# Minimal model which is necessarily singular

I was told during a summer school on the MMP a nice example (which I have also mentioned here on MO) that I'm not able to figure out anymore.

The example (due, I think, to Miles Reid) is a smooth compact threefold $X$ such that the number of sections of $\mathcal{O}_X(m K_X)$ grows like $m^3/4$ (if I recall correctly). The nice thing about this example is the following.

Assume $X$ is birational to a smooth variety $Y$ such that $K_Y$ is nef. Then sections of $K_Y$ grow in the same fashion, in particular $K_Y$ is big, so by Kawamata-Viehweg vanishing we have $h^0 (Y, \mathcal{O}_Y(m K_Y)) = \chi(Y, \mathcal{O}_Y(m K_Y)) \sim m^3/4$, and by Riemann-Roch we find $K_Y^3 = 3/2$, which is not possible, since that number must be integer.

So, if one wants to have a minimal model for $X$, one has to allow singular varieties into the picture.

Can anyone tell me how the variety $X$ is obtained (or another example in a similar flavour)?

EDIT: As I wrote in the comments to VA's answer, I'm looking for an elementary example, where $\mathcal{O}_X(m K_X)$ can be computed and compared to the Riemann-Roch expansion, in order to have a completely intersection-theoretic argument. In particular I'd like to avoid using the concepts of canonical and terminal singularities, since I view this example mainly as a motivation to introduce exactly those concepts. It would also be nice if one could directly find a smooth $X$, rather then using Hironaka to resolve a singular variety with fractional $K_Y^3$ (which one secretly knows is terminal).

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Minimal models of smooth varieties of general type differ by flops, and there are finitely many of them. In dimension 3, a flop does not change the type of the singularity. So if one minimal model is singular then all of them are.

So take any singular 3-fold with terminal singularities and ample $K_X$, and that would be an example of a minimal model which is always singular. E.g., how about starting with $X'=$ the hypersurface $x_1x_2-x_3x_4=0$ in $\mathbb P^4$, and take $X$ to be the double cover of $X'$ ramified in a smooth divisor $D$ of degree $>6$. Let $Y$ be a smooth model of $X$, for example its desingularization.

To obtain an example with fractional $K_X^3$, start with a singular 3-fold $X'$ with terminal singularities with fractional $K_{X'}^3$ and let $X$ be a double cover, as the above. (You want $2K_{X'}^3$ to be fractional in this case, since $K_X^3 = 2(K_{X'}+\frac12 D)^3$.)

There are many choices for $X'$. For example, an appropriate weighted projective space works. If you are lazy to do it yourself, search for Reid's list of 95 Fano 3-folds which are complete intersections in weighted projective spaces. Most of them have fractional $K_{X'}^3$.

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I don't think this answer is in the same spirit, since it assumes one already knows about the singularities which appear in the MMP. Instead, the example I'm thinking of can actually be used to motivate the idea that one cannot do the MMP in the smooth setting (but not all singularities can be accepted, since one needs some intersection theory to say that the canonical is nef). –  Andrea Ferretti Apr 12 '10 at 18:13
I agree, knowledge is a dangerous thing to have. –  VA. Apr 12 '10 at 20:39
I sincerely don't understand your ironic comment. I have asked for an intersection-theoretic argument and you gave a different one. So I have asked for something different. There must be some motivation to introduce canonical, terminal singualrities and so on. So it would be nice to see this example BEFORE learning the machinery you mention (which, by the way, I more or less know). As I said above, it is not even a priori obvious that the right setting for the MMP must include singular varieties, for instance for surfaces this is not true. –  Andrea Ferretti Apr 13 '10 at 8:33
I think VA's accepts it is dangerous to "know the singularities which appear," not that his knowledge of the theory is dangerous. –  quim Apr 13 '10 at 9:19
Andrea: yes, my comment was ironic. I think I understand what you are after: an example of a smooth 3-fold for which $h^0(nK_Y)$ grows like $dn^3/3!$ with $d=1/2$ or some other fractional number. I gave you such an example indirectly; you want something where the statement could be checked easily, without 3-fold theory. Oh well, a search of some didactic papers by Miles Reid may yield such an example. Good luck. –  VA. Apr 13 '10 at 16:37

@Andrea: Perhaps this example would be a good motivation (see also this post). I am referring to your comment above to VA's answer.

I think this is example is due to Iitaka, or someone from his school, in any case it predates Miles Reid. Take a $3$-dimensional abelian variety $A$ and mod out by the involution $(−1)$. Resolve the resulting $64$ double points and call the result X. Then it is relatively easy to prove that $X$ is not birational to a smooth projective variety with a nef canonical bundle (the point is that you have to blow down the exceptional divisors over these $64$ double points, so the minimal model will be $A$ mod $(-1)$). I believe that at the time this example was thought of as proof that minimal models did not exist in higher dimensions, but then Reid and Mori realized that it only means that minimal models need not be smooth.

I think the right way to think about this is that minimal models have no worse than terminal singularities. It turns out that terminal singularities are smooth in codimension 2, so in particular a 2-dimensional terminal singularity is actually smooth. So, one could argue that even minimal models of surfaces have terminal singularities, that is, that's the natural class of singularities for a minimal model. It just so happens that in dimension 2, these singularities are indistinguishable from smooth points.

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Thank you very much, this example looks much more explicit. I will try to work out the details. –  Andrea Ferretti Oct 15 '10 at 14:25