Harmonic Analysis on the Positive Rationals. Computation of Character Sums

TL;DR

This paper explores harmonic analysis techniques on positive rationals, focusing on representing rationals through specific product and quotient forms, and introduces a novel two-dimensional representation of rational pairs.

Contribution

It provides a detailed account of representing rationals via product/quotient forms and introduces a new simultaneous two-dimensional rational representation.

Findings

Representation of rationals using $(an+b)/(An+B)$ forms

A non-intuitive two-dimensional rational pair representation

Insights into harmonic analysis on positive rationals

Abstract

The authors review results implicit in their recent paper [2] on the product/quotient representation of rationals by rationals of the type $( an + b )/ ( An+ B )$ and give a detailed account of a particular related non-intuitive simultaneous ( i.e. two dimensional ) representation of pairs of rationals.

Equations

$$r=\prod _{n_0<n\le k}{\left(\frac{an+b}{An+B}\right)}^{{\epsilon }_n},{\epsilon }_n=0,\pm 1,$$

$$g\left(\frac{an+b}{An+B}\right)=c\ne 0$$

$${\chi }_D\left(\frac{u_{1}k+v_1}{u_{2}k+v_2}\right)=c\ne 0,$$

$$\eta (\beta ,\gamma )={\ell }^{-\alpha }\sum _{u(\text{mod}{\ell }^{\alpha +\max (\beta ,\gamma )})}\chi \left(\frac{a_{1}u+b_1}{{\ell }^{\beta }}\right)\overline{\chi }\left(\frac{A_{1}u+B_1}{{\ell }^{\gamma }}\right)$$

$$g_1(an+b)g_2(An+B)=c_1\ne 0,n>n_0,$$

$$g_1(aBn+1)g_2(Abn+1)=c_2\ne 0.$$

$$g_1(aBn+1)g_2(Ab[(aB+1)n+1]+1)=c_3.$$

$$g_2(Ab(aB+1)n+Ab+1){\overline{g}}_2(Abn+1)=c_4\ne 0.$$

$$\det\left(\begin{array}[]{cc}Ab(aB+1)&Ab+1\\ Ab&1\end{array}\right)$$

$$a=\prod _{n_0<n\le k}(5n+1{)}^{{\epsilon }_n},b=\prod _{n_0<n\le k}(5n-1{)}^{{\epsilon }_n},{\epsilon }_n=0,\pm 1$$

$$26=\prod _{j=1}^k(5n_j+1{)}^{{\epsilon }_j},26=\prod _{j=1}^k(5n_j-1{)}^{{\epsilon }_j}$$

$$g_1\otimes g_2:({\mathbb{Q}}^{\ast }{)}^2\to ({\mathbb{Q}}^{\ast }{)}^2/\Gamma \to \text{unit circle},$$

$$\chi (5n+1)\tilde{\chi }(5n-1)=c\ne 0$$

$$\chi (2)\tilde{\chi }(5k+3^{-1}(5t-1))$$

$$\chi (2)\tilde{\chi }(5k+2^{-3}(5t-1))$$

$$w((5n+1{)}^u(5n-1{)}^v)$$

$$(5n+1{)}^u(5n-1{)}^v\equiv (-1{)}^v[1+5(u-v)n](\text{mod}{5}^2),$$

$$\chi (5n+1)\tilde{\chi }(5n-1)=c_{0}w(5(u-v)n+1).$$

$$w(s)={\theta }^{-1}\sum _{r=1}^{5^2}\overline{w}(r)\exp (2\pi i{5}^{-2}rs).$$

$$\sum _{n=1}^{5^2}w(5(u-v)n+1)={\theta }^{-1}\sum _{r=1}^{5^2}\overline{w}(r)\exp (2\pi i{5}^{-2}r)\sum _{n=1}^{5^2}\exp (2\pi i{5}^{-1}r(u-v)n).$$

$$1=g_1(5n+1)g_2(5n-1)=\tilde{\chi }(5n+1)\tilde{\chi }(5n-1)=\tilde{\chi }((5n{)}^2-1);$$

$$\chi \left(p^{-2r}(a_{1}t+b_1)+a_{1}k\right)\overline{\chi }\left(p^{-r}(A_{1}t+B_1)\right).$$

$$g\left(\frac{a_1{\delta }_{0}n+a_{1}s+b_1}{A_1{\delta }_{0}n+A_{1}s+B_1}\right)={\chi }_{{\delta }_0}\left(\frac{a_{1}s+b_1}{A_{1}s+B_1}\right)$$

$$a=\prod _{n_0<n\le k}(3n+1{)}^{{\epsilon }_n},b=\prod _{n_0<n\le k}(5n+2{)}^{{\epsilon }_n},\epsilon =0,\pm 1,$$

