I cried a lot. I spent many sleepless nights. Still her memories are chasing me, still I am crying. She left me alone. Months passed. I moved to my national capital of Delhi. Still it hurts" says an Indian guy Prasanth who donated love notes and winter coat.
"Love is a violent recreational sport. Proceed at your own risk. Helmets, armor, and steel-toe boots are required by law."
– H.C. Paye
You move on and the next exhibit says "Given to me by an American "boyfriend" when I was 17 and inscribed "for _____ who charmed the savage wolf". I didn't know that he would hound my parents for years and would eventually have a sex change and steal their name for his new persona."
And here's an "Ex-Axe" story or to be more detailed a story of two lesbians and an axe being promoted to a therapy instrument. After being left for another woman by her partner, this woman from Berlin decided to chop her ex's furniture. "Two weeks after she left, she came back for the furniture. It was neatly arranged into small heaps and fragments of wood. She took the trash and left my apartment for good."
"Love is an invited deal for losing peace."
― Seema Gupta
A Linksys router with a message "We tried. Not compatible."
Not every museum will give you an emotional, thoughtful and funny experience like this one and that's the reason I totally recommend it if you are in Zagreb.
And there's something we shouldn't forget;
"There is always love, hope and life after a broken heart."
― Jelord Klinn Cabresos
The best example for this are Olinka Vištica and Dražen Grubišić who founded this museum after their own relationship broke up!
Which object from your past relationship would you leave in the museum?


text:
DU Home » Latest Threads » Forums & Groups » Main » General Discussion (Forum) » Salma Hayek gets knighted...
Thu Jan 5, 2012, 10:14 AM
Sarah Ibarruri (21,043 posts)
Salma Hayek gets knighted??? For WHAT? For marrying a mega-rich pal of right wingnut Sarkozy?
Clearly, knighting in France has just become one huge joke. With so many remarkable women in France, and this actress who has done nearly NOTHING gets KNIGHTED? No wonder the French are angry.
Actress Salma Hayek gains French knighthood
Film actress Salma Hayek has been awarded one of France's top honours by President Nicolas Sarkozy.
Mexican-born Hayek will become a Knight of the Legion d'Honneur at a ceremony in Paris later this year.
The 45-year-old's businessman husband, Francois-Henri Pinault, is an associate of Mr Sarkozy, while his father was also honoured in the New Year's list.
Some readers of French newspaper L'Express have said honours should be awarded to worthier recipients.
123 replies, 17524 views
123 replies Author Time Post
Salma Hayek gets knighted??? For WHAT? For marrying a mega-rich pal of right wingnut Sarkozy? (Original post)
Sarah Ibarruri Jan 2012 OP
WTF do you care? nt
Dreamer Tatum Jan 2012 #1
We already know what you think about most topics. All one has to do is do a search with your name
Sarah Ibarruri Jan 2012 #2
haha... good one
fascisthunter Jan 2012 #38
:) nt
Sarah Ibarruri Jan 2012 #41
Is there an objective and peer-reviewed list of things we're supposed to care about...?
LanternWaste Jan 2012 #55
Ah, well, she certainly looks decorative as well as being smart. n/t
Vogon_Glory Jan 2012 #3
Decorative, yup. Smart? For marrying a mega-rich dude who got her knighted? nt
She's Not A Stupid Woman
Vogon_Glory Jan 2012 #13
"Not being a stupid woman" is not a reason to knight someone. Knighting is for REMARKABLE women
A Mexican was knighted in France?
Iggo Jan 2012 #5
"Only in America," as Yogi Berra would say....
msanthrope Jan 2012 #8
Marrah_G Jan 2012 #54
Someone who has done almost nothing in life was knighted in France, instead of a woman who has
Not that I'm a huge fan of her
Dorian Gray Jan 2012 #103
You're talking to the wrong person. I'm Latin. Want a list of Latin women worthy of this award?
So? I'm Mexican-American. Big whoop.
Iggo Jan 2012 #37
Well I'm not. I'm for the title going to a LATIN WOMAN that has done MUCH MORE. But the 1% awards
I'm sorry. I guess I wasn't clear.
Sure, sure, right.
What the fuck are you talking about?
Really? Re-read your posts. nt
HappyMe Jan 2012 #6
Don't know what they use as standards, still I'd not say she's done nothing
Bluenorthwest Jan 2012 #7
There are countless AMAZING women in France.
Did I say otherwise? No, I did not. I said she has done many things, you say 'nearly
Bluenorthwest Jan 2012 #14
My point is that this is not a remarkable woman. There ARE MANY remarkable women on this planet
Money buys power, and titles.
liberalmuse Jan 2012 #9
Money does buy power and titles, and rights. nt
I think she's a "Dame"
Blue_Tires Jan 2012 #58
That might be only in England. The "Dame" thing, that is.
