# Friedrich extension of an elliptic operator

I'm working with elliptic operators on Riemannian manifolds.

In the case where the metric $g$ on $(M,g)$ is not complete, I understand that, for example, the Hodge laplacian may be not essentially self-adjoint on $C^\infty(\Omega^p(M))$; I want to obtain self-adjoint extensions.

Now, the Hodge laplacian factors as $\Delta = D\cdot D$, where $D=d+\delta$ is the sum of the de Rham differential and the codifferential obtained through the metric. If we define the maximal domain of $D$ as $$\mathcal D_{max}(D)=\{u\in L^2\Omega(M):\ Du\in L^2\Omega(M)\}$$ and the minimal domain as $$\mathcal D_{min}(D)=\{u\in D_{max}(D):\ \exists\ u_j\in C^\infty\Omega(M)\ s.t.\ u_j\to u\text{ and }Du_j\to Du\text{ in }L^2\Omega^p(M)\}$$ Then the Friedrichs extension of $\Delta$ may be characterised as $\Delta^{Fr}=D_{max}D_{min}$. The article I'm reading this from (Mazzeo-Vertman, "Analytic torsion on manifolds with edges") doesn't really explain how one could get to this conclusion.

I'm interested in elliptic operators which factor in a similar fashion, and I'd hope the same property holds for this larger class. Could anybody tell me whether this holds, or propose some good reference on the topic?

Edit: the comment I received seemed to imply it's possible to prove this from scratch. Here's my try: the domain of the Friedrichs extension is the completion of the domain of the operator (for us,the core domain $C^\infty(\Omega(M))$) with respect to the form $$Q(\xi,\eta)=(\xi,D^2\eta) + (\xi,\eta)$$ since both $\xi$ and $\eta$ are assumed to lie in the domain, and $D=D^*$ on the core domain, for the first part $$(\xi,D^2\eta)=(D\xi,D\eta)$$ the elements which I obtain by including limits of converging Cauchy sequences - in the graph norm of $D$, say - are exactly those in $\mathcal D_{min}(D)$, or am I mistaken here? If this were true, then the only requirement left would be that $D^2u$ is still square integrable (for, I assume, we want to be able to write $(D^2\xi,\eta)$). Then the extension would really be $D_{max}D_{min}$.

• Thanks for your comment. I'm new to the subject, so I first wanted to know if this were some kind of big theorem, or something one could try to prove from scratch. Your comment seemed to imply the second, so I gave it a try (see edit). If you have time, could you take a look at it? I'm new to the business, and I could have written obscenities. – nelv Mar 22 '17 at 19:06
• Yes, that's basically the idea. You have determined the form domain to be the graph space of $D_{min}$. Then the final condition for the an element to be in the domain of the Friedrichs extension is that $D_{min}f$ be in the domain of $D_{max}$. I believe that's correct; at least that's how I remember it. – Disintegrating By Parts Mar 23 '17 at 16:09