prop@alttheorem lem@alttheorem coroll@alttheorem defi@alttheorem
quest@alttheorem fact@alttheorem rem@alttheorem exa@alttheorem

# Intrinsic Lipschitz sections of no-linear quotient maps

Daniela Di Donato

###### Abstract.

Le Donne and the author introduced the so-called intrinsically Lipschitz
sections of a fixed quotient map $\pi$ in the context of metric spaces.
Moreover, the author introduced the concept of intrinsic Cheeger energy when
the quotient map is also linear. In this note we investigate about the no-
linearity of $\pi$. In particular, we find a Leibniz formula for the intrinsic
slope when $\pi$ satisfies a weaker condition. After that, we focus our
attention on Carnot groups and using the properties of intrinsic dilations we
show that the dilation of a Lipschitz section is so too. Finally, in Carnot
groups of step 2, we give a suitable additional condition in order to get the
sum of two intrinsically Lipschitz sections is so too.

###### Contents

  1. 1 Introduction
  2. 2 A weaker linearity condition for $\pi$
  3. 3 The intrinsic dilations vs quotient map
  4. 4 The sum of intrinsic Lipschitz sections is so too in Carnot groups of step 2

##  1\. Introduction

Find out a good notion of rectifiability in subRiemannian Carnot groups
[ABB19, BLU07, CDPT07] is unanswered questions which have continued to perplex
many mathematicians for decades. Starting from a negative result by Ambrosio
and Kirchheim [AK00], Franchi, Serapioni and Serra Cassano introduced the
notion of intrinsically Lipschitz maps in [FSSC01, FSSC03b, FSSC03a] (see also
[SC16, FS16]) in order to answer this question. Recently, Le Donne and the
author [DDLD22] generalize this concept in metric spaces and prove some
relevant properties like Ahlfors-David regularity, Ascoli-Arzelá Theorem,
Extension theorem, etc. for the so-called intrinsically Lipschitz sections.

Our setting is the following. We have a metric space $X$, a topological space
$Y$, and a quotient map $\pi:X\to Y$, meaning continuous, open, and
surjective. The standard example for us is when $X$ is a metric Lie group $G$
(meaning that the Lie group $G$ is equipped with a left-invariant distance
that induces the manifold topology), for example a subRiemannian Carnot group,
and $Y$ if the space of left cosets $G/H$, where $H<G$ is a closed subgroup
and $\pi:G\to G/H$ is the projection modulo $H$, $g\mapsto gH$.

######  Definition .

We say that a map $\varphi:Y\to X$ is a section of $\pi$ if

(1) $\pi\circ\varphi=\mbox{id}_{Y}.$

Moreover, we say that a map $\varphi:Y\to X$ is an intrinsically Lipschitz
section of $\pi$ with constant $L$, with $L\in[1,\infty)$, if in addition

(2) $d(\varphi(y_{1}),\varphi(y_{2}))\leq
Ld(\varphi(y_{1}),\pi^{-1}(y_{2})),\quad\mbox{for all }y_{1},y_{2}\in Y.$

Here $d$ denotes the distance on $X$, and, as usual, for a subset $A\subset X$
and a point $x\in X$, we have $d(x,A):=\inf\\{d(x,a):a\in A\\}$.

A first observation is that this class is contained in the class of continuous
maps (see [DDLD22, Section 2.4]) and in the case $\pi$ is a Lipschitz quotient
or submetry [BJL+99, VN88], being intrinsically Lipschitz is equivalent to
biLipschitz embedding, see Proposition 2.4 in [DDLD22]. Moreover, since
$\varphi$ is injective by (1), the class of Lipschitz sections not include the
constant maps.

Conversely to Lipschitz classical notion, the intrinsically Lipschitz sections
cannot be uniformly continuous; an easy example is as follows.

######  Example  ($\pi$ is not linear).

Let $X=(0,1),Y=(1,+\infty)$ and $\pi:(0,1)\to(1,+\infty)$ defined as
$\pi(x)=\frac{1}{x},$ for every $x\in(0,1).$ It is easy to see that $\pi$ is a
quotient map and $\varphi:(1,+\infty)\to(0,1)$ defined as

$\varphi(y)=\frac{1}{y},\quad\forall y\in(1,+\infty),$

is an intrinsic $1$-Lipschitz section of $\pi$. On the other hand, it is well-
known that $\varphi$ is not uniformly continuous.

The long-term objective is to obtain a Rademacher type theorem á la Cheeger
[Che99] (see also [KM16, Kei04]) for the intrinsically Lipschitz sections.

This project fits to an active line of research [Pan89, Vit20, FMS14] on
Rademacher type theorems, but it also studies connections to different notions
and mathematical areas like Cheeger energy, Sobolev spaces, Optimal transport
theory. The reader can see [AGS14a, AGS14b, Stu06, LV09, Vil09, Ha96, Sha00,
JPNJ15].

A first step of this research is given in [DD22b, Proposition 3.1-3.2] and
[DD22a] as follows.

######  Proposition .

Let $\pi:X\to Y$ be a linear and quotient map between a normed space $X$ and a
topological space $Y.$

  1. (1)

the set of all sections is a vector space over $\mathbb{R}$ or $\mathbb{C}.$

  2. (2)

If $\varphi:Y\to X$ is an intrinsically Lipschitz section of $\pi,$ then for
any $\lambda\in\mathbb{R}-\\{0\\}$ the section $\lambda\varphi$ is also
intrinsic Lipschitz for $1/\lambda\pi$ with the same Lipschitz constant.

  3. (3)

If $X=\mathbb{R}^{s}$, $Y$ is a metric space and each fiber is contained in a
straight line and (3) holds, then the sum of two intrinsically Lipschitz
sections is so too.

Here it is fundamental that $\pi$ is a linear map. The aim of this paper is to
prove the last result when $\pi$ does not satisfy this property. We focus our
attention on that property because it is not true in the context of Carnot
groups. More precisely, the main results are

  1. (1)

Proposition 2.2: we give a Leibniz formula for the intrinsic slope when $\pi$
satisfies a weaker linearity condition.

  2. (2)

Proposition 3 and Corollary 3: we focus our attention on the well-known
properties of intrinsic dilations in Carnot groups and we show that the
dilation of a Lipschitz section is so too.

  3. (3)

Proposition 4: an open question is whether the sum of two Lipschitz maps in
the Franchi, Serapioni and Serra Cassano sense is so too. Our idea of proof in
[DD22a] does not work to solve this open question in general case; on the
other hand, if we just consider the linearity of $\pi$ valued in the points of
the two sections, then it is possible to get a positive answer in the context
of Carnot groups of step 2 [BLU07, DD20].

##  2\. A weaker linearity condition for $\pi$

In this section, we study the properties proved in [DD22a] when $\pi:X\to Y$
is not linear. Regarding the definition of the intrinsic Lipschitz constants
the reader can see [DD22a, Section 3]. Here, we ask that there exists
$\lambda>0$ such that

(3) $\pi(\alpha x_{1}+\beta
x_{2})=\alpha^{\lambda}\pi(x_{1})+\beta^{\lambda}\pi(x_{2}),\quad\mbox{ for
any }x_{1},x_{2}\in X,$

and for any $\alpha,\beta\in\mathbb{R}$ with $(\alpha,\beta)\neq(0,0).$

######  Proposition .

Let $\varphi,\psi:Y\to X$ be sections of $\pi.$ Then, for any
$\alpha,\beta\in\mathbb{R}$ with $(a,b)\neq(0,0)$ the map
$\eta:=\alpha\varphi+\beta\psi:Y\to X$ is a section of
$(\frac{1}{\alpha^{\lambda}+\beta^{\lambda}})\pi.$

###### Proof.

Fix $\alpha,\beta\in\mathbb{R}$ with $(\alpha,\beta)\neq(0,0).$ The thesis
follows because
$\pi(\eta(y))=\pi(\alpha\varphi(y)+\beta\psi(y))=(\alpha^{\lambda}+\beta^{\lambda})y$
for any $y\in Y.$ ∎

######  Proposition .

Let $X$ be a normed space and $\varphi:Y\to X$ be an intrinsically
$L$-Lipschitz section of $\pi.$ Then, for any $\alpha\in\mathbb{R}-\\{0\\}$
the map $\alpha\varphi:Y\to X$ is an intrinsically Lipschitz section of
$(\frac{1}{\alpha^{\lambda}})\pi$ with the same Lipschitz constant.