Buns_of_Fire Jan 2012 #87
how else could he get close to her without seeming obvious ???
madrchsod Jan 2012 #16
How else could Sarkozy get friendlier with that mega-millionaire? :) nt
sounds like something out of the Onion. wow, thanks.
inna Jan 2012 #17
It does! Unfortunately, it's true. More and more bs goes on. The world is sounding like The Onion
That's what I thought at first.
Catherina Jan 2012 #92
I'd love to know what your aunt said! Sarkozy trips over his own feet worshipping billionaires
Sarah Ibarruri Jan 2012 #108
Occulus Jan 2012 #120
Seems like an awful lot of effort on Sarkozy's part
Codeine Jan 2012 #18
lol Yes, and it's a waste of an award except for Sarkozy, who is now even tighter with her hubby
More evidence of the control wielded by the 1%.
snagglepuss Jan 2012 #23
It is contemptible. And you're right, the 1% does it again. nt
SaintPete Jan 2012 #25
There are PLENTY of Latin women who do that precise thing and deserve this award FAR MORE than
Sounds like the problem
No. The problem is that knighting is something the 1% do for the 1%, no others need apply. nt
OK. So what?
Thank for those facts.
JNelson6563 Jan 2012 #85
When France gets attacked by a dragon and Hayek armors up and kicks that dragon's ass...
Dr. Strange Jan 2012 #26
LOL!! Yeah. :) nt
because you know no one has ever been knighted for totally stupid reasons
Johonny Jan 2012 #30
Yeah I know. The 1% knight their own, and Salma Hayek became the 1% when she married that
somehow she couldn't possibly be in that 1% on her own
TorchTheWitch Jan 2012 #98
The fact is, her husband's father, a multi-billionaire, is going to be knighted as well. Her
Being named a Knight of the Legion d'Honneur is a minor honor
onenote Jan 2012 #32
You're right. Someone else mentioned that. The 1% knight their own, and not because they're worthy
The Puss and Boots actress is not in the same league as DeNIro. Many in France
Were they thrilled with Putin?
She is an activist and champian of human rights.
EC Jan 2012 #40
But we know that's not why she received it. Many French women and men who've done much more
It looks to me like she is using her good fortune
So that's why she turned me down.
bikebloke Jan 2012 #43
uh, have you seen the way that women looks. I would knight her in a
okieinpain Jan 2012 #44
I have a list for you to knight. :) In fact, turn on the TV or watch any movie. nt
I'm all for it....
DataException Jan 2012 #45
That does it. I vote for you to knight all the Barbie females on TV and movies! :) nt
Come on, be honest....who HASN'T had fantasies of Salma in a suit of armor wielding a longsword?
tjwash Jan 2012 #46
No, no it's not just you. n/t
A Simple Game Jan 2012 #49
Well, why not knight the frikkin' Kardashians, while we're at it? :) It'd be just as good. nt
There isn't a foundry in Europe worthy to make her breastplate, though
lots of celebrities get knighted
Motown_Johnny Jan 2012 #48
That's true. BUT RONALD REAGAN WAS KNIGHTED? For what? For being an a-h??? nt
dunno, I just copied the whole list
Scary if Raygun was knighted! nt
I like Salma Hayek
Cali_Democrat Jan 2012 #56
Performers are usually made a knight in the Order of Arts and Letters, there.
Norrin Radd Jan 2012 #60
Sure would've.
I thought knighting people went out with the revolution anyway.
dameocrat67 Jan 2012 #62
Apparently not. nt
I promise, I'll be outraged next week.
tabasco Jan 2012 #63
I'm going to make you outraged right now lol :) Read Salma's latest comment about her implants:
Looks like Jesus answered her prayers to me....lol (n/t)
EX500rider Jan 2012 #94
Dr. Jesus :) nt
Well Jesus works in mysterious ways! lol (n/t)
EX500rider Jan 2012 #112
Sure does! I wonder if Dr. Jesus has a website and offers financing for lipo. lol nt
I didn't see the interview in question
France no longer has a monarchy yet the titles still exist.
Dawson Leery Jan 2012 #64
Yup. This award was not even given to her. It was given BECAUSE she married a multi-billionaire.
And here is an article from New Zealand on the issue...
That's disgusting but so is everything about Sarkozy.
Yup. Sarkozy is right wing. They have no idea what meritorious means.
Imperial honors, titles of nobility and heredetary monarchies can fuck off.
Poll_Blind Jan 2012 #70
It is. However, we DO have nobility. Occupy calls it the 1%. nt
knights are different in France
Enrique Jan 2012 #73
LOL! Knights in France have implants. nt
You seem overly obsessed with this whole thing
Beaverhausen Jan 2012 #78
Proof positive that Repukes read DU (or infiltrate it). My original post made it on a wingnut site
This message was self-deleted by its author
Obamanaut Jan 2012 #80
Why shouldn't she be a French knight? She's already a Mayan Queen.
Monk06 Jan 2012 #82
She's friends to pythons .....
