Periods in action Pierre Lairez Inria Saclay MEGA 2017 Mthodes - PowerPoint PPT Presentation

Periods in action Pierre Lairez Inria Saclay MEGA 2017 Méthodes efgectives en géométrie algébrique 16 June 2017, Nice

What is a period? rational coefgicients. “Rational coefgicients” may mean This is what we will be interested in. Etymology A period is the integral on a closed path of a rational function in one or several variables with • coefgicients in Q • coefgicients in C ( t ) , the period is a function of t . • 2 π is a period of the sine. ∫ z d x • arcsin( z ) = � 1 − x 2 0 � � 1 − x 2 = 1 d x d x d y • 2 π = � • • y 2 − (1 − x 2 ) π i − 1 1 ∞

Periods with a parameter Liouville (1834) Not expressible in terms of elementary functions Complete elliptic integral since then Many applications in algebraic geometry (Gauß-Manin connection) An ellipse 1 eccentricity t t F ′ O F major radius 1 perimeter E ( t ) √ � 1 − t 2 x 2 • • • • E ( t ) = 1 − x 2 d x − 1 1 ∞ Euler (1733) ( t − t 3 ) E ′′ + (1 − t 2 ) E ′ + tE = 0 geometry of the cycles ↔ analytic properties of the periods

Content 1 Computing periods 2 Multiple binomial sums 3 Volume of semialgebraic sets

Computing periods

Difgerential equations as a data structure (also I An example by Bostan, Chyzak, van Hoeij, and Pech (2011) Representation of D-finite functions (+ root location) (+ init. cond.) Representation of algebraic numbers explicit � � � √ 5 + 2 6 2 + 3 ) implicit x 4 − 10 x 2 + 1 = 0 � ( ) � 27 w (2 − 3 w ) 1/3 2/3 � 2 F 1 ∫ t 2 (1 − 4 w ) 3 explicit 1 + 6 · d w (1 − 4 w )(1 − 64 w ) 0 implicit t ( t − 1)(64 t − 1)(3 t − 2)(6 t + 1) y ′′′ + (4608 t 4 − 6372 t 3 + 813 t 2 + 514 t − 4) y ′′ + 4(576 t 3 − 801 t 2 − 108 t + 74) y ′ = 0

Difgerential equations as a data structure II What can we compute? • addition, multiplication, composition with algebraic functions • power series expansion • equality testing , given difgerential equations and initial condtions • numerical analytic continuation with certified precision (D. V. Chudnovsky and G. V. Chudnovsky 1990; van der Hoeven 1999; Mezzarobba 2010) sage: from ore_algebra import * sage: dop = (zˆ2+1)*Dzˆ2 + 2*z*Dz sage: dop.numerical_solution(ini=[0,1], path=[0,1]) [0.78539816339744831 +/- 1.08e-18] sage: dop.numerical_solution(ini=[0,1], path=[0,i+1,2*i,i-1,0,1]) [3.9269908169872415 +/- 4.81e-17] + [+/- 4.63e-21]*I

The Picard-Fuchs equation Back to the periods with polynomial coefgicients One equation fits all cycles, the Picard-Fuchs equation . R ( t , x 1 ,..., x n ) a rational function γ ⊂ C n a n -cycle ( n -dim. compact submanifold) which avoids the poles of R , for t ∈ U ⊂ C � define y ( t ) ≜ R ( t , x 1 ,..., x n )d x 1 ··· d x n , for t ∈ U γ wanted a difgerential equation a r ( t ) y ( r ) +···+ a 1 ( t ) y ′ + a 0 ( t ) y = 0 ,

A computational handle Computational proof Perimeter of an ellipse R ( t , x , y ) �√ � �� � � 1 − t 2 x 2 1 1 recall E ( t ) = 1 − x 2 d x = d x d y 1 − t 2 x 2 2 π i 1 − ( 1 − x 2 ) y 2 Picard-Fuchs equation ( t − t 3 ) E ′′ + (1 − t 2 ) E ′ + tE = 0 ( t − t 3 ) ∂ 2 R ∂ t 2 + (1 − t 2 ) ∂ R ∂ t + tR = ( 2 t ( − 1 + t 2 ) x ( 1 + x 3 ) y 3 ( ) ) − t ( − 1 − x + x 2 + x 3 ) y 2 ( − 3 + 2 x + y 2 + x 2 ( − 2 + 3 t 2 − y 2 )) ∂ + ∂ 2 2 ∂ x ( − 1 + y 2 + x 2 ( t 2 − y 2 )) ∂ y ( − 1 + y 2 + x 2 ( t 2 − y 2 ))

