Vanishing of $w_2$ for orientable 3-manifolds

Question

Let $M$ be oriented manifold: this happens if and only if $w_1(M)=0$ (the first Stiefel Whitney class, being an element in $H^1(M,\mathbb{Z}_2)$. There is a result that if $M$ is three dimensional then from the vanishing of $w_1$ automatically follow that also $w_2=0$. This should be rather easy consequence of properties of Wu classes and Steenrod squares however I don't see how to proceed in the calculation. I expressed the total Stiefel Whitney class $w$ of $M$ as $Sq(v)$ where $v$ is the total Wu class. After evaluating $Sq(v)$ I arrived, using the fact that $M$ is 3 dimensional (if I'm not mistaken) at the following: $1+w_1+w_2+w_3=1+v_1+v_2++v_1 \cup v_1$. From $w_1=0$ we infer $v_1=0$ therfore also $v_1 \cup v_1=0$ and the question boils down to showing $v_2=0$ but I don't see how to prove this.

Answer 1

From the relation $Sq(v)=w$ we get that $v_2=w_2+w_1^2$. More specifically By Thm 11.14 (which is just an extended form of the above equality) in M-S you have $w_1=v_1$ and $w_2=Sq^1(v_1)+v_2=v_1^2+v_2$. Plug them together you obtain my first claim.

Recall we have a nice characterisation of the (second) Wu-class, as $$Sq^{2}(x)=v_2\smile x$$ for any $x \in H^{n-k}(M;\Bbb Z_2)$ (See Milnor Stasheff page 132). Since in our case $n=3$, we get that for any $x\in H^1(M;\Bbb Z_2)$, $0=v_2\smile x$, which gives you $v_2=0$.

To see this last equality, assume $v_2 \neq 0$. We would have $$1=\langle v_2, (v_2)_*\rangle=\langle v_2, [M]\frown x\rangle=0$$ Where $(v_2)_* \in H_2(M;\Bbb Z_2)$ is the dual of $v_2$ and Poincaré Duality (mod 2) tells us that there exists an element $x \in H^1(M; \Bbb Z_2)$ such that $[M]\frown x=(v_2)_*$

Therefore you have $0=w_2+w_1^2$, which proves your claim