# Triangular arrangements on the projective plane

###### Abstract

In this work we study line arrangements consisting in lines passing through
three non aligned points. We call them triangular arrangements. We prove that
any combinatorics of a triangular arrangement is always realized by a Roots-
of-Unity-Arrangement which is a particular class of triangular arrangements
obtained by deleting lines from the full monomial arrangement ${\cal
A}^{3}_{3}(N)$, for $N$ big enough. Among these Roots-of-Unity-Arrangements we
characterize the free ones. Finally, we give two triangular arrangements
having the same weak combinatorics (that means the same number $t_{i}$ of
points with multiplicity $i,i\geq 2$), such that one is free but the other one
is not.

Simone Marchesi
Universitat de Barcelona
Departament de Matemàtiques i Informàtica
Gran Via de les Corts Catalanes, 585
08007 Barcelona, Spain
e-mail<EMAIL_ADDRESS>

Jean Vallès
Université de Pau et des Pays de l’Adour
LMAP-UMR CNRS 5142, Avenue de l’Université - BP 1155
64013 Pau Cedex, France
e-mail<EMAIL_ADDRESS>
tel: +33559407517

Mathematics Subject Classification (2000): 52C35, 14F05, 32S22

Acknowledgements. We would like to thank the University of Campinas and the
Université de Pau et des Pays de l’Adour for the hospitality and for providing
the best working conditions (in 2017 in Pau and in 2018 in Campinas).

The first author was partially supported by Fundação de Amparo à Pesquisa do
Estado de São Paulo (FAPESP) grant 2017/03487-9, by CNPq grant 303075/2017-1
and by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior -
Brasil (CAPES) - Finance Code 001.

The second author was partially supported by Fundação de Amparo à Pesquisa do
Estado de São Paulo (FAPESP), grant 2018/08524-2.

##  1 Introduction

A line arrangement ${\cal A}=\\{l_{1},\ldots,l_{n}\\}$ in ${\mathbb{P}}^{2}$
is a finite set of distinct lines. The union of these lines forms a divisor
defined by an equation $f=\prod_{i}f_{i}=0$, where $f_{i}=0$ is the equation
defining $l_{i}$. The sheaf ${\cal T}_{{\cal A}}$ of vector fields tangent to
this arrangement can be defined as the kernel of the Jacobian map, which
means:

$\begin{CD}0@>{}>{}>{\cal T}_{\mathcal{A}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}^{3}@>{\nabla
f}>{}>\mathcal{J}_{f}(n-1)@>{}>{}>0,\end{CD}$

where $\mathcal{J}_{f}$ is the ideal sheaf generated by the three partial
derivatives $\nabla f=(\partial_{x}f,\partial_{y}f,\partial_{z}f)$. This
ideal, called Jacobian ideal, defines the Jacobian scheme supported by the
singular points of the arrangement; for instance when ${\cal A}$ is generic
(i.e. it consists of $n$ lines in general position) then $\mathcal{J}_{f}$
defines $\binom{n}{2}$ distinct points. The sheaf ${\cal T}_{{\cal A}}$, which
is a reflexive sheaf over ${\mathbb{P}}^{2}$ and therefore a vector bundle, is
a basic tool to study the link between the geometry, the topology and the
combinatorics of ${\cal A}$.

When ${\cal T}_{{\cal A}}={\cal O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-b)$ the arrangement ${\cal A}$ is called free
arrangement and the integers $(a,b)$ are its exponents; this notion of
freeness was first introduced for a reduced divisor by Saito [13] and by Terao
[14] for hyperplane arrangements.

In [12] Terao conjectures that freeness depends only on the combinatorics of
${\cal A}$, where the combinatorics is described by the intersection lattice
$L({\cal A})$, that is the set of all the intersections of lines in ${\cal
A}$. More precisely, if two arrangements ${\cal A}_{0}$ and ${\cal A}_{1}$
have the same combinatorics (a bijection between $L({\cal A}_{0})$ and
$L({\cal A}_{1})$) and one of them is free then the other one is also free (of
course with the same exponents). This conjecture, despite all the efforts, is
proved, for line arrangements, only up to $13$ lines (see [3]).

A weaker problem concerns the weak combinatorics. The weak combinatorics of a
given arrangement of $n$ lines is given by the knowledge of the integers
$t_{i}$, $i\geq 2$, which are the number of points with multiplicity exactly
$i$ of the arrangement. Let us mention the following formula, proved by
Hirzebruch in [9], involving these numbers (when $t_{n}=t_{n-1}=t_{n-2}=0$):

$t_{2}+t_{3}\geq n+\sum_{i\geq 1}it_{i+4}.$ (1)

In this paper we consider triangular arrangements meaning that there exist
three non aligned points such that every line of the arrangement passes
through one of these points; for instance, reflection arrangements described
in [12] and again in [9] belong to this family.

The paper is organized in the following way.
In Section 2 we will prove preliminary results on triangular arrangement and
introduce the concept of Roots-of-Unity-Arrangement (RUA), which is, in our
opinion, a central family to be studied in the context of arrangements, as
confirmed by the following results.
Indeed, in Section 3 we prove a key result, which states that for any
triangular arrangement it is possible to find a RUA with exactly the same
combinatorics.
Sections 4 and 5 are devoted to the characterization of freeness for RUA,
which we prove it depends on the combinatorics and therefore such family
represents an evidence to Terao’s conjecture.
In Section 6, we show that Terao’s conjecture does not extend to the
assumption of weak-combinatorics. Indeed we prove the following.

######  Theorem.

There exist pairs of arrangements possessing the same weak combinatorics such
that one is free and the other is not.

Finally, in Section 7, we discuss Terao’s conjeture for triangular
arrangements in general.

##  2 The inner triple points of a triangular arrangement

In this section we will explicit the importance of the set $T$ of the triple
points, defined by the triangular arrangement, which are not the vertices of
the triangle. We will describe in particular the case where $T$ is either
empty or a complete intersection.

Let $A,B,C$ be three points not aligned. A line arrangement such that any of
its line passes through $A,B$ or $C$ is called triangular arrangement. If one
of the three side lines is missing we will say that the triangular arrangement
is uncomplete. The set of triangular arrangements consisting in $a+1$ lines
through $A$, $b+1$ lines through $B$, $c+1$ lines through $C$ and the three
side lines will be denoted by $\mathrm{Tr}(a,b,c)$; these arrangements posses
$a+b+c$ lines.

######  Proposition 2.1.

Let $\mathcal{A}\in\mathrm{Tr}(a,b,c)$ then, there is an exact sequence

$\begin{CD}0@>{}>{}>{\cal T}_{\mathcal{A}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c)@>{}>{}>\mathcal{J}_{T}(-1)@>{}>{}>0,\end{CD}$

where $T$ is the smooth finite set of inner triple points (i.e. $A,B,C\notin
T$).

###### Proof.

Let $Z\subset\check{\mathbb{P}}^{2}$ the finite set of points corresponding by
projective duality to the lines of ${\cal A}$. Since $Z$ is contained in a
triangle $\Delta$, this induces the following exact sequence:

$\begin{CD}0@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}(-2)@>{}>{}>\mathcal{J}_{Z}(1<EMAIL_ADDRESS>
The vertices of $\Delta$ belong to $Z$, then it implies that

$\mathcal{J}_{Z/\Delta}(1)={\cal O}_{L_{1}}(-a)\oplus{\cal
O}_{L_{2}}(-b)\oplus{\cal O}_{L_{3}}(-c),$

where $L_{1},L_{2},L_{3}$ are the edges of $\Delta$ corresponding by duality
to the vertices $A,B,C$. Let us consider the incidence variety
${\mathbb{F}}=\\{(x,l)\in{\mathbb{P}}^{2}\times\check{\mathbb{P}}^{2}\mid x\in
l\\}$ and the projection maps $p:{\mathbb{F}}\rightarrow{\mathbb{P}}^{2}$ and
$q:{\mathbb{F}}\rightarrow\check{\mathbb{P}}^{2}$. According to [7, Theorem
1.3] ${\cal T}_{\cal A}=p_{*}q^{*}(\mathcal{J}_{Z}(1))$ and the Fourier-Mukai
transform $p_{*}q^{*}$ applied to the above exact sequence gives:

$\begin{CD}0@>{}>{}>{\cal T}_{\mathcal{A}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c)@>{}>{}>{\cal O}_{{\mathbb{P}}^{2}}(-1)\\\
@V{}V{}V\\\ 0@<{}<{}<\mathrm{R}^{1}p_{*}q^{*}{\cal
O}_{L_{1}}(-a)\oplus\mathrm{R}^{1}p_{*}q^{*}{\cal
O}_{L_{2}}(-b)\oplus\mathrm{R}^{1}p_{*}q^{*}{\cal
O}_{L_{3}}(-c)@<{}<{}<\mathrm{R}^{1}p_{*}q^{*}\mathcal{J}_{Z}(1),\end{CD}$

The sheaf $\mathrm{R}^{1}p_{*}q^{*}\mathcal{J}_{Z}(1)$ is supported on the
scheme of triple points defined by $\mathcal{A}$, while, the last sheaf of the
sequence is supported on the vertices of the triangle $(ABC)$. Therefore the
kernel of the last map is the structural sheaf of the set of triple inner
points $T$. This implies that we have the following exact sequence

$\begin{CD}0@>{}>{}>{\cal T}_{\mathcal{A}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c<EMAIL_ADDRESS>
∎

######  Remark 2.2.

By the hypothesis on $\mathcal{A}$, the set $T$ is smooth. Its length is
related to the second Chern class of ${\cal T}_{\mathcal{A}}$, more precisely
we have that $c_{1}({\cal T}_{\mathcal{A}})=1-a-b-c$ and

$c_{2}({\cal
T}_{\mathcal{A}})=\binom{a+b+c-1}{2}-\binom{a}{2}-\binom{b}{2}-\binom{c}{2}-\mid
T\mid=(ab+bc+ac-a-b-c+1)-\mid T\mid.$ (2)

######  Theorem 2.3.

The bundle ${\cal T}_{\mathcal{A}}$ is free with exponents $(a+b-1,c)$ if and
only if $T$ is a complete intersection $(a-1,b-1)$.

######  Remark 2.4.

This is the most unbalanced splitting that is allowed for ${\cal
A}\in\mathrm{Tr}(a,b,c).$ Indeed, $|T|$ cannot be bigger than $(a-1)(b-1)$.

###### Proof.

Assume that $T$ is a complete intersection $(a-1,b-1)$. Since $T$ is the locus
of inner triple points, the curve defined by $(a-1)$ lines passing through $A$
contains $T$ and the curve defined by $(b-1)$ lines passing through $B$
contains also $T$. These two curves generate the ideal defining $T$, which
implies that the kernel of the last map of the exact sequence

$\begin{CD}0@>{}>{}>{\cal T}_{\mathcal{A}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c)@>{}>{}>\mathcal{J}_{T}(-1)@>{}>{}>0,\end{CD}$

is ${\cal O}_{{\mathbb{P}}^{2}}(-a-b+1)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-c)$.
This proves that ${\cal T}_{\mathcal{A}}$ is free with exponents $(a+b-1,c)$.

Reciprocally, if ${\cal T}_{\mathcal{A}}={\cal
O}_{{\mathbb{P}}^{2}}(-a-b+1)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-c)$ then we
have ${\rm H}^{0}(\mathcal{J}_{T}(a-1))\neq 0$ and ${\rm
H}^{0}(\mathcal{J}_{T}(a-2))=0$. Moreover, the length of $T$, given by the
numerical invariant of the above exact sequence, is $(a-1)(b-1)$ and this
proves that $T$ is a complete intersection $(a-1,b-1)$. ∎

######  Remark 2.5.

If $c\geq a+b-1$ the splitting type of ${\cal T}_{\cal A}$ along the lines
joining $A$ to $C$ or $B$ to $C$ is fixed and it is ${\cal
O}_{l}(1-a-b)\oplus{\cal O}_{l}(-c)$; this is a consequence of [15, Theorem
3.1]. Therefore, under the condition $c\geq a+b-1$ the arrangement is free if
and only if

${\cal T}_{{\cal A}}={\cal O}_{{\mathbb{P}}^{2}}(1-a-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c).$

That’s why, if we want to describe all the possible splitting types of free
triangular arrangements with $a+b+c$ lines ($a+1$ by $A$, $b+1$ by $B$ and
$c+1$ by $C$), we can assume that $c\leq a+b-2$. Then the biggest possible gap
$|a+b-1-c|$ is realized by the complete intersection $(a-1)(b-1)$, in
particular it could be described with a Roots-of-Unity-Arrangement (see the
definition below): let $\zeta$ be a primitive $(c-1)$-root of unity, the
arrangement

$xyz\prod_{i=0}^{a-2}(x-\zeta^{i}y)\prod_{j=0}^{b-2}(y-\zeta^{j}z)\prod_{k=0}^{c-2}(x-\zeta^{k}z)=0$

belongs to $\mathrm{Tr}(a,b,c)$ and it is free with exponents $(a+b-1,c)$.

######  Definition 2.6.

A triangular arrangement $\mathcal{A}$ of $a+b+c$ lines, defined by an
equation

$xyz\prod_{i=1}^{a-1}(x-\alpha_{i}y)\prod_{j=1}^{b-1}(y-\beta_{j}z)\prod_{k=1}^{c-1}(x-\gamma_{k}z)=0,$

is called a Roots-of-Unity-Arrangement (RUA for short) if the coefficients
$\alpha_{i},\beta_{j}$ and $\gamma_{k}$ can all be expressed as powers of a
$n$-root of unity $\zeta$.

