text:
\section{Introduction}

Let $\CA  = (\CA,V)$ be a complex hyperplane arrangement and let
$L(\CA)$ denote its intersection lattice.
We say that $\CA$ is \emph{supersolvable}, provided $L(\CA)$
is supersolvable, see Definition \ref{def:super}.
Thanks to \cite[Prop.\ 3.2]{stanley:super} (see Corollary \ref{cor:stanley}), for $X$ in $L(\CA)$, the
restriction $\CA^X$ of a supersolvable arrangement $\CA$ is itself
again supersolvable.

Now suppose that $W$ is a finite, unitary
reflection group acting on the complex
vector space $V$.
Let $\CA = (\CA(W),V)$ be the associated
hyperplane arrangement of $W$.
In \cite[Thm.\ 1.2]{hogeroehrle:super}, we classified all
supersolvable reflection arrangements.
Extending this earlier work,
the aim of this note is to
classify all supersolvable restrictions
$\CA^X$ for $\CA$ a reflection arrangement.
Since supersolvability is a rather strong condition,
not unexpectedly, there are only
very few additional instances
apart from the obvious
restrictions of supersolvable reflection arrangements.

Moreover, similar to the case of supersolvable reflection arrangements,
we are able to characterize
the irreducible arrangements in this class merely
by the presence
of modular elements of dimension $1$ in their intersection lattice
(see Theorem \ref{super-restriction-rank2}).
This in turn leads to the unexpected, remarkable fact
that reflection arrangements as well as their
restrictions are
supersolvable if and only if they are strictly linearly fibered
(see Corollary \ref{cor:linearlyfiberd}).

The classification
of the irreducible, finite, complex reflection groups $W$ due to
Shephard and Todd \cite{shephardtodd}
states that each such group belongs to one of two types.
Namely, either $W$ belongs to the infinite
three-parameter family $G(r,p,\ell)$ of monomial groups, or else is one of
an additional 34 exceptional groups,
simply named $G_4$ up to $G_{37}$.
As a result, proofs of properties
of $W$ and its arrangement $\CA(W)$ frequently also do
come in two flavors: conceptional, uniform arguments for the
infinite families on the one hand, and adhoc and
mere computational techniques for the exceptional instances,
on the other,
e.g.\ see \cite[\S 6, App.\ B, App.\ C]{orlikterao:arrangements}
and \cite{orliksolomon:unitaryreflectiongroups}.
This dichotomy also prevails the statements
and proofs of this paper.

First we recall the main result from \cite[Thm.\ 1.2]{hogeroehrle:super}:

\begin{theorem}
\label{super}
For $W$ a finite
complex reflection group,
$\CA(W)$ is
supersolvable if and only if
any irreducible factor of $W$ is
of rank at most $2$, or
is isomorphic either to
a Coxeter group
of type $A_\ell$ or $B_\ell$ for $\ell \ge 3$, or to a
monomial group $G(r,p,\ell)$
for $r, \ell \ge 3$  and $p \ne r$.
\end{theorem}

It is easy to see that any central
arrangement of rank at most $2$ is supersolvable,
cf.\ \cite[Rem.\ 2.3]{hogeroehrle:super}.
Thus we focus in the sequel on restrictions $\CA^X$ with $\dim X \ge 3$.

In \cite[Thm.\ 1.2]{amendhogeroehrle:indfree},
we classified all inductively free restrictions $\CA^X$ of
reflection arrangements $\CA$. Since a
supersolvable arrangement is inductively free,
thanks to work of Jambu and Terao  \cite[Thm.\ 4.2]{jambuterao:free}
(see Theorem \ref{thm:superfree}),
the supersolvable restrictions $\CA^X$ of
reflection arrangements $\CA$ form a natural subclass of
the inductively free restrictions.

In order to state \cite[Thm.\ 1.2]{amendhogeroehrle:indfree}, we
require a  bit more notation.
For fixed $r, \ell \ge 2$ and
$0 \le k \le \ell$, we denote by $\CA_\ell^k(r)$ the
sequence of intermediate arrangements that interpolate
between the two reflection arrangements
$\CA(G(r,r,\ell)) =: \CA_\ell^0(r)$ and $\CA(G(r,1,\ell)) =: \CA_\ell^\ell(r)$,
defined in \cite[\S 2]{orliksolomon:unitaryreflectiongroups}
(see also \cite[\S 6.4]{orlikterao:arrangements}).
Note that $\CA_\ell^k(r)$ is not a reflection arrangement for $k \ne 0, \ell$.
Also note that each of the arrangements
does arise as the restriction $\CA(W)^X$ for suitable
$W$ and $X$, e.g.\ see \cite[Prop.\ 6.84]{orlikterao:arrangements}.
Here is \cite[Thm.\ 1.2]{amendhogeroehrle:indfree}.

\begin{theorem}
\label{thm:indfree}
Let $W$ be a finite, irreducible, complex
reflection group with reflection arrangement
$\CA = \CA(W)$ and let $X \in L(\CA)$.
The restricted arrangement $\CA^X$ is inductively free
if and only if one of the following holds:
\begin{itemize}
\item[(i)]
$\CA$ is inductively free;
\item[(ii)]
$W = G(r,r,\ell)$ and
$\CA^X \cong \CA^k_p(r)$, where $p = \dim X$ and $p - 2 \leq k \leq p$;
\item[(iii)]
$W$ is one of $G_{24}, G_{27}, G_{29}, G_{31}, G_{33}$, or $G_{34}$ and $X \in L(\CA) \setminus \{V\}$ with  $\dim X \leq 3$.
\end{itemize}
\end{theorem}

Because the pointwise stabilizer $W_X$ of $X$ in $L(\CA)$ is itself a
complex reflection group, by \cite[Thm.~1.5]{steinberg:differential},
following \cite[\S 3, App.]{orliksolomon:unitaryreflectiongroups} (cf. \cite[\S 6.4, App.\ C]{orlikterao:arrangements}),
we may  label the $W$-orbit
of $X \in L(\CA)$ by the type $T$ say,
of $W_X$; we thus frequently denote such a restriction $\CA(W)^X$
simply by the pair $(W,T)$.

Thanks to \cite[Prop.\ 2.5]{hogeroehrle:super}
and the factorization property of restrictions
\eqref{eq:restrproduct} below,
the question of supersolvability of  $\CA^X$ reduces
to the case when $\CA$ is irreducible.
Thus,  we may assume that  $W$ is irreducible.
In light of Theorem \ref{thm:indfree},
we can now state the main result of our paper.

\begin{theorem}
\label{super-restriction}
Let $W$ be a finite, irreducible, complex
reflection group with reflection arrangement
$\CA = \CA(W)$ and let $X \in L(\CA)$ with $\dim X \ge 3$.
Then the restricted arrangement
$\CA^X$ is supersolvable
if and only if
one of the following holds:
\begin{itemize}
\item[(i)] $\CA$ is supersolvable;
\item[(ii)] $W = G(r,r,\ell)$ and
$\CA^X \cong \CA_p^p(r)$ or $\CA_p^{p-1}(r)$,
where $p = \dim X$;
\item[(iii)] $\CA^X$ is $(E_6,A_3)$, $(E_7, D_4)$, $(E_7, A_2^2)$,
or $(E_8, A_5)$.
\end{itemize}
\end{theorem}

While Theorem \ref{super-restriction} gives a complete classification
of supersolvable restrictions
of reflection arrangements,
as far as isomorphism types
of such restrictions are concerned,
there is quite a bit of redundancy in its statement.
For, the last two restrictions in part (iii) of Theorem \ref{super-restriction}
are isomorphic to each other while the
first two restrictions
are isomorphic to ones in part (ii)
(see Lemma \ref{lem:isoms} below)
and $\CA_p^p(r)$ is
isomorphic to the reflection arrangment of $G(r,1,p)$,
already covered in part (i).
Thus,
apart from supersolvable reflection arrangements themselves,
there is only one additional family of
supersolvable restrictions one for each dimension,
and a single exceptional case,
up to isomorphism.

\begin{corollary}
\label{cor:super-restriction}
Let $W$ be a finite, irreducible, complex
reflection group with reflection arrangement
$\CA = \CA(W)$ and let $X \in L(\CA)$ with $\dim X \ge 3$.
Then $\CA^X$ is supersolvable
if and only if  either
$\CA^X$ is isomorphic to a supersolvable reflection arrangment,
$\CA^X \cong \CA_p^{p-1}(r)$ for some $p \ge 3$,
or $\CA^X \cong (E_7, A_2^2)$.
\end{corollary}

The definition of
supersolvability of $\CA$ entails the existence of
modular elements in $L(\CA)$ of any possible rank;
see \S \ref{ssect:supersolve} for the notion of
modular elements.
In our second main result we  show that
irreducible,
supersolvable restrictions of
reflection arrangements
are characterized merely by the presence
of a modular element of  dimension $1$.

\begin{theorem}
\label{super-restriction-rank2}
For $W$ a finite, irreducible
complex reflection group of rank at least $4$,
let $\CA = \CA(W)$.
Let $X \in L(\CA)$ with $\dim X \ge 3$.
Then $\CA^X$
is supersolvable if and only if
there exists a modular element of dimension $1$
in its lattice $L(\CA^X)$.
\end{theorem}

\begin{remark}
\label{rem:super-restriction-rank2}
(i).
Note that
Theorem \ref{super-restriction-rank2}
also applies to the case $X = V$ and thus gives
a new characterization of
supersolvable
reflection arrangements,
cf.~\cite[Thm.\ 1.3]{hogeroehrle:super}.

