Some hypergeometric functions in positive characteristic and transcendence

C. R. Acad. Sci. Paris, Ser. I 357 (2019) 317–322

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Number theory

Some hypergeometric functions in positive characteristic and transcendence Quelques fonctions hypergéométriques en caractéristique positive et transcendance Mao-Sheng Li Department of Mathematical Sciences, Tsinghua University, Beijing 100084, PR China

a r t i c l e

i n f o

Article history: Received 26 January 2019 Accepted after revision 26 March 2019 Available online 12 April 2019 Presented by the Editorial Board

a b s t r a c t In this work, we study some special hypergeometric functions in positive characteristic, introduced by D. S. Thakur. We shall establish functional relationships among them, and deduce, with the help of the function field version of the Schneider–Lang theorem obtained by J. Yu, that at least one of their values at nonzero algebraic arguments is transcendental. © 2019 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

r é s u m é Nous étudions dans ce travail certaines fonctions hypergéométriques spéciales en caractéristique positive, introduites par D. S. Thakur. Nous allons établir des relations fonctionnelles auxquelles elles satisfont, et déduire, à l’aide de la version pour les corps de fonctions du théorème de Schneider–Lang obtenue par J. Yu, qu’au moins l’une de leurs valeurs aux arguments algébriques non nuls est transcendante. © 2019 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

Version française abrégée Désignons par Fq le corps fini à q éléments, par Fq [ T ] l’anneau intègre des polynômes en T à coefficients dans Fq et par k := Fq ( T ) le corps de fractions de Fq [ T ]. Pour tous les P , Q ∈ Fq [ T ] avec Q = 0, nous définissons | P / Q |∞ := qdeg P −deg Q , et appelons |·|∞ la valeur absolue ∞-adique sur Fq ( T ). Désignons par Fq (( T −1 )) le complété topologique de Fq ( T ) pour |·|∞ , et par C∞ le complété topologique d’une clôture algébrique fixée de Fq (( T −1 )). Finalement, notons k¯ la clôture algébrique de k dans C∞ .

E-mail address: [email protected]. https://doi.org/10.1016/j.crma.2019.03.006 1631-073X/© 2019 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

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Posons D 0 := 1, et D n :=

n −1 k=0

n

q−(a−1)

k

( T q − T q ) pour tout entier n  1. Définissons (a)n = D n+a−1 , pour tout a ∈ Z>0 .

Soient r , s  0 des entiers. Pour tous les entiers ai , b j > 0 (1  i  r , 1  j  s), la fonction hypergéométrique r F s (a1 , . . . , ar ; b 1 , . . . , b s ; z) est définie par r F s (a1 , . . . , ar ; b 1 , . . . , b s ; z ) :=

+∞  (a1 )n · · · (ar )n qn z . D n (b1 )n · · · (b s )n n =0

Elle a été introduite pour la première fois par D. S. Thakur [9] afin d’imiter sa contrepartie dans le cas réel, et elle partage de nombreuses propriétés avec cette dernière. Pour en savoir plus sur cette fonction ainsi que sa motivation, le lecteur intéressé peut consulter [9] (voir aussi [10]). Si r = s = 0, nous obtenons l’exponentielle de Carlitz 0 F 0 ( z)

= e C ( z) :=

+∞ qn  z n =0

Dn

.

Si r = 0, s = 1, et b1 = m + 1, nous obtenons la fonction de Bessel–Carlitz

J m ( z) :=

+∞ 

zq

m+n

q−m

, et 0 F 1 (−; m + 1; z) = J m .

qm n=0 D m+n D n

Notons que la normalisation ici est légèrement différente de celle de [6]. Nous inspirant du travail de L. Denis [6], nous considérons la fonction hypergéométrique spéciale suivante 0 F s (−; b 1 , . . . , b s ; z )

:=

+∞  n =0

zq

n

D n (b1 )n · · · (b s )n

,

et nous étudions ses valeurs spéciales aux arguments algébriques non nuls. Voici les résultats principaux. Théorème 1. Soient s  1 et 0 < b1  b2  · · ·  b s des entiers. Pour tout γ ∈ k non nul, nous avons





tr.degk F 0 (γ ), F 0 ( T γ ), . . . , F 0 ( T s γ )  1, où nous posons F 0 ( z) = 0 F s (−; b1 , b2 , . . . , b s ; z). Théorème 2. Soient s  1 et 0 < b1  b2  · · ·  b s des entiers. Pour tout γ ∈ k non nul, nous avons





tr.degk F 0 (γ ), F 1 (γ ), . . . , F s (γ )  1, où nous posons F j ( z) = 0 F s (−; b1 , . . . , b j −1 , b j + 1, . . . , b s + 1; z), pour 1  j  s. 1. Statements of the main results Let Fq be the finite field with q element, k := Fq ( T ) the rational function field over Fq with T as the indeterminate, and k∞ := Fq (( T1 )) the field of formal Laurent series in 1/ T over Fq . The latter is also the topological completion of k for the canonical ∞-adic absolute value. Finally, we let C∞ denote the topological completion of a fixed algebraic closure of k∞ , and k¯ denote the algebraic closure of k in C∞ . Set D 0 := 1, and D n :=

n −1 k=0

n

k

q−(a−1)

( T q − T q ) for all integers n  1. Set (a)n = D n+a−1 , for all a ∈ Z>0 . Let r , s  0 be integers.

