Kavita Dhanda and Alan Haynes · Journal of Number Theory 268 (2025) · my read of their paper
Take an irrational number $\alpha$ and multiply it by $1, 2, 3, \dots$ Each multiple $q\alpha$ lands some distance from the nearest whole number $p$, and that distance $q\alpha - p$ is the whole subject of Diophantine approximation. The classical question is how small it gets. This paper asks a different question: multiply the error by a power $|q|^\eta$ to level the playing field, and look at every value $|q|^\eta(q\alpha - p)$ at once, for every $q$ and every $p$. Where do those values pile up?
Kavita and Alan are my coauthors on the Peck's theorem paper, and this is the paper that paper grew out of: it's where the dual lattice, the dominant units, and the trick of writing an approximation as $s u$ for a lattice element $s$ and a unit $u$ were first set up. Its answer comes in two halves. For a typical $\alpha$, the values pile up everywhere, as long as $\eta$ isn't too big, and the paper measures exactly how the exceptional $\alpha$ fail. For an algebraic $\alpha$, the values pile up on something you can draw: a discrete set of points on the line, or a nest of ellipses or hyperbolas in the plane, scaled by the norms of a number field. The missing rings in those pictures are the inert primes. Every figure on this page is computed from the paper's own examples, with exact integer arithmetic wherever the integers get large (and they get to $10^{15}$ fast).
Everything on this page is built from one set. Vectors are written in bold in the paper; here $\alpha \in \mathbb{R}^d$ and $p \in \mathbb{Z}^d$ are vectors, $q$ is an integer, and $|x| = \max_i |x_i|$ is the sup norm.
For $\alpha \in \mathbb{R}^d$ and $\eta \in \mathbb{R}$, the set of $\eta$-normalized approximations to $\alpha$ is
$$\mathrm{NA}_\eta(\alpha) = \left\{ |q|^\eta (q\alpha - p) : q \in \mathbb{Z} \setminus \{0\},\ p \in \mathbb{Z}^d \right\}.$$For $d = 1$ it's a set of real numbers: the error of each multiple $q\alpha$, scaled by $|q|^\eta$. Here it is for $\alpha = \sqrt{2}$ and $\eta = 1$. Each row is one $q$; the short arrow is the raw error $q\sqrt{2} - p$ to the nearest whole number, the long one is the same error stretched by $q$, and the stretched values drop onto the line at the bottom.
The dots at the bottom are not landing just anywhere. That's the first hint of the second half of the paper, and we'll come back to it. First, the words for "where the values pile up."
$\mathcal{A}_\eta(\alpha)$ is the set of all accumulation points of $\mathrm{NA}_\eta(\alpha)$: the points $\gamma \in \mathbb{R}^d$ such that every neighborhood of $\gamma$ contains infinitely many elements of $\mathrm{NA}_\eta(\alpha)$.
Unwinding the definition, $\gamma \in \mathcal{A}_\eta(\alpha)$ exactly when for every $\epsilon > 0$ there are $q \neq 0$ and $p$ with
$$0 < \left| q\alpha - \frac{\gamma}{|q|^\eta} - p \right| < \frac{\epsilon}{|q|^\eta}.$$Read it as: $q\alpha$ comes within $\epsilon/|q|^\eta$ of the shifted lattice $\mathbb{Z}^d + \gamma/|q|^\eta$. The whole paper is about two sets of $\alpha$ built from $\mathcal{A}_\eta$.
$\alpha$ is $\eta$-well approximable, written $\alpha \in \mathcal{W}(\eta)$, if $0 \in \mathcal{A}_\eta(\alpha)$: the normalized errors come arbitrarily close to $0$.
$\alpha \in \mathcal{D}(\eta)$ if $\mathcal{A}_\eta(\alpha) = \mathbb{R}^d$: every point of $\mathbb{R}^d$ is an accumulation point of the normalized approximations.
$\mathcal{W}(\eta)$ is the classical object: $\alpha \in \mathcal{W}(\eta)$ means $|q|^\eta \|q\alpha\|$ gets as small as you like, where $\|x\|$ is the distance to the nearest integer vector. $\mathcal{D}(\eta)$ is the new one, and it's stronger: $\mathcal{D}(\eta) \subseteq \mathcal{W}(\eta)$ for every $\eta$. Try it: pick $\alpha$ and slide $\eta$. The strip shows every normalized approximation with $q \leq 10^5$ that lands in the window; when $\alpha$ is $\sqrt{2}$ or the golden ratio and $\eta = 1$, the orange ticks are the places the paper says the dots must go.
On the Peck page the whole game was finding one $n \leq T$ whose errors $n\alpha_i - p_i$ are all small, with a logarithm shaved off. Here nobody is looking for the best approximation. Every pair $(q, p)$ gets a dot, the good ones and the terrible ones, and the question is the shape of the whole cloud.
Before the new sets, the old ones, because the paper's Theorems 1–3 are built to sit next to them. Two facts do most of the work. The first is Dirichlet's theorem in $d$ dimensions.
For every $\alpha \in \mathbb{R}^d$ there are infinitely many $q \in \mathbb{N}$ with
$$q^{1/d}\,\|q\alpha\| < 1.$$So $\mathcal{W}(\eta) \supseteq \mathbb{R}^d \setminus \mathbb{Q}^d$ for every $\eta < 1/d$: the exponent $1/d$ is free. (Rational $\alpha$ are excluded because $\|q\alpha\| = 0$ infinitely often there, and the definition of an accumulation point asks for points near $0$, not equal to it.) The second fact says what happens at and beyond the free exponent, for almost every $\alpha$.
Let $\psi \colon \mathbb{N} \to (0, \infty)$ be decreasing. If $\sum_{q=1}^{\infty} (q\psi(q))^d = \infty$, then for almost every $\alpha \in \mathbb{R}^d$ there are infinitely many $q \in \mathbb{N}$ with $\|q\alpha\| \leq q\psi(q)$. If the sum converges, then for almost every $\alpha$ there are only finitely many.
With $\psi(q) = c\,q^{-1-1/d}$ the sum is $c^d \sum 1/q$, which diverges, so for almost every $\alpha$ the inequality $q^{1/d}\|q\alpha\| \leq c$ holds infinitely often for every $c > 0$: almost every $\alpha$ is in $\mathcal{W}(1/d)$. With $\eta > 1/d$ the sum converges and $\mathcal{W}(\eta)$ has measure zero. The exceptional set at the free exponent has a name.
$\alpha \in \mathbb{R}^d \setminus \mathbb{Q}^d$ is badly approximable if $\displaystyle \inf_{q \in \mathbb{N}} q^{1/d}\,\|q\alpha\| > 0$. The badly approximable points are exactly $\mathcal{W}(1/d)^c$.
