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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?

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    This sounds like something I've never done. Why would you wedge the forms from different trivializations? You should be doing this in a single trivialization (or overlap of two, perhaps computing how things transform when you go from one to the other). – 2017-02-21
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    @TedShifrin Sorry, I didn't make this very clear, but I'm interested in looking on the overlap $U_{\alpha\beta}=U_\alpha\cap U_\beta$. – 2017-02-21
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    I see that. But you should be working with the wedge product in $U_\alpha$ and the wedge product in $U_\beta$ and then, perhaps, comparing the two. I can't imagine wedging a Lie-algebra valued $1$-form on $U_\alpha$ with one on $U_\beta$. I'm assuming you're looking at curvature or some structure equations. – 2017-02-21
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    @TedShifrin Hmm, not so sure I understand, we can just restrict the two $1$-forms to the intersection and then consider the wedge product there, no? Yes, I'm working with a bunch of local curvatures (on each trivialisation) and comparing them. – 2017-02-21
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    So it's what I said, You're computing the wedge product in a *particular* trivialization. You're not wedging across trivializations. At any rate, you still have an $r\times r$ matrix. Whether you're doing matrix multiplication or commutator depends on what you're doing. – 2017-02-21

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