I have trying to understand the proof of the lemma (Page 14, Lemma: 2.2.4) which is stated as follows:
Every ribbon graph can be embedded into an oriented surface such that its cyclic orderings are induced from the orientation of the surface.
I have the following confusions:
- In the first paragraph, it is written that : for every vertex $v$ define a subset $U_v \subset \mathbb{R}^2$ with $|E_v|$ (the geometric realization of the star of $v$, see definition 2.2.2) $\textbf{boundary components}$ labelled by the elements of $E_v$ of the form depicted in Figure 2.6 (Only the straight lines of the boundary are part of the surface.)
In the above context I have the following figure.
What are the boundary components here? Are the red lines boundary components labelled by the elements of $E_v$? As written in the first braket. only the red lines are the part of surfaces
- In the second paragraph, it is written that: Similarly we attach to each edge $e$ a strip $V_e$ as depicted in Figure 2.7 whose two boundary components (again only the straight lines are part of the surface) are labelled by $e_+$ and $e_-$.
So, in the following figure, which are the boundary components lebelled by $e_+$ and $e_-$? Also, it is written that "again only the straight lines are part of the surface". Which straight lines are only the part of surface in this case?
Please help.
I'll first answer (2): $e_-$ and $e_+$ are the two horizontal dotted sides. By convention you can take the bottom one to be $e^+$ and the top $e_-$. Do a simple example: look at the ribbon graph with 2 vertices $v$, $w$ and just one edge $e=\{v,w\}$. Represent the vertices as closed disks. The boundary of these disks has an orientation, for example counter-clockwise, which has to be the same for all vertices. Now take the edge $e$ represented as a ribbon and attach it to these disks. For the orientation to make sense, this ribbon must have one horizontal side oriented towards $v$ and the other horizontal side must be oriented away from $v$. These are your $e_-$ and $e_+$.
Regarding (1) and boundary components: Pick a vertex and an edge. Trace the outgoing arrow out of the edge until you return to where you started from. The path that you traced is one boundary component. To find a different boundary component you would pick a different vertex-edge pair that hasn't already been traced. When you trace out all the vertices and edges, you are done.
The simple example from above of two vertices and one edge has just one boundary component. For a more elaborate example, take a triangle represented as a ribbon graph. You will have 2 boundary components, the "inner" and the "outer" sides of the ribbons.
The definition that you're working with is very formal and very difficult and obsucres what otherwise is a relatively simple concept. I think that you would benefit by understanding first a simpler approach.
An orientable cellularly embedded graph (which is equivalent to a ribbon graph) is a graph $G$ embedded (drawn) on an orientable closed surface $X$ such that $X\setminus G$ is a disjoint union of topological disks. This means that
(i) vertices are points on the surface,
(ii) edges are curved segments between vertices,
(iii) crossings happen only at the vertices and
(iv) $X\setminus G=F_1\cup\dots\cup F_k$ for some $k\in \mathbb N$ and each $F_i$ is homeomorphic to an open disk. The sets $F_i$ are called the faces of $G$ (to be more precise, the faces of a particular embedding of $G$).
Let $G$ be cellularly embedded on $X$. We can represent this embedding as follows: let $e=\{v,w\}$. To each edge assign two labels called half-edges in the following way: if you're standing on the vertex $v$ and looking towards $w$ along $e$, then on the left side of $e$ (where left is defined by the orientation of your surface) place a label $i$. And if you're standing on the vertex $w$ and looking towards $v$ along $e$, place a label $j$ on the left side.
In this way you have represented an edge $e$ as a pair of half-edges $i$ and $j$. This is written as the transposition $(i~j)$. Hence every edge is represented by such a transposition. Now each vertex can be represented by a permutation cycle $(i_1~\cdots i_k)$ which you obtain just by reading the half-edges incident to it in a preferred order (e.g. counter-clockwise).
So what does this have to do with your questions? The $e_-$ and $e_+$ correspond to the half edges $(i~j)$ that we are using to represent an edge. Hence in (2) the $e_-$ and $e_+$ can also be considered as the two vertical dotted sides of an edge which you glue to a vertex (represented as a disk) so that the orientation makes sense. The boundary components are identified with the faces $F_i$ which can be read simply by reading the labels inside a particular face (and if you've done your labeling according to the convention that I've described before, you should have the same number of labels in a face as there are edges bordering it).
Now to obtain a ribbon graph from a cellularly embedded graph, you "inflate" the vertices into closed disks and edges into ribbons and remove the faces. Conversely, to obtain a cellularly embedded graph from a ribbon graph, you glue in the faces, contract the disks into points and ribbons into curved segments.