section[0pt] *[8pt].

# On the Dimension of Kakeya Sets in the First Heisenberg Group
00footnotetext:  2010 Mathematics Subject Classification: Primary 28A75;
Secondary 28A78 $\cdot$ 28A80 Key words and phrases: Kakeya set, Hausdorff
dimension, Heisenberg group. J. L. is supported by the Academy of Finland via
the projects: Quantitative rectifiability in Euclidean and non-Euclidean
spaces, Grant No. 314172, and Singular integrals, harmonic functions, and
boundary regularity in Heisenberg groups, Grant No. 328846.

Jiayin Liu

Abstract.We define Kakeya sets in the Heisenberg group and show that the
Heisenberg Hausdorff dimension of Kakeya sets in the first Heisenberg group is
at least 3. This lower bound is sharp since, under our definition, the
$\\{xoy\\}$-plane is a Kakeya set with Heisenberg Hausdorff dimension 3.

##  1 Introduction

The study of Kakeya sets in Euclidean space is one of the central topics in
geometric measure theory. A set $E\subset{\mathbb{R}}^{n}$ is a Kakeya set if
for every $e\in S^{n-1}$ there exists a unit line segment $I_{e}$ parallel to
$e$ such that $I_{e}\subset E$.

A natural question is to determine the least Hausdorff dimension of Kakeya
sets.

The answer is known in 2-dimensional Euclidean space. Indeed, Kakeya sets in
${\mathbb{R}}^{2}$ turn out to be of Hausdorff dimension equal to 2 which can
be shown by multiple ways. See [7, 9, 15]. However, for Kakeya sets in higher
dimensional Euclidean space, the sharp lower bound is not known. We would like
to remark some progress: Bourgain used two different methods to provide lower
bounds [3, 4], which were further improved by Wolff [18] and Katz-Tao [11]
respectively. Recently, Katz-Zahl [13] and Guth-Zahl [10] enhanced the results
of [18] in ${\mathbb{R}}^{3}$ and ${\mathbb{R}}^{4}$ respectively, which show
that the best known lower bound is $5/2+\epsilon_{0}$ in ${\mathbb{R}}^{3}$
with $\epsilon_{0}$ an absolute constant and $3+1/40$ in ${\mathbb{R}}^{4}$.
In ${\mathbb{R}}^{n}$ with $n\geq 5$, the best known lower bound
$3+(2-\sqrt{2})(n-4)$ was established in [12] by Katz-Tao.

As an analogy to Euclidean Kakeya sets, we can define Kakeya sets in the
Heisenberg group. In this paper, we denote by ${{\mathbb{H}}^{n}}$ the $n$-th
Heisenberg group. When $n=1$, we write ${\mathbb{H}}$ instead of
${\mathbb{H}}^{1}$ for simplicity.

######  Definition 1.1.

A set $E\subset{\mathbb{H}}^{n}$ is a Kakeya set if for every unit line
segment $I\subset{\mathbb{R}}^{2n}\times\\{0\\}$ centred at the origin, there
exists $q\in{{\mathbb{H}}^{n}}$ such that $qI\subset E$.

Here and in what follows, by a unit line segment we mean an isometric copy of
the unit open interval $(0,1)$. Moreover, by Heisenberg Hausdorff measure we
mean the one induced by the Korányi metric on the first Heisenberg group. For
the definition of the Korányi metric, we refer the readers to Section 2.

The Heisenberg Hausdorff dimension of Euclidean Kakeya sets has been studied
in [16] where the author showed a lower bound on the dimension of Kakeya sets.
In [17], the author studied Kakeya sets for general metric spaces in axiomatic
sense.

According to Definition 1.1, it is not hard to show that the Euclidean
Hausdorff dimension of Kakeya sets in ${\mathbb{H}}$ is at least 2. This can
be done as follows. Under orthogonal projection to the $\\{xoy\\}$-plane,
every Kakeya set $E$ in ${\mathbb{H}}$ becomes a Kakeya set $E^{\prime}$ in
${\mathbb{R}}^{2}$. Hence the known lower bound of the Euclidean Hausdorff
dimension of $E^{\prime}$ is also the one of $E$ since orthogonal projection
in ${\mathbb{R}}^{3}$ is Lipschitz. Moreover, 2 is sharp since the
$\\{xoy\\}$-plane is a Kakeya set in ${\mathbb{H}}$ with Euclidean Hausdorff
dimension 2. However, the Heisenberg Hausdorff dimension of the
$\\{xoy\\}$-plane is 3 and the orthogonal projection from ${\mathbb{H}}$ to
the $\\{xoy\\}$-plane is no longer Lipschitz with respect to the Korányi
metric. Hence to calculate a lower bound of Heisenberg Hausdorff dimension of
Kakeya sets seems to be a nontrivial problem.

