A tensor is invariant under diffeomorphism.

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.

• Since $\nabla$ and $\Delta$ are intrinsic, doesn't that mean $\nabla \nabla f - \frac{1}{2}\Delta f g$ is intrinsic? Nov 29, 2023 at 16:51
• 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? Nov 29, 2023 at 17:40
• 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 $\phi_t(x)=x$.) Nov 29, 2023 at 18:35
• You are right, i will edit it. And I doubt gradf is a killing vector field. Nov 29, 2023 at 19:11
• @Kakashi I want to ask what can i do if it is intrinsic Nov 30, 2023 at 17:43