Relation between Rogers Ramanujan continued fraction and $j$-invariant

Question

While going through this answer I found an interesting but slightly complicated relation between Rogers-Ramanujan continued fraction and the j-invariant. I would like to know an elementary proof of the same.

Before proceeding let me define all the necessary terms and symbolism to set the proper context. Let $\tau$ be a complex number with positive imaginary part and $q=\exp(2\pi i\tau) $ so that $|q|<1$. Below I define functions and the relations which I am aware of. The Rogers-Ramanujan continued fraction is given by $$R(q) =\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\dots}}}}\tag{1}$$ Ramanujan studied this function in great detail and obtained the following fundamental identities $$\frac{1}{R(q)}-1-R(q)=\frac{\eta(q^{1/5})} {\eta(q^5)}\tag{2}$$ and $$\frac{1}{R^5(q)}-11-R^5(q)=\left(\frac{\eta(q)}{\eta(q^5)}\right)^6\tag{3}$$ where $\eta(q) $ is Dedekind eta function defined by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{4}$$ To define the $j$-invariant we need to introduce Ramanujan's version of Eisenstein series denoted by $L, M, N$ (symbols $P, Q, R$ are typically used but we want to avoid conflict with Rogers-Ramanujan continued fraction $R(q) $) \begin{align} L(q) &= 1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{5a}\\ M(q)&=1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\tag{5b}\\ N(q) &=1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\tag{5c} \end{align} It should be observed that $L$ is related to $\eta$ via $$L(q) =24q\frac{d}{dq}(\log\eta(q))\tag{6}$$ The $j$-invariant is defined as $$j(q) =\frac{1728M^3(q)}{M^3(q)-N^2(q)}\tag{7}$$ Ramanujan obtained a system of differential equations connecting $L, M, N$: \begin{align} q\frac{dL(q) } {dq} &=\frac{L^2(q)-M(q)}{12}\tag{8a}\\ q\frac{dM(q)}{dq}&=\frac{L(q)M(q)-N(q)}{3}\tag{8b}\\ q\frac{dN(q)} {dq} &=\frac{L(q) N(q) - M^2(q)}{2}\tag{8c} \end{align} Using $(6)$ and $(8a)$ it is evident that $M(q) $ can also be expressed in terms of $\eta(q) $. On the other hand the above differential equations allow us to prove that $$M^3(q)-N^2(q)=1728\eta^{24}(q)\tag{9}$$ and thus we have some expression for $j(q) $ in terms of $\eta(q) $.

The following complicated relation holds between Rogers Ramanujan continued fraction $R(q)$ and $j(q) $ : $$ R^5 (R^{10}+11 R^5-1)^5j+(R^{20}-228 R^{15}+494 R^{10}+228 R^5+1)^3 = 0\tag{10}$$ I checked Wikipedia and found that this is derived from another identity $$j(q) =\frac{(x^2+10x+5)^3} {x} \tag{11}$$ where $$x=125\left(\frac {\eta(q^5)}{\eta(q)}\right)^6\tag{12}$$ Using $(11),(12)$ and $(3)$ we can deduce $(10)$ with a little algebra.

Thus the problem boils down to a proof of equation $(11)$. I don't know if this can be derived using algebraic manipulation of the identities given above or does it need some specific modular equation.

Any proofs or suggestions for proof are welcome. I don't understand the machinery of modular forms properly and would prefer an approach more in the spirit of Ramanujan. The question is however tagged "modular-forms" to get the attention of experts from that tag.

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Update: I have finally managed to give a proof based on modular equation of degree $5$ given by Ramanujan and posted it as an answer. The proof is more of a verification and a more natural proof utilizing some transformation formula of eta function is desired.

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There is a related question which assumes $(10),(11)$ and proves $(3)$, but the approach uses Mathematica.

Answer 1

An approach based on modular equations of degree $5$ works out fine with some tedious algebra. The identity $(11) $ in question will be proved by replacing $q$ with $q^2$.

Let $k, l$ denote the elliptic modulus and $K, L$ the complete elliptic integrals of first kind corresponding to nomes $q, q^5$ respectively and further let $\alpha=k^2,\beta =l^2$. Also let $m=K/L$ denote the multiplier.

