11institutetext: Department of EEE, Imperial College London, London, SW7 2AZ,
United Kingdom.
<EMAIL_ADDRESS>
# Sum-Product Bounds & an Inequality for the Kissing Number in Dimension 16

Andrew Mendelsohn

###### Abstract

We obtain an inequality for the kissing number in 16 dimensions. We do this by
generalising a sum-product bound of Solymosi and Wong for quaternions to a
semialgebra in dimension 16. In particular, we obtain the inequality

$k_{16}\geq\frac{\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|}{\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|}-1,$

where $k_{16}$ is the 16-dimensional kissing number, and $S_{x}$ and
$\mathcal{R}$ are sets defined below. Along the way we also obtain a sum-
product bound for subsets of the octonions which are closed under taking
inverses, using a similar strategy to that used for the quaternions. We use
the fact that the kissing number in eight dimensions is 240 to achieve the
result. Namely, we obtain the bound

$\operatorname{max}(|\mathcal{A}+\mathcal{A}|,|\mathcal{A}\mathcal{A}|)\geq\frac{|\mathcal{A}|^{4/3}}{(1928\cdot\lceil\operatorname{log}|\mathcal{A}|\rceil)^{1/3}},$

where $\mathcal{A}$ is a finite set of octonions such that if
$x\in\mathcal{A}$, $x^{-1}\in\mathcal{A}$ also.

##  1 Introduction

Given an arbitrary subset $\mathcal{A}$ of a ring, let
$\mathcal{A}\mathcal{A}$ be the set of products of elements of $\mathcal{A}$,
and $\mathcal{A}+\mathcal{A}$ the set of sums of elements of $\mathcal{A}$;
how large can both $|\mathcal{A}\mathcal{A}|$ and $|\mathcal{A}+\mathcal{A}|$
be? In [3], Erdős and Szemerédi showed that for subsets of the integers,
$\max(|\mathcal{A}+\mathcal{A}|,|\mathcal{A}\mathcal{A}|)\gg|\mathcal{A}|^{1+\delta}$
for some $\delta>0$. For the quaternions, Chang [4] showed that $\delta\geq
1/54$, and Solymosi and Wong [1] showed $\delta$ can be arbitrarily close to
$1/3$. The state of the art for the exponent for subsets of the quaternions is
the work of Basit and Lund [6], in which the authors show $\delta>1/3$.
In the first part of this work, we observe that the method of [1] adapts to
the octonions, for certain conditioned subsets $\mathcal{A}$. In particular,
for a subset $\mathcal{A}\subset\mathbb{O}\setminus\\{0\\}$, let
$e_{1},...,e_{|\mathcal{A}/\mathcal{A}|}$ be the elements of
$\mathcal{A}/\mathcal{A}:=\mathcal{A}\mathcal{A}^{-1}\cap\mathcal{A}^{-1}\mathcal{A}$,
$a_{i}$ be the number of ways of multiplying an element of $\mathcal{A}$ with
an element of $\mathcal{A}^{-1}$ (in that order) to obtain $e_{i}$, and
$b_{i}$ be the number of ways of multiplying an element of $\mathcal{A}^{-1}$
with an element of $\mathcal{A}$ (in that order) to obtain $e_{i}$, for
$i=1,...,|\mathcal{A}/\mathcal{A}|$. Suppose the $a_{i}$ and $b_{i}$ are such
that
$\sum_{i}a_{i}b_{i}\geq\operatorname{min}(\sum_{i}a_{i}^{2},\sum_{i}b_{i}^{2})$,
and say without loss of generality
$\operatorname{min}(\sum_{i}a_{i}^{2},\sum_{i}b_{i}^{2})=\sum_{i}a_{i}^{2}$.
Then

$\operatorname{max}\left(|\mathcal{A}+\mathcal{A}|,|\mathcal{A}\mathcal{A}^{-1}|\right)\gg\frac{|\mathcal{A}||\mathcal{A}/\mathcal{A}|^{1/4}}{\lceil\operatorname{log}|\mathcal{A}|\rceil^{1/4}}.$

For subsets $\mathcal{A}$ satisfying closure under inversion, we obtain

$\operatorname{max}\left(|\mathcal{A}+\mathcal{A}|,|\mathcal{A}\mathcal{A}|\right)\gg\frac{|\mathcal{A}|^{4/3}}{\lceil\operatorname{log}|\mathcal{A}|\rceil^{1/3}},$

