text:
\section*{Introduction}
In 1878/1879, F. Klein \cite{Kle79} found and studied in detail
a remarkable complex algebraic curve,
which nowadays bears his name. It can be defined
in the complex projective plane
by the homogeneous polynomial
$$\Phi_{4}: x^{3}y + y^{3}z +z^{3}x,$$
and its exceptional properties follow from the fact that it
has an automorphism group of the maximal possible order
according to Hurzwitz's bound \cite{Hur}, namely
$$|\ensuremath{\operatorname {Aut}}(\Phi_{4})| = 84(g-1)= 168.$$
Because $\Phi_{4}$ is a smooth quartic, its group of automorphisms $\ensuremath{\operatorname {Aut}}(\Phi_{4})$ is realized by a subgroup $G\subset\ensuremath{\operatorname {Aut}}(\ensuremath{\mathbb{P}}^2_\ensuremath{\mathbb{C}})=\operatorname{PGL}(3,\ensuremath{\mathbb{C}})$.

In the theory of complex arrangements of lines, there is a particularly symmetric arrangement of 21 lines well known for its properties, extremal in many senses; it was discovered by Klein, and it is the unique set of 21 lines invariant under the action of $G$.
In this work, following the recent trend to extend the study of arrangements of lines to arrangements of curves of higher degree, we study an arrangement of lines and conics closely related to Klein's arrangement of lines, which displays a similarly rich geometry, with some features extreme among arrangements of lines and conics.

Every sufficiently general smooth quartic has 21 reducible polars
(a fact already known by E.~Bertini in 1896, \cite{Bertini}) and in the case of Klein's quartic, each of them splits as the union of one of the lines in Klein's arrangement plus a smooth conic.
Taken together, they form the arrangement $\mathcal{K}$ of 21 lines and 21 conics which we are going to describe.
Our main contribution is the description of all singularities of the arrangement $\mathcal{K}$, namely their position with respect to Klein's quartic and the action of the group $G$. The arrangement has $42$ nodes, $252$ ordinary triple points and $189$ ordinary quadruple points.
We additionally show the following.
\begin{enumerate}
	\item $\mathcal{K}$ is neither free nor nearly free \cite{Dimca}.
	\item The logarithmic Chern slope of the arrangement of 21 conics is $E\approx 2.25$, which is the largest known value in the class of arrangements of curves of degree 2.
	\item The Harbourne index of $\mathcal{K}$ is $h(\mathcal{K})\approx -3.087$, the second most negative known value for reduced curves with ordinary singularities after Wiman's arrangement of lines, which has $h(\mathcal{W})\approx-3.358$.
	\item The ideal $I_2$ (respectively, $I_3$) of the set of all singular points (respectively, of points of multiplicity at least 3) in $\mathcal{K}$ provides examples of \emph{failure of containment} \cite{SzSz}, in the sense that $\displaystyle{I_m^{(3)}\not\subseteq I_m^{2}}$ with $m \in\{2,3\}$.
\end{enumerate}

The equations of the involved curves are all explicit, and all claims we make can be verified by a Computer Algebra System such as Singular \cite{DGPS}. In the Appendix we include scripts to do so. In the text, we give conceptual proofs for all claims except the failure of the containment for $I_2$ due to very involving computations.

Additionally, we study further configurations of curves of higher degree invariant under the action of the Klein group $G$, and also a remarkably analogous configuration of reducible polars of the plane sextic curve invariant under $A_6$, which contains Wiman's arrangement of lines $\mathcal{W}$ mentioned above. For the latter arrangement, after proving its existence we compute its singularities (some of which are not ordinary) and Harbourne index, which turns out to be $-3.38$.

The paper is structured as follows. In Section \ref{sec:klein-lines}, we recall classical results on Klein's quartic and properties of Klein's arrangement of lines from a viewpoint of the Klein group $G$. In Section \ref{sec:construction}, we describe the singularities of Klein's arrangement $\mathcal{K}$ of $21$ conics and $21$ lines. As we shall see, all singularities of the arrangement are ordinary. Starting from Section 3, we focus on possible applications of our arrangement. We compute the Harbourne index of the arrangement $\mathcal{K}$ (Section \ref{sec:h-index}) and also for its further iterations (Section \ref{sec:iteration}). In Section \ref{sec:freeness}, we show that Klein's conic-line arrangement is neither free nor nearly-free, in Section \ref{sec:chern} we put our configuration into the context of log-surfaces and we compute its Chern numbers. In Sections \ref{sec:containment} \& \ref{sec:iteration}, we focus on the containment problem for symbolic and ordinary powers of ideals, showing that certain sets of singular points of $\mathcal{K}$ and other invariant arrangements of higher degree provide new counterexamples to the containment $I^{(3)} \subset I^{2}$. Finally, in Section \ref{sec:wiman}, we study the analogous arrangement of reducible quintics invariant under $A_6$.

We work over the complex numbers.

\section{Klein's arrangement of lines}\label{sec:klein-lines}

Klein's quartic has been extensively studied (the book
\cite{KLBook} contains a plethora of information and references
on this beautiful subject, including a translation of Klein's
original paper) both for its intrinsic properties as a Riemann
surface and as an algebraic plane curve, which is the point of view
we are interested in.

Among the $168$ automorphisms of Klein's quartic,
one can find exactly $21$ involutions, i.e.,
$\alpha \in \ensuremath{\operatorname {Aut}}(\Phi_{4})$ such that $\alpha \neq 1$ and
$\alpha^{2} = 1$. Each of these $21$ involutions,
seen as an automorphism of $\mathbb{P}^{2}$,
is a \emph{harmonic homology} (see \cite[5.7.23]{Cas14}), with a line of fixed points
(its \emph{axis}) and an isolated fixed point (its \emph{center}).
The resulting arrangement of $21$ lines (the axes)
has exactly $21$ quadruple points (at the centers)
and $28$ triple points (which by duality correspond to the 28 bitangents
to the quartic).

We next recall the construction and properties of Klein's
arrangement of lines, following N.~Elkies \cite{Elkies}
(see also \cite{SymmetricBU} by T.~Bauer, S.~Di~Rocco, B.~Harbourne, J.~Huizenga, A.~Seceleanu, and T.~Szemberg).

Let $G= \ensuremath{\operatorname {PSL}}(2,7)\cong\ensuremath{\operatorname {Aut}}(\Phi_{4})$ be the unique simple
group of order $168$. Denoting $\zeta$ a primitive 7-th root of 1,
$G$ has an irreducible $3$-dimensional representation $\rho$
over $\mathbb{Q}(\zeta)$, given by three generators $g,h,i$
as follows:
$$\rho(g) = \begin{pmatrix}
\zeta^4  & 0 & 0\\
0 & \zeta^2 & 0 \\
0 & 0 & \zeta
\end{pmatrix}, \quad
\rho(h) = \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}$$
and
$$\rho(i)=\frac{2\zeta^4+2\zeta^2+2\zeta+1}{7}\begin{pmatrix}
\zeta-\zeta^6 & \zeta^2-\zeta^5 & \zeta^4-\zeta^3 \\
\zeta^2-\zeta^5 & \zeta^4-\zeta^3 & \zeta-\zeta^6 \\
\zeta^4-\zeta^3 & \zeta-\zeta^6 & \zeta^2-\zeta^5
\end{pmatrix}.$$
Note that all three matrices have determinant 1 and the element $i$ has order $2$.
This representation gives an embedding of $G$ into $\ensuremath{\operatorname {SL}}_3(\mathbb{Q}(\zeta))$.  By projectivizing, $G$ acts on $\mathbb{P}^2$ and on its dual,
$\check{\mathbb{P}}^2$. Note that a matrix $M\in \rho(G)\subset \ensuremath{\operatorname {SL}}_3(\mathbb{Q}(\zeta))$ takes the point $p=[a:b:c]$ to the point $[a':b':c']$ determined by
$(a',b',c')=(M(a,b,c)^T)^T$, but if $[A:B:C]\in\check{\mathbb{P}}^2$ is the point dual to the line $L=V(Ax+By+Cz)$, then $M$ takes $L$ to $L'=V(A'x+B'y+C'z)$, where
$(A',B',C')=((M^{-1})^T(A,B,C)^T)^T$; i.e., $(A',B',C')=(A,B,C)M^{-1}$.

\begin{remark}\label{rmk:transpose}
	Observe that the transposed matrix of each generator
	$\rho(g)$, $\rho(h)$, $\rho(i)$ belongs to the group
	they generate, $\rho(G)$ which, abusing the notation, we
	call simply $G$.
	Therefore the projectivity
	$\iota:\mathbb{P}^2\rightarrow\check{\mathbb{P}}^2$
	given in coordinates $[a:b:c]$ and $[A:B:C]$ by the
	identity matrix, even though it is not equivariant,
	maps orbits to orbits. More precisely, points
	$[a:b:c]$ and $[a':b':c']$ lie in the same orbit
	under the action of $G$
	if and only if the lines $V(ax+by+cz)$ and
	$V(a'x+b'y+c'z)$ lie in the same orbit under the dual action.
\end{remark}
Consider the action of $G$ on the homogeneous coordinate ring
$S = \mathbb{C}[x,y,z]$ of $\mathbb{P}^2$.
The ring $S^G$ of polynomials invariant under the action of $G$
is well-known since Klein's work \cite{Kle79}.
It is generated by four polynomials
$\Phi_{4},\Phi_{6},\Phi_{14}$, and $\Phi_{21}$, where $\Phi_d$ has degree $d$.  The invariant $\Phi_{21}$ defines the line arrangement $\mathcal{K}_1=V(\Phi_{21})$.  The polynomials $\Phi_{4},\Phi_{6},\Phi_{14}$ are algebraically independent, and $\Phi_{21}^2$ belongs to the ring generated by $\Phi_{4},\Phi_{6},\Phi_{14}$.

The geometric significance of the invariants $\Phi_d$ is explained in
\cite{Elkies}. The polynomials $\Phi_{4}$ and $\Phi_{6}$ are uniquely
determined up to a constant factor;
we take $\Phi_{4}$ to be Klein's quartic above, and $\Phi_{6}$ to be
$$\Phi_{6} = -\frac{1}{54} H(\Phi_{4}) = xy^5 + yz^5+zx^5-5x^2y^2z^2,$$
where $H(\Phi_{4})$ is the Hessian determinant
$$H(\Phi_{4}) := \begin{vmatrix} \partial^2 \Phi_{4}/\partial x^2&\partial^2 \Phi_{4}/\partial x \partial y&\partial^2 \Phi_{4}/\partial x \partial z\\
\partial^2 \Phi_{4}/\partial y \partial x&\partial^2 \Phi_{4}/\partial y^2&\partial^2 \Phi_{4}/\partial y \partial z\\
\partial^2 \Phi_{4}/\partial z\partial x&\partial^2 \Phi_{4}/\partial z \partial y&\partial^2 \Phi_{4}/\partial z^2\end{vmatrix}.$$
The degree $14$ part of $S^G$ is spanned by $\Phi_{14}$
and $\Phi_{4}^2\Phi_{6}$, so the invariant $\Phi_{14}$ is
not uniquely determined up to constants.
One possible choice is
$$\Phi_{14} = \frac{1}{9}BH(\Phi_{4},\Phi_{6}),$$ where $BH(\Phi_{4},\Phi_{6})$ is the \emph{bordered Hessian} $$BH(\Phi_{4},\Phi_{6}) := \begin{vmatrix} \partial^2 \Phi_{4}/\partial x^2&\partial^2 \Phi_{4}/\partial x \partial y&\partial^2 \Phi_{4}/\partial x \partial z & \partial \Phi_{6} / \partial x\\
\partial^2 \Phi_{4}/\partial y \partial x&\partial^2 \Phi_{4}/\partial y^2&\partial^2 \Phi_{4}/\partial y \partial z & \partial \Phi_{6}/\partial y\\
\partial^2 \Phi_{4}/\partial z\partial x&\partial^2 \Phi_{4}/\partial z \partial y&\partial^2 \Phi_{4}/\partial z^2 & \partial \Phi_{6}/\partial z\\
\partial \Phi_{6}/\partial x & \partial \Phi_{6}/\partial y & \partial \Phi_{6}/\partial z & 0 \end{vmatrix}.$$
Finally, the polynomial $\Phi_{21}$, which splits
as the product of the lines in Klein's arrangement, can
be defined by the following Jacobian determinant
$$\Phi_{21} =\frac{1}{14} J(\Phi_{4},\Phi_{6},\Phi_{14}) = \frac{1}{14} \begin{vmatrix} \partial \Phi_{4}/\partial x & \partial \Phi_{4}/\partial y & \partial \Phi_{4}/\partial z\\
\partial \Phi_{6}/\partial x & \partial \Phi_{6}/\partial y & \partial \Phi_{6}/\partial z\\
\partial \Phi_{14}/\partial x & \partial \Phi_{14}/\partial y & \partial \Phi_{14}/\partial z\\ \end{vmatrix}.$$

