Kavita Dhanda, Josh Flynn, and Alan Haynes · University of Houston · in preparation
Take a number field, say $\mathbb{Q}(\sqrt[3]{2})$, and a basis of it that starts with $1$: here $1, \sqrt[3]{2}, \sqrt[3]{4}$. How well can one integer $n$ bring all of $n\sqrt[3]{2}$ and $n\sqrt[3]{4}$ close to whole numbers at once? In 1955 Cassels and Swinnerton-Dyer showed that the product $n\,\|n\alpha_1\|\,\|n\alpha_2\|$ gets as small as you like. In 1961 Peck squeezed in an extra factor of $\log n$, in any degree. Peck's proof is short but famously opaque.
This page is the paper I'm writing with Kavita Dhanda and Alan Haynes, rebuilt with moving pictures. The paper recasts Peck's proof as a picture in which the whole result follows from a single application of Minkowski's convex body theorem, in the lattice of logarithms of units. The same picture, run inside a congruence subgroup, gives denominators in arithmetic progressions and an analogue of the $p$-adic Littlewood conjecture; and changing the shape of the convex body makes progress on a conjecture Peck raised in the same paper, about handing out the logarithmic savings unequally among the coordinates. Every figure here is computed from the paper's own example fields, $\mathbb{Q}(\sqrt[3]{2})$, $\mathbb{Q}(2\cos\frac{2\pi}{7})$, and $\mathbb{Q}(\sqrt{2}, \sqrt{5})$, with exact arithmetic wherever the numbers get big.
Everything here is measured with one piece of notation.
For $x \in \mathbb{R}$, the distance to the nearest integer is
$$\|x\| = \min_{p \in \mathbb{Z}} |x - p|,$$and for a vector $x \in \mathbb{R}^d$, $\|x\| = \min_{p \in \mathbb{Z}^d} |x - p|$ with $|\cdot|$ the sup norm, so $\|x\| = \max_i \|x_i\|$. So $\|n\alpha\|$ is small exactly when $n\alpha$ is close to a whole number $p$, which is the same as $\alpha$ being close to the fraction $p/n$.
The cornerstone result goes all the way back to the 1840s.
For every $\alpha \in \mathbb{R}$ and every $Q \in \mathbb{N}$ there are integers $p, q$ with $1 \leq q \leq Q$ and
$$\left| \alpha - \frac{p}{q} \right| \leq \frac{1}{q(Q+1)}.$$Equivalently, $\displaystyle\min_{1 \leq q \leq Q} \|q\alpha\| \leq \frac{1}{Q+1}$.
The idea is a pigeonhole. Cut $[0, 1)$ into $Q$ boxes and drop in the $Q + 1$ fractional parts of $0, \alpha, 2\alpha, \dots, Q\alpha$. Two share a box, and their difference is a $q \leq Q$ with $\|q\alpha\| < 1/Q$. The slightly sharper $1/(Q+1)$ in the theorem takes a little more care; the lattice picture in the next section gives it for free.
Letting $Q$ run off to infinity: for every irrational $\alpha$ there are infinitely many $q$ with $q\,\|q\alpha\| \leq 1$. The constant can be improved, and Hurwitz found the best one in 1891.
For every irrational $\alpha$ there are infinitely many $q \in \mathbb{N}$ with
$$q\,\|q\alpha\| < \frac{1}{\sqrt{5}},$$and $\sqrt{5}$ cannot be replaced by anything bigger: for the golden ratio $\varphi = [1; 1, 1, 1, \dots]$, $\displaystyle\liminf_{q \to \infty} q\,\|q\varphi\| = \frac{1}{\sqrt{5}}$.
Numbers with bounded partial quotients are the hardest to approximate. They get a name.
A number $\alpha \in \mathbb{R}$ is badly approximable if there is a constant $C > 0$ with $q\,\|q\alpha\| > C$ for every $q \in \mathbb{N}$. We write
$$\mathbf{BAD} = \{\text{badly approximable numbers}\}.$$Here's what that looks like. Each bar is $q\,\|q\alpha\|$; a number is in $\mathbf{BAD}$ when the bars have a floor they never drop under.
Every quadratic irrational is badly approximable, because its continued fraction repeats; almost every real number is not (Borel and Bernstein: $\mathbf{BAD}$ has measure zero). For two numbers at once, the central question is still open.
For all $\alpha = (\alpha_1, \alpha_2) \in \mathbb{R}^2$,
$$\liminf_{n \to \infty}\, n\,\|n\alpha_1\|\,\|n\alpha_2\| = 0.$$Without the $\|n\alpha_2\|$ the statement is false: that's what $\mathbf{BAD}$ means. Since $\|n\alpha_2\| \leq \frac{1}{2}$ always, the second factor can only push the product down, and the conjecture says it always pushes it all the way. Colloquially: no pair of real numbers is "multiplicatively badly approximable."
The natural place to hunt for a counterexample is a pair like $\sqrt[3]{2}, \sqrt[3]{4}$, which together with $1$ form a basis of a cubic number field. Such a pair is simultaneously badly approximable in the classical sense: $\max(\|n\alpha_1\|, \|n\alpha_2\|) \gg n^{-1/2}$. Whether that badness survives when you multiply the two errors instead of taking the larger one is exactly Littlewood's question, and for these pairs the answer is known.
If $1, \alpha_1, \alpha_2$ form a basis of a cubic number field, then $\displaystyle\liminf_{n \to \infty}\, n\,\|n\alpha_1\|\,\|n\alpha_2\| = 0$.
So the cubic candidates are not counterexamples; they confirm the conjecture. Peck, using a different method, proved more, and in every degree. This is the theorem the whole page is about.
Let $d \geq 2$ and suppose that $1, \alpha_1, \dots, \alpha_d$ is a basis for a real algebraic number field. Then there is a constant $C = C(\alpha_1, \dots, \alpha_d)$ such that, for all $T \geq 2$, there is a positive integer $n \leq T$ with
$$\|n\alpha_1\| \leq C n^{-1/d}, \qquad \|n\alpha_i\| \leq C n^{-1/d} (\log T)^{-1/(d-1)} \quad (2 \leq i \leq d).$$In particular $\displaystyle\liminf_{n \to \infty}\, n \log n\,\|n\alpha_1\|\,\|n\alpha_2\| \cdots \|n\alpha_d\| < \infty$, which for $d = 2$ is stronger than Littlewood's conjecture.
