Discrete geometry from an algebraic point of view · Discrete geometry from an algebraic point of view Valculescu Claudiu ... Discrete geometry is concerned with the study of ﬁnite

Discrete geometry from an algebraic point of view

Valculescu Claudiu

4 September 2014

Valculescu Claudiu Discrete geometry from an algebraic point of view 1 / 25

Preliminaries

DefinitionDiscrete geometry is concerned with the study of finite (discrete) setsof geometric objects and their properties.

Some topics:Packings, coverings and tilingsFor example: What is the maximum density of a packing of unitcircles in the plane?


Preliminaries


Some topics:Packings, coverings and tilings

For example: What is the maximum density of a packing of unitcircles in the plane?


Preliminaries




Preliminaries




PreliminariesSome topics:

Incidence structuresFor example: Given an arbitrary set P of n points and an arbitraryset L of m lines, both in the real plane, what is the maximum numberof point-line incidences, i.e. the maximum cardinality of the set

I(P, L) = (p, l) : p ∈ l , p ∈ P, l ∈ L?

l1

l3

l2l4p1

p4p2

p3

(p4, l1)(p3, l2)

(p4, l2)

(p1, l3)(p1, l4)(p2, l4)

In particular, we are interested in the asymptotic behaviour asm, n→∞.



Incidence structures

For example: Given an arbitrary set P of n points and an arbitraryset L of m lines, both in the real plane, what is the maximum numberof point-line incidences, i.e. the maximum cardinality of the set

I(P, L) = (p, l) : p ∈ l , p ∈ P, l ∈ L?

l1

l3

l2l4p1

p4p2

p3

(p4, l1)(p3, l2)

(p4, l2)

(p1, l3)(p1, l4)(p2, l4)





I(P, L) = (p, l) : p ∈ l , p ∈ P, l ∈ L?

l1

l3

l2l4p1

p4p2

p3

(p4, l1)(p3, l2)

(p4, l2)

(p1, l3)(p1, l4)(p2, l4)





I(P, L) = (p, l) : p ∈ l , p ∈ P, l ∈ L?

l1

l3

l2l4p1

p4p2

p3

(p4, l1)(p3, l2)

(p4, l2)

(p1, l3)(p1, l4)(p2, l4)





I(P, L) = (p, l) : p ∈ l , p ∈ P, l ∈ L?

l1

l3

l2l4p1

p4p2

p3

(p4, l1)(p3, l2)

(p4, l2)

(p1, l3)(p1, l4)(p2, l4)



PreliminariesSome other topics:

Geometric graph theoryFor example: Planarity, crossing number, ...

Topological combinatoricsFor example: Fair division problems, equipartitions



Geometric graph theory

For example: Planarity, crossing number, ...









Topological combinatorics

For example: Fair division problems, equipartitions






Preliminaries


Preliminaries

Some other domains in which discrete geometry has applications:

Theoretical computer science

EngineeringFor example: Robotics;

Computer Imagery and Image processingFor example: Modeling image aggregation processes;

ChemistryFor example: Cristalography;

Other fields of mathematicsFor example: Numerical analysis.


Preliminaries








Preliminaries








Preliminaries








Preliminaries








Preliminaries








Big O and Ω notations

Definition (Big O notation)Let f and g be two functions defined on some subset of the real numbers.One writes

f (x) = O(g(x)),

if and only if there is a positive constant M and x0 > 0, such that

|f (x)| ≤ M|g(x)| for all x ≥ x0.

Definition (Ω notation)Let f and g be two functions defined on some subset of the real numbers.One writes

f (x) = Ω(g(x)),


|f (x)| ≥ M|g(x)| for all x ≥ x0.




f (x) = O(g(x)),


|f (x)| ≤ M|g(x)| for all x ≥ x0.


f (x) = Ω(g(x)),


|f (x)| ≥ M|g(x)| for all x ≥ x0.




f (x) = O(g(x)),


|f (x)| ≤ M|g(x)| for all x ≥ x0.


f (x) = Ω(g(x)),


|f (x)| ≥ M|g(x)| for all x ≥ x0.


Point-line incidencesRemember the point-line incidence question...