(ii).
The condition of irreducibility in
Theorem \ref{super-restriction-rank2}
is necessary, see \cite[Rem.\ 2.6]{hogeroehrle:super}.
\end{remark}

In view of \cite[Cor.~5.112, Thm.~5.113]{orlikterao:arrangements},
Theorem \ref{super-restriction-rank2}
readily implies the following.

\begin{corollary}
\label{cor:linearlyfiberd}
For $W$ a finite, irreducible complex reflection group let $\CA = \CA(W)$ and let $X \in L(\CA)$.
Then $\CA^X$ is strictly linearly fibered
if and only if it is of fiber type.
\end{corollary}

\begin{remark}
\label{rem:linearlyfiberd}
See \cite[Defs.~5.10, 5.11]{orlikterao:arrangements}
for the notions of \emph{strictly linearly fibered} and
\emph{fiber type} arrangements.
Corollary \ref{cor:linearlyfiberd}
is rather striking as the
latter notion is considerably stronger than the first.
In particular, it follows that for $W$ a finite, complex reflection group,
the reflection arrangement $\CA(W)$ is supersolvable if and only if it is
strictly linearly fibered.
This new fact underlines the very special role played by
the class of
reflection arrangements and their restrictions among all arrangements.
\end{remark}

The paper is organized as follows.
In \S \ref{sect:recol} we recall the required notation and
facts about supersolvability of arrangements
and reflection arrangements from
\cite[\S 4, \S6]{orlikterao:arrangements}
and prove some preliminary results.
Theorems \ref{super-restriction} and \ref{super-restriction-rank2} are proved in \S \ref{sect:proof1} and \S \ref{sect:proof2},
respectively.
The dichotomy of the classification of the irreducible
complex reflection groups $W$ into the three
parameter family $G(r,p,\ell)$ and an additional 34 exceptional cases
(see \cite[Table B.1]{orlikterao:arrangements}),
mentioned above, descends to an analogous
dichotomy of the  restrictions $\CA(W)^X$ into the three-parameter family
$\CA^k_\ell(r)$ and a small finite set of exceptional cases.
While our proofs for the restrictions that are isomorphic to
$\CA^k_\ell(r)$ are conceptual and uniform, the statements of
Theorem \ref{super-restriction} and
Corollary \ref{cor:super-restriction} indicate that there cannot
be a uniform argument in general.
For $W$ of exceptional type,
our arguments consist of a case-by-case analysis based
on some technical general lemmas for $3$- and $4$-arrangements
provided in Section \ref{ssect:supersolve}.
Despite the fact that the statement of
Theorem \ref{super-restriction-rank2}
is uniform, still two instances required
some computer calculations.

For general information about arrangements and reflection groups we refer
the reader to \cite{bourbaki:groupes},
\cite{orliksolomon:unitaryreflectiongroups} and
\cite{orlikterao:arrangements}.
Throughout, we use the naming scheme
of the irreducible finite complex reflection groups due to
Shephard and Todd, \cite{shephardtodd}.

\section{Recollections and Preliminaries}
\label{sect:recol}
\subsection{Hyperplane Arrangements}
\label{ssect:hyper}

Let $V = \BBC^\ell$
be an $\ell$-dimensional complex vector space.
A \emph{hyperplane arrangement} is a pair
$(\CA, V)$, where $\CA$ is a finite collection of hyperplanes in $V$.
Usually, we simply write $\CA$ in place of $(\CA, V)$.
We only consider central arrangements, i.e. the origin is contained in the center $T := \bigcap_{H \in \CA}H$ of $\CA$.
The empty arrangement in $V$ is denoted by $\Phi_\ell$.

The \emph{lattice} $L(\CA)$ of $\CA$ is the set of subspaces of $V$ of
the form $H_1\cap \dotsm \cap H_n$ where $\{ H_1, \ldots, H_n \}$ is a subset
of $\CA$.
For $X \in L(\CA)$, we have two associated arrangements,
firstly the subarrangement
$\CA_X :=\{H \in \CA \mid X \subseteq H\}$
of $\CA$ and secondly,
the \emph{restriction of $\CA$ to $X$}, defined by
$\CA^X := \{ X \cap H \mid H \in \CA \setminus \CA_X\}$.

The lattice $L(\CA)$ is a partially ordered set by reverse inclusion:
$X \le Y$ provided $Y \subseteq X$ for $X,Y \in L(\CA)$.
We have a \emph{rank} function on $L(\CA)$ defined by $r(X) := \codim_V(X)$.
The \emph{rank} $r(\CA)$ of $\CA$ is the rank of a maximal element in $L(\CA)$ with respect
to the partial order.
With this definition $L(\CA)$ is a \emph{ranked geometric lattice},
\cite[\S 2]{orlikterao:arrangements}.
The $\ell$-arrangement $\CA$ is called \emph{essential} provided $r(\CA) = \ell$.

Note that the restriction $\CA^X$ is also central,
so that $L(\CA^X)$ is again a geometric lattice.
Let $\CA$ be central and let
$X, Y \in L(\CA)$ with $X < Y$.
We recall the following sublattices of $L(\CA)$ from
\cite[Def.\ 2.10]{orlikterao:arrangements},
$L(\CA)_Y :=\{ Z \in L(\CA) \mid Z \le Y\}$,
$L(\CA)^X :=\{ Z \in L(\CA) \mid X \le Z\}$,
and the \emph{interval}
$[X,Y] := L(\CA)_Y \cap L(\CA)^X = \{ Z \in L(\CA) \mid X \le Z \le Y\}$.

\begin{lemma}
\label{lem:interval}
Let $\CA$ be central and let
$X, Y \in L(\CA)$ with $X < Y$.
Then $L(\CA^Y)$ is a sublattice in $L(\CA^X)$.
In particular, the former is an interval in the latter.
\end{lemma}

\begin{proof}
It follows from  \cite[Lem.\ 2.11]{orlikterao:arrangements}
that
$L(\CA^Y) = L(\CA)^Y
\subseteq
L(\CA)^X = L(\CA^X)$
are both sublattices of $L(\CA)$ and that
$L(\CA^Y) = L((\CA_T)^Y) = [Y, T]$ is an interval in $ L(\CA^X)$.
\end{proof}

Let $\CA = \CA_1 \times \CA_2$ be the
\emph{product}
of the two arrangements $\CA_1$ and $\CA_2$.
With the partial order defined on
$L(\CA_1) \times L(\CA_2)$ by
$(X_1,X_2) \le (Y_1,Y_2)$ provided $X_1 \le Y_1$ and $X_2 \le Y_2$,
there is a lattice isomorphism
$ L(\CA_1) \times L(\CA_2) \cong L(\CA)$ given by
$(X_1, X_2) \mapsto X_1 \oplus X_2$,
\cite[\S 2]{orlikterao:arrangements}.
It is easily seen that
for $X =  X_1 \oplus X_2 \in L(\CA)$, we have
\begin{equation}
\label{eq:restrproduct}
\CA^X = \CA_1^{X_1} \times \CA_2^{X_2}.
\end{equation}

Note that
$\CA \times \Phi_0 = \CA$
for any arrangement $\CA$.
If $\CA$ is of the form $\CA = \CA_1 \times \CA_2$, where
$\CA_i \ne \Phi_0$ for $i=1,2$, then $\CA$
is called \emph{reducible}, else $\CA$ is
\emph{irreducible},
\cite[Def.\ 2.15]{orlikterao:arrangements}.

\subsection{Free Arrangements}
\label{ssect:free}

Let $S = S(V^*)$ be the symmetric algebra of the dual space $V^*$ of $V$.
If $x_1, \ldots , x_\ell$ is a basis of $V^*$, then we identify $S$ with
the polynomial ring $\BBC[x_1, \ldots , x_\ell]$.
By denoting the $\BBC$-subspace of $S$
consisting of the homogeneous polynomials of degree $p$ (and $0$) by $S_p$,
we see that there is a natural $\BBZ$-grading
$S = \oplus_{p \in \BBZ}S_p$, where
$S_p = 0$ for $p < 0$.