The following results concerning the arrangements ${\cal A}^{0}_{3}(n)$,
${\cal A}^{1}_{3}(n)$, ${\cal A}^{2}_{3}(n)$ and ${\cal A}^{3}_{3}(n)$,
defined respectively by the equations $f_{n}=0,xf_{n}=0,xyf_{n}=0$ and
$xyzf_{n}=0$ where $f_{n}=(x^{n}-y^{n})(y^{n}-z^{n})(x^{n}-z^{n})$, are well
known (see in particular in [12, Propositions 6.77 and 6.85] or [9] page 119).
The arrangements ${\cal A}^{3}_{3}(n)$ and ${\cal A}^{0}_{3}(n)$ are
reflection arrangements associated respectively to the full monomial group
$G(n,1,3)$ (see [12, Example 6.29]) and to the monomial group $G(n,3,3)$; we
call them full monomial arrangement and monomial arrangement.

######  Corollary 2.7.

The arrangement $\mathcal{A}_{3}^{3}(n)$ is free with exponents $(n+1,2n+1).$

###### Proof.

The set of inner triple points $T$ is a complete intersection of length
$n^{2}$ defined by the ideal $(x^{n}-y^{n},y^{n}-z^{n})$. ∎

Removing the sides of the triangle and applying the Addition-Deletion theorem
[12, Theorem 4.51], we obtain also:

######  Corollary 2.8.

The arrangements $\mathcal{A}_{2}^{3}(n),\mathcal{A}_{1}^{3}(n)$ and
$\mathcal{A}_{0}^{3}(n)$ are obtained respectively from
$\mathcal{A}_{3}^{3}(n)$, $\mathcal{A}_{2}^{3}(n)$ and
$\mathcal{A}_{1}^{3}(n)$ by deleting one line between two vertices of the
triangle. They are free with exponents respectively equal to $(n+1,2n)$,
$(n+1,2n-1)$ and $(n+1,2n-2)$.

##  3 Roots-of-Unity-Arrangement

This section is dedicated to prove the following result.

######  Theorem 3.1.

Given a triangular arrangement, it is always possible to find a RUA with the
same combinatorics.

###### Proof.

Let us consider the triangular arrangement defined by the following equations

$\left\\{\begin{array}[]{l}x=0\\\ x=\alpha_{i}y\\\ y=0\\\ y=\beta_{j}z\\\
z=0\\\ z=\gamma_{k}x\end{array}\right.$

where $x=y=z=0$ are the lines which compose the triangle, $\alpha_{i}\neq 0$,
$i=1,\ldots,a-1$, and $\alpha_{i_{1}}\neq\alpha_{i_{2}}$ for $i_{1}\neq
i_{2}$, and the same properties hold for the $\beta_{j}$’s and the
$\gamma_{k}$’s, $j=1,\ldots,b-1$ and $k=1,\ldots,c-1$.
Observe that the existence of an inner triple point, defined by three lines
$x=\alpha_{\bar{\imath}}y,y=\beta_{\bar{\jmath}}z$ and $z=\gamma_{\bar{k}}x$
is given by a relation of the following type

$\alpha_{\bar{\imath}}\beta_{\bar{\jmath}}\gamma_{\bar{k}}=1.$

Therefore, we can translate the combinatorics of the arrangement in a family
of equalities

$\alpha_{i_{1}}\beta_{j_{1}}\gamma_{k_{1}}=1$ (3)

for each $i_{1},j_{1},k_{1}$ whose associated lines define an inner triple
point of the arrangement, a family of inequalities

$\alpha_{i_{2}}\beta_{j_{2}}\gamma_{k_{2}}\neq 1$ (4)

for each $i_{2},j_{2},k_{2}$ whose associated lines do not define an inner
triple point of the arrangement, and finally, the inequalities

$\alpha_{i_{1}}\neq\alpha_{i_{2}},\alpha_{i_{1}}\neq
0,\beta_{j_{1}}\neq\beta_{j_{2}},\beta_{j_{1}}\neq
0,\gamma_{k_{1}}\neq\gamma_{k_{2}},\gamma_{k_{1}}\neq 0,$ (5)

for each $i_{1},i_{2}=1,\ldots,a-1$, with $i_{1}\neq i_{2}$,
$j_{1},j_{2}=1,\ldots,b-1$, with $j_{1}\neq j_{2}$, and
$k_{1},k_{2}=1,\ldots,c-1$, with $k_{1}\neq k_{2}$.
Our goal is to find solutions, or at least prove their existence, which
satisfy all the previous relations and that can be expressed as various powers
of a $n$-th root of the unity, for a given $n$.
Let us consider a prime number $p$ and one of its primitive roots $\omega$,
hence, working modulo $p$, we can translate all the relations of type (3) as

$\omega^{v_{i_{1}}}\omega^{w_{j_{1}}}\omega^{t_{k_{1}}}\equiv 1\
(\mathrm{mod}\ p)$

or equivalently, as a family of linear equations

$v_{i_{1}}+w_{j_{1}}+t_{k_{1}}\equiv 0\ (\mathrm{mod}\ p-1).$ (6)

We claim that the linear system defined by the family (6) has always solutions
for any choice of $p$; indeed we have the following result.

######  Lemma 3.2.

The linear system given by the equations

$v_{i_{1}}+w_{j_{1}}+t_{k_{1}}=0,$ (7)

induced by (6) has infinitely many solutions.

###### Proof.

Taking $v_{i}=t_{k}=-1$ and $w_{j}=2$, for each value of $i,j$ and $k$ which
occurs in the linear system, we obtain a non zero solution of an homogeneous
linear system; hence we have infinite solutions. ∎

Since the coefficients of the equations of (7) are integers we get in
particular infinite integer solutions. Therefore, choosing well $p>>0$, we get
as many solutions of the system of linear congruences (6) as we want.

Now, add to our previous linear system, modulo $p$, the associated equality

$v+w+t\equiv 0\ (\mathrm{mod}\ p-1).$

Then either we have less solutions than before or the added condition is a
consequence of the others. Indeed, suppose that for a fixed $p$ we have
exactly the same solutions, this means that the added condition is a linear
combination of the previous ones, i.e.

$v+w+t\equiv\sum_{s}\lambda_{s,p}(v_{i_{s}}+w_{j_{s}}+t_{k_{s}})\
(\mathrm{mod}\ p-1)$

which is equivalent to have

$\omega^{v}\omega^{w}\omega^{t}\equiv\prod_{s}(\omega^{v_{i_{s}}}\omega^{w_{j_{s}}}\omega^{t_{k_{s}}})^{\lambda_{s,p}}\
(\mathrm{mod}\ p)$

and therefore

$\alpha\beta\gamma\equiv\prod_{s}(\alpha_{i_{s}}\beta_{j_{s}}\gamma_{k_{s}})^{\lambda_{s,p}}\
(\mathrm{mod}\ p).$

We conclude noticing that if we have the previous relation for an infinite set
of prime numbers, then we must also have

$\alpha\beta\gamma=\prod_{s}(\alpha_{i_{s}}\beta_{j_{s}}\gamma_{k_{s}})^{\lambda_{s}},$

which implies that the added condition is a consequence to the other
equalities, which is a contradiction because of our hypothesis on the triple
points.

Following the same reasoning as before for the relations expressed in the
family (5), it is possible to find solutions that do also satisfy those
inequalities.

Therefore, as a consequence of what we have said, it is possible to find an
integer $n$ (big if necessary) such that, taking one of its unity roots
$\zeta$, we can associate to each coefficient $\alpha_{i}$, $\beta_{j}$ and
$\gamma_{k}$, respectively, powers $\zeta^{\bar{\alpha}_{i}}$,
$\zeta^{\bar{\beta}_{j}}$ and $\zeta^{\bar{\gamma}_{k}}$ of the unity root.
The arrangement defined by the lines $x=\zeta^{\bar{\alpha}_{i}}y$,
$y=\zeta^{\bar{\beta}_{j}}z$ and $z=\zeta^{\bar{\gamma}_{k}}x$ satisfies all
the conditions expressed in (3), (4) and (5), and has therefore the exact same
combinatorics as the original one. ∎

##  4 Free arrangements obtained by deletion from the full monomial
arrangement’s

In this section we characterize free arrangements in $\mathrm{Tr}(a,b,c)$
obtained by deletion from a full monomial arrangement.

######  Theorem 4.1.

Let $\mathcal{A}$ be a roots of unity arrangement, then $\mathcal{A}$ is free
if and only if is obtained from a full monomial arrangement one, deleting in
each step a line such that the corresponding complementary arrangement has the
miminum possible of triple points.

###### Proof.

Removing inner lines from a full monomial arrangement ${\cal A}^{3}_{3}(N)$ we
find two triangular arrangements, ${\cal A}$ consisting in the remaining lines
and the complementary arrangement ${\cal A}^{c}$, consisting in the deleted
lines. More precisely, when we remove $a^{{}^{\prime}}=N-a+1$,
$b^{{}^{\prime}}=N-b+1$ and $c^{{}^{\prime}}=N-c+1$ lines, with equations
$f_{A}=0,f_{B}=0,f_{C}=0$, respectively from the vertices $A$, $B$ and $C$ of
a full monomial arrangement $\mathcal{A}_{3}^{3}(N)$, these lines form a
triangular subarrangement ${\cal A}^{c}$ defined by the equation
$f_{A}f_{B}f_{C}=0$ and this arrangement possesses, outside the vertices, a
set $T_{\mathrm{rem}}$ of triple inner points (set that can be empty). This
operation induces a commutative diagram

$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-1-N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-1-N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(f_{A},f_{B},f_{C})}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
T}_{{\cal
A}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-a)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(-c)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{\phi}}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{\Gamma}(N+2-a-b-c)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

where ${\cal T}_{\cal A}$ is the logarithmic bundle associated to the
triangular arrangement ${\cal A}\in\mathrm{Tr}(a,b,c)$, $\Gamma$ is the zero
locus of the section of ${\cal T}_{\cal A}$ induced by the syzygy
$(f_{A},f_{B},f_{C})$, $T$ is the set of triple inner points of ${\cal A}$ and
$\mathrm{Sing}(\mathcal{F})=T_{\mathrm{rem}}$. Since ${\cal T}_{\cal A}$ comes
from the full monomial arrangement $\mathcal{A}_{3}^{3}(N)$ we have of course

$\phi=[\frac{x^{N}-y^{N}}{f_{A}},\frac{y^{N}-z^{N}}{f_{B}},\frac{x^{N}-z^{N}}{f_{C}}].$

Moreover, since $T_{\mathrm{rem}}$ and $T$ are disjoint sets of simple points,
we conclude that $\Gamma=T_{\mathrm{rem}}$. This gives
$|T_{\mathrm{rem}}|=c_{2}({\cal T}_{\cal A}(N+1))$. Recall that, from Remark
2.2, we have

$\mid T\mid=(ab+bc+ac-a-b-c+1)-c_{2}({\cal T}_{\mathcal{A}}).$ (8)

Then we obtain the following formula relating $|T|$ and $|T_{\mathrm{rem}}|$:

$|T|=N^{2}-(N+2)(a+b+c-3)-3+ab+ac+bc-|T_{\mathrm{rem}}|.$ (9)

Observe that the equation (8) implies that we will reach the minimum possible
number of inner triple points when the second Chern class of the bundle is as
big as it can be (it means in particular that if the number of inner triple
points of ${\cal A}$ is smaller than this minimum then ${\cal A}$ cannot be
free). This happens exactly when the bundle ${\cal T}_{{\cal A}}$, when
supposed to be free, is as balanced as possible, which means that, depending
on the relations among $a$, $b$ and $c$, the distance between the two
exponents is as little as it can be. More precisely, balanced means that

${\cal T}_{{\cal A}}\simeq{\cal
O}_{{\mathbb{P}}^{2}}\left(\frac{-a-b-c+1}{2}\right)\oplus{\cal
O}_{{\mathbb{P}}^{2}}\left(\frac{-a-b-c+1}{2}\right)$

when $-a-b-c+1$ is even, and

${\cal T}_{{\cal A}}\simeq{\cal
O}_{{\mathbb{P}}^{2}}\left(\frac{-a-b-c}{2}\right)\oplus{\cal
O}_{{\mathbb{P}}^{2}}\left(\frac{-a-b-c}{2}+1\right)$

when $-a-b-c+1$ is odd.

On the other hand, by formula (9), we have that $\mid T\mid$ is as big as
possible, when the complementary arrangement as the fewest triple points. This
happens when we consider a full monomial arrangement with $N^{2}$ inner triple
points, we eliminate the first $a^{{}^{\prime}}=N+1-a$ and
$b^{{}^{\prime}}=N+1-b$ lines respectively to two families, and we choose the
other $c^{{}^{\prime}}=N+1-c$ lines, to be eliminated from the third family,
in order to get the least possible number of triple points in the
complementary arrangement.