They have measure zero but full Hausdorff dimension $d$ (Jarník for $d = 1$, Schmidt for $d \geq 2$), and for $\eta > 1/d$ the set $\mathcal{W}(\eta)$ has dimension $\frac{d+1}{1+\eta}$ (Jarník–Besicovitch). Here is the paper's Table 1, and under it the same story as a picture: the strip of $\mathrm{NA}_\eta$ for a typical $\alpha$ and for $\sqrt{2}$, as $\eta$ sweeps from negative to large.
| $\eta$ | $\mathcal{W}(\eta)$ | $\mathcal{W}(\eta)^c$ |
|---|---|---|
| $\eta < 0$ | $\mathbb{R}^d$ | $\emptyset$ |
| $0 \leq \eta < 1/d$ | $\mathbb{R}^d \setminus \mathbb{Q}^d$ (Dirichlet) | $\mathbb{Q}^d$ |
| $\eta = 1/d$ | $\lambda(\mathcal{W}) = \infty$, $\lambda(\mathcal{W}^c) = 0$ (Khintchine, Groshev) | $\dim_H = d$ (Jarník, Schmidt) |
| $\eta > 1/d$ | $\dim_H = \frac{d+1}{1+\eta}$ (Jarník, Besicovitch) | $\lambda(\mathcal{W}^c) = \infty$, $\lambda(\mathcal{W}) = 0$ |
Two things to watch in that sweep. For $\eta < 0$ the dots are everywhere no matter what $\alpha$ is, because the scale factor $|q|^\eta$ shrinks everything toward $0$ and $\mathbb{Q}$ is dense. Past $\eta = 1$ (that's $1/d$ for $d = 1$) the typical $\alpha$ runs out of dots, and $\sqrt{2}$ ran out even earlier: it's badly approximable, so its dots never get near $0$ at $\eta = 1$.
Now the new set. $\alpha \in \mathcal{D}(\eta)$ asks for much more than $\mathcal{W}(\eta)$: not only does $0$ have to be an accumulation point of the normalized approximations, every $\gamma$ does. Two cases were already known. For $\eta = -1$, $\mathcal{D}(\eta) = \mathbb{R}^d$ by the density of $\mathbb{Q}^d$. For $\eta = 0$, the normalized approximations are just the points $q\alpha - p$, so $\mathcal{A}_0(\alpha) = \mathbb{R}^d$ says the orbit $\{q\alpha\}$ is dense modulo $\mathbb{Z}^d$, and Kronecker's theorem says exactly when that happens.
$K_d$ is the set of $\alpha \in \mathbb{R}^d$ for which $1, \alpha_1, \dots, \alpha_d$ are linearly independent over $\mathbb{Q}$. Kronecker's theorem: $\mathcal{D}(0) = K_d$.
If $\alpha \notin K_d$, the points $q\alpha$ lie on a rational hyperplane, so modulo $\mathbb{Z}^d$ they sit on a proper sub-torus and can never be dense. The same obstruction kills every $\eta \geq 0$: $K_d^c \subseteq \mathcal{D}(\eta)^c$. Here are both cases on the torus for $d = 2$: on the left $\alpha = (\sqrt{2}, \sqrt{3})$, on the right $\alpha_2 = 2\alpha_1 + \frac{1}{3}$, which is stuck on three lines forever.
Apart from $\eta = -1$ and $\eta = 0$, the paper says, "there do not appear to be any non-trivial cases of $\eta$ for which this problem is well understood." So they understood it. The measure theory first.
For $d \in \mathbb{N}$ and $\eta \leq 1/d$,
$$\lambda(\mathcal{D}(\eta)) = \infty \qquad \text{and} \qquad \lambda(\mathcal{D}(\eta)^c) = 0.$$Assume $0 < \eta \leq 1/d$ (smaller $\eta$ is covered by the cases above, since $\mathcal{D}(\eta)$ only grows as $\eta$ shrinks). For $n \in \mathbb{Z}^d$, $q \geq 1$, $\gamma \in \mathbb{R}^d$ and $\epsilon > 0$ let
$$\mathcal{E}_{n,q}(\gamma, \epsilon) = \left\{ x \in n + [0, 1)^d : 0 < \left| q^\eta(qx - p) - \gamma \right| < \epsilon \text{ for some } p \in \mathbb{Z}^d \right\},$$and let $\mathcal{E}_{n,\infty}(\gamma, \epsilon) = \limsup_{q \to \infty} \mathcal{E}_{n,q}(\gamma, \epsilon)$. By the definition of an accumulation point, $\gamma \in \mathcal{A}_\eta(\alpha)$ if and only if $\alpha$ lies in $\mathcal{E}_{n,\infty}(\gamma, \epsilon)$ for every $\epsilon > 0$ (with $n$ the cube containing $\alpha$). So it is enough to show $\lambda(\mathcal{E}_{n,\infty}(\gamma, \epsilon)) = 1$ for every $n$, $\gamma$, $\epsilon$: then for each fixed $\gamma$, almost every $\alpha$ has $\gamma \in \mathcal{A}_\eta(\alpha)$, so (a countable union of null sets) almost every $\alpha$ has $\mathbb{Q}^d \subseteq \mathcal{A}_\eta(\alpha)$, and $\mathcal{A}_\eta(\alpha)$ is closed.
For $\gamma = 0$ this is the Khintchine–Groshev theorem. For $\gamma \neq 0$, unwind the set: up to boundaries, $\mathcal{E}_{n,q}(\gamma, \epsilon)$ is the union over $p$ of the cubes
$$\prod_{i=1}^d \overline{B}\!\left( \frac{p_i}{q} + \frac{\gamma_i}{q^{1+\eta}},\ \frac{\epsilon}{q^{1+\eta}} \right)$$inside $n + [0, 1]^d$: little cubes centered at the rational points $p/q$ shifted by $\gamma/q^{1+\eta}$. Take $\epsilon = 2|\gamma|$. Then each cube contains the unshifted cube $\prod B(p_i/q, |\gamma|/q^{1+\eta})$, and the limsup of those is the set of $x$ with $\|qx\| \leq |\gamma| q^{-\eta}$ infinitely often, which has full measure by Khintchine–Groshev since $\sum (|\gamma| q^{-\eta})^d = \infty$ for $\eta \leq 1/d$. So $\lambda(\mathcal{E}_{n,\infty}(\gamma, 2|\gamma|)) = 1$.
Now shrink $\epsilon$. The cubes of $\mathcal{E}_{n,q}(\gamma, \epsilon)$ sit inside the cubes of $\mathcal{E}_{n,q}(\gamma, 2|\gamma|)$ with a fixed fraction $(\epsilon / 2|\gamma|)^d$ of their volume, so Cassels' lemma (below) says the two limsup sets have the same measure, namely $1$. One technicality: cubes cut by the boundary of $n + [0,1]^d$ aren't cubes, but there are $\ll q^{d-1}$ of them, each of volume $\ll q^{-d(1+\eta)}$, and $\sum q^{d-1-d(1+\eta)} = \sum q^{-1-d\eta} < \infty$, so by Borel–Cantelli throwing them away changes nothing. $\blacksquare$
The lemma that does the shrinking is a higher-dimensional version of one of Cassels', and it's the kind of thing that's obvious once you see the picture.
Let $\{C_n\}$ be hypercubes in $\mathbb{R}^d$ with faces parallel to the coordinate hyperplanes and diameters tending to $0$. If $\epsilon > 0$ and $U_n \subseteq C_n$ are measurable sets with $\lambda(U_n) \geq \epsilon\,\lambda(C_n)$ for all $n$, then
$$\lambda\!\left( \limsup_{n \to \infty} U_n \right) = \lambda\!\left( \limsup_{n \to \infty} C_n \right).$$Shrinking every ball by a fixed factor doesn't change which points are covered infinitely often, up to measure zero. The proof is the Lebesgue density theorem: a point of density of the big limsup set can't keep landing in the parts of the cubes that were thrown away. Here's the proof of Theorem 1 for $d = 1$, row by row: the dashed cubes are the classical ones around $p/q$, the blue ones are shifted by $\gamma/q^{1+\eta}$ and twice as wide, and the orange ones inside them are the $\epsilon$-cubes.