Theorem (Szemeredi-Trotter, 1983)Given a set P of n points and a set L of m lines in the real plane, thenumber of incidences is of order

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

EndreSzemeredi

WilliamTrotter


Point-line incidencesRemember the point-line incidence question...

Theorem (Szemeredi-Trotter, 1983)Given a set P of n points and a set L of m lines in the real plane, thenumber of incidences is of order

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

EndreSzemeredi

WilliamTrotter


Application of Szemeredi-Trotter

DefinitionGiven a finite set A ⊂ R, we define :

A + A = a + a′ : a, a′ ∈ A

A · A = a · a′ : a, a′ ∈ A.

Note the following!If A is an arithmetic progression a + bk, k = 1, ..., n, we have

|A + A| = O(|A|).

If A is a geometric progression a · bk , k = 1, ..., n, we have

|A · A| = O(|A|).

How ”small” can max|A + A|, |A · A| be?




A + A = a + a′ : a, a′ ∈ A

A · A = a · a′ : a, a′ ∈ A.

Note the following!

If A is an arithmetic progression a + bk, k = 1, ..., n, we have

|A + A| = O(|A|).


|A · A| = O(|A|).





A + A = a + a′ : a, a′ ∈ A

A · A = a · a′ : a, a′ ∈ A.


|A + A| = O(|A|).


|A · A| = O(|A|).





A + A = a + a′ : a, a′ ∈ A

A · A = a · a′ : a, a′ ∈ A.


|A + A| = O(|A|).


|A · A| = O(|A|).





A + A = a + a′ : a, a′ ∈ A

A · A = a · a′ : a, a′ ∈ A.


|A + A| = O(|A|).


|A · A| = O(|A|).

How ”small” can max|A + A|, |A · A| be?Valculescu Claudiu Discrete geometry from an algebraic point of view 9 / 25


Gyorgy Elekes

Theorem (Sum-Product, Elekes 1997)Let A ⊂ R be a finite set. Then

max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.

Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).

Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),

where A−1 = x−1, x ∈ A.We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.

We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.We have |L| = |A|2.

Each line in L has at least |A| incidences with P.



Gyorgy Elekes


max|A + A|, |A · A| = Ω(|A|5/4).

Proof.Let P = (A · A)× (A + A).Let L be the set of all lines of the form y = ax + b, with a ∈ A−1 andb ∈ A (we can discard a = 0),where A−1 = x−1, x ∈ A.We have |L| = |A|2.Each line in L has at least |A| incidences with P.



Thus, we have

|A · A|2/3 · |A + A|2/3 · |A|4/3 = Ω(|A|3).

Assume neither |A + A|, nor |A · A| are of order Ω(|A|5/4).

This leads us to a contradiction, and thus

max|A + A|, |A · A| = Ω(|A|5/4),

which completes the proof.

The conjectured bound is ∼ |A|2−ε, for all ε > 0.



Thus, we have

|A · A|2/3 · |A + A|2/3 · |A|4/3 = Ω(|A|3).



max|A + A|, |A · A| = Ω(|A|5/4),





Thus, we have

|A · A|2/3 · |A + A|2/3 · |A|4/3 = Ω(|A|3).



max|A + A|, |A · A| = Ω(|A|5/4),





Thus, we have

|A · A|2/3 · |A + A|2/3 · |A|4/3 = Ω(|A|3).



max|A + A|, |A · A| = Ω(|A|5/4),




Stepping forward...

Definition (Real algebraic curve)We call C ⊂ R2 an real algebraic curve if it isinfinite and there exist a polynomial f ∈ R[x , y ]\0such that

C = (x , y) ∈ R2 : f (x , y) = 0.

Lines, circles, ellipses, hyperbolas, parabolas are algebraic curves.

Definition (Degree of freedom. Multiplicity)Let P ⊆ RD be a set of points and let Γ be a set of curves in RD. We saythat P and 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 any kpoints of P belong to at most M curves in Γ.


Stepping forward...


C = (x , y) ∈ R2 : f (x , y) = 0.




Stepping forward...


C = (x , y) ∈ R2 : f (x , y) = 0.




Stepping forward...

Theorem (Pach-Sharir, 1998)If a set P of points in R2 and a set Γ of algebraic curves in R2 form asystem 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.

