In this section we shall consider a smooth and complete Riemannian manifold with dimension . The Riemannian measure will be denoted by .

The Laplace-Beltrami of will be denoted by . Since is assumed to be complete, as we have seen in the previous section, the operator is essentially self-adjoint on the space . More precisely, there exists an increasing sequence such that on , and , as .

The Friedrichs extension of , which is therefore the unique self-adjoint extension of in will still be denoted by and the domain of this extension is denoted by .

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 in the Hilbert space , we may construct a strongly continuous contraction semigroup in whose infinitesimal generator is ;
- By using the ellipticity of , we may prove that admits a heat kernel, that is: There is a smooth function , , such that for every and , Moreover, the heat kernel satisfies the two following conditions:
- (Symmetry) ;
- (Chapman-Kolmogorov relation) .

- The semigroup is a sub-Markov semigroup: If is a function in , then .
- By using the Riesz-Thorin interpolation theorem, defines a contraction semigroup on , .

Let us now assume that is a compact Riemannian manifold. In that case, it obviously complete. As usual, we denote by the heat semigroup and by the corresponding heat kernel. As a preliminary result, we have the following Liouville’s type theorem.

**Lemma:*** Let such that , then is a constant function.*

**Proof:** From the ellipticity of , we first deduce that is smooth. Then, since is compact, the following equality holds Therefore , which implies that is a constant function

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

**Proposition:*** For , *

**Proof:** Since the constant function is in , by compactness of , we may apply uniqueness of solutions of the heat equation in

It turns out that the compactness of implies the compactness of the semigroup.

**Proposition:** * For the operator is a compact operator on the Hilbert space . It is moreover trace class and
*

**Proof:** We shall provide two proofs of the fact that 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 , the operator is bounded. Moreover, from Rellich’s theorem, the map , is compact. Therefore by composition,

is a compact operator.

The second proof is simpler and more direct. Indeed, from the existence of the heat kernel But from the compactness of , Therefore, the operator

is a Hilbert-Schmidt operator. It is thus in particular a compact operator.

Since , is a product of two Hilbert-Schmidt operators. It is therefore a class trace operator and We conclude then by applying the Chapman-Kolmogorov relation

In this compact framework, we have the following theorem

**Theorem:*** There exists a complete orthonormal basis of , consisting of eigenfunctions of , with having an eigenvalue with finite multiplicity satisfying Moreover, for , , with convergence absolute and uniform for each . *

**Proof:** Let . From the Hilbert-Schmidt theorem for the non negative self adjoint compact operator , there exists a complete orthonormal basis of and a non increasing sequence , such that The semigroup property implies first that for , , The same result is then seen to hold for , and finally for , due to the strong continuity of the semigroup. Since the map is decreasing, we deduce that . Thus, there is a such that . As a conclusion, there exists a complete orthonormal basis of $\mathbf{L}^2_\mu (\mathbb{M},\mathbb{R})$, and a sequence satisfying such that Since , we actually have . Also, if is such that , it is straightforward that and that , so that thanks to Liouville theorem, is a constant function. Therefore .

Since , by differentiating as in , we obtain furthermore that and that . The family forms an orthonormal basis of . We therefore have a decomposition in , Since is the kernel of , it is then straightforward that for , and that . Therefore in , The continuity of , together with the positivity of imply, via Mercer’s theorem that actually, the above series is absolutely and uniformly convergent for

As we stressed it in the statement of the theorem, in the decomposition the eigenvalue is repeated according to its multiplicity. It is often useful to rewrite this decomposition under the form where the eigenvalue is not repeated, that is In this decomposition, is the dimension of the eigenspace corresponding to the eigenvalue and is an orthonormal basis of . If we denote, then is called the reproducing kernel of the eigenspace . It satisfies the following properties whose proofs are let to the reader:

**Proposition:**

- does not depend on the choice of the basis ;
- If , then .

From the very definition of the reproducing kernels, we have

The compactness of also implies the convergence to equilibrium for the semigroup with an exponential rate.

**Proposition:** * Let , then uniformly on , when ,
*

**Proof:** It is obvious from the previous proposition and from spectral theory that in , converges to a constant function that we denote . The convergence is also uniform, because for ,

Moreover, for every , . Therefore Since is constant, we finally deduce the expected result