Homotopy types of complements of 2-arrangements in R^4

LaTeX2e, 25 pages with 5 figures. Revised version, to appear in
  Topology

Abstract:
  We study the homotopy types of complements of arrangements of n transverse
planes in R^4, obtaining a complete classification for n <= 6, and lower bounds
for the number of homotopy types in general. Furthermore, we show that the
homotopy type of a 2-arrangement in R^4 is not determined by the cohomology
ring, thereby answering a question of Ziegler. The invariants that we use are
derived from the characteristic varieties of the complement. The nature of
these varieties illustrates the difference between real and complex
arrangements.

\section{Introduction}
\label{sec:intro}

In~\cite{GM}, Goresky and MacPherson introduced a generalization of the
notion of complex hyperplane arrangement. A {\em $2$-arrangement}
in $\mathbb{R}^{2d}$ is a finite collection $\mathcal{A}$ of codimension~$2$ linear
subspaces so that, for every subset $\mathcal{B}\subseteq \mathcal{A}$, the space
$\bigcap_{H\in \mathcal{B}}\, H$ has even dimension.
The main object of study is the
complement of the arrangement, $X(\mathcal{A})=\mathbb{R}^{2d}\setminus \bigcup_{H\in \mathcal{A}}\, H$.
Goresky and MacPherson computed the cohomology groups of $X$.  Bj\"{o}rner
and Ziegler \cite{BZ} and Ziegler \cite{Z} determined the structure of the
cohomology algebra $H^*(X;\mathbb{Z})$. These results generalize the classical work
of Arnol'd, Brieskorn, and Orlik and Solomon on the cohomology ring of the
complement of a complex hyperplane arrangement, see \cite{OT}.
Unlike the situation obtaining for the Orlik-Solomon algebra, which is
completely determined by the intersection lattice,
there remained an ambiguity in the relations defining $H^*(X;\mathbb{Z})$.
Even in the simplest case of $2$-arrangements in $\mathbb{R}^4$,
a striking phenomenon occurs, showing that this ambiguity
cannot be resolved,~\cite{Z}.

Each $2$-arrangement $\mathcal{A}$ in $\mathbb{R}^4$ is a realization of the uniform
matroid $U_{2,n}$, where $n=|\mathcal{A}|$ is the cardinality of the arrangement.
Thus, the intersection lattice of such an arrangement is uniquely
determined by $n$.  Furthermore, the homology groups of the complement,
$X$, the lower central series quotients of the group $G=\pi_1(X)$,
and the Chen groups of $G$ also depend only on $n$.

On the other hand, the cohomology ring of $X$ is a more subtle invariant.
The relations in $H^*(X;\mathbb{Z})$ depend on, and are determined by the
linking numbers of the associated link.  Ziegler ~\cite{Z} found a pair
of $2$-arrangements of four planes which have non-isomorphic
cohomology rings. His method, which uses an invariant derived
from $H^*(X;\mathbb{Z})$, does not seem, however, to extend beyond $n=4$.

In this paper, we introduce new homotopy-type invariants
of complements of $2$-arrangements.  These invariants, derived
from the Alexander module, work for arbitrary $n$.  As a first step
towards the homotopy classification of $2$-arrangements, we prove
the following (see Corollary~\ref{cor:depth2}).

\begin{thm}  \label{thm:lowerbound}
For every integer $n\ge 1$, there exist at least
$p(n-1) - \lfloor\frac{n-1}{2}\rfloor$
different homotopy types of complements of
$2$-arrangements of $n$~planes in $\mathbb{R}^4$,
where $p( \cdot)$ is the partition
function, and $\lfloor\cdot\rfloor$
is the integer part function.
\end{thm}

At the end of \cite{Z}, Ziegler asks whether the cohomology
ring determines the homotopy type of the complement of a $2$-arrangement,
proposing as a candidate for a negative answer the remarkable pair
of arrangements of $6$ planes found by Mazurovski\u{\i} in \cite{M1}.
Using successive cablings on Mazurovski\u{\i}'s pair, we answer Ziegler's
question, as follows (see Theorem~\ref{thm:mazcables}).

\begin{thm}  \label{thm:homotopytypes}
For every integer $n\ge 6$, there exists a pair of $2$-arrangements
of $n$ planes in $\mathbb{R}^4$, whose complements have isomorphic cohomology rings,
but different homotopy types.
\end{thm}

Rigid isotopy of arrangements implies isotopy of their singularity
links.  The converse is not clear, though, since an isotopy may go outside
the class of such links.  On the other hand, the classification,
up to rigid isotopy, of $2$-arrangements in $\mathbb{R}^4$  is equivalent
to the classification, also up to rigid isotopy, of configurations
of skew lines in $\mathbb{R}^3$.  Such configurations were introduced by
Viro in \cite{V}, and have been intensively studied  since then,
see the survey article by Crapo and Penne~\cite{CP}.  The rigid
isotopy classification of configurations of $n$ skew lines in $\mathbb{R}^3$,
was achieved by Viro~\cite{V} for $n\le 5$, and by
Mazurovski\u{\i}~\cite{M1} for $n=6$.

It is readily seen that rigid isotopy of arrangements implies homotopy
equivalence of their complements.  The converse is not true.  Indeed,
as first noted by Viro, there exist configurations that
are not isotopic to their mirror image.  But clearly, the complements
of mirror pairs are diffeomorphic, and thus homotopy equivalent.
The next result shows that this is the only exception, for $n\le 6$
(see Theorem~\ref{thm:classify6}).

\begin{thm}  \label{thm:lowclassification}
For $2$-arrangements of $n\le 6$ planes in $\mathbb{R}^4$,
the homotopy types of complements are in one-to-one
correspondence with the rigid isotopy types modulo mirror images.
\end{thm}

This theorem recovers Ziegler's classification
of homotopy types of arrangements of $n=4$ planes.
The number of homotopy types from the classification in
Theorem~\ref{thm:lowclassification}, together with the lower bound
from Theorem~\ref{thm:lowerbound}, are tabulated
below.\footnote[2]{Recently, Borobia and Mazurovski\u{\i}~\cite{BM} achieved
the rigid isotopy classification of configurations of $7$ lines.
If the assertion of Theorem~\ref{thm:lowclassification} were
to hold for $n=7$, it would give $37$ distinct homotopy types
of complements of arrangements of $7$ planes.  We have verified
this in the particular case of horizontal arrangements, for which there
are $24$ distinct homotopy types.}

\[
\begin{array}{|l||c|c|c|c|c|c|c|}
\hline
\multicolumn{1}{|c||}{n}
&\multicolumn{1}{|c|}{1}
&\multicolumn{1}{|c|}{2}
&\multicolumn{1}{|c|}{3}
&\multicolumn{1}{|c|}{4}
&\multicolumn{1}{|c|}{5}
&\multicolumn{1}{|c|}{6}
&\multicolumn{1}{|c|}{7}
\\
\hline\hline
\text{Homotopy types}
& 1 & 1 & 1 & 2 & 4 & 11 & ?\\
\hline
\text{Lower bound}
& 1 & 1 & 1 & 2 & 3 & 5 & 8 \\
\hline
\end{array}
\]

The homotopy-type invariants that we use in our classification
of $2$-arrangements are derived from the characteristic varieties of
their complements.  Given a space $X$ with $H_1(X)\cong \mathbb{Z}^n$,
the $k^{\text{th}}$ determinantal ideal of the Alexander module
of $X$ defines a subvariety, $V_k(X)$, of the complex algebraic
torus, $(\mathbb{C}^*)^n$, whose monomial isomorphism type depends
only on the homotopy type of $X$---in fact, only on
$\pi_1(X)$---see~\cite{DF, L1}.
We call $V_k(X)$ the $k^{\text{th}}$ characteristic variety of $X$.
From this variety, we extract in Theorem~\ref{thm:invariant}
the following homotopy-type invariants for the space $X$:
the list $\Sigma_k(X)$ of codimensions of irreducible components,
and the number $\Tors_{p,k}(X)$ of $p$-torsion points.
These numerical invariants are readily computable by standard
methods of geometric topology and commutative algebra, and are
powerful enough to detect all the differences in homotopy
types listed in the above theorems.

The characteristic varieties of complements of divisors in complex
algebraic manifolds have been intensively studied recently,
see~\cite{Ar, L2, Hr, CScv, L3, LY}.
Deep results as to their qualitative nature have
been obtained by Arapura \cite{Ar}, who showed that all the
irreducible components of such characteristic varieties are
(possibly translated) subtori of a complex algebraic torus.
Building on this work, a more precise description of the
characteristic varieties of complex hyperplane arrangements
has emerged.  In all known examples, if $X$ is the complement
of such an arrangement, all positive-dimensional subtori of
$V_k(X)$ pass through the origin $\mathbf{1}$ of the torus.

On the other hand, if $X$ is the complement of a $2$-arrangement
in $\mathbb{R}^4$, we find that the characteristic varieties of $X$ may
contain positive-dimensional subtori that do not pass through~$\mathbf{1}$.
For the non-complex Ziegler arrangement, the variety $V_2$ contains three
subtori of $(\mathbb{C}^*)^4$, one of which is translated by $(1,-1,1,1)$,
see Example~\ref{ex:Zieglercharvar}.  But this is still a rather mild
qualitative difference.  For the indecomposable Mazurovski\u{\i}
arrangements, the variety $V_1$ is not even a union of translated
subtori, see \S\ref{sec:maz}.  These phenomena may be thought
of as manifestations of the non-complex nature of real arrangements.

The paper is organized as follows.

In \S\ref{sec:arrconlin}, we review the basic facts about
$2$-arrangements in $\mathbb{R}^4$, and their associated configurations of
lines and singularity links.
In \S\ref{sec:decomphoriz},
we look in detail at some special classes of arrangements:
the decomposable ones, and the horizontal ones.
In \S\ref{sec:braids}, we associate several braids to a
$2$-arrangement, and use these braids to compute the fundamental
group of the complement.
In \S\ref{sec:detideals}, we review Alexander modules and
define numerical homotopy-type invariants from the associated
characteristic varieties.
In \S\ref{sec:bottomvar}, we study the bottom
characteristic varieties $V_{n-2}$, obtaining a complete characterization
for depth~$2$, completely decomposable arrangements.
In \S\ref{sec:topvar}, we study the top characteristic varieties $V_1$,
and their torsion points.
In \S\ref{sec:maz}, we study in detail the Mazurovski\u{\i}
arrangements, and their cablings.
Using the results and techniques from
\S\S\ref{sec:bottomvar}--\ref{sec:maz}, we complete the homotopy-type
classification of $2$-arrangements of $6$ planes or less in
\S\ref{sec:classification}.

\medskip
\noindent
{\bfseries Acknowledgment.}   This work started from an illuminating
conversation with G\"{u}nter Ziegler, who introduced us to~\cite{Z}.
The computational part of the work was greatly aided by
{\it Mathematica${^{\circledR}}$}, and by the commutative
algebra package {\it Macaulay~2}.  Thanks are due to the referee,
for many valuable suggestions that have improved both the substance
and the style of the paper.

\section{Arrangements, Line Configurations, and Links}
\label{sec:arrconlin}

In this section we collect some facts about arrangements of transverse
planes in $\mathbb{R}^4$, and the corresponding configurations of skew lines
in $\mathbb{R}^3$ and links in $\mathbb{S}^3$.

\subsection{}  \label{defpoly}
We start by defining our basic objects of study in a concrete way.

\begin{defn}  \label{def:2arr}
A {\em $2$-arrangement} in $\mathbb{R}^4$ is a finite collection
$\mathcal{A}=\{H_1,\dots ,H_n\}$ of pairwise transverse $2$-dimensional
vector subspaces of $\mathbb{R}^{4}$.  The {\em union} of the arrangement
is $U(\mathcal{A})=\bigcup_{H\in \mathcal{A}} H$.  The {\em complement} of the arrangement
is $X(\mathcal{A})=\mathbb{R}^4\setminus U(\mathcal{A})$.  The {\em link} of the arrangement is
$L(\mathcal{A})=\mathbb{S}^{3}\cap U(\mathcal{A})$.
\end{defn}

Each plane $H_i\in \mathcal{A}$ can be written as
$H_i=\ker\lambda_i \cap \ker\lambda'_i$, for some linear forms
$\lambda_i, \lambda'_i:\mathbb{R}^4\to \mathbb{R}$.
The transversality condition means that $H_i\cap H_j=\{0\}$,
for all $i\ne j$.  That is,
$\det(\l_i,\l'_i,\l_j,\l'_j)\ne 0$ for $i\ne j$.

Alternatively, identifying $\mathbb{R}^4$ with $\mathbb{C}^2=\{(z,w)\}$,
each plane in $\mathcal{A}$ can be written as $H_i= \{f_i=0\}$, where
$f_i(z,w)=a_iz+b_i\bar{z}+c_iw+d_i\bar{w}$,
for some $a_i,b_i,c_i,d_i\in\mathbb{C}$.
In terms of real coordinates
$x=\operatorname{Re} z$, $y=\operatorname{Im} z$, $u=\operatorname{Re} w$, $v=\operatorname{Im} w$, we have
$\lambda_i(x,y,u,v)=\operatorname{Re} f_i(x+\ii y, u+\ii v)$ and
$\lambda'_i(x,y,u,v)=\operatorname{Im} f_i(x+\ii y, u+\ii v)$, where $\ii=\sqrt{-1}$.

With notation as above, let $f:\mathbb{C}^2\to \mathbb{C}$ be the polynomial
map in $z, \bar{z}, w, \bar{w}$ given by $f=f_1\cdots f_n$.  We say
that $f$ is a {\em defining polynomial} for the arrangement $\mathcal{A}$.
Obviously, the union of the arrangement is the zero locus of the
defining polynomial.

\begin{exmp}  \label{complexlines}
The most basic example of a $2$-arrangement is a {\em complex} arrangement.
Such an arrangement consists of complex lines through the origin of $\mathbb{C}^2$.
 Any two complex arrangements differ by an
$\mathbb{R}$-linear change of variables, and thus have diffeomorphic complements.
We denote the complex arrangement of $n$ lines by $\mathcal{A}_n$,
and take its defining polynomial to be $f_n(z,w)=(z-w)\cdots (z-nw)$.
The link $L(\mathcal{A}_n)$ is the $n$-component Hopf link.
The {\em trivial} arrangement is $\mathcal{A}_1$.
\end{exmp}

\subsection{}  \label{singlink}
Let $\mathcal{A}=\{H_1,\dots ,H_n\}$ be a $2$-arrangement in $\mathbb{R}^4$.
Its link, $L=\{L_1,\dots , L_n\}$, consists of $n$ unknotted circles
in $\mathbb{S}^3$.  The complement of the arrangement, $X(\mathcal{A})$, is homotopy
equivalent to the complement of the link, $Y(L)=\mathbb{S}^{3}\setminus L$,
via radial deformation.  Using this observation, we can compute
homotopy-type invariants of $X=X(\mathcal{A})$ by methods of knot theory.

The homology groups of $X$ depend only on the number
of planes in the arrangement:  $H_0=\mathbb{Z}$, $H_1=\mathbb{Z}^n$, $H_2=\mathbb{Z}^{n-1}$,
$H_k=0$ for $k>2$.  The cohomology ring of $X$, on the other hand,
also depends on the linking numbers $l_{i,j}=\lk(L_i,L_j)$.
Specifically,
\begin{equation*}  \label{eq:cohomology}
H^{*}(X; \mathbb{Z}) = \left.\sideset{}{^*}\bigwedge\mathbb{Z}^n \right\slash \left(
l_{ij}e_ie_j + l_{jk}e_je_k + l_{ki}e_ke_i = 0 \right),
\end{equation*}
where $\bigwedge^{*} \mathbb{Z}^n$ is the exterior algebra on $e_1, \dots , e_n$.
As noted by Ziegler~\cite{Z}, one can compute the linking numbers of
$L(\mathcal{A})$ directly from the defining equations of $\mathcal{A}$.  Indeed, if
$H_i=\{\lambda_i=\lambda'_i=0\}$, then
$l_{i,j}=\sgn(\det(\lambda_i,\lambda'_i,\lambda_j,\lambda'_j))$.

As shown by Ziegler~\cite{Z}, the complement $X$ fibers over
$\mathbb{C}^*=\mathbb{C}\setminus \{0\}$, with fiber $\mathbb{C}\setminus \{n-1\ \text{points}\}$,
and thus $X$ is a $K(G,1)$ space.  Alternatively, since all the linking
numbers are non-zero, the link $L$ is non-split, and thus $Y(L)$
is aspherical, see~\cite{BuZ}.  It follows that the homotopy type of $X$
is determined by the isomorphism class of its fundamental group $G$.

As we shall see in Proposition~\ref{prop:bundle}, the monodromy
of the bundle $X\to \mathbb{C}^*$ is a certain (pure) braid automorphism
$\check{\beta}\in P_{n-1}$, and so $G$ is a semidirect product of
free groups, $G=\mathbb{F}_{n-1}\rtimes_{\check{\beta}} \mathbb{F}_1$.  Since
$\check{\beta}$ acts trivially on homology, a result of Falk and
Randell~\cite{FR} implies that the lower central series quotients
of $G$ depend only on $n$, being equal to those of the product
$\Gamma=\mathbb{F}_{n-1}\times \mathbb{F}_1$.  In fact, since all the linking
numbers of $L$ are equal to $\pm 1$, a result of Massey and Traldi~
~\cite{MT} shows that the lower central series quotients of both
$G$ and $G/G''$ are equal to the corresponding quotients of
$\Gamma$ and $\Gamma/\Gamma''$.

