text:
\section{Introduction}

The lines of curvature of surfaces in $\mathbb R^3$ have been a subject of interest for several centuries, since G. Monge described
the case of the triaxial ellipsoid by direct integration
of the differential equation of these lines. This is the first example of a foliation with singularities
on a compact surface \cite{Monge}.
Despite the fact that many studies have focused on different properties of local and global nature of the lines of curvature of surfaces in $\mathbb R^3$,
(see for instance \cite{Soto}, \cite{Bruce}, \cite{S-X}, \cite{Hopf}, \cite{Spi}),
a complete analytic description of them has not been achieved for the case of an oriented compact surface of genus higher than
one. A reason for this may be that, contrasting with the ellipsoid case which is a quadric surface,
embeddings of compact surfaces of higher genus are quartic or even higher degree algebraic surfaces, \cite{Griffits, Hirsch} and consequently, their
lines of curvature are described by rather complicated differential equations.
The simplest case in this class of surfaces is the double torus.

In the present article we present a description of the foliation of lines of curvature
of an embbeding of the double torus into $\mathbb R^3$.
In order to avoid the problem arising from the degree of the algebraic surface in $\mathbb R^3$
we consider an embedding of the double torus into $\mathbb R^4$ as the link of the real part of the Milnor fibration of a polynomial with isolated singularity at the origin.
Specifically, this surface is embedded into the sphere $\mathbb S_r^3 \subset \mathbb R^4$ of radius $r$ with center at the origin for which
we analyze the lines of curvature with respect to the gradient $\nu$, referred to in our research as $\nu$-lines of curvature, of this polynomial
on the link.

The $\nu$-lines of curvature of surfaces, where $\nu$ is a normal vector field,
have been studied in $\mathbb R^4$ in \cite{F} and  \cite{Ga-S}, where the expression of the differential equation is given in the Monge chart, in which
the surface is defined as the graph of a differentiable function.
In this article, we introduce an equation of these lines according to the description of the surface as the intersection of the loci of two polynomials,
which simplifies the analysis. This equation, together with the symmetries of the link, allow us to apply elements from both algebraic geometry and dynamical systems
to describe the foliation completely. Then, we apply the stereographic projection which transforms the $\nu$-lines of curvature of this spherical embedding into the lines of curvature of the image of the link in $\mathbb R^3$.

The article is organized as follows: in section 2 we present some preliminaries and the formulation of the
differential equation of $\nu$-lines of curvature that will be used in the study (Proposition \ref{3.2} and Corollary \ref{equationframe}). In section 3, we analyze the embbeding of the double torus into $\mathbb R^4$. Namely, we determine a decomposition of the surface
parameterized with convenient coordinate charts which allow us to use the symmetries of the surface to simplify the study. In section 4, we prove the main theorem
which provides the description of the foliation of the $\nu$-lines of curvature on the double torus embbeded into $\mathbb R^4$ as a transversal intersection (Theorem \ref{Maintheorem}).
As a corollary we get the description of the lines of curvature of the image of this embedding under the stereographic projection in $\mathbb R^3$
(Corollary \ref{torusinR3}).

\section{Preliminaries}

\subsection{The $\nu$-lines of curvature of a surface in $\R^4$.}

Let $M$ be a smooth oriented surface immersed in $\R^4$ with the Riemannian
metric
induced by the standard Riemannian metric of $\R^4$. For each $p \in M$
consider
the decomposition $T_p \R^4 =T_pM \oplus N_pM$, where $N_pM$ is the
orthogonal complement
of $T_pM$ in $\R^4$.
Let $\bar \nabla$ be the Riemannian connection of $\R^4$. Given local
vector fields
$X, \ Y$ on $M$, let $\bar X, \ \bar Y$ be local extensions to $\R^4$.
The tangent component of the Riemannian connection in $\R^4$ is the Riemannian
connection of $M: \nabla_XY=(\bar \nabla_{\bar X}\bar Y)^{\top}.$

Let ${\mathcal X}(M)$ and ${\mathcal N}(M)$ be the space of the smooth
vector fields tangent to $M$ and
the space of the smooth vector fields normal to $M$, respectively.
Consider the {\it second fundamental form},
$$II: {\mathcal X}(M)\times {\mathcal X}(M) \rightarrow {\mathcal N} M,
\ II(X,Y) = \bar \nabla_{\bar X}\bar Y - \nabla_XY.$$
This map is symmetric and bilinear. So, at each point $p$, $II(X,Y)$ only depends on
$X(p)$ and $Y(p)$.

Suppose the $\nu$ is a unitary vector field in ${\mathcal N}(M)$. {\it The $\nu$-second fundamental form} of $M$ at $p$ is the quadratic form,
$$II_{\nu}: T_pM \rightarrow \R,\ II_{\nu}(X)= \langle II_p(X,X), \nu \rangle.$$

Recall the {\it $\nu$-shape operator}
$$S_{\nu}: T_pM \rightarrow T_pM,
\ S_{\nu}(X)=-(\bar \nabla_{\bar X}\bar \nu)^{\top} ,$$
\noindent where $\bar \nu$ is a local extension to $\R^4$ of the normal
vector field $\nu$
at $p$ and $\top$ means the tangent component. This operator is
self-adjoint and, for
any $X,\ Y \in T_pM$, satisfies the following equation:
$$<S_{\nu}(X),Y>=<II(X,Y), \nu>.$$

Thus, for each $p \in M$, there exists an
orthonormal
basis of eigenvectors of $S_{\nu} \in T_pM$, for which the restriction of
the second
fundamental form to the unitary vectors, $II_{\nu}|_{S^1}$, takes its
maximal and
minimal values. The corresponding eigenvalues $k_1,\ k_2$ are the {\it maximal}
and {\it minimal $\nu$-principal curvatures}, respectively. The point $p$ is a
$\nu$-umbilic if the $\nu$-principal curvatures coincide.
Let ${\mathcal U}_{\nu}$ be the set of $\nu$-umbilics in $M$.
For any $p \in M \backslash {\mathcal U_{\nu}}$  there are two $\nu$-principal
directions
defined by the eigenvectors of $S_{\nu}$. These fields of directions are
smooth and
integrable, and therefore they define, in  $M \backslash {\mathcal U_{\nu}}$, two families of orthogonal curves, its integrals,
which are called the  {\it $\nu$-principal lines of curvature}, one
maximal and the other
one minimal. The $\nu$-umbilics are considered as the singularities of these foliations.
The differential equation of $\nu$-lines of curvature is
\begin{eqnarray*}
S_{\nu}(X(p))=\lambda(p)X(p)
\end{eqnarray*}

\subsection{Equation of $\nu$-lines of curvature of surfaces defined implicitly in 4-space}

We first determine a characterization of a $\nu$-principal direction at a point of a surface
provided with a normal frame $\{\nu, \mu \}$.

\begin{prop}
Let $M$ be a surface immersed in $\mathbb R^4$.
Assume that $\{\nu, \mu \}$ is a frame of the normal
bundle on $M$.
Consider the unitary vector field $\tilde \nu= \frac{\nu}{||\nu||}$.
If $p \in M$, the vector $X \in T_pM$ is tangent to a $\tilde \nu$- principal direction, if and only if
the following equation holds at $p$:

\begin{eqnarray}\label{equation1}
\langle \bar \nabla_X \nu\wedge X \wedge \mu,\ \nu \rangle=0.
\end{eqnarray}

\end{prop}

\noindent{\bf Proof.} We decompose the covariant derivative as the sum of the following
vector fields:
$$\bar \nabla_X \nu=(\bar \nabla_X\nu)^{\top}+p_{\mu}(X)\mu + p_{\nu}(X)\nu,$$
\noindent where  $p_{\mu},\ p_{\nu}: T_pM \rightarrow \R$ are the projections of
$\bar \nabla_X \nu$ on the lines determined by the normal frame $\{\nu, \mu\}$ at $p$.

Therefore,
\begin{eqnarray}
\langle \bar \nabla_X \nu\wedge X \wedge \mu, \nu \rangle & = & \langle (\bar \nabla_{X} \nu)^{\top} +
p_{\mu}(X)\mu + p_{\nu}(X)\nu\wedge X \wedge \mu, \nu \rangle \nonumber \\
& = & \langle (\bar \nabla_{X} \nu)^{\top} \wedge X \wedge \mu, \nu \rangle.
\end{eqnarray}

Observe that the last expression vanishes if and only if $(\bar \nabla_{X} \nu)^{\top}$ and $X$ are
linearly dependent.

If $\nu$ is unitary, $-(\bar \nabla_{X} \nu)^{\top}=S_{\nu}(X)$ is the $\nu$-shape operator. Then, equation $(\ref{equation1})$ holds
if and only if $S_{\nu}(X)$ and $X$ are linearly dependent; this is equivalent to the fact that $X$ must be an eigenvector of the $\nu$-shape operator.

On the other hand, if $\nu$ is not unitary, then,

\begin{eqnarray*}
(\bar \nabla_{X} \tilde \nu)^{\top} & = & \left(\bar \nabla_{X} \frac{\nu}{||\nu||} \right)^{\top} \\
& = & \left (X \left(\frac{1}{||\nu||}\right)\nu + \frac{1}{||\nu||}(\bar \nabla_{X}\nu) \right)^{\top} \\
& = & \frac{1}{||\nu||}\left(\bar \nabla_{X}\nu \right)^{\top}.
\end{eqnarray*}

\noindent Therefore, from this equation we have
\begin{eqnarray*}
\langle \bar \nabla_X \nu\wedge X \wedge \mu, \nu \rangle & = &
\langle (\bar \nabla_{X} \nu)^{\top} \wedge X \wedge \mu, \nu \rangle \\
& = &- \langle ||\nu || S_{\tilde \nu}(X) \wedge X \wedge \mu, \nu \rangle.
\end{eqnarray*}

\noindent Thus, we conclude that equation $(\ref{equation1})$ holds if and only if $X$ is an eigenvector of the $\tilde \nu$- shape operator. $\hfill\Box$

\begin{rem}
Observe that this proposition allows us to consider any non-unitary normal vector field $\nu$ on $M$ in order
to define the $\tilde \nu$- principal directions of curvature.
This fact will be used in the setting when we consider non-unitary gradient vector fields to define the lines of curvature.
\end{rem}

Consider now a surface $M$ defined as the transversal intersection of the inverse images of regular values of two differentiable functions,
there is a natural frame of the normal bundle of $M$, given by the gradient vector fields
of both functions. Thus, $M$ is endowed with a natural pair of lines of curvature.
A direct computation that uses the multilinearity properties of the left hand side of $(\ref{equation1})$
provides the following expression for the equation of these lines of curvature:

\begin{prop}\label{3.2}
Let $F,G:\mathbb R^4 \rightarrow \mathbb R$ be a pair of differentiable functions. Suppose that $M$ is a surface defined as the transversal intersection of
the inverse image of two regular values of these functions.
Let $\nu$ and $\mu$ be the gradient vector fields of $F$ and $G$, respectively.
Then, the differential equation of the $\nu$-lines of curvature of $M$ is given by:

\begin{eqnarray}
\left( dx_{1} ,\ dx_{2},\  dx_{3},\ dx_{4} \right)
\left(\begin{array} {cccc}
    \Omega_{11} & \Omega_{12} & \Omega_{13} & \Omega_{14}  \\
     \Omega_{12} & \Omega_{22} & \Omega_{23} & \Omega_{24} \\
      \Omega_{13} & \Omega_{23} & \Omega_{33} & \Omega_{34} \\
\Omega_{14} & \Omega_{24} & \Omega_{34} & \Omega_{44} \\
\end{array} \right)\left(\begin{array}{c} dx_1 \\ dx_2 \\ dx_3 \\ dx_4 \end{array}\right)=0, \quad
\label{curvaturelines}
\end{eqnarray}

\noindent where $\{x_1,x_2,x_3,x_4\}$ is the system of parameters of $\mathbb R^4$ and

  $$
\Omega_{ij}= \frac{1}{2}\left(\langle \bar \nabla_{\frac{\partial}{\partial x_i}} \nu
\wedge \frac{\partial}{\partial x_j}\wedge \mu, \nu \rangle +
\langle \bar \nabla_{\frac{\partial}{\partial x_j}}
\wedge \frac{\partial}{\partial x_i} \wedge \mu, \nu \rangle\right).
  $$
\end{prop}

\begin{rem}
This proposition still holds if $p$, $q$ or both are singular values of any of the two functions, but only if $M= F^{-1}(p) \cap G^{-1}(q)$
lies in the complement of the
corresponding critical points. The other hypothesis remain under assumption.
\end{rem}

In case we have a frame of the tangent bundle of $M$, equation $( \ref{curvaturelines})$ has a simpler expression.

\begin{coro}\label{equationframe}
Assume that $V_1, V_2$ is a frame of the tangent bundle of $M$. Then, the expression of the differential equation $( \ref{curvaturelines})$ is:
\begin{eqnarray}
l_1^2 \Omega_{11} + l_1 l_2 \left( \Omega_{12} +  \Omega_{21} \right)+ l_2^2  \Omega_{22}=0
\end{eqnarray}

\noindent where $\Omega_{ij}=\langle \nabla_{V_i} \nu \wedge V_j \wedge \mu, \nu \rangle, \ \ i,j \in \{ 1,2\} $
and $l_i$ is the $i$-component of the vectorial solution $X$ of the equation, namely: $X= l_1V_1 + l_2V_2$.
\end{coro}

\section{An embbeding of the double torus into $\mathbb R^4$}

Let us analyze a family of examples where the $\nu$-lines of curvature can be described using the
expressions presented above.  In \cite{Seade},
J. Seade studied the topology of a class of surfaces obtained as the projections of certain complex hypersurfaces
onto a real line through the origin of $\mathbb C$. Let us consider an element of that class:

  $$
\bar F:\mathbb C^2 \rightarrow \mathbb C,\ \ \bar F(z_1,z_2)= z_1^p + z_2^q,\ \ 1< p,q \in \mathbb N.
  $$

\noindent The hypersurface defined by the real part of $\bar F:\ Re\ \bar F= z_1^p+ z_2^q+ \bar z_1^p + \bar z_2^q$
is diffeomorphic to the cone over the link $L_r= Re\ \bar F^{-1}(0)\cap \mathbb S^3_{r},\ r>0$. Moreover,
$L_r$ is a closed oriented surface of genus $(p-1)(q-1)$
in $\mathbb S^3_r \subset \mathbb R^4$.
For instance, consider the case $p=2,q=3$ in which $L_r$ is a double torus in $\mathbb S^3_r \subset \mathbb R^4$.
We provide an elementary proof of this fact in the following proposition.