In this note, we will show the following

######  Theorem 1.2.

In the first Heisenberg group ${\mathbb{H}}$ equipped with the Korányi metric,
every Kakeya set has Heisenberg Hausdorff dimension at least 3 and this lower
bound is sharp.

In the following, $E$ will denote a Kakeya set in the first Heisenberg group.

Our method to show Theorem 1.2 is based on the idea of [2, 9]. We first encode
each horizontal line segment in $E$ by a quadruple in ${\mathbb{R}}^{4}$
forming a subset $L(E)\subset{\mathbb{R}}^{4}$. Then we transfer the
computation for dimensions of each intersection of $E$ and a plane belonging
to a one parameter family to that of a subset in ${\mathbb{R}}^{3}$ obtained
by certain projections acting on $L(E)$. This can be seen as a duality
principle. Finally we use a recent Marstrand-type projection theorem in
${\mathbb{R}}^{3}$ by Käenmäki-Orponen-Venieri [14] and a co-area inequality
by Eilenberg-Harrold, Jr. [8] to conclude the proof.

Since every Kakeya set in the first Heisenberg group has Heisenberg Hausdorff
dimension at least 3, a further question that may be asked is to find a lower
bound of 3-dimensional Heisenberg Hausdorff measure among all Kakeya sets in
the first Heisenberg group. Unlike the Euclidean Kakeya set, which may have
$n$-dimensional Lebesque measure zero in every ${\mathbb{R}}^{n}$ (for
example, see [1]), it is not easy to show a counterpart for Kakeya sets in
Heisenberg group. Hence we would like to ask the following question

######  Problem 1.3.

Does there exist a Kakeya set in the first Heisenberg group with zero
3-dimensional Heisenberg Hausdorff measure?

The paper is organised as follows. In section 2, we recall some background in
Heisenberg groups, the Marstrand-type projection theorem in ${\mathbb{R}}^{3}$
and the co-area inequality. In section 3, we prove Theorem 1.2.

Acknowledgement

J. L. would like to thank K. Fässler and T. Orponen for many meaningful
discussions.

##  2 Preliminaries

The first part of this section is dedicated to a brief introduction to the
first Heisenberg group $\mathbb{H}$. For a detailed one, we refer the readers
to [5].

The first Heisenberg group $\mathbb{H}$ is $\mathbb{R}^{3}$, equipped with the
group multiplication, for any $w=(x,y,t)$ and
$w^{\prime}=(x^{\prime},y^{\prime},t^{\prime})$, as follows

$ww^{\prime}=\bigg{(}x+x^{\prime},y+y^{\prime},t+t^{\prime}+\frac{1}{2}[xy^{\prime}-x^{\prime}y]\bigg{)}.$
(2.1)

We introduce the Korányi metric on the first Heisenberg group. This is the
left invariant metric given by

$d_{\mathbb{H}}(w,w^{\prime}):=\|(w^{\prime})^{-1}\cdot w\|_{\mathbb{H}}$
(2.2)

where $\|\cdot\|_{\mathbb{H}}$ is defined as

$\|(x,y,t)\|_{\mathbb{H}}=((x^{2}+y^{2})^{2}+16t^{2})^{1/4}.$

We define horizontal lines in the first Heisenberg group ${\mathbb{H}}$ as
lines which can be obtained by a left translation of some line passing through
the origin and lying in the $\\{xoy\\}$-plane.

By the definition of horizontal lines, we know that for any $b\in{\mathbb{R}}$
and $q\in{\mathbb{H}}$, $qI_{b}$ and $qJ_{b}$ are horizontal line and
horizontal line segment respectively where

$I_{b}(\tau)=(\tau,b\tau,0),\ \tau\in{\mathbb{R}}$ (2.3)

and

$J_{b}(\tau)=(\tau,b\tau,0),\
\tau\in(-\frac{1}{2\sqrt{b^{2}+1}},\frac{1}{2\sqrt{b^{2}+1}}).$

The following observation is needed in the proof of Theorem 1.2.

