The cohomology ring of an oriented closed $3$-manifold with $\pi_1=\mathbb Z$

Question

Let $M$ be a closed, orientable, connected manifold of dimension $3$, such that $\pi_1(M)=\mathbb Z$. Find its cohomology ring $H^*(M;\mathbb Z)$.

Clearly $H^0=H^3=\mathbb Z$. Now since connected implies path connected on a manifold and we know $H_1=\pi_1/[\pi_1,\pi_1]=\mathbb Z$, and by Poincaré duality we have $H^2=\mathbb Z$. Since there is no torsion in $H_0$, then $H^1(M)=\hom(H_1,\mathbb Z)=\mathbb Z$. So the cohomology ring has one generator in each positive degree, let's call them $\alpha\in H^1,\beta\in H^2,\gamma\in H^3$. The only nontrivial relations to find are $\alpha\cup\beta$ and $\alpha\cup\alpha$.

By singularity of the cup product pairing $H^1\times H^2\to \mathbb Z$ with $(\phi,\psi)\mapsto(\phi\cup\psi)[M]$, and $[M]$ the fundamental class, we have that $\alpha$ is a generator if and only if $\alpha\cup n\beta=\gamma$ for some integer $n$. If it were true that $\alpha\cup \alpha=\beta$, then we could conclude that $n=\pm1$, but I don't know how to show (or not show) the latter.

Answer 1

You can finish easily using some basic algebraic properties of the cup product: namely, it is bilinear and graded commutative. I encourage you to see how you can finish on your own using these properties before reading the solution hidden below.

First, the cup product is bilinear, so $\alpha\cup n\beta=n(\alpha\cup \beta)$ and so if $n$ were not $\pm1$ then there would be no possible value of $\alpha\cup\beta$. Thus $n=\pm1$, and without loss of generality $n=1$ by changing your generators if necessary.

As for $\alpha\cup\alpha$, since $\alpha$ has degree $1$ and the cup product is graded-commutative, $\alpha\cup\alpha=-\alpha\cup\alpha$ which immediately implies $\alpha\cup\alpha=0$ since $H^2$ is torsion-free.