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This is a really basic question, but I cannot find a explicit answer on any book. Given a Riemannian manifold $(M,g)$, the metric induces an inner product on vectors at any point, but it also induces an inner product on $p$-forms for $1\leq p\leq \dim(M)$.

I am just looking for confirmation that for 0-forms it is given by the obvious one, i.e. for $f,g\in \Omega^0(M) = C^\infty(M)$ $$\langle f,g\rangle_{\Omega^0} = f \cdot g$$ and that I am not missing any factor (say volume - determinant of the metric).

topolosaurus
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    this inner product doesn't depend at all on the metric, so it's up to you to define it however you wish (btw you're using $g$ in two places with different meanings; as the metric and as a function). – peek-a-boo May 20 '21 at 21:59
  • There’s a volume factor missing. The metric on forms is defined using the Hodge star operator and always produces some scalar product times the volume. – Gunnar Þór Magnússon May 21 '21 at 06:16
  • The definition on $\Omega^0(M)$ is usually left out but is implicitly understood to be indeed $\left< f, g \right>(p) = f(p)g(p)$. Note that $\left< f, g\right>$ is not the "inner product" of $f$ and $g$ but the "pointwise inner product" (because the result is a function, not a scalar). The inner product of $f,g$ is usually taken to be the scalar $\int_{M} f \cdot g , d\operatorname{Vol}_M$ (assuming $M$ is oriented and compact) which is the integral of the "pointwise inner product" against a volume form. – levap May 21 '21 at 11:18

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