Poisson Summation

$\displaystyle \sum_{n=-\infty}^{\infty} f(n)=\sum_{m=-\infty}^{\infty} \hat{f}(m)$

where

$\displaystyle \hat{f}(\xi) =\int_{\mathbb{R}} f(x) e^{2\pi ix\xi} dx$

Poisson summation is one of the most beautiful and powerful identities in mathematics. It’s a trace-formula that relates a sum over integer lattice to it’s sum over the dual lattice of integers. In general, it relates a sum over a lattice to a sum over the dual lattice.

$\displaystyle F(x) =\sum_{n=-\infty}^{\infty} f(x+n)$

is periodic function , hence defined on ${\mathbb{R}/\mathbb{Z}}$ .

The summation formula now can be seen as a specialization of the equality of trace computed in different basis- (one in the standard basis as a sum (integral) over the diagonal, and the other as a sum over eigenvalues). The relavant operator is averaging(convolution) operator

$\displaystyle T_{f} (g) = F\star g= \int_{\mathbb{R}/\mathbb{Z}} F(x-y) g(y) dy$

The integral over the diagonal is

$\displaystyle \text{Trace} (T_f)=\int_{\mathbb{R}/\mathbb{Z}} F(0)dy = F(0) =\sum_{n=-\infty}^{\infty} f(n)$

$\displaystyle T_f(e^{2\pi inx)}) =\int_{\mathbb{R}/\mathbb{Z}} F(x-y) e^{2\pi iny}dy = \hat{F}(n) e^{2\pi inx}$

${e^{2\pi inx}}~~~\quad \quad \quad$ are all the eigenfunctions with eigenvalues ${\hat{F}(n).}$

$\displaystyle \begin{aligned} \hat{F}(m) &=\int_{\mathbb{R}/\mathbb{Z}} F(x) e^{2\pi i m y}dy \\ &=\int_{0}^{1} \sum_{n=-\infty}^{\infty} f(y+n) e^{2\pi i m y}dy \\ &=\sum_{n} \int_{0}^{1} f(y+n) e^{2\pi i my}dy\\ &=\sum_{n} \int_{n}^{n+1} f(y) e^{2\pi i m y}dy \\ &=\int_{\mathbb{R}}f(y)e^{2\pi i m y}dy \\ &=\hat{f}(m) \end{aligned}$

Therefore sum over eigenvalues

$\displaystyle \text{Trace} (T_f)= \sum_{n} \hat{F} (n) =\sum_{n} \hat{f} (n)$

So we have,

$\displaystyle \sum_{n=-\infty}^{\infty} f(n)=\sum_{m=-\infty}^{\infty} \hat{f}(m)$

For a smooth function which doesn’t oscillate that much- we expect sampling over integers to capture the coarse scale behavior and hence the sum over integers is expected to be close to the integral over the real line which is the zeroth Fourier coefficient. Poisson summation captures this idea in a quantitative and more precise sense. For a nice function supported on a scale ${X}$ , the Fourier transform is essentially supported near ${0}$ on a scale of ${1/X.}$ And so the Poisson summation helps us to transform sums supported on a length of ${X}$ to length of ${1/X}$ . -Because of this property, Poisson summation is a a very fundamental tool in analysis and number theory.

One can think of Poisson summation as the Plancherel identity applied to the distribution

$D=\sum_{n}\delta(x-n)$

because the Fourier transform of the distribution is itself.

$\displaystyle \sum_{n=-\infty}^{\infty} f(n)=\langle f, D\rangle = \langle \hat{f}, \hat{D}\rangle = \langle \hat{f}, {D}\rangle =\sum_{m=-\infty}^{\infty} \hat{f}(m)$

More general identities (but in fact, special cases of the main Poisson formula):

$\displaystyle \sum_{n=-\infty}^{\infty} f(nt)=\frac{1}{t}\sum_{m=-\infty}^{\infty} \hat{f}(\frac{m}{t})$

$\displaystyle \sum_{n=-\infty}^{\infty} f(x+n)=\sum_{m=-\infty}^{\infty} \hat{f}(m) e^{2\pi imx}$

Applications:

