## Lecture 12. The heat semigroup on a compact Riemannian manifold

In this section we shall consider a smooth and complete Riemannian manifold $(\mathbb{M},g)$ with dimension $n$. The Riemannian measure will be denoted by $\mu$.

The Laplace-Beltrami of $\mathbb{M}$ will be denoted by $L$. Since $\mathbb{M}$ is assumed to be complete, as we have seen in the previous section, the operator $L$ is essentially self-adjoint on the space $\mathcal{C}_c(\mathbb{M},\mathbb{M}$. More precisely, there exists an increasing sequence $h_n\in C_c(\mathbb{M},\mathbb{R})$ such that $h_n\nearrow 1$ on $\mathbb{M}$, and $||\Gamma(h_n,h_n)||_{\infty} \to 0$, as $n\to \infty$.

The Friedrichs extension of $L$, which is therefore the unique self-adjoint extension of $L$ in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ will still be denoted by $L$ and the domain of this extension is denoted by $\mathcal{D}(L)$.

We shall not repeat the whole theory of diffusion semigroups on manifolds, since many of the results that were proved before are easily extended to manifolds. In particular, we may prove:

• By using the spectral theorem for $L$ in the Hilbert space $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, we may construct a strongly continuous contraction semigroup $(\mathbf{P}_t)_{t \ge 0}$ in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ whose infinitesimal generator is $L$;
• By using the ellipticity of $L$, we may prove that $(\mathbf{P}_t)_{t \ge 0}$ admits a heat kernel, that is: There is a smooth function $p(t,x,y)$, $t \in (0,+\infty), x,y \in \mathbb{M}$, such that for every $f \in \mathbf{L}_{\mu}^2 (\mathbb{M},\mathbb{R})$ and $x \in \mathbb{M}$ , $\mathbf{P}_t f (x)=\int_{\mathbb{M}} p(t,x,y) f(y) d\mu (y).$ Moreover, the heat kernel satisfies the two following conditions:
• (Symmetry) $p(t,x,y)=p(t,y,x)$;
• (Chapman-Kolmogorov relation) $p(t+s,x,y)=\int_{\mathbb{M}} p(t,x,z)p(s,z,y)d\mu(z)$.
• The semigroup $(\mathbf{P}_t)_{t \ge 0}$ is a sub-Markov semigroup: If $0\le f \le 1$ is a function in $\mathbf{L}_{\mu}^2 (\mathbb{M},\mathbb{R})$, then $0 \le \mathbf{P}_t f \le 1$.
• By using the Riesz-Thorin interpolation theorem, $(\mathbf{P}_t)_{t \ge 0}$ defines a contraction semigroup on $\mathbf{L}_{\mu}^p (\mathbb{M},\mathbb{R})$, $1 \le p \le \infty$.

Let us now assume that $(\mathbb{M},g)$ is a compact Riemannian manifold. In that case, it obviously complete. As usual, we denote by $(\mathbf{P}_t)_{t \ge 0}$ the heat semigroup and by $p(t,x,y)$ the corresponding heat kernel. As a preliminary result, we have the following Liouville’s type theorem.

Lemma: Let $f \in \mathcal{D}(L)$ such that $Lf=0$, then $f$ is a constant function.

Proof: From the ellipticity of $L$, we first deduce that $f$ is smooth. Then, since $\mathbb{M}$ is compact, the following equality holds $-\int_{\mathbb{M}} f Lf d\mu=\int_{\mathbb{M}} \Gamma(f,f) d\mu.$ Therefore $\Gamma(f,f)=0$, which implies that $f$ is a constant function $\square$

In the compact case, the heat semigroup satisfies the so-called stochastic completeness (or Markov) property.

Proposition: For $t \ge 0$, $\mathbf{P}_t 1=1.$

Proof: Since the constant function $1$ is in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, by compactness of $\mathbb{M}$, we may apply uniqueness of solutions of the heat equation in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ $\square$

It turns out that the compactness of $\mathbb{M}$ implies the compactness of the semigroup.

