Holonomy reduction from constant spinors Let $(M,g)$ be a $d$-dimensional Riemannian oriented, spin, manifold, and let us denote by $S$ the corresponding spinor bundle. The Levi-Civita connection $\nabla$ on $(M,g)$ lifts to a unique spin connection $\nabla^{S}$ on $S$.
Now, suppose there is a no-where vanishing parallel spinor $\epsilon\in\Gamma(S)$ on $M$, namely $\nabla^{S}\epsilon =0$. It is clear then that $Hol(\nabla^{S})\subseteq G$, where $G\subseteq Spin(d)$ is the stabilizer of the spinor under the spin group action. My question is, what is then the holonomy of $\nabla$? Intuitively, it should be $\rho(G)\subseteq SO(d)$, where $\rho\colon Spin(d)\to SO(d)$ is the double covering map, but I do not know how to prove this claim.
Thanks. 
 A: You are right in principle, but a bit of care is needed to make this precise. The spinor representation decomposes into orbits under the action of the spin group (in a rather complicated way). These orbits can be thought of as the possible "algebraic types" of spinors (in a point!), simlar to the (point-wise) rank of a symmetric $\binom02$-tensor field. (I don't know to what extent the algebraic side of the problem is understood in general, a well known example of such a type are pure spinors.) Each such orbit comes with a stabilizer, which is a subgroup in the spin group determined up to conjugacy.
As you expect, the holonomy group will then be contained in the image of this subgroup in $SO(d)$. (This holonomy group is only determined up to conjugacy which is a usual phenomenon. Of course, the "true" holnomy group can be strictly smally than that.) My favorite way to prove this would be to observe that if you take a cuvrve in $M$ and lift it horizontally to the principal Spin(d)-bundle, then the projection of the lift to the orthonormal frame bundle is a horizontal lift of the initial curve. The result then follows from the definition of holonomy in terms of lifts of closed curves to the orthonormal frame bundle.   