Let $\Der(S)$ be the $S$-module of $\BBC$-derivations of $S$
and for $i = 1, \ldots, \ell$
define $D_i := \partial/\partial x_i$ to be the $i\th$ partial derivation of $S$.
Now $D_1, \ldots, D_\ell$ is an $S$-basis of $\Der(S)$ and
we call $\theta \in \Der(S)$
\emph{homogeneous of polynomial degree $p$}
provided
$\theta = \sum_{i=1}^\ell f_i D_i$,
where $f_i \in S_p$ for each $1 \leq i \leq \ell$.
In this case we write $\pdeg \theta = p$.
By defining $\Der(S)_p$ to be the $\BBC$-subspace of $\Der(S)$ consisting
of all homogeneous derivations of polynomial degree $p$,
we see that $\Der(S)$ is a graded $S$-module:
$\Der(S) = \oplus_{p\in \BBZ} \Der(S)_p$.

Following \cite[Def.~4.4]{orlikterao:arrangements},
we define the $S$-submodule $D(f)$ of $\Der(S)$ for $f \in S$ by
$D(f) := \{\theta \in \Der(S) \mid \theta(f) \in f S\}$.

If $\CA$ is an arrangement in $V$,
then for every $H \in \CA$ we may fix $\alpha_H \in V^*$ with
$H = \ker(\alpha_H)$.
We call $Q(\CA) := \prod_{H \in \CA} \alpha_H \in S$
the \emph{defining polynomial} of $\CA$.

The \emph{module of $\CA$-derivations} is the $S$-submodule of $\Der(S)$
defined by
\[
D(\CA) := D(Q(\CA)).
\]
The arrangement $\CA$ is said to be \emph{free} if the module of $\CA$-derivations
$D(\CA)$ is a free $S$-module.

Note that  $D(\CA)$
is a graded $S$-module $D(\CA) = \oplus_{p\in \BBZ} D(\CA)_p$,
where $D(\CA)_p = D(\CA) \cap \Der(S)_p$, see
\cite[Prop.~4.10]{orlikterao:arrangements}.
If $\CA$ is a free $\ell$-arrangement,
then by \cite[Prop.~4.18]{orlikterao:arrangements}
the $S$-module $D(\CA)$ admits a basis of $\ell$ homogeneous derivations $\theta_1, \ldots, \theta_\ell$.
While these derivations are not unique, their polynomial
degrees $\pdeg \theta_i$ are unique (up to ordering).
The set of \emph{exponents} of the free arrangement $\CA$ is the multiset
\[
\exp\CA := \{\pdeg \theta_1, \ldots, \pdeg \theta_\ell\}.
\]
For the stronger notion of  \emph{inductive freeness}, see
\cite[Def.~4.53]{orlikterao:arrangements}.

\subsection{Supersolvable Arrangements}
\label{ssect:supersolve}

Let $\CA$ be an arrangement.
Following \cite[\S 2]{orlikterao:arrangements}, we say
that $X \in L(\CA)$ is \emph{modular}
provided $X + Y \in L(\CA)$ for every $Y \in L(\CA)$.
Let $\CA$ be a central (and essential) $\ell$-arrangement.
The next notion is due to Stanley \cite{stanley:super}.

\begin{defn}
\label{def:super}
We say that $\CA$ is
\emph{supersolvable}
provided there is a maximal chain
\[
V = X_0 < X_1 < \cdots < X_{\ell-1} < X_\ell = T
\]
 of modular elements $X_i$ in $L(\CA)$,
cf.\ \cite[Def.\ 2.32]{orlikterao:arrangements}.
\end{defn}

It is easy to see that
every $2$-arrangement is supersolvable,
\cite[Rem.\ 2.3]{hogeroehrle:super}.
The next result is also straightforward,
\cite[Lem.\ 2.4]{hogeroehrle:super}.

\begin{lemma}
\label{lem:super-rank2}
A $3$-arrangement $\CA$ is supersolvable if and only if
there exists a modular element in $L(\CA)$ of
dimension $1$.
\end{lemma}

Thanks to \cite[Prop.\ 2.5]{hogeroehrle:super},
supersolvable  arrangements behave well with respect to
the  product construction for arrangements from \S \ref{ssect:hyper}.

Supersolvable arrangements are always free,
\cite[Thm.~4.58]{orlikterao:arrangements}:

\begin{theorem}
\label{thm:superfree}
Let $\CA$ be a supersolvable $\ell$-arrangment with
maximal chain
\[
V = X_0 < X_1 < \cdots < X_{\ell-1} < X_\ell = T
\]
of modular elements $X_i$ in $L(\CA)$.
Define $b_i = |\CA_{X_i}\setminus \CA_{X_{i-1}}|$ for $1 \le i \le \ell$.
Then $\CA$ is inductively free with
$\exp \CA = \{b_1, \ldots, b_\ell\}$.
\end{theorem}

Note that it was first proven in \cite[Thm.~4.2]{jambuterao:free} that every supersolvable arrangement is inductively free.
We proceed with some further preliminary results.

\begin{lemma}
\label{lem:modularrestrictions}
Let $X \in L(\CA)$ be modular and $Z \in L(\CA)$. Then $X\cap Z$ is
modular in $L(\CA^Z)$.
\end{lemma}

\begin{proof}
Let $Y \in L(\CA^Z) = L(\CA)^Z \subseteq L(\CA)$.
Since $X$ is modular in $L(\CA)$, we
have $X+Y \in L(\CA)$ and a direct calculation shows
$(Z \cap X) + Y = Z \cap (X + Y)\in L(\CA^Z)$.
\end{proof}

The following
immediate consequence of
Lemma \ref{lem:modularrestrictions}
is due to
Stanley, \cite[Prop.\ 3.2]{stanley:super}.

\begin{corollary}
\label{cor:stanley}
Let $X < Y$ in $L(\CA)$.
If $\CA$ is supersolvable, then so is the
interval $[X,Y]$.
\end{corollary}

Here is a further consequence of
Lemma \ref{lem:modularrestrictions}.

\begin{corollary}
\label{cor:dim4notsupbycounting}
Let $\CA$ be a $4$-arrangement and let $Z \in L(\CA)$ be of
dimension $1$.
Suppose $\vert \CA_Z \vert > \vert \{H \in \CA \mid \CA^H
\text{ is supersolvable } \} \vert $. Then $Z$ is not modular in $L(\CA)$.
\end{corollary}

\begin{proof}
By assumption on $\CA_Z$, there is a $H \in \CA$ with $Z \in L(\CA^H)$ such that $\CA^H$ is not
supersolvable. Therefore, since $\CA^H$ is
a $3$-arrangement, $Z$ is not modular in $L(\CA^H)$,
by Lemma \ref{lem:super-rank2}.
We conclude that $Z$ is not modular in $L(\CA)$,
thanks to Lemma \ref{lem:modularrestrictions}.
\end{proof}

\begin{lemma}
\label{lem:countres}
Let $\CA$ be an arrangement. Then for $X \in L(\CA)$, we have
$\vert \CA_X \vert > \vert (\CA^H)_X \vert$ for all $H \in \CA_X$.
\end{lemma}

\begin{proof}
Note that
$\vert(\CA^H)_X\vert
= \vert \{H \cap H' \mid H' \in \CA_X \setminus \{H\}\}\vert < \vert \CA_X\vert$.
\end{proof}

The next result gives a useful numerical criterion for the non-supersolvability of a free,
irreducible $3$-arrangement $\CA$ in terms of the exponents of $\CA$.

\begin{lemma}
\label{lem:dim3notsuper}
Let $\CA$ be a free, irreducible $3$-arrangement.
Let $\exp \CA = \{1, b_1, b_2\}$ with $b_1 \le b_2$.
If $|\CA_X|\le b_1$ for all $X \in L(\CA)$ of
dimension $1$,
then $\CA$ is not supersolvable.
\end{lemma}

\begin{proof}
Note that since $\CA$ is irreducible, $b_1 > 1$,
\cite[Thm.~4.29(3)]{orlikterao:arrangements}.
Assume $|\CA_X|\le b_1$ and suppose that
$\CA$ is supersolvable with $V < H < X < 0$ a maximal chain of
modular elements in $L(\CA)$. Observe that
$|\CA_H \setminus \CA_V| = |\{H\} \setminus \emptyset| = 1$.
By Theorem \ref{thm:superfree}, we have
$|\CA_X \setminus \CA_H| \in \{b_1,b_2\}$. However,
$|\CA_X \setminus \CA_H| = |\CA_X| -1 < b_1 \le b_2$,
a contradiction.
\end{proof}