Read the two inequalities carefully. Dirichlet's theorem in $d$ dimensions already gives $\|n\alpha_i\| \ll n^{-1/d}$ for every $i$, so the first coordinate gets nothing new; the savings, a factor $(\log T)^{-1/(d-1)}$, go to the other $d - 1$ coordinates, and multiplied out they are the one $\log$ in the product. The notation $f \ll g$ means $f \leq c\,g$ for some constant $c$ depending only on the $\alpha_i$, and $f \asymp g$ means both $f \ll g$ and $g \ll f$.
Instead of thinking about the inequality $|\alpha - p/q| \leq \psi(q)$ one fraction at a time, collect all the points
$$M_\alpha \mathbb{Z}^2 = \left\{ \begin{pmatrix} q \\ q\alpha + p \end{pmatrix} : p, q \in \mathbb{Z} \right\}, \qquad M_\alpha = \begin{pmatrix} 1 & 0 \\ \alpha & 1 \end{pmatrix}.$$The columns $\mathbf{v}_1 = (1, \alpha)$ and $\mathbf{v}_2 = (0, 1)$ of $M_\alpha$ generate a lattice of determinant $1$. The $x$-coordinate of a lattice point is the denominator $q$, and its $y$-coordinate is $q\alpha + p$, so the error $\|q\alpha\|$ is the vertical distance from the point to the $x$-axis. A good rational approximation is a lattice point that wanders very close to the horizontal axis. The tool that finds such points is the one the whole paper runs on.
Let $\Gamma \subseteq \mathbb{R}^d$ be a lattice (a discrete subgroup with compact quotient) with covolume $V$, and let $B \subseteq \mathbb{R}^d$ be compact, convex, and symmetric about the origin. If $\operatorname{vol}(B) \geq 2^d V$, then $B$ contains a nonzero point of $\Gamma$.
Dirichlet's theorem is the case $d = 2$: draw a box centered at the origin, $2Q$ wide and $2/Q$ tall. Its area is exactly $4 = 2^2 \cdot 1$, so it contains a nonzero lattice point, and that point's coordinates are a $q \leq Q$ with $|q\alpha + p| \leq 1/Q$.
Why is Minkowski's theorem true? The argument fits in one loop. Shrink the body by half, so it has exactly one cell's worth of area; copies of it at every lattice point can't all fit without overlapping; and the difference of two points in an overlap is a lattice point that convexity and symmetry put back inside the original body.
Try to beat it. The box below has area $4$ and the lattice has covolume $1$. Stretch it, turn it, and try to lose every nonzero lattice point; then cheat with the red slider and see exactly where the threshold is.
There is a second way to read Dirichlet's stretching box, and although the paper never needs it, it is the picture number theorists carry around. Keep the box a fixed $2 \times 2$ square and stretch the lattice instead: the matrix $g_t = \operatorname{diag}(e^{-t}, e^{t})$ squeezes $M_\alpha\mathbb{Z}^2$ horizontally and stretches it vertically, preserving areas, and Minkowski's theorem applied to $g_t M_\alpha\mathbb{Z}^2$ and the square is Dirichlet's theorem with $Q = e^{t}$. Each lattice $g_t M_\alpha\mathbb{Z}^2$ is a point of the space of unimodular lattices $X_2 = SL_2(\mathbb{R})/SL_2(\mathbb{Z})$, which can be drawn as the keyhole $\mathcal{F} = \{\tau \in \mathbb{H} : |\tau| \geq 1, |\operatorname{Re}\tau| \leq \frac{1}{2}\}$ (a lattice's shape is the ratio $\tau$ of two basis vectors, read in $\mathbb{C}$), and a point high in the keyhole is a lattice with a very short vector. The golden ratio's orbit is a closed loop, because its continued fraction repeats; $\sqrt[3]{2}$'s climbs toward the cusp whenever a partial quotient is large.
Hold on to the shape of that argument: a box that gets longer in one direction and thinner in another, with its area pinned, always catches a lattice point. Peck's theorem is the same move, in a cleverer lattice, where "longer" means "a bigger unit" and "thinner" means "a more balanced one."
Let $K \subseteq \mathbb{R}$ be a number field of degree $[K : \mathbb{Q}] = d + 1$, with $r + 1$ real and $2z$ non-real embeddings into $\mathbb{C}$, so $d + 1 = (r + 1) + 2z$. Label the embeddings $\sigma_0, \dots, \sigma_d$ so that $\sigma_0, \dots, \sigma_r$ are the real ones, $\sigma_0$ is the identity on $K$, and $\sigma_{r+j} = \overline{\sigma_{r+z+j}}$ for $1 \leq j \leq z$. A cubic field has either $(r, z) = (0, 1)$, one real embedding and a complex pair, or $(r, z) = (2, 0)$, totally real.
Identifying $\mathbb{C}$ with $\mathbb{R}^2$, the Minkowski embedding is the injective map $\sigma \colon K \to \mathbb{R}^{d+1}$,
$$\sigma(\beta) = \big( \sigma_0(\beta), \dots, \sigma_{r+z}(\beta) \big)^{t}.$$For $K = \mathbb{Q}(\sqrt[3]{2})$ that's $\sigma(x) = (\sigma_0(x), \operatorname{Re}\sigma_1(x), \operatorname{Im}\sigma_1(x)) \in \mathbb{R}^3$, where $\sigma_1(\sqrt[3]{2}) = \sqrt[3]{2}\,e^{-2\pi i/3}$. Here is $\sigma(\mathcal{O}_K)$ for the ring of integers $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{2}]$.
The end of that loop is the point: the complex coordinate on its own is not discrete. It's the real coordinate $\sigma_0$, growing while $\sigma_1$ shrinks, that keeps the lattice points apart, and that tension is the whole proof in miniature. An additive subgroup of $\mathbb{R}^{d+1}$ is a lattice if it is discrete and cocompact, and we say $\Lambda \subseteq K$ is a lattice in $K$ if $\sigma(\Lambda)$ is one. A basic fact: if $\alpha_0, \dots, \alpha_d \in K$ are $\mathbb{Q}$-linearly independent, then $\alpha_0\mathbb{Z} + \cdots + \alpha_d\mathbb{Z}$ is a lattice in $K$. The Minkowski embedding shows the additive structure of $K$; to see the multiplicative structure of its units, take logarithms.
The logarithmic embedding is the homomorphism $\varphi \colon K^\times \to \mathbb{R}^{r+z+1}$,
$$\varphi(\beta) = \big( \log|\sigma_0(\beta)|, \dots, \log|\sigma_{r+z}(\beta)| \big)^{t}.$$A unit $u \in \mathcal{O}_K^\times$ is totally positive if $\sigma_k(u) > 0$ for $0 \leq k \leq r$.
