Quadratic Reciprocity via Poisson Summation

Quadratic reciprocity compares two apparently unrelated questions. For distinct odd primes $p$ and $q$ , one may ask whether $p$ is a square modulo $q$ , and whether $q$ is a square modulo $p$ . The quadratic reciprocity law says that these answers agree except in one case: when both primes are $3$ modulo $4$ . In Legendre-symbol notation, the theorem is

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

The proof below is built around one object, the quadratic Gauss sum. Its strength is that it explains the origin of the sign. The Chinese remainder theorem will tell us how a Gauss sum modulo $pq$ factors into Gauss sums modulo $p$ and $q$ . But that factorization alone does not tell us whether a sign should appear. The sign comes from somewhere else: from Poisson summation, or more precisely from the Fourier transform of a complex Gaussian.

This is the essential point of the proof. The finite arithmetic of $\mathbf Z/N\mathbf Z$ contributes the Legendre symbols. The real Fourier transform contributes a phase, either $1$ or $i$ . Quadratic reciprocity says that these two kinds of phase bookkeeping fit together perfectly.

Let

$\displaystyle G_N:=\sum_{x\bmod N}\exp\left(\frac{2\pi i x^2}{N}\right).$

We begin with the analytic fact that, for every odd positive integer $N$ ,

$\displaystyle G_N=\varepsilon_N\sqrt{N},\qquad \varepsilon_N=\begin{cases}1,&N\equiv1\pmod4,\\ i,&N\equiv3\pmod4.\end{cases}$

The rest of the proof of quadratic reciprocity will be short once this formula is available. Thus the central task is to understand why Poisson summation produces the two phases $1$ and $i$ .

Poisson summation

At first sight, $G_N$ is a purely finite sum. Poisson summation enters by embedding it into a smoothed infinite sum. The formal series

$\displaystyle \sum_{n\in\mathbf Z}\exp\left(\frac{2\pi i n^2}{N}\right)$

does not converge, since its terms all have modulus $1$ . We therefore insert a small Gaussian damping factor and define, for $t>0$ ,

$\displaystyle \Theta_N(t):=\sum_{n\in\mathbf Z}\exp\left(\frac{2\pi i n^2}{N}\right)e^{-\pi t n^2/N}.$

The parameter $t$ makes the sum rapidly convergent. We will let $t$ tend to zero only after identifying its main term.

There are two ways to read $\Theta_N(t)$ . The first sees the finite Gauss sum $G_N$ . The second applies Poisson summation and sees the Fourier transform of a complex Gaussian. Comparing these two readings evaluates $G_N$ .

First, separate the integers into residue classes modulo $N$ . Write $n=a+kN$ , with $a$ modulo $N$ and $k\in\mathbf Z$ . Since $\displaystyle \frac{(a+kN)^2}{N}=\frac{a^2}{N}+2ak+Nk^2,$ the oscillating factor $\displaystyle \exp\left(\frac{2\pi i n^2}{N}\right)$ depends only on $a$ . Hence

$\displaystyle \Theta_N(t)=\sum_{a\bmod N}\exp\left(\frac{2\pi i a^2}{N}\right)\sum_{k\in\mathbf Z}e^{-\pi Nt(k+a/N)^2}.$

The inner sum is a shifted Gaussian lattice sum. For small $t$ , it is well approximated by the corresponding Gaussian integral. More precisely, Poisson summation for the ordinary real Gaussian gives

$\displaystyle \sum_{k\in\mathbf Z}e^{-\pi h(k+\alpha)^2}=\frac{1}{\sqrt h}+O\left(h^{-1/2}e^{-\pi/h}\right)$

as $h\downarrow0$ , uniformly for $\alpha$ in a bounded set. Taking $h=Nt$ , we obtain

$\displaystyle \Theta_N(t)=\frac{G_N}{\sqrt{Nt}}+o(t^{-1/2}).$

This is the first important interpretation. The finite Gauss sum is the coefficient of the leading blow-up of the smoothed theta sum. The factor $1/\sqrt{Nt}$ is geometric: it comes from the spacing $N$ between the integers in any fixed residue class modulo $N$ .

