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I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!

The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$

The chain groups are

$$ C_0(T^2) = \{ k e^0 | k \in \mathbb{Z} \} = \mathbb{Z}$$

$$ C_1(T^2) = \{ k_1 e^1_a + k_2 e^1_b | k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$

$$ C_2(T^2) = \mathbb{Z}$$

$$ C_k(T^2) = \{ 0 \} , k > 2$$

Now the homology groups:

$$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$

where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)

$$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$

where $\ker \partial_1 = \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.

$im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.

I don't want to use Hurewicz to do $H_1$.

$$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.

Thanks for your help.


I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them...

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I'm not entirely convinced by your computations and arguments. You forgot to say a word about the attaching maps of your cell decomposition of $T^2$. What are they and what maps do they induce on the chain level? –  t.b. May 25 '11 at 14:58
The argument is the right one, still I found you're explanation non very convincing. Could you compute explicitly the maps $\partial_1$ and $\partial_2$? –  Giacomo d'Antonio May 25 '11 at 14:58
Yes, you're on the right track. The thing is that you need to specify how you glue the cells together, so to speak. The attaching maps determine the boundary maps, so if you speak of "comput[ing] the boundary maps", you need to specify the attaching maps first. –  t.b. May 25 '11 at 15:09
@Mariano: Don't be too harsh on Matt! He just learned about the very existence of cellular homology a few hours earlier and I guess this was his first try at a computation :) –  t.b. May 25 '11 at 18:07
@amWhy: My pleasure! Matt is some sort of protegé of mine on this site... I know that it is you, I noticed a new busy editing bee around here...with suspiciously similar gravatar :) At the latest, the ellipses in this comment would have given you away... –  t.b. May 25 '11 at 19:13

1 Answer 1

up vote 5 down vote accepted

Ok, so here are the boundary homomorphisms:

$$ \dots \rightarrow 0 \xrightarrow{d_3} \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^2 \xrightarrow{d_1} \mathbb{Z} \xrightarrow{d_0} 0$$

$d_0 = 0$

$d_1 = 0$ because the attaching map ($f = const.$) as there is one $0$-cell.

$d_2 = c_1 e_1^1 + c_2 e_2^1= 0$ because $f = ab a^{-1}b^{-1}$ so the coefficients are $c_i = +1 - 1$ respectively.

The homology groups as claimed above in the question follow from these maps.

Is this right? Many thanks for your help!

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@Theo Buehler and @Mariano Suárez-Alvarez: 2 up votes mean this is correct, I take it? –  Matt N. May 27 '11 at 15:39
Not quite. Using your notation, since $f=aba^{-1}b^{-1}$, it must be that $d_2=e_1^1+e_2^1-e_1^1-e_2^1=0$. –  wckronholm May 27 '11 at 22:10
@wckronholm: That's how I read it when I upvoted this answer, I read $d_2 = (+1-1)e_{1}^1 + (+1-1)e_{2}^1$, at least after a short moment of puzzlement. But I certainly agree that your way is a much clearer way of putting it :) @Matt: Note that we don't see an @-ping when we haven't commented on the post (answer). Also, only one such ping per comment works (one ping goes to the poster and one to the one addressed by @user, so Mariano wouldn't have been notified had you commented on the comment thread below your question. Be that as it may, I think you can safely accept your answer! –  t.b. May 28 '11 at 6:20
Now that you managed the torus: try to do the compact orientable surface $\Sigma_g$ of genus $g$ (the thing with $g$ holes)! Take a $4g$-gon and identify the sides appropriately. Compute its homology cellularly, then calculate its Euler characteristic $\chi(\Sigma_g)$ and verify that $\chi(\Sigma_g) = 2 - 2g$ in this way. –  t.b. May 28 '11 at 6:27

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