I am trying to do the following exercise :
Show that there does not exist a simply connected closed manifold of dimension $6$ with $H_2(M)=0$ and $\chi(M)=1$.
First thing we notice is that since $M$ is simply connected then it's orientable. Now using Poincare-Duality and the universal coefficients theorem we know that $\mathrm{rk} (H_i)= \mathrm{rk}(H_{6-i}(M))$. From the fact that $\chi(M)=1$, and that $H_2(M)=0=H_1(M)$ we obtain that $\mathrm{rk}(H_3(M))=1$. Now I don't know how to go from here.
Any hint is appreciated. Thanks in advance.
Best Answer
You have shown that $\text{rank}(H^3(M))=\text{rank}(H_3(M))=1$. Thus $H^3(M)$ module torsion is a copy of $\mathbb{Z}$ generated by a cocycle $\alpha\in H^3(M)$. Let $H^3_f(M)$ denote the subgroup of $H^3(M)$ generated by $\alpha$. The cup product pairing $$ H^3_f(M)\times H^3_f(M) \overset{(\varphi,\psi)\mapsto(\varphi\smile \psi)[M]}{\longrightarrow} \mathbb{Z} $$ is non-singular and sends $(\alpha,\alpha)$ to a generator, and by graded commutativity of the cohomology ring of $M$, we get that this is a skew-symmetric form. Any skew-symmetric non-singular form can only exists for even-dimensional rank. This implies that $H^3_f(M)$ is even-dimensional which is a contradiction.