MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the cohomological incarnation, the Riemann hypothesis part of the Weil conjectures for a smooth proper scheme of finite type over a finite field with q elements says that: the eigenvalues of Frobenius acting on the $H^i_\mathrm{et}(X, Q_\ell)$ are algebraic integers with complex absolute value q^{i/2}.

For smooth proper curve C over F_q of genus g, the Riemann hypothesis is often stated in another way as $|N- q - 1| \leq 2g \sqrt(q)$ where N is the number of rational points on C.

Why are these two statements equivalent? Is(Are) there any corresponding inequality(ies) which are equivalent to the Riemann hypothesis in higher dimensions?

share|cite|improve this question
up vote 12 down vote accepted

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I wrote out the details of the argument some years ago (in an unpublished book on zeta and $L$-functions), so it's easy to cut and paste it here.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 $z = q^{-s}$ and the usual version of the Riemann hypothesis for $L(q^{-s})$ is equivalent to the statement that the reciprocal roots of $L(z)$ all have absolute value $\sqrt{q}$. 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: $$ L(z) = (1 - \alpha_1z)\cdots(1-\alpha_dz), \ \ \ \alpha_j \not= 0. $$

Let $L*(z)$ be the polynomial with complex-conjugate coefficients to those of $L(z)$, so $$ L*(z) = (1 - \overline{\alpha_1}z)\cdots(1-\overline{\alpha_d}z). $$ (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 $$ L(1/qz) = \frac{W}{z^d}L*(z) $$ for some constant $W$. If you compare coefficients of the same powers of $z$ on both sides, this functional equation implies the mapping $\alpha \mapsto q/\alpha$ sends reciprocal roots of $L(z)$ to reciprocal roots of $L*(z)$ (and $W$ has absolute value $q^{d/2}$).

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

i$)$ the reciprocal roots of $L(z)$ have absolute value $\sqrt{q}$ (RH for $L(z)$),

ii$)$ the reciprocal roots of $L(z)$ have absolute value $\leq \sqrt{q}$.

Proof. We only need to show ii implies i. Assuming ii, let $\alpha$ be any reciprocal root of $L(z)$, so $|\alpha| \leq \sqrt{q}$. By the functional equation, $q/\alpha$ is a reciprocal root of $L*(z)$, so $q/\alpha = \overline{\beta}$ for some reciprocal root $\beta$ of $L(z)$. Then $|q/\alpha| = |\overline{\beta}| = |\beta| \leq \sqrt{q}$ and thus $\sqrt{q} \leq |\alpha|$. Therefore $|\alpha| = \sqrt{q}$ and i follows. QED

This lemma reduces the proof of the Riemann hypothesis for $L(z)$ from the equality $|\alpha_j| = \sqrt{q}$ for all $j$ to the upper bound $|\alpha_j| \leq \sqrt{q}$ 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 $|\alpha_j|$'s in part ii of Lemma 1 is equivalent to a $O$-estimate on sums of powers of the $\alpha_j$'s which superficially seems weaker.

We will be interested in the sums $$ \alpha_1^n + \cdots + \alpha_d^n, $$ 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) $$L(z) = \exp\left(\sum_{n \geq 1}N_n z^n/n\right)$$ and then logarithmic differentiation shows $$ N_n = -(\alpha_1^n + \dots + \alpha_d^n) $$ for all $n \geq 1$.

Lemma 2. For nonzero complex numbers $\alpha_1,\dots,\alpha_d$ and a constant $B > 0$, the following are equivalent:

i$)$ For some $A > 0$, $|\alpha_1^n + \dots + \alpha_d^n| \leq AB^n$ for all $n \geq 1$.

ii$)$ For some $A > 0$ and positive integer $m$, $|\alpha_1^n + \dots + \alpha_d^n| \leq AB^n$ for all $n \geq 1$ with $n \equiv 0 \bmod m$.