\subsection{}  \label{config}
Now let $H$ be an affine hyperplane in $\mathbb{R}^4$, generic with respect
to $\mathcal{A}$.  The {\em configuration} of $\mathcal{A}$ corresponding to $H$ is
the configuration of skew lines in $\mathbb{R}^{3}$ defined as
$\mathcal{C}_{H}(\mathcal{A})=H\cap U(\mathcal{A})$.  Conversely, given a configuration $\mathcal{C}$
of skew lines, one obtains a $2$-arrangement, $\mathcal{A}_p=\mathcal{A}(\mathcal{C})$, by
coning at a generic point $p$, and translating $p$ to $0$.

\begin{exmp} \label{ex:ziegler}
Let $\mathcal{A}^{+}$ and $\mathcal{A}^{-}$ be the pair of $2$-arrangements considered
by Ziegler in~\cite{Z}.  The arrangement $\mathcal{A}^{+}$ is the complex
arrangement $\mathcal{A}_4$, with defining polynomial
$f^{+}(z,w)=(z-w)(z-2w)(z-3w)(z-4w)$.
The arrangement $\mathcal{A}^{-}$ has defining polynomial
$f^{-}(z,w)=(z-\bar{w})(z-2\bar{w})(z-3w)(z-4w)$.
Projecting onto the hyperplane $\{v=1\}$, we get configurations
$\mathcal{C}^{\pm}=\{\ell_1^{\pm},\ell_2^{\pm},\ell_3,\ell_4\}$,
with equations
\begin{alignat*}{2}  \label{eq:4lines}
\ell_1^{\pm}&=\{x-u=y\mp 1=0\},
&\qquad \ell_2^{\pm}&=\{x-2u=y\mp 2=0\},\\
\ell_3&=\{x-3u=y-3=0\},
&\qquad \ell_4&=\{x-4u=y-4=0\}.
\end{alignat*}
\end{exmp}
\noindent
The two configurations are pictured in Figure~\ref{fig:zpair}.

\begin{figure}
\centerline{\hfill\hskip -1.28 in
\begingraph<0.5cm,0.5cm>[0.35pt]
\Label {$1$} at (1, 1)
\Label {$2$} at (1, 2.5)
\Label {$3$} at (2.55, 4.6)
\Label {$4$} at (3.85, 4.7)
\edge from (4, 4) to (4.5, 0.5)
\edge from (3, 4) to (3.9, 3.1)
\edge from (4.4, 2.6) to (6.5, 0.5)
\edge from (2, 2.5) to (4, 2.125)
\edge from (4.5, 2.03125) to (4.8, 1.975)
\edge from (5.4, 1.8625) to (10, 1)
\edge from (2, 1) to (4.1, 1.13125)
\edge from (4.7, 1.16875) to (5.5, 1.21875)
\edge from (6, 1.25) to (7.4, 1.3375)
\edge from (8.4, 1.4) to (10, 1.5)
\Label {$1$} at (13, 1)
\Label {$2$} at (13, 2.5)
\Label {$3$} at (14.55, 4.6)
\Label {$4$} at (15.85, 4.7)
\edge from (16, 4) to (16.5, 0.5)
\edge from (15, 4) to (15.9, 3.1)
\edge from (16.4, 2.6) to (18.5, 0.5)
\edge from (14, 2.5) to (16, 2.125)
\edge from (16.5, 2.03125) to (16.8, 1.975)
\edge from (17.4, 1.8625) to (19.3, 1.50625)
\edge from (20.5, 1.28125) to (22, 1)
\edge from (14, 1) to (16.1, 1.13125)
\edge from (16.7, 1.16875) to (17.5, 1.21875)
\edge from (18, 1.25) to (22, 1.5)
\endgraph
\hfill}
\caption{Ziegler's pair:
The configurations $\mathcal{C}^{+}$ (left) and $\mathcal{C}^{-}$ (right).}
\label{fig:zpair}
\end{figure}

\begin{figure}
\centerline{\hfill  \hskip -0.1 in
\begingraph<0.5cm,0.5cm>[0.35pt]
\Label {$1$} at (1, 1)
\Label {$2$} at (1, 2)
\Label {$3$} at (1, 3)
\Label {$4$} at (1, 4)
\edge from (2, 1) to (3, 1)
\edge from (2, 2) to (3, 2)
\edge from (2, 3) to (3, 3)
\edge from (2, 4) to (3, 4)
\edge from (3, 1) to (4, 1)
\edge from (3, 2) to (4, 2)
\edge from (3, 3) to (3.3, 3.3)
\edge from (3.7, 3.7) to (4, 4)
\edge from (3, 4) to (4, 3)
\edge from (4, 1) to (5, 1)
\edge from (4, 2) to (4.3, 2.3)
\edge from (4.7, 2.7) to (5, 3)
\edge from (4, 3) to (5, 2)
\edge from (4, 4) to (5, 4)
\edge from (5, 1) to (5.3, 1.3)
\edge from (5.7, 1.7) to (6, 2)
\edge from (5, 2) to (6, 1)
\edge from (5, 3) to (6, 3)
\edge from (5, 4) to (6, 4)
\edge from (6, 1) to (7, 1)
\edge from (6, 2) to (7, 2)
\edge from (6, 3) to (6.3, 3.3)
\edge from (6.7, 3.7) to (7, 4)
\edge from (6, 4) to (7, 3)
\edge from (7, 1) to (8, 1)
\edge from (7, 2) to (7.3, 2.3)
\edge from (7.7, 2.7) to (8, 3)
\edge from (7, 3) to (8, 2)
\edge from (7, 4) to (8, 4)
\edge from (8, 1) to (9, 1)
\edge from (8, 2) to (9, 2)
\edge from (8, 3) to (8.3, 3.3)
\edge from (8.7, 3.7) to (9, 4)
\edge from (8, 4) to (9, 3)
\edge from (9, 1) to (10, 1)
\edge from (9, 2) to (10, 2)
\edge from (9, 3) to (10, 3)
\edge from (9, 4) to (10, 4)
\Label {$1$} at (13, 1)
\Label {$2$} at (13, 2)
\Label {$3$} at (13, 3)
\Label {$4$} at (13, 4)
\edge from (14, 1) to (15, 1)
\edge from (14, 2) to (15, 2)
\edge from (14, 3) to (15, 3)
\edge from (14, 4) to (15, 4)
\edge from (15, 1) to (16, 1)
\edge from (15, 2) to (16, 2)
\edge from (15, 3) to (15.3, 3.3)
\edge from (15.7, 3.7) to (16, 4)
\edge from (15, 4) to (16, 3)
\edge from (16, 1) to (17, 1)
\edge from (16, 2) to (16.3, 2.3)
\edge from (16.7, 2.7) to (17, 3)
\edge from (16, 3) to (17, 2)
\edge from (16, 4) to (17, 4)
\edge from (17, 1) to (17.3, 1.3)
\edge from (17.7, 1.7) to (18, 2)
\edge from (17, 2) to (18, 1)
\edge from (17, 3) to (18, 3)
\edge from (17, 4) to (18, 4)
\edge from (18, 1) to (19, 1)
\edge from (18, 2) to (19, 2)
\edge from (18, 3) to (18.3, 3.3)
\edge from (18.7, 3.7) to (19, 4)
\edge from (18, 4) to (19, 3)
\edge from (19, 1) to (20, 1)
\edge from (19, 2) to (19.3, 2.3)
\edge from (19.7, 2.7) to (20, 3)
\edge from (19, 3) to (20, 2)
\edge from (19, 4) to (20, 4)
\edge from (20, 1) to (21, 1)
\edge from (20, 2) to (21, 2)
\edge from (20, 3) to (21, 4)
\edge from (20, 4) to (20.3, 3.7)
\edge from (20.7, 3.3) to (21, 3)
\edge from (21, 1) to (22, 1)
\edge from (21, 2) to (22, 2)
\edge from (21, 3) to (22, 3)
\edge from (21, 4) to (22, 4)
\endgraph
\hfill}
\caption{Ziegler's pair:
The half-braids $\a^{+}$ (left) and $\a^{-}$ (right).}
\label{fig:zbraids}
\end{figure}

\subsection{} \label{rigidisotop}
Finally, let us consider the natural isotopy relation between arrangements,
modeled on the similar notion for configurations.

\begin{defn}  \label{def:isotopic}
Two arrangements $\mathcal{A}$ and $\mathcal{A}'$ are called {\em rigidly isotopic} if there is
an isotopy of $\mathbb{R}^{4}$ connecting $\mathcal{A}$ to $\mathcal{A}'$ through arrangements.
\end{defn}

The rigid isotopy class of $\mathcal{C}_H(\mathcal{A})$ does not
depend on $H$, and the rigid isotopy class of $\mathcal{A}_p(\mathcal{C})$ does not
depend on $p$.  Therefore, we will denote them simply by $\mathcal{C}(\mathcal{A})$ and
$\mathcal{A}(\mathcal{C})$, respectively.  Moreover, rigid isotopy classes of configurations
are in one-to-one correspondence with rigid isotopy classes of $2$-arrangements.
See Crapo and Penne~\cite{CP} for details and references.

\begin{rem} \label{rem:topmove}
Given an arrangement, we can deform it by means of a rigid isotopy
so that one of the planes has linking number $+1$ will all other planes.
The analogous procedure for bringing one of the lines of a configuration
on top of all others is explained in Penne~\cite{P}.
\end{rem}

\section{Decomposable and Horizontal Arrangements}
\label{sec:decomphoriz}

In this section we look at arrangements that can
be obtained by a sequence of cabling operations from simpler
arrangements, and also at arrangements whose corresponding
configurations are ``horizontal".  We consider in more
detail the subclass of completely decomposable arrangements,
and obtain a normal form for those of depth~$2$.

\subsection{}
\label{cables}
Let us start by recalling the following notion from knot theory
(see~\cite{BZ}).  Let $L=L_1 \cup \cdots \cup L_n$ be a link in $\mathbb{S}^3$.
The $(a,b)$-cable of $L$ about the $k^{\text{th}}$ component
is the link $L\{a,b\}=L\cup K(a,b)$, where $K(a,b)$ is an
$(a,b)$-torus link contained in the boundary
of a tubular neighborhood of $L_k$.

Now let $\mathcal{A}$ be a $2$-arrangement of $n$ planes in $\mathbb{R}^4$, with
defining polynomial $f=f_1\cdots f_n$.  Fix an index $1\le k\le n$,
a positive integer $r$, and a number $\epsilon=\pm 1$.  Given these
data, we define the {\em $\epsilon r$-cable} about the $k^{\text{th}}$
component of $\mathcal{A}$ to be the arrangement $\mathcal{A}^{k}\{\epsilon r\}$
with defining polynomial
\begin{equation*}  \label{def:cabling}
f(f_k+g_1)\cdots (f_k+g_r),
\end{equation*}
where each $g_j$ is a linear form in $z, \bar{z}, w, \bar{w}$, whose
coefficients are sufficiently small with respect to those of $f$, and
such that $\sgn(\det(f_k,g_j))=\epsilon$, for $j=1,\dots, r$.

The cabling operation is well-defined up to rigid isotopy of arrangements.
The reverse operation is called decabling.  It is readily seen that the
link of $\mathcal{A}^{k}\{\pm r\}$ is the $(r, \pm r)$-cable about
the $k^{\text{th}}$ component of $L(\mathcal{A})$.

\begin{defn}  \label{def:compdecomp}
A $2$-arrangement for which no decabling is possible is called
{\it indecomposable}; otherwise, it is called {\it decomposable}.
If $\mathcal{A}$ is connected to the trivial arrangement $\mathcal{A}_1$ by a finite
sequence of cabling moves, then $\mathcal{A}$ is called {\em completely decomposable}.
\end{defn}

\begin{exmp} \label{ex:compdecomp}
The complex arrangement $\mathcal{A}_n$ is the $(n-1)$-cable of $\mathcal{A}_1$,
and thus is completely decomposable.  Its link is the corresponding
cable about the unknot.  The arrangement $\mathcal{A}^{-}$ from Example~\ref{ex:ziegler}
is the $(-1)$-cable of $\mathcal{A}_3$, and thus is also completely decomposable.
\end{exmp}

\subsection{}
We now define $2$-arrangements in $\mathbb{R}^4$ corresponding to
special collections of skew lines in $\mathbb{R}^3$, variously called
join configurations \cite{V}, horizontal configurations \cite{M1},
or spindle configurations \cite{CP}.

\begin{defn}
A configuration is called {\em horizontal} if it is rigidly isotopic
to a configuration whose lines are stacked one over another in distinct
planes, all parallel to a fixed (horizontal) plane.  A $2$-arrangement
which admits an associated horizontal configuration is called {\em horizontal}.
\end{defn}

A horizontal configuration $\mathcal{C}$ of $n$ lines determines a permutation
$\tau=\tau(\mathcal{C})$ on $\{1,\dots , n\}$, as follows.  Project perpendicularly
all lines onto a fixed horizontal plane.  Order these $n$ lines
in decreasing order of their (necessarily distinct) slopes.
Order the $n$ horizontal planes containing the lines in increasing order
of their vertical heights.  For every $i\in\{1,\dots, n\}$, put $\t_{i}=k$
if the $i^{\text{th}}$ line is contained in the $k^{\text{th}}$ horizontal plane.
This defines the permutation $\tau\in S_n$.

Conversely, every permutation $\t\in S_n$ determines a horizontal
configuration $\mathcal{C}(\t)$ (see \cite{DV, M1}), and thereby a
horizontal arrangement $\mathcal{A}(\t)$.  Explicitly,  $\mathcal{A}(\t)$ may
be defined as follows.

\begin{prop} \label{prop:horizontal}
Let $\t\in S_n$. Choose real numbers
$a_i, b_i$, $1\le i\le n$, so that
$a_1 < \dots < a_n$ and $b_{\t_{1}} < \dots < b_{\t_{n}}$.
Then the polynomial
\begin{equation}  \label{eq:horizpoly}
f(z,w)=\prod_{i=1}^{n}{(z-\frac{a_i+b_i}{2}w-\frac{a_i-b_i}{2}\bar{w})},
\end{equation}
defines a horizontal $2$-arrangement, whose associated
permutation is $\t$.
\end{prop}

For horizontal arrangements, the linking numbers have a
particularly simple interpretation.  Namely, if $\mathcal{A}=\mathcal{A}(\t)$,
then $l_{i,j}=\sgn(\t_{i}\t_{j})$.

\begin{exmp}  \label{ex:zieghoriz}
In Example~\ref{ex:ziegler}, pick the vertical coordinate to be
$y=\operatorname{Im} z$. Then the lines of the configurations $\mathcal{C}^{\pm}$ are
contained in horizontal planes, parallel to the plane $y=0$.
In each case, the ordering given by the slopes is $(1,2,3,4)$.
The ordering given by the vertical heights is
$(1,2,3,4)$ for $\mathcal{C}^{+}$, and $(2,1,3,4)$ for $\mathcal{C}^{-}$.
Thus $\mathcal{A}^{+}=\mathcal{A}(1234)$ and $\mathcal{A}^{-}=\mathcal{A}(2134)$.  The defining
polynomials corresponding to the choices $a=(1,2,3,4)$,
$b=(\pm 1, \pm 2, 3, 4)$ in \eqref{eq:horizpoly} are
the polynomials $f^{\pm}$ from Example~\ref{ex:ziegler}.
All linking numbers $l_{i,j}^{\pm}$ are equal
to $+1$, except for $l_{1,2}^{-}=-1$.
\end{exmp}

The permutation $\t$ associated to a horizontal arrangement $\mathcal{A}=\mathcal{A}(\t)$
is not unique.  The following result of Mazurovski\u{\i}~\cite{M2}
lists various ways in which uniqueness is known to fail.

\begin{prop}
\label{prop:maz}
Two horizontal arrangements, defined by permutations $\t$ and $\t'$ in $S_n$,
are rigidly isotopic if:
\begin{alphenum}
\item \label{move-a} $\t'=\sigma\t\sigma'$, where
$\sigma$ and $\sigma'$ are circular permutations of $(1,\dots , n)$;  or
\item \label{move-b} $\t'=\t^{-1}$;  or
\item \label{move-c}
$\t'=(\t_1, \dots, \t_{i-1}, \t'_{i}, \dots, \t'_{i+s}, \t_{i+s+1}, \dots, \t_n)$, where
$(\t_{i}, \dots, \t_{i+s})$ is a permutation of $\{m+1,\dots ,m+s+1\}$, and
 $(\t'_{i}-m, \dots ,\t'_{i+s}-m)=(s+1, \dots, 1)(\t_{i}-m, \dots,
\t_{i+s}-m)(s+1, \dots, 1)$.
\end{alphenum}
\end{prop}

\begin{rem} \label{rem:horizconj}
We do not know whether any two rigidly isotopic horizontal arrangements can
be connected by a finite sequence of moves of type (\ref{move-a}), (\ref{move-b}),
(\ref{move-c}).  There is another set of moves, introduced by Crapo and Penne,
which is conjectured to be complete for horizontal configurations, see~\cite{CP}, p.~80.
At any rate, the precise enumeration of the cosets of $S_n$  modulo the
equivalence relation generated by either set of moves seems to be a
challenging combinatorial problem.
\end{rem}

\begin{exmp} \label{rem:fixmove}
We can use moves of type~(\ref{move-a}) to realize the rigid isotopy
from Remark~\ref{rem:topmove} in the case of horizontal arrangements.
Indeed, if $\mathcal{A}=\mathcal{A}(\tau)$ for some $\t\in S_n$ with $\t_k=n$, then we
can replace $\t$ by $\t'=\t(k+1,\dots , n,1,\dots , k)$.
This yields a new arrangement, $\mathcal{A}'=\mathcal{A}(\t')$, for which
$\t'_n=n$, and $l'_{i,n}=1$ for all $i<n$.
\end{exmp}

\begin{exmp} \label{rem:blockmove}
An important example of move~(\ref{move-c}) is as follows.
Suppose the block $B=(\t_i,\dots, \t_j)$ is obtained  by concatenation
of two blocks, $B_1$ and $B_2$, of consecutive integers,  each block
in either increasing order (a {\em positive block}), or in decreasing
order (a {\em negative block}), and so that $\min B_2=\max B_1+1$.
Then the block $B'=(\t'_i, \dots, \t'_j)$ is also a concatenation
of two blocks  of consecutive integers, $B'_1$ and $B'_2$. Moreover,
$B'_1$ is $B_2$  shifted down by $|B_1|$ and $B'_2$ is $B_1$ shifted up by
$|B_2|$. In essence, the move $(B_1 B_2)\to(B'_1 B'_2)$ allows us to flip-and-shift
adjacent blocks of consecutive integers, provided that $\min B_2=\max B_1+1$.
\end{exmp}

