The heat kernel on the torus

1 Heat kernel on 𝕋

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

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

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

$$g_t(x)=2\pi \sum _{k\in \mathbb{Z}}{k}_t(x+2\pi k),x\in ℝ,$$

which one checks indeed converges for all $x\in ℝ$. 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 $𝕋$, where $𝕋=ℝ/2\pi \mathbb{Z}$.

Let $m$ be Haar measure on $𝕋$: $dm(x)={(2\pi )}^{-1}dx$, and so $m(𝕋)=1$. With ${\parallel f\parallel }_1={\int }_{𝕋}|f|𝑑m$ for $f:𝕋\to ℂ$, we have, because $g_t>0$,

$${\parallel g_t\parallel }_1=\sum _{k\in \mathbb{Z}}{\int }_{𝕋}{k}_t(x+2\pi k)𝑑x={\int }_{ℝ}{k}_t(x)𝑑x=1.$$

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

${\widehat{g}}_t(\xi )$ $=$ ${\displaystyle {\int }_{𝕋}}{g}_t(x)e^{-i\xi x}𝑑m(x)$

$=$ ${\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{𝕋}}{k}_t(x+2\pi k)e^{-i\xi x}𝑑x$

$=$ ${\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{𝕋}}{k}_t(x+2\pi k)e^{-i\xi (x+2\pi k)}𝑑x$

$=$ ${\displaystyle {\int }_{ℝ}}{k}_t(x)e^{-i\xi x}𝑑x$

$=$ ${\widehat{k}}_t\left({\displaystyle \frac{\xi }{2\pi }}\right)$

$=$ $e^{-{\xi }^{2}t}.$

Lemma 1.

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

$$g_t(x)=\sqrt{\frac{\pi }{t}}\exp \left(-\frac{x^2}{4t}\right)\left(1+2\sum _{k\ge 1}\exp \left(-\frac{{\pi }^2{k}^2}{t}\right)\cosh \left(\frac{\pi kx}{t}\right)\right).$$

Proof.

Using the definition of $g_t$,

$g_t(x)$ $=$ $2\pi {\displaystyle \sum _{k\in \mathbb{Z}}}{k}_t(x+2\pi k)$

$=$ $2\pi {\displaystyle \sum _{k\in \mathbb{Z}}}{(4\pi t)}^{-1/2}\exp \left(-{\displaystyle \frac{{(x+2\pi k)}^2}{4t}}\right)$

$=$ $\sqrt{{\displaystyle \frac{\pi }{t}}}\exp \left(-{\displaystyle \frac{x^2}{4t}}\right){\displaystyle \sum _{k\in \mathbb{Z}}}\exp \left(-{\displaystyle \frac{\pi kx}{t}}\right)\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)$

$=$ $\sqrt{{\displaystyle \frac{\pi }{t}}}\exp \left(-{\displaystyle \frac{x^2}{4t}}\right)$

$\left(1+{\displaystyle \sum _{k\ge 1}}\left(\exp \left({\displaystyle \frac{\pi kx}{t}}\right)+\exp \left(-{\displaystyle \frac{\pi kx}{t}}\right)\right)\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)\right),$

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

Definition 2.

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

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

Theorem 3.

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

$$\exp \left(-\frac{{\parallel x\parallel }^2}{4t}\right)g_t(0)\le g_t(x)\le \exp \left(-\frac{{\parallel x\parallel }^2}{4t}\right)\left(\sqrt{\frac{\pi }{t}}+g_t(0)\right).$$

Proof.

Let $x=2\pi m+\theta$ with $|\theta |\le \pi$, so that $\parallel x\parallel =\parallel \theta \parallel =|\theta |$, and $g_t(x)=g_t(\theta )$. Using Lemma 1 and the fact that $\cosh y\ge 1$, we get

$$g_t(\theta )\ge \exp \left(-\frac{{\theta }^2}{4t}\right)\sqrt{\frac{\pi }{t}}\left(1+2\sum _{k\ge 1}\exp \left(-\frac{{\pi }^2{k}^2}{t}\right)\right)=\exp \left(-\frac{{\theta }^2}{4t}\right)g_t(0),$$

hence

$$g_t(x)\ge \exp \left(-\frac{{\parallel x\parallel }^2}{4t}\right)g_t(0),$$

the lower bound we wanted to prove.

Write

$$S=1+2\sum _{k\ge 1}\exp \left(-\frac{{\pi }^2{k}^2}{t}\right)\cosh \left(\frac{\pi k\theta }{t}\right).$$

For any $k\ge 1$, using $|\theta |\le \pi$,

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

Hence

$S$ $\le$ $1+{\displaystyle \sum _{k\ge 1}}\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)\left(1+\exp \left({\displaystyle \frac{{\pi }^{2}k}{t}}\right)\right)$

$=$ $1+{\displaystyle \sum _{k\ge 1}}\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)+\exp \left(-{\displaystyle \frac{{\pi }^{2}k(k-1)}{t}}\right)$

