Algebraic techniques in combinatorial geometry

Valculescu Adrian-Claudiu

17.06.2014

Valculescu Adrian-Claudiu Algebraic techniques in combinatorial geometry

Preliminaries on Discrete (Combinatorial) Geometry

What do we study?

Finite (discrete) sets of geometric objects (points, lines,polygons, polytopes, circles, planes) and their properties.

What do we study?

Some concrete topics :Packings, coverings of the plane (or of higher-dimensionalspaces), Incidence problems, Matroids, Geometric graphtheory, ...

Some examples:

Problem (Erdos)What is the maximum number of times the unit distance can occuramong a set of n points in the plane?

Theorem (Szemeredi-Trotter)Given n points and m lines in the plane, the number of incidencesbetween the set of points and the set of lines is of order

O(n2/3m2/3 + m + n).

Some concrete topics :

Packings, coverings of the plane (or of higher-dimensionalspaces), Incidence problems, Matroids, Geometric graphtheory, ...

Some examples:

O(n2/3m2/3 + m + n).

Some examples:

O(n2/3m2/3 + m + n).

Some examples:

O(n2/3m2/3 + m + n).

Some examples:

O(n2/3m2/3 + m + n).

Some examples:

O(n2/3m2/3 + m + n).

Why algebraic geometry?

Problem (Erdos)Prove that any set of n points in the plane determines at leastΩ(n/

√log n) distinct distances!

Solved...after more than 50 years! - 2010 - L.Guth,N.Katz - using tools from Algebraic geometry.

Tools from Algebraic Geometry...

Definition (Algebraic curve). We call C an algebraic curveif it is infinite and there exists a polynomial f ∈ R[x , y ]\0such that

C = (x , y) ∈ R2 : f (x , y) = 0.

Lines and circles are algebraic curves!

C = (x , y) ∈ R2 : f (x , y) = 0.

The degree of the polynomial f defined above is called degreeof the curve C .

Theorem (Bezout’s inequality)Two algebraic curves of degrees d1 and d2 have at most d1 · d2intersection points, unless they have a common component.

The curve C is called irreducible if the polynomial f isirreducible over R.

...and some applications in Discrete Geometry

Example: We want to bound from above the number ofincidences between a set of n points and a set of m lines (orcircles) in the plane.More generally...... incidences between a set of n points and m algebraiccurves (again in the plane) or varieties (in higher dimensions)under certain assumptions.How to do this?

Example: We want to bound from above the number ofincidences between a set of n points and a set of m lines (orcircles) in the plane.

More generally...... incidences between a set of n points and m algebraiccurves (again in the plane) or varieties (in higher dimensions)under certain assumptions.How to do this?

Example: We want to bound from above the number ofincidences between a set of n points and a set of m lines (orcircles) in the plane.More generally...

... incidences between a set of n points and m algebraiccurves (again in the plane) or varieties (in higher dimensions)under certain assumptions.How to do this?

Example: We want to bound from above the number ofincidences between a set of n points and a set of m lines (orcircles) in the plane.More generally...... incidences between a set of n points and m algebraiccurves (again in the plane) or varieties (in higher dimensions)under certain assumptions.

How to do this?

Example: We want to bound from above the number ofincidences between a set of n points and a set of m lines (orcircles) in the plane.More generally...... incidences between a set of n points and m algebraiccurves (again in the plane) or varieties (in higher dimensions)under certain assumptions.How to do this?

Given a polynomial f , we will denote by Z (f ) the zero set off , namely

Z (f ) = (x , y) ∈ R2 : f (x , y) = 0.Each connected component of R2\Z (f ) will be called cell.

Definition (Partitioning polynomial)Given a set P of n points in the plane, and 1 < r ≤ n a parameter,we say that f is an r -partitioning polynomial for P if noconnected component of R2\Z (f ) contains more than n/r pointsof P.

Z (f ) = (x , y) ∈ R2 : f (x , y) = 0.

Each connected component of R2\Z (f ) will be called cell.

TheoremFor every r > 1, every finite point set P ⊂ R2 admits anr-partitioning polynomial f of degree at most O(

√r).

The idea of polynomial partitioning method for boundingincidences between points and curves: separately boundthe number of incidences between the set of curves and thepoints from each connected component of R2\Z (f ), andrespectively between the set of curves and the points on Z (f ).

Note that, for a given cell Ci , when counting incidences, weonly have to consider the curves intersecting this cell.

√r).

Erdos’ distinct distance problem

Problem (Erdos): Prove that any set of n points in the planedetermines at least Ω(n/

√log n) distinct distances.

Let’s see the idea behind the proof...

Let’s see the idea behind the proof...Valculescu Adrian-Claudiu Algebraic techniques in combinatorial geometry

Idea of proof : Transform the problem to an incidenceproblem between points and curves (G.Elekes).

Use polynomial partitioning.