$\varphi(\mathcal{O}_K^\times)$ is a lattice of rank $r + z$ in the hyperplane of $\mathbb{R}^{r+z+1}$ defined by
$$x_0 + \cdots + x_r + 2x_{r+1} + \cdots + 2x_{r+z} = 0.$$The hyperplane is just the statement that a unit has norm $\pm 1$: the product of all $d + 1$ embeddings has absolute value $1$, and each complex embedding appears twice. For $\mathbb{Q}(\sqrt[3]{2})$ the rank is $1$: every unit is $\pm u_1^k$ for the fundamental unit $u_1 = 1 + \sqrt[3]{2} + \sqrt[3]{4} \approx 3.85$ (its inverse is $\sqrt[3]{2} - 1$). Multiplying by $u_1$ is a leap along a curve in Minkowski space and a fixed step along a line in log space.
Now fix $\alpha_0 = 1$ and a basis $\alpha_0, \alpha_1, \dots, \alpha_d$ of $K$ over $\mathbb{Q}$, write $\alpha = (\alpha_1, \dots, \alpha_d)^t$, and let $\Lambda = \mathbb{Z} + \alpha_1\mathbb{Z} + \cdots + \alpha_d\mathbb{Z}$. The first key observation of the paper, implicit in Peck's proof, is that the lattice dual to $\Lambda$ naturally encodes simultaneous approximations to $\alpha$. The duality is with respect to the trace form $\langle \beta, \gamma \rangle = \operatorname{Tr}(\beta\gamma)$, which is non-degenerate.
There are unique $\mathbb{Q}$-linearly independent $\alpha_0^*, \dots, \alpha_d^* \in K$ with
$$\operatorname{Tr}(\alpha_i \alpha_j^*) = \delta_{ij} \qquad (0 \leq i, j \leq d).$$The dual lattice of $\Lambda$ is $\Lambda^* = \alpha_0^*\mathbb{Z} + \cdots + \alpha_d^*\mathbb{Z}$.
Writing an element $s \in \Lambda^*$ in the dual basis is just taking traces: $s = \sum_i \operatorname{Tr}(\alpha_i s)\, \alpha_i^*$, and the coefficients are integers. The element to watch is $\alpha_1^*$: since $\alpha_0 = 1$, it has $\operatorname{Tr}(\alpha_1^*) = \operatorname{Tr}(\alpha_0\alpha_1^*) = 0$. Here is the example that runs through the page.
Find the dual basis of $1, \alpha_1, \alpha_2$ for $\alpha_1 = \sqrt[3]{2}$, $\alpha_2 = \sqrt[3]{4}$.
The trace of $\alpha_1^k$ is $3$ when $3 \mid k$ and $0$ otherwise (the conjugates of $\alpha_1$ are $\alpha_1, \alpha_1\omega, \alpha_1\omega^2$ with $\omega = e^{2\pi i/3}$, and $1 + \omega^k + \omega^{2k} = 0$ unless $3 \mid k$). So the matrix of the trace form in the basis $1, \alpha_1, \alpha_1^2$ is
$$\big( \operatorname{Tr}(\alpha_1^{i+j}) \big)_{i,j} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 6 \\ 0 & 6 & 0 \end{pmatrix}, \qquad \text{inverse } \begin{pmatrix} 1/3 & 0 & 0 \\ 0 & 0 & 1/6 \\ 0 & 1/6 & 0 \end{pmatrix}.$$The rows of the inverse are the dual basis: $\alpha_0^* = \frac{1}{3}$, $\alpha_1^* = \frac{\alpha_1^2}{6} = \frac{\sqrt[3]{4}}{6}$, $\alpha_2^* = \frac{\alpha_1}{6} = \frac{\sqrt[3]{2}}{6}$. Check: $\operatorname{Tr}(\alpha_1\alpha_1^*) = \operatorname{Tr}(\alpha_1^3)/6 = \operatorname{Tr}(2)/6 = 1$ and $\operatorname{Tr}(\alpha_1^*) = \operatorname{Tr}(\alpha_1^2)/6 = 0$. So $\Lambda^* = \tfrac{1}{3}\mathbb{Z} + \tfrac{\sqrt[3]{4}}{6}\mathbb{Z} + \tfrac{\sqrt[3]{2}}{6}\mathbb{Z}$, and $\alpha_1^* = \sqrt[3]{4}/6$ is the trace-zero element.
To turn dual lattice elements into approximations we need a change of coordinates. Let $\Sigma^*$ be the real $(d+1) \times (d+1)$ matrix with columns $\sigma(\alpha_0^*), \dots, \sigma(\alpha_d^*)$, and for $\beta \in \mathbb{R}^d$ let $u_\beta = \begin{pmatrix} 1 & 0^t \\ \beta & I_d \end{pmatrix} \in SL_{d+1}(\mathbb{R})$.
The matrix $R = u_{-\alpha}(\Sigma^*)^{-1}$ has first column $e_0 = (1, 0, \dots, 0)^t$. Consequently, for $s = q\alpha_0^* + p_1\alpha_1^* + \cdots + p_d\alpha_d^* \in \Lambda^*$,
$$R\sigma(s) = (q,\ p - q\alpha)^t, \qquad \text{and in particular} \qquad R\sigma(\alpha_1^*) = (0, e_1)^t,$$and $R$ has the block form $R = \begin{pmatrix} 1 & c^t \\ 0 & \tilde{R} \end{pmatrix}$ with $c \in \mathbb{R}^d$ and $\tilde{R}$ a $d \times d$ matrix.
The first identity is a one-liner: $(\Sigma^*)^{-1}\sigma(s)$ is the coordinate vector $(q, p)^t$ of $s$ in the dual basis, and $u_{-\alpha}$ subtracts $q\alpha$ from the last $d$ coordinates. So every element of $\Lambda^*$ with $q = \operatorname{Tr}(s) \neq 0$ is a simultaneous approximation to $\alpha$ with denominator $q$ and error vector $p - q\alpha$. The block form says that the error vector is a fixed linear function $\tilde{R}$ of the last $d$ coordinates of $\sigma(s)$, the conjugates $\sigma_1(s), \dots, \sigma_d(s)$, and does not depend on $s = \sigma_0(s)$ itself. For our cubic example, $p_i - q\alpha_i = \operatorname{Tr}(\alpha_i s) - \operatorname{Tr}(s)\alpha_i = 2\operatorname{Re}\big(\sigma_1(s)(\sigma_1(\alpha_i) - \alpha_i)\big)$, which is $\tilde{R}$ written out.