\subsection{}  \label{subsec:shift}
We come now to a special class of horizontal arrangements,
that can be constructed inductively from $\mathcal{A}_1$ by a sequence
of cabling moves.  Let $\mathcal{A}=\mathcal{A}(\t)$, where $\t\in S_n$, and $k=\t_{\ell}$.
An $\epsilon r$-cabling move on the $k^{\text{th}}$ component of $\mathcal{A}$
yields a new horizontal arrangement, $\mathcal{A}(\t')$, where $\t'\in S_{n+r}$
is given by
\begin{equation*}  \label{eq:shift}
\t'_{i}=
\begin{cases}
\t_{i} & \text{if } i\notin\{\ell,\dots,\ell+r\} \text{ and } \t_{i}<k\\
\t_{i}+r& \text{if } i\notin\{\ell,\dots,\ell+r\} \text{ and } \t_{i}>k\\
k+\epsilon(\ell-i)+\frac{1-\epsilon}{2} r& \text{if } i\in\{\ell,\dots,\ell+r\}.
\end{cases}
\end{equation*}
In other words, an $\epsilon r$-cabling move on $k$
shifts all the numbers in $\t$ greater than $k$ by $r$ and replaces $k$ by
$(k,\dots, k+r)$ if $\epsilon=1$ or by $(k+r,\dots, k)$ if $\epsilon=-1$.

\begin{defn}
A horizontal arrangement is called {\em completely decomposable} if
the associated permutation can be obtained from the
identity permutation $(1)$ by a finite sequence of cabling moves.
\end{defn}

It is apparent from the definitions that the link of
a completely decomposable arrangement is obtained from the unknot
by successive $(1,\pm 1)$-cablings.

\begin{exmp}  \label{ex:decomp}
All $2$-arrangements of up to $5$~planes are completely decomposable,
except for $\mathcal{A}(31425)$, which is indecomposable.  Among arrangements
of $6$~planes, for example, $\mathcal{A}(K)=\mathcal{A}(341256)$ is completely decomposable,
$\mathcal{A}(314256)$ is decomposable but not completely so, and $\mathcal{A}(241536)$
is indecomposable.
\end{exmp}

\subsection{}
We now introduce a measure of the complexity of a completely
decomposable arrangement $\mathcal{A}$.  Pick a permutation $\t$
such that $\mathcal{A}=\mathcal{A}(\t)$.  Construct a sequence of permutations
connecting $\t$ to the identity permutation,
$\t=\t_0 \to \t_1 \to \cdots \to \t_d=(1)$, as follows.  At each step,
partition the current permutation into blocks of consecutive integers,
either in increasing order (positive cablings), or in decreasing order
(negative cablings), and contract each block to a single number
(via decabling moves), renumbering accordingly.   Let $d(\t)=d$.

\begin{defn}  \label{def:depth}
The {\em depth} of a completely decomposable arrangement $\mathcal{A}$ is
\[
\depth(\mathcal{A}) = \min_{\t}\, \{d(\t)\mid \mathcal{A} \text{ is rigidly isotopic to } \mathcal{A}(\t)\}.
\]
\end{defn}

\begin{exmp}  \label{ex:depth}
The only arrangement of depth~$0$ is the trivial arrangement
$\mathcal{A}_1=\mathcal{A}(1)$.   The arrangements of depth~$1$ are
the complex arrangements $\mathcal{A}_n=\mathcal{A}(1 \cdots n)$, with $n>1$.
The arrangement $\mathcal{A}(21435)$ is completely decomposable, via
the sequence of permutations $(21435)\to (123)\to (1)$,
and so has depth $2$.
\end{exmp}

For arrangements of depth~$2$, we single out the following type.

\begin{defn} \label{defn:normalform}
Let $\mathcal{A}(\t)$ be a depth~$2$, completely decomposable arrangement.
We say that $\mathcal{A}(\t)$ is in {\em normal form} if $\t=(I_1, \dots, I_r, J)$,
where $I_1, \dots, I_r$ are negative blocks, $J$ is a positive block,
and the following conditions hold:
\begin{romenum}
\item $I_1 < \cdots < I_r < J$,   \label{condition1}
\item $2 \leq |I_1| \leq \cdots \leq |I_r|$,   \label{condition2}
\item $|I_1| \le |J|$ if $r=1$.  \label{condition3}
\end{romenum}
\end{defn}

\begin{prop}  \label{prop:normalform}
Every arrangement of depth~$2$ is rigidly isotopic to a unique
arrangement in normal form.
\end{prop}

\begin{proof}
Let $\mathcal{A}$ be a depth~$2$ arrangement of $n$ planes.  Up to rigid isotopy, we
may assume that $\mathcal{A}=\mathcal{A}(\t)$, where $d(\t)=2$.  Applying the type~(\ref{move-a})
move of Example~\ref{rem:fixmove}, we may further assume that $n$ is fixed
by $\t$. Then the permutation sequence of $\mathcal{A}=\mathcal{A}(\tau)$ has the following form:
$\t\to (1,\dots, r)\to(1)$.  Applying  repeatedly the type~(\ref{move-c}) move of
Example~\ref{rem:blockmove}, we can push all the positive blocks of $\t$
(including singletons) to the right, packing all of them into a single
positive block (that will contain $n$), and also arrange the negative blocks
in increasing order of their sizes from left to the right.  In this way, we
arrive at the normal form $\mathcal{A}(I_1, \dots, I_r, J)$ for $\mathcal{A}$.  The uniqueness
is guaranteed by the conditions imposed on $I_1, \dots, I_r$ and $|J|$.
\end{proof}

Thus, we may refer to {\em the} normal form of an arrangement of depth~$2$.
As we shall see in \S\ref{sec:bottomvar}, the normal form is a complete
homotopy type invariant for complements of such arrangements.

\section{Braids and Fundamental Groups}
\label{sec:braids}

In this section, we associate to a $2$-arrangement of $n$ planes
several braids on $n$ strings, and use these braids to find
presentations for the fundamental group of the complement.

\subsection{}  \label{braids}
Let $B_n$ be Artin's braid group on $n$ strings, with generators
$\sigma_1, \sigma_2, \dots , \sigma_{n-1}$ and relations $\sigma_i\sigma_{j}\sigma_i=
\sigma_{j}\sigma_i\sigma_{j}$ for $|i-j|=1$ and $\sigma_i\sigma_j=\sigma_j\sigma_i$ for $|i-j|>1$,
see~\cite{Bi}.  Also, let
$\Delta_n=(\sigma_{n-1} \cdots \sigma_1)(\sigma_{n-1} \cdots \sigma_2) \cdots
(\sigma_{n-1}\sigma_{n- 2})(\sigma_{n-1})\in B_n$ be
``Garside's braid"---the half-twist on $n$ strings.

Consider a configuration $\mathcal{C}=\{\ell_1,\dots ,\ell_n\}$
of $n$ skew-lines in $\mathbb{R}^3$.   Associated to $\mathcal{C}$,
there is a braid on $n$ strings, $\a=\a(\mathcal{C})\in B_n$, see
Mazurovski\u{\i}~\cite{M1} and Crapo and Penne~\cite{CP}.
The procedure that takes $\mathcal{C}$ to $\a$
is illustrated in Figures~\ref{fig:zpair} and \ref{fig:zbraids}.
Set $\b=\a\Delta_n\a\Delta_n^{-1}$.  We call $\a$ and $\b$, the {\em half-braid},
respectively the {\em full-braid} of the configuration $\mathcal{C}$
(or of the arrangement $\mathcal{A}=\mathcal{A}(\mathcal{C})$).  As is well-known,
conjugation by $\Delta_n$ is the involution $\sigma_i \mapsto \sigma_{n-i}$. Thus,
the braid $\b$ is obtained by concatenating $\a$ with another
copy of $\a$, rotated by $180^{\circ}$, see Figure~\ref{fig:braids}.
Clearly, $\b$ is a pure braid in $P_n$.

\begin{figure}
\[
\begingraph<0.5cm,0.5cm>[0.35pt]
\edge from (0, 2) to (1, 2)
\edge from (0, 3) to (1, 3)
\edge from (0, 4) to (1, 4)
\edge from (1, 2) to (2, 2)
\edge from (1, 3) to (1.3, 3.3)
\edge from (1.7, 3.7) to (2, 4)
\edge from (1, 4) to (2, 3)
\edge from (2, 2) to (2.3, 2.3)
\edge from (2.7, 2.7) to (3, 3)
\edge from (2, 3) to (3, 2)
\edge from (2, 4) to (3, 4)
\edge from (3, 2) to (4, 2)
\edge from (3, 3) to (4, 4)
\edge from (3, 4) to (3.3, 3.7)
\edge from (3.7, 3.3) to (4, 3)
\edge from (4, 2) to (5, 2)
\edge from (4, 3) to (5, 3)
\edge from (4, 4) to (5, 4)
\edge from (8, 2) to (9, 2)
\edge from (8, 3) to (9, 3)
\edge from (8, 4) to (9, 4)
\edge from (9, 2) to (10, 2)
\edge from (9, 3) to (9.3, 3.3)
\edge from (9.7, 3.7) to (10, 4)
\edge from (9, 4) to (10, 3)
\edge from (10, 2) to (10.3, 2.3)
\edge from (10.7, 2.7) to (11, 3)
\edge from (10, 3) to (11, 2)
\edge from (10, 4) to (11, 4)
\edge from (11, 2) to (12, 2)
\edge from (11, 3) to (12, 4)
\edge from (11, 4) to (11.3, 3.7)
\edge from (11.7, 3.3) to (12, 3)
\edge from (12, 2) to (12.3, 2.3)
\edge from (12.7, 2.7) to (13, 3)
\edge from (12, 3) to (13, 2)
\edge from (12, 4) to (13, 4)
\edge from (13, 2) to (14, 2)
\edge from (13, 3) to (13.3, 3.3)
\edge from (13.7, 3.7) to (14, 4)
\edge from (13, 4) to (14, 3)
\edge from (14, 2) to (15, 3)
\edge from (14, 3) to (14.3, 2.7)
\edge from (14.7, 2.3) to (15, 2)
\edge from (14, 4) to (15, 4)
\edge from (15, 2) to (16, 2)
\edge from (15, 3) to (16, 3)
\edge from (15, 4) to (16, 4)
\endgraph
\]
\caption{The braids $\a$ and $\b$ associated to $\mathcal{A}(213)$.}
\label{fig:braids}
\end{figure}

The following result of Mazurovski\u{\i}~\cite{M1}
and Crapo and Penne~\cite{CP} establishes the direct connection
between the link and the braid of an arrangement.  First recall
the classical theorem of Alexander, according to which every link
in $\mathbb{S}^3$ is isotopic to the closure of a braid (see~\cite{Bi}).

\begin{prop} \label{prop:closebraid}
Let $\mathcal{A}$ be a $2$-arrangement in $\mathbb{R}^4$ and $L=L(\mathcal{A})$ its link.
Let $\mathcal{C}=\mathcal{C}(\mathcal{A})$ be the associated configuration of skew lines in $\mathbb{R}^3$
and $\b=\b(\mathcal{C})$ its full-braid.  Then $L$ is isotopic to the
closure of $\b$.
\end{prop}

Let $X$ be the complement of the arrangement $\mathcal{A}$, and
$G=\pi_1(X)$ its fundamental group.  Recall that $X$ is
homotopy equivalent to the complement $Y$ of the link $L$.
Since $L$ is the closure of $\b$, the group $G$ has
Artin presentation
\begin{equation} \label{eq:artinpres}
G=\langle x_1,\dots , x_n \mid  \b(x_i) = x_i,
\: i=1,\dots , n\rangle,
\end{equation}
see~\cite{Bi, BZ}.

\subsection{}
As mentioned in Remark~\ref{rem:topmove}, we can bring one of the
lines of $\mathcal{C}$, say $\ell_n$, on top of all the other ones.
Discarding $\ell_n$, we get a configuration $\check{\mathcal{C}}$
of $n-1$ skew lines, so that $\mathcal{C}=\mathcal{C}'\cup\{\ell_n\}$.
It follows that $L=\check{L}\cup L_n$, where $\check{L}$
is the closure of $\check{\b}=\b(\check{\mathcal{C}})\in P_{n-1}$.
Furthermore, it is readily seen that the half-braid of
$\check{\mathcal{C}}$ is given by
$\iota(\check{\a})=\sigma_1^{-1}\cdots \sigma_{n-1}^{-1}\a$,
where $\iota:B_{n-1}\hookrightarrow B_n$ is the standard
inclusion $\iota(\sigma_i)=\sigma_{i+1}$.
We call $\check{\a}$ and $\check{\b}$, the {\em reduced half-braid},
respectively the {\em reduced full-braid} of the arrangement $\mathcal{A}=\mathcal{A}(\mathcal{C})$.

It is now apparent that the complement of $L$ in $\mathbb{S}^3$
is homotopy equivalent to the complement of $\check{L}$
in the solid torus $\mathbb{S}^1\times D^2 = \mathbb{S}^3 \setminus (L_n\times D^2)$.
These geometric considerations lead to the following:

\begin{prop} \label{prop:bundle}
The complement $X(\mathcal{A})$ of a $2$-arrangement of $n$ planes in $\mathbb{R}^4$
is homotopy equivalent to the total space of a bundle over the circle,
with fiber $D^2\setminus \{n-1 \text{ points}\}$, and monodromy
the braid automorphism $\check{\b}$.
\end{prop}

Thus, $X$ is a $K(G,1)$, with fundamental group a semidirect product
of free groups, $G = \mathbb{F}_{n-1} \rtimes_{\check{\b}} \mathbb{F}_1$.
The Artin representation of $\check{\b}$ provides a presentation for $G$,
corresponding to this split extension:
\begin{equation*} \label{eq:bpres}
G=\langle x_1,\dots , x_n \mid x_n^{-1} x_i x_n = \check{\b}(x_i),
\: i=1,\dots , n-1\rangle.
\end{equation*}

\begin{exmp}\label{ex:complex}
For the complex arrangement $\mathcal{A}_n$, the half-braid is the
half-twist $\a=\Delta_n$, and the full-braid is the full-twist
$\beta=\Delta_n^2$.  Since $\check{\b}=\Delta_{n-1}^2$
acts on $\mathbb{F}_{n-1}$ by conjugation by $x_1\cdots x_{n-1}$,
the group $G$ is isomorphic to $\mathbb{F}_{n-1}\times \mathbb{F}_1$, where
$\mathbb{F}_1=\langle x_1\cdots x_n\rangle$.
\end{exmp}

For a non-complex $2$-arrangement, the group $G$ is in general
{\em not} isomorphic to a direct product, as we shall later see.
Nevertheless, we may still use the underlying idea of
Example~\ref{ex:complex}, and simplify the presentation of $G$,
by cutting off a full twist from $\check{\b}$.

\begin{prop}  \label{prop:xi2}
Let $\mathcal{A}$ be a $2$-arrangement of $n$ planes, with reduced
half-braid $\Check{\a}$.  Set $\xi =\Delta_{n-1}{\check{\a}}^{-1}$.
Then, the fundamental group $G$ of $\mathcal{A}$ is
isomorphic to $\mathbb{F}_{n-1} \rtimes_{\xi^2} \mathbb{F}_1$, and has presentation
\begin{equation}  \label{eq:xipres}
G=\langle x_1,\dots , x_n \mid x_n x_i x_n^{-1} = \xi^2(x_i),
\: i=1,\dots , n-1\rangle.
\end{equation}
\end{prop}
\begin{proof}
Recall that $G_1\rtimes_{\phi} G_2\cong G_1\rtimes_{\phi'} G_2$
if $\phi'=\gamma \phi^{\pm 1}$, where $\gamma\in\Inn(G_1)$.
Thus, it suffices to show that $\check{b}$ differs from $\xi^{-2}$ by an
inner automorphism of $\mathbb{F}_{n-1}$.  This follows from the fact that
$\Delta_{n-1}^2\in \Center(B_{n-1})\cap \Inn(\mathbb{F}_{n-1})$:
\begin{equation*}
\hspace{0.6in}
\check{b}= \check{a} \Delta_{n-1} \check{a} \Delta_{n-1}
   = \xi^{-1} \Delta_{n-1}^2 \xi^{-1}
   = \Delta_{n-1}^2 \xi^{-2}.
\hspace{0.6in}\qed
\end{equation*}
\renewcommand{\qed}{}
\end{proof}

\begin{rem} \label{rem:conjugate}
Recall also that $G_1\rtimes_{\phi} G_2\cong G_1\rtimes_{\phi'} G_2$
if $\phi'=\psi^{} \phi \psi^{-1}$.  We can use this observation
to further simplify the above presentation,
by conjugating $\xi\in P_{n-1}$ by a suitable automorphism of $\mathbb{F}_{n-1}$.
In practice, this will be achieved by either changing the basis
of $\mathbb{F}_{n-1}$, or by conjugating $\xi$ by a suitable braid
$\delta\in B_{n-1}$.
\end{rem}

\subsection{}
We now identify the braids $\a\in B_{n+1}$ and $\xi\in P_{n}$
associated to a horizontal arrangement of $n+1$ planes in terms of the
generators $\sigma_i$ of the braid group and of the generators
$A_{i,j}=\sigma_{j-1} \cdots \sigma_{i+1}\sigma_{i}^{2}\sigma_{i+1}^{-1}
\cdots \sigma_{j-1}^{-1}$ of the pure braid group, respectively.

\begin{prop}  \label{prop:combed}
Let $\mathcal{A}(\t)$ be a horizontal $2$-arrangement of $n+1$ planes.  Then:
\begin{alphenum}
\item \label{al}
The half-braid $\a$ has the form
\[
\a=(\sigma_{n}^{l_{n,n+1}} \cdots \sigma_1^{l_{1,n+1}})(\sigma_{n}^{l_{n-1,n}}
\cdots \sigma_2^{l_{1,n}}) \cdots
(\sigma_{n}^{l_{2,3}}\sigma_{n-1}^{l_{1,3}})(\sigma_{n}^{l_{1,2}}),
\]
where $l_{i,j}$ is the sign of the permutation $(\t_{i}\, \t_{j})$.
\item \label{xi}
The pure braid $\xi$ can be combed as
$\xi=\xi_2 \cdots \xi_{n}$, where
\begin{equation*}  \label{eq:combed}
\xi_j=\prod_{i=1}^{j-1} A_{i,j}^{e_{i,j}} \quad \text{ and } \quad
e_{i,j}=
\begin{cases}
1 &\text{ if } \t_{i} > \t_{j},\\
0 &\text{ otherwise.}
\end{cases}
\end{equation*}
\end{alphenum}
\end{prop}

\begin{proof}
Part \eqref{al} follows from the definitions of $\mathcal{C}=\mathcal{C}(\t)$ and
$\a=\a(\mathcal{C})$, and the fact that the linking numbers of $L(\mathcal{A}(\t))$
are given by $l_{i,j}=\sgn(\t_{i}\t_{j})$.

