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Let $(X,\Delta)$ be a log pair (we assume the coefficients of $\Delta\leq 1$, but could be negative rationals), and I use the definitions in the book of "Birational Geometry of Algbebraic varieties" (Page 56, Def. 2.34; Page 58, Def. 2.37) for "terminal, canonical, klt, plt, lc, dlt" singularities.

My first question is: which of these six singularities do not depend on the resolution of singularities?

To be precise, this means: if one happen to have a log resolution: $Y \to X$, and the discrepancy for this resolution satisfy the requirement for the corresponding singularities, then for all the log resolutions, the discrepancy satisfy the requirement.

I try to use the standard argument that one dominant two resolutions by the third one. However, I always feel I might miss something (because the argument goes through when the coefficients of $\Delta > 1$, in which case, all the statement are false), below is what I did:

Suppose we try to "show" lc doesn't depend on the resolution. Let $\pi: Y_1 \to X$ be the log resolution, and the coefficients of the exceptional divisors $E_1$ are all $\geq -1$ in the log reamification formula. Let $\theta: Y_2 \to X$ be another log resolution. Finally, let $p:W \to Y_2, q: W \to Y_1 $ be a common resolution where the exceptional divisors and the strict transform of $\Delta$ (which I denote by $\Delta_W$) have snc. Then, we have $$K_W + \Delta_W \sim q^*\pi^*(K_X+\Delta) + q^* E_1 + F_1,$$ where $E_1, F_1$ are some exceptional divisors of $\pi, q$ respectively. We can write $$q^*E_1 = q_*^{-1}E_1 + Exc_q(q^*E_1)$$ where $ q_*^{-1}E_1$ is the strict transform of $E_1$, and $Exc_q(q^*E_1)$ belongs to the exceptional divisor of $q$. One can do the same thing for $W \to Y_2 \to X$, and putting them together we have $$q_*^{-1}E_1 + Exc_q(q^*E_1) + F_1 \sim \theta_*^{-1}E_2 + Exc_p(p^*E_1) + F_2.$$ Because every divisor involved in the above expression are exceptional divisors of $p \circ \theta = q \circ\pi$, the numerically equivalence implies equality of divisors. Moreover, because the coefficients on the left are all $\geq -1$ (I am a little worry about $Exc_q(q^*E_1)$, but I think the assumption on the snc of divisors would imply its coefficients $\geq -1$ ), this implies the same thing on the right, and particularly for $E_2$.

My second question is:

Assume the coefficients of $\Delta$ is in $[0,1]$, then what are the differences between "plt" and "dlt"? I didn't realize the subtlety of $dlt$ compared to other type of singularities...

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I guess I can take a stab at this. Certainly we are talking about log resolutions, ie that $\pi : Y \to X$ is proper and birational, $Y$ is regular, and $\text{exc}(pi) \cup \pi^{-1}_* \Delta$ has simple normal crossings. Then:

  • KLT is independent of the log resolution

  • LC is independent of the log resolution.

  • PLT is independent of the log resolution as long as you assume that the resolution also separates the birational transform of all the components of $\lfloor \Delta \rfloor$. In other words it's independent of the resolution as long as that resolution is also an embedded resolution of the components of $\Delta$.

  • DLT is definitely not independent of the resolution, however see the recent book on singularities by Kollár (with contributions by Kovács), for discussions of types of resolutions that one can read off DLT from, also see the work of Szábo.

  • Canonical is not independent of the resolution. Consider the pair $(X, \Delta)$ where $X = \mathbb{A}^2$ and $\Delta = (1-\varepsilon) L_1 + (1 - \varepsilon) L_2$. Then $(X, \Delta)$ is its own resolution, and so you might suspect it is canonical (or terminal) but of course blowing up the origin gives you a divisor with discrepancy $1 - 2 \cdot ( 1 - \varepsilon) = -1 + 2 \varepsilon < 0$ for $0 < \varepsilon \ll 1$. Of course, in the case that $\Delta$ is an integral Cartier divisor and $K_X$ is Cartier, then it is independent of the choice of resolution. Now, what is true is that there are only finitely valuations corresponding to divisors with discrepancy $\leq 0$ for a KLT (and also canonical) pair. Hence it is always enough to go to log resolutions where every divisor, exceptional or not, with discrepancy $\leq 0$ is disjoint from every other divisor with discrepancy $\leq 0$.