\begin{proposition}[\cite{Kle79}, see also \cite{Elkies},\cite{SymmetricBU}]\label{pro:orbits}
	Every point $p\in \ensuremath{\mathbb{P}}^2$ has an orbit
	of size 168, except:

\begin{enumerate}
	\item The triple points of $\mathcal{K}_1$ form an orbit
	$\ensuremath{\mathbf{O}}_{28}$ of size $28$. The point $[1:1:1]$ lies in this orbit.

	\item The quadruple points of $\mathcal{K}_1$ form an orbit $\ensuremath{\mathbf{O}}_{21}$
	of size $21$.

	\item The invariant curves $V(\Phi_{4})$ and $V(\Phi_{6})$ meet in
	an orbit $\ensuremath{\mathbf{O}}_{24}$ of $24$ points.
	The point $[1:0:0]\in \ensuremath{\mathbb{P}}^2$ lies in this orbit.

	\item The invariant curves $V(\Phi_{4})$ and $V(\Phi_{14})$ meet
	in an orbit $\ensuremath{\mathbf{O}}_{56}$ of $56$ points.
	The point $[\omega^2:\omega:1]\in \ensuremath{\mathbb{P}}^2$ lies in this orbit, where $\omega$ is a primitive cube root of 1.

	\item The invariant curves $V(\Phi_{6})$ and $V(\Phi_{14})$ are tangent at an orbit $\ensuremath{\mathbf{O}}_{42}$ of $42$ points belonging to $\mathcal{K}_1$.

	\item Any point on $\mathcal{K}_1$ not mentioned above has an orbit of size $84$.
\end{enumerate}
Moreover, the orbits $\ensuremath{\mathbf{O}}_{21}$ and $\ensuremath{\mathbf{O}}_{28}$ are not contained in either
$V(\Phi_{4})$, $V(\Phi_{6})$ or $V(\Phi_{14})$. The orbit $\ensuremath{\mathbf{O}}_{24}$
is not contained in either $V(\Phi_{14})$ or $\mathcal{K}_1=V(\Phi_{21})$.
The orbit $\ensuremath{\mathbf{O}}_{56}$ is not contained in either
$V(\Phi_{6})$ or $\mathcal{K}_1=V(\Phi_{21})$.
\end{proposition}

By Remark \ref{rmk:transpose}, the orbits of lines follow
the same pattern as the orbits of points.
In particular, there is a unique orbit of 21 lines,
which are the components of $\Phi_{21}$ (and by the resulting duality,
each of the 21 lines contains four points of the orbit of size 21).
Recall that for each line in
$\Phi_{21}$ there is an involution $\alpha\in G$ which fixes it
pointwise --- for a general point $p$ in the line, $\alpha$
generates the stabilizer of $p$, and so $p$ belongs to one of the
orbits of size 84 described in the last item. The collection of centers
of the 21 involutions is also an orbit, namely $\ensuremath{\mathbf{O}}_{21}$.

Similarly, there is an orbit of 28 lines, made up exactly of the 28
bitangents of $\Phi_{4}$ (every smooth quartic has 28 bitangents).
The orbit $\ensuremath{\mathbf{O}}_{56}$ consists of the tangency points of $\Phi_4$ with its bitangents.
The joint arrangement of 21+28 lines and points is in itself
a beautiful object represented combinatorially by the so-called
\emph{Coxeter graph}; the following description is taken from
\cite[Theorem 6.1]{Bit}.

\begin{theorem}\label{thm:49}
Let $P$ be the set of points and $L$ the set of lines
of the Fano plane $\ensuremath{\mathbb{P}}^2_{\ensuremath{\mathbb{F}}_2}$.
There is a labelling of the $49$ singular points of $\Phi_{21}$, i.e.,
in $\ensuremath{\mathbf{O}}_{21} \cup \ensuremath{\mathbf{O}}_{28}$, by the elements of $P \times L$ and a
labelling of the 21 lines in $\Phi_{21}$ and the $28$ bitangents
by the elements of $L \times P$ in such a way that
\begin{enumerate}
\item $\langle p, l \rangle  \in P \times L$ represents a point in
$\ensuremath{\mathbf{O}}_{21}$ iff $p \in l, \langle l, p \rangle \in L \times P$
represents a component of $\Phi_{21}$ iff $l \ni p$,
\item the involution which fixes the line $\langle p, l \rangle$
has center $\langle l, p \rangle$,
\item the point $\langle p, l \rangle $ is on the bitangent $\langle m, q \rangle$ iff $p \in m$ and  $q \in l$,
\item the point $\langle p, l \rangle\in \ensuremath{\mathbf{O}}_{21}$ is on the component
$\langle m, q \rangle$ of $\Phi_{21}$
iff $p=q$ and $l\neq m$, or $p\neq q$ and $l=m$,
\item the point $\langle p, l \rangle\in \ensuremath{\mathbf{O}}_{28}$ is on the component
$\langle m, q \rangle$ of $\Phi_{21}$
iff $p \in m$ and $q \in l$.
\end{enumerate}
There are no other incidences between the $49$ points and the $49$ lines.
\end{theorem}

\section{The arrangement of 21 conics and 21 lines}\label{sec:construction}
It is known that a general smooth quartic has 21 reducible polar curves, namely the polars with respect to the 21 nodes of its Steinerian curve.
The Steinerian curve of a given curve $C$, introduced by J.~Steiner in \cite{Steiner}, is the locus of points $p$ such that the polar of $C$ with respect to $p$ is singular. A modern presentation of the Steinerian and its role in the polarity theory was given by I.~Dolgachev in \cite[1.1]{Dolgachev}.

It is well known (at least since 1896 by Bertini \cite{Bertini}) that the Steinerian of a general quartic has 21 nodes and 24 cusps; this is also the case for Klein's quartic (see \cite[Example~6.1.1]{DK93}, \cite[\S15]{Adler} for modern proofs and references to the history of the subject). Additionally, the Steinerian is invariant under the action of $G$, so its 21 nodes are invariant, hence they are the points of $\ensuremath{\mathbf{O}}_{21}$.
Note also that, since the Steinerian is an invariant curve of degree 12, its equation must be a combination of $\Phi_{4}^3$ and $\Phi_{6}^2$.
The only linear combination of these forms which vanishes at $\ensuremath{\mathbf{O}}_{21}$ is $4\Phi_{4}^3+\Phi_{6}^2$, so this is the equation of the Steinerian.

The polar of $\Phi_{4}$ with respect to each point $p\in \ensuremath{\mathbf{O}}_{21}$ splits as a smooth conic and a line meeting transversely at two points.
To describe in detail the singularities of the arrangement given by the 21 reducible polars, we shall use the gradient map $\nabla(\Phi_{4})$ given
by the partial derivatives of Klein's quartic equation, namely
$$\mathbb{P}^{2} \ni [x:y:z] \overset{9:1}{\longmapsto} [u:v:w]=\left[3x^{2}y + z^{3}: 3y^{2}z + x^{3}: 3z^{2}x+y^{3} \right] \in \check{\mathbb{P}}^{2}.$$
By definition, the polar $P_p(\Phi_4)$ with respect to the point $p=[a:b:c]\in \ensuremath{\mathbb{P}}^2$ is the preimage under $\nabla(\Phi_{4})$ of the line
dual to $p$, $V(au+bv+cw)\subset \check{\mathbb{P}}^{2}$.
Since Klein's quartic is non-singular, the partial derivatives
never vanish simultaneously, so the gradient map
is defined everywhere, i.e., it is a morphism.
Equivalently, the net of polar curves to $\Phi_{4}$,
which are the pullbacks of the lines
by $\nabla(\Phi_{4})$, has no base points.

The morphism $\nabla(\Phi_{4})$ is $9:1$, and it is
ramified along the degree $6$ smooth Hessian curve $V(\Phi_{6})$.
The image of Klein's quartic is its dual, and the image of the Hessian is the discriminant, or branch curve, of the morphism, which we denote $\Delta=\nabla(\Phi_{4})(V(\Phi_{6}))$.
By definition, the Steinerian is the dual of the discriminant $\Delta$ (see also \cite[1.1]{Dolgachev}).

It is well-known that the gradient map
is covariant under the representation $\rho$, respectively its dual ---
in fact this follows from the observation
that the linear space spanned by the partials
is invariant under the action of $\rho$ on $S$.
Hence $\nabla(\Phi_{4})$ maps orbits for the representation $\rho$ to orbits for the dual representation.
For convenience we identify $\ensuremath{\mathbb{P}}^2$ with its dual via the isomorphism $\iota:\ensuremath{\mathbb{P}}^2\rightarrow \check{\mathbb{P}}^{2}$ defined in Remark \ref{rmk:transpose} and given in coordinates by the identity matrix; recall that with this identification,
the orbits by $\rho$ and its dual agree, and we shall simply say that $\nabla(\Phi_{4})$ maps orbits to orbits.

\begin{proposition}\label{pro:reducible_polars}
	$\Phi_{4}$ has exactly $21$ reducible polars, and each consists of a line
	and a conic meeting transversely at two points.
	They are the polars with
	respect to the $21$ quadruple points of $\Phi_{21}$, and
	the preimages by $\nabla(\Phi_{4})$ of the $21$ lines making up
	$\Phi_{21}$. The $21$ lines which are components of reducible
	polars are also the components of $\Phi_{21}$, whereas each of the
	$21$ conics intersects the Klein curve at the eight points of contact
	of four bitangents. The nodes of the 21 reducible polars
	are all distinct and form the orbit $\ensuremath{\mathbf{O}}_{42}$.
\end{proposition}

\begin{proof}
	The fact that the $21$ reducible polars in the net are the polars with respect to the 21 nodes of the Steinerian, and that they split as transverse conic-line pairs, was shown in \cite[Example~6.1.1]{DK93}. As we already mentioned, their invariance under $G$ implies that the 21 nodes of the Steinerian are
	the points in $\ensuremath{\mathbf{O}}_{21}$.
	By definition, the polars of the $21$ points coincide with the
	pullbacks by $\nabla(\Phi_4)$ of the $21$ lines. Also, the $21$ line components of the
	reducible polars form an orbit under the action of $G$,
	and the only orbit of size $21$ is made up of the components
	of $\Phi_{21}$.

