Character Sums and the Riemann Hypothesis

We prove that an innocent looking inequality implies the Riemann Hypothesis and show a way to approach this inequality through sums of Legendre symbols.


Let
where λ is the Liouville lambda-function1 .Since |λ(n)| = 1, this series is absolutely convergent for real x, so that f is continuous, odd and periodic with period 1 on R.Here is a plot of f (x) for 0 x 1 using 1000 terms of the series defining f : Theorem 1.If f (x) 0 for 0 x 1/4, then the Riemann Hypothesis is true.
Theorem 1 is deceptive in that it looks like it should be a a simple matter to prove that f (x) is non-negative.A problem is that it is not clear whether f (x) is differentiable or not, and even if it is it would be difficult to estimate the derivative.So, proving that f (x) > 0 at some point doesn't immediately tell us about f (x) at nearby points.
The "1/4" in Theorem 1 can be replaced by any positive constant.So the real issue is trying to prove that f (x) > 0 for small positive x.

Note that
so that if for some x there is an N such that then, it must be the case that f (x) > 0. We will use this idea a little later.
We can give an "explicit formula" for f in terms of the zeros ρ = β + iγ of ζ: Theorem 2. Assuming the Riemann Hypothesis, ) .
Here ℓ(n) is defined through its generating function for ℜs > 1.Also, X(s) is the factor from the functional equation for ζ(s) which can be defined by = 2(2π) −s Γ(s) cos πs 2 .Note that if the zeros of ζ(s) are simple, then the term with the sum over the zeros of ζ becomes Theorem 2 is nearly a converse to Theorem 1 in the sense that if RH is true and all the zeros are simple and so that the inequality (2) seems plausible.
Finally we remark that the formula of Theorem 2 for f (x) hides very well the fact that f (x) is periodic with period 1!

Prior results
There has been quite a lot of work connecting partial weighted sums of the Liouville and the Riemann Hypothesis.We refer to [BFM] for a nice description of past work.In this paper the authors prove that the smallest value of x for which

Character sums
A possible approach to proving that f (x) > 0 for small x > 0 lies in the fact that λ is completely multiplicative and takes the values ±1.This scenario resembles quadratic Dirichlet characters (for simplicity think Legendre symbols) except that Dirichlet characters can also take the value 0. By the Chinese Remainder Theorem, for any N there is a prime number q such that λ(n) = n q for all n N where .q is the Legendre symbol 2 mod q.As an example: n 163 for all n 40, but they differ at n = 41.Let be the Fourier sine series with λ(n) replaced by n q .If f q (x) 0 for 0 x 1/4 for a sufficiently large set of q, then it must also be the case that f (x) 0 for 0 x 1/4.(The proof is that if f (x 0 ) < 0 for some 0 < x 0 < 1/4, then we can find a q such that n q = λ(n) for all n N where N is chosen so large that |f (x 0 )| > 1/N; then it must be the case by the analog of (1) for f q that f q (x 0 ) < 0.) The same assertion but with q restricted to primes congruent to 3 mod 8 is also valid, since the Legendre symbols for these q can also imitate λ(n) for arbitrarily long stretches 1 n N. We can express this as follows: Theorem 3. If f q (x) 0 for all 0 x 1/4 and all primes q congruent to 3 mod 8, then the Riemann Hypothesis is true.
2 n q = 0 if (n, q) > 1; n q = +1 if n is a square mod q; and n q = −1 if n is not a square modulo q.
Remark 1.We could just as well have stated this theorem for q ≡ 3 mod 4.However, the intention is that we are interested in q for which χ q imitates λ.Insisting that χ q (2) = −1 leads to the condition that q ≡ 3 mod 8.
The sums f q (x) still have the same problem in that it is tricky to prove for sure that they are positive for small positive x.However, the analogue of Theorem 2 above is much simpler, is unconditional, and leads to a straightforward way to check, for any given fixed q, that f q (x) 0 for 0 x 1/4.
Theorem 4. Let x 0. Let q ≡ 3 mod 8 be squarefree.Then n .Now Dirichlet's class number formula enters the picture.Let K = Q( √ −q) be the imaginary quadratic field obtained by adjoining √ −q to the rationals Q.Let h(q) be the class number3 of K. Dirichlet's formula is for squarefree q ≡ 3 mod 4 and q > 3; (see [D] or [IK]).Thus, the Theorem above can be rephrased in terms of h(q).Moreover, we can express L q (1) as a finite character sum Since n q is an odd function of q we also have where Corollary 1.Let q > 3 be squarefree with q ≡ 3 mod 8. Then Here is a plot of We can use the corollary to prove that f 163 (x) 0 for 0 x 1/2 and consequently that f (x) 0 for 1/4 > x 0.043 as follows..05 it follows from (1) that f (x) 0 for 0.25 x 7 163 = 0.043.Corollary 2. f (x) 0 for 0.043 x 0.25.
It seems clear that for any given ǫ > 0 we could replace 0.043 by ǫ in this inequality with enough computation time.Also, if we use Euler products instead of Dirichlet series we can show that f (x) 0 for 1/4 x 0.011.
The following conjecture seems surprising.
Conjecture 1.If q ≡ 3 mod 8 is squarefree, then f q (x) 0 for 0 x 1/2.Remark 2. J. Bober has checked that this inequality is true for all primes q ≡ 3 mod 8 up to 10 9 .Now we turn to the proofs.

