= Poisson summation formula for a Euclidean lattice
{c}
{title2=$\sum_{\lambda\in\Lambda}f(\lambda)=m(\Lambda)^{-1}\sum_{\mu\in\Lambda^\vee}\widehat f(\mu)$}
Let $\Lambda=B\mathbb Z^d$ be a full-rank <Euclidean lattice>, with <covolume> $m(\Lambda)=|\det B|$ and <dual lattice> $\Lambda^\vee=B^{-\mathsf T}\mathbb Z^d$. For a <Schwartz function> $f\in\mathcal S(\mathbb R^d)$ use the <Fourier transform> convention
$$
\widehat f(y)=\int_{\mathbb R^d}f(x)e^{-2\pi i\langle x,y\rangle}\,dx.
$$
Then the <Poisson summation formula> becomes
$$
\boxed{\sum_{\lambda\in\Lambda}f(\lambda)
=\frac1{m(\Lambda)}\sum_{\mu\in\Lambda^\vee}\widehat f(\mu).}
$$
Both <series> have <absolute convergence>. To see the normalization, apply integer-lattice <Poisson summation> to $g(u)=f(Bu)$. The <change of variables formula> gives $\widehat g(k)=|\det B|^{-1}\widehat f(B^{-\mathsf T}k)$, yielding both the reciprocal <covolume> factor and the <dual lattice> argument. The factor is the physical lattice-point density. In particular, for spacing $h>0$,
$$
\sum_{n\in\mathbb Z}f(hn)=\frac1h\sum_{k\in\mathbb Z}\widehat f(k/h).
$$
For the <Gaussian function> $f(x)=e^{-\pi a\|x\|^2}$ with $a>0$, the product of the one-dimensional <Gaussian Fourier transforms> gives
$$
\sum_{\lambda\in\Lambda}e^{-\pi a\|\lambda\|^2}
=\frac{a^{-d/2}}{m(\Lambda)}\sum_{\mu\in\Lambda^\vee}e^{-\pi\|\mu\|^2/a}.
$$
Here $\|\cdot\|$ is the <Euclidean norm>. As a normalization check, take $\Lambda=\operatorname{diag}(2,3)\mathbb Z^2$. Its <covolume> is $6$, and its <dual lattice> is $\operatorname{diag}(1/2,1/3)\mathbb Z^2$, so
$$
\sum_{m,n\in\mathbb Z}e^{-\pi a(4m^2+9n^2)}
=\frac1{6a}\sum_{p,q\in\mathbb Z}e^{-\pi(p^2/4+q^2/9)/a}.
$$
This also follows by multiplying two <Gaussian Poisson summation identities> with parameters $4a$ and $9a$. If one instead uses an angular <Fourier transform> kernel $e^{-i\langle x,\omega\rangle}$, the reciprocal <frequencies> are $2\pi\Lambda^\vee$, while the prefactor remains $m(\Lambda)^{-1}$.
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