###### Proof.

Fix $\alpha\in\mathbb{R}-\\{0\\}$ and $y,z\in Y.$ By Proposition 2, we have
that $\alpha\varphi$ is a section of $(\frac{1}{\alpha^{\lambda}})\pi.$ On the
other hand, noting that if $x\in\pi^{-1}(z),$ then

(4) $\alpha
x\in\pi^{-1}(\alpha^{\lambda}z)=\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(z);$

indeed,

$\pi(\alpha x)=\alpha^{\lambda}\pi(x)=\alpha^{\lambda}z,$

and so $\frac{1}{\alpha^{\lambda}}\pi(\alpha x)=z.$ Now we choose
$x\in\pi^{-1}(z)$ such that

$d\left(\alpha\varphi(y),\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(z)\right)=d(\alpha\varphi(y),\alpha
x).$

Consequently,

$\displaystyle d(\alpha\varphi(y),\alpha\varphi(z))$
$\displaystyle=|\alpha|d(\varphi(y),\varphi(z))\leq
L|\alpha|d(\varphi(y),\pi^{-1}(z))$ $\displaystyle\leq
L|\alpha|d(\varphi(y),x)=Ld\left(\alpha\varphi(y),\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(z)\right).$

∎

######  Theorem 2.1.

Let $X=\mathbb{R}^{s}$, $Y$ be a metric space and $\pi:X\to Y$ be a quotient
map such that each fiber is contained in a straight line. If $\pi$ satisfies
(3), then the sum of two intrinsically Lipschitz sections is so too.

###### Proof.

The proof is exactly as in [DD22a, Theorem 4.1], noting that the fiber of the
sum is $(1/2\pi)^{-1}$ thanks to Proposition 2. ∎

Now we want to obtain the Leibniz formula for the intrinsic Lipschitz
constants using an additional suitable condition on the fibers as in [DD22a].

######  Theorem 2.2 (Leibniz formula).

Let $Y$ be a metric space and $\pi:X\to Y$ be a quotient map such that either
$X=\mathbb{R}$ or $X=\mathbb{R}^{s}$ and each fiber is contained in a straight
line. Assume also that $\pi$ satisfies (3) and $\varphi$ and $\psi$ are
intrinsically $L$-Lipschitz sections of $\pi$ such that there is $c\geq 1$
satisfying

(5)
$d(\pi^{-1}(y),\pi^{-1}(z))\geq\frac{1}{c}d(f(y),\pi^{-1}(z)),\quad\,\forall
y,z\in Y,$

$\mbox{for }f=\varphi,\psi.$ Then, denoting $\eta=\alpha\varphi+\beta\psi$ the
map $Y\to X$ with $\alpha,\beta\in\mathbb{R}-\\{0\\},$ we have that

(6)
$Ils(\eta)(y)\leq\frac{c}{2^{1/\lambda}}(Ils(\varphi)(y)+Ils(\psi)(y)),\quad\forall
y\in Y.$

We need the following lemma which is true in more general setting.

######  Lemma .

Let $X$ be a normed space, $Y$ be a topological space and $\pi:X\to Y$ be a
quotient map satisfies (3). Then

(7)
$|\alpha|d(\pi^{-1}(y_{1}),\pi^{-1}(y_{2}))=d\left(\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(y_{1}),\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(y_{2})\right),\quad\forall
y_{1},y_{2}\in Y,\forall\alpha\in\mathbb{R}-\\{0\\}.$

###### Proof.

Fix $y_{1},y_{2}\in Y.$ If $d(\pi^{-1}(y_{1}),\pi^{-1}(y_{2}))=d(x_{1},x_{2})$
for some $x_{1}\in\pi^{-1}(y_{1})$ and $x_{2}\in\pi^{-1}(y_{2}),$ then as
above (see (4)) we deduce that

$\alpha
x_{i}\in\pi^{-1}(\alpha^{\lambda}y_{i})=\left(\frac{1}{\alpha^{\lambda}}\pi\right)^{-1}(y_{i}),$

for $i=1,2$ and, consequently,

(8) $\displaystyle d(\pi^{-1}(y_{1}),\pi^{-1}(y_{2}))$
$\displaystyle=\|x_{1}-x_{2}\|=\frac{1}{|\alpha|}\|\alpha x_{1}-\alpha
x_{2}\|$
$\displaystyle\geq\frac{1}{|\alpha|}d\left(\left(1/{\alpha^{\lambda}}\pi\right)^{-1}(y_{1}),\left(1/{\alpha^{\lambda}}\pi\right)^{-1}(y_{2})\right).$

On the other hand, if
$d((1/\alpha^{\lambda}\pi)^{-1}(y_{1}),(1/\alpha^{\lambda}\pi)^{-1}(y_{2}))=d(h_{1},h_{2})$
for some $h_{i}\in(1/\alpha^{\lambda}\pi)^{-1}(y_{i})$ for $i=1,2$ then

$\frac{1}{\alpha}h_{i}\in\pi^{-1}(y_{i});$

indeed,

$\pi\left(\frac{1}{\alpha}h_{i}\right)=\frac{1}{\alpha^{\lambda}}\pi(h_{i})=y_{i},\quad\mbox{
for }i=1,2.$

Hence,

(9) $\displaystyle
d((1/\alpha^{\lambda}\pi)^{-1}(y_{1}),(1/\alpha^{\lambda}\pi)^{-1}(y_{2}))$
$\displaystyle=\|h_{1}-h_{2}\|=|\alpha|\left\|\frac{1}{\alpha}h_{1}-\frac{1}{\alpha}h_{2}\right\|$
$\displaystyle\geq|\alpha|d(\pi^{-1}(y_{1}),\pi^{-1}(y_{2})).$

Hence, putting together (8) and (9), we get (7). ∎

###### Proof of Proposition 2.2.

Notice that since $Ils(\varphi)=Ils(\alpha\varphi)$ for any $\alpha\neq 0$
(see Proposition 2), it is sufficient to prove (6) for $\eta=\varphi+\psi.$

We split the proof in two steps. In the first step we prove that $\eta$ is
intrinsic Lipschitz and in the second one we show (6) following a similar
computation in [KM16, Lemma 3.2].

(1). The thesis follows from Theorem 2.1 noting that for $s=1$ it is trivial
that each fiber is contained in a straight line.