So for $\eta \leq 1/d$ almost every $\alpha$ is in $\mathcal{D}(\eta)$, and for $\eta > 1/d$ almost none are (because $\mathcal{D}(\eta) \subseteq \mathcal{W}(\eta)$). That settles the measure. The exceptional sets are where it gets interesting. For $0 < \eta < 1/d$ and $d \geq 2$, the complement $\mathcal{D}(\eta)^c$ contains $K_d^c$ (the rationally dependent $\alpha$, a countable union of hyperplanes), but it contains a lot more, and the extra points are singular.
$\alpha \in \mathbb{R}^d$ is singular if
$$\lim_{R \to \infty} R^d \cdot \min\left\{ \|r \cdot \alpha\| : r \in \mathbb{Z}^d,\ 0 < |r| \leq R \right\} = 0,$$and, for $\omega \geq d$, $E(\omega)$ is the set of $\alpha \in K_d$ for which the same limit with $R^\omega$ in place of $R^d$ is $0$.
Note the switch: this is about the linear form $r \cdot \alpha = r_1\alpha_1 + \cdots + r_d\alpha_d$ being close to an integer, not about $q\alpha$ being close to an integer vector. Dirichlet says $R^d \min \|r \cdot \alpha\| \leq 1$ always; singular means it goes to $0$. For $d = 1$ only rationals are singular, but Khintchine showed that for $d \geq 2$ there are singular points in $K_d$, and Baker, Yavid and Rynne pinned down their size: for $\omega > d$,
$$d - 2 + \frac{d}{\omega} \leq \dim_H E(\omega) \leq d - 2 + \frac{2d + 2}{\omega + 1}.$$Let $0 < \eta < 1/d$. If $d = 1$ then $\mathcal{D}(\eta)^c = K_1^c = \mathbb{Q}$. If $d \geq 2$ then
$$\frac{d}{1 + 1/\eta} \leq \dim_H\!\left( \mathcal{D}(\eta)^c \setminus K_d^c \right) \leq d - 2 + \frac{2(d+1)\big((1+\eta)d - 1\big)}{(1+\eta)d^2}.$$Upper bound. Let $S(\omega)$ be the set of $\alpha \in K_d$ with $Q^\omega \min_{0 < q \leq Q} \|q\alpha\| \to 0$: the $\alpha$ whose uniform approximation is better than Dirichlet's by a power. The key step is Lemma 14: for $\omega < 1/(d\eta + d - 1)$, $\mathcal{D}(\eta)^c \setminus K_d^c \subseteq S(\omega)$. To see it, take $\alpha \in K_d$ with $\alpha \notin S(\omega)$, so $\min_{q \leq Q} \|q\alpha\| \geq c_0 Q^{-\omega}$ for infinitely many $Q$. Cassels' transference principle (Lemma 7: if $\alpha$ is not too well approximable up to $Q$, then its orbit up to $Q' = \max\{Q, \delta^{-d}\}$ is $\varrho$-dense with $\varrho = \max\{\delta, 1/(Q\delta^{d-1})\}$) turns that into: the points $q\alpha + p$ with $Q_0 - Q_1 < q \leq Q_0$ are $(c_1/Q_1^{\eta'})$-dense in $\mathbb{R}^d$, where $Q_1 = c_2 Q^{d\omega}$ and $\eta' = \frac{1}{d\omega} - \frac{d-1}{d}$. The condition on $\omega$ is exactly $\eta' > \eta$. Now fix $\gamma \neq 0$ and $\epsilon > 0$. As $q$ runs over the window $(Q_0 - Q_1, Q_0]$, with $Q_0 = \rho Q_1/(\rho - 1)$ and $\rho = (1 + \epsilon/|\gamma|)^{1/\eta}$, the target $\gamma/q^\eta$ moves by less than $c_\epsilon/Q_1^\eta$, and for $Q$ large that is more than the density $c_1/Q_1^{\eta'}$ of the window's points, since $\eta' > \eta$. So some $q$ in the window has $\|q\alpha - \gamma/q^\eta\| < \epsilon/q^\eta$, which is $\gamma \in \mathcal{A}_\eta(\alpha)$. Hence $\alpha \in \mathcal{D}(\eta)$.
Then Lemma 10 (from the second transference principle, Lemma 8) gives $S(\omega) \subseteq E(d - 1 + d\omega)$, and the Baker–Yavid–Rynne upper bound with $\omega \to 1/(d\eta + d - 1)$ gives the right-hand side.
Lower bound. Let $E^+(\omega)$ be defined like $E(\omega)$ but with $r$ restricted to non-negative coordinates. Lemma 15: $E^+(1 + 1/\eta) \subseteq \mathcal{D}(\eta)^c \setminus K_d^c$. Suppose instead $\alpha \in E^+(1 + 1/\eta) \cap \mathcal{D}(\eta)$ and pick $\epsilon < \frac{1}{2(2d)^{1 + 1/\eta}}$. Since $(1, \dots, 1) \in \mathcal{A}_\eta(\alpha)$, there is a huge $q$ with $|q\alpha_i - 1/|q|^\eta - p_i| < \epsilon/|q|^\eta$ for all $i$; with $R = |q|^\eta/(2d)$ there is $r \geq 0$, $|r| \leq R$, with $\|r \cdot \alpha\| \leq \epsilon R^{-1-1/\eta}$. Then $(r_1 + \cdots + r_d)/|q|^\eta$ is within $\sum r_i |q\alpha_i - |q|^{-\eta} - p_i| + |q|\,\|r \cdot \alpha\|$ of an integer, which is less than $\frac{1}{2}\cdot\frac{r_1 + \cdots + r_d}{|q|^\eta}$ plus something tiny, impossible because $0 < (r_1 + \cdots + r_d)/|q|^\eta \leq \frac{1}{2}$. (The non-negativity of $r$ is what makes $r_1 + \cdots + r_d \neq 0$.) Lemma 16, a modification of Baker's Cantor-set construction with the mass distribution principle, gives $\dim_H E^+(\omega) \geq d/\omega$; the price of insisting on non-negative $r$ is the weaker bound. $\blacksquare$
The upper bound's mechanism deserves its own picture, because it's the heart of why "not singular" forces "every point is a limit." Below, $\alpha$ is a typical pair, the blue dots are the orbit points $q\alpha \bmod 1$ for $q$ in one window $(Q_0 - Q_1, Q_0]$, and the red disc is the target $\gamma/q^\eta$ with its allowed radius $\epsilon/q^\eta$. The target barely moves across the window, the window's points are finer than the target, so one of them lands inside.
For $\eta > 1/d$ the roles flip: $\mathcal{D}(\eta)$ is the small set, and the paper shows it is exactly as large as $\mathcal{W}(\eta)$.
