On modular solutions of certain modular differential equation and supersingular polynomials

Abstract

We extend the results of Kaneko–Zagier and Baba–Granath on relations of supersingular polynomials and solutions of certain second-order modular differential equations.

1 Introduction

An elliptic curve E over a field K of characteristic $p> 0$ is called supersingular if it has no p-torsion over $\overline{K}$. This condition depends only on the j-invariant of E, and it is known that there are only finitely many supersingular j-invariants, all being contained in $\mathbb {F}_{p^2}$ . We define the supersingular polynomial $ss_{p}(X)$ as the monic polynomial whose roots are exactly all the supersingular j-invariants:

$$\begin{aligned} ss_{p}(X)=\prod _{\begin{array}{c} E/\overline{\mathbb {F}}_{p}\\ E : \text {supersingular} \end{array}} \bigl ( X - j(E) \bigr ). \end{aligned}$$

Because the set of supersingular j-invariants in characteristic p is stable under the conjugation over $\mathbb {F}_{p}$, we have $ss_{p}(X) \in \mathbb {F}_{p}[X]$.

Various lifts of $ss_{p}(X)$ to characteristic 0 are reviewed and studied in Kaneko and Zagier [1]. In particular, they constructed a lift by using a certain differential operator on the space of modular forms. Baba and Granath [2] extended this construction by introducing new differential operators.

In this paper, we unify and generalize these results, by considering a differential operator arising from a product of Eisenstein series $E_{4},E_{6}$, and the discriminant function $\varDelta$. With this operator we construct a second-order differential operator which gives rise to an endomorphism of $M_{k}$. We write an eigenform of this operator explicitly in terms of hypergeometric series. For $k=p-1$, we show that the associated polynomial $\widetilde{F}$ of this eigenform F satisfies

$$\begin{aligned} ss_p (X) = X^{\delta }(X-1728)^{\varepsilon } \widetilde{F}(X) \mod p, \end{aligned}$$

with suitable $\delta ,\varepsilon \in \{ 0, 1 \}$.

2 Modular forms and supersingular polynomials

For positive even integer k, we denote by $M_{k}$ the space of holomorphic modular forms of weight k on $\varGamma = \mathrm {PSL}_{2}(\mathbb {Z})$. Let $E_{k}(\tau )$ be the Eisenstein series of weight k on $\varGamma$ defined by

$$\begin{aligned} E_{k}(\tau ) =1-\frac{2k}{B_{k}}\sum _{n=1}^\infty \biggl ( \sum _{d\mid n}d^{k-1} \biggr ) q^{n} \qquad (q=e^{2\pi i \tau }), \end{aligned}$$

where $\tau$ is a variable in the Poincaré upper half-plane $\mathfrak {H}$ and $B_{k}$ the kth Bernoulli number. For even $k\ge 4$, we have $E_{k}(\tau ) \in M_{k}$. We also define the discriminant function $\varDelta (\tau ) \in M_{12}$ and the elliptic modular function $j(\tau )$, respectively, by

$$\begin{aligned} \varDelta (\tau )= & {} \frac{E_{4}(\tau )^3-E_{6}(\tau )^2}{1728}\\= & {} q \prod _{n=1}^\infty (1-q^n)^{24} = q-24q^2+252q^3-1472q^4+ \cdots \end{aligned}$$

and

$$\begin{aligned} j(\tau ) =\frac{E_{4}(\tau )^3}{\varDelta (\tau )}=\frac{1}{q} + 744 + 196884 q + 21493760 q^2 + \cdots . \end{aligned}$$

The Gauss hypergeometric series is defined by

$$\begin{aligned} {}_{2} F_{1}(\alpha , \beta ; \gamma ; x ) = \sum _{n=0}^\infty \frac{ (\alpha )_{n} (\beta )_{n} }{ (\gamma )_{n}} \frac{x^n}{n!} \quad (|x|<1 ), \end{aligned}$$

where $(\alpha )_{0}=1$ and $(\alpha )_{n}=\alpha (\alpha +1) \cdots (\alpha +n-1) \,\,\, (n\ge 1)$. We note that the series ${}_{2} F_{1}(\alpha , \beta ; \gamma ; x )$ becomes a polynomial when $\alpha$ or $\beta$ is a negative integer and $\gamma$ is not a negative integer.

