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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?

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    $\begingroup$ George de Rham does this in his book Differentiable Manifolds. He develops the theory without restricting to orientable manifolds: he uses the language of "even" and "odd" forms. The Hodge-star of an even $k$-form is odd (and vice versa). $\endgroup$
    – Chris Gerig
    Commented Sep 6, 2015 at 19:50
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    $\begingroup$ And the main point is that the formula for the adjoint of $d$ uses the Hodge-star twice, i.e. $d^\ast = \pm *d*$ (the sign depends on the dimension and the degree), so it goes from even forms to even forms, as it should, so, indeed, orientability is not needed to define the Hodge Laplacian. $\endgroup$
    – Robert Bryant
    Commented Sep 6, 2015 at 20:37
  • $\begingroup$ @Chris Gerig But am I right, that in order to define the Hodge star you need the existence of the volume form (which is equivalent to being orientable)? And also the adjoint is defined abstracly (in terms of Hilbert space operators), why we should care whether we are still able to have formula like $d^*=\pm *d*$? $\endgroup$
    – truebaran
    Commented Sep 6, 2015 at 21:17
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    $\begingroup$ @truebaran: Chris' point is that the Hodge star is well-defined without orientability; it's just that it maps odd forms to even forms (and vice versa). Orientability (when it holds) is used to identify the odd forms with the even forms (and, for this, it is essential). The point of the formula $d^\ast = \pm * d *$ is that, because the Hodge star appears twice, $d^\ast$ maps even forms to even forms, so it's well-defined, even without orientability. $\endgroup$
    – Robert Bryant
    Commented Sep 7, 2015 at 0:17
  • $\begingroup$ @RobertBryant Still I'm only partially convinced: so your point is that it is possible to define Hodge star without orientability. As far as I understand it gives you the grading. But there is a general theory of Hilbert spaces operators and from this theory it follows that once you have a densely defined operator $T$ one can construct its adjoint $T^*$ so in particular you can construct $d^*$ without using Hodge star (provided that you have already scalar product on the space of forms: in this case I suspect that you are forces to use density and not volume form because the volume form is) $\endgroup$
    – truebaran
    Commented Sep 8, 2015 at 13:35

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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.

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