Janos Pach Micha SharirValculescu Claudiu Discrete geometry from an algebraic point of view 13 / 25

Applications of algebraic geometry

Paul Erdos

Problem (Erdos, 1946)Prove that any set of n points in the planedetermines at least n1−ε distinct distances.

Solved - 2010 - Guth, Katz - using tools from algebraic geometry.

Larry Guth Nets Hawk Katz


Applications of algebraic geometry

Paul Erdos

Problem (Erdos, 1946)Prove that any set of n points in the planedetermines at least n1−ε distinct distances.

Solved - 2010 - Guth, Katz - using tools from algebraic geometry.

Larry Guth Nets Hawk KatzValculescu Claudiu Discrete geometry from an algebraic point of view 14 / 25

Variation of Erdos’ distinct distance problem

Theorem (Pach-De Zeeuw, 2013)Given a plane algebraic curve C of degree d with n points on it, thenumber of distinct distances spanned by the set of points is Ωd (n4/3),unless C contains a line or a circle.

Janos Pach Frank de Zeeuw

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


Variation of Erdos’ distinct distance problem

Theorem (Pach-De Zeeuw, 2013)Given a plane algebraic curve C of degree d with n points on it, thenumber of distinct distances spanned by the set of points is Ωd (n4/3),unless C contains a line or a circle.

Janos Pach Frank de Zeeuw

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


Distinct values of bilinear functions on algebraic curves

Definition (Complex algebraic curve)We call C ⊂ C2 a complex algebraic curve if there exist a polynomialf ∈ C[x , y ]\0 such that

C = (x , y) ∈ C2 : f (x , y) = 0.

For all 2× 2 matrices M with complex entries, let BM : C2 × C2 → C bedefined as

BM(p, q) = pT Mq,∀p, q ∈ C2.

DefinitionWe call a curve in C2 special if it is a line or it is linearly equivalent to acurve defined by an equation of the form

xk = y l , k, l ∈ Z, (k, l) = 1.




C = (x , y) ∈ C2 : f (x , y) = 0.


BM(p, q) = pT Mq,∀p, q ∈ C2.


xk = y l , k, l ∈ Z, (k, l) = 1.




C = (x , y) ∈ C2 : f (x , y) = 0.


BM(p, q) = pT Mq,∀p, q ∈ C2.


xk = y l , k, l ∈ Z, (k, l) = 1.




C = (x , y) ∈ C2 : f (x , y) = 0.


BM(p, q) = pT Mq,∀p, q ∈ C2.


xk = y l , k, l ∈ Z, (k, l) = 1.



Theorem (Valculescu-De Zeeuw, 2014)Let C be an irreducible algebraic curve in C of degree d, S a finite set ofpoints on C. If BM is defined as above, then the following holds:

|BM(S)| = Ω(d−5|S|4/3),

where BM(S) = B(p, q) : p, q ∈ S, unless C is special or M is singular.

Remark. If M is singular or C is special, there are constructions where thenumber of distinct values is linear in |S|.

Sketch of proof:Idea: reduce the problem to an incidence problem (introduced by Elekes).The same idea was used by Guth and Katz in their solution for the distinctdistance problem.




|BM(S)| = Ω(d−5|S|4/3),







|BM(S)| = Ω(d−5|S|4/3),



Sketch of proof:

Idea: reduce the problem to an incidence problem (introduced by Elekes).The same idea was used by Guth and Katz in their solution for the distinctdistance problem.




|BM(S)| = Ω(d−5|S|4/3),



Sketch of proof:Idea: reduce the problem to an incidence problem (introduced by Elekes).

The same idea was used by Guth and Katz in their solution for the distinctdistance problem.




|BM(S)| = Ω(d−5|S|4/3),






Let BM(S) = B(pi , qj) : pi , qj ∈ S.

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

For each value a ∈ BM(S), let

Ea = (pi , qs) ∈ S2 : BM(pi , qs) = a.

Using Cauchy-Schwarz, we have

|Q| =∑

a∈BM(S)|Ea|2 ≥

1BM(S)

∑a∈BM(S)

|Ea|

2

= n4

BM(S) .






Ea = (pi , qs) ∈ S2 : BM(pi , qs) = a.


|Q| =∑

a∈BM(S)|Ea|2 ≥

1BM(S)

∑a∈BM(S)

|Ea|

2

= n4

BM(S) .