	If $p\in\ensuremath{\mathbf{O}}_{21}$, by Theorem \ref{thm:49}, four
	bitangents of $V(\Phi_{4})$ meet at $p$, so their eight points of
	tangency to $V(\Phi_{4})$ belong to the polar $(\Phi_{4})_p$, which we know splits as a pair conic and line.
	By Theorem \ref{thm:49} and Proposition \ref{pro:orbits},
	the eight bitangency points, which belong to $\ensuremath{\mathbf{O}}_{56}$, do not lie on the $21$ lines of $\mathcal{K}_1$, so
	they belong to the conic
	component.

	The whole set of singular points of the
	$21$ reducible polars (two points on each) is invariant under
	$G$. So these singularities form either the orbit $\ensuremath{\mathbf{O}}_{42}$
	or the orbit $\ensuremath{\mathbf{O}}_{21}$ (in which case each point would
	be singular for two reducible polars).
	But the latter option leads to a contradiction:
	a reducible polar is the preimage of a line
	by $\nabla(\Phi_{4})$, so every singularity on it must belong to
	the ramification locus $\Phi_{6}$, whereas, by Proposition \ref{pro:orbits},
	$\Phi_{6}$ does not contain $\ensuremath{\mathbf{O}}_{21}$. So the singularities
	of the 21 reducible polars are all distinct, and they are exactly
	the 42 points in $\ensuremath{\mathbf{O}}_{42}$.
\end{proof}

\begin{definition}
	We shall denote by $\Phi_{63}=(\nabla(\Phi_{4}))^{*}(\Phi_{21})$
	the equation of the arrangement of 21 reducible polars.
	It splits as $\Phi_{63}=\Phi_{21}\Phi_{42}$, where $\Phi_{42}$
	defines the arrangement $\mathcal{K}_2=V(\Phi_{42})$ of 21 conics.
	We also denote $\mathcal{K}=\mathcal{K}_1+\mathcal{K}_2=V(\Phi_{63})$.
\end{definition}

Our next goal is to determine the singularities of the reduced arrangement
$\mathcal{K}$. Clearly, these are the singularities of the reducible
polars (which were described in Proposition \ref{pro:reducible_polars})
and their intersection points, which
are the preimages of the singularities of $\mathcal{K}_1$.
These will be determined by showing that they lie off the
ramification curve $V(\Phi_{6})$.

\begin{lemma}\label{lem:conincident}
	If $p$ is a point belonging to more than one reducible polar
	of $\Phi_{4}$, then either $\nabla(\Phi_{4})(p)\in \ensuremath{\mathbf{O}}_{21}$
	and $p$ belongs to exactly $4$ reducible polars, or
	$\nabla(\Phi_{4})(p)\in \ensuremath{\mathbf{O}}_{28}$
	and $p$ belongs to exactly $3$ reducible polars.
\end{lemma}
\begin{proof}
	The reducible polars are exactly the preimages by $\nabla(\Phi_{4})$
	of the lines composing $\mathcal{K}_1$, therefore a point as in
	the statement belongs to the preimage of two such lines.
	This implies that $p$ belongs to the preimage of their unique
	intersection, which is either in $\ensuremath{\mathbf{O}}_{21}$ or $\ensuremath{\mathbf{O}}_{28}$.
	Moreover, the reducible polars to which $p$ belongs
	are precisely the preimages of the line components
	going through $\nabla(\Phi_{4})(p)$.
\end{proof}

\begin{lemma}\label{lem:bitangency_points}
	The points in the orbit $\ensuremath{\mathbf{O}}_{56}$ are ordinary singularities
	of multiplicity 3 in $\mathcal{K}$ and in $\mathcal{K}_2$.
	Moreover $\nabla(\Phi_{4})(\ensuremath{\mathbf{O}}_{56})=\ensuremath{\mathbf{O}}_{28}$.
\end{lemma}
\begin{proof}
	Since $\Phi_{21}$ does not vanish at $\ensuremath{\mathbf{O}}_{56}$
	(Proposition \ref{pro:orbits}), the singularities of
	$\mathcal{K}$ and $\mathcal{K}_2$ at $\ensuremath{\mathbf{O}}_{56}$ are equal.

	$\ensuremath{\mathbf{O}}_{56}$ is the orbit formed by the tangency points
	of the 28 bitangents to Klein's quartic. As was shown in the proof of Proposition \ref{pro:reducible_polars}, each (conic) component of
	$\mathcal{K}_2$ goes through 8 points of $\ensuremath{\mathbf{O}}_{56}$, and there
	are 21 such components, so there are $3=21\times 8 / 56$
	components through each point.

	To see that the triple points
	of $\mathcal{K}$ at each point of
	$\ensuremath{\mathbf{O}}_{56}$ are ordinary, we use the
	gradient map $\nabla(\Phi_{4})$.
	Since $\Phi_{6}$ does not vanish at $\ensuremath{\mathbf{O}}_{56}$,
	$\nabla(\Phi_{4})$ is biholomorphic in a neigbourhood of each
	point of $\ensuremath{\mathbf{O}}_{56}$. Thus the triple points of $\Phi_{63}$
	are analytically isomorphic to the triple points of $\Phi_{21}$, which
	--- being unions of lines --- are ordinary triple points,
	and they are located at $\ensuremath{\mathbf{O}}_{28}$.
\end{proof}

\begin{proposition}\label{preimage_28}
	Each point in $\ensuremath{\mathbf{O}}_{28}$ has $9$ distinct preimages by
	the morphism $\nabla(\Phi_{4})$. The whole preimage
	$\nabla(\Phi_{4})^{-1}(\ensuremath{\mathbf{O}}_{28})$ consists of
	$\ensuremath{\mathbf{O}}_{28}$, $\ensuremath{\mathbf{O}}_{56}$, and two orbits of size $84$.
\end{proposition}

\begin{proof}
	We will describe the preimage of $p=[1:1:1]\in \ensuremath{\mathbf{O}}_{28}$.
	It consists of the base locus of the pencil of cubics obtained
	as preimages of lines through $p$. The equations of the
	lines composing $\Phi_{21}$ are well-known, and the three
	going through $p$ are given by the vanishing of the forms
	\begin{align*}
	\Lambda_1&= x+(\zeta^4+\zeta^3)y-(\zeta^4+\zeta^3+1)z,\\
	\Lambda_2&= -(\zeta^4+\zeta^3+1)x+y+(\zeta^4+\zeta^3)z,\\
	\Lambda_3&= (\zeta^4+\zeta^3)x-(\zeta^4+\zeta^3+1)y+z,
	\end{align*}
	where we note that $\Lambda_1+\Lambda_2+\Lambda_3=0$. Thus we
	are looking for the base points of the pencil generated by
	the preimages of the $\Lambda_i$.

	The point $p$ is mapped to itself by $\nabla(\Phi_{4})$, so the linear
	components of the reducible polars $V(\nabla(\Phi_{4})^{-1}(\Lambda_i))$
	are exactly the lines $V(\Lambda_i)$.
	Since $\Phi_{6}$ does not vanish at $p$, the singularity of
	$\mathcal{K}$ at $p$ is isomorphic to that of $\mathcal{K}_1$,
	i.e. none of the conics goes through $p$.

	The explicit computation
	of $\nabla(\Phi_{4})^*(\Lambda_i)$ shows that $\nabla(\Phi_{4})$
	maps each line $V(\Lambda_i)$ onto itself, and provides the equations
	of the three conics mapping to the lines. The first of these is
	\begin{align*}
	\Gamma_1&=(\zeta^4+\zeta^3)x^2+(\zeta^5+\zeta^2)(\zeta^5+\zeta^2+1)
	xy+(\zeta^5+\zeta^2)y^2\\ & \quad +(-\zeta^5-\zeta^2+1)xz+
	(-\zeta^4-\zeta^3+1)yz+(\zeta^5+\zeta^2)(\zeta^4+\zeta^3-1)z^2,
	\end{align*}
	whereas the remaining two, $\Gamma_2$, $\Gamma_3$ are similar, with
	$x, y,z$ cyclically permuted.

	By Lemma \ref{lem:conincident}, every point in the intersection
	of $V(\Lambda_2)$ with $V(\Gamma_1)$ is a triple point of
	$\mathcal{K}$ mapping to $p$, and since it is different from $p$
	it must belong to $V(\Gamma_3)$. More generally

	\[ V(\Lambda_i)\cap V(\Gamma_j)=V(\Lambda_i) \cap V(\Gamma_k), \quad
	\{i,j,k\}=\{1,2,3\}, \]
	and each of these intersection points is a triple point of $\mathcal{K}$.
	If we prove that the intersection $V(\Lambda_2)\cap V(\Gamma_1)$
	consists of two distinct points, it will follow that
	$p$ has 9 distinct preimages as claimed, namely $p$ itself, two points of
	$\ensuremath{\mathbf{O}}_{56}$ (which by Lemma \ref{lem:bitangency_points}
	must be shared by the three conics $V(\Gamma_i)$) and two
	distinct points on each line $V(\Lambda_i)$.

	Plugging the value of $y$ from $\Lambda_2=0$ into $\Gamma_1=0$
	we can explicitly determine the two intersection points as
	the solution of a quadratic form in $x,z$ with nonvanishing
	discriminant.

	So the two intersection points are indeed distinct, which means
	that the conics $V(\Gamma_j)$, $V(\Gamma_k)$ intersect the line
	$V(\Lambda_i)$ transversely. Moreover, at these two points the conics
	$V(\Gamma_j)$, $V(\Gamma_k)$ cannot be tangent, because they
	intersect in two additional points of $\ensuremath{\mathbf{O}}_{56}$.
	Therefore, the three components $V(\Gamma_j)$, $V(\Gamma_k)$,
	$V(\Lambda_i)$ meet at two ordinary triple points of $\mathcal{K}$.

	Each linear component $\Lambda$ of $\Phi_{21}$
	vanishes at 4 points of $\ensuremath{\mathbf{O}}_{28}$, and we just
	showed that each of them has two preimages
	on $V(\Lambda)$ distinct from $p$ itself. This makes for a total of
	$4\cdot 2 \cdot 21=168$ points; by Proposition \ref{pro:orbits}
	these comprise two orbits of size $84$.
\end{proof}
\begin{proposition}
\label{pro:preimage21}
	Each point in $\ensuremath{\mathbf{O}}_{21}$ has $9$ distinct preimages
	by the morphism $\nabla(\Phi_{4})$. The whole preimage
	$\nabla(\Phi_{4})^{-1}(\ensuremath{\mathbf{O}}_{21})$ consists of $\ensuremath{\mathbf{O}}_{21}$
	and two orbits of size $84$.
\end{proposition}

\begin{proof}
    We will argue analogously to Proposition \ref{preimage_28},
    and describe the preimage of
		\[p=[1:\zeta^4+\zeta^3:-(\zeta^4+\zeta^3+1)]\in \ensuremath{\mathbf{O}}_{21}.\]
	The equations of the
	lines composing $\Phi_{21}$ are well-known, and the four
	going through $p$ are given by the vanishing of the forms
	\begin{align*}
	\Lambda_1&= x+(\zeta^5+\zeta)y+(\zeta^5+\zeta^4+\zeta^2+1)z, \\
	\Lambda_2&= x-(\zeta^5+\zeta^4+\zeta^3+\zeta+1)y+(\zeta^5+\zeta^3+\zeta^2+1)z,\\
	\Lambda_3&= x+(\zeta^5+1)y-(\zeta^3+\zeta^2+\zeta)z,\\
	\Lambda_4&= x+(\zeta^2+1)y+(\zeta^3+\zeta^2+\zeta+1)z.
	\end{align*}
	We are looking for the base points of the pencil generated by
	the preimages of the $\Lambda_i$.