For even $k\ge 4$, we can write k uniquely in the form

$$\begin{aligned} k=12m+4\delta +6\varepsilon \quad \text {with} \quad m\in \mathbb {Z}_{\ge 0},\ \delta \in \{ 0,1,2 \} ,\ \varepsilon \in \{ 0,1 \} . \end{aligned}$$

(1)

Under this notation, any modular form $f(\tau ) \in M_{k}$ can be written uniquely as

$$\begin{aligned} f(\tau )= E_{4}(\tau )^{\delta } E_{6}(\tau )^{\varepsilon } \varDelta (\tau )^{m} \widetilde{f} \bigl ( j(\tau ) \bigr ) , \end{aligned}$$

(2)

where $\widetilde{f}$ is a polynomial of degree less than or equal to m. We call $\widetilde{f}$ the associated polynomial of f.

The following representation of $ss_{p}(X)$ is essentially due to Deuring [3].

Lemma 1

Let $p\ge 5$ be a prime number and write $p-1$ in the form $12m+4\delta +6\varepsilon \; ( m\in \mathbb {Z}_{\ge 0},\ \delta \in \{ 0,1,2 \} ,\ \varepsilon \in \{ 0,1 \})$. Then

$$\begin{aligned} ss_p (X) = X^{m+\delta }(X-1728)^{\varepsilon } {}_{2}F_{1}\left( -m ,\frac{5}{12}-\frac{2\delta -3\varepsilon }{6};1;\frac{1728}{X} \right) \mod p. \end{aligned}$$

(3)

Proof

We define the monic polynomial $U_{n}^{\varepsilon }(X)$ of degree $n\ge 0$ by

$$\begin{aligned} X^{n}\, {}_{2}F_{1}\left( \tfrac{1}{12} ,\tfrac{5}{12};1;\tfrac{1728}{X} \right)&= U_{n}^{0}(X) + O \left( \tfrac{1}{X} \right) , \\ X^{n-1} (X-1728)\, {}_{2}F_{1}\left( \tfrac{7}{12} ,\tfrac{11}{12};1;\tfrac{1728}{X} \right)&= U_{n}^{1}(X) + O \left( \tfrac{1}{X} \right) . \end{aligned}$$

By [1, Proposition 5], we have $ss_{p}(X) = U_{m +\delta +\varepsilon }^{\varepsilon }(X) \mod p$. The first two parameters of the hypergeometric series in (3) reduce modulo p to

$$\begin{aligned} \left( -m ,\frac{5}{12}-\frac{2\delta -3\varepsilon }{6} \right) \equiv {\left\{ \begin{array}{ll} (\tfrac{1}{12},\tfrac{5}{12}) \pmod {p} &{} \text {if }p \equiv 1 \pmod {12}, \\ (\tfrac{5}{12},\tfrac{1}{12}) \pmod {p} &{} \text {if }p \equiv 5 \pmod {12}, \\ (\tfrac{7}{12},\tfrac{11}{12}) \pmod {p} &{} \text {if }p \equiv 7 \pmod {12}, \\ (\tfrac{11}{12},\tfrac{7}{12}) \pmod {p} &{} \text {if }p \equiv 11 \pmod {12}. \end{array}\right. } \end{aligned}$$

Since ${}_{2}F_{1}(a,b;c;x) = {}_{2}F_{1}(b,a;c;x)$, we see that $U_{m +\delta +\varepsilon }^{\varepsilon }(X)$ is congruent to the left-hand side of (3) modulo p. $\square$