Useful Lemmas
Lemma 1.For y > 0 we have Summing these leads to the desired formula.See also [T]; the above is the integral of the formula (7.9.5).
x s s(s + 1) This lemma is well-known and is easy to verify.

Proofs of theorems
Proof of Theorem 1.This assertion is a consequence of Landau's Theorem: "If g(n) 0 then the right-most singularity of ∞ n=1 g(n)n −s is real."This is theorem 10 of [HR] and Theorem 1.7 of [MV1].What we actually need is an integral version of this theorem: "If g(x) 0 then the right-most singularity of ∞ 1 g(x)x −s dx is real."The proof of this version is essentially the same as that of the first version (see Lemma 15.1 of [MV1]).The application to our situation is slightly subtle.We argue as follows.Since where 0 < c < 1.The integral is absolutely convergent for 0 < c < 1/2.By Mellin inversion we have We split the integral into two integrals at x = 4 so that say.The integral defining I 1 (s) is absolutely convergent for σ > 1 and the second integral is absolutely convergent for σ < 1.Using the periodicity of f we can show that the second integral converges for σ < 2. Indeed, let Then F (n) = 0 for all integers n and F is bounded.Therefore, This integral converges for ℜs < 2. So, we now have I 2 analytic for ℜs < 2. Clearly, I 1 + I 2 is analytic for ℜs > max{−1/2, ρ − 1} i.e. for ℜs > 0. (The pole of X(1 − s) at s = 0 is canceled by the zero of 1/ζ(s + 1) at s = 0.) It follows that I 1 (s) = (I 1 (s) + I 2 (s)) − I 2 (s) is analytic for ℜs > 0. It follows that I 2 (s) is also analytic for ℜs > 0, and since we already knew it was analytic for ℜs < 2 it follows that I 2 (s) is entire.Now, we can write I 1 as Recall we have assumed that f (1/x) 0 for x 4. Therefore, by Landau's theorem, the rightmost singularity of I 1 (s) is real.Since I 2 is entire, it follows that the rightmost pole of I 1 (s) + I 2 (s) must also be real.But the rightmost real pole of is at s = −1/2.This must be the rightmost pole.Therefore the poles at ρ − 1 must all have their real parts less than or equal to -1/2.In particular, ℜρ 1/2, which is RH.
Proof of Theorem 2. We start again from where 0 < c < 1/2.The integrand has poles only at s = − 1 2 and at s = ρ − 1 where ρ is a complex zero of ζ(s) and nowhere else in the s-plane.The residue at Assuming that the zeros are simple, the residue at s = ρ − 1 is We (carefully) move the path of integration to (c) where −2 < c < −1.To do this we have to cross through a field of poles arising from the zeros of the zeta function.To do this we use Theorem 14.16 of [T1] (see also [R]) to find a path on which 1/ζ(s + 1) ≪ T ǫ where we can safely cross.Using the bounds |X(1 − s)| ≪ T σ−1/2 and ζ(2s + 2) ≪ T − 1 2 −σ we can get the sum of the residues arising from the zeros up to height T together with an error term that tends to 0 as T → ∞.Thus, assuming the zeros are simple, If the zeros are not simple we modify the sum over zeros appropriately.We make the change of variable s → −s in the integral.Using the functional equation for the ζ-function and functional relations for the Γ-function, we see that the new integrand is Then Theorem 2 follows.
Proof of Theorem 4. We denote χ q (n) = n q .By Lemma 1, where 0 < c < 1.Since χ q is odd, we find that the integrand has a pole at s = 0 and nowhere else in the complex plane.We move the path of integration to (c) where c < −1 to see that Now let s → −s in the integral and use the functional equation (see [D], [IK] or [MV1]) L(1 − s, χ q ) = 2q s− 1 2 (2π) −s Γ(s) sin πs 2 L(s, χ q ).After simplification, the integral above is .
By Lemma 3, this integral is The proof of Theorem 4 is complete.
Remark 3. Note that the non-negativity, for 0 < x < 1/4, of the right-hand side of (3) implies the Riemann Hypothesis.This condition only involves Dirichlet L-functions with quadratic characters.Thus, information solely about Dirichlet L-functions potentially gives the Riemann Hypothesis.This example shows that different L-functions somehow know about each other.

