Let $X$ be a compact Riemannian manifold and let $\mathcal{M}(X)$ be the space of regular finite Borel measures with the total variation as norm. 


The Laplace-Belrami-Operator $\Delta$ on $X$ with domain $C^2(M)$ is not densely defined, as $C^2$ is not dense in $\mathcal{M}(X)$. 


**Question:** Can the Laplacian be realized as a densely defined, closed operator on $\mathcal{M}(X)$ that generates an operator semigroup $e^{t\Delta}$?


More specifically, does there exist a subset 
$$\mathrm{dom}(\Delta) \subseteq \{ \mu \in \mathcal{M}(X) \mid \Delta \mu \in \mathcal{M}(X)\}$$ 
(where $\Delta \mu$ is the distributional derivative) such that $\mathrm{dom}(\Delta)$ is dense in $\mathcal{M}(X)$ such that $\Delta$ is closed on $\mathrm{dom}(\Delta)$?



Laplacian on space of measures
The semigroup exists --- it is the adjoint of the heat semigroup on $C(X)$. However, it is *not* strongly continuous at $t=0$ and thus does not have a densely defined generator. The semigroup for $t>0$ is given by 
$$e^{t\Delta}\mu(E) \ = \ \int\_{E} H\_t(x,y) d V(x) d\mu(y) , $$
where $H\_t(x,y)$ is the heat kernel on $X$ and $V$ is the volume measure on $X$. Note that $e^{t\Delta}\mu$ is absolutely continuous with respect to $V$ for $t>0$. It follows that $e^{t\Delta}\mu$ is discontinuous at $t=0$ whenever $\mu$ is not absolutely continuous with respect to $V$. 


Note that $e^{t\Delta}M(X) \subset L^1(d V)$ as soon as $t>0$ and on $L^1$ the semigroup *is* strongly continuous. Thus $e^{t\Delta}$ is strongly continuous for $t>0$. In some sense the semigroup has a *non densely defined* generator, which is $\Delta$ on $L^1(dV)$.


