When people discuss elliptic curve invariants, it is not clear to me what changes are being considered that these invariants are invariant against.
Given an elliptic curve in "standard" form:
$$y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$$
there are some common convenience variables:
$$\begin{aligned}
b_2 &= a_1^2 + 4a_2, \\
b_4 &= a_1 a_3 + 2a_4, \\
b_6 &= a_3^2 +4a_6, \\
b_8 &= a_1^2 a_6 − a_1 a_3 a_4 + 4a_2 a_6 + a_2 a_3^2 − a_4^2,
\end{aligned}$$
and there are the invariants,
$$\begin{aligned}
c_4 &= b_2^2− 24b_4, \\
c_6 &= −b_2^3 + 36 b_2 b_4 − 216 b_6, \\
\Delta &= −b_2^2 b_8 – 8b^3_4 − 27b_6^2 + 9 b_2 b_4 b_6.
\end{aligned}$$
I found this ref which lists those equations and finally doesn't just gloss over the the transformations. But then it proceeds to explain how the invariants change under those transformations, which seems to go against the whole point of an invariant.
Let's look at an example. Consider this curve, chosen to let me do a transformation a little weirder than listed in that reference:
$$y^2 = 4x^3 + q^2(4k+1)$$
where $q,k \in \mathbb{Z}$ and non-zero.
Okay, this is not in standard form. One way to proceed is to multiply by 16, and make a new equation using the variable change $Y=4y$,$X=4x$:
$$Y^2 = X^3 + 16q^2(4k+1).$$
Alternatively, I can look at the initial equation and see that $y\equiv q\ (\text{mod }2)$, so we can write $Y=(y-q)/2$,$X=x$:
$$(2Y+q)^2 = 4Y^2 + 4Yq + q^2 = 4X^3 + q^2(4k+1)$$
$$Y^2 + qY = X^3 + q^2k$$
These two curves do not have the same invariants, despite starting from the same curve and making simple linear transformations. If even linear transformations lead to problems, then how restrictive is this!? I am so confused.
So can someone please help me understand?
I'd very much appreciate a clear statement of what transformations are the invariants actually invariant to, and if there are more general "invariants" if all I care about is whether two curves have solutions I can match up to each other.
Best Answer
Write $$\pi = \frac{\mathrm{d}x}{2y+a_1x+a_3}=\frac{\mathrm{d}y}{3x^2+2a_2x+a_4-a_1y}\text{.}$$ Then $$\pi'=u\pi\text{,}$$ so that the expressions that are actually invariant are certain so-called "modular differential forms", viz., modular forms like
$$\begin{align} \gamma_4 &= c_4 \pi^{\otimes 4} \\ \gamma_6 &= c_6 \pi^{\otimes 6} \\ \gamma_{12} &= \Delta \pi^{\otimes 12} \end{align}$$
and quasi-modular forms like $$\begin{align} \gamma_2 &=(12x+b_2)\pi^{\otimes 2} \\ \gamma_3 &=(2y+a_1x+a_3)\pi^{\otimes 3}\text{.} \end{align}$$
And while it's true that the coefficients of these expressions alone are not genuinely invariant under the law given for the Weierstrass form of curves, they are still "invariant up to powers": if, for two Weierstrass forms $E$, $E'$ over $\mathbb{Q}$, we calculate $c_4$ and $c'_4$, and we find that $c_4/c'_4$ is not of the form $u^4$ for any $\mathbb{Q}$, then we can conclude that $E$ and $E'$ are not isomorphic over $\mathbb{Q}$. To put it another way, we can imagine the "true invariants" of curves over $k$ to be orbits under certain actions of the multiplicative group $k^{\times}$:
$$\begin{align}[a_1]&\in (k\mod 2)/k^{\times}\\ [b_2]&\in(k\mod 12)/(k^{\times})^3 \\ [c_4]&\in k/(k^{\times})^4 \\ [c_6]&\in k/(k^{\times})^6 \\ [\Delta]&\in k/(k^{\times})^{12} \\ \end{align}$$