0
$\begingroup$

Given a map $f:\mathbb{CP^1} \leftarrow \mathbb{CP^1}$ by $f(z)=\frac{4z^2(z-1)^2}{(2z-1)^2}$ Find all branching points and their degrees.

If my calculation is correct I got 4 branching points and in each point degree of $f$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $f$ is 4. But if I directly compute total branching number I get 4. Where did I make a mistake?

$\endgroup$
1
$\begingroup$

$f(z_0) = \infty$ is in fact a branch point. There will be six ramification points $z_0$: $$\begin{matrix} z_0 & f(z_0) \\ \hline 0, 1 & 0 \\ \frac {1 \pm i} 2 & -1 \\ \frac 1 2, \infty & \infty \end{matrix},$$ each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $\pm 2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.