Combinatorial and Additive Number

Theory 2016

Additive combinatorics methods in

associative algebras

Vincent Beck

Université d'Orléans

7 janvier 2016

Credits

Thanks to the organizers

Joint work with Cédric Lecouvey of Université de Tours

Preprint : arXiv:1504.02287

Supported by ANR-12-JS01-0003-01 ACORT grant

Credits

Thanks to the organizers

Joint work with Cédric Lecouvey of Université de Tours

Preprint : arXiv:1504.02287

Supported by ANR-12-JS01-0003-01 ACORT grant

Kneser's and Diderrich's Theorems

Theorem Let G be a group and A,B �nite non empty subsets

of G . Assume A is a commutative subset of G (that is to say

aa′ = a′a for every a, a′ in A) and let H := {g ∈ G , gAB = AB}.We have

|AB| > |A|+ |B| − |H|.

Main idea of the proof

Dyson e-transform : �x e ∈ B and let A′ = A ∩ Be−1 and

B ′ = Ae ∪ B . We have A′B ′ ⊂ AB and |A′|+ |B ′| = |A|+ |B|.

Motivations

Is it possible to replace groups by more complex algebraic

structures (�elds, associative algebras over a �eld) ?

Group→→→ �eld, algebra

Combinatorics→→→ linear algebra

In the rest of the talk, k will be an in�nite �eld

Motivations

Is it possible to replace groups by more complex algebraic

structures (�elds, associative algebras over a �eld) ?

Group→→→ �eld, algebra

Combinatorics→→→ linear algebra

In the rest of the talk, k will be an in�nite �eld

Motivations

Is it possible to replace groups by more complex algebraic

structures (�elds, associative algebras over a �eld) ?

Group→→→ �eld, algebra

Combinatorics→→→ linear algebra

In the rest of the talk, k will be an in�nite �eld

Motivations

Is it possible to replace groups by more complex algebraic

structures (�elds, associative algebras over a �eld) ?

Group→→→ �eld, algebra

Combinatorics→→→ linear algebra

In the rest of the talk, k will be an in�nite �eld

Field extensions

Theorem X.D.Hou, K.H.Leung, Q.Xiang (2002)-Kainrath(2005). Let K be a �eld extension of k . Assume every algebraic

element in K is separable over k . Let A and B be two nonempty

�nite subsets of K ∗. Then

dimk(AB) > dimk(A) + dimk(B)− dimk(H) (1)

where H := {h ∈ K | h〈AB〉k−vs ⊆ 〈AB〉k−vs} and dimk(C ) standsfor dimk(〈C 〉k−vs) for every �nite subset C of K .

Main ideas of the proof

Dyson e-transform, Vandermonde determinant, pigeonhole principle

Field extensions

Theorem X.D.Hou, K.H.Leung, Q.Xiang (2002)-Kainrath(2005). Let K be a �eld extension of k . Assume every algebraic

element in K is separable over k . Let A and B be two nonempty

�nite subsets of K ∗. Then

dimk(AB) > dimk(A) + dimk(B)− dimk(H) (1)

where H := {h ∈ K | h〈AB〉k−vs ⊆ 〈AB〉k−vs} and dimk(C ) standsfor dimk(〈C 〉k−vs) for every �nite subset C of K .

Main ideas of the proof

Dyson e-transform,

Vandermonde determinant, pigeonhole principle

Field extensions

Theorem X.D.Hou, K.H.Leung, Q.Xiang (2002)-Kainrath(2005). Let K be a �eld extension of k . Assume every algebraic

element in K is separable over k . Let A and B be two nonempty

�nite subsets of K ∗. Then

dimk(AB) > dimk(A) + dimk(B)− dimk(H) (1)

where H := {h ∈ K | h〈AB〉k−vs ⊆ 〈AB〉k−vs} and dimk(C ) standsfor dimk(〈C 〉k−vs) for every �nite subset C of K .

Main ideas of the proof

Dyson e-transform, Vandermonde determinant,

pigeonhole principle

Field extensions

Theorem X.D.Hou, K.H.Leung, Q.Xiang (2002)-Kainrath(2005). Let K be a �eld extension of k . Assume every algebraic

element in K is separable over k . Let A and B be two nonempty

�nite subsets of K ∗. Then

dimk(AB) > dimk(A) + dimk(B)− dimk(H) (1)

where H := {h ∈ K | h〈AB〉k−vs ⊆ 〈AB〉k−vs} and dimk(C ) standsfor dimk(〈C 〉k−vs) for every �nite subset C of K .

