TITLE: Spin-H structures
QUESTION [10 upvotes]: 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?

REPLY [13 votes]: 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).
Added Later: In this paper, Aleksandar Milivojevic and I prove the following:

The primary obstruction to the existence of a spin$^h$ structure on an orientable manifold $M$ is $W_5(M)$, the fifth integral Stiefel-Whitney class of $M$. It is worth noting that in the spin$^c$ case the primary (and only) obstruction is $W_3(M)$.
Every orientable manifold of dimension $\leq 7$ is spin$^h$.
In every dimension $\geq 8$, there exist infinitely many homotopy types of closed, simply connected manifolds which are not spin$^h$.

An explicit example of an orientable manifold which is not spin$^h$ is the ten-dimensional manifold $(SU(3)/SO(3))\times(SU(3)/SO(3))$.