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The follow problem appears in the setting of $L^2$-differential forms on manifolds with boundary. An abstracted operator theoretic problem is given below.

Suppose $M$ is a smooth Riemannian manifold with boundary. We have an inner product and the Hodge star on differential forms.

$\langle \omega , \eta \rangle = \int_M \omega \wedge \star \eta$

We have the Cartan derivative $d$ as an unbounded operator, with its domain being suitably regular differential forms. For $\omega, \eta$ smooth and compactly supported we now have

$\langle d\omega,\eta \rangle = \langle \omega, \delta \eta \rangle$

with $\delta := \star d \star$. Now, with algebraic properties of the exterior derivative, we can show more general for $\omega \in dom(d)$ and $\eta \in dom(\delta)$

$\int_M d\omega\wedge\star\eta = \int_M \omega\wedge\star \delta\eta + \int_{\partial M} \operatorname{Tr}\omega \wedge \star \operatorname{Tr}\eta$

so in general a trace term appears. In particular, $\delta$ is not the hermitian adjoint of $d$. We can fix this and define the unbounded operator $d^\ast$ with the same action as $\delta$ but smaller domain, namely $dom(d^\ast) = \{ \eta \in L^2\Lambda \mid \operatorname{Tr}\star\eta = 0 \}$

Assuming I have not committed any serious fallacies, now to my problem.

This difference between $d^\ast$ and $\delta$ does not appear in the setting of manifolds with boundary, and literature on this and the $L^2$ exterior calculus is not as ubiquitous as literature on the fully smooth setting without boundary. However, both $d^\ast$ and $\delta$ are used.

So, whereas $d^\ast$ is the adjoint, what is $\delta$?

I am interested to understand this from a purely operator theoretic point of view (i.e. functional analysis). Of particular interst for me is whether the adjoint of a linear unbounded operator with respect to a pairing may be extended in a canonical way such that "defects" which disturb the adjointness-relation appear ( just as the trace terms above )?

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Already in very simple situations such issues can be examined: on $L^2[a,b]$ the Laplacian on test functions supported strictly in the interior has the obvious continuum of distinct self-adjoint extensions, depending on choices of boundary conditions to make the extension genuinely self-adjoint (as opposed to having leftover integration-by-parts terms as above), and these extensions are mutually incompatible. Perhaps this trivial (very accessible) example suggests something useful to you...? – paul garrett Jul 21 '11 at 15:23
I agree the essence of the problem can be formulated within a functional analytic setting without the analysis on manifolds. This should be more accessible. – shuhalo Jul 21 '11 at 20:13

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