Local galois involution

Question

Let $x(z):=-\frac{(z+1)^3}{z}$ is a meromorphic function from $CP^1\rightarrow \mathbb{C}$. At the point $z=1/2$ and $z=-1$, $dx=0$ hence these are ramifications point. I want to study the Galois involution around this point. Galois involution is maps which are often locally defined around the ramified point it might not be globally defined. Let $s_{1/2}(z)$ is a Galois cover if $x(s_{1/2}(z)=x(z)$. $s_{1/2}$ is locally defined around some open cover of $1/2$ and $s_{1/2}(1/2)=1/2$ Similarly for $s_{-1}(z)$.

I have the following two involutions first around $1/2$. $$ s_{1/2}(z) := \frac12\Bigg(-z-3+\sqrt{z+3)^2+4/z} \Bigg) $$

around 1 $$ s_{-1} (z):= \frac12\Bigg(-z-3-\sqrt{z+3)^2+4/z} \Bigg) $$

It can be checked that that it has all the property defined above. I am having trouble with the involution $s_{-1}(z)$. At the point $z=-1$ $x(z)$ look like $a^3$ in some coordinate which can be checked by Taylor series expansion around $z=-1$. So the Galois involution $s_{-1}(z)$ locally would look like $\omega z $ where $\omega$ is the cube root of unity. When I try to expand to see how $s_{-1}(z)$ look around $z=-1$ it says that it has no Taylor series expansion which worries me. I was expecting locally it should look like $\omega z $.

Please explain what is going wrong?

Answer 1

Your definition of $s_a(z)$ is unclear. If $x(z)$ is holomorphic then locally around $a$ there is $n \ge 1$ such that

$$x(a+z) = x(a)+ z^n P(z), \qquad P(0) \ne 0$$ Let $$\varphi(z) = z P(z)^{1/n} $$ which is locally biholomorphic. Since $z= \varphi^{-1}(z) P(\varphi^{-1}(z))^{1/n}$ then $$x(a+\varphi^{-1}(z)) = x(a)+z^n = x(a)+(\zeta_n z)^n = x(a+\varphi^{-1}(\zeta_n z))$$ And hence $$x(a+z) = x(a+s_a(z)), \qquad s_a(z) = \varphi^{-1}(\zeta_n \varphi(z))$$

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With $a= -1$ and $x(z) = \frac{(z+1)^3}{z}$ you get $$x(-1+z) = z^3P(z), \qquad P(z) = \frac{-1}{1-z}, \qquad \varphi(z) = z \left(\frac{-1}{1-z}\right)^{1/3}$$ See WA for $\varphi^{-1}$'s closed form, and the (local) involution is $$s_{-1}(z) = \varphi^{-1}(\zeta_3 z \left(\frac{-1}{1-z}\right)^{1/3}),\qquad x(-1+z) = x(-1+s_{-1}(z))$$

Answer 2

Let $z = -1+w$ and consider small values of $w$.

Then
$(z+3)^2+4/z = (w+2)^2- \frac 4{1-w} \\ = (w^2+4w+4)-4\frac{(1-w)(1+w+w^2)+w^3}{1-w} \\ = (w^2+4w+4)-4(1+w+w^2)-4\frac {w^3}{1-w} \\ = -3w^2-4\frac {w^3}{1-w} \\ = -3w^2(1+ \frac 43 \frac w{1-w}) \\ $

Let $g(w)$ be the square root of $(1+ \frac 43 \frac w{1-w})$ defined near $w=0$ for which $g(0) = +1$. That is, $g(w) = 1 + O(w)$.

Then the expression above has two square roots, $r_+(w) = +i\sqrt 3g(w)$ and $r_-(w) = -i\sqrt 3 g(w)$

Then $\frac 12(-(-1+w)-3+r_\pm(w)) = -1 + \frac 12(-w+r_\pm(w)) = -1+\frac12(-1\pm i\sqrt 3)w + O(w^2)$

And this is exactly what you would expect. And of course this is not an involution near $-1$.