# If $n$ is not a power of 2 then the dual Stiefel-Whitney class $\bar{w}_{n-1} = 0$

Stiefel-Whitney classes are invertible and for $$w$$, the Stiefel-Whitney class of the tangent bundle of $$M$$, we have its inverse $$\bar{w}$$. I want to prove that if $$n$$ is not a power of 2 then the dual Stiefel-Whitney classes $$\bar{w}_{n-1} = 0$$.

For the Steenrod square $$\mathrm{Sq} : H^\Pi (M) \rightarrow H^\Pi(M)$$ we can define its inverse $$\bar{Sq}$$. Then we can see easily that $$\langle \bar{Sq}(x), \mu \rangle = \langle \bar{w} \cup x , \mu \rangle$$. Set $$x$$ as suitable element in $$H^0(M)$$ we can prove that $$\bar{w}_n = 0$$. And I also think of proving $$\bar{w}_{n-1} = 0$$ by letting $$x$$ be Poincaré dual to $$\bar{w}_{n-1}$$. However I can't check that $$\langle \bar{Sq}(x) , \mu \rangle = 0$$.

• The proof is in Massey's "On the Stiefel-Whitney classes of a manifold" American Journal of Mathematics 1960; see Corollary 1: jstor.org/stable/pdf/2372878.pdf. Apr 24 '20 at 2:00