For $d \in \mathbb{N}$ and $\eta > 1/d$,
$$\dim_H \mathcal{D}(\eta) = \frac{d + 1}{1 + \eta}.$$The upper bound is Jarník–Besicovitch, since $\mathcal{D}(\eta) \subseteq \mathcal{W}(\eta)$. For the lower bound, build a Cantor set $\mathcal{C} \subseteq \mathcal{D}(\eta)$. Enumerate $\mathbb{Q}^d = \{\gamma_i\}$ and let $(i_k) = (1, 1, 2, 1, 2, 3, 1, 2, 3, 4, \dots)$, so every rational target comes up infinitely often. Level $0$ is $[0, 1)^d$ with $Q_0 = 1$. Given the level $k - 1$ cubes $C_r$ of diameter $\delta_{k-1}$, shrink each to the concentric cube $C_r'$ of diameter $\delta_{k-1}/3$, and use Baker's lemma (Lemma 17: for $Q$ large, a cube of diameter $\delta$ contains $\geq c_0 \delta^d Q^{d+1}$ rational points $a/q$ with $q \leq Q$, pairwise $Q^{-(d+1)/d}$ apart, avoiding any given null set) to choose $Q_k$ and at least $m_k = c_0 \delta_{k-1}^d Q_k^{d+1}$ rational points $\beta = a/q$ in $C_r'$ with denominators in $(R, Q_k]$, separated by $\epsilon_k = Q_k^{-(d+1)/d}$. The level $k$ cubes inside $C_r$ are
$$B\!\left( \beta - \frac{\gamma_{i_k}}{q^{1+\eta}},\ \frac{1}{Q_k^{1+\eta} \log Q_k} \right):$$the rational points, shifted toward the current target. Any $\alpha$ in one of them satisfies $0 < |q\alpha - \gamma_{i_k}/q^\eta - p| < 1/(q^\eta \log q)$ for that $q$, so every $\alpha \in \mathcal{C}$ has every $\gamma_i$ as an accumulation point: $\mathbb{Q}^d \subseteq \mathcal{A}_\eta(\alpha)$, and $\mathcal{A}_\eta(\alpha)$ is closed, so $\mathcal{C} \subseteq \mathcal{D}(\eta)$.
Give $\mathcal{C}$ the mass distribution $\mu$ that splits each cube's mass evenly among its children. A set $B$ with $\epsilon_k \leq \operatorname{diam} B < \epsilon_{k-1}$ meets one level $k-1$ cube and at most $\min\{m_k, (\operatorname{diam} B/\epsilon_k)^d\} \leq m_k^{1 - s/d} (\operatorname{diam} B/\epsilon_k)^s$ level $k$ cubes, so $\mu(B) \ll \operatorname{diam}(B)^s / (m_1 \cdots m_{k-1} m_k^{s/d} \epsilon_k^s)$. Choosing each $Q_k$ large enough relative to $Q_{k-1}$ makes the denominator bounded below for every $s < \frac{d+1}{1+\eta}$, and the mass distribution principle gives $\dim_H \mathcal{C} \geq \frac{d+1}{1+\eta}$. $\blacksquare$
The construction, for $d = 1$ and $\eta = \frac{3}{2}$, with denominators small enough to see. Each level picks rational points in the middle third of the previous cube, slides each one toward the current target $\gamma_{i_k}$ by $\gamma/q^{1+\eta}$, and keeps a small interval around it; then the picture zooms in and does it again. The real proof needs $Q_k$ astronomically larger than $Q_{k-1}$, so this is the shape, not the scale.
Here is the paper's Table 2 next to the dimension bounds, drawn. The band for $0 < \eta < 1/d$ is where the singular points live: it starts at dimension $0$ and reaches $d$ at $\eta = 1/d$, consistent with the badly approximable set (dimension $d$) sitting inside $\mathcal{D}(1/d)^c$. Past $1/d$ the curve is Theorem 3.
| $\eta$ | $\mathcal{D}(\eta)$ | $\mathcal{D}(\eta)^c$ |
|---|---|---|
| $\eta < 0$ | $\mathbb{R}^d$ | $\emptyset$ |
| $\eta = 0$ | $K_d$ (Kronecker) | $K_d^c$ |
| $0 < \eta < 1/d$ | $\lambda(\mathcal{D}) = \infty$ (Theorem 1) | $0 < \dim_H(\mathcal{D}^c \setminus K_d^c) < d$ for $d \geq 2$ (Theorem 2) |
| $\eta = 1/d$ | $\lambda(\mathcal{D}) = \infty$, $\lambda(\mathcal{D}^c) = 0$ (Theorem 1) | $\dim_H = d$ ($\mathcal{W}^c \subseteq \mathcal{D}^c$) |
| $\eta > 1/d$ | $\dim_H = \frac{d+1}{1+\eta}$ (Theorem 3) | $\lambda(\mathcal{D}^c) = \infty$, $\lambda(\mathcal{D}) = 0$ |
Turn the dials yourself: $d$ and $\eta$ in, both tables out.
Now the half of the paper I find beautiful. Suppose $\alpha \in K_d$ but $\alpha \notin \mathcal{D}(\eta)$. In general there's no reason to expect the set $\mathcal{A}_\eta(\alpha)$ to have any shape at all. But the first examples anyone would try, $\eta = 1/d$ and $\alpha$ with $1, \alpha_1, \dots, \alpha_d$ a basis of a number field of degree $d + 1$, are badly approximable by a classical result of Perron, so they're in $K_d \cap \mathcal{D}(1/d)^c$. And for those, the paper gives a complete description of the accumulation set in dimensions $1$ and $2$. The one-dimensional answer first.
Let $\alpha = a + b\sqrt{D} \in \mathbb{R}$ with $a, b \in \mathbb{Q}$, $b \neq 0$, $D \geq 2$ squarefree, and $K = \mathbb{Q}(\alpha)$. Then
$$\mathcal{A}_1(\alpha) = \left\{ \pm 2b\sqrt{D}\,N(s) : s \in \Lambda^* \setminus \{0\} \right\},$$where $\Lambda^*$ is the lattice in $K$ dual to $\Lambda = \mathbb{Z} + \alpha\mathbb{Z}$ with respect to the trace form, and $N$ is the norm from $K$ to $\mathbb{Q}$.
So the accumulation set is discrete, and it's a list of norms. For $\alpha = \sqrt{2}$ it works out to $\pm \frac{\sqrt{2}}{4} m$ where $m$ runs over the norms $|a^2 - 2b^2|$ of elements of $\mathbb{Z}[\sqrt{2}]$: $1, 2, 4, 7, 8, 9, 14, 16, 17, \dots$ There is no element of norm $3$, $5$ or $6$ (those primes are inert in $\mathbb{Z}[\sqrt{2}]$), so nothing ever lands at $\frac{3\sqrt{2}}{4}$, $\frac{5\sqrt{2}}{4}$ or $\frac{6\sqrt{2}}{4}$. Watch the dots $q(q\sqrt{2} - p)$ arrive, $q$ up to $10^5$, and sort themselves into the slots.
The smallest slot, $\frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$, is the classical one: for the convergents of $\sqrt{2}$, $q(q\sqrt{2} - p) \to \pm\frac{1}{2\sqrt{2}}$. The others come from the worse approximations, the ones nobody usually looks at. Pick a target and see how the two kinds of number treat it.
In dimension two the answer is a curve, and which curve depends on the field's embeddings.
