Let $$E\quad\colon \quad y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ be an elliptic curve, and let $b_4 = a_1a_3+2a_4$. Serre in Propriétés galoisiennes des points d'ordre fini des courbes elliptiques (Section 5.3, b), p. 305) gives the following formula $$\Delta^{1/3}=b_4-3(x_i x_j+x_kx_l),\label{1}\tag{1}$$ where the $x_i$ are the $x$-coordinates of points of $E$ of order $3$, and $\{1,2,3,4\}=\{i,j\}\cup\{k,l\}$. This formula can be rewritten as \begin{equation*} j^{-1/3}=\frac{\frac{1}{2}\frac{27j}{j-12^3}+3(x_ix_j+x_kx_l)}{\frac{27j}{j-12^3}}.\label{2}\tag{2} \end{equation*} The relation \eqref{2} is a relation between modular forms, therefore its proof comes for free. Questions:

  1. Is there an algebraic reason why \eqref{1} should be true?

  2. Is there an algebraic proof of \eqref{1}?

  3. Are there similar formulas for $\Delta^{1/d}$ where $d$ is a divisor of $24$?

  4. Are there similar formulas for $\Delta^{1/d}$ if $d$ is not a divisor of $24$?

  5. What is the algebraic significance of the number $24$ in this context ($\Delta$ is the $24$-th power of the Dedekind eta function, but what does this mean algebraically)?

  • $\begingroup$ Just an idea for 1): one could use a little bit of classical invariant theory of ternary cubics and compute the Hessian of the cubic plane curve $E$. The 8 points of order three together with the point at infinity are the intersections of the curve with its Hessian (i.e., the nine inflection points). $\endgroup$ May 10, 2019 at 18:15
  • $\begingroup$ I should also add, essentially you would be computing the ternary resultant of two cubics and a line which reduces to a binary resultant of two binary cubics and expressible by a $6\times 6$ Sylvester determinant. I think it should be doable by hand. $\endgroup$ May 10, 2019 at 18:24
  • $\begingroup$ Do the free variables $i$, $j$, $k$, and $l$ in \eqref{1} (currently (1)) indicate summation? Is the $j$ on the left-hand side of \eqref{2} (currently (2)) the same as the index $j$ on the right-hand side? $\endgroup$
    – LSpice
    May 12, 2019 at 20:23
  • $\begingroup$ @LSpice, no, they just indicate the four different $x$-coordinates in some order. In the second relation, the $j$ is used for both the $j$ invariant and an index, $j$ is the $j$ invariant whenever it is not a subscript. $\endgroup$
    – Shimrod
    May 12, 2019 at 20:32
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    $\begingroup$ @LSpice, The three cube roots correspond to the three partitions of the four element set (sorry for late answer). $\endgroup$
    – Shimrod
    May 28, 2019 at 7:09

2 Answers 2


Here is an algebraic proof, without a ton of insight. To make life easier, let's put our curve into reduced Weierstrass form, $y^2 = x^3+a_4 x + a_6$. We put $f(x) = x^3+a_4 x + a_6$. The $3$-torsion points are the flexes, meaning the points where $\tfrac{d^2 y}{(dx)^2}=0$. We compute $$\frac{d^2 y}{(dx)^2} = \frac{d^2 }{(dx)^2} f(x)^{1/2} = (1/2) f'' f^{-1/2} - (1/4) (f')^2 f^{-3/2}=\frac{2 f'' f - (f')^2}{4 f^{3/2}}.$$ So the $x$-coordinates of the $3$-torsion points are the roots of $$2 f'' f - (f')^2 = 3 x^4 + 6 a_4 x^2 + 12 a_6 x - a_4^2.$$ We deduce that the elementary symmetric polynomials in $(x_1, x_2, x_3, x_4)$ take the values $$e_1(x)=0,\ e_2(x)=2 a_4,\ e_3(x) = -4 a_6,\ e_4(x) = -a_4^2/3 . (\ast)$$

