Proof. Necessity. By contradiction, assume that $\theta $ is parallel to an element w in $\mathcal {Z}$ . Recall that $\mathcal {Z}$ is discrete and centro-symmetric and does not contain the origin. Hence, there exists a unique element $w_{0}\in \mathcal {Z}$ , with $\angle w_{0}\theta =0$ , which is parallel to w and has minimal length. As a result, $\|w_{0}\|\sin \angle w_{0}\theta =0$ so that $w_{0}$ is a $\mathcal {Z}$ -convergent to $\theta $ . It is sufficient to show that

$$ \begin{align*} \|v\|\cos\angle v\theta\leq\|w_{0}\| \end{align*} $$

for any $\mathcal {Z}$ -convergent v to $\theta $ , which contradicts the assumption that $\theta $ is $\mathcal {Z}$ -minimal. In fact, if there is a $\mathcal {Z}$ -convergent u to $\theta $ such that $\|u\|\cos \angle u\theta>\|w_{0}\|$ , then the definition of $\mathcal {Z}$ -convergents implies that

$$ \begin{align*} \|u\|\sin\angle u\theta<\|w_{0}\|\sin\angle w_{0}\theta=0, \end{align*} $$

which cannot happen.

Sufficiency. Assume by contradiction that $\theta $ is not $\mathcal {Z}$ -minimal. Then there is a $\mathcal {Z}$ -convergent v to $\theta $ such that

$$ \begin{align*} \|v\|\cos\angle v\theta\geq\|u\|\cos\angle u\theta \end{align*} $$

for any convergent u to $\theta $ . Since $\mathcal {Z}$ has a finite Minkowski constant, $\|v\|\sin \angle v\theta =0$ . Otherwise, the infinite strip

$$ \begin{align*} \{w\in\mathbb{R}^{2}:\|w\|\sin\angle w\theta<\|v\|\sin\angle v\theta\} \end{align*} $$

contains no point in $\mathcal {Z}$ and has infinite area. As a result, $\theta $ is parallel to $v\in \mathcal {Z}$ .

Proof. From the definition of $\mathcal {Z}$ -convergents, for any $k\in \mathbb {Z}$ , the system of inequalities

$$ \begin{align*} \begin{cases} \|v\|\,|\kern-1.4pt\sin\angle v\theta|<\|v_{k}\|\sin\angle v_{k}\theta\\ \|v\|\,|\kern-1.4pt\cos\angle v\theta|<\|v_{k}\|\cos\angle v_{k}\theta \end{cases} \end{align*} $$

admits no solution in $\mathcal {Z}$ . Since $v_{k+1}$ is the next convergent after $v_{k}$ ,

$$ \begin{align*} \|v\|\sin\angle v\theta\geq\|v_{k}\|\sin\angle v_{k}\theta \end{align*} $$

for any $v\in \mathcal {Z}$ satisfying

$$ \begin{align*} \|v_{k}\|\cos\angle v_{k}\theta<\|v\|\cos\angle v\theta<\|v_{k+1}\|\cos\angle v_{k+1}\theta. \end{align*} $$

Therefore, the rectangle

$$ \begin{align*} \{w\in\mathbb{R}^{2}:\|w\|\,|\kern-1.4pt\cos\angle w\theta|<\|v_{k+1}\|\,|\kern-1.2pt\cos\angle v_{k+1}\theta|,\ \|w\|\,|\kern-1.4pt\sin\angle w\theta|<\|v_{k}\|\,|\kern-1.4pt\sin\angle v_{k}\theta|\} \end{align*} $$

contains no point in $\mathcal {Z}$ and has area

$$ \begin{align*} \|v_{k}\|\sin\angle v_{k}\theta\,\|v_{k+1}\|\cos\angle v_{k+1}\theta \end{align*} $$

(see Figure 2). Together with the definition of the Minkowski constant $\mu (\mathcal {Z})$ , this gives the right-hand inequality of (3.1). The left-hand inequality of (3.1) follows from the fact that the rectangle with four vertices $\pm v_{k}$ and $\pm v_{k+1}$ has area

$$ \begin{align*} 2|v_{k}\times v_{k+1}|=2\|v_{k}\|\,\|v_{k+1}\|\sin\angle v_{k}v_{k+1} \end{align*} $$

and is contained in the rectangle constructed above.

