Double a manifold with boundary

Question

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?

Answer 1

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.

Answer 2

You can give $D(M)$ a $C^k$ structure, $k=0,\ldots,\infty,\omega$ if $M$ has one.

Take a $C^k$ function $f$ on $M$ that is $1$ on the boundary with positive normal derivative, and $0\leq f<1$in the interior.

Then you can take $D(M)=\{(x,t)\in M\times\mathbb{R}: f(x)+t^2=1\}$.

Do you want $D(M)$ to be a riemannian $C^k$ manifold ? With isometric copies of $(M,g)$ this will not be possible in general for $k>2$, but sometimes it will, for instance in the locally symmetric case.