# Diophantine vectors

Jordan Bell
March 31, 2016

## 1 Dirichlet’s approximation theorem

Let $m\geq 1$, and for $v\in\mathbb{R}^{m}$ write

 $|v|_{\infty}=\max\{|v_{j}|:1\leq j\leq m\}.$

For a positive integer $r$, let

 $V_{r}=\{k\in\mathbb{Z}^{m}:0<|k|_{\infty}\leq r\},$

which has $N_{r}=(2r+1)^{m}-1$ elements. For any $k\in V_{r}$,

 $|\left\langle v,k\right\rangle|\leq m|k|_{\infty}|v|_{\infty}\leq mr|v|_{% \infty},$

Let $I_{1},\ldots,I_{N_{r}-1}$ be consecutive closed intervals with

 $[0,mr|v|_{\infty}]=\bigcup_{j=1}^{N_{r}-1}I_{j}.$

Then there is some $j$ and some $k^{\prime},k^{\prime\prime}\in V_{r}$, $k^{\prime}\neq k^{\prime\prime}$, with $|\left\langle v,k^{\prime}\right\rangle|,|\left\langle v,k^{\prime\prime}% \right\rangle|\in I_{j}$. If $\left\langle v,k^{\prime}\right\rangle,\left\langle v,k^{\prime\prime}\right\rangle$ have the same sign, then $k=k^{\prime}-k^{\prime\prime}$ satisfies $|\left\langle v,k\right\rangle|\leq|I_{j}|$, and if $\left\langle v,k^{\prime}\right\rangle,\left\langle v,k^{\prime\prime}\right\rangle$ have different signs then $k=k^{\prime}+k^{\prime\prime}$ satisfies $|\left\langle v,k\right\rangle|\leq|I_{j}|$. In either case, $k\in V_{2r}$, and $k$ satisfies

 $|\left\langle v,k\right\rangle|\leq|I_{j}|=\frac{mr|v|_{\infty}}{N_{r}-1}=% \frac{mr|v|_{\infty}}{(2r+1)^{m}-2}.$

## 2 Diophantine vectors

For real $\tau,\gamma>0$, let $D(\tau,\gamma)$ be the set of those $v\in\mathbb{R}^{m}$ such that for any nonzero $k\in\mathbb{Z}^{m}$,

 $|\left\langle v,k\right\rangle|\geq\gamma|k|_{\infty}^{-\tau}.$

In other words,

 $D(\tau,\gamma)=\bigcap_{k\in\mathbb{Z}^{m}\setminus\{0\}}\{v\in\mathbb{R}^{m}:% |\left\langle v,k\right\rangle|\geq\gamma|k|_{\infty}^{-\tau}\}=\bigcap_{k\in% \mathbb{Z}^{m}\setminus\{0\}}D(\tau,\gamma,k).$

Each $D(\tau,\gamma,k)$ is closed, so $D(\tau,\gamma)$ is closed. Let

 $D(\tau)=\bigcup_{\gamma>0}D(\tau,\gamma).$

If $\gamma_{1}\geq\gamma_{2}$ and $v\in D(\tau,\gamma_{1})$, let $k\in\mathbb{Z}^{m}\setminus\{0\}$. Then $|\left\langle v,k\right\rangle|\geq\gamma_{1}|k|_{\infty}^{-\tau}\geq\gamma_{2% }|k|_{\infty}^{-\tau}$, so $v\in D(\tau,\gamma_{2})$, i.e.

 $D(\tau,\gamma_{1})\subset D(\tau,\gamma_{2}),\qquad\gamma_{1}\geq\gamma_{2}.$

Therefore

 $D(\tau,N_{1}^{-1})\subset D(\tau,N_{2}^{-1})\qquad N_{1}\leq N_{2},$

and

 $D(\tau)=\bigcup_{N\geq 1}D(\tau,N^{-1}),$

showing that $D(\tau)$ is an $F_{\sigma}$ set.

