# Partial difference sets with Denniston parameters and cyclotomy

James A. Davis  Department of Math & Computer Science, 212 Jepson Hall, 221
Richmond Way, University of Richmond, VA 23173, USA<EMAIL_ADDRESS>,
Sophie Huczynska  School of Mathematics and Statistics, Mathematical
Institute, University of St Andrews, North Haugh, St Andrews, Fife, KY16 9SS,
Scotland, UK<EMAIL_ADDRESS>,  Laura Johnson  School of Mathematics
and Statistics, Mathematical Institute, University of St Andrews, North Haugh,
St Andrews, Fife, KY16 9SS, Scotland, UK<EMAIL_ADDRESS>and  John
Polhill  Department of Mathematics, Computer Science and Digital Forensics,
Commonwealth University, Bloomsburg, PA, USA<EMAIL_ADDRESS>
###### Abstract.

Denniston constructed partial difference sets (PDSs) with the parameters
$(2^{3m},(2^{m+r}-2^{m}+2^{r})(2^{m}-1),2^{m}-2^{r}+(2^{m+r}-2^{m}+2^{r})(2^{r}-2),(2^{m+r}-2^{m}+2^{r})(2^{r}-1))$
in elementary abelian groups of order $2^{3m}$ for all $m\geq 2,1\leq r<m$. In
this paper, we show that PDSs with the Denniston parameters
$(p^{3m},(p^{m+r}-p^{m}+p^{r})(p^{m}-1),p^{m}-p^{r}+(p^{m+r}-p^{m}+p^{r})(p^{r}-2),(p^{m+r}-p^{m}+p^{r})(p^{r}-1))$
exist in all elementary abelian groups of order $p^{3m}$ for all $m\geq
2,r\in\\{1,m-1\\}$ where $p$ is an odd prime, and present a construction. Our
approach uses PDSs formed as unions of cyclotomic classes.

###### Key words and phrases:

Partial difference sets, Denniston parameters, strongly regular graphs,
cyclotomy

###### 2020 Mathematics Subject Classification:

05E30, 05B10, 11T22

##  1\. Background and motivation

A $k$-subset $D$ of a finite group $G$ (written additively with identity
$0_{G}$) of order $v$ is a $(v,k,\lambda,\mu)$ partial difference set (PDS) if
the multiset of differences $\Delta:=\\{d_{1}-d_{2}:d_{1},d_{2}\in D\\}$
contains every element of $D-\\{0_{G}\\}$ precisely $\lambda$ times and every
element of $G-D-\\{0_{G}\\}$ precisely $\mu$ times. A PDS is equivalent to a
strongly regular graph with a regular automorphism group (a sharply transitive
automorphism group). A PDS can be thought of as a symmetric association scheme
with two classes [1], and there are connections to projective two-weight codes
[4]. Many infinite families have been constructed, both in abelian and
nonabelian groups (see [10] for a good survey). One general family of PDSs is
the so-called Negative Latin Square type PDSs, with parameters
$(n^{2},r(n+1),-n+r+3r^{2},r^{2}+r)$. We note that the order of the group must
be a square in order for the PDS to be Negative Latin Square type.

The focus of this paper will be on a family originally due to Denniston [8];
the parameters are
$(2^{3m},(2^{m+r}-2^{m}+2^{r})(2^{m}-1),2^{m}-2^{r}+(2^{m+r}-2^{m}+2^{r})(2^{r}-2),(2^{m+r}-2^{m}+2^{r})(2^{r}-1))$
for $m\geq 2,1\leq r<m$, and Denniston provided a construction in the
elementary abelian group ${\mathbb{Z}}_{2}^{3m}$ for all of these
possibilities. Davis and Xiang [6] used Galois Rings to show that
${\mathbb{Z}}_{4}^{m}\times{\mathbb{Z}}_{2}^{m}$ contains a PDS for the
$r\in\\{1,m-1\\}$ cases for all $m\geq 2$. Brady [2] used a computer search to
find examples in the $m=2,r=1$ case for $51$ groups of order $64$, and Smith
[15] extended this to an exhaustive search that found examples in $73$ out of
$267$ nonisomorphic groups of order $64$. We note that PDSs with Denniston
parameters will never be of Negative Latin Square type when $m$ is odd; when
$m$ is even there will be one intermediate value of $r$ corresponding to a
Negative Latin Square type PDS.

We examine the question of whether, when $p$ is an odd prime, PDSs exist with
Denniston parameters, i.e. whether there exist PDSs with parameters
$(p^{3m},(p^{m+r}-p^{m}+p^{r})(p^{m}-1),p^{m}-p^{r}+(p^{m+r}-p^{m}+p^{r})(p^{r}-2),(p^{m+r}-p^{m}+p^{r})(p^{r}-1))$.
We answer this question in the affirmative for the $m\geq 2;r\in\\{1,m-1\\}$
cases in the group ${\mathbb{Z}}_{p}^{3m}$.

The paper is organized as follows. Section 2 introduces quadratic forms and
uses these to construct a PDS in $GF(p^{2m})$. Section 3 reframes the main
result of Section 2 in terms of cyclotomic classes. Section 4 defines
characters of the group ${\mathbb{Z}}_{p}^{3m}$ and demonstrates how they can
be used to prove that a subset of a finite abelian group is a PDS. Section 5
contains the construction of the Denniston PDSs in ${\mathbb{Z}}_{p}^{3m}$ for
all $m\geq 2$. Finally, Section 6 contains questions to be considered in the
future.

##  2\. Quadratic forms

For the rest of the paper let $p$ be an odd prime, and consider $GF(p^{2m})$
as a degree-$2$ field extension of $GF(p^{m})$, which is in turn a degree-$m$
extension of the prime field $GF(p)$. Each of $GF(p^{2m})$ and $GF(p^{m})$ can
be viewed as vector spaces over $GF(p)$ (of dimension $2m$ and $m$
respectively).

As an intermediate step to our main result, we use quadratic forms to form a
PDS in $GF(p^{2m})$, which can then be viewed as a union of cyclotomic classes
with indices specified by a Singer difference set. We note that a cyclotomic
“lifting construction” for forming PDSs via subdifference sets of Singer
difference sets appears in a strongly-regular graph setting in [11] and [12]
and could be applied in our case to establish this intermediate step. However,
to guarantee that our main construction is clear to the reader, we present a
self-contained and direct proof of this intermediate PDS construction.

We have the following definition of a quadratic form (see Ma [10]).

######  Definition 2.1.

Let $V$ be an $n$-dimensional vector space over a field $F$. A function
$Q:V\rightarrow F$ is a quadratic form if it satisfies:

  * (i)

For $a\in F$, $Q(ax)=a^{2}Q(x)$.

  * (ii)

The function $R:V\times V\rightarrow F,R(u,v):=Q(u+v)-Q(u)-Q(v)$ is bilinear,
i.e. linear in each argument.

Further, $Q$ is nondegenerate if $R$ is nondegenerate; i.e. $R(u,v)=0$ for all
$v\in V$ implies $u=0$.

We present the following quadratic form which we will use in this paper.

######  Proposition 2.2.

Let $Q:GF(p^{2m})\rightarrow GF(p^{m})$ be given by $Q(x)=x^{p^{m}+1}$. Then
$Q(x)$ is a nondegenerate quadratic form.

###### Proof.

We show that the above requirements are satisfied for $V=GF(p^{2m})$,
$F=GF(p^{m})$ and $Q(x)=x^{p^{m}+1}$. First, let $a\in GF(p^{m})$; then
$a^{p^{m}}=a$. So

$Q(ax)=(ax)^{p^{m}+1}=a^{p^{m}+1}x^{p^{m}+1}=a^{2}x^{p^{m}+1}=a^{2}Q(x).$

We define the Frobenius automorphism $\phi_{p}:GF(p^{2m})\rightarrow
GF(p^{2m})$ by $\phi_{p}(x)=x^{p}$, and the fact that this is an automorphism
[9] implies that $(a+b)^{p^{t}}=a^{p^{t}}+b^{p^{t}}$ for $a,b\in
GF(p^{2m}),t\geq 1$. This leads to the following computation.

$\displaystyle R(u,v)=Q(u+v)-Q(u)-Q(v)$ $\displaystyle=$
$\displaystyle(u+v)^{p^{m}+1}-u^{p^{m}+1}-v^{p^{m}+1}$ $\displaystyle=$
$\displaystyle(u+v)^{p^{m}}(u+v)-u^{p^{m}+1}-v^{p^{m}+1}$ $\displaystyle=$
$\displaystyle(u^{p^{m}}+v^{p^{m}})(u+v)-u^{p^{m}+1}-v^{p^{m}+1}$
$\displaystyle=$ $\displaystyle u^{p^{m}}v+uv^{p^{m}}$

Clearly the expression for $R$ is symmetric in $u$ and $v$. We have

$\displaystyle R(u_{1}+u_{2},v)$ $\displaystyle=$
$\displaystyle(u_{1}+u_{2})^{p^{m}}v+(u_{1}+u_{2})v^{p^{m}}$ $\displaystyle=$
$\displaystyle(u_{1}^{p^{m}}+u_{2}^{p^{m}})v+(u_{1}+u_{2})v^{p^{m}}$
$\displaystyle=$ $\displaystyle
u_{1}^{p^{m}}v+u_{1}v^{p^{m}}+u_{2}^{p^{m}}v+u_{2}v^{p^{m}}$ $\displaystyle=$
$\displaystyle R(u_{1},v)+R(u_{2},v)$

For $a\in GF(p^{m})$, we have

$\displaystyle R(au,v)$ $\displaystyle=$
$\displaystyle(au)^{p^{m}}v+(au)v^{p^{m}}$ $\displaystyle=$ $\displaystyle
a^{p^{m}}u^{p^{m}}v+auv^{p^{m}}$ $\displaystyle=$ $\displaystyle
au^{p^{m}}v+auv^{p^{m}}$ $\displaystyle=$ $\displaystyle aR(u,v).$

So $Q$ satisfies both properties and hence is a quadratic form. Finally, to
prove it is nondegenerate: suppose $R(u,v)=0$ for all $v\in GF(p^{2m})$. Then,
for all $v$, $u^{p^{m}}v+uv^{p^{m}}=0$. Taking $v=u$ we see that
$2u^{p^{m}+1}=0$, ie $u^{p^{m}+1}=0$ and hence $u=0$. ∎

Next, let $Tr_{m/1}:GF(q^{m})\rightarrow GF(q)$ be the trace map, given by
$Tr_{m/1}(x)=x+x^{q}+x^{q^{2}}+\cdots+x^{q^{m-1}}$. We have the following
result of Brouwer [3]:

######  Lemma 2.3.

[Brouwer] Let $F=GF(q^{m})$ and $F_{0}=GF(q)$ ($q$ a prime power and $m>1$).
Let $V$ be a vector space of dimension $d$ over $F$ and write $V_{0}$ for the
same vector space over $F_{0}$ (we will take $V=GF(q^{dm})$).

  * (i)

If $Q:V\rightarrow F$ is a quadratic form on $V$, then $Q_{0}=Tr\circ Q$ is a
quadratic form on $V_{0}$.

  * (ii)

$Q_{0}$ is nondegenerate if and only if $Q$ is nondegenerate and either $q$ is
odd or $d$ is even.

Here we will take $F=GF(p^{m})$, $F_{0}=GF(p)$ and $V=GF(p^{2m})$, so $d=2$.
Let

$Q_{0}=Tr_{m/1}(Q(x))=Tr_{m/1}(x^{p^{m}+1})=x^{p^{m}+1}+x^{p^{m+1}+p}+x^{p^{m+2}+p^{2}}+\cdots+x^{p^{2m-1}+p^{m-1}}.$

Part (i) of Lemma 2.3 implies that $Q_{0}$ is a quadratic form on
$V_{0}=GF(p)$, and part (ii) tells us that $Q_{0}$ is nondegenerate (as $Q$ is
nondegenerate and $q$ is odd).

Next, we state the following useful result (Theorem 2.6 of [10]):

######  Theorem 2.4.

Let $Q_{0}:\mathbb{F}_{q}^{2m}\rightarrow\mathbb{F}_{q}$ be a nondegenerate
quadratic form. Then

$X=\\{x\in\mathbb{F}_{q}^{2m}\setminus\\{0\\}:Q_{0}(x)=0\\}$

is a regular $(q^{2m},q^{2m-1}+\epsilon q^{m-1}(q-1)-1,q^{2m-2}+\epsilon
q^{m-1}(q-1)-2,q^{2m-2}+\epsilon q^{m-1})$-PDS in the additive group of
$\mathbb{F}_{q}^{2m}$, where $\epsilon=\pm 1$ and depends on the choice of
$Q_{0}$.