For all integers ai , b j > 0 (1  i  r , 1  j  s), the hypergeometric function r F s (a1 , . . . , ar ; b1 , . . . , b s ; z) is defined as r F s (a1 , . . . , ar ; b 1 , . . . , b s ; z )

:=

+∞  (a1 )n · · · (ar )n qn z . D n (b1 )n · · · (b s )n n =0

It was introduced firstly by D. S. Thakur [9] to imitate its classical counterpart in the real case, and shares many properties with the latter. For more details on this function and its motivation, the interested reader can consult [9] (see also [10]). If r = s = 0, then we obtain the Carlitz exponential 0 F 0 ( z)

= e C ( z) :=

+∞ qn  z n =0

Dn

.

M.-S. Li / C. R. Acad. Sci. Paris, Ser. I 357 (2019) 317–322

319

If r = 0, s = 1, and b1 = m + 1, then we get the Bessel–Carlitz function

J m ( z) :=

+∞ 

zq

m+n

q−m

qm n=0 D m+n D n

, and 0 F 1 (−; m + 1; z) = J m .

Note that the normalization here is slightly different from that of [6]. As for the classical real case, a natural question is to determine the algebraic nature of the values of the above functions ¯ and at nonzero algebraic arguments. L. I. Wade showed in [13] that e C (γ ) is transcendental over k for all nonzero γ ∈ k, his method is now well known under his name. This result is the exact analog of the classical Hermite–Lindemann theorem ¯ His method (see, for example, [8, p. 37]). L. Denis showed in [6] that J m (γ ) is transcendental over k for all nonzero γ ∈ k. is a part of the well-known motivic method (see, for example, [18,4] for more details). For the general case, D. S. Thakur, Z.-Y. Wen, J.-Y. Yao, and L. Zhao obtained in [12] (see also [11]) a new transcendence criterion by Diophantine approximation that unifies Wade’s method, and showed that r F s (a1 , . . . , ar ; b1 , . . . , b s ; γ ) is transcendental over k, if r < s + 1 and γ ∈ C∞ is nonzero and algebraic such that the separable degree [k(γ ), k]s < q. The last result has been extended recently by J.-Y. Yao [15] to fractional hypergeometric functions, defined by D. S. Thakur [10, p. 226]. Besides the above three methods, the automata method introduced by J.-P. Allouche [1] is also important in the transcendence theory over function fields. For more details on this subject, see also [5,1,2,7,3]. Inspired by the work of L. Denis [6], we shall consider the following special hypergeometric function 0 F s (−; b 1 , . . . , b s ; z ) :=

+∞  n =0

zq

n

D n (b1 )n · · · (b s )n

,

and study its special values at algebraic nonzero arguments. Our main results are the following, which rely heavily on the function field version of the Schneider–Lang theorem obtained by J. Yu [17]. Theorem 1. Let s  1 and 0 < b1  b2  · · ·  b s be integers. For all nonzero γ ∈ k, we have





tr.degk F 0 (γ ), F 0 ( T γ ), . . . , F 0 ( T s γ )  1, where we set F 0 ( z) =

0 F s (−; b 1 , b 2 , . . . , b s ; z).

Theorem 2. Let s  1 and 0 < b1  b2  · · ·  b s be integers. For all nonzero γ ∈ k, we have





tr.degk F 0 (γ ), F 1 (γ ), . . . , F s (γ )  1, where we set F j ( z) = 0 F s (−; b1 , . . . , b j −1 , b j + 1, . . . , b s + 1; z), for 1  j  s. 2. Some preliminary results b s −1

bs

In this section, as above, let s  1 and 0 < b1  b2  · · ·  b s be integers. Set f 0 ( z) = ( F 0 ( z))q , and f j ( z) = ( F j ( z))q for 1  j  s. Inspired by the work of L. Denis [6] about the Bessel–Carlitz function, we obtain the following result, which can be deduced directly from the work of D. S. Thakur [9,10]. Lemma 1. The functions f j ( z) (0  j  s) satisfy the following functional equation:

⎡

⎡

⎤

⎢ ⎢ . ⎣ ..

