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Let us define a Spin-H structure as a reduction of a SO(n)-bundle by the group: $$Spin^H (n)=Spin(n) \times SU(2)/\{ 1,-1\}$$ The Spin-H structures are analogous to the well-known Spin-C structures but for the Hamilton numbers instead of the complex numbers. Can we prove that an oriented riemannian manifold admits a Spin-H structure if and only if it is Spin?

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  • $\begingroup$ I think you meant to write $(Spin(n)×SU(2))/\{1,−1\}$ or $\frac{Spin(n)×SU(2)}{\{1,−1\}}$, yes ? $\endgroup$ – wonderich Aug 16 '18 at 20:35
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As mentioned in Arun Debray's answer, a closed orientable smooth manifold $M$ is spin${}^h$ if and only if there is a principal $SO(3)$-bundle (or equivalently, an orientable real rank three bundle) $P$ such that $w_2(M) = w_2(P)$.

  • If $M$ is spin, then $w_2(M) = 0$. Taking $P$ to be the trivial bundle, we see that $w_2(M) = 0 = w_2(P)$, so $M$ is spin${}^h$.
  • If $M$ is spin${}^c$, then $w_2(M)$ has an integral lift $c \in H^2(M; \mathbb{Z})$. There is a complex line bundle $L$ with $c_1(L) = c$ and hence $w_2(L) = w_2(M)$; one choice of such an $L$ is the complex line bundle associated to a spin${}^c$ structure. Now taking $P = L \oplus \varepsilon^1$, we see that $w_2(M) = w_2(L) = w_2(L\oplus\varepsilon^1) = w_2(P)$, so $M$ is spin${}^h$. For $M = \mathbb{CP}^2$ with its standard spin${}^c$ structure, this is the example given in Arun Debray's answer.
  • If $M$ is a closed oriented Riemannian four-manifold, then $w_2(M) = w_2(\Lambda^+) = w_2(\Lambda^-)$ so taking $P = \Lambda^+$ or $\Lambda^-$ shows that $M$ is spin${}^h$. Note, this can also be deduced from the previous point as all closed orientable smooth four-manifolds are spin${}^c$.
  • If $M$ is the Wu manifold, $SU(3)/SO(3)$, let $P$ be the principal $SO(3)$-bundle $SO(3) \to SU(3) \to SU(3)/SO(3)$. Then $w_2(M) = w_2(P)$, so $M$ is spin${}^h$. This is a notable example as $M$ is not spin${}^c$. I learnt this from Xuan Chen, a former student of Blaine Lawson.

In conclusion, every spin manifold is spin${}^c$, but not conversely (as $\mathbb{CP}^2$ demonstrates), and every spin${}^c$ manifold is spin${}^h$, but not conversely (as $SU(3)/SO(3)$ demonstrates).

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  • $\begingroup$ Based on your answer it sounds like spin^h manifolds are interesting to you outside of just this question. How do they come up in what you're interested in? $\endgroup$ – Arun Debray Jul 7 '18 at 22:59
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    $\begingroup$ I learnt about spin${}^h$ manifolds in Xuan Chen's thesis defense. Beyond that, I haven't really thought much about them. $\endgroup$ – Michael Albanese Jul 8 '18 at 6:31
  • $\begingroup$ @MichaelAlbanese, thanks for mentioning Xuan Chen's thesis. It's short and studies an interesting question! $\endgroup$ – Igor Khavkine Jul 8 '18 at 8:05
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SpinH-structures were studied by Shiozaki-Shapourian-Gomi-Ryu for applications to condensed-matter physics. They prove in Lemma D.9 that a closed manifold $M$ admits a spinH-structure iff it's orientable and there's a principal $\mathrm{SO}_3$-bundle $P\to M$ such that $w_2(P) = w_2(TM)$.

$\newcommand{\CP}{\mathbb{CP}}$This implies $\CP^2$ is spinH but not spin. If $x\in H^2(\CP^2)$ denotes the generator Poincaré dual to a hyperplane and $\bar x$ denotes its reduction in mod 2 cohomology, which is nonzero, then

$$c(\CP^2) = (1+x)^3 = 1 + 3x + 3x^2$$

and therefore $w_2(\CP^2) = \bar x$, so $\CP^2$ isn't spin.