######  Lemma 2.1.

For any $b\in{\mathbb{R}}$ and $q=(q_{1},q_{2},q_{3})\in{\mathbb{H}}$,

  1. (1)

$qI_{b}$ and $qJ_{b}$ can be parameterised as

$qI_{b}(s)=(s,bs+a,-\frac{as}{2}+d),\ s\in{\mathbb{R}}$ (2.4)

and

$qJ_{b}(s)=(s,bs+a,-\frac{as}{2}+d),\
s\in(\epsilon,\epsilon+\frac{1}{\sqrt{b^{2}+1}})$ (2.5)

where $a=q_{2}-bq_{1}$, $d=q_{3}+\frac{1}{2}aq_{1}$ and
$\epsilon=q_{1}-\frac{1}{2\sqrt{b^{2}+1}}$.

  2. (2)

If we denote

$l_{(a,b,d)}:=\\{(s,bs+a,-\frac{as}{2}+d)\in{\mathbb{H}}\ |\
s\in{\mathbb{R}}\\}$ (2.6)

and

$l^{\epsilon}_{(a,b,d)}:=\left\\{(s,bs+a,-\frac{as}{2}+d)\in{\mathbb{H}}\ |\
s\in(\epsilon,\epsilon+\frac{1}{\sqrt{b^{2}+1}})\right\\}.$

Then $l_{(a,b,d)}^{\epsilon}$ has length $1$ with respect to $d_{\mathbb{H}}$
for every $a,b,d,\epsilon$.

  3. (3)

If $b\in(-\sqrt{3},\sqrt{3})$, then the orthogonal projection of
$l_{(a,b,d)}^{\epsilon}$ to the $x$-axis has Euclidean length greater than
$\frac{1}{2}$.

###### Proof.

  1. (1)

Let $I_{b}$ be parameterised as in (2.3). Then by the Heisenberg
multiplication law (2.1), we have

$qI_{b}=\\{(q_{1}+\tau,q_{2}+b\tau,q_{3}+\frac{1}{2}(q_{1}b\tau-q_{2}\tau))\
|\ \tau\in{\mathbb{R}}\ \\}.$

Letting $s=q_{1}+\tau$, $a=q_{2}-bq_{1}$ and $d=q_{3}+\frac{1}{2}aq_{1}$, we
arrive at (2.4). In addition, letting
$\epsilon=q_{1}-\frac{1}{2\sqrt{b^{2}+1}}$, we verify that (2.5) holds.

  2. (2)

Using the definition of $d_{\mathbb{H}}$ and the fact that left translation is
an isometry with respect to $d_{\mathbb{H}}$, we deduce the result.

  3. (3)

From (2.5), the orthogonal projection of $l_{(a,b,d)}^{\epsilon}$ to the
$x$-axis is the interval $(\epsilon,\epsilon+\frac{1}{\sqrt{b^{2}+1}})\subset
x$-axis. Hence when $b\in(-\sqrt{3},\sqrt{3})$, the length of the interval is
greater than $\frac{1}{2}$.

∎

######  Remark 2.2.

In the sense of sub-Riemannian geometry, there exists more general horizontal
curves in ${\mathbb{H}}$ besides horizontal lines. Indeed, associated to the
group operation (2.1), we can define the left invariant vector fields

$\displaystyle X$ $\displaystyle=\frac{\partial}{\partial
x}-\frac{y}{2}\frac{\partial}{\partial t},\quad Y=\frac{\partial}{\partial
y}+\frac{x}{2}\frac{\partial}{\partial t}.$

A Lipschitz curve
${\gamma}=({\gamma}_{1},{\gamma}_{2},{\gamma}_{3}):[a,b]\to{\mathbb{H}}$ is
said to be horizontal if
$\dot{{\gamma}}(s)\in\text{Span}\\{X({\gamma}(s)),Y({\gamma}(s))\\}$, i.e.
$\dot{{\gamma}}(s)=a(s)X({\gamma}(s))+b(s)Y({\gamma}(s))$, for almost every
$s\in[a,b]$. One can check that horizontal lines are indeed horizontal curves
under the this definition. For more information from the sub-Riemannian point
of view, we refer readers to [5, Chapter 2].