that is, $\delta$ is arbitrarily close to $1/3$. To our knowledge this is the
first such bound for subsets of the octonions.
Extending this, in the second half we consider a doubling process applied to
the octonions which yields a division left semialgebra111Where a left
semialgebra is a vector space over a field with a binary operation for which
left multiplication is linear., named the ‘16-ons’ to distinguish it from the
16-dimensional algebra obtained from the Cayley-Dickson doubling process, and
other ‘sedenionic’ constructions over the reals (see, for instance, [7]). This
space is ‘division’ in the sense that every non-zero element has a
multiplicative inverse.
We extend the argument we applied to the octonions to hold in this space for
sets subject to some conditions, which yields an inequality involving the
kissing number in 16 dimensions, $k_{16}$, which rearranges to give an
abstract lower bound for $k_{16}$ in terms of the ratio of sizes of particular
sets of 16-ons, defined below. The result of the first part, giving a sum-
product bound for octonions, can then be seen as a ‘warm-up’ demonstrating
that the work of [1] adapts to a nonassociative setting.
In particular, we find that, when $\mathcal{A}$ is a finite set of 16-ons
which have only the first nine entries possibly non-zero and
$x\in\mathcal{A}/\mathcal{A}$, letting $\ell(x)$ denote the number of
representatives of $x$ in $\mathcal{A}\mathcal{A}^{-1}$, and defining

$\mathcal{R}:=\operatorname{max}_{I}\left\\{x\in\mathcal{A}/\mathcal{A}:\ell(x)\geq
r(x)\text{ and }2^{I}\leq\ell(x)<2^{I+1}\right\\}$

and

$\mathcal{S}_{x}:=\left\\{(a+c,b+d)\in(\mathcal{A}+\mathcal{A})\times(\mathcal{A}+\mathcal{A}):ab^{-1}=x\text{
and }cd^{-1}=\phi(x)\right\\},$

where $x\in\mathcal{R}$ and $\phi:\mathcal{R}\rightarrow\mathcal{R}$ maps an
element of $\mathcal{R}$ to the closest (distinct) element in $\mathcal{R}$,
then

$k_{16}\geq\frac{\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|}{\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|}-1.$

We have not used this to concretely lower bound $k_{16}$. Unfortunately, the
doubling of the 16-ons, the ‘32-ons’, lose necessary algebraic properties for
the argument to hold, so we are unable to use this method for higher
dimensional bounds.

##  2 Preliminaries

Given a finite set $\mathcal{A}$ in a real division algebra $\mathbb{A}$, we
define

$\mathcal{A}+\mathcal{A}=\\{a+b\text{ : }a,b\in\mathcal{A}\\}\text{ and
}\mathcal{A}\mathcal{A}=\\{a\cdot b\text{ : }a,b\in\mathcal{A}\\}.$

Similarly, we let $\mathcal{A}^{-1}$ be the set of the inverses of elements of
$\mathcal{A}$. In addition, we will make reference to what has been called the
ratioset of $\mathcal{A}$:

$\mathcal{A}/\mathcal{A}:=\mathcal{A}\mathcal{A}^{-1}\cap\mathcal{A}^{-1}\mathcal{A}.$

We also need the notion of the multiplicative energy of a set:

######  Definition 1.

Let $\mathcal{A}$ be a finite set in $\mathbb{A}$. Then the multiplicative
energy of $\mathcal{A}$ is defined

$E(\mathcal{A})=|\\{(a,b,c,d)\in\mathcal{A}^{4}\text{ : }ca=db\\}|.$

We are also interested in the following related set:

$E^{\prime}(\mathcal{A})=|\\{(a,b,c,d)\in\mathcal{A}^{4}\text{ :
}ab^{-1}=c^{-1}d\\}|.$

Note that when $\mathcal{A}$ is a subset of an associative ring, these two
sets coincide, and $E(\mathcal{A})=E^{\prime}(\mathcal{A})$. However, this is
not prima facie the case for subsets of the octonions $\mathbb{O}$.
There is the following bound (using the Cauchy-Schwartz inequality) for
associative $\mathbb{A}$:

$E(\mathcal{A})\geq\frac{|\mathcal{A}|^{4}}{|\mathcal{A}\mathcal{A}|}.$

We desire a similar style bound for $E^{\prime}(\mathcal{A})$ with
$\mathcal{A}\subset\mathbb{O}$; we have the following lemma:

######  Lemma 1.

Let $\mathcal{A}\subset\mathbb{O}$ and $0\not\in\mathcal{A}$. Let
$e_{1},...,e_{|\mathcal{A}/\mathcal{A}|}$ be the elements of
$\mathcal{A}/\mathcal{A}$, $a_{i}$ be the number of ways of multiplying an
element of $\mathcal{A}$ with an element of $\mathcal{A}^{-1}$ (in that order)
to obtain $e_{i}$, and $b_{i}$ be the number of ways of multiplying an element
of $\mathcal{A}^{-1}$ with an element of $\mathcal{A}$ (in that order) to
obtain $e_{i}$, for $i=1,...,|\mathcal{A}/\mathcal{A}|$. Suppose the $a_{i}$
and $b_{i}$ are such that
$\sum_{i}a_{i}b_{i}\geq\operatorname{min}(\sum_{i}a_{i}^{2},\sum_{i}b_{i}^{2})$.
Then

