On the nonexistence of linear actions of $PGL(k)$ on $\mathbb C^n$

Question

Let our field be $\mathbb{C}$ and let $PGL(k)=GL(k)$ for $k>1$. I want to show the nonexistence of a linear action of $PGL(k) \curvearrowright \mathbb{C}^n$.

$\newcommand{\Z}{\mathbb{Z}}$ Clearly $H_2(PGL(k),\Z)=H_2(B(PGL(k)),\Z)=H_2(BGL(k),\Z) \neq 0$. According to wikipedia this should be an obstruction to a homomorphism from $PGL(k) \to GL(n)$. Where is a good place to find this fact that is probably standard?

Answer 1

$\newcommand{\C}{\mathbb{C}}$ As always we have the adjoint representation $PGL(k+1) \to GL(Lie(PGL(k+1)))$ where $Lie(PGL(k+1))=gl(k+1)/\C\langle Id \rangle \cong sl_{k+1}$.

It sends $A \in PGL(k+1)$ to $ad_{\tilde A}$ where $\tilde A$ is any lift to $GL(k+1)$. Thus $n$ can be chosen to be $(k^2+k)^2$. So these actions do exist.