# Inner product of $k$-forms

I'm working on the following problem from Lee's Introduction to Riemannian Manifolds:

Let $$(M,g)$$ be a Riemannian $$n$$-manifold. show that for each $$k=1,\ldots, n$$, there is a unique fiber metric $$\langle \cdot, \cdot \rangle_g$$ on the bundle $$\Lambda^k T^*M$$ that satisfies $$\left\langle \omega^1 \wedge \cdots \wedge \omega^k, \eta^1 \wedge \cdots \wedge \eta^k \right\rangle = \det \left( \left\langle \omega^i, \eta^j \right\rangle_g \right)$$ whenever $$\omega^1, \ldots, \omega^k$$, $$\eta^1, \ldots, \eta^k$$ are covectors at a point $$p \in M$$.

The problem comes with the following hint: define the inner product locally by declaring the set of $$k$$-covectors $$$$\mathcal B := \left\{\varepsilon^{i_1} \wedge \cdots \wedge \varepsilon^{i_k}\big|_p : i_1 < \cdots < i_k \right\}$$$$ to be an orthonormal basis for $$\Lambda^k\left(T^*_pM\right)$$ whenever $$\left( \varepsilon^i\right)$$ is a local orthonormal coframe for $$T^*M$$, and proving the resulting inner product satisfies the above equation and is coframe-independent.

But it's hard for me to see how $$\mathcal B$$ being an orthonormal basis implies the determinant formula. I know one can prove the determinant formula from the definition of wedge products $$\omega \wedge \eta = \frac{(k+l)!}{k!l!} \mathrm{Alt}(\omega \otimes \eta)$$ (with $$\omega$$ a $$k$$-form and $$\eta$$ an $$l$$-form) and by using the canonical inner product on $$T^k T^*M$$ given by $$\left\langle \omega^1 \otimes \cdots \otimes \omega^k, \eta^1 \otimes \cdots \otimes \eta^k \right\rangle = \prod_{i=1}^k \langle \omega^i, \eta^i \rangle$$ but this results in $$\frac 1{k!} \det\left(\left\langle \omega^i, \eta^j\right\rangle_g\right)$$. Besides, for my own understanding of the algebra of alternating tensors, I'd like to use a strategy that involves the algebra of wedge products directly. Any suggestions?

EDIT: There appears to be some connection with this inner product and the Gram determinant of (co)vectors, so maybe I can prove that analogous multilinearity properties have to hold for both determinants and this inner product?

• not as much is implied as you think. Given favorite orthonormal bases, one defines the inner product. Then you show that this is "well-defined," meaning different bases give the same thing. Several steps; first orthonormal bases only, then all bases. – Will Jagy May 4 at 16:18
• No that part I understand. I was having trouble proving the equation claimed in the problem from the orthonormal basis declaration. But I figured it out. – D Ford May 4 at 19:29