# Local boundary symmetrisation of Riemannian metrics by coordinate changes

Assume we have a smooth Riemannian metric $g$ on a small one-sided neighborhood $U$ of $0$ on the plane, say $U_\epsilon=\lbrace(x, y): x^2+y^2<\epsilon, y\geq 0\rbrace$.

When does there exist a diffeomorphism $f$ of ( may be smaller) one-sided neighborhood of zero $U_\delta$ such that the resulting Riemannian metric $f_\star g$ can be smoothly extended $\textbf{by relflection}$ along the axis $y=0$.

In particular( but not only) I am mostly interested in cases:

$a)$ $g=\rho(x,y) g_0$, where $\rho(x,y)$ is a conformal factor and $g_0$ is standard Euclidean metric.

$b)$ Case $a$ with flat $g$, i.e. $\Delta\ln\rho=0$

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Assume we cut a surface by a curve $\gamma$ and take doubling of one piece. When the doubling has smooth metric?
Once you reformulate it this way, the answer is evident. First note that $\gamma$ has to be geodesic otherwise the doubling has infinite curvature along the gluing.
Further, you may write the metric near $\gamma$ in semigeodesic coordinates; one coordinate, say $u\ge 0$, is the distance to the geodesic and the other, say $v$, is the parameter of its foot point. In this coordinates the metric has form $(\begin{smallmatrix}1&0\\\ 0&a\end{smallmatrix})$, where $a=a(u,v)$ is a positive function.
The metric on the doubling is $C^\infty$-smooth iff the function $b(u,v)=a(|u|,v)$ is $C^\infty$-smooth near the $v$-axis.