If $0\leq\tau and $\gamma>0$, suppose by contradiction that there is some $v\in D(\tau,\gamma)$. Now, by Dirichlet’s theorem, for each positive integer $r$ there is some $k_{r}\in V_{2r}$ satisfying $|\left\langle v,k_{r}\right\rangle|\leq m|v|_{\infty}2^{-m}r^{-m+1}$. Then, as $|k_{r}|_{\infty}\leq 2r$,

 $m|v|_{\infty}2^{-m}r^{-m+1}\geq|\left\langle v,k_{r}\right\rangle|\geq\gamma|k% _{r}|_{\infty}^{-\tau}\geq\gamma(2r)^{-\tau}=\gamma(2r)^{\tau-m+1}(2r)^{m-1},$

hence

 $(2r)^{-\tau+m-1}\geq\frac{2}{cm|v|_{\infty}}.$

As $\tau, taking $r\to\infty$ yields a contradiction. Therefore

 $D(\tau)=\emptyset,\qquad 0\leq\tau

## 3 Measures of sets

Denote by $\mu$ Lebesgue measure on $\mathbb{R}^{m}$. Let $e_{1},\ldots,e_{m}$ be the standard basis for $\mathbb{R}^{m}$, so

 $|v|_{1}=\sum_{j=1}^{m}|v_{j}|=\sum_{j=1}^{m}|\left\langle v,e_{j}\right\rangle|.$

Let $C=\{v\in\mathbb{R}^{m}:|v|_{\infty}\leq 1\}$. Let $A_{m}$ be the supremum of the $(m-1)$-dimensional Hausdorff measure of the intersection of an $(n-1)$-dimensional affine subspace of $\mathbb{R}^{m}$ and $C$.

We calculate the following.11 1 Dmitry Treschev and Oleg Zubelevich, Introduction to the Perturbation Theory of Hamiltonian Systems, p. 166, Theorem 9.3.

###### Theorem 1.

For $\tau>m-1$ and $\gamma>0$,

 $\mu(C\setminus D(\tau,\gamma))\leq 4\gamma mA_{m}3^{m-1}\zeta(\tau+2-m).$
###### Proof.

Let $k\in\mathbb{Z}^{m}\setminus\{0\}$, and for $t\in\mathbb{R}$, let

 $P_{k,t}=\{x\in\mathbb{R}^{m}:\left\langle x,k\right\rangle=t\},$

and let

 $U_{k}=\left\{x\in\mathbb{R}^{m}:|\left\langle x,k\right\rangle|<\gamma|k|_{% \infty}^{-\tau}\right\}.$

$U$ is the set of points between the hyperplanes $P_{k,-\gamma|k|_{\infty}^{-\tau}}$ and $P_{k,\gamma|k|_{\infty}^{-\tau}}$. The distance between the hyperplanes $P_{k,s}$ and $P_{k,s}$ is $\frac{|s-t|}{|k|_{2}}$, so the distance between the hyperplanes $P_{k,-\gamma|k|_{\infty}^{-\tau}}$ and $P_{k,\gamma|k|_{\infty}^{-\tau}}$ is $d_{k}=\frac{2\gamma|k|_{\infty}^{-\tau}}{|k|_{2}}$. And $|x|_{2}\geq|x|_{\infty}$, so $d_{k}\leq 2\gamma|k|_{\infty}^{-\tau-1}$. But $\mu(C\cap U_{k})\leq dA_{m}$, so

 $\mu(C\cap U)\leq 2\gamma|k|_{\infty}^{-\tau-1}A_{m}.$

Now, $U_{k}=\mathbb{R}^{m}\setminus D(\tau,\gamma,k)$, so

 $C\setminus D(\tau,\gamma)=C\setminus\bigcap_{k\in\mathbb{Z}^{m}\setminus\{0\}}% D(\tau,\gamma,k)=\bigcup_{k\in\mathbb{Z}^{m}\setminus\{0\}}(C\cap U_{k}).$

We remind ourselves that for $r$ a positive integer, the set $V_{r}=\{k\in\mathbb{Z}^{m}:0<|k|_{\infty}\leq r\}$ has $N_{r}=(2r+1)^{m}-1$ elements. Therefore

 $\{k\in\mathbb{Z}^{m}\setminus\{0\}:|k|_{\infty}=r\}=V_{r}\setminus V_{r-1}$

has

 $N_{r}-N_{r-1}=(2r+1)^{m}-(2r-1)^{m}\leq 2m(2r+3)^{m-1}$

elements, using $a^{m}-b^{m}=(a-b)(a^{m-1}+a^{m-2}b+\cdots+ab^{m-2}+b^{m-1})$. Therefore

