Lets consider the following functional analytic definition of the fractional Laplacian: Consider a (complete, connected, oriented) Riemannian manifold $(M,g)$ with corresponding Laplacian $\Delta_{g}$. This operator can naturally be viewed as an unbounded operator $$\Delta_{g}:\mathcal{D}(\Delta_{g})\to L^{2}(M)$$ with domain $\mathcal{D}(\Delta_{g}):=C^{\infty}_{c}(M)$, where $L^{2}(M)$ denotes the Hilbert space of $L^{2}$-functions w.r.t. to the volume measure. On complete manifolds, this operator is essentially self-adjoint and hence, its closure $\overline{\Delta_{g}}$ is the unique self-adjoint extension. Using the spectral theorem, we get for each $s\in [0,\infty)$ a well-defined self-adjoint operator
$$\overline{\Delta_{g}}^{s}:\mathcal{D}(\overline{\Delta_{g}}^{s})\to L^{2}(M)$$
Now, my question is the following: The operator $\overline{\Delta_{g}}^{s}$ can actually be defined on a larger domain, namely, using distribution theory: Let $H^{2s}(M)\subset L^{2}(M)$ be the set of all $\omega\in L^{2}(M)$ such that $\overline{\Delta_{g}}^{s}\omega\in L^{2}(M)$ in the distributional sense, i.e. there exists a $z\in L^{2}(M)$ such that
$$\langle \omega,\overline{\Delta_{g}}^{s}\varphi\rangle_{L^{2}}=\langle z,\varphi\rangle_{L^{2}} $$
for all test functions $\varphi\in C^{\infty}_{c}(M)$. We then write $\overline{\Delta_{g}}^{s}\omega:=z$. This is one possible way to define Sobolev spaces on manifolds, hence the notation $H^{2s}(M)$.
Is the operator $\overline{\Delta_{g}}^{s}\omega$ still self-adjoint on $H^{2s}(M)$, i.e. $$\langle \overline{\Delta_{g}}^{s}\omega,\psi\rangle_{L^{2}}=\langle \omega,\overline{\Delta_{g}}^{s}\psi\rangle_{L^{2}}$$ for all $\omega,\psi\in H^{2s}(M)$?
Of course, clearly $\mathcal{D}(\overline{\Delta_{g}}^{s})\subset H^{2s}(M)$. Furthermore, for $s=1$, clearly $\mathcal{D}(\overline{\Delta_{g}})=H^{2}(M)$, by uniqueness of self-adjoint extensions.
