how to calculate that the second homology group for orientable surface of genus $g$ is $\mathbb{Z}$?
by calculating I mean that find $ker \partial_2$ in chain complex,for example for torus of two hole:
also I know that the first homology group is $\mathbb{Z}^{2g}$ but except torus with one hole I can't do its calculation.
I can think geometrically about that as I study Intuition of the meaning of homology groups but one time I want to see its calculation,also I have the same problem with non-orientable one.for $\mathbb{RP}^2$ and klein bottle I know the algebrical works,but more than it I couldn't go far.
any hint or references will be great,thanks a lot.
Best Answer
I think this is easiest to see using cellular homology. There is only one $2$-cell $U$, so $C_2 \cong \mathbb{Z}$. The boundary of this $2$-cell is the product of the commutators of the $2g$ $1$-cells: $[a_1, b_1] \cdots [a_g, b_g]$ where $[a_i, b_i] = {a_i}^{-1} {b_i}^{-1} a_i b_i$. For instance, if you traverse the boundary of the octagon you've drawn to represent a genus $2$ surface, you'll see that we traverse each edge twice, once in the positive direction, and once in the negative direction. See here for a reference. However, since the homology groups are abelian groups, these commutators are trivial, i.e., $$ \partial_2(U) = -a_1 - b_1 + a_1 + b_1 + \cdots + -a_g - b_g + a_g + b_g = 0 \, . $$ Since $\partial_2$ maps the generator of $C_2$ to $0$, then $\partial_2 = 0$, hence $\ker \partial_2 = C_2 \cong \mathbb{Z}$.