Applying Theorem 2.4 to our quadratic form $Q_{0}$ yields the following:

######  Theorem 2.5.

Let $Q_{0}:\mathbb{F}_{p}^{2m}\rightarrow\mathbb{F}_{p}$,
$Q_{0}(x)=Tr_{m/1}(x^{p^{m}+1})$. Then

$X=\\{x\in GF(p^{2m})\setminus\\{0\\}:Tr_{m/1}(x^{p^{m}+1})=0\\}.$

is a regular
$(p^{2m},p^{2m-1}-p^{m-1}(p-1)-1,p^{2m-2}-p^{m-1}(p-1)-2,p^{2m-2}-p^{m-1})$-PDS
in the additive group of $\mathbb{F}_{p}^{2m}$.

###### Proof.

We apply Theorem 2.4. Using our choice of quadratic form
$Q_{0}:GF(p^{2m})\rightarrow GF(p)$, $Q_{0}(x)=Tr_{m/1}(x^{p^{m}+1})$, our set
$X$ is defined by:

$X=\\{x\in GF(p^{2m})^{*}:Q_{0}(x)=0\\}=\\{x\in
GF(p^{2m})^{*}:Tr_{m/1}(x^{p^{m}+1})=0\\}.$

In our setting, $\epsilon$ as defined in Theorem 2.4 takes the value $-1$. ∎

We observe that $X$ is of Negative Latin Square type (with $n=p^{m}$ and
$r=p^{m-1}-1$).

##  3\. Cyclotomy

In this section, we introduce the notion of cyclotomy, and show how Theorem
2.5 may be expressed in terms of cyclotomic classes. This is useful as it will
provide a natural way to express our main construction for the PDS with
Denniston parameters.

######  Definition 3.1.

Let $GF(q)$ be a finite field of order $q=p^{n}=ef+1$ (where $p$ is prime and
$e>1$). Let $\alpha$ be a primitive element of GF$(q)$. The _cyclotomic
classes_ of order $e$ are defined:

$C_{i}^{e,n}=\alpha^{i}\langle\alpha^{e}\rangle$

for $0\leq i\leq e-1$. Each cyclotomic class is of size $f$.

The study of cyclotomic classes originates with Gauss, and they have been used
to construct difference sets and related objects since the work of Paley [14].
There is particular interest in the question of when individual cyclotomic
classes, or unions of such classes, form difference sets and PDSs; for a
comprehensive recent survey, see [13].

In our setting, we take the field $GF(p^{2m})$ (where $p$ is an odd prime),
and let $\alpha$ be a primitive element of this field. Our classes have the
form $C_{i}^{e,2m}$ where $e|p^{2m}-1$. Note that
$C_{i}^{e,2m}=\alpha^{i}C_{0}^{e,2m}$.

We now reframe the results of the previous section and show that the preimages
under $Q$ of the powers of the generator of $GF(p^{m})$ correspond to certain
cyclotomic classes. This yields a cyclotomic intepretation of the PDS $X$ from
Theorem 2.5.

######  Theorem 3.2.

Let $GF(p^{2m})^{*}=\langle\alpha\rangle$ and
$GF(p^{m})^{*}=\langle\omega\rangle$, where $\omega=\alpha^{p^{m}+1}$. As in
Section 2, let $Q:GF(p^{2m})\rightarrow GF(p^{m})$ be given by
$Q(x)=x^{p^{m}+1}$, let $Q_{0}=Tr_{m/1}(Q(x))$ and let
$X=\\{x\in\mathbb{F}_{q}^{2m}\setminus\\{0\\}:Q_{0}(x)=0\\}$.

  * (i)

The cyclotomic class $C_{i}^{p^{m}-1,2m}$ is the preimage under $Q$ of
$\omega^{i}$ in $GF(p^{m})^{*}$, where $0\leq i\leq p^{m}-2$.

  * (ii)

Let $I=\\{i:Tr_{m,1}(\omega^{i})=0,\quad 0\leq i\leq p^{m}-2\\}$. Then

$X=\bigcup_{i\in I}C_{i}^{p^{m}-1,2m}.$

###### Proof.

For (i), note $C_{i}^{p^{m}-1,2m}=\\{\alpha^{(p^{m}-1)t+i}:0\leq
t\leq(p-1)(p^{m}+1)-1\\}$ ($0\leq i\leq p^{m}-2$) and
$1=\alpha^{p^{2m}-1}=\alpha^{(p^{m}-1)(p^{m}+1)}$. Then

$Q(\alpha^{(p^{m}-1)t+i})=(\alpha^{(p^{m}-1)t+i})^{p^{m}+1}=\alpha^{(p^{m}-1)(p^{m}+1)t}\cdot\alpha^{(p^{m}+1)i}=1\cdot(\alpha^{p^{m}+1})^{i}=\omega^{i}.$

Part (ii) is immediate from the definition of $X$ combined with part (i): the
set $X$ consists of precisely those $C_{i}^{p^{m}-1,2m}$ such that
$\omega^{i}$ are the trace-$0$ elements of $GF(p^{m})$. ∎

Observe that $|I|=p^{m-1}-1$: the trace map $Tr_{m/1}$ is a linear map from
$GF(p^{m})$ with a kernel of size $p^{m-1}$ (the image has size $p$ and the
map is onto), and the zero element is excluded from $I$.

Next, we will demonstrate a connection to Singer difference sets, and obtain a
simpler expression for $X$. Consider $GF(p^{m})$ as a dimension-$m$ vector
space over $GF(p)$ with basis $\\{1,\omega,\omega^{2}\ldots\omega^{m-1}\\}$.
We will later be working in projective space, i.e. we will take the non-zero
$m$-tuples and identify two $m$-tuples if they differ by a scalar factor.

For $GF(p^{m})$, the zero-trace condition defines a hyperplane in the vector
space. A well-known result of Singer (Construction 18.28, [5]) tells us that,
modulo the scalars, its elements form a multiplicative
$(\frac{p^{m}-1}{p-1},\frac{p^{m-1}-1}{p-1},\frac{p^{m-2}-1}{p-1})$-difference
set in the cyclic group of order $\frac{p^{m}-1}{p-1}$ induced by $\omega$ on
the equivalence classes. Let $e=\frac{p^{m}-1}{p-1}$. Taking the set of powers
$i$ such that $\omega^{i}$ is in this difference set, will yield a
corresponding difference set in the additive group $(\mathbb{Z}_{e},+)$.
Observe that $GF(p)\setminus\\{0\\}$ is in fact the cyclotomic class of order
$e$ in $GF(p^{m})$, so we have factored by $\langle w^{e}\rangle=C_{0}^{e,m}$.

Let $T$ be the set of nonzero trace-$0$ elements of $GF(p^{m})$. Since
$|T|=p^{m-1}-1$, there are $\frac{p^{m-1}-1}{p-1}$ elements
$S=\\{\omega^{i}:Tr_{m/1}(\omega^{i})=0,0\leq i\leq e-1\\}$. Taking the
corresponding set of powers yields the set $D=\\{i:\omega^{i}\in S,0\leq i\leq
e-1\\}$, an $(e,\frac{p^{m-1}-1}{p-1},\frac{p^{m-2}-1}{p-1})$ Singer
difference set in $(\mathbb{Z}_{e},+)$. We record the following, which will be
useful in what follows:

######  Lemma 3.3.

The set of nonzero elements $T$ of $GF(p^{m})$ with trace $0$ is

$T=\cup_{j\in
D}\\{\omega^{j},\omega^{e+j},\ldots,\omega^{(p-2)e+j}\\}=\cup_{j\in
D}w^{j}\langle w^{e}\rangle.$

The set $I$ of indices of $T$ is

$I=\cup_{j\in D}\\{j,e+j,\ldots,(p-2)e+j\\}.$

Finally, consider our set $X=\bigcup_{i\in I}C_{i}^{p^{m}-1,2m}.$ We can
rewrite this as

$X=\bigcup_{j\in D}C_{j}^{p^{m}-1,2m}\cup C_{e+j}^{p^{m}-1,2m}\cup\cdots\cup
C_{(p-2)e+j}^{p^{m}-1,2m}$

Observe that since $e=\frac{p^{m}-1}{p-1}$, $e|p^{m}-1$ and so

$C_{0}^{p^{m}-1,2m}\cup C_{e}^{p^{m}-1,2m}\cup\cdots\cup
C_{(p-2)e}^{p^{m}-1,2m}=C_{0}^{e,2m}$

and more generally $C_{j}^{p^{m}-1,2m}\cup C_{e+j}^{p^{m}-1,2m}\cup\cdots\cup
C_{(p-2)e+j}^{p^{m}-1,2m}=C_{j}^{e,2m}$ for $0\leq j\leq p-2$. So the $p-1$
cyclotomic classes of order $p^{m}-1$ corresponding to a fixed $j\in D$
partition a single cyclotomic class of order
$e=\frac{p^{m}-1}{p-1}=p^{m-1}+p^{m-2}+\cdots+p+1$. Summarising, we have the
following result:

######  Theorem 3.4.

Let $p$ be an odd prime and $m>1$. Let $D$ be the
$(\frac{p^{m}-1}{p-1},\frac{p^{m-1}-1}{p-1},\frac{p^{m-2}-1}{p-1})$ Singer
difference set in $(\mathbb{Z}_{\frac{p^{m}-1}{p-1}},+)$ defined by the
trace-0 elements of $GF(p^{m})$. Denote by $C_{i}^{e,2m}$ the $i$-th
cyclotomic class of order $e$ in $GF(p^{2m})$. Then

  * (i)

$X=\bigcup_{j\in D}C_{j}^{\frac{p^{m}-1}{p-1},2m}$

is a regular
$(p^{2m},p^{2m-1}-p^{m-1}(p-1)-1,p^{2m-2}-p^{m-1}(p-1)-2,p^{2m-2}-p^{m-1})$-PDS
in $GF(p^{2m})$.

  * (ii)

$X$ is precisely the PDS
$\\{x\in\mathbb{F}_{q}^{2m}\setminus\\{0\\}:Q_{0}(x)=0\\}$ of Theorem 2.5.

As noted at the start of Section 2, this result could also be obtained as a
special case of the “lifting construction” of [12] (using the fact that
$\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle\cong GF(p)\setminus\\{0\\}$).

We end this section with the following technical corollary which will be used
later:

######  Corollary 3.5.

For $0\leq k\leq\frac{p^{m}-1}{p-1}-1$, let $(-k)+D=\\{(-k+i:i\in D\\}$, and
define

$X_{k}=\bigcup_{j\in(-k)+D}C_{j}^{\frac{p^{m}-1}{p-1},2m}.$

Then $X_{k}$ is a a regular
$(p^{2m},p^{2m-1}-p^{m-1}(p-1)-1,p^{2m-2}-p^{m-1}(p-1)-2,p^{2m-2}-p^{m-1})$-PDS
in $GF(p^{2m})$.

###### Proof.

Immediate since $X=\bigcup_{j\in D}\alpha^{j}C_{0}^{\frac{p^{m}-1}{p-1},2m}$
and

$X_{k}=\bigcup_{j\in(-k)+D}\alpha^{j}C_{0}^{\frac{p^{m}-1}{p-1},2m}=\bigcup_{i\in
D}\alpha^{-k+i}C_{0}^{\frac{p^{m}-1}{p-1},2m}=\alpha^{-k}\bigcup_{i\in
D}\alpha^{i}C_{0}^{\frac{p^{m}-1}{p-1},2m}=\alpha^{-k}X.$

Since $X$ is a PDS comprising a union of cyclotomic classes of $GF(p^{2m})$,
then $\alpha^{-k}X$ is a PDS with the same parameters. ∎

##  4\. Character theory

Let $G$ be a finite abelian group. A character on $G$, often denoted $\chi$,
is a homomorphism from $G$ to the group of complex numbers under
multiplication. The fact that $\chi$ is a homomorphism implies that for all
$g\in G,\chi(g)$ must be an $n^{th}$ root of unity for $n$ an integer
satisfying $ng=0_{G}$. The principal character $\chi_{0}$ is the character
that maps all elements of $G$ to $1\in{\mathbb{C}}$; all other characters are
called nonprincipal.

Consider the field $F=GF(p^{m})$, and let $\omega$ be a primitive element
(i.e. a generator of the multiplicative group of $F$). All nonprincipal
(additive) characters of $F$ have the form $\chi_{k}(x)=(e^{2\pi
i/p})^{Tr_{m/1}(\omega^{k}x)}$ for $0\leq k\leq p^{m}-2$. Similarly, if
$K=GF(p^{2m})$ has primitive element $\alpha$, then all nonprincipal
(additive) characters of $K$ will have the form $\psi_{\ell}(x)=(e^{2\pi
i/p})^{Tr(\alpha^{\ell}x)}$ for $0\leq\ell\leq p^{2m}-2$.

Character theory has been one of the primary tools used to investigate PDSs
because of the following theorem (see [10]).

