text:
Tour Recap: Greenpoint Manufacturing & Design Center
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A dignified old brick structure in the long-time industrial stronghold of East Williamsburg, Brooklyn, tells an interesting—and hopeful—story about the transformation of urban manufacturing over the course of New York City's history. Built as a rope factory in the mid-19th century (and subsequently expanded a number of times), the building at 221 McKibbin Street came to house a furniture manufacturer in the latter half of the 20th. After that business moved its offices to Long Island and sent a hundred manufacturing jobs to Asia, the building was purchased by the Greenpoint Manufacturing and Design Center, a non-profit industrial developer best known for its flagship facility further north along the Newtown Creek, which renovated and re-opened the space in 2009. Since then, the GMDC has used its unique development model to revitalize the facility, sub-dividing the 72,000 square feet of usable space for a dozen smaller manufacturers. Today, once again, the site hosts just shy of a hundred industrial jobs (95, to be exact).
On June 20th, the GMDC team hosted a tour of their McKibbin Street building as part of Open House New York's (OHNY) and New York City Economic Development Corporation's (NYCEDC) Making it Here series. GMDC's CEO, Brian Coleman, started off by speaking with participants about how and why his organization has taken on the task of acting as a landlord for manufacturers in a city where less and less space is available for industrial activity, never mind affordable.
"If we didn't know that we'd be able to stay here," explained Twoseven Inc.'s Franco Götte, "we never would have invested so much in our facility." (Photo: Nicolas Lemery Nantel)
"There are three main differences in the GMDC model," Coleman explained. "The first is that our rents are 15-25% below market. The second is our lease terms: we have a minimum of five years, for new leases, with an option for five more, which creates real estate permanency. The third thing is that we are mission-driven; we're here because we care about these businesses. People have a hard time wrapping their head around why a non-profit would be helping for-profit companies. The reason that GMDC is in this is that those businesses create good jobs."
The average annual wage in GMDC's buildings (there are five scattered across northern Brooklyn) is $47,000/year, which is considerably higher than wages in other sectors where jobs are available for people without a four-year degree. This echoes what many of the experts that have been involved in the MIH series so far have said, time and again, about the importance of manufacturers in creating living wage jobs for working class families and, by extension, supporting stable working class neighborhoods. 92% of the people who work in GMDC buildings are New York City residents, according to Coleman, and 70% walk, bike, or take transit to work. (For more information and stats on GMDC, click here to download a copy of their latest Annual Report). While the challenge of housing affordability within the city is more frequently (and loudly) discussed, in terms of preserving New York's socioeconomic diversity, the importance of providing space for the kinds of businesses that create good-paying working class jobs can't be understated.
92% of the people who work in GMDC buildings are New York City residents. (Photo: Nicolas Lemery Nantel)
That's exactly where GMDC comes in. When renovations were just beginning on McKibbin Street, for example, a woman from the neighboring NYCHA development stopped by and spoke with Coleman about the building, and the recent changes in the neighborhood. East Williamsburg and neighboring Bushwick, which begins a just few blocks south of the GMDC building, went from a relatively unknown industrial corner of the city to a white-hot hub of Brooklyn's exploding arts and cultural scene in just a couple of years, as the development of Williamsburg-proper has pushed artists to move a few stops further down on the L-train. Changes along McKibbin Street have been particularly intense, as the GMDC building sits on the line where the area's Industrial Business Zone ends, and residential zoning begins, allowing as-of-right conversions of older industrial buildings into lofts for new residents.
"'Another condo?' she asked me, sounding kind of sad about it," Coleman said, relating the story. "And I said, 'Nope, it's still going to be manufacturing space.' She was thrilled. 'You mean my brother can get a job here?' I told her I couldn't promise he would get a job here, but that it was going to be a place where he could, potentially, depending on what the tenants needed. That's not an uncommon reaction. When we develop new buildings, we voluntarily go to the Community Boards to tell them what we're doing, and people generally welcome us with open arms because we're bringing jobs into the community, or keeping jobs in the community, which is not the norm these days."
On the day of the tour, workers at Twoseven Inc. could be seen fabricating pieces of a new window display bound for a Manhattan department store. (Photo: Nicolas Lemery Nantel)
Indeed, many of the businesses in the GMDC's McKibbin Street building take on seasonal and contract help depending on their workload; the number of jobs on-site goes up and down depending on the season, but there is plenty of room for more than a hundred people to work here at any given time. The largest tenant on-site is Twoseven Inc., a design/build firm that specializes in the creation of store window displays, retail interiors, and showrooms for high-profile fashion and cosmetics companies around the city. One of the co-founders, Franco Götte, led tour participants on a walk around the factory, explaining that it was a slow period, since a number of jobs had just shipped out. Even still, more than a dozen workers could be seen fabricating pieces of a new window display bound for a Manhattan department store.
OHNY tour participants explore the Twoseven Inc. factory floor. (Photo: Nicolas Lemery Nantel)
Twoseven signed its lease at McKibbin Street fairly soon after the building opened, at the height of the recession, in 2009. Franco noted that that the availability of adjacent space encouraged he and his partner to expand their factory space, allowing them to grow their business—all of which would have been unlikely had they not found their way to a GMDC-owned facility. "If we didn't know that we'd be able to stay here," he said, "we never would have invested so much in our facility."
Still, it is not common for businesses that locate within GMDC facilities to be in expansion mode, as Coleman explained it. "We're sort of the opposite of an incubator. The average age of one of our tenants is 16-17 years. GMDC buildings generally attract mature businesses that have trouble finding space elsewhere in the city where they know they can stay put."
The average age of a business in one of GMDC's buildings is 16-17 years. "We're sort of the opposite of an incubator," says CEO Brian Coleman.
Upstairs, tour participants had the chance to peek inside the Woodwrights, a woodworking shop run by Wyeth Hunnable, whom the GMDC team refers to as "Tenant #1," as he was the first to sign a lease at McKibbin Street. The high, airy space was mostly occupied by three large workstations where Wyeth was making custom wooden panels for an artist with a studio nearby. Many of the Woodwrights' clients are artists, and Wyeth's space reflects the adjacency of his business to the art world. Far from the sawdust-coated room one might imagine upon hearing the term "woodworking shop," the Woodwrights space is painted in bright colors, from the yellow and green loft structure that provides additional storage space, to the pastel mural stretching across the floor, from wall to wall.
The Woodwrights space in the McKibbin Street building reflects the adjacency of his business to the art world, with its brightly painted interior. (Photo: Nicolas Lemery Nantel)
The final stop on the tour was Alchemy Paintworks, a fine art finishing business that works on paint finishing projects for metal sculpture, as well as the repair and restoration of large scale works of art. "Most people don't realize that a lot of artists don't actually do all of the work by hand anymore," said Alchemy's James Terrani, in describing what the company does. "Artists are job creators, now, as well. They employ lots of other creative workers. In the US, artists come to New York; this is where most of the talent is, so this is where the work is, for our company.
Both Alchemy and the Woodwrights are part of a robust ecosystem of manufacturing businesses that play an integral role in supporting New York's world-renowned arts community. Like many of the niche manufacturers that pay a premium to locate and work within the five boroughs, their business models respond to unique market conditions created by New York's exceptionally dense, diverse urban fabric. "A lot of the tenants in our buildings have arts backgrounds," noted GMDC Senior Project Manager Cassandra Smith, near the end of the tour. "Many of their businesses exist because they were able to find commercial applications for their arts skills. We are the Greenpoint Manufacturing and Design Center, and our tenants tend to do some of both of those things."
"Most people don't realize that a lot of artists don't actually do all of the work by hand anymore," said Alchemy's James Terrani (center). (Photo: Nicolas Lemery Nantel)
Today, manufacturers in New York City tend to be smaller, more nimble, and more integrated with design, the arts, and other creative industries. Factories aren't necessarily just places where objects are made; they are places where new products are dreamed up, prototyped, and then manufactured, all within the same facility. If the GMDC offers any indication, there is a bright future for these types of hybrid manufacturing businesses, if the city is willing to make room for them: as of the tour date, GMDC's five buildings, which together contain almost 600,000 square feet of space, are 100% leased.
"We've heard the argument that, since the land our buildings are on is so valuable, we should sell it off and use the proceeds to develop new industrial properties farther out from the core," said Smith. "But people like our tenants want to live and work in New York City, and we think it's good economic policy to make space for manufacturing so that they're making their money here, and spending their money here. So we like to think we're standing on the right side of economic development policy."
"We like to think we're standing on the right side of economic development policy," explained GMDC's Cassandra Smith (left). (Photo: Nicolas Lemery Nantel)
OHNY and NYCEDC thank the Greenpoint Manufacturing and Design Center, Twoseven Inc., Alchemy Paintworks, and the Woodwrights for welcoming participants into their spaces for this tour.
July 1, 2014 Brooklyn, East Williamsburg, Greenpoint Manufacturing & Design Center, Retrofitting the City, Tour Recap 0
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Open House New York (OHNY) broadens public awareness by exposing diverse audiences to distinctive examples of architecture, engineering and design.
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\section{Introduction}
Consider  Anosov diffeomorphisms $f$ and $g$ of a compact smooth
manifold $M$ that are conjugate by a homeomorphism $h$:
$$
h\circ f=g\circ h.
$$
 It is
well known and easy to show that $h$ is in fact H\"older continuous.
When we say that the conjugacy is H\"older or that two
diffeomorphisms are H\"older conjugate we mean that the conjugacy
and its inverse are H\"older continuous.

It is natural to ask the following converse question.
\begin{q}
Is every diffeomorphism that is H\"older conjugate to an Anosov diffeomorphism itself Anosov?
\end{q}

This question was asked by A.~Katok. His motivation came from
differentiable rigidity of higher rank Anosov actions. For example,
a popular object of study is a $\mathbb Z^k$-action which contains
Anosov elements and which is conjugate to an algebraic action for
which Anosov elements are dense. If the answer to the question above
were positive then we would immediately get that Anosov elements are
dense in the original action. Moreover, the Weyl chamber picture in
$\mathbb R^k$ for non-algebraic action would be the same as the one
for the algebraic action. Normally this information is unavailable
or only available through difficult means otherwise. See upcoming
book~\cite{KN} for an introduction to rigidity of Anosov actions.

