Let $\mathfrak{g}$ be a finite dimensional semisimple Lie algebra and $V$ a finite dimensional $\mathfrak{g}$-module. Consider their semidirect sum $\mathfrak{g}\ltimes V$. I want to show that $$H^2(\mathfrak{g}\ltimes V,\mathbb{R})=0\,\,\text{if}\,\,(\wedge^2V^*)^\mathfrak{g}=\{v\in\wedge^2V^*\colon\mathfrak{g}v=0\}=0.$$ This is proven in Theorem 14.1 of https://www.staff.science.uu.nl/~ban00101/lecnotes/repq.pdf, but I don't get what is going on in the second part of the proof.
Let $\omega:(\mathfrak{g}\ltimes V)\times(\mathfrak{g}\ltimes V)\to\mathbb{R}$ be a closed alternating two form. We have to prove that $\omega$ is exact. Since $H^2(\mathfrak{g})=0$, it follows that $\omega_{\mathfrak{g}}=\omega|_{\mathfrak{g}\times\mathfrak{g}}$ is exact, hence there exists $\lambda\in\mathfrak{g}^*$ such that $\omega_\mathfrak{g}=d\lambda$. We extend $\lambda$ to an element in $(\mathfrak{g}\ltimes V)^*$ by setting $\lambda(X,v)=\lambda(X)$. Then we define $\omega_1=\omega-d\lambda$ as a closed alternating two form on $\mathfrak{g}\ltimes V$. In particular, we note that $[\omega]=[\omega_1]$ in $H^2(\mathfrak{g}\ltimes V)$.
Now my question is what is going on next. The author takes a specific look at the map given by $$ \mathfrak{g}\times V\to\mathbb{R},(X,v)\mapsto\omega_1(X,v). $$ However, how is this possible, since $\omega_1$ should take 2 arguments of $\mathfrak{g}\times V$ instead of 1. Later, the author talks about $[X,v]$, but how is this defined, which Lie bracket does he mean overhere? Could anyone help me figure out what is going on?
I believe the author is implicitly employing the embeddings $i_\mathfrak{g}:\mathfrak{g}\to\mathfrak{g}\ltimes V$ and $i_V:V\to \mathfrak{g}\ltimes V$ here; in other words, really the map is $$ (X,v)\mapsto \omega_1((X,0),(0,v)). $$ Likewise, by $[X,v]$ they mean $[(X,0),(0,v)]_{\mathfrak{g}\ltimes V} = (0,Xv)$. It also seems there's a typo in the definition of $\rho$; it should be $$ \rho(X)(t,v) = (\omega_1(X,v),Xv). $$ Then we can use to closedness of $\omega_1$ (in the second equality) to see that: \begin{align} \rho([X,Y])(t,v) &= \big(\omega_1([X,Y],v),[X,Y]v\big) \\ &= \big(\omega_1([X,v],Y)-\omega_1([Y,v],X),XYv-YXv\big) \\ &= \big(\omega_1(Xv,Y)-\omega_1(Yv,X),XYv-YXv\big) \\ &= \big(\omega_1(X,Yv),XYv\big)-\big(\omega_1(Y,Xv),YXv\big)\\ &= \big(\rho(X)\rho(Y)-\rho(Y)\rho(X)\big)(t,v) \end{align} This proves that $\rho:\mathfrak{g}\to\operatorname{End}(\mathbb{R}\oplus V)$ is a Lie algebra homomorphism.