@@ -222,7 +222,7 @@ In other words $\benbrace*{D,g}$ is acting via the following element of $C_c^\in

The computation in the proof of \cref{lem:symbol_welldefined} now shows:

\begin{lemma}

If $D$ is an operator of order $1$, the symbol can be computed as

If $D$ is an operator of order $1$, the symbol can be computed as\marginnote{again, depending on notation the $i$-term might not come up}

\[

\sigma_1(D)(\mathd g) u = i \,\benbrace*{D,g} u

\]

...

...

@@ -230,7 +230,7 @@ The computation in the proof of \cref{lem:symbol_welldefined} now shows:

The symbol gives raise to an extremly important class of differential operators:

\begin{definition}

\begin{definition}\label{def:elliptic}

Let $D \in\DiffOp^k(E_0,E_1)$ and $x \in M$.

Then we say, that $D$ is \Index{elliptic at}$x$ if for each $\xi\in\Tang_x^* M$, $\xi\neq0$ the homomorphism $\sigma_k(D)(\xi)\colon(E_0)_x \to(E_1)_x$ is invertible.

$D$ is \Index{elliptic}, if it is elliptic at each point $x \in M$.

...

...

@@ -267,13 +267,13 @@ Before applying the techniques of \cref{sec:unbounded_operators} to them, we fir

for $u$ and $v$ compactly supported.

\end{definition}

\begin{theorem}

Each differential operator $D \in\DiffOp^k(E_0,E_1)$ has a formal adjoint in$D^\dagger\in\DiffOp^k(E_1,E_0)$ and the formal adjoint is uniquely defined.

Each operator $D \in\DiffOp^k(E_0,E_1)$ has a formal adjoint$D^\dagger\in\DiffOp^k(E_1,E_0)$ and the formal adjoint is uniquely defined.

Furthermore the symbol of the formal adjoint can be computed pointwise:\marginnote{depending on conventions regarding the symbol, there might be a minus sign here, cf.\ \cite[Prop.~10.1.4]{higson2000analytic}}