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?
Extension of Riemannian Metric to Higher Forms
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2I'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. – 2011-02-17
1 Answers
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 $\Lambda^k V$.
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0I mean "inner product", i.e., a symmetric and positive-definite bilinear form. – 2018-09-22