Let $M$ be a Riemannian manifold with totally geodesic boundary $\partial M$. We let $\check{M}$ be its double, i.e. the disjoint union of $M$ with itself under identification of corresponding boundary points. It is then well known that $\check{M}$ is a smooth manifold. Can you tell me whether the metric is still smooth along the boundary or else what is its degree of regularity?
Can you maybe also tell me what it means for the boundary to be an isoperimetric surface for $M$?



Regularity of metric of the double of a Riemannian manifold
You should take a look at some of the answers to previous questions here that address this point, especially [this one](https://mathoverflow.net/questions/67809/the-double-of-a-smooth-manifold-with-boundary) and [this one](https://mathoverflow.net/questions/82270/reference-request-gluing-manifolds-along-pieces-of-boundary). In short, if you double a manifold with totally geodesic boundary you might not have a smooth metric immediately (although you do get a smooth atlas on the double). One can get a smooth metric by either guaranteeing that all the appropriate derivatives vanish, or by using collars (i.e., require the metric is a product near the boundary). These techniques are described in some detail in the above links. 


Regarding isoperimetric hypersurfaces, these are hypersurfaces that minimize area among those that enclose a fixed volume. They are automatically stable constant mean curvature hypersurfaces (although the reverse implication generally fails). Also, in general, the totally geodesic boundary $\partial M$ is not isoperimetric in the double of $M$. For a nice introduction to the subject, take a look at [A. Ros' survey](http://www.ugr.es/~aros/isoper.pdf).


