Let $Y$ be a closed oriented $2$-dimensional manifold, $G$ a Lie group and $Q \to Y$ a principal $G$-bundle with a given section $q.$ Denote by $\mathcal{A}_Q$ the space of connections on $Q,$ and by $L_Q \to \mathcal{A}_Q$ the Chern-Simons line bundle. Suppose we have an Ad-invariant symmetric bilinear form $$\langle -,- \rangle: \mathfrak{g} \times \mathfrak{g} \to \mathbb{C}.$$

Given this data, we can then define the $1$-form $\theta_q$ on $\mathcal{A}_Q$ by $$(\theta_q)_\eta(\dot{\eta}) = 2\pi i \int_Y q^*\langle \eta \wedge \dot{\eta} \rangle, \: \: \:\eta \in \mathcal{A}_Q, \: \dot{\eta} \in T_\eta \mathcal{A}_Q.$$

I have seen (for example, in the paper Classical Chern-Simons Theory, Part I by Freed, pg. 27) the claim that $$\dfrac{i}{2\pi} d(\theta_q)(\dot{\eta_1},\dot{\eta_2}) = -2 \int_Y q^*\langle \dot{\eta_1},\dot{\eta_2} \rangle.$$ Here $d$ denotes the exterior derivative. Is this true? I do not see where the factor of two arises from. I would expect this to follow if we could show that $$d\langle \eta,\dot{\eta} \rangle(\dot{\eta_1},\dot{\eta_2}) = 2 \langle \dot{\eta_1},\dot{\eta_2} \rangle.$$ However, this is not what I get. I would expect the exterior derivative to act as $$d\langle \eta,\dot{\eta} \rangle = \langle d(\eta),\dot{\eta} \rangle + \langle \eta,d(\dot{\eta}) \rangle.$$ Now, further, I would think that $d(\eta) = \dot{\eta},$ so that since $d^2=0,$ I get, when evaluating this on $(\dot{\eta_1},\dot{\eta_2})$ just $\langle \dot{\eta_1},\dot{\eta_2} \rangle.$

**So I would like to ask:**

Is the claimed equality true? If so, why is it true? What is wrong with my proposed way of going about calculating it?