# Quantities $g_2$, $g_3$, $\Delta$

This question is somewhat related to this one. Let $\lambda$ be the modular lambda function. Greenhill (Elliptic Functions, p. 57) states that we may put $$g_2 = \frac{1 - \lambda + \lambda^2}{12}, \quad g_3 = \frac{(1 + \lambda)(1 - 2\lambda)(2 - \lambda)}{432}, \quad \Delta = \frac{\lambda^2 (1 - \lambda)^2}{256}, \tag{\ast}$$ where $g_2$, $g_3$ are the invariants, and $\Delta$ is the discriminant, of the Weierstrass cubic $4x^3 - g_2 x - g_3$. I believe this convention is due to Klein. However, my question is why would one adopt such a convention and can one truly treat them as equalities?

In general, if $X = ax^4 + 4bx^3 + 6cx^2 + 4dx + e$ then we define $\Delta(X) = g_2^3 - 27g_3^2$, where $$g_2(X) = ae - 4bd + 3c^2 \quad \text{and} \quad g_3(X) = ace + 2bcd - ad^2 - b^2 e - c^3.$$

Also, I find it intriguing that the quantities $g_2$, $g_3$, $\Delta$ associated with the Legendre cubic $x(x - 1)(x - \lambda)$ are precisely $(\ast)$. I know that the two elliptic curves $$E_1 : y^2 = 4x^3 - g_2 x - g_3 \quad \text{and} \quad E_2 : y^2 = x(x - 1)(x - \lambda)$$ are isomorphic (because their $j$-invariant is the same), but is there something deeper going on here?

• Could you please explain what you find intriguing? Your general definition with $X=ax^4 + 4bx^3 + 6cx^2 + 4dx + e$ specializes to the first set of formulas when $X=x(x-1)(x-\lambda)$. Surely that's not surprising. What do you mean by the quantities being 'precisely the ones above'? Commented Sep 20, 2014 at 3:02
• @Prometheus No, that is not surprising. Greenhill claims that we may use $(\ast)$ to define $g_2$, $g_3$, $\Delta$. But then $g_2 = g_2(E_2)$, $g_3 = g_3(E_2)$, $\Delta = \Delta(E_3)$, which I think is peculiar. Commented Sep 20, 2014 at 3:14
• I shouldn't believe that it's peculiar. $y^2 = x(x - 1)(x-\lambda)$ is a standard normal form for elliptic curves. [look up "legendre normal form"] Commented Sep 20, 2014 at 3:30
• @BalarkaSen Yes, I know. I think the "modern" formulation of Greenhill's line of thought is the following: Put the Weierstrass elliptic curve $E_1$ into the Legendre form $E_2$, calculate $g_2$, $g_3$, $\Delta$, which are $(\ast)$, and then simply use $(\ast)$ to define $g_2$, $g_3$, $\Delta$. Would you agree? Commented Sep 20, 2014 at 4:19
• Yes, I do believe that is the construction, @glebovg. Commented Sep 20, 2014 at 4:58

If you consider $g_2, g_3, \Delta, \lambda$ as functions of the period ratio $\tau$, then you have to add certain factors to get true equalities. This is obvious from the fact that the modular forms $g_2, g_3, \Delta$ have nonzero weight: Those functions get multiplied with powers of $\tau$ when replacing $\tau$ with $-\tau^{-1}$. On the other hand, $\lambda$ just changes to $1-\lambda$, thus the right-hand sides of (*) have weight zero.

The factor in question is a suitable power of a weight-1 modular form. To derive it, use the known formulae \begin{align} g_2 &= 60 \operatorname{G}_4 & g_3 &= 140 \operatorname{G}_6 \\ \operatorname{G}_4 &= 2\zeta(4)\operatorname{E}_4 = \frac{\pi^4}{45}\operatorname{E}_4 & \operatorname{G}_6 &= 2\zeta(6)\operatorname{E}_6 = \frac{2\pi^6}{945}\operatorname{E}_6 \\ \operatorname{E}_4 &= \frac{1}{2} \left(\varTheta_{00}^8 + \varTheta_{01}^8 + \varTheta_{10}^8\right) & \operatorname{E}_6 &= \frac{1}{2} (\varTheta_{00}^4 + \varTheta_{01}^4) (\varTheta_{00}^4 + \varTheta_{10}^4) (\varTheta_{01}^4 - \varTheta_{10}^4) \\ \lambda &= \frac{\varTheta_{10}^4}{\varTheta_{00}^4} & 1 - \lambda &= \frac{\varTheta_{01}^4}{\varTheta_{00}^4} \\ \Delta &= g_2^3 - 27 g_3^2 = (2\pi)^{12} \operatorname{D} & \operatorname{D} &= \frac{\operatorname{E}_4^3-\operatorname{E}_6^2}{1728} = \frac{1}{256}\varTheta_{00}^8 \varTheta_{01}^8 \varTheta_{10}^8 \end{align} where $\varTheta_{00}, \varTheta_{01}, \varTheta_{10}$ are Jacobi Thetanull functions (of $\tau$, as everything else here), and $\operatorname{G}_k$ and $\operatorname{E}_k$ are Eisenstein series of weight $k$, with $\operatorname{E}_k$ normalized to yield $\lim_{\Im\tau\to\infty}\operatorname{E}_k(\tau) = 1$.