Full text

Proof. For some real $\theta$, $\sum_{j=1}^{k}c_{j}=ke^{i\theta}$. Then $\sum_{j=1}^{k}(1-\textup{Re}\,c_{j}e^{-i\theta})=0$ and the summands are non-negative.

Consideration of the case $\eta(0,0)$ shows that any Dirichlet character $\,(\textup{mod}\,{\ell^{\alpha}})$ candidate for a character on $\mathbb{Q}^{*}/\Gamma$ will have a constant value on the reduced fractions $(a_{1}u+b_{1})/(A_{1}u+B_{1})$, $u\,(\textup{mod}\,{l^{\alpha}})$ and after the updated [2] Lemma 6, satisfy $\ell^{\alpha}\mid 6\Delta_{1}$.

Remark. A closer reading of [2] Lemma 6 shows that either $\ell^{\alpha}\mid A_{1}$, or $p\leq 3$, $p\nmid u_{1}u_{2}$ and $p^{t-1}\mid\Delta_{1}$.

The improvement from $\delta=6(a,A)(aA)^{2}\Delta^{3}$ to $\delta=6\Delta_{1}$ in [2] Theorem 2 follows at once.

Reduction of simultaneous product representations

For integers $a>0$, $A>0$, $b$, $B$ that satisfy $\Delta=aB-Ab\neq 0$, let $(\mathbb{Q}^{*})^{2}$ denote the direct product of two copies of $\mathbb{Q}^{*}$, and $\Gamma$ its subgroup generated by the elements $(an+b)\otimes(An+B)$, $n$ an integer exceeding a given $n_{0}$.

A character on the quotient group $(\mathbb{Q}^{*})^{2}/\Gamma$ extends to a pair of completely multiplicative functions $g_{j}$, $j=1,2$, on $\mathbb{Q}^{*}$ that satisfy $g_{1}(an+b)g_{2}(An+B)=1$, $n>n_{0}$; c.f. Elliott [1], Chapter 15.

The reduction argument given in [2] Section 5 has a misprint in the last line; a correct version may be found in Elliott [1], Chapter 19.

We begin with

[TABLE]

where the coefficients $a$, $b$, $A$, $B$ as positive.

Replacing $n$ by $bBn$,

[TABLE]

Replacing $n$ by $(aB+1)n+1$

[TABLE]

Eliminating between these relations,

[TABLE]

Note that the corresponding discriminant

[TABLE]

has the value $Ab\Delta$. Any Dirichlet character that represents $g_{2}$ in the manner of [2] Theorem 2 will be to a modulus $\delta$ that divides $6Ab\Delta$.

The requirement that $b$, $B$ be positive may be obviated as follows.

Replacing the variable $n$ by $n+k$ for a positive integer $k$ moves the coefficient quartet $a,b,A,B$ to $a$, $ak+b$, $A$, $Ak+B$. This does not affect that values of $(a,b)$, $(A,B)$, or the discriminant. For all sufficiently large $k$, $\delta\mid 6A(ak+b)\Delta$. Consecutive values of $k$ show that $\delta\mid 6Aa\Delta$; choosing $k$ to be a multiple of $\delta$, that $\delta\mid 6Ab\Delta$. Hence $\delta\mid 6A(a,b)\Delta$.

In the notation of Constraints $g_{2}$ is essentially represented by a Dirichlet character to a modulus dividing $6A_{1}\Delta_{1}$.

Likewise a Dirichlet character representing $g_{1}$ will be to a modulus dividing $6a_{1}\Delta_{1}$.

Remark. If $g_{j}$ is a character on the whole group of rationals and $aA<0$, we may employ $g_{j}(-1)$ to replace $a$, $A$ by $-a$, $-A$, as necessary. The outcome of the above argument is then formally the same.

We give an example.

