I'm trying to understand something that is written in Baez & Wise paper "Teleparallel Gravity as a Higher Gauge Theory". In section 4, they discuss Poincaré connection and, first of all, split Poincaré lie algebra - $\mathfrak{io(p,q)}$ - into a direct sum of lorentz - $\mathfrak{o(p,q)}$ - and translation - $\mathbb{R^{p,q}}$ - lie algebras: $$ \mathfrak{io(p,q)} = \mathfrak{o(p,q)} \oplus \mathbb{R^{p,q}} $$
Then, if we have a $IO(p,q)$ principal bundle $\pi:P \rightarrow M$ and a connection $A \in \Omega^1(P,\mathfrak{io(p,q)})$ defined there, it can be decomposed as a sum of two connections: $$ A = \omega + e $$ where $\omega \in \Omega^1(P,\mathfrak{o(p,q)})$ and $e \in \Omega^1(P,\mathbb{R}^{p,q})$.
Proceeding to calculate the curvature of connection $A$, we have naturally: $$ F = d_AA = dA + \frac{1}{2}[A\wedge A] $$
where $[\omega \wedge \eta](X,Y) = [\omega(X) , \eta(Y)] - [\omega(Y) , \eta(X)]$, for $\omega, \eta \in \Omega^1(P,\mathfrak{g})$.
I'm trying to calculate this explicitly:
$$ F = dA + \frac{1}{2}[A \wedge A] = d(\omega + e) + \frac{1}{2}[(\omega + e)\wedge(\omega + e)] $$ $$ = d\omega + de + \frac{1}{2}[\omega \wedge \omega] + \frac{1}{2}[\omega \wedge e] + \frac{1}{2}[e \wedge \omega] + \frac{1}{2}[e \wedge e] $$
Since translation group is abelian, $[e \wedge e] = 0$, and we alson have that $[\omega \wedge e] = [e \wedge \omega] $. Then:
$$ F = d\omega + \frac{1}{2}[\omega \wedge \omega] + de + [\omega \wedge e] $$
But the paper says it easy to check that
$$ F = d_\omega \omega + d_\omega e $$
where $d_\omega \omega = d \omega + \frac{1}{2}[\omega \wedge \omega]$ and $d_\omega e = d e + \frac{1}{2}[\omega \wedge e]$.
In my calculation there is a factor $\frac{1}{2}[\omega \wedge e]$ more. Where's the error?
