I denote the space of all $V$-valued differential $k$-forms on $M$ with $\mathcal A^k(M,V)$. Let $\omega\in \mathcal A^k(M,V)$ and $\eta\in \mathcal A^l(M,W)$, where $V,W$ are finite real vector spaces. Then, we know that the usual wedge is a map $\mathcal A^k(M,V)\times \mathcal A^l(M,W)\to \mathcal A^{k+l}(M,V\otimes W)$. Let $i$ run from 1 to $\dim V$. Let $\{v_i\}$ be a basis for $V$ and $\{w_i\}$ a basis for $W$. Then, $\omega=\omega^iv_i$ and $\eta=\eta^iw_i$ (summation convention), where $\omega^i,\eta^i$ are usual forms, meaning they are elements of $\mathcal A^k(M,\mathbb R)=\mathcal A^kM,\mathcal A^lM$ respectively. Therefore, $\omega\wedge\eta=(\omega^i\wedge \eta^j)v_i\otimes w_j$. Let's assume I want to construct a new product, namely $\curlywedge$, that is a map $\mathcal A^k(M,V)\times \mathcal A^l(M,W)\to \mathcal A^{k+l}(M, W)$.

Following the method of the product $[\omega\wedge\eta]$ for two Lie algebra-valued forms, I can think of two ways, that $V\otimes W\to W$. Both rely on the universal property of the tensor product. The first way is to define a composite map: $$v\otimes w\mapsto (v,w)\stackrel{\langle \sigma,\;\rangle\times id}{\to}(\langle\sigma,v\rangle,w)\to \langle \sigma,v\rangle w\in W$$

where $\langle\;,\;\rangle$ is the pairing product, $v\in V$, $w\in W$ and $\sigma\in V^*$

Alternatively, we can use a homomorphism $\phi:V\to End W$. I'll restrict the case now. Let $\mathfrak g$ be the Lie algebra of the Lie Group $G$ and $(\psi,V)$ be a finite representation for $\mathfrak g$, where $\psi:\mathfrak g\to End V$ is a Lie algebra homomorphism (V is the rep space). Let $\omega\in \mathcal A^1(P,\mathfrak g)$ and $\theta\in \mathcal A^1(P,V)$, namely the connection and solder form for a principal $G$-bundle $P\to M$. We have that $\omega\wedge\theta=\omega^i\wedge\theta^jX_i\otimes v_j\in \mathcal A^2(P,\mathfrak g\otimes V)$, where $\{X_i\},\{v_i\}$ bases for $\mathfrak g,V$ respectively. We want a map $\mathfrak g\otimes V\to V$, so that we can define a product $\mathcal A^k(P,\mathfrak g)\times \mathcal A^l(P,V)\to \mathcal A^{k+l}(M,V)$. It seems reasonable to choose the composite map $X\otimes v\to (X,v)\to\psi(X)v\in V$. From this, one can construct the above desired product, let it be $\curlywedge$ and then it can be proven that $\Theta=d\theta+\omega\curlywedge \theta$ is the usual expression for the torsion 2-form. Bianchi identity for $\Theta$ can also be proven. One sees, that when $V=\mathfrak g$, then $\psi=ad$ and the former $X\curlywedge v$ is nothing more than $[X\wedge v]$.

My problem is that I cannot explicitly write down an exchange rule for $\omega\curlywedge \theta$. I believe it should follow the exchange rule of $[\omega\wedge\eta]$, namely $[\omega\wedge\eta]=(-1)^{kl+1}[\eta\wedge\omega]$, i.e $\omega\curlywedge \theta=\theta\curlywedge \omega$, but I am not able to show this. Instead I find that the exchange rule is of the usual forms.

  • $\begingroup$ The usual situation (where you get the $(-1)^{k\ell+1}$ factor) is with $V$-valued forms, with a Lie bracket on $V$. Then if $\omega = \sum\omega^i\otimes v_i$ and $\eta = \sum \eta^j\otimes v'_j$, we define $[\omega\wedge\eta] = \sum (\omega^i\wedge\eta^j) \otimes [v_i,v'_j]$. I don't see any natural skew-symmetry for you when you switch the order of $\mathfrak g$ and $V$. $\endgroup$ – Ted Shifrin Jan 5 '18 at 19:13
  • $\begingroup$ I agree. But for example, let us have $\Omega\in \mathcal A^2(P,\mathfrak g)$ and $\theta\in \mathcal A^1(P,\mathfrak{g/h})$, where $\mathfrak{g/h}$ is not necessarily a Lie subalgebra of $\mathfrak g$. Then, some authors seem to take $[\Omega\wedge\theta]\in \mathcal A^3(P,\mathfrak{g/h})$ with the usual exchange rules for Lie algebra valued forms. I don't really get how. $\endgroup$ – kospall Jan 6 '18 at 23:16

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