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

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.

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}
• Note that $r^*(dV)=\epsilon\,dV$ isn’t quite right if we treat $dV$ as the volume form because we’re not assuming connectedness (eg think of $M$ as a disjoint union of two copies of $\Bbb{R}$, and $r$ acting as identity on one component and minus identity on the other). If we think of it as a density, then things are fine and orientation doesn’t even cause any issues. Commented May 14, 2023 at 17:59
• Yes, good point. I was implicitly assuming that $M$ is connected. As the joke goes: a manifold with two components is not a manifold -- it's two manifolds instead! Commented May 14, 2023 at 19:27