5
$\begingroup$

A projective smooth variety $X$ is weak Fano if $-K_X$ is nef and big. We say that $X$ is log Fano is there exists a divisor $D$ such that $-(K_X+D)$ is ample and $(X,D)$ is Kawamata log terminal.

Is it true that weak Fano implies log Fano? What is an example of a variety which is log Fano but not weak Fano?

$\endgroup$
4
  • $\begingroup$ Blowing up of the plane at three collinear points? $\endgroup$ Apr 5, 2014 at 22:43
  • $\begingroup$ It seems to me that if $X$ is the blow-up of $\mathbb{P}^2$ at three collinear points, $-K_X$ is nef and big. So $X$ is weak Fano. On the other hand $Cox(X)$ is finitely generated and since $-K_X$ is big we conclude that $X$ is log Fano. I guess that taking $D$ to be the strict transform of the the line through the three points $-(K_X+\epsilon D)$ is ample for $0<\epsilon\ll 1$. $\endgroup$
    – user49214
    Apr 5, 2014 at 23:26
  • $\begingroup$ That is a good point. $\endgroup$ Apr 6, 2014 at 12:14
  • 1
    $\begingroup$ A smooth hypersurface of degree $n+1$ in $\mathbb{P}^n$ is Calabi-Yau, it is certainly not log Fano! $\endgroup$
    – abx
    Apr 8, 2014 at 6:36

3 Answers 3

2
$\begingroup$

That is true. Basically it is a consequence of the following fact:

Let $D$ be a nef and big divisor on an irreducible projective variety $X$. Then there exist an effective divisor $E$ and a rational number $0 <\epsilon\ll 1$ such that $D-\epsilon E$ is ample.

Proof: Let $D$ be a nef and big divisor. Since $D$ is big, by \cite[Corollary 2.2.6]{La}, there exist an ample divisor $A$, an effective divisor $E$, and a positive integer $k$ such that $kD\equiv A+E$. If $h>k$ we can write $hD\equiv (h-k)D+A+E$. The divisor $D^{'} = (h-k)D+A$ is a sum of a nef and an ample divisor. Therefore $D^{'}$ is ample. If $\epsilon = \frac{1}{h}$ we get that $$D-\epsilon E\equiv \epsilon D^{'}$$ is ample.

Now, it is enough to apply this to $D = -K_X$. What you get in the end is the follwing:

Let $X$ be an irreducible, projective variety with at most klt singularities. If $X$ is weak Fano then $X$ is log Fano.

$\endgroup$
4
$\begingroup$

As pointed out by the other answers, every smooth weak Fano is certainly log Fano. But you also asked for an example of a log Fano variety that is not weakly Fano.

Take any toric variety for which $-K_X$ is not nef.

More explicitly, rational ruled surfaces (ruled surfaces over $\mathbb{P}^1$) can have $-K_X$ big but not necessarily nef. For instance, consider the rational ruled surface with respect to $O_{\mathbb{P}^1} \otimes O_{\mathbb{P}^1}(e)$ for $e \gg 0$, this will not be weakly Fano (I think any $e > 2$ works actually). Check out the section in Hartshorne (or Lazarsfeld's Positivity book if you need background on ruled surfaces).

It's a fun exercise to explicitly work out what the $\Delta$ divisor is (can be) that makes $(X, \Delta)$ log Fano in the ruled surface case. You can also see explicitly why ruled surfaces over curves that aren't $\mathbb{P}^1$ cannot be log Fano.

$\endgroup$
1
  • $\begingroup$ Thank you very much. Yes, you are right, $e>2$ works. Here is the computation: the anticanonical divisor is $-K_{X_e} = -2C_0-(2+e)F$, where $C_0$ is the section and $F$ is the fiber. Therefore $-K_{X_e}\cdot C_0 = 2C_0^2+2+e = -e+2$, and $-K_{X_e}$ is not nef for $e>2$. $\endgroup$
    – user49214
    Apr 9, 2014 at 13:31
2
$\begingroup$

I believe that all weak Fano varieties are log Fano. Basically, you can find an effective divisor $D$ such that $-K_X-\epsilon D$ is ample for arbitrarily small $\epsilon$. For a reference, see the proof of proposition 2 of https://www2.bc.edu/dawei-chen/AC.pdf.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.