If the genus of the curve is large relative to $q$, then the main term in the formula $\# X(k) = q - \sum \omega _ i + 1$ is not $q$ but the second term $\sum \omega _ i$ which can (a priori) have size about $2g_ X\sqrt{q}$. In the paper [Drinfeld-number] the authors Drinfeld and Vladut show that this maximum is (as predicted by Ihara earlier) actually at most about $g\sqrt{q}$.

Fix $q$ and let $k$ be a field with $k$ elements. Set

where $X$ runs over geometrically irreducible smooth projective curves over $k$. With this definition we have the following results:

**Proof.**
Given $X$ let $w_1, \ldots , w_{2g}$ and $g = g_ X$ be as before. Set $\alpha _ i = \frac{w_ i}{\sqrt{q}}$, so $|\alpha _ i| = 1$. If $\alpha _ i$ occurs then $\overline{\alpha }_ i = \alpha _ i^{-1}$ also occurs. Then

\[ N = \# X(k) \leq X(k_ r) = q^ r + 1 - (\sum _ i \alpha _ i^ r) q^{r/2} \]

Rewriting we see that for every $r \geq 1$

\[ -\sum _ i \alpha _ i^ r \geq Nq^{-r/2} - q^{r/2} - q^{-r/2} \]

Observe that

\[ 0 \leq |\alpha _ i^ n +\alpha _ i^{n-1} +\ldots +\alpha _ i +1|^2 = (n + 1) + \sum _{j = 1}^ n (n + 1 - j) (\alpha _ i^ j + \alpha _ i^{-j}) \]

So

\begin{align*} 2g(n+1) & \geq - \sum _ i \left(\sum _{j = 1}^ n (n+1-j)(\alpha _ i^ j +\alpha _ i^{-j})\right)\\ & =-\sum _{j = 1}^ n (n+1-j)\left(\sum _ i\alpha _ i^ j +\sum _ i\alpha _ i^{-j}\right) \end{align*}

Take half of this to get

\begin{align*} g(n+1)& \geq - \sum _{j = 1}^ n (n+1-j)(\sum _ i\alpha _ i^ j)\\ & \geq N\sum _{j = 1}^ n (n+1-j)q^{-j/2}-\sum _{j = 1}^ n (n+1-j)(q^{j/2}+q^{-j/2}) \end{align*}

This gives

\[ \frac{N}{g}\leq \left(\sum _{j = 1}^ n \frac{n+1-j}{n+1}q^{-j/2} \right)^{-1} \cdot \left( 1 + \frac{1}{g} \sum _{j = 1}^ n \frac{n + 1 - j}{n + 1}(q^{j/2} + q^{-j/2}) \right) \]

Fix $n$ let $g\to \infty $

\[ A(q)\leq \left(\sum _{j = 1}^ n \frac{n+1-j}{n+1}q^{-j/2}\right)^{-1} \]

So

\[ A(q)\leq \mathop{\mathrm{lim}}\nolimits _{n\to \infty }(\ldots ) = \left(\sum _{j = 1}^\infty q^{-j/2}\right)^{-1}=\sqrt{q}-1 \]

$\square$
## Comments (3)

Comment #15 by Emmanuel Kowalski on

Comment #22 by Johan on

Comment #23 by Emmanuel Kowalski on