######  Theorem 4.1.

Let $G$ be a finite abelian group of order $v$, and let $D\subset G$ have $k$
elements. Then $D$ is a $(v,k,\lambda,\mu)$ PDS in $G$ if and only if for all
nonprincipal characters $\chi$ on $G$ we have
$\chi(D)\in\\{1/2((\lambda-\mu)\pm\sqrt{(\mu-\lambda)^{2}+4(k-\mu)})\\}$.

When we combine Theorems 2.4 and 3.5, we see that the sets $X_{k}$ defined in
Corollary 3.5 will have nonprincipal character values of either $p^{m-1}-1$ or
$-[(p-1)p^{m-1}+1]$. We will use this fact in the next section to prove our
main construction.

When working with PDSs in abelian groups, Delsarte [7] introduced the dual
PDS. The set of characters $G^{*}:=\\{\chi:G\rightarrow{\mathbb{C}}|\chi\mbox{
is a homomorphism}\\}$ is a group under multiplication, and a simple
computation demonstrates that $G^{*}$ is isomorphic to $G$. If $D$ is a
$(v,k,\lambda,\mu)$ PDS in $G$, we define $D^{*}:=\\{\chi\in
G^{*}:\chi(D)=1/2((\lambda-\mu)+\sqrt{(\mu-\lambda)^{2}+4(k-\mu)})\\}$.

######  Theorem 4.2.

If $D$ is a $(v,k,\lambda,\mu)$ PDS in the abelian group $G$ and
$D^{*}=\\{\chi\in
G^{*}:\chi(D)=1/2((\lambda-\mu)+\sqrt{(\mu-\lambda)^{2}+4(k-\mu)})\\}\subset
G^{*}$, then $D^{*}$ is a $(v,k^{*},\lambda^{*},\mu^{*})$ PDS in $G^{*}$ for

$\displaystyle\beta$ $\displaystyle=$ $\displaystyle\lambda-\mu$
$\displaystyle\gamma$ $\displaystyle=$ $\displaystyle k-\lambda$
$\displaystyle\Delta$ $\displaystyle=$ $\displaystyle\beta^{2}+4\gamma$
$\displaystyle k^{*}$ $\displaystyle=$
$\displaystyle[(\sqrt{\Delta}-\beta)(v-1)-2k]/(2\sqrt{\Delta})$
$\displaystyle\lambda^{*}$ $\displaystyle=$
$\displaystyle(v-2k+\beta-\sqrt{\Delta})/\sqrt{\Delta}+[4k^{*}-v^{2}/\sqrt{\Delta}+(v-2k+\beta-\sqrt{\Delta}/\sqrt{\Delta})^{2}]/4$
$\displaystyle\mu^{*}$ $\displaystyle=$
$\displaystyle[4k^{*}-v^{2}/\Delta+(v-2k+\beta-\sqrt{\Delta})^{2}/(\Delta)]/4.$

In our situation, we want to construct a PDS in ${\mathbb{Z}}_{p}^{3m}$ with
the following parameters

$(p^{3m},(p^{m+1}-p^{m}+p)(p^{m}-1),p^{m}-p+(p^{m+1}-p^{m}+p)(p-2),(p^{m+1}-p^{m}+p)(p-1)).$

This is the Denniston $r=1$ case. If we are able to construct this PDS, then
Theorem 4.2 implies that the dual PDS will be the Denniston $r=m-1$ case,
namely

$(p^{3m},(p^{2m-1}-p^{m}+p^{m-1})(p^{m}-1),p^{m}-p^{m-1}+(p^{2m-1}-p^{m}+p^{m-1})(p^{m-1}-2),$
$(p^{2m-1}-p^{m}+p^{m-1})(p^{m-1}-1)).$

For the $r=1$ Denniston case, the character values from Theorem 4.1 are
$p^{m}-p$ and $-((p-1)(p^{m}+1)+1)$. In the next section we define a set that
will have one of these two values for all nonprincipal characters.

##  5\. Main construction

We now turn our attention to the general case, by which we mean the following:
our group will be $GF(p^{m})^{+}\times
GF(p^{2m})^{+}\cong{\mathbb{Z}}_{p}^{3m}$, and we will use $\omega$ as a
primitive element of $GF(p^{m})$ and $\alpha$ as a primitive element of
$GF(p^{2m})$. We define

${\mathcal{D}}:=\displaystyle\bigcup_{i=0}^{\frac{p^{m}-1}{p-1}-1}(\omega^{i}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle)\times(C_{i}^{\frac{p^{m}-1}{p-1},2m}\cup\\{0_{GF(p^{2m})}\\}).$

The next theorem is the main result in the paper, and it demonstrates that
Denniston PDSs exist in the $p$ odd case for $r=1$.

######  Theorem 5.1.

The set ${\mathcal{D}}$ is a
$(p^{3m},(p^{m}-1)\cdot((p-1)(p^{m}+1)+1),p^{m}-p+(p^{m+1}-p^{m}+p)(p-2),(p^{m+1}-p^{m}+p)(p-1))$
PDS in $GF(p^{m})^{+}\times GF(p^{2m})^{+}$.

###### Proof.

In this proof we will write $C_{i}^{(p^{m}-1)/(p-1),2m}$ as
$C_{i}^{(p^{m}-1)/(p-1)}$ for clarity.
Every character $\Phi$ of our group $GF(p^{m})^{+}\times GF(p^{2m})^{+}$ can
be written as $\Phi=\chi_{k}\times\psi_{\ell}$ for $\chi_{k}$ a character of
$GF(p^{m})^{+}$ and $\psi_{\ell}$ a character of $GF(p^{2m})^{+}$. We break
the proof into three cases based on the characters of the group.

Case 1: $\chi_{k}$ principal, $\psi_{\ell}$ nonprincipal:

In this case,
$\chi_{k}(\omega^{i}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle)=p-1$ for
$0\leq i\leq\frac{p^{m}-1}{p-1}-1$. Thus,

$\displaystyle(\chi_{k}\times\psi_{\ell})({\mathcal{D}})$ $\displaystyle=$
$\displaystyle(p-1)\displaystyle\sum_{i=0}^{\frac{p^{m}-1}{p-1}-1}\psi_{\ell}(C_{i}^{\frac{p^{m}-1}{p-1}-1}\cup\\{0_{GF(p^{2m})}\\})$
$\displaystyle=$
$\displaystyle(p-1)\psi_{\ell}(\sum_{i=0}^{\frac{p^{m}-1}{p-1}-1}C_{i}^{\frac{p^{m}-1}{p-1}-1})+(p-1)\cdot\frac{p^{m}-1}{p-1}$
$\displaystyle=$ $\displaystyle(p-1)(-1)+p^{m}-1=p^{m}-p.$

Case 2: $\chi_{k}$ nonprincipal, $\psi_{\ell}$ principal:

In this case,
$\psi_{\ell}(C_{i}^{\frac{p^{m}-1}{p-1}}\cup\\{0_{GF(p^{2m})}\\})=(p-1)(p^{m}+1)+1$
for $0\leq i\leq\frac{p^{m}-1}{p-1}-1$. Thus,

$\displaystyle(\chi_{k}\times\psi_{\ell})({\mathcal{D}})$ $\displaystyle=$
$\displaystyle\displaystyle((p-1)(p^{m}+1)+1)\sum_{i=0}^{\frac{p^{m}-1}{p-1}-1}\chi_{k}(\omega^{i}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle)$
$\displaystyle=$ $\displaystyle-((p-1)(p^{m}+1)+1)$

Case 3: $\chi_{k}$ nonprincipal, $\psi_{\ell}$ nonprincipal:

The kernel of $\chi_{k}$ is

$Ker(\chi_{k})=\\{x\in GF(p^{m}):Tr(\omega^{k}x)=0_{GF(p^{m})}\\},$

and the size of the kernel is $p^{m-1}$. Now, $x\in Ker(\chi_{k})$ precisely
if $x=0_{GF(p^{m})}$ or $\omega^{k}x\in T$. So, by Lemma 3.3, we have
$Ker(\chi_{k})=\omega^{-k}T\cup\\{0_{GF(p^{m})}\\}=\cup_{i\in
D}\omega^{-k+i}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle\cup\\{0_{GF(p^{m})}\\}$.

Let $\overline{k}\equiv k\mod\frac{p^{m}-1}{p-1}$; then
$Ker(\chi_{k})=\cup_{j\in(-\overline{k})+D}\omega^{j}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle\cup\\{0_{GF(p^{m})}\\}$
where $(-\overline{k})+D=\\{(-\overline{k}+i:i\in D\\}$. A consequence of this
is the following:

$\chi_{k}(\omega^{j}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle)=\begin{cases}p-1&j\in-\overline{k}+D\\\
-1&\mbox{otherwise}\par\end{cases}$

Similarly, the set $\cup_{j\in(-\overline{k})+D}C_{j}^{\frac{p^{m}-1}{p-1}}$
(in other words, the union of classes in $GF(p^{2m})$ corresponding to the
indices of the elements of $Ker(\chi_{k})$) is precisely $X_{\overline{k}}$ as
defined in Corollary 3.5, and combining that with Theorem 2.4 we get

$\psi_{\ell}(X_{\overline{k}})=\begin{cases}-[(p-1)p^{m-1}+1]&\\\
p^{m-1}-1&\end{cases}$

Using the fact that $\psi_{\ell}(GF(p^{2m}))=0$, we then have

$\psi_{\ell}(GF(p^{2m})-X_{\overline{k}}-\\{0_{GF(p^{2m})}\\})=\begin{cases}(p-1)p^{m-1}&\\\
-p^{m-1}&\end{cases}$

Combining these results, we obtain

$(\chi_{k}\times\psi_{\ell})({\mathcal{D}})=\sum_{i=0}^{\frac{p^{m}-1}{p-1}-1}\chi_{k}(\omega^{i}\langle\omega^{\frac{p^{m}-1}{p-1}}\rangle)\psi_{\ell}(C_{i}^{(p^{m}-1)/(p-1)}\cup\\{0_{GF(p^{2m})}\\})$

$=(p-1)(\psi_{\ell}(X_{\overline{k}})+\frac{p^{m-1}-1}{p-1})+(-1)\left(\psi_{\ell}(GF(p^{2m})-X_{\overline{k}}-\\{0_{GF(p^{2m})}\\})+(\frac{p^{m}-1}{p-1}-\frac{p^{m-1}-1}{p-1})\right)$

$=\begin{cases}(p-1)(-[(p-1)p^{m-1}+1]+\frac{p^{m-1}-1}{p-1})+(-1)[(p-1)p^{m-1}+(\frac{p^{m}-1}{p-1}-\frac{p^{m-1}-1}{p-1})]&\\\
(p-1)(p^{m-1}-1+\frac{p^{m-1}-1}{p-1})+(-1)(-p^{m-1}+(\frac{p^{m}-1}{p-1}-\frac{p^{m-1}-1}{p-1}))&\end{cases}$

$=\begin{cases}-p^{m}(p-1)-p&\\\ p^{m}-p&\end{cases}$

Since every nonprincipal character $\chi_{k}\times\psi_{\ell}$ has a sum over
${\mathcal{D}}$ of either $-p^{m}(p-1)-p$ or $p^{m}-p$, Theorem 4.1 implies
that ${\mathcal{D}}$ is a PDS as claimed. ∎

Finally, an application of Theorem 4.2 yields:

######  Theorem 5.2.

The dual of ${\mathcal{D}}$ is a
$(p^{3m},(p^{2m-1}-p^{m}+p^{m-1})(p^{m}-1),p^{m}-p^{m-1}+(p^{2m-1}-p^{m}+p^{m-1})(p^{m-1}-2),(p^{2m-1}-p^{m}+p^{m-1})(p^{m-1}-1))$-PDS
in $GF(p^{m})^{+}\times GF(p^{2m})^{+}$.

##  6\. Future directions

In this paper we have expanded the groups containing PDSs with the Denniston
parameters to include odd prime powers. This opens several possible areas for
further exploration.

  * •

In elementary abelian $p$-groups of order $p^{3m}$ ($p$ an odd prime),
construct PDSs with Denniston parameter sets for $2\leq r\leq m-2$.

  * •

Do there exist PDSs with Denniston parameters in non-elementary abelian
$p$-groups when $p$ is an odd prime? As indicated in the Introduction,
constructions exist in the $p=2$ case.

  * •

If there are new PDSs, examine the graphs to see whether they are isomorphic
to known graphs with those parameters.

  * •

Can we find Denniston-like PDSs in groups that are not prime powers?

  * •

What other conclusions can we draw about unions of cyclotomic classes?

##  7\. Acknowledgements

We thank Ken Smith for assistance with computation. The research of Sophie
Huczynska is supported by EPSRC grant EP/X021157/1. Laura Johnson’s research
visit to the University of Richmond was funded by LMS Early Career Research
Travel Grant ECR-2223-10. John Polhill is grateful for the support from the
Gaines Chair at the University of Richmond.

