Wave kernel for the circle $\mathbb{S}^1$ - Poisson Summation Formula

Question

Question : How could I compute the (wave) kernel from the fact I have already found (wave) trace on unit circle?

The definitions are related to the page $25$ of the following pdf.

As the Spectrum$(S^1)=\{n^2 : n\ \in \mathbb{N}^*\}$, the trace (It this relevant for the question?) as distribution is simply $$w(t)=\sum_{k \geq 1} e^{it \sqrt{- \lambda_k}}=\int_{-\infty}^{\infty}W(t,x,x)dx=\frac{1}{e^t-1}.$$

From this fact, I would like to compute the kernel $$W(t,x,y)= \sum_{k \geq 1} e^{it \sqrt{- \lambda_k}} \mu_k(x) \mu_k(y),$$ where $\mu_k$ is the eigenfunctions of the eigenvalues $\lambda_k$, and I found $\mu_k (t) = a_k \cos kt + b_k \sin kt$, $k \in \mathbb{Z}$.

I think we have to use the Poisson Summation Formula, but it is unclear.

(The Poisson Summation Formula) Let $f(x)$ be any piecewise continuous function, defined for each $x$, $-\infty < x < \infty$, such that the sum $F(x)=\sum_{k=-\infty}^{\infty} f(x+2kL)$ converge (absolutely) to a continuous and piecewise $C^1$ function $F(x)$. Assume that this is uniform for $-L \leq x \leq L$. So that $F(x)$ is periodic of the period $2L$, and equals its Fourier series.

I don't understand how to do this, knowing that the exponential function isn't periodic. To help you better understand the concept, I think you can look at a case like the Heat Kernel on the following links #1 and #2.

Thanks!

P.S. Please, be aware that it is not for an homework, but rather for research work.

Answer 1

First, some comments about notation and normalization: If the wave kernel is given by $$W(t,x,y)= \sum_{k \geq 1} e^{it \sqrt{- \lambda_k}} \mu_k(x) \mu_k(y)$$ and the wave trace satisfies $$w(t)=\sum_{k \geq 1} e^{it \sqrt{- \lambda_k}}=\int_{-\infty}^{\infty}W(t,x,x)dx$$ then we must have $\int_{-\infty}^\infty \mu_k(x)^2\,dx=1$. This is natural for the setting cited in the linked thesis, where the functions were taken over the real line. But this is impossible for any eigenfunction periodic on $[0,2\pi)$. Hence I will instead assume that the range of integration in this case is $[0,2\pi)$; thus $\int_0^{2\pi} \mu_k(x)^2\,dx=1$ and $w(t)=\int_0^{2\pi} W(t,x,x)\,dx.$

With this in mind, the normalized eigenfunctions (assuming periodic boundary conditions on $[0,2\pi)$) are $\left\{\dfrac1{\sqrt{2\pi}},\dfrac1{\sqrt{\pi}}\cos x,\dfrac1{\sqrt{\pi}}\sin x,\dfrac1{\sqrt{\pi}}\cos 2x,\dfrac1{\sqrt{\pi}}\sin 2x,\cdots\right\}$ with eigenvalues $\{0,1,1,2,2,\cdots\}$. This basis gives series of the form $\displaystyle \frac{1}{\sqrt{2\pi}}a_0+\frac{1}{\sqrt{\pi}}\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$, as expected of a Fourier expansion.

Proceeding to the wave trace and wave trace, we obtain

\begin{align} w(t) &= 1+2\sum_{k=1}^\infty e^{-k t}=1+\dfrac{2e^{-t}}{1-e^{-t}}=\dfrac{e^t+1}{e^t-1},\\ W(t,x,y) &=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{k=1}^\infty e^{-k t}\cos kx\cos k y+\frac{1}{\pi}\sum_{k=1}^\infty e^{-k t}\sin kx\sin ky \\ &=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{k=1}^\infty e^{-k t}\cos k(x-y).\\ \end{align} All that remains is to simplify the last sum, a task I leave to the interested reader.

Answer 2

I think the kernel is \begin{align*} W\left(t,x,y\right) & =\sum_{n\geq1}e^{-tn}e^{in\left(x-y\right)}\\ & =\frac{1}{e^{\left(t-i\left(x-y\right)\right)}-1},\quad t>0. \end{align*}

Looking at pg 25 of the linked pdf, I think the following makes more sense: \begin{align*} W\left(t,x,y\right) & =\sum_{n=1}^{\infty}\cos\left(nt\right)\sin\left(nx\right)\sin\left(ny\right),\;\text{and}\\ w\left(t\right) & =\sum_{n=1}^{\infty}\cos\left(nt\right) \end{align*}