Eliminate $\varTheta_{01}, \varTheta_{10}$ in favor of $\varTheta_{00}$ and $\lambda$. Then you will arrive at \begin{align} g_2 &= f^4 \frac{1 - \lambda + \lambda^2}{12} \\ g_3 &= f^6 \frac{(2-\lambda)(1+\lambda)(1-2\lambda)}{432} \\ \Delta &= f^{12} \frac{(1 - \lambda)^2 \lambda^2}{256} \\\text{with}\quad f &= 2\pi\varTheta_{00}^2 \end{align}

That said, in elliptic curve equations, those powers of $f$ can be transformed away, which is why Greenhill can do what you have described.

Addendum. Suppose instead you use the homogenized 2-parameter lattice form for $g_2, g_3, \operatorname{G}_4, \operatorname{G}_6, \Delta$, e.g. $$\operatorname{G}_k(\omega_1,\omega_2) = \omega_2^{-k}\operatorname{G}_k(\tau,1) \quad\text{with}\quad \tau = \frac{\omega_1}{\omega_2}$$

Set $\omega_2 = f(\tau)$ and adjust $\omega_1 = \tau\omega_2$ accordingly. Then the above-mentioned result can be rewritten as \begin{align} g_2(\omega_1,\omega_2) &= \frac{1 - \lambda + \lambda^2}{12} \\ g_3(\omega_1,\omega_2) &= \frac{(2-\lambda)(1+\lambda)(1-2\lambda)}{432} \\ \Delta(\omega_1,\omega_2) &= \frac{(1 - \lambda)^2 \lambda^2}{256} \\\text{with}\quad \omega_2 &= 2\pi\varTheta_{00}^2 \end{align} which is what Greenhill states. Note however that this requires a specifically normalized lattice; the normalization is not to $\omega_2=1$ but to $\omega_2=f(\tau)$.

Indeed, for $\tau$ in a certain subset of $\mathbb{H}$, we can identify $$f = 4K(\sqrt{\lambda})$$ where $K$ is a complete elliptic integral. This yields one of the canonical periods used for Jacobian elliptic functions.

• Thank you so much. I truly appreciate your time and effort. Regarding the Addendum, I discussed something similar with O.L. (see the comments). If we normalize the periods a la Klein: $\Omega_1 = \omega_1 \Delta^{1/12}$, $\Omega_2 = \omega_2 \Delta^{1/12}$, then one can show that $g_2$, $g_3$, $\Delta$ all reduce to $(\ast)$. This is discussed in Greenhill (p. 271). Commented Sep 22, 2014 at 6:04
• I have removed the misunderstandable parts of the addendum and clarified the rest (I hope). Sure about the $\Delta^{1/12}(\omega_1,\omega_2)$? From that I get $\Omega_2 = 2\pi\eta^2$ instead of $\Omega_2 = 2\pi\varTheta_{00}^2 = 2\pi\eta^2\mathfrak{f}^4$ where $\mathfrak{f}$ is a Weber function. Commented Sep 22, 2014 at 6:37
• I am pretty sure. If $\Omega_1 = \omega_1 \Delta^{1/12}$, $\Omega_2 = \omega_2 \Delta^{1/12}$, then $\omega_1 = 4K$, $\omega_2 = 4iK$, which corresponds with your results. By the way, I am using a different convention: $\omega_1$ is the real period and $\omega_2$ is the imaginary period, i.e., $\tau = \omega_2/\omega_1$. Commented Sep 22, 2014 at 9:19
• I have taken a look at Greenhill now. On p. 270, he implies that normalizing with $\Delta^{1/12}$ turns $g_2$ to $\gamma_2$ and $g_3$ to $\gamma_3$. In his notation, this is correct. Nowadays, we write $\gamma_2/12$ and $\gamma_3/216$ for that, as we relate to $j = 1728J$ instead of $J$ and absorb the occasional factor 27 into $\gamma_3^2$ so that we have $j=\gamma_2^3$, $j-1728=\gamma_3^2$. Commented Sep 22, 2014 at 23:08
• Cf. Wolfram's formulae (7) ... (17) for the modular Weber functions. Note that in our notation, \begin{align} \varTheta_{00} &= \mathfrak{f}^2\eta &\varTheta_{01} &= \mathfrak{f}_1^2\eta &\varTheta_{10} &= \mathfrak{f}_2^2\eta \\ E_4 &= \gamma_2\eta^8 & E_6 &= \gamma_3\eta^{12} & \operatorname{D} &= \eta^{24} \end{align} where $\eta$ is the Dedekind eta function. Commented Sep 22, 2014 at 23:09