Homogeneous binary quartic forms to elliptic curves

• Broadly I would like to know what is the connection between homogeneous binary quartic forms and elliptic curves.

I see that the invariants on the first space under the action of $SL(2,\mathbb{C})$ are precisely the functions $g_2$ and $g_3$. Also given such a quartic form on $\mathbb{C}^2$, if one thinks its 0-set as the intersection of two conics in $\mathbb{C}^3$ then the determinant of the complex symmetric matrix representing the most general conic passing through this intersection is precisely the "RHS" of the elliptic curve equation.

• What does this determinant mean geometrically?

More specifically consider the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) and define $U = x^2$, $V = 2xy$ and $W=y^2$.

If the action is by the matrix $\left [ \begin{array}{c c } a & b \\ c & d \\ \end{array}\right ]$ ($ad-bc=1$) of $SL(2,\mathbb{C})$ then the following transformation is affected on $\mathbb{C}^2$,

$$\left [ \begin{array}{c} x\\ y\\ \end{array} \right ] \rightarrow \left [ \begin{array}{c} ax + by\\ cx + dy\\ \end{array} \right ]$$

This in turn implies the following transformation on $\mathbb{C}^3$ (defining the matrices, $X$ and $A$),

$$X = \left [ \begin{array}{c} U\\ V\\ W \end{array} \right ] \rightarrow A.X = \left [ \begin{array}{c c c} a^2 & ab & b^2\\ 2ac & (ad+bc) & 2bd \\ c^2 & cd & d^2\\ \end{array} \right ] \left [ \begin{array}{c} U\\ V\\ W \end{array} \right ]$$

$4UW-V^2$ being identically equal to $0$ is obviously kept invariant by the above transformation.

The invariance of $4UW-V^2$ can be thought of as $SL(2,\mathbb{C})$ preserving an inner-product on $\mathbb{C}^3$ with signature, $\left [ \begin{array}{c c c} & & 2 \\ & -1 & \\ 2 & & \\ \end{array} \right ]$

• But $A$ does not seem to be an orthogonal matrix though it satisfies the property of $det(A)=1$. Then why is it being said in the books that the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) induces an action of $SO(3)$ on $\mathbb{C}^3$ (spanned by $U$, $V$ and $W$) with respect to the inner product $\left [ \begin{array}{c c c} & & 2 \\ & -1 & \\ 2 & & \\ \end{array}\right ]$ ?
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It is orthogonal with respect to the given inner product:

sage: R.<a,b,c,d> = QQ['a,b,c,d']
sage: R2 = R.quotient_ring(a*d-b*c-1)
sage: A = Matrix(R2,[[a^2,a*b,b^2],[2*a*c, a*d+b*c, 2*b*d],[c^2,c*d,d^2]])
sage: J = Matrix([[0,0,2],[0,-1,0],[2,0,0]])
sage: A.transpose()*J*A == J
True

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I didn't understand that code you wrote. By a matrix being in $SO(3)$ I would think that it is a $3\times 3$ matrix which satisfies the equation $A^{T}A=AA^{T}=I$. – Anirbit Jan 3 '11 at 18:18
@Anirbit: That's orthogonal with respect to the standard quadratic form. There is an orthogonal group associated with any quadratic form: if the form is given by $x^tJy$, then the orthogonal matrices are those that satisfy $A^tJA=J$. – simplequestions Jan 3 '11 at 18:27
But in that case will $A$ be an element of $SO(3)$? If you change the quadratic form then shouldn't the Lie group also change? – Anirbit Jan 5 '11 at 11:41