# Stiefel Whitney numbers of a $3$- manifold are $0$

In Milnor's book on Characteristic Class he asks to prove that (Problem 11-D) all Stiefel Whitney numbers of a $3$- manifold are $0$. I can show $w_{3}=0$ as $\chi (M)=0$. From dimension consideration only non zero Wu class is $v_1$ and $w_1^3=w_2w_1=v_1^3$. I am stuck how to show $v_1^3=0.$

Can somebody help me?

• do you know that every 3-manifold bounds? May 16, 2015 at 23:25
• Yes I know the fact but I think I am not supposed to use this fact to prove that exercise. May 17, 2015 at 7:45

This problem is an application of the properties of Wu classes to the tangent bundle of a smooth 3 manifold.

In Milnor's notation, the k'th Wu class satisfies the conditions, $<v_k\cup x,\mu> = <Sq^k(x),\mu>$ for all (n-k)-dimensional mod 2 cohomology classes,x, and $\omega_k = \Sigma_{i+j=k}Sq^i(v_j)$ where $\omega_k$ is the k'th Stiefel-Whitney class. Here, $\mu$ is the fundamental mod 2 homology class of the manifold.

The calculations are

• $\omega_1 = Sq^1(v_0) + Sq^0(v_1)$ = $0 + v_1$ by the properties of the Steenrod squares. Thus $v_1 = \omega_1$

• $\omega_2$ = $Sq^2(v_0) + Sq^1(v_1) + Sq^0(v_2)$ = $\omega_1\cup\omega_1 + v_2$. But $v_2 = 0$ since for all one dimensional cohomology classes,$x$, $<v_2\cup x,\mu> = <Sq^2(x),\mu>$ and $Sq^2(x) = 0$. Thus $\omega_2 = \omega_1\cup\omega_1$.

• $\omega_3 = Sq^0(v_3) + Sq^2(v_1) + Sq^1(v_2) + Sq^3(v_0)$ = $v_3 + 0 + 0 + 0$ = $v_3$ = $0.$ Thus $\omega_3 = 0$.

• $<Sq^1(\omega_2),\mu> = <v_1\cup \omega_2,\mu> = <\omega_1\cup\omega_2,\mu> = <Sq^1(\omega_1\cup\omega_1),\mu> = <0,\mu>$ since $Sq^1(\omega_1\cup\omega_1) = Sq^1(\omega_1)\cup\omega_1 + \omega_1\cup Sq^1(\omega_1) = 2\omega_1^3 = 0$. Thus $\omega_1\cup\omega_2 = 0$.

• $\omega_1^3 = 0$ because $\omega_1^2 =\omega_2$