$E^{\prime}(\mathcal{A})\geq\frac{|\mathcal{A}|^{4}|\mathcal{A}/\mathcal{A}|}{\operatorname{min}\left(|\mathcal{A}\mathcal{A}^{-1}|,|\mathcal{A}^{-1}\mathcal{A}|\right)^{2}}.$

When $\mathcal{A}$ is closed under taking inverses,
$E^{\prime}(\mathcal{A})\geq\frac{|\mathcal{A}|^{4}}{|\mathcal{A}\mathcal{A}|}$.

###### Proof.

Without loss of generality, suppose
$\operatorname{min}(\sum_{i}a_{i}^{2},\sum_{i}b_{i}^{2})=\sum_{i}a_{i}^{2}$.
Let $m:=|\mathcal{A}/\mathcal{A}|$. Observe
$\sum_{i=1}^{m}a_{i}=\frac{|\mathcal{A}||\mathcal{A}^{-1}||\mathcal{A}/\mathcal{A}|}{|\mathcal{A}\mathcal{A}^{-1}|}=\frac{|\mathcal{A}|^{2}|\mathcal{A}/\mathcal{A}|}{|\mathcal{A}\mathcal{A}^{-1}|}$.
Then
$E^{\prime}(\mathcal{A})=\sum_{i=1}^{m}a_{i}b_{i}\geq\sum_{i=1}^{m}a_{i}^{2}=\|(a_{1},...,a_{m})\|^{2}\|(1/\sqrt{m},...,1/\sqrt{m})\|^{2}\geq\frac{(\sum_{i=1}^{m}a_{i})^{2}}{m}=\frac{|\mathcal{A}|^{4}|\mathcal{A}/\mathcal{A}|^{2}}{|\mathcal{A}/\mathcal{A}||\mathcal{A}\mathcal{A}^{-1}|^{2}}=\frac{|\mathcal{A}|^{4}|\mathcal{A}/\mathcal{A}|}{|\mathcal{A}\mathcal{A}^{-1}|^{2}}$,
by Cauchy-Schwartz. When $\mathcal{A}=\mathcal{A}^{-1}$, the inequality
simplifies to the stated result. ∎

###  2.0.1 Complex quadrants, quaternionic hexadecants, and octonionic
hexapentacontadictants

One can divide up the complex plane into quadrants, and consider the case when
a given complex number $a+ib$ has both its coefficients positive, or both its
coefficients negative. This corresponds to the complex number lying in the top
right or bottom left quadrant of the plane. Moreover, one can say that given
two complex numbers, both lie in the same quadrant if they both have all
positive or all negative coefficients. Then the norm of the sum of any two
such complex numbers is larger than the norm of either number alone.
In a similar way one can divide up the quaternions, depending on whether a
given quaternion has all its coefficients positive or all negative (or not).
In this case, following [1], one says that two quaternions lie in the same
hexadecant (there being $16=2^{4}$ combinations of the vector of signs of the
coefficients for a given quaternion) if they both have all positive, or all
negative, coefficients.
Finally, we will make the same requirement for the octonions: two octonions
lie in the same hexapentacontadictant if they both have all positive or all
negative coefficients. Then the norm of the sum of any two such octonions is
larger than the individual norms of the elements summed.

###  2.0.2 The 16-ons

The Cayley-Dickson process is a method to take an algebra $\mathbb{A}_{n-1}$,
and create a new algebra $\mathbb{A}_{n}$ with twice the dimension of the old.
When applied to $\mathbb{A}_{0}=\mathbb{R}$, we successively get
$\mathbb{A}_{1}=\mathbb{C}$, $\mathbb{A}_{2}=\mathbb{H}$, and
$\mathbb{A}_{3}=\mathbb{O}$, iterating three times. These are normed division
algebras. Hurwitz showed these are the only normed real division algebras;
later, Pfister [9] gave an identity for writing the product of two sums of 16
squares as the sum of 16 rational squares, essentially showing that norms on
real algebras exist in higher dimensions. In particular, in any power of two
dimension, there exists an algebra with a norm (which satisfies properties
sometimes referred to as being ‘nicely normed’). However, $\mathbb{A}_{n}$ is
not a division algebra once $n>3$, so the norm is not multiplicative: in
particular, since zero-divisors exist, one might have
$\|ab\|=0\neq\|a\|\cdot\|b\|$.
It was observed by Conway and Smith that, tweaking the Cayley-Dickson
construction, one can create a doubling process that yields multiplicative
norms. However, since the only normed division algebras have $n<4$, once at
$n=4$ one starts paying for multiplicativity of the norm with other
properties: for example, alternativity (that is, $x(xy)=x^{2}y)$ is lost at
this point.
The algebras obtained from this adjusted Cayley-Dickson process have been
studied in depth by Smith [5] (it seems with some collaboration with Conway;
this process has been referred to as the ‘Conway-Smith doubling process’ in
[8]), and called $2^{n}$-ons. Given two elements of $\mathbb{A}_{n-1}$, we
define $\mathbb{A}_{n}$ as having additive group
$\mathbb{A}_{n-1}\times\mathbb{A}_{n-1}$ equipped with the multiplication rule

