# The heat kernel on the torus

Jordan Bell
October 7, 2014

## 1 Heat kernel on π

For $t>0$, define $k_{t}:\mathbb{R}\to(0,\infty)$ by11 1 Most of this note is my working through of notes by Patrick Maheux. http://www.univ-orleans.fr/mapmo/membres/maheux/InfiniteTorusV2.pdf

 $k_{t}(x)=(4\pi t)^{-1/2}\exp\left(-\frac{x^{2}}{4t}\right),\qquad x\in\mathbb{% R}.$

For $t>0$, define $g_{t}:\mathbb{R}\to(0,\infty)$ by

 $g_{t}(x)=2\pi\sum_{k\in\mathbb{Z}}k_{t}(x+2\pi k),\qquad x\in\mathbb{R},$

which one checks indeed converges for all $x\in\mathbb{R}$. Of course, $g_{t}(x+2\pi k)=g_{t}(x)$ for any $k\in\mathbb{Z}$, so we can interpret $g_{t}$ as a function on $\mathbb{T}$, where $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$.

Let $m$ be Haar measure on $\mathbb{T}$: $dm(x)=(2\pi)^{-1}dx$, and so $m(\mathbb{T})=1$. With $\|f\|_{1}=\int_{\mathbb{T}}|f|dm$ for $f:\mathbb{T}\to\mathbb{C}$, we have, because $g_{t}>0$,

 $\|g_{t}\|_{1}=\sum_{k\in\mathbb{Z}}\int_{\mathbb{T}}k_{t}(x+2\pi k)dx=\int_{% \mathbb{R}}k_{t}(x)dx=1.$

Hence $g_{t}\in L^{1}(\mathbb{T})$. For $\xi\in\mathbb{Z}$, we compute

 $\displaystyle\hat{g}_{t}(\xi)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{T}}g_{t}(x)e^{-i\xi x}dm(x)$ $\displaystyle=$ $\displaystyle\sum_{k\in\mathbb{Z}}\int_{\mathbb{T}}k_{t}(x+2\pi k)e^{-i\xi x}dx$ $\displaystyle=$ $\displaystyle\sum_{k\in\mathbb{Z}}\int_{\mathbb{T}}k_{t}(x+2\pi k)e^{-i\xi(x+2% \pi k)}dx$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}}k_{t}(x)e^{-i\xi x}dx$ $\displaystyle=$ $\displaystyle\hat{k}_{t}\left(\frac{\xi}{2\pi}\right)$ $\displaystyle=$ $\displaystyle e^{-\xi^{2}t}.$
###### Lemma 1.

For $t>0$ and $x\in\mathbb{R}$,

 $g_{t}(x)=\sqrt{\frac{\pi}{t}}\exp\left(-\frac{x^{2}}{4t}\right)\left(1+2\sum_{% k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)\cosh\left(\frac{\pi kx}{t}% \right)\right).$
###### Proof.

Using the definition of $g_{t}$,

 $\displaystyle g_{t}(x)$ $\displaystyle=$ $\displaystyle 2\pi\sum_{k\in\mathbb{Z}}k_{t}(x+2\pi k)$ $\displaystyle=$ $\displaystyle 2\pi\sum_{k\in\mathbb{Z}}(4\pi t)^{-1/2}\exp\left(-\frac{(x+2\pi k% )^{2}}{4t}\right)$ $\displaystyle=$ $\displaystyle\sqrt{\frac{\pi}{t}}\exp\left(-\frac{x^{2}}{4t}\right)\sum_{k\in% \mathbb{Z}}\exp\left(-\frac{\pi kx}{t}\right)\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)$ $\displaystyle=$ $\displaystyle\sqrt{\frac{\pi}{t}}\exp\left(-\frac{x^{2}}{4t}\right)$ $\displaystyle\left(1+\sum_{k\geq 1}\left(\exp\left(\frac{\pi kx}{t}\right)+% \exp\left(-\frac{\pi kx}{t}\right)\right)\exp\left(-\frac{\pi^{2}k^{2}}{t}% \right)\right),$

which gives the claim, using $\cosh y=\frac{e^{y}+e^{-y}}{2}$. β

###### Definition 2.

For $x\in\mathbb{R}$, let $\|x\|=\inf\{|x-2\pi k|:k\in\mathbb{Z}\}$.

