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?)

  • 5
    $\begingroup$ I don't understand why someone voted me down immediately in 1 min after my post, without leaving me any comment or without possibly reading the entire question (?). Can someone elaborate what is going on? $\endgroup$
    – wonderich
    Sep 11, 2018 at 22:53
  • 7
    $\begingroup$ I didn't downvote, and I think this question is fine, but it might have to do with the frequency of your questions on MSE and MO. One might think, certainly someone who posts an average of around one question per day hasn't properly thought the questions through for themselves. Perhaps you could draw an analogy with the content of Gerhard Paseman's answer here meta.mathoverflow.net/questions/3846/… $\endgroup$ Sep 12, 2018 at 0:20
  • $\begingroup$ Thanks, Aleksandar Milivojevic. Just make sure, do MO discourage asking a question per day? Similarly, does arXiv discourage us to post a paper every week or every month? How do we judge the question is thoughtful or the paper is good in quality or not, simply by its appearing frequency? (I mean, I read and think about math texts/questions every day, for long hours, so it is like a diary for me just to clean up my thought/question.) I remember there is even a badge to encourage asking many questions. Thank you! $\endgroup$
    – wonderich
    Sep 12, 2018 at 2:09
  • $\begingroup$ We can continue this discussion privately (send me an email if that's ok with you). $\endgroup$ Sep 12, 2018 at 2:59

1 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.

  • $\begingroup$ Thanks +1... I have to think more carefully then... $\endgroup$
    – wonderich
    Sep 14, 2018 at 21:52
  • 1
    $\begingroup$ Is the claim true for connected four-manifolds? $\endgroup$ Mar 6, 2019 at 17:31
  • $\begingroup$ @MichaelAlbanese I don't think so -- typically in these kinds of arguments, one can replace disjoint unions with connect sums, right? Is there a reason to expect that not to work for these twisted pin structures? $\endgroup$ Mar 6, 2019 at 21:01
  • $\begingroup$ @ArunDebray: I don't know anything about cohomology with twisted coefficients, so I don't know what to expect. $\endgroup$ Mar 6, 2019 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.