If $S$ denotes the tautological bundle, then $c(S) = 1-x$, so $w_1(S) = 0$ and $w_2(S) = \bar x$. Therefore $S\oplus\underline{\mathbb R}$ is a rank-3 orientable real vector bundle, so its bundle of orthonormal frames admits a reduction to a principal $\mathrm{SO}_3$-bundle $P\to\CP^2$. Since Stiefel-Whitney classes are stable, $w_2(P) = w_2(S) = \bar x$, and therefore $\CP^2$ is spinH.

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  • $\begingroup$ I considered the inclusion map $[Spin(n).Spin(3)/ \{ 1,-1\} ] \hookrightarrow Spin (n+3)$ so that the manifold is Spin-H if and only if there is a $Spin(3)$-fiber bundle $P$ so that $\omega_2 (TM \oplus P)=0,$ $\omega_2 (M)= \omega_2 (P)$ if and only if $\omega_2 (M)=0$ as $Spin_3= S^3$...? I must take a $SO(3)$-fiber bundle? $\endgroup$ – Antoine Balan Jul 7 '18 at 16:52
  • $\begingroup$ @AntoineBalan the condition you mention, that there's a $\mathrm{Spin}_3$-bundle $P$ with $w_2(TM) = w_2(P)$, is data of a $\mathrm{Spin}_n\times\mathrm{Spin}_3$-structure, but a spin^h structure is a $(\mathrm{Spin}_n\times\mathrm{Spin}_3)/\{\pm 1\}$-structure, which is different. The story is very similar to the difference between spin^c structures vs $\mathrm{Spin}_n\times\mathrm U_1$-structures. $\endgroup$ – Arun Debray Jul 7 '18 at 17:31
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No, it is possible construct Spin-H structures on manifolds that don't have spin structures. For instance, the case of a 4-dimensional manifold $M$ was considered in detail in

Avis, S. J. & Isham, C. J. Generalized spin structures on four dimensional space-times. Communications in Mathematical Physics 72, 103-118 (1980).
http://dx.doi.org/10.1007/bf01197630

In fact, they considered a general Lie group $G$ (and corresponding generalized Spin structures), instead of only $G = SU(2)$, and found that the obstruction to the existence of a generalized spin structure was obstructed by a characteristic class valued in $H^3(M,\pi_1(G))$, cf. their equation (4.14). Hence, at least in this case, since $\pi_1(SU(2)) = 0$, there are no obstructions to the existence of Spin-H structures on 4-manifolds. I suspect the situation becomes more complicated in higher dimensions, but I'm not competent to say how.

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  • $\begingroup$ It seems that if I take $G=S^1$ there is a problem because it exists $M$ with $H^3(M, {\bf Z})\neq 0$...Is it false? $\endgroup$ – Antoine Balan Jul 7 '18 at 14:21
  • $\begingroup$ @AntoineBalan, this is exactly the remark that Avis & Isham make on p.113 of their paper. As I'm sure you know, $G=S^1=U(1)$ corresponds to Spin-C structure. $\endgroup$ – Igor Khavkine Jul 7 '18 at 15:28
  • $\begingroup$ In our case, it seems that the group is $G=Spin(n).SU(2)$ so that the obstruction is to be Spin in fact... $\endgroup$ – Antoine Balan Jul 7 '18 at 16:03
  • $\begingroup$ The problem is that if I take $G=\{ e \}$, there is no obstruction? $\endgroup$ – Antoine Balan Jul 7 '18 at 16:10
  • $\begingroup$ @AntoineBalan, I think you should have a look at the discussion in Avis & Isham to see more precisely how the corner cases fit into the general pattern. I think that setting $G=\{e\}$ just gives you back the frame bundle, since $Spin(n) \times_{\mathbb{Z}_2} \{e\} \cong SO(n)$. So indeed, there should be no obstruction. You need $G=\mathbb{Z}_2$ to get the usual Spin structure. I think that Avis & Isham may be working with the hidden assumption that their Lie group $G$ is connected, so that $G=\mathbb{Z}_2$ has to be a special case. $\endgroup$ – Igor Khavkine Jul 7 '18 at 17:58

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