Symbol of a (non linear) differential operator I am interested in knowing whether there is a definition for the symbol of a PDO which is NOT linear.
In Wikipedia and in the book I am reading (An Introduction to Partial Differential Equations by Renardy-Rogers) I only found the definition for linear PDOs.
Here is the Wikipedia link:
http://en.wikipedia.org/wiki/Symbol_of_a_differential_operator
 A: The symbol of a nonlinear differential operator is defined as the symbol of its linearization.
A: See Definition of the principal symbol of a differential operator on a real vector bundle..
For an example, consider the Ricci curvature operator:
\begin{align}
\mathsf{Ricc}:\Gamma(S^2_+M)&\rightarrow\Gamma(S^2M)\\
g&\mapsto\mathsf{Ricc}(g).
\end{align}
The linearisation of the Ricci operator at a given metric $g\in\Gamma(S^2_+M)$ is just the directional derivative of the operator at $g$, and is the map
\begin{align}
D\mathsf{Ricc}|_g:S^2M&\rightarrow S^2M\\
h&\mapsto D\mathsf{Ricc}|_gh=\frac{\text{d}}{\text{d}t}\Big|_{t=0}\mathsf{Ricc}(g+th).
\end{align}
The hard part is calculating it.  Let $(U,\mathsf{x})$ be a chart on $M$ and $\omega\in\Gamma(T^*M)$ be a covector field.  Books on the Ricci flow (Topping, Chapter 2 or Chow & Knopf, Chapter 3) show that locally the principal symbol of the Ricci operator is
\begin{align}
[\hat{\sigma}_\mathsf{Ricc}(\omega)h]_{ij}=\frac{1}{2}g^{st}(\omega_s\omega_ih_{jt}+\omega_s\omega_jh_{it}-\omega_s\omega_th_{ij}-\omega_i\omega_jh_{st}).
\end{align}
It is then easy to show that the Ricci operator is not elliptic since if we set $h_{ij}=\omega_i\omega_j\neq0$, then $\hat{\sigma}_\mathsf{Ricc}(\omega)h=0$.