Expanding $(y-x_1 x_2 - x_3 x_4)(y-x_1 x_3 - x_2 x_4)(y-x_1 x_4 - x_2 x_3)$ gives a polynomial in $y$ whose coefficients are elementary symmetric polynomials in $(x_1, x_2, x_3, x_4)$. By the fundamental theorem of symmetric polynomials, we can write the coefficients of this cubic as polynomials in the $e_j(x)$, and then plug in the formulas from $(\ast)$. (If you use Mathematica, the SymmetricReduction command will do this for you.) I get that this cubic is $$y^3 - 2 a_4 y^2 + \tfrac{4}{3} a_4^2 y - \tfrac{8}{3} a_4^3 - 16 a_6^2$$ $$=y^3 - b_4 y^2 + \tfrac{1}{3} b_4^2 y - \tfrac{1}{3} b_4^3 - 16 a_6^2 = (y-b_4/3)^3 - \tfrac{8}{27} b_4^3 - 16 a_6^2.$$

So the values of $x_i x_j + x_k x_{\ell}$ are $$\tfrac{b_4}{3} + \sqrt[3]{\tfrac{8}{27} b_4^3 - 16 a_6^2} = \frac{b_4 + \sqrt[3]{64 a_4^3 - 432 a_6^2}}{3} = \frac{b_4 + \sqrt[3]{\Delta}}{3}$$ or $$\sqrt[3]{\Delta} = - b_4 + 3(x_i x_j + x_k x_{\ell}).$$ In an earlier draft I said that $64 a_4^3 - 432 a_6^2 = 16 (4 a_4^3 - 27 a_6^2)$ was $16 \Delta$, but apparently the $\Delta$ that shows up in modular forms is $16$ times the classical discriminant of the cubic equation. Thanks to Sylivan JULIEN for pointing this out.

Here is a conceptual explanation for a big piece of this. For any $x_1$, $x_2$, $x_3$, $x_4$, note that $$\frac{(x_1 x_2 + x_3 x_4) - (x_1 x_3 + x_2 x_4)}{(x_1 x_2 + x_3 x_4) - (x_1 x_4 + x_2 x_3)} = \frac{(x_1 - x_4)(x_2 - x_3)}{(x_1 - x_4)(x_2 - x_3)}$$ which is the cross ratio $c(x_1, x_2 : x_3, x_4)$. We want to show that this ratio is a cube root of unity, so we want to show that the cross ratio of $x_1$, $x_2$, $x_3$, $x_4$ is a cube root of unity.

This computation turns out to be easiest when the cube is not in Weierstrass form but Hessian form: $X^3+Y^3+Z^3 = a XYZ$. The flexes of this curve are the $9$ points with homogenous coordinates $(1:-\zeta:0)$, $(0:1:-\zeta)$ and $(-\zeta:0:1)$ with $\zeta^3=1$. If we take $(1:-1:0)$ to be the origin of our curve, then negation is $(X:Y:Z) \mapsto (Y:X:Z)$ and we can take the quotient by negation to be given by the rational map $(X:Y:Z) \mapsto \tfrac{Z}{X+Y}$ (this also takes $(1:-1:0)$ to $\infty$, so it might be something like the $x$-coordinate). The $8$ non-identity flexes map to $0$ and to the $3$ cube roots of $-1$, whose cross ratio is as required.

The next part of this answer addresses the question of whether similar formulas exist for other powers of $\Delta$. I will interpret "similar" as "a polynomial in the $x$ and $y$-coordinates of the $N$-torsion points".

For a modular form $\phi$ of weight $k$, and $g = \left[ \begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \right]$ in $SL_2(\mathbb{Z})$, let $(g^{\ast} \phi)(z) = (cz+d)^{-k} \phi\left( \tfrac{az+b}{cz+d} \right)$.

