ag.algebraic geometry - Possible singularities of the base of a Mori fiber space

Dear Eugene,

I think I can answer one of your questions, namely about $Z$ having rational singularities. It does and here is why:

$Z$ is indeed normal by construction.

Since $f$ is a Mori-contraction, $-K_X$ is $f$-nef and $f$-big. (In fact, since the relative Picard number is $1$ it is $f$-ample). I claim that $R^if_*mathcal O_X=0$ for all $i>0$. If $f$ were birational, this would be a simple consequence of Kawamata-Viehweg vanishing. In this case we need to work a little more.

First, by Kollár's torsion-freeness theorem (see [Yujiro Kawamata, Katsumi Matsuda, and Kenji Matsuki. Introduction to the minimal model problem. In Algebraic geometry, Sendai, 1985, volume 10 of Adv. Stud. Pure Math., pages 283-360. North-Holland, Amsterdam, 1987.] for the format needed here and see the references there for proofs) these sheaves are torsion-free. (Not in general, just in this case!).

Second, let $Usubseteq Z$ be a non-empty subset over which $f$ is flat. The fibers over $U$ are Fano varieties and hence $H^i(F,mathcal O_F)=0$ for $i>0$ for a fiber $F$. I probably should have taken a resolution to start with and then one would not have to worry about singularities, but at least for a general $F$ one can say that $(F,Delta_{|F})$ is klt and hence Kawamata-Viehweg vanishing applies, so we get the above vanishing. Anyway, then by Grauert's theorem [Hartshorne, III.12] it follows that $(R^if_*mathcal O_X)_{|U}=0$ for all $i>0$. However, we have already established that these sheaves are torsion-free, so if they are zero on a non-empty open set then they are zero everywhere.

OK, so we get that $R^if_*mathcal O_X=0$ for all $i>0$. Now we are only almost there because the original definition of rational singularities would require this for a birational morphism and $f$ is decidedly not birational. But this still implies that $Z$ has rational singularities by SJK: A characterization of rational singularities.

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Now, regarding whether $Z$ can be klt, canonical, etc. One simplification I can suggest is that you do not need the $Delta'$ there. If you find a $Delta'$ such that $(Z,Delta')$ is klt, dlt, lc, etc, then so is $Z$ since by the $mathbb Q$-factoriality $Delta'$ will be $mathbb Q$-Cartier. See [Kollár-Mori, 2.35].