Let’s see some details...that we will also use later

Let S be the set of points (That means, |P| = n).Let D(S) = d(pi , qj) : pi , qj ∈ S, where d(a, b) denotes thedistance between a and b.Let Q = (p, p′, q, q′) ∈ S4 : d(p, q) = d(p′q′).

For each a ∈ D(S), let Ea = (pi , qs) ∈ S2 : d(pi , qs) = a.Then, using Cauchy-Schwarz we have

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

Reduce bounding Q to bounding incidences between points andlines, which gives |Q| ≤ n3 log n.

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

Let S be the set of points (That means, |P| = n).

Let D(S) = d(pi , qj) : pi , qj ∈ S, where d(a, b) denotes thedistance between a and b.Let Q = (p, p′, q, q′) ∈ S4 : d(p, q) = d(p′q′).

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

Let S be the set of points (That means, |P| = n).Let D(S) = d(pi , qj) : pi , qj ∈ S, where d(a, b) denotes thedistance between a and b.

Let Q = (p, p′, q, q′) ∈ S4 : d(p, q) = d(p′q′).

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

For each a ∈ D(S), let Ea = (pi , qs) ∈ S2 : d(pi , qs) = a.

Then, using Cauchy-Schwarz we have

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

|Q| =∑

a∈D(S)|Ea|2 ≥

1|D(S)|

∑a∈D(S)

|D(S)| .

Variations of Erdos’ distinct distance problem

Theorem (Sharir-Sheffer-Solymosi, 2013)Given two lines in R2, with n points each, the number of distinctdistances between pairs of points (each point from a line) is

Ω(n4/3),

unless the lines are parallel or orthogonal.

Remark: If the two lines are orthogonal or parallel, there areconstructions for which the number of distances is linear in n.

Ω(n4/3),

Theorem (Pach-De Zeeuw, 2013)Let C be a plane algebraic curve of degree d that does not containa line or a circle. Then any set of n points on C determines atleast Ωd (n4/3) distinct distances.

Remark: If the curve contains a line or a circle, then there areconstructions for which the number of distances spanned is linearin n.

Distinct ”pinned” triangle areas

In the plane...Definition. We say that a triangle is ”pinned” if it has one vertexfixed at the origin.

Theorem (Iosevich-Roche Newton-Rudnev, 2011)A set of n points in R2 determines Ω(n/log(n)) distinct pinnedtriangle areas, unless the points lie on a line passing through theorigin.

Distinct ”pinned” triangle areasIn the plane...

Definition. We say that a triangle is ”pinned” if it has one vertexfixed at the origin.

Distinct ”pinned” triangle areasIn the plane...Definition. We say that a triangle is ”pinned” if it has one vertexfixed at the origin.

On algebraic curves...

Theorem (Charalambides, 2013)Given n points on an irreducible algebraic curve of degree d in R2,there are Ωd (n5/4) distinct pinned triangle areas, unless the curveis a line, ellipse centred at the origin, or hyperbola centred at theorigin.

Distinct ”pinned” triangle areasOn algebraic curves...

Theorem (Valculescu-De Zeeuw, 2014)Let C be an irreducible algebraic curve in R2 of degree d and S aset of n points on C. If C is not of a special form, then thenumber of distinct pinned triangle areas spanned by S is Ωd (n4/3).

Special forms for C : lines, ellipses centered at the origin, andhyperbolas centered at the origin

Remark: If the curve contains a line or a circle, then there areconstructions for which the number of pinned triangle areasspanned is linear in n!

Sketch of proof.

Let A(S) = A(Opi qj) : pi , qj ∈ S, where O(0, 0) is the origin.

Let Q = (p, p′, q, q′) ∈ S4 : A(Opq) = A(Op′q′).

For each a ∈ A(S), let Ea = (pi , qs) ∈ S2 : A(Opi qs) = a.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Sketch of proof.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Sketch of proof.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Sketch of proof.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Sketch of proof.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Sketch of proof.

|Q| =∑

a∈A(S)|Ea|2 ≥

1|A(S)|

∑a∈A(S)

|A(S)| .

Thus, we have obtained

|A(S)| ≤ |Q|.

We reduce to an incidence problem between points and curves.

For this, we define:

Cij = (q, q′) ∈ C2 : A(Opi q) = A(Opjq′).

Cst = (p, p′) ∈ C2 : A(Opqs) = A(Op′qt).

Let Γ be the set of all the curves Cij .

A point (qs , qt) ∈ P lies on Cij iff (pi , pj , qs , qt) ∈ Q.

Distinct ”pinned” triangle areasThus, we have obtained

|A(S)| ≤ |Q|.

A point (qs , qt) ∈ P lies on Cij iff (pi , pj , qs , qt) ∈ Q.Valculescu Adrian-Claudiu Algebraic techniques in combinatorial geometry

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

We need an upper bound for I(P, Γ) !

We use the following result...