## References

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  * [15] K.W. Smith  personal communication, 2023.


$\displaystyle\lesssim\frac{\min\left\\{1,s_{\lambda}^{-1}(1+|\xi|)\right\\}^{M}}{\left(1+s_{\lambda}^{-1}|\xi|\right)^{N_{1}}(1+s_{\lambda}^{-\alpha}\left|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi\right|_{[d-1]})^{N_{2}}}.$

∎

The expression on the right hand side of (42) can further be estimated by the
function

$\displaystyle
S_{\lambda,M,N_{1},N_{2}}(\xi):=\frac{\min\left\\{1,s_{\lambda}^{-1}(1+|\xi|)\right\\}^{M}}{(1+s_{\lambda}^{-1}|\xi|)^{N_{1}}(1+s_{\lambda}^{1-\alpha}\left|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}(\xi/|\xi|)\right|_{[d-1]})^{N_{2}}},\quad\xi\in\mathbb{R}^{d}.$
(43)

As already discussed in [28], this function can be separated into angular and
radial components, allowing us to treat these parts independently in the
integration later. Since the notation in this article differs slightly from
the one in [28], we choose to state the lemma once more here, but refer to
said article for a proof.

######  Lemma 6.6.

([28, Lemma 6.4]) Assume that $s_{\lambda}\geq c>0$ for all
$\lambda\in\Lambda$. For every $M,N_{1},N_{2},K\in\mathbb{N}_{0}$ such that
$K\leq N_{2}$ we have with respect to $\lambda\in\Lambda$ and
$\xi\in\mathbb{R}^{d}$ the uniform estimate

$\displaystyle\frac{\min\left\\{1,s_{\lambda}^{-1}(1+|\xi|)\right\\}^{M}}{\left(1+s_{\lambda}^{-1}|\xi|\right)^{N_{1}}(1+s_{\lambda}^{-\alpha}\left|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi\right|_{[d-1]})^{N_{2}}}\lesssim
S_{\lambda,M-K,N_{1},K}(\xi).$

Next, we want to estimate the scalar product of two functions of the form
(43). Before the actual result, Lemma 6.10, we need some preparation. This is
the part of the proof of Theorem 2.5 which differs the most from the situation
in two dimensions.

######  Lemma 6.7.

Let $a\geq a^{\prime}>0$, $d\in\mathbb{N}$, $d\geq 2$, and $N>1$. Then we have
uniformly for $y\in\mathbb{R}$

$\displaystyle\int_{\mathbb{R}}\frac{\left|x\right|^{d-2}dx}{(1+a\left|x\right|)^{N+d-2}(1+a^{\prime}\left|x-y\right|)^{N+d-2}}\lesssim
a^{-(d-1)}(1+a^{\prime}\left|y\right|)^{-N}.$

###### Proof.

Utilizing the following result from Grafakos [22][Appendix K.1]

$\int_{\mathbb{R}}\frac{dx}{(1+a\left|x\right|)^{N}(1+a^{\prime}\left|x-y\right|)^{N}}\lesssim\max\\{a,a^{\prime}\\}^{-1}(1+\min\\{a,a^{\prime}\\}\left|y\right|)^{-N}$

we can estimate

$\displaystyle\int_{\mathbb{R}}$
$\displaystyle\frac{\left|x\right|^{d-2}dx}{(1+a\left|x\right|)^{N+d-2}(1+a^{\prime}\left|x-y\right|)^{N+d-2}}=a^{-(d-2)}\int_{\mathbb{R}}\frac{\left|ax\right|^{d-2}dx}{(1+a\left|x\right|)^{N+d-2}(1+a^{\prime}\left|x-y\right|)^{N+d-2}}$
$\displaystyle\leq
a^{-(d-2)}\int_{\mathbb{R}}\frac{(1+\left|ax\right|)^{d-2}dx}{(1+a\left|x\right|)^{N+d-2}(1+a^{\prime}\left|x-y\right|)^{N+d-2}}=a^{-(d-2)}\int_{\mathbb{R}}\frac{dx}{(1+a\left|x\right|)^{N}(1+a^{\prime}\left|x-y\right|)^{N+d-2}}$
$\displaystyle\lesssim
a^{-(d-2)}\max\\{a,a^{\prime}\\}^{-1}(1+\min\\{a,a^{\prime}\\}\left|y\right|)^{-N}=a^{-(d-1)}(1+a^{\prime}\left|y\right|)^{-N}.$

∎

We can immediately deduce the following corollary.

######  Corollary 6.8.

Let $a\geq a^{\prime}>0$, $d\in\mathbb{N}\backslash\\{1\\}$, and $N>1$. Then
we have uniformly for $\theta_{0}\in\mathbb{R}$

$\displaystyle\int_{0}^{\pi}\frac{|\sin^{d-2}(\theta)|\,d\theta}{(1+a\left|\sin(\theta)\right|)^{N+d-2}(1+a^{\prime}\left|\sin(\theta-\theta_{0})\right|)^{N+d-2}}\lesssim
a^{-(d-1)}\big{(}1+a^{\prime}\left|\left\\{\theta_{0}\right\\}\right|\big{)}^{-N}.$

###### Proof.

Let us call the integral to be estimated $\mathcal{S}$. Since the integrand on
the left hand side is $\pi$-periodic, we may change the domain of integration
to $[-\pi/2,\pi/2]$. Applying Lemma 6.1, we can further conclude

$\displaystyle\mathcal{S}\asymp\int\limits_{-\pi/2}^{\pi/2}\frac{\left|\theta\right|^{d-2}d\theta}{(1+a\left|\theta\right|)^{N+d-2}(1+a^{\prime}\left|\left\\{\theta-\theta_{0}\right\\}\right|)^{N+d-2}},$

Since $|\\{\theta_{0}\\}|\leq\frac{\pi}{2}$ we can estimate

$\displaystyle\mathcal{S}\lesssim\sum_{\vartheta\in\\{-\pi,0,\pi\\}}\int_{\mathbb{R}}\frac{\left|\theta\right|^{d-2}d\theta}{(1+a\left|\theta\right|)^{N+d-2}(1+a^{\prime}\left|\theta-(\\{\theta_{0}\\}+\vartheta)\right|)^{N+d-2}},$

We now use Lemma 6.7 to estimate this by

$\displaystyle\mathcal{S}\lesssim\sum_{\vartheta\in\\{-\pi,0,\pi\\}}a^{-(d-1)}(1+a^{\prime}\left|\\{\theta_{0}\\}+\vartheta\right|)^{-N}\lesssim
a^{-(d-1)}(1+a^{\prime}\left|\left\\{\theta_{0}\right\\}\right|)^{-N}.$

∎

This result is used to estimate the integral of the angular parts of (43) over
the sphere $\mathbb{S}^{d-1}$.

######  Lemma 6.9.

Let $a,a^{\prime}>0$, $d\in\mathbb{N}$, $d\geq 2$,
$\theta_{\lambda},\theta_{\mu}\in[0,\pi]\times[-\frac{\pi}{2},\frac{\pi}{2}]^{d-3}$,
$\varphi_{\lambda},\varphi_{\mu}\in[0,2\pi]$ and $N>1$. Further, let $d\sigma$
denote the standard surface measure on the sphere $\mathbb{S}^{d-1}$. We then
have the estimate

$\displaystyle\int_{\mathbb{S}^{d-1}}\frac{d\sigma(\eta)}{(1+a|R_{\theta_{\mu}}R_{\varphi_{\mu}}\eta|_{[d-1]})^{N+d-2}(1+a^{\prime}\left|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\eta\right|_{[d-1]})^{N+d-2}}$
$\displaystyle\lesssim\max\\{a,a^{\prime}\\}^{-(d-1)}\big{(}1+\min\\{a,a^{\prime}\\}|\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\\}|\big{)}^{-N},$

where $e_{\lambda}=R^{T}_{\varphi_{\lambda}}R^{T}_{\theta_{\lambda}}e_{d}$ and
$e_{\mu}=R^{T}_{\varphi_{\mu}}R^{T}_{\theta_{\mu}}e_{d}$.

###### Proof.

Note the symmetry of the statement with respect to interchanging the entities
$a,a^{\prime}$ and $\lambda,\mu$. Without loss of generality we can therefore
restrict to the case $a\geq a^{\prime}>0$.

Since the mapping $R_{\theta_{\mu}}R_{\varphi_{\mu}}$ is an isometry, the
integral is equal to

$\displaystyle\mathcal{S}:=\int_{\mathbb{S}^{d-1}}\frac{d\sigma(\eta)}{(1+a\left|\eta\right|_{[d-1]})^{N+d-2}(1+a^{\prime}|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}R^{T}_{\varphi_{\mu}}R^{T}_{\theta_{\mu}}\eta|_{[d-1]})^{N+d-2}}.$

For the integration we parameterize the sphere $\mathbb{S}^{d-1}$ by standard
spherical coordinates, i.e. coordinates
$(\theta_{1},\dots\theta_{d-2},\varphi)\in[0,\pi]^{d-2}\times[0,2\pi)$ such
that for $\eta\in\mathbb{S}^{d-1}$

$\displaystyle\eta(\theta,\varphi)=\begin{pmatrix}\sin(\theta_{1})\cdots\cdots\sin(\theta_{d-2})\cos(\varphi)\\\
\sin(\theta_{1})\cdots\cdots\sin(\theta_{d-2})\sin(\varphi)\\\
\sin(\theta_{1})\cdots\sin(\theta_{d-3})\cos(\theta_{d-2})\\\ \vdots\\\
\cos(\theta_{1})\end{pmatrix}.$

Observe that $\langle\eta,e_{d}\rangle=\cos(\theta_{1})$ and thus
$\theta_{1}=d_{\mathbb{S}}(\eta,e_{d})$. Also note
$|\eta|_{[d-1]}=|\sin(\theta_{1})|$. Letting
$\theta_{0}:=d_{\mathbb{S}}(e_{\lambda},e_{\mu})\in[0,\pi]$ denote the angle
between $e_{\lambda}$ and $e_{\mu}$ we have
$\theta_{0}=d_{\mathbb{S}}(e_{d},R_{\theta_{\lambda}}R_{\varphi_{\lambda}}R^{T}_{\varphi_{\mu}}R^{T}_{\theta_{\mu}}e_{d})$.
Since
$R_{\theta_{\lambda}}R_{\varphi_{\lambda}}R^{T}_{\varphi_{\mu}}R^{T}_{\theta_{\mu}}\in\mathcal{O}(d,\mathbb{R})$
we can apply Lemma 6.4 to estimate
$|R_{\theta_{\lambda}}R_{\varphi_{\lambda}}R^{T}_{\varphi_{\mu}}R^{T}_{\theta_{\mu}}\eta|_{[d-1]}$.
We obtain

$\displaystyle\mathcal{S}$
$\displaystyle\leq\int_{0}^{2\pi}\int_{0}^{\pi}\dots\int_{0}^{\pi}\frac{\sin^{d-2}(\theta_{1})\sin^{d-3}(\theta_{2})\dots\sin(\theta_{d-2})d\theta_{1}d\theta_{2}\dots
d\theta_{d-2}d\varphi}{(1+a|\sin(\theta_{1})|)^{N+d-2}(1+a^{\prime}\min\\{\left|\sin(\theta_{1}+\theta_{0})\right|,\left|\sin(\theta_{1}-\theta_{0})\right|\\})^{N+d-2}}$
$\displaystyle\lesssim\int_{0}^{\pi}\frac{\sin^{d-2}(\theta_{1})\,d\theta_{1}}{(1+a|\sin(\theta_{1})|)^{N+d-2}(1+a^{\prime}\min\\{\left|\sin(\theta_{1}+\theta_{0})\right|,\left|\sin(\theta_{1}-\theta_{0})\right|\\})^{N+d-2}}$
$\displaystyle\leq\sum_{\epsilon\in\\{-1,1\\}}\int_{0}^{\pi}\frac{|\sin(\theta_{1})|^{d-2}\,d\theta_{1}}{(1+a|\sin(\theta_{1})|)^{N+d-2}(1+a^{\prime}\left|\sin(\theta_{1}-\epsilon\theta_{0})\right|)^{N+d-2}}.$

Using Corollary 6.8 we finally arrive at
$\mathcal{S}\lesssim\max\\{a,a^{\prime}\\}^{-(d-1)}\big{(}1+\min\\{a,a^{\prime}\\}\left|\left\\{\theta_{0}\right\\}\right|\big{)}^{-N}$.
∎

With this estimate for the angular components in our toolbox, we proceed to
prove the main result concerning the correlation of functions of the form
(43).