— If $N-a^{{}^{\prime}}-b{{}^{\prime}}-c{{}^{\prime}}\geq 0$ (which is
equivalent to the condition $2N\leq a+b+c-3$), then it is possible to
eliminate this quantity of lines such that no triple points arise in the
complementary arrangement. In such a case the minimal number of triple point
in the complementary arrangement is $0$, that means
$T_{\mathrm{rem}}=\emptyset$ implying ${\rm H}^{1}_{*}({\cal T}_{\cal A})=0$
and by Horrocks’ criterion [10], ${\cal A}$ is free with exponents
$(N+1,a+b+c-N-2)$. On the contrary, if $T_{\mathrm{rem}}\neq\emptyset$ (i.e.
if we delete a line containing a triple point in the complementary when it was
possible to delete a line with no triple point in the complementary) then the
splitting type of ${\cal T}_{\cal A}$ on any line through $\Gamma$ is
different from the generic one, proving that ${\cal A}$ is not free.
The theorem is therefore proven for $2N\leq a+b+c-3$.

— If $N-a^{{}^{\prime}}-b{{}^{\prime}}-c{{}^{\prime}}<0$ (which is equivalent
to the condition $2N>a+b+c-3$) then necessarily $|T_{\mathrm{rem}}|\neq 0$. In
order to minimize this number, we must minimize
$r:=a^{{}^{\prime}}+b{{}^{\prime}}+c{{}^{\prime}}-N=2N+3-a-b-c$. The way to do
it is the following: the first $a^{{}^{\prime}}$ and $b^{{}^{\prime}}$ lines
being deleted, we delete $c^{{}^{\prime}}-r$ lines such that no triple point
arise in the complementary arrangement. The last $r-1$ lines are chosen such
that the number of triple points created in the complementary are

$1+1+2+2+\cdots+(s-1)+(s-1)+s=\frac{r^{2}}{4}$

when $r=2s$, and

$1+1+2+2+\cdots+s+s=\frac{r^{2}-1}{4}$

when $r=2s+1$.

Remark that $r=2N+3-a-b-c$ and $c_{1}({\cal T}_{\cal A})=1-a-b-c$ have the
same parity.

Consider the case when $r$, or identically, when $c_{1}({\cal T}_{\cal
A})=1-a-b-c$ is even.
Then, $|T_{\mathrm{rem}}|=\frac{r^{2}}{4}=\frac{(2N+3-a-b-c)^{2}}{4}.$ From
(8) we have that

$\mid T\mid_{min}=(ab+bc+ac-a-b-c+1)-\frac{1}{4}\left(-a-b-c+1\right)^{2},$

while from (9), we get that

$\mid T\mid_{max}=N^{2}-(N+2)(a+b+c-3)-3+ab+ac+bc-\frac{(2N+3-a-b-c)^{2}}{4}.$

Through direct computation, we get that $\mid T\mid_{max}=\mid T\mid_{min}$
and this implies that ${\cal T}_{{\cal A}}$ is free if and only if it is
constructed eliminating, from a full monomial arrangement, lines minimizing
the triple points in the complementary arrangement.
Analogously, we get the same equality when $-a-b-c+1$ is odd.
The theorem is therefore proven for $2N>a+b+c-3$.

∎

##  5 Free arrangements in the non complete triangle

All the free triangular arrangements coming from full monomial arrangement’s
ones by deletion, as explained in the above construction, contain the three
sides of the triangle, i.e. the three lines joining the three vertices. In
this section we will complete the study of triangular arrangements, proving
that the only free uncomplete triangular arrangements have always the maximal
number of inner triple points, more precisely:

######  Theorem 5.1.

Let ${\cal A}_{0}\in\mathrm{Tr}(a,b,c)$ and $(AB),(AC),(BC)$ its three sides.

  1. 1.

If we remove one side, then

     * •

${\cal A}_{0}\setminus(AB)$ is free if and only if its set of inner triple
points is a complete intersection $(a-1,b-1)$. Then its exponents are
$(c,a+b-2)$.

     * •

${\cal A}_{0}\setminus(AC)$ is free if and only if $b=c$ and its set of inner
triple points is a complete intersection $(a-1,b-1)$. Then its exponents are
$(b,a+b-2)$.

     * •

${\cal A}_{0}\setminus(BC)$ is free if and only if $a=b=c$ and its set of
inner triple points is a complete intersection $(a-1,a-1)$. Then its exponents
are $(a,2a-2)$.

  2. 2.

If we remove two sides, then

     * •

${\cal A}_{0}\setminus[(AB)\cup(AC)]$ is free if and only if its set of inner
triple points is a complete intersection $(a-1,b-1)$. Then its exponents are
$(c,a+b-3)$.

     * •

${\cal A}_{0}\setminus[(AB)\cup(BC)]$ is free if and only if $b=c$ and its set
of inner triple points is a complete intersection $(a-1,b-1)$. Then its
exponents are $(b,a+b-3)$.

     * •

${\cal A}_{0}\setminus[(AC)\cup(BC)]$ is free if and only if $a=b=c$ and its
set of inner triple points is a complete intersection $(a-1,a-1)$. Then its
exponents are $(a,2a-3)$.

  3. 3.

If we remove three sides, then ${\cal A}_{0}\setminus[(AB)\cup(AC)\cup(BC)]$
is free if and only if $a=b=c$ and its set of inner triple points is a
complete intersection $(a-1,a-1)$. Then its exponents are $(a,2a-4)$, or
$a=b=c=2$ with no triple point and its exponents are $(1,1)$.

###### Proof.

We divide the proof in three parts: ${\cal A}$ is a triangular arrangement
with one side removed (Part I), two sides removed (Part II) and three sides
removed (Part III) from a complete triangular arrangement ${\cal
A}_{0}\in\mathrm{Tr}(a,b,c)$.

I) Let us assume first that ${\cal A}$ contains exactly two sides and let $L$
be the missing one. Then ${\cal A}\cup L={\cal A}_{0}\in\mathrm{Tr}(a,b,c)$.
It is clear that ${\cal A}$ and ${\cal A}\cup L$ have the same set $T$ of
inner triple points. Let us denote by $t_{{\cal A}}$ and $t_{{\cal A}\cup L}$
the number of triple points of ${\cal A}$ and ${\cal A}\cup L$ counted with
multiplicities. Then we have

  * •

$t_{{\cal A}\cup L}=|T|+\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$.

  * •

$t_{{\cal A}}=|T|+\binom{a}{2}+\binom{b-1}{2}+\binom{c-1}{2}$, if $L=(BC)$;

  * •

$t_{{\cal A}}=|T|+\binom{a-1}{2}+\binom{b}{2}+\binom{c-1}{2}$, if $L=(AC)$;

  * •

$t_{{\cal A}}=|T|+\binom{a-1}{2}+\binom{b-1}{2}+\binom{c}{2}$, if $L=(AB)$.

The difference $(t_{{\cal A}\cup L}-t_{{\cal A}})$ is $b+c-2$, $a+c-2$ or
$a+b-2$ respectively when $L=(BC),(AC)$ or $(AB)$.

— Assume first that $L=(AB)$. By [8, Proposition 5.1] we have an exact
sequence

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}\cup L}@>{}>{}>{\cal T}_{{\cal
A}}@>{}>{}>{\cal <EMAIL_ADDRESS>
– If $c\leq a+b-2$, the surjection of ${\cal T}_{{\cal A}}$ on ${\cal
O}_{L}(2-a-b)$ induces $({\cal T}_{{\cal A}})_{\mid L}={\cal
O}_{L}(-c)\oplus{\cal O}_{L}(2-a-b)$ and, if we suppose ${\cal A}$ to be free,
we have that ${\cal T}_{\cal A}={\cal O}_{{\mathbb{P}}^{2}}(-c)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(2-a-b)$. By Addition-Deletion theorem [12, Theorem
4.51], $\mathcal{T}_{{\cal A}\cup L}={\cal
O}_{{\mathbb{P}}^{2}}(-c)\oplus{\cal O}_{{\mathbb{P}}^{2}}(1-a-b)$ and,
according to Theorem 2.3, the set of inner triple points is a complete
intersection $(a-1,b-1)$.

– If $c\geq a+b-1$, we consider the multi-arrangement given by the restriction
of ${\cal A}$ on the side $(CB)$. The multiplicities of this multi-arrangement
are $(c,1,\ldots,1,b-1)$ (with $a-1$ multiplicities of type “$1$”). This
implies that the splitting type of $\mathcal{T}_{{\cal A}}$ on the side
$l=(BC)$ is ${\cal O}_{l}(-c)\oplus{\cal O}_{l}(2-a-b)$ by [15, Theorem 3.1].
Then ${\cal A}$ is free if and only if ${\cal T}_{\cal A}={\cal
O}_{{\mathbb{P}}^{2}}(-c)\oplus{\cal O}_{{\mathbb{P}}^{2}}(2-a-b)$. By
Addition-Deletion again $\mathcal{T}_{{\cal A}\cup L}={\cal
O}_{{\mathbb{P}}^{2}}(-c)\oplus{\cal O}_{{\mathbb{P}}^{2}}(1-a-b)$ and the set
of inner triple points is a complete intersection $(a-1,b-1)$.

— When $L=(AC)$ or $L=(BC)$ the proofs are analogous. The additional
equalities $b=c$ or $a=b=c$ come from the fact that permuting the role of
$a,b,c$ the gaps $|a+c-1-b|$ or $|b+c-1-a|$ would be strictly bigger than the
maximal one which is $|a+b-1-c|$ by Remark 2.4.

II) Let us assume now that ${\cal A}$ contains only one side and let $L$ and
$L^{{}^{\prime}}$ the missing sides. Then ${\cal A}\cup L\cup
L^{{}^{\prime}}={\cal A}_{0}\in\mathrm{Tr}(a,b,c)$. Let $t_{{\cal A}}$,
$t_{{\cal A}\cup L}$ and $t_{{\cal A}\cup L\cup L^{{}^{\prime}}}$ be the
number of triple points of ${\cal A}$, ${\cal A}\cup L$ and ${\cal A}\cup
L\cup L^{{}^{\prime}}$, respectively, counted with multiplicities. Then we
have,

  * •

$t_{{\cal A}\cup L\cup
L^{{}^{\prime}}}=|T|+\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$;

  * •

$t_{{\cal A}}=|T|+\binom{a-2}{2}+\binom{b-1}{2}+\binom{c-1}{2}$, if $L=(AB)$
and $L^{{}^{\prime}}=(AC)$;

  * •

$t_{{\cal A}}=|T|+\binom{a-1}{2}+\binom{b-2}{2}+\binom{c-1}{2}$, if $L=(BC)$
and $L^{{}^{\prime}}=(AB)$;

  * •

$t_{{\cal A}}=|T|+\binom{a-1}{2}+\binom{b-1}{2}+\binom{c-2}{2}$, if $L=(AC)$
and $L^{{}^{\prime}}=(BC)$;

from which we get that the difference $(t_{{\cal A}\cup L}-t_{{\cal A}})$ is
equal to $a+b-3$ when $L=(AB)$ and $L^{{}^{\prime}}=(AC)$, $a+c-3$ when
$L=(AC)$ and $L^{{}^{\prime}}=(BC)$, and $b+c-3$ when $L=(BC)$ and
$L^{{}^{\prime}}=(AB)$.

— Assume first that $L=(AB)$ and $L^{{}^{\prime}}=(AC)$. By [8, Proposition
5.1] we have an exact sequence

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}\cup L}@>{}>{}>{\cal T}_{{\cal
A}}@>{}>{}>{\cal <EMAIL_ADDRESS>
– If $c\leq a+b-3$ the surjection of ${\cal T}_{{\cal A}}$ on ${\cal
O}_{L}(3-a-b)$ induces $({\cal T}_{{\cal A}})_{\mid L}={\cal
O}_{L}(-c)\oplus{\cal O}_{L}(3-a-b)$ and if we suppose ${\cal A}$ to be free,
it gives ${\cal T}_{\cal A}={\cal O}_{{\mathbb{P}}^{2}}(-c)\oplus{\cal
O}_{{\mathbb{P}}^{2}}(3-a-b)$. By Addition-Deletion [12, Theorem 4.51], ${\cal
A}\cup L$ is free with exponents $(c,a+b-2)$ and, according to Part I), the
set of inner triple points is a complete intersection of type $(a-1,b-1)$.

– If $c\geq a+b-2$ we consider the other exact sequence given by [8,
Proposition 5.1], i.e.

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}\cup L^{{}^{\prime}}}@>{}>{}>{\cal
T}_{{\cal A}}@>{}>{}>{\cal <EMAIL_ADDRESS>(10)

If $a=2$, considering the multiplicities of the multiarrangement obtained by
restricting ${\cal A}$ onto the unique line through $A$ we find $|T|$ double
points and $c+b-1-2|T|$ simple points. Notice that, since $|T|\leq b-1$, the
disequality

$2(c+b-1-|T|)-1\geq b+c$

is always satisfied, except when $|T|=b-1$ and $b=c$. If this last two
conditions hold, $|T|$ is a complete intersection of type $(1,b-1)$ and the
corresponding arrangement is free with exponents $(b,b-1)$ (see [1, section
2.2] for instance).

Let us now suppose that either $|T|<b-1$ or $b<c$. If ${\cal A}$ is free, then
it must have exponents $(1+|T|,c+b-2-|T|)$ by [15, Theorem 3.1].

If $b<c$ but $|T|=b-1$, then $|T|$ is again a complete intersection, and
relating as before the bundle ${\cal T}_{{\cal A}}$ with the vector bundle
${\cal T}_{{\cal A}\cup L\cup L^{\prime}}$ through addition deletion, we get
the required exponents.

