# Kawamata-Log-Terminal pairs

Let $p_1,...,p_n\in\mathbb{P}^3$ be general points, and let $\Delta\subset\mathbb{P}^3$ be a general surface of degree $d$ with points of multiplicity $m_i$ at $p_i$ for $i = 1,...,n$.

Consider the blow-up $X$ of $\mathbb{P}^{3}$ at $p_1,...,p_n$ and the strict transform $\tilde{\Delta}$ of $\Delta$. For which $d,m_1,...,m_n,$ is $(X,\tilde{\Delta})$ a klt pair?

For instance $m_1 = ... = m_2 = 0,1$ will work, and I guess $m_1 = ... = m_2 = 2$ as well.

• What's the coefficient of $\Delta$? Obviously the pair will never be KLT if the coefficient is $\geq 1$. You could ask for PLT or LC. Mar 16, 2014 at 2:02

I'm not sure what you mean by the pair being klt here: if $\tilde{\Delta}$ appears with coefficient $1$, then it's not. On the other hand, it's sensible to ask whether you can find a boundary $\mathbb Q$-divisor $D$ numerically equivalent to $\tilde{\Delta}$ for which $(X,D)$ is klt. Or you could ask for $(X,\tilde{\Delta})$ to be dlt/lc/plt, as Karl Schwede suggests. In any case, I doubt you can really say much in general -- it's not even possible to say for which $d$ and $m_i$ the class is effective (cf. the Nagata conjecture on $\mathbb P^2$). I guess if the strict transform is basepoint free you're in business, but this probably isn't easy to read off from the numbers.
I'm not sure about your examples: even with $d=1$ and $m_1=m_2=1$ I don't think it's klt -- no matter how you write $\Delta$, it's a sum $\sum a_i P_i$, with $P_i$ a plane through the points and $\sum a_i = 1$, and the divisor you get when you blow up the line has discrepancy $-1$.
• You are right. My question is very vague. I will try to be more precise. Assume $\Delta$ is a surface with ordinary singular points of multiplicities $m_i$ at $p_i$. Then the strict transform $\tilde{\Delta}\subset X$ should be smooth. If $\epsilon$ is a small positive real number the pair $(X,\epsilon\tilde{\Delta})$ should be klt. In this case $Id_{X}$ is a log-resolution and we may write $K_X = Id_X^*(K_X+\epsilon\tilde{\Delta})-\epsilon\tilde{\Delta}$. Therefore the discrepancy is $-\epsilon > -1$. Is this correct? Thank you.
• @ggelli: If $\tilde\Delta\subset X$ is a smooth Cartier divisor, then $(X,\varepsilon\tilde\Delta)$ is klt for any $1>\varepsilon>0$ essentially by the argument you're providing. On the other hand, if you want to change or refine your question, you should do it by editing the question and not by posting a comment to an answer. Mar 16, 2014 at 22:37