(2). Fix $\bar{y}\in Y$ and $r>0.$ Suppose we are given $\varepsilon>0.$ For
any $y\in B(\bar{y},r)$ we have

(10)
$\displaystyle\frac{d(\varphi(\bar{y}),\varphi(y))}{d(\varphi(\bar{y}),\pi^{-1}(y))}\leq
Ils(\varphi)(\bar{y})+\varepsilon\qquad{and}\mbox{\qquad}\frac{d(\psi(\bar{y}),\psi(y))}{d(\psi(\bar{y}),\pi^{-1}(y))}\leq
Ils(\psi)(\bar{y})+\varepsilon.$

On the other hand, thanks to point (1), we can find $z\in B(\bar{y},r)$ so
that

$\displaystyle
Ils(\varphi+\psi)(\bar{y})\leq\frac{d(\varphi(\bar{y})+\psi(\bar{y}),\varphi(z)+\psi(z))}{d(\varphi(\bar{y})+\psi(\bar{y}),(1/2\pi)^{-1}(z))}+\varepsilon,$

and so

$\displaystyle Ils(\varphi+\psi)(\bar{y})$
$\displaystyle\leq\frac{d(\varphi(\bar{y}),\varphi(z))+d(\psi(\bar{y}),\psi(z))}{d(\varphi(\bar{y})+\psi(\bar{y}),(1/2\pi)^{-1}(z))}+\varepsilon$
$\displaystyle\leq\frac{d(\varphi(\bar{y}),\varphi(z))+d(\psi(\bar{y}),\psi(z))}{d((1/2\pi)^{-1}(\bar{y}),(1/2\pi)^{-1}(z))}+\varepsilon$
$\displaystyle=\frac{d(\varphi(\bar{y}),\varphi(z))+d(\psi(\bar{y}),\psi(z))}{2^{1/\lambda}d(\pi^{-1}(\bar{y}),\pi^{-1}(z))}+\varepsilon$
$\displaystyle\leq\frac{c}{2^{1/\lambda}}\frac{d(\varphi(\bar{y}),\varphi(z))}{d(\varphi(\bar{y}),\pi^{-1}(z))}+\frac{c}{2^{1/\lambda}}\frac{d(\psi(\bar{y}),\psi(z))}{d(\psi(\bar{y}),\pi^{-1}(z))}+\varepsilon$
$\displaystyle\leq\frac{c}{2^{1/\lambda}}(Ils(\varphi)(\bar{y})+\varepsilon)+\frac{c}{2^{1/\lambda}}(Ils(\psi)(\bar{y})+\varepsilon)+\varepsilon$

where in the first equality we used the triangle inequality, and in the second
one we used the simple fact $d((1/2\pi)^{-1}(\bar{y}),(1/2\pi)^{-1}(z))\leq
d(\varphi(\bar{y})+\psi(\bar{y}),(1/2\pi)^{-1}(z)),$ noting that
$\varphi(\bar{y})+\psi(\bar{y})\in(1/2\pi)^{-1}(\bar{y})$. In the first
equality we used Lemma 2 for $\alpha^{\lambda}=2$ and in the last two
inequalities we used (5) and (10).

Hence, by the arbitrariness of $\varepsilon,$ the proof is complete. ∎

##  3\. The intrinsic dilations vs quotient map

In this section the general setting is a group $G$ together with a closed
subgroup $H$ of $G$ in such a way that the quotient space $G/H:=\\{gH:g\in
G\\}$ naturally is a topological space for which the map $\pi:g\mapsto gH$ is
continuous, open, and surjective: it is a quotient map.

A section for the map $\pi:G\to G/H$ is just a map $\varphi:G/H\to G$ such
that $\varphi(gH)\in gH$, since we point out the trivial identity
$\pi^{-1}(gH)=gH$. To have the notion of intrinsic Lipschitz section we need
the group $G$ to be equipped with a distance which we assume left-invariant.
We refer to such a $G$ as a metric group. The reader can see [DDLD22, Section
6].

Following the Carnot structure, we present a result where the additional
hypothesis are compatibility conditions between the distance on $G$, the
quotient map and another suitable maps which corresponds to so-called
intrinsic dilations in Carnot groups.

######  Proposition .

Let $G=H_{1}\cdot H_{2}$ be a metric group and, for any fixed $\lambda>0,$
$\delta_{\lambda}:G\to G$ such that

  1. (1)

$\pi_{H_{1}}$ is $k$-Lipschitz at $1;$

  2. (2)

$d(\delta_{\lambda}g,\delta_{\lambda}q)=\lambda d(g,q)$ for any $g,q\in G;$

  3. (3)

$\delta_{\lambda}(\delta_{1/\lambda}(g))=g$ for any $g\in G;$

  4. (4)

$\pi_{H_{1}}(\delta_{\lambda}g)=\delta_{\lambda}\pi_{H_{1}}(g)$ for any $g\in
G.$

Then $\delta_{\lambda}\circ\varphi:G/H_{2}\to G$ is an intrinsically
$L\lambda$-Lipschitz section of $\delta_{1/\lambda}\circ\pi_{H_{1}}$, if
$\varphi:G/H_{2}\to G$ is so too for $\pi_{H_{1}}$ with Lipschitz constant
$L$.

###### Proof.

The fact that $\varphi$ is a section of $\delta_{1/\lambda}\pi_{H_{1}},$
follows from $(3)$ and

$\pi_{H_{1}}(\delta_{\lambda}\varphi(g))=\delta_{\lambda}\pi_{H_{1}}(\varphi(g))=\delta_{\lambda}g,\quad\forall
g\in G.$

Moreover, by $(4)$ we have that for every $a,b\in G/H_{2}$

$d(\delta_{\lambda}\varphi(a),\delta_{\lambda}\varphi(b))=\lambda
d(\varphi(a),\varphi(b))\leq L\lambda d(\varphi(a),\pi_{H_{1}}^{-1}(b)).$

∎

######  Corollary .

Let $G=N\rtimes H$ be a Carnot group and we consider $\delta_{\lambda}:G\to G$
the usual intrinsic dilations $($see, for instance, [SC16]$)$. Then
$\delta_{\lambda}\varphi$ is an intrinsically $L\lambda$-Lipschitz section of
$\delta_{1/\lambda}\pi_{N}$, if $\varphi$ is so too for $\pi_{N}$ with
Lipschitz constant $L$.

###### Proof.

The only non trivial fact is $(4)$ in Proposition 3 but it is true for the
homogeneity of the polynomials that appear in the group operation.

∎

######  Example  (Heisenberg groups).

The _$n$ -th Heisenberg group $\mathbb{H}^{n}$_ is the model case of Carnot
groups. It is the set $\mathbb{R}^{2n+1}$ with the group product $\cdot$ given
by

$(x_{1},\ldots,x_{2n},t)\cdot(x_{1}^{\prime},\ldots,x_{2n}^{\prime},t^{\prime})=\left(x_{1}+x_{1}^{\prime},\ldots,x_{2n}+x_{2n}^{\prime},t+t^{\prime}+\tfrac{1}{2}\sum_{i=1}^{n}x_{i}x_{n+i}^{\prime}-x_{i}^{\prime}x_{n+i}\right)$

for
$(x_{1},\ldots,x_{2n},t),(x_{1}^{\prime},\ldots,x_{2n}^{\prime},t^{\prime})\in\mathbb{R}^{2n+1}$.
We equip $\mathbb{H}^{n}$ with the left-invariant metric

(11) $d(p,q):=\|q^{-1}\cdot p\|,\quad p,q\in\mathbb{H}^{n},$

where $\|(x,t)\|:=\max\\{|x|,\sqrt{|t|}\\}$ and $|\cdot|$ denotes the usual
Euclidean norm on $\mathbb{R}^{2n}$ and on $\mathbb{R}$. Here, for any fixed
$\lambda>0,$ the intrinsic dilation
$\delta_{\lambda}:\mathbb{H}^{n}\to\mathbb{H}^{n}$ is an automorphism defined
as

$\delta_{\lambda}(x_{1},\ldots,x_{2n},t)=(\lambda x_{1},\ldots,\lambda
x_{2n},\lambda^{2}t).$

Hence, it is easy to see that $\delta_{\lambda}(\delta_{1/\lambda}(g))=g$ and
$\|\delta_{\lambda}g\|=\lambda\|g\|$ for any $g\in\mathbb{H}^{n}.$ Moreover,
if we consider the splitting $\mathbb{H}^{n}=H\ltimes N$ given by

$H=\\{(x_{1},\ldots,x_{k},0,\ldots,0)\colon(x_{1},\ldots,x_{k})\in\mathbb{R}^{k}\\}$

and

$N=\\{(0,\ldots,0,x_{k+1},\ldots,x_{2n},t)\colon(x_{k+1},\ldots,x_{2n},t)\in\mathbb{R}^{2n+1-k}\\},$

then

$\displaystyle\pi_{N}(\delta_{\lambda}(p))$ $\displaystyle=\pi_{N}(\lambda
x_{1},\ldots,\lambda x_{2n},\lambda^{2}t)$
$\displaystyle=\left(0,\dots,0,\lambda x_{k+1},\ldots,\lambda
x_{2n},\lambda^{2}t-\frac{1}{2}\lambda^{2}\sum_{i=1}^{k}x_{i}x_{n+i}\right).$

On the other hand,

$\displaystyle\delta_{\lambda}(\pi_{N}(p))$
$\displaystyle=\delta_{\lambda}\left(0,\dots,0,x_{k+1},\dots,x_{2n},t-\frac{1}{2}\sum_{i=1}^{k}x_{i}x_{n+i}\right)$
$\displaystyle=\left(0,\dots,0,\lambda x_{k+1},\ldots,\lambda
x_{2n},\lambda^{2}t-\frac{1}{2}\lambda^{2}\sum_{i=1}^{k}x_{i}x_{n+i}\right),$

and so we get

$\pi_{N}(\delta_{\lambda}(p))=\delta_{\lambda}(\pi_{N}(p)).$

Unfortunately, the non-linearity of $\pi$ can already be seen in this case.