A direct computation
with the help of the standard identification between $\mathbb C^2$ and $\mathbb R^4$ given by
  $$
(z_1=x+iy,z_2=u+iv) \longmapsto (x,y,u,v)
  $$
\noindent shows that $Re\ \bar F$ has the following expression as a function defined in $\mathbb R^4$:

\begin{eqnarray}\label{FF}
F:\mathbb R^4 \rightarrow \mathbb R,\ \  F(x,y,u,v)=x^2- y^2 + u^3 - 3 u v^2.
\end{eqnarray}

\noindent Therefore, if we consider the function

\begin{eqnarray}\label{GG}
G_{r}:\mathbb R^4 \rightarrow \mathbb R,\ \ G_{r}(x,y,u,v)=x^2+ y^2 + u^2 + v^2 - r^2,
\end{eqnarray}

\noindent for any positive $r \in \mathbb R$, the intersection $F^{-1}(0) \cap G_{r}^{-1}(0)$ is the link $L_r$. Let us denote it by $T_{r}(2)$.

The transversality of the intersection is determined in terms of the gradient vector fields. Namely,
let $F,G:\mathbb R^{4} \rightarrow \mathbb R$ be a pair of differentiable functions. Assume that
$ p \in F^{-1}(a) \cap G^{-1}(b)$ is a regular point for both functions. The intersection of $ F^{-1}(a)$ and $G^{-1}(b)$
is transversal at $p$ if and only if the gradient vectors $\mu$ and $\nu$ of these functions are linearly independent at $p$. Thus, we state the following:

\begin{prop}
The intersection $T_{r}(2)=F^{-1}(0) \cap G_{r}^{-1}(0), \  r>0$ is a smooth surface.
\end{prop}

\noindent
{\bf Proof.}  We prove that the inverse images of zero $F^{-1}(0)$ and $G_{r}^{-1}(0)$  intersect transversally.
A direct computation shows that the origin is the unique singular point of both functions, $F$ and $G$. Since it does not lie in $G_{r}^{-1}(0)\ ,\ r>0,$
then it lies off $T_{r}(2)$.
Now, we show that the gradient vector fields $\mu(x,y,u,v)= (2x,-2y, 3u^2-3v^2,-6uv)$
and $\nu(x,y,u,v)=2(x,y,u,v)$ are linearly independent along $F^{-1}(0) \cap G_r^{-1}(0)$, if $r >0$.
To do so, we consider the equation:

\begin{eqnarray}
(2x,-2y, 3u^2-3v^2,-6uv) = \lambda(x,y,u,v),\ \lambda \in \mathbb{R}\setminus\{ 0 \} \label{5}
\end{eqnarray}

\noindent Suppose that $x \neq 0$. Then, $\lambda = 2$ and $y=0$. This implies
that $2v=-6uv$. Consequently,
if $v \neq 0$, then $u= -\frac{1}{3}$ and the relation of the third coordinates above
implies that $2u= 3u^2 - 3v^2$, namely, $v^2=\frac{1}{3}$. By assuming that $G_{r}$ vanishes at $(x,0,-\frac{1}{3},\frac{1}{\pm \sqrt{3}})$,
we get $x^2=r^2- \frac{4}{9}$. By supposing that $F$ vanishes at this point,
we get $x^2<0$ which is a contradiction.  So, if $x \neq 0$ then $v=0$ and $y=0$.
This implies that $u=\frac{2}{3}$.
By evaluating $G_r$ at a point of the form $(x,0,\frac{2}{3},0)$, we get again a contradiction
.
Therefore, $x=0$. Analogous considerations derived from assuming that points of the form $(0,y,u,v)$ are in the intersection of the loci of $G_r$ and $F$ provide
straightforward contradictions
. Therefore, the gradient vector fields are linearly independent at any point of $T_r(2)$.
$\hfill\Box$

Now, we consider the symmetries of $T_{r}(2)$.
It is straightforward to verify that this surface is invariant under the action of the group
$\mathcal G$ generated by the following applications:

$$\Gamma_i: \mathbb C^2 \rightarrow \mathbb C^2,\ \ \ i=1,2,3,$$
 $\Gamma_1(z_1,z_2)=(\bar z_1, z_2)$, $\Gamma_2(z_1,z_2)=(-\bar z_1, z_2)$ and $\Gamma_3(z_1,z_2)=(z_1, e^{\frac{2\pi i}{3}} z_2)$.

\medskip

Let us describe this embedding of the double torus.

From the very definition of the functions $F$ and $G$ we have
four charts of $T_r(2)$ given by the projection onto the $uv$-plane of the following hexagonal regions.

\begin{eqnarray}\label{regions}
H_{++} & = & \lbrace (x,y,u,v) \in T_r(2) : x \geq 0, y \geq 0 \rbrace,  \nonumber \\
H_{+-} & = & \lbrace (x,y,u,v) \in T_r(2) : x \geq 0, y \leq 0 \rbrace, \\
H_{-+} & = & \lbrace (x,y,u,v) \in T_r(2) : x \leq 0, y \geq 0 \rbrace, \nonumber \\
H_{--} & = & \lbrace (x,y,u,v) \in T_r(2) : x \leq 0, y \leq 0 \rbrace. \nonumber
\end{eqnarray}

We consider these sets with boundary, since it will be useful to identify corresponding boundaries in the construction below.
The projection $\mathbb{R}^{4} \rightarrow \mathbb{R}^{2}$
\[ (x,y,u,v) \mapsto (u,v) \] defines the hexagon $H_{uv}$ on the $uv$-plane as follows.

\[ H_{uv}=\lbrace (u,v) \in \mathbb{R}^{2} \quad :  \quad
r^{2}-u^{3}+3uv^{2}-u^{2}-v^{2} \geq 0, \quad r^{2}+u^{3}-3uv^{2}-u^{2}-v^{2} \geq 0 \rbrace. \]

Thus, we consider the coordinate functions

$$\varphi_{\pm  \pm}=\left(\pm \frac{\sqrt{r^2-u^2-u^3-v^2+3 u v^2}}{\sqrt{2}}, \pm \frac{\sqrt{r^2- u^2 + u^3- v^2 - 3 u v^2}}{\sqrt{2}},u, v \right),$$
where the domain of these applications is $H_{uv}$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=7cm]{HexagonoB.eps}
\end{center}
\caption{Projection of the double torus on the $uv$-plane.}
\label{Hexagon}
\end{figure}

The boundary of $H_{uv}$ is constituted by
\[ \partial H_{uv} = X^{1} \cup Y^{1} \cup X^{2} \cup Y^{2} \cup X^{3} \cup Y^{3}, \]
where

\begin{eqnarray*}
X^{1} & = & \lbrace (u,v) \quad : \quad r^{2}-u^{3}+3uv^{2}-u^{2}-v^{2}=0, \quad u \geq 0 \rbrace, \\
Y^{1} & = & \lbrace (u,v) \quad : \quad r^{2}+u^{3}-3uv^{2}-u^{2}-v^{2}=0, \quad u \geq 0, \quad v \geq 0 \rbrace, \\
X^{2} & = & \lbrace (u,v) \quad : \quad r^{2}-u^{3}+3uv^{2}-u^{2}-v^{2}=0, \quad u \leq 0, \quad v \geq 0 \rbrace, \\
Y^{2} & = & \lbrace (u,v) \quad : \quad r^{2}+u^{3}-3uv^{2}-u^{2}-v^{2}=0, \quad u \leq 0 \rbrace, \\
X^{3} & = & \lbrace (u,v) \quad : \quad r^{2}-u^{3}+3uv^{2}-u^{2}-v^{2}=0, \quad u \leq 0, \quad v \leq 0 \rbrace, \\
Y^{3} & = &\lbrace (u,v) \quad : \quad r^{2}+u^{3}-3uv^{2}-u^{2}-v^{2}=0, \quad u \geq 0, \quad v \leq 0 \rbrace.
\end{eqnarray*}

\noindent Observe that

\[ T(2)_+ =  T_r(2) \cap \lbrace x \geq 0 \rbrace = H_{++} \cup H_{+-} \]
\[ T(2)_- =  T_r(2) \cap \lbrace x \leq 0 \rbrace = H_{-+} \cup H_{--}. \]

Therefore, $T(2)_+$ is constructed by gluing the closure of two hexagonal regions through the identification of the curves defined by the equation $y=0$.
That is, $T(2)_+$ is a surface homeomorphic to the sphere minus tree discs (see figure $2$).

\begin{figure}[htb]
\begin{center}
\includegraphics[width=7cm]{Esfera3hoyos.eps}
\end{center}
\caption{First identification.}
\label{Esferatreshoyos}
\end{figure}

Since
$T(2)_-$ is the image of $ T(2)_+$ by the reflection with respect to the hyperplane $x=0$, we obtain
$T(2)$ by gluing two surfaces of this type along the boundary of the missing discs.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=7cm]{mediotorodoble.eps}
\end{center}
\caption{A half of a double torus.}
\label{Hexagon}
\end{figure}

\begin{rem}\label{conjugaciontraslacion}
Let us denote by $T_{\pm \pm}: \mathbb C^2 \rightarrow \mathbb C^2$, the translation of the origin to any of the points $\varphi_{\pm  \pm}(0,0)$, respectively.
Then, the conjugation $T_{\pm \pm}^{-1}\circ g  \circ T_{\pm \pm}$, where $g$ belongs to the group generated by $\Gamma_3$, leaves $T(2)$ invariant.
\end{rem}

We describe the action of $\mathcal G$ on $T_r(2)$. Each point in the interior of $H_{uv}$ represents 4 points identified by the action of the
subgroup of reflections of $\mathcal G$ on the torus.
We denote by $I$ the set of the intersections of the curves $X_i$ and $Y_j$, $i,j = 1,2,3.$
Each point in the boundary $H_{uv}\setminus I $ represents 2 points, while, points in $I$ represent only one point of $T_r(2)$.

Observe that it is enough to determine this region only in $H_{++}$, since
any of the hexagonal regions described in $(\ref{regions})$ are obtained from this one by one reflection.

We analyze the action of $\Gamma_3$ on the closure of the hexagonal region $H_{++}$. So,
consider the following subset of $H_{++}$
\begin{eqnarray}\label{pentagono}
P_{++} = \lbrace (x,y,u,v) \in H_{++} | v \geq 0, v \geq - \sqrt{3} u \rbrace.
\end{eqnarray}
This subset is the closure of a pentagonal region.
Let $P_0$ be the image of the projection of $P_{++}$ in the plane $uv$ (see figure \ref{Pentagon}). We describe the boundary of $P_0$ as follows:
The line segment which goes from the origin to the midpoint of $X^{1}$ is denoted by $e_{1}$;
the upper half of $X^{1}$ is denoted by $e_{2}$; the edge $Y^{1}$ is denoted by $e_{3}$;
the upper half of $X^{2}$ is denoted by $e_{4}$ and finally;
$e_{5}$ denotes the line segment which goes from the midpoint of $X^{2}$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=7cm]{Pentagono.eps}
\end{center}
\caption{A pentagonal fundamental region}
\label{Pentagon}
\end{figure}

If we rotate this pentagon $P_{0}$ by $2\pi/3 $ we obtain another pentagon $P_{1}$, and if we rotate $P_{1}$ we obtain a third pentagon $P_{2}$.
The closure of the projection of $H_{++}$ is the union of closure of these three pentagons. We can define analogously, pentagonal regions in
$P_{-+},\ P_{+-}$ and $P_{--}$ contained in the other hexagonal regions.

\section{Lines of curvature of the double torus}

\subsection{The lines of curvature of the double torus in $\mathbb R^4$}

We consider the unitary vector field

$$ \nu= \frac{F}{||F||},\ {\rm where}\
\nabla F=(2 x, -2 y, 3 u^2 - 3 v^2, -6 u v)
  $$
is the gradient vector field of $F$
on $T_{r}(2)$.

