Isometry Invariance of the Inner Product in $L^2(M)$

Question

Let $(M,g)$ is oriented Riemmanian manifold and endowed with a Riemannian metric $g$. Let $dV$ is the Riemannian volume form Let us consider the function space $L^2(M)$. We define the inner product as follows: $\langle f,g\rangle=\int_M fg dV$

Now, let $r$ be an isometry. I am interested in exploring whether the inner product is invariant under this isometry. Specifically, does the following equation hold true?

$\langle f,g\rangle=\langle f\circ r,g\circ r\rangle$.

As a special case of this, can we at least say that the $L^2$ norm (squared) is preserved under the isometry? In other words, does the following hold?

$\|f\|^2=\langle f,f\rangle=\langle f\circ r,f\circ r\rangle$

Any insights or references to related literature would be greatly appreciated.

Answer 1

This should be a direct application of the following general fact: if $F\colon N\to M$ is a diffeomorphism between oriented manifolds and $\omega$ is a top-degree form on $M$, then $\int_N F^*\omega = \varepsilon\int_M \omega$, with $\varepsilon \in \{+1,-1\}$ chosen according to whether $F$ preserves or reverses orientation (Proposition 16.6 on Lee's Introduction to Smooth Manifolds).

Since $r$ is an isometry, we also know that $r^*({\rm d}V) = \varepsilon\, {\rm d}V$, with the value of $\varepsilon$ again according to whether $r$ preserves orientation or not. Now compute $$\begin{align}\langle f\circ r,g\circ r\rangle_{L^2} &= \int_M (f\circ r)(g\circ r){\rm d}V \\ &= \int_M (r^*f)(r^*g) {\color{red}{\,\varepsilon r^*({\rm d}V)}} \\ &= \varepsilon \int_Mr^*(fg\,{\rm d}V) \\ &= \int_M fg\,{\rm d}V \\ &= \langle f,g\rangle_{L^2}\end{align}$$