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](https://arxiv.org/pdf/hep-th/9206021.pdf) 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?



The exterior derivative of a certain differential form on the space of connections of a surface
For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a *symmetric* bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form $\xi\_\omega$ whose value on a tangent vector $Y\in T\_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.


Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual formula for the differential gives
\begin{align\*}
\mathrm d\xi\_\omega(X,Y) &= X(\xi\_\omega(Y)) - Y(\xi\_\omega(X)) - \xi\_\omega([X,Y])\\
&= \omega(X,Y) - \omega(Y,X) - 0\\
&= 2\omega(X,Y)
\end{align\*}


Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T\_XV\cong V$. Then by definition, $\xi\_\omega = \iota\_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\xi\_\omega = \mathcal L\_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.


More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C\_2}\to V\otimes V\to (V\otimes V)\_{C\_2}$ of $GL\_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL\_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).


As to your calculation that $\mathrm d\eta = \dot\eta$, the only way I know how to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is *not* the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.


