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Independence of hyperlogarithms over function fields via algebraic combinatorics

Matthieu Deneufchˆ atel, G´erard H. E. Duchamp, Vincel Hoang Ngoc Minh, and A. I. Solomon

Laboratoire d’Informatique de Paris Nord, Universit´ e Paris 13

67i`emeS´eminaire Lotharingien de Combinatoire, 20 September 2011

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Outline

1 Motivation

2 Main Theorem

3 Examples

Polylogarithms Counterexample Hyperlogarithms

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 2 / 22

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Outline

1 Motivation

2 Main Theorem

3 Examples

Polylogarithms

Counterexample

Hyperlogarithms

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Polyzetas

Riemann ζ function :

ζ (s ) = X

n≥1

1 n s .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 4 / 22

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Polyzetas

Riemann ζ function :

ζ (s ) = X

n≥1

1 n s .

Generalization (in view of multiplications) : Polyzetas

ζ (s) = X

n

1

>···>n

k

>0

1

n 1 s

1

. . . n s k

k

.

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Polyzetas

Riemann ζ function :

ζ (s ) = X

n≥1

1 n s .

Generalization (in view of multiplications) : Polyzetas

ζ (s) = X

n

1

>···>n

k

>0

1 n 1 s

1

. . . n s k

k

.

(Convergent) Polyzetas are values of polylogs (see below) at 1 : ζ(s) = Li s (1).

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 4 / 22

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Polyzetas

Riemann ζ function :

ζ (s ) = X

n≥1

1 n s .

Generalization (in view of multiplications) : Polyzetas

ζ (s) = X

n

1

>···>n

k

>0

1 n 1 s

1

. . . n s k

k

.

(Convergent) Polyzetas are values of polylogs (see below) at 1 : ζ(s) = Li s (1).

Polylogs can be manipulated as shuffles : algebra structure.

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Outline

1 Motivation

2 Main Theorem

3 Examples

Polylogarithms Counterexample Hyperlogarithms

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 5 / 22

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Data

X an alphabet.

( A , d) a commutative differential algebra over the ring k : differential : ∀ a, b ∈ A , d(ab) = d(a)b + ad(b) ;

d is linear over k ;

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Data

X an alphabet.

( A , d) a commutative differential algebra over the ring k : differential : ∀ a, b ∈ A , d(ab) = d(a)b + ad(b) ;

d is linear over k ;

We require that ker(d) = k ( set of constants = k ).

Extension of d to A hhX ii :

∀S ∈ A hhX ii, d(S ) = X

w∈X

d(hS|w i)w .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 6 / 22

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Data

X an alphabet.

( A , d) a commutative differential algebra over the ring k : differential : ∀ a, b ∈ A , d(ab) = d(a)b + ad(b) ;

d is linear over k ;

We require that ker(d) = k ( set of constants = k ).

Extension of d to A hhX ii :

∀S ∈ A hhX ii, d(S ) = X

w∈X

d(hS|w i)w .

Let C be a differential subfield of A (i.e. d( C ) ⊂ C ).

M : a homogeneous series of degree 1 : M = X

x∈X

u x x ∈ C =1 hhX ii.

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Theorem 1 : Linear independence

Suppose that T ∈ A hhX ii is a solution of the differential equation dT = MT ; hT |1 X

i = 1.

The following conditions are equivalent 1 :

i) The family (hT |w i) w∈X

of coefficients of T is free over C . ii) The family of coefficients (hT |y i) y ∈X ∪{1

X

} is free over C . iii) The family (u x ) x∈X is such that, for f ∈ C and α x ∈ k

d(f ) = X

x∈X

α x u x = ⇒ (∀x ∈ X )(α x = 0).

iv) The family (u x ) x∈X is free over k and

d( C ) ∩ span k ((u x ) x∈X ) = {0} .

1Independence of hyperlogarithms over function fields via algebraic combinatorics, M. D., G. H. E.

