Asymptotic Lower Bounds on Sphere Packing Efficiency of Lattices

\( \DeclareMathOperator{\R}{\mathbb{R}}\)\( \DeclareMathOperator{\Z}{\mathbb{Z}}\)\( \DeclareMathOperator{\C}{\mathbb{C}}\)\( \DeclareMathOperator{\F}{\mathbb{F}}\) 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 $0$.

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

Finding $c_d$ is very difficult for a general $d$. To show $c_d \ge K$, we must prove the existence of $g \in SL_d(\R)$ such that the origin centered ball $B$ with $\mu(B) = K$ 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 $SL_d(\R)/SL_d(\Z)$.

One can show this by constructing a "coarse" fundamental domain of the quotient space.

This gives us the probability space on which we want to use the integrality gap argument. Let us now introduce the random variable.

What we are empirically confirming is the following.

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 $d$!

The details of the Siegel's theorem are too lengthy for this talk. But there is another proof of $c_d \ge 2$ that can be proven within the next 15 minutes!

However the lattices in $\mathcal{L}_p'$ are not unit covolume. But each one of them has a covolume of $p^{d-1}$.

So appropriately normalizing, elements of this set become unit covolume lattices. $$\mathcal{L}_p= \{ C_p \pi_p^{-1}(\F_p v) \ | \ v \in \F_p^d \setminus \{ 0 \} \},$$ when $C_p = p^{-\left( 1- \frac{1}{d} \right)} $.

So $\mathcal{L_p} \subseteq X_d$ is a set of unit covolume lattices, with $\#\mathcal{L}_p \to \infty$ as $p \to \infty$.

Now as before let $f:\R^d \to \R$ be a compactly supported Riemann integrable function. Let $\Phi_f:X_d \to \R$ again be the lattice-sums of $f$.

So let us try to find $\Phi_f$ on $\mathcal{L}_p$. We will assume that the prime $p$ is very large. \begin{align} \frac{1}{\#\mathcal{L}_p}\sum_{\Lambda \in \mathcal{L}_p} \Phi_f(\Lambda) \end{align}

So for large $p$, we can ignore the first term. Or, we can assume that the coeffecient behind the first term was $ (\# \mathcal{L}_p)^{-1}$. So we get that \begin{align} \frac{1}{\#\mathcal{L}_p}\sum_{\Lambda \in \mathcal{L}_p} \Phi_f(\Lambda) & = \frac{1}{\#\mathcal{L}_p} \sum_{v \in C_p \Z^d \setminus ( pC_p \Z^d )} f(v) +\frac{1}{\#\mathcal{L}_p} \sum_{v \in (pC_p\Z^d) \setminus \{0\} } f(v) \\ & = \frac{1}{\#\mathcal{L}_p} \sum_{v \in C_p \Z^d \setminus \{0\} } f(v) . \end{align}

And now, shocking fact!
$ ( C_p^d \# \mathcal{L_p} )\to 1$ as $p \to \infty$ (Easy combinatorics exercise.)

So in fact, as $p \to \infty$, the sum on the right converges to the Riemann integral of $f$ on $\R^d$. $$ \frac{1}{\#\mathcal{L}_p} \left( \sum_{v \in (\# \mathcal{L}_p)^{\frac{1}{d}} \Z^d \setminus \{0\} } f(v) \right) \to \int_{\R^d} f(x) dx . $$

What we have managed to show (with proof!) is

After using the integrality gap lemma, this is a constructive proof of $c_d \ge 2$.

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.

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 $d=960$.

These bounds don't perform well in low dimensions!

However, efforts of constructing explicit lattice have not been so fruitful and already in dimensions $d > 1000$, one of the mentioned asymptotic bounds is the best possible known lower bound.

Thanks to all the background that we have built, we can actually show Venkatesh's 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.

The following is Venkatesh's construction.

Start with $K = \mathbb{Q}[\mu_n]$ , is the $n$th cyclotomic number field. Then $K_{\R}= K\otimes_{\mathbb{Q}}\R$ is a $\varphi(n)$-dimensional real vector space. Set $d = 2 \varphi(n)$

Take $\R^{d} \simeq K_{\R}^{\oplus 2}$. Identify $\Z^{2\varphi(n)}$ with $\mathcal{O}_K^{\oplus 2}$, two copies of the ring of integers in this vector space. Take $G = \Z/n\Z$ acting on this lattice via $\langle \mu_n \rangle$, multiplication by the $n$th roots of unity.

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

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

Again, with this bijection, we will make $Y_d$ a topological space. But does there exists a probability measure on $Y_d$?

Fortunately, the following theorem exists.

So $Y_d$ now has 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 $\Phi_f(\Lambda) \in n \Z$ for $\Lambda \in Y_d$. Such a quadratic form always exists by averaging!

Venkatesh, then proves the following analogue of Siegel's theorem.

This along with the integrality lemma gives $c_{2\varphi(n)} \ge n$.

But what if I told you there is a much more elegant constructive proof?

This approach was developed by Moustrou (2017). It is perfect analogue to Rogers (1947).

The proof of this theorem is exactly the same as Rogers ('47) with very little changes.

Set $r=2$, $K=\mathbb{Q}[\mu_n]$ and one can have a growing collection of finitely many lattices $\mathcal{L}_\mathcal{P}$ such that one of them will have packing efficiency $\ge n$, in space of dimension $2\varphi(n)$.