Unfortunately the answer is negative. We will provide a concrete
counterexample of a $C^{1+Lip}$ diffeomorphism of the 2-torus
$\mathbb T^2$ H\"older conjugate to Anosov but not Anosov itself. In
fact, the counterexample can be constructed to be $C^r$ for any
$r\in(1,3)$ (see remark after Theorem 1 below).

The basic method to produce a non-Anosov diffeomorphism that is
topologically conjugate to Anosov one is to start with an Anosov
diffeomorphism and isotope it pushing stable eigenvalues at a fixed
point to the unit circle. This can be done so that stable and
unstable foliations persist. They remain mutually transversal
everywhere but not uniformly contracting and expanding. The new
system is topologically conjugate to the original Anosov map.
See~\cite{K} for the detailed construction and the proof. Another
similar example was considered in~\cite{L}. It has an additional
feature: stable and unstable manifolds at the fixed point are
tangent.

Looking at the behavior of orbits approaching a fixed point along
the stable manifold we have an exponentially fast approach for the
Anosov map conjugated to a much slower sub-exponential approach. A
H\"older continuous conjugacy would necessarily preserve the
exponential speed, only changing the exponent. This shows that these
diffeomorphisms are far from being H\"older conjugate. Note that in
the meantime the conjugacy or its inverse may turn out to be even
Lipschitz.

Another way to produce such a diffeomorphism is to start with an
Anosov diffeomorphism and ``bend" unstable manifold of a
heteroclinic point $R$ until stable and unstable manifolds at $R$
become tangent. This isotopy can be done locally in the neighborhood
of $R$. The result is a diffeomorphism with stable and unstable
foliation being transverse everywhere but along the orbit of $R$.
Along this orbit stable and unstable manifolds exhibit a tangency.
If we isotope inside of $\mathrm{Diff}^\infty(M)$ then the tangency
is at least cubic: it cannot be quadratic since the stable and
unstable foliations are topologically transverse. This bifurcation
at the boundary of Anosov systems was independently studied by
H.~Enrich~\cite{E} and Ch.~Bonatti, L.~Diaz, F.~Vuillemin
\cite{BDV}. It was shown in~\cite{E} that the new system is
conjugate to the original Anosov map. All periodic points remain
hyperbolic. Hence, unlike in the previous situation, there is a hope
that the topological conjugacy is in fact H\"older continuous.

\begin{figure}[htbp]
\begin{center}

\begin{picture}(0,0)%
\includegraphics{1_tangency.eps}%
\end{picture}%
\setlength{\unitlength}{3947sp}%
\begingroup\makeatletter\ifx\SetFigFont\undefined%
\gdef\SetFigFont#1#2#3#4#5{%
  \reset@font\fontsize{#1}{#2pt}%
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}%
  \selectfont}%
\fi\endgroup%
\begin{picture}(5274,3849)(289,-3223)
\put(3301,-361){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$R$}%
}}}}
\end{picture}%

\end{center}
\caption{Heteroclinic tangency}\label{1_tangency}
\end{figure}

We look at the simplest bifurcation of the type described above.
Consider the arc $f_t$, $t\in[0,1]$, that starts with a linear
hyperbolic automorphism $L=f_0$ and ends with diffeomorphism $f=f_1$
with tangency at a heteroclinic point $R$. The difference is that
instead of creating a cubic tangency we create a transverse
quadratic tangency. The price we pay is that $f_1$ is only
$C^{1+Lip}$ smooth. Higher order derivatives do not exist at point
$L^{-1}(R)$.

\begin{theorem}
\label{main} Suppose that $L$ and $f$ are as in the previous
paragraph. The conjugacy between $L$ and $f$ and its inverse are
H\"older continuous with exponents equal to $1/2-\delta$ and
$1/4-\delta$. Number $\delta$ can be made arbitrarily small by an
appropriate choice of $L$ and $f$.
\end{theorem}

\begin{remark}
Number $1/4$ is not a sharp bound for the exponent. It clearly can
be improved. For a diffeomorphism with a heteroclinic tangency of
order $1+\alpha$, $0<\alpha<2$ our arguments imply that the
conjugacy and its inverse are H\"older continuous with exponents
$\frac1{1+\alpha}-\delta$ and $\frac{2-\alpha}{2(1+\alpha)}-\delta$.
Clearly such a diffeomorphism is only $C^{1+\alpha-\varepsilon}$. We
stick to the case $\alpha=1$ mainly to avoid cumbersome notation.
Notice that if $\alpha$ is close to zero then both exponents are
close to $1$. In the smoothness class $C^3$ and higher our arguments
fail.
\end{remark}

In the next section we point out that this construction also
provides an example of a system for which Mather spectrum differs
from the periodic one.

We heavily rely on the results in~\cite{E} as well as~\cite{BDV}
and~\cite{C}. Thus in the Section 3 we formulate results that are
relevant to our goal. In Section 4 we prove that the conjugacy is
H\"older continuous.

Finally in the last section we present a very short proof of a
positive result from the 2006 Ph.D. thesis of Travis Fisher that
complements ours.

 \begin{theorem}[\cite{F}]
 \label{fisher}
 A $C^{1+Lip}$ diffeomorphism that is conjugate to an Anosov one via a
H\"older conjugacy $h$ is Anosov itself provided that the product of
H\"older exponents for $h$ and $h^{-1}$ is greater than $1/2$.
 \end{theorem}

\begin{remark}
This result holds for any hyperbolic set as well. The proof is the
same. Also we remark that we have removed an unnecessary condition
that was present in the formulation of the result in~\cite{F}.
\end{remark}

\medskip

{\bfseries Acknowledgements.} Anatole Katok suggested the author to
look at the system with a heteroclinic tangency since this is the
simplest situation when one can hope to get a counterexample. The
author would like to thank A.~Katok for discussions, encouragement
and useful comments on the text itself. He also would like to thank
M.~Guysinsky for explaining the idea of the proof of
Theorem~\ref{fisher}.

\section{Periodic spectrum versus Mather spectrum}

Recall the definition of Mather spectrum.
Denote by $\Gamma(TM)$ the set of continuous vector fields with supremum norm. Given a diffeomorphism $f\colon M\to M$ define $f_*\colon\Gamma(TM)\to\Gamma(TM)$
$$
f_*v(\cdot)=Df\left(v(f^{-1}(\cdot))\right).
$$
The specrum $Q_f$ of the complexification of $f_*$ is called Mather spectrum of $f$.

\begin{theorem}[\cite{Math}]
\label{mather}
If non-periodic points of $f$ are dense then any connected component of $Q_f$ is an annulus centered at $0$. Diffeomorphism $f$ is Anosov if and only if $1\notin Q_f$.
\end{theorem}

Define periodic spectrum of a diffeomorphism. Given a periodic point
$x$ of period $p$ denote by $\{\lambda_1(x)^p,\ldots
\lambda_d(x)^p\}$ the set of absolute values of eigenvalues of
$Df^p(x)$. Then
$$
P_f\stackrel{\mathrm{def}}{=}\overline{\bigcup_{x \in
\textup{Per}(f)}\{\lambda_1(x),\ldots ,\lambda_d(x)\}}.
$$

The following is easy to prove.
\begin{prop}
Let $f$ be an Anosov diffeomorphism of $\mathbb T^2$. Then $P_f=Q_f\cap\mathbb R_+$.
\end{prop}

In contrast to above Theorems~\ref{main} and~\ref{mather} imply.
\begin{cor}
Diffeomorphism $f$ from Theorem~\ref{main} provide an example of a
diffeomorphism with dense set of periodic points such that $P_f\neq
Q_f\cap\mathbb R_+$.
\end{cor}

\section{First heteroclinic tangency at the boundary of Anosov systems}

Here we describe some results of~\cite{E},~\cite{BDV} and~\cite{C} that we need.

Let $L$ be hyperbolic automorphism of $\mathbb T^2$. Denote by $e_u$
and $e_s$ the eigenvectors of $L$ and by $\lambda>1$ the unstable
eigenvalue, $Le_u=\lambda e_u$. Let $P$ and $Q$ be two different
fixed points of $L$ and $R$ an intersection of the stable manifold
of $P$ and unstable manifold of $Q$. We may assume that distances to
$R$ from $P$ and $Q$ are equal. Also we assume that the size of a
ball containing $\{P,Q,R\}$ is much smaller than the size of
$\mathbb T^2$. Let $(x,y)$ be coordinates in the neighborhood of $R$
that make stable foliation horizontal and unstable foliation
vertical. Let $B$ be a small ball of radius $r$ centered at $R$.

Define $f_t=\theta_t\circ L$, $t\in[0,1]$ where $\theta_t\colon\mathbb T^2\to\mathbb T^2$ is identity outside of $B$ and given by the following formula on $B$
$$
\theta_t(x,y)=
\left(
  \begin{array}{cc}
    \;\;\;\cos(t\gamma(\rho)) & \sin(t\gamma(\rho)) \\
    -\sin(t\gamma(\rho)) & \cos(t\gamma(\rho)) \\
  \end{array}
\right)
\left(
  \begin{array}{c}
    x \\
    y \\
  \end{array}
\right)
$$
where $\rho=\sqrt{x^2+y^2}$ and $\gamma\colon[0,\infty)\to[0,\pi/2]$ is a $C^\infty$ map satisfying $\gamma(0)=\pi/2$; $\gamma(\rho)=0$ for $\rho\ge r$; $\gamma$ is strictly decreasing on $[0,r]$. Thus on every circle centered at $R$ $\theta_t$ is a rotation by an angle no greater than $\pi/2$. Value $\pi/2$ is achieved for $\rho=0$ and $t=1$. Let $v\in T_R\mathbb T^2$ be the unit vertical vector. From the definition of $f=f_1$ we have
$$
\lim_{n\to\pm\infty}(Df^n)v=0.
$$
Hence $f$ is not Anosov.

