# Extension of Riemannian Metric to Higher Forms

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?

• 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

## 1 Answer

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 $\Lambda^k V$.

• Why does this give an inner product? Do you mean "bilinear form" instead of inner product? – Nathanael Schilling Sep 15 '18 at 18:37
• I mean "inner product", i.e., a symmetric and positive-definite bilinear form. – AlexE Sep 22 '18 at 8:48