Q:

Spin of vector boson in higher dimension

A vector boson $V^{\mu}, \mu = 0,...,3$ has spin 1. To my understanding (correct me if I'm wrong) this is because it transforms as a 4-vector under Lorentz transformation $SO(1,3)$. So the $\mu = 1,2,3$ components will form an irreducible representation of $SO(3)$ subgroup which is isomorphic to spin 1 representation of $SU(2)$, hence $V^{\mu}$ has spin 1.
What if I have more than just 1+3 dimensions? let just say I'm in $1 + N$ dimensions? My vector should now be $V^{M}$ where $M = 0,1,...,N$ and Lorentz transformation is $SO(1,N)$. The logic above will not work anymore because the space components will now transforms under $SO(N)$ for pure rotation. So my question is: What can I say about the spin of $V^M$? Is it still 1? Or something else and why?

A:

The spin of a vector boson in any dimension is spin 1.
What changes with the number of dimensions is the number of degrees of freedom associated with a given spin. A massless vector in four dimensions has two independent degrees of freedom, which can be seen from the rank of what's called the "little group" in the literature. It is the subgroup of the space-time symmetry under which an exemplary momentum for the representation stays invariant. For the massless vector in 4D this is a euclidean group of rank 2. You can also just try to figure out the possible polarization states of a photon.
In 5D, the little group enhances to a rank 3 group, i.e. the 5D photon has three independent degrees of freedom (after gauge fixing and everything). It still has spin 1 though.
Now what spin are we talking about? This spin is w.r.t. the rotation group on 5D Minkowski space. If you perform a dimensional reduction of this representation, you will see that in 4D it decomposes as
$$ \mathbf 1_5 \mapsto \mathbf 1_4 + \mathbf 0_4$$
meaning that a 5D vector looks in 4D like a vector and a scalar. The scalar degree of freedom is exactly when the "photon" is polarized into the extra dimension.