Proposition: For $t > 0$ the operator $\mathbf{P}_t$ is a compact operator on the Hilbert space $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$. It is moreover trace class and $\mathbf{Tr} (\mathbf{P}_t)=\int_\mathbb{M} p(t,x,x)\mu(dx).$

Proof: We shall provide two proofs of the fact that $\mathbf{P}_t$ is a compact operator. You may observe that the first proof does not rely on the existence result of the heat kernel. The first proof stems from the local regularity theory of elliptic operators. Indeed, for $t > 0$, the operator $\mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathcal{H}^0_1(\mathbb{M})$ is bounded. Moreover, from Rellich’s theorem, the map $\iota: \mathcal{H}^{0}_1 (\mathbb{M}) \rightarrow \mathcal{H}^{0}_0 (\mathbb{M})=\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, is compact. Therefore by composition,
$\mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ is a compact operator.

The second proof is simpler and more direct. Indeed, from the existence of the heat kernel $\mathbf{P}_t f (x)=\int_{\mathbb{M}} p(t,x,y) f(y) d\mu (y).$ But from the compactness of $\mathbb{M}$, $\int_{\mathbb{M}}\int_{\mathbb{M}} p(t,x,y)^2 d \mu(x) d\mu(y) < +\infty.$ Therefore, the operator $\mathbf{P}_t: \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R}) \rightarrow \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$
is a Hilbert-Schmidt operator. It is thus in particular a compact operator.

Since $\mathbf{P}_t=\mathbf{P}_{t/2} \mathbf{P}_{t/2}$, $\mathbf{P}_t$ is a product of two Hilbert-Schmidt operators. It is therefore a class trace operator and $\mathbf{Tr} ( \mathbf{P}_t)=\int_{\mathbb{M}}\int_{\mathbb{M}} p(t/2,x,y)p(t/2,y,x) d \mu(x) d\mu(y).$ We conclude then by applying the Chapman-Kolmogorov relation $\square$

In this compact framework, we have the following theorem

Theorem: There exists a complete orthonormal basis $(\phi_n)_{n \in \mathbb{N}}$ of $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, consisting of eigenfunctions of $-L$, with $\phi_n$ having an eigenvalue $\lambda_n$ with finite multiplicity satisfying $0=\lambda_0 < \lambda_1\le \lambda_2 \le \cdots \nearrow +\infty.$ Moreover, for $t > 0$, $x,y \in \mathbb{M}$, $p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y),$ with convergence absolute and uniform for each $t > 0$.

Proof: Let $t > 0$. From the Hilbert-Schmidt theorem for the non negative self adjoint compact operator $\mathbf{P}_t$, there exists a complete orthonormal basis $(\phi_n(t) )_{n \in \mathbb{N}}$ of $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ and a non increasing sequence $\alpha_n(t) \ge 0$, $\alpha_n(t) \searrow 0$ such that $\mathbf{P}_t \phi_n(t)=\alpha_n(t) \phi_n (t).$ The semigroup property $\mathbf{P}_{t+s}=\mathbf{P}_t \mathbf{P}_s$ implies first that for $k \in \mathbb{N}$, $k \ge 1$, $\phi_n(k)=\phi_n(1), \alpha_n(k)=\alpha_n (1)^k.$ The same result is then seen to hold for $k \in \mathbb{Q}$, $k > 0$ and finally for $k \in \mathbb{R}$, due to the strong continuity of the semigroup. Since the map $t \to \| P_t \|_2$ is decreasing, we deduce that $\alpha_n (1) \le 1$. Thus, there is a $\lambda_n \ge 0$ such that $\alpha_n(1)=e^{-\lambda_n}$. As a conclusion, there exists a complete orthonormal basis $(\phi_n)_{n \in \mathbb{N}}$ of $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, and a sequence $\lambda_n$ satisfying $0 \le \lambda_0 \le \lambda_1\le \lambda_2 \le \cdots \nearrow +\infty$ such that $\mathbf{P}_t \phi_n =e^{-\lambda_n t} \phi_n.$ Since $\mathbf{P}_t 1=1$, we actually have $\lambda_0=0$. Also, if $f \in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$ is such that $\mathbf{P}_t f=f$, it is straightforward that $f \in \mathcal{D}(L)$ and that $Lf=0$, so that thanks to Liouville theorem, $f$ is a constant function. Therefore $\lambda_1 > 0$.

