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For an oriented $d$-manifold $M$, we can ask whether the manifold admits a Spin structure, say, if the transition functions for the tangent bundle, which take values in $SO(d)$, can be lifted to $\operatorname{Spin}(d)$ while preserving the cocycle condition on triple overlaps of coordinate charts.

For an unoriented $d$-manifold $M$, we can ask whether the manifold admits Pin$^+$ or Pin$^−$ structures (that is, lifts of transition functions to either $\operatorname{Pin}^+(d)$ or $\operatorname{Pin}^−(d)$ from $O(d)$. This is analogous to lifting the transition functions to $\operatorname{Spin}(d)$ from $SO(d)$ for the spin manifold).

If the manifold $M$ is orientable, then the conditions for Pin$^+$ or Pin$^−$ structures are the same. So it reduces to the old condition that $M$ admits a Spin structure or not.

Every 2-dimensional manifold admits a Pin$^−$ structure, but not necessarily a Pin$^+$ structure.

Every 3-dimensional manifold admits a Pin$^-$ structure, but not necessarily a Pin$^+$ structure.

Question: For some any other $d$, say $d = 0, 1, 4,$ etc. are there statements like: every $d$-dimensional manifold admits a Pin$^-$ structure? Or, every $d$-dimensional manifold admits a Pin$^+$ structure? Or are there useful conditions like SW classes $w_2+w_1^2=0$ like the case for $d = 2, 3$? What are these conditions in other dimensions?

P.S. The original post on MSE received almost no attention for a week.

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  • $\begingroup$ Why does every three-manifold admit a Pin$^-$ structure? I know that every closed three-manifold admits a Pin$^-$ structure. $\endgroup$ Commented Dec 24, 2016 at 0:11

2 Answers 2

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The condition for having a $Pin^+$-structure is the vanishing of $w_2$, and for having a $Pin^-$-structure is the vanishing of $w_2 +w_1^2$, for manifolds of any dimension. This is because the Lie group $Pin^\pm(d)$ is a central extension $$\mathbb{Z}/2 \longrightarrow Pin^\pm(d) \longrightarrow O(d)$$ so there are homotopy fibre sequences $$BPin^\pm(d) \overset{p^\pm}\longrightarrow BO(d) \overset{k^\pm}\longrightarrow K(\mathbb{Z}/2,2)$$ for $k^\pm \in H^2(BO(d) ; \mathbb{Z}/2)$ the class classifying the central extension, which convention dictates is $k^+ = w_2$ and $k^- = w_2 + w_1^2$.

If if $\pi : E \to B$ is a $d$-dimensional vector bundle (which need not be the tangent bundle of a manifold) classified by a map $f : B \to BO(d)$, then $f$ lifts along $p^\pm : BPin^\pm(d) \to BO(d)$ if and only if $k^\pm \circ f : B \to K(\mathbb{Z}/2,2)$ is nullhomotopic, i.e. if and only if $f^*k^\pm =0 \in H^2(B;\mathbb{Z}/2)$.

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  • $\begingroup$ If Pin+-structure is the vanishing of w2, isn't that the same as the Spin structure is the vanishing of w2? What are the differences of criteria for Pin+-structure and Spin-structure ? $\endgroup$
    – miss-tery
    Commented Dec 23, 2016 at 17:45
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    $\begingroup$ A Spin structure exists if both $w_1$ and $w_2$ vanish, whereas a $Pin^+$ structure only requires the latter. $\endgroup$ Commented Dec 23, 2016 at 17:56
  • $\begingroup$ I admit I am a bit of an outsider in this discussion, but I am very confused by the following: Why do you write $w_1^2$ instead of just $w_1$? I though these are $\mathbb{Z}_2$ numbers. $\endgroup$ Commented Apr 8, 2020 at 18:41
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    $\begingroup$ @TomasBzdusek: They are not elements of $\mathbb{Z}_2$, they are cohomology classes: $w_1 \in H^1(BO(d); \mathbb{Z}_2)$ and $w_1^2 \in H^2(BO(d); \mathbb{Z}_2)$. $\endgroup$ Commented Jul 5, 2020 at 1:02
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For $d \geq 4$, let $M_d = (S^1)^{d-4}\times\mathbb{CP}^2$. As tori are parallelisable, $w(M_d) = w(\mathbb{CP}^2)$, in particular $w_1(M_d) = 0$ and $w_2(M_d) \neq 0$, so $M_d$ does not admit a Spin, Pin$^+$, or Pin$^-$ structure.

Therefore, there is no $d \geq 4$ such that every $d$-dimensional manifold admits a Pin$^+$/Pin$^-$ structure.

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