# Lattice packings through division algebras

Presentation by

### Nihar P. Gargava

Doctoral student,
Chair of Number Theory,
Section of Mathematics,
École Polytechnique Fédérale de Lausanne,

## Introduction

$$\DeclareMathOperator{\R}{\mathbb{R}}$$$$\DeclareMathOperator{\Z}{\mathbb{Z}}$$$$\DeclareMathOperator{\C}{\mathbb{C}}$$$$\DeclareMathOperator{\F}{\mathbb{F}}$$ Suppose we have $\{a_1, a_2, ... , a_n \} \in m \Z$ with $a_i \ge 0$ and suppose that $$\frac{1}{n}\sum_{i=1}^{n} a_i < m.$$

What can we conclude about the numbers $a_i$?

Answer: At least one of them is $0$.

#### Lemma

(Integrality gap)
Let $X$ be a probability space and $\Phi: X\to \R^{\ge 0}$ be a random variable that takes values only in $m \Z$ for some integer $m$.
If $\int_X \Phi < m$, then there exists some $x \in X$ such that $\Phi(x) = 0$.

For our application, we will set $X$ to be a set of lattices.

## What is a lattice packing?

Consider $\R^d$ with the standard inner product. A lattice $\Lambda\subseteq \R^d$ is a discrete subgroup such that the quotient space $\R^d / \Lambda$ has a finite induced volume.

Given a lattice $\Lambda$, choose $r > 0$ and consider the open balls $\{ B_{r}\left( v \right)\}_{v \in \Lambda}$ such for any $v_1, v_2 \in \Lambda$, $B_{r}(v_1) \cap B_{r}(v_2) \neq \emptyset \Rightarrow v_1 = v_2$.

Then $(\Lambda,r)$ is called a lattice sphere packing, or simply lattice packing inside $(\R^d, \langle \ , \ \rangle)$.

Packing density of a lattice packing is: $\lim_{R \rightarrow \infty} \frac{\mu\left( B_{R}(0) \cap \left( \bigsqcup_{v \in \Lambda} B_{r}(v) \right) \right)}{\mu\left( B_{R}(0) \right)} = \frac{\mu( B_{r}(0) )}{ \mu(\R^d / \Lambda)}$ It is always in the interval $]0,1[$.

#### Visualizing in $\R^2$

However, note that $2r$ can be at most $$m(\Lambda) = \min_{v \in \Lambda \setminus \{0\}} ||v|| ,$$ otherwise some balls will begin to intersect.

The goal is to maximize packing density. So take $r = \frac{1}{2}m(\Lambda)$. In that case packing density will be equal to $$\frac{\mu( B_{m(\Lambda)/2}(0) )}{ \mu(\R^d / \Lambda)},$$ and is independent of scaling.

To maximize this over all $\Lambda$, it is sufficient to maximize over unit covolume lattices (i.e. $$\mu(\R^d/\Lambda)$$ = 1)

#### Visualizing in $\R^2$

So we somehow find $\sup_{\substack{\Lambda \subseteq \R^d \\ \mu(\R^d/\Lambda)=1}} \mu\left(B_{m(\Lambda)/2}(0) \right)$

$= \sup_{ g \in SL_d(\R)} \frac{1}{2^d} \mu\left(B_{m(g \Z^d)}(0) \right)$

which we use to define our dimensional constant $c_d$ $\colon= \frac{1}{2^d} c_d$

Putting it together, we have $c_{d} = \sup\left\{ \mu\left( B_{r}(0)\right) \ | \ r> 0,\exists~g \in SL_d(\R) \text{ and } B_{r}(0) \cap g\Z^d = \{ 0\}\right\}.$

Clearly, $$~c_d \in~]0,2^d[$$, so supremum exists!

## What is known about $~c_d$?

The exact value of the constant $c_d$ is known only for $d = \{ 1,2,3,4,5,6,7,8,24\}$. For other $d$, we want to understand the asymptotic behaviour. In this talk, we will only focus on lower bounds.

