Every unorientable 4-manifold has a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure

Question

The precise statement on J. W. Morgan's "The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds (MN-44)" that 4-manifold $X$ admits a Spinc structure (Lemma 3.1.2) seems to be that every 4-orientable manifold X admits a $Spin^c$ structure. This means that we impose the $w_1(X)=0$ for the 4-orientable manifold X.

However, for 4-unorientable manifold $M$, we may modify the statement to different structures.

In addition to, $Spin^c=\frac{(Spin \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$

we can have: $$Pin^c=\frac{(Pin^+ \times U(1))}{\mathbb{Z}/2\mathbb{Z}}=\frac{(Pin^- \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$Pin^{\tilde c+}=\frac{(Pin^+ \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$Pin^{\tilde c-}=\frac{(Pin^- \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}}.$$

See for example this Ref, Annals of Physics 394, 244-293 (2018) and References therein.

It seems that I can improve John Morgan's statement to show that

Every unorientable 4-manifold has either a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure. (?)

e.g. My approach is based on improving the map

\begin{equation*} H^1(X;Pin^c) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2), \end{equation*}

\begin{equation*} H^1(X;Pin^{\tilde c+}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2), \end{equation*}

\begin{equation*} H^1(X;Pin^{\tilde c-}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2). \end{equation*} The last maps of all three need to have appropriate constraints between $c_1$ and $w_2(M)$, $w_1(M)$ and $w_1^2(M)$.

Question: I wonder whether there exists any math literature show the similar results like mine above? Or are my statements obviously true? (Or obviously wrong?)

Answer 1

$\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$ The conjecture is false: $\RP^4\amalg(\RP^2\times\RP^2)$ is an unorientable 4-manifold that has no pin^$c$, pin^{$\tilde c-$}, or pin^{$\tilde c+$} structure.

It suffices to find three unorientable 4-manifolds $A$, $B$, and $C$, such that $A$ isn't pin^$c$, $B$ isn't pin^{$\tilde c-$}, and $C$ isn't pin^{$\tilde c+$}. Then $A\amalg B\amalg C$ doesn't admit any of the three structures: a $G$-structure is a reduction of the principal bundle of frames, so a $G$-structure on a manifold induces a $G$-structure on each connected component.

First, $A = \RP^2\times\RP^2$ doesn't have a pin^$c$ structure; this is discussed here.

Then, $B = \RP^4$ has no pin^{$\tilde c-$}-structure. Let $H^*(-; \Z_{w_1})$ denote integer cohomology twisted by the orientation bundle. From Shiozaki-Shapourian-Gomi-Ryu, Lemma D.7, we learn that a manifold $M$ has a pin^{$\tilde c-$}-structure iff $w_2 + w_1^2\in H^2(M;\Z/2)$ admits a lift across the mod 2 reduction map

$$\tag{$*$} H^2(M;\Z_{w_1})\longrightarrow H^2(M; (\Z/2)_{w_1}) = H^2(M;\Z/2).$$

Twisted Poincaré duality tells us that if $M$ is a closed $n$-manifold, there are isomorphisms $H_k(M;\Z)\cong H^{n-k}(M;\Z_{w_1})$, so $H^2(\RP^4;\Z_{w_1})\cong H_2(\RP^4;\Z) = 0$. A quick computation shows that if $x\in H^1(\RP^4;\Z/2)$ is the generator, then $w_2(\RP^4) + w_1(\RP^4)^2 = x^2$; in particular, it's nonzero, so it can't lift to $H^2(\RP^4;\Z_{w_1})$. Therefore $\RP^4$ admits no pin^{$\tilde c-$}-structure.

Finally, $C = \RP^2\times\RP^2$ doesn't have a pin^{$\tilde c+$}-structure. Lemma D.8 of the Shiozaki-Shapourian-Gomi-Ryu paper tells us that a manifold $M$ has a pin^{$\tilde c+$}-structure iff $w_2\in H^2(M;\Z/2)$ admits a lift across ($*$).

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 ($*$) for $M = \RP^2$ 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 ($*$), and therefore $\RP^2\times\RP^2$ has no pin^{$\tilde c+$}-structure.