Riemann surface associated with complete analytic function of $(z^2-1)^{1/3}$ I'm trying to define an analytic function on '$\mathbb{C}$' of the form $f(z)=(z^2-1)^{1/3}$, i.e. I first remove two semi-infinite rays $l_1$ and $l_2$, one going from $1$ to $\infty$ along the positive reals, one from $-1$ to $\infty$ along the negative reals, ending up with the domain $U=\mathbb{C}\setminus(l_1\cup l_2)$.
Then I have three choices for $f$, and they will differ by factors of $e^{2\pi i/3}$ and $e^{4\pi i/3}$. Since $U$ is simply-connected, such a function is well-defined after making the initial choice.
First question: This is pretty vague. How can I justify this?
Second question: If I now want to describe the Riemann surface associated with the complete analytic function of $(z^2-1)^{1/3}$, how can I do so using sort of a gluing procedure? I start with three copies of $U$, but how can I glue them together?
 A: First question: On a simply connected domain $G$, every holomorphic function $f$ without zeros has a holomorphic logarithm, that is, there is a $g \in \mathscr{O}(G)$ with $f(z) = e^{g(z)}$ for all $z\in G$. For the logarithmic derivative of $f$ is holomorphic, and since $G$ is simply connected, it has a primitive $h$ on $G$. Then
$$\frac{d}{dz}\left(f(z)e^{-h(z)}\right) = f'(z)e^{-h(z)} - f(z)h'(z)e^{-h(z)} = e^{-h(z)}\left(f'(z) - f(z)\frac{f'(z)}{f(z)}\right) = 0,$$
so $f(z) = c\cdot e^{h(z)}$ with $c\neq 0$. Hence $g(z) = h(z)+\log c$ is - for every choice of $\log c$ - a holomorphic logarithm of $f$.
In our situation, the function $h(z) = z^2-1$ has no zeros in $U$, and therefore a holomorphic logarithm $g$. Then $e^{g(z)/3}$ is a branch of $\sqrt[3]{z^2-1}$ on $U$.
It is clear that if $f$ is a branch of $\sqrt[3]{z^2-1}$ on any domain then $g(z) = e^{2\pi i/3}f(z)$ and $h(z) = e^{4\pi i/3}f(z)$ define two other branches. Since every $w \neq 0$ has exactly three distinct cube roots, any branch of $\sqrt[3]{z^2-1}$ must agree with one of the three at a given point $z_0$ in the domain, and by continuity, in a neighbourhood of $z_0$. By the identity theorem, it must therefore be one of $f,g$ or $h$.
Second question: Let's call the three copies of $U$ by the names of $U_0,U_1$ and $U_2$. Let $f$ be the branch of $\sqrt[3]{z^2-1}$ on $U$ with $f(0) = -1$, and on $U_k$ choose the branch $f_k(z) = e^{2k\pi i/3}\cdot f(z)$.
Then we need to glue the $U_k$ together along the branch cuts, and we must determine which sheet must be glued to which along each cut. For that, we must find the argument of $f_k(z)$ as $z$ approaches the branch cuts from the upper respectively from the lower half-plane. On the sheet $U_0$, we have $f_0(it) < 0$ for all $t\in \mathbb{R}$. As $z$ traverses a quarter-circle from the imaginary axis to a branch-cut, $z^2-1$ traverses a half-circle, and $\sqrt[3]{z^2-1}$ a third of a half-circle, that is, an angle of $\frac{\pi}{3}$. Thus the argument when approaching the branch-cut on the positive ray from the upper half-plane or the branch-cut on the negative ray from the lower half-plane ($f_0$ is an even function) is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$. The argument when approaching the branch-cuts from the other half-plane is $\pi + \frac{\pi}{3} = \frac{4\pi}{3}$. Writing the arguments in the order of quadrants, and noting that all arguments are increased by $\frac{2\pi}{3}$ when going from one sheet to the next, we get
$$\begin{matrix}
U_0 & \frac{2\pi}{3} & \frac{4\pi}{3} & \frac{2\pi}{3} & \frac{4\pi}{3}\\
U_1 & \frac{4\pi}{3} & 0 & \frac{4\pi}{3} & 0\\
U_2 & 0 & \frac{2\pi}{3} & 0 & \frac{2\pi}{3}
\end{matrix}$$
so we must glue


*

*the second quadrant of $U_0$ to the third quadrant of $U_1$,

*the fourth quadrant of $U_1$ to the first quadrant of $U_2$,

*the second quadrant of $U_2$ to the third quadrant of $U_0$,

*the fourth quadrant of $U_0$ to the first quadrant of $U_1$,

*the second quadrant of $U_1$ to the third quadrant of $U_2$,

*the fourth quadrant of $U_2$ to the first quadrant of $U_0$.

A: A more direct way, without any glueing, is to look directly at the locus $E$ of $y^3=z^2-1$ in $\mathbf C^2$. This equation defines a smooth affine curve, and the projection $(y,z) \mapsto z$ is locally an isomorphism away from the points $(0,1)$ and $(0,-1)$. Moreover this projection is generically three-sheeted. On $E$, the function $y$ satisfies $y^3=z^2-1$, i.e., morally $y$ 'is' $(z^2-1)^{1/3}$, and the function $y$ restricted to each "sheet" gives the three branches of your function.