#### Some asymptotic lower bounds for large dimensions

Lower bound Contribution of Dimensions covered
$c_d \ge 1$ Minkowksi (1896) $\forall ~d \ge 1$
$c_d \ge 2(d-1)$ Ball (1992) $\forall ~d \ge 1$
$c_{4n} \ge 8.8n$ Vance (2011) $d=4n , n \ge 1$
$c_{2\varphi(n)} \ge n$ Venkatesh (2013) $d=2\varphi(k)$ for some $k$

Since $\lim \inf \left( \frac{\varphi(n)}{n} \log \log n\right) = e^{- \gamma}$, the last bound is the best lower bound (among these, and overall) on $c_d$ in infinitely many dimensions. The first dimension where it outperforms all others in this list is $d=960$.

Explicit lattice constructions can help us only upto $d \le 300$.

(Ball, 1992) , the lower bound is as indicated.

(Vance, 2011) , using a probabilistic argument on lattices that lie in vector-space over quaternion division algebra. Works only when $4 \mid d$.

(Venkatesh, 2013) , using a probabilistic argument on lattices that lie in $(K \otimes_\mathbb{Q} \mathbb{R})^2$, $K$ is a cyclotomic field.

(G., 2021) , using a probabilistic argument on lattices that lie in $(D \otimes_\mathbb{Q} \mathbb{R})^2$, $D$ is a division algebra over $\mathbb{Q}$.

#### Theorem

(G. 2021)
Let $D$ be a finite-dimensional division algebra over $\mathbb{Q}$. Let $\mathcal{O} \subseteq D$ be an order in $D$ and $G_{0} \subseteq \mathcal{O}$ be a finite group embedded in the multiplicative group of $D$. Then if $d=2\dim_{\mathbb{Q}}D$, then \begin{align} c_{d} \ge \# G_{0}. \end{align}

To recover Venkatesh's result, set $D=\mathbb{Q}(\mu_n)$, $\mathcal{O} = \mathbb{Z}[\mu_n]$ and $G_0 = \langle \mu_n \rangle$. Hence, this gives \begin{align} c_{2 \varphi(n)} \ge n \end{align}

The cherrypicked sequence of Venkatesh achieves an asymptotic growth of $O( d \log \log d )$. This is achieved by setting $K$ as the $n$th cyclotomic field where $n=\prod_{p < N} p$.

The new result provides more freedom to cherrypick sequences. Instead of choosing a sequence of cyclotomic fields, we can now choose sequences of $\mathbb{Q}$-divison algebras. However, no such sequence will be able to give an asymptotic result strictly better than $O( d \log \log d )$. Improvements in individual dimensions is still possible, as shown before.

## The probabilistic method

Recall:
To show $c_d \ge M$, we must prove the existence of $g \in SL_d(\R)$ such that the origin centered ball $B$ with $\mu(B) = M$ has $g \Z^d \cap B = \{0\}$

This is an optimization problem on the space $X_d:= \{ \Lambda \subset \mathbb{R}^{d} | \ \mu(\mathbb{R}^{d} / \Lambda) = 1 \} = \{ g\Z^d \ | \ g \in SL_d(\R) \}$

$\simeq SL_d(\mathbb{R})/SL_d(\mathbb{Z}).$

A priori, this is a bijection of sets. But now we can pull back the topology and the measure from $SL_d(\R)/SL_d(\Z)$.

$SL_d(\R)$ has the topology of a locally compact group.

$SL_d(\R)$ is unimodular. $SL_d(\Z)$ is a discrete subgroup inside $SL_d(\R)$ and therefore there is a unique left $SL_d(\R)$-invariant measure on $SL_d(\R)/SL_d(\Z)$.

#### Proposition

There exists a unique (upto scaling) natural measure on $SL_d(\R)/SL_d(\Z)$, left-invariant under $SL_d(\R)$ action on cosets.
Furthermore, $SL_d(\mathbb{R})/SL_d(\mathbb{Z})$ under this has a bounded total measure.

This gives us a probability space. Hence we can talk about random unit covolume lattices.

## Lattice-sum function

Consider a bounded measurable function with compact support $f:\R^d \rightarrow \R$.
e.g. the indicator function of a ball.

With this, we can now construct the lattice-sum function $\Phi_f(\Lambda): X_d \rightarrow \R$, given as $\Phi_f(\Lambda) = \sum_{v \in \Lambda \setminus \{ 0\}}^{} f(v).$

Since we can generate random lattices, we can talk about the expected value of $\Phi_f(\Lambda)$.

