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:

  • $H_0(\Sigma_g) = ker d_0 / im d_1 = \mathbb{Z} / 0 = \mathbb{Z}$
  • $H_1(\Sigma_g) = ker d_1 / im d_2 = \mathbb{Z}^{2g} / 0 = \mathbb{Z}^{2g}$
  • $H_2(\Sigma_g) = ker d_2 / im d_3 = \mathbb{Z} / 0 = \mathbb{Z}$
  • 2
    $\begingroup$ You say you're using $g$ one-cells, but your attaching map for $e^2$ involves $2g$ of them. $\endgroup$ Jun 20, 2011 at 10:09
  • $\begingroup$ There is something wrong. Thank you! $\endgroup$ Jun 20, 2011 at 10:13
  • 1
    $\begingroup$ It's a $4g$-gon, not a $2g$-gon! Thanks!! $\endgroup$ Jun 20, 2011 at 10:17
  • $\begingroup$ Ok, maybe someone could write something as an answer, then I can accept it and this question is resolved. Thanks! $\endgroup$ Jun 20, 2011 at 10:52

1 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.

  • $\begingroup$ I don't know if this is the approach you wanted, but I think it may give you a kind of quick-and-dirty way of getting the homology without having to do the whole thing ground-up all the time (tho it is a good idea to do ground-up a few times) $\endgroup$
    – gary
    Jun 20, 2011 at 14:42
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    $\begingroup$ I tried to fix your formatting and TeX. Note: the sign $\times$ is obtained by using \times. If you want more than one letter as superscript use curly braces: $\mathbb{Z}^{2g}$ is obtained by \mathbb{Z}^{2g}. You can see what I did by clicking on the 'edited xx time' ago over my name. $\endgroup$
    – t.b.
    Jun 20, 2011 at 14:52
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    $\begingroup$ @Matt: they are a meridian and a parallel; two representatives of non-trivial cycles that do not bound. The class (m,n) then goes m times around a meridian and n times around a parallel. $\endgroup$
    – gary
    Jun 20, 2011 at 14:58
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    $\begingroup$ @gary: Okay, I see. By the way: add an @-sign in front of your comments, then the corresponding user gets notified. $\endgroup$
    – t.b.
    Jun 20, 2011 at 15:01
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    $\begingroup$ @Matt, another comment: the choice of the two curves has to see with the fact that if we were to cut or remove the curves, the remaining space would be connected; in that sense, they do not bound, so they are non-trivial curves. Hope this is not too trivial of a comment; it helped me when I learnt about it. $\endgroup$
    – gary
    Jun 20, 2011 at 15:02

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