Let us suppose now that $|T|<b-1$; it implies $-1-|T|\geq 1-c$ because $b\leq
c$. If we have a strict inequality, then we have that $2-c-b+|T|=1-c$ thanks
of the surjective map (10) which implies that $|T|=b-1$ and hence a
contradiction. If the equality holds, then we must have that $|T|=b-2$ and
$b=c$, which means that if ${\cal A}$ is free, then ${\cal T}_{{\cal A}}={\cal
O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal O}_{{\mathbb{P}}^{2}}(1-b)$. This would
imply, using addition deletion, that ${\cal T}_{{\cal A}\cup L^{\prime}}={\cal
O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal O}_{{\mathbb{P}}^{2}}(-b)$ which leads to
condradict Part I) because $T$ is not a complete intersection $(1,b-1)$.

Let us assume now that $a\geq 3$.
Since $a\geq 3$ it is clear that $b\leq a+c-3$ and the surjection of ${\cal
T}_{{\cal A}}$ on ${\cal O}_{L^{{}^{\prime}}}(3-a-c)$ induces $({\cal
T}_{{\cal A}})_{\mid L^{{}^{\prime}}}={\cal
O}_{L^{{}^{\prime}}}(-b)\oplus{\cal O}_{L^{{}^{\prime}}}(3-a-c)$ and if we
suppose ${\cal A}$ to be free, we obtain ${\cal T}_{\cal A}={\cal
O}_{{\mathbb{P}}^{2}}(-b)\oplus{\cal O}_{{\mathbb{P}}^{2}}(3-a-c)$. Then, by
Addition-Deletion [12, Theorem 4.51], ${{\cal A}\cup L^{{}^{\prime}}}$ is free
with exponents $(b,a+c-2)$. According to Part I), this occurs if and only if
$b=c$ and $T$ is a complete intersection $(a-1,b-1)$.

— If we remove the lines $(AC)$ and $(BC)$ or the lines $(AB)$ and $(BC)$ the
proofs are analogous. The additional equalities $b=c$ or $a=b=c$ come from the
fact that permuting the role of $a,b,c$ the gaps $|a+c-2-b|$ or $|b+c-2-a|$
would be strictly bigger than the maximal one which is $|a+b-2-c|$ by Remark
2.4.

III) Let us assume now that ${\cal A}$ contains no side of the triangle and
let $L=(AB)$, $L^{{}^{\prime}}=(AC)$ and $L^{{}^{\prime\prime}}=(BC)$ denote
the missing sides. Then ${\cal A}\cup L\cup L^{{}^{\prime}}\cup
L^{{}^{\prime\prime}}={\cal A}_{0}\in\mathrm{Tr}(a,b,c)$. Consider the exact
sequence

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}\cup L^{{}^{\prime\prime}}}@>{}>{}>{\cal
T}_{{\cal A}}@>{}>{}>{\cal
<EMAIL_ADDRESS>
– If $a\leq b+c-4$ then, by the same technique used before, ${\cal A}$ is free
with exponents $(a,b+c-4)$ implying that ${\cal A}\cup L^{{}^{\prime\prime}}$
is free with exponents $(a,b+c-3)$. Since the gap for this last splitting is
bigger than the gap given by the exponents $(c,a+b-3)$, this proves, according
to the Part II), that ${\cal A}$ is free if and only if $a=b=c$ and $T$ is a
complete intersection $(a-1,a-1)$.

– If $a\geq b+c-3$ the possibilities are $(a,b,c)=(2,2,2)$, $(a,b,c)=(2,2,3)$
or $(a,b,c)=(3,3,3).$ The first case corresponding to $3$ lines is free with
exponents $(1,1)$ if there is no triple point (but this case correspond to an
arrangement in $\mathrm{Tr}(1,1,1)$), $(0,2)$ if there is one triple point.
The second case corresponding to $4$ lines is free with exponents $(1,2)$ if
and only if there is one triple point, i.e. $T$ is a complete intersection
$(1,1)$ ; it cannot be free with exponents $(0,3)$ because the $|T|$ would be
$3$ which is impossible for this arrangement. The third case corresponding to
$6$ lines is free with exponents $(2,3)$ if and only if $|T|=4$ is a complete
intersection $(2,2)$ ; it cannot be free with exponents $(1,4)$ because $|T|$
would be $6$ which is impossible for this arrangement. ∎

As a direct consequence of the previous result, we can explicit the following
cases, which describe when we have the same number of lines passing through
each vertex.

######  Corollary 5.2.

Up to a linear change of coordinates, the following holds:

  * •

the only free triangular arrangement of $3n-1$ lines passing through three
points, with $n-1$ inner lines through each vertex, plus two sides, is ${\cal
A}^{2}_{3}(n)$;

  * •

the only free triangular arrangement of $3n-2$ lines passing through three
points, with $n-1$ inner lines through each vertex, plus one side, is ${\cal
A}^{1}_{3}(n)$;

  * •

the only free arrangement of $3n-3$ lines passing through three points, with
$n-1$ inner lines through each vertex, and no side, is ${\cal A}^{0}_{3}(n)$.

##  6 Weak combinatorics

The combinatorics of ${\cal A}$ is determined by the set $L({\cal A})$ of all
the intersections of lines in ${\cal A}$. There is a partial order on this set
corresponding to the inclusion of points in lines. Two line arrangements
${\cal A}_{0}$ and ${\cal A}_{1}$ have the same combinatorics if and only if
there is a bijection between $L({\cal A}_{0})$ and $L({\cal A}_{1})$
preserving the partial order. In [12] Terao conjectures that if two
arrangements have the same combinatorics and one of them is free then the
other one is also free. This problem posed in any dimension and on any field
is still open even on the projective plane and seems far from being proved,
probably because relatively few free arrangements are known.

In this section we will show that if we only suppose the weak combinatorics
hypothesis, the conjecture does not hold. Indeed, we get the following result.

######  Theorem 6.1.

There exist pairs of arrangements possessing the same weak combinatorics such
that one is free and the other is not.

###### Proof.

We prove it by describing an example. We will explain next how to produce a
family of examples of the same kind.

We construct two triangular arrangements ${\cal A}_{0}$ which is free with
exponents $(7,7)$ and ${\cal A}_{1}$ which isn’t free, both in
$\mathrm{Tr}(5,5,5)$ with the same numbers of multiple points
$t_{3}=12,t_{4}=t_{5}=0$, $t_{6}=3$ and $t_{i}=0$ for $i>6$ (the number of
double points is given by the combinatorial formula
$\binom{15}{2}=t_{2}+3t_{3}+\binom{6}{2}t_{6}$).

— Let us construct ${\cal A}_{0}$: it is obtained by removing the six lines
$x=z$, $x=\zeta z$, $y=z$, $y=\zeta z$, $x=\zeta^{2}y$ and $x=\zeta^{4}y$ from
the full monomial arrangement $xyz(x^{6}-y^{6})(y^{6}-z^{6})(x^{6}-z^{6})=0$
as represented in the following picture:

This arrangement is free because the syzygy of degree $2$, that is

$\psi=[(x-z)(x-\zeta z),(y-z)(y-\zeta z),(x-\zeta^{2}y)(x-\zeta^{4}y)],$

has no zero. Indeed this syzygy gives for

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}_{0}}@>{}>{}>{\cal
O}^{3}_{{\mathbb{P}}^{2}}(-5)@>{\phi}>{}>\mathcal{J}_{T}(-1)@>{}>{}>0\end{CD}$

a non zero section ${\cal O}_{{\mathbb{P}}^{2}}(-7)\longrightarrow{\cal
T}_{{\cal A}_{0}}$, being

$\phi=[\frac{x^{6}-z^{6}}{(x-z)(x-\zeta z)},\frac{y^{6}-z^{6}}{(y-z)(y-\zeta
z)},\frac{x^{6}-y^{6}}{(x-\zeta^{2}y)(x-\zeta^{4}y)}].$

This induces a commutative diagram:

$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-7)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-7)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
T}_{{\cal
A}_{0}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
O}^{3}_{{\mathbb{P}}^{2}}(-5)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{\Gamma}(-7)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

where the singular locus of the rank two sheaf $\mathcal{F}$ is the zero set
of $\psi$. Since this zero set is empty, $\mathcal{F}$ is a vector bundle.
That proves $\mathcal{E}xt^{1}(\mathcal{F},{\cal O}_{{\mathbb{P}}^{2}})=0$ and
then $\Gamma=\emptyset$.

Another argument can be used to establish the freeness: the $12$ inner triple
points are distributed as a partition $3+3+3+3$ along the vertical lines and
$3+3+3+3$ along the horizontal lines but $2+3+4+3$ along the diagonal (this
means that this example is very closed to have the same combinatorics than the
non free case: indeed in the following example we’ll see that the partition
along vertical, horizontal and diagonal lines is always $3+3+3+3$); thanks to
[15, Theorem 3.1], the bundle restricted to the line containing only $2$ inner
triple points has the splitting $(7,7)$ which proves the freeness according to
[6, Corollary 2.12].

— Let us construct now ${\cal A}_{1}$: it is obtained by removing the three
lines $x=z$, $y=z$, $x=y$ from the full monomial arrangement
$xyz(x^{5}-y^{5})(y^{5}-z^{5})(x^{5}-z^{5})=0$ as it appears on the picture:

Beginning with ${\cal A}_{3}^{3}(5)$ which is free with exponents $(6,11)$ and
removing the first line we obtain, by Addition-Deletion theorem, a free bundle
with exponents $(6,10)$. Removing the second line we find again a free
arrangement with exponents $(6,9)$. Removing the third line, we don’t find a
free bundle (with splitting $(7,7)$) but a nearly free bundle (defined in [4]
and studied by the authors in [11]) with generic splitting $(6,8)$. The
jumping point is the intersection point of the three removed lines. The three
partitions appearing along the horizontal, vertical and diagonal lines are
$12=3+3+3+3$.

Let us make it more explicit. We found a syzygy of degree $1$, which is

$\psi=[x-z,y-z,x-y],$

and which induces a non zero section ${\cal
O}_{{\mathbb{P}}^{2}}(-6)\longrightarrow{\cal T}$ where

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}_{1}}@>{}>{}>{\cal
O}^{3}_{{\mathbb{P}}^{2}}(-5)@>{\phi}>{}>\mathcal{J}_{T}(-1)@>{}>{}>0\end{CD}$

and

$\phi=[\frac{x^{5}-z^{5}}{x-z},\frac{y^{5}-z^{5}}{y-z},\frac{x^{5}-y^{5}}{x-y}].$

This syzygy admits a common zero $p=(1,1,1)$ and induces a commutative
diagram:

$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-6)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{{\cal
O}_{{\mathbb{P}}^{2}}(-6)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
T}_{{\cal
A}_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\cal
O}^{3}_{{\mathbb{P}}^{2}}(-5)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{\Gamma}(-8)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{J}_{T}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

where the singular locus of the rank two sheaf $\mathcal{F}$ is the zero set
of $p$. Since $p\notin T$ then $p\in\Gamma$ (actually $\Gamma=\\{p\\}$) and
${\cal T}_{{\cal A}_{1}}$ cannot be free.

This example proves that the arrangement consisting in these $15$ lines is
nearly free with the same weak-combinatorics (same numbers
$t_{2},t_{3},\ldots$) and quite the same combinatorics (only one partition
along the diagonals differs) than the one described just before. This shows
that we cannot replace the term combinatorics by weak-combinatorics in the
hypothesis of Terao’s conjecture. ∎

######  Remark 6.2.

In the famous Ziegler’s example of two arrangements ($9$ lines with $6$ triple
points) with the same combinatorics but with different free resolutions, the
situation was explained by the existence of a smooth conic containing the $6$
triple points. Here the situation can be geometrically explained by the
existence of a cubic containing the $12$ inner triple points. Indeed, since
the bundle ${\cal T}_{{\cal A}_{1}}$ described in the previous example is the
kernel of the following exact sequence

$\begin{CD}0@>{}>{}>{\cal T}_{{\cal A}_{1}}@>{}>{}>{\cal
O}_{{\mathbb{P}}^{2}}^{3}(-5)@>{}>{}>\mathcal{J}_{T}(-1)@>{}>{}>0,\end{CD}$

(where $|T|=12$) it gives ${\rm H}^{0}(\mathcal{J}_{T}(3))={\rm H}^{1}({\cal
T}_{{\cal A}_{1}}(4)).$ Moreover, the following non zero global section

$\begin{CD}0@>{}>{}>{\cal O}_{{\mathbb{P}}^{2}}(-6)@>{}>{}>{\cal T}_{{\cal
A}_{1}}@>{}>{}>\mathcal{J}_{p}(-8)@>{}>{}>0,\end{CD}$

where $p$ is the jumping point associated to the nearly free arrangement,
proves that

${\rm h}^{1}({\cal T}_{{\cal A}_{1}}(4))={\rm h}^{1}(\mathcal{J}_{p}(-4))={\rm
h}^{0}({\cal O}_{p})=1.$

######  Remark 6.3.

It is possible to generalize the described examples, and find a family of them
in the following way: consider triangular arrangements in the family
$\mathrm{Tr}(n,n,n)$. The multiplicity of each vertex is $n+1$. Assume that
$n=2k+1$. For arrangements of this family, the maximal possible number for the
inner triple points is $|T|=4k^{2}$ (then the arrangement is ${\cal
A}_{3}^{3}(n-1)$). For a general triangular arrangement, the set of inner
triple points $T$ is empty but if we want to consider free arrangements, $T$
must contain at least $3k^{2}$ points. This minimal number corresponds to the
balanced free arrangement with exponents $(3k+1,3k+1)$. We construct a nearly
free arrangement with generic splitting ${\cal O}_{l}(-3k)\oplus{\cal
O}_{l}(-2-3k)$ and $3k^{2}$ triple inner points in the same family
$\mathrm{Tr}(n,n,n)$. This generic splitting is also the one of the free
arrangement with $3k^{2}+1$ triple inner points constructed by removing
$(k-1)$ (inner) lines from each vertex of the arrangement
$xyz(x^{3k-1}-y^{3k-1})(y^{3k-1}-z^{3k-1})(x^{3k-1}-z^{3k-1})=0.$ In the last
step, instead of removing a line with $k-1$ triple inner points we remove a
line with $k$ triple points. It is always possible by choosing a line of the
third direction passing through a intersection point $\\{p\\}$ of the previous
removing lines in the two other directions. This construction induces and
exact sequence:

$0\rightarrow{\cal O}_{{\mathbb{P}}^{2}}(-3k)\longrightarrow{\cal T}_{\cal
A}\longrightarrow\mathcal{J}_{p}(-2-3k)\rightarrow 0,$

being, one again, $p$ the points associated to the Nearly free arrangement.