Now comes the decisive use of Poisson summation. Rewrite the summand as

$\displaystyle \exp\left(\frac{2\pi i n^2}{N}\right)e^{-\pi t n^2/N}=\exp\left(-\pi\frac{t-2i}{N}n^2\right).$

Thus $\Theta_N(t)=\sum_{n\in\mathbf Z}f_t(n)$ , where

$\displaystyle f_t(x)=\exp\left(-\pi\frac{t-2i}{N}x^2\right).$

For complex $A$ with positive real part, the Gaussian Fourier transform is

$\displaystyle \widehat{e^{-\pi A x^2}}(\xi)=A^{-1/2}e^{-\pi\xi^2/A}.$

The square root is chosen by continuation from positive real $A$ . This choice matters: it is where the eventual eighth-root-of-unity phase enters.

Applying the formula with $A=(t-2i)/N$ , Poisson summation gives

$\displaystyle \Theta_N(t)=\sqrt{\frac{N}{t-2i}}\sum_{m\in\mathbf Z}\exp\left(-\pi\frac{Nm^2}{t-2i}\right).$

This identity is the heart of the argument. The original sum was organized by residue classes modulo $N$ ; the transformed sum is organized by Fourier frequencies $m$ . The transition between them is precisely Poisson summation.

To understand the right-hand side as $t\to0$ , write

$\displaystyle \frac{1}{t-2i}=\frac{t}{t^2+4}+i\frac{2}{t^2+4}.$

Thus the $m$ -th summand contains a decaying Gaussian factor and an oscillating factor. For the frequencies that materially contribute to the sum, one has $m^2$ of size about $t^{-1}$ . On this range, the summand may be replaced, up to an error negligible compared with $t^{-1/2}$ , by

$\displaystyle e^{-\pi Nt m^2/4}e^{-\pi iNm^2/2}.$

The reason this replacement is harmless is worth noting. The difference between the exact phase coefficient and $N/2$ is $O(t^2)$ . After multiplication by $m^2=O(t^{-1})$ , the resulting phase error is only $O(t)$ . Summing against the Gaussian produces an error smaller than the main term.

Now the arithmetic content of the Fourier side appears. Since $N$ is odd, $m^2\equiv0\pmod4$ for even $m$ , and $m^2\equiv1\pmod4$ for odd $m$ . Therefore the phase $e^{-\pi iNm^2/2}$ is $1$ on even frequencies and $e^{-\pi iN/2}$ on odd frequencies. The even and odd Gaussian sums have the same leading size, each contributing $1/\sqrt{Nt}$ . Consequently,

$\displaystyle \sum_{m\in\mathbf Z}\exp\left(-\pi\frac{Nm^2}{t-2i}\right)=\frac{1+e^{-\pi iN/2}}{\sqrt{Nt}}+o(t^{-1/2}).$

This is the exact place where the distinction between $N\equiv1\pmod4$ and $N\equiv3\pmod4$ is born. After Fourier transformation, the even and odd frequencies carry different phases. Their sum determines the value of the Gauss sum.

There is one more phase from the prefactor. As $t\downarrow0$ ,

$\displaystyle \sqrt{\frac{N}{t-2i}}\longrightarrow\sqrt N\frac{e^{i\pi/4}}{\sqrt2}.$

Combining the preceding asymptotics yields

$\displaystyle \frac{G_N}{\sqrt N}=\frac{e^{i\pi/4}}{\sqrt2}\left(1+e^{-\pi iN/2}\right).$

If $N\equiv1\pmod4$ , then $e^{-\pi iN/2}=-i$ , and the right-hand side is $1$ . If $N\equiv3\pmod4$ , then $e^{-\pi iN/2}=i$ , and the right-hand side is $i$ .

Poisson summation has therefore done much more than evaluate a finite sum. It has detected the real, or archimedean, phase of the quadratic form $x^2$ . The phase $1$ or $i$ is the real-place contribution that will later become the sign in quadratic reciprocity.

