Diagonals of rational functions Main Conference of Chaire J Morlet Artin approximation and infinite dimensional geometry 27 mars 2015 Pierre Lairez TU Berlin
Diagonals: definitions and properties Binomial sums Computing diagonals Diagonals: definitions and properties Binomial sums Computing diagonals
Diagonal of a power series Définition f = a i1,,i n x i1 1 x i n n Q x 1,,x n i 1,,i n N n diag f def = a i,,i t i i 0
A combinatorial problem Counting rook paths y 2 1 3 6 4 5 6 5 3 4 (7, 10) 7 a i,j def = nb of rook paths from (0,0) to (i,j) Easy recurrence: a i,j = a k,j + k <i k <j a i,k (0,0) 2 1 x What about a n,n? asymptotic? existence of a recurrence?
Recurrence relations for rook paths dimension 2 9nu n (14 + 10n)u n+1 + (2 + n)u n+2 = 0 dimension 3 192n 2 (1 + n)(88 + 35n)u n +(1 + n)(54864 + 100586n + 59889n 2 + 11305n 3 )u n+1 (2 + n)(43362 + 63493n + 30114n 2 + 4655n 3 )u n+2 +2(2 + n)(3 + n) 2 (53 + 35n)u n+3 = 0 dimension 4 5000n 3 (1 + n) 2 (2705080 + 3705334n + 1884813n 2 + 421590n 3 + 34983n 4 )u n (1 + n) 2 (80002536960 + 282970075928n + + 6386508141n 6 + 393838614n 7 )u n+1 +2(2 + n)(143370725280 + 500351938492n + + 2636030943n 7 + 131501097n 8 )u n+2 (3 + n) 2 (26836974336 + 80191745800n + 100381179794n 2 + + 44148546n 7 )u n+3 +2(3 + n) 2 (4 + n) 3 (497952 + 1060546n + 829941n 2 + 281658n 3 + 34983n 4 )u n+4 = 0
Differential equation for diagonals a i,j = a k,j + a i,k a i,j x i y j = k <i k <j i,j 0 n 0 a n,n t n = diag 1 1 x 1 x y 1 y 1 1 x 1 x y 1 y Theorem (Lipshitz 1988) diagonal differentially finite If R Q(x 1,,x n ) Q x 1,,x n, then diag R satisfies a linear differential equation with polynomial coefficients c r (t)y (r ) + + c 1 (t)y + c 0 (t)y = 0
More properties of diagonals Theorem (Furstenberg 1967) algebraic diagonal If f (t) = a n t n is an algebraic series (ie P(t, f (t)) = 0 for some P Q[x,y]), then it is the diagonal of a rational power series Theorem (Furstenberg 1967) diagonal algebraic mod p If a n t n Q t is the diagonal of a rational power series, then it is an algebraic series modulo p for almost all prime p
Algebricity modulo p Example f = n ( (3n)! n! 3 tn = diag 1 1 x y z ) is not algebraic f (1 + t) 1 4 mod 5 f ( 1 + 6t + 6t 2) 1 6 mod 7 f ( 1 + 6t + 2t 2 + 8t 3) 1 10 mod 11 Besides, ( 27t 2 t ) f + (54t 1)f + 6f = 0
Proof of algebricity modulo p F q, the base field Slicing operators For r Z, E r i a i t i def = i a qi+r t i and E r I a I x I def = a qi+(r,,r ) x I I We check diag E r = E r diag ; x i E r (F) = E r (x q i F) ; G(x)E r (F) = E r (G(x) q F), because G(x q ) = G(x) q, where x q = x q 1,,xq n; If f (t) = i a i t i, then f (t) = 0 r <q t r i a qi+r t qi
Proof of algebricity modulo p F q, the base field Slicing operators For r Z, E r i We check a i t i def = diag E r = E r diag ; x i E r (F) = E r (x q i F) ; i a qi+r t i and E r I a I x I def = G(x)E r (F) = E r (G(x) q F), because G(x q ) = G(x) q, where x q = x q 1,,xq n; If f (t) = i a i t i, then a qi+(r,,r ) x I I f (t) = 0 r <q t r i a qi+r t i q
Proof of algebricity modulo p F q, the base field Slicing operators For r Z, E r i a i t i def = i a qi+r t i and E r I a I x I def = a qi+(r,,r ) x I I We check diag E r = E r diag ; x i E r (F) = E r (x q i F) ; G(x)E r (F) = E r (G(x) q F), because G(x q ) = G(x) q, where x q = x q 1,,xq n; If f (t) = i a i t i, then f (t) = 0 r <q t r E r (f ) q
Proof of algebricity modulo p Let R = A F F q(x), d = max(deg A,deg F) and the F q -vector space V = { diag ( ) P deg P d } ( ( F F q t dim V d+n )) n 1 Operators E r stabilize V Proof ( ) P PF E r diag = diag E q 1 F r F q = diag E r (PF q 1 ) V F