The following technical lemma is our key tool in order to show that there are
no modular elements of dimension $1$ in certain non-supersolvable,
irreducible $4$-arrangements.

\begin{lemma}
\label{lem:show4nonmodular1}
Let $\CA$ be a free, irreducible $4$-arrangement.
Assume there is, up to lattice isomorphism, only one supersolvable restriction
$\CB:= \CA^H$ to
a hyperplane $H \in \CA$.
Let $1 < b_1 \le b_2$ be the exponents of $\CB$ and let
$c = \vert\{ H' \in \CA \mid L(\CA^{H'}) \cong L(\CB)\} \vert$.
If $c \le b_1 + 1$, then there are no modular
elements of dimension $1$ in $L(\CA)$.
\end{lemma}

\begin{proof}
Suppose that $X \in L(\CA)$ is modular of dimension $1$. Then $X$
is also modular in $\CA^H$ for all $H \in \CA_X$,
by Lemma \ref{lem:modularrestrictions},  and since $\CA^H$ is a
$3$-arrangement, it is supersolvable, by Lemma \ref{lem:super-rank2}.
Lemma \ref{lem:dim3notsuper} implies that $|(\CA^H)_X| > b_1$ and Lemma
\ref{lem:countres} shows that $|\CA_X| > b_1 + 1 \ge c$. This
is a contradiction to Corollary  \ref{cor:dim4notsupbycounting}.
\end{proof}

\subsection{Reflection Groups and Reflection Arrangements}
\label{ssect:refl}
The irreducible finite complex reflection groups were
classified by Shephard and Todd, \cite{shephardtodd}.
Let $W  \subseteq \GL(V)$ be a finite complex reflection group.

The \emph{reflection arrangement} $\CA = \CA(W)$ of $W$ in $V$ is
the hyperplane arrangement
consisting of the reflecting hyperplanes of the elements in $W$
acting as reflections on $V$.

The intermediate arrangements $\CA^k_\ell(r)$ (with $0 \leq k \leq \ell$) interpolate between the reflection
arrangements of $G(r,r,\ell)$ and $G(r,1,\ell)$.
Let $H_i := \ker(x_i)$ and $H_{ij}(\zeta) := \ker(x_i - \zeta x_j)$, then $\CA^k_\ell(r) = \{H_1, \ldots, H_k, H_{ij}(\zeta) \mid
1 \leq i < j \leq \ell, \zeta^r = 1\}$. Note that $H_{ij}(\zeta) = H_{ji}(\zeta\inverse)$. We now recall
\cite[Prop.~2.11]{orliksolomon:unitaryreflectiongroups} (see also \cite[Prop.~6.82]{orlikterao:arrangements}):

\begin{proposition}
\label{prop:intermediate}
Let $\CA = \CA^k_\ell(r)$ and let $H \in \CA$. The type of $\CA^H$ is given in Table \ref{intermediate-table}.
\end{proposition}

\begin{table}[ht!b]
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{llll}\hline
 $k$ & \multicolumn{2}{l}{$H$} & Type of $\CA^H$\\ \hline
 $0$ & arbitrary & & $\CA^1_{\ell - 1}(r)$\\
 $1, \ldots, \ell - 1$ & $H_{ij}(\zeta)$ & $1 \leq i < j \leq k < \ell$ & $\CA^{k - 1}_{\ell - 1}(r)$\\
 $1, \ldots, \ell - 1$ & $H_{ij}(\zeta)$ & $1 \leq i \leq k < j < \ell$ & $\CA^k_{\ell - 1}(r)$\\
 $1, \ldots, \ell - 1$ & $H_{ij}(\zeta)$ & $1 < k < i < j \leq \ell$ & $\CA^{k + 1}_{\ell - 1}(r)$\\
 $1, \ldots, \ell - 1$ & $H_i$ & $1 \leq i \leq \ell$ & $\CA^{\ell - 1}_{\ell - 1}(r)$\\
 $\ell$ & arbitrary & & $\CA^{\ell - 1}_{\ell - 1}(r)$\\ \hline
\end{tabular}
\medskip
\caption{Restriction types of $\CA^k_\ell(r)$}
\label{intermediate-table}
\end{table}

Let $\CA = \CA^k_\ell(r)$, fix $H \in \CA$
and let $(\CA, \CA' = \CA\setminus\{H\}, \CA'' = \CA^H)$ be the triple with
respect to $H$.
In Table \ref{restriction-table}, we consider the map
$\CA' \to \CA''$ given by $H' \mapsto H' \cap H$,
where restricting to $H_i$ results in the substitution $x_i \to 0$, and
restricting to $H_{ij}(\zeta)$ results in the substitution $x_i \to \zeta x_j$.

\begin{table}[ht!b]
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{lll}\hline
$H'$ & restricts to & $\tilde{H} \in \CA^{H_n}$\\ \hline\hline
$H_i$ & $i \neq n$ & $H_i$\\ \hline
$H_{ij}(\zeta)$ & $i, j \neq n$ & $H_{ij}(\zeta)$\\
& $i = n$ & $H_j$\\
& $j = n$ & $H_i$\\ \hline
& \\
& \\
\end{tabular}
\qquad
\begin{tabular}{lll}\hline
$H'$ & restricts to & $\tilde{H} \in \CA^{H_{mn}(\eta)}$\\ \hline\hline
$H_i$ & $i \neq m$ & $H_i$\\
& $i = m$ & $H_n$\\ \hline
$H_{ij}(\zeta)$ & $i, j \neq m$ & $H_{ij}(\zeta)$\\
& $i = m, j \neq n$ & $H_{nj}(\eta\inverse\zeta)$\\
& $i \neq n, j = m$ & $H_{in}(\zeta\eta)$\\ \hline
$H_{mn}(\zeta)$ & $\zeta \neq \eta$ & $H_n$\\ \hline
\end{tabular}
\medskip
\caption{Restrictions of $\CA = \CA^k_\ell(r)$ to hyperplanes $H_n$ and $H_{mn}(\eta)$}
\label{restriction-table}
\end{table}

\section{Proof of Theorem \ref{super-restriction}}
\label{sect:proof1}

We maintain the notation from the previous sections.
Note again that $\CA_\ell^k(r)$ is not a reflection arrangement for $k \ne 0, \ell$.

\begin{lemma}
\label{lem:intermediates}
For $r \geq 2$, $\ell \ge 3$ and  $0 \le k \le \ell$,
the arrangement $\CA_\ell^k(r)$
is supersolvable if and only if $\ell - 1 \le k \le \ell$ or $(k,\ell,r) = (0,3,2)$.
\end{lemma}

\begin{proof}
First note that $\CA_3^0(2)$ is just the Coxeter arrangement of type $D_3 = A_3$.
Thus $\CA_3^0(2)$ is supersolvable, by Theorem \ref{super}.

Else for $k = 0$ and $k= \ell$, the result follows from
Theorem \ref{super} and for
$1 \le k \le \ell-3$, the result is a consequence of
Theorems \ref{thm:indfree}(ii)
and \ref{thm:superfree}.
For $k = \ell-1$, this follows from the case for $k = \ell$,
the proof of \cite[Thm.~1.2(iii)]{hogeroehrle:super} and
\cite[Lem.\ 2.62]{orlikterao:arrangements}.
Finally, for $k =\ell -2$, we argue by induction on $\ell$ as follows.
If $\CA$ is any (central) arrangement and $H$ is in $\CA$, then $L(\CA^H)$ is an intervall
in $L(\CA)$. By Corollary \ref{cor:stanley},  if $\CA^H$ is not supersolvable, then neither is $\CA$.
Suppose $\CA = \CA^{\ell - 2}_\ell(r)$ and $H = \ker(x_i - \zeta x_j) \in \CA^{\ell - 2}_\ell(r)$ for some $1 \leq i < j \leq \ell - 2$ and
some $r\th$ root of unity $\zeta$. By Proposition \ref{prop:intermediate}, we have $\CA^H \cong \CA^{\ell - 3}_{\ell - 1}(r)$,
so it suffices to show that $\CA^1_3(r)$ is not supersolvable.
So, let $\CA = \CA^1_3(r)$ and let
$X \in L(\CA)$ be of rank $2$. Then  $\vert\CA_X\vert \in \{2, 3, r, r + 1\}$.
Since $\exp\CA = \{1, r + 1, 2r - 1\}$,  Lemma \ref{lem:dim3notsuper} implies that
$\CA^1_3(r)$ is not supersolvable.
\end{proof}

Note that it was already observed in \cite[Ex.\ 5.5]{jambuterao:free}
that $\CA_3^1(2)$ is not supersolvable.

