Landsberg-Schaar relation

The Landsberg–Schaar relation is a reciprocity formula for quadratic exponential sums. It transforms a finite sum with quadratic phase $qn^2/p$ into another finite quadratic sum in which the roles of $p$ and $q$ are essentially reversed. The transformation includes the striking phase factor $\exp\left(\frac{\pi i}{4}\right)$ . This factor is not incidental: it is the same analytic phase that appears in the evaluation of quadratic Gauss sums and, ultimately, in quadratic reciprocity.

The relation is as follows.

Theorem. For all positive integers $p$ and $q$ ,

$\displaystyle \frac{1}{\sqrt p}\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i qn^2}{p}\right)=\frac{\exp\left(\frac{\pi i}{4}\right)}{\sqrt{2q}}\sum_{m=0}^{2q-1}\exp\left(-\frac{\pi i pm^2}{2q}\right).$

No coprimality assumption is required. The formula is already a reciprocity law: a quadratic sum of length $p$ is transformed into a quadratic sum of length $2q$ .

The proof has one central idea. Rather than evaluating either finite sum directly, we embed the left-hand sum into a Gaussian-damped infinite theta series. This theta series can then be analyzed in two ways. First, its periodic coefficients recover the finite sum on the left-hand side. Second, Poisson summation transforms it into a dual theta series whose periodic coefficients recover the finite sum on the right-hand side. The phase $\exp\left(\frac{\pi i}{4}\right)$ arises from the Fourier transform of a complex Gaussian.

Gaussian Averaging

We begin with a standard consequence of Poisson summation. For $h>0$ and $\alpha\in\mathbb R$ ,

$\displaystyle \sum_{k\in\mathbb Z}e^{-\pi h(k+\alpha)^2}=\frac{1}{\sqrt h}\sum_{\ell\in\mathbb Z}e^{-\pi \ell^2/h}e^{2\pi i\ell\alpha}.$

As $h\downarrow0$ , every term with $\ell\ne0$ is exponentially small. Hence

$\displaystyle \sum_{k\in\mathbb Z}e^{-\pi h(k+\alpha)^2}=\frac{1}{\sqrt h}+o(h^{-1/2}).$

In other words, a very broad Gaussian samples many integers, so its discrete sum is asymptotic to its integral. We will apply this observation after grouping integers into residue classes.

Theta Series

For $\varepsilon>0$ , define the Theta series $\displaystyle \Theta_\varepsilon:=\sum_{n\in\mathbb Z}\exp\left(\frac{2\pi i qn^2}{p}\right)e^{-\pi\varepsilon n^2/p}.$

The Gaussian factor makes the sum absolutely convergent. As $\varepsilon\to0^+$ , the damping weakens, and the leading term of $\Theta_\varepsilon$ reveals a finite quadratic sum.

The phase $\exp\left(\frac{2\pi i qn^2}{p}\right)$ is periodic modulo $p$ , since replacing $n$ by $n+p$ changes $qn^2/p$ by an integer. Writing $n=r+kp$ , with $0\le r\le p-1$ , gives

$\displaystyle \Theta_\varepsilon=\sum_{r=0}^{p-1}\exp\left(\frac{2\pi i qr^2}{p}\right)\sum_{k\in\mathbb Z}e^{-\pi p\varepsilon(k+r/p)^2}.$

For each fixed $r$ , the inner sum has the form above, with $h=p\varepsilon$ . Therefore,

$\displaystyle \sum_{k\in\mathbb Z}e^{-\pi p\varepsilon(k+r/p)^2}=\frac{1}{\sqrt{p\varepsilon}}+o(\varepsilon^{-1/2}),$

uniformly in $r$ . Multiplying by $\sqrt\varepsilon$ , we obtain

$\displaystyle \lim_{\varepsilon\downarrow0}\sqrt\varepsilon\Theta_\varepsilon=\frac{1}{\sqrt p}\sum_{r=0}^{p-1}\exp\left(\frac{2\pi i qr^2}{p}\right).$

This is the first interpretation of the theta sum: the finite quadratic sum on the left-hand side of the Landsberg–Schaar relation is the coefficient of the dominant $\varepsilon^{-1/2}$ growth.

