Spinor reps in $\mathbb{R}^{1,3}\times{}B$ space-times I am considering spinors in a space-time which is $\mathbb{R}^{1,3}\times{}B$ being $B$ a compact manifold of $D$ dimensions. 
I know that in ordinary 4 dimensional space-time spinors are representations of $O(1,3)$. Now, in my case, are spinors expected to be reps of $O(1,3+D)$?
Does the compactness of $B$ impose some restrictions to this? I have the feeling that we have to expect spinors to be reps of $O(1,3)\times{}O(D)$ since I don't feel that making a boost in the compact space is allowed but I am not sure.
Any clarification on the reps spinors are in in the mentioned space-time will be greatly appreciated.
 A: In 3+1 dimensions spinors do not transform under representations of $O(1,3)$, but under representations of the covering group $\operatorname{Spin}(1,3)$, which has the same Lie algebra. The structure group of a semi-Riemannian manifold is determined by the metric signature. Thus if the metric is such that $B$ is spacelike, spinors would transform under $\operatorname{Spin}(1,3+D)$.
(To define what I mean by the structure group: it is always possible to, locally, find a set of vector fields such that the metric is diagonal. This is since with respect to a local basis of vector fields the metric at every point in spacetime is a symmetric matrix. The structure group is the group of linear transformations that preserves this form of the metric. Indeed the Lorentz group is often defined as the group that preserves $\operatorname{diag}(1,-1,-1,-1)$. By Sylvester's law of inertia the structure group is well-defined.)
A boost along the compact directions is allowed, because locally the space looks like $\mathbb{R}^{1,3+D}$ (for a spacelike $B$). It may be that the metric can be cast in a form $$ds^2 = ds_0^2 + dB^2$$
where $ds_0^2$ is non-zero only for vectors tangent to $\mathbb{R}^{1,3}$ and $dB^2$ is non-zero only for vectors tangent to $B$. Then $O(1,3) \times O(D)$ is the group that preserves this form, but the full structure group is larger, since we can for example mix coordinates on $\mathbb{R}^{1,3}$ and on $B$.
The compactness, or rather the topology of $B$, can enter only in that there is a topological obstruction to consistently defining spinors on a manifold. This condition is global and so if the space $B$ is not nice enough, $R^{1,3} \times B$ may not even have spinors. The technical formulation of the condition is that the second Stiefel-Whitney class should vanish.