For part \eqref{xi}, it is enough to show that, for all
$k$ with $1 \leq k \leq n-1$,
\begin{equation} \label{eq:iden1}
\begin{split}
&\Delta_n (\sigma_{n-1}^{-l_{1,2}}) (\sigma_{n-2}^{-l_{1,3}}\sigma_{n-1}^{-l_{2,3}}) \cdots
(\sigma_{n-k}^{-l_{1,k+1}} \cdots \sigma_{n-1}^{-l_{k,k+1}}) = \\
&\qquad (A_{1,2}^{e_{1,2}}) \cdots (A_{1,k+1}^{e_{1,k+1}} \cdots
A_{k,k+1}^{e_{k,k+1}})
\ (\sigma_{k+1} \cdots \sigma_{n-1})\cdots (\sigma_2 \cdots \sigma_{n-k}) \Delta_{n-k} ,
\end{split}
\end{equation}
Indeed, the identity~\eqref{eq:iden1} for $k=n-1$ yields
the desired combed form of $\xi$.

The proof of \eqref{eq:iden1} is by induction on $k$, using the
braid relations.  The step $k=1$ is as follows:
\begin{align*}
\Delta_n \sigma_{n-1}^{\pm 1}
&=(\sigma_1 \cdots \sigma_{n-1}) \cdots (\sigma_1\sigma_2)(\sigma_1) \sigma_{n-1}^{\pm 1} \\
&=(\sigma_1)(\sigma_2\sigma_1) \cdots (\sigma_{n-2}\sigma_{n-3})(\sigma_{n-1}\sigma_{n-2}) \sigma_{n-1}^{\pm 1}
(\sigma_1 \cdots \sigma_{n-3}) \cdots (\sigma_1\sigma_2)(\sigma_1) \\
&=(\sigma_1)(\sigma_2\sigma_1) \cdots (\sigma_{n-2}\sigma_{n-3}) \sigma_{n-2}^{\pm 1} (\sigma_{n-1}\sigma_{n-2})
(\sigma_1 \cdots \sigma_{n-3}) \cdots (\sigma_1\sigma_2)(\sigma_1)  \\
&\quad\cdots \\
&=(\sigma_1) \sigma_1^{\pm 1} (\sigma_2\sigma_1) \cdots (\sigma_{n-2}\sigma_{n-3})(\sigma_{n-1}\sigma_{n-2})
(\sigma_1 \cdots \sigma_{n-3})  \cdots  (\sigma_1\sigma_2)(\sigma_1) \\
&=A_{1,2}^{e} (\sigma_2 \cdots \sigma_{n-1}) \Delta_{n-1},
\end{align*}
where $e=\frac{1\pm 1}{2}$. The induction step is similar but tedious,
and will be omitted.
\end{proof}

\begin{figure}
\[
\begingraph<0.5cm,0.5cm>[0.35pt]
\edge from (2, 1) to (3, 1)
\edge from (2, 2) to (3, 2)
\edge from (2, 3) to (3, 3)
\edge from (2, 4) to (3, 4)
\edge from (2, 5) to (3, 5)
\edge from (3, 1) to (4, 1)
\edge from (3, 2) to (4, 2)
\edge from (3, 3) to (4, 3)
\edge from (3, 4) to (3.3, 4.3)
\edge from (3.7, 4.7) to (4, 5)
\edge from (3, 5) to (4, 4)
\edge from (4, 1) to (5, 1)
\edge from (4, 2) to (5, 2)
\edge from (4, 3) to (4.3, 3.3)
\edge from (4.7, 3.7) to (5, 4)
\edge from (4, 4) to (5, 3)
\edge from (4, 5) to (5, 5)
\edge from (5, 1) to (6, 1)
\edge from (5, 2) to (5.3, 2.3)
\edge from (5.7, 2.7) to (6, 3)
\edge from (5, 3) to (6, 2)
\edge from (5, 4) to (6, 4)
\edge from (5, 5) to (6, 5)
\edge from (6, 1) to (6.3, 1.3)
\edge from (6.7, 1.7) to (7, 2)
\edge from (6, 2) to (7, 1)
\edge from (6, 3) to (7, 3)
\edge from (6, 4) to (7, 4)
\edge from (6, 5) to (7, 5)
\edge from (7, 1) to (8, 1)
\edge from (7, 2) to (8, 2)
\edge from (7, 3) to (8, 3)
\edge from (7, 4) to (8, 5)
\edge from (7, 5) to (7.3, 4.7)
\edge from (7.7, 4.3) to (8, 4)
\edge from (8, 1) to (9, 1)
\edge from (8, 2) to (9, 2)
\edge from (8, 3) to (8.3, 3.3)
\edge from (8.7, 3.7) to (9, 4)
\edge from (8, 4) to (9, 3)
\edge from (8, 5) to (9, 5)
\edge from (9, 1) to (10, 1)
\edge from (9, 2) to (9.3, 2.3)
\edge from (9.7, 2.7) to (10, 3)
\edge from (9, 3) to (10, 2)
\edge from (9, 4) to (10, 4)
\edge from (9, 5) to (10, 5)
\edge from (10, 1) to (11, 1)
\edge from (10, 2) to (11, 2)
\edge from (10, 3) to (11, 3)
\edge from (10, 4) to (10.3, 4.3)
\edge from (10.7, 4.7) to (11, 5)
\edge from (10, 5) to (11, 4)
\edge from (11, 1) to (12, 1)
\edge from (11, 2) to (12, 2)
\edge from (11, 3) to (11.3, 3.3)
\edge from (11.7, 3.7) to (12, 4)
\edge from (11, 4) to (12, 3)
\edge from (11, 5) to (12, 5)
\edge from (12, 1) to (13, 1)
\edge from (12, 2) to (13, 2)
\edge from (12, 3) to (13, 3)
\edge from (12, 4) to (13, 5)
\edge from (12, 5) to (12.3, 4.7)
\edge from (12.7, 4.3) to (13, 4)
\edge from (13, 1) to (14, 1)
\edge from (13, 2) to (14, 2)
\edge from (13, 3) to (14, 3)
\edge from (13, 4) to (14, 4)
\edge from (13, 5) to (14, 5)
\edge from (18, 1) to (19, 1)
\edge from (18, 2) to (19, 2)
\edge from (18, 3) to (19, 3)
\edge from (18, 4) to (19, 4)
\edge from (19, 1) to (19.3, 1.3)
\edge from (19.7, 1.7) to (20, 2)
\edge from (19, 2) to (20, 1)
\edge from (19, 3) to (20, 3)
\edge from (19, 4) to (20, 4)
\edge from (20, 1) to (21, 1)
\edge from (20, 2) to (21, 2)
\edge from (20, 3) to (21, 3)
\edge from (20, 4) to (21, 4)
\edge from (21, 1) to (21.3, 1.3)
\edge from (21.7, 1.7) to (22, 2)
\edge from (21, 2) to (22, 1)
\edge from (21, 3) to (22, 3)
\edge from (21, 4) to (22, 4)
\edge from (22, 1) to (23, 1)
\edge from (22, 2) to (23, 2)
\edge from (22, 3) to (22.3, 3.3)
\edge from (22.7, 3.7) to (23, 4)
\edge from (22, 4) to (23, 3)
\edge from (23, 1) to (24, 1)
\edge from (23, 2) to (24, 2)
\edge from (23, 3) to (24, 3)
\edge from (23, 4) to (24, 4)
\edge from (24, 1) to (25, 1)
\edge from (24, 2) to (25, 2)
\edge from (24, 3) to (24.3, 3.3)
\edge from (24.7, 3.7) to (25, 4)
\edge from (24, 4) to (25, 3)
\edge from (25, 1) to (26, 1)
\edge from (25, 2) to (26, 2)
\edge from (25, 3) to (26, 3)
\edge from (25, 4) to (26, 4)
\endgraph
\]
\caption{The braids $\a$ and $\xi$ associated to $\mathcal{A}(21435)$.}
\label{fig:alphaxi}
\end{figure}

\begin{exmp}
\label{ex:horizbraids}
The complex arrangement $\mathcal{A}_n$ is horizontal, with corresponding
permutation the identity $\tau=(1\cdots n)$.  Thus
$\a=\Delta_n$ and $\xi=1$.  The arrangement $\mathcal{A}=\mathcal{A}^{-}$
from Example~\ref{ex:ziegler} is also horizontal,
with permutation $\tau=(2134)$.  Thus $\a=\sigma_3\sigma_2\sigma_1\sigma_3\sigma_2\sigma_3^{-1}$
and $\xi=A_{1,2}$.  The braids
$\a=\sigma_4\sigma_3\sigma_2\sigma_1\sigma_4^{-1}\sigma_3\sigma_2\sigma_4\sigma_3\sigma_4^{-1}$
and $\xi=A_{1,2}A_{3,4}$ associated to the horizontal arrangement
$\mathcal{A}(21435)$ are illustrated in Figure~\ref{fig:alphaxi}.
\end{exmp}

\section{Determinantal Ideals and Characteristic Varieties}
\label{sec:detideals}

We start this section with a review of the determinantal ideals
of the Alexander module of a space, following Hillman~\cite{Hi} and
Turaev~\cite{T1, T2}.  From the varieties defined by
these ideals, we extract numerical homotopy-type invariants,
that will be used for the rest of this paper.

\subsection{}
Let $X$ be a connected, finite CW-complex, with basepoint $*$, and
fundamental group $\pi_1(X,*)$.  Let $p:\widetilde{X}\to X$ be the
universal abelian cover, corresponding to the abelianization
homomorphism $\ab:\pi_1(X,*)\to H_1(X;\mathbb{Z})$. The relative
homology group $A(X)=H_1(\widetilde{X},p^{-1}(*); \mathbb{Z})$ has
the structure of a (left) module over the group ring $\mathbb{Z}[H_1(X; \mathbb{Z})]$,
and is known as the {\em Alexander module} of $X$.

Now assume that $H_1(X,\mathbb{Z})$ is isomorphic to
$\mathbb{Z}^n$, the free abelian group on $t_1,\dots ,t_n$.
A choice of isomorphism, $\psi: H_1(X) \xrightarrow{\simeq} \mathbb{Z}^n$,
identifies $\mathbb{Z}[H_1(X)]$ with $\L=\mathbb{Z}[t_1^{\pm 1}, \dots , t_n^{\pm 1}]$,
the ring of Laurent polynomials in $n$ variables, and defines a
$\Lambda$-module structure on the Alexander module of $X$,
which we will denote by $A(X,\psi)$.  From a presentation of
the fundamental group,
$\pi_1(X)=\langle x_1,\dots , x_q \mid r_1,\dots , r_s\rangle$,
one gets a presentation for the Alexander module,
\[
\L^s \xrightarrow{M} \L^q\to A(X,\psi)\to 0,
\]
where $M=\begin{pmatrix} \partial r_i/ \partial x_j \end{pmatrix}^{\ab}$
is the abelianized Jacobian matrix of Fox derivatives.

Define the {\em $k^{\text{th}}$ determinantal ideal} of $A_{\psi}(X)$
to be the ideal $E_k(X,\psi)$ generated by the codimension~$k$ minors
of the Alexander matrix $M$.  Clearly, $E_{k}(X,\psi)\subseteq E_{\ell}(X,\psi)$
if $k\le \ell$.  The determinantal ideals depend only on the homotopy
type of $X$ (in fact, only on its fundamental group), and on the
identification $\psi: H_1(X)\to\mathbb{Z}^n$.

If $\pi_1(X)$ has positive deficiency (i.e., admits a presentation
with more generators than relations), then $E_1(X,\psi)$ is of the form
$I \cdot (\Delta_{X,\psi})$, where $I$ is the augmentation ideal of $\L$,
and $\Delta_{X,\psi}\in \L$ is the (multi-variable) {\em Alexander polynomial}
of $X$, see~\cite{EN}.

\subsection{}
We now associate to $X$ subvarieties $V_k(X,\psi)$ of the algebraic
torus $(\mathbb{C}^{*})^n$, defined by the determinantal ideals $E_{k}(X,\psi)$,
following \cite{DF, L1}.
The coordinate ring of $(\mathbb{C}^{*})^n$ is $\L_{\mathbb{C}}=\L\otimes \mathbb{C}$, the ring
of Laurent polynomials with complex coefficients.  Then, for each $k\ge 0$,
we set
\[
V_k(X,\psi)=\{  (t_1,\dots ,t_n) \in (\mathbb{C}^*)^n \mid g(t_1,\dots, t_n)=0,
\text{ for all } g\in \sqrt{E_{k}(X,\psi)\otimes \mathbb{C}}  \},
\]
where $\sqrt{\mathfrak{a}}$ denotes the radical of an ideal $\mathfrak{a}$.
Clearly, $V_{k}(X,\psi)\supseteq V_{\ell}(X,\psi)$ if $k\le \ell$.

\begin{defn}    \label{def:monomisom}
Two algebraic subvarieties $V$ and $V'$ of $(\mathbb{C}^{*})^n$
are said to have the same {\it monomial isomorphism type} if there exists
an automorphism $\phi_A:(\mathbb{C}^{*})^n\to (\mathbb{C}^{*})^n$ of the form
\[
\phi_A(t_i)=t_1^{a_{i_1}} \cdots t_n^{a_{i_n}}, \quad 1 \leq i \leq n,
\]
for some matrix $A=(a_{i,j})\in \GL_n(\mathbb{Z})$, which maps $V$ into $V'$.
\end{defn}

\begin{prop}   \label{prop:towerinvariant}
The monomial isomorphism type of the subvariety $V_k(X,\psi)$ of the
algebraic torus $(\mathbb{C}^{*})^n$ depends only on the isomorphism type of
$\pi_1(X)$, and not on the identification $\psi:H_1(X)\to \mathbb{Z}^n$.  We call
$V_k(X)=V_k(X,\psi)$ the {\em $k^{\text{th}}$ characteristic variety} of $X$.
\end{prop}
\begin{proof}
Let $X$ and $Y$ be connected, finite CW-complexes, and let
$h:\pi_1(X) \to \pi_1(Y)$ be an isomorphism.  Let
$h_*:H_1(X) \to H_1(Y)$ be the abelianization of $h$,
and set $\bar{h}=\psi_{Y}^{} h_* \psi_{X}^{-1}:\mathbb{Z}^n\to \mathbb{Z}^n$.
The extension of $\bar{h}$ to $\L_{\mathbb{C}}=\mathbb{C}\mathbb{Z}^n$ restricts
to an isomorphism $E_k(X,\psi_X)\otimes\mathbb{C} \to E_k(Y,\psi_Y)\otimes\mathbb{C}$,
for each $k\ge 0$.  Now let $\phi$ the (monomial) automorphism
of $(\mathbb{C}^{*})^n$ induced by $\bar{h}$.  Clearly, $\phi$ restricts
to an isomorphism $V_k(X,\psi_X) \to V_k(Y,\psi_Y)$.
\end{proof}

In other words, for each $k\ge 0$, the monomial isomorphism type
of $V_k(X)$ is an isomorphism type of $\pi_1(X)$, and thus, a
homotopy-type invariant of $X$.  Furthermore, if $\pi_1(X)$ has
positive deficiency, the Alexander polynomial $\Delta_{X}=\Delta_{X,\psi}$
is well-defined up to a monomial change of basis in $(\mathbb{C}^{*})^n$,
and up to multiplication by a unit
$c t_1^{i_1}\cdots t_n^{i_n}\in \Lambda_{\mathbb{C}}$.  Note that
$V_1(X)=\mathbf{1}\cup \{\Delta_{X}=0\}$, where $\mathbf{1}=(1,\dots ,1)$
is the origin of the complex torus $(\mathbb{C}^*)^n$.