Let us try to do this experimentally! Let us sample over a set $S \subseteq X_d$ of lattices.

So we see that it is almost the integral.

#### Visualizing in $\R^2$

What we are empirically confirming is the following.

#### Theorem

(Siegel, 1945)
Suppose $f:\mathbb{R}^{d} \rightarrow \mathbb{R}$ is a compactly supported bounded measurable function. Then, the following holds. \begin{align} \int_{X_d} \Phi_f = \int_{SL_d(\mathbb{R})/SL_d(\mathbb{Z})}^{} \left( \sum_{v \in g \mathbb{Z}^{d} \setminus \{ 0\}}^{}f(v) \right) dg = \int_{\mathbb{R}^{d}}^{} f(x) dx, \end{align} where the $dx$ on the right-hand side is the usual Lebesgue measure on $\mathbb{R}^{d}$ and $dg$ is the unique $SL_d(\mathbb{R})$-invariant probability measure on $SL_d(\mathbb{R})/SL_d(\mathbb{Z})$.

But as you saw that for any $\Lambda \in X_d$, when $f$ is the indicator of a ball, we must have $\Phi_f(\Lambda) \in \{0,2,4,6,...\}$. That's because balls are symmetric. $$v \in \text{supp}(f) \cap ( \Lambda \setminus \{0 \}) \Rightarrow -v \in \text{supp}(f) \cap ( \Lambda \setminus \{0 \}).$$

If $f$ is the indicator function of a ball of volume $2- \varepsilon$, then this tells us that for any dimension $d$ $$\int_{X_d} \Phi_f = 2 - \varepsilon$$

Conclusion: There exists some lattice $\Lambda \in X_d$ such that $\Phi_f(\Lambda) = 0$. That is, there is some lattice of unit covolume that intersects trivially with an origin centered ball of volume $2 - \varepsilon$.

Another conclusion: $c_d \ge 2$ for all dimensions $d$!

Here we used a symmetry of $\{\pm 1\}$ under which the ball and the lattice are invariant. Both Venkatesh and the new result use this idea.

Venkatesh's approach was to work with larger a larger group, namely $\Z/n\Z$. For the new result, the symmetries are even bigger and non-commutative.

## Venkatesh's lower bound

The idea is to take a nice enough subcollection $Y_d \subseteq X_d$ of lattices and average the lattice-sum function $\Phi_f$ over them.

Define the set $Y_d$ as $$Y_d = \left\{ \left[ \begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right] \mathcal{O}_K^2 \ | \ a,b,c,d \in K_{\R} = K\otimes \R , ad-bc =1_{K_{\R}} \right\}.$$

In conventional notation, this is just $$Y_d = \{ g ( \mathcal{O}_K^{\oplus 2} ) \ | \ g \in SL_2(K_{\R}) \} \simeq SL_2(K_\mathbb{R} ) /SL_2(\mathcal{O}_K).$$

Again, with this bijection, we will make $Y_d$ a topological space. With a bit of work, we can also show that $Y_d$ has a finite measure with respect to the natural left $SL_2(K_{\R})$-invariant measure.

$Y_d$ can be give probability measure. But why make such a $Y_d$?

Note that whenever $a,b,c,d \in {K}_{\R}$, we have that $$\left[ \begin{matrix} a & b \\ c & d \end{matrix}\right]\left[ \begin{matrix} \mu_n & 0 \\ 0 & \mu_n \end{matrix}\right]=\left[ \begin{matrix} \mu_n & 0 \\ 0 & \mu_n \end{matrix}\right]\left[ \begin{matrix} a & b \\ c & d \end{matrix}\right]$$

And if $x,y \in \mathcal{O}_K$ and if $r \notin n \Z$, we have that $$\left[ \begin{matrix} \mu_n^r & 0 \\ 0 & \mu_n^r \end{matrix}\right]\left[ \begin{matrix} x \\ y \end{matrix}\right]=\left[ \begin{matrix} x \\ y \end{matrix}\right] \Rightarrow x=y=0.$$

Putting the two together implies that for each $g\in SL_2(K_{\mathbb{R}})$, $g(\mathcal{O}_K^{\oplus 2})$ is invariant under the cyclic group action of $\langle \mu_n \rangle$, and the action has full orbits on the non-zero points of the lattice.

