Given \(A,B,C\in\mathbb C\), prove that there is a unique quadratic differential \[ q=q(z)\,dz^2 \] on \(\widehat{\mathbb C}\) having double poles at \(0\), \(1\), and \(\infty\), residues \(A\), \(B\), and \(C\) there respectively, and no other singularities.
Here, if \[ q=\left(\frac{a}{z^2}+O\!\left(\frac1z\right)\right)dz^2 \] near \(0\), then \(\operatorname{Res}_0(q)=a\).
Solution
The unique differential is \[ \boxed{ q= \left( \frac{A}{z^2} +\frac{B}{(z-1)^2} +\frac{C-A-B}{z(z-1)} \right)dz^2. } \] Here ``double poles’’ must be understood as poles of order at most two: if one of \(A,B,C\) vanishes, the corresponding quadratic term may disappear, as is unavoidable for arbitrary prescribed residues.
1. The displayed differential has no finite singularities except possibly at \(0\) and \(1\), and its residues there are \(A\) and \(B\).
Near \(z=0\), \[ \frac{B}{(z-1)^2}=O(1), \qquad \frac{C-A-B}{z(z-1)}=O\!\left(\frac1z\right). \] Thus \[ q= \left( \frac{A}{z^2}+O\!\left(\frac1z\right) \right)dz^2, \] so \[ \operatorname{Res}_0(q)=A. \]
Similarly, with \(u=z-1\), \[ \frac{A}{z^2}=O(1), \qquad \frac{C-A-B}{z(z-1)}=O\!\left(\frac1u\right), \] and hence \[ q= \left( \frac{B}{u^2}+O\!\left(\frac1u\right) \right)du^2. \] Therefore \[ \operatorname{Res}_1(q)=B. \]
2. The residue at infinity is \(C\).
As \(z\to\infty\), \[ \frac{1}{(z-1)^2} =\frac1{z^2}+O(z^{-3}), \qquad \frac{1}{z(z-1)} =\frac1{z^2}+O(z^{-3}). \] Consequently \[ q =\left( \frac{A+B+(C-A-B)}{z^2}+O(z^{-3}) \right)dz^2 =\left( \frac{C}{z^2}+O(z^{-3}) \right)dz^2. \]
To translate this into the local coordinate \(w=1/z\) at infinity, note that \[ dz^2=w^{-4}dw^2. \] Thus \[ q =\left( \frac{C}{w^2}+O\!\left(\frac1w\right) \right)dw^2, \] so \[ \operatorname{Res}_\infty(q)=C. \]
3. Any quadratic differential with the stated singularity restrictions has the form\[ q=\frac{P(z)}{z^2(z-1)^2}\,dz^2, \qquad \deg P\le2. \]
Since the only finite poles are at \(0\) and \(1\), each of order at most two, \[ z^2(z-1)^2q(z) \] is an entire function. Because \(q\) has at most a double pole at infinity, its coefficient in the \(z\)-coordinate satisfies \[ q(z)=O(z^{-2}) \qquad(z\to\infty). \] Therefore \[ z^2(z-1)^2q(z)=O(z^2). \] An entire function of growth \(O(|z|^2)\) is a polynomial of degree at most two. Hence the asserted form follows.
4. The three prescribed residues determine \(P\) uniquely.
Suppose a differential of the form in 3 has all three residues equal to zero. Write \[ P(z)=az^2+bz+c. \] Near \(0\), the coefficient of \(z^{-2}\) is \[ P(0)=c, \] so residue zero at \(0\) gives \(c=0\).
Near \(1\), the coefficient of \((z-1)^{-2}\) is \[ P(1)=a+b+c, \] so residue zero at \(1\) gives \[ a+b+c=0. \]
At infinity, \[ \frac{P(z)}{z^2(z-1)^2} =\frac{a}{z^2}+O(z^{-3}), \] so residue zero at infinity gives \(a=0\). Hence \(c=0\), \(a=0\), and then \(b=0\). Thus the only differential with all three residues zero is the zero differential.
The difference of any two solutions with residues \(A,B,C\) has zero residues, so it must vanish. This proves uniqueness, while <1>1–<1>2 prove existence.