	As with Proposition \ref{preimage_28},
	none of the conics goes through $p$.
	The explicit computation
	of $\nabla(\Phi_{4})^*(\Lambda_i)$ shows that $\nabla(\Phi_{4})$
	exchanges the lines $V(\Lambda_i)$ in sets of two,
	$\Lambda_1\leftrightarrow\Lambda_2$,
	$\Lambda_3\leftrightarrow\Lambda_4$, and provides the equations
	of the four conics mapping to the lines. The first of these is given by
	\begin{gather*}
	\Gamma_1= (\zeta^5+\zeta)x^2-(\zeta^4+\zeta^3-1)xy+(\zeta^5+\zeta^3)y^2+
	\\(-\zeta^3+\zeta^2-\zeta)xz+(\zeta^6-\zeta^4-\zeta)yz+(\zeta^5+\zeta^4)z^2	.
	\end{gather*}
	By Lemma \ref{lem:conincident}, every point in the intersection
	of $V(\Lambda_2)$ with $V(\Gamma_1)$ is a quadruple point of
	$\mathcal{K}$ mapping to $p$, and since it is different from $p$,
	it must belong to the other two conics $V(\Gamma_3)$ and
	$V(\Gamma_4)$. More generally
	\[ V(\Lambda_i)\cap V(\Gamma_j)=V(\Lambda_i) \cap V(\Gamma_k)=
	V(\Lambda_i)\cap V(\Gamma_{\ell}), \quad
	\{i,j,k,\ell\}=\{1,2,3,4\}, \]
	and each of these intersection points is a quadruple
	 point of $\mathcal{K}$.
	If we prove that the intersection $V(\Lambda_2)\cap V(\Gamma_1)$
	consists of two distinct points, it will follow that
	$p$ has $9$ distinct preimages as claimed, namely $p$ itself and two
	distinct points on each line $V(\Lambda_i)$.

	Plugging the value of $y$ from $\Lambda_2=0$ into $\Gamma_1=0$,
	we can explicitly determine the two intersection points as
	the solution of a quadratic form in $x,z$ with non-vanishing
	discriminant.
	So the two intersection points are indeed distinct, which means
	that the conics $V(\Gamma_j)$, $V(\Gamma_k)$ intersect the line
	$V(\Lambda_i)$ transversely. Moreover, at these two points the conics
	$V(\Gamma_j)$, $V(\Gamma_k)$ cannot be tangent, because they
	intersect in two additional points of $V(\Lambda_{\ell})$.
	Therefore, the four components $V(\Gamma_j)$, $V(\Gamma_k)$,
	$V(\Gamma_{\ell})$ and $V(\Lambda_i)$ meet at two ordinary
	quadruple points of $\mathcal{K}$.

	Each linear component $\Lambda$ of $\Phi_{21}$
	vanishes at 4 points of $\ensuremath{\mathbf{O}}_{21}$, and we just
	showed that each of them has two preimages
	on $V(\Lambda)$ distinct from $p$ itself. This makes for a total of
	$4\cdot 2 \cdot 21=168$ points; by Proposition \ref{pro:orbits}
	these comprise two orbits of size $84$.
\end{proof}

\begin{corollary}
	The curve $\mathcal{K}$
	is reduced and has only transversal intersection
	points as singularities, namely
	$189 = 9 \cdot 21$ quadruple points,
	$252 = 9 \cdot 28$ triple points,
	and $42 = 2 \cdot 21$ nodes.
\end{corollary}

\begin{corollary}
	The curve $\mathcal{K}_2$
	is reduced and has only transversal intersection
	points as singularities, namely
	$224 = 8 \cdot 21 + 56$ triple points,
	and $168 = 8 \cdot 21$ nodes.
\end{corollary}

\begin{remark}\label{rmk:cusps_discriminant}
	The discriminant $\Delta$ of the morphism $\nabla(\Phi_{4})$ has degree 18, as it is the image of a sextic under $\nabla(\Phi_{4})$.
	Since $\Delta$ is the dual of the Steinerian, which is a curve of degree 12 with exactly 21 nodes and 24 cusps as singularities,
	Plücker's second formula gives the number of cusps of $\Delta$, namely 42.
	A priori these 42 cusps could coalesce into a smaller number of more complicated singularities, but they must form a union of orbits; thus either $\Delta$ has exactly 42 ordinary cusps at the points of $\ensuremath{\mathbf{O}}_{42}$, or 21 singularities at $\ensuremath{\mathbf{O}}_{21}$.
	On the other hand, by Proposition \ref{pro:preimage21}, we know that the discriminant does not pass through the points of $\ensuremath{\mathbf{O}}_{21}$, so $\Delta$ has ordinary cusps at the points of $\ensuremath{\mathbf{O}}_{42}$.

	We shall check in \ref{rmk:cusps_K1} that the two cusps of $\Delta$ at the two singular points of each reducible polar have the same tangent line, namely the linear component through them.
	This gives additional information on the Steinerian of the Klein curve: its 42 inflectional branches meet in pairs to form its 21 nodes.
\end{remark}

\section{Harbourne indices}\label{sec:h-index}

The arrangement of reducible polars of Klein's quartic turns out to be relevant in the context of the Bounded Negativity Conjecture and $H$-indices.

\begin{conjecture}[BNC] \label{con:BNC}
	Let $X$ be a smooth complex projective surface. There exists a positive integer $b(X) \in \mathbb{Z}$ such that for all \emph{reduced} curves $C \subset X$ one has $C^{2} \geq -b(X)$.
\end{conjecture}

This conjecture is widely open, so much so that it is not even known whether the blowing-up of the complex projective plane along $r\geq 10$ points has bounded negativity.
However, in all \emph{known} cases of reduced curves in the blowing up of $\ensuremath{\mathbb{P}}^2$ at $r$ points, the self-intersection is bounded by $-4r$.
This prompted the introduction of \emph{H-constants} \cite{BdRHHLPSz} and \emph{Harbourne indices}.
\begin{definition}
	Let $C \subset \mathbb{P}^{2}_{\mathbb{C}}$ be a reduced curve of degree $d$, with ordinary singularities. The Harbourne index of $C$ is
	$$h(C) = \frac{d^{2} - \sum_{p\in \ensuremath{\operatorname {Sing}}(C)} \ensuremath{\operatorname {mult}}_{p}(C)^{2}}{|\ensuremath{\operatorname {Sing}}(C)|}.$$
\end{definition}
\begin{definition}
	The global linear Harbourne index of $\mathbb{P}^{2}$ is defined as
	$$H_{1}(\mathbb{P}^{2}) := \inf_{\mathcal{L}}h(\mathcal{L}),$$
	where the infimum is taken over all \emph{reduced} line arrangements $\mathcal{L} \subset \mathbb{P}^{2}$.
\end{definition}

With these definitions in hand, the main result of \cite[Theorem 3.13]{BdRHHLPSz} can be formulated as follows.
\begin{theorem}\label{thm:linear} One has
	$$H_{1}(\mathbb{P}^{2}) \ge -4.$$
\end{theorem}
It is natural to ask whether the above bound is sharp. The most negative known example is  \emph{Wiman's arrangement of lines} \cite{wiman}, the line arrangement of $45$ lines with $120$ triple points, $45$ quadruple points, and $36$ quintuple points. Easy computations reveal that for Wiman's arrangement $\mathcal{W}_1$ one has
$$h(\mathcal{W}_1) = -\frac{225}{67} \approx -3.358.$$
It is worth pointing out that Klein's arrangement of lines $\mathcal{K}_1$ delivers the second most negative known value of a linear Harbourne index, equal to $-3$.

Now we would like to focus on conic-line arrangements in the projective plane having only ordinary singularities. If $\mathcal{CL}$ is a conic-line arrangement consisting of $l$ lines and $k$ conics, the following combinatorial equality holds:
$$ {l \choose 2} + 2kl + 4 {k \choose 2} = \sum_{r\geq 2} {r \choose 2} t_{r},$$
where by $t_{r}$ we denote the number of $r$-fold intersection points, i.e., points where exactly $r$ curves from the arrangement meet, and we are using a convention that ${0 \choose 2} = {1 \choose 2} = 0$.
The above equality provides
$$(2k+l)^{2} -\sum_{r\geq 2}r^{2}t_{r} = 4k+l - \sum_{r\geq 2}rt_{r},$$
and shows that the Harbourne index of a conic-line arrangement $\mathcal{CL}$ with ordinary singularities satisfies
$$h(\mathcal{CL}) = \frac{4k+l - \sum_{r\geq2} rt_{r}}{| {\rm Sing}(\mathcal{CL})|}.$$

Not much is known about $H$-indices of conic-line arrangements. In a very recent paper \cite{Pokora2017}, the first author proved the following result.
\begin{theorem}(\cite[Theorem~2.1]{Pokora2017})\label{thm:pokora17}
	Let $\mathcal{CL} = \{L_{1}, ..., L_{l}, C_{1}, ..., C_{k}\}$ be an arrangement of $l$ lines and $k$ conics such that $t_{r} = 0$ for $r > \frac{2(l+2k)}{3}$. Then one has
	$$t_{2} + \frac{3}{4}t_{3} + (4k+2l-4)k \geq l + \sum_{r\geq 5}\bigg(\frac{r^{2}}{4}-r\bigg)t_{r}.$$
\end{theorem}
Theorem \ref{thm:pokora17} does not lead to any effective lower bound on Harbourne indices of conic-line arrangements. This leads to the following question.
\begin{question}
	How negative Harbourne indices for conic-line arrangements can be?
\end{question}
The Klein arrangement of conics and lines $\mathcal{K}$ has $21$ conics, $21$ lines, and $t_{2} = 42$, $t_{3} = 252$, and $t_{4} = 189.$
Therefore
$$h(\mathcal{K}) =-\frac{71}{23}\approx -3.0865,$$
which is the second smallest known value of Harbourne indices for arrangements of curves in the complex projective plane having transversal intersection points (here we exclude Cremona transform-arrangements constructed with use of line arrangements, for details please consult \cite{PTG}).
Note that the arrangement of conics $\mathcal{K}_2$, on its own,
has Harbourne index equal to $-33/14 \approx -2.3571$.

\section{Freeness of arrangements}\label{sec:freeness}
A fundamental object associated to an arrangement of plane curves $\mathcal{F}$ is the module $D(\mathcal{F})\subset {\rm Der}_{\mathbb{C}}(S)$ of derivations tangent to the arrangement,
$$D(\mathcal{F}) = \{ \theta : \theta(\Phi) \in \langle \Phi \rangle \text{ for all } \Phi \text{ such that } V(\Phi) \in \mathcal{F}\}.$$
In his famous paper \cite{Terao}, Terao showed the link between Poincar\'{e} polynomials for line arrangements $\mathcal{L}$ and the freeness of the module $D(\mathcal{L})$, and it is notoriously difficult to predict whether the freeness of $D(\mathcal{L})$ is combinatorial in nature, which leads to Terao's conjecture.

Next we focus on the freeness of conic-line arrangements, and in particular of the $G$-invariant arrangement $\mathcal{K}$, mostly in the spirit of \cite{SchT}.
One might expect that $\mathcal{K}$ be free, given that
Klein's arrangement of lines $\mathcal{K}_1$ is free as a reflection arrangement \cite[Chapter~6]{OT92}, $\mathcal{K}$ is its pull-back under the gradient map, and the whole construction is invariant under $G$. Note also that all singularities of $\mathcal{K}$ are quasihomogeneous.
Indeed, nodes are always quasihomogeneous, whereas for all triple and quadruple points of $\mathcal{K}$ we showed in Section \ref{sec:construction} that they are biholomorphic to the triple and quadruple points of Klein's arrangement of lines, which are obviously homogeneous.
However, it will turn out that $\mathcal{K}$ \textbf{is not free}.

