Homology of surface of genus $g$

Question

This is a homework question given to me by someone of the community here and it's a generalisation of this. I was wondering if you could have a look and tell me if it's right. Thanks for your help!

Task: Compute the homology of a surface of genus $g$, $\Sigma_g$.

My calculations:

(i) The cell decomposition:

• $1$ two-cell $e^2$ (a $4g$-gon)
• $2g$ one-cells $e^1_i$
• $1$ zero-cell $e^0$

(ii) The attaching map of $e^2$:

• $f_2 = a_1b_1a_1^{-1}b_1^{-1} \dots a_gb_ga_g^{-1}b_g^{-1}$

  The attaching map of $e^1$:

• $f_1 = e^0$

(iii) The chain groups:

• $C_0(\Sigma_g) = \mathbb{Z}$
• $C_1(\Sigma_g) = \mathbb{Z}^{2g}$
• $C_2(\Sigma_g) = \mathbb{Z}$
• $C_k(\Sigma_g) = 0$, $k>2$

(iv) The boundary homomorphisms:

$$\dots \xrightarrow{d_3} C_2(\Sigma_g) \xrightarrow{d_2} C_1(\Sigma_g) \xrightarrow{d_1} C_0(\Sigma_g) \xrightarrow{d_0} 0$$

• $d_0 = 0$
• $d_1 = 0$, because $f_1$ has degree $0$
• $d_2(e^2) = 0$, because each coefficient is $0$

(v) The homology groups:$\DeclareMathOperator{\im}{im}$

• $H_0(\Sigma_g) = \ker d_0 / \im d_1 = \mathbb{Z} / 0 = \mathbb{Z}$
• $H_1(\Sigma_g) = \ker d_1 / \im d_2 = \mathbb{Z}^{2g} / 0 = \mathbb{Z}^{2g}$
• $H_2(\Sigma_g) = \ker d_2 / \im d_3 = \mathbb{Z} / 0 = \mathbb{Z}$

Answer 1

You can get the genus $g$-surface by doing the connected sum of $g$ tori $T=S^1 \times S^1$, i.e.,

$$S_g := T\,\#\,T\,\#\,\cdots\,\#\,T \qquad (g\text{ times}).$$

Assuming you're working over $ \mathbb{Z}$.

If you know the homology of $T$, and how to find that of the connected sum, done.

If not, or if you prefer a different approach, you can: i) Find the homology of $S^1$, then ii) Find the homology of the product $S^1\times S^1$, and iii) Find the homology of the connected sum of $g$ copies in ii):

1. $H_1(S^1) = \mathbb{Z}$

2. How to find the homology of a product space, (e.g., Künneth's formula) it is $\mathbb{Z}^2$

3. Finding the homology of connected sums; it is the direct sum of the respective homologies; the basis curves are pairwise disjoint, so the homology is the direct sum (what happens in one Torus, stays in that Torus) You ultimately get: $$H_1(S_g)=\mathbb{Z}^{2g}$$ you are done.