AA Look at Chowla’s Problem
Andean Medjedovic
Abstract
In this paper we look at the history behind Chowla’s problem on the solutions to L (1 , f ) = 0 for periodic f . We focus on the results given by Baker, Birch and Wirsing onthe topic. We briefly discuss recent results due to Chatterjee, Murty and Pathak whichgive a full solution when combined with the work of Baker, Birch and Wirsing. a r X i v : . [ m a t h . N T ] O c t able of Contents ii Dirichlet’s Theorem
In 1831, Peter Gustav Lejeune Dirichlet introduced his namesake character in order to provethat the arithmetic progression an + b , for natural numbers a, b and n with gcd( a, b ) = 1,takes on prime values infinitely often. The associated L -series of the characters wereinstrumental in his study. During his exploration, Dirichlet managed to demonstrate thevalue of such L -series at 1 is non-zero. This result later proved to be crucial to Chowla’sinvestigation. We begin with a quick review of the relevant details. Definition 1.
We say a function χ : Z → C is a Dirichlet character modulo k if: • χ is completely multiplicative, χ ( nm ) = χ ( n ) χ ( m ) for any n, m ∈ N • χ is k -periodic, χ ( n + k ) = χ ( n ) for any n ∈ N • χ ( n ) (cid:54) = 0 for ( n, k ) = 1 and χ ( n ) = 0 for ( n, k ) (cid:54) = 1Furthermore, a character is principal if χ ( n ) ∈ { , } for all n ∈ N . Notice that if l isa multiple of k , a character on k is also a character on l . We say a character (modulo l ) isprimitive if it is not a character of modulo k for k < l . We use X ( k ) to denote the set ofDirichlet characters modulo k . Proposition 1.
The Dirichlet characters satisfy an orthogonality relation. If χ is a non-principal character modulo k then (cid:88) a ∈ Z /k Z χ ( a ) = 0 (cid:88) χ ∈ X ( k ) χ ( a ) = 0 Where a (cid:54) = 1 in the second sum. If a = 1 then the second sum is φ ( a ) . Definition 2.
Given a character we define the L -series of that character to be the analyticcontinuation of the sum: L ( s, χ ) = ∞ (cid:88) n =1 χ ( n ) n where s ∈ C . One can check as an exercise the any non-principal L -series is holomorphicfor real part of > s > .
1e now prove the aforementioned theorem of Dirichlet[5].
Theorem 1. If χ is a non-principal character then: L (1 , χ ) (cid:54) = 0 Proof.
Let X ( k ) be the set of all Dirichlet characters of modulus k and consider the product ζ k ( s ) = (cid:89) χ ∈ X ( k ) L ( s, χ )as s → + . We first show that ζ k ( s ) has a singularity at s = 1. Re-writing the above as aEuler product we see that, ζ k ( s ) = (cid:89) p (cid:45) k − p | p | s ) φ ( k ) | p | where | p | is the order of p in Z /k Z . We now invoke a theorem of Landau stating that ifthe coefficients of a L -series are at least zero, the real point on the abscissa is a singularityof the L -series. Suppose ζ k ( s ) does not have a singularity at s = 1. Since the L -series ofeach character is holomorphic at s = 1, the abscissa must be at some σ ≤
0. But this isridiculous, as we can see that for s = φ ( k ) the product is unbounded, we have ζ k ( s ) = (cid:89) p (cid:45) k − p | p | s ) φ ( k ) | p | ≥ (cid:89) p (cid:45) k ∞ (cid:88) i =0 p φ ( k ) si ≥ (cid:88) p (cid:45) k p And Euler showed that the sum of the reciprocal of primes diverges.So ζ k ( s ) indeed has a singularity. If χ is a principal character then L ( χ, s ) has a simplepole at s = 1, If L ( χ, s ) = 0 for any other (non-principal) character then it follows that ζ k ( s ) would not have a pole at s = 1. A contradiction, thus L ( χ, s ) (cid:54) = 0. Definition 3.
Cyclotomic PolynomialsWe say the minimal polynomial over Q of any primitive root of n , say e πi/n , is the n th cyclotomic polynomial, Φ n . One can check that it is also possible to define it with thefollowing: Φ n ( x ) = (cid:89) ≤ k ≤ n ( k,n )=1 ( x − e πik/n )2 Chowla’s Question
Sarvadaman Chowla asked the following natural generalization of Dirichlet’s theorem toCarl Ludwig Siegel in 1949[4]. Let f be a non-zero q -periodic function from Z to Q thattakes 0 for values not relatively prime to q (that is f ( p ) = 0 if ( p, q ) > L (1 , f ) = ∞ (cid:88) n =1 f ( n ) n = 0? Lemma 1.