Recall that $\Delta^{1/24}$ is the Dedekind $\eta$ function, which is a modular form of weight $1/2$. I find $1/2$ integer weights confusing, so I'll only look at even powers of $\eta$. Thus, I'll be asking whether $\Delta^{k/12} = \eta^{2k}$ could be a polynomial in $x$ and $y$-coordinates. Note that $\Delta^{k/12}$ has weight $k$.

For any $g \in SL_2(\mathbb{Z})$, we have $g^{\ast} \eta^2 = \chi(g) \eta^2$, where $\chi$ is a character from $SL_2(\mathbb{Z})$ to the twelfth roots of unity, an explicit formula for which can be found in the Wikipedia article linked before. So $g^{\ast} \Delta^{k/12} = \chi(g)^k \Delta^{k/12}$. Note that $\chi$ factors through the quotient $SL_2(\mathbb{Z}/12 \mathbb{Z})$.

Lemma For $1 \leq k \leq 12$, let $\phi$ be a cusp form of weight $k$ (and some level) obeying $g^{\ast} \phi = \chi(g)^k \phi$. Then $\phi$ is a scalar multiple of $\Delta^{k/12}$.

Proof: Since $\Delta^{1/12}$ is nowhere vanishing on the upper half plane, the ratio $\phi / \Delta^{k/12}$ is holomorphic, and is invariant for $SL_2(\mathbb{Z})$. Therefore, it is a polynomial in $j$, and we have $\phi = \Delta^{k/12} \sum_{e=0}^d c_e j^e$ for some polynomial in $j$. But, looking at $q$ series, the leading term of the right hand side is $c_d q^{k/12 - d}$, and the leading power of $q$ on the left hand side is positive, so $d$ must be $0$. $\square$

Now, let $\psi$ be a modular form of weight $1 \leq k \leq 11$. Let $N$ be the LCM of $12$ and the level of $\psi$. Define $$R \psi = \sum_{g \in SL_2(\mathbb{Z}/N \mathbb{Z})} \chi(g)^k g^{\ast} \psi.$$ (Since $12$ divides $N$, it makes sense to evaluate $\chi$ on $SL_2(\mathbb{Z}/N \mathbb{Z})$; since the level of $\psi$ divides $N$, it makes sense to talk about $g^{\ast} \psi$ similarly.) Then $R \psi$ will be a modular form of the same weight $k$, obeying $g^{\ast} \phi = \chi(g)^k \phi$. I claim furthermore that it will be a cusp form. Proof: Let $\Gamma'$ be the kernel of $\chi$. A set of coset representatives $SL_2(\mathbb{Z}/N \mathbb{Z})/\Gamma'$ is given by $\left[ \begin{smallmatrix} 1 & x \\ 0 & 1 \\ \end{smallmatrix} \right]$ for $0 \leq x \leq 11$. So write the sum as $$\sum_{x=0}^{11} \chi\left( \begin{smallmatrix} 1 & x \\ 0 & 1 \\ \end{smallmatrix} \right)^k \left[ \begin{smallmatrix} 1 & x \\ 0 & 1 \\ \end{smallmatrix} \right]^{\ast} \sum_{g \in \Gamma'/\Gamma(N)} g^{\ast} \phi.$$ The inner sum is a modular form for $\Gamma'$, which has only one cusp. The action of $ \left[ \begin{smallmatrix} 1 & x \\ 0 & 1 \\ \end{smallmatrix} \right]$ takes that cusp to itself. So the value of the whole sum at that cusp is $\sum_{x=0}^{11} \chi\left( \begin{smallmatrix} 1 & x \\ 0 & 1 \\ \end{smallmatrix} \right)^k$ times the value of the inner sum at the cusp, and that inner sum is $0$.

So, if $\psi$ is any modular form of weight $1 \leq k \leq 11$, then $R \psi$ is a scalar multiple of $\Delta^{k/12}$. Of course, that scalar might be $0$, but we can hope!