##  7 On Terao’s conjecture

In this final section we use the characterization, given in Theorem 4.1, of
free arrangements obtained from the full monomial arrangement to discuss
Terao’s conjecture for triangular arrangements.

######  Proposition 7.1.

Let ${\cal A}$ be a triangular arrangement and $\mathcal{B}$ its associated
RUA. If $\mathcal{B}$ is free, then ${\cal A}$ is free as well.

###### Proof.

We will prove the result by induction on the number of lines.
It is shown in [3] that the conjecture holds up to 13 lines for any
arrangement, and in particular it holds for triangular ones with such number
of lines.
Suppose that it holds for triangular arrangements constructed with $n-1$ lines
and let us assume that ${\cal A}$ has $n$ lines. Being $\mathcal{B}$ free,
then, thanks again to Theorem 4.1, we can delete a line in order to get
$\mathcal{B}^{\prime}$ again free, and we delete the associate line to ${\cal
A}$, in order to get ${\cal A}^{\prime}$. This arrangement is free because of
the induction hypothesis, and due to the Addition-Deletion Theorem, ${\cal A}$
must be free as well. ∎

## References

  * [1] Abe, T., Faenzi, D., Vallès, J.: Logarithmic bundles of deformed Weyl arrangements of type $A2$. Bull. Soc. Math. France 144, no. 4, 745–761 (2016).
  * [2] Dimca, A.: Hyperplane arrangements. An introduction. Universitext, Springer International Publishing 2017
  * [3] Dimca, A., Ibadula, D., Macinic, A.: Freeness and near freeness are combinatorial for line arrangements in small degrees. arXiv:1712.04400 (arXiv preprint).
  * [4] Dimca, A., Sticlaru, G.: Free and nearly free curves vs. rational cuspidal plane curves. Publ. RIMS Kyoto Univ. 54, 163–179 (2018)
  * [5] Dolgachev, I., Kapranov, M.: Arrangements of hyperplanes and vector bundles on ${\mathbb{P}}_{n}$. Duke Math. J. 71, 633–664 (1993)
  * [6] Elencwajg, G., Forster, O.: Bounding cohomology groups of vector bundles on ${\mathbb{P}}_{n}$. Math. Ann. 246, no. 3, 251–270 (1979)
  * [7] Faenzi, D., Matei, D., Vallès, J.:  Hyperplane arrangements of Torelli type.  Compos. Math., 149(2):309–332 (2013)
  * [8] Faenzi, D., Vallès, J.: Logarithmic bundles and Line arrangements, an approach via the standard construction. J. Lond. Math. Soc. (2) 90 no. 3, 675–694 (2014)
  * [9] Hirzebruch, F.:  Arrangements of lines and algebraic surfaces.  In Arithmetic and geometry, Vol. II, volume 36 of Progr. Math., pages 113–140. Birkhäuser Boston, Mass. 1983
  * [10] G. Horrocks, G.: Vector bundles on the punctured spectrum of a ring. Proc. London Math. Soc. (3) 14, 689–713 (1964)
  * [11] S. Marchesi, S., Vallès, J.: Nearly free arrangements, a vector bundle point of view. arXiv:1712.04867 (arXiv preprint).
  * [12] Orlik, P., Terao, H.: Arrangements of hyperplanes. Grundlehren der Mathematischen Wissenschaften, 300, Springer-Verlag, Berlin 1992
  * [13] Saito, K.: Theory of logarithmic differential forms and logarithmic vector fields. J. Fac. Sci. Univ. Tokyo, Sect. IA Math. 27, no. 2, 265–291 (1980)
  * [14] Terao, H.: Arrangements of hyperplanes and their freeness. I.J. Fac. Sci. Univ. Tokyo, Sect. IA Math. 27, no. 2, 293–312 (1980)
  * [15] Wakefield, M., Yuzvinsky, S.: Derivations of an effective divisor on the complex projective line. Trans. Amer. Math. Soc. 359, no. 9, 4389–4403 (2007)
  * [16] Ziegler, G.: Multiarrangements of hyperplanes and their freeness. Singularities (Iowa City, IA, 1986), Contemp. Math. 90, Amer. Math. Soc., Providence, RI, 345–359 (1989)


# Moduli of Sub-Laplacians on the second Heisenberg group

Sebastiano Nicolussi Golo  currently unafiliated, Italy
<EMAIL_ADDRESS>and  Benjamin Warhurst  Institute of Mathematics,
University of Warsaw, ul. Banacha 2, 02-097 Warsaw, Poland
<EMAIL_ADDRESS>
###### Abstract.

We solve the contact equivalence problem for generalised sub-Laplacians on
$\mathbb{H}^{2}$ and show that the family of sub-Laplacians on
$\mathbb{H}^{2}$ modulo contact equivalence, is parameterised by
$\mathbb{R}^{+}$.

###### Key words and phrases:

Jet spaces; stratified Lie groups; Carnot groups; Laplacian; equivqlence;

###### 2010 Mathematics Subject Classification:

58A20; 22E25; 35R03;

S. N. G. has been supported by the Academy of Finland (grant 328846, “Singular
integrals, harmonic functions, and boundary regularity in Heisenberg groups”,
grant 322898 “Sub-Riemannian Geometry via Metric-geometry and Lie-group
Theory”, grant 314172 “Quantitative rectifiability in Euclidean and non-
Euclidean spaces”), and by the University of Padova STARS Project “Sub-
Riemannian Geometry and Geometric Measure Theory Issues: Old and New”.
S. N. G. and B. W. are grateful for the support provided by the grant of the
National Science Center, Poland (NCN), UMO-2017/25/B/ST1/01955.
We also would like to thank Prof Ian Anderson, Utah State University, for his
insight and tips with MAPLE computation.

###### Contents

  1. 1 Introduction
  2. 2 Horizontal jets over stratified groups
  3. 3 Heisenberg Groups
  4. 4 Sub-laplacians on $\mathbb{H}^{2}$

##  1\. Introduction

Each positive definite $A\in GL(\mathbb{R}^{n})$ determines the second order
operator

$\Delta_{A}=\nabla\cdot A\nabla=\sum_{ij}a_{ij}\frac{\partial^{2}}{\partial
x_{i}\partial x_{i}}.$

Since $A$ is positive definite, it has a symmetric square root $C$, which is
to say $C$ is symmetric and $C^{2}=A^{-1}$. If $f(x)=Cx$, then

$\Delta_{A}(u\circ f)=(\Delta u)\circ f.$

Let $G$ be a connected, simply connected, stratified nilpotent Lie group, with
Lie algebra $\mathfrak{g}=\mathbb{V}_{1}\oplus\dots\oplus\mathbb{V}_{s}$,
where $\mathbb{V}_{i+1}=[\mathbb{V}_{1},\mathbb{V}_{i}]$ and
$[\mathfrak{g},\mathbb{V}_{s}]=0$. Let $d_{i}={\rm dim}\,\mathbb{V}_{i}$ and
$N={\rm dim}\,\mathfrak{g}$. If $e_{i}$ denotes an element of the standard
basis of $\mathbb{R}^{N}$, then we choose $\mathbb{V}_{1}={\rm
span}\\{e_{1},\dots,e_{d_{1}}\\}$, $\mathbb{V}_{2}={\rm
span}\\{e_{d_{1}+1},\dots,e_{d_{2}}\\}$ and so on. The left invariant vector
fields are framed by the set
$\mathscr{F}=\\{\tilde{X}_{1},\dots,\tilde{X}_{N}\\}$ where

$\tilde{X}_{i}u(p)=\frac{d}{dt}u(p\exp(te_{i}))|_{t=0}$

and the horizontal bundle, denoted by $\mathscr{H}$, is the subbundle of $TG$
framed by $\\{\tilde{X}_{1}\dots,\tilde{X}_{d_{1}}\\}$. The quadratic form on
$\mathscr{H}$ which makes $\\{\tilde{X}_{1},\dots,\tilde{X}_{d_{1}}\\}$
orthonormal is left invariant and denoted by $g_{0}$. If $\tilde{V}$,
$\tilde{W}$ are $\mathscr{H}$ valued left invariant vector fields, with
coefficients $v_{i}$ and $w_{i}$ relative to $\mathscr{F}$, then we have that

$g_{0}(\tilde{V},\tilde{W})=(v_{1},\dots,v_{d_{1}})\cdot(w_{1},\dots,w_{d_{1}}).$

A local diffeomorphism $F:U\subset G\to G$, where $U\subset G$ is open, which
preserves the horizontal bundle in the sense that
$dF_{p}(\mathscr{H}_{p})=\mathscr{H}_{F(p)}$, will be referred to as a contact
mapping, despite the fact that the distributions are not necessarily
associated with contact structures.

The horizontal gradient of a real valued function $u$ on $G$ is define to be
the $\mathscr{H}$-valued vector field

${\nabla}_{\scriptscriptstyle\mathscr{H}}u=\sum_{i=1}^{d_{1}}(\tilde{X}_{i}u)\tilde{X}_{i}.$

In coordinates we have
${\nabla}_{\scriptscriptstyle\mathscr{H}}=(\tilde{X}_{1},\dots,\tilde{X}_{d_{1}})$
and the canonical sub-Laplacian is defined to be

$\mathcal{L}u={\nabla}_{\scriptscriptstyle\mathscr{H}}\cdot{\nabla}_{\scriptscriptstyle\mathscr{H}}u=\sum_{i=1}^{d_{1}}\tilde{X}_{i}^{2}u.$

Given a positive-definite symmetric matrix $M=(m_{ij})\in
GL(\mathbb{R}^{d_{1}})$ we define a generalised sub-Laplacian
$\mathcal{L}_{M}$ by setting

$\mathcal{L}_{M}u={\nabla}_{\scriptscriptstyle\mathscr{H}}\cdot
M{\nabla}_{\scriptscriptstyle\mathscr{H}}u=\sum_{ij=1}^{d_{1}}m_{ij}\tilde{X}_{i}\tilde{X}_{j}u.$

One can then ask if there exists a local contact diffeomorphism $f:G\to G$
such that

(1) $\displaystyle\mathcal{L}_{M}(u\circ f)=\mathcal{L}u\circ f.$

For free groups, all sub-Laplacians are equivalent, as shown in [1] (Theorem
16.1.2 page 625) where matters are simplified by the fact that any element of
$GL(\mathbb{R}^{d_{1}})$ extends to a Lie algebra automorphism. The first
Heisenberg group $\mathbb{H}^{1}$ is free, and so $\mathbb{H}^{2}$ is perhaps
the simplest case not covered by the results in [1]. In this study, we deal
with the equivalence problem for sub-Laplacians by applying the theory of jet
spaces over stratified groups, see [4].

##  2\. Horizontal jets over stratified groups

The horizontal Taylor expansion of a smooth real valued function $u$ at a
point $p\in G$, is based on the usual Taylor expansion around zero of the
function $h(t)=u(p\exp(tV))$ when $V\in\mathbb{V}_{1}$. If
$V=\sum_{i=1}^{d_{1}}v_{i}e_{i}$ and

$\tilde{V}^{k}u(p)=\frac{d^{k}}{dt^{k}}u(p\exp(tV))|_{t=0}$

then the $m$-th order Taylor polynomial of $h$ around zero, denoted
$T^{m}_{0}h$, has the form

$\displaystyle T^{m}_{0}h(t)$
$\displaystyle=\sum_{k=0}^{m}\tilde{V}^{k}u(p)\frac{t^{k}}{k!}$
$\displaystyle=\sum_{k=0}^{m}\,\sum_{i_{1},\dots
i_{k}=1}^{d_{1}}v_{i_{1}}\dots
v_{i_{k}}\tilde{X}_{i_{1}}\dots\tilde{X}_{i_{k}}u(p)\frac{t^{k}}{k!}.$

Setting $t=1$ determines the horizontal Taylor polynomial $T^{m}_{p}u$ of
order $m$, defined by

$\displaystyle T^{m}_{p}u(V)$
$\displaystyle=\sum_{k=0}^{m}\frac{1}{k!}\tilde{V}^{k}u(p)$ (2)
$\displaystyle=\sum_{k=0}^{m}\frac{1}{k!}\sum_{i_{1},\dots,i_{k}=1}^{d_{1}}\,\,\,\tilde{X}_{i_{1}}\dots\tilde{X}_{i_{k}}u(p)\,v_{i_{1}}\dots
v_{i_{k}},$

which approximates $V\to u(p\exp(V))$, $V\in\mathbb{V}_{1}$.

