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.



Properties of non-integer powers of the Hodge Laplacian
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



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> *Wolf, Joseph A.*, [**Essential self adjointness for the Dirac operator and its square**](http://dx.doi.org/10.1512/iumj.1973.22.22051), Indiana Univ. Math. J. 22, 611-640 (1973). [ZBL0263.58013](https://zbmath.org/?q=an:0263.58013).
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