Poisson Summation

We now analyze the same theta sum by Fourier methods. The definition of $\Theta_\varepsilon$ can be rewritten as

$\displaystyle \Theta_\varepsilon=\sum_{n\in\mathbb Z}\exp\left(-\pi\frac{\varepsilon-2iq}{p}n^2\right).$

For a complex number $A$ with ${\text{Re}}(A)>0$ , the Fourier transform of a Gaussian is

$\displaystyle \int_{\mathbb R}e^{-\pi A x^2}e^{-2\pi i\xi x}dx=A^{-1/2}e^{-\pi\xi^2/A}.$

The square root is chosen by continuous extension from positive real values of $A$ . This branch choice is responsible for the phase $\exp\left(\frac{\pi i}{4}\right)$ .

Applying Poisson summation with $\displaystyle A=\frac{\varepsilon-2iq}{p},$ we obtain

$\displaystyle \Theta_\varepsilon=\sqrt{\frac{p}{\varepsilon-2iq}}\sum_{m\in\mathbb Z}\exp\left(-\pi\frac{pm^2}{\varepsilon-2iq}\right).$

This is the decisive step. Poisson summation has replaced the original quadratic phase $qn^2/p$ by a dual quadratic phase.

Since $\displaystyle \frac{1}{\varepsilon-2iq}=\frac{\varepsilon}{\varepsilon^2+4q^2}+i\frac{2q}{\varepsilon^2+4q^2},$ the $m$ -th term in the transformed sum is

$\displaystyle \exp\left(-\pi\frac{p\varepsilon}{\varepsilon^2+4q^2}m^2\right)\exp\left(-\pi i\frac{2pq}{\varepsilon^2+4q^2}m^2\right).$

As $\varepsilon\downarrow0$ ,

$\displaystyle \frac{p\varepsilon}{\varepsilon^2+4q^2}=\frac{p\varepsilon}{4q^2}+O(\varepsilon^3),\qquad \frac{2pq}{\varepsilon^2+4q^2}=\frac{p}{2q}+O(\varepsilon^2).$

A comparison on the effective range $|m|\ll\varepsilon^{-1/2}$ , together with Gaussian tail estimates, yields

$\displaystyle \sum_{m\in\mathbb Z}\exp\left(-\pi\frac{pm^2}{\varepsilon-2iq}\right)=\sum_{m\in\mathbb Z}e^{-\pi p\varepsilon m^2/(4q^2)}\exp\left(-\frac{\pi i pm^2}{2q}\right)+o(\varepsilon^{-1/2}).$

The oscillatory factor $\displaystyle \exp\left(-\frac{\pi i pm^2}{2q}\right)$ is periodic modulo $2q$ . Indeed, replacing $m$ by $m+2q$ changes $pm^2/(4q)$ by an integer. Thus we may group the sum by residue classes modulo $2q$ . Writing $m=r+2qk$ , we have

$\displaystyle e^{-\pi p\varepsilon m^2/(4q^2)}=e^{-\pi p\varepsilon(k+r/(2q))^2}.$

Applying the Gaussian averaging principle again gives

$\displaystyle \sum_{m\in\mathbb Z}\exp\left(-\pi\frac{pm^2}{\varepsilon-2iq}\right)=\frac{1}{\sqrt{p\varepsilon}}\sum_{r=0}^{2q-1}\exp\left(-\frac{\pi i pr^2}{2q}\right)+o(\varepsilon^{-1/2}).$

Finally, $\displaystyle \sqrt{\frac{p}{\varepsilon-2iq}}\longrightarrow\exp\left(\frac{\pi i}{4}\right)\sqrt{\frac{p}{2q}}\quad(\varepsilon\downarrow0).$

The phase $\exp\left(\frac{\pi i}{4}\right)$ appears because $-2iq$ has argument $-\pi/2$ , so its inverse square root has argument $\pi/4$ .

