A have a question on the definition of the Sobolev spaces with real degree. Consider a (complete, connected, oriented) Riemannian manifold $(M,g)$ with Laplacian $\Delta_{g}$. For simplicity, we can also think specifically about $\mathbb{R}^{d}$. This operator maps $C^{\infty}_{c}(M)$ to $C^{\infty}_{c}(M)$ and hence can be understood from the functional analystic point of view 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 $L^{2}(M)=\overline{C^{\infty}_{c}(M)}$ denotes the Hilbert space of $L^{2}$-functions defined using the volume form on $(M,g)$. Now, it is well-known that this operator is essentially self-adjoint, which means that it admits a self-adjoint extension (i.e. its closure) $\overline{\Delta}_{g}$
$$\overline{\Delta_{g}}:\mathcal{D}(\overline{\Delta_{g}})\to L^{2}(M)$$
where the domain $\mathcal{D}(\overline{\Delta_{g}})$ is the usual one of the closure of an operator. Now, by self-adjointness, we can define operators $E^{s}:=\overline{\Delta_{g}}^{s}\:\mathcal{D}(E^{s})\to L^{2}(M)$ for any positive $s\in\mathbb{R}$, where the domain $\mathcal{D}(E^{s})$ is the usual one defined by means of the spectral theorem.
Using this, Sobolev spaces can be defined as follows:
$H^{s}(M):=\{f\in L^{2}(M)\mid E^{s/2}f\in L^{2}\}$, where $ E^{s/2}f\in L^{2}$ has to be understood in the weak sense.
Now, my question is the following:
Question: Doesn't the definition imply that $H^{s}(M)=\mathcal{D}(E^{s/2})$, i.e. the Sobolev space is given by the domain of the fractional Laplacian defined via the spectral theorem?
The inclusion $\mathcal{D}(E^{s/2})\subset H^{s}(M)$ is clear, since for any $f\in\mathcal{D}(E^{s/2})$ clearly $E^{s/2}f\in L^{2}(M)$. For the other direction, note that "$E^{s/2}f\in L^{2}$ for $f\in L^{2}(M)$ in the weak sense", by definition means that there exists a $g\in L^{2}(M)$ such that $$\forall\varphi\in C^{\infty}_{c}(M):\,\,\langle f,E^{s/2}\varphi\rangle_{L^{2}}=\langle g,\varphi\rangle_{L^{2}}$$
But now, if the above is true, it is also true more generally $\forall\varphi\in\mathcal{D}(E^{s/2})$, since $C^{\infty}_{c}(M)$ is dense in $\mathcal{D}(E^{s/2})$. In particular, we see that $f\in\mathcal{D}((E^{s/2})^{\ast})=\mathcal{D}(E^{s/2})$, since the above is nothing else than the definition of the domain of the adjoint of $E^{s/2}$, which is a self-adjoint operator. We conclude that $H^{s}(M)\subset\mathcal{D}(E^{s/2})$.
Is there anything wrong in my argument? I think its a bit weird that the Sobolev space coincides with the domain of the Laplacian. If this is the case, why then writing the definition in terms of weak derivatives, isn't it just tautological then?