Dual space of exterior power and exterior power of dual space

Solution 1:

An isomorphism is given by the non-degenerate pairing $\Phi \colon \Lambda^k(V^{*}) \times \Lambda^k(V) \rightarrow \mathbb{F}$ defined by

$$ \Phi(\varphi^1 \wedge \ldots \wedge \varphi^k, v_1 \wedge \ldots \wedge v_k) = \det (\varphi^i(v_j))_{i,j=1}^n $$

and extended bilinearly. Sometimes, when working over $\mathbb{R}$ or $\mathbb{C}$, people use a slightly different pairing $\Phi' = \frac{1}{k!} \Phi$ which differs from $\Phi$ by a constant factor.

Solution 2:

The natural isomorphism $I$ between these spaces is defined by $$ (I(\alpha_1\wedge\ldots\wedge\alpha_k))(v_1\wedge\ldots\wedge v_k):=\sum_\pi \mathrm{sgn}(\pi)\alpha_1(v_{\pi(1)})\ldots \alpha_k(v_{\pi(k)}). $$