I am trying to understand Kloosterman sums and their estimates (e.g. from [1], which does not prove)

$$ \Big| S(m,n;c) \Big| = \Big| \sum_{(x,c) = 1} e\big( \frac{mx + nx^{-1}}{c}\big) \Big| < 2\, c^{3/4} $$

Another resource [2] has a different version of the Kloosterman sum bound:

$$ \bigg| \sum_{x \in \mathbb{F}_q^\ast} \chi(x) \, e \big( \frac{ax + bx^{-1}}{q} \big) \bigg| \leq 2 \sqrt{q} $$

Yet another place has that if $m= n$ a different bound occurs. Perhaps I'm mixing up notations:

$$ K(n,n; c) \leq (n,c)^{1/2} \sqrt{c} \, \tau(c) $$

where $\tau(c)$ is the number of divisors of $c$ (and $K$ is just the same as $S$). This could be a refinement of the first two.

The proofs of these results are very complicated. Is there a group action involed? Is there an interpretation via group theory of this inequality? Looks like there could be a hyperbola involved and that behaves well under fractional linear transformations like $x \mapsto x + 1$ or $x \mapsto - \frac{1}{x}$ in $\mathbb{F}_q^\ast$.

  • $\begingroup$ In a way, this question is a lead-in to other more complicated Kloosterman-type problems that I will save for another time. $\endgroup$ – john mangual Apr 28 '17 at 17:51

You can find a concise but self-contained proof of Weil's bound for Kloosterman sums of prime moduli in these notes. Note that prime moduli constitute the really hard case.

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The different bounds you have given carry with them different assumptions.

The first bound is stated for $\gcd(m,n,c)=1$. The second bound is stated when $a$ and $b$ are both prime to $p$ (though it is trivially true if only one is prime to $p$). The third bound, however, lacks this condition. If $n$ is prime to $c$, clearly we can drop the $(n,c)$ term, making it appear more similar to the second one.

The second bound is a sum over a finite field. This is both more and less general than the other two, which are sums over the integers mod $c$. The intersection of the two is the case when $c=q$ is a prime, so $\tau(c)=2$. In this case the second and third bounds are equal, but the first one is weaker.

The first bound was proved by Kloosterman, the second was proved by Weil, and the third was either proved by Weil or is an easy corollary of his work (the second bound). The method's used in Weil's proof are significantly more elementary, which accounts for the stronger bound. (Recently more elementary arguments have been found using Stepanov's method, but they are still far more difficult than Kloosterman's orginal proof)

Kloosterman's method fundamentally uses the symmetry underlying the problem, which you can interpret as a group action. In the case where $c=p$ is prime, one first estimates $\sum_{m,n\in \mathbb F_p} |S(m,n;c)|^4$ by expanding the sum and using orthogonality of characters. One then observes that $S(m,n;c)$ is preserved by the symmetry $(m,n) \mapsto (am,a^{-1}n)$ so if any $S(m,n;c)$ were large, $p-1$ of them are large. It seems impossible to generalize this method to exponential sums without this symmetry, see Terry's question.

On the other hand, Weil's method does not use any of the obvious symmetry or structure of the problem at all! His method applies equally well to the exponential of an almost arbitrary function over the solution set of an arbitrary algebraic equation (within reason - one can't give a nontrivial bound if the function is constant, for instance). However, he uses a lot of hidden structure of the problem. Of course some of this is related to group theory, although I wouldn't say group theory is doing any of the heavy lifting in the proof. The key step is a bound on the zeroes of the $L$-function of a character of the idele class group of the field $\mathbb F_q(t)$. Weil's proof is here.

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  • $\begingroup$ why does Terry use the pejorative word "cheap"? Also I suspect Kloosterman could have even more symmetry than the ones you mention. $\endgroup$ – john mangual Apr 28 '17 at 21:06
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    $\begingroup$ @johnmangual "cheap" is only a pejorative word if you're not paying for it. $\endgroup$ – Will Sawin Apr 28 '17 at 21:25
  • $\begingroup$ @johnmangual Maybe it does have more symmetry, though not all the symmetries you mention are there (I don't see any involving the transformation $x \to x+1$). Still, you asked whether the symmetries are involved in the proofs, and I have explained to what extent that is true - quite a bit, for Kloosterman's proof, which is also not as complicated you suggest, but not so much for Weil's. $\endgroup$ – Will Sawin Apr 28 '17 at 21:41
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    $\begingroup$ Instead of "significantly more elementary" you meant "significantly less elementary". $\endgroup$ – GH from MO Apr 29 '17 at 0:40
  • $\begingroup$ the Stepanov method with the line and point coincidences is amazing. it's complicated yet it appeals to my geometric intuition. $\endgroup$ – john mangual Apr 29 '17 at 3:09

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