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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.

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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.

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Quasicoherent gave a brief but nice explanation about homogeneous coordinates and I think that Wikipedia's page about it is a good place to begin. To answer why $\mathrm{PSL}(n, \mathbb{F}_q)$ act just like this, I'm going to take a group theoretic approach.

Let $\mathbb{K}$ be a field and $\mathbb{V}$ a $\mathbb{K}-$vector space. Let $u, v \in \mathbb{V}\setminus\{0\}$, we say they are equivalent and denote $u \sim v$ if there is $\lambda \in \mathbb{K}^{\times} = \mathbb{K}\setminus{0}$ such that $u = \lambda v$. This is a equivalence relation. We define the projective space associated with $\mathbb{V}$ as the quotient set $\mathbf{P} \mathbb{V} = (\mathbb{V}\setminus\{0\})/\sim$. Fine, but note that equivalente classes are just the orbits of $\mathbb{K}^{\times}$ in $\mathbb{V}\setminus\{0\}$ in que obvious action $$:\mathbb{K}^{\times} \times (\mathbb{V} \setminus\{0\}) \to \mathbb{V} \setminus\{0\}: (\lambda, v) \mapsto \lambda v$$ So, we have $$\mathbf{P} \mathbb{V} = \frac{\mathbb{V}\setminus\{0\}}{\mathbb{K}^{\times}}$$ Yeah, it's a nice picture.

We defined what this strange $\mathbf{P}$ do with vector spaces (objects). Of course we want this to be a functor: we can define the category of projective spaces based on a incidence structure with collineations; this is the synthetic point of view. Ok, let's go back to real world and define this $\mathbf{P}$ to linear maps.

Let $\mathbb{W}$ other $\mathbb{K}-$vector space and $T \in \mathrm{Mon}(\mathbb{V}; \mathbb{W})$ a one-to-one linear map (we need the 1-1 condition to eliminate de null vector). Define $$\mathbf{P}T: \mathbf{P}\mathbb{V} \to \mathbf{P}\mathbb{W} \,: [v] \mapsto \mathbf{P}T[v] = [Tv]$$ We can check it's well defined: $$\mathbf{P}T[\lambda v] = [T(\lambda v)] = [\lambda Tv] = [Tv] =\mathbf{P}T[v]$$ Finally, we define $$\mathbf{P}GL(\mathbb{V}) = \mathbf{P}(GL(\mathbb{V}))$$ (I love this definition, it's so funny) Of course you can do the same to $SL(\mathbb{V})$ and to $U(\mathbb{V})$ and $SU(\mathbb{V})$ if $\mathbb{V}$ is a euclidian space (i.e., have a inner product).

This is why the action in homogeneous coordinates is just like you said: take a representative (a vector in the $\mathbb{K}^{\times}-$orbit), act your map like a linear operator (matrix multiplication) and then take the homogeneous coordinates. This is exactly what Quasicoherent did and this is the way to obtain the fractional maps you want to recover. Note that I said nothing about the field $\mathbb{K}$ or about the $\mathbb{K}-$dimension of $\mathbb{V}$, it's really general. In fact, you can do the same thing in modules over division rings (we need division to make $\sim$ transitive).

Let's continue just a little bit. The natural question is: let $T, Q \in \mathrm{Mon}(\mathbb{V}; \mathbb{W})$, if $\mathbf{P}T = \mathbf{P}Q$, then what? It's easy to prove that $\mathbf{P}T = \mathbf{P}Q$ iff there is $\lambda \in \mathbb{K}^{\times}$ such that $T = \lambda Q$. YEAH, ORBITS. We got $$\mathbf{P}GL(\mathbb{V}) \simeq \frac{GL(\mathbb{V})}{\mathbb{K}^{\times}}$$ In another point of view, this is just a central extension $$1 \to \mathbb{K}^{\times} \to \mathbf{P}GL(\mathbb{V}) \to GL(\mathbb{V}) \to 1$$ It is really important to study projective representation. If you prefer group quotients in the usual sense (by a normal subgroup and so on ...), see that $\mathbb{K}^{\times}$ is isomorphic with the center of $GL(\mathbb{V})$, so $$\mathbf{P}GL(\mathbb{V}) \simeq \frac{GL(\mathbb{V})}{Z(GL(\mathbb{V}))}$$ And this is the way to generalize this to modules.

You can check that $\mathbf{P}$ is a (covariant) functor.

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  • $\begingroup$ Probably I got carried away and it was really abstract, but this group stuff together with homogeneous coordenates is a good way to understand what is going on. $\endgroup$ – Thadeu Henrique Costa Oct 5 '17 at 1:29

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