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Fix an odd prime $p$ and set $\zeta=e^{2\pi i/p}$. We define the quadratic Gauss sum as

$$g=\sum_{n=0}^{p-1} \zeta^{n^2}.$$

It's pretty easy to show that

$$g^2= \begin{cases} p & \textrm{if } p\equiv 1 \mod 4 \\\ -p & \textrm{if } p\equiv 3 \mod 4, \end{cases}$$

and from this we can deduce quadratic reciprocity; it's harder to determine the modulus. We can actually find an explicit formula for $g$, namely:

$$g= \begin{cases} \sqrt{p} & \textrm{if } p\equiv 1 \mod 4 \\\ i\sqrt{p} & \textrm{if } p\equiv 3 \mod 4. \end{cases}$$

This is the result I refer to for the remainder of the question.


Can we categorify this result?

By categorification, I mean the opposite of decategorification, and by decategorification, I mean the process of removing structure by e.g. taking the cardinality of a set or the dimension of a vector space. (Thus an example of categorification would be interpreting some combinatorial identity of positive integers as a bijection between sets.) This is intentionally vague, because there are plenty of people who have a much better idea of what constitutes categorification than I do, so feel free to interpret "categorification" liberally.


Gauss's original proof of our result uses q-binomial coefficients. (A modern exposition of this proof can be found in "The determination of Gauss sums" by Bruce C. Berndt and Ronald J. Evans.)

Now, $q$-binomial coefficients can be categorified by Grassmannian varieties. What I mean by that is: the $q$-binomial coefficient $\binom{n}{k}_q$ is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the finite field $\mathbb{F}_q$, i.e. the cardinality of the Grassmannian $\textrm{Gr}(n,k)$. Basically, I'm wondering if there is some way this can be connected to the formula for the quadratic Gauss sum, seeing as how the formula is clearly related to the properties of $q$-binomial coefficients.

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The paper, ""; makes precise and proves the following principle, "if an arithmetic statement about the objects can be proved by pretending that they are complex numbers, then there also exists an honest proof". – Spice the Bird Jan 3 '13 at 18:02
Thanks for your comment. I've heard of that result before somewhere, but never managed to track down a reference. It's very interesting, but doesn't really answer my question. If I had some result about objects that decategorified to this Gauss sum identity, then this paper might give me a proof "for free"; the problem is, I don't know what to prove about the objects yet! – Aleksandar Bahat Jan 5 '13 at 0:48

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