BOHICA12 Jan 2012 #91
Pythons need friends and she made Quentin Taratino suck beer off her toes.
I'm too outraged at stuff in THIS country to worry about who France knights.
Bake Jan 2012 #86
True. After the 3 decades of Repukes destroying our country, this is
Your dismissal would ring slightly more true had you not posted repeatedly to this thread ...
11 Bravo Jan 2012 #116
Well ... For Dusk til Dawn ... enough said nt
French Knights? Is no one going to say it?
FSogol Jan 2012 #93
You silly English Kniggetts.........
He's REPRESSING me! You saw it! n/t
eridani Jan 2012 #101
She has magical boobies that can feed the world.
aikoaiko Jan 2012 #97
That's easy for her to say....
FSogol Jan 2012 #102
Post removed
Post removed Jan 2012 #99
Post removed Jan 2012 #100
She deserves it for giving us 'Ugly Betty'.
roody Jan 2012 #111
for some perspective
cemaphonic Jan 2012 #114
Who the hell is that man??? nt
Good heavens! He looks utterly different! nt
look at it this way Sarah - if the aristocracy has respawned in France ....
marasinghe Jan 2012 #119
lol Too true. nt
This thread is awesome.
flvegan Jan 2012 #121
If she's an American she's in a lot of legal trouble.
Quartermass Jan 2012 #122
Response to Sarah Ibarruri (Original post)
Dreamer Tatum (10,925 posts)
1. WTF do you care? nt
Response to Dreamer Tatum (Reply #1)
2. We already know what you think about most topics. All one has to do is do a search with your name
You need not have responded, you know.
Response to Sarah Ibarruri (Reply #2)
fascisthunter (29,381 posts)
38. haha... good one
Response to fascisthunter (Reply #38)
Thu Jan 5, 2012, 12:19 PM
41. :) nt
LanternWaste (37,748 posts)
55. Is there an objective and peer-reviewed list of things we're supposed to care about...?
Is there an objective and peer-reviewed list of things we're supposed to care about and things we're not supposed to care about?
Or are you simply projecting your own lack of mindfulness of a topic onto others?
Vogon_Glory (8,638 posts)
3. Ah, well, she certainly looks decorative as well as being smart. n/t
Response to Vogon_Glory (Reply #3)
4. Decorative, yup. Smart? For marrying a mega-rich dude who got her knighted? nt
13. She's Not A Stupid Woman
Salma Hayek is not a stupid woman. From what I've read about and listened to, she's not an empty-brained airhead with poor thinking skills. Whether or not she actually DID anything worthy of being made a member of the Legion of Honor is another matter altogether.
Response to Vogon_Glory (Reply #13)
15. "Not being a stupid woman" is not a reason to knight someone. Knighting is for REMARKABLE women
Not an ordinary, non-stupid actress who has done a little charity work, but happened to marry a pal of Sarkozy.
Iggo (46,277 posts)
5. A Mexican was knighted in France?
I'm outraged!
Response to Iggo (Reply #5)
msanthrope (37,549 posts)
8. "Only in America," as Yogi Berra would say....
Seriously. That's what he said on being told that the mayor of Dublin was Jewish.
Response to msanthrope (Reply #8)
Marrah_G (28,581 posts)
12. Someone who has done almost nothing in life was knighted in France, instead of a woman who has
accomplished a great deal for humanity.
Response to Sarah Ibarruri (Reply #12)
Fri Jan 6, 2012, 07:38 AM
Dorian Gray (12,951 posts)
103. Not that I'm a huge fan of her
or think she deserved to be knighted, but she's not done "nothing in life" either. She's an advocate against domestic violence, has been a spokesperson for vaccine campaigns and for AIDS research. (Among other charitable endeavors.) Just saying.
http://www.looktothestars.org/celebrity/116-salma-hayek
She's probably been a bigger advocate for international causes than you'd realize.
28. You're talking to the wrong person. I'm Latin. Want a list of Latin women worthy of this award?
Salma Hayek??? Give me a break!
37. So? I'm Mexican-American. Big whoop.
A country not my own gave an honorary title of nobility to an actor. I'm okay with that.
Response to Iggo (Reply #37)
42. Well I'm not. I'm for the title going to a LATIN WOMAN that has done MUCH MORE. But the 1% awards
titles and awards to its own and not to the rest.
And btw, as an ASIDE, if you THINK there's no 1% in Latin America, you're sooooo mistaken. Latin countries are the prototype of poverty-stricken countries where the 1% sucks the lifeblood out of the rest.
61. I'm sorry. I guess I wasn't clear.
My "So? I'm Mexican-American. Big Whoop" comment was meant to be in direct reply to your "I'm Latin" comment.
In other words: "You're Latin? Big deal. I'm Mexican-American. So what? Doesn't change a thing."
