# Before starting

• Survey of past and current work
• Partly joint with J.I. Cogolludo

# Two points determine a line.

$$a_{0,0}z+a_{1,0} x + a_{0,1}y$$

# One point determines a pencil of lines

$$a_{0,0}z+a_{1,0} x + a_{0,1}y$$

# Five (generic) points determine a conic.

$$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$

# Four points determine a pencil of conics

$$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$

# Four points determine a pencil of conics

$$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$

# Nine points determine a cubic

$$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$

# Eight points determine a pencil of cubics

$$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$

# Eight points determine a pencil of cubics?

$$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$

Theorem (Cayley-Bacharach): - If a cubic passes through eight intersections points of two other cubics, it also passes through the ninth.

# Question:

• When is there a pencil of curves?:
• Two curves always generate a pencil
• More than three are in a pencil iff every three of them are.
• So the question really is:
• What conditions should three curves satisfy to be in a pencil?

# First result

Noether fundamental theorem ($$af+bg$$):

Let $$f,g,h$$ be three homogenous polynomials in $$\mathbb{C}[x,y,z]$$. Being $$f$$ and $$g$$ coprime. Then $$h\in (f,g)$$ if and only if $$h\in (f,g)_p \forall p\in \mathbb{P}^2$$. Where $$(f,g)_p$$ is the ideal generated by $$f$$ and $$g$$ in the localization of $$\mathbb{C}[x,y,z]$$ in the maximal ideal corresponding to $$p$$.

• The condition is trivially satisfied for the non-intersection points.
• It reduces the problem to studying the problem locally at the base points.
• The local conditions are algebraic.

# Combinatorialization of the Fundamentalsatz

First case: line arrangements

# Combinatorialization of the Fundamentalsatz

• Choose base points
• Construct the incidence matrix
• $\\ \\ J = \left(\begin{array}{rrrrrrrrr} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \end{array}\right)$

# Combinatorialization of the Fundamentalsatz

Construct the fundamental matrix:

$M = J^t\cdot J - U = \left(\begin{array}{rrrrrrrrr} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 2 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 2 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & -1 \\ 0 & -1 & 0 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 2 & 0 \\ -1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{array}\right)$

# Combinatorialization of the Fundamentalsatz

Reordering lines to get a diagonal sum of boxes:

$M = \left(\begin{array}{rrrrrrrrr} 2 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 2 \end{array}\right)$

# Combinatorialization of the Fundamentalsatz

The boxes give us the components. The kernels of the submatrices the weights:

$ker \left(\begin{array}{rrr} 2 & 0 & -1 \\ 0 & 2 & -1 \\ -1 & -1 & 1 \end{array}\right) = (1, 1, 2)$

# Combinatorialization of the Fundamentalsatz

• Definition: Given a line arrangement $$\mathcal{L}=\{l_0,\ldots, l_n\}$$, a partition $$\mathcal{L} = \mathcal{L}_0 \coprod \mathcal{L}_1 \coprod \cdots \coprod\mathcal{L}_m$$ and an exponent function $$d:\mathcal{L}\mapsto \mathbb{Z}^+$$ is said to form a combinatorial pencil if at each intersection point, one of these conditions hold:

• The point is "monochromatic" (all the lines lie in the same $$\mathcal{L}_i$$)
• All the components have the same multiplicity in the point ($$\sum_{l\in\mathcal{L}_i}d(l) = \sum_{l\in\mathcal{L}_j}d(l)$$)
• Theorem (Falk-Yuzvinsky, M.): The previous method gives all (primitive) combinatorial pencils.

• Theorem (Falk-Yuzvinsky): A line arrangement is a union of three or more curves in a pencil if and only if it admits a combinatorial pencil.

# Generalization: arbitrary curves

Different types of singularities:

# Algebraic invariants of a singularity

Definition: Given a local branch $$f=0$$ at the origin, its multiplicity is defined as the maximum $$p$$ for which $$f\in \mathfrak{m}^p$$

Definition: Given two local branches $$f=0,g=0$$ at the origin, its intersection multiplicty is defined as $dim\frac{\mathcal{O}_p}{(f,g)}$

# Tool for understanding singularities: blowup

Locally:

• Consider a point $$p$$.
• The set of lines through $$p$$ form a $$\mathbb{P}^1$$.
• We have a map $$\pi:\mathbb{L}\times\mathbb{P}^1\mapsto \mathbb{A}^2$$
• $$\pi$$ is bijective outside of $$p$$
• $$\pi^{-1}(p)=\mathbb{P}^1$$

# Properties of the blowup

The blowup "smoothens" the singularity.