The second ingredient is multiplication by a unit. For $u \in K$ let $M_u$ be the real $(d+1) \times (d+1)$ matrix representing multiplication by $u$ in Minkowski coordinates, $\sigma(\beta u) = M_u\sigma(\beta)$. It is block diagonal: $1 \times 1$ blocks $\sigma_k(u)$ for the real embeddings, and $2 \times 2$ blocks $\begin{pmatrix} \operatorname{Re}\sigma_{r+j}(u) & -\operatorname{Im}\sigma_{r+j}(u) \\ \operatorname{Im}\sigma_{r+j}(u) & \operatorname{Re}\sigma_{r+j}(u) \end{pmatrix}$ for the complex pairs, each one a scaling-and-rotation of $\mathbb{C} \cong \mathbb{R}^2$. Which units act on $\Lambda^*$? Those of the multiplier ring $\mathcal{Z}_\Lambda = \{\gamma \in \mathcal{O}_K : \gamma\Lambda \subseteq \Lambda\}$.
Let $\Lambda \subseteq K$ be a lattice. Then $\gamma\Lambda^* \subseteq \Lambda^*$ for every $\gamma \in \mathcal{Z}_\Lambda$, and $u\Lambda^* = \Lambda^*$ for every $u \in \mathcal{Z}_\Lambda^\times$. Furthermore $\mathcal{Z}_\Lambda^\times$ is a subgroup of $\mathcal{O}_K^\times$ of maximal rank $r + z$, so $\varphi(\mathcal{Z}_\Lambda^\times)$ is a lattice in the hyperplane of Dirichlet's unit theorem.
The first claim is one line: $\operatorname{Tr}(\gamma s \lambda) = \operatorname{Tr}(s \cdot \gamma\lambda) \in \mathbb{Z}$ for $s \in \Lambda^*$ and $\lambda \in \Lambda$. When $\Lambda = \mathcal{O}_K$, as in the cubic examples, every unit is in $\mathcal{Z}_\Lambda^\times$; the multiplier ring is what lets the paper avoid assuming anything about $\Lambda$ (for instance that $a$ and $b$ are squarefree in the biquadratic theorem below).
Here is the proof in a paragraph. Start at $\alpha_1^*$, whose error vector is exactly $e_1$: the first error is $1$, all the others are $0$. Multiply it by a unit $u \in \mathcal{Z}_\Lambda^\times$; the product is again in $\Lambda^*$, so it is an approximation with denominator $q = \operatorname{Tr}(\alpha_1^* u) \approx \alpha_1^* u$ and error vector $\tilde{R}$ applied to the conjugates $\sigma_1(\alpha_1^*u), \dots, \sigma_d(\alpha_1^*u)$. If the conjugates $\sigma_1(u), \dots, \sigma_d(u)$ were all equal, the error vector of $\alpha_1^*u$ would be a scalar multiple of the error vector of $\alpha_1^*$, which is $e_1$: all the errors but the first would vanish. They can't all be equal, but Minkowski's theorem, applied in the lattice of logarithms of units, produces units below any given height for which they are nearly equal, and the logarithm in Theorem 1 measures how nearly.
In the complex cubic case $|\sigma_1(u)| = u^{-1/2}$ for every unit (norm $1$), so the only thing that can move $\alpha_1^*u$ off the line $\ell$ through $\sigma_1(\alpha_1^*)$ is the argument $\theta(u) = \operatorname{Arg}\sigma_1(u)$. Here is the paper's Figure 1, in motion: the spiral in the $\sigma_1$-plane on the left, and on the right the normalized errors $n^{1/2}(n\alpha_i - p_i)$ for $n = \operatorname{Tr}(\alpha_1^*u_1^k)$, with Dirichlet's square and a band of half-height $1/\log n$ around the image $\ell'$ of $\ell$.
In the totally real cubic case nothing rotates. A unit scales $(\sigma_1, \sigma_2)$ by two real numbers whose product is $1/u$, so each image of $\alpha_1^*$ lies on a prescribed hyperbola, and what moves it off $\ell$ is the imbalance $\delta(u) = \log|\sigma_1(u)| - \log|\sigma_2(u)|$.
The lemma below is where Minkowski's theorem is used, and it is the main gear of the proof: units $u > 1$ whose conjugates are all close to $u^{-1/d}$ can be found below every sufficiently large bound. For a positive unit $u$ and $1 \leq k \leq r + z$ write
$$\lambda_k(u) = \log|\sigma_k(u)| + \tfrac{1}{d}\log u,$$so that $\lambda_k(u) = 0$ exactly when $|\sigma_k(u)| = u^{-1/d}$. The unit-theorem hyperplane says $\lambda_1 + \cdots + \lambda_r + 2\lambda_{r+1} + \cdots + 2\lambda_{r+z} = 0$, so the last one is determined by the others, and in a cubic field with one complex pair it is $0$ automatically. The lattice to use is
$$\Gamma = \big\{ \big( \log u,\ \lambda_1(u), \dots, \lambda_{r+z-1}(u),\ \theta \big) : u \in P,\ \theta \in \mathbb{R}^z,\ \theta_j \equiv \operatorname{Arg}\sigma_{r+j}(u) \bmod 2\pi \big\} \subseteq \mathbb{R}^d,$$where $P$ is the set of positive units in $\mathcal{Z}_\Lambda^\times$. Each coordinate is a homomorphism in $u$, so $\Gamma$ is a subgroup; it is discrete because small coordinates force $\varphi(u)$ small, hence $u = 1$, hence $\theta \in 2\pi\mathbb{Z}^z$ small, hence $\theta = 0$; and it contains $\{0\} \times 2\pi\mathbb{Z}^z$ with quotient $P \cong \mathbb{Z}^{r+z}$, so its rank is $(r + z) + z = d$. It is a lattice in $\mathbb{R}^d$; write $V$ for its covolume.
Let $d \geq 2$. There is a constant $T_1 = T_1(\Lambda)$ such that, for all $T \geq T_1$, there exists $u \in \mathcal{Z}_\Lambda^\times$ with $1 < u \leq T$ and
$$\big| u^{1/d}\sigma_k(u) - 1 \big| \ll (\log T)^{-1/(d-1)} \qquad (1 \leq k \leq d).$$Let $\varepsilon = \frac{1}{2}\big( 2^d V / \log T \big)^{1/(d-1)}$, and take $T_1$ large enough that $\varepsilon \leq 1/(4d)$ for $T \geq T_1$. The box
$$B = \big\{ x \in \mathbb{R}^d : |x_0| \leq \tfrac{1}{2}\log T,\ |x_k| \leq \varepsilon \text{ for } 1 \leq k \leq d - 1 \big\}$$is compact, convex, symmetric about the origin, and has volume $\log T \cdot (2\varepsilon)^{d-1} = 2^d V$. By Minkowski's theorem it contains a nonzero point of $\Gamma$, say $(\log v, \lambda_1(v), \dots, \lambda_{r+z-1}(v), \theta)$ with $v \in P$. Replacing the point by its negative if necessary (which replaces $v$ by $v^{-1}$ and $\theta$ by $-\theta$) we may assume $\log v \geq 0$, and in fact $\log v > 0$: if $v = 1$ then every $\lambda_k(v) = 0$, and $|\theta| \leq \varepsilon < 2\pi$ with $\theta \in 2\pi\mathbb{Z}^z$ forces $\theta = 0$, so the point would be the origin. Therefore $1 < v \leq T^{1/2}$.