The $k$-th term of $T^{m}_{p}u$ is a homogeneous polynomial of degree $k$, and
can be expressed in "polar" type form, by which we mean
$T^{m}_{p}u(V)=\sum_{k=0}^{m}\frac{1}{k!}A^{k}_{u,p}(V,\dots,V)$, where for
$k>0$, $A^{k}_{u,p}$ is the $k$-linear map defined on $\mathbb{V}_{1}^{k}$ by

(3) $\displaystyle A^{k}_{u,p}\left({{V}}_{1},\dots,{{V}}_{k}\right)$
$\displaystyle=\tilde{V}_{k}\dots\tilde{V}_{1}u(p)$

and $A^{0}_{u,p}=u(p)$. The horizontal $m$-jet of $u$, denoted $j^{m}_{p}u$,
is the formal sum

$j^{m}_{p}u=\sum_{k=0}^{m}\frac{1}{k!}A^{k}_{u,p}.$

The horizontal $m$-jet of a vector valued function $u:G\to\mathbb{W}$ is then
given by $(j^{m}_{p}u_{1},\dots,j^{m}_{p}u_{N})$ where $u_{1},\dots,u_{N}$ are
the coordinate functions relative to a basis of the target vector space
$\mathbb{W}$.

For $k\in\mathbb{N}$, we denote by $\mathtt{HD}^{k}(\mathfrak{g};\mathbb{W})$
the vector subspace of $\mathtt{Lin}^{k}(\mathbb{V}_{1};\mathbb{W})$
consisting of all $k$-multilinear maps of the form $A^{k}_{u,p}$. For $k=0$,
we set $\mathtt{HD}^{0}(\mathfrak{g};\mathbb{W})=\mathbb{W}$. For each
positive integer $m$, we define

$\mathtt{HD}^{\leq
m}(\mathfrak{g};\mathbb{W})=\bigoplus_{k=0}^{m}\mathtt{HD}^{k}(\mathfrak{g};\mathbb{W}).\quad(\mathtt{HD}^{\leq
m}\text{ for short})$

The $m$-th order horizontal jet space, denoted by
$\mathtt{J}^{m}(G,\mathbb{W})$, is the vector bundle $G\times\mathtt{HD}^{\leq
m}\to G$.

The Lie group structure on $\mathtt{J}^{m}(G,\mathbb{W})$ is given by an anti-
semidirect product which we now briefly review. If $\mathfrak{g}$ and
$\mathfrak{h}$ are Lie algebras, then an antimorphism
$\psi:\mathfrak{g}\to{\rm Der}(\mathfrak{h})$ is a linear map with the
property that

$\psi([x,y])=-[\psi(x),\psi(y)]).$

An antimorphism induces a Lie bracket on $\mathfrak{g}\times\mathfrak{h}$ by
the formula

$[(x,X),(y,Y)]=([x,y],-[X,Y]+\psi(y)X-\psi(x)Y).$

We denote this Lie algebra by $\mathfrak{g}\ltimes_{\psi}\mathfrak{h}$ (anti-
semidirect product).

Let $G$ and $H$ be the connected simply connected Lie groups corresponding to
$\mathfrak{g}$ and $\mathfrak{h}$. The antimorphism $\psi$ induces an
antimorphism $\phi:G\to{\rm Aut}(H)$, by which we mean

$\displaystyle\phi(ab)$ $\displaystyle=\phi(b)\phi(a)$ $\displaystyle
d\phi|_{\exp(x)}$ $\displaystyle=e^{\psi(x)}\in{\rm
Aut}(\mathfrak{h})\quad\forall x\in\mathfrak{g}.$

The set $G\times H$ and the product given by

$(a,A)(b,B)=(ab,B\phi(b)A)$

define a Lie group $G\ltimes_{\phi}H$ with Lie algebra
$\mathfrak{g}\ltimes_{\psi}\mathfrak{h}$. When $\mathfrak{h}$ is abelian, the
algebra and group products simplify to the following:

$\displaystyle[(x,X),(y,Y)]$ $\displaystyle=([x,y],\psi(y)X-\psi(x)Y),$
$\displaystyle(a,A)(b,B)$ $\displaystyle=(ab,B+\phi(b)A).$

If $(a,A)\in G\ltimes_{\phi}H$ and
$(x,X)\in\mathfrak{g}\ltimes_{\psi}\mathfrak{h}$, then the left-invariant
vector field generated by $(x,X)$ evaluated at $(a,A)\in G\ltimes_{\phi}H$ is

$\widetilde{(x,X)}(a,A)=(\tilde{x}(a),\psi(x)A+X)$

where $\tilde{x}(a)$ is the left-invariant vector field generated by $x$
evaluated at $a$.

If $k\geq 2$ and $A\in\mathtt{HD}^{k}(\mathfrak{g};\mathbb{W})$, then for
$V\in\mathbb{V}_{1}$, the right contraction of $A$ by $V$, denoted
$V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A$, is the element in
$\mathtt{HD}^{k-1}(\mathfrak{g};\mathbb{W})$ given by

$V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A(V_{1},\dots,V_{k-1})=\,A(V_{1},\dots,V_{k-1},V).$

Moreover, if $k=1$ then
$V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A=A(V)$, and if $k=0$
then $V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A=0$.

Given $V\in\mathbb{V}_{1}$, we extend the definition of the right contraction
by $V$ to a map
$V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}:\mathtt{HD}^{\leq
m}\to\mathtt{HD}^{\leq m}$, where for each $A\in\mathtt{HD}^{\leq m}$, we set
$(V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A)^{k}=V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A^{k+1}$
for $k<m$ and $(V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A)^{m}=0$.
The map $\mathbb{V}_{1}\to{\rm End}(\mathtt{HD}^{\leq m})$, $V\mapsto
V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}$, extends uniquely (by
formula (3)) to a Lie algebra anti-morphism $\psi:\mathfrak{g}\to{\rm
End}(\mathtt{HD}^{\leq m})$, i.e.,

(4)
$\displaystyle[V,W]\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}=-[V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}},W\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}].$

Since $\mathtt{HD}^{\leq m}$ is an abelian Lie algebra, ${\rm
Der}(\mathtt{HD}^{\leq m})={\rm End}(\mathtt{HD}^{\leq m})$. Hence,
$\psi(V)=V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}$ is a Lie
algebra antimorphism from $\mathfrak{g}$ to ${\rm Der}(\mathtt{HD}^{\leq m})$.
We obtain a Lie algebra

$\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W}):=\mathfrak{g}\ltimes_{\psi}\mathtt{HD}^{\leq
m},$

where

(5)
$\displaystyle[(V,A),(W,B)]=([V,W],W\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A-V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}B).$

The Lie algebra $\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W})$ is stratified of
step $\max\\{s,m+1\\}$, where $s$ is the step of $\mathfrak{g}$. The $k$-th
layer of $\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W})$ is

$\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W})_{k}:=\mathbb{V}_{k}\times\mathtt{HD}^{m+1-k}(\mathfrak{g};\mathbb{W})$

where $\mathbb{V}_{k}=\\{0\\}$ if $k>s$ and
$\mathtt{HD}^{m+1-k}(\mathfrak{g};\mathbb{W})=\\{0\\}$ if $k>m+1$.

The simply connected Lie group corresponding to the Lie algebra
$\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W})$ is the semidirect product
$G\ltimes\mathtt{HD}^{\leq m}(\mathfrak{g};\mathbb{W})$ with group law

$(\exp(V),A)(\exp(W),B)=(\exp(V)\exp(W),B+e^{(W\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}})}A).$

Hence $\mathtt{J}^{m}(G,\mathbb{W})$ together with the afore mentioned product
is a stratified Lie group withe stratified Lie algebra
$\mathfrak{j}^{m}(\mathfrak{g};\mathbb{W})$. The underlying manifold structure
of $\mathtt{J}^{m}(G,\mathbb{W})$ is that of the vector bundle
$G\times\mathtt{HD}^{\leq m}\to G$.

The equivalence problem defined by (1) leads to an equivalence by contact maps
of two stratified subgroups of the second horizontal jet space over
$G=\mathbb{H}^{2}$. The subgroups are submanifolds
$\mathtt{J}\mathcal{L},\mathtt{J}\mathcal{L}_{M}\subset\mathtt{J}^{2}(\mathbb{H}^{2},\mathbb{R})$
where $\mathtt{J}\mathcal{L}$ consists of $2$-jets of $\mathcal{L}$-harmonic
functions and $\mathtt{J}\mathcal{L}_{c}$ consists of $2$-jets of
$\mathcal{L}_{c}$-harmonic functions, see (10) for the definition of
$\mathcal{L}_{c}$.

Thus it needs to be established that a contact map of an open set
$\Omega\subseteq G$ can be prolonged to a contact map of and open subset of
the second horizontal jet space. To this end we show directly that we can
prolong from $G$ to $\mathtt{J}^{1}(G,\mathbb{W})$ and then invoke Theorem 5.2
in [4] to get a prolongation to $\mathtt{J}^{2}(G,\mathbb{W})$.

When $m=0$ we have $\mathtt{J}^{0}(G,\mathbb{W})=G\times\mathbb{W}$ where
$(a,A)(b,B)=(ab,A+B)$ and $\mathscr{H}$ is given by the left translation of
$\mathbb{V}_{1}\times\mathbb{W}$. A contact map $f:\Omega\subseteq G\to G$
lifts to a contact map $F$ on $\Omega\times\mathbb{W}$ by setting
$F=(f\circ\pi_{G},\pi_{\mathbb{W}})$. Indeed if
$(a,A)\in\mathtt{J}^{0}(G,\mathbb{W})=G\times\mathbb{W}$ and
$(V,B)\in\mathfrak{j}^{0}(\mathfrak{g},\mathbb{W})=\mathbb{V}_{1}\times\mathbb{W}$,
then

$\gamma(t)=(a,A)(\exp(tV),tB)=(a\exp(tV),A+tB)$

represents a horizontal tangent at $(a,A)$. Moreover

$\displaystyle F(a,A)^{-1}F\circ\gamma(t)$
$\displaystyle=(f(a)^{-1}f(a\exp(tV)),tB)$

represents a horizontal tangent at the identity since

$\displaystyle\frac{d}{dt}F(a,A)^{-1}F\circ\gamma(t)|_{t=0}$
$\displaystyle=\big{(}\frac{d}{dt}f(a)^{-1}f(a\exp(tV))|_{t=0},B\big{)}\in\mathbb{V}_{1}\times\mathbb{W}$

where $f$ being a contact map implies

(6) $\displaystyle
N_{f,a}(V):=\frac{d}{dt}f(a)^{-1}f(a\exp(tV))|_{t=0}\in\mathbb{V}_{1}.$

When $m=1$ we have $\mathtt{J}^{1}(G,\mathbb{W})=G\times\mathtt{HD}^{\leq 1}$
where

$\displaystyle(a,A)(b,B)$
$\displaystyle=(ab,B+A+\log(b)\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A^{1})$

and $\mathscr{H}$ is given by the left translation of
$\mathbb{V}_{1}\times\mathtt{HD}^{1}$. Moreover,

$(b,B)^{-1}=(b^{-1},\log(b)\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}B^{1}-B)$

If $\pi_{0}:\mathtt{J}^{1}(G,\mathbb{W})\to\mathtt{J}^{0}(G,\mathbb{W})$ is
the natural projection, then a contact map $\hat{F}$ of
$\tilde{U}\subseteq\mathtt{J}^{1}(G,\mathbb{W})$ is a prolongation of a
contact map $F$ of $\mathtt{J}^{0}(G,\mathbb{W})$ if
$\pi_{0}\circ\hat{F}=F\circ\pi_{0}$, which when we write
$\hat{F}=(\hat{F}_{G},\hat{F}^{0}+\hat{F}^{1})$, implies that
$\hat{F}_{G}=F_{G}\circ\pi_{0}$ and $\hat{F}^{0}=F^{0}\circ\pi_{0}$. If we
assume $F$ is a prolongation, then $F_{G}=f\circ\pi_{G}\circ\pi_{0}$ and
$\hat{F}^{0}=\pi_{\mathbb{W}}\circ\pi_{0}$, in summary

$\hat{F}(a,A)=(f(a),A^{0}+\hat{F}^{1}(a,A))$

and

$\hat{F}(a,A)^{-1}=(f(a)^{-1},\log(f(a))\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A)-A^{0}-\hat{F}^{1}(a,A))$

If $(a,A)\in\mathtt{J}^{1}(G,\mathbb{W})=G\times\mathtt{HD}^{\leq 1}$ and
$(V,B)\in\mathfrak{j}^{1}(\mathbb{V}_{1},\mathtt{HD}^{1})=\mathbb{V}_{1}\times\mathtt{HD}^{1}$,
then

$\gamma(t)=(a,A)(\exp(tV),tB)=(a\exp(tV),A+t(B+V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A))$

represents a horizontal tangent at $(a,A)$. It follows that

$\displaystyle\hat{F}_{G}\circ\gamma(t)$ $\displaystyle=f(a\exp(tV))$
$\displaystyle\hat{F}^{0}\circ\gamma(t)$
$\displaystyle=A^{0}+t\,V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A^{1}$
$\displaystyle\hat{F}^{1}\circ\gamma(t)$
$\displaystyle=\hat{F}^{1}\big{(}a\exp(tV),A+t(B+V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A)\big{)}$

and

  1. (1)