For $k\in\mathbb{Z}$, $\|x+2\pi k\|=\|x\|$, so it makes sense to talk about $\|x\|$ for $x\in\mathbb{T}$.

###### Theorem 3.

For $t>0$ and $x\in\mathbb{R}$,

 $\exp\left(-\frac{\|x\|^{2}}{4t}\right)g_{t}(0)\leq g_{t}(x)\leq\exp\left(-% \frac{\|x\|^{2}}{4t}\right)\left(\sqrt{\frac{\pi}{t}}+g_{t}(0)\right).$
###### Proof.

Let $x=2\pi m+\theta$ with $|\theta|\leq\pi$, so that $\|x\|=\|\theta\|=|\theta|$, and $g_{t}(x)=g_{t}(\theta)$. Using Lemma 1 and the fact that $\cosh y\geq 1$, we get

 $g_{t}(\theta)\geq\exp\left(-\frac{\theta^{2}}{4t}\right)\sqrt{\frac{\pi}{t}}% \left(1+2\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)\right)=\exp% \left(-\frac{\theta^{2}}{4t}\right)g_{t}(0),$

hence

 $g_{t}(x)\geq\exp\left(-\frac{\|x\|^{2}}{4t}\right)g_{t}(0),$

the lower bound we wanted to prove.

Write

 $S=1+2\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)\cosh\left(\frac{% \pi k\theta}{t}\right).$

For any $k\geq 1$, using $|\theta|\leq\pi$,

 $2\cosh\left(\frac{\pi k\theta}{t}\right)\leq 2\cosh\left(\frac{\pi^{2}k}{t}% \right)=\exp\left(\frac{\pi^{2}k}{t}\right)+\exp\left(-\frac{\pi^{2}k}{t}% \right)\leq 1+\exp\left(\frac{\pi^{2}k}{t}\right).$

Hence

 $\displaystyle S$ $\displaystyle\leq$ $\displaystyle 1+\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)\left(1+% \exp\left(\frac{\pi^{2}k}{t}\right)\right)$ $\displaystyle=$ $\displaystyle 1+\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)+\exp% \left(-\frac{\pi^{2}k(k-1)}{t}\right)$ $\displaystyle\leq$ $\displaystyle 1+\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)+\exp% \left(-\frac{\pi^{2}(k-1)^{2}}{t}\right)$ $\displaystyle=$ $\displaystyle 2+2\sum_{k\geq 1}\exp\left(-\frac{\pi^{2}k^{2}}{t}\right)$ $\displaystyle=$ $\displaystyle 1+\sqrt{\frac{t}{\pi}}g_{t}(0).$

But $g_{t}(\theta)=\sqrt{\frac{\pi}{t}}\exp\left(-\frac{\theta^{2}}{4t}\right)S$, so

 $g_{t}(\theta)\leq\exp\left(-\frac{\theta^{2}}{4t}\right)\left(\sqrt{\frac{\pi}% {t}}+g_{t}(0)\right)=\exp\left(-\frac{\|x\|^{2}}{4t}\right)\left(\sqrt{\frac{% \pi}{t}}+g_{t}(0)\right),$

the upper bound we wanted to prove. β

Applying Lemma 1 with $x=0$ gives $g_{t}(0)\geq\sqrt{\frac{\pi}{t}}$, and using this with the above theorem we obtain

 $g_{t}(x)\leq 2\exp\left(-\frac{\|x\|^{2}}{4t}\right)g_{t}(0).$ (1)
###### Theorem 4.

For $t>0$,

 $\sqrt{\frac{\pi}{t}}\leq g_{t}(0)\leq 1+\sqrt{\frac{\pi}{t}}$

and

 $2e^{-t}\leq g_{t}(0)-1\leq\frac{2e^{-t}}{1-e^{-t}}.$
###### Proof.