Proof of algebricity modulo p Let R = A F F q(x), d = max(deg A,deg F) and the F q -vector space V = { diag ( ) P deg P d } ( ( F F q t dim V d+n )) n 1 Operators E r stabilize V 2 Let f 1,, f s be a basis of V There exist c ij F q [t] such that i, f i = c ij f q j Proof f i = 0 r <q ( = j t r E r (f i ) q = 0 r <q j b ij t r ) f q j 0 r <q t r ( j b ij f j ) q
Proof of algebricity modulo p Let R = A F F q(x), d = max(deg A,deg F) and the F q -vector space V = { diag ( ) P deg P d } ( ( F F q t dim V d+n )) n 1 Operators E r stabilize V 2 Let f 1,, f s be a basis of V There exist c ij F q [t] such that i, f i = c ij f q j j 3 All the elements of V are algebraic Proof i, f q i = j cq ij f q2 j, etc { Thus, over F q (t), Vect (R) } qk 0 k s } Vect {f qk i 0 k s, 1 i s Vect { f qs i 1 i s}
Characterization of diagonals? Conjecture (Christol 1990) integer coefficients + convergent + diff finite diagonal If a n t n Z t, has radius of convergence > 0, and satisfies a linear differential equation with polynomial coefficients, then it is the diagonal of a rational power series A hierarchy of power series For f Q t, let N(f ) be the minimum number of variables x 1,,x N(f ) such that f = diag R(x 1,,x N(f ) ), with R rational power series, if any N(f ) = 1 f is rational N(f ) = 2 f is algebraic irrational ( ) N n (3n)! t n = 3 n! 3 Question : Find a f such that 3 < N(f ) <
Diagonals: definitions and properties Binomial sums Computing diagonals Diagonals: definitions and properties Binomial sums Computing diagonals
Binomial sums Examples 2n k=0 n ( ) 2 ( ) 2 n n + k = k k k=0 n i=0 j=0 ( ) 3 2n ( 1) k = ( 1) n (3n)! k (n!) 3 k=0 ( n k ) ( ) k n + k k j=0 n ( ) 2 ( ) i + j 4n 2i 2j = (2n + 1) i 2n 2i ( 1) ( ) ( n+r +s n n r s r 0 s 0 ) ( n+s s ) ( n+r r (Dixon) ( ) 3 k (Strehl) j ( ) 2 2n n ) ( 2n r s ) ( n = n 4 k) k 0
Binomial sums Further examples Number theory n 3 u n + (n 1) 3 u n 2 = (34n 3 51n 2 + 27n 5)u n 1 n ( ) 2 ( ) 2 n n + k avec u n = (Apéry) k k k=0 Important step in proving that ζ(3) = n 1 1 n 3 Analysis of algorithm Q [50] Develop computer programs for simplifying sums that involve binomial coefficients Exercise 12663 The Art of Computer Programming Knuth (1968)
Definition δ : Z Q is a binomial sum (δ 0 = 1 et δ n = 0 si n 0) n Z a n Q is a binomial sum for all a Q (n,k) Z 2 ( n k) Q is a binomial sum If u,v : Z p Q are bs, then u + v and uv are bs If u : Z p Q are bs and λ : Z p Z q is an affine map, then u λ is a bs If u : Z p Q is a bs, then n 1 (n 1,,n p ) Z p k=0 u k,n2,,n p Q is a bs
Diagonals binomial sums Theorem (Bosta, Lairez, Salvy 2014) A sequence (u n ) n 0 is a binomial sum if and only if its generating function u n t n is the diagonal of a rational series Example n n 0 k=0 ( n k ) 2 ( n+k k ) 2 ( t n = diag 1 (1 x 1 )(1 x 2 )(1 x 3 ) x 4 (x 1 +x 2 x 3 x 1 x 2 x 3 ) )
Corollaries Corollary 1 binomial sum recurrence If (u n ) n 0 is a binomial sum, then it satisfies a linear recurrence with polynomial coefficients Corollary 2 algebraic gf binomial sum If u n t n is an algebraic series then (u n ) n 0 is a binomial sum Corollary 3 binomial sum algebraic gf mod p If (u n ) n 0 is a binomial sum, then u n t n is an algebraic series modulo p for almost all prime p Conjecture integral + exp bounded + recurrence binomial sum If (u n ) n 0 Z N grows at most exponentially and satisfies a linear recurrence with polynomial coefficients, then it is a binomial sum
