# Do all Fano threefolds have effective $c_2$?

Let $X$ be a smooth complex projective Fano threefold. Then the class $c_1(X)$ can be realised as an effective divisor in $X$. It is it true that the class $c_2(X)$ can be realised as an effective curve?

(Note that $c_3$ cannot, since the Euler characteristics of the cubic Fano $3$-fold is negative)

## 1 Answer

By a theorem of Miyaoka (Theorem 6.1 in Y. Miyaoka The Chern classes and Kodaira dimension of an algebraic variety', 1987), we have that for a Fano variety $X$, $c_2(X)\cdot H\ge 0$ for any ample divisor $H$. By duality of the cone of curves and the nef cone, it follows that $c_2(X)$ is at least pseudoeffective (i.e., a limit of effective classes). On the other hand, since $X$ is Fano, the Cone theorem shows that the cone of curves is rational polyhedral, spanned by finitely many classes of rational curves, so $c_2(X)$ is in fact effective.

EDIT: As C. Jiang correctly points out, to apply Miyaoka, one needs also the fact that the tangent bundle of a Fano 3-fold is generically nef. This was proved by Kollár--Miyaoka--Mori--Takagi, and later Peternell.

See also Theorem 1.2 in Q. Xie On Pseudo-Effectivity of the Second Chern Classes for Terminal Threefolds'.

• Actually, for fano 3-folds, this is a theorem of Kollar—Miyaoka—Mori—Takagi in 2000. Commented Apr 8, 2018 at 8:43
• Yes, you are right, it goes beyond much further. I'll update. Commented Apr 8, 2018 at 9:33
• That is not correct. You need to prove that the tangent sheaf is generically nef in order to use Miyaoka’s theorem. That is proved by KMMT for canonical Fano threefolds. Also Peternell showed smooth case, and Wenhao Ou for general cases. Commented Apr 8, 2018 at 12:40