Apply the theorems of Hopf and Borel to show that for an H-space \(X\) that is a connected finite CW complex with \(\tilde{H}_*(X; \mathbb{Z}) \neq 0\), the Euler characteristic \(\chi(X)\) is \(0\).
Solution
Let \[ P_F(t)=\sum_i \dim_F H^i(X;F)t^i. \] Since \(X\) is a finite CW complex, \(P_F\) is a polynomial and \[ \chi(X)=P_F(-1) \] for every field \(F\).
First take \(F=\mathbb Q\). Hopf’s theorem says that the connected graded Hopf algebra \(H^*(X;\mathbb Q)\) is, as an algebra, a tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators. Because \(X\) is finite, \(H^*(X;\mathbb Q)\) is finite-dimensional, so no polynomial generator can occur. Hence \[ H^*(X;\mathbb Q)\cong\Lambda_{\mathbb Q}(x_1,\dots,x_r) \] with all \(|x_i|\) odd. If \(r>0\), then \[ P_{\mathbb Q}(-1)=\prod_{i=1}^r(1+(-1)^{|x_i|})=0, \] so \(\chi(X)=0\).
It remains to rule out \(r=0\) under the hypothesis \(\widetilde H_*(X;\mathbb Z)\ne0\). If \(r=0\), then \(H^*(X;\mathbb Q)=\mathbb Q\), so \(\chi(X)=1\) and all positive-dimensional integral homology is torsion. Choose a prime \(p\) for which \(\widetilde H^*(X;\mathbb F_p)\ne0\); such a prime exists by the universal coefficient theorem.
Borel’s theorem expresses the finite-dimensional Hopf algebra \(H^*(X;\mathbb F_p)\) as a tensor product of single-generator factors, each either an exterior algebra on an odd-dimensional generator or a truncated polynomial algebra \[ \mathbb F_p[y]/(y^{p^k}) \] with \(|y|\) even when \(p\) is odd. If an exterior factor occurs, its Poincare polynomial vanishes at \(-1\), giving \(\chi(X)=0\), contradicting \(\chi(X)=1\). If there are no exterior factors, every nontrivial truncated factor contributes \[ 1+1+\cdots+1=p^k \] to \(P_{\mathbb F_p}(-1)\), so \(\chi(X)\) is divisible by \(p\), again contradicting \(\chi(X)=1\). Therefore \(r=0\) is impossible.
Hence necessarily \[ \boxed{\chi(X)=0.} \]