##  4\. The sum of intrinsic Lipschitz sections is so too in Carnot groups of
step 2

In this section we consider Carnot groups of step 2 as follow.

######  Definition .

We say that $\mathbb{G}:=(\mathbb{R}^{m+n},\cdot,\delta_{\lambda})$ is a
Carnot group of step 2 if there are $n$ linearly independent, skew-symmetric
$m\times m$ real matrices $\mathcal{B}^{(1)},\dots,\mathcal{B}^{(n)}$ such
that for all $p=(p^{1},p^{2})$ and
$q=(q^{1},q^{2})\in\mathbb{R}^{m}\times\mathbb{R}^{n}$ and for all $\lambda>0$

(12) $p\cdot
q=(p^{1}+q^{1},p^{2}+q^{2}+\frac{1}{2}\langle\mathcal{B}p^{1},q^{1}\rangle)$

where
$\langle\mathcal{B}p^{1},q^{1}\rangle:=(\langle\mathcal{B}^{(1)}p^{1},q^{1}\rangle,\dots,\langle\mathcal{B}^{(n)}p^{1},q^{1}\rangle)$
and $\langle\cdot,\cdot\rangle$ is the inner product in $\mathbb{R}^{m}$ and

$\delta_{\lambda}p:=(\lambda p^{1},\lambda^{2}p^{2}).$

When $H$ is one dimensional, without loss of generality (see [BLU07, Chapter
3.4]), we consider the following splitting

$H:=\\{(x_{1},0\dots,0)\,:\,x_{1}\in\mathbb{R}\\},\qquad
N:=\\{(0,x_{2},\dots,x_{m+n})\,:\,x_{i}\in\mathbb{R}\\}.$

Unfortunately, in the context of Carnot groups of step two $\pi_{N}$ is non
linear (see the model case as in Example 3). However, with the additional
hypothesis which is a sort of a compatibility condition between two Lipschitz
sections it is possible to deduce the following proposition in Carnot groups
of step 2.

######  Proposition .

Let $G=N\rtimes H$ be a Carnot group of step 2 with $H$ 1-dimensional and
$\varphi,\psi:G/H\to G$ be intrinsically Lipschitz sections of $\pi_{G/H}$
such that if we put

$\displaystyle\mathcal{A}_{1,\ell}:=\langle\mathcal{B}^{(\ell)}p^{1},q^{1}\rangle-\langle\mathcal{B}^{(\ell)}(0,p_{2}+q_{2},\dots,p_{m}+q_{m}),(p_{1}+q_{1},0\dots,0)\rangle,$
$\displaystyle\mathcal{A}_{2,\ell}:=\langle\mathcal{B}^{(\ell)}(0,p_{2},\dots,p_{m}),(0,q_{2},\dots,q_{m})\rangle+\langle\mathcal{B}^{(\ell)}(0,p_{2},\dots,p_{m}),(p_{1},0)\rangle$
$\displaystyle\quad+\langle\mathcal{B}^{(\ell)}(0,q_{2},\dots,q_{m}),(q_{1},0)\rangle,$

$\forall\ell=1,\dots,n$ where $p_{i}=\varphi_{i}(a)$ and $q_{i}=\psi_{i}(b)$
for any $i=1,\dots,m+n,$ we want that

(13)
$\displaystyle\mathcal{A}_{1,\ell}=\mathcal{A}_{2,\ell},\quad\forall\ell=1,\dots,n.$

Then the sum of $\varphi$ and $\psi$ is intrinsic Lipschitz.

###### Proof.

We want to apply Corollary 4.6 in [DD22a]. The only hypothesis that we need is
the linearity of $\pi$ which we don’t have in general. However, it is
sufficient to show that

$\pi(\varphi(y_{1})+\psi(y_{2}))=\pi(\varphi(y_{1}))+\pi(\psi(y_{2})),\quad\forall
y_{1},y_{2}\in G/H,$

where $\varphi$ and $\psi$ satisfy (13). For simplicity, we identify $G/H$
with $N,$ and $\pi_{G/H}$ with $\pi_{N}$ and we just write $\pi.$ We prove

(14) $\displaystyle\pi(p\cdot q)=\pi(p)\cdot\pi(q),$

for $p=(p^{1},p^{2})=(p_{1},\dots,p_{m+n})$ and
$q=(q^{1},q^{2})=(q_{1},\dots,q_{m+n})$ when
$\mathcal{A}_{1,\ell}=\mathcal{A}_{2,\ell}.$ By an easy computation, we get

$\displaystyle\pi(p)=\left(0,p_{2},\dots,p_{m},[\pi(p)]_{m+1},\dots,[\pi(p)]_{m+n}\right)$

where

m+ℓ
$\displaystyle=p_{m+\ell}+\frac{1}{2}\langle\mathcal{B}^{(\ell)}(0,p_{2},\dots,p_{m}),(p_{1},0\dots,0)\rangle,$

for $\ell=1,\dots,n.$ Consequently,

$\pi(p)\cdot\pi(q)=(0,p_{2}+q_{2},\dots,p_{m}+q_{m},[\pi(p)\cdot\pi(q)]_{m+1},\dots,[\pi(p)\cdot\pi(q)]_{m+n}),$

where

m+ℓ
$\displaystyle=\frac{1}{2}\langle\mathcal{B}^{(\ell)}(0,p_{2},\dots,p_{m}),(0,q_{2},\dots,q_{m})\rangle+[\pi(p)]_{m+\ell}+[\pi(q)]_{m+\ell}.$

On the other hand,

$\displaystyle\pi(p\cdot q)=\left(0,p_{2}+q_{2},\dots,p_{m}+q_{m},[\pi(p\cdot
q)]_{m+1},\dots,[\pi(p\cdot q)]_{m+n}\right),$

where

m+ℓ
$\displaystyle=p_{m+\ell}+q_{m+\ell}+\frac{1}{2}\langle\mathcal{B}^{(\ell)}p^{1},q^{1}\rangle$
$\displaystyle\quad-\frac{1}{2}\langle\mathcal{B}^{(\ell)}(0,p_{2}+q_{2},\dots,p_{m}+q_{m}),(p_{1}+q_{1},0\dots,0)\rangle,$

for $\ell=1,\dots,n.$ Hence we get the thesis using the hypothesis
$\mathcal{A}_{1,\ell}=\mathcal{A}_{2,\ell}.$ ∎

######  Example  (Heisenberg groups).

Let $G=\mathbb{H}^{n}=N\rtimes H$ as in Example 3. By an easy computation, we
get an explicit formula of (13) (here $\ell=1$), i.e., either

$\varphi_{1}\equiv 0$

or

$\psi_{n+1}\equiv 0.$

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# The Image of the Gassner Representation of the Pure Braid Subgroup has Pair-
wise Free Generators

G. Makenzie Cosgrove

###### Abstract

While much is known about the faithfulness of the Burau representation, the
problem remains open for the Gassner representation for every $B_{n}$ with
$n\geq 4$. We first find the definition of the Colored-Burau representation in
[3] and we show that this is equivalent, when restricted to the pure braid
subgroup, to the Gassner representation. The methods of Abdulrahim [2] and
Knudson [9] require analysis within the lower central series of a free
subgroup of the pure braid group. However, Lipschutz’s work [10] gives a
method for analyzing the Gassner representation and the Colored-Burau
structure reduces this analysis to basic linear algebra.

