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I'm working on something that involves tori, and specifically I'm looking for the first homology group of the $n$-torus with coefficients in $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$. However, the Universal Coefficient Theorem and the algebraic tools needed (tensor products of modules, the Tor functor etc.) werent't covered in my algebraic topology class last year, so I'm just asking if you could check and see if I've understood things properly.

With $n$-torus I mean $\mathbb{T}^n := (S_1)^n$. Then we know that $$H_m (\mathbb{T}^n,\mathbb{Z}) = \mathbb{Z}^{\binom{n}{m}}$$ (right?), so in particular $$H_1 (\mathbb{T}^n,\mathbb{Z} ) = \mathbb{Z}^n.$$

Now we consider the exact sequence in the universal coefficient theorem; I believe the Tor term $\mathrm{Tor}(\mathbb{Z}^m,\mathbb{Z}_2)$ vanishes for any $m$, since $\mathbb{Z}^m$ is free. Correct?

This then yields an isomorphism $$H_1(\mathbb{T}^n,\mathbb{Z}_2) = H_1(\mathbb{T}^n,\mathbb{Z}) \otimes \mathbb{Z}_2 = (\mathbb{Z}_2)^n,$$ since $\mathbb{Z}^m \otimes \mathbb{Z}_2 = (\mathbb{Z}_2)^m$ for any $m$. Is this also correct?

Thanks for your time!

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«I believe the Tor term vanishes since $\mathbb Z^m$ is free. Correct?» Yes. – Mariano Suárez-Alvarez Jun 6 '11 at 16:56
«Is this also correct?» Yes, too. – Mariano Suárez-Alvarez Jun 6 '11 at 16:57
Following Mariano's style: "right?" Yes. You can see this for instance using the Künneth formula. – Raeder Jun 6 '11 at 16:59
@Mariano: thanks for your reply; if you post it as an answer, I'll be happy to accept it. – Gerben Jun 6 '11 at 17:04
@Raeder: I've only seen the Künneth formula in the context of cohomology, but I'll look into the chapter of Hatcher. – Gerben Jun 6 '11 at 17:04
up vote 4 down vote accepted

This is all correct, but it is conceptually more complicated than it needs to be to compute the first homology group of the $n$-torus. The $n$-torus is a product of $n$ circles, and therefore has a cell structure consisting of one vertex, $n$ edges, etc., with $\displaystyle\binom{n}{k}$ cells in dimension $k$. (This is the $n$-fold product of the cell structure on the circle that has one vertex and one edge.) Each of these cells has trivial boundary, so we get the following chain complex: $$ \mathbb{Z}_2 \;\;\xrightarrow{\;\;0\;\;}\;\; \cdots \;\;\xrightarrow{\;\;0\;\;}\;\; \mathbb{Z}_2^\binom{n}{2} \;\;\xrightarrow{\;\;0\;\;}\;\; \mathbb{Z}_2^n \;\;\xrightarrow{\;\;0\;\;}\;\; \mathbb{Z}_2 $$ Therefore, $H_1(\mathbb{T}^n,\mathbb{Z}_2) \cong \mathbb{Z}_2^n$, and more generally $H_k(\mathbb{T}^n,A) \cong A^\binom{n}{k}$ for any abelian group $A$.

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Thanks for the less formal point of view! – Gerben Jun 6 '11 at 17:40
I don't that Jim's answer is any less formal, but it certainly uses less abstract nonsense. – jd.r Jun 6 '11 at 17:49

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