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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