Presentation by

IRMA, Strasbourg

4 March 2024

\(\DeclareMathOperator{\R}{\mathbb{R}}\)\(\DeclareMathOperator{\Z}{\mathbb{Z}}\)\(\DeclareMathOperator{\C}{\mathbb{C}}\)\( \DeclareMathOperator{\F}{\mathbb{F}}\)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!

Visualizing in $\R^2$

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. In this talk, we will only focus on lower bounds.

Lower bound | Contribution of | Dimensions covered |
---|---|---|

$c_d \ge 1$ | Minkowksi-Hlawka (~1896-1940) | $\forall ~d \ge 2$ |

$c_d \ge 2(d-1)$ | 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., 2022) , using a probabilistic argument on lattices that lie in $(D \otimes_\mathbb{Q} \mathbb{R})^2$, $D$ is a division algebra over $\mathbb{Q}$.

Comparison of bounds

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 division algbera construction gives 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.

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 $SL_d(\R)/SL_d(\Z)$.

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

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

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.

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 $d$!

Both Venkatesh and the division algebra lattices use this idea. What we want is to find expectation of the lattice sum over a smaller subset of lattices that have a larger group of symmetries.

For Venkatesh, the group of symmetries is always a cyclic group. For the new results, the symmetries are non-commutative.

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.

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 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 $$Y_d = \{ g ( \mathcal{O}_K^{\oplus 2} ) \ | \ g \in SL_2(K_{\R}) \} \simeq SL_2(K_\mathbb{R} ) /SL_2(\mathcal{O}_K). $$

$Y_d$ can be given probability measure.

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

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 $n$!

What about other homogeneous spaces of lattices? All of the above integral formulas were on spaces $\mathcal{G}(\mathbb{R})/ \Gamma$ for some algebraic $\mathbb{Q}$-group $\mathcal{G}$ and an arithmetic subgroup $\Gamma$.

If the group $\mathcal{G}$ is semisimple, we are automatically guaranteed that $\mathcal{G}(\mathbb{R})/ \Gamma$ will be a probability space under the natural left $\mathcal{G}(\mathbb{R})$-invariant Haar measure

Contribution of | Algebraic group | Arithmetic group | Lower bound |
---|---|---|---|

Siegel (1945) | $SL_d(\mathbb{R})$ | $SL_d(\mathbb{Z})$ | $c_d \ge 2$ |

Rogers (1947) | $SL_d(\mathbb{R})$ | $SL_d(\mathbb{Z})$ | $c_{t} \ge 2 \tfrac{t}{e}$ |

Vance (2011) | $SL_t(\mathbb{H})$ | $SL_t(\mathcal{O})$,
$\mathcal{O}=$ Hurwitz-Integers |
$c_{4t} \ge 24 \tfrac{t}{e} $ |

Venkatesh (2013) | $SL_2(K_\mathbb{R})$
$K= \mathbb{Q}(\sqrt[n]{1})$ |
$SL_2(\mathcal{O})$,
$\mathcal{O}=$ Integers |
$c_{2\varphi(n)} \ge n$ |

G. (2022) | $SL_2(D_\mathbb{R})$
$D=$ division ring containing finite group $G_0$ |
$SL_2(\mathcal{O})$,
$\mathcal{O}=$ order in $D$ |
$c_{2\dim(D)} \ge \# G_0$ |

G.-Serban (2022) | $SL_t(D_\mathbb{R})$
$D=$ division ring containing finite group $G_0$ |
$SL_t(\mathcal{O})$,
$\mathcal{O}=$ order in $D$ |
$c_{t\dim(D)} \ge \# G_0 \tfrac{t}{e} $ |

AndrĂ© Weil discussed a much more general formula in 1965.

His formula is an an "adelic" setting, but it can be used to derive integration all the integration formulas that have been used so far!

In particular, one can also derive the higher-moment formulas due to Rogers.

\(\DeclareMathOperator{\vol}{vol}\)For any $\mathbb{Q}$-subspace $ S \subseteq \mathbb{Q}^{n}$, we can define a height $H(S) $ as \begin{equation} H(S) = \vol\left( \frac{S \otimes_\mathbb{Q} \mathbb{R}}{ S \cap \mathbb{Z}^{n}} \right), \end{equation} the volume taken with respect to natural measure for subpaces of $\mathbb{R}^{n}$.

For $n < d$, the higher moment converges absolutely and is equal to the following \begin{align} \int_{SL_d(\mathbb{R})/SL_d(\mathbb{Z})} \left(\sum_{v \in g\mathbb{Z}^{d} \setminus \{0\} } f(v)\right)^{n} = \sum_{\text{subspaces}~S \subseteq \mathbb{Q}^{n}} \frac{1}{H(S)^{d}} \int_{S^{\perp}(\mathbb{R})} f^{\otimes n}(v) dv, \end{align} where \begin{equation} S^{\perp}(\mathbb{R}) = \{ (v_1,v_2,\dots,v_n) \in (\mathbb{R}^{d})^{n } \ | \ \sum \alpha_i v_i = 0~\text{for each}~(\alpha_1,\dots,\alpha_n) \in S\}. \end{equation} and the integral is with respect to the natural measure on it as a subspace in $\mathbb{R}^{d \times n}$.

