The Hasse invariant of the Tate normal form $E_7$ and the supersingular polynomial for the Fricke group $\Gamma _0^*(7)$

Abstract

A formula is proved for the number of linear factors and irreducible cubic factors over ${\mathbb {F}}_l$ of the Hasse invariant ${\hat{H}}_{7,l}(a)$ of the elliptic curve $E_7(a)$ in Tate normal form, on which the point (0, 0) has order 7, as a polynomial in the parameter a, in terms of the class number of the imaginary quadratic field $K={\mathbb {Q}}(\sqrt{-l})$. Conjectural formulas are stated for the numbers of quadratic and sextic factors of $\hat{H}_{7,l}(a)$ of certain specific forms in terms of the class number of ${\mathbb {Q}}(\sqrt{-7l})$, which are shown to imply a recent conjecture of Nakaya on the number of linear factors over ${\mathbb {F}}_l$ of the supersingular polynomial $ss_l^{(7*)}(X)$ corresponding to the Fricke group $\Gamma _0^*(7)$.

Appendix

Let

$$\begin{aligned} \eta (\tau )=q^{1/24} \prod _{n \ge 1}{(1-q^n)}, \ \ q = e^{2\pi i \tau }, \end{aligned}$$

be the Dedekind $\eta$-function. Also, let

$$\begin{aligned} h(\tau ) = q^{-1} \prod _{n \ge 1}{\frac{(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-2})^2(1-q^{7n-5})^2}{(1-q^{7n-1})^3(1-q^{7n-6})^3}}, \end{aligned}$$

a modular function for $\Gamma _1(7)$. See [10, p. 156].

Theorem 10

For $\tau$ in the upper half-plane,

$$\begin{aligned} \left( \frac{\eta (\tau )}{\eta (7\tau )}\right) ^4=\frac{h^3-8h^2+5h+1}{h(h-1)}, \ \ h=h(\tau ). \end{aligned}$$

(8.1)

Also, if $d(\tau )=h(\frac{-1}{7\tau })$, then

$$\begin{aligned} 49\left( \frac{\eta (7\tau )}{\eta (\tau )}\right) ^4=\frac{d^3-8d^2+5d+1}{d(d-1)}, \ \ d=d(\tau ). \end{aligned}$$

(8.2)

It is clear that (8.2) follows from (8.1), using the transformation formula

$$\begin{aligned} \eta \left( \frac{-1}{\tau }\right) = \sqrt{\frac{\tau }{i}} \eta (\tau ). \end{aligned}$$

Theorem 11

The function $h(\tau )$ above satisfies the transformation formulas:

$$\begin{aligned} h\left( \frac{3\tau - 1}{7\tau - 2}\right) =&\ \frac{1}{1-h(\tau )}, \end{aligned}$$

(8.3)

$$\begin{aligned} h\left( \frac{2\tau -1}{7\tau -3}\right) =&\ \frac{h(\tau )-1}{h(\tau )}. \end{aligned}$$

(8.4)

Note that the map $A(\tau )=\frac{2\tau -1}{7\tau -3}$ has order 3 and $A^2(\tau )= \frac{3\tau - 1}{7\tau - 2}$. We first prove (8.4). We use the notation $e(x) = exp(2 \pi i x)$.

Proof

Start with the formulas from [10, p. 157]:

(8.5)

(8.6)

From the infinite products in (8.5) and (8.6) we get

$$\begin{aligned} s(\tau ) t^2(\tau ) =&\ q^{-1} \prod _{n \ge 1}{\frac{(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-2})^2(1-q^{7n-5})^2}{(1-q^{7n-1})^3(1-q^{7n-6})^3}}\\ =&\ h(\tau ). \end{aligned}$$

Hence, the theta function representations in (8.5) and (8.6) yield

To prove the formula for $h\left( \frac{2\tau -1}{7\tau -3}\right) = h(A(\tau ))$ we compute the transforms

First, we have from [10, p. 143, (4.5)], using the mapping $B(\tau ) = \frac{2\tau -7}{\tau -3}$, that

where

$$\begin{aligned} \kappa _1 = e\left( \frac{-27}{4}+\frac{37}{56}\right) \kappa _0 = e\left( \frac{-5}{56}\right) \kappa _0, \end{aligned}$$

and $\kappa _0$ is a fixed 8-th root of unity depending only on the mapping B. Now use [10, (4.3)] with $\ell = 0$ and $m=9$ and the lower sign, according to which

This gives that

Next, we have

where

$$\begin{aligned} \kappa _3 = e\left( \frac{-39}{4}+\frac{81}{56}\right) \kappa _0 = e\left( \frac{39}{56}\right) \kappa _0. \end{aligned}$$

Using [10, (4.3)] with $\ell = 0$ and $m=8$ and the lower sign gives

from which we obtain

Finally, for the third theta function we have

where

$$\begin{aligned} \kappa _5 = e\left( \frac{-51}{4}+\frac{141}{56}\right) \kappa _0 = e\left( \frac{-13}{56}\right) \kappa _0. \end{aligned}$$

Now using [10, (4.3)] with $\ell =0, m=-6$ and the upper sign gives

from which we obtain

Putting this all together gives

Now using the product formula [10, (4.8)] for the theta functions yields that

$$\begin{aligned} h\left( \frac{2\tau -1}{7\tau -3}\right) = \prod _{n \ge 1}{\frac{(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n-3})^2(1-q^{7n-4})^2}{(1-q^{7n-2})^3(1-q^{7n-5})^3}}. \end{aligned}$$

(8.7)