Proposition \ref{3.2} provides the differential equation of $\nu$-lines of curvature
of the double torus $T_r(2)$ immersed in $\mathbb S^3 \subset \mathbb R^4$.

\begin{coro}\label{EcuaDiff}
The differential equation of the $\nu$-lines of curvature of $T_r(2)$ has the following expression:
{\small
\begin{eqnarray}
dX
\left(\begin{array} {cccc}
    0 & 6v(v^2-3u^2) & -yv(2-18u+9\rho^2) & y(9(u^2-v^2)+u(2+9\rho^2))  \\
     * & 0 & xv(2+18u+9\rho^2) & x(9(u^2-v^2)-u(2+9\rho^2)) \\
      * & * & -24xyv & -24xyu \\
* & * & * & 24xyv\\
\end{array} \right)dX^T=0, \quad
\end{eqnarray}}

\noindent where $dX=\left( dx ,\ dy,\  du,\ dv \right)$, $\rho^2=u^2+v^2$ and the
values of $*$ are determined by the symmetry of the matrix.

\end{coro}

We provide the description of the foliation of $\nu$-lines of curvature of the double torus
$T_r(2),\ r>0$.
As is common, we consider {\it the index of an isolated $\nu$-umbilic point} as the index of an isolated singularity
of the field of principal directions on the surface, \cite{Hopf}.
According to the classification provided in \cite{Soto}, (see also \cite{Bruce})
we refer to a {\it $\nu$-umbilic point of type $D_3$} as a generic $\nu$-umbilic point with three separatrices and index $-\frac{1}{2}$.
We define a {\it pathwise separatrix of the $\nu$-principal foliation} as a smooth curve constituted by the union
of separatrices, some of them of maximal $\nu$-principal curvature and the other of minimal $\nu$-principal curvature.
In the theorem below, we will omit the prefix $\nu$ for the elements of the foliation of $\nu$-lines of curvature, such as
$\nu$-umbilic points, $\nu$-separatrices, etc.

\begin{teo}\label{Maintheorem}
The foliation of $\nu$-lines of curvature of the double torus $T_r(2)$ has the following structure:

\noindent i) It only has 4 umbilic points, each one of type $D_3$.

\noindent ii) It has $3$ pathwise separatrices; each one is an immersion of $\mathbb S^1$
containing the whole set of umbilics.

\noindent iii) The first homology group of the double torus $T_r(2)$ is generated by
1-cycles that are pathwise separatrices.

\noindent iv) The complement of the closure of the set of separatrices is foliated by cycles, that is, closed curvature lines diffeomorphic
to $\mathbb S^1$.

\end{teo}

We begin by proving statement $i)$ in the following two propositions.

\begin{prop}\label{4.1}
Let $r \in \mathbb R$ such that $0<r<\frac{1}{\sqrt{10}} $.
The foliation of $\nu$-lines of curvature of $T_r(2)$ has 4 umbilic points.
\end{prop}

\noindent{\bf Proof.} Let us consider the open subset $T_r(2)^1= \{(x,y,u,v)\in T_r(2)| xy \neq 0  \}$ of $T_r(2)$. We define on $T_r(2)^1$ the frame:

\begin{eqnarray*}
V_1^1(x,y,u,v) & = & -\left(\frac{v-3uv}{2x}\right)\frac{\partial}{\partial x}-
\left(\frac{v+3uv}{2y}\right)\frac{\partial}{\partial y}+
\frac{\partial}{\partial v}, \\
V_2^1(x,y,u,v) & = & -\left(\frac{2u+ 3u^2-3v^2}{4x}\right)\frac{\partial}{\partial x}-
\left(\frac{2u- 3u^2+3v^2}{4y}\right)\frac{\partial}{\partial y}+
\frac{\partial}{\partial u}.
\end{eqnarray*}

\noindent After multiplying
the coefficients of the differential equation in Theorem $\ref{equationframe}$ of $\nu$-lines of curvature on $T_r(2)^1$ by the monomial $xy$,
they have the expression:

\begin{eqnarray*}
\Omega_{11}&=& v (-18 u^3 (x^2 - y^2) + 2 u (1 + 18 v^2) (x^2 - y^2) +
   27 u^4 (3 v^2 + x^2 + y^2) - \\
  && 3 u^2 (9 v^4 + x^2 + y^2 + v^2 (3 - 9 x^2 - 9 y^2)) +
   3 (v^4 + 8 x^2 y^2 + 3 v^2 (x^2 + y^2)))\\
\Omega_{12}&=& \frac{1}{2}(36 u^4 (x^2 - y^2) - 54 u^2 v^2 (x^2 - y^2) +
 2 v^2 (-2 + 9 v^2) (x^2 - y^2) \\
&&- 27 u^5 (3 v^2 + x^2 + y^2) +
 6 u^3 (-3 v^2 + 18 v^4 - 2 (x^2 + y^2)) - \\
&& 3 u (9 v^6 + 16 x^2 y^2 + 4 v^2 (x^2 + y^2) -
    v^4 (2 + 9 x^2 + 9 y^2)) )\\
\Omega_{21}&=&\frac{1}{2}(-81 u^5 v^2 + 9 u^4 (x^2 - y^2) - 4 u^2 (-1 + 27 v^2) (x^2 - y^2)+ \\
&& 9 v^4 (-x^2 + y^2) +
 6 u^3 (18 v^4 - 2 (x^2 + y^2) - 3 v^2 (1 + 3 x^2 + 3 y^2)) - \\
&& 3 u (9 v^6 + 16 x^2 y^2 + 4 v^2 (x^2 + y^2) +
    2 v^4 (-1 + 9 x^2 + 9 y^2)))\\
\Omega_{22}&=& -\frac{v}{4} (-81 u^6 + 9 u^4 (4 + 21 v^2 - 6 x^2 - 6 y^2) -
   72 u^3 (x^2 - y^2) +\\
&& 8 u (1 + 18 v^2) (x^2 - y^2) -
 3 u^2 (4 v^2 + 45 v^4 - 20 (x^2 + y^2)) + 3 (9 v^6 + \\
&& 32 x^2 y^2 +
4 v^2 (x^2 + y^2) + 18 v^4 (x^2 + y^2))).
\end{eqnarray*}

Taking into account that conditions $x^2+y^2=-(u^2+v^2) + r^2$ and $x^2-y^2=-(u^3-3uv^2)$ hold when restricted to $T_r(2)$,
these coefficients can be expressed only in terms of $u,v$ as follows:

\begin{eqnarray}\label{sys5}
\Omega_{11}& = & v (6 r^4 - 15 u^6 + 3 u^2 v^2 + u^4 (7 - 27 v^2) +
  3 r^2 (9 u^4 - v^2 + \nonumber \\ && u^2 (-5 + 9 v^2))), \nonumber \\
\Omega_{12}+ \Omega_{21} & = & \frac{u}{2} (-24 r^4 - 3 r^2 (9 u^4 + v^2 (-8 + 9 v^2) + 2 u^2 (-4 + 9 v^2))  \\
 &&+ \ 2 u^2 (3 u^4 + v^2 (-10 + 9 v^2) + u^2 (-2 + 36 v^2))), \nonumber \\
\Omega_{22} & = & \frac{v}{4} (-24 r^4 - 21 u^6 + u^4 (8 - 27 v^2) - 3 u^2 v^2 (-4 + 9 v^2) + \nonumber \\ &&
   3 v^4 (-4 + 9 v^2) + 6 r^2 (-2 u^2 + 9 u^4 + 6 v^2 - 9 v^4)).  \nonumber
\end{eqnarray}

\noindent Therefore, points in $T_r(2)$ where $u=v=0$ are $\nu$-umbilic, namely, points in the intersection
of the plane curves $x^2-y^2=0$ and $x^2+y^2- r^2=0$. These points are:
  $$
\frac{r}{\sqrt 2}(1,1,0,0),\ \frac{r}{\sqrt 2}(-1,1,0,0),\
\frac{r}{\sqrt 2}(1,-1,0,0),\frac{r}{\sqrt 2}(-1,-1,0,0).   $$

A simple analysis, which is omitted, shows that there are no umbilic points in the subset $T_r(2)\setminus T_r^1(2)$.

$\hfill\Box$
\begin{prop}\label{teoindex}
The foliation of $\nu$-lines of curvature of $T_r(2)$ only has 4 umbilic points.
\end{prop}

\noindent {\bf Proof.}
To prove this property we will fix the radius $r=\frac{1}{\sqrt{10}}$ in order to get convenient computations.
Once we prove the property for the double torus $ T_{\frac{1}{\sqrt{10}}}(2)$, the general case
follows the fact that the
$\nu$-principal foliations are invariant under homotheties of $\mathbb R^4$ centered at the origin.

The $\nu$-umbilic points determined in Proposition \ref{4.1} in this case are:

$$
\left \{\frac{1}{\sqrt{20}}(1,1,0,0), \frac{1}{\sqrt{20}}(-1,1,0,0), \frac{1}{\sqrt{20}}(1,-1,0,0),\frac{1}{\sqrt{20}}(-1-,1,0,0) \right\}.
$$

Now we show that these points are the unique $\nu$-umbilics on $T_{\frac{1}{\sqrt{10}}}(2)$.

We assume that $uv \neq 0$ and denote the polynomials
$\frac{1}{v}\Omega_{11}$, $\frac{1}{u}(\Omega_{12}+ \Omega_{21})$ and $\frac{1}{v}\Omega_{22}$, by $\Omega_1$, $\Omega_2$, $\Omega_3$, respectively.
Namely,

\begin{eqnarray}\label{omega}
\Omega_1 &=& 6 r^4 - 15 u^6 + 3 u^2 v^2 + u^4 (7 - 27 v^2) +
  3 r^2 (9 u^4 - v^2 + u^2 (-5 + 9 v^2)), \nonumber \\
\Omega_2 &=&  \frac{1}{2} (-24 r^4 - 3 r^2 (9 u^4 + v^2 (-8 + 9 v^2) + 2 u^2 (-4 + 9 v^2)) + \\
 &&  2 u^2 (3 u^4 + v^2 (-10 + 9 v^2) + u^2 (-2 + 36 v^2))), \nonumber \\
\Omega_3 &=& \frac{1}{4} (-24 r^4 - 21 u^6 + u^4 (8 - 27 v^2) - 3 u^2 v^2 (-4 + 9 v^2) + \nonumber \\ &&
   3 v^4 (-4 + 9 v^2) + 6 r^2 (-2 u^2 + 9 u^4 + 6 v^2 - 9 v^4)).  \nonumber
\end{eqnarray}

Straightforward considerations show that $\nu$-umbilic points satisfy the assumption.
Thus, $p \in T_{\frac{1}{\sqrt{10}}}(2)$
is umbilic if and only if $p$ lies in the intersection of the zero loci of $\Omega_i,\ i=1,2,3.$
Consequently, to determine this intersection we define a reduction
of these polynomials regarding the following lemma whose proof is straightforward, and is therefore omitted.

\begin{lem}\label{lema1}
Let $\rho: \mathbb R^n \rightarrow \mathbb R^m,\ (x_1,...,x_n) \mapsto (y_1,...y_m)$ be a polynomial mapping.
Assume that the induced application
$\rho ^*: \mathbb R[y_1,...y_m] \rightarrow \mathbb  R[x_1,...x_n] $, defined by $\rho^*(Q)= P$, where $P=Q \circ \rho$,
implies that $\rho({\mathcal L}(P)) \subset {\mathcal L}(Q)$, where ${\mathcal L}(P)$ and ${\mathcal L}(Q)$ are the loci of $P$ and $Q$, respectively.
\end{lem}

Let
$$\rho:\mathbb R^2 \rightarrow \mathbb R^2, \ \
(u,v) \mapsto (w=u^2,z=v^2).$$

Thus, the polynomials

\begin{eqnarray*}
\Theta_1(w,z) & = &\frac{3}{50} - \frac{3w}{2} + \frac{97 w^2}{10} - 15 w^3 - \frac{3 z}{10} + \frac{57 w z}{10} -
 27 w^2 z,  \\
\Theta_2(w,z) & = &-\frac{6}{25} - \frac{6 w}{5} + \frac{67 w^2}{5} - 21 w^3 + \frac{18 z}{5} + 12 w z -
 27 w^2 z -
\\ &&\frac{87 z^2}{5} - 27 w z^2 + 27 z^3, \\
\Theta_3(w,z) & = &-\frac{6}{25} + \frac{12 w}{5} - \frac{67 w^2}{10} + 6 w^3 + \frac{12 z}{5} - \frac{127 w z}{5} +
 72 w^2 z -  \\
&& \frac{27 z^2}{10} + 18 w z^2,
\end{eqnarray*}
satisfy $\rho ^*(\Theta_i)= \Omega_i,\ i=1,2,3$.

