What's the relationship between Gauss sums and the normal distribution? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:58:17Z http://mathoverflow.net/feeds/question/11053 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11053/whats-the-relationship-between-gauss-sums-and-the-normal-distribution What's the relationship between Gauss sums and the normal distribution? Qiaochu Yuan 2010-01-07T18:18:22Z 2010-01-13T11:34:16Z <p>Let $p$ be an odd prime and $\left( \frac{a}{p} \right)$ the Legendre symbol. The <strong>Gauss sum</strong></p> <p>$\displaystyle g_p(a) = \sum_{k=0}^{p-1} \left( \frac{k}{p} \right) \zeta^{ak},$</p> <p>where $\zeta_p = e^{ \frac{2\pi i}{p} }$, is a periodic function of period $p$ which is sometimes invoked in proofs of quadratic reciprocity. <a href="http://sbseminar.wordpress.com/2008/10/11/the-sign-of-the-gauss-sum/" rel="nofollow">As it turns out</a>, $g_p(a) = \left( \frac{a}{p} \right) i^{ \frac{p-1}{2} } \sqrt{p}$, so $g_p(a)$ is essentially an eigenfunction of the discrete Fourier transform. Now, if $(a, p) = 1$, we can write</p> <p>$\displaystyle g_p(a) = \sum_{k=0}^{p-1} \zeta^{ak^2}$</p> <p>so Gauss sums are some kind of finite analogue of the normal distribution $e^{-\pi x^2}$, which is itself well-known to be an eigenfunction of the Fourier transform on $\mathbb{R}$. I remember someone claiming to me once that the two are closely related, but I haven't been able to track down a reference. Does anyone know what the precise connection is? Is there a theory of self-dual locally compact abelian groups somewhere out there?</p> http://mathoverflow.net/questions/11053/whats-the-relationship-between-gauss-sums-and-the-normal-distribution/11138#11138 Answer by GS for What's the relationship between Gauss sums and the normal distribution? GS 2010-01-08T12:44:12Z 2010-01-13T11:34:16Z <p>It's true (as the answer below and some of the commenters note) that it's easy to interpret this question in a way that makes it seem trivial and uninteresting. I'm quite sure, however, that pursuing typographical similarity between $e^{x^2}$ and $\zeta^{m^2}$ leads to interesting mathematics, and so here's a more serious attempt at propoganda for some of Ivan Cherednik's work.</p> <p>Pages 6,7,8 and 9 of Cherednik's <a href="http://arxiv.org/abs/math/0110024" rel="nofollow">paper</a> "Double affine Hecke algebras and difference Fourier transforms" explain how to interpolate'' between integral formulas relating the Gaussian to the Gamma function and (a certain generalization of) Gauss sums. </p> <p>More explicitly, he shows that the formula (for many people, it's really just the definition of the Gamma function)</p> <p>$$\int_{-\infty}^{\infty} e^{-x^2} x^{2k} dx=\Gamma \left( k+\frac{1}{2} \right)$$</p> <p>(for $k \in \mathbb{C}$ with real part $>-1/2$) and the Gauss-Selberg sum</p> <p>$$\sum_{j=0}^{N-2k} \zeta^{(k-j)^2/4} \frac{1-\zeta^{j+k}}{1-\zeta^k} \prod_{l=1}^j \frac{1-\zeta^{l+2k-1}}{1-\zeta^l}=\prod_{j=1}^k (1-\zeta^j)^{-1} \sum_{m=0}^{2N-1} \zeta^{m^2/4}$$</p> <p>(where $N$ is a positive integer, $\zeta=e^{2\pi i/N}$ is a prim. $N$th root of $1$, and $k$ is a positive integer at most $N/2$) can both be obtained as limiting cases of the same $q$-series identity. The common generalization of the Gaussian and the function $k \mapsto \zeta^{k^2}$ is the function $x \mapsto q^{x^2}$, and the measures weighting the integral and sum get replaced by Macdonald's measure---essentially the same one that shows up in the constant term conjecture for $A_1$, and that produces the Macdonald polynomials and kick-started the <a href="http://mathoverflow.net/questions/6517/double-affine-hecke-algebras-and-mainstream-mathematics" rel="nofollow">DAHA</a>. The Fourier transform is deformed along with everything else to produce the "Cherednik-Fourier" transform.</p> <p>I don't know how much of the roots of unity story generalizes to higher rank root systems.</p> <p>Note: In the Gauss-Selberg sum, replacing $k$ by the integer part of $N/2$ and manipulating a little (as in the nice exposition by David Speyer linked to in the question above) gives the usual formula for the Gauss sum.</p> http://mathoverflow.net/questions/11053/whats-the-relationship-between-gauss-sums-and-the-normal-distribution/11228#11228 Answer by Idoneal for What's the relationship between Gauss sums and the normal distribution? Idoneal 2010-01-09T11:27:33Z 2010-01-12T16:28:42Z <p>I don't think there is anything deep going on here. The Fourier analysis on finite abelian group is fairly straightforward.</p> <p>Gauss sums are the Fourier coefficients you get when you expand an additive character $k \rightarrow e^{\frac{2\pi iak}{p}}$ with respect to the basis of multiplicative characters (i.e. those that give rise to Dirichlet characters when our group is $\mathbb{Z}/n\mathbb{Z}$). A Gauss sum is a sum of the product of an additive and a multiplicative characters and as such can be thought of as a finite group analogue of the Gamma function. Recall that the Gamma function is the integral on $\mathbb{R}^{>0}$ of the product of $e^{-x}$ (additive character on the reals) and $x^s$ (a multiplicative character on $\mathbb{R}^{>0}$) with respect to the Haar measure $\frac{dx}{x}$ on $\mathbb{R}^{>0}$.</p> <p>You are probably thinking ${\zeta_p}^{ak^2}$ as the finite analogue of the Gaussian $e^{-\pi x^2}$, but as you have written yourself,</p> <p>$g_p(a)=\sum_{k=0}^{p-1} {\zeta_p}^{ak^2}$,</p> <p>a Gauss sum is a sum of `things' that look like the Gaussian and there is no reason why a Gauss sum itself should be something like the Gaussian. </p>