Properties of non-integer powers of the Hodge Laplacian Consider a  complete smooth Riemannian manifold $(M,g)$.
I think that it is not difficult to prove that the $k$ Hodge Laplacian is essentially selfadjoint in the relevant $L^2$ space of $k$ forms, when defining it on the smooth compactly supported $k$ forms.
The standard argument by Chernoff should apply to this case, too.
I am interested in the validity of identities like this one (maybe on some specific  domain) $$\delta_{k+1}(\Delta_{k+1})^{\alpha} = (\Delta_k)^{\alpha} \delta_{k+1}\tag{1}$$
where $\delta$ is the closure of the usual formal adjoint  of $d$ (both defined on the smooth compactly supported forms) and $\Delta$ is the analogous closure (i.e., the unique selfadjoint extension) of the Hodge Laplacian (defined again on smooth compactly supported forms). $\alpha$ is any real. The powers are defined by spectral calculus using the fact that the spectrum of the operator is positive.
For reasons arising from physics, I am quite confident that (1) holds at least when applying both sides to smooth $k+1$-forms with compact support. However I did not manage to find  books or papers on these issues, barring some quite old articles. I am not an expert so I suppose that, simply, I have not searched thoroughly.
A modern reference on these issues would be very appreciated.
 A: This is a nice question. I think there is a trick that makes it obvious enough not to need a special reference. First, if both sides are to be applied to $k+1$-forms, to be consistent your desired identity needs to be
$$
  \delta_{k+1} (\Delta_{k+1})^{\alpha} = (\Delta_k)^{\alpha} \delta_{k+1} . \tag{1}
$$
The trick consists of starting with the operator $D = d + \delta$, aka the Hodge-Dirac operator, which maps the space of all forms to itself. $D$ is obviously symmetric on smooth compactly supported forms. The Hodge-de Rham Laplacian is then $\Delta = D^2$. By block decomposition, $d_k$ is the block of $D$ that maps $k$-forms to $k+1$-forms, $\delta_k$ is the block of $D$ that maps $k$ forms to $k-1$-forms, while $\Delta_k$ is the block of $\Delta$ that maps $k$-forms to $k$-forms. If you can show that $D$ is essentially self-adjoint on the core of smooth forms with compact support, then the desired identity (1) falls out of the block decomposition of the automatic functional-calculus identity
$$
  D |D|^{2\alpha} = |D|^{2\alpha} D .
$$
But $D$ is essentially self-adjoint because $\Delta$ is, since the resolvent of one can be used to construct the resultant of the other by the formula
$$
  \frac{1}{\lambda - D} = \frac{\lambda + D}{\lambda^2 - D^2} .
$$
To make the argument precise, note that the identity $(\lambda - D) \frac{\lambda + D}{\lambda^2 - D^2} = \mathrm{id}$ on smooth compactly supported forms shows that for any non-real $\lambda$ the range of $\lambda - D$ is dense in $L^2$ forms. Hence, by Corollary to Thm.VIII.3(c) in Reed-Simon v.1, $D$ is essentially self-adjoint.
I hope I didn't make a mistake in the last argument. In any case, the essential self-adjointness of $D$ can probably be shown in a more classical way, as it was done for the usual spinor Dirac operator, for instance in

Wolf, Joseph A., Essential self adjointness for the Dirac operator and its square, Indiana Univ. Math. J. 22, 611-640 (1973). ZBL0263.58013.