For the record, I don't have an ethnic (or any other kind of) rooting interest in who gets awarded a French knighthood. That's not what I was saying. What I meant in my post that you responded to with "I'm Latin" was that Americans being pissed that a Mexican actor got a French knighthood seems more than a little silly. That's it. Sorry I wasn't clear.
As for your aside, I don't see where you get that I think there are no billionaires or poverty south of the American border. That's a huuuuuuge stretch to get to an incorrect conclusion.
71. Sure, sure, right.
You figured I was an anglo or a non-Latin and were going to hit me with the bigot thing, until my admission that I am Latin.
As for your assertion that Americans are angry that a Mexican got selected, my responses were spot on. This has nothing to do with Mexicans, and everything to do with the fact that someone got picked for this merely because she's married to a French billionaire who is a pal of Sarkozy, and no other reason. Excellent Latin women who have spent their entire lives working for all manner of causes would never have been considered. Why? Simple, because they simply were not married to the billionaire who is the REAL object of this award.
The world is reacting to this knighting of Salma Hayek with anger or as a huge joke.
76. What the fuck are you talking about?
You brought race into it. Not me.
My post was about nationality.
77. Really? Re-read your posts. nt
HappyMe (20,277 posts)
6. Oh the humanity!
Bluenorthwest (45,319 posts)
7. Don't know what they use as standards, still I'd not say she's done nothing
She has produced many excellent projects that otherwise might not exist, she's an Academy Award nominee as an actress, and she does some good work speaking for immigrant's rights, opposing violence against women and is an advocate for several other issues as well.
Not nearly nothing. A great deal. Sorry. I know, she's like a woman, and all pretty and stuff, so it is fun to say her large set of accomplishments are 'nearly nothing'.
Should she get knighted? Don't know. Has she done lots of things in the world? Yes she has indeed.
Not into the trashing of artists. She's an accomplished person. That's just how it is.
Response to Bluenorthwest (Reply #7)
11. There are countless AMAZING women in France.
This is a waste.
14. Did I say otherwise? No, I did not. I said she has done many things, you say 'nearly
nothing' and I do not agree that she's done nearly nothing. I strongly disagree with that statement. That's all I said. Also, what you said was not 'there are amazing women in France' you said this one woman has done nearly nothing. Which is just incorrect.
See I am taking issue with your statement, as it is inaccurate. Regarding Ms Hayek. More than one male, non French actor has gotten this award. This is the case for many years. So it is not new nor different to award an actor, nor a non French person.
Response to Bluenorthwest (Reply #14)
19. My point is that this is not a remarkable woman. There ARE MANY remarkable women on this planet
MANY! An endless number. This one is not one.
104. Yep
The diminishing of a woman because she's beautiful and an actress into a "she's done nothing" is a problem. A quick internet search shows that she's not done "nothing."
I have NO problem with questioning how she was knighted, but I don't like the dismissal of her because she's Salma Hayek, actress, either.
liberalmuse (18,581 posts)
9. Money buys power, and titles.
But what I want to know is, what do knights do these days? Do we now call her 'Sir Salma Hayek'? I'm pretty ignorant on the whole modern day knight thing.
Response to liberalmuse (Reply #9)
10. Money does buy power and titles, and rights. nt
Blue_Tires (55,445 posts)
58. I think she's a "Dame"
Response to Blue_Tires (Reply #58)
Buns_of_Fire (15,838 posts)
87. That might be only in England. The "Dame" thing, that is.
But that's as far as I'll carry it. Too many opportunities to make a misstep. Nope. Not gonna do it. Wouldn't be prudent.
madrchsod (58,162 posts)
16. how else could he get close to her without seeming obvious ???
Response to madrchsod (Reply #16)
22. How else could Sarkozy get friendlier with that mega-millionaire? :) nt
inna (8,809 posts)
17. sounds like something out of the Onion. wow, thanks.
Response to inna (Reply #17)
21. It does! Unfortunately, it's true. More and more bs goes on. The world is sounding like The Onion
Isn't it?
Catherina (35,568 posts)
92. That's what I thought at first.
Some relatives in France called me shortly after I reading. Of course I brought it up. They're still pissed.
If I wrote what my aunt said, a jury would hide my post.
Imagine someone giving the Medal of Freedom to the foreign wife of Bush Sr's best buddy and saying it was because she sat on a few charities.
Sarkozy just can't keep himself from cheapening the entire Presidency while he screws over the 99% in France.
Response to Catherina (Reply #92)
108. I'd love to know what your aunt said! Sarkozy trips over his own feet worshipping billionaires
He's just an ordinary right winger, so we shouldn't expect anything but the most crass behavior.
And so, from Sarkozy's point of view, it was either knight the multi-billionaire Frenchman's wife (Hayek), or kneel before the multi-billionaire and kiss his private parts.
He chose the former.
Fri Jan 6, 2012, 11:19 PM
Occulus (20,599 posts)
120. meta
"If I wrote what my aunt said, a jury would hide my post. "
I didn't think about it until you said that, but... isn't that a fairly major flaw in the system, that it would give us cause to completely omit another person's words to us?
