# Why is a neighbouthood of a Conical singularity isometric with $\mathbb{C}$?

I'm starting to study flat surfaces with conical singularities and I have some trouble.. In particular I'm reading "On the moduli space of singular euclidean surfaces" by Marc Troyanov. In (1.7) he states that given a piecewise flat surface $S$ with riemannian metric $m$, in a neighbourhood of a conical singularity $p$ of total angle $\theta$, with polar coordinates $r\in\mathbb{R}$, $\varphi\in \mathbb{R}/\theta\mathbb{Z}$, the metric reads $m=dr^2+r^2d\varphi^2$.

He states that $S$ near $p$ is isometric to $\mathbb{C}$ with metric $m_{\beta}=|z|^{2\beta}|dz|^2$ , $\beta=\frac{\theta}{2\pi}-1$, the isometry being given by $f(r,\varphi)=z=\frac{1}{\beta+1}(re^{i\varphi})^{\beta+1}$.

The problem is that, according to my calculations, this is not an isometry. In particular I tried to verify that $m_{(r,\varphi)}(\frac{\partial}{\partial r},\frac{\partial}{\partial r})=f^*{m_{\beta}}_{(r,\varphi)}(\frac{\partial}{\partial r},\frac{\partial}{\partial r})$

But it results:

$|z|^{2\beta}=(\frac{2\pi}{\theta})^{(\frac{\theta}{\pi}-2)}r^{(\frac{\theta^2}{2\pi^2}-\frac{\theta}{\pi})}$

$dz=r^{(\frac{\theta}{2\pi}-1)}((cos(\frac{\theta}{2\pi}\varphi)dr-rsin(\frac{\theta}{2\pi}\varphi)d\varphi)+i(sin(\frac{\theta}{2\pi}\varphi)dr-rcos(\frac{\theta}{2\pi}\varphi)d\varphi))$

$f^*(|z|^{2\beta}|dz|^2)=(\frac{2\pi}{\theta})^{(\frac{\theta}{\pi}-2)}r^{(\frac{\theta^3}{2\pi^3}-2\frac{\theta^2}{\pi^2}+2\frac{\theta}{\pi})}(dr^2+r^2d\varphi)$

and $m_{(r,\varphi)}(\frac{\partial}{\partial r},\frac{\partial}{\partial r})=1\neq f^*m_{\beta}(\frac{\partial}{\partial r},\frac{\partial}{\partial r})$

I checked many times but I can't understand where I'm wrong.. I know that asking you to check for my errors is not a very good question, but I would be infinitely grateful if you could give me a hint.

Let $\tilde{z} = re^{i\varphi}$. You can check that $|d\tilde{z}|^2 = dr^2 + r^2d\varphi^2$. In these coordinates, we have $z = f(\tilde{z}) = \frac{1}{\beta+1} \tilde{z}^{\beta+1}$. As you calculated, $$|dz|^2 = r^{2\beta} (dr^2 + r^2 d\varphi^2) = |\tilde{z}|^{2\beta} |d\tilde{z}|^2.$$
So, $f$ is an isometry.
The part the may be confusing is that $\tilde{z}$ is actually the coordinate on $\mathbb{C}$ and $z$ is the coordinate on the cone since $f$ must be well-defined. For instance, consider $\tilde{z} = re^{i0}=r \sim re^{i2\pi}$ on $\mathbb{C}$. For the map to be well-defined, we must have that $f(r) \sim f(re^{2\pi i})$. This means that $$\frac{1}{\beta +1} r^{\beta+1} \sim \frac{1}{\beta+1}r^{\beta+1}e^{i\theta}.$$ In other words, we need to identify the angles $0$ and $\theta$ in the image, which describes the cone.