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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?

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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

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    $\begingroup$ 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 at 18:34
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    $\begingroup$ 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. $\endgroup$ – Scott Taylor May 30 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. $\endgroup$ – Scott Taylor May 30 at 20:50

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