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Consider $N\subset M$ a submanifold of a Riemannian manifold $M$. I'm interested on the focal points of $N$ into $M$. Naturally for $S^{n-1}\subset S^{n}$ we have that the north and south poles are focal points of $S^{n-1}$ into $S^{n}$.

Changing a little bit the ambient space and submanifold, taking $M= \mathbb{RP}^3$ (the real projective space) and $N=T^2$ (the minimal Clifford torus) what we can talk about the focal points $T^2$ into $\mathbb{RP}^3$? For example, who are them? Are there finite or infinite focal points for this case?

EDIT:

Remark that $q \in M$ is a focal point of $N$ into $M$ if only if $q$ is a singular value of $\exp^\perp$, where $$\exp^\perp : T(N)^\perp\to M.$$

Anderson Lima
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1 Answers1

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Consider $S^3 =\bigg\{ (z_1,z_2)\in \mathbb{C}^2 \bigg| |z_1|^2 +|z_2|^2 =1\bigg\}$. Then, we define $$T:=\bigg\{ (a_1,a_2)\in S^2\bigg| |a_1|=|a_1| = \frac{1}{\sqrt{2}} \bigg\}$$

EXE : Then the focal set of $T$ contains $$S_2 :=\bigg\{ (0,z_2)\in S^3\bigg||z_2|=1\bigg\} $$

Proof : i) If $$(a_1,a_2)\in T,\ c(t):= \sqrt{2}(\cos\ t\cdot a_1,\sin\ t\cdot a_2),\ t\in [0,\pi ] ,$$ then $|c(t)| = 1$ so that $c$ is a curve in $S^3$ between $(\sqrt{2}a_1,0)$ and $(-\sqrt{2}a_1,0)$.

Since $c$ has a length $\pi$, so $c$ is a geodesic.

Hence we have a family of geodesics $\overline{c}_t$ s.t. it starts at $(e^{it}a_1,a_2)\in T,\ |t|<\varepsilon$ and ends at $(0,\sqrt{2}a_2) \in S_2$.

ii) If $S_1=\bigg\{ (z_1,0)\in S^3\bigg||z_1|=1\bigg\}$, then there is a geodesic of length $\frac{\pi}{2}$ between any two points $p_i\in S_i$ attaining a distance between $S_1$ and $S_2$.

Proof : In i), $c$ is a minimizing so that $c|[0,\frac{\pi}{2}]$ is a minimizing.

iii) To complete the proof, we have a claim that $\overline{c}_t'(0)\in (T_{\overline{c}_t(0)} T)^\perp$

Proof : In i), $c$ passes through $T$ and $c$ is a curve attaining distance between $S_1$ and $S_2$. So $\overline{c}_t$ is a curve attaining a distance between $T$ and $S_2$.

EXE : If $\pi :S^3\rightarrow \mathbb{R}P^3$, then the focal set of $\pi(T)$ contains $\pi(S_2)$.

HK Lee
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