iii$)$ $|\alpha_j| \leq B$ 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 $B = \sqrt{q}$.) 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 $|\alpha_j| = |\overline{\alpha_j}|$) iii implies i. To show ii implies iii, we use a cute analytic trick. Assuming ii, the series $$ \sum_{n \equiv 0 \bmod m} (\alpha_1^n + \dots + \alpha_d^n)z^n $$ 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 $|z| < 1/|\alpha_j|$ for all $j$, the series can be computed to be $$ \sum_{j=1}^{d} \frac{\alpha_j^mz^m}{1-\alpha_j^mz^m} = \sum_{j=1}^{d}\frac{1}{1-\alpha_j^mz^m} - d, $$ so the rational function $\sum_{j=1}^{d} 1/(1-\alpha_j^mz^m)$ is holomorphic on the disc $|z| < 1/B$. Therefore the poles of this rational function must have absolute value $\geq 1/B$. Each $1/\alpha_j$ is a pole, so $|\alpha_j| \leq B$ for all $j$. QED

Theorem. The following are equivalent:

i$)$ $L(z)$ satisfies the Riemann hypothesis ($|\alpha_j| = \sqrt{q}$ for all $j$),

ii$)$ $N_n = O(q^{n/2})$ as $n \rightarrow \infty$,

iii$)$ for some $m \geq 1$, $N_n = O(q^{n/2})$ as $n \rightarrow \infty$ through the multiples of $m$.

Proof. Easily i implies ii and ii implies iii. Assuming iii, we get $|\alpha_j| \leq \sqrt{q}$ 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. :)

share|cite|improve this answer
Thanks Keith, very good point about the functional equation, I only realized how it got used after I gave a talk on this stuff for some graduate students. – Brandon Levin Jan 17 '10 at 18:04

The first step is to reformulate the zeta function in terms of l-adic cohomology. It is defined to be $Z(u) := exp(\sum_m N_m*u^m/m)$, where $N_m$ is the number of points over F_{q^m}. So $N_m$ is #{fixed points of Frob^m}, and by the analogue of the Lefschetz fixed point theorem for l-adic cohomology, this implies $N_m = \sum_i (-1)^i Tr Frob^m*|H^i$. We can substitute this in the expression for Z(u) and rearrange to get

$$Z(u) = \prod_{i=0}^{2n} [exp(\sum_m Tr Frob^m*|H^i * u^m/m)]^{(-1)^i}$$

where $n$ is the dimension of the variety. The inner exponentials are actually $det(1-u Frob)|H^i$, which relates zeros and poles of the zeta function to the eigenvalues of Frob* acting on H^i.

Both H^0 and H^2n are 1-dimensional spaces, and we know the eigenvalues of Frob* acting on them: on H^0 Frob* acts as the identity, so the eigenvalue is 1, and on H^{2n} Frob* is multiplication by q^n because Frob is a finite map of degree q^n.

For curves, n=1 and the expression above for Z(u) becomes $Z(u)=[(1-u)(1-qu)]^{-1} P(u)$, where $P(u)$ is a polynomial of degree dim H^1 whose zeros \alpha_i (the eigenvalues of Frob* acting on H^1) have absolute value $|\sqrt{q}|$. Then it's not hard to show (by logarithmically differentiating $Z(u)$ and rearranging) that $N_m = 1 + q - \sum_j \alpha_j^m$. If we assume dim H^1 = 2g (to prove this I guess you need some way to compare Zariski and etale cohomology), this shows that the bound on point counting is the same as the bound on the size of the eigenvalues of Frob on H^1.

For higher dimensions, you could play the same sorts of games, but the expressions wouldn't be as nice.

share|cite|improve this answer
This shows that Weil conjectures implies that bound but I don't see the reverse direction. Is the reverse direction true for curves? – Brandon Levin Oct 13 '09 at 20:23

I believe a consequence of the Riemann hypothesis in terms of number of points is that an n-dimensional smooth projective variety X will have $q^n+O(q^{n-1/2})$ points over $\mathbb{F}_q$, and the constant in front of $q^{n-1/2}$ is bounded above by the dimension of the dimension of $H^{2n-1}(X)$. However, this is not equivalent; the Riemann hypothesis gives you much better information about the number of points in relation to the cohomology.

share|cite|improve this answer

Brandon, The Weil conjecture can easily be deduced from the point-counting estimate by way of the following lemma: If a(j) are some finite set of complex numbers such that | \sum{j} a(j)^n | \leq A*B^n for all n large, then each a(j) satisfies |a(j)| \leq B. A one-line proof of this is given in Iwaniec-Kowalski on page 288. Anyway, let the a(j) be the roots of the characteristic polynomial on H^1 over F_q, and then take B=2g and A=sqrt(q).

Cheers, Dave

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.