Let a nonlinear differential equation is invariant under all diffeomorphisms, then we get lack of parabolicity?



Lack of parabolicity of PDE due to invariancy under diffeomorphisms?
I'll (try to) answer the question specifically in the case of Ricci flow. View the Ricci flow on a closed manifold $M$ as an initial value problem on the positive cone of positive definite symmetric two-tensors, where given an initial metric $g(0),$ we wish to find a path of metrics $g(t)$ solving,
\begin{align}
\frac{\partial{g(t)}}{\partial{t}}=-2\text{Ric}\_{g(t)}(g(t)).
\end{align}
Suppose we have a solution $g(t)$ and let $f:M\rightarrow M$ be any diffeomorphism. Then by the naturality of these tensors $f^{\*}g$ is a solution as well. If the Ricci flow was a parabolic equation, then (part of) elliptic regularity (using compactness of M) would guarantee that the solution space is finite dimensional. But, as the diffeomorphism group is infinite dimensional, we have exhibited an infinite dimensional space of solutions, hence the equation is not parabolic. More directly, calculating the principal symbol of the linearization of the second order differential operator,
\begin{align}
\Gamma(S\_{>0}^{2}(T^\*M))&\rightarrow \Gamma(S^{2}(T^\*M))\\
g &\mapsto Ric(g),
\end{align}
one can explicitly cook up cotangent directions where the principal symbol is non-invertible, also showing that the equation is not parabolic. Someone more well versed in PDE may correct me if I've said something wrong, but this is the gist of the argument. 


The "DeTurck trick" in these cases is to find a way to "break" this diffeomorphism invariance by changing the metric by a carefully chosen diffeomorphism (which results in an elliptic equation) and then showing that one can back-solve to get an honest solution of the Ricci flow.


You can find calculations of the principal symbol and a discussion of these ideas in this honors thesis <https://math.stanford.edu/theses/Stetler%20Honors%20Thesis.pdf> and follow the references therein for more details. 


