MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a little problem to fully understand the next thing:

Let $F:\mathbb{P}^2 \dashrightarrow \mathbb{P}$ a rational map. Let $\sigma:\bar{\mathbb{P}}^2 \rightarrow \mathbb{P}^2$ be a composition of $\sigma$-processes resolving the indeterminacy points of the rational map $F$, so that $\bar{F} = F\circ \sigma$ is a morphism.

We denote by $\bar{Y}_z$ the fibers of the morphism $\bar{F}$. Let $\sigma^{-1}(L_{\infty}) = \bar{S}_{\infty} \cup \bar{Y_{\infty}} \cup(\bigcup L_i)$, where $\infty = \mathbb{P}^1 \setminus \mathbb{C}^1$, the curves $L_i$ are components of the fibers of $\bar{F}$, and the morphism $\bar{F}$ maps each of the irreducible components of the curve $\bar{S}_{\infty}$ onto $\mathbb{P}^1$.

My question is. What are exactly the sets $\bar{S}_{\infty}$, $\bar{Y_{\infty}}$ and $L_i$?.

[Edited] I forgot to define $L_{\infty} = \mathbb{P}^2 \setminus \mathbb{C}^2$

Thank you in advance.

share|cite|improve this question
Does not seem to be clear like this. Would be good to say what is $L_\infty$. – Jérémy Blanc Aug 29 '13 at 8:38
up vote 0 down vote accepted

It seems to me that the only way this can happen is if $L_\infty\subset \mathbb P^2$ maps to $\infty\in\mathbb P^1$.

So, $\overline Y_\infty$ consists of $\sigma^{-1}_*L_\infty$ (a.k.a., the strict transform of $L_\infty$) and possibly a few $\sigma$-exceptional curves that map to $L_\infty$ via $\sigma$.

The rest of the $\sigma$-exceptional curves that map to $L_\infty$ via $\sigma$ are contained in $\overline S_\infty$ and $\bigcup L_i$. The "essential" ones, that is, those who really cause $F$ to be not defined somewhere, i.e., those who map to the entire $\mathbb P^1$ are in $\overline S_\infty$, but these may lie over infinitely near points, so in order to "reach" them you might have to blow up points whose corresponding exceptional divisors get contracted by $\overline F$. Based on the little information you're giving, I don't think one can say more precisely which ones are in which group.

There is one more group of curves, the fibers of $F$ that intersect $L_\infty$ in a point (or points) where $F$ is not defined. The strict transforms of these also end up among the $L_i$.

share|cite|improve this answer
Thanks for the answer. An additional condition is to assume that the generic fiber $Y_z$ is connected. – Geri Aug 30 '13 at 3:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.