It's known that all oriented 4-manifolds admit a $Spin^c$ structure, ie. a spin structure on $TX\oplus\mathcal{L}$ for some complex line bundle $\mathcal{L}$. 


A usual generalization of this structure to unorientable manifolds is to ask for a spin structure on $TX\oplus\mathcal{L}\oplus\mathcal{E}$ where $\mathcal{L}$ is again a complex line and now $\mathcal{E}$ is a real line bundle (which the whole bundle being oriented will force to be the orientation line). This is called a $Pin^c$ structure. In cohomology, it amounts to an integral lift of $w\_2$.


Unfortunately, not all 4-manifolds admit a $pin^c$ structure, eg. $\mathbb{RP}^2 \times \mathbb{RP}^2$. This is easy to see by a computation of $w\_2$.


There is another generalization I'll call a $Pin^{\tilde c}$ structure. In this version, $\mathcal{L}$ and $\mathcal{E}$ combine into a real 2-plane bundle. In cohomology it amounts to a *twisted* integral lift of $w\_2$.


So, do all 4-manifolds admit a twisted integral lift of $w\_2$?


Here are some edits in response to Qiaochu's comment. An integral lift of $w\_2$ is an element of $H^2(X,\mathbb{Z})$ mapping to $w\_2$ in $H^2(X,\mathbb{Z}/2)$. A twisted integral lift lives instead in $H^2(X,\mathbb{Z}^{w\_1})$, with local coefficients in the orientation line.



Are all 4-manifolds $Pin^{\tilde{c}}$?
$\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$No, $\RP^2\times\RP^2$ isn't pin$\tilde c$. This came up while thinking about [another MathOverflow
question](https://mathoverflow.net/q/310373/97265), but I'll rewrite the argument here.


What you call a pin$\tilde c$-structure has been studied in physics, following
[Metlitski](https://arxiv.org/abs/1510.05663) and [Freed-Hopkins](https://arxiv.org/abs/1604.06527), where it's
also called a pin$\tilde c+$-structure. (This is because one can also ask about pin$\tilde
c-$-structures, which are twisted integral lifts of $w\_2 + w\_1^2$.) These structures are discussed in
a little more detail by [Shiozaki-Shapourian-Gomi-Ryu](https://arxiv.org/abs/1710.01886) in Appendix D.


Here's the argument for $\RP^2\times\RP^2$: let $x$ denote the generator of $H^1(-;\Z/2)$ of the first $\RP^2$ and
$y$ be that for the second $\RP^2$. Using the usual CW structure on $\RP^2$ and the product CW strucure on
$\RP^2\times\RP^2$, you can check that $H\_2(\RP^2\times\RP^2;\Z)\cong\Z/2$ and the reduction mod 2 map
$H\_2(\RP^2\times\RP^2;\Z)\to H\_2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H\_2(\RP^2\times\RP^2;\Z)$ to
the Poincaré dual of $x+y$.


The Poincaré duality isomorphisms $H\_2(\RP^2\times\RP^2;\Z)\cong H^2(\RP^2\times\RP^2;\Z\_{w\_1})$ and
$H\_2(\RP^2\times\RP^2;\Z/2)\cong H^2(\RP^2\times\RP^2;\Z/2)$ are natural with respect to change-of-coefficients,
which means that the reduction mod 2 map $\rho\_2\colon H^2(\RP^2\times\RP^2;\Z\_{w\_1})\to
H^2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H^2(\RP^2\times\RP^2;\Z\_{w\_1})\cong\Z/2$ to $x+y$.
However, $w\_2(\RP^2\times\RP^2) = x^2+xy+y^2$, so it's not in the image of $\rho\_2$, and therefore
$\RP^2\times\RP^2$ has no pin$\tilde c+$-structure.


