Refer to: To complete the proof that $\operatorname{PSL}(2,\Bbb F_5)\cong A_5$. I'm starting to understand the group action of a projective linear group of degree $2$ on $\Bbb P^1(\Bbb F_q)$. But that's because I've already learnt about Möbius transformations, i.e. actions of $\operatorname{PSL}(2,\Bbb C)$ on $\Bbb{\widehat C}$, when I was studying non-Euclidean geometry before. So I just need to change the field in question from $\Bbb C$ to a finite one. I want to know more about the structure of projective linear groups of higher degree, e.g. $\operatorname{PGL}(3,\Bbb F_q)$ and $\operatorname{PSL}(3,\Bbb F_q)$.
I have no prior knowledge in projective geometry (at least I think so). Can someone give a quick explanation to me what is $\Bbb P^k(\Bbb F_q)$, or how does it look like? https://en.wikipedia.org/wiki/Projective_space gives a definition like $\Bbb P^k(\Bbb F_q):=(\Bbb F_q^{k+1}-\{(0,0,\dots,0)\})/\sim$ which they say identifies points which lie on the same line passing through origin as in the same equivalence class. That's fine. But I can't visualize it, especially in the context of finite fields. Also https://en.wikipedia.org/wiki/PSL(2,7) says that $$\text{For}\space\gamma=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\in\operatorname{PSL}(3,\Bbb F_2)\space\text{and}\space\mathbf x=\begin{pmatrix}x\\y\\z\end{pmatrix}\in\Bbb P^2(\Bbb F_2),\space\gamma.\mathbf x=\begin{pmatrix}ax+by+cz\\dx+ey+fz\\gx+hy+iz\end{pmatrix}$$
Why does $\gamma$ acts on $\mathbf x$ like this?? I can only see why the elements of $\Bbb P^2(\Bbb F_2)$ are represented as three-dim. vectors. And now I have another serious problem: I can't recover the way I interpret the action of $\operatorname{PSL}(2,*)$ on $*\cup\{\infty\}$ from what I've learnt today! I know that (from what I've learned in Möbius geometry) for $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{PSL}(2,\Bbb F_q)$ and $x\in\Bbb F_q\cup\{\infty\}$, $\gamma.x=\frac{ax+b}{cx+d}$. Now I realized that $\Bbb P^1(\Bbb F_q)$ is actually $\Bbb F_q\cup\{\infty\}$. Why can the $x$ not be represented as a two-dim. vector, but just a scalar here? and the $\frac{ax+b}{cx+d}$: neither does it look like a two-dim vector. I'm sure that what I learnt about Möbius transformations can't be wrong, neither can the description of linear fractional transformations as an element of $\operatorname{PSL}(2,*)$ acting on $x$ by $x\mapsto\frac{ax+b}{cx+d}$ be. Just how to reconcile my intepretation with the new one derived from the definition of a projective line?
I think I'm writing too much. I must stop here.
Best Answer
Here's an answer to your last question. In homogeneous coordinates, $\newcommand{\P}{\mathbb P} \newcommand{\F}{\mathbb F} \P^1(\F_q)$ can be written as the set of all pairs $[X_0:X_1]$ with $X_0, X_1 \in \F_q$ not both $0$. The point $[X_0:X_1]$ corresponds to the line through the origin $X_0 x + X_1 y = 0$ in $\mathbb{A}^2(\F_q)$. Since the equations $X_0 x + X_1y = 0$ and $\lambda X_0 x + \lambda X_1 y = 0$ define the same line for any $\lambda \neq 0$, then in terms of our coordinates we have $[X_0:X_1] = [\lambda X_0 : \lambda X_1]$ for all $\lambda \in \F_q^\times$. (It is in this sense that the coordinates are homogeneous.)
Note that if $X_1 \neq 0$ we can write $[X_0:X_1] = [X_0/X_1 : 1]$ and the set $$ \{[X_0:X_1] \in \P^1(\F_q) \mid X_1 \neq 0\} = \{[z:1] \mid z \in \F_q\} $$ can be identified with $\F_q$. Thus we have the stratification $$ \P^1(\F_q) = \{[z:1] \mid z \in \F_q\} \cup \{[1:0]\} = \F_q \cup \{\infty\} \, . $$
A matrix $\gamma = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in \mathrm{PSL}_2(\F_q)$ then acts by multiplication as you stated: $$ \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} X_0\\ X_1\end{pmatrix} = \begin{pmatrix} aX_0 + bX_1\\ cX_0 + dX_1\end{pmatrix} $$ or to emphasize the fact that we're still using homogeneous coordinates: $$ \begin{pmatrix} a & b\\ c & d \end{pmatrix} [X_0 : X_1] = [aX_0 + bX_1 : cX_0 + dX_1] \, . $$ But when $X_1 \neq 0$ we have $[X_0:X_1] = [z:1]$ where $z = X_0/X_1$, so this expression can be rewritten \begin{align*} \begin{pmatrix} a & b\\ c & d \end{pmatrix} [X_0:X_1] = [aX_0 + bX_1 : cX_0 + dX_1] = [a(X_0/X_1) + b : c(X_0/X_1) + d] = [az + b : cz + d] \, . \end{align*} If in addition we have $cz + d \neq 0$, then we can divide through $$ [az + b : cz + d] = \left[\frac{az + b}{cz + d} : 1 \right] $$ which, under the identification mentioned above, can identified with $\frac{az+b}{cz+d}$, recovering your original definition.
I may add something later about how locally $\P^k$ looks like $\mathbb{A}^k = \mathbb{F}_q^k$, whose elements are just $k$-tuples, but hopefully this at least answers your last question.