Following \cite[Section 1.4]{SchT}, we know that if $\mathcal{CL}$ is a reduced conic-line arrangement, then
$$D(\mathcal{CL}) \simeq E \oplus D_{0}(\mathcal{CL}),$$
where $E$ is the Euler derivation and $D_{0}(\mathcal{CL})$ corresponds to the module of syzygies on the Jacobian ideal of the defining polynomial of $\mathcal{CL}$; the properties of $D_{0}(\mathcal{CL})$ determine freeness of the arrangement $\mathcal{CL}$.

We now prove that $\mathcal{K}$ is not free following an idea kindly suggested to us by the referee. Denote by $r$ the minimal degree such that the homogeneous component $D_{0}(\mathcal{K})_{r}$ is non-zero.
By \cite[Collorary 1.2]{DimcaTjurina}, $\mathcal{K}$ is a free arrangement if and only if $\tau(\mathcal{K})= 62^{2} - r(62-r)$, where $\tau(\mathcal{K})$ is the total Tjurina number of $\mathcal{K}$. As we observed a few paragraphs earlier all singularities of $\mathcal{K}$ are quasihomogeneous, so the total Tjurina number is equal to the total Milnor number, i.e.,
$$\tau(\mathcal{K}) =  \mu(\mathcal{K}) = \sum_{p \in {\rm Sing}(\mathcal{K})} ({\rm mult}_{p} - 1)^{2} = 1 \cdot 42 + 4 \cdot 252 + 9 \cdot 189 = 2751.$$
This leads to the equation
$$r^2 - 62r + 1093=0$$
which does not have real roots, so $\mathcal{K}$ cannot be free.

In a recent paper \cite{Dimca}, Dimca and Sticlaru introduced a larger class of arrangements, called \textit{nearly free curves}, which can be defined as follows.
\begin{definition}
\label{def:near}
	Let $\mathcal{F}$ be an arrangement of plane curves, then $\mathcal{F}$ is called \emph{nearly free} if the Jacobian module $N := I_{\mathcal{F}} / J_{\mathcal{F}}$
	does not vanish and its homogeneous pieces $N_k$ satisfy $\dim N_{k} \leq 1$ for any $k$.
\end{definition}
Following the same approach as for freeness suggested by the referee, and continuing to denote $r$ the minimal degree such that the homogeneous component $D_{0}(\mathcal{K})_{k}$ is non-zero, $\mathcal{K}$ is nearly free if and only if
	$$\tau(\mathcal{K}) = 62^{2} - r(62-r) - 1,$$
by \cite[Theorem 1.3]{DimcaTjurina}.
Since the equation again has no real roots, $\mathcal{K}$ cannot be nearly free.

\begin{remark}
\label{Dimca1}
Going further in the analysis, it is natural to ask how far $\mathcal{K}$ is from being free; to this end Dimca introduced the so called \textit{defect of freeness}, defined as
	$$\nu (\mathcal{CL}) = {\rm max}_{k} \{ {\rm dim} N_{k}\}.$$
Checking whether a certain arrangement $\mathcal{CL}$ is free can be reduced to one of the following to conditions:
	\begin{itemize}
		\item ${\rm pdim}(S/J_{\mathcal{CL}}) = 2$, where $J_{\mathcal{CL}}$ denotes
		the Jacobian ideal and $S/J_{\mathcal{CL}}$ the Milnor algebra of
		the defining polynomial of $\mathcal{CL}$;
		\item the Jacobian module $N := I_{\mathcal{CL}} / J_{\mathcal{CL}}$ vanishes, where $I_{\mathcal{CL}}$ denotes the the saturation of $J_{\mathcal{CL}}$ with respect to the irrelevant ideal $(x,y,z)$.
	\end{itemize}
	In the case of our Klein invariant conic-line arrangement $\mathcal{K}$, using a Singular script, we can compute the minimal resolution of the Milnor algebra $S/J_{\mathcal{K}}$, which has the following form:
	\begin{equation}\label{eq:nonfree}
	0 \rightarrow S(-121) \rightarrow S(-115) \oplus S(-99) \oplus S(-93)\rightarrow S^{3}(-62) \rightarrow S,
	\end{equation}
	Theorem 2.8 ii) in \cite{Dimca} describes the resolutions of Milnor algebras for nearly free curves; denoting by $(d_1,d_2,d_3)$ the degrees of the relations, a necessary condition is $d_1+d_2=\deg \mathcal{CL}$.

	This way we obtain an alternative computational check that $\mathcal{K}$ is neither free (because ${\rm pdim}(S/J_{\mathcal{K}}) = 3$) nor almost free (because $(d_{1},d_{2},d_{3}) = (31,37,53)$) thanks to \eqref{eq:nonfree}.
	Additionally, we can compute the defect of freeness.
	The resolution of the Milnor algebra (\ref{eq:nonfree}) gives $r = 93 - 62 = 31$. Then by a recent result by Dimca \cite[Theorem 1.2 (2)]{DimcaV} we can calculate that
$$\nu(\mathcal{K}) = \bigg\lceil \frac{3}{4}(d-1)^{2} \bigg\rceil - \tau(\mathcal{K}) =  \bigg\lceil \frac{3}{4}\cdot62^{2} \bigg\rceil - 2751 = 132,$$
so $\mathcal{K}$ is far away from being free.
\end{remark}
\begin{remark}
\label{Dimca2}
If $E$ is the rank two vector bundle associated to the graded  $S$-module $D_{0}(\mathcal{K})$, then using a recent result by Abe and Dimca \cite[Theorem 1.1 (2)]{AbeDimca} we can show that the generic splitting type of $E$ (after we restrict $E$ to a generic line) is $(31,31)$.
\end{remark}
\section{Chern slopes}\label{sec:chern}
Let $X$ be a smooth projective surface and $D$ be a simple normal crossing divisor. A log-surface is a smooth surface $U :=X \setminus D$. We refer to $U$ as the pair $(X,D)$. The logarithmic Chern numbers of $(X,D)$ are defined as follows:
$$\overline{c}_{1}^{2}(X,D) = (K_{X}+D)^{2} \text{ and } \overline{c}_{2}(X,D) = e(X) - e(D).$$
We consider log-surfaces defined by plane curve arrangements, i.e., pairs $(\mathbb{P}^{2},\mathcal{F})$, where $\mathcal{F}$ is a arrangement of $k\geq 3$ smooth curves, each having degree $d \geq 1$, such that all singularities of $\mathcal{F}$ are ordinary and there is no point where all curves meet. Consider the blowing up $\pi : X \rightarrow \mathbb{P}^{2}$ along singular points of $\mathcal{F}$ having multiplicities $\geq 3$, and denote by $\mathcal{F}'$ the reduced total transform of $\mathcal{F}$ under $\pi$. Then $(X,\mathcal{F}')$ is a log-surface. We can compute the logarithmic Chern numbers of $(X,\mathcal{F}')$, namely:
$$\overline{c}_{1}^{2}(X,\mathcal{F}') = 9 + (d^{2}-6d)k + \sum_{r \geq 2}(3r-4) t_{r},$$
$$\overline{c}_{2}(X,\mathcal{F}') = 3 +(d^{2}-3d)k + \sum_{r\geq 2}(r-1)t_{r}.$$
In \cite[Theorem 3.1]{Pokora}, the second author proved the following result.
\begin{theorem}
	Let $\mathcal{F} \subset \mathbb{P}^{2}$ be a curve arrangement defined as above such that each irreducible component has degree $d \geq 2$, then for the associated log-surface $(X,\mathcal{F}')$ one has
	$$\overline{c}_{1}^{2}(X,\mathcal{F}') < \frac{8}{3} \overline{c}_{2}(X,\mathcal{F}').$$
\end{theorem}
However, it is not known whether the above bound is sharp. Unfortunately, we could not find in the literature interesting examples which could potentially lead to logarithmic Chern slopes $E(X,\mathcal{F}') :=\overline{c}_{1}^{2}(X,\mathcal{F}') / \overline{c}_{2}(X,\mathcal{F}')$ close to $8/3$. Focusing on the case of curve arrangements $\mathcal{F} \subset \mathbb{P}^{2}$ having degree $d=2$, Klein's arrangement $\mathcal{K}_2$ consists of $21$ conics with $224$ triple and $168$ double points as the intersections. Therefore
$$E(X, \mathcal{K}_2') \approx 2.25,$$
and this logarithmic Chern slope, to the best of our knowledge, is the largest known value in the class of curve arrangements of degree $2$.

\section{On the containment problem}\label{sec:containment}
In this section, we would like to briefly discuss the containment problem for symbolic and ordinary powers of homogeneous ideals of finite sets of points in the projective plane. Here we consider only a very special case of that problem, for a general introduction we refer to the survey \cite{SzSz}. Let $\mathcal{P} = \{p_{1}, ...,p_{s}\} \subset \mathbb{P}^{2}$ be a finite set of mutually distinct points and we define the following radical ideal
$$I=I(\mathcal{P}) = I(p_{1}) \cap ... \cap I(p_{s}).$$
Then the $m$-th symbolic power of $I$ can be defined as
$$I^{(m)} = I^{m}(p_{1}) \cap ... \cap I^{m}(p_{s}).$$
Geometrically speaking, the $m$-th symbolic power is the set of homogeneous forms vanishing along $p_{i}$'s with multiplicity $\geq m$.
In recent years, and spurred by the work of Ein--Lazarsfeld--Smith \cite{ELS02}, Hochster and Huneke \cite{HH02} and Harbourne and Huneke \cite{HH}, the following problem has attracted considerable attention.
\begin{problem}
Determine the pairs $(m,r) \in \mathbb{Z}_{> 0}^{2}$ such that the containment
$$I^{(m)} \subset I^{r}$$
holds.
\end{problem}
\begin{theorem}[Ein--Lazarsfeld--Smith]\label{thm:els}
Let $I = I(p_{1}) \cap ... \cap I(p_{s})$ be a saturated ideal of a reduced finite set of points in $\mathbb{P}^{2}$, then the containment
$I^{(2r)} \subset I^{r}$ holds for all $r\ge 0$.
\end{theorem}

It is worth noting that the groundbreaking and elegant Theorem \ref{thm:els} holds regardless of the position of points.
It is natural to ask whether the above result is sharp. In this direction, Huneke proposed the following problem.
\begin{question}[Huneke]\label{qu:huneke}
Let $I = I(p_{1}) \cap ... \cap I(p_{s})$ be a saturated ideal of a reduced finite set of points in $\mathbb{P}^{2}$. Does the containment
$I^{(3)} \subset I^{2}$
hold?
\end{question}
Most sets of points do satisfy Huneke's containment. However, in \cite{DSTG}, a first counterexample was found: the radical ideal of $12$ triple points of the dual-Hesse arrangement of $9$ lines. By \cite{BdRHHLPSz}, the radical ideal of the singular points of Klein's arrangement of lines $\mathcal{K}_1$ is also a counterexample to Huneke's question.

S. Akesseh proved in \cite{Ake17} that the pullback of the ideal $I$ of a set of points in $\ensuremath{\mathbb{P}}^2$ by a morphism $\ensuremath{\mathbb{P}}^2 \rightarrow \ensuremath{\mathbb{P}}^2$, like our gradient map $\nabla(\Phi_{4})$, satisfies the same containment relations as $I$.
By Propositions \ref{preimage_28} and \ref{pro:preimage21}, the pullback by $\nabla(\Phi_{4})$ of the radical ideal of the singular points of $\mathcal{K}_1$ is the radical ideal $I_3$ of the set of points of multiplicity at least 3 in the arrangement $\mathcal{K}$ of lines and conics, and this is a counterexample to Huneke's question. In fact, $\Phi_{63} \in I_3^{(3)} \setminus I_3^{2}$.

