TITLE: Hodge de Rham operator and orientability
QUESTION [12 upvotes]: Let $(M,g)$ be a Riemannian manifold. One can consider the exterior algebra bundle $\Lambda(T^*M)$. The sections of this bundle are differential forms, to be noted by $\Omega^k(M)$. One can consider de Rham differential $d:\Omega^*(M) \to \Omega^{*+1}(M)$. In order to define the so called Hodge-de Rham operator one needs the adjoint of this differential: this requiers some choice of scalar product on the space of all forms. If $M$ is orientable there is natural notion of integration which is defined via volume form (nonvanishing top form)-this notion is absent in the nonorientable situation. However there is also the notion of density: as far as I know, with this notion one can define scalar products on forms as well and so we are able to define the adjoint of $d$ and therefore the Hodge de Rham operator. My question is: 

Is there a flaw in above argument? If not, why we consider Hodge-de Rham operator only for orientable manifolds?

REPLY [8 votes]: The comments above almost say it all. Let me just add one small point of motivation. There are several notions of integration on manifolds. If you have a measure, you can integrate functions. A Riemannian metric is one way of getting a measure. If you have a top degree form, you need an orientation.
To combine both, consider odd top forms, that is, elements of $$\Omega^k(M;o(TM))=\Lambda^kT^*M\otimes o(TM)\;,$$ where $n=\dim M$ and  $o(TM)\cong\Lambda^nTM$ denote the bundle of local orientations. The integral over forms in $\Omega^n(M;o(TM))$ is well-defined because they "carry an orientation with them". These are the densities in the question, and they correspond to (signed) measures on $M$. A typical example is the Euler form of the Levi-Civita connection on $M$.
The Hodge star relates integration of functions with integration of forms. For example $$\int_M\|\alpha\|^2\,d\mathrm{vol}_g=\int_M\alpha\wedge *\alpha$$ on oriented Riemannian manifolds $(M,g)$. If you drop the orientation, you can still define $*\colon\Omega^k(M)\to\Omega^{n-k}(M,o(TM))$ and the formula above still works. You can also define $D=d+*^{-1}d*$ if you regard the second $d$ as acting on $\Omega^\bullet(M,o(TM))$. Because the local formula for both $d$s is the same, you won't even notice a difference. And with the formula above and Stokes theorem (for odd forms), you can check that $*^{-1}d*$ is the formal adjoint of $d$.
Edit To address the last question: the only reason I see why one would regard the Hodge-de Rham operator on oriented manifolds only is related to the decomposition of $H^{2k}(M)$ of an oriented $4k$-dimensional manifold into positive and negative forms. This is needed for example to define the signature, or to talk about antiselfdual metrics in dimension 4.