Schur’s Proof of Sign of Gauss Sum

Consider the quadratic Gauss sum

$\displaystyle g_p=\sum_{n=0}^{p-1} e^{2 \pi i n^{2} / p}.$

It’s easy to see that

$\displaystyle g_p^2 = \left( \frac{-1}{p}\right)p$

which implies that

$\displaystyle g_{p}=\pm \sqrt{p} \text { if } p \equiv 1~(\bmod 4) \quad \text { and } g_{p}=\pm i \sqrt{p} \text { if } p \equiv 3 ~(\bmod 4).$

But how do we determine the sign?

There are many ways to prove the result (Poisson summation, theta series, Reciprocity by Contour integrals, Esterman evaluation etc). Here we show a linear algebra proof by Schur.

Schur’s proof:

Let ${S}$ be the ${p \times p}$ matrix whose ${ij}$ th entry is ${\zeta^{ij}}$ where ${\zeta =e^{\frac{2\pi i}{p}}}$ is the ${p}$ -th root of unity. $S$ is the Fourier matrix. That is

$\displaystyle S f(x) = \hat f(x) .$

The Gauss sum is the trace of $S$ and hence we compute the determinant and eigenvalues in order to evaluate the trace.

We have $\displaystyle S\bar S = pI_p$ because

$\displaystyle \sum_{j=0}^{p-1}\zeta^{ij}\zeta^{-jk} = \begin{cases} p \quad \text{ if } i=k, \\ 0 \quad \text{ otherwise.}\end{cases}$

Therefore ${\frac{S}{\sqrt p}}$ is a unitary matrix.

Because ${S}$ is the Fourier matrix and hence we have by Fourier inversion (or just computation) that

$\displaystyle S^2 f(x) = pf(-x).$

Thus a basis of eigenvectors for ${S^2}$ is given by

${ \delta_0}$ and ${\delta(x) +\delta(-x)}, x= 1,2 , \cdots p-1/2$ for the eigenvalue ${\sqrt p}$ and

${\delta(x)-\delta(-x)}, x= 1,2,3, \cdots p-1/2$ for the eigenvalue ${-\sqrt p}$ .

From these eigenvalues above, we can compute the determinant of ${S^2}$ to be

$\displaystyle (-1)^{{p \choose 2}} p^p.$

Therefore the determinant of ${S/\sqrt{p}}$ has to be ${\pm i^{p \choose 2}.}$

We will now determine the sign of this determinant (This is where the main work is!)

The determinant of ${S}$ is a vandermonde determinant given by

$\displaystyle {\det}\left(S\right)=\prod_{0 \leq s<r \leq p-1}\left(\zeta^{r}-\zeta^{s}\right)$

$\displaystyle = \prod_{0 \leq s <r \leq p-1} e^{\frac{\pi i (r+s)}{p}} \left( e^{\frac{\pi i (r-s)}{p}} - e^{\frac{\pi i (s-r)}{p}} \right)$

$\displaystyle = \prod_{0 \leq s <r \leq p-1} e^{\frac{\pi i (r+s)}{p}} \prod_{0 \leq s <r \leq p-1} \left(2 i \sin \frac{(r-s) \pi}{p}\right)$

${ \prod_{0 \leq s <r \leq p-1} e^{\frac{\pi i (r+s)}{p}}}$ is equal to ${1}$ . Therefore

$\displaystyle \det (S)= i^{{p \choose 2}} \prod_{0 \leq s <r \leq p-1} \left(2 \sin \frac{(r-s) \pi}{p}\right)$

But the quantities ${\sin \frac{(r-s) \pi}{p}}$ are positive, therefore we determined the sign of the determinant and we have

$\displaystyle \det(S) = i^{p \choose 2} p^{p/2}.$

Because ${ S^4 =p^2 I,}$ we get that the eigenvalues of ${S}$ are ${\pm \sqrt p, \pm i \sqrt p.}$

If ${x, y, z, t}$ are the mutliplicities of these eigenvalues we get

$\displaystyle x+y =\frac{p+1}{2}, z+t =\frac{p-1}{2}$

because ${\sqrt p}$ and ${-\sqrt p}$ are the eigenvalues for ${S^2.}$

$\displaystyle \det (S) = i^{p \choose 2} p^{p/2} =p^{p/2}(-1)^{y+t} i^{z+t}$

$\displaystyle Trace(S) = x-y +(z-t)i.$

Now we already know the value of the trace up to sign, using these relations we can deduce that the sign of the Gauss sum in all the cases has to positive!

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