Double of Riemannian manifold.

Question

Let $M$ be a Riemannian manifold with totally geodesic boundary $\partial M$. We consider its double $\check{M}$, i.e. the disjoint union of $M$ with itself under identification of corresponding boundary points. It is 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 $M$?

Answer 1

Here is an example of a smooth metric whose double is not $\mathcal C^3$:

Consider the smooth manifold $M = ]-1,1[ \times \mathbb R$ with global coordinate system $(t,s)$, where $t\in ]-1,1[$ and $s \in \mathbb R$. We can define a smooth metric on $M$ via $$g(s,t) = dt^2 + (t^3 + 1)^2ds^2.$$ Smoothness of $g$ follows since $(t^3 + 1)$ is smooth and does not vanish on $]-1,1[$. Now one can show that $s \mapsto (0,s)$ defines a geodesic of $(M,g)$ (this might not be too trivial, but is essentially due to the fact that $(t^3 + 1)$ has a saddle point at $t = 0$). Consider $N := [0,1[ \times \mathbb R \subset M$, which defines a smooth submanifold with boundary of $M$. Then $(N,g_N)$ is a smooth Riemannian manifold with boundary. Let $(\tilde N,\tilde g)$ be the riemannian double. We claim that $\tilde g$ is not $\mathcal C^3$:

To see this let $K(t,s)$ be the sectional curvature of $g$ at $(t,s)$. For a metric $dt^2 + f^2(t)ds^2$ there is the well known formula $$K(s,t) = \frac{f''(t)}{f(t)}.$$ In our case $$K(t,s) = K(t) = \frac{t}{t^3 + 1}$$ and $$K'(t) = \frac{1}{(t^3 + 1)^2}.$$ In particular $K'(0) = 1$. Now consider the double metric $\tilde g$ of the double $\tilde N$ of $(N,g_N)$. Again we have $\tilde N \cong ]-1,1[ \times \mathbb R$ with coordinates $(t,s)$ as above.

Now assume the metric $\tilde g$ is $\mathcal C^3$. Then $t \mapsto \tilde K(t)$ is $\mathcal C^1$. On the other hand the map $(t,s) \mapsto (-t,s)$ is an isometry of $\tilde N$. This implies that $\tilde K'(0) = 0$. But we have $\tilde K(t) = K(|t|)$ for $t \neq 0$, so $\lim_{t \to 0}\tilde K'(t) = \lim_{t \to 0} K'(t) = 1$, a contradiction.

In general i would guess, that the double metric is always $\mathcal C^2$, due to the vanishing of the second fundamental form at $\partial M$, but not necessarily three times differentiable.