In this paper, we denote by $\mathcal{H}_{{\mathbb{H}}}^{s}$ (resp.
$\mathcal{H}_{{\mathbb{R}}}^{s}$) the $s$-dimensional Hausdorff measure
induced by the Korányi metric (resp. Euclidean metric) and by
$\dim_{\mathcal{H}}^{{\mathbb{H}}}$ (resp.
$\dim_{\mathcal{H}}^{{\mathbb{R}}}$) the Hausdorff dimension of sets induced
by Korányi metric (resp. Euclidean metric). In addition, given any set
$A\subset{\mathbb{H}}$, we denote

$L(A):=\\{(a,b,d,\epsilon)\in{\mathbb{R}}\times(-\sqrt{3},\sqrt{3})\times{\mathbb{R}}\times{\mathbb{R}}\
|\ l^{\epsilon}_{(a,b,d)}\subset A\\}$ (2.7)

and

$L(A,c):=\\{(a,b,d,\epsilon)\in L(A)\ |\
l^{\epsilon}_{(a,b,d)}\cap\\{x=c\\}\neq\emptyset\\}.$ (2.8)

Next, we recall a version of a Marstrand-type projection theorem [14, Theorem
1.2]:

######  Theorem 2.3.

Suppose that
${\gamma}:[0,2\pi)\to{\mathbb{R}}^{3},\theta\mapsto{\gamma}(\theta)=\frac{1}{\sqrt{2}}(\cos\theta,\sin\theta,1)$.
If $K\subset{\mathbb{R}}^{3}$ is Borel set, then
$\dim_{\mathcal{H}}^{{\mathbb{R}}}\rho_{{\gamma}(\theta)}(K)=\min\\{\dim_{\mathcal{H}}^{{\mathbb{R}}}K,1\\}$
for almost every $\theta\in[0,2\pi)$.

Here, and in what follows, for any $x\in{\mathbb{R}}^{3}\setminus\\{0\\}$,
$\rho_{x}:{\mathbb{R}}^{3}\to\text{Span}(x)$ denotes the Euclidean orthogonal
projection to the straight line passing through the origin and $x$.

We also need the following co-area inequality [8, Theorem 1]:

######  Theorem 2.4.

Let $X$ be an arbitrary metric space, $0\leq\alpha<\infty$ be real numbers and
$F\subset X$ be any subset. Then, for any $1$-Lipschitz map
$f:X\to{\mathbb{R}}$ we have

$\int_{\mathbb{R}}^{\ast}\mathcal{H}^{\alpha}_{X}(F\cap
f^{-1}(y))\,dy\leq\mathcal{H}^{\alpha+1}_{X}(F).$ (2.9)

Here, $\int^{\ast}_{\mathbb{R}}g\,dy$ is the upper integral of
$g:{\mathbb{R}}\to[0,+\infty)$. That is

$\int^{\ast}_{\mathbb{R}}g(y)\,dy=\inf\int_{\mathbb{R}}h(y)\,dy$

where the infimum is taken over all measurable functions
$h:{\mathbb{R}}\to[0,+\infty)$ satisfying $0\leq g(y)\leq h(y)$ for a.e.
$y\in{\mathbb{R}}$.

##  3 Proof of Theorem 1.2

###### Proof of Theorem 1.2.

Since every set is contained in a $G_{\delta}$-set of the same dimension, we
may assume $E$ to be $G_{\delta}$.

_Step 1. Properties of $L(E,c)$_
First, we need

Claim I: _For every $c\in{\mathbb{R}}$, $L(E,c)$ is a $G_{\delta}$ set in
${\mathbb{R}}^{4}$._

###### Proof of Claim I.

Recalling (2.7),

$L(E,c)=\\{(a,b,d,\epsilon)\in L(E)\ |\
l^{\epsilon}_{(a,b,d)}\cap\\{x=c\\}\neq\emptyset\\}.$

Since $E$ is a $G_{\delta}$ set, we can find a sequence of open sets
$\\{E_{i}\\}_{i\in{\mathbb{N}}}$ such that $E_{i}\supset E_{i+1}$ for each
$i\in{\mathbb{N}}$ and

$E=\bigcap_{i\in{\mathbb{N}}}E_{i}.$

Consider the sets

$L(E_{i},c)=\\{(a,b,d,\epsilon)\in L(E_{i})\ |\
(l^{\epsilon}_{(a,b,d)}\cap\\{x=c\\})\neq\emptyset\\}.$

We assert $L(E_{i},c)$ is open for any $c\in{\mathbb{R}}$ and
$i\in{\mathbb{N}}$. Consider an arbitrary quadruple $(a,b,d,\epsilon)\in
L(E_{i},c)$, i.e.

