TITLE: Vanishing of Aronhold S-invariant on the cubic forms on $H^2(X, \mathbb Q)$
QUESTION [8 upvotes]: I am considering several examples of compact complex threefolds $X$ such that $rk H^2(X)=3$.
Note that we have a cubic form on $H^2(X, \mathbb Q)$ which comes from the cup product.
I calculated the Aronhold S-invariants for those cubic forms and found that they are all zero,
which is very curious to me.
So let me ask some questions:
If S-invariants for a ternary cubic form is zero, what does it mean to the cubic form?
Is it a general phenomenon that S-invariants for cubic forms on such $X$'s are zero?
If it not,  what does zero S-invariant possibly mean to $X$?

REPLY [3 votes]: Here are some references that may be relevant. In the case $b_2=3$ the cubic from coming from the cup product gives rise to a plane cubic curve -- take the vanishing locus of the ternary cubic form on the projective plane $\mathbb{P}({\rm H}^2(X,\mathbb{Q}))$. Some statements concerning (non-)realizability of different types of cubics can be found in Sections 4.3 and 5.3 of 

C. Okonek and A. van de Ven. Cubic forms and complex 3-folds. L'Enseignement Math. 41 (1995), 297-333. 

To connect this with the question, Proposition 5.13.2 (p. 251) of the following paper gives a characterization of the types of plane cubic curves coming from ternary cubics with Aronhold invariant $S=0$: 

I. Dolgachev and V. Kanev. Polar covariants of plane cubics and quartics. Adv. Math. 98 (1993), 216-301. 

(In particular, the types 2,4,6,7 in the Okonek-van de Ven paper have nontrivial Aronhold invariants.)