Denote by $\mathcal U$ and $\mathcal V$ neighborhoods of segments
$PR$ and $Qf^{-1}(R)$ that contain $B$ and $f^{-1}(B)$ respectively.

\begin{theorem}[\cite{E}, \cite{BDV}, \cite{C}]
There exist $r$ small enough and function $\gamma$
such that corresponding arc $f_t$ as defined above satisfies the following.
\begin{enumerate}
\item Diffeomorphisms $f_t$ are Anosov for $t<1$.
\item Diffeomorphism $f$ possesses invariant contracting and expanding foliations $W^s$ and $W^u$. The leaves of $W^s$ and $W^u$ are $C^1$ immersed curves. Denote by $E^s$ and $E^u$ distribution tangent to these foliations.
\item \label{cubic} Foliations $W^s$ and $W^u$ are transverse everywhere but along the orbit of $R$. At the point $R$ they have cubic tangency. Namely, there is $\tau>1$ such that for $S(x,y)\in B$
\begin{equation}
\tan\measuredangle(E^u(S),e_s)\ge\tau(x^2+y^2).
\end{equation}
Analogous inequality holds for distribution $E^s$.
\item \label{Eu_horizontal} For any $S\notin\mathcal U$
$$
\tan\measuredangle(E^u(S),e_u)<\varepsilon.
$$
Number $\varepsilon$ can be made arbitrarily small by the choice of
$L$, $f$ and $\mathcal U$. If we let
\begin{multline*}
\qquad\qquad
B_\infty=\{x\in\mathbb T^2:\;\exists i>0\;\;\;\mbox{such that}\\
 f^{-i}(x)\in B,\;\{f^{-i}(x), f^{-i+1}(x),\ldots x\}\subset\mathcal U\},
\end{multline*}
 then for any $S\notin B_\infty$
$$
\tan\measuredangle(E^u(S),e_u)<1.
$$
Analogous statement holds for $E^s$ and $\mathcal V$.
\item \label{top_conj} Diffeomorphism $f$ is conjugate to $L$ by a homeomorphism $h$, $h\circ f=L\circ h$.
\end{enumerate}
\end{theorem}

\begin{remark}
Technical statements~(T\ref{cubic}) and~(T\ref{Eu_horizontal}) are not stated explicitly in the papers quoted but they follow from the cone constructions that are carried out there.
\end{remark}

It may seem that since the size of $B$ is small $E^u$ is almost
vertical inside of $B_\infty$ and almost horizontal outside. In
fact, the transition through the boundary of $B_\infty$ is
continuous. Parameter $\tau$ increases when $r$ goes to zero.

We will be working with exactly the same construction, but $\theta$
must be chosen differently. Function $\gamma$ can be chosen
differently with $\gamma^{(r)}(0)=0$ for $r<2$ and $\gamma''(0)<0$.
Then $\theta\in C^{1+Lip}$ and the tangency is quadratic. This way
instead of~(T\ref{cubic}) we have
\begin{equation}
\label{quadratic}
 \tan\measuredangle(E^u(S),e_s)\ge\tau\sqrt{x^2+y^2},\;\; \tau>1.
\end{equation}

We outline proofs of~(T\ref{Eu_horizontal}) and~(\ref{quadratic}) at
the end of this section.

\begin{remark}
For conservative systems the theorem above was established
in~\cite{C}. Original proof~\cite{E} required that product of the
eigenvalues at $P$ is greater than $1$ while the product of the
eigenvalues at $Q$ is less than $1$. Assumption on the eigenvalues
at $P$ and $Q$ in~\cite{BDV} is even more restrictive. The main
motivation of~\cite{C} was to extend the example to systems with
homoclinic tangency. Our proof works for homoclinic intersection as
well. We work with a heteroclinic intersection only for convenience.
Also we would like to remark that our proof of H\"older continuity
works for the original construction in~\cite{E}. One needs to start
the isotopy with a $C^1$ small perturbation of $L$ that satisfies
above assumptions on eigenvalues at $P$ and $Q$ instead of starting
with $L$.
\end{remark}

Let us recall the proof of~(T\ref{top_conj}) from~\cite{E} since
this is the statement that we strengthen.

\begin{proof}
The main tool here is the following result of P.~Walters.
\begin{theorem}[\cite{W}]
Let $g\colon M\to M$ be an Anosov diffeomorphism. Then there exists an $\varepsilon_0>0$ such that for each $0<\varepsilon<\varepsilon_0$ there exists $\delta>0$ such that if $\tilde g\colon M\to M$ is a homeomorphism and if $d(g,\tilde g)<\delta$, then there exists a unique continuous map $h$ of $M$ onto $M$ with $h\circ \tilde g=g\circ h$ and $d(h,id)<\varepsilon$.
\end{theorem}
Apply the theorem for $g=L$ and $\tilde g=f$ to get semiconjugacy $h$ with
$d_{C^0}(h,id)<\varepsilon$. Note that $d_{C^0}(L,f)\to 0$ as $r\to 0$.
We have to take $r$ small enough so that $\varepsilon$ is smaller than constant associated
to the local product structure of $W^s$ and $W^u$. This guarantees that $h$ is injective.
Indeed if $h$ glues together some points on, say, unstable manifold then iterating forward
we get that $h$ glues together some points fixed distance apart. This is impossible since $h$ is close to identity.

Hence, by the invariance of domain theorem, $h$ is a homeomorphism.
\end{proof}

\begin{proof}[Sketch of proof of~(T\ref{Eu_horizontal})]
Cone constructions in~\cite{E} and~\cite{C} imply that $\forall
S\notin B_\infty$
$$
\tan\measuredangle(E^u(S), e_u)<1.
$$
Then given a small number $\varepsilon$ there exists $N$ such that
$\forall S\notin\cup_{i=0}^Nf^i(B_\infty)$
$$
\tan\measuredangle(E^u(S), e_u)<\varepsilon.
$$
It is possible to fatten $\mathcal U$ so that set $\tilde B_\infty$
that corresponds to new fattened neighborhood $\tilde{\mathcal U}$
of $PR$ contains $\cup_{i=0}^Nf^i(B_\infty)$ as shown on the
Figure~\ref{2_fatten_U}.

\begin{figure}[htbp]
\begin{center}

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\includegraphics{2_fatten_U.eps}%
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\end{picture}%

\end{center}
\caption{The hatched set is $B_{\infty}$. Distance $|QR|$ is much
bigger than $f^N(B)=f^2(B)$. Hence it is possible to fatten
$\mathcal U$ to $\tilde{\mathcal U}$ so that unstable distribution
outside $\tilde B_\infty$ is $\varepsilon$-close to horizontal
vector $e_u$.}\label{2_fatten_U}
\end{figure}

For that we need to make sure that $f^N(B)$ is small compared to the
distance $|QR|$. This can be achieved by appropriate choice of
automorphism $L$, $P$ and $Q$.

We fix a hyperbolic matrix $L$ that induces an automorphism of
$\mathbb T^2=\mathbb R^2/\mathbb Z^2$ with $d$ fixed points. We fix
size $r$ of the ball $B$ and the map $\theta|_B$. The trick now is
to choose the torus $\mathbb T^2$ to be ``big" when compared to
eigenvalue $\lambda$ of $L$ and $r$.

Linear map $L$ induces a hyperbolic automorphism of $\mathbb
T^2=\mathbb R^2/k\mathbb Z^2$, where $k$ is a big integer. This
automorphism is a finite cover of the automorphism of $\mathbb
T^2=\mathbb R^2/\mathbb Z^2$. It also has $d$ fixed points.
Obviously the distances between those fixed points are big now.
Hence $P$ and $Q$ can be chosen so that $|QR|$ is big.
\end{proof}
\begin{remark}
We will use the fact that $|QR|$ can be chosen big independently of
$r$ and $\lambda$ several times in the course of the proof of
H\"older continuity.
\end{remark}

\begin{prop}
\label{propos} Given a point $S\in\mathcal U$. Denote by $d(S)$ the
distance to $PR$. Then
\begin{equation}
\label{cubic_for E^s} \tan\measuredangle(E^s(S),e_s)\le
\kappa\,d(S)^2.
\end{equation}
where $\kappa$ is a number that depends on $\varepsilon$
from~(T\ref{Eu_horizontal}).
\end{prop}

\begin{proof}[Proof of Proposition]
Let $N$ be the smallest positive integer such that
$f^N(S)\notin\mathcal U$. Then $\lambda^nd(S)\approx1$ and
$\tan\measuredangle(E^s(f^N(S)),e_s)\le\varepsilon$
by~(T\ref{Eu_horizontal}) since $f^N(S)\notin\mathcal V$ as well.
$$
\tan\measuredangle(E^s(S),e_s)=\lambda^{-2n}\tan\measuredangle(E^s(f^N(S)),e_s)\le
C\varepsilon d(S)^2.
$$
\end{proof}

\begin{proof}[Sketch of proof of~(\ref{quadratic})]
Denote by $A$ and $\tilde A$ points of intersection of the line $QR$
and boundary $\partial B$ as shown on Figure~\ref{3_q_variation}. It
follows follows from the definition of $f$ that straight segment
$[Q,A]$ is inside of $W^u(Q)$ while straight segment $[P,R]$ is
inside $W^s(P)$. The shape of $W^u(Q)$ between $A$ and $\tilde A$ is
completely determined by $\theta$. Namely, it is the image of the
segment $[A, \tilde A]$ under $\theta$. We remark that on the other
hand the shape of $W^u(Q)$ between $A$ and $R$ determines $\theta$
since map $\theta$ is a rigid rotation on every circle around $R$.

\begin{figure}[htbp]
\begin{center}

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\end{center}
\caption{Establishing linear variation of the angle on the circle
$\EuScript C$.}\label{3_q_variation}
\end{figure}

First let us establish linear variation of the
angle~(\ref{quadratic}) along $\theta[A,R]$. It follows from the
choice of $\theta$. Let $C\in[A,R]$ be the point such that
$\measuredangle(E^u(\theta(C)),e_s)=\pi/2$. Then for any
$S\in\theta[C,R]$ we have
$$
\tan\measuredangle(E^u(S),e_s)\ge\tau\sqrt{x^2+y^2}
$$
since the tangency is quadratic. For $S\in\theta[A,C]$
$$
\tan\measuredangle(E^u(S),e_s)\ge c
$$
where $c$ is a big constant, $c\gg 1$.