The following example shows that the supersolvable
arrangements from Lemma \ref{lem:intermediates}
do actually occur as restrictions of the reflection
arrangement of $W = G(r,r,\ell)$.

\begin{example}
Let $ p,r \ge 2$, $\ell = 2p-1$, and
$\zeta$ is an $r$th root of unity.
Let $W = G(r,r,\ell)$ and
for $1 \le i \ne j \le \ell$
let $H_{i,j}(\zeta) = \ker(x_i - \zeta x_j)$ be a hyperplane in
$\CA = \CA(W)$.
Then $X = \cap_{i=1}^{p-1} H_{2i-1,2i}(\zeta)$
belongs to $L(\CA)$ with $\dim X = \ell - (p-1) = p$.
According to
the description of the orbit structure of $W$ on $L(\CA)$ in
\cite[Prop.~2.14]{orliksolomon:unitaryreflectiongroups}
(cf. \cite[Prop.\ 6.84]{orlikterao:arrangements}), we have
$\CA^X \cong \CA_p^{p-1}(r)$
which is supersolvable, by Lemma \ref{lem:intermediates}.

If $Y$ is in $L(\CA)$ with $\dim Y = p$ such that $Y$ is contained in
at least one coordinate hyperplane $\ker(x_i)$, then again by
\cite[Prop.~2.14]{orliksolomon:unitaryreflectiongroups}
(\cite[Prop.\ 6.84]{orlikterao:arrangements}),
we see that
$\CA^Y \cong \CA_p^p(r) = \CA(G(r,1,p))$
which is supersolvable, by Theorem \ref{super}.
\end{example}

As indicated in the Introduction,
we use the convention
to label the $W$-orbit
of $X \in L(\CA)$ by the type $ T$ which is
the Shephard-Todd label \cite{shephardtodd}
of the complex reflection group $W_X$.
We then denote the restriction $\CA^X$ simply by the pair
$(W,T)$.
The following result is due to Orlik and Terao,
\cite[App.\ D]{orlikterao:arrangements}.

\begin{lemma}
\label{lem:isoms}
We have the following lattice isomorphisms of
$3$-dimensional restrictions:
\begin{itemize}
\item [(i)] $(E_6,A_3) \cong \CA_3^2(2)$;
\item [(ii)] $(E_7, D_4) \cong \CA_3^3(2)$;
\item [(iii)] $(F_4,\tilde{A_1}) \cong (F_4,A_1) \cong (E_7,A_1^4) \cong (E_7,(A_1A_3)') \cong (E_8,A_1 D_4) \cong (E_8,D_5)$;
\item [(iv)] $(E_6,A_1 A_2) \cong (E_7,A_4)$;
\item [(v)] $(E_7,A_2^2) \cong (E_8,A_5)$;
\item [(vi)] $(E_8,A_1^2 A_3) \cong (E_8,A_2 A_3)$;
\item [(vii)] $(G_{26},A_0) \cong (G_{32}, C(3)) \cong (G_{34}, G(3,3,3))$.
\end{itemize}
\end{lemma}

Armed with Lemma
\ref{lem:dim3notsuper},
we can now determine all $3$-dimensional
supersolvable restrictions for an ambient
irreducible, non-supersolvable
reflection arrangement.

\begin{lemma}
\label{lem:dim3}
Let $\CA = \CA(W)$ be an irreducible, non-supersolvable
reflection arrangement of exceptional type.
Let $X \in L(\CA)$ with $\dim X = 3$.
Then $\CA^X$ is supersolvable if and only if
$\CA^X$ is
$(E_6,A_3)$, $(E_7, D_4)$, $(E_7, A_2^2)$,
or $(E_8, A_5)$.
\end{lemma}

\begin{proof}
Using Theorem \ref{super}, the tables of all orbit types
for the irreducible reflection groups of exceptional type in
\cite[App.]{orliksolomon:unitaryreflectiongroups} (see also \cite[App.\ C]{orlikterao:arrangements})
and some
explicit computer aided calculations (cf. Remark \ref{rem:computations}),
we check which of the $3$-dimensional
restrictions satisfies the hypothesis of Lemma
\ref{lem:dim3notsuper}.
In Table \ref{table:rank3restrictions}
we present in each of the 3-dimensional restrictions $\CB = \CA(W)^X$
of the exceptional cases all values
$|\CB_Y|$, where $Y \in L(\CB)$ is of dimension $1$
(disregarding the lattice isomorphisms from Lemma \ref{lem:isoms}).
Each of the cases labelled ``false'' satisfies the condition from
Lemma  \ref{lem:dim3notsuper} and is therefore not
supersolvable.

\begin{longtable}{l|l|c|c}
$\CB :=\CA^X$ & $\exp \CB$ & $\{|\CB_{Y}|\mid \operatorname{rk}(Y) = 2\}$
 & supersolvable\\\endfirsthead
\endhead\toprule
\label{table:rank3restrictions}
$(F_4,A_1)$        &  1,5,7        & \{2,3,4\}        &false\\
$(F_4,\tilde A_1)$ &  1,5,7        & \{2,3,4\}        &false\\
$(G_{29},A_1)$     &  1,9,11       & \{2,3,4,5\}      &false\\
$(H_4,A_1)$        &  1,11,19      & \{2,3,5,6\}      &false\\
$(G_{31},A_1)$     &  1,13,17      & \{2,3,6\}        &false\\
$(G_{32},C(3))$    &  1,7,13       & \{2,4,5\}        &false\\
$(G_{33},A_1^2)$   &  1,7,9        & \{2,3,4,5\}      &false\\
$(G_{33},A_2)$     &  1,6,7        & \{2,3,4,5\}      &false\\
$(G_{34},A_1^3)$   &  1,13,19      & \{2,3,4,5,8\}    &false\\
$(G_{34},A_1A_2)$  &  1,13,16      & \{2,3,4,5,7\}    &false\\
$(G_{34},A_3)$     &  1,11,13      & \{2,3,4,6\}      &false\\
$(G_{34},G(3,3,3))$ &  1,7,13       & \{2,4,5\}        &false\\
$(E_6,A_1^3)$      &  1,4,5        & \{2,3,4\}        &false\\
$(E_6,A_1A_2)$     &  1,4,5        & \{2,3,4\}        &false\\
$(E_6,A_3)$        &  1,3,4        & \{2,3,4\}        & {\bf true}\\
$(E_7,A_1^4)$      &  1,5,7        & \{2,3,4\}        &false\\
$(E_7,A_1^2A_2)$   &  1,5,7        & \{2,3,4,5\}      &false\\
$(E_7,A_2^2)$      &  1,5,7        & \{2,3,4,6\}      & {\bf true}\\
$(E_7,(A_1A_3)')$  &  1,5,7        & \{2,3,4\}        &false\\
$(E_7,(A_1A_3)'')$ &  1,5,5        & \{2,3,4\}        &false\\
$(E_7,A_4)$        &  1,4,5        & \{2,3,4\}        &false\\
$(E_7,D_4)$        &  1,3,5        & \{2,3,4\}        & {\bf true}\\
$(E_8,A_1^3A_2)$   &  1,7,11       & \{2,3,4,5,6\}    &false\\
$(E_8,A_1A_2^2)$   &  1,7,11       & \{2,3,4,6\}      &false\\
$(E_8,A_1^2A_3)$   &  1,7,9        & \{2,3,4,6\}      &false\\
$(E_8,A_2A_3)$     &  1,7,9        & \{2,3,4,6\}      &false\\
$(E_8,A_1A_4)$     &  1,7,8        & \{2,3,4,5\}      &false\\
$(E_8,A_5)$        &  1,5,7        & \{2,3,4,6\}      & {\bf true}\\
$(E_8,A_1D_4)$     &  1,5,7        & \{2,3,4\}        &false\\
$(E_8,D_5)$        &  1,5,7        & \{2,3,4\}        &false\\
\hline
\caption{The supersolvable 3-dimensional restrictions of the exceptional groups}
\end{longtable}

Next we argue that the 4 cases labelled ``true'' in
Table \ref{table:rank3restrictions} are indeed supersolvable:

The restrictions
$(E_6,A_3)$ and $(E_7,D_4)$ are supersolvable, thanks to Lemmas
\ref{lem:intermediates}  and \ref{lem:isoms}(i),~(ii).
Moreover, by Lemma \ref{lem:isoms}(v), $(E_7,A_2^2) \cong (E_8,A_5)$, so we only have
to check the case $(E_7,A_2^2)$.

