2
$\begingroup$

In 1995 Ros proved that minimal surfaces in the round three-sphere $S^3$ enjoy a two-piece property: any hyperplane $\Pi \subset \mathbf{R}^4$ divides every minimal surface $\Sigma \subset S^3$ into two connected components. (The same result in fact holds in any dimension.) This is (to me) almost miraculous, because it does not assume that $\Pi$ intersects $\Sigma$ transversally. I have been trying to get a better understanding on this; my question arose in this context.

Let $\Pi \subset \mathbf{R}^4$ be a plane that is orthogonal to a minimal surface $\Sigma \subset \mathbf{R}^3$ at some point $p$ say. Does $\Pi$ meet $\Sigma$ transversally, or can they be tangent at some other point?

$\endgroup$

1 Answer 1

0
$\begingroup$

In your question do you mean hyperplane in place of plane and $\Sigma \subset S^3$ in place of $\Sigma \subset \mathbf{R}^3$? Making this (perhaps mistaken) interpretation, you can have tangency at another point. For example take a Clifford torus $T \subset S^3$, a point $p \in T$, and a great circle (one of two) $C \subset T$ through $p$. Let $S$ be the totally geodesic sphere (the intersection of a hyperplane $\Pi$ with $S^3$) containing $C$ and orthogonal at $p$ to $T$. Then $T \cap S=C \cup C'$, with $C'$ another great circle intersecting $C$ orthogonally at two antipodal points $\pm q$ (distance $\pi/2$ along $C$ from $p$), at each of which $T$ and $S$ are tangent.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .