Arguments where the $j$-invariant has non-trivially real values — and a conjectured duality

Question

The $j$-function (a.k.a. $j$-invariant up to a factor $1728$) is most often only considered for arguments where it yields real values.

It is well known that for $a,b,n\in\mathbb N$, $j\left(\dfrac {a+\sqrt{-n }}{b}\right)$ is algebraic (not necessary real), and that for $a=0$, $b=1$ or $b=2$ it is real. The literature seems to deal essentially with those latter cases.

Now while usually, $n$ is kept constant and the underlying algebraic structure of the field $\mathbb Q(\sqrt{-n})$ is involved, I have taken a somewhat transversal approach and looked for the $n$'s (not only integers) that yield real values of $j(\cdot)$ for fixed $a,b\in\mathbb N$ where $b\geqslant3$. By periodicity, it is sufficient to look at $1\leqslant a<b$ (and because of $j(z)=\overline{j(1-\bar z)}$ even only at $ a<\frac b2$). Note that $b$ is more "salient" for the patterns than $a$.

It turns out that for $b\geqslant3$, there are only a finite number of those what I'd call, for lack of a better term, "reality roots". Denote their set by $R_{a,b}$.
For instance, $R_{1,3}$ consists of $7$ such roots, viz. $n=\frac18,\frac15,\frac12,1,2,5,8$. Some (all?) of those sets contain progressions with typically "modular" appearances, e.g. for $a=3$, $b=7$ there are $12$ reality roots, which may be written as $$R_{3,7}=\Bigl\lbrace{\frac1{24},\frac2{20},\frac3{16},\frac4{12},\frac5{8},\frac64,\frac83,\frac{10}2,\frac{12}1, 3\cdot11,4\cdot10,\frac1{5\cdot9} \Bigr\rbrace}.$$ How to prove the following conjectures for $b\geqslant3$?

• All reality roots are rational. [Seems to be obvious at second glance, but why?]
• The cardinality $|R_{a,b}|$ is at most $2b+1$ (and I'd guess from numerical evidence that it's always at least $b$).
• For $a=1$, the set $R_{1,b}$ is explicitly given by $$R_{1,b}=\Bigl\lbrace{\frac1{b^2-1},\frac1{2b-1},\frac2{2b-2},\frac3{2b-3},\dots,\frac{2b-2}2,\frac{2b-1}1, b^2-1\Bigr\rbrace}.$$
• The sets $R_{a_1,b}$ and $R_{a_2,b}$ are "reciprocal" to each other (i. e. $\alpha\in R_{a_1,b}\Longleftrightarrow\frac1\alpha\in R_{a_2,b}$) if $a_1a_2\equiv\pm 1\pmod b$, so the structure of the multiplicative group $(\mathbb{Z}/b\mathbb{Z})^\times$ carries over.
  In fact, for $a_1a_2\equiv- 1\pmod b$, it seems like we have altogether for all $n\in\mathbb R^+$ a kind of general duality $$\color{blue}{j\biggl(\dfrac {a_1+\sqrt{-\frac1n }}{b}\biggr)=j\biggl(\dfrac {a_2+\sqrt{-n }}{b}\biggr)},$$
  while for $a_1a_2\equiv+ 1\pmod b$, we have the conjugate $$\color{blue}{j\biggl(\dfrac {a_1+\sqrt{-\frac1n }}{b}\biggr)=\overline{j\biggl(\dfrac {a_2+\sqrt{-n }}{b}\biggr)}}.$$ .

For the last (conjectured) statement, remember that $R_{a,b}=R_{-a,b}=R_{b-a,b}$. It implies furthermore that $R_{a,b}$ is "self-reciprocal" whenever $a^2\equiv\pm 1\pmod b$; also note that the "if" might well be an "iff".

Answer 1

Recall that the $j$-invariant identifies $\Bbb{H}/{\rm{PSL}}_2(\Bbb{Z})$ with $\Bbb{C}$; and a fundamental domain for the action of ${\rm{PSL}}_2(\Bbb{Z})$ on $\Bbb{H}$ is $D:=\left\{z\in\Bbb{H}\,\big|\, |z|>1,|\Re(z)|<\frac{1}{2}\right\}$. The $j$-function admits an expansion in terms of $q:={\rm{e}}^{2\pi{\rm{i}}z}$ with positive coefficients. Under the transformation $z\mapsto-\bar{z}$ of $\Bbb{H}$, $q$ becomes $\bar{q}$. Hence $j(z)=\overline{j(-\bar{z})}$. Thus $j(z)$ is real iff $-\bar{z}\in{\rm{PSL}}_2(\Bbb{Z})\cdot z$. In the fundamental domain $D$, $z=-\bar{z}$ if $z$ lies on the vertical axis. If $-\bar{z}$ is obtained from $z$ via applying a non-trivial element of ${\rm{PSL}}_2(\Bbb{Z})$, we should look at the boundary $\partial D$ of the fundamental domain: $-\bar{z}=z\pm 1$ if $\Re(z)=\mp\frac{1}{2}$, and $-\bar{z}=-\frac{1}{z}$ if $|z|=1$. We conclude that, up to the action of ${\rm{PSL}}_2(\Bbb{Z})$, a point $z$ with $j(z)\in\Bbb{R}$ lies on $\Re(z)=\frac{1}{2},0$ or on the unit circle. ($z\mapsto z+1$ takes $\Re(z)=-\frac{1}{2}$ to $\Re(z)=\frac{1}{2}$, so we only keep the former.) But under the element $z\mapsto\frac{1}{1-z}$ of ${\rm{PSL}}_2(\Bbb{Z})$, the unit circle is mapped to $\Re(z)=\frac{1}{2}$. Conversely, on either of $\Re(z)=0$ or $\Re(z)=\frac{1}{2}$ the value of $j$ is real because $q={\rm{e}}^{2\pi{\rm{i}}z}\in\Bbb{R}$. All in all: $$ j(z_0)\in\Bbb{R}\iff z_0\in {\rm{PSL}}_2(\Bbb{Z})\cdot\left\{z\in\Bbb{H}\,\big|\, \Re(z)\in\left\{0,\frac{1}{2}\right\}\right\}. $$

