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I am looking at the following Riemann surface (let's call it $M$), \begin{equation} y^n=\frac{(x-x_1)(x-x_3)}{(x-x_2)(x-x_4)} \end{equation} which is a Riemann surface of genus $n-1$. It can be thought of as a quotient of the complex plane by a Schottky group $\Sigma$, \begin{equation} M\cong\mathbb{C}'/\Sigma \end{equation} where $\mathbb{C}'$ is the domain of discontinuity of $\Sigma$. Then, we can look at the covering, \begin{equation} \pi_{\Sigma}:\mathbb{C}'\rightarrow \mathbb{C}'/{\Sigma} \end{equation} which takes $z$ to $[z]$. The inverse map $w=\pi^{-1}_{\Sigma}$ is multivalued on $\mathbb{C}'$, \begin{equation} w\sim\gamma w \end{equation} where $\gamma\in\Sigma$. However, the Schwarzian derivative of $w$ in some coordinate patch $z$ is single valued, \begin{equation} \{w,z\}=\frac{w'''}{w'}-\frac{3}{2}\left(\frac{w''}{w'}\right)^2. \end{equation} On overlapping coordinate patches, it transforms like, \begin{equation} \{w',z'\}=\left(\frac{dz}{dz'}\right)^2\{w,z\}+\{z',z\} \end{equation} which means it is a projective connection on $M$. Also, near the $z_i$s, since we can use the coordinate, $y^n\sim(z-z_i)$, \begin{equation} \{w',y\}\sim\{z,y\}=-\frac{1}{2}\frac{n^2-1}{y^2}+\cdots. \end{equation} According to Faulkner - The Entanglement Renyi Entropies of Disjoint Intervals in AdS/CFT, these properties are enough to fix the Schwarzian derivative up to $3g-3$ unknown accessory parameters, \begin{equation} \{w,z\}=\Delta\left(\sum_{i=1}^4\frac{1}{(z-z_i)^2}+\frac{-z_3+z_1+z_2+z_4-2z}{(z-z_1)(z-z_2)(z-z_4)}\right)+\sum_{s=1}^{3(n-2)}p_s\omega_s \end{equation} where $\omega$s are a basis for the quadratic differentials on $M$ and $\Delta=1/2(1-1/n^2)$. I understand that we can add any linear combination of quadratic differntials on $M$ to the form of the Schwarzian because they transform uniformly, but the first term in the brackets is not clear to me. It should transform like a projective connection. Does anyone have a justification for including the first term multiplying $\Delta$?

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First of all, you should take care about which coordinates you take. Namely, Faulkner uses the coordinate $x$ given by the defining equation $$y^n=\frac{(x-x_1)(x-x_2)}{(x-x_3)(x-x_4)}$$ as a branched projective structure (he denotes this by $z$ in his paper!). He then computes the Schwarzian derivative of $w$ - the Schottky uniformisation - with respect to $x$. As $x$ is a well-defined branched projective structure $${w,x}(dx)^2$$ is a well-defined meromorphic differential on $M$, having prescribed singularities at $x_1,\dots,x_4$. He derives in equation (4.10) in his paper that the quadratic residues must be $\Delta$. This is well-known since Mandelbaum.

The second term in the brackets is needed as ${w,x}(dx)^2$ has no pole over $x=\infty.$ It might look a bit unnatural as it has no pole at $z_3$, but in fact $\frac{(dx)^2}{(x-x_1)\dots(x-x_4)}$ is a holomorphic quadratic differential.

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  • $\begingroup$ Is there any way I can check that the term multiplying $\Delta$ is a projective connection? The last terms are quadratic differentials, so for the whole thing to be a projective connection, the first term needs to be one. $\endgroup$ Sep 25, 2021 at 8:55
  • $\begingroup$ I don't know exactly what you mean by your question. The term $q=\{w,x\}$ is not a connection but a meromorphic quadratic differential (considered on $M$ by pull-back), the Schwarzian of $w$ with respect to the branched projective structure $x.$ But you can use $q$ this to define a (smooth) complex projective structure on $M$. Consider any complex structure, e.g. the uniformisation. With respect to this the projective structure $x$ is given by a meromorphic quadratic differential $Q$. Its quadratic residues are the negatives of the quadratic residues of $q$. $\endgroup$
    – Sebastian
    Sep 25, 2021 at 10:41
  • $\begingroup$ Moreover, it can be checked by computation that $q+Q$ is holomorphic. Hence, $q$ defines a smooth complex projective structure on $M.$ The corresponding flat connection is given by adding $q+Q$ (in the usual way) to the flat connection giving the chosen projective structure (e.g. uniformisation). Note that in certain special cases (e.g. $x_k$ being 4th roots of 1) , one can uniquely characterize the uniformisation projective structure, i.e. determine the accessory parameters $\sum p_s\omega_s$. It might also work for Schottky uniformisation, by using additional symmetries. $\endgroup$
    – Sebastian
    Sep 25, 2021 at 10:48

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