######  Lemma 6.10.

Let $\alpha\in[0,1]$, $d\in\mathbb{N}$, $d\geq 2$, and
$M,N_{1},N_{2}\in\mathbb{N}_{0}$. Further, let $(\Lambda,\Phi_{\Lambda})$ and
$(\Delta,\Phi_{\Delta})$ be parametrizations with
$(s_{\lambda},e_{\lambda},x_{\lambda})=\Phi_{\Lambda}(\lambda)$ and
$(s_{\mu},e_{\mu},x_{\mu})=\Phi_{\Delta}(\mu)$ for $\lambda\in\Lambda$,
$\mu\in\Delta$, such that $c\leq s_{\lambda}$ and $c\leq s_{\mu}$ for a fixed
constant $c>0$. Then for $A>0$ and $B>1$ satisfying

$\displaystyle N_{1}>\frac{d}{2},\quad M+d>N_{1}\geq
A+\frac{1+\alpha(d-1)}{2},\quad\text{and}\quad N_{2}\geq B+d-2$

the following estimate holds true with an implicit constant independent of
$\lambda\in\Lambda$ and $\mu\in\Delta$,

$\displaystyle(s_{\lambda}s_{\mu})^{-\frac{1+\alpha(d-1)}{2}}\int\limits_{\mathbb{R}^{d}}S_{\lambda,M,N_{1},N_{2}}(x)S_{\mu,M,N_{1},N_{2}}(x)\,dx\lesssim\max\left\\{\frac{s_{\lambda}}{s_{\mu}},\frac{s_{\mu}}{s_{\lambda}}\right\\}^{-A}\left(1+\min\\{s_{\lambda},s_{\mu}\\}^{1-\alpha}|\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\\}|\right)^{-B}.$

###### Proof.

Without loss of generality we subsequently assume $s_{\lambda}\leq s_{\mu}$.
The strategy is to separate the integration into an angular and a radial part
and estimate these independently. For the estimate of the angular part we can
use Lemma 6.9, which yields

$\displaystyle(s_{\lambda}s_{\mu})^{-\frac{(1+\alpha(d-1)}{2}}\int_{0}^{\infty}\int_{\mathbb{S}^{d-1}}S_{\lambda,M,N_{1},N_{2}}(\eta,r)S_{\mu,M,N_{1},N_{2}}(\eta,r)r^{d-1}\,d\sigma(\eta)dr$
$\displaystyle\lesssim(s_{\lambda}s_{\mu})^{-\frac{1+\alpha(d-1)}{2}}s_{\mu}^{-(1-\alpha)(d-1)}s_{\mu}^{d}\big{(}1+s_{\lambda}^{1-\alpha}|\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\\}|\big{)}^{-B}\cdot\,\mathcal{S}$

with a remaining radial integral

$\mathcal{S}:=s_{\mu}^{-d}\int_{0}^{\infty}\frac{\min\left\\{1,s_{\lambda}^{-1}(1+r)\right\\}^{M}}{(1+s_{\lambda}^{-1}r)^{N_{1}}}\frac{\min\left\\{1,s_{\mu}^{-1}(1+r)\right\\}^{M}}{(1+s_{\mu}^{-1}r)^{N_{1}}}r^{d-1}\,dr.$

Note that for the estimate we used the assumptions $s_{\lambda}\leq s_{\mu}$,
$B>1$ and $N_{2}\geq B+d-2$. It remains to verify the relation
$(s_{\mu}s_{\lambda})^{-(1+\alpha(d-1))/2}s_{\mu}^{-(1-\alpha)(d-1)}s_{\mu}^{d}\cdot\mathcal{S}\lesssim(s_{\mu}/s_{\lambda})^{-A}$,
or equivalently

$\displaystyle\mathcal{S}\lesssim\Big{(}\frac{s_{\mu}}{s_{\lambda}}\Big{)}^{-A-\frac{1+\alpha(d-1)}{2}}.$

To prove this, we split the integration of $\mathcal{S}$ into three parts
$\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{3}$ corresponding to the
integration ranges $0\leq r\leq 1$, $1\leq r\leq s_{\mu}$, and $s_{\mu}\leq r$
respectively.

#####  $0\leq r\leq 1$:

Here we estimate $\min\left\\{1,s_{\lambda}^{-1}(1+r)\right\\}^{M}\leq
s_{\lambda}^{-M}(1+r)^{M}\leq 2^{M}s_{\lambda}^{-M}$ and
$(1+s_{\lambda}^{-1}r)^{N_{1}}\geq 1$, and similarly for the index $\mu$.
Hence, the integral over this part can be estimated by

$\displaystyle\mathcal{S}_{1}\lesssim
s_{\mu}^{-d}s_{\lambda}^{-M}s_{\mu}^{-M}\int_{0}^{1}r^{d-1}\,dr\asymp
s_{\mu}^{-(M+d)}s_{\lambda}^{-M}\lesssim\left(\frac{s_{\mu}}{s_{\lambda}}\right)^{-(M+d)},$

where the last inequality holds because of the uniform lower bound $0<c\leq
s_{\lambda}$ for $\lambda\in\Lambda$. Finally observe that the assumed
inequalities imply $M+d>A+\frac{1+\alpha(d-1)}{2}$.

##### $1\leq r\leq s_{\mu}$

We estimate the terms involving $\mu$ as follows:
$(1+s_{\mu}^{-1}r)^{N_{1}}\geq 1$ and $(r+1)\leq 2r$. Hence

$\displaystyle\min\left\\{1,s_{\mu}^{-1}(1+r)\right\\}^{M}\leq
s_{\mu}^{-M}(1+r)^{M}\leq s_{\mu}^{-M}(r+r)^{M}\leq 2^{M}s_{\mu}^{-M}r^{M}.$

For the terms with $\lambda$’s, we have $(1+s_{\lambda}^{-1}r)^{N_{1}}\geq
s_{\lambda}^{-N_{1}}r^{N_{1}}$ and
$\min\left\\{1,s_{\lambda}^{-1}(1+r)\right\\}^{M}\leq 1$. The integral
$\mathcal{S}_{2}$ hence satisfies

$\displaystyle\mathcal{S}_{2}\lesssim
s_{\mu}^{-d}s_{\lambda}^{N_{1}}s_{\mu}^{-M}\int_{1}^{s_{\mu}}r^{M-N_{1}+d-1}\,dr\lesssim
s_{\mu}^{-(M+d)}s_{\lambda}^{N_{1}}s_{\mu}^{M+d-N_{1}}=\Big{(}\frac{s_{\mu}}{s_{\lambda}}\Big{)}^{-N_{1}},$

where it was used that $M+d>N_{1}$, which implies $M+d-N_{1}-1>-1$, for the
integration. By assumption $N_{1}\geq A+\frac{1+\alpha(d-1)}{2}$, giving the
desired result.

##### $s_{\mu}\leq r$

We estimate both terms like the $\lambda$-terms above to obtain

$\displaystyle\mathcal{S}_{3}\lesssim
s_{\mu}^{-d}s_{\lambda}^{N_{1}}s_{\mu}^{N_{1}}\int_{s_{\mu}}^{\infty}r^{d-1-2N_{1}}\,dr\lesssim
s_{\mu}^{N_{1}-d}s_{\lambda}^{N_{1}}s_{\mu}^{d-2N_{1}}\lesssim\Big{(}\frac{s_{\mu}}{s_{\lambda}}\Big{)}^{-N_{1}}.$

The integral converges since $N_{1}>\frac{d}{2}$. Since $N_{1}\geq
A+\frac{1+\alpha(d-1)}{2}$ the proof is finished. ∎

###  6.3 Cancellation Estimate

Theorem 2.5 provides estimates for the scalar products of $\alpha$-molecules.
To derive them we evaluate these scalar products on the Fourier side, where we
can take advantage of cancellation phenomena. Technically, the method is based
on a clever integration by parts involving the following differential
operator, depending on $\lambda\in\Lambda$, $\mu\in\Delta$,

$\displaystyle\mathscr{L}_{\lambda,\mu}:=\mathcal{I}-s_{0}^{2\alpha}\Delta-\frac{s_{0}^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\langle
e_{\lambda},\nabla\rangle^{2},$ (44)

where $s_{0}=\min\\{s_{\lambda},s_{\mu}\\}$, $\mathcal{I}$ is the identity
operator, $\nabla$ the gradient and $\Delta$ the standard Laplacian.

Lemma 6.11 shows how $\mathscr{L}_{\lambda,\mu}$ acts on products of functions
$a_{\lambda}$, $b_{\mu}$ which satisfy (7).

######  Lemma 6.11.

Let $a_{\lambda}$ and $b_{\mu}$ satisfy (7) for every multi-index
$\rho\in\mathbb{N}_{0}^{d}$ with $|\rho|_{1}\leq L$ and assume
$s_{\lambda},s_{\mu}\geq c>0$. Then we can write the expression

$\displaystyle\mathscr{L}_{\lambda,\mu}\left(a_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)b_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)\right)$

as a finite linear combination of terms of the form

$\displaystyle
p_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)q_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)$

with functions $p_{\lambda},q_{\mu}$, which satisfy (7) for all multi indices
$\rho\in\mathbb{N}_{0}^{d}$ with $|\rho|_{1}\leq L-2$.

###### Proof.

For convenience we introduce the operators
$O_{\lambda}:=A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}$
and $O_{\mu}:=A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}$.
Further, we define the functions
$\widetilde{a}_{\lambda}(\xi):=a_{\lambda}(O_{\lambda}\xi)$ and
$\widetilde{b}_{\mu}(\xi):=b_{\mu}(O_{\mu}\xi)$. We also abbreviate
$\xi_{\lambda}:=O_{\lambda}\xi$ and $\xi_{\mu}:=O_{\mu}\xi$. Taking into
account $s_{\lambda}\gtrsim 1$, we observe $\|O_{\lambda}\|_{2\rightarrow
2}=\|A^{-1}_{\alpha,s_{\lambda}}\|_{2\rightarrow
2}=\max\\{s_{\lambda}^{-\alpha},s_{\lambda}^{-1}\\}\lesssim
s_{\lambda}^{-\alpha}.$ Analogously, it holds $\|O_{\mu}\|_{2\rightarrow
2}=\|A^{-1}_{\alpha,s_{\mu}}\|_{2\rightarrow 2}\lesssim s_{\mu}^{-\alpha}$.
Finally, we introduce the ‘transfer’ matrix

$\displaystyle
T_{\lambda,\mu}:=R_{\theta_{\mu}}R_{\varphi_{\mu}}R^{T}_{\varphi_{\lambda}}R^{T}_{\theta_{\lambda}}\in\mathcal{O}(d,\mathbb{R}).$
(45)

After these remarks we turn to the proof, where we treat the components of
$\mathscr{L}_{\lambda,\mu}$ separately.

##### $\mathcal{I}$

This term causes no pain.

##### $s_{0}^{2\alpha}\Delta$

By the product rule we have

$\displaystyle\Delta(\widetilde{a}_{\lambda}\widetilde{b}_{\mu})=\underbrace{2\langle\nabla\widetilde{a}_{\lambda},\nabla\widetilde{b}_{\mu}\rangle}_{\bf
A}+\underbrace{\widetilde{a}_{\lambda}\Delta\widetilde{b}_{\mu}+\widetilde{b}_{\mu}\Delta\widetilde{a}_{\lambda}}_{\bf
B}.$

In the following we first treat part A and then part B.

##### A

The chain rule yields
$\nabla\widetilde{a}_{\lambda}(\xi)=O^{T}_{\lambda}\nabla
a_{\lambda}(\xi_{\lambda})$ for every $\xi\in\mathbb{R}^{d}$ and an analogous
formula for $\widetilde{b}_{\mu}$. Thus we obtain

$\big{\langle}\nabla\widetilde{a}_{\lambda}(\xi),\nabla\widetilde{b}_{\mu}(\xi)\big{\rangle}=\big{\langle}O^{T}_{\lambda}\nabla
a_{\lambda}(\xi_{\lambda}),O^{T}_{\mu}\nabla
b_{\mu}(\xi_{\mu})\big{\rangle}=\big{\langle}O_{\mu}O^{T}_{\lambda}\nabla
a_{\lambda}(\xi_{\lambda}),\nabla b_{\mu}(\xi_{\mu})\big{\rangle}.$

The expression $\big{\langle}O_{\mu}O^{T}_{\lambda}\nabla a_{\lambda},\nabla
b_{\mu}\big{\rangle}$ is a linear combination of the products
$\partial_{i}a_{\lambda}\partial_{j}b_{\mu}$, where $i,j\in\\{1,\ldots,d\\}$,
with the entries of the matrix $O_{\mu}O^{T}_{\lambda}$ as coefficients. The
functions $\partial_{i}a_{\lambda}$ and $\partial_{j}b_{\mu}$ clearly satisfy
(7) for every $\rho\in\mathbb{N}_{0}^{d}$ with $|\rho|_{1}\leq L-1$. Moreover,
the entries of the matrix $O_{\mu}O^{T}_{\lambda}$ are bounded in modulus by
$\|O_{\mu}O^{T}_{\lambda}\|_{2\rightarrow 2}$, which in turn obeys the
estimate

$\|O_{\mu}O^{T}_{\lambda}\|_{2\rightarrow
2}=\|A^{-1}_{\alpha,s_{\mu}}T_{\lambda,\mu}A^{-1}_{\alpha,s_{\lambda}}\|_{2\rightarrow
2}\leq\|A^{-1}_{\alpha,s_{\mu}}\|_{2\rightarrow
2}\|A^{-1}_{\alpha,s_{\lambda}}\|_{2\rightarrow
2}\lesssim(s_{\mu}s_{\lambda})^{-\alpha}\leq s_{0}^{-2\alpha},$

where $s_{0}=\min\\{s_{\lambda},s_{\mu}\\}$. This shows that the function
$s_{0}^{2\alpha}{\bf A}$ can be written as claimed.