⎢ ⎢0 ⎢ .. ⎥=⎢ ⎥ ⎢ . ⎦ ⎢ ⎣0

f 0 ( T z) ⎢ f 1 ( T z) ⎥ ⎥ ⎢ ⎢ f 2 ( T z) ⎥

f s ( T z)

Tq

0

b s −1

1 T

.. . 0 0

qb s −b1

⎤⎡ ... 0 ⎥ ... 0 ⎥⎢ ⎥⎢ .. ⎥ ⎢ ⎢ . ⎥ ⎥⎢ ⎣ ... 1 ⎦ ... T

⎤

f 0 ( z) f 1 ( z) ⎥ ⎥ f 2 ( z) ⎥

⎡

0 ⎢0

⎤⎡

⎤q

0 ... 0 f 0 ( z) 0 . . . 0 ⎥ ⎢ f 1 ( z) ⎥ ⎥ ⎥⎢ .. .. ⎥ ⎢ f 2 ( z) ⎥

⎢ ⎢. ⎥ + ⎢ .. . ⎥ . ⎢ . ⎥ ⎥ ⎢ ⎥ ⎥⎢ . .. . ⎦ ⎦ ⎦ ⎣ ⎣ . . 0 0 ... 0 1 0 ... 0 f s ( z) f s ( z)

Proof. For all functions f ( z) and for a ∈ Z, define

 f ( z) := f ( T z) − T f ( z), and a f ( z) := f ( T z) − T q

−a

f ( z).

By the formulas in [9] with r = 0 (see also [10, pp. 229–230]), for all integers c j > 0, we have

( 0 F s (−; c 1 , ..., c s ; z)) = ( 0 F s (0; c 1 + 1, ..., c s + 1; z))q , c j −1 ( 0 F s (−; c 1 , ..., c j , ..., c s ; z)) = 0 F s (−; c 1 , ..., c j − 1, ..., c s ; z), if c j > 1,

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from which and by definition, we deduce

f 0 ( T z) = ( 0 F s (−; b1 , . . . , b s ; T z))q

= Tq = Tq

b s −1

( 0 F s (−; b1 , . . . , b s ; z))q

b s −1

= Tq

b s −b j b s −b j

= = Tq

+ ( 0 F s (−; b1 + 1, . . . , b s + 1; z))q

bs

 −b j qbs = T q F j ( z) + b j F j ( z)

bs

bs

( F j ( z))q + ( F j +1 ( z))q

bs

f j ( z) + f j +1 ( z), for 1  j  s − 1,

f s ( T z) = ( 0 F s (−; b1 , . . . , b s + 1; T z))q



b s −1

f 0 ( z) + f 1 ( z),

f j ( T z) = ( F j ( T z))q

= Tq

 qbs −1 = T ( 0 F s (−; b1 , . . . , b s ; z)) + ( 0 F s (−; b1 , . . . , b s ; z))

b s −1

−b s

bs

qbs ( 0 F s (−; b1 , . . . , b s + 1; z)) + bs ( 0 F s (−; b1 , . . . , b s + 1; z)) bs

= T ( 0 F s (−; b1 , . . . , b s + 1; z))q + ( 0 F s (−; b1 , . . . , b s ; z))q

bs

= T f s ( z) + ( f 0 ( z))q . 2

Hence the desired functional equation holds.

Now let | · |∞ be the canonical ∞-adic absolute value over C∞ such that | T |∞ = q. For all Let f ( z) =

∞

j =0

α ∈ C∞ , set d(α ) :=

log|α |∞ . log q

α j z j be an entire function on C∞ . Define 



Mr ( f ) = max d(α j ) + r j , and ρ f = lim sup j 0

log Mr ( f )

r →+∞

log q

The latter is called the order of f . If f is Fq -linear, i.e., f ( z) =

ρ f = lim sup j →+∞

. ∞ j =0

j

c j zq , then we obtain (see [16]),

−jqj . d(c j )

(1)

Let K be a finite extension of k. For all

α ∈ K , define 

size(α ) = max d(β) : β ∈ k¯ is a conjugate of α . According to J. Yu [17], an Fq -linear entire function f of finite order is an E q -function with respect to K , if one can write f ( z) =

∞

j =0

j

c j zq , where c j ∈ K and ∃C > 0 such that size(c j )  C for all integers j  0, and there exists a sequence