Then we have the standard formulas \begin{align} M(q^2)&=\left(\frac{2K}{\pi}\right)^4(1-\alpha(1-\alpha))\tag{1}\\ \eta(q^2)&=2^{-1/3}\sqrt {\frac {2K}{\pi}}(\alpha(1-\alpha))^{1/12}\tag{2}\\ \eta(q^{10})&=2^{-1/3}\sqrt{\frac{2L}{\pi}}(\beta(1-\beta))^{1/12}\tag{3} \end{align} Dr. Bruce C. Berndt and his collaborators analyzed Ramanujan's modular equations of degree $5$ and gave the following relations between $\alpha, \beta, m$: \begin{align} (\alpha^3\beta)^{1/8}&=\frac{\rho+3m-5}{4m}\tag{4}\\ (\alpha\beta^3)^{1/8}&=\frac{\rho+m^2-3m}{4m}\tag{5}\\ \rho&=\sqrt{m^3-2m^2+5m}\tag{6} \end{align} From the above equations we get $$\alpha=\frac{(\alpha^3\beta)^{3/8}}{(\alpha\beta^3)^{1/8}}=\frac{(\rho+3m-5)^3}{16m^2(\rho+m^2-3m)}$$ Multiplying numerator and denominator by $\rho-m^2+3m$ and using some algebra we get $$\alpha=\frac{(\rho+3m-5)^2(2m+\rho)} {16m^3(m-1)} $$ Next we replace $\alpha, \beta, m$ by $1-\beta,1-\alpha,5/m$ to get $$((1-\alpha)(1-\beta)^3)^{1/8}=\frac{\rho+3m-m^2}{4m}$$ and $$((1-\alpha)^3(1-\beta))^{1/8}=\frac{\rho+5-3m}{4m}$$ and thus $$1-\alpha=\frac{(\rho+5-3m)^2(2m-\rho)} {16m^3(m-1)} $$ and hence $$\alpha(1-\alpha)=\frac{(m-1)(5-m)^5}{256m^5} \tag{7}$$ A similar calculation leads to the relation $$\left(\frac{\beta(1-\beta)}{\alpha(1-\alpha)} \right)^{1/2}=\frac{m^2(m-1)^2}{(5-m)^2}\tag{8}$$ Let us now observe that if $$x=125\left(\frac{\eta (q^{10})}{\eta(q^2)}\right)^6$$ then by the identities $(2),(3),(8)$ we get $$x=\frac{125(m-1)^2}{m(5-m)^2}$$ On the other hand we have $$j(q^2)=\frac{M^3(q^2)}{\eta^{24}(q^2)}=\frac{(1-\alpha(1-\alpha)) ^3}{2^{-8}(\alpha(1-\alpha))^2}=\frac{(256m^5-(m-1)(5-m)^5)^3}{m^5(m-1)^2(5-m)^{10}}$$ And the expression $(x^2+10x+5)^3/x$ equals $$\frac{(3125(m-1)^4+250m(m-1)^2(5-m)^2+m^2(5-m)^4)^3}{m^5(m-1)^2(5-m)^{10}}$$ Our job is thus complete if we can show that $$3125(m-1)^4+250m(m-1)^2(5-m)^2+m^2(5-m)^4=256m^5-(m-1)(5-m)^5$$ Both sides are polynomials of degree $6$ and hence we need to verify the identity for at most $7$ values of $m$. The verification is trivial for $m=0,1,5$ and one may try to verify it for small values like $m=-1,-2, 2,3$.

There is another approach to rewrite the desired identity so that both sides can be factorized easily. Thus the desired identity is equivalent to $$250m(m-1)^2(5-m)^2+3125(m-1)^4=256m^5+(m-1)(m-5)^5-m^2(m-5)^4$$ or $$250m(m-1)^2(m-5)^2+3125(m-1)^4=3125-6250m+3750m^2-1000m^3+125m^4+250m^5$$ or $$(m-1)^2(2m(m-5)^2+25(m-1)^2)=25-50m+30m^2-8m^3+m^4+2m^5$$ It can be checked now that each side factors as $$(m-1)^2(2m^3+5m^2+25)$$

Answer 2

(A very extended comment.)