##### B

Due to symmetry it suffices to treat the term
$\widetilde{b}_{\mu}\Delta\widetilde{a}_{\lambda}$. Since
$\widetilde{b}_{\mu}(\xi)=b_{\mu}(\xi_{\mu})$ for $\xi\in\mathbb{R}^{d}$ and
since $b_{\mu}$ fulfills condition (7) for every $\rho\in\mathbb{N}_{0}^{d}$
with $|\rho|_{1}\leq L$, the function $b_{\mu}$ is a suitable first factor
with the required properties. Let us investigate the second factor
$\Delta\widetilde{a}_{\lambda}$.

The second derivative of $\widetilde{a}_{\lambda}$ is at each
$\xi\in\mathbb{R}^{d}$ a bilinear mapping
$\mathbb{R}^{d}\times\mathbb{R}^{d}\to\mathbb{R}$, which by the chain rule
satisfies for $v,w\in\mathbb{R}^{d}$

$\displaystyle\widetilde{a}^{\>\prime\prime}_{\lambda}(\xi)[v,w]=a^{\prime\prime}_{\lambda}(\xi_{\lambda})[O_{\lambda}v,O_{\lambda}w].$

Thus, we have the expansion

$\displaystyle\Delta\widetilde{a}_{\lambda}(\xi)=\sum_{i=1}^{d}\widetilde{a}^{\prime\prime}_{\lambda}(\xi)[e_{i},e_{i}]=\sum_{i=1}^{d}a^{\prime\prime}_{\lambda}(\xi_{\lambda})[O_{\lambda}e_{i},O_{\lambda}e_{i}].$

Let $\rho\in\mathbb{N}_{0}^{d}$ be a multi-index with $|\rho|_{1}\leq L-2$.
Then the partial derivative with respect to $\rho$ of the function
$\xi\mapsto\sum_{i=1}^{d}s_{0}^{2\alpha}a^{\prime\prime}_{\lambda}(\xi)[O_{\lambda}e_{i},O_{\lambda}e_{i}]$
clearly exists. It remains to prove the frequency localization (7).

In view of
$\partial^{\rho}(a_{\lambda}^{\prime\prime})=(\partial^{\rho}a_{\lambda})^{\prime\prime}$
we can estimate for every $i\in\\{1,\ldots,d\\}$ and every
$\xi\in\mathbb{R}^{d}$

$\displaystyle
s^{2\alpha}_{0}|\partial^{\rho}a^{\prime\prime}_{\lambda}(\xi)[O_{\lambda}e_{i},O_{\lambda}e_{i}]|\leq
s^{2\alpha}_{0}{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\partial^{\rho}a_{\lambda}^{\prime\prime}(\xi)\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}\|O_{\lambda}\|_{2\rightarrow
2}^{2}\lesssim{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\partial^{\rho}a_{\lambda}^{\prime\prime}(\xi)\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}.$

The norm of the bilinear mapping is given by
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\partial^{\rho}a_{\lambda}^{\prime\prime}(\xi)\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}=\sup_{|v|,|w|=1}|\partial^{\rho}a^{\prime\prime}_{\lambda}(\xi)[v,w]|$.
This is equal to the spectral norm of the corresponding Hesse matrix.
Therefore we can deduce
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\partial^{\rho}a_{\lambda}^{\prime\prime}(\xi)\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}\lesssim\sup_{|\beta|_{1}=2}\left|\partial^{\beta}\partial^{\rho}a_{\lambda}(\xi)\right|$.
The functions $\partial^{\beta}\partial^{\rho}a_{\lambda}$ satisfy (7) for
every $\beta\in\mathbb{N}_{0}^{d}$ with $|\beta|_{1}=2$ due to the assumption
on $a_{\lambda}$. The required frequency localization follows.

#####
$s_{0}^{2}(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2})^{-1}\langle
e_{\lambda},\nabla\rangle^{2}$

First we put $w_{1}:=s_{0}^{2}$,
$w_{2}:=s_{0}^{2\alpha}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{-2}$,
and
$w_{3}:=s_{0}^{1+\alpha}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{-1}$
and notice that the pre-factor satisfies

$\displaystyle
s_{0}^{2}(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2})^{-1}\leq\min\\{w_{1},w_{2},w_{3}\\}.$
(46)

The first two estimates are obvious. For the third, recall that $1+t^{2}\geq
2t$ for all $t\in\mathbb{R}$. Hence,

$\displaystyle
s_{0}^{2}(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2})^{-1}\leq\textstyle{\frac{1}{2}}s_{0}^{2}(s_{0}^{1-\alpha}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|)^{-1}\leq
s_{0}^{1+\alpha}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{-1}.$

We begin with the product rule, which yields

$\displaystyle\langle
e_{\lambda},\nabla\rangle^{2}\big{(}\widetilde{a}_{\lambda}\widetilde{b}_{\mu}\big{)}=\widetilde{b}_{\mu}\langle
e_{\lambda},\nabla\rangle^{2}\widetilde{a}_{\lambda}+2(\langle
e_{\lambda},\nabla\rangle\widetilde{a}_{\lambda})(\langle
e_{\lambda},\nabla\rangle\widetilde{b}_{\mu})+\widetilde{a}_{\lambda}\langle
e_{\lambda},\nabla\rangle^{2}\widetilde{b}_{\mu}.$ (47)

Recall that
$e_{\lambda}=R^{T}_{\varphi_{\lambda}}R^{T}_{\theta_{\lambda}}e_{d}$. We
calculate with the chain rule for $\xi\in\mathbb{R}^{d}$

$\displaystyle\big{\langle}e_{\lambda},\nabla\widetilde{a}_{\lambda}(\xi)\big{\rangle}=\big{\langle}O_{\lambda}e_{\lambda},\nabla
a_{\lambda}(\xi_{\lambda})\big{\rangle}=\big{\langle}A^{-1}_{\alpha,s_{\lambda}}e_{d},\nabla
a_{\lambda}(\xi_{\lambda})\big{\rangle}=s_{\lambda}^{-1}\partial_{d}a_{\lambda}(\xi_{\lambda}),$

where we used $O_{\lambda}e_{\lambda}=A^{-1}_{\alpha,s_{\lambda}}e_{d}$. Using
the ‘transfer’ matrix $T_{\lambda,\mu}$ from (45), we similarly obtain

$\displaystyle\big{\langle}e_{\lambda},\nabla\widetilde{b}_{\mu}(\xi)\big{\rangle}=\big{\langle}O_{\mu}e_{\lambda},\nabla
b_{\mu}(\xi_{\mu})\big{\rangle}=\big{\langle}A^{-1}_{\alpha,s_{\mu}}T_{\lambda,\mu}e_{d},\nabla
b_{\mu}(\xi_{\mu})\big{\rangle}.$

Next, we note that $\langle
e_{\lambda},\nabla\rangle^{2}\widetilde{a}_{\lambda}(\xi)=\widetilde{a}_{\lambda}^{\prime\prime}(\xi)[e_{\lambda},e_{\lambda}]$.
Together with the chain rule, this implies

$\displaystyle\langle
e_{\lambda},\nabla\rangle^{2}\widetilde{a}_{\lambda}(\xi)=a_{\lambda}^{\prime\prime}(\xi_{\lambda})[O_{\lambda}e_{\lambda},O_{\lambda}e_{\lambda}]=a^{\prime\prime}_{\lambda}(\xi_{\lambda})[A^{-1}_{\alpha,s_{\lambda}}e_{d},A^{-1}_{\alpha,s_{\lambda}}e_{d}]=s^{-2}_{\lambda}\partial^{2}_{d}a_{\lambda}(\xi_{\lambda}).$

We also obtain

$\displaystyle\langle
e_{\lambda},\nabla\rangle^{2}\widetilde{b}_{\mu}(\xi)=b^{\prime\prime}_{\mu}(\xi_{\mu})[O_{\mu}e_{\lambda},O_{\mu}e_{\lambda}]=\big{(}\langle
A^{-1}_{\alpha,s_{\mu}}T_{\lambda,\mu}e_{d},\nabla\rangle^{2}b_{\mu}\big{)}(\xi_{\mu}).$

Let us henceforth use the abbreviation
$\eta:=T_{\lambda,\mu}e_{d}\in\mathbb{S}^{d-1}$. Plugging the above
calculations into (47) leads to the following expression for $\langle
e_{\lambda},\nabla\rangle^{2}\big{(}\widetilde{a}_{\lambda}\widetilde{b}_{\mu}\big{)}(\xi)$
at $\xi\in\mathbb{R}^{d}$

$\displaystyle
s_{\lambda}^{-2}b_{\mu}(\xi_{\mu})\cdot\partial_{d}^{2}a_{\lambda}(\xi_{\lambda})+2s_{\lambda}^{-1}\partial_{d}a_{\lambda}(\xi_{\lambda})\cdot\langle
A^{-1}_{\alpha,s_{\mu}}\eta,\nabla
b_{\mu}(\xi_{\mu})\rangle+a_{\lambda}(\xi_{\lambda})\cdot\big{(}\langle
A^{-1}_{\alpha,s_{\mu}}\eta,\nabla\rangle^{2}b_{\mu}\big{)}(\xi_{\mu}).$ (48)

For the first summand of (48) we consider the product of the functions
$s^{-2}_{\lambda}\partial^{2}_{d}a_{\lambda}$ and $b_{\mu}$. Since
$s_{\lambda}^{-2}\leq s_{0}^{-2}$ and in view of (46) the pre-factor $w_{1}$
is compensated. Due to the assumptions on $a_{\lambda}$ and $b_{\mu}$ the
product is thus of the desired form.

Let us put $\eta_{[d-1]}:=(\eta_{1},\ldots,\eta_{d-1},0)^{T}\in\mathbb{R}^{d}$
and $\eta_{[d]}:=(0,\ldots,0,\eta_{d})^{T}\in\mathbb{R}^{d}$ and observe that

$A^{-1}_{\alpha,s_{\mu}}\eta=A^{-1}_{\alpha,s_{\mu}}(\eta_{[d-1]}+\eta_{[d]})=s_{\mu}^{-\alpha}\eta_{[d-1]}+s_{\mu}^{-1}\eta_{[d]}.$

The second summand of (48) then becomes – up to the factor $2$ –

$\partial_{d}a_{\lambda}(\xi_{\lambda})\cdot\big{(}s_{\lambda}^{-1}s_{\mu}^{-\alpha}\langle\eta_{[d-1]},\nabla
b_{\mu}(\xi_{\mu})\rangle+s_{\lambda}^{-1}s_{\mu}^{-1}\eta_{d}\partial_{d}b_{\mu}(\xi_{\mu})\big{)}.$

We choose the function $\partial_{d}a_{\lambda}$ as the first factor, which
clearly has the required properties, and the function

$\displaystyle\xi\mapsto
s_{\lambda}^{-1}s_{\mu}^{-\alpha}\langle\eta_{[d-1]},\nabla
b_{\mu}(\xi)\rangle+s^{-1}_{\lambda}s_{\mu}^{-1}\eta_{d}\partial_{d}b_{\mu}(\xi).$

as the second factor. The second component of this function causes no problems
because $|\eta_{d}|\leq 1$ and the pre-factor $w_{1}$ is compensated due to
$(s_{\lambda}s_{\mu})^{-1}\leq s^{-2}_{0}$. To deal with the other term,
notice that by Lemma 6.2
$|\eta_{[d-1]}|=|\eta|_{[d-1]}\asymp\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|$.
Thus

$s^{-1}_{\lambda}s^{-\alpha}_{\mu}|\langle\eta_{[d-1]},\nabla
b_{\mu}\rangle|{\lesssim}s^{-1}_{\lambda}s^{-\alpha}_{\mu}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right||\nabla
b_{\mu}|.$

The fact that $\partial_{i}b_{\mu}$, $i\in\\{1,\ldots,d\\}$, satisfy (7) by
assumption, and that
$s^{-1}_{\lambda}s^{-\alpha}_{\mu}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|$
compensates $w_{3}$, implies that also the first component satisfies the
required properties.