$l^{\epsilon}_{(a,b,d)}\cap\\{x=c\\}\neq\emptyset\quad\text{and}\quad
l^{\epsilon}_{(a,b,d)}\subset E_{i},$ (3.1)

Thanks to the openness of $E_{i}$ and the interval
$(\epsilon,\epsilon+\frac{1}{\sqrt{b^{2}+1}})$, we deduce that for
$(a^{\prime},b^{\prime},d^{\prime},\epsilon^{\prime})$ close enough to
$(a,b,d,\epsilon)$, we have

$l^{\epsilon^{\prime}}_{(a^{\prime},b^{\prime},d^{\prime})}\cap\\{x=c\\}\neq\emptyset\quad\text{and}\quad
l^{\epsilon^{\prime}}_{(a^{\prime},b^{\prime},d^{\prime})}\subset E_{i}$

and hence $(a^{\prime},b^{\prime},d^{\prime},\epsilon^{\prime})\in
L(E_{i},c)$, which implies $L(E_{i},c)$ is open.

We are left to show

$L(E,c)=\bigcap_{i\geq 1}L(E_{i},c).$ (3.2)

Note that the direction “$\subset$” of (3.2) is obvious. We show the direction
“$\supset$”. If there exists $(a,b,d,\epsilon)\in\bigcap_{i\geq 1}L(E_{i},c)$,
then we infer that

$l^{\epsilon}_{(a,b,d)}\cap\\{x=c\\}\neq\emptyset\quad\text{and}\quad
l^{\epsilon}_{(a,b,d)}\subset E_{i},\text{ for all }i.$ (3.3)

On the other hand, $E=\cap_{i\in{\mathbb{N}}}E_{i}$ implies

$\\{x=c\\}\cap\left(\bigcap_{i\in{\mathbb{N}}}E_{i}\right)=\\{x=c\\}\cap E.$
(3.4)

Combining (3.3) and (3.4) we conclude

$l^{\epsilon}_{(a,b,d)}\subset E,$

which verifies (3.2). ∎

Furthermore, we have the following

Claim II: _There exists at least one $c_{0}\in{\mathbb{R}}$ satisfying_

$\mathcal{H}^{1}_{\mathbb{R}}(\pi_{123}(L(E,c_{0})))>0$ (3.5)

_where $\pi_{123}$ is the orthogonal projection from ${\mathbb{R}}^{4}$ to the
subspace spanned by the first three coordinates._

###### Proof of Claim II.

Since $E$ is a Kakeya set in ${\mathbb{H}}$, by Definition 1.1 and recalling
(2.7), we have

$\pi_{2}(L(E))\supset(-\sqrt{3},\sqrt{3})$

where $\pi_{2}$ is the orthogonal projection from ${\mathbb{R}}^{4}$ to the
subspace spanned by the second coordinate, which implies

$\mathcal{H}^{1}_{\mathbb{R}}(\pi_{2}(L(E)))>0.$ (3.6)

By observing that

$L(E)=\bigcup_{c\in\mathbb{Q}}L(E,c)$

and using (3.6), we infer that there exists $c_{0}$ such that

$\mathcal{H}^{1}_{\mathbb{R}}(\pi_{2}(L(E,c_{0})))>0.$ (3.7)

Noting that

$\mathcal{H}^{1}_{\mathbb{R}}(\pi_{123}(L(E,c)))\geq\mathcal{H}^{1}_{\mathbb{R}}(\pi_{2}(L(E,c))),\
\forall c\in{\mathbb{R}},$

we conclude the proof. ∎

We end step 1 with the following

Claim III: _We can find a Borel set $B\subset\pi_{123}(L(E,c_{0}))$ with_

$\mathcal{H}_{{\mathbb{R}}}^{1}(B)>0$ (3.8)