Theorem: Every plane curve can be resolved to a normal crossing divisor by a finite number of blowup at points.

Lemma: Let $$f,g$$ be local branches in a point $$p$$. Let $$\bar{f},\bar{g}$$ be their strict transforms. Then $(\bar{f},\bar{g}) = (f,g) - m_p(f)\cdot m_p(g)$

# Generalization of the incidence lattice

After resolving to normal crossings, we have:

• The strict transforms of the original components.
• The exceptional divisors that appeared during the blowup process
• An incidence between them given by the multiplicity

# Generalization of the process

• Choose base exceptional divisors
• Construct the incidence matrix
• Proceed as before

# Example

$(- x^{3} + y^{3} - y^{2}) \cdot x \cdot (y - 1) \cdot y$

# Example

$\left(\begin{array}{rrrrr} 2 & 1 & 1 & 0 & -1 \end{array}\right)$

# Example

$\left(\begin{array}{rrrrrr} 2 & 1 & 1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 0 & 1 & -1 \end{array}\right)$

# Example

$\left(\begin{array}{rrrrrrr} 2 & 1 & 1 & 0 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & -1 \end{array}\right)$

# Example

$J = \left(\begin{array}{rrrrrrrrrr} 2 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & -1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1 \end{array}\right)$

# Choose base components

$J = \left(\begin{array}{rrrrrrrr} 2 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right)$

# Degrees matrix

$D = \left(\begin{array}{rrrrrrrr} 9 & 3 & 3 & 3 & 0 & 0 & 0 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$

# Fundamental matrix

$J^t\cdot J - D = \left(\begin{array}{rrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 & -1 & 0 & -1 & 0 \\ 0 & -1 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 3 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 \end{array}\right)$

# After reordering

$\left(\begin{array}{rrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & -1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 \end{array}\right)$

The decomposition in boxes gives us the partition.

The kernels of the boxes give us the powers of each component:

$(- x^{3} + y^{3} - y^{2}), x^3 , (y - 1) \cdot y^2$

# Main result

• Definition: Given a plane curve, with irreducible components $$\mathcal{C}=\{C_1,\ldots C_m\}$$ a combinatorial pencil is a partition $$\mathcal{C}=\mathcal{C}_1\coprod\mathcal{C}_2\coprod \cdots \coprod \mathcal{C}_n$$ and an exponent function $$d:\mathcal{C}\mapsto \mathbb{Z}^+$$ such that at any singular point $$p$$ of the curve, one of the following two options hold:

• All components that go through $$p$$ live in the same $$\mathcal{C_i}$$
• For each local branch $$b$$ at $$p$$, $$b\in \mathcal{C}_i$$, and every $$j,k\neq i$$, the following equality holds:

$\sum_{b'\in \mathcal{C}_j}d(b')\cdot (b,b')_p = \sum_{b'\in \mathcal{C}_k}d(b')\cdot (b,b')_p$

• Theorem (Cogolludo, M.): A Curve allows a combinatorial pencil if and only if it is the union of three or more fibers of a pencil of curves. Moreover the fibers are given by the partition and exponents of the combinatorial pencil.

# Higher dimensions

• Question: Given $$n+1$$ hypersurfaces in $$\mathbb{C}\mathbb{P}^n$$, when do they belong to a linear system of dimension $$n-1$$?

• Still open

• Definition (Libgober): A hyperplane arrangement is said to be an isolated non normal crossing (INNC) if the intersection of $$i$$ hyperplanes has codimension $$i$$ except maybe in isolated points.
• Theorem (M.): An INNC arrangement in $$\mathbb{C}\mathbb{P}^n$$ where at most $$n+1$$ hyperplanes meet at a point is a union of $$n+1$$ fibers of a linear system of dimension $$n$$ if and only if it admits a partition such that every $$INNC$$ point is the intersection of one hyperplane in each element of the partition.