Diophantine vectors

1 Dirichlet’s approximation theorem

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

$${|v|}_{\mathrm{\infty }}=\max \{|v_j|:1\le j\le m\}.$$

For a positive integer $r$, let

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

$$|⟨v,k⟩|\le m{|k|}_{\mathrm{\infty }}{|v|}_{\mathrm{\infty }}\le mr{|v|}_{\mathrm{\infty }},$$

Let $I_1,\mathrm{\dots },I_{N_r-1}$ be consecutive closed intervals with

$$[0,mr{|v|}_{\mathrm{\infty }}]=\bigcup _{j=1}^{N_r-1}{I}_j.$$

Then there is some $j$ and some $k^{\prime },k^{\prime \prime }\in V_r$, $k^{\prime }\ne k^{\prime \prime }$, with $|⟨v,k^{\prime }⟩|,|⟨v,k^{\prime \prime }⟩|\in I_j$. If $⟨v,k^{\prime }⟩,⟨v,k^{\prime \prime }⟩$ have the same sign, then $k=k^{\prime }-k^{\prime \prime }$ satisfies $|⟨v,k⟩|\le |I_j|$, and if $⟨v,k^{\prime }⟩,⟨v,k^{\prime \prime }⟩$ have different signs then $k=k^{\prime }+k^{\prime \prime }$ satisfies $|⟨v,k⟩|\le |I_j|$. In either case, $k\in V_{2r}$, and $k$ satisfies

$$|⟨v,k⟩|\le |I_j|=\frac{mr{|v|}_{\mathrm{\infty }}}{N_r-1}=\frac{mr{|v|}_{\mathrm{\infty }}}{{(2r+1)}^m-2}.$$

2 Diophantine vectors

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

$$|⟨v,k⟩|\ge \gamma {|k|}_{\mathrm{\infty }}^{-\tau }.$$

In other words,

$$D(\tau ,\gamma )=\bigcap _{k\in {\mathbb{Z}}^m\setminus \{0\}}\{v\in {ℝ}^m:|⟨v,k⟩|\ge \gamma {|k|}_{\mathrm{\infty }}^{-\tau }\}=\bigcap _{k\in {\mathbb{Z}}^m\setminus \{0\}}D(\tau ,\gamma ,k).$$

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

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

If ${\gamma }_1\ge {\gamma }_2$ and $v\in D(\tau ,{\gamma }_1)$, let $k\in {\mathbb{Z}}^m\setminus \{0\}$. Then $|⟨v,k⟩|\ge {\gamma }_1{|k|}_{\mathrm{\infty }}^{-\tau }\ge {\gamma }_2{|k|}_{\mathrm{\infty }}^{-\tau }$, so $v\in D(\tau ,{\gamma }_2)$, i.e.

$$D(\tau ,{\gamma }_1)\subset D(\tau ,{\gamma }_2),{\gamma }_1\ge {\gamma }_2.$$

Therefore

$$D(\tau ,N_1^{-1})\subset D(\tau ,N_2^{-1})\mathit{\hspace{1em}\hspace{1em}}{N}_1\le N_2,$$

and

$$D(\tau )=\bigcup _{N\ge 1}D(\tau ,N^{-1}),$$

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

If 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 $|⟨v,k_r⟩|\le m{|v|}_{\mathrm{\infty }}{2}^{-m}{r}^{-m+1}$. Then, as ${|k_r|}_{\mathrm{\infty }}\le 2r$,

$$m{|v|}_{\mathrm{\infty }}{2}^{-m}{r}^{-m+1}\ge |⟨v,k_r⟩|\ge \gamma {|k_r|}_{\mathrm{\infty }}^{-\tau }\ge \gamma {(2r)}^{-\tau }=\gamma {(2r)}^{\tau -m+1}{(2r)}^{m-1},$$

hence

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

As , taking $r\to \mathrm{\infty }$ yields a contradiction. Therefore

3 Measures of sets

Denote by $\mu$ Lebesgue measure on ${ℝ}^m$. Let $e_1,\mathrm{\dots },e_m$ be the standard basis for ${ℝ}^m$, so

$${|v|}_1=\sum _{j=1}^m|v_j|=\sum _{j=1}^m|⟨v,e_j⟩|.$$

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

We calculate the following.

Theorem 1.

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

$$\mu (C\setminus D(\tau ,\gamma ))\le 4\gamma m{A}_m{3}^{m-1}\zeta (\tau +2-m).$$

Proof.

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

$$P_{k,t}=\{x\in {ℝ}^m:⟨x,k⟩=t\},$$

and let

$U$ is the set of points between the hyperplanes $P_{k,-\gamma {|k|}_{\mathrm{\infty }}^{-\tau }}$ and $P_{k,\gamma {|k|}_{\mathrm{\infty }}^{-\tau }}$. The distance between the hyperplanes $P_{k,s}$ and $P_{k,s}$ is $\frac{|s-t|}{{|k|}_2}$, so the distance between the hyperplanes $P_{k,-\gamma {|k|}_{\mathrm{\infty }}^{-\tau }}$ and $P_{k,\gamma {|k|}_{\mathrm{\infty }}^{-\tau }}$ is $d_k=\frac{2\gamma {|k|}_{\mathrm{\infty }}^{-\tau }}{{|k|}_2}$. And ${|x|}_2\ge {|x|}_{\mathrm{\infty }}$, so $d_k\le 2\gamma {|k|}_{\mathrm{\infty }}^{-\tau -1}$. But $\mu (C\cap U_k)\le d{A}_m$, so

$$\mu (C\cap U)\le 2\gamma {|k|}_{\mathrm{\infty }}^{-\tau -1}{A}_m.$$

Now, $U_k={ℝ}^m\setminus D(\tau ,\gamma ,k)$, 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 has $N_r={(2r+1)}^m-1$ elements. Therefore

