# Can this integral be made non-positive?

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 $$c_{p_0}$$.

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