Let $V$ be a finite-dimensional vector space.
Is there an isomorphism between $\Lambda^k(V^\ast)$ and $\left(\Lambda^k(V)\right)^\ast$?
I was able to prove this with the additional requirement of an inner product on $V$ (and thus subsequently on $\Lambda^k(V)$) via
$$
\require{AMScd}
\begin{CD}
\left(\Lambda^k(V)\right)^\ast @>\mathcal{J}^{-1}>> \Lambda^k(V) @>\Lambda^kJ>> \Lambda^k(V^\ast)
\end{CD}
$$
where $J: V \to V^\ast$ and $\mathcal{J}: \Lambda^k(V) \to \left(\Lambda^k(V)\right)^\ast$ are the isomorphisms given by the Riesz representation theorem and $\Lambda^kJ$ is the map given by $v_1\wedge \cdots \wedge v_k \mapsto J(v_1) \wedge \cdots \wedge J(v_k)$.
Is there another way to identify these two spaces without the requirement of an inner product on $V$? I read Qiaochu Yuan's comment to his answer on a similar question but did not really understand it I fear.
Thank you very much.
Best Answer
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.