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There are two inner product on differential forms:

  1. $\langle \alpha,\beta\rangle$ induced from Riemannian metric $g$ by defining on 1-forms as dual of vector fields then extending to all differential forms i.e. $\langle e_{i_1}\wedge \dots\wedge e_{i_k}, e_{j_1}\wedge \dots\wedge e_{j_k}\rangle =\det[\langle e_{i_s}, e_{j_s}\rangle]$.
  2. On compact oriented Riemannian manifold $(M,g)$ $$(\alpha,\beta)=\int_M\alpha\wedge\star\beta.$$

Q1: Are these two inner products on differential forms equal?

Q2: If the answer to Q1 is "NO" then is it important to notice that two operators are adjoint of each other (or an operator is symmetric or self-adjoint) w.r.t. which metric? e.g. $d$ and $\delta$ that are adjoint w.r.t. second inner product but I don't know it is w.r.t. other one.

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    $\begingroup$ How do you extend the first one to differential form? $\endgroup$
    – Didier
    Mar 14 at 17:31
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    $\begingroup$ No, the first definition is not on differential forms; it's pointwise or at best local. I suggest, however, that you relate $\alpha\wedge\star\beta$ at a point to the definition in 1. $\endgroup$ Mar 14 at 21:31
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    $\begingroup$ I think what @TedShifrin meant was that (1) is defined as an inner product on $\Lambda^p_xM$ for $x\in M$, while (2) is defined as an inner product on $\Gamma(\Lambda^pM)$. The relation between (1) and (2) is that (2) is the global version of (1), obtained by integrating (1). $\endgroup$
    – Didier
    Mar 15 at 9:34
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    $\begingroup$ $\Gamma(\Lambda^pM)$ is the set of sections of the vector bundle $\Lambda^pM$, and is usually written as $\Omega^p(M)$ yes. Its elements are differential forms of degree $p$, and are global objects. $\endgroup$
    – Didier
    Mar 15 at 10:31
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    $\begingroup$ What I said is that $(2)$ is not a Riemannian metric. It is an inner product on the vector space of $p$-forms. An analog would be the inner product on $L^2[0,1]$ defined by $\langle f,g\rangle = \int_0^1 fg$. There is no such thing as $\left(\int_0^1f^2\right)(x)$, because $\int_0^1f^2$ is a number, not a function. $\endgroup$
    – Didier
    Mar 19 at 11:29
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Disclaimer: the comment section is overgrowing so here is an answer that I hope will erase all your doubts.

First part: As said in the comment section, $(1)$ is a inner product on the vector space $\Lambda^p_xM$ while $(2)$ is an inner product on the vector space $\Omega^p(M)$. The link between them is that $(2)$ is obtained by integrating $(1)$ over the whole manifold $M$.

An analog is this: consider the set $M=[0,1]$. Then for each $x\in [0,1], T_x[0,1] = \mathbb{R}$ and one can define an inner product on $T_x[0,1] = \mathbb{R}$ by $\langle a,b \rangle_x = a\times b$. This is $(1)$.

A vector field on $[0,1]$ is just a smooth function $f:[0,1] \to \mathbb{R}$, and $(2)$ is here $$ \langle f,g\rangle = \int_0^1 f(x) g(x) \mathrm{d}x = \int_0^1 \langle f, g\rangle_x \mathrm{d}x. $$

Second part: If $V$ is a vector space and $\langle\cdot,\cdot\rangle$ is an inner product, one can create a Riemannian metric on $V$, thought as a manifold the following way. As a vector space, the tangent bundle of $V$ is trivial: $$TV = V\times V$$ and one can define the Riemannian metric $g_v = \langle\cdot,\cdot\rangle$ for $v\in V$. It is a constant Riemannian metric because the canonical trivialization makes $g_v$ to be a function independant of $v \in V$. Take $V = \Omega^p(M)$ and $\langle \alpha,\beta\rangle = \int_M \alpha \wedge \star \beta$. Now, forget that $V$ and $\langle\cdot,\cdot\rangle$ is defined thanks to a Riemannian manifold $(M,g)$ and just look at its structure: it is a vector space with an inner product. Hence, for this inner product $\|\alpha\|$ is a number.

If you really want to think of this construction as a Riemannian manifold, like in the first paragraph, then $\|\alpha\|$ will be a function: $$ \|\alpha\| : \beta \in \Omega(M)^p \mapsto \|\alpha\|(\beta) = \|\alpha\|\in \mathbb{R} $$ which is constant and does not take points of $M$ as entries.

Comment: if you really do not understand what I said, here is just a question for you: for $x \in M$, how would you define $\left(\int_M \alpha\wedge \star \beta\right)(x)$?

This is the exact same thing as this question: how would you define $\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$?

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