Below you can find the original question, but here is the main underlying calculation that seems bizarrely impossible to find written down anywhere:
$$
\begin{pmatrix}
a&b\\c&d
\end{pmatrix}
\wedge
\begin{pmatrix}
e&f\\g&h
\end{pmatrix}
\overset{?}{=}
ah-fc
$$
The bounty on this question is really for this one calculation, and the original question below is just for some (historical) context.
Let $X$ be a complex $n$-manifold with a rank-$r$ vector bundle $E$ and open cover $\{U_\alpha\}$ such that $E|_{U_\alpha}\cong\mathbb{C}^r$.
Say we have two $r\times r$-matrices $P,Q$ of differential $1$-forms, i.e.
- $P\in\Omega^1_{U_\alpha}\otimes\operatorname{End}(E|_{U_\alpha})$;
- $Q\in\Omega^1_{U_\beta}\otimes\operatorname{End}(E|_{U_\beta})$.
Question 1: What is the wedge product $P\wedge Q$? I believe it should just be matrix multiplication, but using the wedge product component-wise, i.e. $$(P\wedge Q)_{ij}=\sum_{k}(P_{ik}\wedge Q_{kj})$$ but (having read the article on vector-valued differential forms on the infallible Wikipedia) we should have that $$P\wedge Q\in\Omega^2_{U_{\alpha\beta}}\otimes\operatorname{End}(E|_{U_\alpha})\otimes \operatorname{End}(E|_{U_\beta}).$$ Then, using the fact that the $E|_{U_\alpha}$ are finite-dimensional vector spaces, we see that $P\wedge Q$ should be an $r^2\times r^2$-matrix of differential $2$-forms (since $\operatorname{End}(V)\otimes\operatorname{End}(W)\cong\operatorname{End}(V\otimes W)$ for f.d. vector spaces). In this case, the wedge product should be given by something like the Kronecker product of $P$ and $Q$.
Question 2: Now let $M$ be an $r\times r$ matrix of holomorphic functions (i.e. $0$-forms). What can we say about $P\wedge MQ$? It seems clear that we should always have
- $(MP)\wedge Q=M(P\wedge Q)$;
- $P\wedge MQ=PM\wedge Q$;
- $P\wedge Q=-(Q^t\wedge P^t)^t$ (or something similar, depending on the answer to Question 1).
However, using these two 'facts' it doesn't seem possible to 'pull out the $M$' from an expression of the form $P\wedge MQ$. Is there a way of doing so? Matrix non-commutativity gets in the way here, but are there restrictions we can place on $M$, $P$, or $Q$ to get some nice result?
Best Answer
The calculation as stated is wrong. This is because, given $A,B\in\operatorname{End}(V)$, and $v,w\in V$, the "good" definition of $A\wedge B\in\operatorname{End}(V\wedge V)$ is given by $$(A\wedge B)(v\wedge w)= Av\wedge Bw-Aw\wedge Bv$$ which is not necessarily equal to just $Av\wedge Bw$, which is what I was using. (Note, for example, that we need the result to be alternating)
In particular, the notation is very confusing, since it is the good definition to have $$(A\wedge A)(v\wedge w) = Av\wedge Aw$$ and so this $\wedge$ operator is not given by the same equation in the case where $A\neq B$, i.e. the notation "$\wedge$" is not used consistently between these two (different) cases!
Finally, to give the correct version of the specific calculation in the question, we have that
$$ \begin{pmatrix} a&b\\c&d \end{pmatrix} \wedge \begin{pmatrix} e&f\\g&h \end{pmatrix} = ah-fc+bg-de. $$