# Diffeomorphism Invariance of frobenius norm

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

We have a isometry $$\phi:(M,\phi^*g) \to (M,g)$$.

One thing I want to know is, how can I efficiently show that intrinsic quantities like $$|\nabla\nabla f|^2$$ and $$(\Delta f)$$ are invariant under the isometry.

I have showed $$|\phi_* A_p|_{g}=|A_p|_{\phi^*g}$$ for any vector field $$A$$ and similar version of equality for tensors.

Yet, I still have no idea to show my problem. Is it calculate everything out in local coordinate rly the only way to solve this problem?

Any help is appreciated, thank you.

• Yeah you can work in coordinates. If something only depends on the metric tensor, then an isometry preserves it. For example, since $\text{div}(X) = \frac{1}{\sqrt{g}}\partial_{x_j}(\sqrt{g}X^j)$, it follows that div is invariant under isometry. Same reasoning applies to $\text{grad}$ and $\Delta$. Given that we are on a Riemannian manifold though, practically everything depends only on the metric ... Dec 3, 2023 at 1:16