\subsection{}
By themselves, the characteristic varieties are not very practical homotopy-type
invariants.  We extract from them several numerical invariants that
are powerful enough for our purposes.  For each integer $p\ge 2$,
let
\[\Omega_p^n=\{(\omega_1,\dots ,\omega_n)\in (\mathbb{C}^*)^n \mid
\omega_i\: \text{ is a $p^{\text{th}}$ root of unity}\}\]
be the set of $p$-torsion points of $(\mathbb{C}^*)^n$.

\begin{thm} \label{thm:invariant}
The following are isomorphism type invariants of $\pi_1(X)$:
\begin{alphenum}
\item \label{inv1}
The list $\Sigma_k(X)$ of codimensions of irreducible
components of $V_k(X)$;

\item \label{inv2}
The list $\Sigma_{\mathbf{1},k}(X)$ of codimensions of irreducible
components of $V_k(X)$ passing through $\mathbf{1}$;

\item \label{inv3}
The number $\Tors_{p,k}(X)=\left|{\Omega_p^n\cap V_k(X)}\right|$
of $p$-torsion points of $V_k(X)$.

\end{alphenum}
\end{thm}

\begin{proof}
By Proposition~\ref{prop:towerinvariant}, an isomorphism of fundamental
groups determines a monomial isomorphism of the corresponding
characteristic varieties.  Part~\eqref{inv1} follows from the fact
that an isomorphism of algebraic varieties sends irreducible
components to irreducible components of the same codimension.
Part~\eqref{inv2} follows from Part~\eqref{inv1}, and the fact that
a monomial isomorphism fixes $\mathbf{1}$.  Part~\eqref{inv3} follows
from the fact that a monomial isomorphism preserves the set of
$p$-torsion points.
\end{proof}

\subsection{}  \label{alexmod}
Now let $X=X(\mathcal{A})$ be the complement of a $2$-arrangement of
$n$ planes in $\mathbb{R}^4$.  Recall that $X$ has the homotopy type
of a $2$-complex (modeled on the Artin presentation of its
fundamental group $G$), and that $H_1(X)=\mathbb{Z}^n$.  Thus, we
can define the {\em $k^{\text{th}}$ characteristic variety of $\mathcal{A}$}
to be $V_k(\mathcal{A})=V_k(X)$.  As we shall see, the descending tower of
characteristic varieties has the form $(\mathbb{C}^{*})^n=V_0 \supseteq V_1 \supseteq
\dots \supseteq V_{n-2} \supseteq V_{n-1}\supseteq V_{n}=\emptyset$,
with $V_1$ being a hypersurface in $(\mathbb{C}^{*})^n$, if $n\ge 3$,
and $V_{n-1}$ consisting of the single point $\mathbf{1}$, if $n\ge 2$.
We will focus on the nontrivial ends of the tower, namely
$V_{1}$ and $V_{n-2}$, which we shall call the {\em top},
respectively the {\em bottom} characteristic variety of $\mathcal{A}$.

In order to find explicit equations for the characteristic varieties,
we need to choose a particular presentation for $G=\pi_1(X)$.
Unless otherwise specified, we shall use
the presentation \eqref{eq:xipres} associated to the semidirect
product structure $G=\mathbb{F}_{n-1}\rtimes_{\xi^2} \mathbb{F}_1$ from
Proposition~\ref{prop:xi2}.  This presentation yields an
identification $\psi_{\xi}:H_1(X)\to \mathbb{Z}^n$.
Let $A=A(X,\psi_{\xi})$ be the corresponding Alexander module.
A presentation matrix for $A$ is the $(n-1)\times n$
(Alexander) matrix
\begin{equation*}
\label{alexmat}
M=\begin{pmatrix}
t_n\cdot \id-\Theta(\xi^2) & d_1
\end{pmatrix},
\end{equation*}
where $d_1=\left( 1-t_1 \ \cdots \ 1-t_{n-1}\right)^{\top}$ and
$\Theta: P_{n-1} \to \GL_{n-1}(\L)$ is the Gassner representation
of the pure braid group, see Birman~\cite{Bi}.

The $k\times k$ minors of $M$ generate the determinantal ideal $E_k$,
whose radical, $\sqrt{E_k}$, defines the $k^{\text{th}}$
characteristic variety $V_k=V_k(\mathcal{A})$.  Note that $E_{n-1}=I$ and
$E_{n}=\Lambda$, and so $V_{n-1}=\mathbf{1}$ and $V_{n}=\emptyset$.

Now recall that a link group has deficiency~$1$, see e.g.~\cite{BZ}.
Thus we may define the {\em Alexander polynomial of $\mathcal{A}$} to be
$\Delta_{\mathcal{A}}=\Delta_{X,\psi_{\xi}}$. For $n=1$, we have
$\Delta_{\mathcal{A}}=1$.  For $n>1$, we have
\[
\Delta_{\mathcal{A}}(t_1,\dots ,t_n) = \frac{1}{t_n-1}
\det \left( t_n\cdot \id-\Theta(\xi^2) \right),
\]
see Penne~\cite{P}.  Thus $\Delta_{\mathcal{A}}=1$ for $n=2$.
For $n\ge 3$, the triviality of
the Gassner representation evaluated at $\mathbf{1}$
implies that $\mathbf{1}\in V_1(\mathcal{A})$ and so
$V_1(\mathcal{A})=\{\Delta_{\mathcal{A}}(t_1,\dots ,t_n)=0\}$.

\begin{rem}  \label{rem:aplink}
For certain purposes, it is more natural to start from the
Artin presentation \eqref{eq:artinpres} associated to the semidirect
product structure $G=\mathbb{F}_{n-1}\rtimes_{\b} \mathbb{F}_1$. The resulting
presentation, $A(X,\psi_{\b})$, for the Alexander module coincides
with the usual presentation of the Alexander module of the
link $L(\mathcal{A})$.  We will denote the associated Alexander polynomial
by $\Delta_{L(\mathcal{A})}=\Delta_{X,\psi_{\b}}$.
\end{rem}

\begin{exmp}

\label{ex:Cxcharvar}
Let $\mathcal{A}_n$ be the arrangement of $n\ge 3$ complex lines through the
origin of $\mathbb{C}^2$.  Recall that $\xi=1$ and $\beta=\Delta_n^2$ in this case.
It is readily seen that $E_k=I \cdot (t_{n}-1)^{n-k-1}$.
Thus $V_1=\dots =V_{n-2}=\{t_n-1=0\}$, and
$\Delta_{\mathcal{A}_n}=(t_n-1)^{n-2}$, whereas
$\Delta_{L(\mathcal{A}_n)}=(t_1\dots t_n-1)^{n-2}$.
\end{exmp}

\begin{exmp}
\label{ex:Zieglercharvar}
Let $\mathcal{A}$ be the arrangement $\mathcal{A}^{-}=\mathcal{A}(2134)$. Recall that
$\xi=A_{1,2}$. The Artin representation of $\xi:\mathbb{F}_3\to \mathbb{F}_3$ is given
by $\xi(x_1)=x_1x_2x_1x_2^{-1}x_1^{-1}$, $\xi(x_2)=x_1x_2x_1^{-1}$,
$\xi(x_3)=x_3$.  Consider the new basis $y_1=x_1$, $y_2=x_1x_2$, $y_3=x_3$
for $\mathbb{F}_3$.  In this basis, $\xi(y_1)=y_2y_1y_2^{-1}$, $\xi(y_2)=y_2$,
$\xi(y_3)=y_3$, and so the Alexander matrix is:
\[
M= \begin{pmatrix}
t_4-t_2^2 & (t_2+1)(t_1-1) & 0 & 1-t_1 \\
0 & t_4-1 & 0 & 1-t_2 \\
0 & 0 & t_4-1 &
1-t_3
\end{pmatrix}.
\]
The determinantal ideals are $E_1=I \cdot (t_4-1)(t_4-t_2^2)$,
and $E_2=I \cdot (t_4-1,t_2^2-1) + \L \cdot(t_2+1)(t_1-1)(t_3-1)$.
The characteristic varieties are
\begin{align*}  \label{eq:avz}
\begin{split}
V_1&=\{t_4-1=0\} \cup \{t_4-t_2^2=0\}, \\
V_2&=\{t_4-1=t_2+1=0\} \cup \{t_4-1=t_2-1=t_1-1=0\} \\
& \qquad\cup \{t_4-1=t_2-1=t_3-1=0\}.
\end{split}
\end{align*}

Now let $\mathcal{A}$ be the arrangement $\mathcal{A}^{+}=\mathcal{A}(1234)$.
We know from Example~\ref{ex:Cxcharvar} that its characteristic
varieties are $V_1=V_2=\{t_4-1=0\}$.  We plainly see that
the characteristic varieties of $\mathcal{A}^{+}$ and $\mathcal{A}^{-}$ have a
different number of components, and so are not isomorphic.
Thus, $X^{+}\not\simeq X^{-}$.  In fact, $H^*(X^{+})\ncong H^*(X^{-})$,
as was shown by Ziegler~\cite{Z}.
\end{exmp}

\section{Bottom Characteristic Varieties}
\label{sec:bottomvar}

In this section we study the bottom characteristic varieties $V_{n-2}(\mathcal{A})$
of arrangements of $n$ planes that are obtained from the trivial
arrangement by a sequence of cabling operations.  We obtain a
complete characterization of these varieties when the sequence
has length~$2$.

\subsection{}
Let $\mathcal{A}$ be a depth~$2$ arrangement, and $\mathcal{A}(I_1,\dots ,I_r,J)$ its
normal form, as introduced in Definition~\ref{defn:normalform}.
By Proposition~\ref{prop:combed},
the pure braid $\xi$ can be taken to be $\xi=A_{I_1} \cdots A_{I_r}$,
where $A_I$ is the full twist on the strings $I$, with the convention
that whenever a negative block appears as a subindex of a braid generator,
it will be understood as a set of integers in increasing order.

Let us first consider a full twist $A_I$ corresponding to the
indexing set $I=\{j, j+1, \dots, k\}$.
The Artin representation of $A_I$ is given by $A_I(x_i)=x_Ix_ix_I^{-1}, i \in I$
and $A_I(x_i)=x_i$ otherwise, where $x_I=x_j \cdots x_k$.
The change of basis $y_k=x_I$ and $y_i=x_i$ for $i \neq k$
simplifies $A_I$ to $A_I(y_i)=y_ky_iy_k^{-1}, i \in I$ and $A_I(y_i)=y_i$
otherwise.  The Gassner representation of $A_I^2$ in this new basis
is given by:

\begin{equation}  \label{eq:gasstwist}
\Theta(A_I^2)=
\begin{pmatrix}
t_k^2  & \cdots & 0       & (t_k+1)(1-t_j)     \\
\vdots & \ddots &  \vdots & \vdots             \\
0      & \cdots & t_k^2   & (t_k+1)(1-t_{k-1}) \\
0      & \cdots & 0       & 1
\end{pmatrix}
\end{equation}

Now notice that the factors $A_{I_1}, \dots, A_{I_r}$ of $\xi$ braid on
mutually disjoint groups of strings.  Therefore, we can change
the basis in the free group for each block separately, as above.
Hence, the Gassner representation $\Theta(\xi^2)$ is a block-diagonal matrix,
with blocks as in~\eqref{eq:gasstwist}.  The Alexander matrix is:

\begin{equation}  \label{eq:alexdepth2}
M=
\begin{pmatrix}
t_n-\Theta(A_{I_1}^2)  & \cdots & 0                  & 0               & d_1(I_1)  \\
\vdots             & \ddots &  \vdots            & \vdots          & \vdots     \\
0                  & \cdots & t_n-\Theta(A_{I_r}^2)  & 0               & d_1(I_r)  \\
0                  & \cdots & 0                  & t_n-\id_{|J|-1} & d_1(J)
\end{pmatrix}
\end{equation}
where $d_1(I)$ is the column vector whose entries are $t_i-1$,
for $i\in I$.

The radical $\sqrt{E_{n-2}}$ of the ideal of $2 \times 2$-minors of $M$
is generated by
\[
t_n-1; \quad t_{k_p}^2 -1; \quad (t_{k_p}+1)(t_{i_p}-1)(t_{j_p}-1),
\]
where $k_p=\max I_p$,  for all $i_p \in I_p \setminus \{k_p\}$,
$j_p \not\in I_p$, and $1\le p \le r$.

The ideal $\sqrt{E_{n-2}}$ defines the characteristic variety $V_{n-2}$.
In order to describe this subvariety of $(\mathbb{C}^*)^n$, we need some notation.
Given a subset $I$ of $[1,n]=\{1,\dots,n\}$, let $\Bar{I} = [1,n]\setminus I$,
and $\Check{I} = \Bar{I} \cup \{\max I\}$.  Also, let $T(I)$ be the
subtorus of $(\mathbb{C}^*)^n$ given by $T(I)=\{t_i-1=0 \mid i\in I\}$,
and $\Bar{T}(I)$ the translated subtorus given by
$\Bar{T}(I)=\{t_i+1=0 \mid i\in I\}$.

\begin{prop}  \label{prop:bottomdepth2}
Let $\mathcal{A}$ be a depth~$2$, completely decomposable arrangement of $n$ planes, with
normal form $\mathcal{A}(I_1,\dots ,I_r,J)$.  The bottom characteristic variety $V_{n-2}(\mathcal{A})$
has the following irreducible components:
\begin{alphenum}
\item  Subtori passing through $\mathbf{1}$:
\begin{align*}
&T({\Check{I}_p}), \ \: \text{for}\ \: p\in [1,r] \quad \text{and}\quad
T({\Check{J}}), \ \: \text{if}\ \: |J|>1.
\intertext{
\item
Translated subtori:
}
&T({\cup_{p\notin P} I_{p}}\cup \{n\}) \cap
\overline{T}({\cup_{p\in P} \{\max I_p\}}),
\ \: \text{for}\ \: \emptyset \ne P\subseteq [1,r].
\end{align*}
\end{alphenum}
\end{prop}

\begin{exmp}  \label{ex:comps6}
The arrangement $\mathcal{A}=\mathcal{A}(214356)$ from Example~\ref{ex:depth}
is in normal form, with $I_1=\{2,1\}$, $I_2=\{4,3\}$, and $J=\{5,6\}$.
The components of $V_4(\mathcal{A})$ are:
\begin{align*}
&\{t_6-1=t_4+1=t_2+1=0\},\\
&\{t_6-1=t_4+1=t_2-1=t_1-1=0\},\\
&\{t_6-1=t_4-1=t_3-1=t_2+1=0\},\\
&\{t_6-1=t_5-1=t_4-1=t_3-1=t_2-1=0\},\\
&\{t_6-1=t_5-1=t_4-1=t_2-1=t_1-1=0\},\\
&\{t_6-1=t_4-1=t_3-1=t_2-1=t_1-1=0\}.
\end{align*}
\end{exmp}

\subsection{}
Let $\mathcal{A}$ be a completely decomposable arrangement of
depth~$2$, with normal form $\mathcal{A}(I_1,\dots ,I_r,J)$.  Recall that
$|I_1|\le \dots \le |I_r|$.  Define
\begin{equation*} \label{eq:sofa}
S(\mathcal{A})=\{|I_1|,\dots ,|I_r|\}.
\end{equation*}
Since we also have $I_1\le \dots \le I_r<J$, the ordered list $S(\mathcal{A})$,
together with the number of planes, $n=|J|+\sum_{k=1}^r |I_k|$,
determines the normal form.

Let $\Sigma=\Sigma_{n-2}(\mathcal{A})$ be the list of codimensions of
irreducible components of $V_{n-2}(\mathcal{A})$, and
$\Sigma_{\mathbf{1}}=\Sigma_{\mathbf{1},n-2}(\mathcal{A})$
be the sublist corresponding to components passing
through $\mathbf{1}$.   From Proposition~\ref{prop:bottomdepth2}, we see that
\begin{subequations}  \label{eq:comps}
\begin{alignat}{2}
\Sigma_{\mathbf{1}}&=
\{n+1-|I_p|\}_{p=1,\dots, r}  && \text{if } |J|=1 \label{eq:c11}\\
\Sigma_{\mathbf{1}}&=
\{n+1-|I_p|\}_{p=1,\dots, r} \cup \{n+1-|J|\} &\qquad&
\text{if } |J|>1 \label{eq:c12}\\
\Sigma \setminus \Sigma_{\mathbf{1}} &=
\{r+1+\sum_{p\notin P}(|I_p| -1)\}_{\emptyset\subsetneq P\subseteq [1,r]}
&\qquad& \label{eq:cminusc1}
\end{alignat}
\end{subequations}
The lists $\Sigma_{\mathbf{1}}$ and $\Sigma$ have lengths
\begin{equation}  \label{eq:length}
d_1=r+\epsilon_J+1, \qquad d=2^r+r+\epsilon_J,
\end{equation}
where $\epsilon_J=0$ if $|J|>1$ and $\epsilon_J=-1$ if  $|J|=1$.

\begin{thm}  \label{thm:depth2}
The list $S(\mathcal{A})$ is a complete homotopy-type invariant
for depth~$2$, completely decomposable arrangements $\mathcal{A}$
of $n$ planes.
\end{thm}

\begin{proof}
Let $\mathcal{A}'$ be another completely decomposable arrangement of depth~$2$,
with normal form  $\mathcal{A}(I'_1, \dots, I'_r, J')$.  Assume $X(\mathcal{A})\simeq X(\mathcal{A}')$.
Then, by Theorem~\ref{thm:invariant}, $\Sigma(\mathcal{A})=\Sigma(\mathcal{A}')$ and
$\Sigma_{\mathbf{1}}(\mathcal{A})=\Sigma_{\mathbf{1}}(\mathcal{A}')$.  We want to show that
$S(\mathcal{A})=S(\mathcal{A}')$.  There are four cases
to consider, according to the sizes of $J$ and $J'$:

\begin{itemize}
\item $|J|=1, |J'|>1$.  Then, by \eqref{eq:length},
the system of equations $d=d'$, $d_1=d_1'$ has no solution.

