Let \(D=\{w\in\mathbb C:|w|<1\}\), and let \(g:D\to D\) be a surjective holomorphic map with \(g(0)=0\). Suppose that \(g\) is a two-sheeted branched covering, with its only ramification point at \(w=0\). An example is \(g(w)=w^2\).
Express the given conditions explicitly in terms of \(g\) and its derivatives.
Show that \(|g(w)|\leq|w|^2\) for all \(|w|<1\).
Suppose that \(g(1/2)=i/4\). What is the strongest statement about \(g(w)\) that follows from (b)? Explain.
Solution
1. Part (a): derivatives determine local multiplicity, while fiber cardinalities specify the two sheets.
For a holomorphic map \(g:D\to D\), the explicit conditions are \[ g(0)=0,\qquad g'(0)=0,\qquad g''(0)\ne0,\qquad g'(w)\ne0\quad(0<|w|<1), \] together with \[ g^{-1}(0)=\{0\},\qquad \#g^{-1}(z)=2\quad(z\in D\setminus\{0\}). \] The fibers over nonzero values consist of distinct points. Equivalently, every value in \(D\) has two preimages counted with multiplicity, and the only multiple point of a fiber is the double zero at zero. The fiber conditions include surjectivity; derivative conditions alone would not specify the number of sheets.
To verify the equivalence, a two-sheeted holomorphic branched covering has local multiplicity two at its ramification point and multiplicity one elsewhere. The Taylor series translates these assertions into the displayed derivative conditions. The double preimage over zero uses both sheets, so there can be no other preimage of zero; a nonbranch value has exactly two simple preimages.
Conversely, suppose the displayed conditions hold. At each of the two preimages of a nonzero value, the holomorphic inverse function theorem gives a local inverse [@SS03]. Choose disjoint inverse neighborhoods and intersect their image neighborhoods. Every value of this intersection already has its two preimages in these neighborhoods; the fiber count excludes any others. Thus the intersection is evenly covered by two sheets.
Near zero write \(g(w)=w^2h(w)\), where \(h\) is holomorphic and \(h(0)=g''(0)/2\ne0\). On a small disk, \(h\) is nonzero and has a holomorphic square root \(s\) [@SS03]. The local coordinate \(v=ws(w)\) has nonzero derivative at zero and puts \(g\) in the form \(v^2\). A sufficiently small punctured target disk has both its preimages in this coordinate neighborhood; the fiber count again excludes any others. This gives precisely the required single quadratic branch.
2. Part (b): division by the double zero gives the bound.
Since \(g(0)=g'(0)=0\), the function \(H(w)=g(w)/w^2\) for \(w\ne0\) extends holomorphically across zero by the Taylor series, with \(H(0)=g''(0)/2\). For \(0<r<1\), on \(|w|=r\) one has \(|H(w)|=|g(w)|/r^2\leq r^{-2}\), because \(g(D)\subset D\). The maximum modulus principle therefore gives \(|H(w)|\leq r^{-2}\) throughout \(|w|\leq r\) [@SS03]. For a fixed \(w\in D\), let \(r\uparrow1\) with \(r>|w|\). It follows that \(|H(w)|\leq1\), and hence \[ |g(w)|=|w|^2|H(w)|\leq|w|^2. \] This argument uses only the disk bound and the double zero, not the stronger global covering hypotheses.
3. Part (c): the extra value determines the entire map.
The assumed value gives \(H(1/2)=(i/4)/(1/4)=i\). Thus the holomorphic function \(H\), bounded by one, attains modulus one at an interior point. The maximum modulus principle makes \(H\) constant [@SS03], and that value fixes the constant as \(i\). Consequently \[ \boxed{g(w)=iw^2\quad\text{for all }w\in D.} \] This function satisfies every hypothesis: it has only the double zero at zero, nonzero derivative elsewhere, and exactly two distinct disk preimages of each nonzero disk value. It also has the prescribed value at \(1/2\).