Further remarks
Since h(q) ≫ ǫ q 1/2−ǫ we see that f q (x) 0 for a ≪ x ≪ q −1/2−ǫ .In particular, f q (a/q) 0 for a ≪ q 1/2−ǫ .But this doesn't give information about f (x).Also, the Polya-Vinogradov inequality tells us that max N N n=1 χ q (n) ≪ q 1/2 log q and the work of Montgomery and Vaughan [MV] shows that the Riemann Hypothesis for L(s, χ) implies that max Moreover, it is known that the right hand side here can not be replaced by any function that goes to infinity slower.It is also known, assuming the Riemann Hypothesis for L(s, χ), that L(1, χ) ≪ log log q.
Our desired inequality can be expressed in terms of L(1, χ) as max It appears that both sides of this inequality can be as big as √ q log log q.
A question is whether the converse of Theorem 1 is true.It might be possible to approach this by showing that the " 3 2 " derivative of f (x) is positive at x = 0 so that there is a small interval to the right of 0 for which f (x) ≥ 0. This method, or trying to prove (2) directly, would involve explicit estimates (assuming RH) for 1/ζ(s) in the critical strip; see [MV1] section 13.2 for a good approach to such explicit estimates.
Finally, we mention that f (x) can be evaluated at a rational number x = a/q as an average involving Dirichlet L-functions L(s, χ) where χ is a character modulo q.
6. Evaluation of f q (a/p).
A couple of formulas may help us move forward here.One is that if θ 1 and θ 2 are characters with coprime moduli m 1 and m 2 respectively, then (see [IK,(3.16)]) The other is that for a character θ modulo m and a positive integer r where B r is the rth Bernoulli polynomial (see [Wa,Theorem 4.2]).Recall the functional equation (see [D]) for a primitive character θ mod m: It follows that for an even θ = χ q ψ, with q ≡ 3 mod 4 and ψ an odd character modulo p, Therefore, ℑ{τ (ψ)ψ(a)L(2, χ q ψ)} = ℜ{ 2π 2 χ q (p)ψ(aq) We sum this equation over the odd characters modulo p using Thus, we have Theorem 5.For primes p and q both congruent to 3 modulo 4 and 1 ≤ a < p/2 we have for all p < q which are primes congruent to 3 modulo 8 and all 0 < a < p/2, then the Riemann Hypothesis follows.
We note that by these techniques one can show Theorem 6.
When this formula is compared with our earlier formula we deduce the identity a 3 for q ≡ 3 mod 4.
Now we indicate another possible direction.
Proposition 1.If f q (x) = 0 then x is a rational number.
Proof.By Corollary 1, f q (x) = 0 implies that S q (q/2) − S q (qx) = 0.But S q (q/2) = h(q) is an integer.So f q (x) = 0 implies that S q (qx) is a rational number.Now This has the shape integer − integer qx which can only be rational if x is a rational number.
So, it suffices to show that f q (x) has no rational zeros; perhaps a congruence argument could work.However, Theorem 5 is not much use here because the hypothetical x for which f q (x) = 0 would likely have a denominator that is divisible by q, so the conditions of Theorem 5 don't hold.
We remark that there are rational values of x for which the numerator of f q (x) is congruent to 0 modulo q; for example These examples, which all seem to have an x with denominator divisible by q, might be worth studying further.
Here is one final formula that may or may not be useful.Suppose that f q (x) = 0. Let y = xq.Then either n≤y χ q (n) = h(q) and n≤y nχ q (n) = 0 or else y satisfies .
The first alternative seems unlikely as in that case there would be an interval on which f q (x) would be identically 0.

Conclusion
Conjecture 1 has been checked for primes up to 10 9 and it holds for those primes.However, probabilistic grounds call into question it's truth for all primes q ≡ 3 mod 8.Of course, one only needs it's truth for a set of characters χ q for which χ q (n) = λ(n) for all n ≤ N q where N q → ∞ with q.Presumably something like this is correct (and should be equivalent to RH), but it is not clear how to proceed.But the results of section 6, suggest a slightly alternative way forward which may have a more arithmetic flavor.
a number with absolute value at most 1, not necessarily the same at each occurrence.Now for a an integer, S 163 (163x) is constant for x in the interval [ a a 163 ) 0.0095 0.0095 0.019 0.038 0.047 0.066 0.076 0.095 0.12 0.14 Since Θ 20