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Suppose $(M,g)$ is a compact Riemannian manifold with total geodesic boundary. The double of it is given by $$ D(M) = M\cup_fM $$ where $f:\partial M\to\partial M$ is an identity map. In general $D(M)$ is not a $C^\infty$ manifold. But I wonder if $D(M)$ is a $C^2$ manifold or not.

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2 Answers 2

up vote 16 down vote accepted

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.

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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.

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