Fix a circle $\EuScript C$ of radius $\rho$ centered at $R$. Let
$D=\EuScript C\cap\theta[A,R]$. We know that
estimate~(\ref{quadratic}) holds for $D$ and we would like to
establish it for other points on $\EuScript C$.

Consider distribution $\tilde E^u\subset T_{\EuScript C}\mathbb T^2$
given by the formula $\tilde E^u=DL\,E^u$. Then $E^u$ on $\EuScript
C$ is given by $E^u=Df E^u=D\theta\tilde E^u$.

If we denote by $v$ and $u$ normal and tangent vector fields to
$\EuScript C$ then $D\theta$ with respect to bases $(v(\cdot),
u(\cdot))$ and $(v(\theta(\cdot)),u(\theta(\cdot)))$ is given by the
shear matrix
\begin{equation}
\label{d_teta}
 D\theta=\left(
  \begin{array}{cc}
    1 & \alpha \\
    0 & 1 \\
  \end{array}
\right)\!,\; \alpha=\alpha(\rho)>0.
\end{equation}

According to Proposition~\ref{propos}
$$
\tan\measuredangle(\tilde E^u(S),e_u)\le\kappa\rho^2,\;\; S\in
\EuScript C.
$$
Meanwhile
$$
\tan\measuredangle(E^u(D),e_s)\ge\tau\rho.
$$
These inequalities together with the formula for $D\theta$ above
imply linear variation of angle~(\ref{quadratic}) for all
$S\in\EuScript C$.
\end{proof}

\section{Topological conjugacy is H\"older}

Here we prove that the conjugacy $h$ and its inverse are H\"older continuous.

We mimic the standard proof of H\"older continuity of conjugacy
between two Anosov systems (e.~g. Section 19.1 in~\cite{KH}). The
conjugacy maps stable and unstable foliations of $f$, $W^s$ and
$W^u$, into stable and unstable foliations of $L$, $W_L^s$ and
$W_L^u$, respectively.
\begin{step}
A restriction of $h$ to a leaf of $W^u$ is H\"older continuous with the exponent
equal to $1-\delta$. A restriction of $h^{-1}$ to a leaf of $W_L^u$ is H\"older continuous with the exponent equal to $1/4-\delta$. Number $\delta$ depends on the choice of $L$ and $r$ and can be made arbitrarily small.
 \end{step}
\begin{remark}
The distances with respect to which we show H\"older continuity are
induced Riemannian distances
 on the leaves of $W^u$ and $W_L^u$.
\end{remark}

Analogously, $h$ and $h^{-1}$ are H\"older continuous when
restricted to stable leaves with exponents $1-\delta$ and
$1/4-\delta$ respectively. We immediately get the following.
\begin{prop}
Homeomorphism $h^{-1}$ is H\"older continuous with exponent
$1/4-\delta$.
\end{prop}
This follows from a standard argument that utilizes uniform
transversality of $W_L^s$ and $W_L^u$.

To conclude that $h$ is H\"older as well one needs to have $W^s$ and $W^u$ to be uniformly transversal. In our situation $W^s$ and $W^u$ are uniformly transversal only outside a neighborhood of the orbit of $R$. This leads to further loss of the exponent by factor of $1/2$.
\begin{step}
Conjugacy $h$ is H\"older continuous with exponent equal to $1/2-\delta$.
\end{step}
Heuristically it is clear that the loss of the exponent at the second step is inevitable. In the second step we ``straighten out" the quadratic tangency. Thus the exponent is no greater than $1/2$.

Together with~(\ref{quadratic}) Proposition~\ref{propos} gives us
control on the angle between $E^s$ and $E^u$ in the neighborhood of
the orbit of $R$ which is crucial for carrying out estimates in Step
2.

Throughout the proof we denote by $d^u$, $d^s$, $d_L^u$ and $d_L^s$
induced Riemannian distances along the leaves of corresponding
foliations.

\begin{proof}[Proof of Step 1]
Uniform continuity of $h$ implies that $\exists C_1>0$ such that
\begin{multline*}
\frac{1}{C_1}\, d_L^u(h(a),h(b))\le d^u(a,b)\le C_1d_L^u(h(a),h(b))\;\;\;\;\;
\mbox{whenever}\;\;\;\;\; d^u(a,b)\ge\frac {r}{10}
\end{multline*}
To prove H\"older estimates for close-by points $a$ and $b\in
W^u(a)$ we need to have exponential estimates on expansion along
$E^u$. Given a point $a\in\mathbb T^2$ let
$$
D^u(a)=\|Df(a)v^u\|,\;\;\;v^u\in E^u, \|v^u\|=1.
$$
If $a\notin L^{-1}(B)$ then, obviously, $\lambda^{-1}\le
D^ua\le\lambda$. If $a\in L^{-1}(B)$ then $D^ua$ can be bigger.
Still there exists $D$ such that $D^ua\le D$ for $a\in L^{-1}(B)$.
In fact, using estimates on $\alpha$ from~(\ref{d_teta}) one can
show that $D=2\lambda$ works but we will not use it. Keeping $r$ and
$\theta$ fixed choose $L$ (e.~g. pass to a finite cover as in proof
of~(T\ref{Eu_horizontal})), $P$ and $Q$ so that $|PR|$ is large and
hence first return time $m$ to $L^{-1}(B)$ is big. Thus $\forall
a\in\mathbb T^2$
\begin{equation}
\label{product}
\prod_{i=1}^{m-1}D^uf^i(a)\le\lambda^{m-1}D\le(\lambda^{1+\delta})^m,
\end{equation}
where $\delta=\delta(m, D)$ is a small number. It follows that
\begin{equation}
\exists C_2 :\;\; \forall n>0\; \|Df^nv^u\|\le C_2 (\lambda^{1+\delta})^n\|v^u\|,\;v^u\in E^u.
\end{equation}

Now we use standard argument to prove H\"older continuity. Take
$a\in W^u(b)$ close to $b$. Let $N$ be the smallest number such that
$d^u(f^N(a),f^N(b))\ge r/10$. Then
\begin{multline*}
d_L^u(h(a),h(b))=\frac{1}{\lambda^N}d_L^u\left(L^N(h(a)),L^N(h(b))\right)\le\frac{C_1}{\lambda^N}d^u(f^N(a),f^N(b))\\
\le\frac{C_1C_3}{\lambda^N}d^u(f^N(a),f^N(b))^{1-\delta}\le\frac{C_1C_2^{1-\delta}C_3}{\lambda^N}(\lambda^{1+\delta})^{N(1-\delta)}d^u(a,b)^{1-\delta}\le Cd^u(a,b)^{1-\delta}.
\end{multline*}

To show that $h^{-1}$ is H\"older along $W_L^u$ we need an estimate
on the product in~(\ref{product}) from below. According
to~(T\ref{Eu_horizontal})  if $a\notin \mathcal U\cup L^{-1}(B)$
then $D^u(a)\ge\mu$, where $\mu=\mu(\varepsilon)$ and
$\mu\nearrow\lambda$ when $\varepsilon\to 0$ (we remark that the
choices we do to make $\varepsilon$ small do not affect $\lambda$).
If $a\in \mathcal U\backslash B_\infty$ then another inequality
from~(T\ref{Eu_horizontal}) provide an estimate on expansion.

If $a\in B_{\infty}$ then, obviously, $D^u(a)\ge\lambda^{-1}$ but we
need to have a better control. Split $B_{\infty}$ into its connected
components
$$B_{\infty}=\bigcup_{i\ge 0}B_i$$ (see Figure~\ref{2_fatten_U}). Let $(x^i,y^i)$ be coordinates in $B_i$ obtained by parallel
transport of $(x,y)$ from $R$ to $f^i(R)$. Consider rectangles
\begin{equation}
\label{bar1}
\bar B_i=\{(x^i,y^i) : |x^i|\le r\lambda^{-i}, |y^i|\le r\lambda^{-3i} \}.
\end{equation}
Also let
\begin{equation}
\label{bar2}
\bar B=\bigcup_{i\ge 0}\bar B_i.
\end{equation}

Let $\xi=\tau r$ where $\tau$ is from~(\ref{quadratic}).
\begin{lemma}
\label{lemma}
Consider a point $S(x^n,y^n)\in B_n$,  $S\notin \bar B_n$. Then
$$
\tan\measuredangle(E^u(S),e_s)\ge\xi.
$$
\end{lemma}
\begin{proof}
Note that this follows from~(\ref{quadratic}) if $n=0$.

$$
\tan\measuredangle(E^u(S),e_s)=\lambda^{2n} \tan\measuredangle(E^u(f^{-n}(S)),e_s).
$$
Clearly $(\lambda^{-n}x^n,\lambda^ny^n)$ are $(x,y)$ coordinates of $f^{-n}(S)$. If $|x^n|\ge r\lambda^{-n}$ then
  $$
  \lambda^{2n} \tan\measuredangle(E^u(f^{-n}(S)),e_s)\ge \lambda^{2n}\tau(\lambda^{-n}x^n)\ge\tau r.
  $$
If $|y^n|\ge r\lambda^{-3n}$ then
   $$
   \lambda^{2n} \tan\measuredangle(E^u(f^{-n}(S)),e_s)\ge
   \lambda^{2n}\tau(\lambda^{n}y^n)\ge\tau r.
   $$                                                           \end{proof}

Take an arbitrary point $a\notin\bar B$. Then
$\tan\measuredangle(E^u(a),e^s)\ge\xi$ and hence there exists $m>0$
such that $D^uf^m(a)>\mu$. Therefore contraction along unstable
mostly happens only inside $\bar B$. After the point leaves $\bar B$
its unstable direction ``recovers" after $m$ steps even before the
point leaves $\mathcal U$.

Now we will be doing estimates from below on expansion along $E^u$
by analyzing itinerary of a point.

