$SU(N)$ is the $N$-fold cover of $PSU(N)$. They share the same Lie algebra, so the Yang-Mills action would look identical locally. The center of $SU(N)$ is just $Z_N$. At the level of representations, the fundamental representation of $SU(N)$ is a projective representation of $PU(N)$, and only the adjoint ones are linear representations of $PU(N)$.

If the matter fields all transform in the adjoint representation, then it makes sense to say that the gauge group is actually $PU(N)$. A simple explanation is that by taking tensor product of adjoint representations you never get the fundamental ones, so the Hilbert space is restricted.

Because $PSU(N)=SU(N)/Z_N$, the global topology of $PU(N)$ is nontrivial. For example, the fundamental group $\pi_1(PU(N))=Z_N$, so there are nontrivial "vortex lines" in the scalar matter field, around which you pick up a holonomy in the center $Z_N$. These topological excitations themselves are one-dimensional objects, and have "codimension" 2.

Quarks in $SU(3)$ QCD transform as the fundamental representation.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Meng Cheng