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Let $X$ be a finite CW complex and $\chi(X)$ its Euler characteristic (defined using integer coefficients). When is it true that $\chi(X)=\sum (-1)^i \dim H_i(X;F)$, where $F$ is a field?

I thought it would be true for all fields, but I noticed that for $X$ the Klein bottle and $F=\mathbb{Z}/2$ this is false! In fact Bredon (Geometry and topology) claims it to be true, but it isn't..

Najib Idrissi
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Angry Euler
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    Could you give a more precise reference to Bredon? – Hoot Jul 10 '15 at 16:49
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    How is it not true for the Klein bottle? $\chi(X) = 0$ in any case. Bredon is correct that it doesn't depend on the choice of field - use the universal coefficient theorem. –  Jul 10 '15 at 16:50
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    Mike Miller is right. I made a mistake. If you calculate carefully, then, over a field of characteristic $2$, you should get $\chi(\text{Klein bottle})=1-2+1=0$, while you would get $\chi(\text{Klein bottle})=1-1+0=0$ for other fields. Hence, the Klein bottle is not a counterexample that you claimed to be. – Batominovski Jul 10 '15 at 17:03

2 Answers2

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$\newcommand{\Z}{\mathbb{Z}}$The Klein bottle has the following integral homology groups: $$H_*(K; \Z) = (\Z, \Z \oplus \Z/2\Z, 0, 0, \dots).$$ This gives an Euler characteristic $\chi_\Z = 1-1 = 0$. Over $\Z/2\Z$, the universal coefficient theorem yields: $$H_*(K; \Z/2\Z) = (\Z/2\Z, \Z/2\Z^2, \Z/2\Z, 0, 0, \dots)$$ and so the Euler characteristic is $\chi_{\Z/2\Z} = 1-2+1 = 0$. This agrees with the previous computation.

In general, suppose $X$ is a space with finite integral homology (i.e. a finite number of nonzero homology groups, and these groups are all finitely generated), for example a finite CW-complex. Then its Euler characteristic is well-defined. Let $n$ be the top dimension of nonvanishing homology.

  1. First suppose that $\newcommand{\F}{\mathbb{F}}\F$ is a field of characteristic zero. Let $b_k$ be the $k$th Betti number of $X$, ie. $$H_k(X;\Z) = \Z^{b_k} \oplus T$$ where $T$ is the torsion subgroup. Then $$\chi_\Z = b_0 - b_1 + b_2 - \dots = \sum_k (-1)^k b_k.$$ Now the universal coefficient theorem says $$H_k(X;\F) = H_k(X;\Z) \otimes \F \oplus \operatorname{Tor}(H_{k-1}(X;\Z), \F).$$ Since the field is of characteristic zero, the $\operatorname{Tor}$-term vanishes, and you're left with $H_k(X;\F) = \F^{b_k}$. It follows that $\chi_\F = \chi_\Z$.
  2. Suppose now that $\F$ is a field of characteristic $p$. Suppose also that $H_k(X; \Z) = \Z^{b_k} \oplus (\Z/p\Z)^{c^p_k} \oplus T^p_k$, where $T^p_k$ is the torsion part which is not $p$-torsion. The universal coefficient theorem gives: $$H_k(X; \F) = \begin{cases} \F^{b_0 + c^p_0} & k = 0 \\ \F^{b_k + c^p_k + c^p_{k-1}} & 1 \le k \le n \\ \F^{c^p_n} & k = n+1 \end{cases}$$ (in this formula, I write $c^p_0$ knowing fully well that in fact $c^p_0 = 0$. It does not matter.) Then the Euler characteristic becomes: $$\chi_\F = (b_0 + c^p_0) - (b_1 + c^p_1 + c^p_0) + \dots + (-1)^n (b_n+c^p_n+c^p_{n-1}) + (-1)^{n+1}c^p_n$$ Each $c^p_k$ cancels with the one in the next factor, so all is left is $$\chi_\F = b_0 - b_1 + \dots = \chi_\Z.$$
Najib Idrissi
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  • don't you think it would be more elegant to prove it regardless of the characteristic, without a case distinction? – Mister Benjamin Dover Jul 11 '15 at 11:59
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    Sure @BenDover. Do you happen to have such a proof? I would be glad to read it. – Najib Idrissi Jul 11 '15 at 12:37
  • Well, my proof by cellular homology has no case study – Michael Jul 11 '15 at 12:39
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    But it only works for finite CW complexes, @Michael, and requires some homological algebra lemma (I'm not saying it's a bad proof, just that there are pros and cons to each approach). In general I don't know if any space with finite homology is weakly equivalent to a finite CW complex (it might very well be the case, though instinctively I would say no). – Najib Idrissi Jul 11 '15 at 12:41
  • Ok, that's a point – Michael Jul 11 '15 at 12:45
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    @NajibIdrissi we have $\dim_F \mathrm{Tor}(H_{k-1}(X),F)=\dim_F \mathrm{Tor}(H_{k-1}(X){\mathrm{tor}},F)=\dim_F H{k}(X)_{\mathrm{tor}}\otimes F$, where the last follows from the exact sequence $0\rightarrow\mathrm{Tor}(F,\mathbb{Z}/n)\rightarrow F\rightarrow F\rightarrow F\otimes\mathbb{Z}/n\rightarrow 0$. – Mister Benjamin Dover Jul 11 '15 at 14:24
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    @NajibIdrissi: Pick some absurdly big (really, not finitely presented suffices) perfect group $G$; for instance, $G = SL_2(\mathbb R)$. Build a CW complex with $\pi_1(X) = G$. Now add higher cells to kill off higher homology. If you require $\pi_1(X)$ finitely presented I think it might be true that if $H_i(X)$ are all finitely generated, $X$ is homotopy equivalent to a finite complex. –  Jul 11 '15 at 16:34
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There is also a way showing the independence of $\chi (X)$ without using the universal coefficient theorem, but using cellular homology:

Let $a_n$ denote the number of cells of $X$ in dimension $n$. Then we get the cellular chain complex $$\ldots \longrightarrow F^{a_k} \longrightarrow F^{a_{k-1}} \longrightarrow \ldots $$ Using that the Euler characteristic of a chain complex equals that of its Homology (which is proved using the dimension formula for vector spaces), we get $$\chi_F=\sum (-1)^i a_i$$ and $$\chi_F=\sum (-1)^i \dim H_i(X;F)$$ which is then independent of the field.

Michael
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  • this does not explain the relation to integral homology. – Mister Benjamin Dover Jul 11 '15 at 11:58
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    Well, the definition of the integral euler characteristic goes via the rank of the integral groups, which is the dimension of the corresponding rational homology. Thus by definition rational and integral euler characteristic agree. – Michael Jul 11 '15 at 12:37