Second Stiefel-Whitney Class of a 3 Manifold

This is exercise 12.4 in Characteristic Classes by Milnor and Stasheff.

The essential content of the exercise is to show that $w_2(TM)=0$, where $M$ is a closed, oriented 3-manifold, $TM$ its tangent bundle, and $w_2$ the second Stiefel-Whitney class. The hint is to use Wu's formula (11.14), which in this instance says that $$w_2(TM)=v^2,$$

where $v\in H^1(M,\mathbb{Z}_2)$ is such that, for any $x\in H^2(M,\mathbb{Z}_2)$, $vx=Sq^1(x)$. Now I can show from this that $v^3=0$, but why does $v^2=0$? I don't think I know enough about Steenrod squares to finish!

I should mention that I know there are other ways of showing $w_2(TM)=0$ (spin structures, etc.), but I am trying to follow the hint!

• If $TM$ is orientable then $w_1(TM)$ vanishes, which implies that $v^2 = 0$. – user27126 Jan 10 '13 at 20:38

$\newcommand{\Sq}{\operatorname{Sq}}$ For this exercise, it is useful to recall the following property of the Steenrod squares $$\Sq^k: H^n(M; \Bbb Z/2) \longrightarrow H^{n+k}(M; \Bbb Z/2):$$

$(\ast)$ If $a \in H^n(M)$, then $\Sq^0(a) = a$, $\Sq^n(a) = a \smile a$, and $\Sq^k(a) = 0$ for all $k > n$.

The axiomatic description of the Steenrod squares, which includes the above property as an axiom, is given by Milnor and Stasheff in Chapter 8, where they use the Steenrod squares to construct Stiefel-Whitney classes.

We can use $(\ast)$ to compute all the Stiefel-Whitney classes of a compact, oriented $3$-manifold $M$. In what follows, we write $v_k$ for the degree $k$ part of the total Wu class $$v = 1 + v_1 + v_2 + v_3.$$

Since $M$ is oriented, we know that $w_1 = 0$. By Wu's formula, $$w_1 = \Sq^1(1) + \Sq^0(v_1) = v_1,$$ where we used $(\ast)$ twice. Hence $v_1 = 0$ in this case.

Now let us show that $w_3$ is zero. First, by the definition of the Wu class, $$v_2 \smile x = \Sq^2(x)$$ for all $x \in H^1(M; \Bbb Z/2)$. But since $1 < 2$, we have that $$v_2 \smile x = 0$$ for all $x \in H^1(M; \Bbb Z/2)$, and hence $$v_2 = 0.$$ Similarly, $$v_3 = 0.$$ Now by Wu's formula, $$w_3 = \Sq^3(1) + \Sq^2(v_1) + \Sq^1(v_2) + \Sq^0(v_3) = 0$$ by using $(\ast)$ on the first two terms and the fact that $v_2 = v_3 = 0$ on the last two terms.

Finally, by Wu's formula, $$w_2 = \Sq^2(1) + \Sq^1(v_1) + \Sq^0(v_2) = 0$$ by using $(\ast)$ on the first term and the fact that $v_1 = v_2 = 0$ on the last two terms.

We conclude that $w_1 = w_2 = w_3 = 0$.

• Sorry for being MIA for a few days. Anyway, this is indeed the best answer. Simple and to the point. Here I am fiddling with Bocksteins, and it never occured to me to look at what Wu's formula says about $w_1$. Thanks again! – user641 Jan 15 '13 at 6:12