Using Lemma 1 we have

 $g_{t}(0)\geq\sqrt{\frac{\pi}{t}}.$

For each $x\in\mathbb{R}$ we have

 $g_{t}(x)=\sum_{k\in\mathbb{Z}}\hat{g}_{t}(k)e^{ikx}=\sum_{k\in\mathbb{Z}}e^{-k% ^{2}t}e^{ikx}=1+2\sum_{k\geq 1}e^{-k^{2}t}\cos(kx).$

Writing $\phi(t)=\sum_{k\geq 1}e^{-k^{2}t}$, we then have

 $g_{t}(0)=1+2\phi(t).$

But as $e^{-x^{2}t}$ is positive and decreasing, bounding a sum by an integral we get

 $\phi(t)\leq\int_{0}^{\infty}e^{-x^{2}t}dx=\frac{1}{\sqrt{t}}\int_{0}^{\infty}e% ^{-x^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{t}},$

hence

 $g_{t}(0)=1+2\phi(t)\leq 1+\sqrt{\frac{\pi}{t}}.$

Moreover, because $\phi(t)\geq e^{-t}$ (lower bounding the sum by the first term), we have

 $g_{t}(0)=1+2\phi(t)\geq 1+2e^{-t}.$

Finally, because $e^{-tk^{2}}\leq e^{-tk}$ for $k\geq 1$,

 $\phi(t)\leq\sum_{k\geq 1}e^{-tk}=e^{-t}\frac{1}{1-e^{-t}},$

thus

 $g_{t}(0)\leq 1+\frac{2e^{-t}}{1-e^{-t}}.$

β

Taking $t\to 0$ and $t\to\infty$ in the above theorem gives the following asymptotics.

###### Corollary 5.
 $g_{t}(0)\sim\sqrt{\frac{\pi}{t}},\qquad t\to 0$

and

 $g_{t}(0)-1\sim 2e^{-t},\qquad t\to\infty.$

## 2 Heat kernel on πn

Fix $n\geq 1$, and let $\mathscr{A}=(a_{1},\ldots,a_{n})$, $a_{i}$ positive real numbers. We define $g_{t}^{\mathscr{A}}:\mathbb{R}^{n}\to(0,\infty)$ by

 $g_{t}^{\mathscr{A}}(x)=\prod_{k=1}^{n}g_{a_{k}t}(x_{k}),\qquad x=(x_{1},\ldots% ,x_{n})\in\mathbb{R}^{n}.$

For $x\in\mathbb{R}^{n}$ and $\xi\in\mathbb{Z}^{n}$ we have

 $g_{t}^{\mathscr{A}}(x+2\pi\xi)=\prod_{k=1}^{n}g_{a_{k}t}(x_{k}+2\pi\xi_{k})=% \prod_{k=1}^{n}g_{a_{k}t}(x_{k})=g_{t}^{\mathscr{A}}(x),$

so $g_{t}^{\mathscr{A}}$ can be interpreted as a function on $\mathbb{T}^{n}$.

Let $m_{n}$ be Haar measure on $\mathbb{T}^{n}$:

 $dm_{n}(x)=\prod_{k=1}^{n}dm(x_{k})=\prod_{k=1}^{n}(2\pi)^{-1}dx_{k}=(2\pi)^{-n% }dx,$

which satisfies $m_{n}(\mathbb{T}^{n})=1$. Define $\mu_{t}^{\mathscr{A}}$ to be the measure on $\mathbb{T}^{n}$ whose density with respect to $m_{n}$ is $g_{t}^{\mathscr{A}}$:

 $d\mu_{t}^{\mathscr{A}}=g_{t}^{\mathscr{A}}dm_{n}.$

We now calculate the Fourier coefficients of $g_{t}^{\mathscr{A}}$. For $\xi\in\mathbb{Z}^{n}$,