So if $f$ is the indicator function of an origin-centered ball with respect to a quadratic form that is invariant under $\langle \mu_n \rangle$, we have that $\Phi_f(\Lambda) \in \{0,n,2n,3n,4n,\ldots\}$ for $\Lambda \in Y_d$. Such a quadratic form always exists by averaging!

Venkatesh proves the following analogue of Siegel's theorem.

#### Theorem

(Venkatesh 2013)
Let $d = 2\varphi(n)$. Suppose $f:K_{\mathbb{R}}^2 \rightarrow \mathbb{R}$ is a compactly supported bounded measurable function. Then, the following holds. \begin{align} \int_{Y_d} \Phi_f = \int_{SL_2(K_\mathbb{R})/SL_2(\mathcal{O}_K)}^{} \left( \sum_{v \in g \mathcal{O}_K^{\oplus 2} \setminus \{ 0\}}^{}f(v) \right) dg = \int_{\mathbb{R}^{d}}^{} f(x) dx, \end{align} where the $dx$ on the right-hand side is that Lebesgue measure on $\mathbb{R}^{d}$ that makes $\mathcal{O}_K^{\oplus 2}$ of unit covolume and $dg$ is the unique $SL_2(K_\mathbb{R})$-invariant probability measure on $Y_d$.

Conclusion: By setting $f$ as the indicator function of a ball in a suitable quadratic form, we conclude that there exists some lattice $\Lambda \in Y_d$ such that $\Phi_f(\Lambda) = 0$. That is, there is some lattice of unit covolume that intersects trivially with an origin centered ball of volume $n - \varepsilon$.

Another conclusion: $c_{2 \varphi(n)} \ge n$ for all $n$!

## Using division algebras

The division algebra case is on similar lines. Let $D$ be a finite-dimensional division algebra over $\mathbb{Q}$. Let $\mathcal{O} \subseteq D$ be an order in $D$. We work in $d = 2 \dim_{\mathbb{Q}} D$ dimensions. Define $D_{\R}= D \otimes_\mathbb{Q} \R$. $$Y_d = \{ g ( \mathcal{O}^{\oplus 2} ) \ | \ g \in SL_2(D_{\R}) \} \simeq SL_2(D_\mathbb{R} ) /SL_2(\mathcal{O}).$$
Here $$SL_2(D_{\R}) = \left\{ \left[ \begin{matrix} a & b \\ c & d \end{matrix}\right] \ | \ \left[ \begin{matrix} x \\ y \end{matrix}\right] \mapsto \left[ \begin{matrix} ax + by \\ cx + dy \end{matrix}\right] \text{ is a measure preserving map on } D_{\R}^{\oplus 2} \right\}$$

$Y_d$ consists of lattices that are invariant under diagonal right-multiplication by units in $\mathcal{O}$ $$g ( \mathcal{O}^{\oplus 2} ) = g ( \mathcal{O}^{\oplus 2} ) \left[ \begin{matrix} \mu & 0 \\ 0 & \mu \end{matrix}\right] , \text{ for any } \mu \in \mathcal{O}^{*}$$

And if $x,y \in \mathcal{O}$ and if $\mu \in \mathcal{O}^{*} \setminus \{ 1_{D}\}$, we have that $$\left[ \begin{matrix} x \\ y \end{matrix}\right] \left[ \begin{matrix} \mu & 0 \\ 0 & \mu \end{matrix}\right]=\left[ \begin{matrix} x \\ y \end{matrix}\right] \Rightarrow x=y=0.$$

#### Theorem

(G. 2021)
Let $d = 2\varphi(n)$. Suppose $f:D_{\mathbb{R}}^2 \rightarrow \mathbb{R}$ is a compactly supported bounded measurable function. Then, the following holds. \begin{align} \int_{Y_d} \Phi_f = \int_{SL_2(D_\mathbb{R})/SL_2(\mathcal{O})}^{} \left( \sum_{v \in g \mathcal{O}^{\oplus 2} \setminus \{ 0\}}^{}f(v) \right) dg = \int_{\mathbb{R}^{d}}^{} f(x) dx, \end{align} where the $dx$ on the right-hand side is that Lebesgue measure on $\mathbb{R}^{d}$ that makes $\mathcal{O}^{\oplus 2}$ of unit covolume and $dg$ is the unique $SL_2(D_\mathbb{R})$-invariant probability measure on $Y_d$.