Computing periods Theory and practice (People who wrote a paper that solves the problem.) Oaku, Salvy, Singer, Takayama, Wilf, Zeilberger, etc. Kauers, Koutschan, Lairez, Lipshitz, Movasati, Nakayama, Nishiyama, algorithms Almkvist, Apagodu, Bostan, Chen, Christol, Chyzak, van Hoeij, existence Grothendieck (1966), Monsky (1972), etc. . Problem (mostly) solved! given R ( t , x 1 ,..., x n ) , a rational function find a 0 ,..., a r ∈ Q [ t ] , with a r ̸= 0 and r minimal C 1 ,..., C n ∈ Q ( t , x 1 ,..., x n ) with poles( C i ) ⊆ poles( R ) , such that a r ( t ) ∂ r R n ∂ t r +···+ a 1 ( t ) ∂ R ∂ C i ∑ ∂ t + a 0 ( t ) R = ∂ x i i = 1 see also Picard (1902) for n � 3

Multiple binomial sums joint work with Alin Bostan and Bruno Salvy

What are binomial sums? (Strehl) Examples (Dixon) ( ) 3 2 n 2 n = ( − 1) n (3 n )! ∑ ( − 1) k ( n !) 3 k k = 0 ( ) 2 ( ) 2 ( )( ) ( ) 3 n k n n + k n n + k k ∑ ∑ ∑ = k k k k j j = 0 k = 0 k = 0 ( ) 2 ( ) ( ) 2 n n i + j 4 n − 2 i − 2 j 2 n ∑ ∑ = (2 n + 1) i 2 n − 2 i n i = 0 j = 0 ( )( )( )( )( ) ( ) 4 n n n + s n + r 2 n − r − s n ∑ ∑ ∑ ( − 1) n + r + s = r s s r n k r � 0 s � 0 k � 0

What are binomial sums? Conjectured by Brent, Ohtsuka, Osborn, and Prodinger (2014) Conjectured by Le Borgne More examples Both proved using periods! ( )( ) ( ) 2 � = 2 n 4 ( n − 1)(3 n 2 − 6 n + 2) 2 n 2 n 2 n ∑ ∑ � � � i 3 j 3 ( i 2 − j 2 ) n + i n + j (2 n − 3)(2 n − 1) n i j 1 + F − 1, − 1 + 2 F 0,0 − F 0,1 + F 1,0 − 3 F 1,1 + F 1,2 − F 3,1 + 3 F 3,2 n n n n n n n n ( n + 2 )( n + 2 )( n + 2 ) n ∑ m m + 1 m + 2 − F 3,3 − 2 F 4,2 + F 4,3 − F 5,2 , = n n n n ( n + 2 )( n + 2 ) m = 0 1 2 n − 1 d − a )(( n + d + 1 − 2 a − 2 c + 2 b )) ∑ ∑ ( d − a − c )( ) ( n + d + 1 − 2 a − 2 c + 2 b where F a , b n = − n c d − a − c n − a − c + b n + 1 − a − c + b c = 0 d = 0

What are binomial sums? Definition The not so formal grammar of binomial sums → integer linear combination of the variables ( ) → → Cst ∑ → → + → · n =

Computing binomial sums with periods basic block Example product conclusion Generating functions of binomial sums are periods! ( ) 3 2 n 2 n ( − 1) n (3 n )! ∑ ( − 1) k = ? n ! 3 k k = 0 � (1 + x ) n ( ) n 1 d x = k 2 π i x k x � (1 + x 1 ) 2 n ( ) 3 (1 + x 2 ) 2 n (1 + x 3 ) 2 n 2 n d x 1 d x 2 d x 3 1 = (2 π i ) 3 k x k x k x k x 1 x 2 x 3 1 2 3 ( i = 1 (1 + x i ) 2 ) x 1 x 2 x 3 − t ∏ 3 � d x 1 d x 2 d x 3 1 summation y ( t ) = ( i = 1 (1 + x i ) 2 )( i = 1 (1 + x i ) 2 ) 3 − t ∏ 3 1 − t ∏ 3 (2 i π ) 3 x 2 1 x 2 2 x 2 where y ( t ) is the generating function of the l.h.s. � 1 x 1 x 2 d x 1 d x 2 simplification y ( t ) = (2 i π ) 2 x 2 1 x 2 2 − t (1 + x 1 ) 2 (1 + x 2 ) 2 (1 − x 1 x 2 ) 2 integration t (27 t + 1) y ′′ + (54 t + 1) y ′ + 6 y = 0 , i.e. 3(3 n + 2)(3 n + 1) u n + ( n + 1) 2 u n + 1 = 0