$(a,b)(c,d)=\left(ac-\overline{b\bar{d}},\overline{\bar{b}\bar{c}}+\overline{\bar{b}\overline{\bar{a}\overline{\overline{b^{-1}}\bar{d}}}}\right),$

where $\overline{\cdot}$ is conjugation on $\mathbb{A}_{n-1}$. For the
$2^{n}$-ons, conjugation can be written $\overline{(a,b)}=(\overline{a},-b)$,
with $a,b\in\mathbb{A}_{n-1}$. When $n=4$, that is, in the case of the 16-ons,
the above simplifies to

$(a,b)(c,d)=\left(ac-d\bar{b},cb+\bar{a}b^{-1}\cdot bd\right)$

When $b=0,$ the multiplication is defined $(a,0)(c,d)=(ac,\bar{a}d)$. We list
the basic properties of the $2^{n}$-ons.

######  Proposition 1.

[5, Theorem 15] If $a,b,x,y\in\mathbb{A}_{n}$, $n\geq 0$, and
$s,t\in\mathbb{R}$, then

  1. 1.

$A$ unique $2^{n}$-on 1 exists, with $1x=x1=x$.

  2. 2.

$A$ unique $2^{n}$-on 0 exists with $0+x=x+0=x$ and $0x=x0=0.$

  3. 3.

$x+y=y+x,(x+y)+z=x+(y+z)$; $-x$ exists with $x+(-x)=x-x=0$.

  4. 4.

There exists a norm $\|\cdot\|$ such that $\|x\|^{2}=x\bar{x}=\bar{x}x$ with
$\|x\|^{2}\|y\|^{2}=\|xy\|^{2}$.

  5. 5.

$s\cdot xy=sx\cdot y=xs\cdot y=x\cdot sy=x\cdot ys$.

  6. 6.

‘Weak linearity’ holds: $(x+s)y=xy+sy$ and $x(y+s)=xy+xs$.

  7. 7.

Left distributivity holds: $x(y+z)=xy+xz$.

  8. 8.

If $x\neq 0$, a unique $x^{-1}$ exists satisfying $x^{-1}x=xx^{-1}=1$, equal
to $x^{-1}=\bar{x}\|x\|^{-2}$. Inversion commutes with complex conjugation:
$\overline{x^{-1}}=\bar{x}^{-1}$.

  9. 9.

Left alternativity holds: $x\cdot xy=x^{2}y$.

  10. 10.

Left cancellation holds: $x\cdot x^{-1}y=y$.

Note the one-sidedness of many of the above properties. In particular, the
16-ons form a division left semialgebra, since all non-zero elements have an
inverse but they lack the full distributivity properties required to be an
algebra.
We define $\mathcal{A}+\mathcal{A}$, $\mathcal{A}\mathcal{A}$,
$\mathcal{A}\mathcal{A}^{-1}$, $E(\mathcal{A})$, and $E^{\prime}(\mathcal{A})$
for finite subsets of the 16-ons identically as in the octonionic case. We
denote the 16-ons by $\mathbb{S}$. The lemma bounding
$E^{\prime}(\mathcal{A})$ when $\mathcal{A}\subset\mathbb{O}$ adapts to the
case of $\mathcal{A}\subset\mathbb{S}$ immediately, and we state it below.

######  Lemma 2.

Let $\mathcal{A}\subset\mathbb{S}$ and $0\not\in\mathcal{A}$. Let
$e_{1},...,e_{|\mathcal{A}/\mathcal{A}|}$ be the elements of
$\mathcal{A}/\mathcal{A}$, $a_{i}$ be the number of ways of multiplying an
element of $\mathcal{A}$ with an element of $\mathcal{A}^{-1}$ (in that order)
to obtain $e_{i}$, and $b_{i}$ be the number of ways of multiplying an element
of $\mathcal{A}^{-1}$ with an element of $\mathcal{A}$ (in that order) to
obtain $e_{i}$, for $i=1,...,|\mathcal{A}/\mathcal{A}|$. Suppose the $a_{i}$
and $b_{i}$ are such that
$\sum_{i}a_{i}b_{i}\geq\operatorname{min}(\sum_{i}a_{i}^{2},\sum_{i}b_{i}^{2})$.
Then

$E^{\prime}(\mathcal{A})\geq\frac{|\mathcal{A}|^{4}|\mathcal{A}/\mathcal{A}|}{\operatorname{min}\left(|\mathcal{A}\mathcal{A}^{-1}|,|\mathcal{A}^{-1}\mathcal{A}|\right)^{2}}.$

When $\mathcal{A}$ is closed under taking inverses,
$E^{\prime}(\mathcal{A})\geq\frac{|\mathcal{A}|^{4}}{|\mathcal{A}\mathcal{A}|}$.