To get packing bounds, fix a finite subgroup $G_0 \subseteq \mathcal{O}^*$ to act diagonally on the right of $\mathcal{O}^{\oplus 2}$. Choose $f$ to be the indicator function of a ball invariant under the right action of $G_0$. Then $\Phi_f(\Lambda) \in \{0, |G_0|, 2|G_0|, \dots\}$ for all $\Lambda \in Y_d$.

Conclusion: We get the bounds $$c_{2 \dim_{\mathbb{Q}} D} \ge \# G_0.$$

Let $(G_{0},\mathcal{O}, D )$ be as finite subgroup inside an order inside a $\mathbb{Q}$-division algebra respectively. Then there exists a lattice packing in dimensions $d= 2\dim_{\mathbb{Q}} D$ whose packing efficiency is at least $\frac{1}{2^{d}} ( \#G_{0} )$.

Note that the following tightening can be done once we have a tuple $(G_{0},\mathcal{O},D)$. When $D$ is a $\mathbb{Q}$-division algebra, the $\mathbb{Q}$-span of $G_{0}$ in $D$ is also a division algebera. Let $\mathbb{Z}\langle G_{0}\rangle \subseteq \mathcal{O}$ be the $\mathbb{Z}$-span of $G_{0}$, then we get that $(G_{0},\mathbb{Z}\langle G_{0}\rangle ,\mathbb{Q}\langle G_{0}\rangle )$ is another tuple that gives us a lower bound.

So we only need to find finite subgroups $G_0$ that live in some $\mathbb{Q}$-division algebra $D$ and we can get the rest.

Fortunately, there exists a complete classification of such finite subgroups due to Amitsur, 1955.    ## Summary of Amitsur's result

The following construction $D=(E/F,\sigma,\gamma)$ is called a cyclic division algebra.

• $\DeclareMathOperator{\Gal}{Gal}$ $\DeclareMathOperator{\N}{N}$ $F$ is a number field over $\mathbb{Q}$,
• $E/F$ is a cyclic extension of degree $n$, i.e. the field extension $E/F$ is Galois and the Galois group is cyclic,
• $\sigma$ is a generator of the cyclic group $\Gal(E/F)$ and
• $\gamma \in F^{*}$, with the property that the multiplicative order of $\gamma$ in the group $K^{*}/ N_{F}^{E}(E^{*})$ is exactly $n$. That is, $\gamma^{k} \notin N_{F}^{E}(E^{*})$ for any $k \in \{ 1,2,\dots,n-1\}$ and $\gamma^{n} = N_{F}^{E}(x)$ for some $x\in E^{*}$.
When this happens we say that $\gamma \in F^{*}$ is a non-norm element.

Consider a formal element $b$ that does not commute with $E$ and satisfies $b^{n} = \gamma$. $D$ is now defined as per the isomorphism \begin{align} \label{eq:iDentify} D \simeq E \oplus E b \oplus E b^{2} \oplus \dots \oplus E b^{n-1},\end{align} with the rule that \begin{align} bl = \sigma(l)b \text{ for all }l \in E.\label{eq:ruleofD} \end{align}

If $\gamma$ does not satisfy the non-norm condition, then we call $(E/F,\sigma,\gamma)$ just a cyclic algebra.

All division algebras over $\mathbb{Q}$ are cyclic division algebras!

The dimension of $D=(E/F,\sigma,\gamma)$ over $\mathbb{Q}$ is $\dim_{\mathbb{Q}}D = [F:\mathbb{Q}]n^2 = [E:\mathbb{Q}]n$.

For $1 \le r \le m$ and $(m,r)=1$, consider the integers

• $\DeclareMathOperator{\ord}{ord} n = \ord_{m} r$ is the multiplicative order or $r$ modulo $m$, that is the smallest postive integer $k$ such that $m \mid r^{k}-1$,
• $s = \gcd(r-1,m)$,
• $t = m/s$.