_and at least one of the following holds:_

$B\subset\pi_{123}(L(E,c)),\quad\forall c\in[c_{0}-\frac{1}{4},c_{0}]$ (3.9)

_or_

$B\subset\pi_{123}(L(E,c)),\quad\forall c\in[c_{0},c_{0}+\frac{1}{4}].$ (3.10)

###### Proof of Claim III.

Using Lemma 2.1(3), we observe that if
$l^{\epsilon}_{(a,b,d)}\cap\\{x=c_{0}\\}\neq\emptyset$, then either

$l^{\epsilon}_{(a,b,d)}\cap\\{x=c_{0}-\frac{1}{4}\\}\neq\emptyset,$

or

$l^{\epsilon}_{(a,b,d)}\cap\\{x=c_{0}+\frac{1}{4}\\}\neq\emptyset.$

Figure 1: A line segment that intersects both $\\{x=c_{0}\\}$ and
$\\{x=c_{0}+\frac{1}{4}\\}$.

Hence we have

$L(E,c_{0})\subset[L(E,c_{0})\cap L(E,c_{0}-\frac{1}{4})]\cup[L(E,c_{0})\cap
L(E,c_{0}+\frac{1}{4})].$

We conclude

$\pi_{123}(L(E,c_{0}))\subset\pi_{123}(L(E,c_{0})\cap
L(E,c_{0}-\frac{1}{4}))\cup\pi_{123}(L(E,c_{0})\cap L(E,c_{0}+\frac{1}{4})).$
(3.11)

From the above inclusion, we can assume, without loss of generality, that

$\mathcal{H}^{1}_{\mathbb{R}}\left(\pi_{123}(L(E,c_{0})\cap
L(E,c_{0}+\frac{1}{4}))\right)\geq\frac{1}{2}\mathcal{H}^{1}_{\mathbb{R}}(L(E,c_{0}))>0$

where the last inequality results from Claim II.

On the other hand, if $(a,b,d,\epsilon)\in L(E,c_{0})\cap
L(E,c_{0}+\frac{1}{4})$, then for any $c\in[c_{0},c_{0}+\frac{1}{4}]$, we have
$(a,b,d,\epsilon)\in L(E,c)$, which indicates

$L(E,c_{0})\cap L(E,c_{0}+\frac{1}{4})\subset L(E,c)\text{ for any
}c\in[c_{0},c_{0}+\frac{1}{4}].$

By Claim I, for any $c\in{\mathbb{R}}$, we know that $L(E,c)$ is a
$G_{\delta}$ set and hence $\pi_{123}(L(E,c_{0})\cap L(E,c_{0}+\frac{1}{4}))$
is an analytic set. Hence we can apply Corollary 2 in [6] to choose $B$ to be
a closed subset of $\pi_{123}(L(E,c_{0})\cap L(E,c_{0}+\frac{1}{4}))$ with
$\mathcal{H}^{1}_{\mathbb{R}}(B)>0$. Therefore $B$ satisfies the assumption of
the claim. ∎

Claim III enables us to choose $c_{0}\in{\mathbb{R}}$ such that, without loss
of generality, there exists a Borel set $B\subset\pi_{123}(L(E,c_{0}))$
satisfying (3.8), i.e.

$\mathcal{H}_{{\mathbb{R}}}^{1}(B)>0.$

and (3.10). Therefore, we infer that

$\dim_{\mathcal{H}}^{{\mathbb{R}}}(B)\geq 1.$ (3.12)

_Step 2. Establish a duality principle._
Recall the definition of $l_{(a,b,d)}$ in (2.6). For every
$c\in[c_{0},c_{0}+1/4]$, we consider $E_{c}\subset\\{x=c\\}\cap
E\subset{\mathbb{H}}$ defined by

$\displaystyle E_{c}:$ $\displaystyle=\\{l_{a,b,d}\cap\\{x=c\\}\ |\ (a,b,d)\in
B\\}$ $\displaystyle=\\{(c,bc+a,-\frac{ac}{2}+d)\ |\ (a,b,d)\in B\\}.$ (3.13)

Use left translation $T_{(-c,0,0)}$ to translate $E_{c}$ to $\\{yot\\}$-plane,
which means $E_{c}^{1}:=T_{(-c,0,0)}(E_{c})$ lies in $\\{yot\\}$-plane and has
same dimension as $E_{c}$. See the above Figure 2.

