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Let $X_n = Bl_{p_1,...,p_n}\mathbb{P}^2$ be the blow-up of $\mathbb{P}^2$ in $n$ general points $p_1,...,p_n\in\mathbb{P}^2$.

Let $f_i:\mathbb{P}^{2}\dashrightarrow\mathbb{P}^1$ be the linear projection from the point $p_i\in\mathbb{P}^2$. Then $f_i$ lifts to a fibration $\widetilde{f}_i:X_n\rightarrow\mathbb{P}^1$.

So we get $n$ fibrations $\widetilde{f}_i:X_n\rightarrow\mathbb{P}^1$ for $i=1,...,n$. Now let $g:X_n\rightarrow\mathbb{P}^1$ be a fibration of $X_n$ on $\mathbb{P}^1$. Is it true that $g$ must be one of the $\widetilde{f}_i$'s?

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  • $\begingroup$ No. If $n=2$, you can blow down the line $L$ joining the two points to get a $\mathbb{P}^1\times\mathbb{P}^1$ and thus get two maps to $\mathbb{P}^1$. You can see that since $L$ goes to a point, these can not be one of the $\tilde{f}$'s. $\endgroup$
    – Mohan
    Commented Jul 28, 2015 at 23:47
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    $\begingroup$ I do not understand your argument. In the case $n=2$ there are just two fibrations to $\mathbb{P}^1$. The strict transform of $L$ is just a common irreducible component of special fibers of the $\widetilde{f}_i$'s. It is easy too see this fact thinking $\mathbb{P}^2$ blown-up in two points as $\mathbb{P}^1\times\mathbb{P}^1$ blown-up in one point. $\endgroup$
    – Puzzled
    Commented Jul 29, 2015 at 0:28
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    $\begingroup$ @S_Z_S: Your picture looks familiar, but didn't you have a different username before? $\endgroup$
    – Jason Starr
    Commented Jul 29, 2015 at 1:56
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    $\begingroup$ You know, Polizzi and I did give correct answers to one of your earlier questions, yet you did not accept either of our answers. $\endgroup$
    – Jason Starr
    Commented Jul 29, 2015 at 2:59
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    $\begingroup$ I think another nice counterexample comes from cubic surfaces. Any line on a cubic gives a fibration onto $\mathbf P^1$. Fixing 6 skew lines $L_1,\ldots,L_6$ identifies the surface as a blowup of $\mathbf P^2$ with the $L_i$ as exceptional curves. Each of the $L_i$ is a section for the fibration coming from any of the other $L_j$. But now let $L$ be a line which intersects $L_1$ say: then $L_1$ is contained in a fibre of the fibration corresponding to $L$, so that fibration must be different from the other 6. $\endgroup$
    – Lazzaro Campeotti
    Commented Jul 29, 2015 at 19:08

2 Answers 2

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As Mohan indicated, this is not always going to be true. Let $n$ be $4$. The group $\text{Aut}(\mathbb{P}^2)$ acts transitively on the set of ordered $4$-tuples $(p_0,p_1,p_2,p_3)$ such that no $3$ of these points are collinear. Thus, up to projective linear transformation, assume the points are $p_0=[1,1,1]$, $p_1=[1,0,0]$, $p_2=[0,1,0]$, and $p_3=[0,0,1]$.

Denote the homogeneous coordinates on $\mathbb{P}^2$ by $[X,Y,Z]$. Then consider the rational function $$g = X(Z-Y)/Z(Y-X).$$ This gives a rational map $\mathbb{P}^2\dashrightarrow \mathbb{P}^1$ that extends to a regular, projective, flat morphism $$g:X_4\to \mathbb{P}^1.$$ None of the $4$ exceptional divisors is contained in a fiber of $g$, thus it equals none of $f_1$, $f_2$, $f_3$, nor $f_4$.

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As in Jason's answer take $n = 4$. The fibrations $\widetilde{f}_i$'s for $i=1,..,4$ are induced by the linear systems $\mathcal{L}_i$ of the lines through $p_i$ for $i=1,...,4$.

Now, take the linear system of $\mathcal{C}_{1,2,3,4}$ of conics through $p_1,...,p_4$. This is a pencil and therefore induces a rational map $g:\mathbb{P}^2\dashrightarrow\mathbb{P}^1$. Once you blow-up the four base points you get a morphism $\widetilde{g}:X_4\rightarrow\mathbb{P}^1$. Since $\widetilde{g}$ and the $\widetilde{f}_i$'s are induced by non-equivalent linear systems we have $\widetilde{g}\neq \widetilde{f}_i$ for any $i=1,...,4$.

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    $\begingroup$ You are saying the same thing as what I said . . . $\endgroup$
    – Jason Starr
    Commented Jul 29, 2015 at 17:29
  • $\begingroup$ Yes, I guess it's another way to say the same thing. Indeed, your $g$ is given by conics. $\endgroup$
    – Puzzled
    Commented Jul 29, 2015 at 18:14

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