# Possible number of components of anticanonical sections of projective manifolds

Let $$M$$ be a connected projective complex manifold with a smooth anticanonical divisor $$D$$ ($$D \sim -K_M$$). Let $$k$$ be the number of components of $$D$$.

Some cheap thoughts give:

If $$M$$ is a Fano manifold of dimension higher than one, $$k=1$$.

If $$M=\mathbb P^1$$ or $$M= \mathbb P^1 \times X$$, where $$X$$ is a projective manifold with trivial canonical class, then $$k=2$$.

Is it possible to have $$k$$ bigger than two? Is $$k$$ bounded?

• I have this idea involving a smooth plane cubic (i.e. anticanonical divisor in $\mathbb P^2$) and blowing up a bunch of points not on the cubic, but every time I try to compare this to the usual thing where the points are on the cubic I get turned around. But I'm guessing you've already thought of this and I'm wrong, hence the need for the question, because otherwise you can achieve any $k$ by blowing up $k-1$ points. But I thought I should mention this anyway, just in case. – Tabes Bridges May 1 at 6:35

(Let me turn my comment into an answer to give a correct attribution.)

Yes, in the setup you describe, the maximum number of connected components of the divisor $$D$$ is 2. This is related to the so-called connectedness principle in birational geometry.

The statement that you want follows from Proposition 5.1 of Log canonical singularities are Du Bois by Kollár and Kovács. They prove in particular that if $$(Y,\Delta)$$ is an lc pair such that $$K_Y + \Delta$$ is $$\mathbf Q$$-linearly equivalent to 0, then the non-klt locus $$\operatorname{nklt}(Y,\Delta)$$ has 1 or 2 components.

In your case, taking $$\Delta=D$$, then $$\operatorname{nklt}(Y,\Delta)$$ is just (the support of) $$D$$, so this set has at most 2 connected components.

Here is how one can prove that the number of components is at most $$2$$ in the case of complex projective surfaces. So let $$X$$ be such a surface and let $$D_1\cup \ldots\cup D_k$$ be a smooth anti-canonical divisor with $$k>0$$.

(i) Since $$-K_X$$ is effective, we see that $$X$$ has Kodaira dimension $$-\infty$$. I.e. it is either a rational surface or an irrational ruled surface.

(ii) All components $$D_i$$ are elliptic curves (by adjunction formula, for example).

(iii) Suppose first that we are in the irrational ruled case and let $$F$$ be a fiber. Since through every point of $$X$$ a fiber passes, and all $$D_i$$'s are ellipitic curves, we see that $$D_i\cdot F>0$$. At the same time $$D\cdot F=-K_X\cdot F=2$$. So we have at most two divisors $$D_i$$.

(iv) Suppose now we are in the rational case. Then either $$X$$ is $$\mathbb CP^2$$ and $$D$$ is a smooth elliptic curve or again $$X$$ admits the structure of a $$\mathbb CP^1$$ fibration (over $$\mathbb CP^1$$), where we can reason as in (iii).

Case of higher dimension? What follows is speculative, since my Mori theory knowledge is close to zero. It seems to me that the above reasoning has chances to succeed in higher dimensions as well. At least, in case $$X$$ is a Mori fiber space, it looks to me that $$D$$ cuts out on a generic (Mori) fiber a divisor in the anti-canonical class with the same number of components. So we might proceed by induction.