Since Lemma \ref{lema1} implies that $\bigcap_{i=1}^3 \rho({\mathcal L}(\Omega_i)) \subset \bigcap_{i=1}^3{\mathcal L}(\Theta_i)$,
it is enough to show that $\bigcap_{i=1}^3{\mathcal L}(\Theta_i)= \emptyset$ to complete the proof
.
We fix the radio of the sphere and compute using Wolfram Mathematica to get a set of generators of the ideal defined by $\Theta_i,\ i=1,2,3.$ Then, we state

\begin{lem} Let us asumme that $r=\frac{1}{\sqrt{10}}$. Then,
the ideal generated by the polynomials $\Theta_i, i=1,2,3$ has the following Groebner basis:
\Small{
\begin{eqnarray*}
G_1(w,z)&=& \Gamma_1(z)=2457 + 355500 z - 16427260 z^2 + 227180200 z^3 - 1418476000 z^4 +
 \\
&&  4524012000 z^5 - 7251120000 z^6 + 4665600000 z^7 \\
G_2(w,z)&=& -272440432411875 + 651925265295213 w + 3881340837877779 z + \\
&& 6558340381939640 z^2 - 389422787482597000 z^3 +
 2318821632923662800 z^4  \\
&& - 5434755866556408000 z^5 +
 4628749229159040000 z^6.
\end{eqnarray*}}
\end{lem}

The polynomial $G_1$ depends only  on $z$ and its  roots are computable. The positive ones are

\begin{eqnarray*}
z_1& =& \frac{1}{20}, \ \  {\rm and} \\
z_2& = &\frac{1}{240}\left(67 - \left(\frac{13^5}{\sigma}\right)^{\frac{1}{3}} - (13 \sigma )^{\frac{1}{3}}\right),
\end{eqnarray*}
where $\sigma=1289-216 \sqrt{35}$.
Thus, the intersection of the loci of $G_1$ and $G_2$ is obtained by substituting these roots in the polynomial
$G_2$ and solving for $w$. That is, for $z_1$ we get
$w_1=3/20$, while for
$z_2$ we get
\begin{eqnarray*}
w_2&=&\frac{\sigma^{-\frac{5}{3}}}{720}(13^{\frac{1}{3}}(556848 \sqrt{35}-3294481 )+13^{\frac{2}{3}}\sigma^{\frac{1}{3}}(2808 \sqrt{35}-16757) - \\
&& \sigma^{\frac{2}{3}}(14472 \sqrt{35}-86363 )).
\end{eqnarray*}
\medskip
 By a direct substitution we see that none of these points lies on the sphere of radius $r=\frac{1}{\sqrt{10}}$. This completes the proof under the hypothesis
$xy \neq 0$. If $x$ or $y$ vanishes, a similar analysis can be applied. It is simpler because most of the equations above involve easier expressions, so it is omitted.
$\hfill\Box$

\medskip

Let us analyze the $\nu$-lines of curvature near the umbilic points in order to determine the umbilic separatrices and their indexes.

Taking the umbilic point $\dfrac{r}{\sqrt 2}(1,1,0,0)$ as the origin, we can introduce, for the local analysis, a system of coordinates $X,Y,Z,W,$ in a neighborhood of this
point, where the double torus can be parameterized as the graph of a differentiable function $h:U \subset \mathbb R^2\to \mathbb R^2.$ The parametrization has the form
${\bf x}(u,v)=(f(u,v),g(u,v),u,v),$ where:

\begin{align*}
  f(u,v)=&\dfrac{r}{\sqrt 2}-\dfrac{\sqrt{r^2-u^2-u^3-v^2+3uv^2}}{\sqrt 2},\\
  g(u,v)=& \dfrac{r}{\sqrt 2}- \dfrac{\sqrt{r^2-u^2+u^3-v^2-3uv^2}}{\sqrt 2}.
\end{align*}

\begin{lem}
The separatrices at the umbilic point $\dfrac{r}{\sqrt 2}(1,1,0,0)$ have slope $p=0,\;\pm \sqrt 3.$
\end{lem}
\noindent {\bf Proof.}
In \cite{F} is shown that if we have a local immersion of a surface into $\mathbb R^4$ with a parameterization of the form
\begin{align*}
  {\bf x}(u,v)&=( \frac{k}{2}(u^2+v^2)+\frac{a}{6}u^3+\frac{b}{2}uv^2+\frac{c}{6}v^3+R_1,\\
  &\frac{\alpha}{2}u^2+\beta uv+ \frac{\gamma}{2}v^2+\frac{\delta}{6}u^3+\frac{{\varepsilon}}{2}u^2v+\frac{\zeta}{2}uv^2+\frac{\eta}{6}v^3+R_2,u,v),
\end{align*}
then the slopes of the separatrices at the umbilic point are given by the solutions of the equation:
  $$
(b+\beta n)p^3+(\beta m-c-(\gamma-\alpha)n)p^2-(2b-a+(\gamma-\alpha)m+\beta n)p-\beta m=0,
  $$
where $p=dv/du$ is the variable to determine,
and $k,\;a,\;b,\;c,\;\alpha,\;\beta,\;\gamma,\;\delta,\;{\varepsilon},\;\zeta,\;\eta$ are the local parameters of the immersion.

We use the Taylor expansion of $f(u,v)$ and $g(u,v)$ up to order three, to express the parameterization as follows:
\begin{align*}
  {\bf x}(u,v)=& (\dfrac{1}{2\sqrt 2r}(u^2+v^2)+\dfrac{3}{6\sqrt 2 r}u^3-\dfrac{3}{2\sqrt 2r}uv^2,\\
  & \dfrac{1}{2\sqrt2r}(u^2+v^2)-\dfrac{1}{2\sqrt2r}u^3+\dfrac{9}{6\sqrt2r}uv^2,u,v).
\end{align*}
By solving the previous equation we obtain that $p=0,\;\pm \sqrt 3.$ $\Box$

Then, by applying the action of the group $\mathcal G$ on $T_r(2)$ we conclude the following:

\begin{prop}\label{teoindex}
The umbilic points of $T_r(2)$ are of type $D_3$.
\end{prop}

\medskip
Now we prove statement ii).

\begin{prop}\label{circ1}
  There exists a pathwise separatrix, homeomorphic to a circle, which goes through all the umbilic points.
\end{prop}
\begin{proof}
If we consider a curve $S_1(t):I\to \mathbb R^4$ given by $S_1(t)=(h_1(t), h_2(t),$ $h_3(t), h_4(t)),$ it is easy to show that it satisfies the
equation of the curvature lines when the last coordinate vanishes, i.e., for $h_4(t)=0.$ Also, if we consider that the curve is in the double torus,
 $S_1(t)$ should simultaneously satisfy conditions $F(S_1)=0$ and $G_r(S_1)=0$. Consequently, we obtain a curve defined in $4$ parts given by:
\begin{align*}
  S_{11}(t)&=(-\frac{\sqrt{r^2-t^2 -t^3}}{\sqrt 2},-\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},t,0),\\
  S_{12}(t)&=(\frac{\sqrt{r^2-t^2 -t^3}}{\sqrt 2},-\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},t,0),\\
  S_{13}(t)&=(\frac{\sqrt{r^2-t^2 -t^3}}{\sqrt 2},\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},t,0),\\
  S_{14}(t)&=(-\frac{\sqrt{r^2-t^2 -t^3}}{\sqrt 2},\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},t,0).
\end{align*}
We have that $U_i \in S_{1i}(t),\;i=1,2,3,4$ where $U_1=(-\frac{r}{\sqrt 2},-\frac{r}{\sqrt 2},0,0),$
$U_2=(\frac{r}{\sqrt 2},-\frac{r}{\sqrt 2},0,0),$ $U_3=(\frac{r}{\sqrt 2},\frac{r}{\sqrt 2},0,0),$ $U_4=(-\frac{r}{\sqrt 2},\frac{r}{\sqrt 2},0,0)$
.

We have that $(\Gamma_1\circ \Gamma_2 \circ \Gamma_1\circ \Gamma_2)(S_{11})= (\Gamma_1\circ \Gamma_2 \circ \Gamma_1)(S_{12})=(\Gamma_1\circ \Gamma_2)(S_{13})=\Gamma_1(S_{14})=S_{11}.$ In addition,
 the matrices $\Gamma_2$ and $\Gamma_1$ have as fixed points the points obtained taking $x=0$ for $\Gamma_2$ and $y=0$ for $\Gamma_1.$
 Taking this in the curvature line, we obtain $4$ fixed points $p_i,\; i=1,2,3,4$ where $p_1=S_{12}\cap S_{13},$
$p_2=S_{11}\cap S_{14},$ $p_3=S_{11}\cap S_{12}$ and  $p_4=S_{13}\cap S_{14}$.
Thus, since $S_1(t)$ is a differentiable curve and $S_{1i},\;i=1,2,3,4$ are connected arcs with just one point in common each two of them,
and since we can generate $S_{1i},\; i=2,3,4$ with the action of $\mathcal G$ on $S_{11}$, we conclude that  $S_1(t)$ is homeomorphic to a circle.
\end{proof}

\begin{prop}\label{circ2}
There exist two pathwise separatrices obtained by rotating $S_1(t)$ $2\pi/3$ and $4\pi/3$ with center at an umbilic point.
\end{prop}
\begin{proof} We know that the curvature lines are invariant under rotations. By applying $\Gamma_3$ to the separatrix $S_1$,
we obtain a new curvature line $S_2:I\to \mathbb R^4$ associated to $p=\sqrt 3$ given by:

\begin{align*}
  S_{21}(t)&=(-\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, -\frac{\sqrt{r^2-t^2+t^3}}{\sqrt 2},-\frac{t}{2},\frac{\sqrt 3}{2}t),\\
  S_{22}(t)&=(\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, -\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},-\frac{t}{2},\frac{\sqrt 3}{2}t),\\
  S_{23}(t)&=(\frac{\sqrt{r^2-t^2-t^3)}}{\sqrt 2}, \frac{\sqrt{r^2- t^2 + t^3}}{\sqrt 2},-\frac{t}{2},\frac{\sqrt 3}{2}t),\\
  S_{24}(t)&=(-\frac{\sqrt{r^2-t^2-t^3}}{\sqrt 2}, \frac{\sqrt{r^2-t^2+t^3}}{\sqrt 2},-\frac{t}{2},\frac{\sqrt 3}{2}t),
\end{align*}

where $U_i \in S_{2i}(t),\;i=1,2,3,4.$

In the same way, applying $T_{-U_3}\circ \rho_{4\pi/3}\circ T_{U_3}$ we obtain a new curvature line $S_3:I\to \mathbb R^4$ associated to $p=-\sqrt 3$ and given by:

\begin{align*}
  S_{31}(t)&=(-\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, -\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},-\frac{t}{2},-\frac{\sqrt 3}{2}t),\\
  S_{32}(t)&=(\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, -\frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},-\frac{t}{2},-\frac{\sqrt 3}{2}t),\\
  S_{33}(t)&=(\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, \frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},-\frac{t}{2},-\frac{\sqrt 3}{2}t),\\
  S_{34}(t)&=(-\frac{\sqrt{r^2-t^2 - t^3}}{\sqrt 2}, \frac{\sqrt{r^2-t^2 + t^3}}{\sqrt 2},-\frac{t}{2},-\frac{\sqrt 3}{2}t),
\end{align*}

where $U_i \in S_{3i}(t),\;i=1,2,3,4.$
\end{proof}

\medskip

Proof of statement $iii)$. We begin by describing a $CW$ complex structure  which includes the $\nu$-lines of curvature
and the $\nu$-umbilics as cells of $T_r(2)$.

\begin{lem}\label{CW-structure}
The set $P_{++} \subset T_r(2)$ admits the following $CW$-complex structure.

\noindent i) Five 0-cells, one of them is an umbilic point.

\noindent ii) Five 1-cells, all of them are lines of curvature. Among them, two are separatrices containing the umbilic point.

\noindent iii) One 2-cell.
\end{lem}

\begin{proof}
We begin by defining the 2-cell as the interior of $P_{++}$, see (\ref{pentagono}).
Then, the 1-cells are described as follows:
\begin{eqnarray*}
a_1& = & \{ (x,y,u,v) \in P_{++}; v=0\}, \ \ a_2=\{ (x,y,u,v) \in P_{++}; x=0, \ u>0\} \\
a_3& = &  \{ (x,y,u,v) \in P_{++}; y=0\}, \ \ a_4=\{ (x,y,u,v) \in P_{++}; x=0, \ v>0\} \\
&& \ \ \ \ \ \ \ a_5 =    \{ (x,y,u,v) \in P_{++}; v=-\sqrt{3}u\}.
\end{eqnarray*}
Observe that since the projection of $P_{++}$ onto the $uv$-plane is $P_0$,
the projections of these 1-cells  are $e_1, ... , e_5$, respectively.
We point out that, according to the differential equation of $\nu$-lines of curvature of
Corollary $\ref{EcuaDiff}$, a straightforward computation implies that $a_i,\ i=1,...,5$
are $\nu$-lines of curvature of $T_r(2)$. Moreover, the curves $S_{13}$ defined in Proposition $\ref{circ1}$ and $S_{23}$ defined
in Proposition $\ref{circ2}$
are  parameterizations of $a_1$ and $a_5$, respectively. Thus, these 1-cells are separatrices. Finally, the 0-cells are defined
as the preimages of the vertices of the pentagonal region $P_0$ under the projection. Among them, the one corresponding to the origin
in the $uv$-plane is the umbilic $(\frac{r}{\sqrt{2}}, \frac{r}{\sqrt{2}},0,0)$.
\end{proof}

The $CW$-complex structure of $P_{++}$ in this lemma allows us to state the following:

\begin{prop}
The double torus $T_r(2)$ admits a $CW$-complex structure such that the 0-skeleton includes the umbilic points and the 1-skeleton
the separatices.
\end{prop}
\begin{proof}
The cells of the $CW$-complex structure of $T_r(2)$
are the images of the cells of the pentagonal region $P_{++}$
under the action of the group $\mathcal G$. We describe them as follows:
The 0-skeleton is constituted by sixteen 0-cells. Namely, four of the orbit of $v_0$,
six for the orbit of $v_1$,
and three of the orbit $v_i, i=2,3$. Observe that the umbilic points belong to the orbit of $v_0$.
The 1-skeleton is constituted by thirty 1-cells defined by the orbits of the boundary of the pentagonal region. Namely,
the orbit of $a_1$ consists of twelve 1-cells including $a_5$; all of them are separatrices. The orbit of $a_i$ consists of
six 1-cells for each $i=2,3,4$.
The 2-skeleton consists of
twelve 2-cells defined by the orbit of $P_{++}$.
By a direct computation, it can be verified that this decomposition provides the Euler-characteristic of the double torus:
$$12 -30 +16= -2= \chi(T_r(2)).$$
\end{proof}

\begin{coro}
The first homology group of the double torus $T_r(2)$, $H_1(T_r(2))$, with coefficients in $\mathbb Z_2$
has a basis whose representatives are cycles constituted by pathwise separatrices.
\end{coro}

\begin{proof}
We provide a basis $\{\alpha_i\}_{i=1}^4$ of $H_1(T_r(2))$
whose representatives are the cycles $\{c_i\}_{i=1}^4$, which are defined as follows.

