Wednesday, 27 August 2008

ag.algebraic geometry - Equivalent Statements of Riemann Hypothesis in the Weil Conjectures

The reason why that inequality is equivalent to RH (for curves) is the functional equation.



I wrote out the details of the argument some years ago (in an unpublished book on zeta and L-functions), so it's easy to cut and paste it here. The polynomial L(z) I speak about below would arise in practice as the numerator of the zeta-function of the curve, with z=qs and the usual version of the Riemann hypothesis for L(qs) is equivalent to the statement that the reciprocal roots of L(z) all have absolute value sqrtq. That's the form of the Riemann hypothesis I will be referring to in what follows.



Suppose we have a polynomial L(z) over the complex numbers with constant term 1 and degree d, factored over its reciprocal roots:
L(z)=(1alpha1z)cdots(1alphadz),alphajnot=0.



Let L(z) be the polynomial with complex-conjugate coefficients to those of L(z), so
L(z)=(1overlinealpha1z)cdots(1overlinealphadz).
(Sorry, I want to make the asterisk into an exponent in this particular .tex situation, but it isn't working and I don't have time to figure it out.)
Assume L(z) and L(z) are connected by the functional equation
L(1/qz)=fracWzdL(z)
for some constant W. If you compare coefficients of the same powers of z on both sides, this functional equation implies the mapping alphamapstoq/alpha sends reciprocal roots of L(z) to reciprocal roots of L(z) (and W has absolute value qd/2).



Lemma 1. Granting the functional equation above, the following conditions are equivalent:



i) the reciprocal roots of L(z) have absolute value sqrtq (RH for L(z)),



ii) the reciprocal roots of L(z) have absolute value leqsqrtq.



Proof. We only need to show ii implies i.
Assuming ii, let alpha be any reciprocal root of L(z),
so |alpha|leqsqrtq. By the functional
equation, q/alpha is a reciprocal
root of L(z), so q/alpha=overlinebeta
for some reciprocal root beta of L(z).
Then |q/alpha|=|overlinebeta|=|beta|leqsqrtq and thus sqrtqleq|alpha|.
Therefore |alpha|=sqrtq and i follows. QED



This lemma reduces the proof of the Riemann hypothesis for L(z) from the equality |alphaj|=sqrtq for all j to the upper bound |alphaj|leqsqrtq for all j. Of course the functional equation was crucial in explaining why the superficially weaker inequality implies the equality.



Next we want to show the upper bound on the |alphaj|'s in part ii of Lemma 1 is equivalent to a O-estimate on sums of powers of the alphaj's which superficially seems weaker.



We will be interested in the sums
alphan1+cdots+alphand,
which arise from the theory of zeta-functions as coefficients in an exponential generating function: since L(z) has constant term 1, we can write (as formal power series over the complex numbers) L(z)=expleft(sumngeq1Nnzn/nright)
and then logarithmic differentiation shows
Nn=(alphan1+dots+alphand)
for all ngeq1.



Lemma 2.
For nonzero complex numbers alpha1,dots,alphad and a
constant B>0, the following are equivalent:



i)
For some A>0, |alphan1+dots+alphand|leqABn for
all ngeq1.



ii)
For some A>0 and positive integer m,
|alphan1+dots+alphand|leqABn for
all ngeq1 with nequiv0bmodm.



iii) |alphaj|leqB for all j.



Part ii is saying you only need to show part i when n runs through the (positive) multiples of any particular positive integer to know it is true for all positive integers n. It is a convenient technicality in the proof of the Riemann hypothesis for curves, but the heart of things is the connection between parts i and iii. (We'd be interested in part iii with B=sqrtq.) You could set m=1 to make the proof below that ii implies iii into a proof that i implies iii. The passage from i to iii is what Dave is referring to in his answer when he cites the book by Iwaniec and Kowalski.



Proof.
Easily i implies ii and (since |alphaj|=|overlinealphaj|)
iii implies i.
To show ii implies iii, we use a cute analytic trick. Assuming ii, the series
sumnequiv0bmodm(alphan1+dots+alphand)zn
is absolutely convergent for |z|<1/B, so the series defines a holomorphic function on this disc. (The sum is over positive multiples of m, of course.)
When |z|<1/|alphaj| for
all j, the series can be computed to be
sumdj=1fracalphamjzm1alphamjzm=sumdj=1frac11alphamjzmd,
so the rational function sumdj=11/(1alphamjzm) is holomorphic
on the disc |z|<1/B. Therefore the poles of this rational function must have absolute value geq1/B. Each 1/alphaj is a pole, so |alphaj|leqB for all j. QED



Theorem.
The following are equivalent:



i) L(z) satisfies the Riemann hypothesis (|alphaj|=sqrtq for all j),



ii) Nn=O(qn/2) as nrightarrowinfty,



iii) for some mgeq1, Nn=O(qn/2) as nrightarrowinfty through the multiples of m.



Proof.
Easily i implies ii and ii implies iii.
Assuming iii, we get |alphaj|leqsqrtq for all j by
Lemma 2, and this inequality over all j is equivalent to i
by Lemma 1. QED



Brandon asked, after Rebecca's answer, if the inequality implies the Weil conjectures (for curves) and Dave also referred in his answer to the Weil conjectures following from the inequality. In this context at least, you should not say "Weil conjectures" when you mean "Riemann hypothesis" since we used the functional equation in the argument and that is itself part of the Weil conjectures. The inequality does not imply the Weil conjectures, but only the Riemann hypothesis (after the functional equation is established).



That the inequality is logically equivalent to RH, and not just a consequence of it, has some mathematical interest since this is one of the routes to a proof of the Weil conjectures for curves.



P.S. Brandon, if you have other questions about the Weil conjectures for curves, ask your thesis advisor if you could look at his senior thesis. You'll find the above arguments in there, along with applications to coding theory. :)

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