Let $\alpha \in \mathbb{R}^2$ with $\{1, \alpha_1, \alpha_2\}$ a basis for a cubic field $K$, and let $\Lambda^*$ be the lattice dual to $\Lambda = \mathbb{Z} + \alpha_1\mathbb{Z} + \alpha_2\mathbb{Z}$ with respect to the trace form.
The rest of the page is the proof, and the proof is a set of pictures. It takes three sections: the lattices a number field gives you, the lemma that says every good approximation is "a lattice element times a unit," and the lemma that says those products trace out one curve.
Let $K \subseteq \mathbb{R}$ be a number field of degree $d + 1$ with $r + 1$ real embeddings $\sigma_0, \dots, \sigma_r$ and $s$ pairs of complex ones $\sigma_{r+i} = \overline{\sigma_{r+s+i}}$, so $d + 1 = r + 1 + 2s$, and $\sigma_0$ is the identity. (This is the paper's convention; the Number Theory course will write $r_1, r_2$ for these counts.) Two ways to put $K$ in a Euclidean space.
The Minkowski embedding $\sigma \colon K \to \mathbb{R}^{d+1}$ is $\sigma(\alpha) = (\sigma_0(\alpha), \dots, \sigma_{r+s}(\alpha))^t$, with each complex coordinate read as a point of $\mathbb{R}^2$. A subset $\Lambda \subseteq K$ is a lattice if $\sigma(\Lambda)$ is a lattice in $\mathbb{R}^{d+1}$.
The logarithmic embedding $\phi \colon K^\times \to \mathbb{R}^{r+s+1}$ is $\phi(\alpha) = (\log|\sigma_0(\alpha)|, \dots, \log|\sigma_{r+s}(\alpha)|)^t$. It is a homomorphism from multiplication to addition, and it sends the units $\mathcal{O}_K^\times$ into the hyperplane $x_0 + \cdots + x_r + 2x_{r+1} + \cdots + 2x_{r+s} = 0$, where by Dirichlet's unit theorem they form a lattice.
If $\alpha_0, \dots, \alpha_d \in K$ are $\mathbb{Q}$-linearly independent, $\Lambda(\alpha_0, \dots, \alpha_d) = \alpha_0\mathbb{Z} + \cdots + \alpha_d\mathbb{Z}$ is a lattice. The paper always takes $\alpha_0 = 1$ and $\alpha_1, \dots, \alpha_d$ the coordinates of the vector $\alpha$. The lattice's partner comes from the trace.
The trace form on $K$ is $\langle \alpha, \beta \rangle = \mathrm{Tr}(\alpha\beta)$. For a basis $\alpha_0, \dots, \alpha_d$ the matrix $A = (\langle \alpha_i, \alpha_j \rangle)$ is rational and invertible (its determinant is a rational multiple of the discriminant), and the dual basis $\alpha_j^* = \sum_i a^*_{ij}\alpha_i$, built from $A^{-1} = (a^*_{ij})$, satisfies $\langle \alpha_i, \alpha_j^* \rangle = 1$ if $i = j$ and $0$ otherwise. The dual lattice is $\Lambda^* = \Lambda(\alpha_0^*, \dots, \alpha_d^*) = \{\alpha \in K : \langle \alpha, \beta \rangle \in \mathbb{Z} \text{ for all } \beta \in \Lambda\}$.
Why the dual lattice? Because of what the trace does to it. An element $s = q\alpha_0^* + p_1\alpha_1^* + \cdots + p_d\alpha_d^*$ of $\Lambda^*$ has $\mathrm{Tr}(s) = \langle s, \alpha_0 \rangle = q$ and $\mathrm{Tr}(s\alpha_i) = \langle s, \alpha_i \rangle = p_i$. So the integers $(q, p)$ of an approximation are the coordinates of an element of $\Lambda^*$, and the error vector is a trace too: $p_i - q\alpha_i = \mathrm{Tr}(s\alpha_i) - \alpha_i\mathrm{Tr}(s) = \sum_{j \geq 1} \sigma_j(s)\big(\sigma_j(\alpha_i) - \alpha_i\big)$, which only involves the conjugate embeddings of $s$. That identity is the engine of everything below, and it's also why the figures can compute errors for $q \approx 10^{15}$ without losing a digit: nothing is subtracted from anything of the same size.
One more object: the units that are allowed to act. Multiplying $\Lambda^*$ by an arbitrary unit of $\mathcal{O}_K$ might not land back in $\Lambda^*$, so the paper uses the ring of multipliers.
For a lattice $\Lambda \subseteq K$, the multiplier ring is $\mathcal{Z}_\Lambda = \{\gamma \in \mathcal{O}_K : \gamma\Lambda \subseteq \Lambda\}$, a subring of $\mathcal{O}_K$ that is itself a lattice. Its units $\mathcal{Z}_\Lambda^\times$ satisfy $\gamma\Lambda^* \subseteq \Lambda^*$, by $\langle \gamma\alpha, \beta \rangle = \langle \alpha, \gamma\beta \rangle$.
If $\Lambda$ is a lattice in $K$, then $\phi(\mathcal{Z}_\Lambda^\times)$ is a lattice in the unit hyperplane of $\mathbb{R}^{r+s+1}$: the multiplier units have full rank $r + s$.
Pick $n \in \mathbb{N}$ with $n\Lambda \subseteq \mathcal{O}_K$, so $\sigma(n\Lambda)$ is a finite-index subgroup of $\sigma(\mathcal{O}_K)$. There are finitely many subgroups of that index, and multiplication by a unit (a volume-preserving automorphism of $\sigma(\mathcal{O}_K)$) permutes them. So for each basis unit $v_i$ of $\mathcal{O}_K^\times$ some power $v_i^{b_i}$ fixes $n\Lambda$, hence fixes $\Lambda$, hence lies in $\mathcal{Z}_\Lambda^\times$. The $v_i^{b_i}$ generate a full-rank subgroup. $\blacksquare$
Since the unit logarithms form a lattice in the hyperplane, there is a $\kappa_\Lambda > 1$ such that every cube of side $2\log\kappa_\Lambda$ centered on the hyperplane contains one. That gives, for every $t > 0$, a unit $u$ with $\phi(u)$ in the box $[\log t - \log\kappa_\Lambda, \log t + \log\kappa_\Lambda] \times \prod_{i=1}^{r+s} [-\frac{\log t}{d} - \log\kappa_\Lambda, -\frac{\log t}{d} + \log\kappa_\Lambda]$, that is,
$$\kappa_\Lambda^{-1} t \leq |u| \leq \kappa_\Lambda t \qquad \text{and} \qquad \kappa_\Lambda^{-1} t^{-1/d} \leq |\sigma_i(u)| \leq \kappa_\Lambda t^{-1/d} \quad (1 \leq i \leq d).$$A unit $u \in \mathcal{Z}_\Lambda^\times$ satisfying those inequalities for some $t > 1$ is a dominant unit: it is large in the identity embedding and uniformly small, of size about $|u|^{-1/d}$, in every other one.