Figure 2: Translating $E_{c}$ to $\\{yot\\}$-plane

Recalling (3) and the Heisenberg multiplication law (2.1), we deduce that

$\displaystyle E_{c}^{1}$ $\displaystyle=T_{(-c,0,0)}(E_{c})$
$\displaystyle=\\{(-c,0,0)\cdot(c,bc+a,-\frac{ac}{2}+d)\ |\ (a,b,d)\in B\\}$
$\displaystyle=\\{(0,bc+a,-ac-\frac{bc^{2}}{2}+d)\ |\ (a,b,d)\in B\\}.$ (3.14)

Notice that the third coordinate of points in $E_{c}^{1}$ expressed in (3)
takes the form

$-ac-\frac{bc^{2}}{2}+d=\left\langle(-c,-\frac{c^{2}}{2},1),(a,b,d)\right\rangle$

where $\langle,\rangle$ is the Euclidean inner product in ${\mathbb{R}}^{3}$.

By considering the $t$-axis as ${\mathbb{R}}$ and letting

$\varphi:\\{yot\\}\to{\mathbb{H}},\ (0,y,t)\mapsto(0,0,t),$ (3.15)

we can write

$\displaystyle\varphi(E_{c}^{1})$
$\displaystyle=\left\\{\left\langle(-c,-\frac{c^{2}}{2},1),(a,b,d)\right\rangle\
|\ (a,b,d)\in B\right\\}$
$\displaystyle=\left(1+\frac{c^{2}}{2}\right)\rho_{(-c,-\frac{c^{2}}{2},1)}(B).$
(3.16)

Equation (3) implies that $\varphi(E_{c}^{1})$ can be viewed as a Euclidean
projection of $B$ to the one parameter family of lines
$\Gamma=\\{{\gamma}_{c}:{\mathbb{R}}\to{\mathbb{R}}^{3}\ |\
t\mapsto(-ct,-\frac{c^{2}}{2}t,t),c\in[c_{0},c_{0}+1/4]\\}$ up to scalings.
Letting $t=1$, we observe that $c\mapsto(-c,-\frac{c^{2}}{2},1)$ forms a part
of parabola $\chi$ in ${\mathbb{R}}^{3}$. Hence this one parameter family of
lines forms part of a cone $C_{1}$ in ${\mathbb{R}}^{3}$, i.e.

$C_{1}=\\{(x,y,z)\in{\mathbb{R}}^{3}\ |\ x^{2}=-2yz\\}.$

Moreover, the intersection of $\Gamma$ and the unit sphere in
${\mathbb{R}}^{3}$ is contained in a circle and can be parameterised as

$\widetilde{\gamma}(c)=\bigg{\\{}\frac{2}{2+c^{2}}(-c,-\frac{c^{2}}{2},1)\ |\
c\in[c_{0},c_{0}+1/4]\bigg{\\}}.$

We see the arc $\widetilde{\gamma}$ and the parabola $\chi$ are both conical
curves, they can be included in one same cone as Figure 3 depicts.

Figure 3: Parabola $\chi$ and Arc $\widetilde{\gamma}$ can be included in one
same cone

_Step 3. Conclusion._
In Theorem 2.3, the family of lines passing through the origin and
${\gamma}(\theta)=\frac{1}{\sqrt{2}}(\cos\theta,\sin\theta,1)$ also spans a
cone $C_{2}=\\{(x,y,z)\in{\mathbb{R}}^{3}\ |\ x^{2}+y^{2}=z^{2}\\}$ in
${\mathbb{R}}^{3}$.

We observe that the cone $C_{2}$ can be obtained by a rotation
${\mathcal{R}}:(x,y,z)\mapsto(x,\frac{\sqrt{2}}{2}(y+z),\frac{\sqrt{2}}{2}(z-y))$
acting on $C_{1}$ and $\widetilde{\gamma}$ is mapped to an arc of ${\gamma}$,
i.e.