As before (sketch of proof of (T\ref{Eu_horizontal})) we make sure
by the choice of $L$ that after a point leaves $\mathcal U$ it
spends at least time $4m+4$ before it enters $\mathcal V$.

Take a point $a(x,y)\in B$ and  assume that $f^i(a)\in \bar B_i$,
$i=0,\ldots n$, $f^{n+1}(a)\notin \bar B_{n+1}$. By the Lemma
$|x|\le r\lambda^{-2n}$, $|y|\le r\lambda^{-2n}$. Hence
$f^{-1}(a),\ldots f^{-2n}(a)\in\mathcal V$, $D^uf^{-i}(a)\ge\mu$,
$i=2,\dots 2n$. Simple calculation with $D\theta$ shows that
$D^uf^{-1}(a)\ge 1$. Start with $f^{-2n}(a)$ and wait time
$4n+4m+4$. We know for sure that during that time the orbit has not
entered $\mathcal V$ again. Thus
\begin{multline}
\label{period_estimate}
\prod_{i=-2n}^{2n+4m+4}D^uf^i(a)=\prod_{i=-2n}^{-1}D^uf^i(a)
\prod_{i=0}^{n+m}D^uf^i(a)
\prod_{i=n+m+1}^{2n+4m+4}D^uf^i(a)\\
\ge\mu^{2n-1}\lambda^{-n-m-1}\mu^{n+3m+4}=\lambda^{-n-m-1}\mu^{3n+3m+3}.
\end{multline}
This estimate is good enough to get exponent $1/2-\delta$. The only
problem is that it holds only along specific orbit segments
described above. Call them ``cycles".

As before take $a\in W^u(b)$ close to $b$ and let $N$ be the smallest number such that $d^u(f^N(a),f^N(b))\ge r/10$.

Let us first explain the idea informally. We have to study how the
length of $d^u(f^i(a),f^i(b))$ changes as $i=0,\ldots N$. We
decompose this time segment into ``cycles" as
in~(\ref{period_estimate}). These ``cycles" do not overlap. There
might be some ``gaps" between the ``cycles" that only improve the
estimate since the time spent in the ``bad" set $\bar B_\infty$ is
inside the ``cycles". The difficulty that we have to deal with is
that at the beginning ``cycle" might be ``incomplete". The ``worst"
situation is when $a$ is close to $R$. Same problem occurs at the
end --- the last ``cycle" might happen to be ``cut" at the end.

Notice that in the description above we ignore returns to $\mathcal
U\backslash B$. The expansion rate inside this set might be less
than $\mu$ but still is greater than 1 according
to~(T\ref{Eu_horizontal}). Therefore these returns are much easier
to take care of. Hence we consider only ``cycles" that correspond to
returns to $B$.

The problem of the ``cut" at the end is easy to deal with. Denote by
$t_j$, $j=1,\ldots l$ the lengths of ``cycles". We count incomplete
``cycles" as well. We also consider numbers $n_j$, $j=1,\ldots l$.
In the notation of~(\ref{period_estimate}) $t_j=4n_j+4m+4$. Assume
that the last ``cycle" number $l$ is incomplete. During the last
``cycle" the segment enters $B$, spends time $n_l$ inside $\bar
B_{\infty}$ and then it recovers to the size of $r/10$ during the
time less than $n_l+4m+4$. When the segment leaves $\bar B_{n_l}$ it
has length of order $\lambda^{-3n_l}$ since that is the vertical
size of $\bar B_{n_l}$ and the unstable foliation is roughly
vertical in $\bar B_{n_l}$. Clearly number $n_l$ is not big since
otherwise the segment cannot recover to the size $r/10$. We can
control $N$ independently of $n_l$ by choosing $a$ and $b$ extremely
close to each other. This may result in big H\"older constant but
the exponent will not be affected. In fact, since $n_l$ is bounded
this difficulty at the end of ``time-window" $i=0,\ldots N$ can be
taken care of by the H\"older constant.

Now assume that the first ``cycle" is incomplete as well. In contrast to above $n_1$ might happen to be big when compared to $N$ since the size of the segment is small at the beginning. Clearly we need to examine the ``worst" situation when $a\in B$. As before we can argue that after time $n_1$ the size of the segment is of the order $\lambda^{-3n_1}$. Number $N-n_1$ is greater than $3n_1$ since during this time the segment grows up to size $r/10$.
Hence, using~(\ref{period_estimate}) we get
\begin{multline}
\label{prelim}
d^u(f^N(a), f^N(b))\approx \lambda^{-n_1}\lambda^{n_1}\lambda^{\frac34(N-2n_1)}\lambda^{-\frac14(N-2n_1)}d^u(a,b)\\
=\lambda^{-\frac12(N-2n_1)}d^u(a,b)=\lambda^{-\frac12(\frac N2+\frac N2-2n_1)}d^u(a,b)
\ge\lambda^{\frac N4}d^u(a,b).
\end{multline}
This clearly good enough to get exponent $\frac14-\delta$.

From now on we will be providing details to the scheme described above. Still we stay away from completely rigorous technical discussion. The technical details are plentiful while the way argument works is fairly transparent.

To get the estimate $N\ge 4n_1$ we need to redefine slightly the set $\bar B$, numbers $n_i$ and $t_i$ accordingly. First fix $\mu$ close to $\lambda$, $\mu<\lambda$.
\begin{equation*}
\bar B_i=\{(x^i,y^i) : |x^i|\le \tilde r\lambda^{-i}, |y^i|\le \tilde r\lambda^{-3i} \},
\end{equation*}
\begin{equation*}
\bar B=\bigcup_{i\ge 0}\bar B_i,
\end{equation*}
where $\tilde r=\tilde r(\mu)<r/20$ is chosen so that $D^u(x)\ge\mu^{-1}$ for any $x\in B\backslash \bar B_0$ and by the arguments of Lemma~\ref{lemma} $D^u(x)\ge\mu^{-1}$ for any $x\notin \bar B_\infty$.

We study lengths of the segment during the time $N$ defined above.
We consider the ``worst" case when $a$ and $b$ are close to $R$. Let
$n_1$ be the smallest integer such that
\begin{equation*}
\frac{d^u(f^{n_1+1}(a),f^{n_1+1}(b))}{d^u(f^{n_1}(a),f^{n_1}(b))}\ge\mu^{-1}.
\end{equation*}
Then it is easy to see that after some fixed number of iterates $m=m(\mu)$ which is independent of $n_1$ we will have
\begin{equation*}
\frac{d^u(f^{n_1+m+1}(a),f^{n_1+m+1}(b))}{d^u(f^{n_1+m}(a),f^{n_1+m}(b))}\ge\mu.
\end{equation*}

\begin{remark}
Notice that any iterate of the segment does not cross more than one
connected component of $\bar B_\infty$. Indeed, the vertical
distance between $\bar B_n$ and $\bar B_{n+1}$ is of order
$\lambda^{-n}$, horizontal sizes of $\bar B_n$ and $\bar B_{n+1}$
are of order $\lambda^{-3n}$ while in the ``gap" between $B_n$ and
$B_{n+1}$ unstable foliation is in horizontal cone field according
to~(T\ref{Eu_horizontal}). It follows that local unstable leaves do
not intersect $\bar B_n$ and $\bar B_{n+1}$.
\end{remark}

Analogously define numbers $n_j$, $j=1,\ldots l$. Then define $t_j=n_j+4m+4$ and corresponding ``cycles" as before. Clearly we have an analogue of~(\ref{period_estimate})
\begin{equation}
\label{period_estimate_new}
d^u(f^{s_j+t_j}(a),f^{s_j+t_j}(b))\ge\lambda^{-n-m-1}\mu^{3n+3m+3}d^u(f^{s_j}(a), f^{s_j}(b)).
\end{equation}
where $s_j$ is the starting time of a full ``cycle". If the
itinerary of the segment has complete first ``cycle" then the same
way as in the proof of H\"older continuity of $h$
using~(\ref{period_estimate_new}) we get
$$
d^u(h^{-1}(a),h^{-1}(b))\le C d^u(a,b)^{\frac12-\delta}
$$
with $\delta=\delta(\mu)\searrow 0$ as $\mu\nearrow\lambda$.

Now we go back to  the ``worst" case when the first ``cycle" starts
near $R$. Definitions of $\bar B_\infty$ and $n_1$ guarantee that at
least half of the segment $[f^{n_1-1}(a),f^{n_1-1}(b)]\subset
W^u(f^{n_1-1}(a))$ lies inside of $\bar B_{n_1-1}$. Recall that
unstable foliation is almost vertical inside $\bar B_\infty$. Hence
$$
d^u(f^{n_1-1}(a), f^{n_1-1}(b))\le \frac r{10}\lambda^{-3(n_1-1)}.
$$
Recall that $D^u(x)\le\lambda$ for any $x\notin L^{-1}(B)$ and $d^u(f^N(a),f^N(b))\ge r/10$. It follows that $N-n_1\ge\Gamma n_1$ where $\Gamma$ depends on frequency of visits to $L^{-1}(B)$ and can be made arbitrarily close to $3$. Now the preliminary estimate~(\ref{prelim}) transforms into the following one

\begin{multline*}
d^u(f^N(a),f^N(b)) \ge\lambda^{-n_1-m} \mu^{n_1+3m+4}\mu^{\frac34(N-2n_1)}
\lambda^{-\frac14(N-2n_1)}\\
\ge\lambda^{-n_1-m} \mu^{n_1+3m+4}(\mu/\lambda)^{\frac14(N-2n_1)}
\mu^{\frac12(N-2n_1)}\\
=\lambda^{-n_1-m}\mu^{n_1+3m+4}(\mu/\lambda)^{\frac14(N-2n_1)} \mu^{\frac12(1-\Delta)(N-2n_1)}\mu^{\frac12\Delta(N-2n_1)}\\
\ge\mu^{\frac12\Delta(N-2n_1)}\ge\mu^{\frac14(N-\delta)},
\end{multline*}
where $\delta=\delta(\Gamma,\Delta)$ is small and $\Delta=\Delta(\lambda,\mu)$ is chosen so that $\mu^{\frac12(1-\Delta)(N-2n_1)}$ compensates the factors in front of it compensates the factor in front of . Number $\Delta\nearrow 1$ as $\mu\nearrow\lambda$. It follows that $\delta$ can be arbitrarily small.
\end{proof}

\begin{proof}[Proof of Step 2]
First of all let us notice that outside of $\mathcal U$ foliations $W^s$ and $W^u$ are uniformly transversal. Then, by the standard argument, H\"older continuity along $W^s$ and $W^u$ implies H\"older continuity of $h$ with exponent $1-\delta$ outside of $\mathcal U$.