\item $|J|=|J'|=1$.  Then, by \eqref{eq:c11},
$\Sigma_{\mathbf{1}}(\mathcal{A})=\Sigma_{\mathbf{1}}(\mathcal{A}')$ implies $S(\mathcal{A})=S(\mathcal{A}')$.

\item $|J|=|J'|>1$.  Then,  \eqref{eq:c12},
$\Sigma_{\mathbf{1}}(\mathcal{A})=\Sigma_{\mathbf{1}}(\mathcal{A}')$ implies $S(\mathcal{A})=S(\mathcal{A}')$.

\item $|J|>|J'|>1$.  Then, by  \eqref{eq:length}, $r=r'$.  If
$r=1$,  equation~\eqref{eq:c12} implies that $\{|I_1|, |J|\}=\{|I'_1|,
|J'|\}$  as {\it unordered} lists.  But condition~\eqref{condition3} from
Definition~\ref{defn:normalform} and  the fact that $|J|>|J'|$ rule out
this possibility.  If $r>1$, equation~\eqref{eq:cminusc1} implies
$|J|=n+r+2-\max( \Sigma \setminus \Sigma_{\mathbf{1}}) -
\min( \Sigma \setminus \Sigma_{\mathbf{1}})$.  Hence $|J|-|J'|=r-r'=0$,
which again is impossible.
\end{itemize}

Conversely, assume $S(\mathcal{A})=S(\mathcal{A}')$.  Then, as noted above,
the normal forms of $\mathcal{A}$ and $\mathcal{A}'$ coincide.
By Proposition~\ref{prop:normalform}, $\mathcal{A}$ and $\mathcal{A}'$ are rigidly
isotopic, and thus $X(\mathcal{A})\simeq X(\mathcal{A}')$.
\end{proof}

\begin{cor}  \label{cor:depth2}
The number of homotopy classes of $2$-arrangements of $n$~planes
which are completely decomposable of depth at most ~$2$ equals
$p(n-1) - \lfloor(n-1)/2\rfloor$, where $p( \cdot)$ is the partition
function, and $\lfloor\cdot\rfloor$ is the integer part function.
\end{cor}

\begin{proof}
Follows from the Theorem by an elementary counting argument.
\end{proof}

\section{Top Characteristic Varieties}
\label{sec:topvar}

In this section we study the top characteristic variety $V_1(\mathcal{A})$ of a
$2$-arrangement $\mathcal{A}$, and the number $\Tors_{p,1}(\mathcal{A})$ of its
$p$-torsion points, for $p$ a prime number.

\subsection{}
Let us start with a completely decomposable arrangement, $\mathcal{A}=\mathcal{A}(\t)$.
Let $\t=\t_0\to\t_1 \to\cdots\to\t_d=(1)$ be the decomposition
sequence for $\t$.  Recall that each permutation in the sequence
is partitioned into blocks of consecutive integers.
We call such a block $B$ {\em essential} if either $|B| \ge 2$, or
$B=\t_{d-1}$ and $|B| > 2$.

\begin{thm}  \label{thm:topvariety}
The top characteristic variety $V_1(\mathcal{A})$ of a completely
decomposable arrangement of $n$ planes is the union of an arrangement
$\mathcal{V}(\mathcal{A})$ of codimension~$1$ subtori of $(\mathbb{C}^*)^n$, all passing
through $\mathbf{1}$.
\end{thm}
\begin{proof}
Choose $\t\in S_n$ so that $\mathcal{A}=\mathcal{A}(\t)$ and $\depth(\mathcal{A})=d(\t)$.
Let $m$ be the number of essential blocks of $\t$.
Recall from \S\ref{subsec:shift} that $L=L(\mathcal{A})$ is an iterated torus link,
obtained by $(1,\pm 1)$-cablings on the unknot.  Thus, it is
a spliced link in the sense of Eisenbud and Neumann~\cite{EN}.
The decomposition sequence of $\t$ corresponds to a minimal
splice diagram of $L$:  The (signed) essential blocks $B_{1},\dots , B_{m}$
of the permutations in the sequence correspond to the (signed) nodes
$v_{n+1},\dots , v_{n+m}$ of the diagram, and the integers $1, \dots, n$
to the arrowheads $v_{1},\dots , v_{n}$.   Then, according to
\cite{EN}, Theorem~12.1, the Alexander polynomial of $L$ is given by:
\[
\Delta_{L}(t_1, \dots, t_n)=\prod_{j=n+1}^{n+m}(t_1^{l_{1,j}}t_2^{l_{2,j}}
\cdots t_n^{l_{n,j}} -1)^{\delta_j -2},
\]
where $l_{i,j}=\pm 1$ is the linking number of $L_i$ with the
``virtual component" corresponding to $v_j$, and $\delta_j$
is the valency of $v_j$.  Thus, each irreducible component of
$V_1(\mathcal{A})=\{\Delta_{L}=0\}$ is a codimension~$1$ subtorus.
(It actually can be shown that $V_1(\mathcal{A})$ has precisely $m$ components.)
\end{proof}

To compute the number of torsion points on $V_1(\mathcal{A})$, we may
now use a result of Bj\"{o}rner and Ekedahl~\cite{BE}.
Indeed, an arrangement $\mathcal{V}$ of codimension~$1$ subtori in
$(\mathbb{C}^*)^n$ defines an arrangement $\mathcal{V}_{p}$ of hyperplanes
in $(\mathbb{Z}_p)^n$:
To a subtorus $t_1^{a_1}\cdots t_n^{a_n}-1=0$ corresponds
the hyperplane $a_1 x_1+\cdots +a_nx_n=0 \mod p$.
Proposition~3.2 of \cite{BE} then implies the following.

\begin{prop}	  \label{prop:p-torsion}
The number of $p$-torsion points on the union
$U=U(\mathcal{V})$ of an arrangement of subtori in $(\mathbb{C}^*)^n$ is given by:
\[
 \Tors_p(U)=-\sum_{x \in L \setminus \{\hat{0}\}}~\mu(\hat{0},x)p^{\dim(x)},
\]
where $L$ is the intersection lattice of the arrangement
$\mathcal{V}_{p}$, with minimal element $\hat{0}=(\mathbb{Z}_p)^n$, and
M\"{o}bius function $\mu$.
\end{prop}

\begin{exmp} \label{ex:depth3}
The arrangement $\mathcal{A}=\mathcal{A}(312546)$ is completely decomposable, of depth $3$.
Its decomposition sequence is $(312546)\to (2134)\to (12)\to (1)$.
Proposition~\ref{prop:combed} gives $\xi=A_{1,3}A_{2,3}A_{4,5}$.
The Artin representation of $\xi$ in the basis
$y_1=x_1$, $y_2=x_1x_2$, $y_3=x_1x_2x_3$, $y_4=x_4$, $y_5=x_4x_5$ is
given by $\xi(y_1)=y_3y_2^{-1}y_1y_2y_3^{-1}$, $\xi(y_2)=y_3y_2y_3^{-1}$,
$\xi(y_3)=y_3$, $\xi(y_4)=y_5y_4y_5^{-1}$, and $\xi(y_5)=y_5$.
The Alexander polynomial is $\Delta_{\mathcal{A}}(t_1,\dots,t_6)=
(t_6-1)(t_6-t_5^{2})(t_6-t_3^{2})(t_6-t_3^{2}t_2^{-2})$.
Proposition~\ref{prop:p-torsion} yields $\Tors_{2,1}(\mathcal{A})=32$
and $\Tors_{3,1}(\mathcal{A})=585$.
\end{exmp}

For an arrangement of depth~$2$, we can give a more precise description
of the top characteristic variety, based on formula \eqref{eq:alexdepth2}
for the Alexander matrix.
\begin{prop}  \label{prop:torsdecomp}
Let $\mathcal{A}$ be a depth~$2$ arrangement of $n$ planes,
with normal form $\mathcal{A}(I_1, \dots, I_r, J)$, and let
$k_q=\max I_q$, for $1 \le q \le r$.  Then:
\vskip 5pt
\begin{alphenum}
\item
$\Delta_{\mathcal{A}}(t_1,\dots,t_n)=
(t_n-1)^{|J|+r-2}\prod_{q=1}^{r}(t_n-t_{k_q}^2)^{|I_q|-1};$
\vskip 4pt
\item
$V_{1}(\mathcal{A})=\{\mathbf{t}\in(\mathbb{C}^*)^n \mid
(t_n-1)\prod_{q=1}^{r}(t_n-t_{k_q}^2) = 0\};$
\vskip 4pt
\item
$\Tors_{2,1}(\mathcal{A})=2^{n-1}$ and
$\Tors_{p,1}(\mathcal{A})=p^{n-r-1}(p^{r+1}-(p-1)^{r+1})$, for $p\ge3$.
\end{alphenum}
\end{prop}

\subsection{}
Let $L$ be a link in $\mathbb{S}^3$.  The Alexander polynomial of a sublink
of $L$, and that of an $(a,b)$-cable about $L$, can be computed from
the Alexander polynomial of $L$, via the following well-known formulae of
Torres and Sumners--Woods, see~\cite{EN, Hi, T1}.

\begin{thm} \label{thm:tsw}
Let $L=L_1 \cup \dots \cup L_n$ be a link in $\mathbb{S}^3$.
Set $T=t_1^{l_1} \cdots t_{n-1}^{l_{n-1}}$, where $l_i=\lk(L_i,L_n)$.
Then:
\begin{subequations} \label{eq:iterated}
\begin{align}
\Delta_L(t_{1}, \dots , t_{n-1}, 1)&=(T-1)\Delta_{L \setminus L_n}
(t_{1}, \dots , t_{n-1}),
\label{eq:torres}\\
\intertext{
Moreover, if $L'=L\{a,b\}$, with $\gcd(a,b)=1$, then:}
\Delta_{L'}(t_{1}, \dots , t_{n}, t_{n+1})&=
(T^{a}t_{n}^{b}t_{n+1}^{b}-1)\Delta_{L}(t_{1}, \dots , t_{n-1}, t_n^{a}t_{n+1}).
 \label{eq:sw}
\end{align}
\end{subequations}
\end{thm}

\begin{cor}  \label{cor:vcable}
Let $\mathcal{A}$ be an arrangement of $n$ planes, and let $\mathcal{A}^k\{r\}$
be an $r$-cable about it.  Then:
\vskip 4pt
\begin{alphenum}
\item  \label{it1}
$V_{1}(\mathcal{A}\{r\})=\{\mathbf{t}\in
(\mathbb{C}^*)^{n+r}\mid t_1\cdots t_n -1=0 \,\text{ or }\,
(t_1,\dots ,t_n)\in V_{1}(\mathcal{A})\};$
\vskip 4pt
\item  \label{it2}
$\Tors_{p,1}(\mathcal{A}\{r\})=p^{r-1}\Tors_{p,1}(\mathcal{A}\{1\}).$
\end{alphenum}
\end{cor}
\begin{proof}
Let $L=L(\mathcal{A})$.
Recall from \S\ref{cables} that $L(\mathcal{A}\{1\})=L\{1,1\}$.
By the Sumners-Woods formula \eqref{eq:sw}, we have
\[
\Delta_{L(\mathcal{A}\{1\})}(t_1, \dots , t_n, t_{n+1})=
(t_1\cdots t_{n+1}-1)\Delta_{L}(t_1, \dots , t_{n-1}, t_nt_{n+1}).
\]
After a monomial change of basis,
this implies part \eqref{it1} for $r=1$.
The general case follows from the same formula, by induction on $r$.
Part \eqref{it2} follows immediately from part \eqref{it1}.
\end{proof}

\subsection{}
We conclude this section with a recursion formula for $\Tors_{2,1}(\mathcal{A})$.
Let $\Delta_{L(\mathcal{A})}$ be the Alexander polynomial of the link
$L(\mathcal{A})$.  Define the {\em single-variable Alexander polynomial} of $\mathcal{A}$ to be
$\Delta_{\mathcal{A}}(t):=(t-1)\Delta_{L(\mathcal{A})}(t,\dots ,t)$.  Furthermore, set
\[
\delta(\mathcal{A})=\begin{cases}
1&\quad\text{if } \Delta_{\mathcal{A}}(-1)=0, \\
0&\quad\text{otherwise}.
\end{cases}
\]
\begin{exmp}  \label{ex:deltacx}
The single-variable Alexander polynomial of the complex arrangement $\mathcal{A}_n$ is
$\Delta_{\mathcal{A}_n}(t)=(t-1)(t^n-1)^{n-2}$.  Thus, $\delta(\mathcal{A}_n)=\frac{1+(-1)^n}{2}$.
\end{exmp}

\begin{thm} \label{thm:2tors}
Let $\mathcal{A}=\{H_1,\dots, H_n\}$ be a $2$-arrangement in $\mathbb{R}^4$.
The number of $2$-torsion points of the top characteristic
variety $V_1(\mathcal{A})$ is given by the following formula:
\begin{equation} \label{eq:deltasum}
\Tors_{2,1}(\mathcal{A})=2^{n-1}-\frac{1+(-1)^n}{2}+\delta(\mathcal{A})+
\sum_{\mathcal{B}\in \Gamma(\mathcal{A})}\delta(\mathcal{B}),
\end{equation}
where $\Gamma(\mathcal{A})$ is the set of all indecomposable, proper
sub-arrangements of $\mathcal{A}$ with an odd number of planes.
\end{thm}
\begin{proof}
 By definition,
\begin{equation} \label{eq:torsum}
\Tors_{2,1}(\mathcal{A}) = \sum_{\omega\in\Omega_2^n} c_{\omega},
\end{equation}
where $\Omega_2^n = \{(\omega_1,\dots ,\omega_n)\in(\mathbb{C}^*)^n \mid \omega_i=\pm 1\}$,
and $c_{\omega}=1$ if $\Delta_{L(\mathcal{A})}(\omega)=0$,
and $c_{\omega}=0$ otherwise.  For $\omega\in\Omega_2^n$, let
$\mathcal{A_{\omega}}=\{ H_i\in\mathcal{A} \mid \omega_i = -1\}$.
There are several cases to consider:

\begin{itemize}
\item If $\omega=(-1,\dots ,-1)$, then $\mathcal{A_{\omega}}=\mathcal{A}$, and so
$c_{\omega}=\delta(\mathcal{A})$.

\item If $\omega\ne (-1,\dots ,-1)$, then $\mathcal{A_{\omega}}$ is a proper
sub-arrangement of $\mathcal{A}$, and so, by repeated application of
Torres's formula~\eqref{eq:torres}, we have
\begin{equation} \label{eq:delomega}
\Delta_{L(\mathcal{A})}(\omega) = ((-1)^{|\mathcal{A_{\omega}}|} - 1)\Delta_{L(\mathcal{A_{\omega}})}(-1,\dots ,-1).
\end{equation}
\begin{itemize}
\item
If $|\mathcal{A_{\omega}}|$ is even, this formula says that
$\Delta_{L(\mathcal{A})}(\omega)=0$, and so $c_{\omega}=1$.
There are $2^{n-1}-\frac{1+(-1)^n}{2}$ such contributions to
the sum \eqref{eq:torsum}.

\item
If $|\mathcal{A_{\omega}}|$ is odd, and $\mathcal{A_{\omega}}$ is decomposable, we may write
$\mathcal{A_{\omega}}=\mathcal{A}_{\omega}'\{\pm 1\}$, with the cabling done about the last component
of $\mathcal{A}_{\omega}'$.  Let $\mathcal{A}_{\omega}''$ be the sub-arrangement obtained by deleting
the last component of $\mathcal{A}_{\omega}'$.  Clearly, $|\mathcal{A}_{\omega}''|=|\mathcal{A_{\omega}}|-2$.
Formulas \eqref{eq:sw}, and \eqref{eq:torres} give
\begin{equation}  \label{eq:recursion}
\Delta_{L(\mathcal{A_{\omega}})}(-1,\dots ,-1)=-2\Delta_{L(\mathcal{A}_{\omega}')}(-1,\dots ,-1,1)
=4\Delta_{L(\mathcal{A}_{\omega}'')}(-1,\dots ,-1).
\end{equation}
Hence, $\delta(\mathcal{A_{\omega}})=\delta(\mathcal{A}_{\omega}'')$.   Iterating this decabling-deletion
procedure, we eventually reach an arrangement $\mathcal{B}$ for which the procedure
must stop.  There are two possibilities:
\begin{itemize}
\item
One is $\mathcal{B}=\mathcal{A}_3$, in which case $c_{\omega}=\delta(\mathcal{A}_3)=0$.