\noindent First, we consider
\begin{eqnarray*}
c'_1= b_1 +b_2,\ c'_2= \Gamma_3(c'_1),
\end{eqnarray*}
where $b_1= a_4 + \Gamma_3(a_2)$ and $b_2= \Gamma_1(b_1)$.
Then, we define
\begin{eqnarray*}
c_1= d_1 +d_2+ \Gamma_1(d_1 +d_2),\ c_2= \Gamma_3(c_1),
\end{eqnarray*}
where, $d_1$ is the separatrix whose projection is diagonal of slope $\sqrt{3},\ u,v \geq0$
and $d_2= \Gamma_3(d_1)$.
Thus, $c_i,\ i=1,2.$ is a pathwise separatrix homologous to $c'_i,\ i=1,2.$
We define $d_3= a_3 + \Gamma_2(a_3)$ and $d_4= \Gamma_3(d_3)$.
Thus, we have defined a basis $\{c_1', c_2', d_3, d_4  \}$  of  $H_1(T_r(2))$ whose cycles are constituted by
$\nu$-lines of curvature and belong to the $1$-skeleton.
Finally, by choosing the representatives
$$c_3=a_1+a_5+ \Gamma _2(a_1+a_5)\ {\rm and}\ c_4=\Gamma_3(c_3),$$
since $c_i$ is homologous to $d_i$, $i=3,4$, we get the desired basis of $H_1(T_r(2))$ whose cycles $\{c_i \}_{i=1}^4$ are
pathwise separatrices.
\end{proof}

Proof of statement $iv)$. In the proof of this statement, we will use some argument involving properties of the flow of a planar vector field.
The Poincar\'e Bendixon theorem will play a relevant role. For the basic notions and the proof of this theorem see \cite{HirschSmale}.

\begin{prop}\label{closedsolutions}
The lines of curvature on the complement of the separatices of $T_r(2)$ are immersions of $\mathbb S^1$.
\end{prop}

\noindent {\bf Proof.} Let $\mathcal L_1$ and $\mathcal L_2$ be the foliations of maximal and minimal $\nu$-lines of curvature, respectively, in the closure of the
region $P_0$  (see figure \ref{Pentagon}). We assume that $e_1$ and $e_5$ are separatrices in $\mathcal L_1$ containing the origin in the $uv$-plane. Thus, $e_2$ and $e_4$ belong to
$\mathcal L_2$ and $e_3$ belongs in turn to $\mathcal L_1$
.
In the interior of $P_0$, the equation of  maximal (respectively minimal) $\nu$-lines of curvature is defined by a vector field $X_1$ without singularities.
Moreover, we can assume that $X_1$ extends to $e_2$ pointing towards the region $P_0$ along $e_2$. Consequently, each solution with initial condition at $e_2$
never returns to this edge of $P_0$
.

\begin{lem}
The $\omega$-limit of the solution $\gamma_p$ of $X_1$ with initial condition $ \gamma_p(0)=p \in e_2$ does not contain points in $e_1 \cup e_3\cup e_5$.
Moreover, this solution intersects $e_4$.
\end{lem}
\noindent {\bf Proof of the lemma.}
We show that there exists a tubular neighborhood of $e_1 \setminus \{(0,0)\}$ which does not intersect $\gamma_p(t),\ t>o.$
Consider the Poincar\'e application

\begin{eqnarray*}
\pi: \Sigma_3^r(0) \rightarrow e_2
\end{eqnarray*}
where $\Sigma_3(0)$ is the minimal $\nu$-line of curvature in $P_0$ extending the separatrix in the complement of the closure of $P_0$, and where
$\Sigma_3^r(0)$ is a neighborhood of the origin in $\Sigma_3(0)$ of ratio $r$. Since this application is continuous, we can choose $\delta>0$ such that
$\pi\left( \Sigma_3^{\delta}(0) \right) \subset e_2^{{\varepsilon}=|p|/2}(p_0)$, where $p_0= e_1\cap e_2$ and $e_2^{{\varepsilon}=|p|/2}(p_0)$ is the neighborhood
of $p_0 \in e_2$ with ratio equal to a half of the distance from $p$ to $p_0$. Since $p \in e_2$ is not in $e_2^{{\varepsilon}=|p|/2}(p_0)$, the
uniqueness of the solutions of $X_1$ implies that $\gamma_p(0)$ never intersects the tubular neighborhood of $e_1$, defined by the solutions of $X_1$
with initial conditions in  $e_2^{{\varepsilon}=|p|/2}(p_0)$. Analogous arguments allow us to extend such a tubular neighborhood to the whole union
$e_1 \cup e_3\cup e_5$.

Now we prove the second statement. Suppose that $\gamma_p$ does not intersect $e_4$. Observe that
the $\omega$-limit of the solution $\gamma_p$ is closed
and contained in $P_0$ and thus bounded without singular points. Then the Poincar\'e-Bendixon theorem implies that $\gamma_p$ is a closed solution.
This implies in turn that there is a singular point in the bounded region determined by this solution. So we get a contradiction.
\qed

\begin{lem}\label{unicidad}
If $p,\ q \in e_2$ and $r \in e_4$ are such that $\gamma_p(t_0)= r= \gamma_q(t_1)$, where $t_i, i=0,1,$ is the first value
of the parameter where the corresponding
solution intersects $e_4$,
then $p=q$ and $t_0=t_1$.
\end{lem}
\noindent {\bf Proof of the lemma.}
\begin{eqnarray*}
\gamma_p:&[0,t_0]& \rightarrow \overline P_0,\ \gamma_p(0)=p, \ \gamma_p(t_0)=r \\
\gamma_q:&[0,t_1]& \rightarrow \overline P_0,\ \gamma_q(0)=q, \ \gamma_q(t_0)=r.
\end{eqnarray*}
We define $i= \gamma_p[0,t_0] \cap \gamma_q[0,t_1]$. Thus, this set
is a non-empty closed subset of each solution. Morever, the flow box theorem implies that $i$ is an open set.
Then, $i = \gamma_p[0,t_0]$ and the proof follows from this equation. \qed

\noindent {\bf Proof of Proposition \ref{closedsolutions}.}

When the curve $\gamma_p(t)$ reaches the side $e_{4}$ at $r$ in the $uv$-plane, the curve $\overline \gamma(t)$ reaches the boundary of $H_{++}$
at the same point. Thus,
in $T_r(2)$ the curve $\overline \gamma_p(t)=(x(t),y(t),u(t),v(t)),\ x(t)\geq 0$ crosses the boundary of $H_{++}$ and arrives to $H_{-+}$, now with $x(t) < 0$.
Under the projection $\pi(x,y,u,v)=(u,v)$ it returns to the interior of the pentagon with negative values in the first entrance.
Note that $\pi \circ \overline \gamma = \gamma$. That is, the projection of $\pi\circ \gamma_p^-(t)$ is contained in
$\gamma([0,t_0])$. Therefore, it returns to the pentagon following the solutions but in the opposite direction.
So, Lemma \ref{unicidad} implies that this solution reaches $e_2$ at $p$ implying that $\overline \gamma_p$  is a closed solution. \qed

\begin{rem}
Observe that each line of curvature of the statement of the proposition is contained in the union
$P_{++} \cup P_{-+}$, where these sets are the pentagonal regions defined in (\ref{Pentagon}), or in another pair of these regions.
\end{rem}

\subsection{The lines of curvature of the double torus in $\mathbb R^3$}

We consider a surface $M$, defined as the transversal intersection of the sphere of radius $r$ centered at the origin, $\mathbb S^3_r \subset \mathbb R^4$,
with the inverse image of a value of a differentiable function $F:\mathbb R^4 \rightarrow \mathbb R$
with an isolated singularity at the origin. Let $\nu$ be a unit vector field parallel to the gradient of $F$.
The lines of curvature transform in a proper way under the stereographic projection. We state the property in a more general setting.

\begin{prop}
Consider a surface $M \subset \mathbb S^3_r$.
The stereographic projection $\sigma:\mathbb S^3_r\setminus{\{p\}} \subset \mathbb R^4 \rightarrow \mathbb R^3$, where $p \in \mathbb S^3_r \setminus M$, provides an immersion
of $M$ into $\mathbb R^3$, which transforms the $\nu$-lines of curvature of $M$ into the lines of curvature of $\sigma(M)$, for any field $\nu$ normal to $M$ not parallel to the radial vector field. That is, a curve
$l \subset M$ is a  $\nu$-line of curvature if and only if $\sigma(l)$ is a line of curvature of $\sigma(M)$.
\end{prop}

\noindent{\bf Proof.}
We begin by recalling that any spherical surface $M \subset \mathbb R^4$ is $R$-umbilic, where $R$ is the {\it radial vector field}, defined as the
unitary vector field parallel to the gradient of the function $G$.
Moreover, any normal field independent to the radial vector field at each point of $M$
defines the same principal configuration (see Lemma $2.1$ in \cite{R-S}). Consider this normal field $\nu$. Its orthogonal projection to
$T_pS^3_r$ defines a non-null field $\hat \nu$ normal to $M$ at each point.
The stereographic projection
as a conformal map transforms
the vector field $\hat \nu$ into one of the normal fields of $\sigma(M)\subset \mathbb R^3$, and the $\hat \nu$-lines of curvature on $M$ into
the lines of curvature $\sigma(M)$. Since the $\nu$-lines of curvature of $M$ coincide with the $\hat \nu$-lines of curvature, the proof is complete.
$\hfill\Box$

We denote the embedding $\sigma(T_r(2)), r= 1$ of the double torus into  $\mathbb R^3$ by $\mathbb T_2$.

\begin{coro}\label{torusinR3}
The principal configuration of the embedded double torus $\mathbb T_2 \subset \mathbb R^3$ has
the same structure of that of $T_r(2)$ provided by Theorem \ref{Maintheorem}.
\end{coro}

We conclude by observing that the approach applied in this article to study
the lines of curvature of the double torus can be extended to describe the
lines of curvature of model embeddings of compact oriented surfaces of
genus higher than two. Specifically, we have used the differential equation
in Corollary 2.5 to describe the $\nu$-lines of curvature of the model defined
in section 3 as the link $L_r = Re \bar F \cap {\mathbb S}^3, r > 0$, of genus $(p-1)(q-1),
p, q \geq 3$, where the vector field $\nu$ is the gradient of
$Re \bar F = z_1^p + z_2^q + \bar z_1^p + \bar z_2^q$.
The symmetries of this model allowed us to determine the
type of umbilic points. Moreover, the topological decomposition of the link
induced a CW-structure which includes umbilic points and separatrices as
some of its cells. Furthermore, arguments of dynamical systems, similar to
those used in this article, could be applied to prove that the lines of
curvature are closed cycles. Finally, as in the case of the double torus,
the stereographic projection defines the embedding of $L_r, r > 0$ into $\mathbb R^3$
which transforms the foliations of $\nu$-lines of curvature into the foliation
of lines of curvature of the image of the link in $\mathbb R^3$. However, certain
problems for the determination of the number of umbilic points appeared in
the case of an arbitrary genus, which need to be solved to complete the
description.

