Suppose that $V$ and $W$ are finite-dimensional vector spaces over $\mathbb{F}$. If $\varphi, \psi \in \hom(V,W)$, there are at least two interpretations of the symbol $\varphi \wedge \psi$:
- It is the wedge product in the exterior algebra of the vector space $\hom(V,W)$.
- It is the image of $\varphi, \psi$ under the functor $\bigwedge$, i.e. a map $V \wedge V \to W \wedge W$. (Edit: Mariano has pointed out to me that this does not make sense in general)
First Question: I gather that these two interpretations have nothing to do with each other? This is kind of confusing when for instance $W=\mathbb{F}$, and we have $\varphi \wedge \psi$ as an element of $(V^* \wedge V^*)$, but under the second interpretation $\varphi \wedge \psi = 0$, because $\mathbb{F} \wedge \mathbb{F} = 0$. (Edit: this question does not make sense, as Mariano points out.)
Second Question: Again in the special case $W=\mathbb{F}$, we can understand the action of $\varphi_1 \wedge \dotsb \wedge \varphi_k$ by taking an isomorphism $\bigwedge^k(V^*) \xrightarrow{\alpha} (\bigwedge^k(V))^*$ given by $\alpha(\varphi_1 \wedge \dotsb \wedge \varphi_n)(v_1 \wedge \dotsb \wedge v_n) = \det(\varphi_i(v_j))$. As far as I can tell, this correspondence is mostly arbitrary. Is this correct?
Third Question: When $W \neq \mathbb{F}$, how can I understand the operation of $\varphi_1 \wedge \dotsb \wedge \varphi_k$ on $\bigwedge^k(V)$? Would we then need some correspondence $\bigwedge^k(V^* \otimes W) \to (\bigwedge^k(V \otimes W^*))^*$?