To analyze $R_{a,b}$, one should check when for some $s>0$ an expression of the form $\frac{\alpha s{\rm{i}}+\beta}{\gamma s{\rm{i}}+\delta}$ or $\frac{\alpha (\frac{1}{2}+s{\rm{i}})+\beta}{\gamma (\frac{1}{2}+s{\rm{i}})+\delta}$ where $\begin{bmatrix} \alpha &\beta\\ \gamma &\delta \end{bmatrix}\in{\rm{SL}}_2(\Bbb{Z})$ is of the form $\frac{a+\sqrt{-n}}{b}$. I assume $a,b\in\Bbb{N}$ and $\gcd(a,b)=1$ (because otherwise the expression can be simplified).

Addressing the first bullet point:

If $b\geq 3$ and $\frac{\alpha s{\rm{i}}+\beta}{\gamma s{\rm{i}}+\delta}$ or $\frac{\alpha (\frac{1}{2}+s{\rm{i}})+\beta}{\gamma (\frac{1}{2}+s{\rm{i}})+\delta}$ is equal to $\frac{a+\sqrt{n}{\rm{i}}}{b}$ for some $n>0$, then $n\in\Bbb{Q}$.

Proof) The real part of $\frac{\alpha s{\rm{i}}+\beta}{\gamma s{\rm{i}}+\delta}$ is given by $\frac{\alpha\gamma s^2+\beta\delta}{\gamma^2s^2+\delta^2}$. The equality $\frac{\alpha\gamma s^2+\beta\delta}{\gamma^2s^2+\delta^2}=\frac{a}{b}$ amounts to \begin{equation*} b\beta\delta-a\delta^2=(a\gamma^2-b\alpha\gamma)s^2. \tag{$\star$} \end{equation*} It suffices to show $s^2\in\Bbb{Q}$. That can only fail if $a\gamma^2-b\alpha\gamma$ and $b\beta\delta-a\delta^2$ are both zero. If $\gamma=0$, then, since $\begin{bmatrix} \alpha &\beta\\ \gamma &\delta \end{bmatrix}\in{\rm{SL}}_2(\Bbb{Z})$, we should have $\alpha=\delta\in\{\pm 1\}$ and the corresponding transformation is $z\mapsto z+\beta$. Therefore, the real part of $\frac{\alpha s{\rm{i}}+\beta}{\gamma s{\rm{i}}+\delta}$ (resp. $\frac{\alpha (\frac{1}{2}+s{\rm{i}})+\beta}{\gamma (\frac{1}{2}+s{\rm{i}})+\delta}$) lies in $\Bbb{Z}$ (resp. $\frac{1}{2}\Bbb{Z}$). But this cannot be the case because the real part is $\frac{a}{b}$ where $\gcd(a,b)=1$ and $b\geq 3$. It remains to address the case of $a\gamma=b\alpha$ and $b\beta\delta-a\delta^2=0$. One cannot have $b\beta=a\delta$ since that along with $a\gamma=b\alpha$ contradicts $\alpha\delta-\beta\gamma=1$. So it remains to address $a\gamma=b\alpha$ and $\delta=0$. If $\delta=0$, then $\gamma\in\{\pm 1\}$. But then $b\mid a$ which is impossible due to $\gcd(a,b)=1, b\geq 3$. Finally, when $\frac{\alpha (\frac{1}{2}+s{\rm{i}})+\beta}{\gamma (\frac{1}{2}+s{\rm{i}})+\delta}$ is considered instead of $\frac{\alpha s{\rm{i}}+\beta}{\gamma s{\rm{i}}+\delta}$, the only difference is that on the LHS of $(\star)$ $\beta$ and $\delta$ must be replaced with $\beta+\frac{\alpha}{2}$ and $\delta+\frac{\gamma}{2}$ respectively. Again, $s^2\in\Bbb{Q}$ unless $$ a\gamma^2-b\alpha\gamma=b\left(\beta+\frac{\alpha}{2}\right)\left(\delta+\frac{\gamma}{2}\right)-a\left(\delta+\frac{\gamma}{2}\right)^2=0. $$ As argued above, $\gamma=0$ cannot occur and hence $a\gamma=b\alpha$. If $\delta+\frac{\gamma}{2}=0$, then $\delta=\pm 1$ due to $\alpha\delta-\beta\gamma=1$ in which case $\gamma=\pm 2$. Thus $b\mid 2a$, again contradicting $\gcd(a,b)=1, b\geq 3$. We end up with $a\gamma=b\alpha$ and $b\left(\beta+\frac{\alpha}{2}\right)=a\left(\delta+\frac{\gamma}{2}\right)$. This is impossible because of $\alpha\delta-\beta\gamma=1$.