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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:$\DeclareMathOperator{\im}{im}$

  • $H_0(\Sigma_g) = \ker d_0 / \im d_1 = \mathbb{Z} / 0 = \mathbb{Z}$
  • $H_1(\Sigma_g) = \ker d_1 / \im d_2 = \mathbb{Z}^{2g} / 0 = \mathbb{Z}^{2g}$
  • $H_2(\Sigma_g) = \ker d_2 / \im d_3 = \mathbb{Z} / 0 = \mathbb{Z}$
ultralegend5385
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Rudy the Reindeer
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1 Answers1

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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.

t.b.
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gary
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  • 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) – gary Jun 20 '11 at 14:42
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    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. – t.b. Jun 20 '11 at 14:52
  • Nice, thanks. What are the basis curves? – Rudy the Reindeer Jun 20 '11 at 14:55
  • PS: I'm doing ground-up to practice ; ) – Rudy the Reindeer Jun 20 '11 at 14:56
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    @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. – gary Jun 20 '11 at 14:58
  • Theo: thanks, I am still working on it. – gary Jun 20 '11 at 14:59
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    @gary: Okay, I see. By the way: add an @-sign in front of your comments, then the corresponding user gets notified. – t.b. Jun 20 '11 at 15:01
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    @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. – gary Jun 20 '11 at 15:02
  • @Theo: I meant to say that I am still working on learning how to do the formatting correctly; thanks for your patience! – gary Jun 20 '11 at 15:25
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    @gary: Thanks. Explanations can never be too trivial ; ) – Rudy the Reindeer Jun 20 '11 at 15:44
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    @gary: I reincorporated my changes. I hope they're fine with you. To learn LaTeX I found Tobi Oetiker's Not so short introduction to LaTeX invaluable (download lshort.pdf there). To learn the syntax for markdown formatting here, here's more than you need to know. – t.b. Jun 20 '11 at 17:45
  • So what's the reason for the downvote? – gary Feb 18 '24 at 00:18