Is there such a thing as an infinity vector?

Question

I've been see the following question on group theory:

Let $p$ be a prime, and let $G = SL2(p)$ be the group of $2 \times 2$ matrices of determinant $1$ with entries in the field $F(p)$ of integers $\mod p$.

(i) Define the action of $G$ on $X = F(p) \cup \{ \infty \}$ by Mobius transformations. [You need not show that it is a group action.]

State the orbit-stabiliser theorem. Determine the orbit of $\infty$ and the stabiliser of $\infty$. Hence compute the order of $SL2(p)$.

I know matrices are isomorphic to Mobius maps, but not how the action of a mobius map can be used to define the action of a matrix (I don't really know what this part means to be honest). I tried the next part, but wasnn't sure whether the consider the vector $(\infty,\infty)$, $(\infty,a)$ or $(b,\infty)$ $(a,b \in F(p))$. Any help would be greatly appreciated!!

(Sorry the question title isn't very related to the question, I just didn't know what to put specifically!)

Answer 1

Use homogeneous coordinates. A homogeneous coordinate vector $[x:y]$ represents the number $x/y$. The vector $[x:0]$ for any $x\neq 0$ represents infinity. Conventionally one would write infinity as $[1:0]$ but the choice of representative is arbitrary.

Here is how the matrix operation relates to the Möbius map:

$$ \begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot\begin{pmatrix}x\\y\end{pmatrix} \qquad \frac xy\mapsto \frac{ax+by}{cx+dy} =\frac{a\frac xy+b}{c\frac xy+d} $$

So your regular $z\mapsto\frac{az+b}{xz+d}$ is modeled nicely in this way, as long as $y\neq 0$.

Answer 2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$ You have $$f(z) = \frac{a z + b}{c z + d},$$ and $$f(\infty) = \frac{a \infty + b}{c \infty + d} = \frac{a}{c}.$$

This is best understood by letting the group act on the projective line $$ \mathcal{P} = \Set{[x, y] : x, y \in F(p), (x, y) \ne (0, 0)}, $$ by $$ f([x, y]) = [a x + b y, c x + d y], $$ where $[x, y] = \Set{\lambda (x, y) : \lambda \in F(p)^{*} }$. In this notation, $\infty = [1, 0]$.

Answer 3

Let $F$ be any field. The group ${\rm GL}_2(F)$ (and thus any of each subgroups) has a natural linear action on the $F$-vector space $F^2$. By linearity, this action descends to an action on the projective line ${\Bbb P}^1(F)$ which is explicitely given by $$ \begin{pmatrix} a & b\\ c & d\end{pmatrix}\cdot z=\frac{az+b}{cz+d} $$ Here $z\in F\cup\{\infty\}$ represents a homothety class of a vector in $F^2$ and $z=\infty$ is nothing but the class of the vector $(1,0)$.