Ea = (pi , qs) ∈ S2 : BM(pi , qs) = a.


|Q| =∑

a∈BM(S)|Ea|2 ≥

1BM(S)

∑a∈BM(S)

|Ea|

2

= n4

BM(S) .






Ea = (pi , qs) ∈ S2 : BM(pi , qs) = a.


|Q| =∑

a∈BM(S)|Ea|2 ≥

1BM(S)

∑a∈BM(S)

|Ea|

2

= n4

BM(S) .



Define the following curves:Cij = (q, q′) ∈ C2 : BM(pi , q) = BM(pj , q′).

Cst = (p, p′) ∈ C2 : BM(p, qs) = BM(p′, 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.That means

n4

|BM(S)| ≤ |Q| = I(P, Γ) ≤?.

An upper bound for|I(P, Γ)| will give us a lower

bound on |BM(S)|.


Distinct values of bilinear functions on algebraic curvesDefine the following curves:

Cij = (q, q′) ∈ C2 : BM(pi , q) = BM(pj , q′).

Cst = (p, p′) ∈ C2 : BM(p, qs) = BM(p′, 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.That means

n4

|BM(S)| ≤ |Q| = I(P, Γ) ≤?.


bound on |BM(S)|.




Cst = (p, p′) ∈ C2 : BM(p, qs) = BM(p′, 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.That means

n4

|BM(S)| ≤ |Q| = I(P, Γ) ≤?.


bound on |BM(S)|.




Cst = (p, p′) ∈ C2 : BM(p, qs) = BM(p′, 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.

That means

n4

|BM(S)| ≤ |Q| = I(P, Γ) ≤?.


bound on |BM(S)|.





n4

|BM(S)| ≤ |Q| = I(P, Γ)

≤?.


bound on |BM(S)|.





n4

|BM(S)| ≤ |Q| = I(P, Γ) ≤?.


bound on |BM(S)|.Valculescu Claudiu Discrete geometry from an algebraic point of view 19 / 25


The desired bound is given by the following generalization of thePach-Sharir incidence theorem.

Theorem (Solymosi-De Zeeuw, 2014)Let A,B ⊂ C2 with |A| = |B|, let P = A× B ⊂ C4, and let Γ be a set ofalgebraic curves of degree at most d2 in C4, with |P| = |Γ| = m2. If anytwo points of P are contained in at most d2 curves of Γ, then we have

I(P, Γ) = O(d10/3m8/3).

The points and curves defined above might not fulfill the conditions of thetheorem...

We have to remove some curves and some points!





I(P, Γ) = O(d10/3m8/3).







I(P, Γ) = O(d10/3m8/3).







I(P, Γ) = O(d10/3m8/3).





What points and curves should we remove?

No, we will remove only the following:

Γ0 = Cij ∈ Γ : ∃Ckl ∈ Γ : |Cij ∩ Ckl | =∞,

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



What points and curves should we remove?

No, we will remove only the following:

Γ0 = Cij ∈ Γ : ∃Ckl ∈ Γ : |Cij ∩ Ckl | =∞,

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



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

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

One can prove that an irreducible algebraic curve of degree d has atmost d7 linear automorphisms, unless it is a special curve, whichcompletes the proof.

What about special curves?





















In fact, yes. Just consider the inner product, given by

BI(p, q) = xpxq + ypyq.

On a special curve, the number of distinct values of BI can be linear in thesize of the set of points. Also, if M is singular, the number of distinctvalues can be linear.



In fact, yes.

Just consider the inner product, given by















Examples of constructions spanning linear number of distinct values:

Let us defineS := (2li , 2ki ) : i = 1, ..., |S|.

ThenBI((2li , 2ki ), (2lj , 2kj)) = (2l )i+j + (2k)i+j .

That means

|BI(S)| = 2|S| − 1.

Thus, the number of distinct values is linear in |S|.






That means

|BI(S)| = 2|S| − 1.







That means

|BI(S)| = 2|S| − 1.







That means

|BI(S)| = 2|S| − 1.







That means

|BI(S)| = 2|S| − 1.



Thank you!


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Discrete geometry from an algebraic point of view · Discrete geometry from an algebraic point of view Valculescu Claudiu ... Discrete geometry is concerned with the study of ﬁnite