${\gamma}(\theta(c))\\!=\\!{\mathcal{R}}\circ\widetilde{\gamma}(c)\\!=\\!\frac{2}{2+c^{2}}\left(\\!-c,\frac{\sqrt{2}}{4}(2-c^{2}),\frac{\sqrt{2}}{4}(2+c^{2})\\!\right)\\!=\\!\frac{1}{\sqrt{2}}\left(\frac{-2\sqrt{2}c}{2+c^{2}},\frac{2-c^{2}}{2+c^{2}},1\right),\,c\in\\![c_{0},c_{0}+\frac{1}{4}],$

where $\theta(c)$ is determined from

$(\cos(\theta(c)),\sin(\theta(c)))=\left(\frac{-2\sqrt{2}c}{2+c^{2}},\frac{2-c^{2}}{2+c^{2}}\right).$

This implies

$\rho_{(-c,-\frac{c^{2}}{2},1)}((x,y,z))=\rho_{{\gamma}(\theta(c))}({\mathcal{R}}(x,y,z)),\quad\forall(x,y,z)\in{\mathbb{R}}^{3}.$
(3.17)

By Claim III and (3.12), we know $B$ is Borel and
$\dim_{\mathcal{H}}^{{\mathbb{R}}}(B)\geq 1$. We use (3.17) and apply Theorem
2.3 to the family of lines passing through the origin and
$\\{{\gamma}(\theta(c))\\}_{c\in[c_{0},c_{0}+1/4]}$ to deduce that

$\dim_{\mathcal{H}}^{{\mathbb{R}}}[\rho_{(-c,-\frac{c^{2}}{2},1)}(B)]=\dim_{\mathcal{H}}^{{\mathbb{R}}}[\rho_{{\gamma}(\theta(c))}({\mathcal{R}}(B))]=1\quad
a.e.\ c\in[c_{0},c_{0}+1/4].$ (3.18)

Recalling the definition of $\varphi$ in (3.15) and according to (2.2), we
know $\varphi$ is $1$-Lipschitz with respect to $d_{\mathbb{H}}$ and for any
set $A\subset t$-axis,

$\dim_{\mathcal{H}}^{\mathbb{H}}(A)=2\dim_{\mathcal{H}}^{\mathbb{R}}(A).$

Hence for any $c\in[c_{0},c_{0}+1/4]$ such that (3.18) holds, combining (3),
(3), (3.18) and the above equality, we conclude

$\displaystyle\dim_{\mathcal{H}}^{{\mathbb{H}}}(\\{x=c\\}\cap E)$
$\displaystyle\geq\dim_{\mathcal{H}}^{{\mathbb{H}}}(E_{c})=\dim_{\mathcal{H}}^{{\mathbb{H}}}(E_{c}^{1})\geq\dim_{\mathcal{H}}^{{\mathbb{H}}}(\phi(E_{c}^{1}))$
$\displaystyle=2\dim_{\mathcal{H}}^{{\mathbb{R}}}(\rho_{(-1,\frac{c}{2},c^{2})}(B))$
$\displaystyle=2$

and for any $0<\alpha<2$, we deduce

$\mathcal{H}_{\mathbb{H}}^{\alpha}(\\{x=c\\}\cap E)=\infty.$

By definition of $d_{\mathbb{H}}$, the map
$f:({\mathbb{H}},d_{\mathbb{H}})\to{\mathbb{R}},(x,y,t)\to(x,0,0)$ is
$1$-Lipschitz. Now letting $X={\mathbb{H}}$, $Y=[c_{0},c_{0}+1/4]$ and
$F=E\cap\\{(x,y,t)\in{\mathbb{H}}\ |\ x\in[c_{0},c_{0}+1/4]\\}$ in Theorem
2.4, for any $0<\alpha<2$, we derive

$\mathcal{H}_{\mathbb{H}}^{\alpha+1}(F)\geq\int^{\ast}_{[c_{0},c_{0}+1/4]}\mathcal{H}_{\mathbb{H}}^{\alpha}(F\cap
f^{-1}(y))\,dy=\infty,$

which implies

$\dim_{\mathcal{H}}^{{\mathbb{H}}}(E)\geq\dim_{\mathcal{H}}^{{\mathbb{H}}}(E\cap\\{(x,y,t)\in{\mathbb{H}}\
|\ x\in[c_{0},c_{0}+1/4]\\})=\dim_{\mathcal{H}}^{{\mathbb{H}}}(F)\geq 3.$

We finish the proof. ∎

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Jiayin Liu

Department of Mathematics and Statistics, University of Jyväskylä.

E-mail<EMAIL_ADDRESS>