 $\displaystyle\mu(C\setminus D(\tau,\gamma))$ $\displaystyle\leq\sum_{k\in\mathbb{Z}^{m}\setminus\{0\}}\mu(C\cap U_{k})$ $\displaystyle\leq\sum_{k\in\mathbb{Z}^{m}\setminus\{0\}}2\gamma|k|_{\infty}^{-% \tau-1}A_{m}$ $\displaystyle=2\gamma A_{m}\sum_{r=1}^{\infty}\sum_{\{k\in\mathbb{Z}^{m}% \setminus\{0\}:|k|_{\infty}=r\}}r^{-\tau-1}$ $\displaystyle\leq 2\gamma A_{m}\sum_{r=1}^{\infty}2m(2r+1)^{m-1}\cdot r^{-\tau% -1}$ $\displaystyle=4\gamma mA_{m}\sum_{r=1}^{\infty}(2r+1)^{m-1}r^{-\tau-1}.$

We estimate

 $\sum_{r=1}^{\infty}(2r+1)^{m-1}r^{-\tau-1}\leq\sum_{r=1}^{\infty}(3r)^{m-1}r^{% -\tau-1}=3^{m-1}\sum_{r=1}^{\infty}r^{m-\tau-2},$

and therefore

 $\mu(C\setminus D(\tau,\gamma))\leq 4\gamma mA_{m}\cdot 3^{m-1}\zeta(\tau+2-m).$

But

 $C\setminus D(\tau)=\bigcap_{N\geq 1}(C\setminus D(\tau,N^{-1})),$

and by Theorem 1, if $\tau>m-1$ then $\mu(C\setminus D(\tau,N^{-1}))\to 0$ as $N\to\infty$. Therefore

 $\mu(C\cap D(\tau))=0,\qquad\tau>m-1.$

## 4 Cohomological equation

Let $\mathbb{T}^{m}=\{z\in\mathbb{C}^{m}:|z_{1}|=1,\ldots,|z_{m}|=1\}$ and write $\nu$ for the Haar measure on $\mathbb{T}^{m}$ for which $\nu(\mathbb{T}^{m})=1$. For $k\in\mathbb{Z}^{m}$ let $\chi_{k}(z)=\prod_{j=1}^{m}z_{j}^{k_{j}}$. Let $\Delta(\tau,\gamma)$ be the set of those $z\in\mathbb{T}^{m}$ such that

 $|\chi_{k}(z)-1|\geq\gamma|k|_{1}^{-\tau},\qquad k\in\mathbb{Z}^{m}\setminus\{0\}.$

Let

 $\Delta(\tau)=\bigcup_{\gamma>0}\Delta(\tau,\gamma),$

and then

 $\Delta=\bigcup_{\tau>0}\Delta(\tau).$

For $\lambda\in\mathbb{T}^{m}$ define $R_{\lambda}:\mathbb{T}^{m}\to\mathbb{T}^{m}$ by

 $R_{\lambda}(z)=\lambda\cdot z=(\lambda_{1}z_{1},\ldots,\lambda_{m}z_{m}).$

In the following theorem, (1) is called a cohomological equation.22 2 Anatole Katok, Combinatorial Constructions in Ergodic Theory and Dynamics, p. 71, Theorem 11.5.

###### Theorem 2.

For $\lambda\in\mathbb{T}^{m}$, $\lambda\in\Delta$ if and only if for any $h\in C^{\infty}(\mathbb{T}^{m})$ there is some $\psi\in C^{\infty}(\mathbb{T}^{m})$ such that

 $h(z)-\int_{\mathbb{T}^{m}}hd\nu=\psi(R_{\lambda}z)-\psi(z),\qquad z\in\mathbb{% T}^{m}.$ (1)
###### Proof.

It is a fact that $\chi_{k}\in\widehat{\mathbb{T}}^{m}$ and that $k\mapsto\chi_{k}$ is an isomorphism of topological groups $\mathbb{Z}^{d}\to\widehat{\mathbb{T}}^{m}$. For $f\in L^{1}(\nu)$, $\widehat{f}:\mathbb{Z}^{m}\to\mathbb{C}$ is defined by

 $\widehat{f}(k)=\int_{\mathbb{T}^{m}}f(z)\overline{\chi_{k}(z)}d\nu(z)=\int_{% \mathbb{T}^{m}}f(z)\chi_{k}(z)^{-1}d\nu(z).$