Since $\mathbf{P}_t \phi_n =e^{-\lambda_n t} \phi_n$, by differentiating as $t \to 0$ in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, we obtain furthermore that $\phi_n \in \mathcal{D}(L)$ and that $L\phi_n=-\lambda_n \phi_n$. The family $(x,y)\to \phi_n(x) \phi_m(y)$ forms an orthonormal basis of $\mathbf{L}^2_{\mu \otimes \mu} (\mathbb{M}\times \mathbb{M},\mathbb{R})$. We therefore have a decomposition in $\mathbf{L}^2_{\mu \otimes \mu} (\mathbb{M}\times \mathbb{M},\mathbb{R})$, $p(t,x,y)=\sum_{m,n \in \mathbb{M}} c_{mn} \phi_m(x) \phi_n(y).$ Since $p(t,\cdot,\cdot)$ is the kernel of $\mathbf{P}_t$, it is then straightforward that for $m \neq n$, $c_{mn}=0$ and that $c_{nn}=e^{-\lambda_n t}$. Therefore in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, $p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y).$ The continuity of $p$, together with the positivity of $\mathbf{P}_t$ imply, via Mercer’s theorem that actually, the above series is absolutely and uniformly convergent for $t > 0$ $\square$

As we stressed it in the statement of the theorem, in the decomposition $p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\lambda_n t} \phi_n (x) \phi_n (y),$ the eigenvalue $\lambda_n$ is repeated according to its multiplicity. It is often useful to rewrite this decomposition under the form $p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\alpha_n t} \sum_{k=1}^{d_n} \phi^n_k (x) \phi^n_k (y),$ where the eigenvalue $\alpha_n$ is not repeated, that is $0=\alpha_0 < \alpha_1 < \alpha_2 < \cdots$ In this decomposition, $d_n$ is the dimension of the eigenspace $\mathcal{V}_n$ corresponding to the eigenvalue $\alpha_n$ and $(\phi^n_k)_{1 \le k \le d_n }$ is an orthonormal basis of $\mathcal{V}_n$. If we denote, $\mathcal{K}_n(x,y)= \sum_{k=1}^{d_n} \phi^n_k (x) \phi^n_k (y),$ then $\mathcal{K}_n$ is called the reproducing kernel of the eigenspace $\mathcal{V}_n$. It satisfies the following properties whose proofs are let to the reader:

Proposition:

• $\mathcal{K}_n$ does not depend on the choice of the basis $(\phi^n_k)_{1 \le k \le d_n }$;
• If $f \in \mathcal{V}_n$, then $\int_{\mathbb{M}} \mathcal{K}_n (x,y) f(y) d\mu(y) =f(x)$.

From the very definition of the reproducing kernels, we have $p(t,x,y)=\sum_{n=0}^{+\infty} e^{-\alpha_n t} \mathcal{K}_n(x,y).$

The compactness of $\mathbb{M}$ also implies the convergence to equilibrium for the semigroup with an exponential rate.

Proposition: Let $f \in \mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, then uniformly on $\mathbb{M}$, when $t \to +\infty$, $\mathbf{P}_t f \rightarrow \frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu.$

Proof: It is obvious from the previous proposition and from spectral theory that in $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, $\mathbf{P}_t f$ converges to a constant function that we denote $\mathbf{P}_\infty f$. The convergence is also uniform, because for $s,t,T > 0$,
$\| \mathbf{P}_{t+T} f -\mathbf{P}_{s+T} f \|_\infty$
$= \sup_{x \in \mathbb{M}} \left| \mathbf{P}_{T} ( \mathbf{P}_{t} f -\mathbf{P}_{s} f) (x) \right|$
$= \sup_{x \in \mathbb{M}} \left|\int_{\mathbb{M}} p(T,x,y) ( \mathbf{P}_{t} f -\mathbf{P}_{s} f) (y) d\mu(y) \right|$
$\le \left( \sup_{x \in \mathbb{M}} \sqrt{ \int_{\mathbb{M}} p(T,x,y)^2 d\mu(y)}\right) \| \mathbf{P}_{t} f -\mathbf{P}_{s} f \|_2.$
Moreover, for every $t \ge 0$, $\int_{\mathbb{M}} \mathbf{P}_t f d\mu =\int_{\mathbb{M}} f d\mu$. Therefore $\int_{\mathbb{M}} \mathbf{P}_\infty f d\mu =\int_{\mathbb{M}} f d\mu.$ Since $\mathbf{P}_\infty f$ is constant, we finally deduce the expected result $\mathbf{P}_\infty f=\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu$ $\square$

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