##  1 Introduction

The faithfulness of the Burau representation has been well-studied. It is
known that the representation is faithful for $n=3$ and unfaithful for $n\geq
5$ [6]. The case $n=4$ is still open, but several attempts have been made to
answer the question. Beridze and Traczyk showed that the faithfulness depends
on the restriction to a free subgroup generated by two elements and that the
cubes of their images generate a free group [5]. Their proof was topological
in nature, however and, more recently, Beridze, Bigelow, and Traczyk gave an
alternate proof using the Ping-Pong Lemma that the same holds modulo $p$ [4].

When it comes to the Gassner representation, it is not known for any $n\geq 4$
whether the representation is faithful. In fact, the faithfulness of the
Gassner representation is dependent only upon its faithfulness when restricted
to free subgroups of the pure braid group. For example, Kundson located the
kernel within the lower central series of a certain free subgroup of the pure
braid group [9]. The Gassner representation is originally constructed as a
Magnus representation using Fox derivatives and the computations can be seen
in [2]. These computations are strenuous and the resulting matrices require
additional rewriting to make use of them.

In this paper, we prefer to use the Colored-Burau representation introduced in
[3], which appears to be computationally favorable to Fox derivatives. Rather
than obtaining a representation via homology computations as in the ordinary
Burau representation, the Colored-Burau representation is simply an imposition
of algebraic structure. In Section 2, we introduce the Colored-Burau
representation, compute the images of the pure braid generators found in [7],
and show these align with the Gassner matrices. In Section 3, we use the
preceding computations to provide a short linear-algebraic proof that the
faithfulness of the Gassner representation depends only upon its restriction
to a free subgroup. Finally, in Section 4, we show, following the idea in
[10], that the Gassner matrices for the generators of the free subgroup are
pairwise free.

##  2 The Colored-Burau Representation

Here we introduce the Colored-Burau representation from [3]. This
representation is a generalization of the famous Burau representation in that
it is defined for the entire braid group. However, we use this represenation
to perform explicit computations on the pure braid subgroup. These computaions
will show that the restriction of the Colored-Burau representation to the pure
braid subgroup is precisely the Gassner representation. In Section 3 and
Section 4 we will leverage this to write short linear-algebraic proofs about
the Gassner representation.

###  2.1 A Convention

Fix an integer $n\geq 2$. For $1\leq i<j\leq n$ and a scalar or variable $t$
we denote by $t_{i\to j}$ the column vector (of implicit size $n$) with $t$ in
positions $i$ up to and including $j-1$ and $0$ elsewhere. On the other hand,
given a length-$n$ column vector ${\bf v}$, we write $c_{s}({\bf v})$ for the
$n\times n$ matrix whose $s$-column is ${\bf v}$ (and all of whose other
entries vanish). So, we could write

$c_{2}([1,2,3]^{T})=\begin{bmatrix}0&1&0\\\ 0&2&0\\\ 0&3&0\end{bmatrix}.$

This example also illustrate the convention that we number rows and columns
starting at $1$ rather than $0$.

###  2.2 The Colored-Burau Map

Fix an integer $n\geq 2$. Denote by $B_{n}$ the Braid group on $n$ strands.
Let $T=(t_{1},\ldots t_{n})$ be a tuple of formal variables. Denote by
$L_{n}:=\mathbb{Z}[t_{1}^{\pm 1},\ldots,t_{n}^{\pm 1}]$, the integral Laurent
polynomial ring in $n$ variables. The Colored-Burau map of $B_{n}$ is

$cb:B_{n}\to GL_{n}(L_{n})$

from the braid group to the set of invertible $n\times n$ matrices with
entries that are Laurent polynomials in the variables $t_{1},\ldots,t_{n}$.
The Colored-Burau map sends the generators $\sigma_{i}$ to the following:

$\sigma_{1}\mapsto\begin{bmatrix}-t_{1}&1&&\\\ &1&&\\\ &&\ddots&\\\
&&&1\end{bmatrix}$

and, for $i\neq 1$,

$\sigma_{i}\mapsto\begin{bmatrix}1&&&\\\ &\ddots&\\\ &&t_{i}&-t_{i}&1&&\\\
&&&&&\ddots&\\\ &&&&&&1\end{bmatrix},$

which is the identity except for the $i^{\text{th}}$ row where it has in
succession $t_{i},-t_{i},1$ with $-t_{i}$ on the diagonal. Using the
convention from Section 2.1, we can write:

$\sigma_{i}\mapsto I_{n}+c_{i-1}((t_{i})_{i-1\to i})+c_{i}((1-t_{i})_{i\to
i+1})+c_{i+1}(1_{i+1\to i+2})$

(when $i=1$, the $c_{i-1}$ term vanishes because we index columns from $1$,
not $0$).

Note that applying the quotient $t_{i}\sim t$ that identifies all $t_{i}$ with
a single indeterminant will recover the Burau representation of $B_{n}$.

###  2.3 The Colored-Burau Group

Let $n\geq 2$ be an integer and $T=(t_{1},\ldots,t_{n})$ be a tuple of formal
variables. The Colored-Burau group is the outer semi-direct product

$G=GL_{n}(L_{n})\rtimes S_{n}$

of the symmetric group on $n$ symbols acting on $L_{n}$. The action is by
permutation of the variables. In particular, there is a homomorphism
$\pi:B_{n}\to S_{n}$ given by $s_{i}:=\pi(\sigma_{i})=(i,i+1)$, so that a
braid $\beta\in B_{n}$ may act on an element of $GL_{n}(L_{n})$ as

${}^{\beta}M:={}^{\pi(\beta)}M$

Accordingly, the multiplication in $G$ is

$(M,s)\star(M^{\prime},s^{\prime}):=(M{}^{s}M^{\prime},ss^{\prime})$

The Colored-Burau representation, therefore, is the map

$CB:B_{n}\to G$

given by

$\sigma_{i}\mapsto(cb(\sigma_{i}),s_{i})$

###  2.4 Matrices for Pure Braid Generators

Among other sources, Birman shows in [7] the Pure Braid subgroup is generated
by

$A_{i,j}=\sigma_{i}\sigma_{i+1}\cdots\sigma_{j-2}\sigma_{j-1}^{2}\sigma_{j-1}^{-1}\cdots\sigma_{i}^{-1}$

together with relations that we do not list here. Moreover, Birman shows there
is a split exact sequence

$1\to F_{n-1}\to P_{n}\to P_{n-1}\to 1$

such that $F_{n-1}\cong\langle A_{1,j}\rangle_{j=2}^{n}$. As such, we describe
explicitly the matrices $cb(A_{i,j})$ and their inverses. The reader can check
that the formula derived here agrees precisely with Abdulrahim’s result (see
Proposition 3.3.3 in [2]), however, our derivation is via computations in the
Colored-Burau group.

######  Lemma 2.1.

Let $n\geq 2$ and

$A_{i,j},\ 1\leq i<j\leq n$

be a pure braid generator for some $i,j$. The matrix

$cb(A_{i,j})\in GL_{n}(L_{n})$

is equal to

$\displaystyle I_{n}$ $\displaystyle+\quad c_{i-1}((-t_{i}t_{j}+t_{i})_{i\to
j})$ $\displaystyle+\quad c_{i}((t_{j}-1)_{i\to j})$ $\displaystyle+\quad
c_{j-1}((t_{i}t_{j}-t_{j})_{i\to j})$ $\displaystyle+\quad
c_{j}((-t_{i}+1)_{i\to j}),$

with the following caveat: When $i=1$ the term $c_{i-1}$ is absent.