The Chinese remainder theorem and the factorization of $G_{pq}$

Let $p$ be an odd prime. For $c\not\equiv0\pmod p$ , define

$\displaystyle G_p(c):=\sum_{x\bmod p}\exp\left(\frac{2\pi i c x^2}{p}\right).$

We need the basic identity

$\displaystyle G_p(c)=\left(\frac{c}{p}\right)G_p(1).$

To see this, let $\chi_p$ be the quadratic character modulo $p$ , extended by $\chi_p(0)=0$ . The congruence $x^2\equiv y\pmod p$ has $1+\chi_p(y)$ solutions. Thus

$\displaystyle G_p(c)=\sum_{y\bmod p}(1+\chi_p(y))\exp\left(\frac{2\pi i c y}{p}\right).$

The part without $\chi_p$ vanishes because

$\displaystyle \sum_{y\bmod p}\exp\left(\frac{2\pi i c y}{p}\right)=0.$

Hence

$\displaystyle G_p(c)=\sum_y\chi_p(y)\exp\left(\frac{2\pi i c y}{p}\right).$

Replacing $y$ by $c^{-1}z$ , we get

$\displaystyle G_p(c)=\chi_p(c)\sum_{z\bmod p}\chi_p(z)\exp\left(\frac{2\pi i z}{p}\right)=\left(\frac{c}{p}\right)G_p(1).$

This is a useful conceptual step. The Legendre symbol is not added artificially at the end of the proof. It appears because the quadratic character is naturally adapted to the finite Fourier transform.

Now let $p\ne q$ be odd primes. Choose integers $A,B$ satisfying $Aq\equiv1\pmod p$ and $Bp\equiv1\pmod q$ . The Chinese remainder theorem identifies a pair $(u,v)$ , with $u\bmod p$ and $v\bmod q$ , with the class

$\displaystyle x\equiv Aqu+Bpv\pmod{pq}.$

Squaring this expression, the cross term is divisible by $pq$ , and therefore disappears modulo $pq$ . Using the defining congruences for $A$ and $B$ , one finds

$\displaystyle \exp\left(\frac{2\pi i x^2}{pq}\right)=\exp\left(\frac{2\pi i A u^2}{p}\right)\exp\left(\frac{2\pi i B v^2}{q}\right).$

Summing over $u$ and $v$ gives

$\displaystyle G_{pq}=G_p(A)G_q(B).$

By the coefficient-change formula,

$\displaystyle G_{pq}=\left(\frac{A}{p}\right)\left(\frac{B}{q}\right)G_pG_q.$

Since $A\equiv q^{-1}\pmod p$ and $B\equiv p^{-1}\pmod q$ , and a Legendre symbol has the same value on a unit and its inverse, this becomes

$\displaystyle G_{pq}=\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)G_pG_q.$

This is the finite arithmetic side of the proof. The Chinese remainder theorem almost factors the Gauss sum perfectly. The correction is exactly the product of the two Legendre symbols.

Comparison

We now evaluate the two sides using the Gauss-sum formula above. We have

$\displaystyle G_{pq}=\varepsilon_{pq}\sqrt{pq},\qquad G_p=\varepsilon_p\sqrt p,\qquad G_q=\varepsilon_q\sqrt q.$

Substituting and cancelling $\sqrt{pq}$ , we obtain

$\displaystyle \left(\frac{q}{p}\right)\left(\frac{p}{q}\right)=\frac{\varepsilon_{pq}}{\varepsilon_p\varepsilon_q}.$

If at least one of $p,q$ is $1\pmod4$ , the quotient on the right is $1$ . If $p\equiv q\equiv3\pmod4$ , then $pq\equiv1\pmod4$ , so $\varepsilon_{pq}=1$ , while $\varepsilon_p\varepsilon_q=i^2=-1$ . The quotient is then $-1$ . Hence

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

This is quadratic reciprocity.

The architecture of the proof is now visible. The Chinese remainder theorem describes how the quadratic phase behaves at the finite primes $p$ and $q$ . Poisson summation describes the same quadratic phase at the real place. The Legendre symbols record the finite correction; the numbers $1$ and $i$ record the real correction. Quadratic reciprocity is the statement that these two phase computations are compatible.

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