Let $M^2$ be an orientable surface with boundary endowed with a Riemannian metric $g$. We know that if we put in the manifold $M \times \mathbb{R}$ the product metric $\overline{g} = g + dt^2$, then the slices $M_t = M \times \{t\}$ are totally geodesic and are free-boundary surfaces, which means that their boundary meet the boundary $\partial (M \times \mathbb{R}) = \partial M \times \mathbb{R}$ orthogonally.

Suppose now that we choose an angle $\theta \in (0, \pi)$. Is there a "natural" Riemannian metric $g_\theta$ on $M \times \mathbb{R}$ such that the slices $M_t$ are totally geodesic, meet the boundary of $M \times \mathbb{R}$ at a constant angle $\theta$ and $g_{\pi/2}$ agrees with the product metric $\overline{g}$? Is there an explicit formula?

Remark: Let $N_t$ be the unit normal to $M_t$ in the metric $g_\theta$, and $\overline{N}$ be the outward unit normal to $\partial M \times \mathbb{R}$ in this metric. We say that the slice $M_t$ meets the boundary $\partial M \times \mathbb{R}$ at an angle $\theta$ if $$g_\theta(N_t, \overline{N}) = \cos \theta$$ along $\partial M_t$.

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  • $\begingroup$ what about $k\cdot t^2\cdot g+dt^2$? $\endgroup$ – Anton Petrunin Jun 5 '19 at 21:20
  • $\begingroup$ @AntonPetrunin what is $k$ and what happens when $t=0$? $\endgroup$ – Eduardo Longa Jun 6 '19 at 17:43
  • $\begingroup$ $k$ is a constant so the metric is defined only for $t>0$, but if you reparametrize it by $\ln t$ you get a metric on the product $M\times \mathbb{R}$. $\endgroup$ – Anton Petrunin Jun 7 '19 at 12:56
  • $\begingroup$ Crossposted from math.stackexchange.com/questions/3237696/… $\endgroup$ – Nate Eldredge Jun 16 '19 at 3:50

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