$\displaystyle\pi_{G}\big{(}\hat{F}(a,A)^{-1}\hat{F}\circ\gamma(t)\big{)}=f(a)^{-1}f(a\exp(tV))$

  2. (2)

$\displaystyle\pi_{\mathtt{HD}^{0}}\big{(}\hat{F}(a,A)^{-1}\hat{F}\circ\gamma(t)\big{)}=-\log\big{(}f(a)^{-1}f(a\exp(tV))\big{)}\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A)+tV\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A^{1}$

  3. (3)

$\displaystyle\pi_{\mathtt{HD}^{1}}\big{(}\hat{F}(a,A)^{-1}\hat{F}\circ\gamma(t)\big{)}=\hat{F}^{1}\circ\gamma(t)-\hat{F}^{1}(a,A).$

Note that in item (2) we use the fact that if $a=\exp(V)$ and $b=\exp(W)$ then

$\log(a^{-1}b)=\log(a)-\log(b)+H(V,W)$

where $H$ is an expression involving brackets of order greater than one which
implies

$\log(a)\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A)-\log(b)\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A)=-\log\big{(}a^{-1}b\big{)}\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A).$

If $\hat{F}$ is to be a contact map, then we require that the expression on
the right in item (2) vanishes which implies

(7) $\displaystyle
V\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}A^{1}=N_{f,a}(V)\raisebox{0.0pt}{\scalebox{1.0}[-1.0]{$\lrcorner$}}\hat{F}^{1}(a,A).$

Since $f$ is a contact map, the set $\\{N_{f,a}(e_{i}):i=1\dots d_{1}\\}$ is a
basis for $\mathbb{V}_{1}$ and $N_{f,a}$ is an invertible linear map of
$\mathbb{V}_{1}$ to itself. It now follows from (7) that

$\hat{F}^{1}(a,A)=A^{1}\circ N_{f,a}^{-1}.$

If

$\eta(g)=A^{1}\circ N_{f,a}^{-1}\circ\log\circ L_{f(a)}^{-1}(g)+A^{0}$

then $\eta(f(a))=A^{0}$ and

$\tilde{V}\eta(f(a))=A^{1}\circ N_{f,a}^{-1}(V),$

hence if follows that

$\hat{F}^{0}(a,A)+\hat{F}^{1}(a,A)=A_{\eta,f(a)}^{0}+A_{\eta,f(a)}^{1}$

is an element of $\mathtt{HD}^{\leq 1}$. The same construction starting with
$f^{-1}$, then prolonging to $F^{-1}$, and then prolonging to $\hat{F}^{-1}$
yields the inverse of $\hat{F}$ and so $\hat{F}$ is a diffeomorphism.

When $m=2$, we apply Theorem 5.2 in [4], see Remark 5.4.

##  3\. Heisenberg Groups

Following [7], the Heisenberg group $\mathbb{H}^{n}$ is given by
$\mathbb{R}^{2n+1}$, where we denote coordinates by $(x,y,t)$,
$x,y\in\mathbb{R}^{n}$ and $t\in\mathbb{R}$, and the product is given by

$(x,y,t)(x^{\prime},y^{\prime},t^{\prime})=(x+x^{\prime},y+y^{\prime},t+t^{\prime}-2x\cdot
y^{\prime}+2y\cdot x^{\prime})).$

If $\\{e_{i}:i=1,\dots,2n+1\\}$ is the standard basis for $\mathbb{R}^{2n+1}$
and $\mathfrak{h}^{n}$ denotes the Lie algebra where the nontrivial Lie
brackets are $[e_{i},e_{n+i}]=-4e_{2n+1}$, $i=1,\dots,n$, then the product
above is the Baker–Campbell–Hausdorff model of $\mathbb{H}^{2}$ built on
$\mathfrak{h}^{n}$.

The left invariant vector fields are framed by

$\displaystyle\tilde{X}_{i}$ $\displaystyle=\frac{\partial}{\partial
x_{i}}+2y_{j}\frac{\partial}{\partial t}\quad i=1,\dots,n$
$\displaystyle\tilde{Y}_{i}$ $\displaystyle=\frac{\partial}{\partial
x_{i}}-2x_{j}\frac{\partial}{\partial t}\quad i=1,\dots,n$
$\displaystyle\tilde{T}$ $\displaystyle=\frac{\partial}{\partial t}$

for which the only nontrivial brackets are

$[\tilde{X}_{i},\tilde{Y}_{i}]=-4\tilde{T}\quad i=1,\dots,n.$

The frame $\\{dx_{i},dy_{i},\theta\\}$, where
$\theta=dt+2\sum_{i}(x_{i}dy_{i}-y_{i}dx_{i})$, is the dual of the frame
$\\{\tilde{X}_{i},\tilde{Y}_{i},\tilde{T}\\}$ and ${\rm
ker\,}\theta=\mathscr{H}$. In particular, $\theta$ is a contact form since

$(d\theta)^{n}\wedge\theta=2^{2n}dx_{1}\wedge\dots\wedge dx_{n}\wedge
dy_{1}\wedge\dots\wedge dy_{n}\wedge dt.$

The left invariant $2$-form

$B=\frac{1}{4}d\theta=\sum_{i}dx_{i}\wedge dy_{i}$

is symplectic on $\mathscr{H}$ and represented by the matrix

$\Omega=\begin{pmatrix}0&I_{n}\\\ -I_{n}&0\end{pmatrix}$

relaltive to the chosen coordinates. Furthermore, Cartan’s formula implies
that

$[\tilde{V},\tilde{W}]=-4B(\tilde{V},\tilde{W})\tilde{T}$

for all left invariant horizontal vector fields $\tilde{V}$ and $\tilde{W}$.
The complex structure $J_{0}=\Omega^{T}$ is compatible with $B$ and $g_{0}$,
that is to say

$g_{0}(\tilde{V},\tilde{W})=B(\tilde{V},J_{0}\tilde{W})$

for all left invariant horizontal vector fields $\tilde{V}$ and $\tilde{W}$.

A local diffeomorphism $f:\mathbb{H}^{n}\to\mathbb{H}^{n}$ preserves
$\mathscr{H}$ if and only $f^{*}\theta=f^{*}\theta(\tilde{T})\theta$ and is
therefore a contact transformation in the strict sense. Left translations are
contact transformations.

An automorphism $\alpha$ of $\mathbb{H}^{n}$ must satisfy
$d\alpha(\tilde{T})=\lambda\tilde{T}$ since $\tilde{T}$ is central, and is
therefore a contact transformation. Two fundamental examples are dilation by
$\lambda>0$

$\displaystyle\delta_{\lambda}(x,y,t)$ $\displaystyle=(\lambda x,\lambda
y,\lambda^{2}t)$

and the reflection

(8) $\displaystyle r(x,y,t)$ $\displaystyle=(x,-y,-t).$

Consequently, every automorphism of $\mathbb{H}^{n}$ is a composition
$\alpha\circ\delta_{\lambda}\circ r$ where $\alpha$ is normalised, that is to
say, $d\alpha(\tilde{T})=\tilde{T}$. Furthermore, a normalised automorphism
$\alpha$ satisfies

$B(d\alpha(\tilde{V}),d\alpha(\tilde{V}))=B(\tilde{V},\tilde{W})$

for all left invariant horizontal vector fields $\tilde{V}$ and $\tilde{W}$,
and is thus a symplectic transformation represented in the chosen coordinates
by an element of $\operatorname{Sp}(2n,\mathbb{R})$. It follows that the group
of grading preserving automorphism of the Lie algebra is

$\mathbb{R}_{+}\times(\operatorname{Sp}(n,\mathbb{R})\cup
dr(\operatorname{Sp}(n,\mathbb{R})))$

where $\mathbb{R}_{+}$ acts by $d\delta_{\lambda}$. The group of "horizontal
orientation" preserving automorphisms of $\mathbb{H}^{n}$ is
$\mathbb{R}_{+}\times{\rm Sp}(2n,\mathbb{R})$ and the action is this case is
given by the representation

$(\lambda,S)\to\begin{pmatrix}\sqrt{\lambda}S&0\\\ 0&\lambda\end{pmatrix}.$

Williamson’s Theorem (Theorem 8.3.1, page 244, [3]) gives the symplectic
version of the well known diagonalisation of a positive-definite symmetric
matrix by the orthogonal group.

######  Theorem 3.1 (Williamson).

Let $M$ be a positive-definite symmetric real $2n\times 2n$ matrix. Then there
exists $S\in\operatorname{Sp}(2n,\mathbb{R})$ such that

$S^{T}MS=\Lambda_{M}=\begin{pmatrix}\Lambda&0\\\ 0&\Lambda\end{pmatrix}$

where $\Lambda={\rm diag}(\lambda_{1},\dots,\lambda_{n})$ and $\pm
i\lambda_{j}$ are the eigenvalues of $\,-\Omega M^{-1}$.

The Williamson diagonal form gives the following factorisation

$\displaystyle\mathcal{L}_{M}u$
$\displaystyle=S^{-1}{\nabla}_{\scriptscriptstyle\mathscr{H}}\cdot\Lambda_{M}S^{-1}{\nabla}_{\scriptscriptstyle\mathscr{H}}.$

Furthermore, we can dilate by $\lambda_{1}^{-1}$ and conclude that for any
positive-definite symmetric real $2n\times 2n$ matrix, we can change
coordinates by a Lie group automorphism $\alpha$ of $\mathbb{H}^{n}$ such that
the frame given by $\bar{X}_{i}=d\alpha(\tilde{X}_{i})$,
$\bar{Y}_{i}=d\alpha(\tilde{Y}_{i})$ and $\bar{T}=d\alpha(\tilde{T})$, is left
invariant and

$\displaystyle\mathcal{L}_{M}(u\circ\alpha)$
$\displaystyle=({\bar{\nabla}}_{\scriptscriptstyle\mathscr{H}}\cdot\Lambda_{M}(1,\lambda_{2}/\lambda_{1},\dots,\lambda_{n}/\lambda_{1}){\bar{\nabla}}_{\scriptscriptstyle\mathscr{H}}u)\circ\alpha$
(9)
$\displaystyle=(\bar{X}_{1}^{2}u)\circ\alpha+(\bar{Y}_{1}^{2}u)\circ\alpha+\sum_{i=2}^{n}\frac{\lambda_{i}}{\lambda_{1}}(\bar{X}_{i}^{2}u)\circ\alpha+\frac{\lambda_{i}}{\lambda_{1}}(\bar{Y}_{i}^{2}u)\circ\alpha$

where ${\bar{\nabla}}_{\scriptscriptstyle\mathscr{H}}$ is the horizontal
gradient operator relative to the frame $\\{\bar{X}_{i},\bar{Y}_{i}\\}$.

##  4\. Sub-laplacians on $\mathbb{H}^{2}$

In this section we consider the simplest nontrivial case of the equivalence
problem for sub-Laplacians, namely $\mathbb{H}^{2}$. As a consequence of (9),
we need only to consider the equivalence problem for the one parameter family

(10)
$\displaystyle\mathcal{L}_{c}=\tilde{X}_{1}^{2}+\tilde{Y}_{1}^{2}+c(\tilde{X}_{2}^{2}+\tilde{Y}_{2}^{2}),\qquad
0<c.$

We will show that for each $c>0$, the existence of a local contact
diffeomorphism $F:\mathbb{H}^{2}\to\mathbb{H}^{2}$ with the property that

$\mathcal{L}_{c}(u\circ F)=\mathcal{L}u\circ F$

requires that the prolongation of $F$ to the second horizontal jet space
$\mathtt{J}^{m}(\mathbb{H}^{2},\mathbb{R})$ , restricts to a Pansu
differentiable map between two stratified subgroups of
$\mathtt{J}^{m}(\mathbb{H}^{2},\mathbb{R})$ which implies that the Lie
algebras of these two groups must be isomorphic. The nonexistence of an
isomorphism between the afore mentioned Lie algebras can easily be verified
with MAPLE using the package LieAlgebras.

