Is there some geometric intuition for the quotient $G/Z(G)$, where $G=GL_n(\mathbb{R})$?

Question

Let $G=GL_n(\mathbb{R})$ be the $n$th general linear group. Its center $Z(G)$ is given by all scalar matrices $aI$ with nonzero determinant. How can I get an intuitive picture of $G/Z(G)$? I know that the equivalence relation is obviously $A\sim B$ iff there exists $\lambda\neq 0$ such that $A=\lambda B$, but I'm having trouble getting a concrete picture.

Answer 1

By definition, this is the projective linear group, sometimes denoted $PGL(n, \Bbb R)$. There are only finitely many elements in $SL(n, \Bbb R)\cap Z(G)$, so the homomorphism $$SL(n, \Bbb R) \hookrightarrow GL(n, \Bbb R) \to PGL(n, \Bbb R)$$ is an immersion and (as Lie groups), $$\dim PGL(n, \Bbb R) = \dim GL(n, \Bbb R) - \dim Z(GL(n, \Bbb R)) = n^2 - 1 = \dim SL(n, \Bbb R) .$$ (The image of $SL(n, \Bbb R)$ under this map is denoted $PSL(n, \Bbb R)$.) Thus, $PGL(n, \Bbb R)$ locally looks like $SL(n, \Bbb R)$, that is, they have the same universal cover, which is usually denoted $\widetilde{SL(n, \Bbb R)}$.

Now, $\det (a I) = a^n$, so if $n$ is odd, the only scalar matrix in $SL(n, \Bbb R)$ is $I$, and it follows that $SL(n, \Bbb R) \cong PGL(n, \Bbb R)$.

On the other hand, if $n$ is even, then the scalar matrices in $SL(n, \Bbb R)$ are $\pm I$, so in this case $SL(n, \Bbb R) \to PSL(n, \Bbb R)$ is a $2$-fold cover, and it follows from some topology that $SL(n, \Bbb R) \not\cong PGL(n, \Bbb R)$.

Answer 2

One might get some if none intuition about a group by knowing where it acts. In this case this group acts naturally on projective space, and in fact transitively and faithfully.