## Lecture 10. The heat semigroup on the circle

In the next few lectures, we will show that the diffusion semigroups theory we developed may actually be extended without difficulties to a manifold setting. As a motivation, we start with a very simple example.

We first study the heat semigroup on the simplest (non Euclidean) Riemannian manifold: the circle $\mathbb{S}^1= \left\{ e^{i\theta}, \theta \in \mathbb{R} \right\}.$ The Laplace operator on $\mathbb{R}^n$, $\Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$ is the canonical diffusion operator on $\mathbb{R}^n$. A natural question to be asked is: in the same way, is there a canonical diffusion operator on $\mathbb{S}^1$. A first step, of course, is to understand what is a diffusion operator on $\mathbb{S}^1$. We characterized diffusion operators as linear operators on the space of smooth functions that satisfy the maximum principle. Once a notion of smooth functions on $\mathbb{S}^1$ is understood, this maximum principle property can be taken as a definition. The circle $\mathbb{S}^1$ may be identified with the quotient space $\mathbb{R} / 2\pi \mathbb{Z}$. More precisely, it is easily shown that a smooth function, $f :\mathbb{R} \rightarrow \mathbb{C}$ which is $2\pi$ periodic, i.e. $f(\theta+2\pi)=f(\theta),$ can be written as $f(\theta)=g\left( e^{i\theta} \right),$ for some function $g: \mathbb{S}^1 \rightarrow \mathbb{C}$. Conversely, any function $g: \mathbb{S}^1 \rightarrow \mathbb{C}$ defines a $2\pi$ periodic function on $\mathbb{R}$ by setting $f(\theta)=g\left( e^{i\theta} \right).$ So, with this in mind, we simply say that $g: \mathbb{S}^1 \rightarrow \mathbb{C}$ is a smooth function if $f$ is. With this identification between the set of smooth $2\pi$ periodic functions on $\mathbb{R}$ and the set of smooth functions on $\mathbb{S}^1$, it then immediate that the canonical diffusion operator $\Delta_{\mathbb{S}^1}$ on $\mathbb{S}^1$ should write, $\Delta_{\mathbb{S}^1} g ( e^{i\theta})=f''(\theta).$ The corresponding diffusion semigroup is also easily computed from the heat semigroup on $\mathbb{R}$. Indeed, a natural computation leads to

$\left( e^{t \Delta_{\mathbb{S}^1} }g \right) ( e^{i\theta}) =\left( e^{t \frac{d^2}{d\theta^2} }f \right) ( \theta)$
$= \int_{-\infty}^{\infty} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy$
$=\sum_{k \in \mathbb{Z}} \int_{2k \pi }^{2k\pi +2\pi} f(y+\theta) \frac{e^{-\frac{y^2}{4t} } }{\sqrt{4\pi t}} dy$
$=\sum_{k \in \mathbb{Z}} \int_{0 }^{2\pi} f(y-2k\pi+\theta) \frac{e^{-\frac{(y-2k\pi)^2}{4t} } }{\sqrt{4\pi t}} dy$
$= \int_{0 }^{2\pi} f(y+\theta) p(t,y) dy$,
where $p(t,y)=\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y-2k\pi)^2}{4t} }$. This allows to define the heat semigroup on $\mathbb{S}^1$ as the family of operators defined by $\mathbf{P}_t g (e^{i\theta} )= \int_{0 }^{2\pi} g(e^{i\nu }) p(t,\theta-\nu) d\nu.$ The natural domain of this operator is $\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ where $\mu$ is the measure on $\mathbb{S}^1$ which is defined through the property $\int_{\mathbb{S}^1} g d\mu = \int_0^{2\pi} f(\theta) d\theta.$ The reader may then check the following properties for this semigroup of operators:

• (Semigroup property) $\mathbf{P}_{t+s} =\mathbf{P}_t \mathbf{P}_s$;
• (Strong continuity) The map $t \rightarrow \mathbf{P}_t$ is continuous for the operator norm on $\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$;
• (Contraction property) $\|\mathbf{P}_t g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) } \le \| g \|_{\mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R}) }$;
• (Self-adjointness) For $g_1,g_2 \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$, $\int_{\mathbb{S}^1} (\mathbf{P}_t g_1) g_2 d\mu=\int_{\mathbb{S}^1} g_1(\mathbf{P}_t g_2) d\mu$
• (Markov property) If $g \in \mathbf{L}^2_\mu(\mathbb{S}^1,\mathbb{R})$ is such that $0 \le g \le 1$, then $0 \le \mathbf{P}_t g \le 1$.

Exercise.

• Prove the Poisson summation formula: If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a smooth and rapidly decreasing function, then $\sum_{m \in \mathbb{Z}} f(m) e^{im \theta}=\sum_{k \in \mathbb{Z}} \hat{f} (\theta -2k\pi).$
• Deduce that the heat kernel on $\mathbb{S}^1$ may also be written $p(t,y)=\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y}.$

Exercise. From the previous exercise, the heat kernel on $\mathbb{S}^1$ is given by $p(t,y) =\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y} =\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y 2k\pi)^2}{4t} }$.

• By using the subordination identity $e^{-\tau | \alpha | } =\frac{\tau}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{\tau^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad \tau \neq 0, \alpha \in \mathbb{R},$ show that for $\tau > 0$, $\frac{1+e^{-2\pi \tau}}{1-e^{-2\pi \tau}} =\frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \frac{2\tau}{\tau^2+n^2}$
• The Bernoulli numbers $B_k$ are defined via the series expansion $\frac{x}{e^x -1}=\sum_{k=0}^{+\infty} B_k \frac{x^k}{k!}.$ By using the previous identity show that for $k \in \mathbb{N}$, $k \neq 0$, $\sum_{n=1}^{+\infty} \frac{1}{n^{2k}} =(-1)^{k-1} \frac{(2\pi)^{2k} B_{2k} }{2(2k)!}.$

Exercise. Show that the heat kernel on the torus $\mathbb{T}^n=\mathbb{R}^n / (2 \pi \mathbb{Z})^n$ is given by $p(t,y) = \frac{1}{(4\pi t)^{n/2}} \sum_{k \in \mathbb{Z}^n} e^{-\frac{\|y+2k\pi\|^2}{4t} }=\frac{1}{(2\pi)^n} \sum_{l\in \mathbb{Z}^n} e^{i l \cdot y -\| l \|^2 t}.$

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