3 Construction of the endomorphism

In this section, we construct an endomorphism $\phi _{g,k}$ of $M_{k}$. Let r, s, t be integers, not all zero, and k be an even integer greater than or equal to 4. Then, for the meromorphic modular form $g(\tau ) = E_{4}(\tau )^r E_{6}(\tau )^s \varDelta (\tau )^t \not \equiv 0$ of weight $u:=4r+6s+12t$ and $f \in M_{k}$, we define the differential operator $\partial _{g}$ by

$$\begin{aligned} \partial _{g}(f)(\tau ) = \partial _{g,k}(f)(\tau ) = f'(\tau ) -\frac{k}{u} \frac{g'(\tau )}{g(\tau ) } f(\tau ) \quad \left( \,\, \prime = \frac{1}{2 \pi i} \frac{d}{d\tau } = q \frac{d}{d q} \right) , \end{aligned}$$

and for $m\in \mathbb {Z}_{\ge 0},\ \delta \in \{ 0,1,2 \}$, and $\varepsilon \in \{ 0,1 \}$ with $k=12m+4\delta +6\varepsilon$, define the operator $\phi _{g,k}$ by

$$\begin{aligned} \phi _{g,k}(f)&= \frac{1}{E_{4}} \left\{ (\partial _{g,k+2}\circ \partial _{g,k}) (f) - \frac{t^2 k(k+2)}{u^2} E_{4} f \right. \nonumber \\&\qquad - \frac{432}{u^2} (s k -u\varepsilon )(s k -u\varepsilon +4(r+2 s+3 t)) \frac{E_{4} \varDelta }{E_{6}^{2}} f \nonumber \\&\qquad \left. + \frac{192}{u^2} (r k -u \delta ) \bigl ( r k -u \delta + 6(r+s+2 t) \bigr ) \frac{\varDelta }{E_{4}^{2}} f \right\} . \end{aligned}$$

(4)

Note that the function $g(\tau )$ is not always a holomorphic modular form. Except for the case of $(r,s,t)=(0,0,1)$, the image of $f \in M_{k}$ under $\partial _{g,k}$ is not holomorphic in general.

Theorem 1

The differential operator $\phi _{g,k}$ is an endomorphism of $M_{k}$.

To prove the theorem, we need two lemmas.

Lemma 2

The operator $\partial _{g}$ is written as

$$\begin{aligned} \partial _{g}(f) = \frac{4 r}{u}\partial _{E_{4}}(f) + \frac{6 s}{u}\partial _{E_{6}}(f) + \frac{12 t}{u}\partial _{\varDelta }(f) = \partial _{\varDelta }(f) + \frac{k}{6 u} \left( 2 r \frac{E_{6}}{E_{4}} + 3 s \frac{E_{4}^2}{E_{6}} \right) f . \end{aligned}$$

(5)

Proof

This is easily computed by using the well-known relation (due to Ramanujan)

$$\begin{aligned} E'_{2}=\frac{E^{2}_{2}-E_{4}}{12},~~E'_{4}=\frac{E_{2}E_{4}-E_{6}}{3},~~E'_{6}=\frac{E_{2}E_{6}-E^{2}_{4}}{2}. \end{aligned}$$

(6)

$\square$

Lemma 3

Put $v= (sk-u \varepsilon )/2$ and $w= (rk - u a)/3$. Then

$$\begin{aligned} u\, \partial _{g,k}(E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c})&= v E_{4}^{a+2} E_{6}^{\varepsilon -1} \varDelta ^{c} + w E_{4}^{a-1} E_{6}^{\varepsilon +1} \varDelta ^{c} , \nonumber \\ u^2\, (\partial _{g,k+2}\circ \partial _{g,k}) (E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c})&=1728 v(v+2(r+2 s+3 t)) E_{4}^{a+1} E_{6}^{\varepsilon -2} \varDelta ^{c+1} \nonumber \\&{} \quad +(v+w)(v+w -2 t) E_{4}^{a+1} E_{6}^{\varepsilon } \varDelta ^{c} \nonumber \\&{} \quad -1728 w(w+2(r+s+2 t)) E_{4}^{a-2} E_{6}^{\varepsilon } \varDelta ^{c+1}. \end{aligned}$$

(7)

