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

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  • $\begingroup$ No pde is invariant under all diffeomorphisms unless it is of zero order. $\endgroup$ Jul 23, 2014 at 18:42

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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.

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  • $\begingroup$ So, you use of elliptic regularity, but you assume the equation is parabolic . I don't undrestabd really $\endgroup$
    – Alon
    Jul 23, 2014 at 21:48
  • $\begingroup$ Also Ricci flow is degenerate parabolic and is not strictly parabolic and this is the main issue $\endgroup$
    – Alon
    Jul 23, 2014 at 21:49
  • $\begingroup$ I think the point is that the "degeneracy" is exactly due to the diffeomorphism invariance of the equation, namely that the degenerate directions of the principal symbol arise from Lie derivatives of the metric. $\endgroup$ Jul 23, 2014 at 22:29
  • $\begingroup$ Still you have not answered to my question. I think a part of your answer need to edit $\endgroup$
    – Alon
    Jul 23, 2014 at 23:51
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What do you mean? For example the nonlinear parabolic PDE $$ u_t = (1 + u^2) u_{xx}$$ is invariant under some diffeomorphisms (e.g. translations).

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  • $\begingroup$ See remark 5.1.2 for instance books.google.com/… $\endgroup$
    – Alon
    Jul 23, 2014 at 16:51
  • $\begingroup$ Ah, so you're talking about invariance under all diffeomorphisms... $\endgroup$ Jul 23, 2014 at 17:35
  • $\begingroup$ Yes, I edited it again. Sorry for that $\endgroup$
    – Alon
    Jul 23, 2014 at 17:59

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