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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 $X$?

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The Aronhold invariant vanishes when the form has border rank $\le 3$, i.e., lies in the Zariski closure of sums of three cubes of linear forms. I don't know how your specific cubic looks like but if you show it is a sum of three cubes that would explain the phenomenon you observed.

If you don't immediately see the three linear forms, one way to get them is to compute the Hessian of your cubic and then try to factor it as a product of three linear forms. More precisely, if you can write your cubic $C$ as $C=L_1^3+L_2^3+L_3^3$ then the Hessian of $C$ should be proportional to the product $L_1 L_2 L_3$. So this should help identify these linear forms $L_i$ if they exist.

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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.)

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  • $\begingroup$ The characterization of $S=0$ in Dolgachev-Kanev is the same as in my answer, since having a polar s-gon with $s\le 3$ means the cubic is a sum of three cubes. $\endgroup$ – Abdelmalek Abdesselam Jan 4 '18 at 18:08

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