Codeine (25,586 posts)
18. Seems like an awful lot of effort on Sarkozy's part
to get Salma Hayak to kneel in front him.
And he'd probably STILL be shorter than her!
Response to Codeine (Reply #18)
20. lol Yes, and it's a waste of an award except for Sarkozy, who is now even tighter with her hubby
the mega-millionaire. Money talks, doesn't it?
snagglepuss (12,704 posts)
23. More evidence of the control wielded by the 1%.
Contemptible that a minor celeb is awarded such a high honor.
Response to snagglepuss (Reply #23)
24. It is contemptible. And you're right, the 1% does it again. nt
SaintPete (533 posts)
25. Activism
Hayek's charitable work includes increasing awareness on violence against women and discrimination against immigrants.
On July 19, 2005, Hayek testified before the U.S. Senate Committee on the Judiciary supporting reauthorizing the Violence Against Women Act.
In February 2006, she donated $25,000 to a Coatzacoalcos, Mexico, shelter for battered women and another $50,000 to Monterrey based anti-domestic violence groups.
Hayek is a board member of V-Day, the charity founded by playwright Eve Ensler.
Since the birth of her daughter, Hayek has worked to help mothers in developing nations worldwide, teaming up with Pampers and UNICEF to help stop the spread of life-threatening maternal and neonatal tetanus. She is a global spokesperson for the Pampers/UNICEF partnership 1 Pack = 1 Vaccine to help raise awareness of the program.
http://en.wikipedia.org/wiki/Salma_Hayek
Does she deserve the knighting? Maybe, maybe not. Who really cares?
Does she deserve your insults and scorn? - not a chance...
Response to SaintPete (Reply #25)
29. There are PLENTY of Latin women who do that precise thing and deserve this award FAR MORE than
this woman. But, oh yeah, I forgot! They didn't marry a PAL of right wing Sarkozy.
35. Sounds like the problem
is yours, not hers.
36. No. The problem is that knighting is something the 1% do for the 1%, no others need apply. nt
39. OK. So what?
why are you upset about this? It affects you not, and is completely ceremonial.
If they want to have their little titles and honors, let them.
Or fight it.
But why let it make you angry?
JNelson6563 (28,151 posts)
85. Thank for those facts.
I thought I had read of her work for these causes in the article you posted. Thanks for posting that. I don't understand the scorn being heaped on her.
Dr. Strange (25,658 posts)
26. When France gets attacked by a dragon and Hayek armors up and kicks that dragon's ass...
I expect to hear some serious apologies from some of you people in this thread.
Response to Dr. Strange (Reply #26)
27. LOL!! Yeah. :) nt
105. Ha ha ha ha ha
Best post in thread.
Johonny (18,078 posts)
30. because you know no one has ever been knighted for totally stupid reasons
in European history before. Pointless titles given to the upper class aren't like exactly a new European tradition.
Response to Johonny (Reply #30)
31. Yeah I know. The 1% knight their own, and Salma Hayek became the 1% when she married that
pal of Sarkozy.
TorchTheWitch (11,065 posts)
98. somehow she couldn't possibly be in that 1% on her own
You know, because she's pretty and all, so she couldn't possibly have any accomplishments of her own, must be dumb as a brick and just couldn't possibly have earned a place in the 1% category through her own efforts without having to marry some dude (and of course, since she's pretty and all she just couldn't possibly have married the dude for any other reason than money because even though she's a well-known and accomplished actress she must be a broke gold-digging hussy).
Your frequent misogynist attacks on attractive women - or for that matter any woman you disagree with on something - is really getting tiresome. But what is so irksome is that for someone who purports to be so against misogyny some of the worst objectifying and dehumanizing of women I've seen on this forum has been coming from you yet you seem entirely oblivious to it.
As far as Salma Hayek receiving a knighthood is concerned, the only thing to be upset about it is that in this day and age there are still European countries that cling to the trappings of class divide and the ridiculous notion of nobility by still having a monarchy, titles for the modern gentry-by-birth and pass out knighthoods... all that's missing is the jousting and a few beheadings.
Response to TorchTheWitch (Reply #98)
106. +1
107. The fact is, her husband's father, a multi-billionaire, is going to be knighted as well. Her
husband's father and her husband are intimate friends with Sarkozy. This is not the first time Sarkozy the right wingnut does something like this. Right wingers never knew an award that couldn't advance their cronyist desires.
As for her accomplishments, ask me to name Latin women who have done a million times more than this woman, and dedicated their whole lives to it. This woman is getting an award from Sarkozy as a pat in the back to her husband, who is his pal.