###### Proof.

When $n=2$, we need only check $A_{1,2}$, which we obtain as the result of a
single group multiplication:

$\displaystyle CB(A_{1,2})$ $\displaystyle=CB(\sigma_{1}^{2})$
$\displaystyle=\left(\begin{bmatrix}-t_{1}&1\\\
0&1\end{bmatrix},(1,2)\right)\star\left(\begin{bmatrix}-t_{1}&1\\\
0&1\end{bmatrix},(1,2)\right)$ $\displaystyle=\left(\begin{bmatrix}-t_{1}&1\\\
0&1\end{bmatrix}\cdot\begin{bmatrix}-t_{2}&1\\\
0&1\end{bmatrix},(1,2)\cdot(1,2)\right)$
$\displaystyle=\left(\begin{bmatrix}t_{1}t_{2}&1-t_{1}\\\
0&1\end{bmatrix},e\right)$

and the matrix component is equal

$\underbrace{\begin{bmatrix}1&0\\\
0&1\end{bmatrix}}_{I_{2}}+\underbrace{\begin{bmatrix}t_{2}-1&0\\\
0&0\end{bmatrix}}_{c_{1}((t_{2}-1)_{1\to
2})}+\underbrace{\begin{bmatrix}t_{1}t_{2}-t_{2}&0\\\
0&0\end{bmatrix}}_{c_{1}((t_{1}t_{2}-t_{2})_{1\to
2})}+\underbrace{\begin{bmatrix}0&1-t_{1}\\\
0&0\end{bmatrix}}_{c_{2}((1-t_{1})_{1\to
2})}=\begin{bmatrix}t_{1}t_{2}&1-t_{1}\\\ 0&1\end{bmatrix}$

as claimed.

Now suppose the result holds for some $B_{k}$ with $k>2$ and for all $1\leq
i<j\leq k$. Set $n=k+1$. By our inductive hypothesis, it suffices to prove the
formula for $j=n$. Thus, we further induct on $\ell=n-i$. When $\ell=1$, we
obtain the result immediately by one group multiplication:

$\displaystyle CB(A_{n-1,n})$
$\displaystyle=CB(\sigma_{n-1}^{2})=CB(\sigma_{n-1})\star CB(\sigma_{n-1})$
$\displaystyle=\left(\begin{bmatrix}I_{n-3}&\\\ &1&0&0\\\
&t_{n-1}&-t_{n-1}&1\\\
&0&0&1\end{bmatrix},(n-1,n)\right)\star\left(\begin{bmatrix}I_{n-3}&\\\
&1&0&0\\\ &t_{n-1}&-t_{n-1}&1\\\ &0&0&1\end{bmatrix},(n-1,n)\right)$
$\displaystyle=\left(\begin{bmatrix}I_{n-3}&\\\ &1&0&0\\\
&t_{n-1}&-t_{n-1}&1\\\ &0&0&1\end{bmatrix}\cdot\begin{bmatrix}I_{n-3}&\\\
&1&0&0\\\ &t_{n}&-t_{n}&1\\\ &0&0&1\end{bmatrix},(n-1,n)\cdot(n-1,n)\right)$
$\displaystyle=\left(\begin{bmatrix}I_{n-3}&\\\ &1&0&0\\\
&t_{n-1}-t_{n-1}t_{n}&t_{n-1}t_{n}&1-t_{n-1}\\\ &0&0&1\end{bmatrix},e\right)$

and the matrix component is equal

$\displaystyle\underbrace{\begin{bmatrix}I_{n-3}&0&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{bmatrix}}_{I_{n}}$
$\displaystyle+\underbrace{\begin{bmatrix}I_{n-3}&0&0&0\\\ 0&1&0&0\\\
0&-t_{n-1}t_{n}+t_{n-1}&1&0\\\
0&0&0&1\end{bmatrix}}_{c_{n-2}((-t_{n-1}t_{n}+t_{n-1})_{n-1\to n})}$
$\displaystyle+\underbrace{\begin{bmatrix}I_{n-3}&0&0&0\\\ 0&1&0&0\\\
0&0&t_{n}-1&0\\\ 0&0&0&1\end{bmatrix}}_{c_{n-1}((t_{n}-1)_{n-1\to n})}$
$\displaystyle+\underbrace{\begin{bmatrix}I_{n-3}&0&0&0\\\ 0&1&0&0\\\
0&0&t_{n}t_{n-1}-t_{n}&0\\\
0&0&0&1\end{bmatrix}}_{c_{n-1}((t_{n}t_{n-1}-t_{n})_{n-1\to n})}$
$\displaystyle+\underbrace{\begin{bmatrix}I_{n-3}&0&0&0\\\ 0&1&0&0\\\
0&0&0&-t_{n-1}+1\\\ 0&0&0&1\end{bmatrix}}_{c_{n}((-t_{n-1}+1)_{n-1\to n})}$
$\displaystyle=\begin{bmatrix}I_{n-3}&\\\ &1&0&0\\\
&t_{n-1}-t_{n-1}t_{n}&t_{n-1}t_{n}&1-t_{n-1}\\\ &0&0&1\end{bmatrix}$

as claimed.

Now suppose our formula holds for some $2\leq\ell\leq n-2$ (we treat the case
$i=1$ separately). Since

$A_{i-1,n}=\sigma_{i-1}A_{i,n}\sigma_{i-1}^{-1},$

we perform two group multiplications. Firstly, since $A_{i,n}$ is a pure
braid, there is no permuting of the variables in $cb(\sigma_{i-1}^{-1})$.
Thus, we multiply

$\displaystyle cb(A_{i,n})cb(\sigma_{i-1}^{-1})$
$\displaystyle=\begin{bmatrix}I_{i-2}&\\\ &1&0&0&0&0\\\
&t_{i}t_{n}+t_{i}-1&t_{n}&0&t_{i}t_{n}-t_{n}&1-t_{i}\\\
&t_{i}t_{n}+t_{i}-1&t_{n}-1&I_{\ell-2}&t_{i}t_{n}-t_{n}&1-t_{i}\\\
&\vdots&\vdots&&\vdots&\vdots\\\
&t_{i}t_{n}+t_{i}-1&t_{n}-1&0&1+t_{i}t_{n}-t_{n}&1-t_{i}\\\
&0&0&0&0&1\end{bmatrix}\cdot\begin{bmatrix}I_{i-2}&&&\\\ &1&0&0\\\
&1&\frac{-1}{t_{i}}&\frac{1}{t_{i}}\\\ &0&0&1\\\ &&&&I_{\ell-1}\end{bmatrix}$

The product is non-unital in precisely the columns $i-1,i,i+1,$ and $n-1$ with
the entries being

$\begin{bmatrix}I_{i-3}&\\\ &1&0&0&0&0&0\\\
&1&\frac{-1}{t_{i}}&\frac{1}{t_{i}}&0&0&0\\\ \\\
&t_{i}(t_{n}-1)&\frac{1-t_{i}+t_{i}t_{n}}{t_{i}}&\frac{t_{i}-1}{t_{i}}&I_{\ell-2}&t_{i}t_{n}-t_{n}&1-t_{i}\\\
&\vdots&\vdots&\vdots&&\vdots&\vdots\\\
&t_{i}(t_{n}-1)&\frac{1-t_{i}+t_{i}t_{n}}{t_{i}}&\frac{t_{i}-1}{t_{i}}&0&1+t_{i}t_{n}-t_{n}&1-t_{i}\\\
&0&0&0&0&0&1\end{bmatrix}.$

The second group multiplication has the effect of permuting every occurrence
of $t_{i}$ in the above to $t_{i-1}$ and subsequently multiplying by
$cb(\sigma_{i-1})$, hence we obtain

