Let \(M\) be a closed orientable \(4\)-manifold whose second homology \(H_2(M; \mathbb{Z})\) has rank \(1\). Show that there does not exist a free action of the group \(\mathbb{Z}_2\) on \(M\).
Solution
1. Let \(b_i=\operatorname{rank}H_i(M;\mathbb Z)\). Since \(M\) is closed and orientable of dimension \(4\), Poincaré duality gives\[ b_0=b_4=1,\qquad b_1=b_3. \]
This is the rank form of integral Poincaré duality.
2. Since \(b_2=1\) by hypothesis,\[ \chi(M)=1-b_1+1-b_3+1=3-2b_1, \]which is odd.
Insert the Betti numbers into the alternating-sum formula for Euler characteristic and use \(b_3=b_1\).
3. If a free \(\mathbb Z/2\)-action existed, the quotient map\[ M\to M/(\mathbb Z/2) \]would be a two-sheeted covering of a finite CW complex.
A free action of a finite group on a closed manifold is properly discontinuous and the quotient is again a compact manifold; the orbit map is a covering.
4. Euler characteristic would then satisfy\[ \chi(M)=2\chi(M/(\mathbb Z/2)), \]which is even.
Euler characteristic multiplies by the degree of a finite covering.
5. This contradicts
2. Therefore\[ \boxed{\text{no free }\mathbb Z/2\text{-action exists on }M.} \]
An integer cannot be both odd and even.