Binomial sums as diagonals Skectch of the proof Proposition All binomial sums are linear combinations of sequences in the form (k 1,,k p ) Z p [1] ( R 0 R k 1 1 R k ) d d, where R 0,,R p Q(x 1,,x d ) With one variable, [1] (RS n ) = [x1 n xn d+1 ]( R ) 1 x 1 x d+1 S } {{ } may not be a power series
Diagonals as binomial sums Skectch of the proof Let R = 1 + a 1 x m 1 + + x m Q(x 1,,x d ) r ( ) R = x m k1 + + k 0 r a k 1 k k N r 1,,k 1 ak r r x k 1m 1 x k r m r r } {{ } ( k1 + +k r k 1,,k r x m 0 ) = ( k1 + +k r binomial sum [x n 1 xn d ]R = k 1 C k ) ( k2 + +k r k 2 k Z e C k 1 Γ (n,k) où ) ( kr ) 1 +k r k r, so Ck is a Γ = { e (n,k) R R e + m 0 + k i m i = (n,,n) } i=1
Diagonals: definitions and properties Binomial sums Computing diagonals Diagonals: definitions and properties Binomial sums Computing diagonals
Algorithmic Lipshitz theorem Theorem (Lipshitz 1988) diagonal differentially finite If R Q(x 1,,x n ) Q x 1,,x n, then diag R satisfies a linear differential equation with polynomial coefficients c r (t)y (r ) + + c 1 (t)y + c 0 (t)y = 0 How to compute the differential equation? Once computer, we get everything we want about the diagonal It would allow to prove identities between binomial sums
Diagonals as integrals Basic fact: C, diag ( x i C x i x j C x j ) = 0 Consider the following transformation T : Q(x 1,,x n ) Q(t,x 1,,x n 1 ) ( ) 1 t R R x 1,,x n 1, x 1 x n 1 x 1 x n 1 T (R) = n 1 i=1 diag R = 1 (2iπ) n C i diag R = 0 x i T (R)dx 1 dx n 1
Computing integrals K a field of characteristic 0 with a derivation δ (usually K = Q(t) and δ = t ) R = a f K(x 1,,x n ) Problem Find c 0,,c r K such that C 1,,C n K(x) c r δ r (R) + + c 1 δ(r) + c 0 (R) = n i=1 C i x i Problem (bis) Compute a basis and normal forms in K(x 1,,x n )/ n i=1 x i K(x 1,,x n )
Computing integrals K a field of characteristic 0 with a derivation δ (usually K = Q(t) and δ = t ) R = a f K(x 1,,x n ) Problem Find c 0,,c r K such that C 1,,C n K(x) c r δ r (R) + + c 1 δ(r) + c 0 (R) = n i=1 C i x i Problem (bis) Compute a basis and normal forms in K[x 1,,x n, f 1 ]/ =: H n Rham (An K n i=1 x i K[x 1,,x n, f 1 ] \ V(f )) Hn 1 Rham (V(f ))
Finiteness of the de Rham cohomology Theorem (Grothendieck 1966) H n Rham (An K \ V(f )) is a finite dimensional K-vector space Theorem (Griffiths 1969) dim K H n Rham (An K \ V(f )) < (deg f + 1)n Corollary The diagonal of R(x 0,,x n ) is solution of a linear differential equation with polynomial coefficients of order at most (d + 1) n, where d is the degree of the denominator of T (R)
Reduction of pole order Homogeneous case f K[x 0,,x n ] homogeneous def V f = { a f homogeneous of degree n 1 } q def Notation : i = x i c Fact : i f q 1 = ic f q 1 (q 1)c i f f q Rewriting rule i c i i f f q 1 q 1 i i c i f q 1 (maps V f V f ) Theorem (Griffiths 1969) If V(f ) P n K is smooth then a f q V a f, f q = c i i f s a f q 0 i
Reduction of pole order Homogeneous singular case Rewriting rules are ambiguous: if i c i i f = 0, then 0 i i c i f q We can add the rules i i c i 0, it still preserves equivalence classes modulo derivatives New reductions 0 rk r b f appear, we add the rules b rk r + 1 q f 0 q f q rg 2 Theorem There exists an r > 0 such that for all a f q V f a f q V f, a f q = i i c i f s a f q rk r 0 Leads to an efficient algorithm for computing rational integrals (Lairez 2015)
An example f = 2xyz(w x)(w y)(w z) w 3 (w 3 w 2 z + xyz) e(q,r) : number of indepedent rational functions a/f q that are not reducible with rules of rank r q 0 1 2 3 4 q > 4 no rule 0 10 165 680 1771 36q 3 e(q, 1) 0 10 86 102 120 18q e(q, 2) 0 10 7 6 6 6 e(q, 3) 0 9 1 0 0 0 dim H 3 (P 3 \ V(f )) = 10