I am currently writing a paper about the Hodge theorem for an algebraic topology course. The specific formulation I am proving can be stated thus. Let $M$ be a compact, orientable n-dimensional Riemannian manifold. We use $\Delta^kM$ to denote the vector space of *elliptic* differential $k$-forms, i.e the differential $k$-forms $\alpha \in \Omega^kM$ such that $\Delta \alpha = 0$ where $\Delta = dd^* + d^*d$ is the Laplace Beltrami operator. We use $H^kM$ to denote the $k$-th de Rham cohomology of $M$. The Hodge thoerem states that:
$$
\Delta^kM \simeq H^kM
$$

A big part of the proof of this theorem is showing that the following equality holds: $$ \Omega^kM = \Delta^kM \oplus \Delta(\Omega^kM) $$

This equality in follows from the following facts;

a) $\Omega^kM$ sits inside some Sobolev space $H$ where $\Delta$ is well defined.

b) Since $\Delta$ is a linear self-adjoint operator on $H$, we have that: $$H = Ker(\Delta) \oplus Im(\Delta^*) = Ker(\Delta) \oplus Im(\Delta)$$ This means that any $\alpha \in \Omega^kM$ can be expressed as $\alpha = \beta + \Delta\gamma$ where $\beta, \gamma \in H$.

c) Now, $\beta$ is a harmonic form and it is not difficult to show that this implies that it is closed (i.e that $d\beta = 0$). This implies that it is smooth, so $\beta \in \Omega^kM$. Thus, $\Delta\gamma$ must be smooth. By elliptic regularity, this implies that $\gamma$ is smooth, leading us to the result that $\gamma \in \Omega^kM$ and thus that $$ \Omega^kM = \Delta^kM \oplus \Delta(\Omega^kM) $$

I have done a fair bit of snooping, but I have not found a clear, short explanation as to what the Sobolev space $H$ looks like and how it constructed. So I guess the best formulation of my question would be:

How does one construct a Sobolev space $H$ containing the space of differential k-forms $\Omega^kM$ of a closed Riemannian manifold $M$ such that the differential Laplace-Beltrami operator $\Delta$ is well-defined?