Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been reading about Riemannian manifolds, and have come across a comment that says that for a metric $g$ on an $N$-dimensional manifold $M$, considered as a bilinear map $$ g:\Omega^1(M) \times \Omega^1(M) \to C^{\infty}(M), $$ there exists a canonically induced bilinear map $$ g_k:\Omega^k(M) \times \Omega^k(M) \to C^{\infty}(M), $$ for all $2 \geq k \leq N$. What is this canonically induced $g_k$ defined?

share|improve this question
3  
Think about the following situation first: If $V$ is a vector space and $g: V^{\ast} \times V^{\ast} \to \mathbb{R}$ is bilinear then there is a canonical way of extending $g$ to a bilinear map $\wedge^{k} V^{\ast} \times \wedge^{k} V^{\ast} \to \mathbb{R}$. Now do this pointwise for each fiber of $\Omega^{1}(M)$. –  t.b. Feb 17 '11 at 20:43
    
I'm sorry but I can't what the extension of $g$ to a bilinear map $g: \Lambda^k V^* \times \Lambda^k V^* \to R$ is. –  MikhailMatrix Feb 17 '11 at 21:03
add comment

1 Answer

up vote 4 down vote accepted

If we have an inner product g on a vector space V, we can define an inner product on $\bigotimes_{i=1}^k V$ via $$g(v_1 \otimes \cdots \otimes v_k, w_1 \otimes \cdots \otimes w_k) := \frac{1}{k!}g(v_1, w_1)\cdots g(v_k, w_k).$$

The factor 1/k! has the following explanation: If the wedge product if defined via $$\omega \wedge \eta := \frac{(r+s)!}{r!s!}\operatorname{Alt}(\omega \otimes \eta),$$ where $\omega$ is an r-form and $\eta$ an s-form, then we get $$g(v_1 \wedge \ldots \wedge v_k, w_1 \wedge \ldots \wedge w_k) = \operatorname{det}(g(v_i, w_j)).$$

This gives the nice statement, that if $\{v_1, \ldots, v_n\}$ is an orthonormal basis of V, then $\{v_{i_1} \wedge \ldots \wedge v_{i_k}: 1 \le i_1 < \ldots < i_k \le n\}$ is an orthonormal basis of $\Lambda^k V$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.