# Curvature of dual connection

Let $d_A \colon \Omega^0(E) \longrightarrow \Omega^1(E)$ be a connection on a bundle $E\longrightarrow M$, and suppose its connection matrix is $\theta=(\theta_{ij})$ for some local frame of $E$. I know that the curvature matrix is given by $$\Theta = d\theta + \theta \wedge \theta.$$

I want to prove that the curvature of the dual connection $d_{A^*}$ vanishes iff the curvature of $d_A$ vanishes.

But, if i take the dual coframe of the previous frame, I know that the connection matrix of $d_{A^*}$ is given by $-\theta^t$. Then the curvature matrix of the dual connection would be $$\Theta^* = - ( d\theta) ^t + (\theta\wedge\theta)^t$$ which needn't vanish. Am I missing something?

Yes, you seem to have forgotten that operation $\wedge$ is anticommutative. As you know, the curvature of a connection in a local frame is expressed as $dA + A\wedge A$, for some matrix of 1-forms $A$. But this is not a „multiplication” of matrices, so it is not true in general that $A^{t}\wedge A^{t} = (A\wedge A)^{t}$! In fact, quite the opposite:
$$A^{t}\wedge A^{t} = -(A\wedge A)^{t}.$$
So, finally you should get that $\Theta^* = -\Theta^{t}$.