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For a full Euclidean lattice and a positive definite real symmetric matrix $A$, the Gaussian theta sum obeysUse the Fourier transform kernel $e^{-2\pi ix\cdot z}$, the identity $\widehat{e^{-\pi x^{T}Ax}}(z)=(\det A{)}^{-1/2}{e}^{-\pi z^T{A}^{-1}z}$, and the Poisson summation formula for a Euclidean lattice. No integrality or self-duality of the Euclidean lattice is assumed.

For a full Euclidean lattice $\Lambda$ and a positive definite real symmetric matrix $A$, the Gaussian theta sum is ${\theta }_{\Lambda }(A)={\sum }_{\lambda \in \Lambda }{e}^{-\pi {\lambda }^{T}A\lambda }$. Its absolute convergence follows from Gaussian decay and the polynomial growth of the number of lattice points in a ball. The Poisson summation formula for a Euclidean lattice gives its anisotropic theta functional equation.

For a full-rank Euclidean lattice, ${\Theta }_{\Lambda }(\tau )={\sum }_{x\in \Lambda }{e}^{\pi i|x{|}^2\tau }$ obeys ${\Theta }_{\Lambda }(\tau )=(-i\tau {)}^{-n/2}m(\Lambda {)}^{-1}{\Theta }_{{\Lambda }^{\vee }}(-1/\tau )$. Apply the Poisson summation formula for a Euclidean lattice and the complex Gaussian Fourier transform. The identity holds without assuming an integral or self-dual lattice.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

For a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$, its dual lattice isIf $\Lambda =B{\mathbb{Z}}^n$, then ${\Lambda }^{\prime }=B^{-\mathsf{T}}{\mathbb{Z}}^n$, $m(\Lambda )=|\det B|$ and $m({\Lambda }^{\prime })=m(\Lambda {)}^{-1}$. The characters of a real torus identify the additive dual lattice with the multiplicative character group byThe map is well defined precisely because $u\in {\Lambda }^{\prime }$, and is injective. For surjectivity, a continuous group homomorphism from the compact torus into ${\mathbb{C}}^{\times }$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb{R},+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to ${\mathbb{R}}^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $⟨u,x⟩$ for a unique $u$. Triviality on $\Lambda$ says $u\in {\Lambda }^{\prime }$, proving surjectivity.

Use the Fourier transform convention $\hat{f}(u)={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi i⟨u,x⟩}dx$, and take $f$ to be a Schwartz function. The periodization of a Schwartz function $P_f(x)={\sum }_{\lambda \in \Lambda }f(x+\lambda )$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character ${\chi }_u$ isThe equality follows by translating each cell and using $e^{2\pi i⟨u,\lambda ⟩}=1$. The rapidly convergent Fourier series can be evaluated at zero, yielding the Poisson summation formula for a Euclidean latticeThis argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.

For $\tau$ in the complex upper half-plane, let $g_{\tau }(x)=e^{\pi i\tau |x{|}^2}$. Scaling the self-dual real Gaussian function gives its Fourier transform at $\tau =it$, $t>0$, and holomorphic continuation in $\tau$ gives the complex Gaussian Fourier transformThe branch is $\exp [-(n/2)\mathrm{Log}(-i\tau )]$ with the logarithm on the right half-plane; it is positive for $\tau =it$. Both the integrals and the lattice sums are locally normally convergent on the complex upper half-plane. Applying the Poisson summation formula proves the lattice theta functional equationNo integrality or self-duality hypothesis on the lattice is needed.

Put $a=n/2$, $m=m(\Lambda )$ and ${\theta }_{\Lambda }(t)={\Theta }_{\Lambda }(it)$. The Epstein zeta function converges absolutely for $\mathrm{Re}s>a$, and its Mellin transform representation isAt infinity, ${\theta }_{\Lambda }(t)-1$ decays exponentially. Define the entire functionOn $(0,1)$, substitute ${\theta }_{\Lambda }(t)=m^{-1}{t}^{-a}{\theta }_{{\Lambda }^{\prime }}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the pole-subtracted theta integral for an Epstein zeta functionThe formula initially holds for $\mathrm{Re}s>a$ and continues the completed Epstein zeta function meromorphically to all $\mathbb{C}$. Apply the same formula to the dual lattice at $a-s$, using $m({\Lambda }^{\prime })=m^{-1}$ and ${\Lambda }^{\prime \prime }=\Lambda$. The entire terms and the two rational terms match, provingFinally $E_{\Lambda }(s)={\pi }^s{\mathcal{E}}_{\Lambda }(s)/\Gamma (s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma (s)$, and $E_{\Lambda }(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.

For a full Euclidean lattice $\Lambda \subseteq {\mathbb{R}}^n$, its dual lattice isFor a Schwartz function, the Poisson summation formula for a Euclidean lattice statesTo prove it, periodize $f$ over $\Lambda$. The resulting function on ${\mathbb{R}}^n/\Lambda$ has Fourier coefficient $m(\Lambda {)}^{-1}\hat{f}(\mu )$ at $\mu \in {\Lambda }^{\vee }$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.

Put $L_{\tau }=y^{-1/2}(\mathbb{Z}\tau +\mathbb{Z})$, a covolume-one lattice, and defineTermwise Mellin transformation in the initial half-plane gives

Split the integral at $t=1$. The integral over $[1,\infty )$ is entire in $s$ because the theta sum decays exponentially. Apply the Poisson summation formula for a Euclidean lattice and the Fourier eigenfunction calculation from part b to the interval $(0,1]$, then substitute $t\mapsto 1/t$. This rewrites the small-time integral as another exponentially convergent integral over $[1,\infty )$ plus explicit elementary Mellin terms. Those terms are meromorphic, but for positive even $k$ their apparent poles are canceled by the zeros of $1/\Gamma (s+k)$. The displayed formula therefore continues holomorphically to every $s\in \mathbb{C}$, proving the analytic continuation of a weight-k real-analytic Eisenstein series.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.