Before starting

• Ask questions!
• 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?

Not as trivial as it might seem:

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 \(p\).

Remarks about the Fundamentalsatz

• 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.

Example:

Example:

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

Example

Example

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

Partial answer

• 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.

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