0
$\begingroup$

Let $\Sigma$ be a closed orientable surface of genus $g$ and let $\alpha = \{\alpha_1,...,\alpha_g\}$ be a set of pairwise nonintertersecting nonseparating simple closed curves on $\Sigma$. Then $\alpha$ determines a way of identifying the boundary of a genus $g$ handlebody $H$ with $\Sigma$. If $\gamma$ is a simple closed curve in $\Sigma$ that does not intersect any of the $\alpha_i$, does it follow that $\gamma$ will bound a properly embedded disk when we identify $\Sigma$ with $\partial H$?

$\endgroup$
  • $\begingroup$ Not sure if I understand. The curves $\alpha_j$ can be contractible? Further, if you have $\alpha_1$, a nontrivial circle on a torus, and $\gamma$ be just close to $\alpha_1$, going "paralelly", then $\gamma$ doesn't bound any disc... $\endgroup$ – Peter Franek Apr 24 '17 at 20:12
  • $\begingroup$ Whoops - forgot to say nonseparating. Parallel is fine - those will bound disks. $\endgroup$ – user101010 Apr 25 '17 at 1:17
1
$\begingroup$

It is probably assumed that the curves are also pairwise non-isotopic (since otherwise (1) the curves do not determine the handlebody attachment uniquely and (2) the answer would be "no" since you could take $g$ parallel copies of the same nonseparating curve in a genus-$g$ surface, and for $g>1$, given any handlebody attachment there are plenty of disjoint curves that don't bound disks).

Identify $\Sigma$ with $\partial H$, and let $D_1,\dots,D_g\subset H$ be a system of disjoint disks such that $\partial D_i=\alpha_i$ for each $i$. If $\gamma$ does not intersect any of the $\alpha_i$, then $\gamma$ is an embedded loop in the boundary of $B=H-\bigcup_i \nu(D_i)$. Since $B$ is a ball, $\gamma$ bounds a disk $D\subset B$, which is also a disk in $H$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.