Finally let $u = v^2$, so $u \in \mathcal{Z}_\Lambda^\times$ and $1 < u \leq T$. For the real embeddings $\sigma_k(u) = \sigma_k(v)^2 > 0$ and $u^{1/d}\sigma_k(u) = e^{\lambda_k(u)}$ with $|\lambda_k(u)| = 2|\lambda_k(v)| \leq 2d\varepsilon$ (the last $\lambda$ is bounded by $d$ times the largest of the others). For the complex pairs $\operatorname{Arg}\sigma_{r+j}(u) = 2\theta_j$ with $|2\theta_j| \leq 2\varepsilon < \pi$, so $u^{1/d}\sigma_{r+j}(u) = e^{\lambda_{r+j}(u) + 2i\theta_j}$ with $|\lambda_{r+j}(u) + 2i\theta_j| \leq 4d\varepsilon$. In all cases $u^{1/d}\sigma_k(u) = e^{\zeta_k}$ with $|\zeta_k| \leq 4d\varepsilon \leq 1$, so $|u^{1/d}\sigma_k(u) - 1| \leq |\zeta_k| e^{|\zeta_k|} \leq 12d\varepsilon \ll (\log T)^{-1/(d-1)}$. $\blacksquare$
For a cubic field $d = 2$, $\Gamma$ is a lattice in the plane, the box is a rectangle $\log T$ wide and $2\varepsilon = 4V/\log T$ tall, and you can watch it work. In $\mathbb{Q}(\sqrt[3]{2})$ the second coordinate is the argument $\theta(u)$, and $\Gamma$ is generated by $(\log u_1, \theta(u_1))$ and $(0, 2\pi)$.
In $\mathbb{Q}(\rho)$, $\rho = 2\cos\frac{2\pi}{7}$, the second coordinate is $\lambda_1(u) = \frac{1}{2}\delta(u)$. The units $\rho$ and $\rho' = \rho^2 - 2 = 2\cos\frac{4\pi}{7}$ (both roots of $x^3 + x^2 - 2x - 1$) give the two generators, and the covolume is much smaller, so the box is thinner.
Drive the box yourself. Click a caught point to see which unit it is, and the approximation that $u = v^2$ gives.
The same rectangle does both cubic cases. Its long side is "how big a unit may be," $\log u \leq \frac{1}{2}\log T$. Its short side is "how balanced the unit must be": in the complex case that is an angle, in the totally real case a logarithmic ratio. The area is pinned at $4V$, so the short side is forced to be about $1/\log T$, and that is the logarithm in Peck's theorem.
For honesty, here is how the hunt for a balanced unit looked in my notes before we saw the box, in the totally real case. Step one: the logarithms of the units form a lattice in the plane, so every point of the plane is within some fixed distance $K$ of a unit; walk the perfectly balanced target $(\log T, 0)$ out along the ray and there is always a unit within $K$ of it, balanced up to a constant factor $e^{\pm K}$. Counting the strip gives about $\log T$ of them below $T$.
Step two: the imbalances $\xi = \lambda_1(u)$ of those $\log T$ units sit in a fixed interval, so two of them are within $1/\log T$ of each other, and the quotient of those two units has imbalance $1/\log T$. That's a correct proof of the cubic case, and it is two pigeonholes in a row. Minkowski's box is both at once: its long side is step one, its short side is step two, and it works in every degree without changing a word.
Suppose $u \in \mathcal{Z}_\Lambda^\times$ satisfies $u > 1$ and
$$\sigma_k(u) = u^{-1/d}(1 + \epsilon_k), \qquad |\epsilon_k| \leq \varepsilon, \qquad 1 \leq k \leq d, \tag{14}$$for some $0 < \varepsilon \leq 1$. Then the multiplication matrix splits into a perfectly balanced part and a small error:
$$M_u = \begin{pmatrix} u & 0^t \\ 0 & u^{-1/d} I_d \end{pmatrix} + u^{-1/d} E,$$where $E$ is block diagonal with blocks $0$, then $\epsilon_1, \dots, \epsilon_r$, then the $2 \times 2$ matrices of multiplication by $\epsilon_{r+1}, \dots, \epsilon_{r+z}$. Each row of $E$ has at most two nonzero entries of size at most $\varepsilon$, so $|Ex| \leq 2\varepsilon|x|$.
With $u$ as in (14), let $s = \alpha_1^* u \in \Lambda^*$ and write $R\sigma(s) = (q, p - q\alpha)^t$. Then
$$q = \alpha_1^* u + O(u^{-1/d}), \qquad |p_1 - q\alpha_1 - u^{-1/d}| \ll \varepsilon u^{-1/d}, \qquad |p_i - q\alpha_i| \ll \varepsilon u^{-1/d} \quad (2 \leq i \leq d), \tag{15}$$and feeding in the unit of Lemma 8 gives Theorem 1.
Write $\sigma(\alpha_1^*) = (\alpha_1^*, w)^t$ with $w \in \mathbb{R}^d$. The identities $R\sigma(\alpha_1^*) = (0, e_1)^t$ and the block form of $R$ give $\alpha_1^* + c^t w = 0$ and $\tilde{R}w = e_1$. Therefore
$$\begin{pmatrix} q \\ p - q\alpha \end{pmatrix} = RM_u\sigma(\alpha_1^*) = R\begin{pmatrix} u\alpha_1^* \\ u^{-1/d}w \end{pmatrix} + u^{-1/d}RE\sigma(\alpha_1^*) = \begin{pmatrix} \alpha_1^*(u - u^{-1/d}) \\ u^{-1/d}e_1 \end{pmatrix} + u^{-1/d}RE\sigma(\alpha_1^*).$$Since $|RE\sigma(\alpha_1^*)| \ll \varepsilon \leq 1$, this is (15). The first error, $p_1 - q\alpha_1$, is pinned near $u^{-1/d}$: it gets no savings. The others are all $\ll \varepsilon u^{-1/d}$.