It is more interesting to notice that the radical ideal $I_2$ of the whole set of singular points of $\mathcal{K}$ is a new counterexample to Question \ref{qu:huneke}.
Indeed, it is clear that $\Phi_{6}\Phi_{63} \in I_2^{(3)}$, and we checked using Singular that $\Phi_{6}\Phi_{63}\notin I_2^2$.

In the next section we obtain an additional counterexample.
\section{Iterated preimages}\label{sec:iteration}

Several of the interesting properties we have shown for the arrangement $\mathcal{K}$ of reducible polars are consequences of its being the pullback of Klein's line arrangement $\mathcal{K}_1$.
This suggests iterating the process, so we consider
the preimage $\Phi_{189}=\nabla(\Phi_4)^*(\Phi_{63})$.
Since $\Phi_{63}=\nabla(\Phi_4)^*(\Phi_{21})=\Phi_{42}\Phi_{21}$, there is a splitting $\Phi_{189}=\Phi_{126}\Phi_{42}\Phi_{21}$, where $\Phi_{126}=\nabla(\Phi_4)^*(\Phi_{42})$ is an invariant arrangement of $21$ sextics. Further iterations lead to invariant arrangements of higher degree:
\[
(\nabla(\Phi_4)^k)^*(\Phi_{21})=\Phi_{14\cdot 3^{k}}\cdots
\Phi_{42}\Phi_{21}.
\]
Our first observation is that each arrangement $V(\Phi_{14\cdot 3^{k}})$ contains the points of $\ensuremath{\mathbf{O}}_{42}$. This follows from the following lemma, and allows us to give a new type of counterexamples to Huneke's question \ref{qu:huneke}.

\begin{lemma}
$\nabla(\Phi_{4})(\ensuremath{\mathbf{O}}_{42})=\ensuremath{\mathbf{O}}_{42}$.
\end{lemma}
\begin{proof}
We saw in Proposition \ref{pro:preimage21} that $\nabla(\Phi_{4})(\ensuremath{\mathbf{O}}_{42})$ is disjoint from $\ensuremath{\mathbf{O}}_{21}$; since the image of $\ensuremath{\mathbf{O}}_{42}$ has to be an orbit of size dividing $42$, it follows that $\nabla(\Phi_{4})(\ensuremath{\mathbf{O}}_{42})=\ensuremath{\mathbf{O}}_{42}$.
\end{proof}

\begin{proposition}
	Let $T=\nabla(\Phi_{4})^{-2}(\ensuremath{\operatorname {Sing}}(\mathcal{K}_1))$.
	The ideal $I_X$ of every reduced set of points $X$ with
	$$ T \subseteq X \subseteq T \cup \ensuremath{\mathbf{O}}_{42}.$$
	satisfies $\Phi_{189}\in I_X^{(3)}\setminus I_X^2$, so  $I_X^{(3)}\not\subseteq I_X^2$.
\end{proposition}
\begin{proof}
	Since $\Phi_{126}=(\nabla(\Phi_{4})^2)^*(\Phi_{21})$, it is clear that $\Phi_{126}$ has multiplicity 3 at every point of $T$.
	By the previous lemma, $V(\Phi_{21})$, $V(\Phi_{42})$ and $V(\Phi_{126})$ pass through all points of $\ensuremath{\mathbf{O}}_{42}$, so $\Phi_{189}=\Phi_{126}\cdot\Phi_{42}\cdot\Phi_{21}$ vanishes at order 3 along $\ensuremath{\mathbf{O}}_{42}$, and $\Phi_{189}\in I_X^{(3)}$ for every $X\subseteq T \cup \ensuremath{\mathbf{O}}_{42}.$

	On the other hand, by the result of Akesseh \cite{Ake17} mentioned above, $\Phi_{189}\notin I_T^2$, and obviously $I_X^2\subseteq I_T^2$.
\end{proof}

It is worth pointing out that in this example, the same polynomial gives failure of containment for the whole set $T\cup \ensuremath{\mathbf{O}}_{42}$ of its points of multiplicity $\ge 3$, but also for sets strictly included in it, a phenomenon not previously described. Note however that nested sets of points whose ideals exhibit failure of containment was already observed in the case of Wiman's arrangement of lines  \cite[Example 3.3.5]{Har16}, where the containment $I^{(3)} \subset I^{2}$ still does not hold if we remove exactly one point of multiplicity $3$ from the set of all $201$ singular points. In the next section we present some details about Wiman's arrangement of lines and its further applications in the context of our work.

\medskip

These iterated pullbacks are interesting examples also for the question of bounded negativity. However, their singularities are not ordinary any longer. For instance, $V(\Phi_{63})$ is tangent to $V(\Phi_{42})$ at the points of $\ensuremath{\mathbf{O}}_{42}$.

\begin{lemma}\label{lem:tangency}
	For every $k\ge 1$, $\Phi_{14\cdot 3^{k}}$ is smooth along $\ensuremath{\mathbf{O}}_{42}$.
	Moreover, $\Phi_{14\cdot 3^{k}}$ is tangent to $\mathcal{K}_2$ at every $p\in \ensuremath{\mathbf{O}}_{42}$.
\end{lemma}
\begin{proof}
	Choose a point $p \in \ensuremath{\mathbf{O}}_{42}$, and denote $p'=\nabla(\Phi_{4})(p)\in \ensuremath{\mathbf{O}}_{42}$.
	Let $L$, respectively $C$, be the component of $\mathcal{K}_1$, respectively $\mathcal{K}_2$, through $p$.
	Choose an analytic neighborhood $U$ at $p$ and local coordinates $(x,y)$ in $U$ such that $L:\{x=0\}$, $C:\{y=0\}$.
	Let $\alpha\in\ensuremath{\operatorname {Aut}}(\Phi_{4})$ be such that $\alpha(p)=p'$.
	Then $\alpha(U)$ is an open neighborhood of $p'$, and $\tilde{x}=x \circ \alpha^{-1}$, $\tilde{y}=y \circ \alpha^{-1}$ are local coordinates, such that
	$\tilde x=0$, resp. $\tilde y=0$, is a local equation of the component of $\mathcal{K}_1$, resp. $\mathcal{K}_2$, through $p'$.
	There is no harm in identifying $U$ with $\alpha(U)$ and $(\tilde{x},\tilde{y})$ with $(x,y)$, and we shall do so in the sequel (in fact, there are points $p \in \ensuremath{\mathbf{O}}_{42}$ fixed by $\nabla(\Phi_{4})$, see the Singular computations in the appendix, and choosing such a $p$ even eliminates the abuse of notation.)

	Shrinking the neighborhood $U$ if necessary, we may assume that $\nabla(\Phi_{4})$ is given locally as $\nabla(\Phi_{4})(x,y)=(xy,g)$, where $g=g(x,y)\in \ensuremath{\mathcal{O}}_{\ensuremath{\mathbb{P}}^2,p}$ is a power series in $x,y$ converging in $U$ to a local equation of $\nabla(\Phi_{4})^*(C)$, and hence of $\Phi_{126}$.
	General polars of the Klein quartic passing through $p$ are smooth, hence the series $g$ is of order 1 (and $\Phi_{126}$ is smooth at $p$).
	On the other hand, since the discriminant $\Delta$ of $\nabla(\Phi_{4})$ is singular at $p'$ by Remark \ref{rmk:cusps_discriminant}, the local intersection multiplicity of $g(x,y)=0$ with the Jacobian of $\nabla(\Phi_{4})$,
	\[J\nabla(\Phi_{4})_p=\begin{vmatrix}
	y & x \\ \frac{\partial g}{\partial x}(x,y) & \frac{\partial g}{\partial x}(x,y)
	\end{vmatrix}\]
	is at least two. Hence the initial form of $g$ is either $ax$ or $ay$ for some $a\in \ensuremath{\mathbb{C}}$.
	But $J\nabla(\Phi_{4})_p$ is a local equation at $p$ of the sextic $V(\Phi_{6})$, and we know that $V(\Phi_{6})$ is \emph{not} tangent to $\mathcal{K}_1$ along $\ensuremath{\mathbf{O}}_{42}$, so the initial form of $g$ is $ay$.
	Therefore $\nabla(\Phi_{4})^*(C)$ is tangent to $C$ at $p$, i.e., $V(\Phi_{126})$ is tangent to $\mathcal{K}_2$.
	Moreover, the preimage of every curve smooth at $p$ and tangent to $C$ has initial form $ay$ and hence will be smooth and tangent to $C$ as well, proving by recurrence that every $\Phi_{14\cdot 3^{k}}$ is tangent to $C$, hence to $\mathcal{K}_2$, at $p$.
\end{proof}

\begin{remark}\label{rmk:cusps_K1}
	It follows from the computation of the Jacobian in the last paragraph of the proof that the singular polar $L+C:\{xy=0\}=\nabla(\Phi_{4})^*(L)$ has local intersection multiplicity 3 with the Jacobian, and therefore $L$ has intersection multiplicity 3 with the discriminant. In other words, $L$ is the tangent line to the cusp of $\Delta$ at $p$, as announced in \ref{rmk:cusps_discriminant}.
\end{remark}

\begin{proposition}
	For every $k\ge 1$, $(\nabla(\Phi_{4})^k)^*(\mathcal{K}_1)$ is reduced.
\end{proposition}
\begin{proof}
	Assume $(\nabla(\Phi_{4})^k)^*(\mathcal{K}_1)$ is non-reduced.
	Then $(\nabla(\Phi_{4})^{k-1})^*(\mathcal{K}_1)$ must contain the (irreducible) discriminant curve $\Delta$.
	But this has a cusp at each point of $\ensuremath{\mathbf{O}}_{42}$ (Remark \ref{rmk:cusps_discriminant}), whereas by the preceding lemma every component of $\mathcal{K}_{k-1}$ is smooth along $\ensuremath{\mathbf{O}}_{42}$, a contradiction.
\end{proof}

With a little more effort it is possible to compute the intersection multiplicities between branches of the $k$-th pullback of the Klein arrangement $\mathcal{K}_1$, and hence determine exactly the singularities at $\ensuremath{\mathbf{O}}_{21}$, $\ensuremath{\mathbf{O}}_{28}$, $\ensuremath{\mathbf{O}}_{42}$ and their preimages.
Thus we can give an upper bound $H_k$ for the Harbourne index of $(\nabla(\Phi_{4})^k)^*(\mathcal{K}_1)$, taking into account multiplicities at infinitely near points. It turns out that $H_k$ is a decreasing sequence converging to $-1283/410 \approx -3.1293$.
We skip the cumbersome details.

\section{Remarks on Wiman's arrangement}\label{sec:wiman}

The largest possible primitive subgroup of $\ensuremath{\operatorname {PGL}}_3$, isomorphic to $A_6$, was found by H.~Valentiner in 1889 \cite{Val89}.
Its ring of invariants was computed by A.~Wiman in 1896 \cite{wimanA}, see also \cite{wiman}; the invariant curve of the smallest degree is a unique sextic which we call \emph{Wiman's smooth sextic}\footnote{Not to be confused with the nodal sextic usually associated with Wiman, whose group of automorphisms is isomorphic to $S_5$.}, and it also comes with an extremely special arrangement of lines, $\mathcal{W}_1$, which consists of 45 lines with exactly 36 quintuple points, 45 quadruple points, and 120 triple points as singularities ---first described by F.~Gerbaldi in 1882 \cite{gerbaldi}.
It has been studied in recent years with regard to Bounded Negativity and the containment problem, yielding for instance the most negative $H$-index known for a curve with ordinary singularities.
It turns out that Wiman's smooth sextic also exhibits a remarkable configuration $\mathcal{W}$ of 45 reducible polars, as we prove below.
Not surprisingly, the theory of polarity and the Steinerian curve of sextics is much less developed and understood than that for quartic curves.
Thus, providing a complete theoretical description of the singularities of the reducible polars of Wiman's smooth sextic, as we did for Klein's quartic, is far beyond the scope of this work.
Instead, we now report on the relevant properties found computationally using Singular (scripts in the Appendix).