Computing binomial sums with periods • Many related works on multiple sums (Chyzak, Egorychev, Garoufalidis, Koutschan, Sun, Wegshaider, Wilf, Zeilberger, etc) algorithmically Theorem + Algorithm (Bostan, Lairez, and Salvy 2016) One can decide the equality between binomal sums. —Wilf and Zeilberger, 1992 • Excellent running times, thanks to simplification and better algorithms for integration • Subtelties in the translation recurrence operators → actual sequences , not handled • “This approach, while it is explicit in principle, in fact yields an infeasible algorithm.”

Binomial sums are diagonals of rational functions Theorem (Bostan, Lairez, and Salvy 2016) The converse does not hold, but... Corollaries of Furstenberg’s theorems (Furstenberg 1967) . (but finitely many). ( u n ) n � 0 is a binomial sum if and only if u n = a n ,..., n , for some rational power series ∑ I a I x I . ) d x 2 � ( ··· d x n 1 t NB. diag R ( x 1 ,..., x n ) = R x 2 ··· x n , x 2 ,..., x n (2 π i ) n − 1 x 2 x n • If ∑ n u n t n is algebraic, then ( u n ) n � 0 is a binomial sum. • If ( u n ) n � 0 is a binomial sum, then ∑ n u n t n is algebraic modulo p for all prime p

Algebricity modulo a prime Apéry’s numbers (Straub 2014) ( ) 2 ( ) 2 n n + k n ∑ ∑ t n y ( t ) ≜ k k n k = 0 1 = diag (1 − x − y )(1 − z − w ) − wxyz y ( t ) is transcendental, however y ( t ) ≡ ( t 2 − 1) − 1 mod 5 2 ) − 1 ( ( t − 1)( t 2 − 1) y ( t ) ≡ 3 mod 7 y ( t ) ≡ (( t + 1)( t + 5)( t + 7)( t + 8)( t + 9)) − 1 mod 11 5 ) − 1 ( ( t 2 + 8 t + 1)( t 2 + 6 t + 1)( t − 1) 6 ( t 2 + 5 t + 1) − 1 y ( t ) ≡ mod 13 12 and of course t 2 ( t 2 − 34 t + 1) y ′′′ + 3 t (2 t 2 − 51 t + 1) y ′′ + (7 t 2 − 112 t + 1) y ′ + ( t − 5) y = 0.

Volume of semialgebraic sets joint work with Mohab Safey El Din

A numeric integral What is the volume of this shape? • Basic question • Few algorihms • Monte-Carlo • Henrion, Lasserre, and Savorgnan (2009) • Exponential complexity with respect to precision • No certification on precision { x 2 + y 2 + z 2 � 1 − 2 10 ( x 2 y 2 + y 2 z 2 + z 2 x 2 )}

Volumes are periods proof Stokes formula + Leray tube map satisfies a Picard-Fuchs equation! not so useful There is no parameter. Proposition For any generic f ∈ R [ x 1 ,..., x n ] , } ≜ ∫ � 1 x 1 ∂ f { vol f � 0 d x 1 ··· d x n = d x 1 ··· d x n . 2 π i f ∂ x 1 { f � 0 } Tube { f = 0 } better say For a generic t , � 1 x 1 ∂ f | x n = t { } { } vol f � 0 x n = t d x 1 ··· d x n − 1 ∩ = 2 π i f | x n = t ∂ x 1 � �� � ∫ ∞ { } { } { } NB. vol f � 0 = vol f � 0 ∩ x n = t d t −∞

The “volume of a slice” function , basis of the solution space of the Picard-Fuchs equation volume of the slice { } y 1 , y 2 z coordinate − 1 1 0 · y 1 + 0 · y 2 1.0792353... · y 1 − 40.100605... · y 2 0 · y 1 + 0 · y 2