...but first we have to define:

DefinitionLet P ⊂ RD and let Γ be a set of curves in RD. We say that Pand Γ form a system with k degrees of freedom and multiplicity Mif any two curves in Γ intersect in at most M points of P, and anyk points of P belong to at most M curves in Γ.

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

That meansn4

|A(S)| ≤ |Q| = I(P, Γ).

Theorem (Pach-Sharir)If a set P of points in R2 and a set Γ of algebraic curves in R2

form a system with 2 degrees of freedom and multiplicity M, then

I(P, Γ) ≤ CM ·max|P|2/3|Γ|2/3, |P|, |Γ|,

where CM is a constant depending only on M.

Theorem (Pach-Sharir)If a set P of points in R2 and a set Γ of algebraic curves in R2

form a system with 2 degrees of freedom and multiplicity M, then

I(P, Γ) ≤ CM ·max|P|2/3|Γ|2/3, |P|, |Γ|,

where CM is a constant depending only on M.

The sets P and Γ defined above may not quite form a system with2 degrees of freedom...

However, we can show that after removing small subsets of pointsand curves they do form such a system.

What points and curves should we remove?

LetΓ0 = Cij ∈ Γ : ∃Ckl ∈ Γ : |Cij ∩ Ckl | =∞.

Similarly, let

P0 = (qs , qt) ∈ P : ∃(qu, qv ) ∈ P : |Cst ∩ Cuv | =∞.

LemmaIf C is not exceptional, then for all Cij ,Ckl ∈ Γ\Γ0 we have

|Cij ∩ Ckl | ≤ d2,

and for all pairs of points in P\P0, there are at most d2 curves inΓ that contain both points.

Similarly, let

|Cij ∩ Ckl | ≤ d2,

Similarly, let

|Cij ∩ Ckl | ≤ d2,

Similarly, let

|Cij ∩ Ckl | ≤ d2,

We want to show that Γ0 and P0 are relatively small!

We do this by showing that if two curves have infinite intersection,then this is related to an automorphism of C (invertible lineartransformation that fixes the curve). In the paper, we prove:

LemmaAn irreducible algebraic curve of degree d has ≤ 4d linearautomorphisms, unless it is a line or linearly equivalent to one ofthe following:- an ellipse or a hyperbola centered at the origin;- a parabola;- a ”pseudohyperbola” (algebraic curve of equation xpyq = 1,p, q > 1) or a ”pseudocusp” (algebraic curve of equationxpy−q = 1, p, q > 1).

Remark. Parabolas, pseudohyperbolas and pseudocusps can beexcluded!

Distinct ”pinned” triangle areasWe want to show that Γ0 and P0 are relatively small!

We do this by showing that if two curves have infinite intersection,then this is related to an automorphism of C (invertible lineartransformation that fixes the curve).

In the paper, we prove:

Explicit constructions on the exceptional curves:

For lines:After rotating and scaling we can assume that C is the line y = 1.Choose

S = (i , 1) : i = 1, ..., n.

The pinned triangle areas are of the form 12 |i − j | - the number is

linear in n.

For lines:

After rotating and scaling we can assume that C is the line y = 1.Choose

S = (i , 1) : i = 1, ..., n.

linear in n.

For lines:After rotating and scaling we can assume that C is the line y = 1.

ChooseS = (i , 1) : i = 1, ..., n.

linear in n.

S = (i , 1) : i = 1, ..., n.

linear in n.

S = (i , 1) : i = 1, ..., n.

linear in n.

For ellipses centered at the origin:After scaling we can assume that C is the unit circle.Choose

S = (cos(2πi/n), sin(2πi/n)) : i = 1, ..., n.

The pinned triangle areas are of the form 12 sin(2π|i − j |/n) - the

number is linear in n.

For ellipses centered at the origin:

After scaling we can assume that C is the unit circle.Choose

For ellipses centered at the origin:After scaling we can assume that C is the unit circle.

Choose

number is linear in n.Valculescu Adrian-Claudiu Algebraic techniques in combinatorial geometry

For hyperbolas centered at the origin:Afer rotating and scaling we can assume C is the hyperbola xy = 1.Choose

S = (2i , 2−i ) : i = 1, ..., n.

The pinned triangle areas are of the form 12 |2

i−j − 2j−i | - thenumber is linear in n.

For hyperbolas centered at the origin:

Afer rotating and scaling we can assume C is the hyperbola xy = 1.Choose

S = (2i , 2−i ) : i = 1, ..., n.

For hyperbolas centered at the origin:Afer rotating and scaling we can assume C is the hyperbola xy = 1.

ChooseS = (2i , 2−i ) : i = 1, ..., n.

S = (2i , 2−i ) : i = 1, ..., n.

Open problems

Extend to higher dimensions.

Extend to distinct values of polynomials on algebraic curves.

Improve the exponent 4/3.

Open problems

Thank you!

Thank you! :)

Algebraic techniques in combinatorial geometry · Algebraic techniques in combinatorial geometry...

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