Compactness of $PGL(n, \mathbb{R})$ and $PGL(n, \mathbb{C})$

Question

I know that the $PGL(n, \mathbb{R})$ or $PGL(n, \mathbb{C})$ is Lie group, because $PGL(n, F) = GL(n, F) / Z(n, F)$, where $Z(n, F)$ - scalar transformation and $F$ is $\mathbb{C}$ or $\mathbb{R}$.

I think, that both of these groups is not compact, but I don't know how to prove it?

Thank you so much!

Answer 1

Let us begin with $PGL(n,\mathbb{R})$. It can be defined as the quotient of $SL(n,\mathbb {R})$ by $\{\pm\operatorname{Id}\}$ and $\{\pm\operatorname{Id}\}$ is compact, since it is finite. But $SL(n,\mathbb{R})$ isn't. Therefore, $SPL(n,\mathbb{R})$ isn't either. (When $G$ is a topological group and $H$ is a closed normal subgroup, then $G$ is compact if and only if both $H$ and $G/H$ are.)

Since $PGL(n,\mathbb{R})$ is a closed subgroup of $PGL(n,\mathbb{C})$ and $PGL(n,\mathbb{R})$ is not compact, $PGL(n,\mathbb{C})$ is not compact either.