That's what the worldwide objection to this is about.
onenote (39,021 posts)
32. Being named a Knight of the Legion d'Honneur is a minor honor
Good grief. Knight (or more correctly "Chevalier" is the lowest rung on the Legion d'Honneur scale. Its a ceremonial honor that gets bestowed on all sorts of people for all sorts of reasons, including thousands of foreigners. Among those getting the nod over the years: Vladmir Putin. Clint Eastwood. Barbra Streisand. Alan Greenspan.
Do you think that there was no one more "qualified" (based on whatever it is that you think consists of the qualifications for this yawn of an honor) than those particular "Knights"?
If you want to criticize the awarding of honorary "knighthoods" by the French to foreign actors, actresses, political leaders, I can understand. But to single out the award to one actress as somehow being that egregious compared to the others?
Response to onenote (Reply #32)
33. You're right. Someone else mentioned that. The 1% knight their own, and not because they're worthy
It's all a thing of the 1%, for the 1%.
34. The Puss and Boots actress is not in the same league as DeNIro. Many in France
aren't happy with this decision.
47. Were they thrilled with Putin?
EC (12,287 posts)
40. She is an activist and champian of human rights.
She promotes the arts and promotes education of the history of South America. Both her and Antonio Bandarras do a lot of work in human rights. I'm just guessing, but maybe that's why?
And on edit: For the work she has done for women's rights and her arts projects I admire her very much.
Oh and her work exposing the South American 1% in different movies etc. I think we should be giving her applause. She has done some really good work showing how the elites took over countries and used the countries riches for personnal gain...much like is happening here now. Even her movie Freda shows how liberals (socialists) were persecuted. I'm glad she made those movies and exposed who the bad guys really are.
Response to EC (Reply #40)
84. But we know that's not why she received it. Many French women and men who've done much more
weren't even considered.
Sarkozy's just giving his Gucci friend another decoration for the mantlepiece.
I have nothing against Hayek, except her ridiculous comment thanking God for the big boobs she prayed for as a child.
Because of your posts, I googled her charity work. I don't know how complete this is. It's good stuff but is it really enough to get another country's top award? If she weren't married to Pinault, there's no way she would have gotten it. If you read about the Pinaults, they're 1% exploiters. Her father-in-law, Sarkozy's pal, is the 74th wealthiest person in the world. http://www.forbes.com/lists/2006/10/FUBG.html
Don't you find that very unfair to poorer people, especially French, who don't have those kind of connections but spend their whole lives helping others either at home or abroad?
Charity biography
Mexican actress Salma Hayek is highly active in raising awareness about violence against women and discrimination against immigrants, and since the birth of her daughter last year, has added campaigns helping children to her list of charity works.
Her own charity, the Salma Hayek Foundation, which previously supported organizations giving aid to and raising awareness for battered women, has now branched out to work with disadvantaged children on the streets of Mexico, helping them to overcome problems such as drugs, violence or lack of education.
As a spokesperson for the Pampers 1 Pack = 1 Vaccine campaign, she helps raise awareness and funds which allow UNICEF to provide tetanus vaccines to pregnant women, protecting the women for up to 15 years, and their babies for several months after the birth.
She says of the program:
"I have long been a supporter of every child's right to health and happiness. As a new mother, I'm thrilled to help Pampers and UNICEF raise awareness of their initiative to provide tetanus vaccines to those at risk."
Hayek took part in the America: A Tribute to Heroes charity telethon for victims of 9/11.
Hayek breastfed a desperate mother's hungry baby during a fact-finding trip to Sierra Leone, Africa, in 2008.
Salma received the Cookie magazine "Smart Cookie" award in 2009 for her active participation in charity since entering the realm of motherhood.
Salma Hayek joins forces with HRW to stop violence towards women
Salma Hayek recently joined the Human Rights Watch (HRW) organization in hope of asking presidential candidates in Mexico to commit to the prevention and punishment of crimes against women. Hayek, HRW, and 65 other social groups in Mexico and U.S. got together to make this call an open letter that was sent to candidates running for office in July.Hayek believes that every Mexican deserves a president that will end all these ridiculous killings against women. Non-government organizations have stated that for the last 13 years, more than 400 women have been murdered or have "disappeared" in Ciudad Juarez, in the state of Chihuahua.
"It's outrageous that Mexican women must live afraid of the killers harassing them on the streets," Hayek stated. Jose Miguel Vivanco, HWR director in the Americas, stated that "despite some efforts done by the government to investigate these brutal crimes, women are still being kidnapped, tortured, and killed in Mexico."
http://www.soulfrito.net/site/index.php?option=com_content&task=section&id=7&Itemid=39
88. It looks to me like she is using her good fortune
to help others... What do you want? She's not forgetting or ignoring what is really going on in the streets of the countries around her and she is doing what she can. As far as the knighthood, I congratulate her and do not begrudge her. Yes, I'm sure there are others that deserve some recongnition and I'm sure they will be rewarded someday, by someone. So don't disparage one, because there are others.
bikebloke (5,258 posts)
43. So that's why she turned me down.
okieinpain (9,396 posts)
44. uh, have you seen the way that women looks. I would knight her in a
heartbeat.
