Suppose $(M,g)$ is a compact Riemannian manifold with totally geodesic boundary. Its *double* is given by
$$
D(M) = M\cup\_fM
$$
where $f:\partial M\to\partial M$ is the identity map. In general $D(M)$ is not a $C^\infty$ manifold. Is $D(M)$ a $C^2$ manifold?



Double a manifold with boundary
It *is* a $C^\infty$ manifold if you define charts properly (e.g. using geodesics normal to the boundary as coordinate lines).


The *metric* of the double is $C^2$ (but not always $C^3$). Indeed, since the boundary is totally geodesic, the normal derivative of the metric tensor (in the above mentioned coordinates) vanishes, and the second normal derivatives on two copies match due to the symmetry.