If such an f exists we must have (cid:80) qn =1 f ( n ) = 0 .Proof. Suppose, without loss of generality that (cid:80) qn =1 f ( n ) = p >
0. Now consider the sumwe are interested up to some large qy , y ∈ N and sum by parts, qy (cid:88) f ( n ) n = qy (cid:88) f ( n ) + (cid:90) qy (cid:0)(cid:80) t f ( n ) (cid:1) t dt = py + (cid:90) qy tq p + O (1) t dt = py + ln( py ) + O (1)So as y increases the sum is unbounded.Chowla was able to prove his conjecture in the case that f was an odd periodic functionwith values in {− , , } and with prime period, q , with ( q −
1) also prime. In a letterto Chowla, Siegel extended the result to include all rational-valued odd periodic functionswith prime period, q . Theorem 2 (Baker, Birch, and Wirsing, 1973[2]) . Suppose f : Z → A is non-zero, (where A are the algebraic numbers) with period q so that:(i) f ( r ) = 0 if < ( r, q ) < q (ii) Φ q is irreducible over Q ( f (1) , · · · , f ( q )) Then L (1 , f ) = ∞ (cid:88) n =1 f ( n ) n = 0 . (1) cannot hold. f so that 1 holds even if we allow f to be even so long as we force f (0) = 0. As one can see for themselves, the theorempresented by Baker et al. is more general still.We mention the necessity of both conditions ( i ) and ( ii ). Suppose we omitted condition( i ), and consider the function, f defined by L ( s, f ) = (1 − p − s ) ζ ( s )where ζ ( s ) is, of course, the Riemann zeta funcation. As s goes to 1, (1 − p − s ) ζ ( s ) goesto 0 since ζ ( s ) has a only a simple pole. It remains to show that f ( n ) is periodic. I claimthe period is p . Indeed expand each term of the RHS:(1 − p − s ) n s = 1 n s − p ( np ) s + p ( np ) s If we define ( a | b ) to be 1 when a | b and 0 otherwise, for integers a, b then we can write out f ( n ) explicitly like so: f ( n ) = 1 − p ( p | n ) + p ( p | n ) n . So f ( n + p ) = f ( n ) and so condition ( i ) is necessary.Condition ( ii ) is also necessary, it is possible for L (1 , f ) to be algebraically dependentfor certain f . One such example is for the 2 dirichlet characters modulo 12 with valuesgiven below. n χ ( n ) 1 − − χ ( n ) 1 − − L (1 , χ ) = π , L (1 , χ ) = π √ and so 2 χ − √ χ = f implies that L (1 , f ) = 0. The12 th cyclotomic is x − x + 1 which is reducible over Q ( √ ii ) is necessary.Immediately from the theorem we get a non-trivial corollary: Corollary 1.
Suppose ( q, φ ( q )) = 1 and let X (cid:48) ( q ) be the set of non-principal charactersmodulo q . Then L (1 , χ ) are linearly independent over Q for χ ∈ X (cid:48) ( k ) . roof. Suppose we have some relation (cid:80) Ni =1 α i L (1 , χ i ) = 0 where the α i ∈ Q . Then wecan define f = (cid:80) Ni =1 α i χ i and get a contradiction to the above theorem, if our f satisfiescondition ( ii ) (it clearly satisfies ( i )). So this cannot happen. We get the corollary by animplication of condition ( ii ). Φ q is irreducible over Q φ ( q ) if [ Q φ ( q ) ( e πi/q ) : Q φ ( q ) ] = φ ( q ).Computing the degree of the LHS gives φ ( q ) φ (( φ ( q ) , q )) = φ ( q )Thus ( φ ( q ) , q ) is one or two. So f will satisfy ( ii ) if ( φ ( q ) , q ) = 1 and then the corollaryholds. We work towards a proof of 2 following the argument given by Baker, Birch and Wirsing.The general idea of the proof is to transform our sum using shifted versions of it to provethat it must be 0 on a orthogonal set. This would imply that f = 0 in the first place. Tobegin, let F q be the set of f : Z → A with period q so that L (1 , f ) = ∞ (cid:88) n =1 f ( n ) n = 0and let G q be the set of functions of form g ( s ) = 1 q q (cid:88) r =1 f ( r ) e − πrs/q for an f ∈ F q . Note that 1 implies g (0) = 0 for any g ∈ G q . Lemma 2. g ∈ G q if and only if q − (cid:88) s =1 g ( s ) log(1 − e πis/q ) = 0 Proof.