So, let $\CB = \{H_0,\ldots,H_{12}\}$ be the arrangement given by $(E_7,A_2^2)$.
Since $\CB$ is a $3$-arrangement, we only need to compute the
intersections of rank $2$.
Each $X \in L(\CB)$ is uniquely given by $\CB_X$. The $X \in L(\CB)$ of rank
$2$ correspond to the following subarrangements $\CB_X$ of $\CB$:

\begin{table}[h!t]
\begin{tabular}{rrrr}
$\{H_0, H_1, H_3, H_6, H_9, H_{12}\}$, & $\{H_0, H_2, H_{10}\}$,    & $\{H_0, H_4, H_5, H_{11}\}$, & $\{H_0, H_7, H_8\}$, \\
$\{H_1, H_2, H_7, H_{11}\}$,           & $\{H_1, H_4, H_{10}\}$,    & $\{H_1, H_5, H_8\}$,         &                      \\
$\{H_2, H_3, H_4\}$,                   & $\{H_2, H_5, H_6\}$,       & $\{H_2, H_8, H_{12}\}$,      & $\{H_2, H_9\}$,      \\
$\{H_3, H_5, H_7\}$,                   & $\{H_3, H_8\}$,            & $\{H_3, H_{10}\}$,           & $\{H_3, H_{11}\}$,   \\
$\{H_4, H_6, H_7\}$,                   & $\{H_4, H_8, H_9\}$,       & $\{H_4, H_{12}\}$,           &                      \\
$\{H_5, H_9, H_{10}\}$,                & $\{H_5, H_{12}\}$,         &                              &                      \\
$\{H_6, H_8, H_{10}, H_{11}\}$,        &                            &                              &                      \\
$\{H_7, H_9\}$,                        & $\{H_7, H_{10}, H_{12}\}$, &                              &                      \\
$\{H_9, H_{11}\}$,                     &                            &                              &                      \\
$\{H_{11}, H_{12}\}$.                  &                            &                              &                      \\
\end{tabular}
\end{table}

Let $Z:=H_0 \cap H_1 \cap H_3 \cap H_6 \cap H_9 \cap H_{12}$.
For all $X \in L(\CB)$ of dimension $1$ we have $X + Z \in L(\CB)$, since $X$ and
$Z$ are subsets of a common hyperplane of $\CB$. Therefore, $Z$ is a modular
element of dimension $1$ in $L(\CB)$ and thus $\CB$ is supersolvable, by Lemma \ref{lem:super-rank2}.
\end{proof}

Now, with the aid of Corollary \ref{cor:stanley} and
Lemma \ref{lem:dim3} we can show that among
irreducible, non-supersolvable
reflection arrangement of exceptional type there are no
$4$-dimensional supersolvable restrictions.

\begin{lemma}
\label{lem:dim4s}
Let $\CA = \CA(W)$ be an irreducible, non-supersolvable
reflection arrangement of exceptional type.
Let $X \in L(\CA)$ with $\dim X = 4$.
Then $\CA^X$ is not supersolvable.
\end{lemma}

\begin{proof}
Thanks to Lemma \ref{lem:dim3}, for every $W$ as in the statement, other than
$W$ of type $E_6$, $E_7$ or $E_8$, already
every $3$-dimensional restriction is not supersolvable.
It thus follows from Corollary \ref{cor:stanley} that
in these cases
no higher-dimensional
restriction is supersolvable either.

So we are left to check the following
restrictions $(E_6,A_1^2)$, $(E_6,A_2)$, $(E_7,(A_1^3)')$,
$(E_7,(A_1^3)'')$, $(E_7,A_1A_2)$,$(E_7,A_3)$, $(E_8,A_1^4)$, $(E_8,A_1^2A_2)$,
$(E_8,A_2^2)$, $(E_8,A_1A_3)$, $(E_8,A_4)$, $(E_8,D_4)$.
Again by Corollary \ref{cor:stanley}, it suffices
to find a restriction to a hyperplane in these cases which is
not supersolvable; in the following table, we exhibit
a suitable restriction in each instance:
\begin{center}
\begin{tabular}{l|l}
  $\CA^X$ & non-supersolvable restriction \\
\hline
$(E_6,A_1^2)$     &$(E_6,A_1A_2)$\\
$(E_6,A_2)$       &$(E_6,A_1A_2)$\\
$(E_7,(A_1^3)')$  &$(E_7,A_1^4)$ \\
$(E_7,(A_1^3)'')$ &$(E_7,A_1^4)$\\
$(E_7,A_1A_2)$    &$(E_7,A_4)$\\
$(E_7,A_3)$       &$(E_7,A_4)$\\
$(E_8,A_1^4)$     &$(E_8,A_1D_4)$\\
$(E_8,A_1^2A_2)$   &$(E_8,A_1A_4)$\\
$(E_8,A_2^2)$     &$(E_8,A_2A_3)$ \\
$(E_8,A_1A_3)$    &$(E_8,A_2A_3)$\\
$(E_8,A_4)$       &$(E_8,A_1A_4)$\\
$(E_8,D_4)$       &$(E_8,A_1D_4)$\\
\end{tabular}
\end{center}
This is readily extracted from the information in the tables of
\cite[App.]{orliksolomon:unitaryreflectiongroups} (\cite[App.\ C]{orlikterao:arrangements}).
\end{proof}

\begin{corollary}
\label{cor:dimge4}
Let $\CA = \CA(W)$ be an irreducible, non-supersolvable
reflection arrangement of exceptional type.
Let $X \in L(\CA)$ with $\dim X \ge 4$.
Then $\CA^X$ is not supersolvable.
\end{corollary}

\begin{proof}
Thanks to Lemma \ref{lem:dim4s},
every restriction of dimension $4$ is non-supersolvable.
The result then follows from Corollary \ref{cor:stanley}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{super-restriction}]
Thanks to Corollary \ref{cor:stanley},
every interval of a supersolvable lattice is itself supersolvable.
Consequently, if $\CA$ is supersolvable, then so is any restriction $\CA^X$,
by Lemma \ref{lem:interval}.
If $\CA^X$ is as in (ii) or (iii), then
$\CA^X$ is supersolvable, thanks to
Lemmas \ref{lem:intermediates} and \ref{lem:dim3} and
Corollary \ref{cor:dimge4}.
This gives the reverse implication.

If $\CA^X$ is supersolvable but  $\CA$ is not,
then $\CA^X$  is as in (ii) or (iii), again
by Lemmas \ref{lem:intermediates} and \ref{lem:dim3} and
Corollary \ref{cor:dimge4}.
This gives the forward implication.
\end{proof}

\section{Proof of Theorem \ref{super-restriction-rank2}}
\label{sect:proof2}

We start by determining all modular elements in the lattice of $\CA^k_\ell(r)$.  We introduce the following notation:
Let $\Bool_k$ be the Boolean arrangement in $\BBC^k$ and define $\CB^k_\ell := \Bool_k \times \Phi_{\ell - k}$ to be the
$\ell$-arrangement consisting of the first $k$ coordinate hyperplanes in $\BBC^\ell$, i.e. $\CB^k_\ell = (\{H_1, \ldots, H_k\},
\BBC^\ell)$. Then clearly $\CB^k_\ell \subseteq \CA^m_\ell(r)$ for all $m \geq k$.

\begin{lemma}
\label{lem:modularelements}
Let $\CA = \CA^k_\ell(r)$ with $\ell \geq 3$, $0 \leq k \leq \ell$ and $r \geq 2$.
Then all elements of $L(\CB^k_\ell) \subseteq
L(\CA)$ are modular in $L(\CA)$.

\begin{proof}
Let $X \in L(\CB^k_\ell)$ with $1 \leq \dim X < \ell - 1$ (i.e.\ $X$ is not trivially modular) and $I = \{1, \ldots ,\ell\}$.
Then there is a subset $\Delta_X \subseteq \{1, \ldots, k\}$ such that $X = \bigcap\limits_{i \in \Delta_X}H_i$.

Let $Y \in L(\CA)$ with $\dim Y = p$. Then using the construction in \cite[\S 2]{orliksolomon:unitaryreflectiongroups}
(cf. \cite[\S 6.4]{orlikterao:arrangements}), we can find a subset $\Delta_Y \subseteq I$ and a partition $\Lambda_Y =
(\Lambda_1, \ldots, \Lambda_p)$ of $I\setminus\Delta_Y$ together with $r\th$ roots of unity $\theta_{ij}$ such that
$$Y = \bigcap\limits_{i \in \Delta_Y}H_i \cap \bigcap\limits_{i, j \in \Lambda_1}H_{ij}(\theta_{ij}) \cap \cdots
\cap \bigcap\limits_{i, j \in \Lambda_p}H_{ij}(\theta_{ij}),$$
where we agree that $H_{ii}(\theta) = \BBC^\ell$ for all $i$ and all $\theta$.
Consequently, in the following we
occasionally omit the intersections corresponding to $\Lambda_i$ with $\vert\Lambda_i\vert = 1$.

