Take a genus $g$ surface $S$ standardly embedded in $\mathbb{R}^3$, by which I mean it is unknotted. Surface $S$ bounds a volume $V$ that deformation retracts on a standardly embedded planar graph $G$ with $\beta_1 = g$, and that only has degree $3$ vertices.

Among the loops on $S$ that are null homotopic in $V$, there is a subset that are boundaries of embedded disks in $V$ that intersect $G$ exactly once for some choice of $G$ as above.

Do these loops (or perhaps close variants) have a name? Do they have an alternate definition?


1 Answer 1


They are often called meridians of $G$. Note that there are many graphs $G$ to which $V$ deformation retracts (most nonplanar); if you are not particular about which graph $G$ then they are called meridians of $V$. $V$ is called a handlebody and $G$ is a spine of the handlebody. See, for example, Scharlemann's "Refilling meridians in a genus 2 handlebody complement" arXiv:math/0603705

  • 1
    $\begingroup$ In a paper I found meridians are defined as essential curves bounding an embedded disk. In particular the condition that it should intersect a (well chosen) spine exactly once is not specified. Does that mean that the condition is always true, or are these meridians more general than what I defined? $\endgroup$
    – alesia
    May 30, 2020 at 18:34
  • 1
    $\begingroup$ They are essentially equivalent ideas. Suppose $D$ is a collection of essential discs in $V$ such that no two are isotopic and such that any other essential disc disjoint from $D$ is isotopic to a disc in $D$. Then cutting $V$ along $D$ produces a collection of 3-balls with 3 scars from $D$ in each boundary. Cone the center of each scar to the center of the 3-ball to get a "tripod". Undoing the cutting along $D$ connects the tripods together (or two edges of a tripod together) to form a graph $G$ to which $V$ deformation retracts. $\endgroup$ May 30, 2020 at 20:42
  • $\begingroup$ Also the graph $G$ and the discs $D$ satisfy your requirements, except that $G$ may not be planar, and given any embedded disc in $V$ with boundary essential in $S$ it is contained in such a collection of discs. (That is little harder to prove, but is a standard innermost circle/outermost arc argument. Probably you can find the argument in Jaco's or Hempel's books on 3-manifolds.) Insisting that $G$ be planar is much more restrictive and the corresponding discs are really associated more to $G$ than to $V$. I would still call them meridian discs though. $\endgroup$ May 30, 2020 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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