\medskip
\textbf{Acknowledgement:}
The work of the second and third authors was supported by CONACYT grant 283017, the
third author was also supported by PAPIIT-DGAPA-UNAM grant IN118217.

redpajama_set_name: RedPajamaArXiv


This program is offered completely on-line - discuss options with your advisor.
Upon completion of the Accounting Degree program, students will be prepared for employment in general accounting occupations. They will be prepared to work in public, private, or governmental agencies as accounting clerks, accounting technicians, bookkeepers, accounting support personnel, or payroll assistants.
Prepare and interpret financial records for a business while applying generally accepted accounting principles and industry standards.
Identify and explain common internal controls necessary in business organizations.
Communicate professionally, both orally and in writing.
Compute payrolls and prepare basic federal and state payroll tax forms.
Prepare basic income tax returns for individuals and businesses.
Analyze the legal, ethical, and practical implications of business decisions.


text:
A FAMILY COMPANY
Go Further With Our Family™
The Johnson Family
Samuel Curtis Johnson
Herbert F. Johnson, Sr.
H.F. Johnson, Jr.
Fisk Johnson
What Family Means
This We Believe
At Work for a Better World
Architecture and Tours
Community Interest Programs
Johnson Family
Samuel Curtis Johnson: The Generous and Determined Founder of a Family Company
This We Believe: Our Company Values Have Guided SC Johnson for Five Generations
SC Johnson: Who We Are
Everything is Chemistry: Why SC Johnson is Tackling Negative Perceptions about Synthetic Chemicals
Global Product Transparency: SC Johnson Offers Ingredient Lists So You Can Make Choices for Your Family
No Secrets: SC Johnson Introduces 100% Transparency for Fragrance Ingredients
Corporate Transparency
Improving Products
Minimizing Our Footprint
2018/2019 Sustainability Report
Ending Plastic Waste: SC Johnson is Committed to Increasing Plastic Reuse and Stopping Plastic Pollution
Ocean Plastic: SC Johnson is Tackling Plastic Pollution in Indonesia While Helping Families
Forest Conservation: SC Johnson Continues Efforts to Stop Deforestation
Statements & Policies
SC Johnson Launches Global Partnership to Fight Ocean Plastic and Poverty
SC Johnson Chairman and CEO Fisk Johnson to Receive Honor for Efforts to Raise Awareness about Human Trafficking
Borough of Bradley Beach Partners with SC Johnson to Launch New Jersey's First Pilot Program for Residential Curbside Plastic Film Recycling
Global Partnership With Plastic Bank Fights Ocean Plastic Pollution and Poverty
Turning the Tide on Ocean Plastic
Leading the Way on Ocean Plastic
SC Johnson Plans Shift to Geothermal Power at Headquarters, Significantly Reducing Energy Use
Company Continues to Further Diversify Clean Energy Portfolio Through 'All of the Above' Approach
RACINE, Wis., April 1, 2019 – SC Johnson today is announcing its intention to use geothermal energy to power its Racine headquarters, a move that will greatly reduce the company's environmental footprint and positions the company as a leader in the private sector in the transition to cleaner, renewable energy sources.
SC Johnson plans to install a GeoExchange* system at its iconic Frank Lloyd Wright-designed campus to provide sustainable heating and cooling throughout the facilities, using the constant temperature of the Earth. The project – which is pending city and state approvals – is expected to reduce energy usage by an estimated 42 percent. There is a 40 percent reduction made up of decreasing consumption from the implementation of a GeoExchange system, including transforming the current boiler facility to a new energy efficient thermal plant, and an additional 2 percent reduction by utilizing PV for renewable energy. Combined with other sustainable projects, the facility will save another 15 to 20 percent in energy usage, resulting in a total facility-wide reduction of 57 to 62 percent of the current energy load.
"As a global company, you think about your footprint in all the places you operate and how to minimize your impact," said Fisk Johnson, Chairman and CEO of SC Johnson. "Leading the industry in an environmentally responsible manner starts at home. For us, that meant taking a look at our operations and finding where we can lessen our impact by reducing greenhouse gas emissions, addressing air quality and increasing the amount of energy offset from renewable resources. Transitioning to geothermal energy at our headquarters goes a long way toward accomplishing those goals."
SC Johnson has for two decades been an industry leader in the use of renewable energy in the production of its products. The company's commitment to lowering its carbon footprint and utilizing a diverse, renewables-heavy energy mix is evident in its across-the-board embrace of wind, solar and biofuel:
At the company's manufacturing plant in Surabaya, Indonesia, waste husks from rice grains are used as a fuel source for heating water for production. The initiative dramatically cut down on the use of fossil fuels, resulting in a reduction of 7,000 metric tons of greenhouse gas emissions annually.
Manufacturing facilities in China, India, Indonesia and Mexico have solar projects for hot water heating.
Three of the company's manufacturing sites – Bay City, Michigan; Mijdrecht, Netherlands; and Gorzow, Poland – run on 100 percent wind energy, while facilities in Mount Pleasant, Wisconsin (Waxdale) and Toluca, Mexico, get a portion of their power from wind.
Waxdale also generates 85 percent of its energy from cleaner-burning natural gas. Methane gas from a nearby public landfill generates 28 percent of the facility's energy.The other 57 percent comes from a second cogeneration system that uses cleaner-burning natural gas.
Altogether, SC Johnson's efforts have resulted in the reduction of greenhouse gas emissions in its worldwide manufacturing sites by 62 percent since the company's 2000 baseline.
"At SC Johnson, we started using renewable energy more than 15 years ago to make products like Windex, Pledge and others," Johnson said. "Today, a third of our global energy use comes from renewable sources. It's an investment, but it's the right thing, now and for the future."
SC Johnson's Administration Building – which remains the only operational corporate headquarters designed by Wright – opened 80 years ago, in 1939. Additionally, roughly 13,000 square feet of photovoltaic (PV) panels (or ~620 panels) will be installed on the building roofs of the company's west campus. A GeoExchange well field will be located under a northwest parking lot, and an existing boiler plant will be converted to a next-generation thermal plant.
The installation of the GeoExchange system – which repurposes energy in a highly efficient manner but is not considered renewable – is the most significant of many efficiency improvements at SC Johnson headquarters. The PV panels will serve the important function of offsetting energy consumption from the grid with self-generated renewable energy. SC Johnson hopes to complete the GeoExchange field installation by the fall of 2019, with the entire campus converted to the new thermal plant by the fall of 2020.
Company-owned wind turbines are in operation at Mijdrecht and Waxdale; the Wisconsin facility has two operational turbines, each standing 415 feet tall, and both requiring approval from the U.S. Federal Aviation Administration for use. At Mijdrecht, the 262-foot turbine generates nearly half the amount of energy needed to run the 3 million-square-foot site, with the remaining energy as purchased wind power.
Though the use of wind turbines in busy urban areas can prove challenging, the positive environmental impacts make it well worth the effort. The Waxdale wind turbines generate 8 million kilowatt-hours of electricity per year – enough to power 770 homes. Using those turbines saves about 6,000 metric tons of carbon emissions compared to what would have been created using fossil fuels.
"SC Johnson has made a commitment to diversifying energy sources and prioritizing the use of renewables whenever possible," Johnson said. "Our work is far from over, and we're excited to do our part at home in Racine with the GeoExchange system at our headquarters. But we also believe that lowering our carbon footprint around the world will make a positive impact on the communities we serve."
* GeoExchange is a registered trademark of Geothermal Heat Pump Consortium Inc. d/b/a Geothermal Exchange Organization.
About SC Johnson
SC Johnson is a family company dedicated to innovative, high-quality products, excellence in the workplace and a long-term commitment to the environment and the communities in which it operates. Based in the USA, the company is one of the world's leading manufacturers of household cleaning products and products for home storage, air care, pest control and shoe care, as well as professional products. It markets such well-known brands as GLADE®, KIWI®, OFF!®, PLEDGE®, RAID®, SCRUBBING BUBBLES®, SHOUT®, WINDEX® and ZIPLOC® in the U.S. and beyond, with brands marketed outside the U.S. including AUTAN®, BAYGON®, BRISE®, KABIKILLER®, KLEAR®, MR MUSCLE® and RIDSECT®. The 133-year-old company, which generates $10 billion in sales, employs approximately 13,000 people globally and sells products in virtually every country around the world. www.scjohnson.com
інформацію про склад мийного засобу
© 2020 S. C. Johnson & Son, Inc. All Rights reserved

redpajama_set_name: RedPajamaCommonCrawl


Snow White celebrates her Diamond Anniversary in 2012 and fans of Snow White and the Seven Dwarfs and Walt Disney will definitely want to add the 2012 calendar to their collection. It's a keepsake calendar designed to help celebrate this special anniversary.
A typical twelve month calendar, this one features large 13 by 15 inch pages, packed with beautiful color images. The pages have been given a linen look adding to the elegant feel of this calendar.
The environmentally conscious will be glad to know that Snow White's calendar is printed on FSC certified paper that is 100% recyclable, using soy and vegetable inks.
What fun to share 2012 with Snow White and the Seven Dwarfs.
Order your Snow White calendar from Amazon.
See Jim Shore's Snow White Figurines.
Snow White and the Seven Dwarfs Blu-ray Movie Review.


Posted January 7, 2014 by Tom & filed under General.
So with a new site, comes a new philosophy. Going forward, all Red Monkey Goo's new websites, be it our clients to our own in house projects, will be responsive by default. We see a lot of businesses still pushing this as a added extra. A nice to have. In this increasingly mobile world we believe it should be standard.


text:
Die Liste der niederländischen Meister im Mountainbike führt die Sieger der Wettkämpfe um die niederländischen Meisterschaften im Mountainbike auf. Diese werden seit 1988 durchgeführt.

Siegerliste Männer

Cross Country

Frauen

Cross Country

Weblinks
Datenbank der Jahresergebnisse

Niederlande
Person (Radsport, Niederlande)
Niederlande, Mountainbike
MountainbikeMeisterschaften, Niederlande

redpajama_set_name: RedPajamaWikipedia


Planning a group trip to one of the world's most exciting capitals? London is a thrilling and entertaining place to visit no matter what you're in to but navigating the city when you're unfamiliar with it can be stressful and time-consuming. So why not make the most out of your time in London and hire one of our vehicles for a hassle-free city tour?
At coachrental.co.uk we're dedicated to finding you the best deals on a wide range of coaches and minibuses, whether you're just planning a short tour of all the capital's top sights or a whole weekend in the city. With our coach hire service, you can tailor your sightseeing trip to your own schedule only visiting the places you want to see and our hire packages are flexible enough to incorporate nights out at the West End, bar and club crawls, special events like Notting Hill Carnival and much more.
We've got a fantastic selection of coaches and minibuses available to hire, from compact minibuses and standard coaches to spacious double deckers with enough room for up to 80 passengers. So whether you're planning an intimate outing with a small group of family and friends or a big trip requiring multiple vehicles, we can design a bespoke hire package to suit your needs.
Hiring one of our vehicles is a great way to discover London without getting stuck in endless traffic or getting lost on the Underground. If you want to find out more about our great deals on coach hire in London, check out coachrental.co.uk today.


Are you seeking legitimate custom sociology papers? We are the ultimate solution that many students are seeking. We are writing organization that specializes in the provision of legitimate custom sociology papers to students. Why fail yet we have been established to assist. Students will find help from us in completing their assignments. Additionally, we assist students with tips and guidelines on writing different academic papers. There are several aspects that make our company stand out from the rest.
We are a writing institution that provides legitimate custom sociology papers that are affordable. Students seek writing assistance from companies that are affordable. The affordability of our writing services means that students can access our services despite their economic ability. Our writing company is the most affordable in the world. The affordability of our services has seen our company attract students from across the globe. We are also loved because we provide legitimate custom sociology papers that are of high quality. An academic writing has to be of high quality before it is categorized as legitimate. Our company is recognized for the cheap prices and the high standards of legitimate custom sociology papers that we provide. Students should be careful when dealing with writing companies that are exorbitant. An exorbitant price of the paper is not a guarantee of quality.
The academic writing company is reachable 24/7. Our clients can testify to the ease of access of our company. We have various means of communication that clients can use. First, our clients can utilize Skype to communicate with the customer support team. Alternatively, clients can send mails to the customer support system. The email can contain details of the assignments that client wants to be completed. Clients can also call the customer support desk to discuss matters related to payment of the writing assignment. The ease of access enables clients to seek urgent writing assistance. Our writers can complete legitimate custom sociology papers that are needed within the shortest duration. The 24/7 access has enabled our company to complete legitimate custom sociology papers that are urgent. Complaints about delayed assignment are unheard of, in our company.
Our company is also recognized for the high level of professionalism it upholds. We aim maintain a high level of professionalism has been possible. This is because of the presence of professional writers. Out writers are well trained. Our writers are knowledgeable on the diverse writing styles needed in academic writing. Our writers know different writing styles e.g. MLA, Harvard and Oxford. Our writers ascertain that legitimate custom sociology papers have been written per the writing style that the client prefers. Our writers also know how to write different academic papers. Our clients write term paper, essay, article review, research work and dissertation. The papers are assigned to writers based on their academic and professional qualifications. The company boasts of writers who are qualified in diverse fields. Our company can complete legitimate custom sociology papers in fields e.g. science, arts, medicine and mathematics.


The 45th | Republican warns of impeachment if Russia probe is halted — United States Studies Centre Subscribe here to receive The 45th in your inbox.
Republican Senator Jeff Flake has told The Washington Post that he would support impeachment proceedings against Donald Trump if the president ends Special Counsel Robert Mueller's investigation into Russian interference in the 2016 presidential election "without cause".
"If [Trump] fires [Mueller] without cause, how different is that from what Nixon did with the 'Saturday Night Massacre'? He left before impeachment came, but that was the remedy then and that would be the remedy now," Flake said.
Flake has been a prominent critic of the president, but his warning came days after similar salvos from Republican elder Senator Orrin Hatch, who said firing Mueller would be a "foolish action" and "the stupidest thing" the president could do.
Trump has stepped up his criticism of Mueller, attacking him by name for the first time on Twitter in the days following the president's decision to fire FBI deputy director Andrew McCabe – 26 hours before his scheduled retirement.


