Generalizing Klein's order 7 formula $y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$ to order 13?

Question

I. Level 7

In Klein's "On the Order-Seven Transformations of Elliptic Functions", he gave two elegant resolvents of degrees 8 and 7 in pages 306 and 313. Translated to more understandable notation, we have,

$$x^8+14x^6+63x^4+70x^2-7 = x\sqrt{j(\tau)-1728}\tag1$$

$$y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j(\tau)\tag2$$

For general $j$, in Magma notation these are 8T43 = PGL(2,7) and 14T16, respectively, both of which have order $2\times168 = 336$ hence generally not solvable in radicals. But I found the octic in $x$ has a nice solution by the septic in $y$. Let $\zeta = e^{2\pi i/7}$, then,

$$x_k = \frac{\alpha}4\left(\pm\sqrt{y_1+\alpha^6}\pm\sqrt{y_2+\alpha^6}\pm\dots\pm\sqrt{y_7+\alpha^6}\,\right)$$

where $\alpha = \zeta+\zeta^2+\zeta^4 = \frac{-1+\sqrt{-7}}2$, a radical in the character table of PSL(2,7), and signs chosen appropriately. This is a similar method used for the 8T25 octic $x^8-x^7+29x^2+29=0$ here.

However, if $j$ is the j-function like $j\Big(\tfrac{1+\sqrt{-163}}2\Big)$, then the order is smaller and is now solvable in radicals.

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II. Level 13

The equation $(1)$ involves level 7 eta quotients. But this has a level 13 counterpart, namely,

$$\frac{(x^2 + 6x + 13)(x^6 + 10x^5 + 46x^4 + 108x^3 + 122x^2 + 38x - 1)^2}{x} = j(\tau)-1728$$

Analogous to its younger sibling, this is 14T39 = PGL(2,13) with order $24\times7\times13=2184$ hence is generally not solvable in radicals. But the order is smaller if $j$ is the j-function and is solvable.

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III. Question

Is there a transitive group of degree $13m$ and order $2184$? The online databases show there are none for $m=1,2$ but is silent for $m>2$.

P.S. The objective is create a formula for the j-function with $13$ as the highest exponent, but with coefficients as algebraic numbers of deg $m$, analogous to formula $(2)$.

Answer 1

The smallest $m$ for which $\mathrm{PGL}(2,13)$ acts faithfully and transitively on a set of $13m$ elements is $m=6$, with point stabiliser a dihedral group of order $28$. I think you may be seeing a manifestation of Camille Jordan's theorem that $\mathrm{PSL}(2,p)$ has a subgroup of index $p$ if and only if $p\leqslant 11$.

Answer 2

The accepted answer is of course fine, but the conclusions drawn in the comment below it are not correct at all, so let me add some more:

The "actual" reason why the octic equation (1) is defined over $\mathbb{Q}$ whereas the septic isn't is that the automorphism group $PGL_2(7)$ of $PSL_2(7)\cong PSL_3(2)$ (the latter being the Galois group of equation (2)) embeds into $S_8$, but not into $S_7$. By the way, introducing a square root in (1) is not necessary for this (if anything, it's a curiosity of this particular case that even after pulling back by the square root one still gets a rational function), there is also the octic equation $j(\tau) = \dfrac{(x^2+5x+1)^3(x^2+13x+49)}{x}$. The group-theoretic reason for the existence of these two rational function expressions is that there is a triple of elements (of orders 2,3 and 7) in $PSL_2(7)$ which is a so-called "genus-$0$ triple" in both the actions on 7 and on 8 points. There are general theoretical criteria for when such an expression can be defined over $\mathbb{Q}$, and one necessary condition is that the full symmetric normalizer $C_7\rtimes C_6$ of the order-$7$ subgroup needs to be in the Galois group. That's not the case in $PSL_3(2)\le S_7$, but in $PGL_2(7)\le S_8$, which is why the field of definition of the octic descends to $\mathbb{Q}$.

For $PGL_2(13)$, the situation is different. The index-$78$ ($=6\cdot 13$) subgroups are maximal both in $PGL_2(13)$ and in its normal subgroup $PSL_2(13)$, so there is no lower degree action of $PSL_2(13)$ of degree e.g. $3\cdot 13$ not extending to $PGL_2(13)$. This means that if one parameterizes the fixed field of these index-$78$ subgroups inside the Galois closure of the given equation, it will be doable by a degree-$78$ equation over $\mathbb{Q}$, and there's no way of dropping the degree from $78$ to something lower at the cost of increasing the field of definition from $\mathbb{Q}$ to a larger number field (even it there were, this larger number field would have to be inside the constant field extension of the Galois closure of the given equation, which is only degree $2=[PGL:PSL]$, not $6$; again, the reason why this worked at all for $PGL_2(7)$ was that the index-$14$ subgroups of $PGL_2(7)$ lie inside $PSL_2(7)$, so over the fixed field of the latter group (i.e., the constant field $\mathbb{Q}(\sqrt{-7})$), one gets an equation of degree $7$).

Also, it depends what is meant by "a (degree-$13$) version of Klein's formula"; one can of course write down an equation $F(x,j)=0$ for the degree-$78$ function field extension of the bottom field $\mathbb{Q}(j)$, as one can do for any extension; but this will not be a rational function expression $j=f(x)$ any more! For this to be possible, the respective triple of permutations associated to the function would have to be still a genus-$0$ triple in the other action; Magma says it's instead genus $3$, so it will be just some equation for a genus $3$ curve here.

Answer 3

(This is an addendum to prevent clutter on the original post.)

Klein's 7th deg formula is,

$$j(\tau) = y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3$$

Of course, one can get rid of the square roots and this is actually a 14T16 with order $2\times168 = 336$. The price to pay is it becomes a quadratic in $j(\tau)$.

However, using Mathematica, I found a transformation such that $j(\tau)$ remains linear. Let,

$$y = \frac{-(z+\sqrt{-7})^3}{z^3-7z^2+7z+7}$$

and we have the rather nice,

$$j(\tau) = \frac{(\Phi_{14})^3}{(\Phi_{6})^7} = \frac{\big(4\,z\,(z^2 + 7)(z^2 - 7z + 14)(5z^2 - \color{red}{14}z - 7)\big)^3}{\big(z^3-7z^2+7z+7\big)^7}$$

which is a 21T20 with same order $2\times168 = 336.$

Note 1: This $7m$-deg formula is the same as in page 92 of Elkies' article "The Klein Quartic in Number Theory" but the red number 14 corrects a small typo of the article which had it as 15.

Note 2: Per Dave Benson's answer, there is a possibility this has a $13m$-deg counterpart for $m=6$. Hopefully, the formula will not be so cumbersome.