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Here's the previous question: Homology of the Klein Bottle It asks what are the homology groups of the Klein bottle.

My question is this: Are we always working over $\mathbb{Z}$? Say we denote by $C_n^{\Delta}(X;F)$ the vector space over a field $F$ with basis the $n$-simplices of some space $X$. Now let $X=K$ the Klein bottle. How does $H_n(K;F)$ differ for $F=\mathbb{R},\mathbb{Q},\mathbb{F}_p,\mathbb{Z}$?

I guess where I might be confused is what we actually do when $C_n^{\Delta}(X;F)$ is over a field $F$ and not just the integers. In the nice square of the Klein bottle, does the $0$-simplex $v$ now generate $\mathbb{R}$? And similarly for the 1-simplices $a,b,c$? Or is it simply that the coefficients of the formal sums $\sum \alpha_n \sigma_{\alpha}^n$ in $C_n^{\Delta}(X;F)$ now can take on real values? I'm still not sure how that would affect the homology groups.

I hope you can help shed some light on the topic! Thanks.

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Numbersandsoon
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1 Answers1

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You can just do the same calculations as with integer coefficients. Using the same notations as in this answer to your previous question, you have $H_1(K;G)=\langle d\rangle/\langle 2d\rangle\oplus\langle c\rangle=G\oplus G/2G$.

The boundary of a chain $g_1U+g_2L$ with coefficients $g_1,g_2\in G$ is $g_1(a+b-c)+g_2(a-b+c)=(g_1+g_2)a+(g_1-g_2)b+(g_2-g_1)c$ which is zero iff $g_1=g_2$ and $g_1+g_2=0$, so $H_2(K;G)=\{gU+gL: 2g=0\}\cong\{g\in G:2g=0\}$.

$H_0$ is simply the direct sum of copies of $G$, one for each path-component, so it is just $G$ here.

The universal coefficient theorem states that there is an exact sequence

$0\to H_n(K)\otimes G\to H_n(K;G)\to H_{n-1}(K)\star G\to 0,$

where $A\star B$ is the so-called "torsion product" of $A$ and $B$. Here are some properties of $\star$:

  • $A\star B=B\star A$,
  • $A\star\mathbb Z=0$,
  • $A\star\mathbb Z_n=\text{ker}(A\overset n\to A)=\{a\in A: na=0\}$,
  • $A\star\bigoplus_{i=1}^nB_i=\bigoplus_{i=1}^n(A\star B_i)$.

Now the universal coefficient theorem gives the exact sequences

$0\to 0\otimes G=0\to H_2(K;G)\overset\cong\to (\mathbb Z\oplus\mathbb Z_2)\star G=\mathbb Z_2\star G\to 0$

and

$0\to(\mathbb Z\oplus\mathbb Z_2)\otimes G=G\oplus G/2G\overset\cong\to H_1(K;G)\to\mathbb Z\star G=0\to 0,$

which confirm the calculations above.

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Benedikt Hunger
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