Action of $\mathbb{Z}/3\mathbb{Z}$ on $P^{1}$ I am reading from the book Topics in Galois theory by Serre.
I have the following question ,
take $G=\mathbb{Z}/3\mathbb{Z}$. The group $G$ acts on $P^1$ by 
$$\sigma x\;=\;1/(1-x)$$
where $\sigma$ is generator of $G$.
Am I interpreting this action correctly. I am thinking of it as following, think $P^1$ as extended complex plane and think x as a complex number. I am not able to interpret this action geometrically thinking of $P^1$ as set of lines.
If we write $T=x+ \sigma x + \sigma^{2} x$. How $T$ gives a map $Y=P^1\rightarrow P^1/G$.
 A: Thinking first of $\mathbb{C}^2$, the action of the group $G = \mathbb{Z}/3\mathbb{Z}$ is generated by the linear transformation 
$$\begin{pmatrix}0 & 1 \\ -1 & 1 \end{pmatrix} \cdot \begin{pmatrix}w \\ z \end{pmatrix} = \begin{pmatrix}z \\ -w+z  \end{pmatrix}
$$ 
Under the projection map $\mathbb{C}^2 - \{0\} \to P^1$, all the points in a single "line" $aw+bz=0$ get mapped to a single element of $P^1$. If $a \ne 0$ then we can rewrite this equation as $w = - \frac{b}{a} z$, set $\zeta = -\frac{b}{a}$ (the "minus slope"), rewrite the equation of the line as $w=\zeta z$, and instead map that line to $\zeta \in \mathbb{C} \subset \mathbb{C} \cup \{\infty\}$. 
If instead $a=0$ we map the line (whose equation may be rewritten $z=0$) to $\infty$. 
When this is done, the point $\frac{w}{z}=\zeta$ is mapped to the point $$\frac{z}{-w+z} = \frac{1}{-\frac{w}{z}+1} = \frac{1}{-\zeta+1}
$$
In other words, the action is generated by $\sigma \zeta = \frac{1}{-\zeta + 1}$, where your argument $x$ has been changed to the "minus slope" variable $\zeta$.
Thus, the $\mathbb{C} \cup \{\infty\}$ model for $P^1$ can be thought of as "minus slope space".
Let me now stick with your variable $x$. Let me explain how to understand the formula
$$\sigma x = \frac{1}{-x+1}
$$
One thing to do is to find the fixed points
$$\sigma x = x
$$
Substituting the formula we obtain
$$\frac{1}{-x+1}=x
$$
and solving we obtain
$$x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i
$$
Now, working back to the $P^1$ model, you can think of the two roots of this equation as the "minus slopes" of the two eigenspaces of the linear transformation $\begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}$. One eigenspace is
$$w = - \bigl(\frac{1}{2} + \frac{\sqrt{3}}{2}i\bigr) \, z
$$
and the other eigenspace is similar but with the slope changed to its complex conjugate.

Now you ask about the map $T = x + \sigma x + \sigma^2 x$. Here, I must admit, I am unsure what you are looking for, but I will make some guesses.
In the $\mathbb{C} \cup \{\infty\}$ model, where $x$ is "minus slope", this formula for $T$ makes no sense: adding "minus slopes", or adding ordinary "slopes" for that matter, has no meaning. 
On the other hand this map does makes sense in $\mathbb{C}^2$ using vector addition, where $x$ is replaced by a vector variable $\vec v = \begin{pmatrix}w \\ z \end{pmatrix}$ and where $\sigma$ is replaced by the matrix $M = \begin{pmatrix}0 & 1 \\ -1 & 1 \end{pmatrix}$. Since $M$ is a linear transformation taking lines to line, it therefore also makes sense in the space of lines $P^1$. 
The point seems to be that $T\sigma = T(M \sigma)$ for all $\sigma$ (do the calculation to verity that $T=TM$). So $T$ can be thought of as a map $T : P^1 \to P^1$ having the property that it maps each $G$-orbit in the domain $P^1$ to a single point in the range $P^1$. Thus the orbit space $P^1 / G$ is shown, using $T$, to be equivalent to $P^1$ itself.
