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Let $M \subset \mathbb{S}^3$ be a closed, connected and orientable embedded (and minimal, if important) surface. Choose a unit normal vector field $\eta: M \to \mathbb{S}^3$ along $M$ and a point $p_0 \in \mathbb{S}^3$ such that $-p_0 \not \in M$, and define a function $c_{p_0} : M \to \mathbb{R}$ by $$c_{p_0}(p) = \frac{\langle \eta(p), p_0 \rangle}{1 + \langle p, p_0 \rangle}.$$

An alternative expression for $c_{p_0}$ is

$$c_{p_0}(p) = \frac{\langle \eta(p), p_0 - \langle p, p_0 \rangle p \rangle}{1 + \langle p, p_0 \rangle} = \left\langle \eta(p), -\tan\left( \frac{d_{p_0}(p)}{2} \right) \nabla d_{p_0}(p) \right\rangle,$$

where $d_{p_0}(p)$ is the distance between $p$ and $p_0$ and $\nabla d_{p_0}$ is the gradient (in the sphere) of this function.

Here is my question: is it possible to choose $p_0 \in \mathbb{S}^3$ and $\eta$ such that:

  1. $-p_0 \not \in M$,
  2. $\int_M c_{p_0} \, \mathrm{d} A \leq 0$,
  3. if $\Omega$ denotes the connected component of $\mathbb{S}^3 \setminus M$ such that $\eta$ points outside $\Omega$, then $-p_0 \not \in \Omega$?

I am interested in this setup because I want to apply the divergence theorem on $\Omega$, but $-p_0$ cannot belong to this region because it is a singularity of $c_{p_0}$.

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  • $\begingroup$ Do I understand correctly that you consider $\mathbb S^3$ embedded in $\mathbb R^4$ and endowed with the induced Riemannian metric? $\endgroup$ – Alex M. Apr 10 at 16:28
  • $\begingroup$ Yes, this is the setup. $\endgroup$ – Eduardo Longa Apr 10 at 16:28

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