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## Extending a Riemannian Metric to one on $k$-Forms [closed]

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 $2 \leq k \leq N$. How is this canonically induced $g_k$ defined?

It seems that an answer to the question should follow from the simpler Euclidean case, which is just a question about linear algebra: If $V$ is a vector space and $g:V^\ast \times V^\ast \to R$ is bilinear then is there a canonical way of extending $g$ to a bilinear map $$\Lambda^k V^\ast \times \Lambda^k V^\ast \to \mathbb{R}?$$ There should be some trick with an anti-symmetry construction, but I can't spot.

Sorry if ths question is too basic for the level of the site.

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This is basic contruction found in many textbooks. One approach is to define $e_{i_1} \otimes \ldots \otimes e_{i_k}$ to be orthonormal basis for your tensor product and extend by linearity. This respects (anti)symmetrization. As an exercise in linear algebra you can find formula for scalar product of two general elements. – robot Feb 19 2011 at 14:46
Indeed, this is not really a question about riemannian geometry, but one about linear algebra. Given an inner product on a vector space $E$, you get an inner product on $\Lambda^k E$ whose value on monomials $v_1 \wedge \cdots \wedge v_k$, $w_1 \wedge \cdots \wedge w_k$ is the determinant of the matrix whose entries are $\langle v_i,w_j\rangle$. You then extended it bilinearly to all of $\Lambda^k E$. (By the way, what you have called $g$ is not what is usually called the metric in Riemannian geometry. Your $g$ is the induced "metric" on the cotangent space.) – JosÃ© Figueroa-O'Farrill Feb 19 2011 at 15:05
Given an inner product on $V$, extend it canonically to one on $V^{\otimes k}$ via $\langle v_1 \otimes \dots \otimes v_k, w_1 \otimes \dots \otimes w_k \rangle = \langle v_1, w_1 \rangle \dots \langle v_k, w_k \rangle$. Using the antisymmetrizer map (see my answer in <a href="mathoverflow.net/questions/54343/… thread</a>) you can embed $\bigwedge^k(V)$ into $V^{\otimes k}$. Restricting the inner product on $V^{\otimes k}$ to the image of the antisymmetrizer, you get the determinant formula Jose mentions. – MTS Feb 19 2011 at 21:43