${\small\begin{bmatrix}I_{i-2}\\\ &1&0&0\\\ &t_{i-1}&-t_{i-1}&1\\\ &0&0&1\\\
&&&&I_{n-i-1}\end{bmatrix}}\cdot\begin{bmatrix}I_{i-3}&\\\ &1&0&0&0&0&0\\\
&1&\frac{-1}{t_{i-1}}&\frac{1}{t_{i-1}}&0&0&0\\\ \\\
&t_{i-1}(t_{n}-1)&\frac{1-t_{i-1}+t_{i-1}t_{n}}{t_{i-1}}&\frac{t_{i-1}-1}{t_{i-1}}&I_{\ell-2}&t_{i-1}t_{n}-t_{n}&1-t_{i-1}\\\
&\vdots&\vdots&\vdots&&\vdots&\vdots\\\
&t_{i-1}(t_{n}-1)&\frac{1-t_{i-1}+t_{i-1}t_{n}}{t_{i-1}}&\frac{t_{i-1}-1}{t_{i-1}}&0&1+t_{i-1}t_{n}-t_{n}&1-t_{i-1}\\\
&0&0&0&0&0&1\end{bmatrix},$

which is non-unital in precisely columns $i-1,i,n-1,$ and $n$ with the entries
being

$\begin{bmatrix}I_{i-2}&\\\ &1&0&0&0&0\\\
&t_{i-1}t_{n}+t_{i-1}-1&t_{n}&0&t_{i-1}t_{n}-t_{n}&1-t_{i-1}\\\
&t_{i-1}t_{n}+t_{i-1}-1&t_{n}-1&I_{\ell-2}&t_{i-1}t_{n}-t_{n}&1-t_{i-1}\\\
&\vdots&\vdots&&\vdots&\vdots\\\
&t_{i-1}t_{n}+t_{i-1}-1&t_{n}-1&0&1+t_{i-1}t_{n}-t_{n}&1-t_{i-1}\\\
&0&0&0&0&1\end{bmatrix}$

and this agrees with the proposed formula, as the reader can verify.

Finally, we handle the case $i=1$. By induction, we have

$cb(A_{2,n})=\begin{bmatrix}1&0&0&0&0\\\
t_{2}t_{n}+t_{2}-1&t_{n}&0&t_{2}t_{n}-t_{n}&1-t_{2}\\\
t_{2}t_{n}+t_{2}-1&t_{n}-1&I_{n-4}&t_{2}t_{n}-t_{n}&1-t_{2}\\\
\vdots&\vdots&&\vdots&\vdots\\\
t_{2}t_{n}+t_{2}-1&t_{n}-1&0&1+t_{2}t_{n}-t_{n}&1-t_{2}\\\
0&0&0&0&1\end{bmatrix}$

and, since $A_{1,n}=\sigma_{1}A_{2,n}\sigma_{1}^{-1}$, we obtain

$cb(A_{1,n})=\begin{bmatrix}t_{n}&0&t_{1}t_{n}-t_{n}&1-t_{1}\\\
t_{n}-1&I_{n-3}&t_{1}t_{n}-t_{n}&1-t_{1}\\\ \vdots&&\vdots&\vdots\\\
t_{n}-1&0&1+t_{1}t_{n}-t_{n}&1-t_{1}\\\ 0&0&0&1\end{bmatrix},$

as the reader can verify. Finally, we note that this agrees with the proposed
formula. ∎

######  Corollary 2.2.

Using the same notation as in Lemma 2.1, the matrix

$cb(A_{ij})^{-1}\in GL_{n}(L_{n})$

is equal to

$I_{n}+c_{i-1}\left(\left(\frac{t_{j}-1}{t_{j}}\right)_{i\to
j}\right)+c_{i}\left(\left(\frac{1-t_{j}}{t_{i}t_{j}}\right)_{i\to
j}\right)+c_{j-1}\left(\left(\frac{1-t_{i}}{t_{i}}\right)_{i\to
j}\right)+c_{j}\left(\left(\frac{t_{i}-1}{t_{i}t_{j}}\right)_{i\to j}\right),$

with the following caveat: When $i=1$ the first term $c_{i-1}$ is absent.

######  Corollary 2.3.

Using the same notation as in Lemma 2.1, taking determinants gives

$\det(cb(A_{i,j}))=t_{i}t_{j}.$

##  3 The Faithfulness of the Gassner Representation

The main result of this section is that the faithfulness of the Gassner
representation is determined completely by its faithfulness on the free
subgroup $F_{n-1}\cong\langle A_{1,j}\rangle_{j=2}^{n}$. The following result
due to Long has provided for such reductions in many other contexts. In fact,
[8] uses a similar argument on the level of Lie Algebras.

######  Lemma 3.1 (Thereom 2.2 in [11]).

Let $A$ be a braid group and $\rho:A\to M$ be a representation. Furthermore,
suppose $\rho$ is faithful on $Z(A)$. Then $\rho$ is faithful on $A$ if and
only if it is faithful on $B\lhd A$ for some non-trivial, non-central, normal
subgroup $B$.

In [9], Knudson uses the restriction to a free subgroup to show the Gassner
representation is faithful if and only if

$\Gamma^{i}F_{n-1}/\Gamma^{i+1}F_{n-1}\to\Gamma^{i}g_{n}(F_{n-1})/\Gamma^{i+1}g_{n}(F_{n-1})$

is injective for all $i$ (here $g_{n}$ is the restriction of Gassner to
$F_{n-1}$). However, the injectivity fails for $i=5$. Nevertheless, he shows
the kernel is contained in $[\Gamma^{3}F_{n-1},\Gamma^{2}F_{n-1}]$.

While the result given in this section is not novel, the proof requires only
the computation of a determinant, which is simple for Colored-Burau matrices.
This allows us to ultimately circumvent the computations in the lower central
series of $F_{n-1}$ by performing linear algebraic operations on the matrices
for the pure braid generators directly. We require only the following result
about centers of braid groups.

######  Lemma 3.2.

For $n\geq 3$, $Z(B_{n})=Z(P_{n})$.

###### Proof.

Certainly any braid $\beta\in Z(B_{n})$ must commute with all pure braids,
hence $Z(B_{n})\subseteq Z(P_{n})$. The reverse containment is a consequence
of Corollary 1.8.4 in [7], where it is shown that $Z(B_{n})$ is generated by
the pure braid $\sigma=A_{1,2}(A_{1,3}A_{2,3})\cdots(A_{1,n}\cdots
A_{n-1,n})$. ∎

######  Theorem 3.3.

The Gassner representation on $P_{n}$ is faithful if and only if it is so on
$F_{n-1}\leq P_{n}$.

###### Proof.

Certainly the Gassner representation is faithful on $F_{n-1}$ if it is so on
$P_{n}$. Thus, it suffices to show the converse, which we do using Lemma 3.1.
Since $F_{n-1}$ is normal, being that it is the kernel of a homomorphism in
the split exact sequence

$1\to F_{n-1}\to P_{n}\twoheadrightarrow P_{n-1}\to 1,$

and because $F_{n-1}$ is non-trivial and non-central in $P_{n}$, we need only
prove that the Gassner representation is faithful on $Z(P_{n})$.

Since any braid in $Z(P_{n})$ must have an image that is central in
$GL_{n}(L_{n})$, the image must be scalar. Using Lemma 3.2, Corollary 2.3, and
the fact that $cb$ and $\det$ are homomorphisms, we obtain

$\det(cb(\sigma))=\prod_{j=2}^{n}t_{j}^{j-1}\prod_{i=1}^{j}t_{i}=\prod_{i=1}^{n}t_{i}^{n-1},$

so that $cb(\sigma)$ has infinite order. Since $Z(B_{n})=Z(P_{n})$ and
$\sigma$ generates the former, $cb$ must be faithful on the latter. By Lemma
2.1, $cb$ is the same as the Gassner representation and this completes the
proof. ∎

##  4 The Main Result

Now we show that the images of the Pure Braid generators $cb(A_{1j})$ are
pairwise free. The proof uses the Ping-Pong Lemma, which we record here for
reference. A proof of this can be found, for example, in [1].