 $\displaystyle\mathscr{F}(g_{t}^{\mathscr{A}})(\xi)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{T}^{n}}g_{t}^{\mathscr{A}}(x)e^{-i\xi\cdot x}dm_{n}% (x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{T}^{n}}\prod_{k=1}^{n}g_{a_{k}t}(x_{k})e^{-i\xi_{1}% x_{1}-\cdots-i\xi_{n}x_{n}}dm_{n}(x)$ $\displaystyle=$ $\displaystyle\prod_{k=1}^{n}\int_{\mathbb{T}}g_{a_{k}t}(x_{k})e^{-i\xi_{k}x_{k% }}dm(x_{k})$ $\displaystyle=$ $\displaystyle\prod_{k=1}^{n}\hat{g}_{a_{k}t}(\xi_{k})$ $\displaystyle=$ $\displaystyle\prod_{k=1}^{n}e^{-\xi_{k}^{2}a_{k}t}$ $\displaystyle=$ $\displaystyle e^{-tq(\xi)},$

where

 $q(\xi)=\sum_{k=1}^{n}a_{k}\xi_{k}^{2},\qquad\xi\in\mathbb{Z}^{n}.$
###### Definition 6.

For $x=(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}$ we define

 $\|x\|_{\mathscr{A}}^{2}=\frac{1}{a_{1}}\|x_{1}\|^{2}+\cdots+\frac{1}{a_{n}}\|x% _{n}\|^{2},$

with $\mathscr{A}=(a_{1},\ldots,a_{n})$.

For $\xi=(\xi_{1},\ldots,\xi_{n})\in\mathbb{Z}^{n}$, because $\|x_{k}+2\pi\xi_{k}\|=\|x_{k}\|$, we have $\|x+2\pi\xi\|_{\mathscr{A}}=\|x\|_{\mathscr{A}}$, so it makes sense to talk about $\|\cdot\|_{\mathscr{A}}$ on $\mathbb{T}^{n}$.

Using Theorem 3 and (1) we get the following.

###### Theorem 7.

For $t>0$ and $x\in\mathbb{R}^{n}$,

 $\exp\left(-\frac{\|x\|_{\mathscr{A}}^{2}}{4t}\right)g_{t}^{\mathscr{A}}(0)\leq g% _{t}^{\mathscr{A}}(x)\leq 2^{n}\exp\left(-\frac{\|x\|_{\mathscr{A}}^{2}}{4t}% \right)g_{t}^{\mathscr{A}}(0).$

Combining this with Theorem 4 we obtain the following. The first inequality is appropriate for $t\to 0^{+}$ and the second inequality for $t\to\infty$.

###### Theorem 8.

For $t>0$ and $x\in\mathbb{R}^{n}$,

 $\exp\left(-\frac{\|x\|_{\mathscr{A}}^{2}}{4t}\right)\prod_{k=1}^{n}\sqrt{\frac% {\pi}{a_{k}t}}\leq g_{t}^{\mathscr{A}}(x)\leq 2^{n}\exp\left(-\frac{\|x\|_{% \mathscr{A}}^{2}}{4t}\right)\prod_{k=1}^{n}\left(1+\sqrt{\frac{\pi}{a_{k}t}}\right)$

and

 $\exp\left(-\frac{\|x\|_{\mathscr{A}}^{2}}{4t}\right)\prod_{k=1}^{n}\left(1+2e^% {-a_{k}t}\right)\leq g_{t}^{\mathscr{A}}(x)\leq 2^{n}\exp\left(-\frac{\|x\|_{% \mathscr{A}}^{2}}{4t}\right)\prod_{k=1}^{n}\left(1+\frac{2e^{-a_{k}t}}{1-e^{-a% _{k}t}}\right).$

## 3 The infinite-dimensional torus

$\mathbb{T}^{\infty}$ with the product topology is a compact abelian group. Let $m_{\infty}$ be Haar measure on $\mathbb{T}^{\infty}$:

 $dm_{\infty}(x)=\prod_{k=1}^{\infty}dm(x_{k}),\qquad x=(x_{1},x_{2},\ldots)\in% \mathbb{T}^{\infty},$

where $m$ is Haar measure on $\mathbb{T}$.

