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?

1$\begingroup$ I am not sure what you call "multiplicity", but by all reasonable definitions a $(1)$curve has multiplicity 1 (if $L=mD$, $L^2=1$ implies $m=1$). $\endgroup$– abxCommented Jul 18, 2022 at 19:10

1$\begingroup$ @abx I think the setting is that the class of the fiber is given by $m_1 D_1+ m_2 D_2 + m_3 D_3 + n L $, where $D_1, D_2, D_3, L$ are irreducible (and reduced), and the goal is to calculate $n$. $\endgroup$– Will SawinCommented Jul 18, 2022 at 19:13
2 Answers
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.

1$\begingroup$ Thanks for your answer @Dori Beijleri. I have a confision though. If the solutions comes to be all zeroes then what does that mean? what about infinitely many solutions? $\endgroup$– totaCommented Jul 18, 2022 at 21:26

1$\begingroup$ @tota If there are no solutions this means that the configuration cannot appear as a configuration of curves in the fiber of a smooth fibered surface. If there are infinitely many solutions it means that we need some extra input to actually solve the problem. I'll add this in another edit. $\endgroup$ Commented Jul 18, 2022 at 23:03

1$\begingroup$ In fact if the system is consistent it will always have infinitely many solutions because if $F$ is the given fiber then $dF$ is also a solution. So if you want to compute the multiplicities for a higher genus fibration you need to know something about the minimal degree of local multisections and what fiber components they can pass through. See for example type $mI_n$ fibers on an elliptic surface. $\endgroup$ Commented Jul 19, 2022 at 2:33


$\begingroup$ @DoriBejleri: Hi, Dori. In this situation it is not possible to have multiple fibers, see my answer below. $\endgroup$– ritaCommented Jan 7 at 17:41
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.