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I. Klein

In "On the Order-Seven Transformation of Elliptic Functions" (pp. 287-331), he discusses in p. 298 what we now call the Klein quartic,

$$\lambda^3\mu+\mu^3\nu+\nu^3\lambda= 0\tag1$$

and in p. 313 introduces what we can call the Klein septic resolvent,

$$z^7-2^2\cdot7^2\big(7\mp\sqrt{-7}\big)z^4+2^5\cdot 7^4\big(5\mp\sqrt{-7}\big)\color{red}z\\ \mp 2^9\cdot3\cdot7^3\sqrt{-7}\frac{g_2}{\sqrt[3]\Delta}=0\tag2$$

(where the red linear $\color{red}z$ is missing in the paper and I assume is a typo). In the same page, he says the roots of $(2)$ in terms of $\lambda,\mu, \nu$ are,

$$z =\frac{\pm 2\sqrt{-7}\Big(P_1+\frac{-1\mp\sqrt{-7}}{2}P_2\Big)}{\sqrt[3]\nabla}\tag3$$

where,

$$P_1 = \gamma^{2x}\lambda^2+\gamma^{x}\mu^2+\gamma^{4x}\nu^2\\ P_2 =\gamma^{6x}\mu\nu+\gamma^{3x}\nu\lambda+\gamma^{5x}\lambda\mu$$

with $\gamma= e^{2\pi i/7}$ as first mentioned in p. 313.

II. Ramanujan

Unbeknownst to Klein (d. 1925), it turns out Ramanujan (d. 1920) found an elegant parameterization to $(1)$. Define the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2}\, b^{n(n-1)/2}$$

then,

$$\lambda = q^{1/56}f(-q^3,-q^4)\\ \mu= \color{red}{-}q^{25/56}f(-q,-q^6)\\ \nu= q^{9/56}f(-q^2,-q^5)\\$$

where $q=e^{2\pi i\tau}$.

III. Question

I tried to implement this in Mathematica. Unfortunately, I couldn't get $(3)$ to be a root of the septic $(2)$.

Q: Was I wrong in assuming that any parametrization to $(1)$ would do? Or is there some typo or confusion of variables in the paper that screwed up my implementation?

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  • $\begingroup$ Isn't the upside-down Delta the Hessian of (20)? Also, $x$ ranges from 0 to 6, cf. (18). $\endgroup$ – literature-searcher Mar 21 at 22:43
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There seems to be a problem with Klein's septic equation $(2)$ combined with the purported roots in $(3)$. Let $\,k\,$ be any integer. Define $$ P_1(k) := \gamma^{k}\mu^2 + \gamma^{2k}\lambda^2 + \gamma^{4k}\nu^2 \tag{1} $$ and $$ P_2(k) := \gamma^{3k}\lambda\nu + \gamma^{6k}\mu\nu + \gamma^{5k}\lambda\mu. \tag{2} $$ Let $$ \,a := 2\sqrt{-7}/\eta(\tau)^2 \; \textrm{ and } \; b := -(7+\sqrt{-7})/\eta(\tau)^2. \tag{3} $$ Define the roots of $(2)$ as $$ r_k := a\,P_1(k) + b\,P_2(k). \tag{4} $$ The polynomial $\, P(z) := (z - r_1)(z - r_2)\cdots (z - r_7)\,$ expands to $$ P(z) = z^7 \!-\! 2^2\! \cdot\! 7^2\, (7+\sqrt{-7})\, z^4 \!+\! 2^5\!\cdot \!7^4\, (5+\sqrt{-7})\,z \!+\! 2^7\! \cdot\! 7^3 \sqrt{-7} \frac{g_2(\tau)}{\eta(\tau)^8}. \tag{5}$$ This was for values of $\,\lambda,\mu,\nu\,$ as suggested by Tito Piezas III.

P.S. Note that $\, \sqrt[3]{\Delta} = \eta(\tau)^8$ is the denominator of the constant term of $P(z).$ Also note that $\, 1728 J(\tau) = j(\tau) = g_2(\tau)^3/\eta(q)^{24}\,$ where $\,J(\tau)\,$ is Klein's invariant. Thus using cube roots $\, 12\sqrt[3]{J(\tau)} = g_2(\tau)/\eta(\tau)^8\,$ and the constant term can be written as $\, 2^9\!\cdot 3 \cdot\! 7^3 \sqrt{-7} \sqrt[3]{J(\tau)} \,$ which is closer to Klein's version. In fact, just before Klein's equation $(49)$ (Tito's equation $(2)$) Klein writes $\ J = g_2^2/\Delta.$

P.P.S. I see that Tito has used a set of roots $\,y_k\,$ differing by a common factor from $\,z_k\,$ to simplify the septic. Perhaps Klein would have used that version, but he preferred $\,J(\tau)\,$ instead of $\,j(\tau).$

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  • $\begingroup$ Beautiful! I verified it with my own revised Mathematica implementation. I knew there may have been a 2nd typo in the paper (after I found the 1st one). And to complicate things, I used the modern definition of the modular discriminant as $$\Delta' = (2\pi)^{12} \eta^{24}(\tau)$$ though Mathworld did point out that older literature might just use, $$\Delta = \eta^{24}(\tau)$$ which is the case in this post. $\endgroup$ – Tito Piezas III Mar 22 at 2:53
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(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y\Big(y^3-\frac{8}{\alpha^3}\sqrt{-7}\Big)\Big(y^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\alpha P_2(k)}{\eta^2(\tau)}\\ \alpha=\frac{-1+\sqrt{-7}}2$$

for $k=1,2\dots7,$ with $P_1(k)$ and $P_2(k)$ as defined by Klein (and Somos).

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  • 2
    $\begingroup$ Note that, $$\frac{5+\sqrt{-7}}2=-\left(\frac{1+\sqrt{-7}}2\right)^3$$ $\endgroup$ – Tito Piezas III Mar 22 at 3:57

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