Wedge product of matrix-valued differential forms

Question

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.

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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?

Answer 1

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

$$ \begin{pmatrix} a&b\\c&d \end{pmatrix} \wedge \begin{pmatrix} e&f\\g&h \end{pmatrix} = ah-fc+bg-de. $$

Answer 2

$\newcommand{\End}{\operatorname{End}}$ I don't fully understand the question. But I would assume that you want something that does not depend on the trivialization. And the best way to do this is to not use the trivialization at all. Also, since this is a point wise calculation, this is really a question about abstract multilinear algebra. So here is an unsuccessful attempt.

Let $T$ and $V$ be vector spaces. Let $P, Q \in T^*\otimes \End(V)$. Their tensor product is $P\otimes Q \in (T^*\otimes\End(V))\otimes (T^*\otimes\End(V))$. So you could define their wedge product to be $$ P\wedge Q = P\otimes Q - Q\otimes P \in (T^*\otimes\End(V))\otimes (T^*\otimes\End(V)) $$ If you choose a basis $(\theta^1, \dots, \theta^n)$ of $T^*$ and a basis $(e_1, \dots, e_r)$ of $E$ with dual basis $(\epsilon^1, \dots, \epsilon^r)$, the we can write \begin{align*} P &= p_{k\alpha}^\beta \theta^k\otimes \epsilon^\alpha\otimes e_\beta\\ Q &= q_{k\alpha}^\beta \theta^k\otimes \epsilon^\alpha\otimes e_\beta\\ P\otimes Q &= p_{k\alpha}^\beta q_{j\eta}^\gamma \theta^k\otimes \epsilon^\alpha\otimes e_\beta\otimes \theta^j\otimes \epsilon^\eta\otimes e_\gamma\\ P\wedge Q &= (p_{k\alpha}^\beta q_{j\eta}^\gamma - q_{k\alpha}^\beta p_{j\eta}^\gamma) \theta^k\otimes \epsilon^\alpha\otimes e_\beta\otimes \theta^j\otimes \epsilon^\eta\otimes e_\gamma \end{align*} But this is probably not what you want. There is, however, a natural map $$ (T^*\otimes\End(V))\otimes (T^*\otimes\End(V)) \rightarrow T^*\otimes T^*\otimes\End(V) $$ It's easiest to define this map using the bases: $$ \theta^k\otimes \epsilon^\alpha\otimes e_\beta\otimes \theta^j\otimes \epsilon^\eta\otimes e_\gamma \mapsto \theta^k\otimes\theta^j(\langle e_\beta,\epsilon^\eta\rangle \epsilon^\alpha\otimes e_\gamma) = \delta^\eta_\beta \theta^k\otimes\theta^j\otimes \epsilon^\alpha\otimes e_\gamma $$ This can be used to define a tensor product $P\otimes Q \in T^*\otimes T^*\otimes \End(V)$, where $$ P\otimes Q = p_{k\alpha}^\beta q_{j\beta}^\gamma \theta^k\otimes \theta^j \otimes \epsilon^\alpha\otimes e_\gamma $$ This corresponds to multiplying the two matrices of $1$-forms. Using this, we can define the wedge product to be \begin{align*} P\wedge Q = P\otimes Q - Q\otimes P &= (p_{k\alpha}^\beta q_{j\beta}^\gamma - q_{k\alpha}^\beta p_{j\beta}^\gamma) \theta^k\otimes \theta^j \otimes \epsilon^\alpha\otimes e_\gamma \in T^*\otimes T^* \otimes\End(V). \end{align*} At this point, I don't see how to proceed further. In particular, I don't see how to turn this into an element in $\bigwedge^2T^*\otimes\End(V)$. But that doesn't mean it can't be done. Or maybe there is some other clever way to define the wedge product.