\item
The other is $\mathcal{B}\in \Gamma(\mathcal{A})$, in which case $c_{\omega}=\delta(\mathcal{B})$.
Clearly, any element of $\Gamma(\mathcal{A})$ can be reached by the above
procedure; thus there are $|\Gamma(\mathcal{A})|$ such contributions to the sum
\eqref{eq:torsum}.
\end{itemize}
\end{itemize}
\end{itemize}
This completes the proof.
\end{proof}

\begin{rem}  \label{rem:lowboundtors}
Note that $ 2^{n-1} -1\le \Tors_{2,1}(\mathcal{A})\le 2^{n}$.  If $\Tors_{2,1}(\mathcal{A}) = 2^{n-1}-1$,
and $n\ge 3$, then the top characteristic variety $V_1(\mathcal{A})$ is {\em not}
the union of translated subtori of $(\mathbb{C}^*)^n$.   For, otherwise,
at least one of the subtori must be of the form $T=\{t_1^{a_1}\cdots t_n^{a_n}-1=0\}$,
since $\Delta_{L(\mathcal{A})}(1,\dots,1)=0$.  But the torus $T$ has $2^{n-1}$ torsion
points of order $2$.
\end{rem}

\begin{cor} \label{cor:almostcompldec}
If all the proper subarrangements of $\mathcal{A}$ are completely decomposable,
then $\Tors_{2,1}(\mathcal{A})=2^{n-1}-\frac{1+(-1)^n}{2}+\delta(\mathcal{A})$.
\end{cor}

\begin{cor} \label{cor:compldec}
If $\mathcal{A}$ is completely decomposable, then $\Tors_{2,1}(\mathcal{A})=2^{n-1}$.
\end{cor}
\begin{proof}
The recursion formula~\eqref{eq:recursion}, together with
Example~\ref{ex:deltacx} imply that $\delta(\mathcal{A})=\frac{1+(-1)^n}{2}$,
and the conclusion follows from the previous corollary.
\end{proof}

\begin{exmp} \label{ex:h1}
The arrangement $\mathcal{A}=\mathcal{A}(31425)$ is horizontal, indecomposable, and all
its subarrangements are completely decomposable.  The (single variable)
Alexander polynomial is $\Delta_{\mathcal{A}}(t)=(t-1)^4(4t^2-t+4)$, and so $\delta(\mathcal{A})=0$.
From Corollary~\ref{cor:almostcompldec}, we get $\Tors_{2,1}(\mathcal{A})=16$.
\end{exmp}

\begin{exmp} \label{ex:h2}
The arrangement $\mathcal{A}=\mathcal{A}(314256)$ is decomposable, but not completely
decomposable, since it has $\mathcal{A}(31425)$ as a subarrangement. We have
$\Delta_{\mathcal{A}}(t)=(t^6-1)(t-1)^3(t+1)(3t^2-2t+3)$, and so $\delta(\mathcal{A})=1$.
From Theorem~\ref{thm:2tors}, we get $\Tors_{2,1}(\mathcal{A})=32$.
\end{exmp}

\begin{exmp} \label{ex:h3}
The arrangement $\mathcal{A}=\mathcal{A}(241536)$ is horizontal, indecomposable, and all
its proper subarrangements are completely decomposable.  We have
$\Delta_{\mathcal{A}}(t)=(t-1)^5(5t^4+6t^2+5)$, and so $\delta(\mathcal{A})=0$.
From Corollary~\ref{cor:almostcompldec}, we get $\Tors_{2,1}(\mathcal{A})=31$.
Hence, $V_1(\mathcal{A})$ is not a union of translated subtori of $(\mathbb{C}^*)^6$.
\end{exmp}

\section{Mazurovski\u{\i}'s arrangements}
\label{sec:maz}

In this section, we study the $2$-arrangements associated to
Mazurovski\u{\i}'s configurations.  Using their associated cablings,
we find infinitely many pairs of arrangements whose complements are
cohomologically equivalent, but not homotopy equivalent.

\subsection{}
In \cite{M1}, Mazurovski\u{\i} introduced a remarkable pair of
configurations of skew lines, $K$ and $L$, which have the same
linking numbers, but are not rigidly isotopic.  Let $\mathcal{K}=\mathcal{A}(K)$ and
$\mathcal{L}=\mathcal{A}(L)$ be the corresponding arrangements of planes.
The arrangement $\mathcal{K}$ is horizontal, with associated permutation
$\tau=(341256)$.  Moreover, $\mathcal{K}$ is completely decomposable, of depth $3$;
a minimal decomposition sequence is $(341256)\to (213)\to (12)\to (1)$.
The arrangement $\mathcal{L}$ is neither horizontal, nor decomposable.
Defining polynomials for $\mathcal{K}$ and $\mathcal{L}$ are given by
\begin{align*}  \label{polyKL}
f_{\mathcal{K}}(z,w)&=f(z,w)\cdot (z-7w),\\
f_{\mathcal{L}}(z,w)&=f(z,w)\cdot (z-\frac{6-7\ii}{2} w-
\frac{3+14\ii}{2}\bar{w}),
\end{align*}
where $f$ is the following defining polynomial for $\mathcal{A}(34125)$:
\begin{equation*}  \label{polyf}
\begin{split}
f(z,w)&=(z-\frac{5-5\ii}{2}w+\frac{3-5\ii}{2}\bar{w})
(z-\frac{7-10\ii}{2}w+\frac{3-10\ii}{2}\bar{w})\\
&\quad\times
(z-\frac{5-14\ii}{2}w-\frac{3+14\ii}{2}\bar{w})
(z-\frac{7-9\ii}{2}w-\frac{3+9\ii}{2}\bar{w})
(z-6w).\\
\end{split}
\end{equation*}

\begin{figure}
\[
\begingraph<0.5cm,0.5cm>[0.35pt]
\Label {$1$} at (1, 8)
\Label {$2$} at (1, 9)
\Label {$3$} at (1, 10)
\Label {$4$} at (1, 11)
\Label {$5$} at (1, 12)
\Label {$6$} at (1, 13)
\edge from (2, 8) to (3, 8)
\edge from (2, 9) to (3, 9)
\edge from (2, 10) to (3, 10)
\edge from (2, 11) to (3, 11)
\edge from (2, 12) to (3, 12)
\edge from (2, 13) to (3, 13)
\edge from (3, 8) to (4, 8)
\edge from (3, 9) to (4, 9)
\edge from (3, 10) to (4, 10)
\edge from (3, 11) to (4, 11)
\edge from (3, 12) to (3.3, 12.3)
\edge from (3.7, 12.7) to (4, 13)
\edge from (3, 13) to (4, 12)
\edge from (4, 8) to (5, 8)
\edge from (4, 9) to (5, 9)
\edge from (4, 10) to (5, 10)
\edge from (4, 11) to (4.3, 11.3)
\edge from (4.7, 11.7) to (5, 12)
\edge from (4, 12) to (5, 11)
\edge from (4, 13) to (5, 13)
\edge from (5, 8) to (6, 8)
\edge from (5, 9) to (6, 9)
\edge from (5, 10) to (5.3, 10.3)
\edge from (5.7, 10.7) to (6, 11)
\edge from (5, 11) to (6, 10)
\edge from (5, 12) to (6, 12)
\edge from (5, 13) to (6, 13)
\edge from (6, 8) to (7, 8)
\edge from (6, 9) to (6.3, 9.3)
\edge from (6.7, 9.7) to (7, 10)
\edge from (6, 10) to (7, 9)
\edge from (6, 11) to (7, 11)
\edge from (6, 12) to (7, 12)
\edge from (6, 13) to (7, 13)
\edge from (7, 8) to (7.3, 8.3)
\edge from (7.7, 8.7) to (8, 9)
\edge from (7, 9) to (8, 8)
\edge from (7, 10) to (8, 10)
\edge from (7, 11) to (8, 11)
\edge from (7, 12) to (8, 12)
\edge from (7, 13) to (8, 13)
\edge from (8, 8) to (9, 8)
\edge from (8, 9) to (9, 9)
\edge from (8, 10) to (9, 10)
\edge from (8, 11) to (9, 11)
\edge from (8, 12) to (8.3, 12.3)
\edge from (8.7, 12.7) to (9, 13)
\edge from (8, 13) to (9, 12)
\edge from (9, 8) to (10, 8)
\edge from (9, 9) to (10, 9)
\edge from (9, 10) to (10, 10)
\edge from (9, 11) to (9.3, 11.3)
\edge from (9.7, 11.7) to (10, 12)
\edge from (9, 12) to (10, 11)
\edge from (9, 13) to (10, 13)
\edge from (10, 8) to (11, 8)
\edge from (10, 9) to (11, 9)
\edge from (10, 10) to (10.3, 10.3)
\edge from (10.7, 10.7) to (11, 11)
\edge from (10, 11) to (11, 10)
\edge from (10, 12) to (11, 12)
\edge from (10, 13) to (11, 13)
\edge from (11, 8) to (12, 8)
\edge from (11, 9) to (11.3, 9.3)
\edge from (11.7, 9.7) to (12, 10)
\edge from (11, 10) to (12, 9)
\edge from (11, 11) to (12, 11)
\edge from (11, 12) to (12, 12)
\edge from (11, 13) to (12, 13)
\edge from (12, 8) to (13, 8)
\edge from (12, 9) to (13, 9)
\edge from (12, 10) to (13, 10)
\edge from (12, 11) to (13, 11)
\edge from (12, 12) to (12.3, 12.3)
\edge from (12.7, 12.7) to (13, 13)
\edge from (12, 13) to (13, 12)
\edge from (13, 8) to (14, 8)
\edge from (13, 9) to (14, 9)
\edge from (13, 10) to (14, 10)
\edge from (13, 11) to (14, 12)
\edge from (13, 12) to (13.3, 11.7)
\edge from (13.7, 11.3) to (14, 11)
\edge from (13, 13) to (14, 13)
\edge from (14, 8) to (15, 8)
\edge from (14, 9) to (15, 9)
\edge from (14, 10) to (15, 11)
\edge from (14, 11) to (14.3, 10.7)
\edge from (14.7, 10.3) to (15, 10)
\edge from (14, 12) to (15, 12)
\edge from (14, 13) to (15, 13)
\edge from (15, 8) to (16, 8)
\edge from (15, 9) to (16, 9)
\edge from (15, 10) to (16, 10)
\edge from (15, 11) to (16, 11)
\edge from (15, 12) to (16, 13)
\edge from (15, 13) to (15.3, 12.7)
\edge from (15.7, 12.3) to (16, 12)
\edge from (16, 8) to (17, 8)
\edge from (16, 9) to (17, 9)
\edge from (16, 10) to (17, 10)
\edge from (16, 11) to (17, 12)
\edge from (16, 12) to (16.3, 11.7)
\edge from (16.7, 11.3) to (17, 11)
\edge from (16, 13) to (17, 13)
\edge from (17, 8) to (18, 8)
\edge from (17, 9) to (18, 9)
\edge from (17, 10) to (18, 10)
\edge from (17, 11) to (18, 11)
\edge from (17, 12) to (17.3, 12.3)
\edge from (17.7, 12.7) to (18, 13)
\edge from (17, 13) to (18, 12)
\edge from (18, 8) to (19, 8)
\edge from (18, 9) to (19, 9)
\edge from (18, 10) to (19, 10)
\edge from (18, 11) to (19, 11)
\edge from (18, 12) to (19, 12)
\edge from (18, 13) to (19, 13)
\Label {$1$} at (1, 1)
\Label {$2$} at (1, 2)
\Label {$3$} at (1, 3)
\Label {$4$} at (1, 4)
\Label {$5$} at (1, 5)
\Label {$6$} at (1, 6)
\edge from (2, 1) to (3, 1)
\edge from (2, 2) to (3, 2)
\edge from (2, 3) to (3, 3)
\edge from (2, 4) to (3, 4)
\edge from (2, 5) to (3, 5)
\edge from (2, 6) to (3, 6)
\edge from (3, 1) to (4, 1)
\edge from (3, 2) to (4, 2)
\edge from (3, 3) to (4, 3)
\edge from (3, 4) to (4, 4)
\edge from (3, 5) to (3.3, 5.3)
\edge from (3.7, 5.7) to (4, 6)
\edge from (3, 6) to (4, 5)
\edge from (4, 1) to (5, 1)
\edge from (4, 2) to (5, 2)
\edge from (4, 3) to (5, 3)
\edge from (4, 4) to (4.3, 4.3)
\edge from (4.7, 4.7) to (5, 5)
\edge from (4, 5) to (5, 4)
\edge from (4, 6) to (5, 6)
\edge from (5, 1) to (6, 1)
\edge from (5, 2) to (6, 2)
\edge from (5, 3) to (5.3, 3.3)
\edge from (5.7, 3.7) to (6, 4)
\edge from (5, 4) to (6, 3)
\edge from (5, 5) to (6, 5)
\edge from (5, 6) to (6, 6)
\edge from (6, 1) to (7, 1)
\edge from (6, 2) to (6.3, 2.3)
\edge from (6.7, 2.7) to (7, 3)
\edge from (6, 3) to (7, 2)
\edge from (6, 4) to (7, 4)
\edge from (6, 5) to (7, 5)
\edge from (6, 6) to (7, 6)
\edge from (7, 1) to (7.3, 1.3)
\edge from (7.7, 1.7) to (8, 2)
\edge from (7, 2) to (8, 1)
\edge from (7, 3) to (8, 3)
\edge from (7, 4) to (8, 4)
\edge from (7, 5) to (8, 5)
\edge from (7, 6) to (8, 6)
\edge from (8, 1) to (9, 1)
\edge from (8, 2) to (9, 2)
\edge from (8, 3) to (9, 3)
\edge from (8, 4) to (9, 4)
\edge from (8, 5) to (8.3, 5.3)
\edge from (8.7, 5.7) to (9, 6)
\edge from (8, 6) to (9, 5)
\edge from (9, 1) to (10, 1)
\edge from (9, 2) to (10, 2)
\edge from (9, 3) to (9.3, 3.3)
\edge from (9.7, 3.7) to (10, 4)
\edge from (9, 4) to (10, 3)
\edge from (9, 5) to (10, 5)
\edge from (9, 6) to (10, 6)
\edge from (10, 1) to (11, 1)
\edge from (10, 2) to (11, 2)
\edge from (10, 3) to (11, 3)
\edge from (10, 4) to (11, 5)
\edge from (10, 5) to (10.3, 4.7)
\edge from (10.7, 4.3) to (11, 4)
\edge from (10, 6) to (11, 6)
\edge from (11, 1) to (12, 1)
\edge from (11, 2) to (12, 2)
\edge from (11, 3) to (11.3, 3.3)
\edge from (11.7, 3.7) to (12, 4)
\edge from (11, 4) to (12, 3)
\edge from (11, 5) to (12, 5)
\edge from (11, 6) to (12, 6)
\edge from (12, 1) to (13, 1)
\edge from (12, 2) to (13, 3)
\edge from (12, 3) to (12.3, 2.7)
\edge from (12.7, 2.3) to (13, 2)
\edge from (12, 4) to (13, 4)
\edge from (12, 5) to (13, 5)
\edge from (12, 6) to (13, 6)
\edge from (13, 1) to (14, 1)
\edge from (13, 2) to (14, 2)
\edge from (13, 3) to (13.3, 3.3)
\edge from (13.7, 3.7) to (14, 4)
\edge from (13, 4) to (14, 3)
\edge from (13, 5) to (14, 5)
\edge from (13, 6) to (14, 6)
\edge from (14, 1) to (15, 1)
\edge from (14, 2) to (15, 2)
\edge from (14, 3) to (15, 3)
\edge from (14, 4) to (15, 4)
\edge from (14, 5) to (15, 6)
\edge from (14, 6) to (14.3, 5.7)
\edge from (14.7, 5.3) to (15, 5)
\edge from (15, 1) to (16, 1)
\edge from (15, 2) to (16, 2)
\edge from (15, 3) to (16, 3)
\edge from (15, 4) to (16, 5)
\edge from (15, 5) to (15.3, 4.7)
\edge from (15.7, 4.3) to (16, 4)
\edge from (15, 6) to (16, 6)
\edge from (16, 1) to (17, 1)
\edge from (16, 2) to (17, 2)
\edge from (16, 3) to (16.3, 3.3)
\edge from (16.7, 3.7) to (17, 4)
\edge from (16, 4) to (17, 3)
\edge from (16, 5) to (17, 5)
\edge from (16, 6) to (17, 6)
\edge from (17, 1) to (18, 1)
\edge from (17, 2) to (18, 2)
\edge from (17, 3) to (18, 3)
\edge from (17, 4) to (18, 4)
\edge from (17, 5) to (17.3, 5.3)
\edge from (17.7, 5.7) to (18, 6)
\edge from (17, 6) to (18, 5)
\edge from (18, 1) to (19, 1)
\edge from (18, 2) to (19, 2)
\edge from (18, 3) to (19, 3)
\edge from (18, 4) to (19, 4)
\edge from (18, 5) to (19, 5)
\edge from (18, 6) to (19, 6)
\endgraph
\]
\caption{Mazurovski\u\i's pair: The half-braids $\a_{\mathcal{K}}$ (top) and
$\a_{\mathcal{L}}$ (bottom).}
\label{fig:KandL}
\end{figure}

The half-braids associated to $\mathcal{K}$ and $\mathcal{L}$ are pictured in
Figure~\ref{fig:KandL}.  We see that the linking numbers of $L(\mathcal{K})$ are
$l_{1,4}=l_{2,4}=l_{1,3}=l_{2,3}=-1$, and all other $l_{i,j}=1$,
whereas the linking numbers of $L(\mathcal{L})$ are
$l_{1,5}=l_{2,5}=l_{1,4}=l_{2,4}=-1$,
and all other $l_{i,j}=1$.  The reordering of the components of $L(\mathcal{L})$
that fixes $1$, $2$, $6$ and permutes $3$, $4$, $5$ to $4$, $5$, $3$
identifies the linking numbers of $L(\mathcal{K})$ and $L(\mathcal{L})$.   Thus,
$H^*(X(\mathcal{K});\mathbb{Z})\cong H^*(X(\mathcal{L});\mathbb{Z})$.