Duchamp, H. N. Minh and A. Solomon,CAI 2011, LNCS 6742, pp. 127–139. Springer, Heidelberg (2011)

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 7 / 22

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“Slices“ of the free monoid

1 X

X 2 X 3

.. .

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Linear independence of the bottom triangle

hT |1i hT |X i

hT |X 2 i hT |X 3 i

.. .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 8 / 22

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Linear independence of the bottom triangle

Linear independence of the whole triangle

hT |1i hT |X i

hT |X 2 i hT |X 3 i

.. .

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Outline

1 Motivation

2 Main Theorem

3 Examples

Polylogarithms Counterexample Hyperlogarithms

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 9 / 22

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Polylogarithms

X = {x 0 , x 1 }. Ω = C \ (]−∞, 0[ ∪ ]1, +∞[). u 0 (z ) = 1

z , u 1 (z) = 1 1 − z .

1 0

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Polylogarithms

X = {x 0 , x 1 }. Ω = C \ (]−∞, 0[ ∪ ]1, +∞[). u 0 (z ) = 1

z , u 1 (z) = 1 1 − z . Definition

∀z ∈ Ω,

Li x

0n

(z ) = ln n (z ) n! . Li x

1

w (z) =

Z z 0

dt

1 − t Li w (t), and, ∀w ∈ X x 1 X ,

Li x

0

w (z) = Z z

0

dt

t Li w (t).

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 10 / 22

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Remark

Let w = x 0 s

1

−1 x 1 . . . x 0 s

k

−1 x 1 ↔ s = (s 1 , . . . , s k ).

It can be shown that the Taylor expansion of these functions is given by Li w (z ) = Li s (z) = X

n

1

>n

2

> ··· >n

k

> 0

z n

1

n 1 s

1

. . . n s k

k

.

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Application of the theorem

A = functions from Ω = C \ (]−∞, 0[ ∪ ]1, +∞[) to C . C = field of functions on Ω (germs of analytic functions).

T = generating series of polylogs : T (z ) = X

w∈X

Li w (z)w.

M (z ) = 1

z x 0 + 1 1 − z x 1 .

Differential equation : Drinfel’d equation d

dz T (z ) = M (z)T (z ).

Consequence : Linear independance of polylogs over C .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 12 / 22

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Let X = {x 0 , x 1 }. u 0 (z ) = 1 and u 1 (z) = 1 z . Encoding integrals : x i

Z z z

i

·u i (s )ds x i = α z z

i

(x i )x i , z 0 = 0, z 1 = 1.

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Let X = {x 0 , x 1 }. u 0 (z ) = 1 and u 1 (z) = 1 z . Encoding integrals : x i

Z z z

i

·u i (s )ds x i = α z z

i

(x i )x i , z 0 = 0, z 1 = 1.

1 Z z

0

x 0 ds = zx 0

Z z 1

ds

s x 1 = ln(z )x 1

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 13 / 22

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Let X = {x 0 , x 1 }. u 0 (z ) = 1 and u 1 (z) = 1 z . Encoding integrals : x i

Z z z

i

·u i (s )ds x i = α z z

i

(x i )x i , z 0 = 0, z 1 = 1.

1 Z z

0

x 0 ds = zx 0

Z z 1

ds

s x 1 = ln(z )x 1 Z z

0

sx 0 ds

s x 1 = zx 0 x 1

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Let X = {x 0 , x 1 }. u 0 (z ) = 1 and u 1 (z) = 1 z . Encoding integrals : x i

Z z z

i

·u i (s )ds x i = α z z

i

(x i )x i , z 0 = 0, z 1 = 1.

1 Z z

0

x 0 ds = zx 0

Z z 1

ds

s x 1 = ln(z )x 1 Z z

0

sx 0 ds

s x 1 = zx 0 x 1

α z 0 (x 0 x 1 n ) ≡ α z 0 (x 0 ).

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 13 / 22

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Let X = {x 0 , x 1 }. u 0 (z ) = 1 and u 1 (z) = 1 z . Encoding integrals : x i

Z z z

i

·u i (s )ds x i = α z z

i

(x i )x i , z 0 = 0, z 1 = 1.