It follows from~(\ref{quadratic}) and Proposition~\ref{propos} that
inside $B$ the angle  $\measuredangle(E^s,E^u)$ varies linearly with
distance to $R$. This allows to show that $h$ is H\"older continuous
with exponent $1/2-\delta$ inside $B$. More work is required to
establish H\"older continuity in the rest of $\mathcal U$. We start
with an observation that allows to reduce our task to establishing
H\"older inequality for points inside of a single $B_n$, $n\ge 0$.

Introduce vertical and horizontal cones
$$
\mathcal C_v(x)=\{v\in T_x\mathbb T^2 : \tan\measuredangle(v,e_s)<\varepsilon\},
$$
$$
\mathcal C_h(x)=\{v\in T_x\mathbb T^2 : \tan\measuredangle(v,e_s)>\xi\}
$$
with $\varepsilon$ as in~(T\ref{Eu_horizontal}) and $\xi$ as in Step
1. These cones have disjoint interiors. Moreover, $E^s(x)\in\mathcal
C_v(x)$ for any $x\in\mathcal U$ by~(T\ref{Eu_horizontal}) and
$E^u(x)\in\mathcal C_h(x)$ for any $x\in \mathcal U\backslash\bar
B_\infty$ by Lemma~\ref{lemma}. Thus $\mathcal C_v$ and $\mathcal
C_h$ provide good control of $E^s$ and $E^u$ in $\mathcal
U\backslash\bar B_\infty$.

Take $a$ and $b$ closeby inside $\mathcal U$. Let $e$ be the
intersection of local unstable manifold $W^u(a,r/10)$ and local
stable manifold $W^s(b,r/10)$.

\begin{lemma}
\label{lemma2}
Assume that some fixed proportion with respect to the length of local unstable manifold connecting $a$ and $e$ lies inside of $\mathcal C_h$ --- meaning that tangent vector is in the cone. Then
$$
d(h(a),h(b))\le Cd(a,b)^{1-\delta}.
$$
\end{lemma}
Recall that we know that local stable manifold is in $\mathcal C_v$. Then the proof of the Lemma is a straightforward adjustment of the standard one when the whole local unstable manifold lies inside of $\mathcal C_h$ as well.

We have remarked in the  course of the proof of Step 1 that local
unstable leaves do not meet different connected components of $\bar
B_\infty$. Together with Lemma~\ref{lemma2} this implies that we are
only left to deal with points $a$ and $b$ such that local unstable
manifold connecting $a$ and $e$ lies almost entirely in $\bar B_n$
for some $n\ge 0$.

Denote by $(x,y)$ the coordinate system centered at $f^n(R)$ with $x$-axis being horizontal. Observe further that it is enough to consider points $a$ and $b$ that have the same $y$-coordinate. Let $a=a(x_1,y_1)$, $b=b(x_2,y_1)$ and $e=e(x_3,y_3)$. We are aiming at proving the estimate
\begin{equation}
\label{beak}
C|x_1-x_2|^{\frac12}\ge|y_1-y_3|.
\end{equation}
Together with Step 1 this would imply that $h$ is H\"older with exponent $\frac12-\delta$.

Given a point $S(x,y)\in B_n$ we have
\begin{multline*}
\tan\measuredangle(E^u(S),e_s)=\lambda^{2n}\measuredangle(E^u(f^{-n}(S)),e_s)\ge
\lambda^{2n}\tau\sqrt{(\lambda^{-n}x)^2+(\lambda^ny)^2}\\
\ge\lambda^{2n}\tau\lambda^{-n}\sqrt{x^2+y^2}\ge\sqrt{x^2+y^2}
\end{multline*}
and by Proposition~\ref{propos}
$$
\tan\measuredangle(E^s(S),e_s)\le\kappa d(S)^2\le\frac{x^2}2.
$$
These inequalities provide control on the angle needed to carry out the estimate~(\ref{beak}).

Denote by $\EuScript B$ the ``beak" formed by the unstable manifold
connecting $a$ and $e$ and the stable manifold connecting $b$ and
$e$. We will consider two representative cases illustrated on
Figure~\ref{4_beaks}.

\begin{figure}[htbp]
\begin{center}

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\includegraphics{4_beaks.eps}%
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\put(577,683){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$W^u(Q)$}%
}}}}
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}}}}
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}}}}
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\end{picture}%

\end{center}
\caption{Beaks.}\label{4_beaks}
\end{figure}

\medskip

{\bfseries Case A. $D\stackrel{\mathrm{def}}{=}dist(\EuScript B, f^n(R))\ge|y_1-y_3|$}.
In this case according to the estimates above we have that $[a,e]$ is tilted at least by $D$ while $[b,e]$ is tilted at most by $D^2$. Hence
$$
|x_1-x_2|\ge|y_1-y_3|(D-D^2)\ge\frac12|y_1-y_3|D\ge\frac12|y_1-y_3|^2.
$$

\medskip

{\bfseries Case B.} We allow $|y_1-y_3|$ to be greater than $dist(\EuScript B, f^n(R))$. Also we make a simplifying assumption that $x_3=0$ and $y_3>0$. Then
\begin{multline*}
|x_1-x_3|\ge\int_{y_3}^{y_1}\sqrt{x^2+y^2}d\,{length}-\int_{y_3}^{y_1}\frac{x^2}2d\,{length}\\
\ge\int_{y_3}^{y_1}(y-y^2/2)d\,{length}\ge\frac13|y_1-y_3|^2.
\end{multline*}

When $x_3\neq 0$ the estimate is similar but first one needs to ``cut" the ``tip of the beak" where $x^2\le y^2$ does not hold. When $y_3<0$ while $y_1>0$ or vice versa modification in the same spirit is required.
\end{proof}

\section{A positive result}

Here we prove Theorem~\ref{fisher}.

Let $M$ be a Riemannian manifold and $d(\cdot,\cdot)$ the distance
function induced by the Riemannian metric. Let $f\colon M\to M$ be
an Anosov diffeomorphism and $g \colon M\to M$ a diffeomorphism
H\"older conjugate to $\tilde f$:
$$
h\circ\tilde f=f\circ h.
$$
Let $\alpha$ be H\"older exponent of $h$ and $\beta$ the H\"older
exponents of $h^{-1}$. Recall that we assume that $\alpha\beta>1/2$.

\begin{definition}
A sequence of points $\{y_i\in M; i\in\mathbb Z\}$ is called
$\varepsilon$-pseudo orbit for $g\colon M\to M$ if
$d(g(y_i),y_{i+1})<\varepsilon$, $i\in\mathbb Z$.
\end{definition}
\begin{definition}
We say that a real orbit $\{f^i(x); i\in\mathbb Z\}$
$\delta$-shadows an $\varepsilon$-pseudo orbit $\{y_i; i\in\mathbb
Z\}$ if $d(f^i(x),y_i)<\delta$.
\end{definition}
\begin{definition}
Diffeomorphism $g\colon M\to M$ is {\it quasi-Anosov} if for all
non-zero $v\in TM$ the sequence $\{\|T^iv\|; i\in\mathbb Z\}$ is
unbounded.
\end{definition}

We will be using the following characterization of Anosov systems.

\begin{theorem}[e.~g.~\cite{M}]
Diffeomorphism $g\colon M\to M$ is Anosov if and only if $g$ is
quasi-Anosov and all dimensions of stable manifolds at periodic
points are the same.
\end{theorem}
Dimensions of stable manifolds at periodic points of $\tilde f$ are
the same since $\tilde f$ is topologically conjugate to $f$. Hence
we only need to show that $\tilde f$ is quasi-Anosov.

Assume that $\tilde f$ is not quasi-Anosov. Then $\exists v\in
T_{\tilde x}M$, $\|v\|=1$, such that $\|T^nv\|\le 1$ for all
$n\in\mathbb Z$. Define $v_n=T^nv$, $n\in\mathbb Z$. For any
sufficiently small $\varepsilon>0$ consider sequence $\{\tilde
x_n=exp(\varepsilon v_n); n\in\mathbb Z\}$. Sequence $\{\varepsilon
v_n\}$ being $\varepsilon$-small and diffeomorphism being
$C^{1+Lip}$ imply that there exists a constant $\tilde c$ that
depends on $\tilde f$ only such that $\{\tilde x_n\}$ is $\tilde
c\varepsilon^2$-pseudo orbit which is obviously $\delta$-shadowed by
the orbit of $\tilde x$ with $\varepsilon/2<\delta<2\varepsilon$.

Let $x=h(\tilde x)$ and $x_n=h(\tilde x_n)$, $n\in\mathbb Z$.
Applying H\"older inequalities we get that $\{x_n\}$ is
$c_1\varepsilon^{2\alpha}$-pseudo orbit for $f$ that is
$\delta$-shadowed by $\{f^n(x)\}$ with
$\delta>c_2\varepsilon^{1/\beta}$. Constants $c_1$ and $c_2$ do not
depend on $\varepsilon$. To make it more transparent denote
$\xi=c_1\varepsilon^{2\alpha}$.