Why is it a finite number?

Let $m < n$ and let $G(m,n)$ be the set of $m$-dimensional subspaces of $\mathbb{Q}^n$. Then \begin{equation} \# \{ S \in G(m,n) \ | \ H(S) \le T \} \asymp T^{n} \end{equation}

The formula basically writes the higher moment as a "height zeta function" value. In fact the fact that the right hand side exists is the equivalent to saying that the height zeta functions on Grassmannian varieties absolutely converge at special values.

Knowing higher moments of Siegel gives more information about random lattices!

When $n \ll \sqrt{d}$, the higher moments start behaving like moments of a Poisson distribution.

Let $\Lambda \subseteq \mathbb{R}^{d}$ be a random unit covolume lattice and let $B$ be a ball at origin of volume $V$. Consider the random variable $\rho(\Lambda) = \left( \# B \cap \Lambda \setminus \{ 0\} \right)$. Then, provided $d\geq \lceil \tfrac{1}{4}n^{2}+3\rceil,$ we have $$ 2^n \cdot m_{n}( \tfrac{V}{2} ) \leq \mathbb{E}[\rho(\Lambda)^n]\leq 2^n\cdot m_{n}( \tfrac{V}{2} ) +E_{n,d}\cdot (V+1)^{n-1},$$ where \begin{equation} m_{n}(\lambda) = e^{-\lambda}\sum_{r=0}^\infty\frac{\lambda^{r}}{r!} r^{n} = \mathbb{E}_{X \sim \mathcal{P}(\lambda) }(X^{n}) \label{eq:def_of_poisson} \end{equation} is the $n$th moment of a Poisson distribution with parameter $\lambda$ and where $E_{n,d}$ is an error term decaying exponentially as $d$ increases: $$E_{n,d} \le 2\cdot 3^{ \lceil \tfrac{n^2}{4}\rceil}\cdot ( \tfrac{\sqrt{3}}{2})^d+21\cdot 5^{ \lceil \tfrac{n^2}{4}\rceil}\cdot (\tfrac{1}{2})^{d}.$$

Incidentally, the $\mathcal{O}_{K}$-lattices considered by Venkatesh are also used in coding theory and cryptography.

We fix $t \in \mathbb{Z}_{\ge 2}$ and look at the following random set of lattices.
$$ \{ g \cdot \mathcal{O}_K^{\oplus t} \ | \ g \in SL_t(K_{\R}) \} \simeq SL_t(K_\mathbb{R} ) /SL_t(\mathcal{O}_K). $$

We call these lattices random $\mathcal{O}_K$ lattices.

Each $\mathcal{O}_K$ lattice $\Lambda \subseteq K_\mathbb{R}^t$ lives in a real space of dimension $t \times [K:\mathbb{Q}]$. We equip $K_\mathbb{R}^t$ with the "normalized trace form" so that each $\Lambda$ has unit covolume. On $K_{\mathbb{R}}$ it is \begin{equation} \langle x,y \rangle = \Delta_K^{-\frac{2}{[K:\mathbb{Q}]}} \text{Tr}(x\overline{y}). \end{equation} On $K_{\mathbb{R}}^t$, it is extended by just summing up "coordinate-wise".

Then does Poisson like behaviour also exist for this specific class of random lattices? We are interested in $t \rightarrow \infty$ and, also the more difficult scenario, $[K:\mathbb{Q}] \rightarrow \infty$.

Let $\Lambda \subseteq K_\mathbb{R}^{t}$ be a random $\mathcal{O}_K$-lattice. and let $B$ be a ball at origin of volume $V$. Consider the random variable $\rho(\Lambda) = \left( \# B \cap \Lambda \setminus \{ 0\} \right)$. Then, provided $t\geq t_0 = \Omega( n^3 \log \log n)$ we have $$ \omega_K^n \cdot m_{n}( \tfrac{V}{\omega_K} ) \leq \mathbb{E}[\rho(\Lambda)^n]\leq \omega_K^n\cdot m_{n}( \tfrac{V}{\omega_K} ) +E_{n,t,K}\cdot (V+1)^{n-1},$$ where $m_{n}(\lambda) = \mathbb{E}_{X \sim \mathcal{P}(\lambda) }(X^{n})$ as before and where $E_{n,t,K}$ is an error term satisfying $$E_{n,t,K} \ll e^{-\varepsilon (t-t_0) [K:\mathbb{Q}]\ } \text{ for some }\varepsilon > 0 ,$$ as long as $n$ is fixed and $t \cdot [K:\mathbb{Q}]$ grows, with $K$ always remaining in a class of number fields satisfying the Bogomolov property.