Proof

One can easily see that the operator $\partial _{\varDelta }$ satisfies the Leibniz rule:

$$\begin{aligned} \partial _{\varDelta ,k+l}(FG) = \partial _{\varDelta , k}(F) G + F \partial _{\varDelta , l}(G) \end{aligned}$$

for $F \in M_{k}$ and $G \in M_{l}$. Hence we can prove the lemma by direct calculation using (5) and the following relations:

$$\begin{aligned} \partial _{\varDelta }(E_{4}) = -\frac{1}{3} E_{6}, \quad \partial _{\varDelta }(E_{6}) = -\frac{1}{2} E_{4}^{2},\quad \partial _{\varDelta }(\varDelta ) =0,\quad E_{4}^3 - E_{6}^2 =1728 \varDelta . \end{aligned}$$

$\square$

Proof of Theorem 1

For even $k\ge 4$, write k in the form $k=12 m +4\delta +6\varepsilon$ as before and assume the numbers a, c satisfy $a\equiv \delta \mod 3 \; (0\le a\le 3m +\delta ), \, 0\le c \le m$, and $k=4a +6\varepsilon +12c$, so that the forms $E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c}$ constitute basis elements of $M_{k}$. We now compute $\phi _{g,k}(E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c})$.

Since $(v+w)(v+w- 2t) = t^2 k(k+2) - 2 t(k+1)uc +u^2 c^2$, we can obtain from (7) the following equation:

$$\begin{aligned}&{} u^2\, (\partial _{g,k+2}\circ \partial _{g,k}) (E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c})- t^2 k(k+2) E_{4} \cdot E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \\&\quad =1728 v(v+ 2(r+2 s+3 t)) E_{4}^{a+1} E_{6}^{\varepsilon -2} \varDelta ^{c+1}\\&\qquad + u^2 c \{ c - 2 t(k+1)/u \} E_{4}^{a+1} E_{6}^{\varepsilon } \varDelta ^{c} -1728 w(w+ 2(r+s+2 t)) E_{4}^{a-2} E_{6}^{\varepsilon } \varDelta ^{c+1}. \end{aligned}$$

Furthermore, by using $1728v(v+ 2(r+2s+3t)) = 432(sk-u\varepsilon )(sk-u\varepsilon +4(r+2s+3t))$, we have

$$\begin{aligned}&u^2\, (\partial _{g,k+2}\circ \partial _{g,k}) (E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c}) - t^2 k(k+2) E_{4} \cdot E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \\&\qquad - 432 (s k-u\varepsilon )(s k-u\varepsilon +4(r+2 s+3 t)) \frac{E_{4} \varDelta }{E_{6}^2} \, E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \\&\quad = u^2 c \left\{ c - \frac{2 t(k+1)}{u} \right\} E_{4}^{a+1} E_{6}^{\varepsilon } \varDelta ^{c}\\&\qquad -1728 w(w+ 2(r+s+2 t)) E_{4}^{a-2} E_{6}^{\varepsilon } \varDelta ^{c+1} . \end{aligned}$$

We define $\lambda (x) = \frac{192}{u^2} (r k - u x )(r k - u x + 6(r+s+2 t))$, then $1728w(w+2(r+s+2t)) = u^2 \lambda (a)$. Adding $u^{2} \lambda (\delta ) E_{4}^{a-2} E_{6}^{\varepsilon } \varDelta ^{c+1}$ to both sides of the above equation and dividing them by $u^{2} E_{4}$, we get

$$\begin{aligned} \phi _{g,k}(E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c})&= \frac{1}{E_{4}} \left\{ (\partial _{g,k+2}\circ \partial _{g,k}) (E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c}) - \frac{t^2 k(k+2)}{u^2} E_{4} \cdot E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \right. \nonumber \\&{} \quad - \frac{432}{u^2} (s k-u\varepsilon )(s k-u\varepsilon +4(r+2 s+3 t)) \frac{E_{4} \varDelta }{E_{6}^2} \, E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \nonumber \\&{} \left. \quad + \frac{48}{u^2} (r k-u \delta )(r k-u \delta + 6(r+s+2 t)) \frac{\varDelta }{E_{4}^2} \, E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} \right\} \nonumber \\&= c \left\{ c - \frac{2 t(k+1)}{u} \right\} E_{4}^{a} E_{6}^{\varepsilon } \varDelta ^{c} -(\lambda (a) -\lambda (\delta )) E_{4}^{a-3} E_{6}^{\varepsilon } \varDelta ^{c+1} . \end{aligned}$$

