What kind of manifolds have the same Stiefel Whitney classes and dual Stiefel Whitney classes? I have been studying Milnor and Stasheff's Characteristic Classes and had the following question. 
Let $M$ be an $n$-dimensional differentiable manifold. Let $w_i(M)$ be the $i^{\text{th}}$ Stiefel Whitney (SW) class of the tangent bundle of $M$. We have $w(M)=1+w_0+w_1+\cdots$ is the total SW class and the total dual SW class $\overline w(M)=1+\overline w_0+\overline w_1+\cdots$ where each $\overline w_i$ is the $i^{\text{th}}$ dual SW class. By Whitney duality theorem $\overline w_i(M)=w_i(\nu)$ where $\nu$ is the normal bundle of $M$.
My question: 

Is it possible for $w(M)=\overline w(M)$ (and hence for all $i$ , $w_i(M)=\overline w_i(M)$)? What class of manifolds would satisfy (if at all) this property?

Thank you.
 A: Let $K$ denote the Klein bottle.
Recall that $K$ is non-orientable, so $w_1(K)\neq 0$. As the top Stiefel-Whitney number of a closed manifold is the mod $2$ reduction of its Euler characteristic (see Corollary $11.12$ of Characteristic Classes by Milnor and Stasheff for the non-orientable case) and $\chi(K) = 0$, we see that $w_2(K) = 0$ so $w(K) = 1 + w_1(K) \neq 1$.
On any closed surface $M$, its second Wu class vanishes so $w_1(M)^2 + w_2(M) = 0$; in particular, $w_1(K)^2 = 0$. So we see that 
$$w(K)(1 + w_1(K)) = (1 + w_1(K))(1+w_1(K)) = 1 + 2w_1(K) + w_1(K)^2 = 1$$ 
and therefore $\overline{w}(K) = 1 + w_1(K) = w(K)$.
By the same argument, connected sums of Klein bottles also have this property; in fact, these are the only closed surfaces with this property. For rather trivial reasons, this property is preserved under products with manifolds $M$ such that $w(M) = 1$. In particular, for any $n \geq 2$, $K\times S^{n-2}$ is an $n$-dimensional example with non-trivial total Stiefel-Whitney class.
All of the above examples are non-orientable. An orientable example is given by $M =  \mathbb{CP}^2\#\mathbb{CP}^2$. As $H^i(M; \mathbb{Z}_2) = 0$ for $i = 1, 3$, $w(M) = 1 + w_2(M) + w_4(M)$. As $M$ is not spin, $w_2(M) \neq 0$ and as $\chi(M) = 4$ is even, $w_4(M) = 0$, so $w(M) = 1 + w_2(M)$ and hence $w(M)^2 = 1 + w_2(M)^2$. On any closed four-manifold, its fourth Wu class vanishes, which in this case tells us that $w_2(M)^2 = 0$ so $w(M)^2 = 1$ and hence $\overline{w}(M) = 1 + w_2(M) = w(M)$. By the same argument, $\mathbb{CP}^2\#\overline {\mathbb{CP}^2}$ is another orientable example.
