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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 $Y=P^1\rightarrow P^1/G$.

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Not sure whether this is what you are after, but one way to give a geometric meaning is the following.

You can think of $P^1{\mathbb C}$ as the "boundary at infinity" of the 3-dimensional hyperbolic space. To every "ideal tetrahedron" (i.e., a tetrahedron in hyperbolic 3-space with vertices at infinity) you can associate the cross ratio of its four vertices.

(The best way to think of the cross ration is probably that the cross ratio of the 4 points $0,1,z,\infty$ is $z$ and that the cross ratio is invariant under the action of $PSL(2,{\mathbb C})$. Since every non-degenerate 4-tuple is of the form $A0, A1, Az, A\infty$ for some $A\in PSL(2,{\mathbb C})$, this determines the cross ratio uniquely.)

So to every non-degenerate tetrahedron with vertices at infinity you get a cross ratio in $P^1{\mathbb C}$. (Actually for a non-degenerate terahedron you only get elements in $P^1{\mathbb C}\setminus\left\{0,1,\infty\right\}$.)

Now, the symmetric group $S_4$ acts by permutation on the four vertices of a tetrahedron. You get an induced action on the possible cross ratios (i.e., on $P^1{\mathbb C}\setminus\left\{0,1,\infty\right\}$) and you can check that for even permutations $\sigma\in A_4$ there are only three possibilities, either $z$ is mapped to itself, or to $$\frac{1}{1-z}$$ or to $$1-\frac{1}{z}.$$(The action extends in the obvious way to $\left\{0,1,\infty\right\}$.) Let $Stab(z)\subset A_4$ be the stabilizer of some $z$, then $Stab(z)\backslash A_4$ is isomorphic to ${\mathbb Z}/3{\mathbb Z}$.

So the natural action of $A_4\subset S_4$ by permuting vertices of tetrahedra factors over an action of ${\mathbb Z}/3{\mathbb Z}$ which is exactly the action you described above.

You'll find more detail in https://en.wikipedia.org/wiki/Cross-ratio#Six_cross-ratios_and_the_anharmonic_group

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  • $\begingroup$ Thank for taking time to answer .Can you give me some hint for second part of my question. $\endgroup$
    – Tensor_Product
    Commented Jul 18, 2016 at 8:10
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    $\begingroup$ What is the point of the second question? Obviously $T$ is invariant under the $Z/3Z$-action, so it gives a well-defined map to $P^1/G$. $\endgroup$
    – ThiKu
    Commented Jul 18, 2016 at 8:13
  • $\begingroup$ SIr , I think there are gaps in my understanding. What I was thinking that each $x\in P^1$ goes to $orbit(x)$ under the map. Then how $T$ plays a role here. I understand that $T$ is $G-invariant$. What I know is that given a group action we have orbit space which is just set of $G-$ orbits and it is denoted by $X/G$ $\endgroup$
    – Tensor_Product
    Commented Jul 18, 2016 at 8:26
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    $\begingroup$ You'd better think of $T$ as a map $P^1/G\to P^1$: all three points $z, \frac{1}{1-z}$ and $1-\frac{1}{z}$ are mapped to the same image. Of course one can compute that image, but I don't see a geometric interpretation of it. $\endgroup$
    – ThiKu
    Commented Jul 18, 2016 at 8:53

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