Multiplicity of irreducible component of a singular fiber of a $\mathbb{P}^1$-fibration

Question

Let $X$ be a smooth projective surface and $f:X\to\mathbb{P}^1$ be a $\mathbb{P}^1$-fibration with a singular fiber consisting of a tree with three irreducible rational ($-2$)-curves $D_1$, $D_2$, $D_3$ and a ($-1$)-curve $L$ intersecting one of $D_i$'s. I want to know is it possible to calculate multiplicity of $L$ in the scheme theoretic fiber of this singular fiber? Is it possible to calculate multiplicity of the singular fiber from above information?

Answer 1

If we assume that the rational curves intersect in nodes, then we can compute it from the data of the dual graph. Since $L$ intersects only one of the $D_i$, then the $D_i$ necessarily form a chain, say $D_1 \cup D_2 \cup D_3$ and without loss of generality, there are two possibilities: 1) $L$ intersects $D_1$, or 2) $L$ intersects $D_2$. With that in mind we can compute the multiplicities of the curves in the fiber using the following basic lemma.

Lemma: If $F = f^*p$ is a scheme theoretic fiber, then $F.E = 0$ for all curves $E$ contained in a fiber of $f$.

Then writing our fiber $F = a_1D_1 + a_2D_2 + a_3D_3 + bL$, we have a system of equations $F.D_i = F.L = 0$. On the other hand, we know the self intersections $D_i^2 = -2$ and $L^2 = -1$, we know that $D_1.D_2 = D_2.D_3 = 1$ and $D_1.D_3 = 0$, and finally we know $L.D_i$ in each of the two cases. Using this we can solve completely for the $a_i$ and $b$ in each of the two cases.

Edit: I think the curves actually have to intersect in nodes under the assumption that $X$ is smooth. This is just a question about the reduced fiber, call it $C$, which is a tree of rational curves on a smooth surface $X$. The idea is to compute the contribution of the $\delta$-invariant of the singularity to the arithmetic genus.

Let's suppose that $p \in C$ is a singular point and let $C' \to C$ be the partial normalization at $p$. If $p$ lies on a unique irreducible component of $C$ then $p_a(C) - p_a(C') = \delta > 0$ which is impossible as $p_a(C) = 0$. Otherwise, $p$ is at the intersection of exactly two components of $C$ (by the assumption that $C$ is a tree). Then $C'$ is a disjoint union of two components $A$ and $B$ so $p_a(C') = p_a(A) + p_a(B) - 1 = -1$ and $p_a(C) - p_a(C') = \delta$ so we conclude that $\delta = 1$. Then its not hard to see that the node is the only plane curve with $2$ branches and $\delta = 1$.

Edit 2: It turns out that in the first case there are no solutions which means that such a configuration cannot appear as the fiber of a genus $0$ fibration on a smooth surface. We can see this geometrically as follows. Since $L$ is a $(-1)$-curve we can contract it to obtain a smooth surface that the fibration factors through, but then $D_1$ becomes a $(-1)$-curve which can contract to another smooth surface through which the fibration factors. Now $D_2$ becomes a $(-1)$ curve so after one more contraction we have a smooth surface with a fibration and a fiber that is set theoretically just $D_3$, but we also have that $D_3^2 = -1$ (after these three contractions) and this is impossible.

In the second case solving the linear system just yields that $a_2 = b = 2a_3$ and $a_1 = a_3$ so we still need one more input. This comes from Tsen's theorem which guarantees (at least over an algebraically closed field) that the fibration $f$ has a section. This means that the fiber $F$ contains at least one reduced component. Since the $a_i$ and $b$ are all positive integers, this gives us the unique solution $a_1 = a_3 = 1$ and $a_2 = b = 2$. We can achieve such a fiber by starting with a fiber that is just $\mathbb{P}^1$, blowing up once, then blowing up again at the new node we produced which gives us a fiber that looks like $D_1 + 2D_2 + D_3$, and finally blowing up a generic point on $D_2$ to obtain the $D_1 + 2D_2 + D_3 + 2L$ fiber.

Answer 2

I think Dori's answer can be simplified using adjunction on $X$. Using Dori's notation, one has $2g(F)-2=K_XF=-b<0$, so the only possibility is that $b=2$ and the general $F$ has genus 0.

This already answers the original question, but, as Dori did, one can describe the fiber precisely. Imposing $LF=0$ one gets $a_2=2$. Now blow down $L$ to get a fibered surface $X'$. The image $D_2'$ of $D_2$ is a $-1$-curve, the images $D'_1$, $D'_3$ of $D_1$ and $D_3$ are again $-2$-curves and $2D'_2+a_1D'_1+a_3D'_3$ is a fiber $F'$ of a fibration of genus 0. Imposing $D'_2F'=0$ one gets $a_1D'_1D'_2+a_2D'_3D'_2=2$. Assuming $D'_1D'_2>0$, there are only two possibilities: (1) $a_1=a_3=D'_1D'_2=D'_3D'_2=1$; (2) $a_1D'_1D'_2=2$, $D'_2D'_3=0$, and in this case $D'_3D'_1>0$ since the fiber $F'$ is connected. Imposing $F'D'_3=F'D'_1=0$ it is easy to see that case (2) leads to a contradiction. So we have exactly the fiber described at the end of Dori's answer.

Finally note that a fibration with rational fibers cannot have a multiple fiber $F=mA$, since one would have $A^2=0$, $m=2$ and $K_XA=-1$ but $A^2+K_XA$ is even, again by the adjunction formula.