Main ideas of the proof

Dyson e-transform, Vandermonde determinant, pigeonhole principle

Applications

Theorem Kneser's theorem. Let G be an abelian group and

A,B �nite non empty subsets of G , and

H := {g ∈ G , gAB = AB}. We have

|AB| > |A|+ |B| − |H|.

Main ideas of the proof

Realize G = 〈A,B〉gr as the Galois group of a �eld extension (of

�eld of fractions) k ↪→ K , de�ne a map from G to K and use

Galois correspondence theorem.

Applications

Theorem Kneser's theorem. Let G be an abelian group and

A,B �nite non empty subsets of G , and

H := {g ∈ G , gAB = AB}. We have

|AB| > |A|+ |B| − |H|.

Main ideas of the proof

Realize G = 〈A,B〉gr as the Galois group of a �eld extension (of

�eld of fractions) k ↪→ K , de�ne a map from G to K and use

Galois correspondence theorem.

The algebra case

Question : Can we replace K by an associative algebra A in Hou,

Leung and Xiang's or Kainrath's theorem ?

Answer : Of course, not ! Because of zero divisors : we can have

AB = 0 for a big subspaces A and B of A.

→→→Hypothesis on A and B : 〈A〉k−vs and 〈B〉k−vs contain an

invertible element.

→→→ unformal Hypothesis on A : "it is easy to construct invertibles

elements in A".

More precisely, we assume that A is of either one of the following

type �nite dimensional algebras, Banach algebras or �nite products

of �eld extensions of k .

In particular, for all a ∈ A there exists λ ∈ k such that a − λ1 is

invertible in A.

The algebra case

Question : Can we replace K by an associative algebra A in Hou,

Leung and Xiang's or Kainrath's theorem ?

Answer : Of course, not ! Because of zero divisors : we can have

AB = 0 for a big subspaces A and B of A.

→→→Hypothesis on A and B : 〈A〉k−vs and 〈B〉k−vs contain an

invertible element.

→→→ unformal Hypothesis on A : "it is easy to construct invertibles

elements in A".

More precisely, we assume that A is of either one of the following

type �nite dimensional algebras, Banach algebras or �nite products

of �eld extensions of k .

In particular, for all a ∈ A there exists λ ∈ k such that a − λ1 is

invertible in A.

The algebra case

Question : Can we replace K by an associative algebra A in Hou,

Leung and Xiang's or Kainrath's theorem ?

Answer : Of course, not ! Because of zero divisors : we can have

AB = 0 for a big subspaces A and B of A.

→→→Hypothesis on A and B : 〈A〉k−vs and 〈B〉k−vs contain an

invertible element.

→→→ unformal Hypothesis on A : "it is easy to construct invertibles

elements in A".

More precisely, we assume that A is of either one of the following

type �nite dimensional algebras, Banach algebras or �nite products

of �eld extensions of k .

In particular, for all a ∈ A there exists λ ∈ k such that a − λ1 is

invertible in A.

The algebra case

Question : Can we replace K by an associative algebra A in Hou,

Leung and Xiang's or Kainrath's theorem ?

Answer : Of course, not ! Because of zero divisors : we can have

AB = 0 for a big subspaces A and B of A.

→→→Hypothesis on A and B : 〈A〉k−vs and 〈B〉k−vs contain an

invertible element.

→→→ unformal Hypothesis on A : "it is easy to construct invertibles

elements in A".

More precisely, we assume that A is of either one of the following

type �nite dimensional algebras, Banach algebras or �nite products

of �eld extensions of k .

In particular, for all a ∈ A there exists λ ∈ k such that a − λ1 is

invertible in A.

The algebra case

Question : Can we replace K by an associative algebra A in Hou,

Leung and Xiang's or Kainrath's theorem ?

Answer : Of course, not ! Because of zero divisors : we can have

AB = 0 for a big subspaces A and B of A.

→→→Hypothesis on A and B : 〈A〉k−vs and 〈B〉k−vs contain an

invertible element.

→→→ unformal Hypothesis on A : "it is easy to construct invertibles

elements in A".

More precisely, we assume that A is of either one of the following

type �nite dimensional algebras, Banach algebras or �nite products

of �eld extensions of k .

In particular, for all a ∈ A there exists λ ∈ k such that a − λ1 is

invertible in A.

Diderrich's theorem for algebras

Theorem B., Lecouvey (2015). Assume A is isomorphic to

a subalgebra of an algebra of the three preceeding types.

