2
$\begingroup$

I have recently taken interest in Mori's Minimal Model Program (MMP) and I struggle to figure out why it stops when the canonical divisor $K_X$ of our variety $X$ is nef.

For now, I have understood the following:

  • If $X$ is a normal and projective variety with terminal singularities whose canonical divisor $K_X$ is not nef, then the Cone Theorem tells us that there are rational curves $C$ that are $K_X$-negative and that generate an extremal ray in the closed cone of curves $\overline{NE}(X)$. Moreover, if $g: X \rightarrow X'$ is a divisorial contraction of such extremal ray, then $X'$ is also a normal and projective variety with terminal singularities.

  • If $X$ is moreover nonsingular and $2$-dimensional, then the contraction $g:X \rightarrow X'$ of a $K_X$-negative extremal ray induces a variety $X'$ that is nonsingular as well.

  • If $Y$ is a (smooth projective) $2$-dimensional variety whose canonical divisor $K_Y$ is nef, then (via the Adjunction Formula) $Y$ has no $(-1)$-curves, and therefore (via Castelnuovo's Theorem) there is no birational morphism $\mu: Y \rightarrow Y'$ (where $Y' \neq Y$ is a smooth projective surface). A (smooth projective) surface with nef canonical divisor is therefore minimal.

My question is the following: In the general case, if $Y$ is a minimal model (i.e. a projective variety with terminal singularities and whose canonical divisor $K_Y$ is nef),then does that mean that there is no birational morphism $\mu: Y \rightarrow Y'$(where $Y' \neq Y$ is a projective variety $Y'$ with terminal singularities)?

$\endgroup$

1 Answer 1

3
$\begingroup$

In general, I think the most we can say is that if $\mu: Y \to Y'$ is such a morphism, then $\mu$ is small; it has no exceptional divisors.

In this situation, $$K_{Y/Y'} := K_Y - \mu^* K_{Y'} = \sum a_E E$$ where this sum ranges over all the exceptional divisors of $\mu$. Moreover $a_E > 0$ for all $E$, since $Y'$ has terminal singularities. This divisor is $\mu$-nef because if $C \subset Y$ is a curve contracted by $\mu$, then $(K_{Y/Y'}).C = K_Y.C \geq 0$, since $Y$ is a minimal model.

By the negativity lemma, we know $- \sum a_E E$ is effective, which can only be the case if $\mu$ has no exceptional divisors.

To be honest, I'm having trouble finding examples of such small morphisms between minimal models but I strongly suspect they exist in general. (If I find one which is easy to describe I'll edit it in.)

However, there are some settings where what you say holds.

For example, suppose there is a curve $C \subset Y$ contracted by $\mu$ so that $K_Y.C > 0$. Then, $K_{Y'}$ is not $\mathbb{Q}$-Cartier. Indeed, if $m > 0$ is so that $m K_Y$ and $mK_{Y'}$ are Cartier, then $\mu^* m K_{Y'}$ and $m K_Y$ agree outside of a codimension $\geq 2$ set, which implies they are linearly equivalent. This is absurd since $(\mu^* m K_{Y'}).C = 0$.

As such, if $K_Y$ is ample then the exceptional set of any such $\mu$ would have to be empty, so $\mu$ would have to be an isomorphism.

$\endgroup$
1
  • $\begingroup$ I apologize for the edit and unedit. I thought I came across an example today but I was mistaken: the example did not have positive canonical bundle. $\endgroup$
    – Daniel
    Commented Jun 2 at 17:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .