1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Does not seem to be clear like this. Would be good to say what is $L_\infty$. $\endgroup$ Aug 29, 2013 at 8:38

1 Answer 1

0
$\begingroup$

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 $L_i$.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. An additional condition is to assume that the generic fiber $Y_z$ is connected. $\endgroup$
    – Geri
    Aug 30, 2013 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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