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Hi,

My question seems to be closely related to that one : What strict resolutions of singularities are needed? (and it is perhaps even included in it, but I am not so sure).

Let $X$ be a Gorenstein projective variety with canonical singularities. Does there exist a sequence :

$ X_N \rightarrow ... X_i \rightarrow X_{i-1} ... \rightarrow X_0 = X$ such that:

i) $X_N$ is smooth,

ii) $X_i \rightarrow X_{i-1}$ is the blow up along a smooth subvariety $Z_{i-1} \subset X_{i-1}$

iii) The exceptional divisor of $X_i \rightarrow X_{i-1}$ has only Log-canonical (or perhaps DuBois?) singularities?

Even if Log-canonical or DuBois is impossible, is there any regularity condition one can impose on these exceptional divisors?

Thanks.

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  • $\begingroup$ Franz, Can I ask why you want ii)? Blowing up a smooth point/subvariety on a singular variety is highly-non-controllable (unlike the case with a smooth ambient space). $\endgroup$
    – Karl Schwede
    Commented Apr 5, 2011 at 15:09
  • $\begingroup$ Well in Fact, I would like to know if Kodaira hold on the exceptional divisor (perhaps should I have formulated my question like this...). I looked a bit in the litterature and, apparently, DuBois is the right thing to have Kodaira. Do you have any idea about that? $\endgroup$
    – Franz
    Commented Apr 6, 2011 at 19:46
  • $\begingroup$ @Schwede. What were you referring to in your comment? I would like to understand. What is the sense in which they are highly non-controllable? $\endgroup$
    – O.R.
    Commented Jun 5, 2011 at 7:11

1 Answer 1

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You can try the following.

EDIT: At this point, I don't think this can work in full generality. However, it will give you some bounds on the singularities, which is better than nothing.

Take a resolution of singularities obtained by blowing-up a sequence of smooth centers $X_N \to X_{N-1} \to \dots \to X_1 \to X_0 = X$. Fix an $i$ you care about. Then by Corollary 1.4.3 of LINK: BCHM, you can contract all the exceptional divisors of $X_N \to X_i$ and obtain $X_i'$, something that agrees with $X_i$ in codim 1 (ie, up to flips/flops).

Then you could look at the largest coefficient you can stick on $E_i$ the (possibly non-prime) exceptional divisor of $X_i' \to X_{i-1}'$ (maybe that's only a rational map, I'm not sure) as you form this relative minimal-model. If you can stick a coefficient of $1-\varepsilon$ (for $\varepsilon > 0$) in this process then that should force $E_i$ to have nice singularities. In particular they should have semi-log canonical singularities and thus Du Bois singularities by LINK: KK (LC-singularities are Du Bois). The point is that a limit of klt singularities should be log canonical, and if you have a log canonical pair with (some) divisors of coefficient 1, then those divisors should be ``semi-log canonical'' for a sufficiently general notion of semi-log canonical. In particular, it certainly follows from LINK: KK that such singularities are automatically Du Bois.

However, even if you can't manage the $1-\varepsilon$ bound, the bounds you can put on the coefficient might be useful to bound the singularities that appear (ie, you would effectively be bounding the log canonical threshold).

Again, this is very vague and I don't know if the details can be made to work.

Maybe Sandor will have some better thoughts.

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  • $\begingroup$ Hi Karl! Thanks for your answer. I will think about it but you could you explain just a bit more the second part of your answer (I am not very used with relative divisorial log terminal model...)? Thanks again for your time. $\endgroup$
    – Franz
    Commented Apr 4, 2011 at 17:31
  • $\begingroup$ Franz, I should have said relative minimal dlt model. There's some discussion of this in the KK paper I link above. In particular, see theorem 3.1 in that paper (on the existence of such things). Anyway, I think I cleaned up my answer a bit. The first thing I was worried about was not really a worry. $\endgroup$
    – Karl Schwede
    Commented Apr 4, 2011 at 22:37
  • $\begingroup$ Thanks very much! I will look at the Kollar-Kovacs' paper very carrefully so! $\endgroup$
    – Franz
    Commented Apr 5, 2011 at 7:05
  • $\begingroup$ After additional thought, I don't think this can work in full generality, but it might give you some tools in the specific case you are looking at. $\endgroup$
    – Karl Schwede
    Commented Apr 5, 2011 at 15:07

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