Schur’s Evaluation of the Sign of Gauss Sum

Consider the quadratic Gauss sum

\displaystyle g_p=\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i n^2}{p}\right),

where p is an odd prime. Writing the expression using the quadratic character modulo p ,

\displaystyle g_p=\sum_{m=0}^{p-1}\left(\frac{m}{p}\right)\exp\left(\frac{2\pi i m}{p}\right).

and squaring gives

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

Thus the magnitude and possible direction of g_p are already known:

\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).

The squaring argument leaves exactly one ambiguity: which sign occurs?

There are several ways to settle this ambiguity. One can use Poisson summation, theta functions, contour integration, or Estermann’s estimate. Schur’s proof is different in character. It treats the Gauss sum as the trace of the finite Fourier transform. The square of that transform reveals its possible eigenvalues, while its determinant fixes the remaining sign.

Let \displaystyle \zeta=\exp\left(\frac{2\pi i}{p}\right), and let V=\mathbb C^{\mathbb F_p} be the vector space of complex-valued functions on the finite field \mathbb F_p . Define the unnormalized Fourier transform

\displaystyle (Sf)(x):=\sum_{y\bmod p}\zeta^{xy}f(y).

In the standard basis, S is the p\times p matrix

\displaystyle S=(\zeta^{xy})_{x,y\bmod p}.

Its diagonal entries are \zeta^{x^2} , so its trace is exactly the Gauss sum:

\displaystyle {\text{tr}}(S)=\sum_{x\bmod p}\zeta^{x^2}=g_p.

Thus the problem of evaluating g_p becomes a spectral problem: determine the trace of the Fourier matrix. It is convenient to normalize the transform: \displaystyle F:=\frac{1}{\sqrt p}S. The orthogonality of additive characters gives

\displaystyle \sum_{y\bmod p}\zeta^{(x-z)y}=\begin{cases}p,&x=z,\\ 0,&x\ne z.\end{cases}

Hence \displaystyle SS^\ast=pI, so F is unitary. In particular, it is diagonalizable and all of its eigenvalues have absolute value 1 . More importantly, Fourier inversion gives a simple formula for the square of S :

\displaystyle (S^2f)(x)=\sum_{y\bmod p}\sum_{z\bmod p}\zeta^{xy+yz}f(z).

Interchanging the sums gives

\displaystyle (S^2f)(x)=\sum_{z\bmod p}\left(\sum_{y\bmod p}\zeta^{y(x+z)}\right)f(z).

The inner sum vanishes unless x+z\equiv0\pmod p , in which case it equals p . Therefore \displaystyle (S^2f)(x)=pf(-x). Let R denote the reflection operator \displaystyle (Rf)(x):=f(-x). Then \displaystyle S^2=pR,\quad F^2=R.

This is the main structural fact. The space V splits into its even and odd parts: \displaystyle V=V^+\oplus V^-, where \displaystyle V^+=\{f(-x)=f(x)\},\quad V^-=\{f(-x)=-f(x)\}. The dimensions are \displaystyle \dim V^+=\frac{p+1}{2},\quad \dim V^-=\frac{p-1}{2}.

On the even subspace, R acts as 1 , so F^2=1 . Thus the eigenvalues of F on V^+ are 1 and -1 . On the odd subspace, R acts as -1 , so F^2=-1 . Thus the eigenvalues of F on V^- are i and -i . Equivalently, the eigenvalues of S=\sqrt pF are among \displaystyle \sqrt p,-\sqrt p, i\sqrt p,-i\sqrt p.

Let m_1,m_{-1},m_i,m_{-i} be the multiplicities of the eigenvalues 1,-1,i,-i of F . The even-odd decomposition gives

\displaystyle m_1+m_{-1}=\frac{p+1}{2},\qquad m_i+m_{-i}=\frac{p-1}{2}.

The trace is \displaystyle {\text{tr}}(F)=m_1-m_{-1}+i(m_i-m_{-i})=\frac{g_p}{\sqrt p}.

At this stage, the spectrum has narrowed the Gauss sum to the expected four directions. To determine which direction occurs, we need one further piece of information: the determinant.

