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Just as for functions $f_1,f_2 \in L^2(\Omega_1)$, with $\Omega_1 \subset \mathbb R ^d$ and the Euclidean metric, we have: $$\langle f_1, f_2 \rangle = \int_{\Omega_1} \overline{f_1 (x)} f_2(x) dx$$

What is the analog inner product on a manifold $(\mathcal M, g)$? Can it be expressed as this?:

$$\langle f_1, f_2 \rangle_g = \int_\mathcal{M} \overline{f_1(x)} f_2(x) g^*(x) dx$$

for some function $g^*(x)$ determined by $g$?

I am rather unfamiliar with basic concepts of Differential Geometry, but I am familiar with introductory Functional Analysis. Most of the descriptions I have found of this involve notation/technical details I am unfamiliar with.

jacktrnr
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    There's something called the volumn form $dV$, locally given by $\sqrt{\det g} dx^1\cdots dx^n$, which is used in the integration. – Arctic Char Jun 02 '22 at 18:00
  • @ArcticChar Thanks! – jacktrnr Jun 02 '22 at 20:19
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    The point is we should somehow make the manifold $M$ into a measure space. So, we need a certain $\sigma$-algebra, and a measure on it. A-priori, this is completely arbitrary, but if you're given a metric tensor field $g$, then you can use this to define a measure $\lambda_g$ (often written $dV_g$) (see surface measure and Gauss-Green theorem for the details). Now that we have a measure space $(M,\mathcal{L}(M),\lambda_g)$, you can carry out all the measure-theoretic stuff with $L^p$ spaces. In particular, $L^2(\lambda_g)$ is a Hilbert space. – peek-a-boo Jun 03 '22 at 13:55

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