Taking $\displaystyle f(x) =e^{-\pi x^{2}}, \hat{f}(\xi) =e^{-\pi \xi^{2}}$ ,we get

$\displaystyle \theta(z) =\sum_{n} e^{-\pi zn^{2}} = \frac{1}{\sqrt{z}} \sum_{n \in \mathbb{Z}} e^{-\pi n^{2} / z} = \frac{1}{\sqrt{z}} \theta(\frac{1}{z}).$

$\displaystyle \sum_{n \in \mathbb{Z}} \frac{1}{(z+n)^{2}}=\frac{\pi^{2}}{\sin ^{2}(\pi z)}$

follows using

$\displaystyle f(x) =\frac{1}{(z+x)^{2}}, \quad \quad \widehat{f}(n)=\left\{\begin{array}{ll} (-2 \pi i)^{2} n e^{2 \pi i n z} & \text { if } n>0 \\ 0 & \text { if } n \leq 0 \end{array}\right.$

and

$\displaystyle \sum_{n=1}^{\infty}(-2 \pi i)^{2} n e^{2 \pi i n z}=\sum_{n=1}^{\infty}-2 \pi i \frac{d}{d z} e^{2 \pi i n z}=-2 \pi i \frac{d}{d z} \frac{e^{2 \pi i z}}{1-e^{2 \pi i z}}=\frac{\pi^{2}}{\sin ^{2}(\pi z)}$

From this we can deduce

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{1}{2} \lim _{x \rightarrow 0}\left(\sum_{n \in \mathbb{Z}} \frac{1}{(x+n)^{2}}-\frac{1}{x^{2}}\right) = \frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\pi^{2}}{\sin ^{2}(\pi x)}-\frac{1}{x^{2}}\right)=\frac{\pi^{2}}{6}.$

2. Starting with

$\displaystyle \sum_{n=-\infty}^{\infty} f(nt)=\frac{1}{t}\sum_{m=-\infty}^{\infty} \hat{f}(\frac{m}{t})$

and taking the Mellin transform with respect to ${t}$ (integrate against multiplicative characters ${t^s}$ ) to get

$\displaystyle \begin{aligned} \int_{0} ^{\infty}\left(\sum_{n=-\infty}^{\infty} f(nt)\right)t^s \frac{dt}{t} &=\sum_{n=-\infty}^{\infty} \int_{0}^{\infty} f(nt) t^s \frac{dt}{t} \\ &=\sum_{n=-\infty}^{\infty}\frac{1}{n^s} \int_{0} ^{\infty}f(t) t^s \frac{dt}{t} \quad \quad \text{change of variables} \quad (t \rightarrow nt)\\ &=\left(\sum_{n=-\infty}^{\infty}\frac{1}{n^s} \right)M(f,s)\\ &=\zeta(s) M(f,s) \end{aligned}$ $\displaystyle \begin{aligned} \int_{0} ^{\infty}\left(\frac{1}{t}\sum_{m=-\infty}^{\infty} \hat{f}(\frac{m}{t}) \right)t^s \frac{dt}{t} &=\sum_{n=-\infty}^{\infty} \int_{0}^{\infty} \hat{f}(m/t) t^s \frac{dt}{t} \\ &=\sum_{n=-\infty}^{\infty}\frac{1}{m^{1-s}} \int_{0} ^{\infty}\hat{f}(t) t^{1-s} \frac{dt}{t} \quad \quad \text{change of variables} \quad (t \rightarrow m/t)\\ &=\left(\sum_{n=-\infty}^{\infty}\frac{1}{n^s} \right)M(\hat {f},s)\\ &=\zeta(1-s) M(\hat{f},1-s) \end{aligned}$

Therefore Poisson summation gives

$\displaystyle \zeta(s) M(f,s) =\zeta(1-s) M(f,1-s)$

3. Using ${f(x) =e^{-\pi x^{2}}}$ and taking the mellin transforms of ${\theta(t)}$ as above we get

$\displaystyle M(f,s) =\int_{0}^{\infty}e^{-\pi t^{2}} t^{s} \frac{dt}{t} = \pi^{-s/2} \Gamma(s/2)$ $\displaystyle \zeta(s) M(f,s) =\zeta(1-s) M(\hat{f},1-s) =\zeta(1-s)M(f,1-s)$

