Wedge product is nondegenerate symmetric bilinear form

Question

Let$$f: \Lambda^k(\mathbb{R}^n) \times \Lambda^{n - k}(\mathbb{R}^n) \to \mathbb{R}, \quad f(\alpha, \beta) = \alpha \wedge \beta.$$How do I see that $f$ is a nondegenerate symmetric bilinear form?

Answer 1

Fix a basis such as $e_1 \wedge e_2, e_1 \wedge e_3, e_1 \wedge e_4, e_2 \wedge e_3, e_2 \wedge e_4, e_3 \wedge e_4$ and work out the Gram matrix $G = (G_{ij})_{ij}$, where $G_{ij}$ is the value of $f$ on the $i$-th and $j$-th basis vector.

It's easy to do because $f$ is just the determinant: for example, $$G_{12} = f(e_1 \wedge e_2, e_1 \wedge e_3) = \mathrm{det}(e_1,e_2,e_1,e_3) = 0.$$ If you think about it you will see that $G$ is a particular permutation matrix.

You need to show that $G$ is invertible and symmetric to prove your claim.

Answer 2

First, you should have the one-dimensional space $\Lambda^n\mathbb R^n$ as the target of your pairing. Second, it does not make much sense to talk about symmetry in this case (and in the naive sense that pairing is not symmetric since $\beta\wedge\alpha=(-1)^{k(n-k)}\alpha\wedge\beta$.) But I guess at any rate the main issue is non-degeneracy.

To see this, recall that the wedge product of $n$-vectors in $\mathbb R^n$ is non-zero if and only if the vectors are linearly independent (and thus form a basis). Taking the standard basis $\{e_i\}$ for $\mathbb R^n$, consider the induced basis of $\Lambda^\ell\mathbb R^n$, which can be written as $e_I=e_{i_1}\wedge e_{i_2}\wedge\dots\wedge e_{i_\ell}$, where $I\subset\{1,\dots,n\}$ is an $\ell$-element subset whose elements are $i_1<i_2<\dots<i_\ell$. Now if take the bases $\{e_I\}$ for $\Lambda^k\mathbb R^n$ and $\{e_J\}$ for $\Lambda^{n-k}\mathbb R^n$. Then consider $e_I\wedge e_J$. If $I\cap J\neq\emptyset$, this vanishes since there are two equal elements in the resulting wedge product of vectors. On the other hand, if $J$ is the complement of $I$ in $\{1,\dots,n\}$, then you just have to re-order factors to see that $ e_I\wedge e_J=\pm e_1\wedge\dots\wedge e_n\neq 0$. For this non-degeneracy follows readily: writing $\alpha\in\Lambda^k\mathbb R^n$ as a linear combination of the $e_I$, vanishing of the wedge product with $e_J$ implies vanishing of the coefficient of $e_{J^c}$, where $J^c$ is the complement of $J$ in $\{1,\dots,n\}$. If this holds for all $J$, you get $\alpha=0$.