1 Z z

0

x 0 ds = zx 0

Z z 1

ds

s x 1 = ln(z )x 1 Z z

0

sx 0 ds

s x 1 = zx 0 x 1

α z 0 (x 0 x 1 n ) ≡ α z 0 (x 0 ).

Problem : Third condition of the theorem.

iii) The family (u x ) x∈X is such that, for f ∈ C and α x ∈ k d(f ) = X

α x u x = ⇒ (∀x ∈ X )(α x = 0).

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Hyperlogarithms

Definition (1928, Lappo-Danilevski) Let a 0 , . . . a k ∈ C . Then

L(a i

n

, . . . , a i

1

|γ ) = Z z

z

0

Z s

n

z

0

. . . Z s

2

z

0

ds 1

s 1 − a i

1

. . . ds n s n − a i

n

with γ : z 0 z a path such that

a j

i

∈ / γ and s i ∈ γ, ∀i ∈ {1, . . . , n} . If z 0 6= a i

1

, the integral converges.

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 14 / 22

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Hyperlogarithms

Definition (1928, Lappo-Danilevski) Let a 0 , . . . a k ∈ C . Then

L(a i

n

, . . . , a i

1

|γ ) = Z z

z

0

Z s

n

z

0

. . . Z s

2

z

0

ds 1

s 1 − a i

1

. . . ds n s n − a i

n

with γ : z 0 z a path such that

a j

i

∈ / γ and s i ∈ γ, ∀i ∈ {1, . . . , n} . If z 0 6= a i

1

, the integral converges.

Our theorem applies as well to families of inputs of the type u i (z) = λ i

z − a i , λ i ∈ C .

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Field of germs

Idea : Field of analytic functions with variable domains.

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 15 / 22

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Field of germs

Idea : Field of analytic functions with variable domains.

Ω : connected, simply connected, analytic domain.

B : a filter basis of Ω of open connected (non void) subsets of Ω :

∀S i , S j ∈ B , ∃S k ∈ B , S k ⊂ S i ∩ S j .

S i

S j

S k

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Field of germs

Idea : Field of analytic functions with variable domains.

Ω : connected, simply connected, analytic domain.

B : a filter basis of Ω of open connected (non void) subsets of Ω ; A correspondence C such that

∀U ∈ B , C [U] is a subring of C ω (U , C ), satisfying :

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 15 / 22

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Field of germs

Idea : Field of analytic functions with variable domains.

Ω : connected, simply connected, analytic domain.

B : a filter basis of Ω of open connected (non void) subsets of Ω ; A correspondence C such that

∀U ∈ B , C [U] is a subring of C ω (U , C ), satisfying :

Inverse : if f ∈ C [U ] \ {0}, ∃W ∈ B such that W ⊂ U − O f and

f −1 (defined on W ) is in C [W ].

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Field of germs

Idea : Field of analytic functions with variable domains.

Ω : connected, simply connected, analytic domain.

B : a filter basis of Ω of open connected (non void) subsets of Ω ; A correspondence C such that

∀U ∈ B , C [U] is a subring of C ω (U , C ), satisfying : Inverse : if f ∈ C [U ] \ {0}, ∃W ∈ B such that W ⊂ U − O f and f −1 (defined on W ) is in C [W ].

Compatibility with restrictions : ∀U , W ∈ B , W ⊂ U, res WU (C [U ]) ⊂ C [W ] .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 15 / 22

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Example

What is the good domain if the inputs are u i (z) = λ i z − a i ?

It is always possible to cut the complex plane with half rays to form a simply connected domain on which the u i ’s are analytic :

0

+ +

+ +

+

+ +

+ +

+

b

a

1

b

a

2

b

a

3

b

a

4 b

a

5

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Encoding integrals with words

X = {x 1 , . . . , x n }.