Assume A and B to be two �nite nonempty subsets of A such that

〈A〉k−vs. ∩ U(A) 6= ∅ and 〈B〉k−vs. ∩ U(A) 6= ∅.

Assume that A is commutative and A(A) admits a �nite number of

�nite-dimensional subalgebras.

Let H := {x ∈ A, x〈AB〉k−vs ⊂ 〈AB〉k−vs}.We have

dimk(AB) > dimk(A) + dimk(B)− dim(H)

Sketch of proof of the theorem

Lemma Assume A is of one of the three preceeding types.

Let A and B be two �nite subsets of A such that A is

commutative, 〈A〉k−vs. ∩ U(A) 6= ∅ and 〈B〉k−vs. ∩ U(A) 6= ∅.

Then, for each a ∈ 〈A〉k−vs ∩ U(A), there exists a (commutative)

�nite-dimensional subalgebra Aa of A such that Aa ⊆ A(A) and a

vector space Va contained in 〈AB〉k−vs such that Va ∩ U(A) 6= ∅,

AaVa = Va, 〈aB〉k−vs ⊆ Va and

dimk(Va) + dimk(Aa) > dimk(A) + dimk(B).

Proof of Lemma e-Dyson transform, recursion on dimension of

〈A〉k−vs ; the type of A ensures us that at each step there are

enough invertible elements in 〈A〉k−vs .

Sketch of proof of the theorem

Lemma Assume A is of one of the three preceeding types.

Let A and B be two �nite subsets of A such that A is

commutative, 〈A〉k−vs. ∩ U(A) 6= ∅ and 〈B〉k−vs. ∩ U(A) 6= ∅.

Then, for each a ∈ 〈A〉k−vs ∩ U(A), there exists a (commutative)

�nite-dimensional subalgebra Aa of A such that Aa ⊆ A(A) and a

vector space Va contained in 〈AB〉k−vs such that Va ∩ U(A) 6= ∅,

AaVa = Va, 〈aB〉k−vs ⊆ Va and

dimk(Va) + dimk(Aa) > dimk(A) + dimk(B).

Proof of Lemma e-Dyson transform, recursion on dimension of

〈A〉k−vs ; the type of A ensures us that at each step there are

enough invertible elements in 〈A〉k−vs .

Sketch of proof of the theorem

Let (x1, . . . , xn) be a basis of 〈A〉k−vs with x1 invertible.

For any α ∈ k , set xα = x1 + αx2 + · · ·+ αn−1xn.

For each α such that xα is invertible (there exists an in�nity of such

α), there exists a �nite-dimensional subalgebra Aα and a subspace

Vα such in the preceeding lemma.

Since A(A) has only a �nite number of subalgebras, there exists n

distincts elements of k , α1, . . . , αn such that

B := Aα1 = · · · = Aαn . Moreover (xα1 , . . . , xαn) is a basis of

〈A〉k−vs .

We have B ⊂ H.

A �rst example

Example Let C the Banach algebra of continuous functions from

[ 0 , 1 ] into R. Let V and W be �nite dimensional subspaces

containing a positive function. We have

dimR(VW ) > dimR(V ) + dimR(W )− 1.

Back to Kneser-Diderrich's theorem

for groups

Let G be a group and A,B �nite non empty subsets of G . Assume

A is a commutative subset of G and let H := {g ∈ G , gAB = AB}.

|AB| > |A|+ |B| − |H|.

Going to algebras. We consider A = C[G ]. This is a subalgebra of a

Banach algebra.

The algebra A(A)C-alg.' C[X±1

1, . . . ,Xn

±1]r has only a �nite number

of �nite dimensional subalgebras.

The stabilizer of 〈AB〉C−vs in C[G ] is C[H].

Diderrich's theorem in C[G ] gives Didderich's theorem for G .

Back to Kneser-Diderrich's theorem

for groups

Let G be a group and A,B �nite non empty subsets of G . Assume

A is a commutative subset of G and let H := {g ∈ G , gAB = AB}.

|AB| > |A|+ |B| − |H|.

Going to algebras. We consider A = C[G ]. This is a subalgebra of a

Banach algebra.

The algebra A(A)C-alg.' C[X±1

1, . . . ,Xn

±1]r has only a �nite number

of �nite dimensional subalgebras.

The stabilizer of 〈AB〉C−vs in C[G ] is C[H].

Diderrich's theorem in C[G ] gives Didderich's theorem for G .