We now want to know that the $x$-coordinates of $N$-torsion points are modular forms of level $N$ and weight $2$, and the $y$-coordinates are modular forms of level $N$ and weight $3$. So we can take a polynomial in $x$'s and $y$'s of appropriate weight, apply the $R$-operator and hope.

If we are going to have a chance, we better make sure the level is high enough. $\Delta^{k/12}$ has level $\tfrac{12}{GCD(k,12)}$, so we should take $N$ divisible by this. The most obvious thing to try is to take $N = \tfrac{12}{GCD(k,12)}$.

We identify the $N$-torsion points with $(\mathbb{Z}/N \mathbb{Z})^2$. For $(a,b) \in (\mathbb{Z}/N \mathbb{Z})^2$, we denote the coordinates of the corresponding torsion point as $(x(a,b), y(a,b))$. Note that the action of $SL_2(\mathbb{Z}/N \mathbb{Z})$ on $x(a,b)$ and $y(a,b)$ is precisely the action on the vectors $(a,b)$. (Row or column vectors? I don't feel like working that hard.)

I'll present cases in order of complexity:

$k=6$: $\Delta^{1/2}$ has level $2$, so we work with $2$-torsion points. We want a polynomial of weight $6$, so we try cubics in the $x$-variables. Applying $R$ to $x(1,0)^2 x(0,1)$, we obtain $$x(1,0)^2 x(0,1) + x(0,1)^2 x(1,1) + x(1,1)^2 x(1,0)-x(0,1)^2 x(1,0) - x(1,1)^2 x(0,1) - x(1,0)^2 x(1,1)$$ $$=(x(1,0) - x(0,1)) (x(1,0) - x(1,1)) (x(0,1) - x(1,1)).$$ Sure enough, $$\Delta = (x(1,0) - x(0,1))^2 (x(1,0) - x(1,1))^2 (x(0,1) - x(1,1))^2.$$ This is far from the shortest way to obtain this identity, but it works.

$k=4$ This is the one the OP started with. This time, $\Delta^{1/3}$ has level $3$, so we work with $3$-torsion points. We want a polynomial of weight $4$, so we try quadratics in the $x$-variables. We have $x(a,b) = x(-a,-b)$, so we index the $x$-variables by the points of $\mathbb{P}^1(\mathbb{F}_3)$, written in homogenous coordinates $x(a:b)$. The action of $SL_2(\mathbb{Z})$ on the $4$ points of $\mathbb{P}^1(\mathbb{F}_3)$ is by the alternating subgroup $A_4$. So we want to take a quadratic monomial, $x(1:0) x(0:1)$ and average it with an order $3$ character of $A_4$. Writing $\omega$ for a cube root of unity, and not working hard enough to figure out which one I mean, we get that $\Delta^{1/3}$ is proportional to $$\left( x(1:0) x(0:1) + x(1:1) x(1:2) \right) + \omega \left( x(1:0) x(1:1) + x(0:1) x(1:2) \right) + \omega^2 \left( x(1:0) x(1:2) + x(0:1) x(1:1) \right)$$ as desired. Something else cute happens here: If we chose the other power of $\omega$, we get $0$. So we can use this to rewrite the formula in simpler ways.

$k=1$ No monomial in $x$'s and $y$'s can have weight $1$.

$k=2$ or $k=3$. So we want linear monomials in $x$'s or $y$'s. However, every torsion point is stabilized by some conjugate of $\left[ \begin{smallmatrix} 1 & \ast \\ 0 & 1 \\ \end{smallmatrix} \right]$, and averaging over this stabilizing subgroup gives $0$, so we just get $0$ if we apply $R$ to any $x(a,b)$ or $y(a,b)$. (François Brunault, in comments, states that something stronger is true: No linear combination of $x$'s or $y$'s is ever a cusp form. I think I've reconstructed the proof, but I'll leave it to him.)