For arbitrarily small $\xi>0$ we have constructed a $\xi$-pseudo
orbit and a true orbit that $\delta$-shadows the pseudo orbit with
$\delta>c\xi^\kappa\stackrel{\mathrm{def}}{=}\xi^{1/2\alpha\beta}$
where $\kappa<1$ by the assumption. Meanwhile it is a well known
simple fact that $\delta$ can be estimated from above $\delta<C\xi$
where $C$ depends only on $f$. The proof is straightforward and
exploits local product structure of stable and unstable foliation of
$f$. For $\xi$ small enough these bounds on $\delta$ contradict each
other. Hence we have arrived at a contradiction.

redpajama_set_name: RedPajamaArXiv


text:
MixerModule::MixerModule(ModularSynth& synth)
	:SynthModule(synth, moduleId, 2, 2, 0)
{
}

MixerModule::~MixerModule()
{
}

void MixerModule::cycle()
{
	float mixing = getInput(1);
	float mixA, mixB;

	if (mixing < 0)
	{
		mixA = 1.0f;
		mixB = std::min(1.0f, std::max(0.0f, 1.0f - fabsf(mixing)));
	}
	else
	{
		mixA = std::min(1.0f, std::max(0.0f, 1.0f - mixing));
		mixB = 1.0f;
	}

	setOutput(0, mixA * getInput(0));
	setOutput(1, mixB * getInput(0));
}

const char * MixerModule::getInputName(int input) const
{
	static const char *names[] = {"Input", "Mixing"};
	return names[input];
}

const char * MixerModule::getOutputName(int output) const
{
	static const char *names[] = {"Output A", "Output B"};
	return names[output];
}

const char * MixerModule::getName() const
{
	return "Mixer";
}

SynthModule * MixerModule::createModule(ModularSynth& synth)
{
	return new MixerModule(synth);
}

redpajama_set_name: RedPajamaGithub


text:
Lightlife urges Impossible Foods and Beyond Meat to clean up their labels
In an ad in several newspapers, the 41-year-old company says its burger made with 11 easily recognized ingredients represents a future that is "cleaner, tastier, and nutritionally superior."
Courtesy of Lightlife
Megan Poinski @meganpoinski
UPDATE: Aug. 26, 2020: Impossible Foods posted a response on Medium that called Lightlife's campaign a "highly misleading" and "desperate attempt" to cast doubt on a product with which "it can't compete on quality or value." Impossible Foods pointed out that the campaign is financed by one of North America's largest animal agriculture companies, and said it was proud of its use of genetic engineering.
"Fortunately, a growing number of consumers recognize meat-industry propaganda and the agenda behind it. And when they go to the grocery store, they pick the product that delivers on taste, nutrition and sustainability. They pick Impossible Burger," the response states.
UPDATE: Aug. 25, 2020: Beyond Meat emailed Food Dive a written statement responding to the campaign. "If Lightlife were clear on our ingredients, they would see that our food is made from simple, plant-based ingredients. With no GMOs. No synthetic additives. No carcinogens. No hormones. No antibiotics. No cholesterol. Our foods are designed to have the same taste and texture as animal-based meat, giving more consumers more options that are better for them and the planet."
About a year ago, Lightlife President Dan Curtin challenged his team to make the cleanest plant-based burger they could create, using as few ingredients as possible, and ensuring they would be recognizable by consumers who flip over the package to see what's inside.
After launching the new version of the Lightlife Burger — which is made with 11 ingredients known to consumers — Curtin is making the same challenge to competitors in the plant-based meat space. An open letter to Impossible Foods and Beyond Meat signed by Curtin is running as an advertisement in major newspapers such as the New York Times and The Wall Street Journal.
"Enough with the hyper-processed ingredients, GMOs, unnecessary additives and fillers, and fake blood," the letter reads. "While we want the same things — a greener planet and a more sustainable food system — at Lightlife, we've chosen a very different way to get there. We're making a clean break from both of you 'food tech' companies that attempt to mimic meat at any cost."
Lightlife, which has been around for 41 years and is one of the original players in the meat-alternative market, has been working to keep abreast of what consumers want in plant-based meat. While the sector is booming — Lightlife sold more than 100 million servings of its plant-based meals in the last 12 months, Curtin told Food Dive — penetration is low. Polling done before the coronavirus pandemic indicated about half of consumers have tried plant-based meat, according to the International Food Information Council.
Curtin said Lightlife hired Boston Consulting Group to examine what consumers thought about plant-based meat. The goal, he noted, was for the company to find out what consumers wanted, and then try to deliver. The company connected with more than 11,500 people and the results were clear.
"The thing that screamed out to us time and time again was ... consumers are a little confused about plant based," Curtin said. "They looked at some of the ingredient decks. They didn't understand why there were so many ingredients in them. There were things in there that they weren't familiar with ... But 98% of the people we spoke to said that their needs were not being met, that they couldn't find what they were looking for."
It was obvious to Curtin what he needed to do next: simplify the ingredients list, minimize the process, ensure Lightlife is made out of things that consumers recognize and get competitors on board.
"They didn't understand why there were so many ingredients in them. There were things in there that they weren't familiar with ... But 98% of the people we spoke to said that their needs were not being met, that they couldn't find what they were looking for."
Dan Curtin
President, Lightlife
Curtin started out with the challenge issued to his team, led by Jitendra Sagili, chief of R&D and food technology officer. Curtin said he was unwilling to deviate from the core principles of what he wanted: Good taste, simple ingredients, a meat-like texture and a high-nutritional profile.
"It was not an easy task," Sagili said. "We took several months, several iterations."
He said Lightlife used fermentation from natural processes to transform some of these ingredients, like using mushrooms to create the extract for umami meat-like notes. Colors came from ingredients that were minimally processed and natural, such as beet powder, while onion powder and garlic provided some of the flavors.
Curtin and Sagili both highlighted Lightlife's manufacturing process. It's fairly easy to follow and understand, Curtin said, and every step is done in one of Lightlife's own plants. While the burgers are well-processed — many things need to be done before plants and other vegetables can become something that looks and performs like ground beef — Sagili said consumers can easily understand what Lightlife does.
"There is always a process involved, right?" Sagili said. "Right from the crop state, all the way to when you put it in food. Process is definitely the way you make it more edible and more presentable in an application."
Sagili said if he had all of the ingredients to make a Lightlife Burger, as well as common kitchen tools like blenders and mixers, he could make the burger in an actual kitchen — no factory required.
In the plant-based meat sector nowadays, it's rare for companies to call one another out. Although there are rivalries between companies, Impossible Foods and Beyond Meat — the two biggest solely plant-based meat companies —say they have just one competitor: the meat industry.
Lightlife looks at things differently. The brand is a division of Greenleaf Foods, the plant-based arm of Canada-based Maple Leaf Foods, which owns several large meat brands sold in that country.
Curtin said Lightlife is a pro-food company that supports consumer choice for protein. Lightlife does not anticipate the conventional meat industry will go away or be completely displaced by plant-based offerings, he said. Lightlife also does not advocate that consumers turn their backs on meat — though their products are suitable for vegans and vegetarians. Curtin said he wants to serve flexitarians — who sometimes eat plant-based foods — and reducetarians — who are trying to cut down on consumption of animal-based food.
"We want to be as encompassing as we can," Curtin said. "We don't want to alienate people because we don't believe that that's the right thing to do with it. Taking a more holistic approach is where we believe it's going to win at the end."
Curtin was not sure how Beyond Meat and Impossible Foods would react to the campaign. In the past, when outside organizations have attacked plant-based meat companies for not-so-clean-label products, Impossible Foods was quick to provide a snarky response. However, those critics were outside the industry, not part of it.
What Curtin hopes to see is other plant-based meat companies following Lightlife's lead for cleaner labels. Though in the end, he said the reaction he cares most about is the one from consumers.
"When they try it, they're going to have an amazing experience," Curtin said. "The products are absolutely delicious. And I think that that's what it's bringing a call to: What's going on and what we're doing. ...The consumers are going to, I think, be incredibly impressed with what they see and then what their experience is, and they'll be coming back for more and more and more."
Follow Megan Poinski on Twitter
Filed Under: Manufacturing Ingredients Corporate Meat / Protein Marketing
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Case note, in Latin, of the Genoese family mentioned in Doc [ID:3371]. Signed by William Batt.
Summary Case note, in Latin, of the Genoese family mentioned in Doc [ID:3371]. Signed by William Batt.
Case of the second son of Antonio Marchionne, who is suffering from epilepsy, as reported by William Batt.


Almost every single lady wants to settle down one day and have a family. But before that there is some things you need to know before you marry and that leads us to the topic for today what every woman should know before courtship.
The Macmillan dictionary defines courtship as a period of time when a man and woman have a romantic relationship with each other before marriage.
courtship can also be defined as when a brother and sister has reached an agreement that God is Leading them into marriage.
You must remember that not all courtship Leads to marriage.
One of the reasons why marriage fail is because of the mistakes people make in courtship.
-Dating is arrangement to meet as part of a sexual or romantic relationship.
– courtship is between two people who are ready for marriage.
Before courtship begins the man must have proposed and you as the lady accepted and your parents and family members know about it and then you both become officially engaged. so don't say you are in courtship without any marriage proposal.
-KNOW GOD: you can think about settling down without having GOD in mind because God is the one who will bring the right man to you.
– SERVE GOD: your service in the house of God is like storing money in the bank for rainy days as you wait for your man to locate you keep busy in the house of God because it pays.
-GET TO KNOW YOU: it is quite funny when you see a lot of young ladies who are preparing themselves for marriage without getting to know themselves, as a young lady you can't think of settling down if you don't know yourself and what you're capable of ,so before starting the courtship make sure you know your weaknesses and your strengths this will help you in the long run.
-PREPARE YOURSELF: as you wait for your partner keep busy and prepare yourself for marriage it is never too early to start attending relationship seminars.
Please note that you can't be locked up in your room and wear horrible clothes and expect to get a marriage proposal,go out: to places you know you will meet responsible men.
Ask him to give you time to pray and seek God's face make sure you introduce him to your parents and pastors.
Don't say yes to the first guy who seeks your hand in marriage because the right one may be the last one so be careful. when many men are asking for your hand in marriage all at the same time you need to pray so as to make the right decision don't for any reason whatsoever feel reluctant and decide to choose who you want to marry by yourself let God, your parents and your pastor help you in your decision REMEMBER THAT MARRIAGE IS A LIFE TIME THING.
NOTE that courtship should last 6 months or 2 years no more no less.
-don't enter into courtship with someone you are way older than unless you want to become a miserable baby sitter.
DON'T ENTER INTO COURTSHIP WITH SOMEONE YOUR PARENTS PASTORS OR EVEN GOD DISAPPROVE OF. REMEMBER LOVE IS BLIND AND MARRIAGE AN EYE OPENER. TRUST ME YOU DON'T WANT TO END UP WITH THE WRONG MAN.
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Top 3 Ebira Youths To Look Forward To in The Year 2019.