Here is all of that for the paper's first example, $K = \mathbb{Q}(\sqrt[3]{2})$, with $\alpha = (\sqrt[3]{2}, \sqrt[3]{4})$ and $\Lambda = \mathbb{Z}[\sqrt[3]{2}] = \mathcal{O}_K$, so $\mathcal{Z}_\Lambda = \mathcal{O}_K$. There's one real embedding and one complex pair, so $\sigma(K) \subseteq \mathbb{R} \times \mathbb{C}$ and the unit hyperplane $x_0 + 2x_1 = 0$ is a line. The fundamental unit is $u = 1 + \sqrt[3]{2} + \sqrt[3]{4} \approx 3.847$, with $\sigma_1(u) = u^{-1/2} e^{i\theta}$ exactly (its norm is $1$), so every power $u^k$ is dominant with $\kappa_\Lambda = 1$. On the left, the powers of $u$ in $\mathbb{R} \times \mathbb{C}$: out along the real axis, spiraling in toward the complex origin. On the right, their logarithms march along the unit line.
Take $\Lambda = \Lambda(1, \alpha_1, \dots, \alpha_d)$. The plan of the paper's Section 6 is two statements: (i) every $(1/d)$-normalized approximation of bounded size comes from an element $su \in \Lambda^*$ with $s$ in a finite set and $u$ a dominant unit; (ii) for fixed $s$, the approximations from $su$ as $u$ runs over the dominant units lie near one fixed surface, scaled by $|N(s)|^{1/d}$. Statement (i) is a pigeonhole in the log picture.
Let $1, \alpha_1, \dots, \alpha_d$ be a $\mathbb{Q}$-basis of a number field and $\Lambda = \Lambda(1, \alpha_1, \dots, \alpha_d)$. For every $C > 0$ there is a finite set $S_C \subseteq \Lambda^*$ such that whenever $q \in \mathbb{N}$ and $p \in \mathbb{Z}^d$ satisfy $|q\alpha - p| \leq C q^{-1/d}$,
$$q\alpha_0^* + p_1\alpha_1^* + \cdots + p_d\alpha_d^* = su$$for some $s \in S_C$ and some dominant unit $u \in \mathcal{Z}_\Lambda^\times$.
Let $\alpha^* = q\alpha_0^* + p_1\alpha_1^* + \cdots + p_d\alpha_d^*$. Writing $B$ for the matrix of the embeddings of $\alpha_0, \dots, \alpha_d$ (with the complex rows doubled and split into real and imaginary parts), the dual basis satisfies $(\sigma(\alpha_0^*) \cdots \sigma(\alpha_d^*)) = B^{-t}$, and a short computation gives
$$\sigma(\alpha^*) = \sum_{i=1}^d (p_i - q\alpha_i)\,\sigma(\alpha_i^*) + (q, 0, \dots, 0)^t.$$So $|\alpha^* - q| \leq C(\alpha) q^{-1/d}$ and $|\sigma_i(\alpha^*)| \leq C(\alpha) q^{-1/d}$ for $i \geq 1$, with $C(\alpha) = C\sum_i |\sigma(\alpha_i^*)|$: the element $\alpha^*$ is about $q$ in the identity embedding and about $q^{-1/d}$ in the others. Now take a dominant unit $u$ with $\kappa_\Lambda^{-1} q \leq |u| \leq \kappa_\Lambda q$ and $|\sigma_i(u)| \asymp q^{-1/d}$. Dividing, $|\sigma(\alpha^* u^{-1})| \leq \kappa_\Lambda(1 + C(\alpha))(1 + q^{-1-1/d})$: bounded, independently of $q$. The set $S_C$ of elements of $\Lambda^*$ with all embeddings bounded by that constant is finite because $\Lambda^*$ is a lattice, and $\alpha^* u^{-1} \in S_C$. $\blacksquare$
So the good approximations, in log coordinates, lie in a strip parallel to the unit line, and multiplying by $u^{-1}$ slides each one along the line into a single window. Only finitely many lattice points fit in the window. Here every dot is a real approximation $(q, p)$ to $(\sqrt[3]{2}, \sqrt[3]{4})$ from the figures below, placed at $(\log q, \log|\sigma_1(\alpha^*)|)$, and then slid home.
Conversely, every $s \in \Lambda^*$ and unit $u$ give an approximation: $q_u = \mathrm{Tr}(su)$ and $p_{u,i} = \mathrm{Tr}(su\alpha_i)$. Statement (ii) says what those approximations look like.
With $\alpha$, $\Lambda$ as above and $s \in \Lambda^*$, for each $u \in \mathcal{Z}_\Lambda^\times$ there are $\gamma_u \in \mathbb{R}$ and $\beta_u \in \mathbb{R}^d$ with
$$|q_u|^{1/d}(q_u\alpha - p_u) = \gamma_u\, M(\alpha)\, \beta_u,$$where $M(\alpha)$ is an invertible real $d \times d$ matrix depending only on $\alpha$, $\beta_u$ lies on the surface
$$|x_1 \cdots x_r| \prod_{i=1}^{s} \left( x_{r+2i-1}^2 + x_{r+2i}^2 \right) = 1,$$and $\gamma_u \to |N(s)|^{1/d}$ as $u \to \infty$ through dominant units.
From $q_u = \mathrm{Tr}(su)$ and $p_{u,i} = \mathrm{Tr}(su\alpha_i)$,
$$q_u\alpha - p_u = \begin{pmatrix} \alpha_1 - \sigma_1(\alpha_1) & \cdots & \alpha_1 - \sigma_d(\alpha_1) \\ \vdots & & \vdots \\ \alpha_d - \sigma_1(\alpha_d) & \cdots & \alpha_d - \sigma_d(\alpha_d) \end{pmatrix} \begin{pmatrix} \sigma_1(su) \\ \vdots \\ \sigma_d(su) \end{pmatrix}.$$The complex embeddings come in conjugate pairs, and $w\bar{z} + \bar{w}z = 2\mathrm{Re}(wz)$, so each pair of complex columns can be replaced by the real columns $2\mathrm{Re}(w_{i,j})$ and $-2\mathrm{Im}(w_{i,j})$, with $w_{i,j} = \alpha_j - \sigma_{r+i}(\alpha_j)$, acting on $(\mathrm{Re}\,z_i, \mathrm{Im}\,z_i)$ with $z_i = \sigma_{r+i}(su)$. That real matrix is $M(\alpha)$. Set
$$\gamma_u = |q_u|^{1/d}\,|\sigma_1(su) \cdots \sigma_d(su)|^{1/d}, \qquad \beta_u = \frac{|q_u|^{1/d}}{\gamma_u}\big(\sigma_1(su), \dots, \sigma_r(su), \mathrm{Re}\,z_1, \mathrm{Im}\,z_1, \dots\big)^t,$$and $\beta_u$ satisfies the surface equation by construction. For the limit,
$$\gamma_u^d = \frac{|q_u|\,|N(su)|}{|\sigma_0(su)|} = |N(s)|\,\frac{|\mathrm{Tr}(su)|}{|su|} = |N(s)| \left| \frac{su}{|su|} + \sum_{i=1}^d \frac{\sigma_i(su)}{|su|} \right|,$$and along dominant units $|\sigma_i(su)|/|su| \to 0$, so $\gamma_u^d \to |N(s)|$. Finally $M(\alpha)$ is invertible because, letting $s$ vary, $q_u$ and $p_u$ take every value in $\mathbb{Z} \times \mathbb{Z}^d$ (take $s = q\alpha_0^* + \sum p_i\alpha_i^*$ and $u = 1$), so the left side spans $\mathbb{R}^d$. $\blacksquare$
For a complex cubic field ($r = 0$, $s = 1$) the surface is the unit circle $x_1^2 + x_2^2 = 1$, and $M(\alpha)$ turns it into an ellipse. For a totally real cubic ($r = 2$) it's $|x_1 x_2| = 1$, a pair of conjugate hyperbolas. Here's the mechanism for one $s$ in $\mathbb{Q}(\sqrt[3]{2})$: on the left $\sigma_1(su^k)$ spirals inward by $u^{-1/2}$ and turns by $\theta \approx 146^\circ$ each step; in the middle the direction $\beta_k$ walks around the unit circle by $\theta$; on the right the actual normalized approximation $\sqrt{q_k}(q_k\alpha - p_k)$ lands on the ellipse, with $\gamma_k$ already within $10^{-6}$ of $|N(s)|^{1/2}$ by $k = 7$.