Wiman's smooth sextic can be defined by the equation
\[\Psi_6=10x^3y^3+9z(x^5+y^5)-45x^2y^2z^2-135z^4xy+27z^6\]

and its group of automorphisms is $\ensuremath{\operatorname {Aut}}(\Psi_6))\cong A_6$ \cite{wiman}.
As for the case of Klein's curve, there is a group $G$ of $3\times 3$ matrices representing all automorphisms of $V(\Psi_6)$, with the particularity that in this case $G$ is a triple cover of $A_6$, as it contains the matrices $\omega^i I_3$, where $\omega$ is a third root of unity.
If coordinates are chosen adequately, also in this case the group of matrices $G$ is closed under transposition  (see \cite{SymmetricBU})  so we shall work with those coordinates in the appendix, although the equation $\overline{\Psi}$ of the sextic is then not as nice as $\Psi_6$, the one given by Wiman.

The ring of invariants of $G$ is generated by $\Psi_6$, its Hessian $\Psi_{12}$, the bordered Hessian $\Psi_{30}$ of $\Psi_6$ and $\Psi_{12}$, and the Jacobian $\Psi_{45}$ of the three first invariants, which is the equation of the arrangement of 45 lines.

Also in this case the group $G$ is generated by involutions $\alpha$, each of which is a harmonic homology with center at one of the 45 quadruple points of the arrangement $\mathcal{W}_1$ and axis at one of the 45 lines.

\begin{lemma}
	If $\alpha\in \ensuremath{\operatorname {PGL}}_3(\ensuremath{\mathbb{C}})$ is a general homology that fixes an irreducible curve $C\subset \ensuremath{\mathbb{P}}^2$ of degree $d>1$, and $C$ intersects the axis of the homology transversely, then the polar of $C$ with respect to the center of $\alpha$ is reducible, and contains the axis of $\alpha$.
\end{lemma}
\begin{proof}
	Let $p$ be the center and $L$ be the axis of $\alpha$. Since $\alpha$ fixes $C$ and $L$, it also fixes the tangent lines to the $d$ points of $C\cap L$. But all lines fixed by $\alpha$ other than $L$ pass through the center $p$. Therefore the $d$ points of $C\cap L$ belong to the polar curve $\partial_p(C)$, which has degree $d-1$. Hence $L$ is a component of $\partial_p(C)$.
\end{proof}

\begin{corollary}
	The polar of $\Psi_6$ with respect to each of the 45 quadruple points of $\mathcal{W}_1$ decomposes as $L+Q$ where $L$ is one of the 45 lines in $\mathcal{W}_1$ and $Q$ is a quartic.
\end{corollary}

A Singular computation shows that the quartic component of each of the 45 reducible polars is smooth, intersecting the linear component at 4 distinct points, disjoint from the set of singularities of $\mathcal{W}_1$.
In contrast with the behavior of the reducible polars of Klein's quartic, here the different quartic curves are tangent at some points, namely each of them passes through 8 of the 72 inflection points, and the five quartics going through the same inflection point are tangent there (the gradient map $\nabla(\Psi_{6})$ is ramified at these points, which map to the quintuple points of $\mathcal{W}_1$).
Therefore the number of proper quintuple points on $\mathcal{W}$ is $5^2\cdot 36-72$, and there are $72$ additional infinitely near quintuple points.
Taking into account all multiple points, proper and infinitely near, the Harbourne index of $\mathcal{W}$ is smaller than that of $\mathcal{W}_1$, and in fact we have $h(\mathcal{W})=-1173/347\approx-3.38$.

We want to remark as well that each set of 4 nodes on one of the 45 reducible polars belongs to the (smooth) Hessian $V(\Psi_{12})$, and they have the same image under the Steiner map to the Steinerian, namely the point $p$ whose polar is singular on them. In particular, the Steinerian (which in this case is a curve of degree 48) has 45 points of multiplicity 4 at the quadruple points of $\mathcal{W}_1$.

\section*{Acknowledgments}
We want to thank I.~Dolgachev for sharing with us facts and references on the Steinerian curve, which were essential for completing this work. We would also like to thank an anonymous referee for crucial comments that allowed to improve Section 4 devoted to the freeness of arrangements, and for suggesting Remarks \ref{Dimca1} and \ref{Dimca2}.

This work was begun during the first author's visit at Universitat Aut\`{o}\-no\-ma de Barcelona, under the financial support of the Spanish MINECO grant MTM2016-75980-P, which also supports the second author.
During the project Piotr Pokora was supported by the programme for young researchers at Institute of Mathematics Polish Academy of Sciences.
In the first phase of the project Piotr Pokora was a member of Institute of Algebraische Geometrie at Leibniz Universit\"at Hannover.
\section*{Appendix}
Here we present our Singular script which allows to verify some of the above claims.

{\scriptsize
\verbatiminput{Singular_Klein_paper.txt}}

{\footnotesize
	\bibliographystyle{plainurl}

redpajama_set_name: RedPajamaArXiv


The Estadio Tecnológico is a stadium located in city of Monterrey, Mexico. It is the property of the Monterrey Institute of Technology and Higher Education. The Monterrey Football Club play their home games there, but only rent the space and are not affiliated with the Institute. The stadium was opened in July 17, 1950 by Mexican president Miguel Alemán Valdés. Originally, it seated 20,000 spectators, but in 1985 it was expanded to 38,000 seats. In the early 90's executive boxes were added which reduced the capacity to its current figure. The Estadio Tecnológico was a venue for the 1986 Mexico FIFA World Cup, the highest competition in worldwide professional soccer. On February 12, 2006, the Irish rock band U2 opened the 4th leg of its Vertigo Tour at the Estadio Tecnológico, to a crowd of 42,000.


I'd like to share my latest faerybeads dragons and mouse. I love these dragons even more than I imagined and I'm so surprised by the mouse. He was going on a different combo but to my delight he can hold his own with dragons. So for me this bracelet represents the balance of sweetness and fierceness, vulnerability and self-protection. Best wishes to you this weekend!
Re: Two dragons and a...mouse?
Plumbun, Elephants are supposed to be scared of mice, so maybe dragons are too !
Looks great, the feet of the mouse are adorable.


text:
Viking expands Twin Otter Series 400 international sales network to include Korea, Baltic States
Viking Air Limited of Victoria, British Columbia, has assigned Hanbaek Aerospace of Seoul, South Korea, as the Twin Otter Series 400 sales representative for the Republic of Korea (ROK). Hanbaek will be responsible for all new commercial and military sales of the Series 400 Twin Otter aircraft in-country. The UK-based Liftec Limited will represent the Twin Otter Series 400 in Eastern Europe, CIS (excluding Russia), and the Baltic States. Liftec will be responsible for all new commercial and military sales of the Series 400 Twin Otter aircraft throughout its region. Read more about Hanbaek's assignment. Read more about Liftec's assignment.
From CTV: An astronaught's-eye view of Earth, captured on video
CTV News reports on AIAC Pacific member UrtheCast Corp. and its first three astronaught's-eye view video clips, featuring views of Boston, London, and Barcelona from the ISS. Read article.
KF Aeropspace announces new domestic and international cargo services
KF Aerospace, based in Kelowna, B.C., highlights new route structures and a number of key strategic partners as it re-enters the domestic and international air cargo services. Read More Français.
Seaplane trainer: Pacific Sky Aviation and Viking Air break ground on world's first Twin Otter full flight simulator
Pacific Sky Aviation Inc. and Viking Air Ltd. broke ground May 26 at the site of the world's first full flight simulator featuring a seaplane configuration for the legendary Twin Otter aircraft. Pictured from left: Mauro Pezzetta, Michael Coughlin, Sherry Brydson, Rob McDonald, David Curtis, and Doug Stroud. Read More
Shoot video of your BBQ or golf game from space
Vancouver's UrtheCast promises a unique summer photo-op — a view from space of your backyard party or golf tournament. Read more.
Minister Fast welcomes Canadian aerospace success story
Cascade Aerospace contract with the Mexican Air Force creates jobs and opportunities for Canadians The Honourable Ed Fast, Minister of International Trade, today congratulated Cascade Aerospace Inc. on the delivery of an upgraded C-130K Hercules aircraft to Mexico's Secretariat of National Defense (SEDENA). Read more.
Cascade Aerospace delivers 100th CC-130 to Royal Canadian Airforce
Cascade Aerospace, an Operating Unit of IMP Aerospace & Defence, today celebrated the 100th delivery of a CC-130 Hercules aircraft from its base in Abbotsford, British Columbia to the Royal Canadian Air Force (RCAF) following scheduled maintenance. The aircraft will be returned to service to continue its role as one of Canada's premier long range Fixed Wing Search and Rescue aircraft. Read More.
Pacific Sky Aviation reaches program milestone towards Twin Otter Series 400 Flight Simulation Centre
In partnership with the National Research Council (NRC) in Ottawa, Pacific Sky has just completed the first of two phases of flight testing on the new Twin Otter Series 400 aircraft manufactured by Viking Air Limited. After four weeks conducting over 800 ground and flight manoeuvres, the aircraft performance data collected by NRC will be submitted to TRU Simulation + Training in Montreal, the simulator manufacturer. TRU Simulation will then use the data to build and prepare the new Series 400 Twin Otter Full Flight Simulator (FFS) for qualification by Transport Canada to the highest regulatory standard currently available. Known as "Level D", this highest standard of FFS will place the new Twin Otter simulator in a league of its own with respect to realism and quality of training for the wheeled version of the airplane. Read More.
Vancouver-based company UrtheCast will film 'epic' from space
UrtheCast Corp. and PepsiCo Inc. partner to re-launch campaign, says Financial Post. Read article here.
Finance Minister Mike de Jong discusses BC budget and aerospace with Abbotsford Chamber of Commerce
Aerospace draws big cheer. "You heard me talk about aerospace. I'm very proud of this partnership – it is an initiative that was born out of Abbotsford. The people involved are from Abbotsford. The leaders who are taking us and establishing ourselves on the national stage are from this part of British Columbia. So we have taken some of your money and we have made a commitment over the next number of years of several million dollars to promote the aerospace sector both nationally and internationally," said de Jong. Read more. Download the PDF.

