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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.

Apply the Poisson summation formula to $h(x)=e^{2\pi i\tau x^2}$, with Fourier kernel $e^{-2\pi ix\xi }$. Its complex Gaussian Fourier transform isChoose the square root holomorphic on the half-plane and positive when $\tau$ is positive imaginary. Summing over integer $\xi$ givesHere the printed theta series of integer squares uses $q=e^{2\pi i\tau }$; the common theta-constant convention instead uses $e^{\pi iz}$.

To keep track of the shifted series, put $z=2\tau$ and defineThus $\varphi (\tau )={\Theta }_4(2\tau )$. The same Poisson calculation with a phase or shifted lattice gives ${\Theta }_4(-1/z)=\sqrt{-iz}{\Theta }_2(z)$ and ${\Theta }_2(-1/z)=\sqrt{-iz}{\Theta }_4(z)$. Translation gives ${\Theta }_4(z+2)={\Theta }_4(z)$ and ${\Theta }_2(z-1)=e^{-\pi i/4}{\Theta }_2(z)$. These are theta-constant inversion and translation laws.

Let $U\tau =\tau /(2\tau +1)$ and put $w=-1/z-1$, so $2U\tau =-1/w$. Raising the preceding identities to the eighth power removes all square-root and phase ambiguities:Also $\varphi (\tau +1{)}^8=\varphi (\tau {)}^8$. The given generators, including their negatives, therefore establish weight four for ${\varphi }^8$ on ${\Gamma }_0(2)$, and weight $4k$ for ${\varphi }^{8k}$.

The defining series is holomorphic and its expansion at infinity has no negative powers. At the other modular cusp,Writing $q_0=e^{\pi i\tau }$ gives ${\Theta }_2(\tau /2)=2q_0^{1/8}{\sum }_{n\ge 0}{q}_0^{n(n+1)/2}$, so the right side begins $16q_0$ and is a holomorphic power series in $q_0$. Taking its $k$th power proves modular cusp holomorphy for every $k>0$. ConsequentlyThe exponent in the shifted series is $n^2$, as printed in the PDF, not the corrupted exponent in the TeX aid.

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.