Global invariance under $SU(N)$ is equivalent to the conservation of $N^2-1$ charges – these charges are nothing else than the generators of the Lie algebra ${\mathfrak su}(N)$ that mix some components of $SU(N)$ multiplets with other components of the same multiplets. These charges don't commute with each other in general. Instead, their commutators are given by the defining relations of the Lie algebra,
$$ [\tau_i,\tau_j] = f_{ij}{}^k \tau_k $$
But these generators $\tau_i$ are symmetries because they commute with the Hamiltonian,
$$[\tau_i,H]=0.$$
None of these charges may be interpreted as the "gluon number". This identification is completely unsubstantiated not only in QCD but even in the simpler case of QED. What is conserved in electrodynamics because of the $U(1)$ symmetry is surely *not* the number of photons! It's the electric charge $Q$ which is something completely different. In particular, photons don't carry any electric charge.

Similarly, this single charge $Q$ – generator of $U(1)$ – is replaced by $N^2-1$ charges $\tau_i$, the generators of the algebra ${\mathfrak su}(N)$, in the case of the $SU(N)$ group.

Also, it's misleading – but somewhat less misleading – to suggest that the conserved charges in the globally $SU(N)$ invariant theories are just the $N$ color charges. What is conserved – what commutes with the Hamiltonian – is the whole multiplet of $N^2-1$ charges, the generators of ${\mathfrak su}(N)$.

Non-abelian algebras may be a bit counterintuitive and the hidden motivation behind the OP's misleading claim may be an attempt to represent $SU(N)$ as a $U(1)^k$ because you may want the charges to be commuting – and therefore to admit simultaneous eigenstates (the values of the charges are well-defined at the same moment). But $SU(N)$ isn't isomorphic to any $U(1)^k$; the former is a non-Abelian group, the latter is an Abelian group.

At most, you may embed a $U(1)^k$ group into $SU(N)$. There's no canonically preferred way to do so but all the choices are equivalent up to conjugation. But the largest commuting group one may embed into $SU(N)$ isn't $U(1)^N$. Instead, it is $U(1)^{N-1}$. The subtraction of one arises because of $S$ (special, determinant equals one), a condition restricting a larger group $U(N)$ whose Cartan subalgebra would indeed be $U(1)^N$.

For example, in the case of $SU(3)$ of real-world QCD, the maximal commuting (Cartan) subalgebra of the group is $U(1)^2$. It describes a two-dimensional space of "colors" that can't be visualized on a black-and-white TV, to use the analogy with the red-green-blue colors of human vision. Imagine a plane with hexagons and triangles with red-green-blue and cyan-purple-yellow on the vertices.

But grey, i.e. color-neutral, objects don't carry any charges under the Cartan subalgebra of $SU(N)$. For example, the neutron is composed of one red, one green, one blue valence quark. So you could say that it has charges $(+1,+1,+1)$ under the "three colors". But that would be totally invalid. A neutron (much like a proton) actually carries no conserved QCD "color" charges. It is neutral under the Cartan subalgebra $U(1)^2$ of $SU(3)$ because the colors of the three quarks are contracted with the antisymmetric tensor $\epsilon_{abc}$ to produce a singlet. In fact, it is invariant under all eight generators of $SU(3)$. It has to be so. All particles that are allowed to appear in isolation must be color singlets – i.e. carry vanishing values of all conserved charges in $SU(3)$ – because of confinement!

So as far as the $SU(3)$ charges go, nothing prevents a neutron from decaying to completely neutral final products such as photons. It's only the (half-integral) spin $J$ and the (highly approximately) conserved baryon number $B$ that only allow the neutron to decay into a proton, an electron, and an antineutrino and that make the proton stable (so far) although the proton's decay to completely quark-free final products such as $e^+\gamma$ is almost certainly possible even if very rare.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Luboš Motl