This paper studies the costs of the transit system of Santiago, Chile, and proposes pricing schedules to finance it considering a fixed amount of subsidy. We estimate a cost function for the firms providing bus services in Santiago and study their industrial structure. We also estimate a demand model using aggregate data, which delivers information on the demand price elasticity and the effect of other variables related to the supply (e.g., capacity and frequency). Finally, we compute four pricing schedules: uniform price with and without subsidy and two-part tariffs with and without subsidy. Our results show that: (i) the industry exhibits economies of scale when considering both demand-oriented and supply-oriented output measures; (ii) measures of technical efficiency are consistent with the observed exit of firms from the market; (iii) the budget-balanced fare without subsidy is higher than the actual fare, suggesting that subsidies are justified; (iv) the budget-balanced fare with subsidy is higher than the actual fare, suggesting that subsidies are insufficient; (v) two-part tariffs either with or without subsidy are the most effective means of increasing bus travel demand and user welfare; and (vi) a menu of tickets with two options (a uniform price and a two-part tariff) is a Pareto-improving pricing schedule that balances the budget and increases bus travel demand.


Being the fast reader that I am, you are sure to be guaranteed with new reviews most days in a week. My reviews come purely out of my heart and how the book has made me feel. If a book does not move me, I do not feel inspired to write about it.
If you want to get any book reviewed, then you can contact me via social media or through postal address. Visit my Contact page for more information.
I am open to paperbacks and hardcovers. Fictions of almost all genres are welcome.
Nothing compares to the happiness I get when my reviews make others want to read that book.


text:
New Blog Tour for "Banged up Heart"
Below you can follow along with my current blog tour. Check out the reviews and info for each of the tour stops!
Each featured post will be available on the date it is done.
Blog Tour Dates & Links
July 8 (today) @ The Muffin
Interview and giveaway.
http://muffin.wow-womenonwriting.com/
July 9 @ Book Santa Fe Crystal Otto
Crystal J. Casavant-Otto shares her review after listening to the touching and inspiring memoir Banged Up Heart by Shirley Melis.
http://www.booksantafe.info/
July 10 @ Lisa Buske
Lisa Buske reviews the Audible version of Banged-Up Heart: Dancing with Love and Loss by Shirley Melis.
https://www.lisambuske.com/
July 11 @ Bring on Lemons with Tara Forst
Wisconsin mother and business woman Tara Forst shares her thoughts after listening to the inspirational audible version of Banged-Up Heart by Shirley Melis.
http://bringonlemons.blogspot.com/
July 12th @ BreakEven Books
Today's Author Spotlight at BreakEven Books is on Shirley Melis and her memoir Banged-Up Heart - now available on Audible!
https://breakevenbooks.com/
July 13 @ Bring on Lemons with Angela Williams
Angela Williams reviews Shirley Melis's Banged Up Heart.
July 14 @ Bring on Lemons with Cathy Hansen
Educator and Entrepreneur Cathy Hansen reads and reviews Banged Up Heart by Shirley Melis. You'll want to stop by Bring On Lemons today for your chance to learn more about this touching and encouraging memoir (now available on Audible).
July 15 @ Lisa Haselton
Shirley Melis is being interviewed today by Lisa Haselton. Stop at Lisa's blog to learn more about Banged Up Heart, a non-fiction, memoir (now available on Audible), about finding the courage and strength to love again.
http://lisahaseltonsreviewsandinterviews.blogspot.com/
July 16 @ Bring on Lemons with Madeline Sharples
Fellow memoir writer Madeline Sharples interviews Shirley Melis about her book Banged Up Heart. (now available on Audible) Don't miss this heart felt interview about courage and love.
http://madelinesharples.com/
July 17 @ Jerry Waxler
Coach and Author Jerry Waxler shares his thoughts after reading Shirley Melis's memoir Banged Up Heart.
www.memorywritersnetwork.com/blog
July 18 @ Deb Blanchard
Teacher Deb Blanchard gives insight into "Banged Up Heart" the touching memoir by Shirley Melis (now available on Audible).
https://www.facebook.com/Reviews-by-Deb-1737715249887520/
July 19 @ Linda Appleman Shapiro
Don't miss today's interview between memoirist Linda Appleman Shapiro and Shirley Melis. Find out more about Melis and her memoir Banged Up Heart (now available on Audible).
http://applemanshapiro.com/
July 20 @ Bring on Lemons with Penny Harrison
Wisconsin business owner and avid reader Penny Harrison shares her thoughts and feelings about Shirley Melis's memoir Banged Up Heart (now available on Audible).
July 22 @ Memoir Writers Journey with Kathleen Pooler
Kathleen Pooler shares her thoughts with readers of Memoir Writer's Journey - find out what Pooler has to say about reading Shirley Melis's memoir Banged Up Heart.
http://krpooler.com/
July 23 @ Bring on Lemons with Pang Papathopolus
Wisconsin professional caregiver and health care worker Pang gives her review of the Audible version of Banged-Up Heart by Shirley Melis.
July 24 @ The Constant Story with David W. Berner
Author and radio personality David W Berner reviews Shirley Melis's book Banged Up Heart and shares his thoughts with readers at The Constant Story.
http://davidwberner.blogspot.com/
July 25@ World of My Imagination
Nicole Pyles shares her review of Banged-Up Heart on Audible - hear what she has to say about this memoir in it's audio form!
https://theworldofmyimagination.blogspot.com/
July 26 @ Bring on Lemons with Michelle DelPonte
Michelle DelPonte shares her review of the Audible version of Shirley Melis's Banged Up Heart today at Bring on Lemons. Don't miss this exciting blog stop and book giveaway.
Home-Grown Authors
By Shirley Melis
"I came here expecting to buy produce, and here I am, buying books!" The tall man in khakis and a blue polo shirt cradled three books in one arm while his wife listened to another author at the table pitch her book.
Home-Grown Authors, sponsored by the New Mexico Book Association, is the brainchild of Maxine Davenport, a local author who writes compelling short stories and novels about strong women. Love Is a Legal Affair is her latest. As many as six local authors can be found indoors at Santa Fe's Farmers Market every Tuesday morning until Thanksgiving. Their works of fiction and non-fiction run the gamut from westerns and murder mysteries to memoir and travel stories.
• Gone to the Dogs is author Tom Lohr's story of his 103-day odyssey to find the best combination of baseball and hot dogs at major league baseball parks in North America.
• Belinda Perry, author of An Old Woman's Lies, also writes westerns, using the name William Luckey. "When I started writing westerns, I figured no one would buy a western written by a woman," Belinda confides.
• Taking the act of walking seriously is author Michael Metras's mission. Germany to Rome in 64 Days: Our Pilgrimage is Michael's book about a walk he took with his wife, whom he met on a walk across northern Spain. On the Germany-to-Rome trek, each went through nine pairs of shoes.
Farmer's Market Author's Table
NM Book Association sponsors the table, exhibitors must be members of NMBA. For further information about exhibiting, please contact Maxine Davenport: maxine@kewa.com.
Authors present on October 10, from left to right (and latest publication):
Maxine Neely Davenport ("Love is a Legal Affair")
Shirley Melis ("Banged-Up Heart")
Roberta Parry ("Killing Time")
Tom Lohr ("Command Decision")
Belinda Perry (pen name: William Luckey, "Time Alone")
PREPARING FOR MY BOOK LAUNCH AT COLLECTED WORKS IN SANTA FE, NM
When owner Dorothy Massey offered me a Reading/Signing at Collected Works, I was euphoric. Reading at Collected Works, Santa Fe's #1 independent bookstore, would be a dream come true. But what parts of my 300-page memoir would I read?
"You need a program," said editor Morgan Farley, who suggested I take a look at some author videos on YouTube. I clicked on a YouTube of one author who impressed me because she looked and sounded spontaneous. Much to my surprise, I found when clicking onto her other YouTube videos that she'd repeated the same "program" time after time. Heartened by the idea that I might put together a reading I could use more than once, I selected passages that followed my story line without revealing the ending.
For a practice reading in front of Morgan, I copied pages from a pdf of my book. Squinting to read without my glasses and rushing through the passages, I could see the disappointment in Morgan's face. "You have time," she said, "to make this good. I've heard you read before; I know you can do it. You can find a recording machine at Best Buy for $100. Get one and listen to yourself."
Over the next few days, I made a few decisions:
No glasses -- To read effectively without wearing my glasses, I typed out the passages with a font size of 14 or 16. I would not be reading from a hand-held book.
Develop patter (connective tissue) between passages – My goal was to be as seamless as possible from the moment I started speaking until I stopped. I had to think about what was happening between passages and how to summarily cover those happenings without revealing everything and at the same time, bridge the gaps without losing the story line.
Find a recording device -- My husband Frank offered me his iPhone with voice memo app I used to record before finding the same app on my own iPhone.
Hearing myself read was enlightening. "Drop yourself into your story," Morgan advised. When I did, I found myself reliving the story with feeling that I hadn't expressed on my first practice read.
For the next few mornings, I practiced my patter aloud during 30-minute treadmill sessions. I wanted to memorize it so that I could look up and out at the audience except when looking down to read the passages. In the afternoons, when nobody was around, I'd tape myself. Eventually, I was satisfied with my reading, my voice inflections and pauses. (Morgan, a poet who reads beautifully, was a great help with this.)
And then there was the question of using a microphone. At Collected Works, I would be on a small stage, a platform three giant steps above the main floor. One afternoon, about a week before my reading, Dorothy arranged for one of her staff to set up the microphone so I could test the sound and determine how close I needed to be to the mouthpiece. She offered a music stand onto which I'd drop my pages as I read.
The day of my reading, I awoke feeling a little nervous but as ready as I could be. That evening, before an audience of 115, I learned that my initial concerns, subsequent decisions and practice paid off.
Choosing a Subtitle
Pressed to come up with a subtitle for Banged-Up Heart -- in time for a poster display at my college reunion -- I quickly settled on A Widow's Story of Star-Crossed Love. It wasn't until many months later, about the time I landed a publisher, that I began to lament my hastily concocted subtitle. A potential reader might think my book was about a widow who falls in love with someone who dumps her, and that was not the case. Why hadn't I seen that earlier? If a subtitle tells the reader what the memoir is about, what was my memoir really about? "Love and loss," I said to my editor, who'd asked the question. She nodded her agreement. But the words "love and loss" alone seemed incomplete. What was it about love and loss that would give the reader insight into the essence of Banged-Up Heart?
My thoughts turned to the Anne Lamott Plan B piece where I'd found my title, and that's when it hit me. The subtitle should say something about dancing. I ran a couple of possibilities by Terra Nova Books editor, Marty Gerber: "How about A Widow Dances with Love and Loss?" "Yes," he said, "you were widowed at the start; yes, you're twice widowed at the end. But in between is the story of two diverse individuals – powered by love—trying valiantly to know each other and find a way together to battle an overwhelming enemy. I simply feel that the 'human-ness' of the tale you tell is so much more than the 'widow-ness' angle." "How about A Dance with Love and Loss?" I asked. "It's not a single dance," he said. "You're dancing with love and loss from beginning to end." And that's how the subtitle, Dancing with Love and Loss, came to be. Sometimes it takes time and distance to be able to re-focus effectively on the essence of what you've written. And having the reaction of someone else, such as editor Marty Gerber, can prove invaluable.
Choosing a Title
"Send your creative works for display at Reunion." My college reunion was three months off but, regrettably, my memoir, my creative work, was not finished. I was still rewriting with no likelihood of completing it in time. But what an opportunity to promote it! I thought. I might even find the name of a literary agent from an author classmate. My editor agreed. With her encouragement, I met with Amiel Gervers, a tall brunette who'd left an advertising career in New York City to live in Santa Fe, New Mexico.
"Shirley, I can design a website for your book, and I can have it finished with posters and postcards in time for your college reunion." Stirring sugar into her tea at the Teahouse on Canyon Road, Amiel told me what she'd need. "First, I want to read your book. I know you haven't finished rewriting it, but that's all right. I want to select quotes to go with some of these photos you've shown me. You'll need to send these to me electronically. Also, I'll need a bio from you and while you at it, something about your "Tahoe Traveler" writing experience. In my twenties, I'd traveled in Europe and the Middle East, writing a column for the Tahoe Daily Tribune.
Within days, Amiel had everything she needed from me with one exception: a title for the book. I'd sent her a photo to use as a cover mock-up -- a burned out forest with nascent green grass denoting regrowth in the foreground. But I was still clueless about a title. Months earlier, I'd discarded a working title, Love + Loss x Two, because I'd narrowed my focus to my recent love and loss.
Hoping for an epiphany, I hunkered down with the last several chapters. I would be discussing one, possibly more, of these with my editor in a few days. In one chapter,
my attention riveted on a selection by writer Anne Lamott that the Rev. Kim Beach read at John's memorial service. That's it! I thought.
Eager to share my find with editor Morgan Farley, I could hardly contain myself as I sat down at the table for our editing session. "I have a title!" I blurted. "So do I," she said. "Banged-Up Heart," I said. "That's what I came up with, too." Morgan beamed with delight. And that's the story of how I found a title for my memoir – in time to promote it at my college Reunion. Since then, the cover has changed but the title remains.