For $t>0$, let $\mu_{t}$ be the measure on $\mathbb{T}$ whose density with respect to Haar measure $m$ is $g_{t}$:

 $d\mu_{t}=g_{t}dm.$

This is a probability measure on $\mathbb{T}$.

Let $\mathscr{A}=(a_{1},a_{2},\ldots)\in\mathbb{N}^{\infty}$. For $t>0$ we define

 $\mu_{t}^{\mathscr{A}}=\prod_{k=1}^{\infty}\mu_{a_{k}t}.$

This is a probability measure on $\mathbb{T}^{\infty}$.22 2 Christian Berg determines conditions on $\mathscr{A}$ and $t$ so that $\mu_{t}^{\mathscr{A}}$ is absolutely continuous with respect to Haar measure $m_{\infty}$ on $\mathbb{T}^{\infty}$: Potential theory on the infinite dimensional torus, Invent. Math. 32 (1976), no. 1, 49β100.

The Pontryagin dual of $\mathbb{T}^{\infty}$ is the direct sum $\bigoplus_{k=1}^{\infty}\mathbb{Z}$, which we denote by $\mathbb{Z}^{(\infty)}$, which is a discrete abelian group. For $\xi\in\mathbb{Z}^{(\infty)}$ and $x\in\mathbb{T}^{\infty}$, we write

 $e_{\xi}(x)=\exp\left(i\sum_{k=1}^{\infty}\xi_{k}x_{k}\right).$

The Fourier transform of $\mu_{t}^{\mathscr{A}}$ is $\mathscr{F}(\mu_{t}^{\mathscr{A}}):\mathbb{Z}^{(\infty)}\to\mathbb{C}$ defined by

 $\mathscr{F}(\mu_{t}^{\mathscr{A}})(\xi)=\int_{\mathbb{T}^{\infty}}e_{-\xi}(x)% dm_{\infty}(x),\qquad\xi\in\mathbb{Z}^{(\infty)},$

which is

 $\displaystyle\int_{\mathbb{T}^{\infty}}e_{-\xi}(x)dm_{\infty}(x)$ $\displaystyle=\int_{\mathbb{T}^{\infty}}\exp\left(-i\sum_{k=1}^{\infty}\xi_{k}% x_{k}\right)d\mu_{t}^{\mathscr{A}}(x)$ $\displaystyle=\int_{\mathbb{T}^{\infty}}\prod_{k=1}^{\infty}\exp(-i\xi_{k}x_{k% })d\mu_{t}^{\mathscr{A}}(x)$ $\displaystyle=\prod_{k=1}^{\infty}\int_{\mathbb{T}}\exp(-i\xi_{k}x_{k})g_{a_{k% }t}(x_{k})dm(x_{k})$ $\displaystyle=\prod_{k=1}^{\infty}\hat{g}_{a_{k}t}(\xi_{k})$ $\displaystyle=\prod_{k=1}^{\infty}\exp(-\xi_{k}^{2}a_{k}t)$ $\displaystyle=\exp\left(-t\sum_{k=1}^{\infty}a_{k}\xi_{k}^{2}\right).$

## 4 Convergence of infinite products

If $c_{k}\geq 0$, then for any $n$,

 $1+\sum_{k=1}^{n}c_{k}\leq\prod_{k=1}^{n}(1+c_{k})\leq\exp\left(\sum_{k=1}^{n}c% _{k}\right).$

Thus, the limit of $\prod_{k=1}^{n}(1+c_{k})$ as $n\to\infty$ exists if and only if

 $\sum_{k=1}^{\infty}c_{k}<\infty.$

For the second inequality in Theorem 8, the limit of $\prod_{k=1}^{n}\left(1+2e^{-a_{k}t}\right)$ as $n\to\infty$ exists if and only if

 $\sum_{k=1}^{\infty}2e^{-a_{k}t}<\infty.$