  • Terminal is not independent of the log resolution (indeed, the same example works). The same argument should also work here in terms of finding sufficiently big resolutions as we did for canonical.

Your computation

I think you are making it too hard. The point for KLT / LC is that if $\pi: Y \to X$ is any birational map from normal $Y$ then if $(X, \Delta)$ is KLT/LC, then $(Y, -K_Y + \pi^*(K_X + \Delta) )$ is also KLT/LC. (KLT and LC are rigged so that this happens, and the converse works too).

This means that if you take a further resolution $\rho : W \to Y$ then we see that the pair $$ \big(W, -K_W + \rho^*(K_Y - K_Y + \pi^*(K_X + \Delta)) \big) = \big(W, -K_W + \rho^* \pi^*(K_X + \Delta) \big) $$ In particular, the fact that $(W, -K_W + \rho^* \pi^*(K_X + \Delta) )$ has KLT/LC singularities implies that $(Y, -K_Y + \pi^*(K_X + \Delta))$ also has KLT/LC singularities. And so it's easy to see that things are independent of the resolution.

PLT vs DLT

A simple normal crossings pair with coefficients in $[0,1]$ always has DLT singularities, it only has PLT singularities if each divisor with coefficient $=1$ is disjoint from every other such divisor.

Basically, the point of DLT is that it is a more controlled version of LC (many objects which you produce via the MMP which a priori have LC singularities also have DLG singularities). In particular, the worst singularities of a DLT variety occur where the pair is a SNC pair, and so while these are technically more severe, they can be very easy to deal with.

Hopefully this helps.

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  • $\begingroup$ Dear Karl: Thank you so much for your constant help!! I think I understand most part of your answer. However, there is one place I was unable to get straight: you claim $(X,\Delta)$ and $(Y, -K_Y + \pi^*(K_X + \Delta))$ would be KLT/LC simultaneously. I guess this is related to Lemma 2.30 in Koll\'ar and Mori's book, however, in that lemma, they do not assume anything about the discrepancy. I know I must make something wrong, because, otherwise, use the same argument as you posted, one can show canonical/terminal singularity does not depend on the resolution. $\endgroup$
    – Li Yutong
    Commented Aug 12, 2014 at 8:11
  • $\begingroup$ Dear Li Yutong, I don't have Kollár-Mori in front of me, what's the idea of 2.30? The reason that KLT/LC are ok is because there is no distinction between exceptional and non-exceptional discrepancies (note the discrepancy of a non-exceptional divisor is just the coefficient of $\Delta$ along it). In particular, what is exceptional over $X$ is not always exceptional over $Y$. $\endgroup$
    – Karl Schwede
    Commented Aug 12, 2014 at 14:28
  • $\begingroup$ The Lemma 2.30 is "Let $Y \to X$ be a proper birational morphism between normal varieties. Let $\Delta_Y, \Delta_X$ be $\mathbb{Q}$-divisor such that $K_Y + \Delta_Y \equiv f^*(K_X + \Delta_X)$ and $f_*{\Delta_Y}=\Delta_X.$ Then, for any divisor $F$ over $X$, the discrepancy $a(F,Y,\Delta_Y) = a(F,X,\Delta_X)$." I don't see where the assumptions rely on KLT/LC particularly. Where did I get wrong? $\endgroup$
    – Li Yutong
    Commented Aug 13, 2014 at 0:10
  • $\begingroup$ Hi, the problem is that for canonical or terminal, you exclude certain divisors, the nonexceptional divisors. As you blowup, more and more valuations become nonexceptional. In particular, if you start with the example I gave you on $\mathbb{A}^2$, the lines have discrepancy $-1 + \varepsilon < 0$. But this doesn't stop things from being canonical since they are non-exceptional. The blowup at the origin shows it isn't canonical. On the other hand, for every KLT $(X, \Delta)$ there is a log resolution $Y \to X$ with $(Y, \Delta_Y = -K_Y + \pi^*(K_X + \Delta))$ having canonical singularities. $\endgroup$
    – Karl Schwede
    Commented Aug 13, 2014 at 0:28
  • $\begingroup$ Dear Karl, Thank you again, I think I understand where I did wrong! $\endgroup$
    – Li Yutong
    Commented Aug 13, 2014 at 8:28

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