text:
listowel fire department
Thinking of the fallen fire fighters who gave their lives to protect listowel today. RIP. Prayers go to family and friends. They were heroes
Our Thoughts and Prayers are going out to the families of the 2 Fire Fighters who lost their lives while fighting a fire in Listowel Ontario.
Two volunteer firefighters were killed in the line of duty in southwestern Ontario, Canada on Thursday listowel fire department battling a commercial department-store fire in Listowel, Ont.,listowel fire department which is 160 kilometres east of Toronto, Ontario。
A short time later, two firefighters were unaccounted for. Firefighters conducted a search of the building and found the two downed firefighters who had succumbed to injuries they suffered while fighting the fire.
No further details about the victims were available at the present time. The firefighters' bodies were still in the building at 20:00 hours., ET, Thursday, and the listowel fire department Fire Marshal's office had taken over the scene. Fire fighter Line of duty deaths is not common in Canada and having a fire in which there is a double LODD is even more unheard of.
sincere condolences to the North Perth fire department and the people of Listowel. It's going to be a difficult day tomorrow.
Today at 15:30 we lost two brave firefighters from Listowel Volunteer Fire Department from (North Perth, Ontario, Canada) They perished in a structure fire .
My thoughts and prayers go to the Listowel Fire Department and to the families of the two volunteer firemen who lost their lives today fight。


Forgiveness of a Chaos Dwarf - Quests :: Tip.It RuneScape Help :: The Original RuneScape Help Site!
Start location: Speak to Commander Veldaban, in north-west Keldagrim.
1. Speak to Commander Veldaban. He tells you people are missing from Keldagrim East, but you are interrupted by a guard that alerts the commander that the Chaos Dwarves are attacking the Black Guard. There is still a chaos dwarf holed up in the guard tower with some special weapon.
2. You will be transported, with a short cutscene, to the guard tower south of Keldagrim. Go upstairs with Veldaban and kill the level 100 Chaos Dwarf Cannoneer, who has a gun equipped. Take the slip it drops and speak to Veldaban.
3.The slip is a grand exchange order slip. Take the mine cart or teleport to the grand exchange, and talk to a clerk to exchange the slip for a package.
4. Return to Veldaban in the bar in eastern Keldagrim. Ask him if red axe is involved, and he asks for a Guardman's stout. Buy him one (at no cost), and then agree to take him back to the headquarters. After a short cutscene where Veldaban embarrasses himself, he will go to sleep.
5. Speak to Veldaban while sleeping, and he will mention Hilda.
6. Return to the guard tower, south of the headquarters, where you killed the chaos dwarf, and use the boot from the package on the footprints.
7. Follow the footprints to the fire pit near the end of the trail, and use the bowl, also from the package, on the fire. Remove the bowl, and use the rod on the now-dead fire. Turn the rod, and it will open up a secret passage in the wall to the west.
8. Walk through the passage, through the cave, and try to operate the controls. Exit the mines and return to Keldagrim. Speak to Commander Veldaban where you originally found him; he will be awake now and follows you back to the secret passage.
9. Operate the controls on the Dwarven machinery and Commander Veldaban will offer to take the stones and man the board while you ride the cart. Now free both hands and you can ride the cart, while Veldaban operates the controls. The order to insert the coloured stones to change direction is: Yellow, Green, Yellow, Green, Green, Yellow, Yellow, Yellow, Yellow.
10. Lower the bridge, and open the prison door. Speak to Fjoila, and the prisoners will escape back to Keldagrim.
11. Turn the switch south of the prison and proceed through the door. Veldaban continues to talk about Hilda. Watch the chaos dwarves convert Brunolt into a chaos dwarf.
12. Warning: Dwarf cannoneers will attack you after the cutscene. Go through the door and speak to Veldaban. After talking to him, use the metal rectangle, then the turnscrew on the sorting machine. Squeeze through the hole in the wall, then ride the mine cart back to Keldagrim. Watch the cut scene, and the quest will be complete!
Your hunter level can be boosted with a Hunter potion, the Oo'glog mud pool, or Spicy stew to reach level 61 to use the boot with the footprints.
Armour with a high range defence bonus is suggested.
The North Keldagrim mining area contains 8 Gold ore, 10 Adamantite ore, and 2 Runite ore rocks.
Chaos Dwarves, Chaos Dwarf Cannoneers, and Dwogres on the battlefield drop the Dragon pickaxe. The cannoneers also drop a hand cannon and ammunition.


Christian & Company Marine Surveyors, Inc. has been in business since 1995 providing high quality pleasure craft marine surveys for both power boats and sailing vessels. Our marine surveyors are individuals of integrity who are active members of The Society of Accredited Marine Surveyors (Kells Christian, AMS ? # 301, George Jarvie, AMS ? #941 and Brad Destache, SA). This is your assurance that our marine surveyors have passed the rigorous testing required for membership.Marine Travelift Christian & Company?s marine surveys are accepted by all insurance companies, financial institutions, and other marine related industries. Our areas of expertise include wooden boats, fiberglass boats, steel boats, and aluminum boats. We also have performed marine surveys on all types of pleasure craft and yachts, commercial vessels, including fishing boats, and also ferro cement vessels. Mechanical surveys are available for most marine engines. If your insurer or lending institution has requested that you provide them with a recent in-the-water or hauled marine survey; or any of our other services, please give us a call. We?re here to help you in any way we can. Our commitment to the boating public is to provide quality, professional service and ensure your boat provides you with the pleasure, recreation, and fun you expect. Our unwavering participation in continued education assures our clients of the most up-to-date knowledge on safety issues, regulations, industry trends, and market conditions in the boating industry.


Over the years, many patterns were produced to knit outfits for Sindy and other teenage dolls. I have collected many of these patterns and I have now set up a new site where you can see all the patterns and they are also available to purchase as a PDF file sent by email. There are already a lot of patterns up and more will be added shortly.
Visit Sindy Knitting Patterns now.
Click here for free Patterns including the one shown below.


text:
Q: Webpack somehow bypasses Express routing completely I am starting from this excellent tutorial: https://www.fullstackreact.com/articles/using-create-react-app-with-a-server/ and trying to extend it by adding a new page to serve through a new route. However after hours of mucking around I am realizing that somehow create-react-app is doing some weird magic (as mentioned in their docs here):
`create-react-app` configures a Webpack development server to run on `localhost:3000`.

This development server will bundle all static assets located under `client/src/`.

All requests to `localhost:3000` will serve `client/index.html` which will include Webpack's `bundle.js`.

The key quote is "All requests to localhost:3000 will serve client/index.html". I have no idea how this happens. So even though i mess around with routes/index.js:
   app.route('/')
        .get(function (req, res) {
            res.sendFile(bipath.join(__dirname, '../public', 'THISCANBEANYRANDOMFILENAME.html'))
        });

it doesnt matter because webpack is somehow directing localhost:3000 to index.html anyway. where and how is it doing this? Bottom line I am trying to modify my routes to serve a new html file and am running into all sorts of filepath issues (yes, even when i use require('path') or sendFile(...,{root: __dirname}).)
So what exactly is going on here and can you give me any hints to help me out?
Edit: this could be from babel as well as webpack - i'm not exactly clear where babel hands off and where webpack starts.

A: I haven't played around with create-react-app, but it seems like instead of using the default npm start, you could create your own server file and run that.
This looks like a good example.
https://medium.com/@patriciolpezjuri/using-create-react-app-with-react-router-express-js-8fa658bf892d#.6y4rrl61q
Alternatively, if you're looking to have routes used as an api, you could proxy them to a different port like shown in the tutorial you linked.

redpajama_set_name: RedPajamaStackExchange


Please Update your bookmarks for the following sites.
The Website for The Children Of The Amulet has moved to www.amulet.csldesigns.net. Please update your bookmarks.
The Website for Shoplifting The Funnierside has moved to www.funnyside.csldesigns.net. Please update your bookmarks.
Also check out the other websites off www.csldesigns.net.


text:
Villa kan syfta på:

 Villa –  ett friliggande bostadshus
 Sommarvilla – ett större enfamiljshus avsett att användas som bostad under sommaren
 Villa (djur) – ett släkte av tvåvingar
 Villa Lidköping BK – en bandyklubb i Sverige
 Aston Villa FC – en fotbollsklubb i Birmingham i Storbritannien

Personer
 David Villa (1981–), en spansk fotbollsspelare som spelar för Atlético Madrid

Geografi

Estland

 Villa, Viljandimaa, en by i Viljandi kommun i landskapet Viljandimaa,
 Villa, Rõuge kommun, en by i Rõuge kommun i landskapet Võrumaa,
 Villa, Võru kommun, en by i Võru kommun i landskapet Võrumaa,

Italien

 Villadossola, kommun, Piemonte, Verbania,
 Villa di Tirano, kommun, Lombardiet, Provincia di Sondrio,

Se även
 Vila (olika betydelser)


The official µTorrent® (uTorrent) torrent client for Windows, Mac, Android and Linux-- uTorrent is the #1 BitTorrent download client on desktops worldwide. Watch Sins of the Past online. Stream Xena: Warrior Princess season 1, episode 1 instantly.


text:
The Kajang 2 Railway station is a KTM Komuter infill train station located in Kajang 2, Kajang, Selangor. The station is on the KTM Komuter's Seremban Line.

Station features and history
According to the Managing Director of the MKH Berhad, Tan Sri Eddy Chen Lok Loi, the company had purchased  of land in Kajang 2 as early as 2003 and had proposed to the government to jointly develop  of railway reserve land through a smart partnership model. The construction of this railway station is an in-kind contribution received by the Railway Assets Corporation (RAC) as a result of leasing the railway land to Srijang Kemajuan Sdn. Bhd. (SKSB) through a Development Cum Lease Agreement (DCLA) on October 12, 2012. This lease privatization project was approved by the Cabinet on February 8, 2012, with a lease period starting from October 8, 2018 until October 7, 2078. MKH Berhad will be responsible for building a new railway station, which will be handed over to the Railway Assets Corporation as a public facility and operated by the Keretapi Tanah Melayu Berhad (KTMB) for commuter train services.

Eddy Chen also mentioned that the two-story Kajang 2 station is scheduled to be completed in March 2018. It is located between  Station and the  station, near the Terminal Kajang and the Jalan Reko rail section. After the station is operational, passengers can take commuter trains to Kajang Station and transfer to the MRT Kajang Line to head north to Kuala Lumpur. The station's architectural design is by Swan & Maclaren Architects.。

This new train station was built to support the leading development of the  Kajang 2 new township by the local developer, MKH Berhad. MKH funded the construction of the station with RM33 million, and the entire development area covers . Additionally, two to three years after completion, the KTM Mall around the station will also be operational. As of 2017, the construction of the station has already been completed to 40%. The surrounding facilities of the Kajang 2 station include the Rafflesia International School, the Kajang 2 Chinese Primary School built by the company, and a proposed private hospital. The company expects an annual passenger rate of 550,000 people to use the new station. Before the station is operational, MKH has allocated 200 parking spaces on the nearby land for future use, and will combine the multi-storey car park that can accommodate 2,000 parking spaces with the future train station mall.

The station was completed in 2020, but due to the pandemic, it was delayed until now to be put into use.The construction cost of the station due to the pandemic-related factors, has exceeded the budget and reached RM70 million.The station has been operational since March 13, 2023, with 46 train services operating on weekdays and 33 train services operating on weekends and public holidays. The KTM Komuter operates on a current timetable, with a frequency of 30 minutes during peak hours and 60 minutes during off-peak hours.

The railway station covers an area of  and is equipped with various facilities, including parking for 500 vehicles, a pedestrian overpass, four escalators, facilities for the disabled, eight retail kiosks ready for rent, security surveillance cameras, a passenger drop-off area, and other basic facilities such as toilets, surau, and covered platforms. Two accessible lifts equivalent to three-storey in height will be installed at both ends of the pedestrian bridge. The parking area will be operated by Srijang Kemajuan Sdn. Bhd. with a maximum fee of RM5.50 per day.
There is another pre-set parking lot with 200 parking spaces located outside the station, which will be open in the future. In order to encourage nearby residents to use the new train station, the parking fee for the first month of opening will be free.

Eddy Chen also mentioned that pedestrian walkways in the residential areas of Kajang 2 will be connected to the train station, and a pedestrian overpass will be built across the railway tracks and Jalan Reko to Bandar Baru Bangi Section 5, allowing pedestrians to walk along a 600-meter shoulder path to the Terminal Kajang, thus realizing the concept of a public transportation hub. Therefore, this is not only convenient for residents of the new township, but also benefits the public.

Bus Services
Rapid KL trunk route  (Kajang - Lebuh Pudu), MRT Feedr Bus routes  (Stadium Kajang - UKM),  (Stadium Kajang - Teras Jernang) and Smart Selangor  (Kajang - Bandar Baru Bangi) serves the station at Jalan Reko's Plaza Perabot bus stop.

Gallery

References

External links

KTM Komuter Routes

Seremban Line
Railway stations in Selangor


1 06:08 Briar Cliff (Iowa) - Freeman, Logann 94 yd pass from Diaz, Jacob. (Ylagan, Noah kick is good).
1 01:35 Dakota Wesleyan - Tate Gale 30 yd field goal.
2 05:02 Briar Cliff (Iowa) - Brown, Elijah 65 yd pass from Diaz, Jacob. (Delia, Steven kick failed).
3 08:27 Dakota Wesleyan - Tate Gale 25 yd field goal.
4 02:36 Briar Cliff (Iowa) - French, Tyler 12 yd pass from Diaz, Jacob. (Delia, Steven kick is good).