######  Lemma 4.1 (Ping-Pong).

Let $G$ be a group acting on a set $X$ and let $H_{1},H_{2}\leq G$ be
subgroups. Let $|H_{i}|>2$ for some $i$. If $X_{1},X_{2}\subseteq X$ are
disjoint and nonempty and $H_{1}(X_{2})\subseteq X_{1}$ and
$H_{2}(X_{1})\subseteq X_{2}$, then $G\cong H_{1}\ast H_{2}$.

With the Ping-Pong Lemma established, we shall now use it to prove that the
matrices $\\{cb(A_{1,j})\\}_{j=2}^{n}$ are pairwise free. Firstly, we observe
that, for $j\neq j^{\prime}$, any relation held by
$cb(A_{1,j}),cb(A_{1,j^{\prime}})$ must hold for all tuples
$T=(t_{1},\ldots,t_{n})$. Thus, if we can find a single tuple
$T=(\tau_{1},\ldots,\tau_{n})$ for which the resulting matrices are free, then
$cb(A_{1,j})$ and $cb(A_{1,j^{\prime}})$ must be free as well. Secondly, the
freeness is independent of basis: if

$\prod_{k=1}^{N}cb(A_{1,i_{k}})^{\alpha_{k}}=I_{n}$

where $i_{k}\in\\{j,j^{\prime}\\}$ for all $k$, then the change of basis
matrix $P$ gives

$\prod_{k=1}^{N}Pcb(A_{1,i})^{\alpha_{k}}P^{-1}=P\left(\prod_{k=1}^{N}cb(A_{1,i_{k}})^{\alpha_{k}}\right)P^{-1}=PP^{-1}=I_{n}.$

Thus, we shall evaluate $cb(A_{1,j}),cb(A_{1,j^{\prime}})$ and let the
resulting matrices in $GL_{n}(\mathbb{Z})$ act on $\mathbb{R}^{n}$ after a
change of basis that reduces the action on a fixed subspace (depending on
$j,j^{\prime}$) to that of the following free matrices.

######  Lemma 4.2.

Let $G=SL_{2}(\mathbb{Z})$ and

$A=\begin{bmatrix}1&2\\\ 0&1\end{bmatrix}\qquad B=\begin{bmatrix}1&0\\\
2&1\end{bmatrix}$

Then $\langle A,B\rangle\cong F_{2}$ is free.

###### Proof.

We use the Ping-Pong Lemma. Let

$H_{1}=\\{A^{n}:n\in\mathbb{Z}\\}=\left\\{\begin{bmatrix}1&2n\\\
0&1\end{bmatrix}:n\in\mathbb{Z}\right\\}\qquad\text{and}\qquad
H_{2}=\\{B^{n}:n\in\mathbb{Z}\\}=\left\\{\begin{bmatrix}1&0\\\
2n&1\end{bmatrix}:n\in\mathbb{Z}\right\\}.$

Consider the standard action of $G$ on $X=\mathbb{R}^{2}$ by linear
transformations. Set

$X_{1}=\left\\{\begin{bmatrix}x\\\
y\end{bmatrix}\in\mathbb{R}^{2}:|x|>|y|\right\\}\qquad\text{and}\qquad
X_{2}=\left\\{\begin{bmatrix}x\\\
y\end{bmatrix}\in\mathbb{R}^{2}:|x|<|y|\right\\}$

Now, if ${\bf v}=\begin{bmatrix}x\\\ y\end{bmatrix}\in X_{2}$, then $A^{n}{\bf
v}=\begin{bmatrix}x+2ny\\\ y\end{bmatrix}\in X_{1}$ since

$|x+2ny|\geq|2ny|-|x|\geq 2|y|-|x|>2|y|-|y|=|y|.$

Likewise, if ${\bf v}=\begin{bmatrix}x\\\ y\end{bmatrix}\in X_{1}$, then
$B^{n}{\bf v}=\begin{bmatrix}x\\\ 2nx+y\end{bmatrix}\in X_{2}$ since

$|2nx+y|\geq|2nx|-|y|\geq 2|x|-|y|>2|x|-|x|=|x|.$

∎

These are the same matrices that appear in blocks in the representation given
in [10] after evaluating at $t_{i}=-1$ for all $i$. In order to describe the
change of basis explicitly, we evaluate the matrices $cb(A_{1,j})$ and
$cb(A_{1,j^{\prime}})$, and compute their eigenspaces. The new basis is then
given in terms of the eigenvectors.

######  Lemma 4.3.

Let $n\geq 4$. Denote by $\\{e_{k}\\}_{k=1}^{n}$ the standard basis for
$\mathbb{R}^{n}$. Define $\tau$ to be the evaluation homomorphism given by
taking $t_{i}=-1$ for all $1\leq i\leq n$. Then the only eigenvalue for
$\tau(cb(A_{1,j}))$ is $\lambda=1$ and the eigenvectors are

  * •

for $j=2$: $e_{k}$ for $k\neq 2$.

  * •

for $j\geq 3$: $e_{k}$ for $k\notin\\{1,j-1,j\\}$, $v_{j}=e_{1}+e_{j}$, and
$w_{j}=\sum\limits_{i=1}^{j-1}e_{i}$.

###### Proof.

This is an immediate consequence of Lemma 2.1. ∎

######  Theorem 4.4.

Let $n\geq 4$. For all $2\leq j<j^{\prime}\leq n$, the group $\langle
cb(A_{1,j}),cb(A_{1,j^{\prime}})\rangle\cong F_{2}$ is free.

###### Proof.

For ease of notation, let $M_{j}:=\tau(cb(A_{1,j}))$, where $\tau$ is the
evaluation homomorphism $t_{i}=-1$ for all $1\leq i\leq n$ and let
$\\{e_{k}\\}_{k=1}^{n}$ be the standard basis for $\mathbb{R}^{n}$. We begin
with the case $j=2$. If $j^{\prime}=3$, consider $A_{1,2}$ and $A_{1,3}$ as
elements of $B_{3}$. Here, the Burau representation is known to be faithful
[6], so the Colored-Burau representation must be likewise. Consequently,
evaluating gives that $M_{2}$ and $M_{3}$ are free. Now assume that
$j^{\prime}>3$. By Lemma 4.3, we can make the following change of basis by
replacing $e_{j^{\prime}-1}$ with $v_{j^{\prime}-1}=e_{1}+e_{j^{\prime}-1}$.
We thereby obtain the following blocks in the upper left $2\times 2$ minors

$X=\begin{bmatrix}1&2\\\ 0&1\end{bmatrix}\qquad Y=\begin{bmatrix}1&0\\\
-2&1\end{bmatrix},$

for $M_{2}$ and $M_{j^{\prime}}$, respectively. The matrices $X$ and $Y^{-1}$
are those from Lemma 4.2. Thus, $M_{2}$ and $M_{j^{\prime}}$ cannot have a
relation as this would induce a relation on $X$ and $Y^{-1}$.

If $j>2$, we may simultaneously replace $e_{j-1}$ with $w_{j}$ and $e_{j}$
with $w_{j^{\prime}}$ to obtain

$\begin{bmatrix}*&0&*\\\ *&X&*\\\ *&0&*\\\
\end{bmatrix}\qquad\text{and}\qquad\begin{bmatrix}*&0&*\\\ *&Y&*\\\ *&0&*\\\
\end{bmatrix}$

for $M_{j}$ and $M_{j^{\prime}}$, respectively, in the $2\times 2$ minor with
the upper left entry at position $(j-1,j-1)$. The same argument from the
preceding paragraph shows that $M_{j}$ and $M_{j^{\prime}}$ must not generate
a relation, hence are free. ∎

## References

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Department of Mathematics, University at Buffalo, Buffalo, NY 14260-2900, USA

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