$\le$ $1+{\displaystyle \sum _{k\ge 1}}\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)+\exp \left(-{\displaystyle \frac{{\pi }^2{(k-1)}^2}{t}}\right)$

$=$ $2+2{\displaystyle \sum _{k\ge 1}}\exp \left(-{\displaystyle \frac{{\pi }^2{k}^2}{t}}\right)$

$=$ $1+\sqrt{{\displaystyle \frac{t}{\pi }}}{g}_t(0).$

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

$$g_t(\theta )\le \exp \left(-\frac{{\theta }^2}{4t}\right)\left(\sqrt{\frac{\pi }{t}}+g_t(0)\right)=\exp \left(-\frac{{\parallel x\parallel }^2}{4t}\right)\left(\sqrt{\frac{\pi }{t}}+g_t(0)\right),$$

the upper bound we wanted to prove. ∎

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

$$g_t(x)\le 2\exp \left(-\frac{{\parallel x\parallel }^2}{4t}\right)g_t(0).$$ (1)

Theorem 4.

For $t\mathrm{>}\mathrm{0}$,

$$\sqrt{\frac{\pi }{t}}\le g_t(0)\le 1+\sqrt{\frac{\pi }{t}}$$

and

$$2e^{-t}\le g_t(0)-1\le \frac{2e^{-t}}{1-e^{-t}}.$$

Proof.

Using Lemma 1 we have

$$g_t(0)\ge \sqrt{\frac{\pi }{t}}.$$

For each $x\in ℝ$ we have

$$g_t(x)=\sum _{k\in \mathbb{Z}}{\widehat{g}}_t(k)e^{ikx}=\sum _{k\in \mathbb{Z}}{e}^{-k^{2}t}{e}^{ikx}=1+2\sum _{k\ge 1}{e}^{-k^{2}t}\cos (kx).$$

Writing $\varphi (t)={\sum }_{k\ge 1}{e}^{-k^{2}t}$, we then have

$$g_t(0)=1+2\varphi (t).$$

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

$$\varphi (t)\le {\int }_0^{\mathrm{\infty }}{e}^{-x^{2}t}𝑑x=\frac{1}{\sqrt{t}}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}𝑑x=\frac{1}{2}\sqrt{\frac{\pi }{t}},$$

hence

$$g_t(0)=1+2\varphi (t)\le 1+\sqrt{\frac{\pi }{t}}.$$

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

$$g_t(0)=1+2\varphi (t)\ge 1+2e^{-t}.$$

Finally, because $e^{-t{k}^2}\le e^{-tk}$ for $k\ge 1$,

$$\varphi (t)\le \sum _{k\ge 1}{e}^{-tk}=e^{-t}\frac{1}{1-e^{-t}},$$

thus

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

∎

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

Corollary 5.

$$g_t(0)\sim \sqrt{\frac{\pi }{t}},t\to 0$$

and

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

2 Heat kernel on 𝕋ⁿ

Fix $n\ge 1$, and let $𝒜=(a_1,\mathrm{\dots },a_n)$, $a_i$ positive real numbers. We define $g_t^{𝒜}:{ℝ}^n\to (0,\mathrm{\infty })$ by

$$g_t^{𝒜}(x)=\prod _{k=1}^n{g}_{a_{k}t}(x_k),x=(x_1,\mathrm{\dots },x_n)\in {ℝ}^n.$$

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

$$g_t^{𝒜}(x+2\pi \xi )=\prod _{k=1}^n{g}_{a_{k}t}(x_k+2\pi {\xi }_k)=\prod _{k=1}^n{g}_{a_{k}t}(x_k)=g_t^{𝒜}(x),$$

so $g_t^{𝒜}$ can be interpreted as a function on ${𝕋}^n$.

Let $m_n$ be Haar measure on ${𝕋}^n$:

$$d{m}_n(x)=\prod _{k=1}^{n}dm(x_k)=\prod _{k=1}^n{(2\pi )}^{-1}d{x}_k={(2\pi )}^{-n}dx,$$

which satisfies $m_n({𝕋}^n)=1$. Define ${\mu }_t^{𝒜}$ to be the measure on ${𝕋}^n$ whose density with respect to $m_n$ is $g_t^{𝒜}$:

$$d{\mu }_t^{𝒜}=g_t^{𝒜}d{m}_n.$$

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

$ℱ(g_t^{𝒜})(\xi )$ $=$ ${\displaystyle {\int }_{{𝕋}^n}}{g}_t^{𝒜}(x)e^{-i\xi \cdot x}𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{𝕋}^n}}{\displaystyle \prod _{k=1}^n}{g}_{a_{k}t}(x_k)e^{-i{\xi }_1{x}_1-\mathrm{\cdots }-i{\xi }_n{x}_n}d{m}_n(x)$

$=$ ${\displaystyle \prod _{k=1}^n}{\displaystyle {\int }_{𝕋}}{g}_{a_{k}t}(x_k)e^{-i{\xi }_k{x}_k}𝑑m(x_k)$

$=$ ${\displaystyle \prod _{k=1}^n}{\widehat{g}}_{a_{k}t}({\xi }_k)$

$=$ ${\displaystyle \prod _{k=1}^n}{e}^{-{\xi }_k^2{a}_{k}t}$

$=$ $e^{-tq(\xi )},$

where

$$q(\xi )=\sum _{k=1}^n{a}_k{\xi }_k^2,\xi \in {\mathbb{Z}}^n.$$

Definition 6.