Combining the transformed theta-sum formula with these asymptotics, we obtain

$\displaystyle \lim_{\varepsilon\downarrow0}\sqrt\varepsilon\Theta_\varepsilon=\frac{\exp\left(\frac{\pi i}{4}\right)}{\sqrt{2q}}\sum_{r=0}^{2q-1}\exp\left(-\frac{\pi i pr^2}{2q}\right).$

The two preceding limits compute the same quantity. Equating them gives

$\displaystyle \frac{1}{\sqrt p}\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i qn^2}{p}\right)=\frac{\exp\left(\frac{\pi i}{4}\right)}{\sqrt{2q}}\sum_{m=0}^{2q-1}\exp\left(-\frac{\pi i pm^2}{2q}\right).$

This is the Landsberg–Schaar relation.

The proof is important for more than the final identity. It begins with a finite quadratic phase modulo $p$ , but it does not evaluate that phase by finite algebra alone. Instead, it embeds the finite sum into a theta function on the real line. Poisson summation then exchanges the integer lattice with its Fourier-dual lattice.

Under this transformation, the quadratic coefficient $\displaystyle \frac{q}{p}$ is replaced by the dual coefficient $\displaystyle -\frac{p}{4q},$ while the Fourier transform contributes the phase $\exp\left(\frac{\pi i}{4}\right)$ . Thus the Landsberg–Schaar relation forms a bridge between finite arithmetic and real Fourier analysis. The finite sums arise because the quadratic phases are periodic, while the phase $\exp\left(\frac{\pi i}{4}\right)$ comes from the square root in the Fourier transform of a complex Gaussian.

Quadratic Reciprocity: The Landsberg–Schaar relation is not itself the quadratic reciprocity law, but it has the same underlying structure. Suppose that $p$ and $q$ are distinct odd primes, and define the quadratic Gauss sum

$\displaystyle G(a;\ell):=\sum_{x\bmod\ell}\exp\left(\frac{2\pi i ax^2}{\ell}\right),\qquad G_\ell:=G(1;\ell).$

For $\ell\nmid a$ , one has the standard change-of-coefficient identity

$\displaystyle G(a;\ell)=\left(\frac{a}{\ell}\right)G_\ell,$

where $\left(\frac{a}{\ell}\right)$ is the Legendre symbol.

The left-hand side of the Landsberg–Schaar relation therefore becomes

$\displaystyle \frac{1}{\sqrt p}G(q;p)=\left(\frac{q}{p}\right)\frac{G_p}{\sqrt p}.$

On the right, splitting the sum over $m\bmod 2q$ into even and odd residue classes gives

$\displaystyle \sum_{m\bmod 2q}\exp\left(-\frac{\pi i pm^2}{2q}\right)=\bigl(1-i^{pq}\bigr)G(-p;q).$

Using the change-of-coefficient identity once more, $\displaystyle G(-p;q)=\left(\frac{-p}{q}\right)G_q.$

Thus the Landsberg–Schaar relation compares the two Legendre symbols $\left(\frac{q}{p}\right)$ and $\left(\frac{p}{q}\right),$ with the remaining sign supplied by the Gaussian phase and the congruence classes of $p$ and $q$ modulo $4$ . After inserting the classical evaluation of quadratic Gauss sums, this comparison becomes

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

This is quadratic reciprocity. The sign in quadratic reciprocity is therefore not an arbitrary parity correction. It is the finite-arithmetic shadow of the phase $\exp\left(\frac{\pi i}{4}\right)$ that appears in the Fourier transform of a complex Gaussian. In this sense, the Landsberg–Schaar relation is a Fourier-analytic reciprocity law from which the classical quadratic reciprocity law naturally emerges.

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