In order to apply the diffusion semigroup theory developed in the first lectures and construct without ambiguity the semigroup associated to the Laplace-Beltrami operator L, we need to know if L is essentially self-adjoint. Interestingly, this property of essential self-adjointness is clodely related to a metric property of the underlying Riemannian structure: The completeness of the associated distance.
Given an absolutely continuous curve , we define its Riemannian length by If , let us denote by the set of absolutely continuous curves such that The Riemannian distance between and is defined by By using reparametrization, we may define the Riemannian distance in a equivalent way by using the notion of sub-unit curve. Let be an absolutely continuous curve. Since the vector fields ‘s form a basis of at each point, we may find continuous functions such that The curve is then said to be sub-unit if for almost every , . By using reparametrization, it is easily seen that for , .
Exercise: Let . Show that for every ,
An important fact is that hence defined is indeed a distance that induces the usual topology of .
Definition: The function defined above is a distance that induces the usual topology of .
Proof: Since any curve can be parametrized backwards and forwards, we have . The triangle inequality is easily proved by using juxtaposition of curves. Plainly , so it remains to prove that if then . Let such that . Let us denote . The closed Euclidean ball is compact, therefore there exist two constants such that for every and , Let now be an absolutely continuous curve such that . Let
As a consequence, we deduce that Therefore is indeed a distance. Moreover, it is shown as above that for every , there are constants such that for every , This implies that induces the usual topology of
As shown in the following proposition, the distance is intrinsically associated to the Laplace-Beltrami operator.
Proposition: For , we have
Proof: Let . We denote Let be a sub-unit curve such that We can find such that and . Let now . From the change of variable formula we have, Therefore, from Cauchy-Schwarz inequality, As a consequence
We now prove the converse inequality which is trickier. The idea is to use the function that satisfies and . However, giving a precise meaning to is not so easy, because it turns out that is not differentiable at . It suggests to use an approximation of the identity to regularize and avoid the discussion of this differentiability issue. More precisely, fix , and for , consider the function where , , , and , , has the property that and for . Since for any , , it is easy to see that , for some constant . Hence
Proposition: The metric space is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.
Proof: It is clear that if closed and bounded sets are compact then the metric space is complete; It comes from the fact that Cauchy sequence are convergent if and only if they have at least one cluster value. So, we need to prove that closed and bounded sets for the distance are compact provided that is complete. To check this, it is enough to prove that closed balls are compact. Let . Observe that if the closed ball is compact for some , then is closed for any . Define Since induces the usual topology of , . Let us assume that and let us show that it leads to a contradiction. We first show that is compact. Since is assumed to be complete, it suffices to prove that is totally bounded: That is, for every there is a finite set such that every point of belongs to a -neighborhood of .
So, let small enough. By definition of , the ball is compact; It is therefore totally bounded. We can find a finite set such that every point of lies in a -neighborhood of . Let now . We claim that there exists such that . If , there is nothing to prove, we may therefore assume that . Consider then a sub-unit curve such that , . Let We have . On the other hand, As a consequence, In every case, there exists therefore such that . We may then pick in such that . From the triangle inequality, we have . So, at the end, it turns out that every point of lies in a -neighborhood of . This shows that is totally bounded and therefore compact because is assumed to be complete. Actually, the previous argument shows more, it shows that if every point of lies in a -neighborhood of a finite , then every point of will lie -neighborhood of , so that the ball is also compact. This contradicts the fact the definition of . Therefore every closed ball is a compact set, due to the arbitrariness of
Checking that the metric space is complete is not always easy in concrete situations. From the Hopf-Rinow theorem, it suffices to prove that the closed balls are compact. The following proposition is therefore useful.
Proposition: Suppose that the vector fields ‘s have globally Lipschitz coefficients . Then the closed ball is compact for every and . As a consequence the metric space is complete.
Proof: By the hypothesis on the ‘s there exists a constant such
that for any . Fix and let , be a sub-unit curve such that Letting we obtain We infer that , for some depending only on . Integrating the latter inequality one has . for some constant . The previous estimate shows in particular, that We conclude that is Euclidean compact. Since the metric and the Euclidean topology coincide, it is also -compact
Completeness of the metric space is related to the essential self-adjointness of the Laplace-Beltrami operator.
Theorem: If the metric space is complete, then the Laplace-Beltrami operator is essentially self-adjoint.
Proof: We know that if there exists an increasing sequence such that on , and , as , then the operator is essentially self-adjoint. We are therefore reduced to prove the existence of such a sequence. Let us fix a base point . We can find an exhaustion function such that By the completeness of and the Hopf-Rinow theorem, the level sets are relatively compact and, furthermore, as . We now pick an increasing sequence of functions such that on , outside , and . If we set , then we have , on as , and