Notice the formula for $\gamma_u^d$ never equals $|N(s)|$ exactly: the sum of the small conjugates is nonzero. So the points never sit on the curve, they only approach it, which is just what an accumulation point needs.
With Lemmas 18 and 19 in hand, Theorem 4 is short.
$\mathcal{A}_1(\alpha) = \{\pm 2b\sqrt{D}\,N(s) : s \in \Lambda^* \setminus \{0\}\}$ for $\alpha = a + b\sqrt{D}$.
Fix $C > 0$ and let $S_C$ be the finite set of Lemma 18. The unit group $\mathcal{Z}_\Lambda^\times$ has rank $1$, so the positive dominant units are a sequence $u_1 < u_2 < \cdots \to \infty$. If $q|q\alpha - p| \leq C$ then $q\alpha_0^* + p\alpha_1^* = \pm su_k$ for some $s \in S_C$, and by Lemma 19 with $d = 1$, $M(\alpha) = \alpha - \sigma_1(\alpha) = 2b\sqrt{D}$ and $\beta = \pm 1$:
$$q_{u_k}(q_{u_k}\alpha - p_{u_k}) = \pm 2b\sqrt{D}\,\gamma_{u_k}, \qquad \gamma_{u_k} \to |N(s)|.$$So every accumulation point is one of the listed values. Conversely, for any $s \neq 0$ the numbers $\gamma_{u_k}$ are never exactly $|N(s)|$ (the formula for $\gamma_u^d$ has the nonzero conjugate term), so the sequence on the left accumulates at at least one of $\pm 2b\sqrt{D}|N(s)|$, and $\mathcal{A}_1(\alpha)$ is symmetric about $0$. $\blacksquare$
For the complex cubic case the same argument gives $\mathcal{A}_{1/2}(\alpha) \subseteq \bigcup_s |N(s)|^{1/2}\mathcal{C}_K$ with $\mathcal{C}_K = M(\alpha)\{x_1^2 + x_2^2 = 1\}$. For the reverse inclusion, write the fundamental unit as $\sigma_1(u) = u^{-1/2} e(\theta)$. The angle $\theta$ must be irrational: otherwise some $\sigma_1(u)^k$ would be real, so $u^k \in \mathbb{Q}$, so $u^k = \pm 1$. Then for fixed $s$ with $\sigma_1(s) = \rho_s e(\psi_s)$, the directions $\beta_k = (\cos(\psi_s + k\theta), \sin(\psi_s + k\theta))$ are dense on the circle, the $\gamma_k$ tend to $|N(s)|^{1/2}$ without ever equaling it, and every point of the ellipse is an accumulation point.
That's the whole left half of the paper's Figure 1, and here it is, computed and in motion. $\Lambda^* = \frac{\sqrt[3]{2}}{6}\mathbb{Z}[\sqrt[3]{2}]$, so $|N(s)| = |N(\beta)|/108$ for $\beta \in \mathbb{Z}[\sqrt[3]{2}]$, and the rings are indexed by the norms $|N(\beta)| = |x^3 + 2y^3 + 4z^3 - 6xyz|$. Each ring's dots are genuine normalized approximations, with $q = \mathrm{Tr}(su^k)$ computed exactly, ring by ring in order of the norm. Ring 7 never comes.
Seven is inert in $\mathbb{Z}[\sqrt[3]{2}]$: $x^3 \equiv 2 \pmod 7$ has no solution, so $(7)$ stays prime, with norm $343$, and no element has norm $7$. The same goes for $13$, $19$, $37$, and every multiple of them that isn't a multiple of their cube. The norm form is simple enough to play with by hand. Find an element of each norm; the one you can't find is the point.
Who decided $x^3 + 2y^3 + 4z^3 - 6xyz$ should factor so nicely? I spent ten minutes typing triples into it instead of reading the totally real case. For the record, $(-1, 1, 1)$ gives $11$ and $(1, 1, 0)$ gives $3$. Still nothing for $7$. There will never be anything for $7$.
When $K$ is totally real the surface is $|x_1 x_2| = 1$, so for each $s$ the approximations lie near $|N(s)|^{1/2}\mathcal{C}_K^\pm$ with $\mathcal{C}_K^\pm = M(\alpha)\{x_1 x_2 = \pm 1\}$, and the sign is the sign of $\sigma_1(su)\sigma_2(su)$. The unit group now has rank $2$, and the paper uses the units $u_a = u_1^{2a_1}u_2^{2a_2}$ (even powers, so they're totally positive and never flip the sign). The hard part of the reverse inclusion is showing that every arc of the hyperbola gets hit, which needs units whose two small conjugates have any prescribed ratio.
For any $0 < c_1 < c_2$, the set
$$\mathcal{U}(c_1, c_2) = \left\{ u_a > 1 : a \in \mathbb{Z}^2,\ c_1 < \frac{\sigma_1(u_a)}{\sigma_2(u_a)} < c_2 \right\}$$is infinite and uniformly discrete in $(0, \infty)$.
In the log embedding, $u_a \in \mathcal{U}(c_1, c_2)$ means $\phi(u_a)$ lies in $R_>(c_1, c_2) = \{x_0 > 0,\ \log c_1 < x_1 - x_2 < \log c_2\}$. The part of that region with $0 < x_0 < c_3$ meets the unit hyperplane $x_0 + x_1 + x_2 = 0$ in a bounded set, so it holds finitely many lattice points; hence the $x_0$ coordinates of lattice points in the region are uniformly discrete, and so are the units themselves (they're spaced at least geometrically).