Now let $T \geq 2$. By the first estimate there is a constant $C' \geq 2$ with $|q| \leq C'u$ whenever $u > 1$ satisfies (14). For $T$ large, apply Lemma 8 with $T/C'$ in place of $T$: it gives $u \in \mathcal{Z}_\Lambda^\times$ with $1 < u \leq T/C'$ satisfying (14) with $\varepsilon \ll (\log(T/C'))^{-1/(d-1)} \ll (\log T)^{-1/(d-1)}$. If $T$ is also large enough that the implied constant in the last estimate of (15) times $\varepsilon$ is less than $1$, then $q \neq 0$: otherwise the integers $p_i = p_i - q\alpha_i$ for $2 \leq i \leq d$ have absolute value less than $1$, so they vanish, and $s = p_1\alpha_1^*$; but $s = \alpha_1^* u$, so $u = p_1$ would be a rational integer which is also a unit, contradicting $u > 1$.
Let $n = |q|$, so $1 \leq n \leq C'u \leq T$, and let $p_i' = \operatorname{sgn}(q)\,p_i$, so that $\|n\alpha_i\| \leq |n\alpha_i - p_i'| = |p_i - q\alpha_i|$. Since $u^{-1/d} \leq C'^{1/d} n^{-1/d}$, the second and third estimates of (15) give
$$\|n\alpha_1\| \leq u^{-1/d}(1 + O(\varepsilon)) \ll n^{-1/d}, \qquad \|n\alpha_i\| \ll \varepsilon u^{-1/d} \ll n^{-1/d}(\log T)^{-1/(d-1)} \quad (2 \leq i \leq d),$$which is Theorem 1 for all $T \geq T_2$, the largest of the thresholds imposed on $T$. For $2 \leq T < T_2$ the bounds hold with $n = 1$ after enlarging $C$, since $\|\alpha_i\| \leq 1/2$. $\blacksquare$
Run it yourself, with the paper's own example, exactly: every $q$ and $p_i$ below is an integer computed from the BigInt powers of $u_1$.
Where exactly is the $\log$? The box in $\Gamma$ has area $2^d V$ no matter what. Its long side is $\log T$, so its $d - 1$ short sides multiply to about $1/\log T$, and each is about $(\log T)^{-1/(d-1)}$. A short side is how far the caught unit's conjugates are from perfectly balanced, and by (15) that imbalance is the error in every coordinate but the first. One convex body, one logarithm.
The same argument, run inside a congruence subgroup of the unit group, controls the denominator $n$ modulo any $m$.
Let $d \geq 2$ and suppose that $1, \alpha_1, \dots, \alpha_d$ is a basis for a real algebraic number field $K$. Then there is a constant $C = C(\alpha_1, \dots, \alpha_d)$ such that, for every positive integer $m$, there are infinitely many positive integers $n$ divisible by $m$ with
$$n^{1/d} \max_{1 \leq i \leq d} \|n\alpha_i\| \leq C. \tag{3}$$Furthermore, for every prime $p$,
$$\liminf_{n \to \infty}\, n^{1/d}(\log n)^{1/(r+z)}\,|n|_p \max_{1 \leq i \leq d} \|n\alpha_i\| < \infty, \tag{4}$$where $r + z$ is the rank of the unit group of $K$ and $|n|_p$ is the $p$-adic absolute value.
The mechanism is two lines. Let $U_m = \{u \in \mathcal{Z}_\Lambda^\times : u \equiv 1 \bmod m\mathcal{Z}_\Lambda\}$, a subgroup of finite index $I_m \leq m^{d+1}$. If $u = 1 + m\gamma \in U_m$ then, since $\operatorname{Tr}(\alpha_1^*) = 0$ and $\alpha_1^*\gamma \in \Lambda^*$ by Lemma 7,
$$q = \operatorname{Tr}(\alpha_1^* u) = m\operatorname{Tr}(\alpha_1^*\gamma) \in m\mathbb{Z}.$$Now run the proof of Lemma 8 with the positive units of $U_m$ in place of $P$. The lattice $\Gamma_m$ it produces is a sublattice of $\Gamma$ of index at most $I_m$, so its covolume is at most $I_m V$, and the proof goes through verbatim: for all $T$ with $\log T \geq c_1 I_m$ there is a unit $u \in U_m$ with $1 < u \leq T$ and $|u^{1/d}\sigma_k(u) - 1| \leq c_2 (I_m / \log T)^{1/(d-1)}$. Feeding it into the proof of Theorem 1 gives an $n \leq T$ divisible by $m$ with $\|n\alpha_i\| \ll n^{-1/d}(I_m/\log T)^{1/(d-1)} \ll n^{-1/d}$, with constants that do not depend on $m$, which is (3). For (4) take $m = p^k$: every element of $(\mathcal{Z}_\Lambda/p^k\mathcal{Z}_\Lambda)^\times$ has order less than $p^{k+d}$ and $\mathcal{Z}_\Lambda^\times$ has $r + z$ generators besides $-1$, so $I_{p^k} \ll_p p^{k(r+z)}$; choosing $T_k = \exp(c_1 I_{p^k})$ gives $n_k \leq T_k$ divisible by $p^k$ with $|n_k|_p \leq p^{-k} \ll I_{p^k}^{-1/(r+z)} \ll (\log n_k)^{-1/(r+z)}$.
Here it is in $\mathbb{Q}(\sqrt[3]{2})$ for $m = 2$ and $m = 3$. The order of $u_1$ modulo $m$ is $4$ and $3$, so $U_m$ is every fourth, then every third, power of $u_1$; the sublattice is sparser, the box must be taller to catch a point, and when it does, the denominator it produces is divisible by $m$ on the nose.
Taking $m = p^k$ in (3) and letting $k \to \infty$ gives $\liminf_n n^{1/d}|n|_p \max_i \|n\alpha_i\| = 0$ for every prime $p$: an analogue for $\alpha$ of the $p$-adic Littlewood conjecture of de Mathan and Teulié. Their Théorème 3.1 gives (4) with exponent $1/d$ on the logarithm; Bugeaud proved (3) with $C$ independent of $m$ by the same algebraic construction but a linear recurrence in place of Minkowski's theorem, and claimed exponent $1$, but de Mathan has pointed out that the argument appears incomplete for totally real cubic fields, where he proves exponent $1/2$. The exponent $1/(r+z)$ here agrees with those results when $d = 2$ and improves on $1/d$ whenever $K$ has a non-real embedding. Whether exponent $1$ holds in general is open; the proof only needs $p^k \mid \operatorname{Tr}(\alpha_1^* u)$, which far more units satisfy than $u \in U_{p^k}$, and numerical experiments in the fields of $x^3 - x - 1$, $x^3 - 3x - 1$, $x^3 - 4x - 1$, and $x^4 - x - 1$ with $p = 2, 3$ are consistent with exponent $1$.