(8)

The right-hand side is an element of $M_{k}$ if $a \ge 3$. If $a < 3$, we have $a=\delta$ (because $a \equiv \delta \pmod {3}$) and the coefficient $\lambda (a) -\lambda (\delta )$ of $E_{4}^{a-3} E_{6}^{\varepsilon } \varDelta ^{c+1}$ vanishes, hence the right-hand side is in $M_{k}$. Thus $\phi _{g,k}$ is an endomorphism of $M_{k}$. $\square$

4 Modular solutions of $\phi _{g,k}(f)=0$ and supersingular polynomials

Throughout this section, we assume $2t(k+1) \not = c u \, (1 \le c \le m )$ for given r, s, t, and $k=12 m +4\delta +6\varepsilon$. By Eq. (8), we see that the matrix representation of $\phi _{g,k}$ in the ordered base $\{ E_{4}^{3m+\delta } E_{6}^{\varepsilon }, \dots , E_{4}^{\delta } E_{6}^{\varepsilon } \varDelta ^{m} \}$ is a triangular matrix and obtain the eigenvalues $c ( c - \tfrac{2t(k+1)}{u} ) , \; 0\le c\le m$ of $\phi _{g,k}$ as diagonal elements. Hence, under the assumption, all eigenvalues of endomorphism $\phi _{g,k}$ are different.

Theorem 2

(i) The following modular form $F_{g,k}(\tau ) = 1+O(q)$ is the unique eigenvector of $\phi _{g,k}$ with eigenvalue 0:

$$\begin{aligned} F_{g,k}(\tau )&= E_{4}(\tau )^{3 m+\delta } E_{6}(\tau )^{\varepsilon } \nonumber \\&\quad \times {}_{2}F_{1} \left( -m ,\, \frac{5}{12}+\frac{(2 r -3 s -6 t)(k+1)}{6 u} -\frac{2\delta -3\varepsilon }{6} \,; 1-\frac{2 t(k+1)}{u} ; \frac{1728}{j(\tau )} \right) . \end{aligned}$$

(9)

(ii) Let $k=p-1$ where $p\ge 5$ is prime and assume that $u \not \equiv 0 \pmod {p}$. Then the associated polynomial $\widetilde{F}_{g,p-1}(X)$ of $F_{g,p-1}(\tau )$ has p-integral coefficients and

$$\begin{aligned} ss_p (X) = X^{\delta }(X-1728)^{\varepsilon } \widetilde{F}_{g,p-1}(X) \mod p. \end{aligned}$$

Proof

(i) By using (5) and (6) to expand the differential equation $\phi _{g,k}(f)=0$, we obtain

$$\begin{aligned}&f''(\tau ) + A(\tau ) f'(\tau ) +B(\tau ) f(\tau ) =0, \nonumber \\&A(\tau ) = -\frac{k+1}{6} E_{2} +\frac{k+1}{3 u} \left( 3 s \frac{E_{4}^{2}}{E_{6}} +2 r \frac{E_{6}}{E_{4}} \right) , \nonumber \\ B(\tau )&= \frac{k(k+1)}{12} E'_{2} - \frac{k(k+1)}{36 u} \cdot \frac{9s E'_{4}E_{4}^{2} +4r E'_{6} E_{6}}{E_{4} E_{6}} \nonumber \\&\quad +\frac{E_{4}^{3} -E_{6}^{2}}{E_{4}^{2} E_{6}^{2}} \left\{ \frac{s \varepsilon (k+1)}{2 u} E_{4}^{3} - \frac{2 r \delta (k+1) -u \delta (\delta -1)}{9 u} E_{6}^{2} \right\} . \end{aligned}$$

(10)