On the other hand, from the relation $s^7(\tau ) = h(\tau ) (h(\tau )-1)^2$ we easily derive the product formula

$$\begin{aligned} h(\tau )-1 = q^{-1} \prod _{n \ge 1}{\frac{(1-q^{7n-3})^3(1-q^{7n-4})^3}{(1-q^{7n-1})^2(1-q^{7n-6})^2(1-q^{7n-2})(1-q^{7n-5})}}. \end{aligned}$$

Dividing by the product formula for $h(\tau )$ yields that

$$\begin{aligned} \frac{h(\tau )-1}{h(\tau )} = \prod _{n \ge 1}{\frac{(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n-3})^2(1-q^{7n-4})^2}{(1-q^{7n-2})^3(1-q^{7n-5})^3}} \end{aligned}$$

and proves the formula. From (8.4) it follows that

$$\begin{aligned} h\left( \frac{3\tau -1}{7\tau -2}\right) = h(A^2(\tau )) = \frac{h(A(\tau ))-1}{h(A(\tau ))} = \frac{1}{1-h(\tau )}, \end{aligned}$$

which is formula (8.3). This proves Theorem 11. $\square$

Remark

Alternatively, we could finish the above proof by noting that $h\left( \frac{2\tau -1}{7\tau -3}\right)$ is a modular function for $\Gamma _1(7)$, since $T \in \Gamma _1(7)$ satisfies

$$\begin{aligned} h(A T(\tau )) = h(SA(\tau )) = h(A(\tau )), \ \ \text {for some} \ S \in \Gamma _1(7). \end{aligned}$$

Since $h(\tau )$ is a Hauptmodul for $\Gamma _1(7)$ and $h(\tau ) \rightarrow h(A(\tau ))$ induces an automorphism of $\textsf {K}_{\Gamma _1(7)}/\textsf {K}_{\Gamma _0(7)}$, it follows that $h(A(\tau ))$ is also a Hauptmodul and therefore equal to a linear fractional expression in $h(\tau )$. Furthermore, the six values of $h(\tau )$ at the cusps of $\Gamma _1(7)$ (namely, $\infty , 1 ,0$ and the roots $r_i$ of $x^3-8x^2+5x+1$, by (4.4)), are permuted by this automorphism (as residues of h modulo the prime divisors of $\textsf {K}_{\Gamma _1(7)}$ at infinity). Since the value $\infty$ corresponding to $\tau = \infty i$ is mapped to $h(A(\tau )) = 1$, by the product formula (8.7), it follows that the value 1 must be mapped to 0. This holds because $h(A(\tau )) = \frac{h+a}{h+b}$ cannot map a root $r_i$ to 0: otherwise $a = -r_i$, and applying the map twice sends 1 to $\infty$ (A has order 3), which would yield that b satisfies the irreducible equation $b^2+b+1-r_i = 0$ over ${\mathbb {Q}}(\zeta _7)$ (the norm of its discriminant is $-43$). But then $\frac{1-r_i}{1+b} = r_j$, for some j, would be impossible and $\frac{h+a}{h+b}$ could not map 1 to one of the $r_j$. Hence, $a = -1, b = 0$ and $h(A(\tau )) = \frac{h-1}{h}$. From the resulting product formula for $(h(\tau )-1)/h(\tau )$ we can derive the equation $s^7(\tau ) = h(\tau ) (h(\tau )-1)^2$.

Now we use the fact that $[\Gamma _0(7): \Gamma _1(7) \cup (-I)\Gamma _1(7)]=3$, from which it follows that $1, A, A^2$ are representatives for the cosets of $\Gamma _1[7] = \Gamma _1(7) \cup (-I)\Gamma _1(7)$ in $\Gamma _0(7)$. It follows that the function

$$\begin{aligned} \nonumber z(\tau )&= h(\tau )+h(A(\tau ))+h(A^2(\tau ))\\&= h(\tau )+\frac{h(\tau )-1}{h(\tau )}+\frac{1}{1-h(\tau )} = \frac{h^3(\tau )-3h(\tau )+1}{h(\tau )(h(\tau )-1)} \end{aligned}$$

(8.8)

is a modular function for $\Gamma _0(7)$ with a simple pole at $\infty i$ and the value $z(\tau ) = z(0) = 8$ at the other cusp of $\Gamma _0(7)$ (since the values of $h(\tau )$ at the cusps of $\Gamma _1(7)$ lying above 0 are the roots $r_i$ of $x^3-8x^2+5x+1$). It is clear that $\left( \frac{\eta (\tau )}{\eta (7\tau )}\right) ^4$ is a Hauptmodul for $\Gamma _0(7)$ [25, pp. 46, 51], and comparing q-expansions gives that

$$\begin{aligned} z(\tau ) = \left( \frac{\eta (\tau )}{\eta (7\tau )}\right) ^4+8. \end{aligned}$$

This shows that

$$\begin{aligned} \left( \frac{\eta (\tau )}{\eta (7\tau )}\right) ^4&= z(\tau )-8 = \frac{h^3(\tau )-3h(\tau )+1}{h(\tau )(h(\tau )-1)}-8\\&= \frac{h^3(\tau )-8h^2(\tau )+5h(\tau )+1}{h(\tau )(h(\tau )-1)}, \end{aligned}$$

which is (8.1). See [11, (4.24), p. 89].

These identities are closely related to several of Ramanujan’s entries in the unorganized material of his Notebooks. See entries 31 and 32 in [1, pp. 174–184]. In particular, the product representation of $h(\tau )-1$ in the above proof is equivalent to Entry 32(ii) of [1, p. 176] (or [2, (1.2)]); and the relation $f_7(h(\tau ),j_7^*(\tau )) = 0$ in Sect. 4 is, assuming Theorem 10, equivalent to Entry 32(iii) of [1, p. 176] (or [2, (1.3)]). Also see [6, Theorem 7.14, p. 440] and [19].