###  2.0.3 Niners

The subset of the 16-ons comprised of elements with the last seven entries
equal to zero form a useful subset of the 16-ons. This is so because they are
the largest additively closed subset to satisfy both left and right
distributivity, and they are also closed under inversion (not, however,
multiplication). Both left and right distributivity appear crucial to the
following argument. We prove this basic property below.

######  Proposition 2.

Let $x,y,z\in\mathbb{S}\setminus 0$ and $x$ be a niner. Then $(y+z)x=yx+zx$.

###### Proof.

Write $x=(x_{1},x_{2})$ with $x_{i}\in\mathbb{O}$, $i=1,2$, and similary for
$y$ and $z$. We have $x_{2}\in\mathbb{R}$. Then

$\displaystyle(y+z)x$ $\displaystyle=(y_{1}+z_{1},y_{2}+z_{2})(x_{1},x_{2})$
$\displaystyle=\left((y_{1}+z_{1})x_{1}-x_{2}(\overline{y_{2}+z_{2}}),x_{1}(y_{2}+z_{2})+(\overline{y_{1}+z_{1}})(y_{2}+z_{2})^{-1}\cdot(y_{2}+z_{2})x_{2}\right)$
$\displaystyle=\left(y_{1}x_{1}+z_{1}x_{1}-x_{2}\overline{y_{2}}-x_{2}\overline{z_{2}},x_{1}y_{2}+x_{1}z_{2}+(\overline{y_{1}}+\overline{z_{1}})x_{2}\right).$

Moreover, we have

$\displaystyle yx+zx$
$\displaystyle=(y_{1},y_{2})(x_{1},x_{2})+(z_{1},z_{2})(x_{1},x_{2})$
$\displaystyle=\left(y_{1}x_{1}-x_{2}\overline{y_{2}},x_{1}y_{2}+\overline{y_{1}}y_{2}^{-1}\cdot
y_{2}x_{2}\right)+\left(z_{1}x_{1}-x_{2}\overline{z_{2}},x_{1}z_{2}+\overline{z_{1}}z_{2}^{-1}\cdot
z_{2}x_{2}\right)$
$\displaystyle=\left(y_{1}x_{1}-x_{2}\overline{y_{2}}+z_{1}x_{1}-x_{2}\overline{z_{2}},x_{1}y_{2}+\overline{y_{1}}y_{2}^{-1}\cdot
y_{2}x_{2}+x_{1}z_{2}+\overline{z_{1}}z_{2}^{-1}\cdot z_{2}x_{2}\right)$
$\displaystyle=\left(y_{1}x_{1}-x_{2}\overline{y_{2}}+z_{1}x_{1}-x_{2}\overline{z_{2}},x_{1}y_{2}+\overline{y_{1}}x_{2}+x_{1}z_{2}+\overline{z_{1}}x_{2}\right).$

∎

##  3 Obtaining the Bound for the Octonions

We follow the method of [1], in order to derive a sum-product bound for the
octonions; the main technical difficulty is re-working the proofs therein to
get around the restrictions imposed by nonassociativity. To obtain the desired
bound, it suffices to prove

$\frac{E^{\prime}(\mathcal{A})}{\log\lceil|\mathcal{A}|\rceil}\ll|\mathcal{A}+\mathcal{A}|^{2}.$

For $x\in\mathcal{A}/\mathcal{A}$, let $\ell(x)$ ($r(x)$ respectively) denote
the number of representatives of $x$ in $\mathcal{A}\mathcal{A}^{-1}$
($\mathcal{A}^{-1}\mathcal{A}$ respectively). We can then express
${E}^{\prime}(\mathcal{A})$ in terms of $\ell(x)$ and $r(x)$:

$E^{\prime}(\mathcal{A})=\sum_{x\in\mathcal{A}/\mathcal{A}}\ell(x)r(x)$

Without loss of generality, assume that

$\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)r(x)\geq\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\
\ell(x)\leq r(x)\end{subarray}}\ell(x)r(x).$

Then we can bound $E^{\prime}(\mathcal{A})$ as follows:

$E^{\prime}(\mathcal{A})=\sum_{x\in\mathcal{A}/\mathcal{A}}\ell(x)r(x)\leq
2\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)r(x)\leq
2\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)^{2}.$

