Entry 70

 III. Level 3. The McKay-Thompson series of class 3A for the Monster (A030197)

$$\begin{align}j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\ &=  \left(\left(\frac{d_1}{d_3}\right)^2+9\Big(\frac{d_9^3}{d_1\,d_3^2}\Big)\right)^6\end{align}$$ The second form shows they may be \(6\)th powers. Examples: 

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -(\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -(2\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &= -(4\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-89/3}}{2}\Big) &= -(10\sqrt3)^6\end{align}$$ Compare to a similar phenomenon for Level 2. These have discriminant \(d=3p\) for prime \(p=5,17,41,89\) and have class number 2. For class number 6,

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-29/3}}{2}\Big) &= -(x\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-113/3}}{2}\Big) &= -(y\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-137/3}}{2}\Big) &= -(z\sqrt3)^6\end{align}$$ where \((x,y,z)\) are the real roots of the cubics

$$x^3 - x^2 - 4x - 5 = 0\\ y^3 - 14y^2 - 4y - 16 =0\\ z^3 - 22z^2 + 44z - 32 = 0$$ and so on. For class number 10, $$j_{3A}\Big(\tfrac{1+\sqrt{-53/3}}{2}\Big) = -(u\sqrt3)^6$$ where \(u\) is the real roof of the solvable quintic $$u^5 - 8u^4 + 19u^3 - 26u^2 + 16u - 11 = 0$$ as well as for other \(d\).

Entry 69

II. Level 2. The McKay-Thompson series of class 2B for the Monster.

$$j_{2B}(\tau) =\left(\frac{d_1}{d_2}\right)^{24}$$ Examples: 

$$\begin{align}j_{2B}\big(\tfrac12\sqrt{-10}\big) & =\, 2^6\,U_{5}^{12} = 2^6\left(\tfrac{1+\sqrt{5}}2\right)^{12}\\ j_{2B}\big(\tfrac12\sqrt{-58}\big) & = \,2^6\,U_{29}^{12} = 2^6\left(\tfrac{5+\sqrt{29}}2\right)^{12}\\  j_{2B}\Big(\tfrac{1+\sqrt{-5}}2\Big) & = -2^6\,U_{5}^6 = -2^6\left(\tfrac{1+\sqrt{5}}2\right)^6\\ j_{2B}\Big(\tfrac{1+\sqrt{-13}}2\Big) & = -2^6\,U_{13}^6 = -2^6\left(\tfrac{3+\sqrt{13}}2\right)^6\\ j_{2B}\Big(\tfrac{1+\sqrt{-37}}2\Big) & = -2^6\,U_{37}^6 = -2^6\big(6+\sqrt{37}\big)^6\end{align}$$ which have class number \(h(-d)=2\).  The quadratic irrationals \(U_n\) are fundamental units, solutions to Pell equations \(x^2-ny^2=-1\) as discussed in Level 1.

Entry 68

II. Level 2. The McKay-Thompson series of class 2A for the Monster

$$\begin{align}j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\ &= \left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4\end{align}$$ The second form shows they may be \(4\)th powers. Examples: 

$$j_{2A}\big(\tfrac12\sqrt{-10}\big)=12^4\\ \; j_{2A}\big(\tfrac12\sqrt{-58}\big)=396^4\\  j_{2A}\Big(\tfrac{1+\sqrt{-5}}2\Big) = -\big(4\sqrt2\big)^4\\ j_{2A}\Big(\tfrac{1+\sqrt{-13}}2\Big) = -\big(12\sqrt2\big)^4\\ j_{2A}\Big(\tfrac{1+\sqrt{-37}}2\Big) = -\big(84\sqrt2\big)^4$$ which have class number \(h(-d)=2\). For class number \(h(-d)=4\)

$$\begin{align}j_{2A}\Big(\tfrac{1+\sqrt{-17}}2\Big) &= -2^{11}\big(4+\sqrt{17}\big)^2 \big({-1}+\sqrt{17}\big)\\ j_{2A}\Big(\tfrac{1+\sqrt{-73}}2\Big) &= -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3\\ j_{2A}\Big(\tfrac{1+\sqrt{-97}}2\Big) &= -2^{11}\cdot3^4\big(59+6\sqrt{97}\big)^4\big({-9}+\sqrt{97}\big)\\ j_{2A}\Big(\tfrac{1+\sqrt{-193}}2\Big) &= -2^{11}\cdot3^4\big(208+15\sqrt{193}\big)^4\big(903+65\sqrt{193}\big)\end{align}$$ All the quadratic irrationals with odd powers are odd fundamental solutions to Pell equations \(x^2-dy^2=-16\). For example, the initial solution to \(x^2-193y^2=-16\) is \((x,y)=(903,\,65)\) which appears above.