redpajama_set_name: RedPajamaCommonCrawl


text:
Stevens Point Schools ponders charter school's slipping enrollment as board weighs high school expansion
Alan Hovorka
STEVENS POINT - Talk to parents at Stevens Point's charter school and you'll find people grateful for a learning space where their child is flourishing instead of floundering in a traditional school setting.
Shiloh Simons said attending Point of Discovery made his son, Aidan, more engaged in school and successful.
"It made him excited about school again," Simons, a Stevens Point resident, said. "I think it's opened his eyes to different avenues he didn't know before."
But five years after it opened, the charter middle school faces a key decision: Point of Discovery leaders want to expand into 10th grade, and potentially into 11th and 12th grades later, but the plan lingers as the Stevens Point School Board weighs the future direction of the school, which now houses sixth- to ninth-graders.
Complicating the proposal is a school board that wants assurances that an expansion into high school grades will have long-term viability given a slight enrollment decline at the charter school in recent years.
Stevens Point Schools Superintendent Craig Gerlach said he expects to give the board a recommendation on whether to approve the expansion during its next meeting at 6:30 p.m. Monday.
In the meantime, charter school parents like Simons wait to know where they will send their high school-aged kids in the 2020-2021 school year. If the expansion doesn't happen, next year's 10th-graders and their parents will need to figure out how to manage the transition from a school of fewer than 100 to a high school with about 1,500 students.
Point of Discovery expansion would set students on a career path early
Point of Discovery's education model differs from a traditional in-class, homework-driven curriculum. It focuses on "expeditionary learning," a concept that aims to teach students through project-based education by using an overarching theme, or topic, to lace all school subjects together. The goal is to teach children to think critically and engage with their community.
The proposal to expand the school into 10th grade would follow the school's existing model of project-based learning but add another layer through personalized learning plans and educational pathways that, in essence, begin to map out a student's post-high school life.
The proposal includes high school education tracks in arts, entrepreneurship, health care, liberal arts, military, STEAM design and vocational. In a traditional public high school, students have more control over their schedules and classes during their junior and senior years. The proposal would give Point of Discovery underclassmen that degree of control sooner.
"We're really uniquely positioned to make it a truly personalized education for students and to support them because of our smallness and the depth of the relationships we have," Point of Discovery Principal Dan Lathrop said.
Point of Discovery students showcase their learning through regular celebrations of learning, the most recent of which took place on Jan. 15. The most recent semester focused on identifying and answering a need in the larger community, which manifested through projects such as a donation box for the Humane Society of Portage County or a cost-effective drone for use in forestry.
Jolene Bellin, a Stevens Point resident, said the model at Point of Discovery helped her son, Nathan, find his own voice to express his ideas sooner and identify how what he learns in school applies to everyday life. Bellin then opted to enroll the rest of her kids in the charter school because of the clear pathway it created for her son.
"They already know how (their education) should be applied. That's a pretty big difference from mainstream schooling," she said.
Point of Discovery opened for the 2015-16 school year with 52 students and nearly doubled the next school year to 96 students. In the 2017-2018 school year, the school added ninth grade and reached 98 students. But last year and this year saw back-to-back enrollment declines resulting in the 86 students enrolled today. The school occupies the former Jackson Elementary School that closed in 2005 and cost more than $1 million to renovate.
Approving the 10th-grade expansion can't be about accommodating this year's freshman students who want to stay at Point of Discovery next year, Gerlach said. Any expansion has to have a long-term goal that meets student needs, he said.
"It has to be because that's what we really believe is the right thing to do," Gerlach said during a January board meeting.
Board members reiterated their support for the school and its educational model but sent the proposal back to administrators, saying they needed more information. Board members noted the proposal seemed to transform the charter school into a mix of elementary, middle and high school.
"What's being proposed is very different (from the school's current structure). I mean it's just fundamentally different," School Board President Meg Erler said in January.
Erler also asked for additional assurances enrollment would turn around and asked Gerlach to make a direct recommendation on whether to approve or deny the expansion.
"I want this option to remain in this district and I frankly want and need our administration to best inform us as a board how to do that," Erler said.
Point of Discovery leaders think expansion will help enrollment
The push for adding high school to Point of Discovery started as a grassroots effort from parents in the second half of the last school year, said Connie Neegard, director of secondary education for Stevens Point schools.
The plans by Point of Discovery would also drop its maximum possible enrollment from 200 to 120 students — even after adding a grade level — to avoid incurring additional costs to the district, which is wrangling with how to correct a $1 million deficit.
Initially, Point of Discovery leaders asked the school board for permission to expand into a full high school by adding 10th, 11th and 12th grades one grade at a time, and then eventually eliminate sixth grade. The board asked for more information before it could decide on that request. Neegard and the school came back with a revised request that asked for a smaller, gradual expansion.
Neegard said the school could still expand to 11th and 12th grades and drop sixth grade in the future but such decisions will likely be contingent on annual progress on increasing enrollment and annual reports to the school board.
The decline in enrollment since 2018 derives from some parents opting not to enroll their children in ninth grade at Point of Discovery, despite its 2017 addition, out of concern their child would struggle transitioning back to a traditional learning environment, Lathrop said during a January board meeting. Some parents already have had one child graduate from Point of Discovery and have talked the struggles their child faces in adjusting to an atmosphere of nearly 1,500 students at Stevens Point Area Senior High.
The charter school also didn't aggressively market itself throughout the district out of fear people would perceive Point of Discovery as siphoning resources and students from other schools such as P.J. Jacobs or Ben Franklin, Lathrop said.
"I don't want people to perceive us as stealing kids. Because that's not our intent. Our intent is to have kids choose us for the right reasons and they love what we do, et cetera," Lathrop said.
The promotional plan the school laid out includes:
Incorporating Point of Discovery into the district sixth grade transition day
Producing promotional videos with interviews with parents and students
Having a team of parents and students conduct a social media campaign
Visiting parochial schools
Holding open houses and informational meetings
Sending news releases
Neegard said the school hunkered down and built a curriculum to prove the school's value to the district after it opened in 2015.
Point of Discovery consistently meets expectations on the Wisconsin Department of Public Instruction's annual state report cards that measure how well schools meet state standards.
"We just wanted to get our school going and have a solid curriculum, make sure our kids are achieving," Neegard said. "But now the district, I think, understands that this is just a different choice and it's not going to pull kids from any particular building in chunks," she said.
Contact reporter Alan Hovorka at 715-345-2252 or ahovorka@gannett.com. Follow him on Twitter at @ajhovorka.


The Holmdel Police Department is currently recruiting volunteers to serve on the Domestic Violence Response Team (DVRT). In affiliation with 180 Turning Lives Around (180), and with the assistance of the response team volunteers, the Holmdel Police Department continues to make available this service to victims of domestic violence during the initial stages of a highly emotional and traumatic experience. 180 will be conducting an intensive free 40-hour mandatory training course for Domestic Violence Response Team (DVRT) volunteers - Mondays, Wednesdays, and Thursdays, April 29 – May 23, 2019 - 6:00pm-9:30pm, in the Courtroom at Hazlet Police Headquarters, 255 Middle Road, Hazlet, NJ 07730. Applications are currently being accepted.
The volunteer advocates with 180's DVRT Program are civilian members of the community who work collaboratively with law enforcement to provide support, information, and referrals to victims of domestic violence at Police Headquarters. Advocates also discuss with victims safety planning and their legal rights in regard to obtaining a Temporary Restraining Order. By providing empathy and crucial perspective of the situation, these highly-trained advocates help to empower victims to make informed decisions for themselves and their families.
Basic requirements for volunteers to apply include that they must be eighteen years of age or older, have access to reliable transportation, possess a valid driver's license, be willing to serve on an on-call shift basis, participate in an interview process, submit to background investigations and fingerprinting, and successfully complete the mandatory training. The Holmdel Police Department and 180 are committed to a culturally diverse team to better serve the community, so bi-lingual capability is helpful. Prior knowledge of domestic violence is not required. The identities of the DVRT volunteers are kept anonymous. Please contact Lieutenant Michael Pigott of the Holmdel Police Department, DVRT DVLO, at 732-946-9690 ext. 1742 or (mpigott@holmdelpolice.org), to obtain an application or for additional information.
For over thirty-six years, 180 Turning Lives Around has been dedicated to providing shelter, counseling, support, prevention, education, and advocacy for residents of Monmouth County who are affected by domestic violence and sexual violence. If you, or someone you know, is a victim of domestic violence or sexual violence, you can call the 180 Turning Lives Around Confidential Hotline at 732-264-4111 or 888-843-9262, 24-hours a day, 7-days a week. For more information, visit www.180nj.org. In an emergency, dial 9-1-1.
For over 30 years, 180-Turning Lives Around has been dedicated to providing shelter, counseling, protection, support, prevention, education, and advocacy for residents of Monmouth County who are affected by domestic violence and sexual assault. If you or someone you know is a victim of domestic violence or sexual violence, you can call the 180-Turning Lives Around Confidential Hotline at 732-264-4111, 24/7. For more information, visit the New Jersey 180 website.
In 2001 the Holmdel Police Department formed a Domestic Violence Response Team (DVRT). The volunteer advocates with the DVRT Program are civilian members of the community who work in collaboration with the police to provide crisis intervention and safety planning to victims of domestic violence at Police Headquarters. The specially-trained advocates also discuss with the victim their legal rights in regard to obtaining a temporary restraining order.
The Holmdel Police Department is currently recruiting volunteers to serve on its Domestic Violence Response Team (DVRT). In conjunction with 180 Turning Lives Around (180), and with the assistance of the response team volunteers, the Holmdel Police Department continues to make available this service to victims of domestic violence. See more information about the Program, or to volunteer, contact Lt. Michael Pigott at 732-946-9690 ext. 1742 or mpigott@holmdelpolice.org.


text:
ERNÄHRUNG, BEWEGUNG, UMWELT » ays to watch The Open live on Sky S
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#1 | ays to watch The Open live on Sky S 10.04.2019 03:17
Yesterday, an article from the Huffington post took a detailed look at the current state of the Silverdome, and as you can see from the images, it isnt pretty. Head to the @UniquelyDetroit Facebook page for a chance to win a seat from the Silverdome! http://t.co/daqLNCHgO2 pic.twitter.com/rZfwmeRQ9w— Uniquely Detroit (@UniquelyDetroit) April 29, 2014 Although the images of the ruins, and the absence of a tenant, would make you think that the Silverdome has been abandoned for over a decade, it actually played host to a soccer game between European giants AC Milan and Panathinaikos only a few years ago. It was also the home field for an American Ultimate Disc team in 2012, the Detroit Mechanix. It seems the lack of a roof, combined with the exposure to a less than forgiving mother nature, has desecrated the insides of the stadium. For more images, and an incredible video taken from inside the Silverdome ruins, check out the full article here. Air Max 270 Sale Australia . And on Sunday against the Houston Astros they were pleased to see his work finally pay off with his first win since May 24. Air Max 270 Cheap Australia . CSKA received the heavier punishment of playing its next European home game behind closed doors, after its supporters displayed racist and far-right symbols in the Czech Republic during a 2-1 loss to Viktoria Plzen on Dec. http://www.airmax270cheapaustralia.com/.com) - John Wall had 15 points, 12 assists and four steals as the Washington Wizards defeated the New Orleans Pelicans 92-85 on Monday night. Air Max 270 Online Shopping .5 million contract, according to a person familiar with the deal. Hatcher agreed to terms Thursday, according to a person who spoke on condition of anonymity because the contract had not been signed. Air Max 270 Cheap Wholesale . "I met her, I think, a week ago. We went to a Norwood hockey store and picked her out some gear.Phil Mickelson leads the way after the opening round of The 145th Open, with a huge number of top shots produced at Royal Troon. Colin Montgomerie got the tournament underway on his home course when he hit the historic live first tee shot at 6.35am, with the Scot producing a couple of impressive strokes on his way to a level-par 71. Open on Sky Ways to watch The Open live on Sky Sports Lee Westwood made a move up the leaderboard with a chip-in eagle, while Patrick Reed got his week off to a fast start with an eagle-two from the third fairway. Louis Oosthuizen fired the weeks first hole-in-one when he nailed his tee shot at the par-three 14th, while a number of plaayers produced hole outs from well off the green.dddddddddddd. A long range birdie from Luke Donald also features, as does a monster blow with the putter from Matt Jones.Click on the video at the top of the page to see the top shots from the opening round! Get a Sky Sports Pass Dont miss The Open. Watch live on NOW TV today for £10.99. No contract. Watch The Open throughout the week on Sky Sports - your proud new home of The Open Also See: Mickelson close to history First round as it happened Ways to watch The Open Royal Troon leaderboard ' ' '
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