A class of number fields $\{ K_i \}_{i \in I}$ satisfies the Bogomolov property if there exists $c>0$ such that for each $\alpha \in \bigcup_i K_i$, either $\alpha$ is a root of unity or $$h(\alpha) = \tfrac{1}{\deg(\alpha)} \sum_{v \in \cal{M}_{\mathbb{Q}(\alpha) } } \max\{0, \log|\alpha|_v\} \geq c.$$

The Bogomolov property is satisfied by the class of cyclotomic number fields, the class of abelian number fields, and most importantly, for a fixed number field (so $t$ can grow and $K$ can be fixed).

But there are classes of number fields where the Bogomolov property fails. In fact, we can show that for one such class of number fields, the second moment is not the Poisson distribution.

First you need a higher moment formula. For this, we can use Weil's work (difficult to read) or some recent preprints by S. Kim or N. Hughes or my doctoral thesis.

For any number $t$ of copies of $K_\mathbb{R}$ let $f:K_{\mathbb{R}}^{t} \rightarrow \mathbb{R}$ be a compactly supported continuous function. Then, putting the Haar probability measure on $SL_{t}(K_\mathbb{R})/SL_{t}(\mathcal{O}_K)$, we have that \begin{equation} \int_{SL_{t}(K_\mathbb{R})/SL_{t}(\mathcal{O}_K)} \left(\sum_{v \in \gamma \mathcal{O}_K^{ t}} f(v) \right)^n d \gamma = \sum_{m=0}^{{n}}\sum_{\substack{D \in M_{m \times n }(K) \\ \text{rank}(D) = m \\ D \text{ is row reduced echelon}}} \mathfrak{D}(D)^{-t} \int_{x \in K_\mathbb{R}^{t \times m }} f^{\otimes n}(x D ) dx, \end{equation} where $\mathfrak{D}(D)$ is the index of the sublattice $\{ C \in M_{1 \times m }(\mathcal{O}_K) \mid C \cdot D \in M_{1 \times n}(\mathcal{O}_K)\}$ in $M_{1 \times m}(\mathcal{O}_K)$.

What follows is many pages of calculations!

We split the matrices into four different types of matrices. One type of matrices are the main terms, the rest are error terms.

It is a sum over integrals like the following with $a_{ij} \in K$. $$\int_{x_1,x_2,\cdots , x_m \in K_{\mathbb{R}}^t} f(a_{11} x_1 + \cdots + a_{1m}x_m ) f(a_{21}x_1 + \cdots + a_{2m}x_m) \\ \cdots f(a_{n1}x_1 + \cdots + a_{nm}x_m ) dx_1 ... dx_m .$$

When $f$ is the indicator function of a ball, this is the volume of intersection of $n$ ellipsoidal cylinders with a subspace.

For some types of matrices (including the main term), we can actually do this integration exactly.

For other types, we use a convex combination argument. Ellipsoid intersection with subspace is another ellipsoid whose volume is possible to calculate! We can then relate the volume of this to the height of the matrix entries.

\(\DeclareMathOperator{\OK}{\mathcal{O}_K}\)Here are some of the calculations for the second moment. Let $f = \mathbb{I}_B$ be the indicator function of an origin-centered ball $B$. The integration formula first gives that our expected value is $$\vol(B)^2+\sum_{\alpha\in K^\times}[\OK:\alpha^{-1}\OK \cap\OK]^{-t}\cdot\vol(B\cap\alpha B).$$

To arrive at our result it suffices to prove exponential decay of the sum \begin{equation} \sum_{\alpha\in (K^\times\setminus\mu_K)/\mu_K}[\OK:\alpha^{-1} \OK \cap\OK]^{-t}\cdot\frac{\vol(B\cap\alpha B)}{\vol(B)}, \end{equation}

The main terms are then contributed by $\alpha \in \langle \mu_n \rangle$.

For the error terms, we write $S_\beta=\sum_{\alpha\in (\OK^\times\setminus\mu_K)/\mu_K}\frac{\vol(B\cap\alpha\beta B)}{\vol(B)}$ and evaluate $$\sum_{\beta \in (K^{\times}\setminus \OK^{\times})/\OK^{\times}} [\OK:\beta^{-1} \OK \cap\OK]^{-t} S_{\beta}.$$

We then approximate $$\frac{\vol(B\cap\alpha\beta B)}{\vol(B)} \leq \frac{1}{\cosh( 2 \cdot h(\alpha \beta))^{t[K:\mathbb{Q}]/2 }}$$ using the convex combinations idea. Here the Bogomolov property helps to prove exponential decay.

And then it remains to put a strict inequality to bound the point counts $$\# \{ \alpha \in \OK^{\times} \mid h(\alpha \beta ) \leq T\}$$ in terms of $\beta,T$. This is done using a packing argument in the log embedding of units.

Then, combine the point counts with exponential decay to get $$ S_{\beta} \ll N(\beta)^{-t \varepsilon_1} \cdot \exp( \varepsilon_2 (t[K:\mathbb{Q}])$$

Keeping track of the constants, we can bound the sum of $S_{\beta}$ over $\beta$ with a special value of Dedekind zeta function.

nihar.gargava@epfl.ch

2107.04844, 2111.03684, 2308.15275

nihargargava.com/leipzig_seminar

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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!