Let $\mathfrak{g}$ be a finite dimensional Lie algebra. The definition of the Koszul differential is given in the article by Kumar and Vergne on equivariant cohomology page 133 as follow:
Let us now consider the Lie group $G$ with lie algebra $\mathfrak{g}$. Recall the definition of the Koszul differential $c_L$ on the space $T^*= L \otimes \bigwedge^*\mathfrak{g}^*$ calculating the cohomology of a $\mathfrak{g}$-module $L$: for $ \alpha \in L \otimes \bigwedge^p\mathfrak{g}^*$, $c_L \alpha \in L \otimes \bigwedge^{p+1}\mathfrak{g}^*$ is defined by $$ (c_L \alpha)(X_1,...,X_{p+1}) = \sum_i (-1)^{i+1}X_i.\alpha(X_1,..., \hat{X_i},...,X_{p+1}) + \sum_{i<j}(-1)^{i+1}\alpha([X_i,X_j]X_1,..., \hat{X_i},...\hat{X_j},...,X_{p+1})$$ where $X_1,...,X_{p+1}$ are elements of $\mathfrak{g}$.
Let $\mathfrak{g}= \mathfrak{k} \oplus \mathfrak{r}$ where $\mathfrak{k} $ and $ \mathfrak{r}$ are Lie subalgebras of $ \mathfrak{g}$, where $ \mathfrak{r}$ is of dimension $n$. Let $K^i $ be a basis of $ \mathfrak{k}$ with dual basis $K_i$. Fix the element $v' \in \bigwedge ^n \mathfrak{r}^*$.
In page 141 (in the proof of lemma 36) I don't understand how to get the following formula $$ c(v') = - \sum_i (\mathrm{Tr}(\mathrm{ad}_\mathfrak{k} K^i))K_i \wedge v'.$$
Thank you!