Proof. As $\theta $ is $\mathcal {Z}$ -minimal, the height of its kth $\mathcal {Z}$ -convergent strictly increases to infinity as k tends to infinity.

Figure 2

A pair of consecutive $\mathcal {Z}$ -convergents.

Sufficiency. Assume that

$$ \begin{align*} \lim\limits_{k\rightarrow+\infty}\|v_{k+1}\|\cos\angle v_{k+1}\theta\,\|v_{k}\|\sin\angle v_{k}\theta=0. \end{align*} $$

Then for any $\delta>0$ , there exists $k_{0}$ such that

$$ \begin{align*} \|v_{k+1}\|\cos\angle v_{k+1}\theta\,\|v_{k}\|\sin\angle v_{k}\theta<\delta \end{align*} $$

for any $k\geq k_{0}$ . Since $\|v_{k}\|\cos \angle v_{k}\theta $ strictly increases to infinity as k increases to infinity, for any $T> \|v_{k_{0}}\|\cos \angle v_{k_{0}}\theta $ , there is a unique $k\geq k_{0}$ such that

$$ \begin{align*} \|v_{k}\|\cos\angle v_{k}\theta<T\leq v_{k+1}\cos\angle v_{k+1}\theta. \end{align*} $$

As a result,

$$ \begin{align*} \|v_{k}\|\sin v_{k}\theta<\frac{\delta}{\|v_{k+1}\|\cos\angle v_{k+1}\theta}\leq\delta/T, \end{align*} $$

which means that $v_{k}$ is a solution of the system of inequalities

$$ \begin{align*} \begin{cases} \|v\|\sin\angle v\theta<\delta/T\\ 0<\|v\|\cos\angle v\theta<T \end{cases} \end{align*} $$

for $T> v_{k_{0}}$ .

Necessity. By the definition of $\omega $ -singularity, for any $\delta>0$ , there exists $k_{0}$ such that for any $k\geq k_{0}$ , the system of inequalities

$$ \begin{align*} \begin{cases} \|v\|\sin\angle v\theta<{\delta}/({\|v_{k+1}\|\cos\angle v_{k+1}\theta})\\ 0<\|v\|\cos\angle v\theta<\|v_{k+1}\|\cos\angle v_{k+1}\theta \end{cases} \end{align*} $$

has solutions $v\in \mathcal {Z}$ . Either

$$ \begin{align*} 0<\|v\|\cos\angle v\theta<\|v_{k}\|\cos\angle v_{k}\theta \end{align*} $$

or

$$ \begin{align*} \|v_{k}\|\cos\angle v_{k}\theta\leq \|v\|\cos\angle v\theta<\|v_{k+1}\|\cos\angle v_{k+1}\theta. \end{align*} $$

The definition of the consecutive pair of $|v_{k}|$ and $|v_{k+1}|$ implies that

$$ \begin{align*} \|v_{k}\|\sin\angle v_{k}\theta\leq\|v\|\sin\angle v\theta \end{align*} $$

for both of the above cases. Therefore,

$$ \begin{align*} \|v_{k+1}\|\cos\angle v_{k+1}\theta\,\|v_{k}\|\sin\angle v_{k}\theta<\delta \end{align*} $$

for any $k\geq k_{0}$ . Since $\delta>0$ is arbitrary,

$$ \begin{align*} \lim\limits_{k\rightarrow+\infty}\|v_{k+1}\|\cos\angle v_{k+1}\theta\,\|v_{k}\|\sin\angle v_{k}\theta=0. \\[-40pt] \end{align*} $$

Proof of Theorem 1.2

To prove the theorem, we first note that the sufficiency is obvious. For the necessity, by contradiction, we assume that $\theta $ is a singular and a minimal direction of $(X,\omega )$ . Combining Lemma 3.2 and Theorem 3.3,

$$ \begin{align*} \lim\limits_{k\rightarrow+\infty}|v_{k}\times v_{k+1}|=0. \end{align*} $$

Since $|v_{k}\times v_{k+1}|\neq 0$ , this yields

$$ \begin{align*} \inf\{|u\times v|\neq0:u,v\in\mathcal{Z}\}=0. \end{align*} $$

Recall that $(X,\omega )$ is a Veech surface. By Lemma 2.5, we get the contradiction.