That finishes the divisors of $12$. Also, $8=4+4$ and $10=4+6$, so $\Delta^{8/12}$ and $\Delta^{10/12}$ are products of things we already have. There are two other cases I find interesting: $k=5$ and $k=9$. (The case $k=7$ seems like just a messier version of $k=5$, and $11 = 5+6$, which is why I don't care so much about them.)

$k=5$: We want $12$-torsion points, and we want weight $5$, so we want products of an $x$ and a $y$. If I didn't screw up, if $(a,b)$ and $(c,d)$ fail to generate $(\mathbb{Z}/12 \mathbb{Z})^2$, then $R x(a,b) y(c,d)=0$. However, if they generate, then the result sure doesn't look like zero! I get that $$\sum_{\left[ \begin{smallmatrix} a&b \\ c&d \\ \end{smallmatrix} \right] \in SL_2(\mathbb{Z}/12 \mathbb{Z}) } \chi \left( \begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \right)^5 x(a,b) y(c,d)$$ should be a scalar multiple of $\Delta^{5/12}$! Does anyone have the computational chops to work out which one?

$k=9$ This time, we can try $4$-torsion points and polynomials of weight $9$. There are a lot of choices, but I think a very natural one to try is $y(1,0) y(0,1) y(3,3)$. Using the identity $y(a,b) = - y(-a, -b)$, there are only $8$ terms, which I think are the following: $$y(1,0) y(0,1) y(1,1)+i y(1,2) y(0,1) y(1,1)-i y(1,0) y(2,1) y(1,1)-y(1,2) y(2,1) y(1,1)-i y(1,0) y(0,1) y(1,3)-y(1,2) y(0,1) y(1,3)+y(1,0) y(2,1) y(1,3)+i y(1,2) y(2,1) y(1,3).$$ So, this is supposed to be a multiple of $\Delta^{3/4}$. Which one?

  • $\begingroup$ very nice............ $\endgroup$ May 10, 2019 at 20:34
  • $\begingroup$ I know nothing to the subject but why do you get an exponent of 2 for $f'$? $\endgroup$ May 10, 2019 at 22:09
  • $\begingroup$ Or maybe this has something to do with the fact that according to wikipedia, the discriminant of the relevant cubic equation is 16 times the modular discriminant (why on earth are both denoted by $\Delta$ while they differ?). $\endgroup$ May 10, 2019 at 22:40
  • $\begingroup$ The quartic polynomial is the $3$-division polynomial whose Galois group over $\mathbb{Q}(a_4,a_6)$ is $\operatorname{PGL}_2(\mathbb{F}_3)\cong S_4$ and your degree 3 polynomial in $y$ is the cubic resolvent. $\endgroup$ May 11, 2019 at 0:53
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    $\begingroup$ Finally for a cool related answer see math.stackexchange.com/questions/1593701/… ! $\endgroup$
    – alpoge
    May 11, 2019 at 11:51

Regarding Q5, the modular form $\Delta$ has weight 12, so it is a section of the line bundle $\omega^{\otimes 12}$ on the modular curve $Y$. Here $\omega$ is the Hodge bundle, which can be defined as $\omega = \pi_* \Omega^1_{E/Y}$ where $\Omega^1_{E/Y}$ is the sheaf of relative Kähler differentials on the universal elliptic curve $\pi : E \to Y$ (note that here $Y=Y(1)$, so there is no universal elliptic curve, to get around this problem we must treat $Y(1)$ as an algebraic stack).

If $d$ is a divisor of $12$ then $\Delta^{1/d} = \eta^{24/d}$ will be a cusp form of weight $12/d$ (I don't recall the precise level), so everything is algebraic. If you want to get to $\eta$ (in other words $d=24$) then you need half-integral weight modular forms. The interpretation of such modular forms uses the metaplectic cover of $\mathrm{SL}_2(\mathbb{R})$, which is not an algebraic group anymore, only a Lie group (see David Loeffler's answer for more details). The point is that the Hodge bundle $\omega$ is not the square of a line bundle, but it becomes a square once you pull back to the metaplectic group.


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