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Let $X$ be a normal scheme ad $D = \sum_id_iD_i\subset X$ be a $\mathbb{Q}$-divisor such thay $K_X+D$ is $\mathbb{Q}$-Cartier. Let $f:Y\rightarrow X$ be a log resolution of the pair $(X,D)$ and let us write: $$K_Y = f^{*}(K_X+D)+\sum_{j}b_jE_j-\widetilde{D}$$ where the $E_j$'s are $f$-exceptional and $\widetilde{D}$ is the total transform of $D$. In the definition of log canonical pair we require that $b_j\geq -1$ for any $j$. My question is the following: do we require $0\leq d_i\leq 1$ as well?

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  • $\begingroup$ My understanding is that, in general, there is no hypothesis about the $d_i$. $\endgroup$ – Jason Starr Dec 8 '14 at 18:34
  • $\begingroup$ I think so. However in this paper arxiv.org/pdf/math/0205009v1.pdf at the end of Problem $7.1$ (pag 32) the author claims that to require that the pair is log canonical implies $0\leq d_i\leq 1$ but I do not see how. $\endgroup$ – user58018 Dec 8 '14 at 18:44
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    $\begingroup$ Jason, in Kollár-Mori, they require that all the $b_j \geq -1$ for all resolutions (ie, it holds on all valuations). This guarantees that the $d_i \leq 1$. $\endgroup$ – Karl Schwede Dec 8 '14 at 23:48
  • $\begingroup$ @KarlSchwede: I was reading Koll'ar's "Singularities of Pairs". The inequality on $d_i$ there is not a hypothesis of the definition. But, as you explain, it seems to be a consequence of the definition. $\endgroup$ – Jason Starr Dec 9 '14 at 22:35
  • $\begingroup$ Yeah, this is confusing and it is probably not emphasized enough when the definitions are made. The point is that every divisor has a discrepancy, and for the ones in the boundary it is $(-1)$-times their coefficient. This also occurs in the definition of klt: the coefficients of the components of the boundary have to be strictly less than $1$. $\endgroup$ – Sándor Kovács Dec 12 '17 at 1:32
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I think at the very least you should require that $d_i \leq 1$ even if this is not clear from the literature.

If you don't require that $d_i \leq 1$, then lots of things don't work.

Problems with the given definition

  1. The definition of log canonical you gave is not independent of the choice of resolution.

Let me give an example. Set $X = \mathbb{A}^2$ and let $D = 3 \text{div}(x)$. Then $(X, D)$ is its own log resolution and so using your definition, the $b_j$'s are all zero and so this pair is log canonical. However, if $Y \to X$ is the blowup of the origin, then there is one exceptional divisor $E$ and $E$'s coefficient in $K_Y - f^*(K_X + D)$ is equal to $1-3 = -2$.

  1. If you require that your condition holds for all log resolutions (or even all birational maps $f : Y \to X$ with $Y$ normal) then that implicitly guarantees that the coefficients of $d_i$ are all $\leq 1$. Just blow up the points on $D_i$ with coefficients $> 1$ repeatedly. Note that in Kollár-Mori, they do require all resolutions / valuations.

  2. One really wants to say that $(X, D)$ is log canonical if and only if $(Y, -K_Y + \pi^*(K_X + D))$ is log canonical for any proper birational map $f : Y \to X$ with $Y$ normal. This also guarantees that $d_i \leq 1$.

  3. You need both $d_i \leq 1$ and $d_i \geq 0$ in order to guarantee the Kodaira-type vanishing theorems you rely on. For instance, if $(X, D)$ is LC and if $L$ is a line bundle such that $L-K_X-D$ is ample, then $H^i(X, L) = 0$ for $i > 0$ if your $d_i$ are in $[0, 1]$.

An alternate definition

Let me say that I prefer a slightly different way to setup the definition. Write

$$K_Y = f^*(K_X + D) + \sum_j b_j E_j$$

without subtracting off the strict transform.

First notice that this means that some of the $E_j$ will be non-exceptional (that's totally ok). Then

Definition $(X, D)$ is log canonical if the $b_j \geq -1$. $(X, D)$ is KLT if the $b_j > -1$.

This definition then directly forces the $d_j \leq 1$ (respectively $d_j < 1$ for KLT). Setup this way, everything is completely independent of the choice of resolution.

Note that I didn't require that the $d_j \geq 0$ (this depends on the application, but based on 3. above, it might be reasonable to allow negative $d_i$ in the definitions. I should note that allowing negative $d_i$ is probably slightly more problematic for LC than it is for KLT, but I've written enough already).

DLT and ?LT singularities

Note that DLT singularities are a bit different (and then the $d_i$ are assumed $\leq 1$). Then they are setup not to be independent of the choice of resolution. See the recent book by Kollár and Kovács on singularities for a discussion of what kind of ``resolutions'' DLT are independent of (or think about Szabo's characterization of DLT singularities).

There are other notions of log terminal in the literature too, which make other weaker requirements about requiring things for all or some resolutions (for instance, weakly log terminal).

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