Suppose I have the following quadratic differential on the Riemann sphere with four punctures:
\begin{equation} q = -\frac{9 t \left(216+t^3\right)}{\left(-27+t^3\right)^2} dt^2 \end{equation}
This has four zeroes, and four second-order poles (one for each of the punctures).
Now, since $q$ is on the sphere, we should be allowed to do an $SL\left(2, \mathbb{C}\right)$ reparameterisation, of the form $t \rightarrow \frac{az+b}{cz+d}$. So, to test, choose $a=c=b=1$, and $d=2$. Then, using the quadratic differential transformation rule that if $q = f\left(z\right)dz^2 = g\left(w\right)dw^2$, then
\begin{equation} f\left(z\right) = g\left(w\left(z\right)\right)\left(\frac{d w\left(z\right)}{dz}\right)^2 , \end{equation}
I find
\begin{equation} q = -\frac{9 \left(1729+4324 z+3894 z^2+1516 z^3+217 z^4\right)}{(2+z)^2 \left(215+321 z+159 z^2+26 z^3\right)^2} dz^2 . \end{equation}
This now has five second-order poles. But it should still be living on a sphere with four punctures, so it looks like it has picked up another pole. This confuses me; I had really hoped the number of poles would be invariant under such reparameterisations.
So, my question is:
Is such an $SL\left(2,\mathbb{C}\right)$ transformation on a quadratic differential on a Riemann sphere with punctures actually legitimate? If not, why not?
If the transformation is legitimate, what is the explanation for the extra pole in the quadratic differential after the transformation?
Thanks in advance for any help!
The answer to the question, as Alan Muniz points out, is that such $SL\left( 2, \mathbb{C} \right)$ transformations of the quadratic differentials are legitimate, and they do preserve the number of poles of the quadratic differential. The problem in the above is that I incorrectly counted five poles in $q\left(z\right)$, when in fact there are only four - since $q\left(z\right)$ is holomorphic at $\infty$.