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 \begin{equation} \mathcal B := \left\{\varepsilon^{i_1} \wedge \cdots \wedge \varepsilon^{i_k}\big|_p : i_1 < \cdots < i_k \right\} \end{equation} 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?

  • $\begingroup$ 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. $\endgroup$ – Will Jagy May 4 at 16:18
  • $\begingroup$ 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. $\endgroup$ – D Ford May 4 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.