# Constructing Riemann surfaces

At the risk of asking a question that has been already answered, I have been trying to figure out how to construct the Riemann surface of slightly more complicated examples, but after reading examples in Alfors; Churchill and Brown; Gamelin; Silverman and several existing examples here, I feel like I have a good intuition of examples such as $w=\sqrt[n]{z}$ or the logarithm, but I am unable to take that understanding and go much further.

For example, I would like to try to find a Riemann surface with an intersecting branch cut. It seems like a nice way to do this is to take a function like

$$f(z) = \sqrt{(z-1)(z+1)(z-i)(z+i)}.$$

From my perspective, $f(z)$ is a four-valued function everywhere except the points $z=1,-1,i,-i,\infty$ where it is only a single valued function. As such, I would need four sheets and from there my ability to go further promptly falls flat on its face.

I have a very strong inclination to construct my branch cut as a "+" shape connecting the four points $i,-i,1,-1$ respectively, but this seems like a poor choice for a few reasons:

1. I am defining a branch cut between -1 and 1, then another cut running between -i and i. The intersection seems bad to me, but I cannot justify whether or not it truly matters.
2. If I do this, it seems like we can traverse from any of the four sheets by passing through the origin. In that sense, if I define my branch cuts in this manner, I think I am introducing a new point where $f$ is four-valued at the origin. I think in that case, I have to then run a third branch cut connecting the origin to the point at infinity.
3. If I make this choice, I have no clue how to actually begin the gluing process for my sheets.

It seems like a smarter way might be to define four rays originating at the four points $i,-i,1,-1$ and sending them each off to infinity. But to remark on item 3 again, even if I do that I am not quite sure how to align these cuts (also I am not sure if we can choose our branch cut so that this makes sense). In the case of this four ray choice I just described, I think I have two options:

1. Construct glue my sheets such that starting on sheet #1, going around any of the points $1,-1,i,-i$ in a closed positively oriented curve will always take me to sheet 2, then sheet 3, then sheet 4. While going in the opposite orientation takes me to sheet 4, then 3, then 2 then 1. That is, from sheet #1, I can only travel to sheet 2 or 4, but I cannot skip to sheet 3 without traversing through 2 or 4 (but then what happens when I try to make a closed curve through the point at infinity?).
2. Alternatively, it seems like it might be possible to glue the sheets in such a way that I can traverse from sheet 1 to any of the 3 remaining sheets without visiting a different sheet to get there.

This motivates why I chose the four root as my example of interest: In the case of a cubic root, we can always traverse from one sheet to another without visiting any intermediary sheets.

Finally, using the Riemann-Hurwitz formula,

$$2-2g^{\prime} = n(2-2g)-\sum_{P}(e_{p}-1)$$

is causing me issues because it is not obvious to me how to determine the value of $g$. I think in the case of a complex plane, this value of $g$ must always be zero since we just have an complex plane which has no "holes".

Riemann surfaces of algebraic functions are exposed in the textbook "Forster, Otto: Riemann surfaces. Chapter I, Sect. 8 Algebraic functions." (I refer to the German edition from 1977 and take over its notations.) Existence of these Riemann surfaces is proved by means of holomorphic coverings. After presenting the general theorem Forster explicitely calculates the example $\sqrt {f(z)}$.

Your question asks for the Riemann surface $Y$ of $\sqrt {f(z)}$ with $f(z) = z^4-1$, notably for its genus $g(Y)$.

Hence we work with compact Riemann surfaces. We start with the Riemann surface $X = \mathbb P^1$ and the polynomial

$$P(T) = T^4 - f(z) \in \mathscr M(X)[T]$$

over the field $\mathscr M(X)$ of meromorphic functions on projective space $\mathbb P^1.$ Note that the holomorphic function $f(z)$ is a polynomial of degree $4$. It can be considered a meromorphic function on $X$ with a pole of order $4$ at $\infty \in X$.

The polynomial $P(T) \in \mathscr M(X)[T]$ is irreducible, because its discriminant $\Delta \in \mathscr M(X)$ does not vanish identically. By the general theory (Theorem 8.9) a compact Riemann surface $Y$, a proper holomorphic covering

$$\pi:Y \longrightarrow X$$

of degree $4$ and a meromorphic function $F \in \mathscr M(Y)$ with $(\pi^*P)(F) = 0$ exist. The latter means

$$F^4(y) = f (p(y)) \ for \ all \ y \in Y.$$

It explains why the triple $(Y,\pi, F)$ is named the Riemann surface defined by $P(T)$ and why the algebraic function $F$ is named the $4$-th root of f. The triple $(Y, \pi, F)$ is uniquely determined up to canonical isomorphy.

We compute the genus $g(Y)$ by the Riemann-Hurwitz formula

$$g(Y) = (b/2) + n * (g(X) - 1) + 1$$

with $n = 4$, $g(X) = 0$ and $b$ the total branching order of $\pi$. The image of the branch divisor of $\pi$ is the set

$$A := \{1,-1, i, -i, \infty\} \subset X.$$

For each $a \in A, a \neq \infty$, on a small simply connected neighbourhood $U_a \subset X$ of $a$ we have $f(z) = (z - a) * g_a(z)$ with

$$g_a(z) = \prod_{x \in A - \{a \} } (z-x)$$

holomorphic without zeroes. Hence a holomorphic function $h_a$ in $U_a$ exists with $g_a(z) = h_a^4(z)$. And for $z \in U_a$ holds

$$f(z) = (z-a) * h_a^4(z).$$

With $U_a^* := U_a - \{a \}$ the restriction $\pi \ | \pi^{-1}(U_a^*) \longrightarrow U_a^*$ is an unbranched 4-fold covering and $\# \pi^{-1}(a) = 1$. The branching order of $\pi$ at $y \in \pi^{-1}(a)$ is $4 -1 = 3$.

On the other hand, on a small simply connected neighbourhood $U_{\infty} \subset X$ of $\infty \in X$ we have $f(z) = z^4*g_{\infty}(z)$ with $g_{\infty}(z)= 1 - (1/z^4)$ holomophic and without zeroes. Hence

$$f(z) = h_{\infty}^4(z)$$

with a meromorphic function $h_{\infty}$ on $U_{\infty}$. The restriction $\pi \ | \ \pi^{-1}(U_{\infty}) \longrightarrow U_{\infty}$ splits into $4$ disjoint coverings, each of degree 1. We obtain $\# \pi^{-1}(\infty) = 4$, and $\pi$ has branching order $0$ at every point $y\in \pi^{-1}(\infty)$.

Summing up we get $b = 4 * 3 = 12$ and $g(Y) = 6 -4 + 1 = 3$.

• Thank you! I will have to read the reference and take careful look at this example. It has been a long time since I looked at this problem because I couldn't get anywhere for weeks. – JessicaK Oct 30 '14 at 6:17