A binary quartic form
$$aX^4+bX^3Y+cX^2Y^2+dXY^3+Y^4$$
decomposes as a product of linear factors $Y-t_jX$, $j=1,\dotsc,4$. I would like to have an explicit formula for symmetrization of the crossratio of $t_j$.
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1$\begingroup$ Do you mean the j-invariant of the elliptic curve $y^2=ax^4+bx^3+cx^2+dx+1$? $\endgroup$– Robin ChapmanCommented Jul 7, 2010 at 13:54
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$\begingroup$ Yes, that is exactly what I am looking for. $\endgroup$– David MarínCommented Jul 7, 2010 at 14:00
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4$\begingroup$ If you have access to a computer algebra system, you can do the following. Let $\xi$ denote a solution to $f(1,\xi)=0$ where $f$ is your quartic. Then $f(X,Y+\xi X)=b'X^3Y+\cdots+Y^4$. The elliptic curve is now isomorphic to $y^2=b'x^3+c'x^2+d'x+1$. Transform it to the usual Weierstrass form and take the $j$-invariant. Note that $b'$ etc. will have $\xi$s in them, but they should all cancel out via the equation $f(1,\xi)=0$ in the final result. $\endgroup$– Robin ChapmanCommented Jul 7, 2010 at 14:23
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2 Answers
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The $j$ invariant is
$$j=\frac{S^3}{S^3-27T^2}$$
where
$$S=a-\frac{bd}{4}+\frac{c^2}{12}$$
and
$$T=\frac{ac}{6}+\frac{bcd}{48}-\frac{c^3}{216}-\frac{ad^2}{16}-\frac{b^2}{16}.$$
For more details see my article "A computational solution to a question by Beauville on the invariants of the binary quintic", J. Algebra 303 (2006) 771-788. The preprint version is here.
answered Jul 7, 2010 at 22:32
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$\begingroup$ Please beware you are calculating J which is j/1728 $\endgroup$– OliverCommented May 24, 2020 at 22:15
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$\begingroup$ There are different conventions for what the proper normalization of $j$ is. Number theorists have their preferred choice. Folks working on moonshine matters have their own. Classical invariant theorists also. The is no universally canonical choice. $\endgroup$ Commented May 24, 2020 at 23:01
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1$\begingroup$ @AbdelmalekAbdesselam It's certainly true that there is no universally used choice. However, up to translation by an integer, there is a canonical choice if one wants a formula that has the correct properties in all characteristics, including characteristic 2 and 3. So that would seem to make the "number theorists'" version canonical, up to (as I said) translation by an integer. OTOH, if you're doing complex analysis, one could take the "$j$-invariant" of $X^3+AX+B$ to be $$\frac{(\pi^e+i\tan(\pi//\sqrt2))\cdot4A^3}{4A^3+27B^2},$$ for example. :) $\endgroup$ Commented May 25, 2020 at 1:18
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The j-invariant of the binary quartic form $aX^4+bX^3Y+cX^2Y^2+dXY^3+eY^4$ can be expressed by the two basic invariants I and J of the binary quartic form. \begin{align*} j&=\frac{I^3}{\Delta}=\frac{I^3}{I^3-27J^2}\\ I&=\begin{vmatrix} a & \frac{c}{6} \\ \frac{c}{6} & e \\ \end{vmatrix} -4 \begin{vmatrix} \frac{b}{4} & \frac{c}{6} \\ \frac{c}{6} & \frac{d}{4} \\ \end{vmatrix}=a e-\frac{b d}{4}+\frac{c^2}{12}\\ J&=\begin{vmatrix} a & \frac{b}{4} & \frac{c}{6} \\ \frac{b}{4} & \frac{c}{6} & \frac{d}{4} \\ \frac{c}{6} & \frac{d}{4} & e \\ \end{vmatrix}=\frac{a c e}{6}-\frac{a d^2}{16}-\frac{b^2 e}{16}+\frac{b c d}{48}-\frac{c^3}{216}\\ \end{align*}
An Introduction to Invariants and Moduli
(Cambridge Studies in Advanced Mathematics, Series Number 81)(Page 27 Proposition 1.25., Page 427 Theorem 11.44.)
https://www.math.ens.psl.eu/~benoist/refs/Mukai.pdf
Invariants of binary and ternary forms (Page 18)
https://ritzenth.pages.math.cnrs.fr/web//TEDI/tedi5-ritzenthaler.pdf