Since $1\leq\ell(x)\leq|\mathcal{A}|$, there exists an index $I$ such that
elements of the set
$\mathcal{R}:=\left\\{x\in\mathcal{A}/\mathcal{A}:\ell(x)\geq r(x)\right.$ and
$\left.2^{I}\leq\ell(x)<2^{I+1}\right\\}$ contribute at least
$1/\left\lceil\log_{2}|\mathcal{A}|\right\rceil$ of the above sum. That is,

$\frac{E^{\prime}(\mathcal{A})}{2\lceil\log|\mathcal{A}|\rceil}\leq\sum_{x\in\mathcal{R}}\ell(x)^{2}<|\mathcal{R}|2^{2I+2}.$

For each $x\in\mathcal{R}$, we define a set $\mathcal{S}_{x}$ by

$\mathcal{S}_{x}=\left\\{(a+c,b+d)\in(\mathcal{A}+\mathcal{A})\times(\mathcal{A}+\mathcal{A}):ab^{-1}=x\text{
and }cd^{-1}=\phi(x)\right\\},$

where $\phi:\mathcal{R}\rightarrow\mathcal{R}$ maps an element of
$\mathcal{R}$ to the closest (distinct) element in $\mathcal{R}$, that is

$\|x-\phi(x)\|\leq\|x-y\|$

for all $x,y\in\mathcal{R}$. We show that
$\bigcup_{s\in\mathcal{R}}\mathcal{S}_{x}$ has sufficiently large size in
$(\mathcal{A}+\mathcal{A})\times(\mathcal{A}+\mathcal{A})$ (i.e. comprises a
proportion larger than $E(\mathcal{A})/\lceil\log|\mathcal{A}|\rceil$,
multiplied by some constant factor).
In the proofs of the next two lemmas, we use the fact that while the octonions
are not associative, they satisfy weaker identities than associativity. In
particular, if $x,y\in\mathbb{O}$, then $y=(yx)x^{-1}=x^{-1}(xy)$ [2]. The
following lemma guarantees that

$\left|\mathcal{S}_{x}\right|\geq\ell(x)\ell(\phi(x))\geq 2^{2I}.$

######  Lemma 3.

Given $p,q\in\mathcal{A}+\mathcal{A}$ and distinct
$x,y\in\mathcal{A}\mathcal{A}^{-1}$, there is at most one quadruple
$(a,b,c,d)\in\mathcal{A}^{4}$ such that

$a+c=p,\quad b+d=q,\quad ab^{-1}=x,\quad cd^{-1}=y.$

###### Proof.

Since $x$ and $y$ are distinct, their difference is invertible. Then solving
the equations

$d=(y-x)^{-1}(p-xq),\quad b=q-d,\quad c=yd,\quad a=xb$

yields a unique tuple $(a,b,c,d)$, if such a solution exists. ∎

######  Lemma 4.

Let $a,b,c$ and $d$ be octonions, and $b$ and $d$ be non-zero and lie in the
same hexapentacontadictant. Then

$\|(a+c)(b+d)^{-1}-ab^{-1}\|\leq\|cd^{-1}-ab^{-1}\|.$

###### Proof.

We have

$\displaystyle\|(a+c)(b+d)^{-1}-ab^{-1}\|$
$\displaystyle=\|a+c-(ab^{-1})(b+d)\|\frac{1}{\|b+d\|}$
$\displaystyle=\|a+c-(ab^{-1})b-(ab^{-1})d\|\frac{1}{\|b+d\|}$
$\displaystyle=\|a+c-a-(ab^{-1})d\|\frac{1}{\|b+d\|}$
$\displaystyle=\|cd^{-1}-ab^{-1}\|\frac{\|d\|}{\|b+d\|}$
$\displaystyle\leq\|cd^{-1}-ab^{-1}\|.$

∎

######  Lemma 5.

Let $P\in\mathbb{R}^{8}$ be contained in $m$ closed balls, denoted
$\mathcal{B}_{i}$, each with center $Q_{i}\in\mathbb{R}^{8}$, for $i=1,...,m$.
Suppose $Q_{i}$ is not contained in the interior of any $\mathcal{B}_{j}$, for
$j\neq i$. Then $m\leq 241$.

###### Proof.

In the same manner as the proof of [1, Lemma 8], one can construct $m-1$ non-
overlapping spheres of equal radius, all touching some central sphere. Since
the maximum number of spheres that can be arranged in this configuration in
dimension 8 is 240 (the so-called kissing number), we arrive at the result. ∎

Combining the above results, we have

$|\mathcal{A}+\mathcal{A}|^{2}\geq\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|\geq\frac{1}{241}\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|\geq\frac{1}{241}|\mathcal{R}|2^{2I}\geq\frac{E^{\prime}(\mathcal{A})}{1928\lceil\log|\mathcal{A}|\rceil},$

which implies the required bound.