The logarithmic savings in Theorem 1 is split equally among $d - 1$ coordinates. Peck asked whether it has to be. In his words (his degree is $n + 1$ where ours is $d + 1$, and his coordinate with no savings is the last one rather than the first):
"A number of interesting problems can be raised in connection with (3). In one direction it can be asked whether $n - 1$ of the inequalities (2) can be improved with factors which are not all the same; e.g., one might conjecture that we can find infinitely many solutions of the inequalities $|q_0\beta_j - q_j\beta_0| < Cq_0^{-1/n}/f_j(q_0)$ $(j = 1, \dots, n-1)$, $|q_0\beta_n - q_n\beta_0| < Cq_0^{-1/n}$, with $f_1(q_0) \cdots f_{n-1}(q_0) = \log q_0$ and $f_j(q_0) \geq 1$."
Functions $f_2, \dots, f_d$ defined for $T \geq 3$ are admissible if $f_i(T) \geq 1$ for each $i$ and $f_2(T) \cdots f_d(T) = \log T$ for every $T \geq 3$.
Nobody seems to have treated the conjecture since 1961. Note that results about the product $n\|n\alpha_1\| \cdots \|n\alpha_d\|$ alone cannot distinguish between different choices of the $f_i$, since the reweighting leaves the product unchanged. The paper proves the inequalities of Peck's conjecture, for every admissible choice, for a family of quartic fields, and for an arbitrary field when the $f_i$ are not too far apart.
Let $a$ and $b$ be positive integers, none of $a$, $b$, $ab$ a perfect square, and set $\alpha_1 = \sqrt{ab}$, $\alpha_2 = \sqrt{a}$, $\alpha_3 = \sqrt{b}$, so that $1, \alpha_1, \alpha_2, \alpha_3$ is a basis for $\mathbb{Q}(\sqrt{a}, \sqrt{b})$. Then there are constants $C$ and $T_0$, depending only on $a$ and $b$, such that for every admissible pair $f_2, f_3$ and every $T \geq T_0$ there is a positive integer $n \leq T$ with
$$\|n\alpha_1\| \leq Cn^{-1/3}, \qquad \|n\alpha_2\| \leq \frac{Cn^{-1/3}}{f_2(T)}, \qquad \|n\alpha_3\| \leq \frac{Cn^{-1/3}}{f_3(T)}. \tag{5}$$Take $a = 2$, $b = 5$, $f_2(T) = \log T$, and $f_3 \equiv 1$. What does Theorem 3 say?
This is Corollary 4 of the paper: there are constants $C$ and $T_0$ such that, for every $T \geq T_0$, there is a positive integer $n \leq T$ with
$$\|n\sqrt{10}\| \leq Cn^{-1/3}, \qquad \|n\sqrt{2}\| \leq \frac{Cn^{-1/3}}{\log T}, \qquad \|n\sqrt{5}\| \leq Cn^{-1/3}.$$All of the logarithm goes to $\sqrt{2}$, and $\sqrt{5}$ gets none. Nothing about $a$ or $b$ being squarefree was assumed; the multiplier ring takes care of that.
Let $d \geq 2$ and suppose that $1, \alpha_1, \dots, \alpha_d$ is a basis for a real algebraic number field. For every $c > 0$ there are constants $C$ and $T_0$, depending only on $\alpha_1, \dots, \alpha_d$ and $c$, such that for all admissible $f_2, \dots, f_d$ and all $T \geq T_0$ satisfying
$$\max_{2 \leq i \leq d} f_i(T) \leq c \min_{2 \leq i \leq d} f_i(T)^2, \tag{6}$$there is a positive integer $n \leq T$ with $\|n\alpha_1\| \leq Cn^{-1/d}$ and $\|n\alpha_i\| \leq Cn^{-1/d}/f_i(T)$ for $2 \leq i \leq d$.
Taking all the $f_i$ equal recovers Theorem 1 for $T \geq T_0$, and (6) holds for all large $T$ whenever the $f_i$ are powers of $\log T$ whose exponents differ by a factor less than two. The proof of Theorem 1 used the unit of Lemma 8 only through the single bound $|\epsilon_k| \leq \varepsilon$, so it bounds every error $p_i - q\alpha_i$, $i \geq 2$, by the same quantity. To distribute the savings unevenly we need two things: an exact description of the errors as a function of the deviation $\epsilon$, and a version of Lemma 8 whose box may have different side lengths in different directions.
The error map. Let $V_{\mathbb{R}}$ be the real vector space of $\xi \in \mathbb{C}^d$ with $\xi_k \in \mathbb{R}$ for $k \leq r$ and $\xi_{r+z+j} = \overline{\xi_{r+j}}$, of dimension $d$, and $H = \{\xi \in V_{\mathbb{R}} : \xi_1 + \cdots + \xi_d = 0\}$, of dimension $d - 1$. Define $\Psi \colon V_{\mathbb{R}} \to \mathbb{R}^d$ by $\Psi(\xi) = \tilde{R}\,\iota\big(\sigma_1(\alpha_1^*)\xi_1, \dots, \sigma_d(\alpha_1^*)\xi_d\big)$, where $\iota$ lists real parts and imaginary parts in the order $\sigma$ uses.
$\Psi$ is a linear isomorphism from $V_{\mathbb{R}}$ onto $\mathbb{R}^d$, with components
$$\Psi_i(\xi) = \sum_{k=1}^{d} \sigma_k(\alpha_1^*)\big(\sigma_k(\alpha_i) - \alpha_i\big)\xi_k \qquad (1 \leq i \leq d). \tag{19}$$In particular $\Psi(\mathbf{1}) = e_1$, and $\Psi' = (\Psi_2, \dots, \Psi_d)$ restricts to a linear isomorphism from $H$ onto $\mathbb{R}^{d-1}$. For a unit $u$ satisfying (14) the estimates of (15) sharpen to the exact identity
$$p_i - q\alpha_i = u^{-1/d}\big( \delta_{i1} + \Psi_i(\epsilon) \big) \qquad (1 \leq i \leq d). \tag{20}$$Up to normalization, the components of $\Psi(\epsilon)$ are the linear forms in the conjugates of $u$ that Peck works with directly. The rectifying frame adds a geometric point of view: choosing a unit amounts to choosing a point $\epsilon$, and the $d - 1$ errors we wish to make small are the coordinates of that point in the system $\Psi_2, \dots, \Psi_d$. For a totally positive unit $u > 1$ write $\zeta(u) \in V_{\mathbb{R}}$ for the vector with $\zeta_k(u) = \lambda_k(u)$ for real $k$ and $\zeta_{r+j}(u) = \lambda_{r+j}(u) + i\operatorname{Arg}\sigma_{r+j}(u)$, so that $u^{1/d}\sigma_k(u) = e^{\zeta_k(u)}$ and $\epsilon_k = e^{\zeta_k(u)} - 1$; the norm condition puts $\zeta(u) \in H$.