The determinant of S is a Vandermonde determinant:

\displaystyle \det(S)=\prod_{0\le s<r\le p-1}(\zeta^r-\zeta^s).

For 0\le s<r\le p-1 ,

\displaystyle \zeta^r-\zeta^s=2i\exp\left(\frac{\pi i(r+s)}{p}\right)\sin\left(\frac{\pi(r-s)}{p}\right).

The sine factors are all positive, because \displaystyle 0<\frac{\pi(r-s)}{p}<\pi.

The exponential factors contribute \displaystyle \exp\left(\frac{\pi i}{p}\sum_{0\le s<r\le p-1}(r+s)\right).

Every integer 0,1,\ldots,p-1 appears in exactly p-1 pairs, so

\displaystyle \sum_{0\le s<r\le p-1}(r+s)=(p-1)\sum_{j=0}^{p-1}j=\frac{p(p-1)^2}{2}.

Therefore the exponential factor is \displaystyle \exp\left(\frac{\pi i(p-1)^2}{2}\right)=1, because p is odd. Thus

\displaystyle \det(S)=i^{\binom p2}\prod_{0\le s<r\le p-1}2\sin\left(\frac{\pi(r-s)}{p}\right).

The remaining product is positive. Also,

\displaystyle |\det(S)|^2=\det(SS^\ast)=\det(pI)=p^p.

Hence \displaystyle \det(S)=i^{\binom p2}p^{p/2}, and after normalization, \displaystyle \det(F)=i^{\binom p2}.

On the other hand, the determinant is the product of the eigenvalues:

\displaystyle \det(F)=(-1)^{m_{-1}}i^{m_i}(-i)^{m_{-i}}.

Since m_i+m_{-i}=(p-1)/2 , this becomes

\displaystyle \det(F)=(-1)^{m_{-1}+m_{-i}}i^{(p-1)/2}.

Comparing this with the Vandermonde evaluation yields

\displaystyle (-1)^{m_{-1}+m_{-i}}=(-1)^{(p-1)/2}.

This parity relation is exactly the missing information.

First suppose that \displaystyle p=4k+1. Then \displaystyle m_1+m_{-1}=2k+1,\quad m_i+m_{-i}=2k. Since we already know that \displaystyle \frac{g_p}{\sqrt p}\in\{-1,1\}, the imaginary part of {\text{tr}}(F) must vanish. Hence \displaystyle m_i=m_{-i}=k.

The real part is either 1 or -1 , so \displaystyle m_{-1}\in{k,k+1}. The determinant condition gives \displaystyle m_{-1}+m_{-i}=m_{-1}+k\equiv0\pmod2. Thus m_{-1}\equiv k\pmod2 , which selects \displaystyle m_{-1}=k,\quad m_1=k+1. Therefore \displaystyle {\text{tr}}(F)=1, and so \displaystyle g_p=\sqrt p.

Now suppose that \displaystyle p=4k+3. Then \displaystyle m_1+m_{-1}=2k+2,\qquad m_i+m_{-i}=2k+1. This time \displaystyle \frac{g_p}{\sqrt p}\in\{-i,i\}, so the real part of the trace vanishes. Hence \displaystyle m_1=m_{-1}=k+1. The imaginary part is 1 or -1 , so \displaystyle m_{-i}\in{k,k+1}.

The determinant condition now gives \displaystyle m_{-1}+m_{-i}=k+1+m_{-i}\equiv1\pmod2. Thus \displaystyle m_{-i}\equiv k\pmod2, which forces \displaystyle m_{-i}=k,\qquad m_i=k+1. Therefore \displaystyle {\text{tr}}(F)=i, and hence \displaystyle g_p=i\sqrt p.

We have proved the classical evaluation

\displaystyle g_p=\begin{cases}\sqrt p,&p\equiv1\pmod4,\\ i\sqrt p,&p\equiv3\pmod4.\end{cases}

The conceptual structure of the proof is worth emphasizing. The identity F^2=R confines the spectrum of the Fourier transform to the four fourth roots of unity. This almost determines the trace, but not its sign. The Vandermonde determinant supplies the missing orientation information: it records the total phase of the Fourier matrix. Once that phase is known, the multiplicities, and therefore the trace, are forced.

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