This is not so much answer but to show that equations $(10)$ and $(11)$ are not isolated results, but part of a beautiful family of simple formulas of order $n$. It would be nice to have all these formulas in one place.

Surprisingly, their simplicity depends on basic arithmetic: whether $n$ integrally divides $24/(n-1)$. Note the Dedekind eta function $\eta(q)$ involves the $24$th root $q^{1/24}$.

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I. Prime $n = 2,3,5,7,13.$

$$j(q) =\frac{(x-16)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt2\,\eta(q^2)}{\eta(q)}\right)^{24}$$ $$\; j(q) =\frac{(x+3)^3(x+27)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt3\,\eta(q^3)}{\eta(q)}\right)^{12}$$ $$j(q) =\frac{(x^2+10x+5)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt5\,\eta(q^5)}{\eta(q)}\right)^6$$ $$j(q) =\frac{(x^2+5x+1)^3(x^2+13x+49)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt7\,\eta(q^7)}{\eta(q)}\right)^4$$ $$j(q) =\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x, \;\text{with}\;\; x=\left(\frac {\sqrt{13}\,\eta(q^{13})}{\eta(q)}\right)^2$$

Their unsigned discriminants $D_n$ are beautifully consistent,

$$ D_2 = 2^2(j-1728)\,j^2\\ D_3 = 3^3(j-1728)^2\,j^2\\ D_5 = 5^5(j-1728)^2\,j^4\\ D_7 = 7^7(j-1728)^4\,j^4\\ D_{13} = 13^{13}(j-1728)^6\,j^8$$

Order $n=2$ is better known in the guise of the Weber modular functions, while the rest of the first set do not seem to have names.

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II. Square $n=4,9,25.$

$$j(q) =\frac{(x^2-48)^3}{x^2-64}, \quad\text{with}\quad x=\left(\frac {\sqrt4\,\eta(q^{4})}{\eta(q)}\right)^8+8$$ $$j(q) =\frac{x^3(x^3-24)^3}{x^3-27}, \quad\text{with}\quad x=\left(\frac {\sqrt9\,\eta(q^{9})}{\eta(q)}\right)^3+3$$ $$j(q) = \frac{-(x^{20}+12 x^{15}+14 x^{10}-12 x^5+1)^3}{x^{25} (x^{10}+11 x^5-1)} , \;\text{with}\;\; x^{-1}-x=\left(\frac {\sqrt{25}\,\eta(q^{25})}{\eta(q)}\right)^1+1$$

Or alternatively, $$j(q) =\frac{(x^2+192)^3}{(x^2-64)^2}, \quad\text{with}\quad x=\left(\frac {\eta(q^{1/2})}{\eta(q^2)}\right)^8+8$$ $$j(q) =\frac{x^3(x^3+216)^3}{(x^3-27)^3}, \quad\text{with}\quad x=\left(\frac {\eta(q^{1/3})}{\eta(q^3)}\right)^3+3$$ $$j(q) = \frac{-(r^{20}-228 r^{15}+494 r^{10}+228 r^5+1)^3}{r^5 (r^{10}+11 r^5-1)^5} , \;\text{with}\;\; r^{-1}-r=\left(\frac {\eta(q^{1/5})}{\eta(q^5)}\right)^1+1$$

Their unsigned discriminants $D_n$,

$$ D_4 = 4^a\,(j-1728)^3\,j^4\\ D_9 = 9^b\,(j-1728)^6\,j^8\\ D_{25} = 25^c\,(j-1728)^{30}\,j^{40}$$

for some integer $a,b,c$.

Generalizing the OP's post, the three square orders can be connected to three kinds of $q$-continued fractions: the first to the octic cfrac (the power $8$ and octahedral symmetry), the second to the cubic cfrac (the power $3$ and tetrahedral symmetry), and the third, of course, to the Rogers-Ramunujan cfrac (and icosahedral symmetry) which are the 3 symmetries of the Platonic solids. Ramanujan studied all three cfracs.