The MUSe team participates at the Nuit des Chercheurs event which takes place at Talence on the September 27th.
We will present our research activities on UAVs, NFC technology and dynamic networks.
It will be possible to pilot a drone with body movements, to live a short interactive tour of Talence through NFC tags deployed in the town. Also, a demonstration of the JBotSim software will be performed and finally, you will be invited to program ground robots based on the Arduino platform.


We've helped many clients with their financial needs. For us, there's no better feeling of helping a client get a great outcome! You can view the below testimonials and more on our Facebook and Google Business page.
I am so happy I chose Kim and her team at Kaboodle Finance to help me with finding a better option for my home loan. Having dealt with Kim in the past I knew I would be getting great service and I wasn't disappointed. From start to finish Kim and her team kept me up to date with the application process, answered all my questions, and were prompt in returning phone calls and/or emails. The service they gave ensured me that my business was important to them and never once did I feel like 'just another customer'. I would highly recommend Kaboodle Finance to anyone.
There is nothing this woman and her team can't do, we've been dealing with Kim for many many years. Everything is a simple straightforward, easy to understand process all the way till you sign the dotted line and beyond, she's never to busy to help or give advice. 5 starts is not enough to rate her service. Thanks Kim!!
Amazing service, they take the stress out of purchasing property. I'm so happy i chose Kaboodle Finance for my home loan. Friendly staff that go above and beyond to make everything so easy and simple. I'd like to thank the staff for answering all my enquiries no matter how trivial and always finding the right information to give me the best advice. It makes everything so easy having a professional who is willing to help, even with the little things. Highly recommend!
After going through a broker who had taken over a week to provide us with lender options (we were never provided them in the end) we were referred to Kim through a friend. She took our call after hrs, provided us with a lender over the phone, we sent her all our docs the following morning and that day our application for a pre-approval was accepted – all in under 24hrs! Kim is HIGHLY recommended. Quick, reliable and patient as we have never been through this process before.
Kim thank you so much!!!! We look forward to working with you further once we begin putting offers in for our first home!
The team at Kaboodle Finance have gone above and beyond to help us through our journey! Their friendly staff have helped us through refinancing and investing! These can be daunting journeys for individuals, families or couples and the team at Kaboodle Finance make it so easy! I can't recommend them highly enough!


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Precision Ag Bytes 6/6
carrie muehling June 6, 2018
Farm Market iD has released the 2018 Annual Update of their database. The update will provide complete details on farmers, crops, livestock and land for the 2017 crop year. The database covers 94 percent of major crop production in the United States going back more than a decade, and is collected from a combination of public and private sources.
As part of a 33-year effort to protect sensitive lands and improve water quality and wildlife habitat on private lands, the U.S. Department of Agriculture will resume accepting applications for the voluntary Conservation Reserve Program. Eligible farmers, ranchers, and private landowners can sign up at their local Farm Service Agency office between June 4 and Aug. 17, 2018.
The Association of Equipment Manufacturers (AEM) has presented Statistics Outstanding Merit Awards to 29 companies for their superior participation in AEM data reporting programs.
Building on its SeedCommand line of planting products, Ag Leader introduces SureForce. SureForce is a new, bidirectional hydraulic down force system with hydraulic uplift. Weight on the gauge wheels of every row is sensed 200 times per second, and force is automatically adjusted up or down to maintain consistent depth in changing soil conditions.
HELM Agro US, Inc. announced that the California Department of Pesticide Regulations has approved HELMET and HELMET SPC herbicides for use on selected crops in that state. HELMET and HELMET SPC are HELM Agro US' newest additions to the metolachlor herbicide portfolio. Both products deliver superior grass and small-seeded broadleaf control with the added benefit of residual activity. One application controls tough-to-manage weeds, including those resistant to glyphosate, PPO and ALS herbicides.
AGCO Corporation announces a development partnership with agricultural technology company AgIntegrated (AGI) to deliver ag information management systems services to AGCO customers. The partnership will enable AGCO customers to connect more easily with their agronomy service providers, create seamless data connectivity, and digitize the farm.
Deere & Company took legal action against Precision Planting LLC and AGCO Corporation, claiming that the companies have infringed on 12 John Deere patents. The Deere patents relate to several of the unique and inventive aspects of John Deere planting equipment, many of which are embodied in the state-of-the-art technology Deere has named ExactEmerge.
Concentric Ag Corporation (formerly Inocucor), a developer and producer of proprietary biological and essential plant nutrient inputs for specialty and broadacre crops, has hired Mark Trostle as vice president of global regulatory affairs. Trostle is an expert in federal, state and international regulations for crop inputs.


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Ian Murray MP Working Hard for Edinburgh South
People's Vote and Best for Britain
Elsie Inglis and the commemoration of women
This month we have been celebrating the life and legacy of Dr Elsie Inglis.
Born in India in 1864, she was the daughter of John Inglis, a chief commissioner in the Indian civil service. She studied medicine at Dr Sophia Jex-Blake's then newly opened Edinburgh School of Medicine for Women and was one of the first women in Scotland to finish higher education, although she was not allowed to graduate. She went on to complete her training under Sir William Macewen at the Glasgow Royal Infirmary.
The now famous exhortation, "My good lady, go home and sit still", was the response that Dr Elsie Inglis received when she asked the War Office whether female doctors and surgeons could serve in frontline hospitals in World War One. At that time and for many years to come, that was the attitude that women faced in making vital contributions to society.
Despite attempts to repress her efforts-and those of many other women-to contribute, Elsie did not, in the words of the exhortation, "sit still". Instead, she persevered, setting up the Scottish women's hospitals on the frontlines. These hospitals were all-female units that played a vital role with Britain's allies, including the French, the Belgians and the Serbs, who regard her as a hero.
This month in Parliament I was honoured to speak at a debate in parliament on Dr Inglis and the contribution of women to World War One. You can watch my speech here:
Earlier in the month I was also privileged to present an award to the inaugural winner of the Dr Elsie Inglis-Scottish Women's Hospitals' Trust Medal and Prize. Thank you to Ian McFarlane and the Trust for organising and preparing me for the debate.
You can also read about why I support the campaign for a statue of Elsie Inglis in Edinburgh: https://www.edinburghnews.scotsman.com/news/health/men-back-statue-of-pioneering-edinburgh-woman-elsie-inglis-1-4620381
31 Minto Street, Edinburgh, UK View on Google Maps
Email: ian.murray.mp@parliament.uk
Promoted by Lewis Wotherspoon on behalf of Ian Murray MP, both at 31 Minto Street, Edinburgh, EH9 2BT


Well, I haven't been updating this much over the last week as nothing very exciting has been happening, so there was no point in boring you. The weekend was pretty good as Chris took the train up from Hamburg and I gave him a tour of Copenhagen. No, that's not true, we just went out drinking but he did get to see the famous statue of the mermaid.
It's amazing to think that I've been back in Copenhagen for three weeks, but already I feel the lust of new places pulling me away again. I've spent much of the last week looking into flights and reading travel books as Nina and I plan how to spend her time away from university. We haven't made any concrete decisions yet, but when we do you'll be the first to know.
When we were in South Africa in January we passed a serious looking motorboat on the quay in Cape Town. It had www.ms-starship.com on the superstructure so I wrote it down with the intention of finding out just what it was up to. Tonight I finally got around to their website, and it is very impressive with daily updates of their position and photographs from the day online. One day I want a satellite uplink for Zamindar! Check it out.


Game of Thrones is one of those great, unlikely Hollywood success stories. Showrunners David Benioff and Dan Weiss took a complicated series of fantasy novels and, under the care of HBO, crafted a cultural phenomenon. At its best, Game of Thrones makes incorporating dense, compelling characters, and a wide scope into a single package look easy.
But making anything this ambitious is never easy. Each season of Game of Thrones is an ordeal and a half to create. And the roadblocks arrived early – it's no secret that the show's original pilot had to be almost entirely reshot. Now, years later, Benioff and Weiss seem comfortable talking about their disastrous first stab at this show and it's the kind of behind-the-scenes story that is equally horrifying and uplifting.
And to their credit, they did go about changing everything. Director Tom McCarthy (a newly minted Oscar nominee for Spotlight) was replaced by television veteran Tim Van Patten and a reported 90% of the pilot was reshot and restructured. The role of Catelyn Stark was recast, with Michelle Fairley replacing Jennifer Ehle. Emilia Clarke replaced Tamzin Merchant in the role of Daenerys Targaryen. Entire sequences – like a flashback to the execution of Ned Stark's father and the onscreen death of Jon Arryn – were cut. More importantly, some vague information was clarified, as it wasn't clear that Jaime and Cersei were siblings in the original cut, making their shocking final scene totally weightless.
I will never forget being invited to the premiere of the first season. I went in just thinking, 'Well, I guess we'll just see how this goes.' I sat there and this show unfolds and I am stunned. Stunned. And I very specifically remember walking out and I said to [Weiss and Benioff], 'That is the biggest rescue in Hollywood history.' Because it wasn't just that they had saved something bad and turned it really good. You had saved a complete piece of sh*t and turned it into something brilliant. That never happens!
And the rest, as they say, is history. If you want to know more of the finer details, the Vanity Fair article digs deeper into other aspects of the original pilot and it is well worth your time. In the meantime, Game of Thrones season 6 is set to premiere on April 24. No full trailers have been revealed yet, but here's all you need to know: everyone is dead or dying.


Could a high INR causes brain damage such as stroke. If yes ,severe headaches could it be the symptoms?
How high are you talking about? This is something that you really need to talk to your Doctor or coumadin clinician about your concerns. It would be better if you got your answer from a professional and not someone on this site as we all will react to coumadin differently or not at all.