Orientable surface of genus \(g\)
- \(g=2\) is given by \(a+b-a-b+c+d-c-d\)
Solution
1. A closed orientable surface of genus \(g\) is obtained from a \(4g\)-gon by the boundary word\[ \boxed{a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}}. \]
For each handle, take the standard square model of a torus with boundary word \(aba^{-1}b^{-1}\). Taking the connected sum of \(g\) such handles gives a polygon whose boundary word is the product of these \(g\) commutators.
2. In the additive arrow notation of the question, the word is\[ a_1+b_1-a_1-b_1+\cdots+a_g+b_g-a_g-b_g. \]
Writing a reversed edge as the negative of its oriented label translates the multiplicative word in 1 into the displayed notation.
3. For \(g=2\) this specializes to\[ a+b-a-b+c+d-c-d, \]as stated.
Take \((a_1,b_1,a_2,b_2)=(a,b,c,d)\) in 2.