[TABLE]

of positive integers $a$, $b$ if and only if $a\equiv 1\,(\textup{mod}\,{5})$, $(ab)^{2}\equiv 1\,(\textup{mod}\,{5^{2}})$.

In particular, there are infinitely many simultaneous representations

[TABLE]

*with $\varepsilon_{j}=\pm 1$, the integers $n_{j}$ exceeding $n_{0}$.

In this case the general character

[TABLE]

coincides on the primes not dividing 30 with a pair of Dirichlet characters to a modulus dividing 300. We may follow in outline the argument given in the two practical sections of [2], the pair $g(an+b)$, $\overline{g}(An+B)$ there replaced by $g_{1}(an+b)$, $g_{2}(An+B)$; in effect reduce ourselves to the consideration of

[TABLE]

where $\chi$, $\widetilde{\chi}$ are possibly distinct Dirichlet characters to each of the moduli $2^{2}$, 3, $5^{2}$. We consider these cases in turn.

Modulus 3. Choose $t$ to satisfy $5t\equiv 1\,(\textup{mod}\,{3})$ and set $n=t+3k$. Corresponding to $\eta(0,1)$ with $\alpha=1$, $\beta=0$, $\gamma=1$, we see that

[TABLE]

has a constant value when no zero. Summing over $k\,(\textup{mod}\,{3})$ shows this not to be tenable unless $\widetilde{\chi}$ is principal.

Likewise, $\chi$ is principal.

Modulus $2^{2}$. Choose $t$ to satisfy $5t\equiv 1\,(\textup{mod}\,{2^{3}})$ and set $n=t+2^{3}k$. Corresponding to $\eta(1,3)$ with $\alpha=2$, $\beta=1$, $\gamma=3$, we see that

[TABLE]

has a constant value when not zero. Summing over $k\,(\textup{mod}\,{2^{2}})$ shows this to be untenable unless $\widetilde{\chi}$ is principal.

Likewise $\chi$ is principal.

Remark. The closer reading of [2] Lemma 6 slightly simplifies these cases.

At this stage, for suitable characters $\chi$, $\widetilde{\chi}$ to the modulus $5^{2}$, $g_{1}\overline{\chi}\otimes g_{2}\overline{\widetilde{\chi}}$ has a constant value on the elements $(5n+1)\otimes(5n-1)$ defining $\Gamma$, and is 1 on all primes save possibly $p=2,3$; by the argument at the end of [2] Lemma 5, on those also.

Modulus $5^{2}$. This case is perhaps the most interesting.

If $w$ is a generator of the Dirichlet character group $\,(\textup{mod}\,{5^{2}})$, there are representations $\chi=w^{u}$, $\widetilde{\chi}=w^{v}$ for positive integers $u$, $v$. The hypothesis that $\chi(5n+1)\widetilde{\chi}(5n-1)$ has a constant value becomes that when not zero

[TABLE]

has a constant value. Since

[TABLE]

for a suitable non-zero constant $c_{0}$,

[TABLE]

The character $w$ is certainly primitive and for a suitable non-zero gaussian sum $\theta$ has a representation

[TABLE]

In particular,

[TABLE]

The summand $\overline{w}(r)=0$ unless $5\nmid r$, in which case the innersum over $n$ is zero unless $5\mid(u-v)$.

The characters $\chi$, $\widetilde{\chi}$ differ multiplicatively by a character $\,(\textup{mod}\,{5})$.

Without loss of generality we assume $\chi=\widetilde{\chi}\chi_{5}$ to be such a representation. Then

[TABLE]

$\widetilde{\chi}(-1)=1$. The character $\widetilde{\chi}$ has order dividing 10; it is the square of an arbitrary character $\,(\textup{mod}\,{5^{2}})$.

Conversely, such a pair $g_{1}=\widetilde{\chi}\chi_{5}$, $g_{2}=\widetilde{\chi}$ satisfies the requirement $g_{1}(5n+1)g_{2}(5n-1)=1$.

This determines the dual group of $(\mathbb{Q}^{*})^{2}/\Gamma$.

The pair $a\otimes b$ belongs to the principal class of $(\mathbb{Q}^{*})^{2}/\Gamma$ if and only if for all pairs $\widetilde{\chi}\chi_{5}\otimes\widetilde{\chi}$ with $\widetilde{\chi}$ of order dividing 10, $\widetilde{\chi}(ab)\chi_{5}(a)=1$.

