Unitary matrices and rotation of n-dimensional torus

Question

I am reading a paper on quantum automata, and within the proof of theorem $6$, there is the following statement (I have added a definition for completeness of the statement).

Let $U$ be a unitary matrix of rank $n$. Then, in its diagonal basis, $U$ rotates $n$ complex numbers on the unit circle by $n$ different angles $\omega_i$, $1 \leq i \leq n$. We can think of this as a rotation of an $n$-dimensional torus. If $V = (c\epsilon)^n$ is the volume of a $n$-dimensional ball of radius $\epsilon$, then $U^k$ is within a distance $\epsilon$ of the identity matrix for some number of iterations $k \leq \frac{1}{V}$.

Could anyone help me to understand the reasoning or give any hint to study and try to understand this?

Answer 1

I think it means this: As $U$ is unitary, it can be unitarily diagonalized as $U=Q^\ast DQ$, where $Q$ is a unitary matrix and $D$ is a diagonal matrix, say, $D=\operatorname{diag}(e^{i\omega_1},\ldots,e^{i\omega_n})$ with $\omega_j\in[0,2\pi)$ for each $j$. Therefore, $U^k=Q^\ast D^kQ$, where $D^k=\operatorname{diag}(e^{ik\omega_1},\ldots,e^{ik\omega_n})$. Now the paper claims that if $V=(c\epsilon)^n$ is the volumn of an $n$-ball with radius $\epsilon$ (so that $c^n\equiv\pi^{n/2}/\Gamma\left(\frac n2+1\right)$, if that matters), then for some natural integer $k\le\frac1V$, we have $|e^{ik\omega_j}-1|<\epsilon$ for every $j$. (Presumably $\epsilon$ is sufficiently small. Otherwise it is easy to find a counterexample.)

As to the torus bit, each $j$-th diagonal entry of $D^k$ represents a rotation on the Argand plane by the angle $k\omega_j$, and the locus of the rotations of the ball $B=\{(z_1,\ldots,z_n)\in\mathbb{C}^n: |z_j-1|<\epsilon\}$ by $D, D^2, D^3, \ldots$ lies on a torus. So, if the above statement on $k$ is true, then the ball will return to within a distance $\epsilon$ from its initial position after $k$ rotations. Yet I don't see how the statement on $k$ is proved in the paper.