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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.

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    $\begingroup$ I can't extract from either of your descriptions what you mean by this. How does a complex line bundle and a real line bundle "combine into a real 2-plane bundle," and/or what is a twisted integral lift of $w_2$? (What is the twist, and what is it twisting?) $\endgroup$
    – Qiaochu Yuan
    Commented Jan 11, 2016 at 18:15
  • $\begingroup$ I've problems as well to follow your description. Could you describe the group $Pin^{\tilde c}$, please? As for $Pin^c$-structures, I guess that restricting the $Pin^c$ action to a $Spin^c$ action, you recover a $Pin^c$ structure on the orientation covering of $X$. Is this correct? $\endgroup$
    – Sebastian Goette
    Commented Jan 12, 2016 at 12:16
  • $\begingroup$ A $Pin^c$ structure is an equivariant $Spin^c$ structure on the orientation double cover, while a $Pin^{\tilde c}$ structure is a twisted equivariant $Spin^c$ structure there. (The Chern class flips sign between the two sheets.) $\endgroup$
    – user404153
    Commented Jan 12, 2016 at 17:59
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    $\begingroup$ You are right, I made a mistake (I momentarily forgot that the Euler class of a non-orientable bundle lives in twisted cohomology...). So $w_2 + w_1^2$ must be the mod 2 reduction of a class in twisted cohomology. Then I think that $X =\mathbb{RP}^4$ provides a counter-example. If I did my back of the envelope calculations right (please check) then $w(TX) = 1 + x + x^4$ and so $w_2 + w_1^2 = x^2$ is non-trivial, but I think $H^2(X, \mathbb{Z}^{w_1}) = 0$, so this class can't be the reduction of a twisted cohomology class. $\endgroup$
    – Chris Schommer-Pries
    Commented Jan 14, 2016 at 12:57
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    $\begingroup$ Oh you're looking at a third structure actually. I'm looking for a twisted integral lift of $w_2$, not $w_2+w_1^2$. These are inequivalent in the twisted case. $\endgroup$
    – user404153
    Commented Jan 16, 2016 at 21:19

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$\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$No, $\RP^2\times\RP^2$ isn't pin$\tilde c$. This came up while thinking about another MathOverflow question, but I'll rewrite the argument here.

What you call a pin$\tilde c$-structure has been studied in physics, following Metlitski and Freed-Hopkins, 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 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.

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  • $\begingroup$ Thanks, Arun! Indeed it seems @wonderich and I are thinking about the same things :) $\endgroup$
    – user404153
    Commented Sep 15, 2018 at 2:11

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