Another cell complex structure of closed orientable genus $g$ surface?

Question

Suppose we have a wedge sum of $2g$ circles, labeled by $a_1,b_1,\dots, a_g,b_g$. It is well known that if we attach a 2-cell with attaching map given by $a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}$ then we get an orientable closed surface of genus $g$. Do we also get the same result if we attach a 2-cell with attaching map given by $a_1b_1a_2b_2\cdots a_gb_g a_1^{-1}b_1^{-1}\cdots a_g^{-1}b_g^{-1}$? The representation of $\pi_1$ is given by $$ \langle a_1,b_1,\dots,a_g,b_g : a_1b_1a_2b_2\cdots a_gb_ga_1^{-1}b_1^{-1}\cdots a_g^{-1}b_g^{-1}\rangle$$ and its abelianization will be $\Bbb Z^{2g}$, the same as the orientable closed surface of genus $g$, so I am confused

Answer 1

Yes, this is the genus-$g$ orientable closed surface.

There's a great way to represent surfaces using "chord diagrams": see these notes by Sergei Chmutov for example. I learned about this in Prof. Chmutov's class in Spring 2020.

I'll use $g=2$ for demonstration. Consider the fundamental polygon.

One approach at this point would be to just notice from this picture is that this surface is orientable, and using your remark about the abelianization, by the classification of surfaces, we have the genus-$g$ orientable surface. But hopefully what follows will be a more satisfying geometric argument.

It's hard to imagine making all of the above identifications simultaneously. But since all vertices are identified, we can imagine puncturing a hole around the vertex, to get a picture like

Our surface is just this punctured surface, with a disk glued into the unique light-blue puncture.

Now it is easier to imagine these identifications simultaneously: we just have strips running across the diagram. We can imagine this as a "chord diagram":

If any of our strips needed to be twisted, we would have to notate that, but luckily none of our strips need to be twisted.

There's now a manipulation that we can do: we can slide the strips along each other. Chmutov's notes have this picture of what we're doing:

Now we'll first slide the yellow band along the red band, along the path indicated by the gray arrow. You might worry that this twists the red band, but it is twisted by a full turn, not a half turn, so this is inconsequential. We continue by sliding the yellow band along the blue band, then along the green band:

Reverting the strips to being identifications and gluing back in the light-blue-bounded disk, we get the standard fundamental polygon for the genus-2 surface:

For larger $g$, the process is much the same. The yellow chord will have to pass over several pairs of chords instead of just the blue and green, but it can do this one at a time in the same way. Once we have one pair of chords separated, we can freely pass the others through it:

By allowing this free-pass-through of the isolated chord pair, we can inductively separate the rest of the chord pairs.