Let $d \geq 2$ and let $Y \colon H \to \mathbb{R}^{d-1}$ be a linear isomorphism. There are positive constants $L = L(Y)$ and $J = J(Y, \Lambda)$ with the following property. Suppose $T > 1$ and $\eta_2, \dots, \eta_d > 0$ satisfy
$$\prod_{i=2}^{d} \eta_i = \frac{J}{\log T} \qquad \text{and} \qquad L \max_{2 \leq i \leq d} \eta_i \leq \frac{1}{4}. \tag{22}$$Then there is a totally positive $u \in \mathcal{Z}_\Lambda^\times$ with $1 < u \leq T$, $|Y_i(\zeta(u))| \leq 2\eta_i$ for $2 \leq i \leq d$, and $|\zeta(u)| \leq 2L\max_i \eta_i \leq 1/2$.
The proof is the proof of Lemma 8 with the box $\{|x_0| \leq \frac{1}{2}\log T,\ |Y_i(\Theta(\lambda, \theta))| \leq \eta_i\}$, whose volume is again $2^d V$ by the first condition in (22); the generalization is only that the $d - 1$ side lengths are now free, subject to their product. With $Y = \Psi'|_H$, the box's sides are the errors themselves, and Theorem 5 follows by the computation of Theorem 1, with one extra wrinkle: $\epsilon = e^{\zeta} - 1$ is $\zeta$ only up to a second-order term $|\zeta|^2$, and hypothesis (6) is exactly what keeps that term below $\eta_i$. Here is the box in the biquadratic field, where $d = 3$, $z = 0$, and $H$ is the plane $\zeta_1 + \zeta_2 + \zeta_3 = 0$ drawn in the coordinates $y_2 = \zeta_1 - \zeta_3$, $y_3 = \zeta_2 - \zeta_3$.
The plane above is a slice. The real body lives in three dimensions, $(\zeta_1, \zeta_2, \log u)$: the slab $|\log u| \leq \frac{1}{2}\log T$ times the parallelogram $\{|y_2| \leq \eta_2, |y_3| \leq \eta_3\}$, a prism of volume $2^3 V$ whatever the split. Here it is turning, with its shape sweeping from all-on-$\sqrt{5}$ to all-on-$\sqrt{2}$ and back, and the units inside it lighting up.
Why does the biquadratic field escape hypothesis (6)? Label the nontrivial embeddings $\sigma_1 \colon (\sqrt{a}, \sqrt{b}) \mapsto (-\sqrt{a}, \sqrt{b})$, $\sigma_2 \colon (\sqrt{a}, \sqrt{b}) \mapsto (\sqrt{a}, -\sqrt{b})$, $\sigma_3 = \sigma_1\sigma_2$. The traces of $\sqrt{a}$, $\sqrt{b}$, $\sqrt{ab}$ all vanish, so the dual basis is $\alpha_0^* = 1/4$ and $\alpha_i^* = 1/(4\alpha_i)$, and substituting into (19) gives
$$\Psi_1(\xi) = \tfrac{1}{2}(\xi_1 + \xi_2), \qquad \Psi_2(\xi) = 2\alpha_1^*\alpha_2(\xi_1 - \xi_3), \qquad \Psi_3(\xi) = 2\alpha_1^*\alpha_3(\xi_2 - \xi_3). \tag{27}$$So with $y_2 = \zeta_1 - \zeta_3$ and $y_3 = \zeta_2 - \zeta_3$, the essential point is that
$$\Psi_2(\epsilon) = 2\alpha_1^*\alpha_2\big(e^{\zeta_1} - e^{\zeta_3}\big) = 2\alpha_1^*\alpha_2\, e^{\zeta_3}\big(e^{y_2} - 1\big), \qquad \Psi_3(\epsilon) = 2\alpha_1^*\alpha_3\, e^{\zeta_3}\big(e^{y_3} - 1\big).$$Each error depends on one of the two variables $y_2, y_3$, and vanishes exactly on the line $y_i = 0$. That circumvents the second-order approximation to $\epsilon$ entirely, and Lemma 10 with $Y = (y_2, y_3)$ and $L = 1$ gives Theorem 3 for every admissible pair, with $\eta_i = \mu_i/f_i(T)$ and the larger $f_i$ getting the smaller box side. Nothing is required of the smaller $f_i$, which may be identically $1$.
The logarithmic savings is genuinely logarithmic. De Mathan has shown, using Baker's theory of linear forms in logarithms, that for these $\alpha$ and any $c > 0$, every $n \geq 2$ with $\max_i \|n\alpha_i\| \leq cn^{-1/d}$ satisfies $\max_{i \geq 2} \|n\alpha_i\| \gg_c n^{-1/d}(\log n)^{-\kappa}$ for some $\kappa = \kappa(c)$. So the factor $(\log T)^{-1/(d-1)}$ cannot be replaced by a positive power of $n^{-1}$, and since that bound constrains only the largest of the errors, it constrains only the smallest of the $f_i$.
One box, three theorems. Left, Lemma 8's box in the lattice of units, growing with $\log T$. Middle, the same box in a congruence sublattice, where it needs four times the area and pays for it with a denominator divisible by $2$. Right, the box of Lemma 10, changing shape at fixed area to hand the savings to whichever coordinate you like.
Peck's theorem says a basis of a number field is approximable just a little better than Dirichlet's theorem promises, by a logarithm spread over $d - 1$ coordinates. The proof here says why in a sentence: the error vector of $\alpha_1^* u$ is the imbalance of the unit $u$, read through a fixed linear map, and Minkowski's theorem finds a unit below any height whose imbalance is about $(\log T)^{-1/(d-1)}$, because the box in the lattice of unit logarithms has a fixed area and a side of length $\log T$.
Three questions stay open. Whether the exponent $1/(r+z)$ in Theorem 2 can be raised to $1$, as the numerics suggest. Whether hypothesis (6) in Theorem 5 can be removed, which is what the biquadratic case manages by having each error depend on a single coordinate of $H$. And Peck's conjecture read literally, with $f_i(n)$ in place of $f_i(T)$: for non-decreasing $f_i$ the two are the same, but without monotonicity the conjecture as written is a stronger statement than what is proved here.