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Consider $(M,g)$ is a Riemannian Manifold (compact).

For a smooth function $f(x,t): M \times \mathbb{R} \to \mathbb{R}$,

It is known that we can induce a one parameter group of diffeomorphism such that $$\frac{\partial}{\partial t} \phi_t(x) = \nabla f (t, \phi_t (x)) $$ And $$\phi_0(x)=x$$

I want to show the trace free part of hessian of $f$ ,i.e. $$\nabla \nabla f -\frac{1}{2}\Delta f g$$ has length that is invariant under the diffeomorphism. Which means take $\phi_t^*g=\overline{g}$, $\phi_t^*\nabla=\overline{\nabla}$ , we have $$|\nabla \nabla f -\frac{1}{2}\Delta f g|^2=|\overline{\nabla}\overline{ \nabla} f -\frac{1}{2}\overline{\Delta} f \overline{g}|^2 $$

Any help will be appreciated.

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  • $\begingroup$ Since $\nabla$ and $\Delta$ are intrinsic, doesn't that mean $\nabla \nabla f - \frac{1}{2}\Delta f g$ is intrinsic? $\endgroup$
    – Kakashi
    Nov 29, 2023 at 16:51
  • $\begingroup$ Yes, but is $\phi_t$ an isometry?(I don't know how to show it), or is there something I overlooked and can prove the result together with intrinsic property? $\endgroup$ Nov 29, 2023 at 17:40
  • $\begingroup$ You should not expect $\phi_t$ to be an isometry in general. If a Killing vector field is the gradient of a function, then it must be parallel and the function harmonic. (Also, you probably meant to write $\phi_0(x)=x$ instead of $\phi_t(x)=x$.) $\endgroup$
    – Ivo Terek
    Nov 29, 2023 at 18:35
  • $\begingroup$ You are right, i will edit it. And I doubt gradf is a killing vector field. $\endgroup$ Nov 29, 2023 at 19:11
  • $\begingroup$ @Kakashi I want to ask what can i do if it is intrinsic $\endgroup$ Nov 30, 2023 at 17:43

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