Frame Independence of Inner Product on Differential Forms I am trying to prove that $\langle \omega^1\wedge\dots\wedge\omega^k,\eta^1\wedge\dots\wedge\eta^k\rangle=det(
\langle(\omega^i)^\#,(\eta^j)^\#\rangle)$ is independent of the choice of frame (where these are all one forms).  I'm pretty uncertain how to do this, but here's my attempt:
Let $(\varepsilon^i)$ be an arbitrary frame dual to $(E_i)$.  We then have that the Riemannian tensor is $g=g_{ij}\varepsilon^{i}\otimes\varepsilon^{j}$ and that $(\varepsilon^i)^\#=g^{\alpha l}\varepsilon^l E_{\alpha}=g^{ii}\varepsilon^i E_i$, since $\varepsilon^i$ is a coframe.  If I can show that $det(\langle(\varepsilon^i)^\#,(\varepsilon^j)^\#\rangle)$ is independent of the choice of frame (by showing that $\langle(\varepsilon^i)^\#,(\varepsilon^j)^\#\rangle$ is independent of choice of frame), then I've proved that the general inner product is independent of the choice of frame.
However, I end up with: $\langle (\varepsilon^i)^\#,(\varepsilon^j)^\#\rangle=g_{ij}\varepsilon^i(g^{ii}\varepsilon^i E_{i})\varepsilon^j(g^{j} \varepsilon^j E_{j})=g_{ij}g^{ii}\varepsilon^ig^{jj}\varepsilon^j=\varepsilon^ig^{jj}\varepsilon^j$, which is not (at least not obviously) independent of the choice of frame... I'm sure there's something wrong with my use of the raising operator, but at this point I'm stuck and frustrated and would appreciate any insight
 A: The problem asks you to show that there exists a unique inner product on $\Lambda^k(V^{*})$ that satisfies that for all $\omega^1, \dots, \omega^k, \eta^1, \dots, \eta^k \in V^{*}$ we have
$$ \left< \omega^1 \wedge \dots \wedge \omega^k, \eta^1 \wedge \dots \wedge \eta^k \right> = \det \left( \left< \left( \omega^i \right)^{\sharp}, \left( \eta^j \right)^{\sharp} \right> \right). $$
This formula doesn't involve any choice of basis for $V$. Now, the point is that not very element of $\Lambda^k(V^{*})$ is of the form $\omega^1 \wedge \dots \wedge \omega^k$ for some $\omega^i \in V^{*}$ and, even worse, an element $u \in \Lambda^k(V^{*})$ might have infinitely many different representations as $u = \omega^1 \wedge \dots \wedge \omega^k$ with different $\omega^i$'s. For example, 
$$ (e^1 + e^2) \wedge (e^1 - e^2) = (-2e^1) \wedge e^2 = e^1 \wedge (-2e^2) $$
are three different representations of the same element in $\Lambda^2(V^{*})$.
This discussion implies that you can't naively define an inner product by the formula because:


*

*The formula doesn't cover all elements of $\Lambda^{k}(V^{*})$. This is not that bad because elements of $\Lambda^{k}(V^{*})$ are linear combinations of elements of the form $\omega^1 \wedge \dots \wedge \omega^k$ and since the inner product must be bilinear, the formula in fact determines how the inner product behaves on the whole of $\Lambda^k(V^{*})$. This also shows that if such an inner product exists, it is unique.

*The more serious problem is that elements might have many different representations which might result in inconsistent results in the right hand side.


One can argue that everything is alright without choosing a basis but this will require appealing to the universal property of the exterior product and I don't remember if Lee covers it in his book. A different approach suggested by Lee is to pick a basis $e^I$ for $\Lambda^k(V^{*})$ where $e^1, \dots, e^n$ is a dual basis of an orthonormal basis for $V$. Then, every element  $\omega \in \Lambda^k(V^{*})$ has a unique representation $\omega = \sum a_I e^I$ and you can define an inner product on $\Lambda^k(V^{*})$ by the formula
$$ \left< \sum a_I e^I, \sum b_J e^J \right>_{(e^i)} := \sum a_I b_J \delta^{I,J}. $$
This amounts to declaring that the elements $e^I$ form an orthonormal basis of $\Lambda^k(V^{*})$ and is clearly well-defined but now it is not clear that this satisfies the property above. If you try and check the property, you might see that it will be useful to prove in advance that if you start with a different basis $f^i$ of $V^{*}$ (which is still required to be dual to an orthonormal basis of $V$) and you define an inner product the same way using $f^i$ instead, you get the same inner product ($\left< \cdot, \cdot \right>_{(e^i)} = \left< \cdot, \cdot \right>_{(f^i)}$). 
