algebraic-topology
I understand the construction of a torus from a square by
pasting opposite edges of a square and also its CW structure of that.
It's easy to imagine.
But how to understand the construction of a surface of genus 2g from
a polygon with 4g sides and its CW structure. Because in the torus case
by folding and pasting the square we get the object. This we can imagine.
But in the 2g genus case how to imagine the procedure.
Please help me to understand. Thanks in advance.
Fortunately, the one you understand can be readily seen in the cell-complex construction.
So, take a rectangle, identify opposite sides. Now, draw a picture of a torus, and draw the rectangle on it. This is very important that you can do this. What does the rectangle look like on the torus? It likes like a sort of figure-8 (sort of). All four corners are the same point. Two pairs of opposite sides are associated, so we get only 2 edges, not 4. Good.
This is how the cell-structure comes, too. Take one point (0-cell). Take 2 1-cells (each loop in the figure 8). And take 1 2-cell. But how do we attach our 2 cell? Well, a 2-cell is just a square. So on the original rectangle that you drew and understood, why don't you just take that to be your 2-cell? Then the attaching maps are precisely those implied by your drawing.
So the cell-structure and the rectangle are, in fact, the exact same. In fact, when I give cell structures for genus-g surfaces, I give them in that fashion.
It all comes down to (in my opinion) finding that figure 8 on the torus itself, to understand what that rectangle is. If this doesn't make sense, comment, and I'll upload an image.
Here, we see the figure 8 and the two loops. All four corners are the same point. Their intersection is the 0-cell, the red and the blue are each 1-cells, and the surface is a single 2-cell is attached with the following attaching map (where a is the red side, b is the blue)
Take the dodecagon at the origin with one pair of edges intersecting the $y$-axis (call them the top and bottom faces) and one pair intersecting the $x$-axis. Cut the polygon along the $x$ axis, and un-identify the left and right faces. This gives two octagons, each with an opposing pair of unmatched edges. Identify these new edges, and you should have what you're looking for.
Best Answer
I do more graph theory, but we touch on a bit of topology there, so assuming that I understand your question, it's basically the same procedure.
Using a diagram such as this one, you paste the edges marked with the same letters together. You line up the two edges with the same letter in such a way that the arrows are pointing in the same direction. If you think about it for a few minutes, you'll see how this forms the handles for an orientable surface, or the crosscaps for a non-orientable surface.
For example, let's look at diagram 8.5.4 in the link. For those following along at home, this is a surface with two handles, so the flat representation is an octagon, with edge sequence
$$aba^{-1}b^{-1}cdc^{-1}d^{-1}$$
where the negative one means that the edge has direction reversed. So we start by pasting $a$ to $a^{-1}$, remembering to keep the arrows pointing in the same direction, which you can imagine forms a sort of cylinder with $b$ as an edge. Now you can paste $b$ to $b^{-1}$, which is a little harder to visualize but if you stare at it, you'll see that you have formed a handle. Now repeat with the other 4 edges, and you have a surface with 2 handles.