Duality in

To fix ideas, let $V$ be an $n$-dimensional real vector space. A $k$-vector is an element of $\bigwedge^k V$ (the exterior algebra) and a $k$-form is an element of $\bigwedge^k V^*$. 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 $\bigwedge^n V$ and $\bigwedge^n V^*$ are of dimension one. We also see (this time by direct computation) that there are nondegenerate bilinear pairings

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 $\bigwedge^n V^*$ and thus gives an isomorphism of that space with the ground field $\mathbb R$. We get a compatible isomorphism of $\bigwedge^n V$ with $\mathbb R$ by considering the dual metric. Composing these isomorphisms with the above nondegenerate pairings induces isomorphisms

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

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 $\mathbb R$, which in turn will identify $\bigwedge^k V^*$ and $\bigwedge^{n-k} V$ 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 $\bigwedge^n V$, 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.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es