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?
 A: 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.
