Finding poles of quadratic differentials

Question

Consider the following quadratic differential (on a Riemann surface):

$$ \phi_1\left(z\right)=\frac{P_4\left(z\right)}{\left(z-1\right)^2\left(z-a\right)^2}\frac{\mathrm{d}z^2}{z^2} $$

Here, $P_4\left(z\right)$ is a quartic polynomial (polynomial subscripts indicate degrees). This quadratic differential has second order poles at $z=0,1,a,\infty$. We can see this for $z=\infty$ by setting $w=1/z$; then $\mathrm{d}z^2/z^2=\mathrm{d}w^2/w^2$. This has poles of order two when $w=0$, i.e. when $z=\infty$.

Now consider the slightly more general quadratic differential with second order poles:

$$\phi_2\left(z\right)=\frac{P_4\left(z\right)}{Q_3^2\left(z\right)}\mathrm{d}z^2$$

The denominator of $\phi_2\left(z\right)$ is the same order as that of $\phi_1\left(z\right)$, but we now allow for the possibility that a factor of $z^2$ cannot be factored out. My question is: how many poles does $\phi_2\left(z\right)$ have, and where are they?

My approach to this question is as follows. Write:

$$\phi_2\left(z\right)=\frac{P_4\left(z\right)}{\left(\sum_{i=0}^3{q_i z^i}\right)^2}\mathrm{d}z^2=\frac{P_4\left(z\right)}{\left(\sum_{i=0}^3{q_i z^{i-1}}\right)^2}\frac{\mathrm{d}z^2}{z^2}$$

Then it seems that we have three second order poles as the roots of $Q_3\left(z\right)$, and as $z\rightarrow\infty$, by applying the same analysis in terms of $w$ as we did for $\phi_1\left(z\right)$, a second order pole at infinity. So we still have four second order poles. Is this correct?

Thanks in advance for any help! Hopefully it's an easy question to answer..

Answer 1

If $R=P/Q$ is a rational function, the quadratic differential $R(z)\,dz$ has:

• pole of order $m$ at $z_0\in \mathbb C$ if $R$ has pole of order $m$ at $z_0$.
• pole of order $m=\deg P-\deg Q + 4$ at $z_0=\infty$, provided $m\ge 1$. The term $+4$ comes from the square of the derivative of $z\mapsto 1/z$, as you observed.

Thus, in case of $R=P_4/Q_3^2$ are are assured to have a pole of order $4-6+4=2$ at infinity.

At a finite point $z_0$, a pole is possible only if $Q(z_0)=0$. But you should consider the possibility that $P_4$ also vanishes at $z_0$. The complete analysis is:

• if $P_4(z_0)\ne 0$, and $Q_3$ has a zero of order $k$ at $z_0$, then the quadratic differential has the pole or order $2k$ at $z_0$.
• if $P_4$ has a zero of order $m$ at $z_0$, and $Q_3$ has a zero of order $k$ at $z_0$, then the quadratic differential has the pole or order $2k-m$ at $z_0$, provided $2k-m\ge 1$.