Let us turn to the last summand of (48). The first factor $a_{\lambda}$ is of
the desired form. For the second factor we expand the function $\langle
A^{-1}_{\alpha,s_{\mu}}\eta,\nabla\rangle^{2}b_{\mu}$ in the form

$s^{-2\alpha}_{\mu}\langle\eta_{[d-1]},\nabla\rangle^{2}b_{\mu}+2s^{-1-\alpha}_{\mu}\eta_{d}\langle\eta_{[d-1]},\nabla\rangle\partial_{d}b_{\mu}+s^{-2}_{\mu}\eta^{2}_{d}\partial^{2}_{d}b_{\mu}.$

Its partial derivatives of order $\rho\in\mathbb{N}_{0}^{d}$ with
$|\rho|_{1}\leq L-2$ clearly exist, and we get the estimate

$|\langle
A^{-1}_{\alpha,s_{\mu}}\eta,\nabla\rangle^{2}\partial^{\rho}b_{\mu}|\lesssim
s_{0}^{-2\alpha}|d_{\mathbb{S}}(e_{\lambda},e_{\mu})|^{2}\sum_{i,j=1}^{d-1}|\partial_{i}\partial_{j}\partial^{\rho}b_{\mu}|+2s_{0}^{-1-\alpha}|d_{\mathbb{S}}(e_{\lambda},e_{\mu})||\nabla\partial_{d}\partial^{\rho}b_{\mu}|+s_{0}^{-2}|\partial^{2}_{d}\partial^{\rho}b_{\mu}|.$

Here we again used that
$|\eta_{[d-1]}|\asymp\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|$
according to Lemma 6.2. This estimate completes the proof, taking into account
the estimate (46) of the pre-factor and the fact that the partial derivatives
of $b_{\mu}$ up to order $L$ satisfy (7). ∎

###  6.4 Proof of Theorem 2.5

At last we have all the tools available to prove Theorem 2.5. Write $\Delta
x=x_{\mu}-x_{\lambda}$. An application of the Plancherel identity yields

$\displaystyle\langle m_{\lambda},p_{\mu}\rangle$
$\displaystyle=\langle\hat{m}_{\lambda},\hat{p}_{\mu}\rangle$
$\displaystyle=(s_{\lambda}s_{\mu})^{-\frac{\alpha(d-1)+1}{2}}\int_{\mathbb{R}^{d}}\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\exp(2\pi
i\langle\xi,\Delta x\rangle)\,d\xi$

for two $\alpha$-molecules $m_{\lambda}$ and $p_{\mu}$ with respective
generators $a_{\lambda}$ and $b_{\mu}$. According to Lemma 6.5 the Fourier
transforms of the generators therefore satisfy (42) for every
$\rho\in\mathbb{N}_{0}^{d}$ with $|\rho|_{1}\leq L$

Next, we want to exploit cancellation. For this we utilize the differential
operator $\mathscr{L}_{\lambda,\mu}$ from (44). First, we observe that partial
integration yields

$\displaystyle\Big{\langle}\mathscr{L}_{\lambda,\mu}^{N}$
$\displaystyle\exp(2\pi i\langle\xi,\Delta
x\rangle),\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\Big{\rangle}$
$\displaystyle=\Big{\langle}\exp(2\pi i\langle\xi,\Delta
x\rangle),\mathscr{L}_{\lambda,\mu}^{N}\big{(}\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\big{)}\Big{\rangle},$

since the boundary terms vanish due to the decay properties (7) of the
generators and its derivatives. Note that we assume $N_{1}>d/2$ and $L\geq
2N$. Second, we calculate for $\xi\in\mathbb{R}^{d}$

$\displaystyle\mathscr{L}_{\lambda,\mu}^{N}\big{(}\exp(2\pi i\langle\xi,\Delta
x\rangle)\big{)}=\left(1+4\pi^{2}s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+\frac{4\pi^{2}s_{0}^{2}\langle e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\right)^{N}\exp(2\pi
i\langle\xi,\Delta x\rangle).$

Consequently, we have

$\displaystyle\langle
m_{\lambda},p_{\mu}\rangle=\left(1+4\pi^{2}s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+\frac{4\pi^{2}s_{0}^{2}\langle e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\right)^{-N}\cdot\,\mathcal{S}_{\lambda,\mu},$

with

$\displaystyle\mathcal{S}_{\lambda,\mu}:=(s_{\lambda}s_{\mu})^{-\frac{\alpha(d-1)+1}{2}}\int\limits_{\mathbb{R}^{d}}\mathscr{L}_{\lambda,\mu}^{N}\big{(}\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\big{)}\exp(2\pi
i\langle\xi,\Delta x\rangle)\,d\xi.$

Since $L\geq 2N$ by assumption, Lemma 6.11 can iteratively be applied $N$
times, and we conclude that

$\displaystyle\mathscr{L}_{\lambda,\mu}^{N}\big{(}\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\big{)}$

can be written as a finite linear combination of terms of the form

$p_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)q_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)),$

where $p_{\lambda}$ and $q_{\mu}$ satisfy (7) (for the multi-index just
containing zeros).

Using Lemma 6.5 and putting $K=2N+d-2\leq N_{2}$ in Lemma 6.6 then yields

$\displaystyle\big{|}\mathscr{L}_{\lambda,\mu}^{N}\big{(}\hat{a}_{\lambda}(A^{-1}_{\alpha,s_{\lambda}}R_{\theta_{\lambda}}R_{\varphi_{\lambda}}\xi)$
$\displaystyle\overline{\hat{b}_{\mu}(A^{-1}_{\alpha,s_{\mu}}R_{\theta_{\mu}}R_{\varphi_{\mu}}\xi)}\big{)}\big{|}$
$\displaystyle\lesssim
S_{\lambda,M-(2N+d-2),N_{1},2N+d-2}(\xi)S_{\mu,M-(2N+d-2),N_{1},2N+d-2}(\xi).$

Due to the assumptions, we can further choose a number $\widetilde{N}\leq
N_{1}$ which satisfies

$\displaystyle(M-(2N+d-2))+d>\widetilde{N}\geq N+\frac{1+\alpha(d-1)}{2}.$
(49)

Since $\widetilde{N}\leq N_{1}$ we have the estimate
$S_{\eta,M-(2N+d-2),N_{1},2N+d-2}\leq
S_{\eta,M-(2N+d-2),\widetilde{N},2N+d-2}$ for $\eta=\lambda$, $\mu$. Hence, we
obtain

$\displaystyle|\mathcal{S}_{\lambda,\mu}|$
$\displaystyle\lesssim(s_{\lambda}s_{\mu})^{-\frac{\alpha(d-1)+1}{2}}\int_{\mathbb{R}^{d}}S_{\lambda,M-(2N+d-2),N_{1},2N+d-2)}(\xi)S_{\mu,M-(2N+d-2),N_{1},2N+d-2}(\xi)\,d\xi$
$\displaystyle\lesssim(s_{\lambda}s_{\mu})^{-\frac{\alpha(d-1)+1}{2}}\int_{\mathbb{R}^{d}}S_{\lambda,M-(2N+d-2),\widetilde{N},2N+d-2}(\xi)S_{\mu,M-(2N+d-2),\widetilde{N},2N+d-2}(\xi)\,d\xi$
$\displaystyle\lesssim\max\left\\{\frac{s_{\lambda}}{s_{\mu}},\frac{s_{\mu}}{s_{\lambda}}\right\\}^{-N}(1+s_{0}^{(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|)^{-2N}.$

Here we used (49) and Lemma 6.10 in the last line (using this $S$ and setting
$\tilde{M}=M-(2N+d-2)$, $A=N$ and $B=2N$ ($B>1$, $A>0$ since $N>1$)).

Altogether, we arrive at the desired estimate

$\displaystyle|\langle m_{\lambda},p_{\mu}\rangle|$
$\displaystyle\lesssim\max\left\\{\frac{s_{\lambda}}{s_{\mu}},\frac{s_{\mu}}{s_{\lambda}}\right\\}^{-N}\left(1+s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+\frac{s_{0}^{2}\langle e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\right)^{-N}\big{(}1+s_{0}^{(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|\big{)}^{-2N}$
$\displaystyle\lesssim\max\left\\{\frac{s_{\lambda}}{s_{\mu}},\frac{s_{\mu}}{s_{\lambda}}\right\\}^{-N}\left(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}+s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+\frac{s_{0}^{2}\langle e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\right)^{-N}$
$\displaystyle\lesssim\omega_{\alpha}(\lambda,\mu)^{-N}.$

For the last estimate observe that the inequality between the arithmetic and
the geometric mean

$\displaystyle\big{(}1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}\big{)}+\frac{s_{0}^{2}\langle
e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\geq
2s_{0}|\langle e_{\lambda},\Delta x\rangle|$

implies

$\displaystyle 1$
$\displaystyle+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}+s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+\frac{s_{0}^{2}\langle e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\geq$
$\displaystyle\frac{1}{2}\left(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}+s_{0}^{2\alpha}\left|\Delta
x\right|^{2}\right)+\frac{1}{2}\left(1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}+\frac{s_{0}^{2}\langle
e_{\lambda},\Delta
x\rangle^{2}}{1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}}\right)$
$\displaystyle\quad\quad\quad\quad\quad\quad\gtrsim
1+s_{0}^{2(1-\alpha)}\left|\left\\{d_{\mathbb{S}}(e_{\lambda},e_{\mu})\right\\}\right|^{2}+s_{0}^{2\alpha}\left|\Delta
x\right|^{2}+s_{0}\left|\langle e_{\lambda},\Delta
x\rangle\right|=1+d_{\alpha}(\lambda,\mu).$

This concludes the proof. ∎

## Acknowledgements

The first author acknowledges support by Deutsche Forschungsgemeinschaft (DFG)
Grant KU 1446/18 - 1 and by the Deutscher Akademischer Austausch Dienst
(DAAD). The second author would like to thank Anton Kolleck for enlightening
discussions on this and related topics. Moreover, both authors would like to
thank Gitta Kutyniok for carefully proofreading the article and providing many
suggestions for enhancing the readability of the paper.

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# Charged excitons or trions in 2D parabolic quantum dots?

Nguyen Hong Quang1,2 and Nguyen Que Huong3 1 Graduate University of Science
and Technology, VAST,
18 Hoang Quoc Viet, Nghia Do, Cau Giay, Hanoi, Vietnam
2 Institute of Physics, Vietnam Academy of Science and Technology (VAST),
18 Hoang Quoc Viet, Nghia Do, Cau Giay, Hanoi, Vietnam
3 Marshall University, One John Marshall Drive, Huntington WV 25701
<EMAIL_ADDRESS>
###### Abstract

So far in the literature the terms “charged exciton” and “trion” are often
confused with each other and mostly considered as the same. In this work we
show this is not the case in 2D quantum dots with a parabolic confinement. By
using the unrestricted Hartree-Fock method the energy and binding energy of
both charged excitons and trions in 2D parabolic quantum dots are calculated
in dependence on the confinements of charge carriers in quantum dot. It is
shown that the binding energies of the charged exciton and the trion behave
differently in regard to the ratios of the confinements between the electron
and hole. The effect of the external magnetic field on the binding energies of
charged excitons has been also considered.
Keywords: charged exciton, trion, parabolic quantum dot, Hartree-Fock method,
binding energy.

##  1 Introduction

Exciton effects have been intensively studied in low-dimensional semiconductor
systems not only theoretically but experimentally in recent years due to their
critical role in electronics, optoelectric devices and quantum information
technology in the future [1, 2, 3].

A trion is a bound state of three charged particles, which can be either two
electrons associated with a hole or two holes associated with an electron.
Trions can also be bound states of an exciton with an electron or an exciton
with a hole. In this case they are also known as negatively charged excitons
or positively charged excitons [4, 5, 6, 7]. In fact, in trion states, like
excitons and other exciton-based systems, such as biexcitons, made up of pairs
of electrons - holes, the charge carriers are correlated with each other
through Coulomb interaction, which has been studied a lot before, but has
attracted great attention recently, especially in atomic-size thin
semiconductor layers based on transition metal dichalcogenide monolayers [8,
9, 10, 11].

Quite recently, new terms “duo”, “trio”, “quatuor” for quasi-particle
consisting of two, three and four charge carriers, respectively has been
introduced in quantum dots by Combescot [12] and “quadron” for four charge
carriers in 2D semiconductor quantum dots by us [13, 14, 15] to distinguish
them from conventional excitons, charged excitons and biexcitons. In
particular, self-organized semiconductor quantum dots are still in special
interest of researchers today, since the binding energies of quasi-particles
in the system are large and the effect of magnetic fields can be studied along
with the symmetry of the quantum dot. [12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22].

