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Selfadjointness of the Dirac operator

If $T$ is a densely defined operator on a Hilbert space $\sH$, its adjoint $T^*$ has domain

\begin{displaymath}
\Dom T^* := \set{\phi \in \sH : \exists  \chi \in \sH
\text...
...{\phi} = \braket{\psi}{\chi}
\text{ for all } \psi \in \Dom T}
\end{displaymath}

and then $T^*\phi := \chi$, of course, so that the formula $\braket{T\psi}{\phi} = \braket{\psi}{T^*\phi}$ holds. If $T$ is symmetric, then clearly $\Dom T \subseteq \Dom T^*$ with $T^* = T$ on $\Dom T$: that is, $T^*$ is an extension of $T$ to a larger domain.

The second adjoint $T^{**} =: \Bar{T}$ is called the closure of $T$ (symmetric operators always have this closure), where the domain of the closure is
\begin{align*}
\Dom\Bar{T} := \set{\psi \in \sH : \exists \phi\in \sH
&\text{ a...
...at } \psi_n \to \psi \text{ and }
T\psi_n \to \phi \text{ in } \sH}
\end{align*}
In other words, the graph of $\Bar{T}$ in $\sH \oplus \sH$ is the closure of the graph of $T$. And then, of course, we put $\Bar{T}\psi := \phi$. When $T$ is symmetric, we get

\begin{displaymath}
\Dom T \subseteq \Dom\Bar{T} \subseteq \Dom T^*.
\end{displaymath}


\begin{defn}
We say that $T$ is \textbf{selfadjoint} if $T = T^*$; thus $T$ is...
...djoint} if it is symmetric and its closure
$\Bar{T}$ is selfadjoint.
\end{defn}


\begin{remark}
Selfadjoint operators have \textit{real spectra}:
$\spec(T) \subs...
...ed
measure'' on Borel subsets of $\bR$ with support in $\spec(T)$.
\end{remark}

The main result of this chapter is that the Dirac operator on a compact Riemannian spin manifold is essentially selfadjoint. This was proved by Wolf in 1973; he actually showed the result also for noncompact manifolds which are complete with respect to the Riemannian distance given by the metric [Wolf]. In his proof, completeness is needed to establish that closed geodesic balls are compact; that proof is also given in the book by Friedrich [Fri]. For simplicity, we deal here only with the compact case.


\begin{thm}
Let $(M,g)$ be a compact boundaryless Riemannian spin manifold. The...
...or $\Dslash$ is essentially selfadjoint on its original
domain $\sS$.
\end{thm}


\begin{proof}
There is a natural norm on $\Dom\Dslash^*$, given by
\begin{displa...
...Dslash^*$ : which establishes that
$\Bar{\Dslash}$ is selfadjoint.
\end{proof}


next up previous contents
Next: The Schrödinger-Lichnerowicz formula Up: Dirac operators Previous: Symmetry of the Dirac   Contents
Pawel Witkowski 2006-03-14