Suppose the region held no lattice point at all. Then neither does the mirror region $R_\leq(c_1, c_2)$ with $x_0 \leq 0$: a lattice point $\lambda$ there could be nudged by a lattice point $\lambda'$ with $|\lambda_1' - \lambda_2'| < \epsilon$ and $\lambda_0' > |\lambda_0|$ (Minkowski's convex body theorem gives one) into $\lambda + \lambda' \in R_>(c_1, c_2)$. So the whole strip $R(c_1, c_2) = \{\log c_1 < x_1 - x_2 < \log c_2\}$ holds no nonzero lattice point. But its intersection with the hyperplane has infinite area, so by Blichfeldt's theorem two of its points differ by a lattice vector $\xi$ with $0 \leq \xi_1 - \xi_2 < \log c_2 - \log c_1$. If $\xi_1 - \xi_2 > 0$, some multiple of $\xi$ lands in the strip, a contradiction; so $\xi_1 = \xi_2$, i.e. $|\sigma_1(u_a)| = |\sigma_2(u_a)|$ for some $a \neq 0$. That can't happen: $u_a$ is a cubic irrational, so $\sigma_1(u_a) \neq \sigma_2(u_a)$ by separability, and $\sigma_1(u_a) = -\sigma_2(u_a)$ would make $\mathrm{Tr}(u_a) = u_a \notin \mathbb{Q}$. $\blacksquare$
Here's the unit lattice of $\mathbb{Q}(2\cos\frac{2\pi}{7})$ in the coordinates $(x_0, x_1 - x_2)$ of the hyperplane, with the strip $\log c_1 < x_1 - x_2 < \log c_2$ and the points it catches, and then Blichfeldt's step: two points of the strip whose difference is a lattice vector, whose multiples march along until one is inside.
Drag the strip's edges yourself. Every unit inside it gives an approximation on a particular arc of the hyperbola, because the ratio $\sigma_1(u)/\sigma_2(u)$ is what decides where on the hyperbola $\beta_u = \big(\sqrt{x_1/x_2}, \sqrt{x_2/x_1}\big)$ lands.
For any $s \in \Lambda^* \setminus \{0\}$ and $C > 0$, the accumulation points of $\{\pm|q_a|^{1/2}(q_a\alpha - p_a) : a \in \mathbb{Z}^2,\ u_a > 1\} \cap [-C, C]^2$ form exactly $|N(s)|^{1/2}\mathcal{C}_K^+ \cap [-C, C]^2$ or $|N(s)|^{1/2}\mathcal{C}_K^- \cap [-C, C]^2$, according to the sign of $\sigma_1(s)\sigma_2(s)$.
Say $\sigma_1(s), \sigma_2(s) > 0$. By Lemma 19, $|q_a|^{1/2}(q_a\alpha - p_a) = \gamma_a M(\alpha)\beta_a$ with $x_{a,j} = \sigma_j(su_a) > 0$, $\gamma_a = |q_a|^{1/2}|x_{a,1}x_{a,2}|^{1/2}$ and $\beta_a = \big((x_{a,1}/x_{a,2})^{1/2}, (x_{a,2}/x_{a,1})^{1/2}\big)^t$. The units whose points land in $[-C, C]^2$ have ratio $\sigma_1(u_a)/\sigma_2(u_a)$ in some $(c_1, c_2)$, so by Lemma 20 they are uniformly discrete in $(1, \infty)$, and by the formula for $\gamma_a^2$ every accumulation point is on $|N(s)|^{1/2}\mathcal{C}_K^+$ and no point of the set is. For the other direction, an arc of the hyperbola corresponds to a ratio interval $(c_1, c_2)$; Lemma 20 gives infinitely many units with that ratio, and letting them tend to infinity pushes the points into any $\epsilon$-neighborhood of the arc. $\blacksquare$
So each $s$ contributes one of the two hyperbolas, $\ell_s$ being the identity when $\sigma_1(s)\sigma_2(s) > 0$. Could the dominant units that aren't of the form $u_a$ (odd powers) add points on the other hyperbola? Only through a unit $\tilde{u}$ with $\sigma_1(\tilde{u})\sigma_2(\tilde{u}) < 0$, and then $\tilde{s} = s\tilde{u}$ is another element of $\Lambda^*$, with the same $|N|$ and the opposite sign, whose own term in the union already accounts for them. That finishes Theorem 5. Here's the right half of the paper's Figure 1: $\alpha = (\rho, \rho^2)$ with $\rho = 2\cos\frac{2\pi}{7}$, the positive root of $x^3 + x^2 - 2x - 1$. The field is $\mathbb{Q}(\zeta_7)^+$, its units are $\pm\rho^a(\rho^2 - 2)^b$, and $\rho^2 - 2$ has conjugates of mixed sign, so every norm shows both hyperbolas, red for $\mathcal{C}^+$ and blue for $\mathcal{C}^-$. The rings are indexed by $49|N(s)|$, the norms of elements of $\mathbb{Z}[\rho]$: $1, 7, 8, 13, 27, 29, \dots$
The paper ends the introduction with a rarer field, found in Ennola and Turunen's tables of totally real cubic fields: $\alpha$ the smallest positive root of $x^3 - 19x + 21$, discriminant $15529 = 53 \cdot 293$. Its special feature is that every positive unit has only positive conjugates. I found its units by a search: $u_1 = 5 - 5\alpha + \alpha^2$ and $u_2 = -5 + 3\alpha + \alpha^2$, and every unit with coefficients up to $3000$ is a product of those two and $-1$, all of them totally positive or totally negative. So multiplying $s$ by a unit never changes the sign of $\sigma_1(s)\sigma_2(s)$: each unit orbit of $\Lambda^*$ paints a single hyperbola, and that's what the paper's Figure 2 shows.
One small correction to the paper's remark here. It says that in this field any two elements of $\Lambda^*$ with the same norm pick the same hyperbola. That's true within a unit orbit, which is what the figure shows, but not across orbits: $s = (0, -1, -6)$ and $\tilde{s} = (-6, -6, -3)$ in the dual basis both have $N = 81/15529$, their ratio has norm $1$ but trace $-14.19$ (so it isn't a unit), and their $\sigma_1\sigma_2$ have opposite signs. Both hyperbolas of that norm appear, just from different ideals. Theorem 5 itself is unaffected.
The first half of the paper puts $\mathcal{D}(\eta)$ on the same footing as the classical $\mathcal{W}(\eta)$: full measure up to the free exponent $1/d$, measure zero beyond it, the same Hausdorff dimension $\frac{d+1}{1+\eta}$ there, and in between, for $d \geq 2$, an exceptional set of singular points whose dimension is pinned between $\frac{d\eta}{1+\eta}$ and a number that climbs to $d$. The gap between those two bounds is the obvious open question, and it's tied to the dimension of singular vectors, which is active research on its own.
The second half says that for the algebraic examples, the cloud of normalized approximations has a skeleton: the surface $|x_1 \cdots x_r|\prod(x_{r+2i-1}^2 + x_{r+2i}^2) = 1$ of Lemma 19, pushed through $M(\alpha)$ and scaled by every norm of the dual lattice. Lemmas 18 and 19 hold in every dimension, so the natural conjecture is that the same description holds for $d \geq 3$, with ellipsoids, hyperboloids, and the mixed surface $|x_1|(x_2^2 + x_3^2) = 1$ for a quartic field with one complex pair. The missing piece is the distribution of units: in rank $2$, Lemma 20 found units with any ratio $\sigma_1/\sigma_2$, and in higher rank one needs units landing in every cone of the unit hyperplane. Here are the two surfaces for $d = 3$, waiting.
For me the memorable line of the paper is the one about ring $7$. A plot of $(q\sqrt[3]{2} - p_1, q\sqrt[3]{4} - p_2)$, scaled by $\sqrt{q}$, knows which primes are inert in $\mathbb{Q}(\sqrt[3]{2})$. The arithmetic of the field is visible in the approximation picture, if you draw enough of it, and drawing enough of it took exactly the machinery that the Peck paper then reused.