##  4 Obtaining a Bound for the 16-ons

In this section we adapt the analysis carried out for the octonions for niner
subsets of the 16-ons. The weaker properties satisfied by the elements of
$\mathbb{S}$ introduce some subtleties into the proofs and force the
restriction to sets of niners (which satisfy good distributive identities),
but the argument proceeds largely as before. Having completed this, we
rearrange the final inequalities to get a result lower bounding $k_{16}$.

###  4.0.1 Hexatriacontapentactapentaliahexacismyriants

We will need to divide the 16-dimensional plane into parts depending on the
coefficients of a given 16-on. Instead of using increasingly lengthy numerical
prefixes, we will simply refer to the portions of the plane containing 16-ons
all of whose coordinates in the vector representation in $\mathbb{R}^{16}$ are
positive, or all negative, as the privileged orthants. That is, for a 16-on
$x=\alpha+\beta i+\gamma j+...\in\mathbb{S}$, we can write
$x=(\alpha,\beta,\gamma,...)\in\mathbb{R}^{16}$, and the privileged orthants
contain the 16-ons with vectors whose entries are all positive, or all
negative.

To obtain the bound for the 16-ons, we restrict to subsets $\mathcal{A}$ of
niners, and it again suffices to prove

$\frac{E(\mathcal{A})}{\log\lceil|\mathcal{A}|\rceil}\ll|\mathcal{A}+\mathcal{A}|^{2}.$

We define $\ell(x)$ and $r(x)$ as before and express
${E}^{\prime}(\mathcal{A})$ as

$E^{\prime}(\mathcal{A})=\sum_{x\in\mathcal{A}/\mathcal{A}}\ell(x)r(x),$

and assume that

$\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)r(x)\geq\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\
\ell(x)\leq r(x)\end{subarray}}\ell(x)r(x).$

The bound on $E^{\prime}(\mathcal{A})$

$E^{\prime}(\mathcal{A})=\sum_{x\in\mathcal{A}/\mathcal{A}}\ell(x)r(x)\leq
2\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)r(x)\leq
2\sum_{\begin{subarray}{c}x\in\mathcal{A}/\mathcal{A}\\\ \ell(x)\geq
r(x)\end{subarray}}\ell(x)^{2}$

follows. Since $1\leq\ell(x)\leq|\mathcal{A}|$, there is an index $I$ such
that elements of $\mathcal{R}:=\left\\{x\in\mathcal{A}/\mathcal{A}:\ell(x)\geq
r(x)\right.$ and $\left.2^{I}\leq\ell(x)<2^{I+1}\right\\}$ contribute at least
$1/\left\lceil\log_{2}|\mathcal{A}|\right\rceil$ of the above sum. Thus

$\frac{E^{\prime}(\mathcal{A})}{2\lceil\log|\mathcal{A}|\rceil}\leq\sum_{x\in\mathcal{R}}\ell(x)^{2}<|\mathcal{R}|2^{2I+2}.$

For $x\in\mathcal{R}$, we define $\mathcal{S}_{x}$ by

$\mathcal{S}_{x}=\left\\{(a+c,b+d)\in(\mathcal{A}+\mathcal{A})\times(\mathcal{A}+\mathcal{A}):b^{-1}a=x\text{
and }d^{-1}c=\phi(x)\right\\},$

where $\phi:\mathcal{R}\rightarrow\mathcal{R}$ is a map sending an element to
a closest element in $\mathcal{R}$, i.e.

$\|x-\phi(x)\|\leq\|x-y\|$

for all $x,y\in\mathcal{R}$. We show that
$\bigcup_{s\in\mathcal{R}}\mathcal{S}_{x}\subset(\mathcal{A}+\mathcal{A})\times(\mathcal{A}+\mathcal{A})$
has size $\gg E(\mathcal{A})/\lceil\log|\mathcal{A}|\rceil$.
In the proofs of the next two lemmas, we use the fact that while the 16-ons
are not associative, they satisfy weaker identities than associativity. In
particular, we have $x^{-1}(xy)=y$. Moreover, left multiplication of sedenions
is a linear operation, i.e. $a(b+c)=ab+ac$ for $a,b,c\in\mathbb{S}$. However,
we note that for arbitrary 16-ons, right multiplication is not linear. This is
the reason we restrict to subsets of niners; by Proposition 2, the equation
$(u+v)w=uw+vw$ holds when $u,v\in\mathbb{S}$ and $w$ is a niner. The following
lemma implies