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

has

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

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

$\mu (C\setminus D(\tau ,\gamma ))$ $\le {\displaystyle \sum _{k\in {\mathbb{Z}}^m\setminus \{0\}}}\mu (C\cap U_k)$

$\le {\displaystyle \sum _{k\in {\mathbb{Z}}^m\setminus \{0\}}}2\gamma {|k|}_{\mathrm{\infty }}^{-\tau -1}{A}_m$

$=2\gamma A_m{\displaystyle \sum _{r=1}^{\mathrm{\infty }}}{\displaystyle \sum _{\{k\in {\mathbb{Z}}^m\setminus \{0\}:{|k|}_{\mathrm{\infty }}=r\}}}{r}^{-\tau -1}$

$\le 2\gamma A_m{\displaystyle \sum _{r=1}^{\mathrm{\infty }}}2m{(2r+1)}^{m-1}\cdot r^{-\tau -1}$

$=4\gamma m{A}_m{\displaystyle \sum _{r=1}^{\mathrm{\infty }}}{(2r+1)}^{m-1}{r}^{-\tau -1}.$

We estimate

$$\sum _{r=1}^{\mathrm{\infty }}{(2r+1)}^{m-1}{r}^{-\tau -1}\le \sum _{r=1}^{\mathrm{\infty }}{(3r)}^{m-1}{r}^{-\tau -1}=3^{m-1}\sum _{r=1}^{\mathrm{\infty }}{r}^{m-\tau -2},$$

and therefore

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

∎

But

$$C\setminus D(\tau )=\bigcap _{N\ge 1}(C\setminus D(\tau ,N^{-1})),$$

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

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

4 Cohomological equation

Let ${𝕋}^m=\{z\in {ℂ}^m:|z_1|=1,\mathrm{\dots },|z_m|=1\}$ and write $\nu$ for the Haar measure on ${𝕋}^m$ for which $\nu ({𝕋}^m)=1$. For $k\in {\mathbb{Z}}^m$ let ${\chi }_k(z)={\prod }_{j=1}^m{z}_j^{k_j}$. Let $\mathrm{\Delta }(\tau ,\gamma )$ be the set of those $z\in {𝕋}^m$ such that

$$|{\chi }_k(z)-1|\ge \gamma {|k|}_1^{-\tau },k\in {\mathbb{Z}}^m\setminus \{0\}.$$

Let

$$\mathrm{\Delta }(\tau )=\bigcup _{\gamma >0}\mathrm{\Delta }(\tau ,\gamma ),$$

and then

$$\mathrm{\Delta }=\bigcup _{\tau >0}\mathrm{\Delta }(\tau ).$$

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

$$R_{\lambda }(z)=\lambda \cdot z=({\lambda }_1{z}_1,\mathrm{\dots },{\lambda }_m{z}_m).$$

In the following theorem, (1) is called a cohomological equation.

Theorem 2.

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

$$h(z)-{\int }_{{𝕋}^m}h𝑑\nu =\psi (R_{\lambda }z)-\psi (z),z\in {𝕋}^m.$$ (1)

Proof.

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

$$\widehat{f}(k)={\int }_{{𝕋}^m}f(z)\overline{{\chi }_k(z)}𝑑\nu (z)={\int }_{{𝕋}^m}f(z){\chi }_k{(z)}^{-1}𝑑\nu (z).$$

If the Fourier series of $f$ converges pointwise,

$$f(z)=\sum _{k\in {\mathbb{Z}}^m}\widehat{f}(k){\chi }_k(z),z\in {𝕋}^m.$$

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

$$|\widehat{f}(k)|\le C_R{|k|}_1^{-R},k\in {\mathbb{Z}}^m\setminus \{0\}.$$

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

$\widehat{\psi \circ R_{\lambda }}(k)$ $={\displaystyle {\int }_{{𝕋}^m}}\psi (\lambda z)\overline{{\chi }_k(z)}𝑑\nu (z)$

$={\displaystyle {\int }_{{𝕋}^m}}\psi (z)\overline{{\chi }_k({\lambda }^{-1}x)}𝑑\nu (z)$

$={\chi }_k(\lambda )\widehat{\psi }(k).$

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

$$\widehat{h}(k)-{\delta }_0(k)\cdot {\int }_{{𝕋}^m}h𝑑\nu ={\chi }_k(\lambda )\widehat{\psi }(k)-\widehat{\psi }(k),k\in {\mathbb{Z}}^m,$$

then, if ${\chi }_k(\lambda )\ne 1$,

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

Now suppose by contradiction that $\lambda \notin \mathrm{\Delta }$. This means that there are ${\tau }_N\to \mathrm{\infty }$ such that for each $N$, there is some ${\gamma }_N>0$ and some $k_N\in {\mathbb{Z}}^m\setminus \{0\}$ such that . Define

$$\widehat{h}(k)={|{\chi }_k(\lambda )-1|}^{1/2}\cdot 1_{\{k_N\}}(k).$$

For $R>0$ let ${\tau }_N\ge 2R$. Then for $k\in {\mathbb{Z}}^m$, either $\widehat{h}(k)=0$ or if $k=k_N$ then

which shows that $h$ is $C^{\mathrm{\infty }}$. There is some $\psi \in C^{\mathrm{\infty }}({𝕋}^m)$ satisfying (1), according to which, for ${\chi }_k(\lambda )\ne 1$,

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

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 $C^{\mathrm{\infty }}$. Therefore $\lambda \in \mathrm{\Delta }$.

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

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