\subsection{} \label{subsection:kandl}
In order to distinguish between the cohomologically equivalent
arrangements $\mathcal{K}$ and $\mathcal{L}$, we turn to their characteristic varieties.
From Figure~\ref{fig:KandL}, we see that the reduced half-braids
of $\mathcal{K}$ and $\mathcal{L}$ are:
\begin{equation*}  \label{halfKL}
\check{\a}_{\mathcal{K}}=\sigma_4\sigma_3\sigma_2\sigma_1\sigma_4\sigma_3^{-1}\sigma_2^{-1}\sigma_4^{-1}\sigma_3^{-1}\sigma_4, \quad
\check{\a}_{\mathcal{L}}=\sigma_4\sigma_2\sigma_3^{-1}\sigma_2\sigma_1^{-1}\sigma_2\sigma_4^{-1}\sigma_3^{-1}\sigma_2\sigma_4.
\end{equation*}
The braids $\xi=\Delta_5 \check{\a}^{-1}\in P_5$ are expressed in terms
of the pure braid generators, as follows.  For $\mathcal{K}$, which is horizontal,
Proposition~\ref{prop:combed} yields $\xi_{\mathcal{K}}=A_{1,3}A_{2,3}A_{1,4}A_{2,4}$.
For $\mathcal{L}$, it is more convenient to work with the conjugate
$\xi'_{\mathcal{L}}=\delta^{-1} \xi_{\mathcal{L}}\delta $, where $\delta=\sigma_1\sigma_3\sigma_4^{-1}$.
Routine combing of the braid yields
$\xi'_{\mathcal{L}}=A_{1,3}A_{2,3}A_{4,5}A_{1,4}A_{4,5}^{-1}A_{2,4}$.

The Artin representation of $\xi=\xi_{\mathcal{K}}$ in the basis
$y_1=x_1$, $y_2=x_3$, $y_3=x_1x_2$, $y_4=x_1x_2x_3x_4$, $y_5=x_5$ is
given by $\xi(y_1)=y_4y_3^{-1}y_1y_3y_4^{-1}$, $\xi(y_2)=y_3y_2y_3^{-1}$,
$\xi(y_3)=y_4y_3y_4^{-1}$, $\xi(y_4)=y_4$, and $\xi(y_5)=y_5$.
The Alexander matrix of $\mathcal{K}$ is:
\begin{equation*}  \label{eq:alexK}
\begin{pmatrix}
t_6-t_4^{2}t_3^{-2}  & 0 &  t_4^{2}t_3^{-2}(t_3+1)(1-t_1) 	& (t_4+1)(t_1-1) & 0 & t_1-1 \\
0             							& t_6-t_3^{2} & (t_4+t_3)(t_2-1)						& (1-t_3)(t_2-1)	& 0 & t_2-1 \\
0             							& 0 & t_6-t_4^{2}  																			& (t_4+1)(t_3-1) & 0 & t_3-1 \\
0             							& 0 & 0                 														&					t_6-1						& 0 & t_4-1 \\
0             							& 0 & 0                 														& 0										& t_6-1 & t_5-1 \\
\end{pmatrix}
\end{equation*}

An elementary computation shows that the bottom variety $V_4(\mathcal{K})$
has $6$ irreducible components---$3$ codimension~$4$ translated
subtori of $(\mathbb{C}^{*})^6$, and $3$ codimension~$5$ subtori passing
through $\mathbf{1}$---given by the following equations:
\begin{align*}
&\{t_6-1=t_4+1=t_3+1=t_2-1=0\},\\
&\{t_6-1=t_4+1=t_3-1=t_1-1=0\},\\
&\{t_6-1=t_5-1=t_4-1=t_3+1=0\},\\
&\{t_6-1=t_5-1=t_4-1=t_3-1=t_1-1=0\},\\
&\{t_6-1=t_5-1=t_4-1=t_3-1=t_2-1=0\},\\
&\{t_6-1=t_4-1=t_3-1=t_2-1=t_1-1=0\}.
\end{align*}

The primary decomposition of the ideal $E_4(\mathcal{L})$ is much harder to find.
The implementation in {\it Macaulay~2}~\cite{GS} of the Eisenbud, Huneke,
and Vasconcelos algorithm yields such a decomposition, and the result is
that $V_4(\mathcal{L})=V_4(\mathcal{K})$.  Thus, the bottom varieties fail to distinguish
between the $\mathcal{K}$ and $\mathcal{L}$ arrangements.

Let us then consider the top varieties.  It is readily seen that
the Alexander polynomial of $\mathcal{K}$ is $\Delta_{\mathcal{K}}(t_1,\dots,t_6)=
(t_6-1)(t_6-t_3^{2})(t_6-t_4^{2})(t_6-t_4^{2}t_3^{-2})$, and so
$V_1(\mathcal{K})$ is the union of $4$ codimension~$1$ subtori
of $(\mathbb{C}^{*})^6$.  Since $\mathcal{K}$ is completely decomposable,
Corollary~\ref{cor:compldec} implies that $\Tors_{2,1}(\mathcal{K})=32$.

The Alexander polynomial of $\mathcal{L}$ may be computed using
{\it Mathematica}~\cite{W}.  The result is too long to be displayed here,
but suffices to say that it is an irreducible polynomial over $\mathbb{Z}$,
consisting of $667$ monomials.   Direct computation shows that the
single variable Alexander polynomial is
$\Delta_{\mathcal{L}}(t)=3(t-1)^5(3t^2 - 2t + 3)^2$.  Hence $\delta({\mathcal{L}})=0$.
Since, as is readily checked, all proper subarrangements of $\mathcal{L}$ are
completely decomposable, Corollary~\ref{cor:almostcompldec}
implies that $\Tors_{2,1}(\mathcal{L})=31$.

Thus, $\Tors_{2,1}(\mathcal{K})\ne\Tors_{2,1}(\mathcal{L})$.  (As noted in
Remark~\ref{rem:lowboundtors}, this arithmetic difference
translates into a geometric  difference: $V_1(\mathcal{K})$ is a
union of subtori, whereas $V_1(\mathcal{L})$ is not  even the union
of translated subtori.)   Appealing now to Theorem~\ref{thm:invariant},
we conclude that the complements of $\mathcal{K}$ and $\mathcal{L}$ are not homotopy
equivalent, although, as mentioned previously, they are cohomologically
isomorphic.  This answers Ziegler's question from \cite{Z}.

\subsection{}
We now use cablings of $\mathcal{K}$ and $\mathcal{L}$ to show that the above phenomenon
happens for arrangements of $n$ planes, for any $n\ge 6$.

\begin{thm}  \label{thm:mazcables}
Let $\mathcal{K}$ and $\mathcal{L}$ be Mazurovski\u{\i}'s arrangements of $6$
transverse planes in $\mathbb{R}^4$. Let $\mathcal{K}\{r\}$ and $\mathcal{L}\{r\}$ be their $r$-cables.
Then, for each $r\ge 0$,
\begin{alphenum}
\item
$H^*(X(\mathcal{K}\{r\});\mathbb{Z})\cong H^*(X(\mathcal{L}\{r\});\mathbb{Z})$;
\vskip 4pt
\item
$X(\mathcal{K}\{r\})\not\simeq X(\mathcal{L}\{r\})$.
\end{alphenum}
\end{thm}

\begin{proof}  As noted above, the links of $\mathcal{K}$ and $\mathcal{L}$
have the same linking numbers.  Hence, the links of $\mathcal{K}\{r\}$ and
$\mathcal{L}\{r\}$ have the same linking numbers.  This implies that the complements
of $\mathcal{K}\{r\}$ and $\mathcal{L}\{r\}$ are cohomologically equivalent.

Although $\mathcal{K}$ and $\mathcal{L}$ are distinguished by their $2$-torsion points,
their cables are not.  Indeed, for $r \ge 1$,
$\Tors_{2,1}(\mathcal{K}\{r\})=\Tors_{2,1}(\mathcal{L}\{r\})=2^{r+5}$,
as can be deduced from Theorem~\ref{thm:2tors} for $r=1$,
and from Corollary~\ref{cor:vcable} for $r>1$.
Hence, we turn to $3$-torsion points.  A {\it Mathematica} computation
shows that $\Tors_{3,1}(\mathcal K\{1\}) = 3^5\cdot 7$ and
$\Tors_{3,1}(\mathcal L\{1\}) = 3^3\cdot 61$.
From Corollary~\ref{cor:vcable}, we get
\[
\Tors_{3,1}(\mathcal K\{r\}) = 3^{r+4}\cdot 7,
\quad\text{and}\quad
\Tors_{3,1}(\mathcal L\{r\}) =  3^{r+2}\cdot 61,
\]
showing that the respective complements are indeed
not homotopy equivalent.
\end{proof}

\subsection{}
Mazurovski\u{\i} introduced in \cite{M1} another interesting configuration
of $6$ lines, which he called $M$.  Like the $L$ configuration,
the $M$ configuration is non-horizontal and indecomposable (they
are the only two such configurations of $6$ lines, up to rigid isotopy
and mirror images).  But, unlike $L$, the $M$ configuration does
not have the linking numbers of any horizontal configuration.
Let $\mathcal{M}=\mathcal{A}(M)$ be the corresponding arrangement.
A defining polynomial for it is:
\begin{equation*}  \label{Mpoly}
\begin{split}
f_{\mathcal{M}}(z,w)&=
(z-(10-\ii)w+
(9-4\ii)\bar{w})
(z-(3-4\ii)w-
(1+4\ii)\bar{w})
\\
&\quad\times
(z-\frac{5-10\ii}{2}w+
\frac{1-10\ii}{2}\bar{w})
(z-(6-5\ii)w+
\frac{1-10\ii}{2}\bar{w})
\\
&\quad\times
(z-\frac{21-29\ii}{4}w+
\frac{1-9\ii}{4}\bar{w})
(z-6w).
\end{split}
\end{equation*}
\noindent
The reduced braids associated to $\mathcal{M}$ are:
\[
\check{\a}_{\mathcal{M}}=\sigma_2\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}\sigma_4^{-1}\sigma_2\sigma_1^{-1}\sigma_2^{-1}
\quad\text{and}\quad
\xi_{\mathcal{M}}=A_{2,4}A_{1,2}A_{3,4}A_{1,5}A_{3,5}.
\]

A {\it Macaulay~2} computation shows that $V_4(\mathcal{M})$ consists
of sixteen $2$-torsion points.
A {\it Mathematica} computation reveals that the Alexander polynomial
of $\mathcal{M}$ is an irreducible polynomial over $\mathbb{Z}$, consisting of $317$
monomials. The single variable Alexander polynomial is
$\Delta_{\mathcal{M}}(t)=(t-1)^5(t^2-t+1)(t^6-5t^5-t^4-6t^3-t^2-5t+1)$,
and so $\delta(\mathcal{M})=0$.  Theorem~\ref{thm:2tors} gives
$\Tors_{2,1}(\mathcal{M})=31$, the same as for $\mathcal{L}$.  (Thus,
the top variety of $\mathcal{M}$ is not the union of translated subtori.)
On the other hand, a computation yields $\Tors_{3,1}(\mathcal{L})=527$ and
$\Tors_{3,1}(\mathcal{M})=421$, showing that $X(\mathcal{L})\not\simeq X(\mathcal{M})$.

\section{Classification of $2$-Arrangements of $n\le 6$ Planes}
\label{sec:classification}

We start by reviewing the rigid isotopy classification
of arrangements of up to $6$ planes.  We then show that
the invariants introduced in \S\ref{sec:detideals} are
powerful enough to classify up to homotopy the complements of
such arrangements.

\subsection{}
As noted in \S\ref{rigidisotop}, rigid isotopy types of
$2$-arrangements in $\mathbb{R}^4$ are in one-to-one correspondence with
rigid isotopy types of skew-lines configurations in $\mathbb{R}^3$.
An important concept introduced by Viro~\cite{V} was that of
a mirror image of a configuration.  We now translate this notion
to arrangements.

\begin{defn}
An arrangement $\mathcal{A}'$ is called a {\em mirror image} of $\mathcal{A}$
if there is a reflection of $\mathbb{R}^4$ sending $\mathcal{A}$ to $\mathcal{A}'$.
The mirror image of $\mathcal{A}$ is unique up to rigid isotopy; we
denote it by $\overline{\mathcal{A}}$.  An arrangement $\mathcal{A}$ which is not isotopic
to $\overline{\mathcal{A}}$ is called {\em non-mirror}.
\end{defn}

As shown by Viro, there exist many non-mirror arrangements.
For example, the complex arrangement, $\mathcal{A}_n$, and its mirror image
under complex conjugation, $\overline{\mathcal{A}}_n$, are not rigidly isotopic
provided $n\ge 3$.  Also, an arbitrary arrangement of $n$ lines is
non-mirror, provided $n\equiv 3 \pmod{4}$.

Viro~\cite{V} and Mazurovski\u{\i}~\cite{M1} classified, up to rigid
isotopy, all configurations $6$ lines or less.  For up to $5$ lines,
linking numbers invariants were used.  For $6$ lines, those invariants
cannot tell apart the $K$ and $L$ configurations.  For that, the Morton
trace of the reduced full-braid is used in~\cite{M1}.  Translated to
arrangements, the complete list of the $33$ rigid isotopy types is
as follows:
\begin{align*}  \label{list6}
n=1: &\quad\mathcal{A}(1)\\
n=2: &\quad\mathcal{A}(12)\\
n=3: &\quad\mathcal{A}(123)^*\\
n=4: &\quad\mathcal{A}(1234)^*,\: \mathcal{A}(2134)\\
n=5: &\quad\mathcal{A}(12345)^*,\: \mathcal{A}(21345)^*,\: \mathcal{A}(21435)^*,\: \mathcal{A}(31425) \\
n=6: &\quad\mathcal{A}(123456)^*,\: \mathcal{A}(213456)^*,\: \mathcal{A}(321456),\:
\mathcal{A}(214356)^*,\: \mathcal{A}(215436)^*, \\
     &\quad\mathcal{A}(312546),\:	\mathcal{A}(341256)^*,\: \mathcal{A}(314256)^*,\:
\mathcal{A}(241536),\: \mathcal{A}(L)^*,\: \mathcal{A}(M)^*
\end{align*}
where $\mathcal{A}^*$ stands for a non-mirror arrangement $\mathcal{A}$ and its mirror
image $\overline{\mathcal{A}}$.

\subsection{}
We now turn to the homotopy classification of complements of arrangements.
Clearly, arrangements that are either rigidly isotopic, or mirror images
of one another, have diffeomorphic (and thus, homotopy equivalent) complements.
Thus, if we delete from the above list the mirror image $\overline{\mathcal{A}}$ from
each pair $\mathcal{A}^*=(\mathcal{A},\overline{\mathcal{A}})$, we are left with a list $20$
arrangements, such that, the complement of any arrangement of $n\le 6$ planes
is homotopy equivalent to the complement of one of the arrangements in
this shorter list.   Table~\ref{tab:classify} shows that there are no
repetitions among the homotopy types of these $20$ arrangements.
Hence, we have the following.

\begin{thm}  \label{thm:classify6}
For $2$-arrangements of $n\le 6$ planes in $\mathbb{R}^4$,
the homotopy types of complements are in one-to-one
correspondence with the rigid isotopy types modulo mirror images.
\end{thm}

\begin{table}
\begin{minipage}{1.02\columnwidth}
\[
\begin{array}{|c||c|c||c|c|c|}
\hline
\multicolumn{1}{|c||}{n}
&\multicolumn{1}{|c|}{\mathcal{A}}
&\multicolumn{1}{|c||}{\depth}
&\multicolumn{1}{|c|}{\Sigma_{n-2}}
&\multicolumn{1}{|c|}{\Tors_{2,1}}
&\multicolumn{1}{|c|}{\Tors_{3,1}}
\\ \hline\hline
1 & \mathcal{A}(1)        & 0 & 0  & 0 & 0 \\
\hline
2 & \mathcal{A}(12)       & 1 & 0  & 1 & 1 \\
\hline
3 & \mathcal{A}(123)      & 1 & 1 & 4 & 9\\
\hline
4 & \mathcal{A}(1234)     & 1 & 1  & 8 & 27 \\
& \mathcal{A}(2134)       & 2 & 2,3_2  & 8 & 45 \\
\hline
5 & \mathcal{A}(12345)    & 1 & 1 & 16 & 81 \\
& \mathcal{A}(21345)      & 2 & 2,3,4 & 16 & 135  \\
& \mathcal{A}(21435)      & 2 & 3,4_4  & 16 & 171 \\
& \mathcal{A}(31425)      & - & 5_{11} & 16 & 141 \\
\hline
6 & \mathcal{A}(123456)   & 1 & 1 & 32 & 243  \\
& \mathcal{A}(213456)     & 2 & 2,3,5  & 32 & 405 \\
& \mathcal{A}(321456)     & 2 & 2,4_2  & 32 & 405 \\
& \mathcal{A}(215436)     & 2 & 3,4_2,5_2  & 32 & 513 \\
& \mathcal{A}(214356)     & 2 & 3,4_2,5_3  & 32 & 513 \\
& \mathcal{K}=\mathcal{A}(341256) & 3 & 4_3,5_3 & 32 & 567 \\
& \mathcal{A}(312546)     & 3 & 4,5_6,6 & 32 & 585 \\
& \mathcal{A}(314256)     & - & 5_6,6_5 & 32 & 495 \\
& \mathcal{A}(241536)     & - & 5_2,6_{13} & 31 & 513 \\
& \mathcal{L}						      & - & 4_3,5_3 & 31 & 527 \\
& \mathcal{M} 			     	  & - & 6_{16} & 31 & 421 \\
\hline
\end{array}
\]
\begin{center}
\caption{Arrangements of $n\le 6$ planes:
Sequence of codimensions of components of $V_{n-2}$---where
$i_k$ stands for $i$ repeated $k$ times---and number of
$2$- and $3$-torsion points on $V_1$.\label{tab:classify}}
\vspace{-2pc}
\end{center}
\end{minipage}
\end{table}