Euler characteristic: dependence on coefficients

Question

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..

Answer 1

$\newcommand{\Z}{\mathbb{Z}}$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.

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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 $$\chi_\F = b_0 - b_1 + \dots = \chi_\Z.$$

Answer 2

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.