α j ∈ Fq [ T ] \ {0} satisfying the following conditions: (1) there exists a constant c > 0 such that d(α j )  c jq j , for all integers j  1; (2) for all integers j , h (0  j  h), αh c j are algebraic integers in K ; (3) if q j 1 + q j 2 + · · · + q j s < qn , then α j 1 α j 2 · · · α j s divides αn in Fq [ T ]. From the above definition, we deduce immediately the following result. Lemma 2. The function f 0 ( z) is an E q -function with respect to k of order s+1 1 . Proof. By definition, f 0 ( z) =

∞ n=0

n+b s −1

zq

qb s −1 qb s −b1 Dn D n+b −1 ... D n+b s −1 1

. Note that d( D n ) = n qn for all integer n  0, hence f 0 ( z) is entire and

additive. Moreover by the formula (1), we have

ρ f 0 = lim

n→∞ qb s −1 n qn

= lim

n→∞

+ qbs −b1 (n

(n + b s − 1) qn+bs −1 + b1 − 1) qn+b1 −1 + · · · + (n + b s − 1) qn+bs −1

n + bs − 1 1 = . (s + 1)n − s + b1 + · · · + b s s+1

To conclude, it suffices to take

αn = D ns+1 , and note that D n =

n

j =1 ( T

qj

− T )q

n− j

.

2

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321

Finally, we recall without proof the following theorem of J. Yu [17].

¯ Theorem 3. Let K be a finite extension of k. Let g 1 , g 2 be E q -functions with respect to K that are algebraically independent over k. Then, there are only finitely many points at which g 1 , g 2 simultaneously assume values in K . 3. Proofs Proof of Theorem 1. By contradiction, suppose that all the F 0 (γ ), F 0 ( T γ ), . . . , F 0 ( T s γ ) were algebraic over k. Set K = k(γ , f 0 (γ ), f 0 ( T γ ), . . . , f 0 ( T s γ )). Then K is a finite extension of k. Set

d0 = qb s −1 , and d j = qb s −b j (1  j  s). By Lemma 1, we have f j ( T z) = T d j f j ( z) + f j +1 ( z) (0  j  s − 1), and f s ( T z) = T ds f s ( z) + ( f 0 ( z))q , from which we deduce directly s 

f j ( T s − j +1 z ) =

j =0

s 

T d j f j ( T s− j z ) +

j =0

and then we obtain f 0 ( T s+1 z) =

s −1 

f j +1 ( T s− j z) + ( f 0 ( z))q ,

j =0 s j =0

T d j f j ( T s− j z) + ( f 0 ( z))q .

Below, by induction on j (0  j  s), we show that, for 0  i  s − j, each f j ( T i z) is a linear combination of f 0 ( T m z) (0  m  s) with coefficients in Fq [ T ]. The case where j = 0 is direct. Now assume that the desired result holds for all integers j (0  j  n), and n < s. Note that f n+1 ( T i z) = f n ( T i +1 z) − T dn f n ( T i z) for 0  i  s − n − 1; hence, by induction, the right-hand side is a linear combination of f 0 ( T m z) (0  m  s) with coefficients in Fq [ T ], so is f n+1 ( T i z). Consequently, we can find P 0 , P 1 , . . . , P s ∈ Fq [ T ] such that

f 0 ( T s+1 z) = P s f 0 ( T s z) + P s−1 f 0 ( T s−1 z) + . . . + P 0 f 0 ( z) + ( f 0 ( z))q .

(2)

Set g 1 ( z) = f 0 ( z), and g 2 ( z) = z. Then g 2 ( z) is an E q -function with respect to k of order 0, and it is algebraically ¯ for g 1 (z) is entire and not a polynomial, hence a transcendental function over C∞ (z) (see independent with g 1 ( z) over k, [14]). Then by Theorem 3, we obtain that there are only finitely many points at which g 1 , g 2 simultaneously assume values in K . However, by the Fq -additivity of g 1 and the equation (2), we have g 1 (Fq [ T ]γ ) ⊆ K . Absurd! So Theorem 1 holds. 2 Proof of Theorem 2. By contradiction, suppose that all the F 0 (γ ), F 1 (γ ), . . . , F s (γ ) were algebraic over k. Set K = k(γ , f 0 (γ ), f 1 (γ ), ..., f s (γ )). Then K is a finite extension of k. By Lemma 1 and the Fq -linearity of f i ( z) (0  j  s), we obtain directly f j (Fq [ T ]γ ) ⊆ K (0  j  s). In particular, we have f 0 (Fq [ T ]γ ) ⊆ K . Absurd, by the same argument as above. So Theorem 2 holds. 2 Acknowledgements The author would like to thank heartily Jia-Yan Yao for interesting discussions on the subject. He would like also to thank the National Natural Science Foundation of China (Grants No. 11371210 and No. 11871295) for partial financial support. Finally he would like to thank heartily the anonymous referee for his careful reading, pertinent comments, and valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

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