It seems well-known that any smooth plane quartic can be written as the vanishing of $Q_0Q_2 -Q_1^2$. Is there a good way to work out these quadratic factors $Q_0,Q_1,Q_2$? For example, given the Klein quartic $X^3Y + Y^3Z + Z^3X = 0$ what would these quadratics be?
If not, is there at least some computational package for this purpose? Thanks for your help!
The condition on your coefficients is an underdetermined system of quadratic equations. So specializing some of them (for instance setting them $0$), you can try to get a system with a finite number of (complex) solutions, and the chances are good that they are actually rational. Such systems can be handled with SageMath (which uses Singular, msolve or other systems as backend). In the specific case, a possible solution is \begin{align} Q_0 &= X^2 - XY + Y^2 + XZ\\\\ Q_1 &= X^2 - XY + YZ\\\\ Q_2 &= X^2 - XZ + YZ + Z^2. \end{align} Added later upon request in the comment: Here is the rather naive and straightforward Sage code which runs 160 seconds on my machine. It shows that a rational solution requires at least $11$ nonzero coefficients.
e2 = [(i, j) for i in range(3) for j in range(3-i)]
n2 = len(e2)
R = PolynomialRing(QQ, 'a', 3*n2)
Rxy.<x, y> = R[]
T0, T1, T2 = [sum(R.gen(m+k*n2)*x^i*y^j for m, (i, j) in enumerate(e2))
for k in range(3)]
f = x^3*y + y^3 + x
l = (T0*T2-T1^2-f).coefficients()
N = R.ngens()
for m in range(1, N):
print('m = ', m)
for s in Subsets(range(N), m):
i = min(s)
if i >= n2:
continue
l0 = l + [R.gen(i)-1] + [R.gen(j) for j in range(N) if
j != i and not j in s]
I = ideal(l0)
if I.dimension() == 0:
V = I.variety()
if len(V) > 0:
break
else:
continue
break
v = V[0]
Q0, Q1, Q2 = [sum(c.subs(v)*xy for c, xy in Q) for Q in [T0, T1, T2]]
print(f'Q0 = {Q0}\nQ1 = {Q1}\nQ2 = {Q2}\n{Q0*Q2-Q1^2 == f}')
Consider the second Veronese embedding $$ v_2 \colon \mathbb{P}^2 \to \mathbb{P}^5 $$ and let $S$ be its image. Then for any quartic curve $C \subset \mathbb{P}^2$ the image $v_2(C) \subset S$ is cut out by a quadric in $\mathbb{P}^5$, and this quadric is well defined modulo quadrics passing through $S$. And to obtain the require presentation of $C$ you need to find a quadric of rank 3 in this space.
For example, if $u_{00},\dots,u_{22}$ are the coordinates in $\mathbb{P}^5$, and $C$ is the Klein quartic, you need a quadric of rank 3 in the space generated by \begin{align} u_{00}u_{01} + u_{11}u_{12} + u_{22}u_{02},\\ u_{00}u_{11} - u_{01}^2,\\ u_{00}u_{22} - u_{02}^2,\\ u_{11}u_{22} - u_{12}^2,\\ u_{01}u_{12} - u_{11}u_{02},\\ u_{12}u_{02} - u_{22}u_{01},\\ u_{02}u_{01} - u_{00}u_{12}, \end{align} but not in the space generated by the last six quadrics.
Here the first equation cuts out $C$ in $S$, while the other six cut out $S$.