Entry 67

Let \(j=j(\tau)\) be the j-function. Define $$j_{1B}(\tau)=432\frac{\sqrt{j}+\sqrt{j-1728}}{\sqrt{j}-\sqrt{j-1728}}=\frac1q-120+10260q-901120q^2+\dots$$ which is (A299954). Then the following values $$\begin{align}j_{1B}\big(\tfrac{1+\sqrt{-19}}2\big) &= -432\,U_{19a}=-432\left(\sqrt{96^3/12^3}+\sqrt{96^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) &= -432\,U_{43b}=-432\left(\sqrt{960^3/12^3}+\sqrt{960^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-67}}2\big) &= -432\,U_{67c}=-432\left(\sqrt{5280^3/12^3}+\sqrt{5280^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-163}}2\big) &= -432\,U_{163c}=-432\left(\sqrt{640320^3/12^3}+\sqrt{640320^3/12^3+1}\right)^2\end{align}$$ are fundamental units, solutions to Pell equations \(x^2-ny^2=\pm1\) and where \((a,b,c,d) = (6, 15, 330, 10005)\). The smaller Heegner numbers like \(d=11\) don't yield fundamental units. And the second one \(U_{43b}=U_{645}\) involves a cube (Wolfram computation), $$j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) = -432\,U_{645}=-432\left(\frac{127+5\sqrt{645}}2\right)^3$$ which seems surprising. Note that $$432\big(\sqrt{-U_{645}}+1/\sqrt{-U_{645}}\big)^2=-960^3$$

Entry 66

Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\).

The McKay-Thompson series of class 1A for the Monster (A007240), disregarding the constant term \(744\), is the well-known j-function \(j(\tau)\). Given the three Weber modular functions \(\mathfrak{f}_n\) 

$$\begin{align}j_{1A}(\tau) &=\left(\frac{\mathfrak{f}(\tau)^{16}+\mathfrak{f}_1(\tau)^{16}+\mathfrak{f}_2(\tau)^{16}}{2}\right)^3\\ &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3\end{align}$$ They tend to be cubes, but not always. Examples: 

$$\begin{align}j_{1A}\big(\tfrac{1+\sqrt{-67}}{2}\big) &= -12^3\big(21^2-1\big)^3 \,=\, -5280^3\\ j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big) &= -12^3\big(231^2-1\big)^3=-640320^3\end{align}$$ while the two highest \(d\) with class number \(h(-d)=2\) are

$$\; j_{1A}\big(\tfrac{1+\sqrt{-403}}{2}\big)=-12^3\big((5301 + 1470\sqrt{13})^2-1\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-427}}{2}\big)=-12^3\big((7215 + 924\sqrt{61})^2-1\big)^3$$ and so on, though if \(d\) is a multiple of \(3\) then it is almost a cube $$\begin{align}j_{1A}\big(\tfrac{1+\sqrt{-15}}2\big) &=-3^3\,U_5^2\,\big(5+4\sqrt{5}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-51}}2\big) &=-48^3\,U_{17}^2\,\big(5+\sqrt{17}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-123}}2\big) &=-480^3\,U_{41}^2\,\big(8+\sqrt{41}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-267}}2\big) &=-240^3\,U_{89}^2\,\big(625+53\sqrt{89}\big)^3\end{align}$$ where \(U_n\) are fundamental units, $$\begin{align}U_5 &= \tfrac{1+\sqrt{5}}2\\ U_{17} &= 4+\sqrt{17}\\ U_{41} &= 32+5\sqrt{41}\\ U_{89} &= 500+53\sqrt{89}\end{align}$$ or fundamental solutions to the Pell equation \(x^2-ny^2 = -1\) with the first as the golden ratio \(\phi\ =U_5\). These will appear later in Class 3B.