Spin-H structures 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?
 A: 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.
A: 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 $(SU(3)/SO(3))\times(SU(3)/SO(3))$.
A: 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.