When $r=1$, we will assume $n=s=1$. In these definitions, we will think of $m$ and $r$ as two parameters and $n,s,t$ will automatically be set as defined above.

With this, consider the cyclic algebra $\mathfrak{U}_{m,r} = (\mathbb{Q}(\mu_{m}), F , \sigma_{r}, \mu_{m}^{t} )$, where $F$ is the subfield of $\mathbb{Q}(\mu_{m})$ fixed by $\sigma_{r}$, and $\sigma_{r}$ is the field automorphism of $\mathbb{Q}(\mu_{m})$ given by $\mu_{m} \mapsto \mu_{m}^{r}$.

A priori, $\mathfrak{U}_{m,r}$ is just a $\mathbb{Q}$-algebra which may not be a division algebra. For this to be a division algebra, we want that $\mu_{m}^{t}$ is a non-norm element of $F$.

#### Theorem

(Amitsur, 1955)
Consider the following conditions on the numbers $m,r,n,s,t$ defined above. Then $\mathfrak{U}_{m,r}$ is a division algebra if and only if both of the following hold.
• One of the following two conditions hold.
• $\gcd(n,t) = 1$. This implies that $\gcd(s,t) = 1$.
• $n=2n', m=2^{\alpha} m',s=2s'$, for some $\alpha \ge 2$ and $m',s',n'$ are odd numbers, such that $\gcd(n,t)=\gcd(s,t)=2$ and $2 ^{\alpha} \mid (r + 1)$.
• One of the following two conditions hold.
• $n=s=2$ and $m \mid (r +1)$.
• For every prime $q \mid n$ there exists a prime $p \mid m$ such that if $m = p^{\alpha} m'$ with $p \nmid m'$, we get $q \nmid \ord_{m'} r$. In addition, at least one of the following must hold regarding $p,q$.
• $p \neq 2$ and $\gcd(q,\frac{ p^{ \delta} - 1}{s} ) = 1$, where $\delta = \ord_{m'} p$.
• $p=q=2$ and $m/4 \equiv \delta \equiv 1 \pmod{2}$, where $\delta$ is as above.

The division algebra $\mathfrak{U}_{m,r}$ contains the following group.

\begin{align} G_{m,r}:= \langle A,B \ | \ A^{m} = 1, B^{n} = A^{t}, BAB^{-1} = A^{r}\rangle. \end{align} We get that $|G_{m,r}|= mn = m \ord_{m}r$. When $r=1$, $G_{m,r}$ is a cyclic group of order $m$.

It embeds as per $i:G_{m,r} \rightarrow \mathfrak{U}_{m,r}^{*}$ defined sending $A \mapsto \mu_{m}$ and $B \mapsto b$.

When $\ord_m 2$ is odd, one can get a slightly better group of size $24 |G_{m,r}|$ embedded in the division algebra $\left(\tfrac{-1,-1}{\mathbb{Q}}\right) \otimes_{\mathbb{Q}} \mathfrak{U}_{m,r}$.

Amitsur then proves that apart from two more sporadic examples, these two are the only families of examples.

#### Theorem

(Amitsur, 1955)
The following is an exhaustive list of finite groups $$G_{0}$$ that can be embedded in some $$\mathbb{Q}$$-division algebra $$D$$.
 Group structure Conditions on the parameters Size of the group Dimension of the smallest division algebra containing the group $$G_{0} \subseteq D^{*}$$ $$\#G_{0}$$ $$\dim_{\mathbb{Q}} \mathbb{Q}\langle G_{0}\rangle$$ $$\mathfrak{D}^{*}$$ $$48$$ $$16$$ $$\mathfrak{I}^{*}$$ $$120$$ $$20$$ $$G_{m,r}$$ $$(m,r) = 1,$$ $$\mathfrak{U}_{m,r}$$ is a division algebra $$m\ord_{m} r$$ $$\varphi(m) \ord_{m} r$$ $$\mathfrak{T}^{*} \times G_{m,r}$$ $$\ord_{m} 2$$ is odd and the rest as above 24 $$m\ord_{m} r$$ 4 $$\varphi(m) \ord_{m} r$$

Here $$\mathfrak{T}^{*}, \mathfrak{D}^{*}, \mathfrak{I}^{*}$$ are the binary tetrahedral group, binary octahedral group and binary icosahedral group respectively. They are finite groups whose respective size is 24,48 and 120.