If the Fourier series of $f$ converges pointwise,

 $f(z)=\sum_{k\in\mathbb{Z}^{m}}\widehat{f}(k)\chi_{k}(z),\qquad z\in\mathbb{T}^% {m}.$

It is a fact that $f\in C^{\infty}(\mathbb{T}^{m})$ if and only if for any $R>0$ there is some $C_{R}$ such that

 $|\widehat{f}(k)|\leq C_{R}|k|_{1}^{-R},\qquad k\in\mathbb{Z}^{m}\setminus\{0\}.$

For $\psi\in L^{1}(\mathbb{T}^{m})$ and $k\in\mathbb{T}^{m}$, because $\nu$ is invariant under multiplication in $\mathbb{T}^{m}$,

 $\displaystyle\widehat{\psi\circ R_{\lambda}}(k)$ $\displaystyle=\int_{\mathbb{T}^{m}}\psi(\lambda z)\overline{\chi_{k}(z)}d\nu(z)$ $\displaystyle=\int_{\mathbb{T}^{m}}\psi(z)\overline{\chi_{k}(\lambda^{-1}x)}d% \nu(z)$ $\displaystyle=\chi_{k}(\lambda)\widehat{\psi}(k).$

Suppose that for every $h$ is $C^{\infty}$ that there is some $\psi\in C^{\infty}(\mathbb{T}^{m})$ satisfying (1). Taking the Fourier transform of (1),

 $\widehat{h}(k)-\delta_{0}(k)\cdot\int_{\mathbb{T}^{m}}hd\nu=\chi_{k}(\lambda)% \widehat{\psi}(k)-\widehat{\psi}(k),\qquad k\in\mathbb{Z}^{m},$

then, if $\chi_{k}(\lambda)\neq 1$,

 $\widehat{\psi}(k)=\frac{\widehat{h}(k)}{\chi_{k}(\lambda)-1}.$

Now suppose by contradiction that $\lambda\not\in\Delta$. This means that there are $\tau_{N}\to\infty$ such that for each $N$, there is some $\gamma_{N}>0$ and some $k_{N}\in\mathbb{Z}^{m}\setminus\{0\}$ such that $|\chi_{k_{N}}(\lambda)-1|<\gamma_{N}|k_{N}|_{1}^{-\tau_{N}}$. Define

 $\widehat{h}(k)=|\chi_{k}(\lambda)-1|^{1/2}\cdot 1_{\{k_{N}\}}(k).$

For $R>0$ let $\tau_{N}\geq 2R$. Then for $k\in\mathbb{Z}^{m}$, either $\widehat{h}(k)=0$ or if $k=k_{N}$ then

 $|\widehat{h}(k)|=|\chi_{k_{N}}(\lambda)-1|^{1/2}<\gamma_{N}^{1/2}|k_{N}|_{1}^{% -\tau_{N}/2}\leq\gamma_{N}^{1/2}|k_{N}|_{1}^{-R}=\gamma_{N}^{1/2}|k|_{1}^{-R},$

which shows that $h$ is $C^{\infty}$. There is some $\psi\in C^{\infty}(\mathbb{T}^{m})$ satisfying (1), according to which, for $\chi_{k}(\lambda)\neq 1$,

 $\widehat{\psi}(k)=\frac{\widehat{h}(k)}{\chi_{k}(\lambda)-1}.$

But $|\widehat{\psi}(k)|$ is either $0$ or if $k=k_{N}$ then $|\chi_{k_{N}}(\lambda)-1|^{-1/2}>\gamma_{N}^{-1/2}|k_{N}|_{1}^{\tau_{N}/2}$. Thus the Fourier coefficients of $\psi$ are unbounded, which contradicts that $\psi$ is $C^{\infty}$. Therefore $\lambda\in\Delta$.

Now suppose that $\lambda\in D$ and let $h\in C^{\infty}(\mathbb{T}^{m})$. Define $\psi$ by

 $\widehat{\psi}(k)=\begin{cases}\frac{\widehat{h}(k)}{\chi_{k}(\lambda)-1}&\chi% _{k}(\lambda)\neq 1\\ 0&\chi_{k}(\lambda)=1.\end{cases}$

The facts that $\lambda\in D$ and that $h$ is $C^{\infty}$ yield that $\psi$ is $C^{\infty}$. It is straightforward from the definition of $\widehat{\psi}(k)$ that $\psi$ satisfies (1). ∎