0
$\begingroup$

The heat kernel on the circle $\mathbb S^1\cong \mathbb{R}/\mathbb{Z}$ is given by $$(*) \qquad k_t(\theta)=\frac{1}{\sqrt{4\pi t}}\sum_{n\in \mathbb Z}e^{-\frac{(\theta-n)^2}{4 t}}, \quad \theta\in \mathbb{R}/\mathbb{Z}.$$

In a PDF, I did not understand the meaning of the following remark: $\, (*)$ is the $\mathbb{Z}$ -average of the heat kernel on $\mathbb R$ given by $$k_t(x)=\frac{1}{\sqrt{4\pi t}} e^{-\frac{x^2}{4 t}}, \quad x\in \mathbb{R}.$$ Also, the technique of unwinding $\mathbb{Z}$-averages?

Can someone explain to me more. Thank you in advance

$\endgroup$
1
$\begingroup$

I wouldn't call it $\mathbb{Z}$-average because "average" means dividing by the total weight, and this isn't happening here. This process is also known as "periodization" or "periodic summation" (Wikipedia). Given a function $f$ on $\mathbb{R}$, we can define a periodic function $g$ by $$ g(x) = \sum_{n\in\mathbb{Z}}f(x+n) \tag1 $$ provided the series converges (which is does when $f$ has sufficiently fast decay at infinity).

"Unwinding $\mathbb{Z}$-average" would mean the opposite: given a periodic function $g$, find $f$ such that (1) holds.

For example, if $g(x) = \sin 2\pi x$, then one can take $$ f(x) = \begin{cases} \sin 2\pi x,\quad &0\le x\le 1, \\ 0 \quad &\text{otherwise}\end{cases} $$ This particular example doesn't seem important. But when $f$ solves some translation-invariant PDE on $\mathbb{R}$ (such as the heat equation), we get a solution of the same PDE on the circle $\mathbb{R}/\mathbb{Z}$ via (1), and that is nice to have.

Closely related: Poisson summation formula, which says that the Fourier series of $g$ can be obtained from the Fourier transform of $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.