Response to okieinpain (Reply #44)
52. I have a list for you to knight. :) In fact, turn on the TV or watch any movie. nt
DataException (38 posts)
45. I'm all for it....
She's hot!!!
Response to DataException (Reply #45)
53. That does it. I vote for you to knight all the Barbie females on TV and movies! :) nt
tjwash (8,219 posts)
46. Come on, be honest....who HASN'T had fantasies of Salma in a suit of armor wielding a longsword?
Or is that just me?
Response to tjwash (Reply #46)
A Simple Game (9,214 posts)
49. No, no it's not just you. n/t
51. Well, why not knight the frikkin' Kardashians, while we're at it? :) It'd be just as good. nt
59. There isn't a foundry in Europe worthy to make her breastplate, though
Motown_Johnny (22,308 posts)
48. lots of celebrities get knighted
Sean Connery, Elton John, Helen Mirren, Judi Dench, Rolf Harris, Henry Cooper, Bob Geldof, Paul McCartney, George Martin, Ian McKellan, Ralph Richardson, Christopher Lee, Roger Moore, Terry Wogan, Mick Jagger, Salman Rushdie, Bono, Nick Faldo, Terry Practchett, Alfred Hitchcock, Edmund Hillary, Barbara Cartland, Ben Kingsley, Kiri Te Kanawa, Alan Sugar, Bob Hope, George Bush, Ronald Reagan, Andre Previn, Bill Gates.
Read more: http://wiki.answers.com/Q/Which_celebrities_have_been_knighted#ixzz1ibiqGdON
Response to Motown_Johnny (Reply #48)
50. That's true. BUT RONALD REAGAN WAS KNIGHTED? For what? For being an a-h??? nt
57. dunno, I just copied the whole list
75. Scary if Raygun was knighted! nt
Cali_Democrat (30,439 posts)
56. I like Salma Hayek
I'm really happy for her
Norrin Radd (4,959 posts)
60. Performers are usually made a knight in the Order of Arts and Letters, there.
I wonder why they didn't give her that one, instead? Would've quelled a lot of angst.
Response to Norrin Radd (Reply #60)
65. Sure would've.
The French are angry at Sarkozy about this. On the other hand, if it gets Sarkozy's right wing behind out of office, it might have been worth it.
http://www.yousaytoo.com/french-anger-of-sarkozy-because-of-salma-hayek/1675555
dameocrat67 (475 posts)
62. I thought knighting people went out with the revolution anyway.
n/t
Response to dameocrat67 (Reply #62)
68. Apparently not. nt
tabasco (22,974 posts)
63. I promise, I'll be outraged next week.
Response to tabasco (Reply #63)
66. I'm going to make you outraged right now lol :) Read Salma's latest comment about her implants:
"FoxNews.com adds a sort of snarky postscript noting that Hayek last month told BBC gabshow host Graham Norton that as a child she prayed to Jesus for large breasts.
"I went to a church that had a saint that was supposed to do a lot of miracles," Hayek said on The Graham Norton Show.
"I put my hands in the holy water and went: Please Jesus give me some boobs."
http://www.philly.com/philly/entertainment/136593351.html
EX500rider (8,512 posts)
94. Looks like Jesus answered her prayers to me....lol (n/t)
Response to EX500rider (Reply #94)
109. Dr. Jesus :) nt
Response to Sarah Ibarruri (Reply #109)
112. Well Jesus works in mysterious ways! lol (n/t)
Response to EX500rider (Reply #112)
113. Sure does! I wonder if Dr. Jesus has a website and offers financing for lipo. lol nt
110. I didn't see the interview in question
but is it possible that was said tongue in cheek? It sounds like something I'd say tongue in cheek if asked on Graham Norton's comedic BBC talk show about my large tatas.
Dawson Leery (19,326 posts)
64. France no longer has a monarchy yet the titles still exist.
This award is given out to powerful people as a symbolic gesture of the alliance between powerful people.
Response to Dawson Leery (Reply #64)
67. Yup. This award was not even given to her. It was given BECAUSE she married a multi-billionaire.
Period, end of story.
69. And here is an article from New Zealand on the issue...
http://www.stuff.co.nz/entertainment/celebrities/6213879/Outrage-over-Hayeks-Knighthood
France's highest order of merit has been branded a joke after the Mexican-born actress Salma Hayek was made a Knight of the Legion of Honour.
News that the 45-year-old was to become a Chevalier among other actors and singers prompted one former minister to refuse the award, while others warned it would see Napoleon ''turning in his grave''.
President Nicolas Sarkozy has the last word on who receives the Legion of Honour, created by Napoleon Bonaparte in 1802 in his attempt to replace nobility with a republican meritocracy.
