Duality in exterior algebras

To fix ideas, let $V$ be an $n$ -dimensional real vector space. A $k$ -vector is an element of $\overset{k}{⋀} V$ (the exterior algebra) and a $k$ -form is an element of $\overset{k}{⋀} V^{*}$ . These spaces are dual to one another but not in a natural way.

Now, the top exterior power of a finite-dimensional vector space is one-dimensional, so both $\overset{n}{⋀} V$ and $\overset{n}{⋀} V^{*}$ are of dimension one. We also see (this time by direct computation) that there are nondegenerate bilinear pairings

\overset{k}{⋀} V \times \overset{n - k}{⋀} V \to \overset{n}{⋀} V and \overset{k}{⋀} V^{*} \times \overset{n - k}{⋀} V^{*} \to \overset{n}{⋀} V^{*}

for all $k$ .

The problem is that while the top exterior powers are of dimension one, we do not have a canonical isomorphism between them. To get one, we fix an inner product $g$ on $V$ . Its determinant (or volume form) is a nonzero element of $\overset{n}{⋀} V^{*}$ and thus gives an isomorphism of that space with the ground field $R$ . We get a compatible isomorphism of $\overset{n}{⋀} V$ with $R$ by considering the dual metric. Composing these isomorphisms with the above nondegenerate pairings induces isomorphisms

\overset{k}{⋀} V ≅ \overset{n - k}{⋀} V^{*} and \overset{k}{⋀} V^{*} ≅ \overset{n - k}{⋀} V .

We finally get the correspondence between $k$ -forms and $(n - k)$ -vectors by following this last non-canonical isomorphism:

\overset{k}{⋀} V^{*} \to \overset{n - k}{⋀} V .

Keep an eye: technically speaking we only need to fix a volume form on the space $V$ to get these isomorphisms, because such a form will induce the necessary isomorphisms of the top exterior powers with $R$ , which in turn will identify $\overset{k}{⋀} V^{*}$ and $\overset{n - k}{⋀} V$ via the (now non-) canonical nondegenerate pairings. Sloppy, awful people sometimes do this silently by fixing a basis of $V$ , which they then either define to be orthonormal or use to define a nonzero element of $\overset{n}{⋀} V$ , which then brings us to this kind of confusion later on.

On the other hand, the inner product let us go further an define the Hodge star operator.