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

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?

• 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 $m=1$).
– abx
Commented Jul 18, 2022 at 19:10
• @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 $n$. Commented Jul 18, 2022 at 19:13

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.

• 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?
– tota
Commented Jul 18, 2022 at 21:26
• @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. Commented Jul 18, 2022 at 23:03
• 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. Commented Jul 19, 2022 at 2:33
• Thanks again!! @Dori Bejleri
– tota
Commented Jul 19, 2022 at 6:03
• @DoriBejleri: Hi, Dori. In this situation it is not possible to have multiple fibers, see my answer below.
– rita
Commented 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.