Hayek is married to the billionaire Francois-Henri Pinault, whose business interests include the Gucci fashion label and Christie's auction house. He is also a close friend of Mr Sarkozy's
81. That's disgusting but so is everything about Sarkozy.
Sarkozy has made this such a joke that
Former minister Henri Torre has turned down receiving the Legion of Honor this year after he learned that Hayek, among others, would also be given the honor. "Too many people have been nominated who do not deserve to be honored," he said. "They have made a mockery of this high distinction by naming any old people."
http://www.sheknows.com/entertainment/articles/851941/does-salma-hayek-deserve-be-knighted-in-france
83. Yup. Sarkozy is right wing. They have no idea what meritorious means.
Look at GW Bush, the poster boy for idiots and cronyism.
Poll_Blind (23,864 posts)
70. Imperial honors, titles of nobility and heredetary monarchies can fuck off.
I really love that in America you do not have to recognize a title of nobility. May seem like a small thing but it's very nice.
Response to Poll_Blind (Reply #70)
72. It is. However, we DO have nobility. Occupy calls it the 1%. nt
Enrique (27,461 posts)
73. knights are different in France
Response to Enrique (Reply #73)
74. LOL! Knights in France have implants. nt
Beaverhausen (24,278 posts)
78. You seem overly obsessed with this whole thing
really, why is this such a big deal to you?
Put some energy toward something more important...say...there are millions of children starving in the world.
79. Proof positive that Repukes read DU (or infiltrate it). My original post made it on a wingnut site
Someone emailed me the link. I burst out laughing when I saw it. It's actually an honor to be bashed by the 'pukes. I think it's a good guideline. If the 'pukes hate it, you know it's a noble and good thing.
Obamanaut This message was self-deleted by its author.
Monk06 (7,675 posts)
82. Why shouldn't she be a French knight? She's already a Mayan Queen.
&feature=results_video&playnext=1&list=PL29EDB12A64D38E12
Response to Monk06 (Reply #82)
BOHICA12 (471 posts)
91. She's friends to pythons .....
Response to BOHICA12 (Reply #91)
95. Pythons need friends and she made Quentin Taratino suck beer off her toes.
Wait a minute QT directed this. So we know where he stands regarding toe sucking and licking Julette Lewis' pussy, which is also a line from this early and quite demented Tarantino flick. On the other hand Tito and Tarantula got featured which is a plus.
Bake (21,977 posts)
86. I'm too outraged at stuff in THIS country to worry about who France knights.
There's only so much outrage I can produce.
Response to Bake (Reply #86)
89. True. After the 3 decades of Repukes destroying our country, this is
literally nothing.
11 Bravo (23,574 posts)
116. Your dismissal would ring slightly more true had you not posted repeatedly to this thread ...
which you fucking initiated.
90. Well ... For Dusk til Dawn ... enough said nt
FSogol (43,517 posts)
93. French Knights? Is no one going to say it?
"You don't frighten us, English pig dogs. Go and boil your bottoms, you sons of a silly person. I blow my nose at you, so-called "Arthur King," you and all your silly English K-nig-hts."
"I don't want to talk to you no more, you empty headed animal food trough wiper. I fart in your general direction. Your mother was a hamster and your father smelt of elderberries."
Response to FSogol (Reply #93)
96. You silly English Kniggetts.........
Go away or will taunt you a second time.
eridani (51,907 posts)
101. He's REPRESSING me! You saw it! n/t
aikoaiko (33,165 posts)
97. She has magical boobies that can feed the world.
On a more serious note, she did speak out well for breastfeeding.
Response to aikoaiko (Reply #97)
102. That's easy for her to say....
roody (10,840 posts)
111. She deserves it for giving us 'Ugly Betty'.
cemaphonic (4,138 posts)
114. for some perspective
You know who else has a knighthood? This guy:
Entertainers sometimes get meaningless honorary titles. Big deal.
Response to cemaphonic (Reply #114)
115. Who the hell is that man??? nt
117. Elton John
I mean, who else would be playing the piano while wearing a duck suit?
118. Good heavens! He looks utterly different! nt
marasinghe (1,253 posts)
119. look at it this way Sarah - if the aristocracy has respawned in France ....
Madame la Guillotine can't be far behind.
plus, i think it's okay her getting it, just for her work in "Desperado"; though Banderas is much more deserving.
having to genuflect -- or whatever asskissing she had to do for the ceremony -- to the Sarkozy piece of shit, is punishment enough.
if you think it isn't, just dwell on the fact that she has to have sex with that rich arrogant bastard she's married to.
Response to marasinghe (Reply #119)
Sat Jan 7, 2012, 12:01 AM
123. lol Too true. nt
flvegan (64,185 posts)
121. This thread is awesome.
Quartermass (457 posts)
122. If she's an American she's in a lot of legal trouble.
It's illegal for Americans to have a title of nobility.
She may have done nothing to deserve it, but so what? Why shouldn't she get knighted, other than for being an American? Is there any other good reason why she shouldn't?