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.

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.

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

  • 1
    $\begingroup$ Do you know how to prove that $f(z)=x^2$ is algebraic over $\Bbb{C}[j]$ ? The polynomial is $\prod_{\gamma\in \Gamma_0(5)\backslash SL_2(\Bbb{Z})} (X-f(\gamma(z)))=\sum_{n=0}^N c_n(z)X^n$, the $c_n(z)$ are holomorphic on $\Bbb{H}$, $SL_2(\Bbb{Z})$ invariant and meromorphic at $i\infty$, thus polynomial in $j(z)$. The polynomial is found from the negative coefficients of the expansion of $c_n(z)$ at $i\infty$. From there we get a polynomial such that $h(R,j)=0$, we can factorize $h(X,Y)=\prod_i h_i(X,Y)\in\Bbb{C}[X,Y]$ and check if $h_i(R,j)=0$, your polynomial (10) is a product of some $h_i$. $\endgroup$
    – reuns
    Jan 26, 2021 at 10:00
  • $\begingroup$ @reuns: I don't have much idea of modular forms so really difficult to make sense of your comment. I know the representation of these functions in terms of elliptic integrals and moduli, but then I don't think getting an expression for eta quotient $\eta(q^5)/\eta(q)$ in terms of $K, k$ would be easy. $\endgroup$
    – Paramanand Singh
    Jan 26, 2021 at 10:21
  • $\begingroup$ @reuns: Well proving that these things are algebraic is not that difficult. They follow from standard theory of transformation of elliptic integrals. What is really tricky is to get explicit algebraic formula for them. That's what I want here and perhaps by a direct algebraic calculation. $\endgroup$
    – Paramanand Singh
    Jan 26, 2021 at 12:29
  • $\begingroup$ @reuns: yeah I am looking for an approach via hand calculation. I really wonder why the techniques of Jacobi, Gauss, Ramanujan were ditched in modern treatments of these ideas. But lets not digress too much here. $\endgroup$
    – Paramanand Singh
    Jan 26, 2021 at 12:34
  • 1
    $\begingroup$ I am saying there is a standard method, which is boring and useless to do by hand, whose algorithm is interesting and has nothing hard, and this is clearly the answer to your question as stated. The algorithm is to compute the negative Fourier coefficients of $f(\gamma(z))$ for each $\gamma\in \Gamma_0(5)\backslash SL_2(\Bbb{Z})$ (the quotient of a group by a subgroup) then to subtract some powers of $j$ to cancel the pole at $i\infty$ (also I meant $f(z)=x^4=a\Delta(z)/\Delta(5z)$). $\endgroup$
    – reuns
    Jan 26, 2021 at 12:37

2 Answers 2


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 $$(m-1)^2(2m^3+5m^2+25)$$

  • $\begingroup$ btw holds good for complex $m$ as well. $\endgroup$
    – Narasimham
    Jan 27, 2021 at 15:04

(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 $q^{1/24}$.

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.

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.


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