The main task that remains is to explicitly construct the two stratified
subgroups of $\mathtt{J}^{m}(\mathbb{H}^{2},\mathbb{R})$ and their Lie
algebras, which can then be analysed using MAPLE. To proceed we need a basis
for $\mathtt{HD}^{\leq 2}$. By the Poincaré–Birkhoff–Witt theorem ([2],
I.2.7), the family

(11)
$\displaystyle\\{\tilde{X}_{1}^{\alpha_{1}}\tilde{X}_{2}^{\alpha_{2}}\tilde{Y}_{1}^{\beta_{1}}\tilde{Y}_{2}^{\beta_{2}}\tilde{T}^{\gamma}:\alpha_{i},\beta_{j},\gamma\in\mathbb{N}\cup\\{0\\}\\}$

is a basis for the left invariant differential operators on $\mathbb{H}^{2}$.
It follows that the left invariant differential operators of homogeneous order
$m\leq 2$ are spanned by the set

(12)
$\displaystyle\\{\tilde{X}_{1},\tilde{X}_{2},\tilde{Y}_{1},\tilde{Y}_{2},\tilde{X}_{1}^{2},\tilde{X}_{1}\tilde{X}_{2},\tilde{X}_{1}\tilde{Y}_{1},\tilde{X}_{1}\tilde{Y}_{2},\tilde{X}_{2}^{2},\tilde{X}_{2}\tilde{Y}_{1},\tilde{X}_{2}\tilde{Y}_{2},\tilde{Y}_{1}^{2},\tilde{Y}_{1}\tilde{Y}_{2},\tilde{Y}_{2}^{2},\tilde{Y}_{1}\tilde{X}_{1}\\}.$

Note that we have replaced the element $\tilde{T}$ that is prescribed by (11)
with $\tilde{Y}_{1}\tilde{X}_{1}$. Since
$[\tilde{X}_{1},\tilde{Y}_{1}]=[\tilde{X}_{2},\tilde{Y}_{2}]=-4\tilde{T}$ we
have

(13)
$\displaystyle\tilde{T}=\frac{1}{4}(\tilde{Y}_{1}\tilde{X}_{1}-\tilde{X}_{1}\tilde{Y}_{1})\quad{\rm
and}\quad\tilde{Y}_{2}\tilde{X}_{2}=\tilde{X}_{2}\tilde{Y}_{2}+\tilde{Y}_{1}\tilde{X}_{1}-\tilde{X}_{1}\tilde{Y}_{1}$

and it follows that (12) is a basis for $\mathtt{HD}^{\leq 2}$.

The two jet has the form $j^{2}_{p}u=\sum_{k=0}^{2}\frac{1}{k!}A_{u,p}^{k}.$
Since $A_{u,p}^{1}(V)=\tilde{V}u(p)$, the set

$\\{A_{X_{1}}=dx_{1},A_{X_{2}}=dx_{2},A_{Y_{1}}=dy_{1},A_{Y_{2}}=dy_{2}\\}$

forms a basis for $\mathtt{HD}^{1}(\mathfrak{g};\mathbb{R})$. The second order
term of $j^{2}_{p}u$ is defined on $\mathbb{V}_{1}^{2}$ by

$A_{u,p}^{2}(V,W)=\tilde{V}\tilde{W}u(p)$

which when expressed relative to (12) becomes

$\displaystyle A_{u,p}^{2}(V,W)=$
$\displaystyle\,v_{1}w_{1}\tilde{X}_{1}^{2}u+(v_{1}w_{2}+v_{2}w_{1})\tilde{X}_{1}\tilde{X}_{2}u+(v_{1}w_{3}-v_{4}w_{2})\tilde{X}_{1}\tilde{Y}_{1}u$
$\displaystyle+(v_{1}w_{4}+v_{4}w_{1})\tilde{X}_{1}\tilde{Y}_{2}u+v_{2}w_{2}\tilde{X}_{2}^{2}u+(v_{3}w_{2}+v_{2}w_{3})\tilde{X}_{2}\tilde{Y}_{1}u$
$\displaystyle+(v_{2}w_{4}+v_{4}w_{2})\tilde{X}_{2}\tilde{Y}_{2}u+v_{3}w_{3}\tilde{Y}_{1}^{2}u+(v_{3}w_{4}+v_{4}w_{3})\tilde{Y}_{1}\tilde{Y}_{2}u$
$\displaystyle+v_{4}w_{4}\tilde{Y}_{2}^{2}u+(v_{3}w_{1}+v_{4}w_{2})\tilde{Y}_{1}\tilde{X}_{1}u.$

and the basis elements of $\mathtt{HD}^{2}(\mathfrak{g};\mathbb{R})$ are:

$\displaystyle A_{X_{1}^{2}}$ $\displaystyle=dx_{1}\otimes dx_{1}$
$\displaystyle A_{X_{1}X_{2}}$ $\displaystyle=dx_{1}\otimes
dx_{2}+dx_{2}\otimes dx_{1}$ $\displaystyle A_{X_{1}Y_{1}}$
$\displaystyle=dx_{1}\otimes dy_{1}-dy_{2}\otimes dx_{2}$ $\displaystyle
A_{X_{1}Y_{2}}$ $\displaystyle=dx_{1}\otimes dy_{2}+dy_{2}\otimes dx_{1}$
$\displaystyle A_{X_{2}^{2}}$ $\displaystyle=dx_{2}\otimes dx_{2}$
$\displaystyle A_{X_{2}Y_{1}}$ $\displaystyle=dy_{1}\otimes
dx_{2}+dx_{2}\otimes dy_{1}$ $\displaystyle A_{X_{2}Y_{2}}$
$\displaystyle=dx_{2}\otimes dy_{2}+dy_{2}\otimes dx_{2}$ $\displaystyle
A_{Y_{1}^{2}}$ $\displaystyle=dy_{1}\otimes dy_{1}$ $\displaystyle
A_{Y_{1}Y_{2}}$ $\displaystyle=dy_{1}\otimes dy_{2}+dy_{2}\otimes dy_{1}$
$\displaystyle A_{Y_{2}^{2}}$ $\displaystyle=dy_{2}\otimes dy_{2}$
$\displaystyle A_{Y_{1}X_{1}}$ $\displaystyle=dy_{1}\otimes
dx_{1}+dy_{2}\otimes dx_{2}.$

As the notation suggests,
$j^{2}_{p}u=\sum_{K\in\mathcal{I}\cup\\{0\\}}\tilde{K}u(p)A_{K}$ where
$\mathcal{I}$ is the ordered list of symbols given by $\eqref{hset}$ with
tilde removed, and $\tilde{K}u(p)=u(p)$ when $K=0$. The ordered set

$\displaystyle E=\\{e_{1},\dots
e_{4},A_{X_{1}^{2}},\dots,A_{Y_{1}X_{1}},e_{5},A_{X_{1}},A_{X_{2}},A_{Y_{1}},A_{Y_{2}},1\\}$

forms a stratified basis for the $21$-dimensional Lie algebra
$\mathfrak{j}^{2}(\mathfrak{g};\mathbb{R})$ where

$\displaystyle\mathfrak{j}^{2}(\mathfrak{g};\mathbb{R})_{1}$
$\displaystyle=\mathbb{V}_{1}\times\mathtt{HD}^{2}(\mathfrak{g};\mathbb{R})={\rm
span}\,\\{e_{1},\dots e_{4},A_{X_{1}X_{2}},\dots,A_{Y_{1}X_{1}}\\}$
$\displaystyle\mathfrak{j}^{2}(\mathfrak{g};\mathbb{R})_{2}$
$\displaystyle=\mathbb{V}_{2}\times\mathtt{HD}^{1}(\mathfrak{g};\mathbb{R})={\rm
span}\,\\{e_{5},A_{X_{1}},A_{X_{2}},A_{Y_{1}},A_{Y_{2}}\\}$
$\displaystyle\mathfrak{j}^{2}(\mathfrak{g};\mathbb{R})_{3}$
$\displaystyle=\mathbb{V}_{3}\times\mathtt{HD}^{0}(\mathfrak{g};\mathbb{R})=\mathbb{R}.$

Note that for $A\in\mathtt{HD}^{2}(\mathfrak{g};\mathbb{R})$, the choice of
basis and (4) require that we define

$[e_{5},A]:=\frac{1}{4}(A(e_{2},e_{1})-A(e_{1},e_{2})).$

The bracket table for the basis $E$ is as follows (we do not include the
trivial row and column generated by $E_{21}$, see [5] for the MAPLE file
J2H2-JLc.mw):

$\displaystyle\begin{array}[]{ccccccccccccccccccccccc}&&&&&&&&&&&&&&&&&&&\\\
0&0&-4E_{16}&0&-E_{17}&-E_{18}&0&-E_{20}&0&0&0&0&0&0&-E_{19}&0&-E_{21}&0&0&0\\\
0&0&0&-4E_{16}&0&-E_{17}&E_{20}&0&-E_{18}&-E_{19}&-E_{20}&0&0&0&-E_{20}&0&0&-E_{21}&0&0\\\
4E_{16}&0&0&0&0&0&-E_{17}&0&0&-E_{18}&0&-E_{19}&-E_{20}&0&0&0&0&0&-E_{21}&0\\\
0&4E_{16}&0&0&0&0&0&-E_{17}&0&0&-E_{18}&0&-E_{19}&-E_{20}&0&0&0&0&0&-E_{21}\\\
E_{17}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
E_{18}&E_{17}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&-E_{20}&E_{17}&0&0&0&0&0&0&0&0&0&0&0&0&\frac{1}{4}E_{21}&0&0&0&0\\\
E_{20}&0&0&E_{17}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&E_{18}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&E_{19}&E_{18}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&E_{20}&0&E_{18}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&E_{19}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&E_{20}&E_{19}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&E_{20}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
E_{19}&E_{20}&0&0&0&0&0&0&0&0&0&0&0&0&0&-\frac{1}{4}E_{21}&0&0&0&0\\\
0&0&0&0&0&0&-\frac{1}{4}E_{21}&0&0&0&0&0&0&0&\frac{1}{4}E_{21}&0&0&0&0&0\\\
E_{21}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&E_{21}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&E_{21}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&E_{21}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
&&&&&&&&&&&&&&&&&&&&\end{array}$

We denote by $\eta_{k}(u,p)$, the coordinates of $A_{u,p}$ relative to $E$,
more specifically:

$\displaystyle A_{u,p}^{1}=\sum_{k=17}^{20}\eta_{k}(u,p)E_{k}\quad{\rm
and}\quad A_{u,p}^{2}=\sum_{k=5}^{15}\eta_{k}(u,p)E_{k}.$

If $u$ is $\mathcal{L}_{c}$-harmonic then

(14) $\displaystyle\eta_{5}+\eta_{12}+c(\eta_{9}+\eta_{14})=0$

and we let $\mathtt{J}\mathcal{L}_{c}$ denote the subbundle of
$\mathtt{J}^{2}(G,\mathbb{R})\to G$ defined by (14). Furthermore, since
$\mathtt{J}\mathcal{L}_{c}$ is obtained by a homogeneous linear equation of
the first layer coordinates, it is a Lie subgroup of
$\mathtt{J}^{2}(G,\mathbb{R})$.

The bracket table for the basis of the Lie algebra of
$\mathtt{J}\mathcal{L}_{c}$ is as follows (we do not include the trivial row
and column generated by $F_{20}$, see [5] for the MAPLE file JL-to-JLc.mw ):

$\displaystyle\begin{array}[]{ccccccccccccccccccc}&&&&&&&&&&&&&&&&&&\\\
0&0&-4\,F_{{15}}&0&-F_{{17}}&0&-F_{{19}}&cF_{{16}}&0&0&F_{{16}}&0&cF_{{16}}&-F_{{18}}&0&-F_{{20}}&0&0&0\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&0&-4\,F_{{15}}&-F_{{16}}&F_{{19}}&0&-F_{{17}}&-F_{{18}}&-F_{{19}}&0&0&0&-F_{{19}}&0&0&-F_{{20}}&0&0\\\
4\,F_{{15}}&0&0&0&0&-F_{{16}}&0&0&-F_{{17}}&0&-F_{{18}}&-F_{{19}}&0&0&0&0&0&-F_{{20}}&0\\\
0&4\,F_{{15}}&0&0&0&0&-F_{{16}}&0&0&-F_{{17}}&0&-F_{{18}}&-F_{{19}}&0&0&0&0&0&-F_{{20}}\\\
F_{{17}}&F_{{16}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&-F_{{19}}&F_{{16}}&0&0&0&0&0&0&0&0&0&0&0&\frac{1}{4}\,F_{{20}}&0&0&0&0\\\
F_{{19}}&0&0&F_{{16}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
-cF_{{16}}&F_{{17}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&F_{{18}}&F_{{17}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&F_{{19}}&0&F_{{17}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
-F_{{16}}&0&F_{{18}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&F_{{19}}&F_{{18}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
-cF_{{16}}&0&0&F_{{19}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
F_{{18}}&F_{{19}}&0&0&0&0&0&0&0&0&0&0&0&0&-\frac{1}{4}\,F_{{20}}&0&0&0&0\\\
0&0&0&0&0&-\frac{1}{4}\,F_{{20}}&0&0&0&0&0&0&0&\frac{1}{4}\,F_{{20}}&0&0&0&0&0\\\
F_{{20}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&F_{{20}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&F_{{20}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
0&0&0&F_{{20}}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\ &&&&&&&&&&&&&&&&&&\end{array}$

Let $F$ be a local contact diffeomorphism on $\mathtt{J}^{2}(G,\mathbb{R})$.
If $\mathcal{L}_{c}(u\circ F)=\mathcal{L}u\circ F$ then $\mathcal{L}u=0$
implies $\mathcal{L}_{c}(u\circ F)=0$ and so the range of
$F|_{\mathtt{J}\mathcal{L}_{c}}$ is
$\mathtt{J}\mathcal{L}:=\mathtt{J}\mathcal{L}_{1}$, moreover
$F|_{\mathtt{J}\mathcal{L}_{c}}$ maps the horizontal bundle of
$\mathtt{J}\mathcal{L}_{c}$ to that of $\mathtt{J}\mathcal{L}$, and so it’s
Pansu derivative is a Lie algebra isomorphism between the Lie algebras of
$\mathtt{J}\mathcal{L}_{c}$ and $\mathtt{J}\mathcal{L}$ (see section 2 and
Theorem 2.9 in [6] for more detail). However, by direct calculation, see [5]
for the MAPLE file JL-to-JLc.mw, we can show that the afore mentioned algebras
are isomorphic only in the case $c=1$.

## References

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