16
$\begingroup$

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:

  • $H_0(\Sigma_g) = ker d_0 / im d_1 = \mathbb{Z} / 0 = \mathbb{Z}$
  • $H_1(\Sigma_g) = ker d_1 / im d_2 = \mathbb{Z}^{2g} / 0 = \mathbb{Z}^{2g}$
  • $H_2(\Sigma_g) = ker d_2 / im d_3 = \mathbb{Z} / 0 = \mathbb{Z}$
$\endgroup$
4
  • 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

15
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.