Note that in the two families, $\# G_0 / \dim_{\mathbb{Q}} \mathbb{Q} \langle G_0 \rangle = m/\varphi(m)$ or $6 m / \varphi(m)$. Looking at this, we can conclude that along no sequence of new constructions will we get $c_d > O(d \log \log d)$.

We can leverage the second family to get a sequence that shows that $c_d \ge O(d (\log \log d)^{7/24})$

$\DeclareMathOperator{\ord}{ord}$ We pick $$m = \prod_{ \substack{ p \text{ is prime} \\ p \le x \\ 2 \nmid \ord_{p} 2} }^{} p$$ and $r=1$. Then observe that with this, we get that $m$ is odd and $\ord_{m}2$ is also odd. Using the classification just given before, we get that $c_{8\varphi(m)} \ge 24m$. This sequence was the one picturized previously.

The following is also an interesting sequence made of new examples.

Let $\{ p_1,p_2,\dots\}$ be the sequence of all primes and suppose $q= 1+\prod_{i=1}^{N}p_{i}$ is a prime for some $N$. Then we can choose an integer $r$ such that $r \equiv 1 \pmod{p_i}$ for each $i$ but has $\ord_{q} r = q-1$, i.e. $r$ is a generator of $\mathbb{F}_{q}^{*}$.

Set $m=q(q-1) = q \prod_{i=1}^{N} p_i$. Then, one can check that $\mathfrak{U}_{m,r}$ is a division algebra so we can have $$c_{2 (q-1)^{2}\varphi(q-1) } \ge q(q-1)^{2}.$$ If there are infinitely many such primes, then this sequence asymptotically approaches $O(d\log \log d)$.

Summary: Amitsur's work allows us to conclude that there are no asymptotic improvements possible when we allow some non-commutative symmetries.

## Effectiveness

These results prove existence of a lattice with a high packing density. Can we actually construct such lattices algorithmically?

#### Theorem

Let $D$ be a finite-dimensional division algebra over $\mathbb{Q}$. Let $\mathcal{O} \subseteq D$ be an order in $D$ and $G_{0} \subseteq \mathcal{O}$ be a finite group embedded in the multiplicative group of $D$. Then in $d=2\dim_{\mathbb{Q}}D$ dimensional real space, there exists a finite time algorithm to construct a lattice $\Lambda \subseteq \mathbb{R}^d$ such that the packing density of $\Lambda$ is at least \begin{align} \tfrac{1}{2^d} \# G_{0} - \varepsilon, \text{ for every }\varepsilon . \end{align}

The approach is based on methods of Moustrou (2017). The idea is that $\int_{Y_d} \Phi_f$ can be approximated by averaging $\Phi_f$ over a growing family of finitely many lattices.

## Thank you for your attention!

Email:
nihar.gargava@epfl.ch

arxiv:
2107.04844

Slides:
nihargargava.com/divalg_seminar

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## Appendix: How to generate random 2-dimensional lattices

To the map $\psi:[\pi/3,2\pi/3]\times ]0,1] \rightarrow \mathbb{H}$ given by $\psi(a,b) = \cos(a) + i \sin(a)/b$ is a measure preserving map!

It maps the rectangle bijectively to a fundamental domain of $\mathbb{H}/SL_2(\mathbb{Z})$.

Using this, the following map randomly generates a lattice. $\psi_1: [0,2\pi]\times [\pi/3,2\pi/3] \times ] 0,1] \to SL_2(\mathbb{R})\\ \psi_1(x,y,z) = \left[ \begin{smallmatrix} 1 & \cos(y) \\ 0 & 1 \end{smallmatrix} \right] \left[ \begin{smallmatrix} \left(\frac{\sin(y)}{z}\right)^{\frac{1}{2}} & 0 \\ 0 & \left({\frac{\sin(y)}{z}}\right)^{-\frac{1}{2}} \end{smallmatrix} \right] \left[ \begin{smallmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{smallmatrix} \right]$

This only works for $d=2$. It is not known how to generalize this to higher dimensions!