First we show that f ( r ) = (cid:80) qs =1 g ( s ) e πirs/q . Indeed, substitute g ( s ) in and sum5ver s : q (cid:88) s =1 g ( s ) e πirs/q = 1 q q (cid:88) s =1 q (cid:88) l =1 f ( l ) e πis ( l − r ) /q = 1 q q (cid:88) s =1 f ( r ) + 1 q (cid:88) l (cid:54) = r ≤ l ≤ q f ( l ) e πi ( r − l ) − e πi ( r − l ) /q − f ( r ) + 0 = f ( r )Now to prove the actual lemma expand the taylor series of the logarithm: q − (cid:88) s =1 g ( s ) og (1 − e πis/q ) = − ∞ (cid:88) k =1 q − (cid:88) s =1 g ( s ) e πisk/q k = − ∞ (cid:88) k =1 f ( k ) k = 0Where the last equality comes from the definition of f .Here we invoke Baker’s theorem on linear forms in logarithms. It turns out we don’tneed the actual bound given by Baker in [1], we just need to show a certain linear form inlogs is zero and so the coefficients in the form must be 0, by Baker. Lemma 3.
Let σ be an automorphism of A and g ∈ G q . Then σg ∈ G q .Proof. Consider the set log(1 − e πis/q ) for s = 1 , · · · , q − Q . Let log( α i ) be a linearlyindependent subset of maximum cardinality i = 1 , · · · , t so they form a basis and:log(1 − e πis/q ) = t (cid:88) i =1 a rs log( α r )for some a rs ∈ Q . Let β r = (cid:80) q − s =1 g ( s ) a rs .Then (cid:80) q − s =1 g ( s ) log(1 − e πis/q ) = 0 implies that t (cid:88) i =1 β i log( α i ) = q − (cid:88) s =1 g ( s ) t (cid:88) r =1 a rs log( α r ) = 0since q − (cid:88) s =1 g ( s ) t (cid:88) r =1 a rs log( α r ) = q − (cid:88) s =1 g ( s ) log(1 − e πis/q ) . t (cid:88) i =1 β i log( α i ) = 0 . Since all of the log( α i ) are linearly independent over Q it follows that each β i is 0. So (cid:80) q − s =1 g ( s ) a rs = 0. Now for any automorphism σ note that q − (cid:88) s =1 σ ( g ( s )) log(1 − e πis/q ) = t (cid:88) r =1 σ ( q − (cid:88) s =1 g ( s ) a rs ) log( α r ) = 0By linearity of σ . The last sum is 0 since each term in the sum over r is 0. By lemma 2, σg ∈ G q . Lemma 4.
Given an automorphism of A , σ , define h to be the unique integer modulo q by σ − e πi/q = e πih/q . Define f (cid:48) ( n ) = σf ( hn ) for all n , then f (cid:48) ∈ F q .Proof. Since f ( r ) and g ( r ) are related by the formula in 2 then there is some g ( s ) ∈ G q sothat σf ( hn ) = σ (cid:32) q − (cid:88) s =1 g ( s ) e πihns/q (cid:33) = q − (cid:88) s =1 σ ( g ( s )) σ ( e πinhs/q )= q − (cid:88) s =1 σg ( s ) e πins/q Thus σf ( hn ) ∈ F q if and only if σg ( s ) ∈ G q by the way G q and F q were constructed.Also by 3, σq ( s ) ∈ G q and so σf ( hn ) ∈ F q . Proof.
We are ready to prove the main result (Theorem 2). The goal is to show thatif the properties ( i ) and ( ii ) are forced on f , f = 0. Recall that ( ii ) state that theminimal polynomial of e πi/q is irreducible over Q adjoined with the values that f takes7n. Accordingly, we may choose some automorphism σ that fixes f ( n ) for all n and takes e πih/q to e πi/q for any h with ( h, q ) = 1. Then σf ( hn ) = f ( hn ) is in F q .Consider now the sum given by q (cid:88) h =1( h,q )=1 ∞ (cid:88) n =1 f ( hn ) n = 0 . It is 0 since each term summing over h is 0. Interchange the sums and group terms withthe same denominator to get ∞ (cid:88) n =1 (cid:80) q h =1( h,q )=1 f ( hn ) n = 0 . The sum (cid:80) q h =1( h,q )=1 f ( hn ) is 0 if 1 < ( n, q ) < q by ( i ). If ( n, q ) = q then the sumis φ ( q ) f (0) as each term is f (0). Finally, if ( n, q ) = 1, then hn takes on all values, p ,modulo q , with ( p, q ) = 1 since we always have ( h, q ) = 1. Thus the sum in question is (cid:80) qs =0 f ( s ) − f (0) = − f (0) by 1.The idea now is to sum over the individual periods of f . Over any period of length q ,say, ( n + 1 , · · · n + q ) we have one multiple of q and φ ( q ) integers coprime to q .0 = ∞ (cid:88) n =1 (cid:80) q h =1( h,q )=1 f ( hn ) n = ∞ (cid:88) m =0 q (cid:88) s =1 (cid:80) q h =1( h,q )=1 f ( h ( mq + s )) mq + s = f (0) ∞ (cid:88) m =0 φ ( q ) q + mq − (cid:88) ≤ s ≤ q ( s,q )=1 s + mq Now each term in the sum over m is less than than 0, φ ( q )) q + mq − (cid:80) ≤ s ≤ q ( s,q )=1 1 s + mq <
