TITLE: Manifolds with $w_1(TM)\cup w_1(TM)=0$ and $w_2(TM)=0$ but $w_1(TM)\neq 0$
QUESTION [5 upvotes]: For a generic dimension $d$, is there an nonorientable manifold $M$ (i.e. $w_1(TM)\neq 0$) with vanishing $w_1(TM)\cup w_1(TM)$ and $w_2(TM)$, i.e.,
$$w_1(TM)\cup w_1(TM)=0, ~~~~~ w_2(TM)=0, ~~~~~w_1(TM)\neq 0?$$
Here $w_i(TM)$ is the $i^{\text{th}}$ Stiefel-Whitney class of the tangent bundle of the manifold $M$. For $d=2$, the Klein bottle is an example.
If such manifolds exist, what kind of structure do they carry? For example, if $w_1(TM)=0$, and $w_2(TM)=0$, then the manifold can be equipped with a spin structure, and we say it is a spin manifold. I'd like to see what is the corresponding structure in the above more complicated case.

REPLY [6 votes]: A smooth manifold $M$ admits a pin$^+$ structure if and only if $w_2(M) = 0$, and a pin$^-$ structure if and only if $w_1(M)^2 + w_2(M) = 0$; see this page for some information on pin structures. The manifolds you are enquiring about satisfy both conditions and hence admit pin$^+$ and pin$^-$ structures. However, as $w_1(M) \neq 0$, they are non-orientable, so they do not admit spin structures.
There exist manifolds of arbitrary dimension which satisfy your requirements. For example, $K\times S^n$ where $K$ is the Klein bottle.