For $x\mathrm{=}\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{)}\mathrm{\in }{\mathrm{R}}^n$ we define

$${\parallel x\parallel }_{𝒜}^2=\frac{1}{a_1}{\parallel x_1\parallel }^2+\mathrm{\cdots }+\frac{1}{a_n}{\parallel x_n\parallel }^2,$$

with $\mathrm{A}\mathrm{=}\mathrm{(}{a}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_n\mathrm{)}$.

For $\xi =({\xi }_1,\mathrm{\dots },{\xi }_n)\in {\mathbb{Z}}^n$, because $\parallel x_k+2\pi {\xi }_k\parallel =\parallel x_k\parallel$, we have ${\parallel x+2\pi \xi \parallel }_{𝒜}={\parallel x\parallel }_{𝒜}$, so it makes sense to talk about $\parallel \cdot {\parallel }_{𝒜}$ on ${𝕋}^n$.

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

Theorem 7.

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

$$\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)g_t^{𝒜}(0)\le g_t^{𝒜}(x)\le 2^n\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)g_t^{𝒜}(0).$$

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

Theorem 8.

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

$$\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)\prod _{k=1}^n\sqrt{\frac{\pi }{a_{k}t}}\le g_t^{𝒜}(x)\le 2^n\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)\prod _{k=1}^n\left(1+\sqrt{\frac{\pi }{a_{k}t}}\right)$$

and

$$\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)\prod _{k=1}^n\left(1+2e^{-a_{k}t}\right)\le g_t^{𝒜}(x)\le 2^n\exp \left(-\frac{{\parallel x\parallel }_{𝒜}^2}{4t}\right)\prod _{k=1}^n\left(1+\frac{2e^{-a_{k}t}}{1-e^{-a_{k}t}}\right).$$

3 The infinite-dimensional torus

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

$$d{m}_{\mathrm{\infty }}(x)=\prod _{k=1}^{\mathrm{\infty }}dm(x_k),x=(x_1,x_2,\mathrm{\dots })\in {𝕋}^{\mathrm{\infty }},$$

where $m$ is Haar measure on $𝕋$.

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

$$d{\mu }_t=g_{t}dm.$$

This is a probability measure on $𝕋$.

Let $𝒜=(a_1,a_2,\mathrm{\dots })\in {\mathbb{N}}^{\mathrm{\infty }}$. For $t>0$ we define

$${\mu }_t^{𝒜}=\prod _{k=1}^{\mathrm{\infty }}{\mu }_{a_{k}t}.$$

This is a probability measure on ${𝕋}^{\mathrm{\infty }}$.

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

$$e_{\xi }(x)=\exp \left(i\sum _{k=1}^{\mathrm{\infty }}{\xi }_k{x}_k\right).$$

The Fourier transform of ${\mu }_t^{𝒜}$ is $ℱ({\mu }_t^{𝒜}):{\mathbb{Z}}^{(\mathrm{\infty })}\to ℂ$ defined by

$$ℱ({\mu }_t^{𝒜})(\xi )={\int }_{{𝕋}^{\mathrm{\infty }}}{e}_{-\xi }(x)𝑑m_{\mathrm{\infty }}(x),\xi \in {\mathbb{Z}}^{(\mathrm{\infty })},$$

which is

${\displaystyle {\int }_{{𝕋}^{\mathrm{\infty }}}}{e}_{-\xi }(x)𝑑m_{\mathrm{\infty }}(x)$ $={\displaystyle {\int }_{{𝕋}^{\mathrm{\infty }}}}\exp \left(-i{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\xi }_k{x}_k\right)𝑑{\mu }_t^{𝒜}(x)$

$={\displaystyle {\int }_{{𝕋}^{\mathrm{\infty }}}}{\displaystyle \prod _{k=1}^{\mathrm{\infty }}}\exp (-i{\xi }_k{x}_k)d{\mu }_t^{𝒜}(x)$

$={\displaystyle \prod _{k=1}^{\mathrm{\infty }}}{\displaystyle {\int }_{𝕋}}\exp (-i{\xi }_k{x}_k)g_{a_{k}t}(x_k)𝑑m(x_k)$

$={\displaystyle \prod _{k=1}^{\mathrm{\infty }}}{\widehat{g}}_{a_{k}t}({\xi }_k)$

$={\displaystyle \prod _{k=1}^{\mathrm{\infty }}}\exp (-{\xi }_k^2{a}_{k}t)$

$=\exp \left(-t{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{a}_k{\xi }_k^2\right).$

4 Convergence of infinite products

If $c_k\ge 0$, then for any $n$,

$$1+\sum _{k=1}^n{c}_k\le \prod _{k=1}^n(1+c_k)\le \exp \left(\sum _{k=1}^n{c}_k\right).$$

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

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