0. Thismeans that f (0) = 0. We want to show that the weighted sum (cid:80) qn =1 χ ( n ) f ( n ) = 0 for8rbitrary χ ∈ X ( q ). By orthogonality, that would mean f ( n ) = 0 for all n , as each sumcan be viewed as a component of the multiplication of the orthogonal matrix M = χ i ( j )with the vector [ f ( i )] ( i,q )=1 for some ordering of the φ ( q ) many χ ∈ X ( q ). By invertabilityof the matrix the result follows.If χ ∈ X ( q ) is principal then the result is true by the lemma 1. Suppose χ isn’t principalan define b ( n ) = q (cid:88) h =1 χ ( h ) f ( hn )so that b (1) χ ( n ) = b ( n ) . To see this note that if hn = m then χ ( m ) χ ( n ) = χ ( h ) by complete multiplicativity. Sothen χ ( m ) χ ( n ) f ( m ) = χ ( h ) f ( hn ) and we can sum over all the numbers coprime to q from1 , · · · , q (or all numbers from 1 , · · · , q since f ( m ) is 0 when ( m, q ) >
1) which yields b (1) χ ( n ) = q (cid:88) m =1 χ ( m ) χ ( n ) f ( m ) = q (cid:88) h =1 χ ( h ) f ( hn ) = b ( n ) . But now b (1) L (1 , χ ) = ∞ (cid:88) n =1 b ( n ) n = q (cid:88) h =1 χ ( h ) ∞ (cid:88) n =1 f ( hn ) n = 0and we have seen that L (1 , χ ) is not 0 so b (1) = (cid:80) qh =1 χ ( h ) f ( h ) = 0. Since χ was arbitrary, f = 0. Which was what was wanted. We briefly review the other theorems in [2] and more recent work done by Chatterjee,Murty, and Pathak [3]. In their paper on the problem Baker et al. were able to dropconditions ( i ) and ( ii ) and classify the solutions to 1 under a much weaker condition.Namely, that f must be odd. 9 heorem 3. Let q ≥ be a natural number. Suppose f is odd, algebraic valued andperiodic modulo q so that holds. Then f is contained in the span (over A ) of the followingdepending on if q is odd or even, respectively: f l ( n ) = (cid:18) sin ( nπ/q ) sin ( π/q ) (cid:19) l l = 3 , · · · , q − f l ( n ) = cos ( nπ/q ) cos ( π/q ) (cid:18) sin ( nπ/q ) sin ( π/q ) (cid:19) l l = 3 , , · , sq − L (1 , f ) = 0 for a q -periodic function f and condition( i ) holds then it was shown that f has to be odd in the same paper. About a year ago,Chatterjee, Murty and Pathak were able to generalize the above result in their paper withthe following theorem. Theorem 4 (Chatterjee, Murty, Pathak[3]) . Let f be algebraically valued with period q .Let f o = f ( x ) − f ( − x )2 f e = f ( x ) + f ( − x )2 be the odd and even parts of f , respectively. Then L (1 , f ) = 0 ⇐⇒ L (1 , f o ) = 0 and L (1 , f e ) = 0Chatterjee et al. then provide a complete characterization of the even f so that L (1 , f ) = 0. Theorem 5.
Consider the set of all f , where f is an even algebraically valued q -periodicfunction so that L (1 , f ) = 0 . This set is exactly the space spanned (over the algebraics) by ˆ F d,c Where d | q , ≤ c ≤ d − and ˆ F d,c is the Fourier transform of F d,c = F (1) d,c − F (2) d,c . Here F (1) d,c and F (2) d,c are defined by F (1) d,c ( n ) = (cid:40) if n ≡ c modulo q otherwise F (2) d,c ( n ) = (cid:40) if n ≡ qcd modulo q otherwise Together with 3, this gives a complete description of the periodic functions with L (1 , f ) =0, solving Chowla’s problem in full generality.10 Acknowledgments
The author would like to thank Cameron Stewart for suggesting the topic and relevantresults. We would also like to thank him for teaching a course on Linear forms in Loga-rithms and exposing us to Baker’s theorem. Finally, we would like to thank the audienceto which this summary was presented, for pointing out several clarifications.11 eferences [1] Alan Baker. “Linear forms in the logarithms of algebraic numbers (IV)”. In:
Mathe-matika
Journal of Number Theory arXiv preprint arXiv:1809.01841 (2018).[4] Sarvadaman Chowla. “The nonexistence of nontrivial linear relations between theroots of a certain irreducible equation”. In:
Journal of Number Theory, vol. 2, no. 1,pp. 120-123