Now let $\tilde{\Lambda}_i := \Lambda_i \cap \Delta_X$ for $i = 1, \ldots, p$. If $\tilde{\Lambda}_i = \emptyset$ for all $i$,
then $Y \subseteq X$ and hence $X + Y = X \in L(\CA)$.
Thus we may assume that $\{i \mid \tilde{\Lambda}_i \neq \emptyset\} = \{1, \ldots, q\}$ for some $1 \leq q \leq p$. Then we get
$$X + Y = \bigcap\limits_{i \in \Delta_X \cap \Delta_Y}H_i \cap \bigcap\limits_{i, j \in \tilde{\Lambda}_1}
H_{ij}(\theta_{ij}) \cap \cdots \cap \bigcap\limits_{i, j \in \tilde{\Lambda}_q}H_{ij}(\theta_{ij}) \in L(\CA),$$
and we are done.
\end{proof}
\end{lemma}

\begin{lemma}
\label{lem:startinduction}
Let $\CA = \CA^k_3(r)$ with $1 \leq k \leq 3$ and $r \geq 2$ and let $X \in L(\CA)$ of dimension $1$.
Then  $X$ is
modular in $L(\CA)$ if and only if $X = H_i \cap H_j$ for some $1 \leq i < j \leq k$.

\begin{proof}
By Lemma \ref{lem:modularelements} every such element is modular, so in the following we assume that $X$ is not of this form. If
$X = H_i \cap H$ for some $H \in \CA$ and $k < i \leq 3$, then a simple argument shows that $X$ is not modular. Thus we may
assume that $X$ is either $X^\eta = H_1 \cap H_{23}(\eta)$ or $X_{\theta_2 \theta_3} = H_{12}(\theta_2) \cap H_{13}(\theta_3)$.
A simple calculation shows that $X^\eta + X_{\theta_2 \theta_3} = \ker((\theta_2\eta + \theta_3)x_1 - \eta x_2 - x_3)$ which is
not contained in $L(\CA)$ whenever $\theta_2\eta + \theta_3 \neq 0$ and hence proves the lemma.
\end{proof}
\end{lemma}

\begin{proposition}
\label{prop:boolsub}
Let $\ell \geq 3$, $0 \leq k \leq \ell$ and $r \geq 2$ and let $\CA = \CA^k_\ell(r) \not\cong \CA^0_3(2)$. Then $X \in L(\CA)$
is modular in $L(\CA)$ if and only if $X \in L(\CB^k_\ell)$ or $X \in \CA \cup \{0, \BBC^\ell\}$.

\begin{proof}
Let $X \in L(\CA)$ with $1 \leq \dim X = p \leq \ell - 2$, that is $X$ is none of the obviously modular elements, cf.\ \cite[Ex.\ 2.28]{orlikterao:arrangements}.
Using the construction in \cite[\S 2]{orliksolomon:unitaryreflectiongroups} (\cite[\S 6.4]{orlikterao:arrangements}),
we can find a subset $\Delta_X \subseteq I$ and a
partition $\Lambda_X = (\Lambda_1, \ldots, \Lambda_p)$ of $I\setminus\Delta_X$ together with $r\th$ roots of unity $\theta_j$
such that
$$X = \bigcap\limits_{j \in \Delta_X}H_j \cap \bigcap\limits_{j \in \Lambda_1\setminus\{m_1\}}H_{m_1j}(\theta_j)\cap
\cdots \cap \bigcap\limits_{j \in \Lambda_p\setminus\{m_p\}}H_{m_pj}(\theta_j),$$
for $m_i := \min\Lambda_i$ ($1 \leq i \leq p$). We may assume that $\vert\Lambda_1\vert \geq \ldots \geq \vert\Lambda_p\vert$.
It is easy to see that $X$ is not modular whenever $\Delta_X \not\subseteq \{1, \ldots, k\}$, so we assume $\Delta_X \subseteq
\{1, \ldots, k\}$. Then we have $\vert\Lambda_1\vert = \ldots = \vert\Lambda_p\vert = 1$ if and only if $X \in L(\CB^k_\ell)$.
By Lemma \ref{lem:modularelements}, the elements of $L(\CB^k_\ell)$ are modular. Thus we assume $\vert\Lambda_1\vert > 1$ and
show that $X$ is not modular. For $H \in \CA$, we know by Proposition \ref{prop:intermediate} that $\CA^H \cong
\CA^{k'}_{\ell - 1}(r)$ with $0 < k' \leq \ell - 1$. Using Lemmas \ref{lem:modularrestrictions},
\ref{lem:startinduction} and induction on $\ell$, it suffices to show that for $\ell \geq 4$ we can choose $H \in \CA$ such that
$X \cap H$ is not modular in $\CA^{k'}_{\ell - 1}(r)$. Now we discuss two cases:

\paragraph{\textbf{Case 1:}} The rank of $X$ is at least $3$. Here we have two subcases:

\subparagraph{\textbf{Case 1a:}}
If $\Delta_X \neq \emptyset$, then choose $i \in \Delta_X$ and set $H := H_i$. Considering Table
\ref{restriction-table}, we see
that in $L(\CA^H)$ the element $X \cap H = X$ is of the form
$$X = \bigcap\limits_{j \in \Delta_X\setminus\{i\}}H_j \cap \bigcap\limits_{j \in \Lambda_1\setminus\{m_1\}}H_{m_1j}(\theta_j)
\cap \cdots \cap \bigcap\limits_{j \in \Lambda_p\setminus\{m_p\}}H_{m_pj}(\theta_j).$$
We still have $\vert\Lambda_1\vert > 1$ and hence $X$ is not contained in the lattice of the Boolean subarrangement
$\CB^{k'}_{\ell - 1}$ of $\CA^H$. As $X$ is clearly neither zero nor a hyperplane in $H$ (nor $H$ itself), it is not
modular in $L(\CA^H)$ by induction hypothesis.

\subparagraph{\textbf{Case 1b:}} If $\Delta_X = \emptyset$, then we may assume $\vert\Lambda_1\vert \geq 3$ or
$\vert\Lambda_1\vert = \vert\Lambda_2\vert = 2$,
because $X$ is not contained in $\CA$. In both cases we set $H := H_{m_1i}(\theta_i)$ for some $i \in \Lambda_1
\setminus\{m_1\}$. Now using Table \ref{restriction-table}, we see that $X = X \cap H \in L(\CA^H)$ is of the form
$$X = \bigcap\limits_{j \in \Lambda_1\setminus\{i, m_1\}}H_{ij}(\theta_i\inverse\theta_j) \cap \cdots \cap \bigcap\limits_{j
\in \Lambda_p\setminus\{m_p\}}H_{m_pj}(\theta_j),$$
where in the case $\vert\Lambda_1\vert = 2$ the first intersection is empty.
Clearly, we have $\vert\Lambda_1\setminus\{i\}\vert > 1$ or $\vert\Lambda_2\vert > 1$ and hence $X$ is not contained in the
lattice of the
Boolean subarrangement $\CB^{k'}_{\ell - 1}$ of $\CA^H$. As $X$ is clearly neither zero nor a hyperplane in $H$ (nor $H$ itself),
it is not modular in $L(\CA^H)$ by induction hypothesis.

\paragraph{\textbf{Case 2:}} Now let $X$ be of rank $2$. We may assume that $X$ is either $X_a = H_1 \cap H_{23}(\theta)$, $X_b
= H_{12}(\theta_2) \cap
H_{13}(\theta_3)$ or $X_c = H_{12}(\theta_2) \cap H_{34}(\theta_4)$.
We set $H := H_{24}(\eta) \in \CA$ and set $\tilde{X}_\gamma := X_\gamma \cap H$ for $\gamma \in \{a, b, c\}$.
Clearly, $\tilde{X}_\gamma \in L(\CA^H)$ ($\gamma = a, b, c$) is none of the trivially modular elements.
Table \ref{restriction-table} shows that in $L(\CA^H)$ we get the following cases
$$\tilde{X}_a = H_1 \cap H_{34}(\theta\inverse\eta), \quad \tilde{X}_b = H_{14}(\theta_2\eta) \cap H_{13}(\theta_3)
\quad\text{and}\quad \tilde{X}_c = H_{14}(\theta_2\eta) \cap H_{34}(\theta_4).$$
Thus $\tilde{X}_\gamma$ ($\gamma = a, b, c$) is not contained in the lattice of the Boolean subarrangement $\CB^{k'}_{\ell - 1}$
of $\CA^H$ and hence it is not modular by induction hypothesis.
\end{proof}
\end{proposition}

We obtain the following immediate consequence of Proposition \ref{prop:boolsub}.

\begin{corollary}
\label{cor:modelem1}
Let $\CA = \CA(G(r,p,\ell))$ and $X \in L(\CA)$. Then $\CA^X$ is supersolvable if and only if $L(\CA^X)$ contains a modular
element of dimension $1$.
\end{corollary}

\begin{proof}
It follows from Lemma \ref{lem:intermediates} and Proposition \ref{prop:boolsub} that  $\CA^k_\ell(r)$ is supersolvable
if and only if there is a modular element of dimension $1$ in $L(\CA^k_\ell(r))$. Now the proofs of
\cite[Prop.~2.5]{orliksolomon:unitaryreflectiongroups} and \cite[Prop.~2.14]{orliksolomon:unitaryreflectiongroups}
(see also \cite[Prop.~6.77 and 6.84]{orlikterao:arrangements}) imply the assertion.
\end{proof}

We record another consequence of  Proposition \ref{prop:boolsub} which together with \cite[Prop.~2.14]{orliksolomon:unitaryreflectiongroups}
(\cite[Prop.~6.84]{orlikterao:arrangements}) gives an alternative proof of Theorem \ref{super-restriction}(ii).

\begin{corollary}

$\CA^k_\ell(r)$ is supersolvable if and only if $k \in \{\ell - 1, \ell\}$ or $(k, \ell, r) = (0, 3, 2)$.