Definition of the iterated integrals α z z

0

(w ) for w ∈ X and z 0 , z ∈ Ω:

α z z

0

(1 X

) = 1;

α z z

0

(x i ) = Z z

z

0

u i (s)ds, x i ∈ X ; α z z

0

(x i w) =

Z z z

0

u i (s)ds α s z

0

(w ), x i ∈ X , w ∈ X .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 17 / 22

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Encoding integrals with words

X = {x 1 , . . . , x n }.

Definition of the iterated integrals α z z

0

(w ) for w ∈ X and z 0 , z ∈ Ω:

α z z

0

(1 X

) = 1;

α z z

0

(x i ) = Z z

z

0

u i (s)ds, x i ∈ X ; α z z

0

(x i w) =

Z z z

0

u i (s)ds α s z

0

(w ), x i ∈ X , w ∈ X . Then if u i (z ) = λ i

z − a i ↔ x i ,

α z z

0

(x j

0

. . . x j

n

) = L(a j

n

, . . . , a j

0

|γ ) with γ : z 0 z in Ω.

Generating series of hyperlogarithms : T (z) := X

w∈X

α z z

0

(w )w .

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Encoding integrals with words

X = {x 1 , . . . , x n }.

Definition of the iterated integrals α z z

0

(w ) for w ∈ X and z 0 , z ∈ Ω:

α z z

0

(1 X

) = 1;

α z z

0

(x i ) = Z z

z

0

u i (s)ds, x i ∈ X ; α z z

0

(x i w) =

Z z z

0

u i (s)ds α s z

0

(w ), x i ∈ X , w ∈ X . Then if u i (z ) = λ i

z − a i ↔ x i ,

α z z

0

(x j

0

. . . x j

n

) = L(a j

n

, . . . , a j

0

|γ ) with γ : z 0 z in Ω.

General idea : Derivating T term by term, we obtain the following non commutative differential equation :

d

dz T (z ) = M (z )T (z ), with M(z ) = X

x

i

∈X

u i (z )x i .

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 17 / 22

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Integrator

Let M (z ) = X

x

i

∈X

u i (z )x i = X

x

i

∈X

1

z − a i x i and z 0 ∈ C . We define the integrator H z

0

:

H z

0

:

A hhX ii → A ≥1 hhX ii S 7→ H z

0

[S ] =

Z z z

0

M(s)S(s )ds

Since ∀S , H z n

0

[S ] ∈ A ≥n hhX ii,

hH z n

0

[S] |w i 6= 0 only for n ≤ |w |.

Therefore, we can define the sum X

w∈X

X

n≥0

hH z n

0

[S ] |w iw = H z

0

[S] = X

n≥0

H z n

0

[S ] .

(38)

(Non commutative) Differential equation

It is clear that

H z

0

= 1 + H z

0

H z

0

Therefore, ∀ S ∈ A hhX ii such that dS = 0 (constant series), d H z

0

[S]

= d S + H z

0

H z

0

[S]

= MH z

0

[S ] , and H z

0

[S ] satisfies the (non commutative) differential equation

dP = MP.

Since

H z

0

[1] = T (z ) = X

w∈X

α z z

0

(w )w we obtain the promised differential equation.

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 19 / 22

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Application of the theorem

X = {x 1 , . . . , x n }.

B : filter basis obtained by cutting the complex plane.

C = Field of germs of functions on B fulfilling condition i) of theorem 1 (for example, the field of rational functions or the field of functions that are inessential at all the points a i ).

T = generating series of hyperlogs : T (z) = X

w∈X

α z z

0

(w )w .

M (z ) =

n

X

i =1

λ i

z − a i x i , λ i 6= 0, ∀i.

Consequence : Linear independance of hyperlogs over C .

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Conclusion and perspectives

Conclusion :

New and simpler proof of known results (without monodromy) ; Generalization of these results to a wider class of algebras.

Perspectives : Implementation.

M. Deneufchˆatel (LIPN - Universit´e Paris 13) Independence of Hyperlogarithms 09/20/2011 21 / 22

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Thank you for your attention!

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