Solving the Bring quintic using the Monster?

Question

I. Method

Hermite's method to solve the Bring quintic by functions that obey $x^8+y^8=1$ implicitly uses octahedral symmetry, while Emil Jann Fiedler's solution by the Rogers-Ramanujan continued fraction $R(q)$ in this MO post uses icosahedral symmetry.

With no details from Fiedler on the latter's derivation, I've been trying to reverse-engineer it hoping that, by using different functions such as Borwein cubic theta functions which obey $x^3+y^3=1$, it will lead to the last Platonic symmetry, the tetrahedral. (Update: The Huber quintic theta functions in this post which obey $x^5+y^5=1$ can solve the Bring as well.)

Given a fifth root of unity $\zeta = e^{2\pi i/5}$, we start with the function,

$$V_n = \frac{2-\big(1-\zeta^n\, a\big)\big(1+\zeta^{2n}\, b\big)}{\zeta^{4n}}$$

Expand $\prod_{n=0}^4\big(x-V_n)$ to get,

$$x^5-5Ax^2-5Bx-C =0$$

which is in principal quintic form and where,

\begin{align} A &= a^2 - b + a^3 b^2 + a b^3\\[4pt] B &= a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4\\[4pt] C &= 1 + a^5 + 10 a^3 b + 10 a b^2 - 10 a^4 b^3 + 10 a^2 b^4 - b^5 + a^5 b^5 \end{align}

By two happy "coincidences", if $a=R(q)$ and $b=R(q^2)$, then $A = 0$ since it is the modular equation between them, while $B = \pmb\square$ is a square (to be seen later). Since $A = 0$, the quintic will be in Bring form. Scale variables $x = B^{1/4}y$ to simplify further,

$$y^5 - 5y - \frac{C}{B^{5/4}}=0$$

Equate this to a generic Bring for some free parameter $h$,

$$y^5 - 5y - h^{1/4}=0$$

and we get a system of $3$ equations in $3$ unknowns $(a,b,j)$,

\begin{align}A &= a^2 - b + a^3 b^2 + a b^3 = 0\\[5pt] h &= \left(\frac{C}{B^{5/4}}\right)^4 = \frac{(1 + a^5 + 10 a^3 b + 10 a b^2 - 10 a^4 b^3 + 10 a^2 b^4 - b^5 + a^5 b^5)^4}{(a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4)^5}\\[5pt] j &=\frac{-(a^{20} - 228a^{15} + 494a^{10} + 228a^5 + 1)^3}{a^5(a^{10} + 11a^5 - 1)^5} \end{align}

where the last is the formula for the j-function in terms of $a=R(q)$. I thought resolving this would result in a horrible mess and a $120$-deg equation in $j$ did pop up. But factored, it was just a nice quadratic in j with a multiplicity of $60$ (the order of icosahedral symmetry),

$$\Big(j^2 - (h^2 - 207h + 3456 )j + (h + 144)^3\Big)^{60}=0$$

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II. Monster

The appearance of the equation,

$$j^2 - (h^2 - 207h + 3456 )j + (h + 144)^3 = 0\tag1$$

was a surprise, as the first two McKay-Thompson series of the Monster satisfy this relation,

\begin{align} j=j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3\\[5pt] &= \left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^{\color{red}3}\\[5pt] &= \frac{1}{q} + 744 + 196884q + 21493760q^2 +\cdots\\[7pt] h=j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2\\[5pt] &=\left(\left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta\big(\frac{\tau}2\big)\,\eta(4\tau)}\right)^{4}-4 \left(\frac{\eta\big(\frac{\tau}2\big)\,\eta(4\tau)}{\eta(\tau)\,\eta(2\tau)}\right)^{4}\right)^{\color{red}4}\\[5pt] &= \frac{1}{q} + 104 + 4372q + 96256q^2 +\cdots\\[5pt] \end{align}

which (without the constant term) are adjacent sequences A007240 and A007241. Note they are generally cubes and fourth powers. For example,

\begin{align} j=j_{1A}\left(\tfrac{\sqrt{-58}}2\right) &= 30^3\big(140989 + 26163\sqrt{29}\big)^3\\[4pt] h=j_{2A}\left(\tfrac{\sqrt{-58}}2\right) &= \,396^4 \end{align}

with the latter used in Ramanujan's famous $1/\pi$ formula. And as was mentioned earlier, if $a=R(q)$ and $b=R(q^2)$, then $B$ in fact is a square,

$$B = a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4 = a^2b^2\left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(5\tau)\,\eta(10\tau)}\right)^2$$

with the eta quotient also a McKay-Thompson series, for class $10C$, or A132041.