\end{corollary}

In the remainder of this section we
consider the case when $W$ is of exceptional type.
We start with a further consequence of Lemma \ref{lem:dim3}.

\begin{corollary}
\label{cor:4dim1mod}
Let $\CA = \CA(W)$ be an irreducible, non-supersolvable
reflection arrangement of exceptional type.
Let $X \in L(\CA)$ with $\dim X = 4$.
Then $\CA^X$ is supersolvable if and only if
$L(\CA^X)$
contains a modular element of dimension $1$.
\end{corollary}

\begin{proof}
Thanks to Lemma \ref{lem:dim3}, for every $W$ as in the statement,
other than $W$ of type $E_6$, $E_7$ or $E_8$,
every $3$-dimensional restriction already fails to be supersolvable.
Consequently, using Lemma \ref{lem:super-rank2},
there are no modular elements of dimension $1$ in $L(\CA)$.

We are left to check the $4$-dimensional restrictions
for  $W$ of type $E_6$, $E_7$ and $E_8$.

Using \cite[App.]{orliksolomon:unitaryreflectiongroups} (cf. \cite[App.\ C]{orlikterao:arrangements}), it is easily seen that
each of
$(E_7,(A_1^3)')$, $(E_8,A_1^4)$, $(E_8,A_1^2A_2)$, and $(E_8,D_4)$ has no
supersolvable restriction to a hyperplane
and therefore none of them admits a modular element of dimension $1$,
by Lemmas \ref{lem:super-rank2} and \ref{cor:stanley}.

The next set of instances we consider are ones which
have
only one type of restriction to a hyperplane which is supersolvable.
We can then use the condition of Lemma \ref{lem:show4nonmodular1} to show
that none of these cases admits a modular element of dimension $1$.
We use the tables of \cite[App.]{orliksolomon:unitaryreflectiongroups}
(\cite[App.\ C]{orlikterao:arrangements}) to determine the number of restrictions
which are supersolvable and denote this number again by $c$, as in Lemma
\ref{lem:show4nonmodular1}:\\
$(E_6,A_1^2)$: the supersolvable restriction has exponents $\{1,3,4\}$ and $c=3$.\\
$(E_7,(A_1^3)'')$: the supersolvable restriction has exponents $\{1,3,5\}$ and $c=1$.\\
$(E_7,(A_1A_2)')$: the supersolvable restriction has exponents $\{1,5,7\}$ and $c=4$.\\
$(E_7,A_3)$: the supersolvable restriction has exponents $\{1,3,5\}$and $c=3$.\\
$(E_8,A_2^2)$: the supersolvable restriction has exponents $\{1,5,7\}$ and $c=6$.\\
$(E_8,A_1A_3)$: the supersolvable restriction has exponents $\{1,5,7\}$ and $c=4$.\\

Unfortunately, Lemma \ref{lem:show4nonmodular1} does not apply for
$(E_6,A_2)$ and $(E_8,A_4)$, since $c$ is too big in these
instances. Using a computer, we checked directly
that there are no modular elements of dimension $1$ in these two cases
(cf.\ Remark \ref{rem:computations}).
\end{proof}

\begin{corollary}
\label{cor:dim1mod}
Let $\CA = \CA(W)$ be an irreducible, non-supersolvable
reflection arrangement of exceptional type.
Let $X \in L(\CA)$ with $\dim X \ge 3$. Then $\CA^X$ is supersolvable if and only if $L(\CA^X)$ contains a modular
element of dimension $1$.
\end{corollary}
\begin{proof}
For $\dim X = 3$,
the assertion follows from
Lemma \ref{lem:super-rank2} and Corollary \ref{cor:4dim1mod} shows that
this also holds for $\dim X = 4$. Since all restrictions $\CA^X$ with $\dim X\ge 4$
are not supersolvable for the exceptional types, the result follows for $\dim X>4$
by Lemma \ref{lem:modularrestrictions}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{super-restriction-rank2}]
The forward implication is clear from Definition \ref{def:super}.

The reverse implication follows
for general central arrangements of rank up to $3$
by \cite[Rem.\ 2.3]{hogeroehrle:super} and Lemma \ref{lem:super-rank2}.
For $W = G(r, p, \ell)$,
the result is  Corollary \ref{cor:modelem1}.
Finally, for $W$ of exceptional type and
$X \in L(\CA)$ with $\dim X \geq 4$
such that $\CA^X$ is not supersolvable,
the result follows from
Corollary \ref{cor:dim1mod}.

Note that
Corollary \ref{cor:modelem1} and
Corollary \ref{cor:dim1mod} also
apply in the case $X = V$.
This gives the equivalence in the statement for the underlying
reflection arrangement $\CA(W)$ itself
(cf.\ Remark \ref{rem:super-restriction-rank2}(i)).
\end{proof}

\begin{remark}
\label{rem:computations}
In order
to establish several of our results
we first use the functionality for complex reflection groups
provided by the   \CHEVIE\ package in   \GAP\
(and some \GAP\ code by J.~Michel)
(see \cite{gap3} and \cite{chevie})
in order to obtain explicit
linear functionals $\alpha$ defining the hyperplanes
$H = \ker \alpha$ of the reflection arrangement
$\CA(W)$ and the relevant restrictions $\CA(W)^X$.

We then use the functionality of
\Sage\  (\cite{sage}) to determine the data in
Table \ref{table:rank3restrictions} and in the proof of Lemma \ref{lem:dim3}.
Moreover, \Sage\ was used
to establish the fact that
the two restrictions
$(E_6, A_2)$ and $(E_8, A_4)$
do not admit modular elements of dimension $1$
in the proof of Corollary \ref{cor:4dim1mod}.
\end{remark}

\bigskip {\bf Acknowledgments}:
We acknowledge
support from the DFG-priority program
SPP1489 ``Algorithmic and Experimental Methods in
Algebra, Geometry, and Number Theory''.

\bigskip

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redpajama_set_name: RedPajamaArXiv


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text:
Last modified 340 days ago (Feb. 14, 2019)
Soil contamination found at transfer station
By PHYLLIS ZORN
Contaminated soil was discovered last week during excavation work at the county's transfer station project.
Refuse director Bud Druse alerted county commissioners to the contamination during Monday's county commission meeting.
Leon Osbourn, project engineer for Kaw Valley Engineering in Junction City, said it appears a small area of diesel fuel was dumped or spilled adjacent to where a new foundation will soon be laid.
The transfer station building was originally a Marion electric plant operated by diesel fuel.
"It looked like maybe some diesel fuel got dumped," Osbourn said. "There might have been an old diesel storage tank there, we don't know. Somebody could have lost a five gallon drum of diesel."
Osbourn said a soil sample has been sent to a laboratory in Kansas City for testing. Test results should be known in a couple weeks.
Although Osbourn doesn't yet know the quantity of dirt with contamination, he said it appears to be a small area.
The size of the contaminated area dictates what will need to be done to remediate the contamination.
Osbourn said another surprise found during excavation is old concrete foundations.
Despite the discoveries, work is continuing at the site.
In other matters at the commission meeting:
Commissioners discussed the need to create a job description for a county engineer to work with the road and bridge department, then tabled the issue to a later meeting.
Discussed the possibility of an agreement with Bergkamp Construction, which is rebuilding a road nearby and has offered to sell the county crushed concrete removed from the job. Road and Bridge supervisor Jesse Hamm said there is no guarantee the crushed concrete will not contain metal.
Talked about trucks working for Diamond Vista wind farm project in the northern section of the county driving on roads not designated as haul routes despite an agreement with the county that trucks are to remain on designated roadways. Hamm said he has contacted the developer several times about the issue and often gets no response from them.
Last modified Feb. 14, 2019

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