Let $(X,D)$ be a pair and $f:Y\rightarrow X$ a log resolution. Write
$$
K\_Y + \widetilde{D} = f^{\*}(K\_X) + \sum\_{i}a\_iE\_i
$$
where $\widetilde{D}$ is the strict transform of $D$. I found the following definition:


the pair $(X,D)$ is $(t,c)$ is $X$ is terminal and $(X,D)$ is canonical meaning that $a\_i\geq 0$ for all $i$.


Now, assume that $X$ is smooth and $D = D\_1 + \dots + D\_r$, where the $D\_i$ are prime divisors, is simple normal crossing. Then the identity is a log resolution and hence $(X,D)$ is $(t,c)$. Is this correct? Or must one interpret the absence of exceptional divisors as the possibility of having arbitrarily negative discrepancies?


I am asking since this does not seem to match the arguments in a paper that I am reading. Are there different definitions of $(t,c)$ pair?


For instance is $\mathbb{P}^2$ with the divisor $D = \{xyz=0\}$ a canonical pair?


Thanks a lot.



Definition of canonical pair
You are using the wrong equation to compute discrepancies. It should be
$$ K\_Y = f^\*(K\_X + D) + \sum a\_E(X,D) E $$
where the $E$ are not all necessarily exceptional.


For example if $(X,D)$ is already smooth and simple normal crossing then
$$ K\_X = \operatorname{id}\_X^\*(K\_X + D) - D $$
and so every irreducible component $D\_i\subset D$ has discrepancy $a\_{D\_i}(X,D)=-1$. Thus the simple normal crossing pair $(X,D)$ is *log canonical*, and not canonical.