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III. Modular lambda function

To find the needed elliptic modulus $k$ and $\tau$, we use the j-function formula in terms of the modular lambda function $\lambda(\tau)$ and elliptic parameter $m=k^2$,

$$j= \frac{256(1 - m + m^2)^3}{(m - m^2)^2}$$

Substitute this into the relation between the two McKay-Thompson series $j_{1A}$ and $j_{2A}$, and it has three quartic factors in $m$. We assume $h = c^4$ and $m=k^2$ and it factors further into two quartics in $k$, the relevant one being the near-palindromic,

$$k^4+\left(\tfrac{c}2\right)^2 k^3+2k^2-\left(\tfrac{c}2\right)^2 k+1 =0$$

The discriminant $D$ of $x^5-5x+c=0$ is $D = 5^5(c^4-256)$. The quartic roots are,

$$k =\sqrt{\frac{\sqrt{2}\,c-\sqrt{c^2+\sqrt{c^4-256}}}{\sqrt{2}\,c+\sqrt{c^2+\sqrt{c^4-256}}}}$$

with the first two by $\sqrt{D/5^5}=\pm\sqrt{c^4-256}$ so it seems expected that $D$ appears.

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IV. Quintic solution

We now have all the information needed. Let,

$$y^5-5y+c=0$$

then the five roots $y_n$ for $n=0,1,2,3,4$ are,

$$y_n = \frac{2-\big(1-\zeta^n\, a\big)\big(1+\zeta^{2n}\,b\big)}{\pm B^{1/4}\;\zeta^{4n}}$$

where,

$$B = a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4$$

$$\,a = R(q),\,\quad b = R(q^2)$$

$$\zeta = e^{2\pi i/5},\quad q=e^{2\pi i\tau}$$

$$\tau = \frac{K'(k)}{K(k)}\sqrt{-\frac14}=\frac{_2F_1\big(\tfrac12,\tfrac12,1,\,1-k^2\big)}{_2F_1\big(\tfrac12,\tfrac12,1,k^2\big)}\sqrt{-\frac14}$$

$$k =\tan\left(\frac14\arcsin\Big(\frac{16}{c^2}\Big)\right)$$

and $K(k)$ is the complete elliptic integral of the first kind for real $c$. (For complex $c$, one may have to use the other roots of the quartic.) It is similar to Emil Jann Fiedler's solution but with the advantage we know how it was derived. (Plus an interesting detour through the Monster along the way.)

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V. Question

Can we use a function similar in form to,

$$V_n = \frac{2-\big(1-\zeta^n\, a\big)\big(1+\zeta^{2n}\, b\big)}{\zeta^{4n}}$$

where $(a,b)$ is related to Ramanujan's cubic continued fraction or the Borwein's cubic theta function to solve the Bring quintic? It would finally add the missing tetrahedral symmetry of the Platonic solids.

Answer 1

(A tentative answer using analogues of Hermite's approach.)

I've been analyzing the situation from Ramanujan's theories of elliptic functions to alternative bases of signature $(2,3,4,6)$. Hermite's solution of the Bring quintic seem to depend on properties peculiar to signature $2$.

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I. Signature 2 and $\color{blue}{x^8+y^8 = 1}$

Given the Dedekind eta function $\eta(\tau)$. Then,

$$\left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8+\left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1$$

which is equivalent to the Jacobi theta functions,

$$\left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4+\left(\frac{\vartheta_4(0,q)}{\vartheta_3(0,q)}\right)^4 = 1$$

Define $A(\tau) =\dfrac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}.$ The modular relation between $A(5\tau)$ and $A(\tau)$ is a sextic,

$$u^6 - v^6 + 5u^2v^2(u^2 - v^2) + 4u v(1 - u^4v^4)=0$$

If $v = A(\tau)$, then the six roots $u$ are,

$$-u = -A(5\tau),\;A\big(\tfrac{\tau}{5}\big),\;A\big(\tfrac{\tau+16}{5}\big),\;A\big(\tfrac{\tau+32}{5}\big),\;A\big(\tfrac{\tau+48}{5}\big),\;A\big(\tfrac{\tau+64}{5}\big)$$

These $u$ can be used to solve the Bring quintic. Define the function,

$$\alpha(\tau) = \Big({-A(5\tau)}-A\big(\tfrac{\tau}{5}\big)\Big) \left(A\big(\tfrac{\tau+m}{5}\big)-A\big(\tfrac{\tau+4m}{5}\big)\right) \left(A\big(\tfrac{\tau+2m}{5}\big)-A\big(\tfrac{\tau+3m}{5}\big)\right)$$

for $\color{blue}{m=16}$. Then expanding,

$$\prod_{n=0}^4\big(x-\alpha(\tau+n)\big)= x^5+c_1x+c_2 = 0$$

Three coefficients vanish and where the $c_i$ are expressions in terms of $A(\tau)$. Equating to a generic Bring, one can then find $\tau$. But the situation is different for signature $3$.