$\left|\mathcal{S}_{x}\right|\geq\ell(x)\ell(\phi(x))\geq 2^{2I}.$

######  Lemma 6.

Given a finite set $\mathcal{A}$ of niners, $p,q\in\mathcal{A}+\mathcal{A}$
and distinct $x,y\in\mathcal{A}^{-1}\mathcal{A}$, there is at most one
quadruple $(a,b,c,d)\in\mathcal{A}^{4}$ such that

$a+c=p,\quad b+d=q,\quad b^{-1}a=x,\quad d^{-1}c=y.$

###### Proof.

Since $x$ and $y$ are distinct, their difference is invertible. Then solving
the equations

$d=(y-x)^{-1}(p-xq),\quad b=q-d,\quad c=dy,\quad a=bx$

yields a unique tuple $(a,b,c,d)$, if such a solution exists. ∎

######  Lemma 7.

Let $a,b,c$ and $d$ be niners such that $(a+c,b+d)\in\mathcal{S}_{x}$ for some
$x$, and $b$ and $d$ be non-zero and lie in the same privileged orthant. Then

$\|(b+d)^{-1}(a+c)-b^{-1}a\|\leq\|d^{-1}c-b^{-1}a\|.$

###### Proof.

We have

$\displaystyle\|(b+d)^{-1}(a+c)-b^{-1}a\|$
$\displaystyle=\|a+c-(b+d)(b^{-1}a)\|\frac{1}{\|b+d\|}$
$\displaystyle=\|a+c-b(b^{-1}a)-d(b^{-1}a)\|\frac{1}{\|b+d\|}$
$\displaystyle=\|d^{-1}c-b^{-1}a\|\frac{\|d\|}{\|b+d\|}.$

∎

The above lemma says that if $(a+c,b+d)\in S_{x}$, there is a 16-on
$(b+d)^{-1}(a+c)$ lying in the closed ball centered at $x$ of radius
$\phi(x)$. The following lemma shows that there cannot be too many 16-ons
lying in any given such closed ball.

######  Lemma 8.

Let $P\in\mathbb{R}^{16}$ be contained in $m$ closed balls, denoted
$\mathcal{B}_{i}$, each with center $Q_{i}\in\mathbb{R}^{16}$, for
$i=1,...,m$. Suppose $Q_{i}$ is not contained in the interior of any
$\mathcal{B}_{j}$, for $j\neq i$. Then $m\leq 7333$.

###### Proof.

As in the proof of [1, Lemma 8], one can construct $m-1$ non-overlapping
spheres of equal radius, all touching some central sphere. Since the maximum
number of spheres that can be arranged in this configuration in dimension 16
is (currently) bounded between 4320 and 7332 [10], we arrive at the result. ∎

As in the octonionic case, we then obtain

$|\mathcal{A}+\mathcal{A}|^{2}\geq\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|\geq\frac{1}{7333}\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|\geq\frac{1}{7333}|\mathcal{R}|2^{2I}\gg\frac{E(\mathcal{A})}{\lceil\log|\mathcal{A}|\rceil}.$
(1)

###  4.0.2 Bounding $k_{16}$

Note that the inequalities $(\ref{equation2})$ can be rewritten

$|\mathcal{A}+\mathcal{A}|^{2}\geq\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|\geq\frac{1}{k_{16}+1}\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|\geq\frac{1}{k_{16}+1}|\mathcal{R}|2^{2I}\gg\frac{E(\mathcal{A})}{(8k_{16}+8)\lceil\log|\mathcal{A}|\rceil}.$

Rearranging, we obtain
$k_{16}\geq\frac{E(\mathcal{A})}{8\lceil\log|\mathcal{A}|\rceil|\mathcal{A}+\mathcal{A}|^{2}}-1$,
and more tightly,

$k_{16}\geq\frac{\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|}{\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|}-1.$

Therefore by choosing sets $\mathcal{A}\subset\mathbb{S}$ carefully, one may
be able to establish concrete lower bounds on $k_{16}$. Stated precisely, we
have

######  Theorem 4.1.

Let $\mathcal{A}\subset\mathbb{S}$ be a finite set of niners such that
$0\not\in\mathcal{A}$. Let $\mathcal{R}$ and $S_{x}$ be defined as above. Then

$k_{16}\geq\frac{\sum_{x\in\mathcal{R}}\left|\mathcal{S}_{x}\right|}{\left|\bigcup_{x\in\mathcal{R}}\mathcal{S}_{x}\right|}-1.$

##  5 Conclusion

We have obtained sum-product bounds, firstly for subsets of the octonions, and
secondly for subsets of niners within the 16-ons, this latter result implying
a bound on the kissing number $k_{16}$. The primary outstanding question from
this work is whether a concrete lower bound on $k_{16}$ can be obtained by
selecting particular sets of niners.

##  6 Acknowledgements

The author would like to thank Edmund Dable-Heath for useful conversations.

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