4. Vander Corput Method: The following formula follows Poisson summation and stationary phase approximation for the Fourier integrals:

$\displaystyle \sum_{n=a}^{b} e(f(n))\approx \sum_{\substack{f^{\prime}(a) \leq r \leq f^{\prime}(b)\\ f^{\prime}\left(x_{r}\right)=r}} \frac{ e\left(f\left(x_{r}\right)-r x_{r}+\frac{1}{8}\right)}{\sqrt{f^{\prime \prime}\left(x_{r}\right)}}$

This method is very useful in estimating exponential sums. Because the identity is a involution, it is applied in conjunction with ideas of Weyl-differencing to get good results. (Look at estimates for zeta function)

5. $\displaystyle \sum_{\vec n \in \mathbb{Z}^{n}} f(\vec n)=\sum_{\vec m \in \mathbb{Z}^{n}} \hat{f}(\vec n)$

is a straight forward generalization to integer lattice in larger dimensions

6. For a lattice ${\Gamma \subset V \equiv \mathbb{R}^n}$ for some ${n.},$ the dual lattice is given by

$\displaystyle \Gamma' =\{\gamma' | \langle \gamma', \gamma \rangle \in \mathbb{Z} ~~\text{for all} ~~\gamma \in \Gamma \}$

The Poisson summation in this case is

$\displaystyle \sum_{\gamma \in \Gamma} f(\gamma)=\frac{1}{\text{Vol}(V / \Gamma)} \sum_{\gamma' \in \Gamma^{\prime}} \hat{f}(\gamma')$

If ${ \Gamma=A{\mathbb{Z}^n}}$ , then by applying Poisson summation for ${F(x) = f(Ax),}$ we get the above formula. The volume terms comes from the Jacobian of the map ${x \rightarrow Ax.}$

7. Modularity of theta function:

$\displaystyle \Theta_{\Gamma}(t)=t^{-n / 2} \frac{1}{\text{Vol}(V / \Gamma)} \Theta_{\Gamma^{\prime}}\left(t^{-1}\right)$

where

$\displaystyle \Theta_{\Gamma}(t)=\sum_{x \in \Gamma} e^{-\pi t \langle x, x\rangle}$

$\displaystyle \sum_{n=0}^{\infty} r_{2}(n) w(n)=\sum_{m=0}^{\infty} r_{2}(m) \int_{0}^{\infty} w(x) \pi J_{0}(2 \pi \sqrt{x m}) \mathrm{d} x$

where

${r_{2}(n)=\#\left\{(a, b) \in \mathbb{Z}^{2}: a^{2}+b^{2}=n\right\}}$

by applying Poisson summation to $\displaystyle f(x,y) = f(x^2+y^2)$

9. Applying Poisson summation for functions of the form ${f(x,y) =f(xy)}$ essentially gives the Voronoi summation formula:

$\displaystyle \sum_{a<n<b} d(n) f(n)=\int_{a}^{b}(\log x+2 \gamma) f(x) d x+\sum_{n=1}^{\infty} d(n) \int_{a}^{b} f(x) \left(4 K_{0}\left(4 \pi \sqrt{n}x^{1 / 2}\right)-2 \pi Y_{0}\left(4 \pi \sqrt{n}x^{1 / 2}\right)\right) d x$

10. We can replace the role of ${\mathbb{Z} \in \mathbb{R}}$ with ${\mathbb{Q} \in \mathbb{A}}$ (rationals inside the ring of adeles), more generally ${K \in \mathbb{A}_{K}}$ (a number field inside its adele ring)to get

$\displaystyle \sum_{k \in K} f(k)=\sum_{k' \in K}^{\infty} \hat{f}(k').$

for a schwarz function ${f : \mathbb{A}_{K} \rightarrow \mathbb{C}}$ Just as before, considering ${f(tx)}$ and taking the mellin transform with ${t}$ for ${t \in \mathbb{A}^{*}}$ gives the functional equation for completed zeta function