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II. Signature 3 and $\color{blue}{x^3+y^3 = 1}$

We have,

$$\left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3+\left(\frac{\left(\frac{\eta(\tau)}{\eta(9\tau)}\right)^3}{\left(\frac{\eta(\tau)}{\eta(9\tau)}\right)^3+9}\right)^3=1$$

which is equivalent to the Borweins' cubic theta identity,

$$\left(\frac{c(q)}{a(q)}\right)^3+\left(\frac{b(q)}{a(q)}\right)^3 = 1$$

Define $B(\tau) = \dfrac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}$. The relation between $B(5\tau)$ and $B(\tau)$ is again a sextic,

$$(u^3 - 1)(v^3 - 1)(u^3 + 75 u v + 330 u^2 v^2 + v^3 + 321 u^3 v^3 + 1) - (1 - u v)^6 = 0$$

If $v = B(\tau)$, then the six roots $u$ are,

$$u = B(5\tau),\;B\big(\tfrac{\tau}{5}\big),\;B\big(\tfrac{\tau+3}{5}\big),\;B\big(\tfrac{\tau+6}{5}\big),\;B\big(\tfrac{\tau+9}{5}\big),\;B\big(\tfrac{\tau+12}{5}\big)$$

Let,

$$\beta(\tau) = \Big(B(5\tau)-B\big(\tfrac{\tau}{5}\big)\Big) \left(B\big(\tfrac{\tau+m}{5}\big)-B\big(\tfrac{\tau+4m}{5}\big)\right) \left(B\big(\tfrac{\tau+2m}{5}\big)-B\big(\tfrac{\tau+3m}{5}\big)\right)$$

for $\color{blue}{m=3}$. Then expanding,

$$\prod_{n=0}^4\big(x-\beta(\tau+n)\big)= x^5+c_1x^3+c_2x+c_3 = 0$$

Only two coefficients vanish (after experimenting with various sign combinations). It is worse for signature $4$.

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III. Signature 4 and $\color{blue}{x^2+y^2 = 1}$

We have,

$$\left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2+\left(\frac{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8+32}\right)^2=1$$

Define $C(\tau) = \dfrac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}$. The relation between $C(5\tau)$ and $C(\tau)$ is still a sextic.

Since the sextic $P(u,v)=0$ is long, suffice to say that if $v = C(\tau)$, then the six roots $u$ are,

$$u = C(5\tau),\;C\big(\tfrac{\tau}{5}\big),\;C\big(\tfrac{\tau+24}{5}\big),\;C\big(\tfrac{\tau+48}{5}\big),\;C\big(\tfrac{\tau+72}{5}\big),\;C\big(\tfrac{\tau+96}{5}\big)$$

Define,

$$\gamma(\tau) = \Big({-C(5\tau)}-C\big(\tfrac{\tau}{5}\big)\Big) \left(C\big(\tfrac{\tau+m}{5}\big)-C\big(\tfrac{\tau+4m}{5}\big)\right) \left(C\big(\tfrac{\tau+2m}{5}\big)\color{blue}{+}C\big(\tfrac{\tau+3m}{5}\big)\right)$$

for $\color{blue}{m=24}$. Then expanding,

$$\prod_{n=0}^4\big(x-\gamma(\tau+n)\big)= x^5+c_1x^3+c_2x^2+c_3x+c_4 = 0$$

Now only one coefficient vanishes (after experimenting with various sign combinations). I didn't bother with signature $6$.

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IV. Some remarks

In summary, using the functions $A(\tau),\,B(\tau),\,C(\tau)$, we can construct quintics where three, two, or just one coefficient(s) vanishes, respectively. So it makes the $A(\tau)$ of signature $2$ "special" since it attains the Bring form. However, there are other functions, but that's for another post. (For more context, this MO post might be informative.)

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V. Tetrahedral symmetry

(As the question looks for a function associated with tetrahedral symmetry, there are some promising leads, but the research is still in progress.)

Answer 2

(New answer):

1. One approach was to express the modular lambda function $\lambda(\tau)=A^8(\tau)$ in terms of the cubic continued fraction $C(\tau)$. For example, define, \begin{align} A(\tau) &= \frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\\[5pt] C(\tau) &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\end{align} and let $\alpha = C(\tau)$ and $\beta = C(2\tau).\,$ Then, $$A^8(6\tau) =\frac{16(\alpha+\alpha^4)\,\beta^4}{(1-4\alpha^2\beta^2)^2}$$ However, it seemed a bit unsatisfactory.
2. An alternative approach was to follow Hermite. He solved the Bring quintic using a modular equation between $A(\tau)$ and $A(5\tau)$. It turns out a modular equation between $C(\tau)$ and $C(5\tau)$ can solve the Bring quintic after all.
3. The new solution is in this MO post as it raises questions of its own, with a familiar octic popping up. (And as usual, two McKay-Thompson series of class 6 for the Monster appear too.)