This is a special case of modular differential equations with regular singularities at elliptic points for $\mathrm {SL}_{2}(\mathbb {Z})$ treated in [4]. More explicitly, the differential equation (10) is expressed as follows using the symbol in [4, Theorem B]:

$$\begin{aligned} \mathcal {D}_{k} \left( \frac{s(k+1)}{u}, \; \frac{2 r(k+1)}{3 u}, \; \frac{s \varepsilon (k+1)}{2 u} ,\; \frac{2 r \delta (k+1) -u \delta (\delta -1)}{9 u} \right) . \end{aligned}$$

Applying [4, Theorem C] to this parameters, we get the hypergeometric representation of $F _{g,k}(\tau )$. We note that the exponent of $E_{6}(\tau )$ is a solution of the following quadratic equation:

$$\begin{aligned} x^{2} - \left( \frac{2s(k+1)}{u} +1 \right) x + \frac{2 s \varepsilon (k+1)}{u} =0. \end{aligned}$$

Since $\varepsilon \in \{ 0,1 \}$, we have $\varepsilon (\varepsilon -1) =0$ and thus the left-hand side of the above equation factors into $(x-\varepsilon )( x- 2s(k+1)/u +\varepsilon -1 )$. As pointed out in [4, Remark 4], we can choose $\varepsilon$ as exponent of $E_{6}(\tau )$. (ii) For $k=p-1$, by (2) and the hypergeometric formula (9), the associated polynomial $\widetilde{F}_{g,p-1}(X)$ of $F_{g,p-1}(\tau )$ is as follows:

$$\begin{aligned} \widetilde{F}_{g,p-1}(X)&= X^{m} {}_{2}F_{1}\left( -m ,\frac{5}{12}+\frac{(2r-3s-6t) p}{6 u} -\frac{2\delta -3\varepsilon }{6};1-\frac{2 t p}{u};\frac{1728}{X} \right) \\&\equiv X^{m} {}_{2}F_{1}\left( -m ,\frac{5}{12} -\frac{2\delta -3\varepsilon }{6};1;\frac{1728}{X} \right) \mod p. \end{aligned}$$

Hence $X^{\delta }(X-1728)^{\varepsilon } \widetilde{F}_{g,p-1}(X)$ is congruent to $ss_{p}(X)$ modulo p by Lemma 1. $\square$

Remark 1

The case of $(r,s,t)=(0,0,1)$ was studied in the paper [1] by Kaneko and Zagier. The corresponding operator

$$\begin{aligned} \partial _{\varDelta ,k}(f)(\tau ) = f'(\tau )-\frac{k}{12} E_{2}(\tau ) f(\tau ) \; : M_{k} \rightarrow M_{k+2} \end{aligned}$$

is called the Ramanujan–Serre derivative. We note that the logarithmic derivative of $\varDelta (\tau )$ is equal to $E_{2}(\tau )$. If $k \not \equiv 2 \pmod {3}$, the function $F_{\varDelta ,k}(\tau )$ coincides with $F_{k}(\tau )$ in [1, Sect. 8] up to a constant multiple. Moreover, Baba and Granath studied the cases of $(r,s,t)=(1,0,0)$ and (0, 1, 0) in [2]. The corresponding operators are given, respectively, by

$$\begin{aligned} \partial _{E_{4},k }(f)(\tau ) = f'(\tau )-\frac{k}{4} \frac{E'_{4}(\tau )}{E_{4}(\tau )} f(\tau ) , \quad \partial _{E_{6},k }(f)(\tau ) = f'(\tau )-\frac{k}{6} \frac{E'_{6}(\tau )}{E_{6}(\tau )} f(\tau ). \end{aligned}$$

Hence, the differential equations $\phi _{E_{4}, k}(f)=0$ and $\phi _{E_{6}, k}(f)=0$ coincide with [2, Eq. (5)] and [2, Eq. (8)], respectively. Consequently, the symbols $F_{E_{4} ,k}(\tau )$ and $F_{E_{6} ,k}(\tau )$ we use are same as theirs, but the definition of our operator $\phi _{g,k}$ and their operator $\phi$ are slightly different.