4
$\begingroup$

Let $\Sigma$ be a compact orientable connected $2$-manifold with a non-empty boundary. Let $\widehat \pi(\Sigma)$ denote the set of free homotopy classes of curves in $\Sigma$. We say $x\in \widehat \pi(\Sigma)$ is peripheral, if there is representative $\alpha\colon \Bbb S^1\to \Sigma$ of $x$ with $\text{im}(\alpha)\subseteq \partial \Sigma$.

Question: Let $x\in\widehat \pi(\Sigma)$ be a non-trivial non-peripheral element. Does there exist a simple closed curve $\beta\subseteq \Sigma$ such that the geometric intersection of $\alpha$ and $\beta$ is non-zero, where $\alpha$ is a representative of $x$.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

This is true for all (compact, connected, oriented) surfaces that admit essential non-peripheral simple closed curves.

The sphere, disk, annulus and pants do not admit essential non-peripheral curves. The sphere and disk have trivial fundamental group. So the answer is "yes" vacuously for the sphere and disk, and "no" for the annulus and pants (as suggested by Johannes’).

Suppose that $S$ is any other compact, connected, oriented surface. Now $S$ admits filling laminations. Any sufficiently good (Hausdorff close) simple closed curve approximation to a filling lamination does what you want. The existence of such laminations is a consequence of Thurston’s theory of pseudo-Anosov maps.

Improve this answer
$\endgroup$
5
$\begingroup$

I don't think so. Let $\Sigma$ be the pair of pants, and $\alpha$ the curve, both pictured below. Then $x=[\alpha]$ is non-trivial since it has non-zero winding number with one of the holes. It is non-peripheral because it has non-zero winding number with two holes. Now, $x$ is trivial in the plane, so any loop $\beta$ in $\Sigma$, simple or not, wil have zero intersection number with any representative of $x$.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ I think your example works, but I'm not so sure about your reasoning. Note that Sam Nead's answer shows that other planar surfaces do have the required property. $\endgroup$
    – HJRW
    Commented Sep 28, 2022 at 9:52
  • 1
    $\begingroup$ Yes, it seems this establishes the algebraic intersection number is always zero. A non-simple $\beta$ may have nonzero geometric intersection number with $\alpha$ (for instance, $\alpha$ itself has nonzero geometric intersection number with $\alpha$). But every simply curve on a pair of pants is peripheral and thus has geometric intersection number zero with every other curve, including $\alpha$. Nice example. $\endgroup$
    – mme
    Commented Sep 28, 2022 at 11:56

You must log in to answer this question.

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