In [12, 13, 14, 15] the authors have demonstrated the essential differences
between ordinary biexciton and “quatuor” [12] or “quadron” [13, 14, 15] in
quantum dots. The main reason for these differences is the confinement impact
on the charged carriers in the quantum dot, where multi-particle pair
interactions between four carriers should be considered equally. In our
previous works [13, 14, 15] we have demonstrated that the confinement and the
Coulomb interaction in a small InAs quantum dot lead to a bound quadron,
rather a conventional biexciton.

To develop further the investigations on this direction, in this work we
consider differences between usual charged excitons and trions in self-
assembled semiconductor quantum dots with 2D parabolic potential of
confinement. By using the Hatree-Fock method, the energies and binding
energies of the charged excitons and trions have been calculated for an entire
range of confinement ratio between the electron and the hole. We will show
that the binding energies of the charged exciton and the trion behave
differently in regard to the ratios of the confinements between the electron
and hole. The magnetic field effect on the binding energy of the charged
excitons has been also presented.

##  2 The model

We consider the system of electrons and holes in a 2D quantum dot with
parabolic potential, and in a magnetic field $\vec{B}\|z$. The total
Hamiltonian of the system of $N$ electrons and $M$ holes ($N=2,M=1$ or
$N=1,M=2$ for charged excitons and trions) can be written in the effective-
mass approximation, as follows [13, 14, 15],

$\widehat{H}=\sum_{i=1}^{N}h(\vec{r}_{i})+\sum_{k=1}^{M}h^{\prime}(\vec{r}_{k})+\sum_{i=1;i<j}^{N}\frac{e^{2}}{\epsilon
r_{ij}}+\sum_{k=1;k<l}^{M}\frac{e^{2}}{\epsilon
r_{kl}}-\sum_{i=1}^{N}\sum_{k=1}^{M}\frac{e^{2}}{\epsilon r_{ik}}\ ,$ (1)

where $h(\vec{r}_{i})$ and $h^{\prime}(\vec{r}_{k})$ are the Hamiltonians of a
single electron and a single hole, given by the formula (2) and (3) and the
other terms describe the electron-electron, the hole-hole and the electron-
hole Coulomb interactions, respectively, and $\epsilon$ is the material’s
dielectric constant.

For a single electron and a single hole in 2D parabolic quantum dots and in
the presence of a magnetic field, the Hamiltonians are:

$\displaystyle h(\vec{r}_{i})$ $\displaystyle=$
$\displaystyle-\frac{\nabla^{2}_{i}}{2m^{*}_{e}}+\frac{m^{*}_{e}}{2}(\omega^{2}_{e}+\frac{1}{4}\omega^{2}_{ce})r^{2}_{i}+\frac{1}{2}\omega_{ce}\hat{L}_{zi}\
,$ (2) $\displaystyle h^{\prime}(\vec{r}_{k})$ $\displaystyle=$
$\displaystyle-\frac{\nabla^{2}_{k}}{2m^{*}_{h}}+\frac{m^{*}_{h}}{2}(\omega^{2}_{h}+\frac{1}{4}\omega^{2}_{ch})r^{2}_{k}+\frac{1}{2}\omega_{ch}\hat{L}_{zk}\
,$ (3)

where the effective mass of the electron (hole) are denoted by $m^{*}_{e}$
($m^{*}_{h}$) and their confinement potentials are denoted by $\omega_{e}$
($\omega_{h}$). The expressions for the cyclotron frequency for the electron
(hole) are $\omega_{ce}=eB/m^{*}_{e}$ ($\omega_{ch}=eB/m^{*}_{h}$). Note that
in (2) and (3) we neglect the Zeeman effect because of its smallness. The last
term relates with the orbital angular momentum of the electron (hole),
respectively $\hat{L}_{zi}$ ($\hat{L}_{zk}$).

In polar coordinates $\vec{r}=(r,\varphi)$ we have the single electron’s
eigenfunction and the eigenvalue in the states $(n,m)$, as follows:

$\displaystyle\chi^{e}_{n,m}(r,\varphi)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2\pi}}e^{im\varphi}\sqrt{\frac{2n!}{(n+|m|)!}}\alpha_{e}(\alpha_{e}r)^{|m|}e^{-(\alpha_{e}r)^{2}/2}L_{n}^{|m|}((\alpha_{e}r)^{2})\
,$ (4) $\displaystyle E^{e}_{n,m}$ $\displaystyle=$
$\displaystyle\Omega_{e}(2n+|m|+1)+\frac{1}{2}m\omega_{c_{e}}\ ,$ (5)

where
$\alpha_{e}=\sqrt{m_{e}^{*}\Omega_{e}},\Omega_{e}=(\omega_{e}^{2}+\frac{1}{4}\omega_{c_{e}}^{2})^{1/2}$,
and $L_{n}^{|m|}(r)$ is generalized Laguerre polynomial. Note that we have the
analogous formulae for the hole, in which the index $h$ for the hole replaces
the index $e$ for the electron.

In [13, 14, 15] we have presented in details the Hartree-Fock-Roothaan
formulation for the multi-particle systems, using the functions (4) as basic
functions, here we only show the final expression for the total energy of the
system in our study:

$\displaystyle E$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{\mu,\nu}\Big{\\{}\delta_{\mu\nu}P_{\mu\nu}^{T}[\Omega_{e}(2n+|m|+1)+m\omega_{c_{e}}]+P_{\mu\nu}^{\alpha}F_{\nu\mu}^{\alpha}+P_{\mu\nu}^{\beta}F_{\nu\mu}^{\beta}\Big{\\}}$
(6) $\displaystyle+$
$\displaystyle\frac{1}{2}\sum_{\mu,\nu}\Big{\\{}\delta_{\mu\nu}P_{\mu\nu}^{\prime\
T}[\Omega_{h}(2n+|m|+1)-m\omega_{c_{h}}]+P_{\mu\nu}^{\prime\
\alpha}F_{\nu\mu}^{\prime\ \alpha}+P_{\mu\nu}^{\prime\
\beta}F_{\nu\mu}^{\prime\ \beta}\Big{\\}},$

where the explicit expressions for $P_{\mu\nu}^{T},P_{\mu\nu}^{\prime\ T}$ ,
$P_{\mu\nu}^{\alpha},P_{\mu\nu}^{\beta}$ , $P_{\mu\nu}^{\prime\ \alpha}$ ,
$P_{\mu\nu}^{\prime\ \beta}$ and
$F_{\mu\nu}^{\alpha,\beta},F_{\mu\nu}^{\prime\ \alpha,\beta}$ are given in
[13].

##  3 The results and discussions

For numerical calculations, we take the parameters for InAs/GaAs quantum dots
with $m^{*}_{e}=0.067m_{o}$, $\omega_{e}=49$ meV, $m^{*}_{h}=0.25m_{o}$,
$\omega_{h}=25$ meV,$\ \epsilon_{s}=12.53$. The units of length is
$a_{B}^{*}=\epsilon_{s}/m^{*}_{e}e^{2}=9.9$ nm, and the units of energy are is
$2Ry^{*}=m^{*}_{e}e^{4}/\epsilon^{2}_{s}=11.61$ meV [13, 14, 15].

In the study of the effect of confinement on the binding energies, for the
reason of comparison we use two parameter sets, namely set 1 with
$\omega_{e}=49$ meV, $m^{*}_{e}=0.067m_{o}$, and $m^{*}_{h}=0.25m_{o}$, and
set 2 with $\omega_{e}=49$ meV, $m^{*}_{e}=0.067m_{o}$, and
$m^{*}_{h}=0.067m_{o}$.

Denote $m_{e}=m^{*}_{e}/m_{o};m_{h}=m^{*}_{h}/m_{o}$, we have for the set 1:
$m_{e}/m_{h}=0.27$, and for the set 2: $m_{e}/m_{h}=1$.

Figure 1: (color online) The comparison of the negatively charged exciton
binding energies with the $m_{e}/m_{h}=0.27$ and $m_{e}/m_{h}=1$ as function
of electron-to-hole confinement.

Figure 2: (color online) The comparison of the negative trion binding energies
with the $m_{e}/m_{h}=0.27$ and $m_{e}/m_{h}=1$ as function of electron-to-
hole confinement ratio.

In Figure 2, Figure 2, Figure 4 and Figure 4 the binding energies of the
negatively charged exciton and the negative trion in the ground state without
magnetic fields have been shown as function of the confinement ratio between
the electron and the hole. To see the impact of masses of carriers, we compare
the results calculated for both parameter sets with $m_{e}/m_{h}=0.27$, and
$m_{e}/m_{h}=1$, respectively. One can see in the Figure 4 and Figure 4 that
for the entire range of the electron-to-hole confinement ratio
$\omega_{e}/\omega_{h}$, the negative trion binding energy is always larger
than that of the negatively charged exciton. We notice that for small values
of the electron-to-hole confinement ratio both negative trions and negatively
charged excitons are bound systems but they become unbound when the
confinement ratio between the electron and the hole increases.

Figure 3: (color online) The comparison of the negative trion and negatively
charged exciton binding energies with the $m_{e}/m_{h}=0.27$ as function of
electron-to-hole confinement ratio.

Figure 4: (color online) The comparison of the negative trion and negatively
charged exciton binding energies with the $m_{e}/m_{h}=1$ as function of
electron-to-hole confinement ratio.

For the positively charged excitons and positive trions we also get the
similar differences as between negatively charged excitons and negative trion.
In Figure 6 and Figure 6 the binding energies of the positively charged
excitons and positive trions in the ground states as function of electron-to-
hole confinement ratio are compared in two parameter sets with
$m_{e}/m_{h}=0.27$ and $m_{e}/m_{h}=1$. Again the positive trion binding
energy is always larger than that of the positively charged excitons.

Figure 5: (color online) The comparison of the positively charged exciton
binding energies with the $m_{e}/m_{h}=0.27$ and $m_{e}/m_{h}=1$ as function
of electron-to-hole confenement ratio.

Figure 6: (color online) The comparison of the positive trion binding energies
with the $m_{e}/m_{h}=0.27$ and $m_{e}/m_{h}=1$ as function of electron-to-
hole confinement ratio.

Figure 7: (color online) The change of the negatively charged exciton binding
energy as function of magnetic fields.

Figure 8: (color online) The change of the positively charged exciton binding
energy as function of magnetic fields.

However, here we observe the transition from anti-binding to binding state of
both positive trion and positively charged exciton when increasing the
electron-to-hole confinement ratio. This situation is opposite for the above
case of negative trions and negatively charged excitons, because the
contributions of the hole-hole interaction is larger than that of electron-
electron interaction to suppress the electron-hole interaction. It is
interesting to note that our new results can help to understand and clarify
sensitive changes in the binding energy of trions in natural ensembles of
InAs/GaAs quantum dots with randomly fluctuating parameters [21].

Figure 8 and Figure 8 show how the magnetic field affect the charged exciton
state. It is seen that magnetic fields increase the binding energy of
positively charged excitons but decrease it for the case of negatively charged
exciton, although the effect is rather small.

In [14, 15] we have analyzed the agreement between our results calculated by
Hartree-Fock method and the experimental data for the biexciton binding energy
[19]. In this work we also see that in InAs/GaAs self-assembled quantum dot we
have the good agreement between our results and the experimental data for the
negatively and positively charged exciton binding energies too. Indeed, the
value of electron-to-hole confinement ratio in InAs/GaAs self-assembled
quantum dot is $\omega_{e}/\omega_{h}=1.96$, that corresponds to the value of
the negatively and positively charged exciton binding energies $0.58\
(2Ry^{*})\approx 6.73$ meV and $-0.62\ (2Ry^{*})\approx-7.2$ meV,
respectively. Our results agree very well with the experimental value of
$6.2\pm 0.4$ meV for the negatively exciton and rather good with the
experimental value interval of [-1 meV:-6 meV] for positively charged exciton,
respectively [19].

##  4 Conclusion

In conclusion, the charged exciton and trion states in 2D parabolic quantum
dots have been studied in the Hartree-Fock-Roothaan formulation. The trion and
charged exciton binding energies have been calculated for the entire range of
the confinement ratio between the electron and the hole. It is shown that the
binding energies of charged excitons and trions can be either negative or
positive, depending on the correlation ratio of the electron-hole confinement
potential, but binding energies of negative or positive trion systems are
generally larger than that of the corresponding negatively or positively
charged excitons. Our theoretical calculation results are in good agreement
with the experimental data on the binding energies of negative and positive
trions in self-organizing small quantum dots InAs/GaAs. It is shown that the
binding energy of positively charged excitons is increased by magnetic fields
while it is decreased in the case of negatively charged exciton, although the
effect is rather small. Our results help further clarify the properties of the
elementary excitations and the differences between charged excitons and trions
in semiconductor quantum dots.

## Acknowledgments

This work is supported by the Vietnam Academy of Science and Technology (VAST)
under project No. NVCC05.12/21-21 and partly by the project No.ICP.2021.01 of
the International Centre of Physics (ICP), Institute of Physics.

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