Since $\chi_{5}$ may be any character $\,(\textup{mod}\,{5})$, with $\widetilde{\chi}$ principal $a\equiv 1\,(\textup{mod}\,{5})$ is required.

Then $\widetilde{\chi}(ab)=1$ for $\widetilde{\chi}\,(\textup{mod}\,{5^{2}})$ that satisfy $\widetilde{\chi}(-1)=1$; $\chi((ab)^{2})=1$ for all $\chi\,(\textup{mod}\,{5^{2}})$; $(ab)^{2}\equiv 1\,(\textup{mod}\,{5^{2}})$.

In this section we note that any prime $p\geq 5$ dividing the parameter $\delta$ that occurs in Theorem 2 necessarily divides $(a_{1},A_{1})$, hence $(a,A)$.

If such a prime satisfies $p^{r}\|\delta$, then $p^{r}\mid\Delta_{1}$. Since any Dirichlet character to a prime-power modulus will induce a character to a modulus that is a higher power of that same prime, without loss of generality we may assume that $p^{r}\|\Delta_{1}$.

If now $p\nmid a_{1}$ and $p^{2r}\mid(a_{1}n+b_{1})$, then $p^{r}$ divides $A_{1}(a_{1}n+b_{1})+\Delta_{1}$, i.e. $a_{1}(A_{1}n+B_{1})$, hence $A_{1}n+B_{1}$. The terms in the sum $p^{r}\eta(2r,r)$ are well defined and when non-zero have a constant value. This remains valid even if the underlying Dirichlet character $\chi\,(\textup{mod}\,{p^{r}})$ is no longer primitive.

Setting $n=t+p^{2r}k$, where $a_{1}t+b_{1}\equiv 0\,(\textup{mod}\,{p^{2r}})$, the corresponding summands become

[TABLE]

Note that $p^{r+1}\nmid(A_{1}t+B_{1})$, otherwise $p^{r+1}\mid\Delta_{1}$.

Summing over a complete residue class system $k\,(\textup{mod}\,{p^{r}})$ we see that $\chi$ must be principal.

With emphasis on $\overline{\chi}$ we may draw the same conclusion if $p\nmid A_{1}$.

A slight modification of the argument shows the conclusion also to be valid if $p=3$ unless $3\|\delta$ and $3\nmid a_{1}A_{1}\Delta_{1}$.

Suppose now that $p^{r}\|\delta$ and the character $\,(\textup{mod}\,{p^{r}})$ component of the representing Dirichlet character $\,(\textup{mod}\,{\delta})$ is principal. Set $\delta_{0}=p^{-r}\delta$. If $\left((a_{1}s+b_{1})(A_{1}s+B_{1}),\delta_{0}\right)=1$, then

[TABLE]

provided the numerator and denominator of the argument of $g$ are not divisible by $p$. In particular, c.f. [2] Lemma 4, second part, $g\overline{\chi}_{\delta_{0}}(p)=1$ so long as $p\nmid(a_{1},A_{1})$.

Without loss of generality we may assume that any prime greater than 3 that does not divide $(a_{1},A_{1})$ also does not divide $\delta$.

Save for the possible exception noted above, this also holds in the case $p=3$.

In the corresponding simultaneous representation via $(a_{1}n+b_{1})\otimes(A_{1}n+B_{1})$ and employing a pair of Dirichlet characters $\chi_{\delta}\otimes\chi_{\gamma}$, $\delta\mid 6a_{1}\Delta_{1}$ and has prime factors exceeding 3 only if they divide $a_{1}$, $\gamma\mid 6A_{1}\Delta_{1}$ and has prime factors exceeding 3 only if they divide $A_{1}$, $\Delta_{1}=a_{1}B_{1}-A_{1}b_{1}$, as before.

In the example $a_{1}=3$, $b_{1}=1$, $A_{1}=5$, $B_{1}=2$ this permits only that $\chi_{\delta}$ may be any Dirichlet character $\,(\textup{mod}\,{3})$, $\chi_{\gamma}$ the principal character $\,(\textup{mod}\,{5})$. Given $n_{0}$, there is a simultaneous representation

[TABLE]

if and only if $a\equiv 1\,(\textup{mod}\,{3})$, $(b,5)=1$.

This accords with the results of [1], Chapter 19.