Are there proofs of the existence of such coordinates near every point of a riemaniann surface $(M,g)$ using Cartan connections and curvature ?
Here is a proof i tried to write down, but i suspect it may be wrong since i dont use complicated theory of PDEs. I write $(G,P)$ for the Euclidean Klein geometry where $G$ is the Euclidean group an $P=O_2(\mathbb{R})$. I write $(e_1,e_2,A)$ for the usual basis of the Lie algebra $\mathfrak{g}=Lie(G)$ of the euclidean group, so $e_1,e_2$ are the infinitesimal translations and $A=\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$ is the usual basis of $\mathfrak{p}=Lie(P)$. The conformal Klein geometry is denoted by $(G',P')$.
Start from the torsionfree euclidean Cartan geometry $(E,\omega)$ corresponding to the metric
Extend it canonically to a conformal Cartan geometry, which means $(E',\omega')$ where $E′$ is an extension of the orthogonal principal bundle E and ω′|E=ω. In particular, the curvature function of this Cartan connection is of the form $k_{\omega'}|_E = k_\omega=fe_1^* \wedge e_2^* \otimes A$ where $A$ is the usual generator of $\mathfrak{p}=Lie(P)$ and $f$ is an $P$-invariant function on $E$.
Construct a new conformal Cartan connection $\omega''$ on $E'$ as follow. Denote by $e_{-1},e_{-2}$ the vectors corresponding to the opposite roots of those associated to $e_1,e_2$ (in a Chevalley basis of $\mathfrak{g}'=Lie(G')$).Then $A=[e_1,e_{-2}] = -[e_2,e_{-1}]$. So we can consider a $P'$-équivariant function $\varphi$ uniquely defined by : $$\varphi|_E = - fe_1^* \otimes (e_{-1}-e_{-2}) + f e_2^* \otimes (e_{-1}-e_{-2}) $$ which statisfies $\partial \varphi = -k_\omega$. We construct $\omega''=\omega'+ \varphi \circ \omega'$.
Using Bianchi identity, we have $d\varphi = 0$. In particular, the change of curvature between $\omega'$ and $\omega''$ is $\partial \varphi$, so $\omega''$ is a flat conformal Cartan connection on $E'$.
$(E',\omega'')$ admits a developpement $\psi : U\longrightarrow G'/P'$ near every point $x\in M$. On a neighboorhood of $\psi(x)$ there exist a flat metric $g_0$, so $g'= \psi^\star g_0$ is a flat metric on a neighboorhood of $x$. By construction, this metric is in the conformal class of $g|_U$.
Do you detect a problem with this proof ?
Thank you
The definition of $\varphi$ in the question doesn't lead to a flat conformal geometry.
However, since we can consider a general section $\varphi = a e_1^* \otimes e_{-1} - b e_2^* \otimes e_{-2}$ (where $a,b$ are $P'$-invariant differentiable functions on $E'$) to construct a general conformal Cartan connection $\omega''$ from $\omega'$, without torsion.
Let's consider the parabolic decomposition $\mathfrak{g}'=\mathfrak{g}_1 \oplus \mathfrak{g}'_0 \oplus \mathfrak{g}'_{-1}$ of $\mathfrak{g}'$ (where $\mathfrak{g}'_0 \oplus \mathfrak{g}'_{-1} = \mathfrak{p}$). If we want $\omega''$ to be flat, then by looking at the $\mathfrak{g}'_0$-component of the new curvature, we must have : $a+b = -f$. So $a=-fc$ and $b=-f(1-c)$ for a differentiable function $c$ on $E'$.
The PDE appears when we look at the $\mathfrak{g}'_{-1}$-component of the new curvature. Let's denote by $Y_1$ and $Y_2$ the $\omega'$-constant vector fields associated with $e_1$ and $e_2$. The condition is that : $$\mathcal{L}_{Y_1}(a) + \mathcal{L}_{{Y}_2}(b) = \mathcal{L}_{Y_1}(fc) + \mathcal{L}_{Y_2}(f(1-c))=0$$ Since $c$ and $f$ are the pullback of diffrentiable functions on $M$, this can be rewritten into a PDE over the base manifold by considering exponential coordinates.
It looks like a Beltrami equation but i can't put it concretly into this form.