Let $V$ be a vector space over the field of real numbers $\mathbb R$. I know that there is an isomorphism $$\Phi:\Lambda^k V^*\longrightarrow (\Lambda^k V)^*,$$ defined by $$\Phi(\varepsilon_1\wedge \ldots \wedge \varepsilon_k)(x_1\wedge \ldots \wedge x_k)=\det(\varphi_i(x_j)).$$ Is it possible to write the inverse explicitly? What would that be?
Thanks.
As far as I know there is not any formula for the inverse (at least without choosing a basis). As evidence for this, note that $\Phi$ is actually not always an isomorphism: it is only an isomorphism if $V$ is finite-dimensional. So if there is a formula for $\Phi^{-1}$, it would have to somehow be a formula that only makes sense if $V$ is finite-dimensional.
One way you can compute $\Phi^{-1}$ explicitly is by choosing a basis $\{e_1,\dots,e_n\}$ for $V$. Given such a basis, let $\{f_1,\dots,f_n\}$ be the dual basis for $V^*$. Note that $\{e_{i_1}\wedge e_{i_2}\wedge\dots e_{i_k}: i_1<i_2<\dots<i_k\}$ is a basis for $\bigwedge^k V$. Let $\{f_{i_1i_2\dots i_k}:i_1<i_2<\dots<i_k\}$ be the dual basis of $\left(\bigwedge^k V\right)^*$. Now we can give a formula for $\Phi^{-1}$ on this basis: $$\Phi^{-1}(f_{i_1i_2\dots i_k})=f_{i_1}\wedge f_{i_2}\wedge\dots f_{i_k}.$$ Indeed, you can easily compute that $\Phi(f_{i_1}\wedge f_{i_2}\wedge\dots f_{i_k})(e_{j_1}\wedge e_{j_2}\wedge\dots e_{j_k})$ is $1$ if $j_m=i_m$ for all $m$ and $0$ otherwise (assuming the $i_m$ and the $j_m$ are both in increasing order), so $\Phi(f_{i_1}\wedge f_{i_2}\wedge\dots f_{i_k})$ is the dual basis element $f_{i_1i_2\dots i_k}$. (Note that this discussion very much requires $V$ to be finite-dimensional, since otherwise the "dual bases" are not actually bases at all.)