On the wikipedia page of Exterior Algebra under the section on alternating multilinear forms, it says:
“... By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $\left( \Lambda^k V \right)^* $. If $V$ is finite-dimensional, then the latter is naturally isomorphic to $ \Lambda ^k \left(V^*\right) $. ...”
I understand the first part, but I don’t see how the dual space of the $k$-th exterior power of $V$ is naturally isomorphic to the $k$-th exterior power of the dual of $V$? I get that they have the same dimension, but what’s the specific natural isomorphism?
I want to know this because I don’t see how alternating multilinear forms correspond to exterior products of covectors.
Usually the natural isomorphism is given via a perfect pairing $$\Lambda^k V^\ast \times \Lambda^k V \rightarrow k,~ (\phi_1\wedge \cdots \wedge \phi_k, v_1\wedge\cdots \wedge v_k)\mapsto \det(\phi_i(v_j))$$ which gives rise to the desired isomorphism. The naturality of this construction is easily verified.