Suppose we know, that the dynamics of theory with chiral fermions (say, left) and gauge field (for simplicity, abelian) leads us to presence of anomalous commutator of canonical momentum E(x):
[Ei(x),Ej(y)]∼Δ(Ai(x),Aj(y)),(1)
where A is canonical coordinate.
Is this information the only one which we need to conclude that corresponding fermion current JLμ isn't conserved, i.e.,
∂μJμL≠0?
In particular, if the answer is "yes", for the left charge QL=∫d3rJ0L there must be
dQLdt∼[H,QL]≠0
due to anomalous commutator.
Another formultion of the question: does the presence of anomalous commutator (1) guarantee the presence of gauge anomaly, i.e., current non-conservation?
An edit
It seems that the answer is obviously yes. First, even if I don't know precise structure of anomalous commutators, it is enough to know that they are non-zero. Thus, in particular, the "Gauss laws" G(x)=∇⋅E−J0 don't commute with each other. This means, that physical states |ψ⟩ of theory no longer satisfy the relation
G(x)|ψ⟩=0
This means violation of the unitarity, and hence the gauge anomaly.
In the case when we know the precise form (1) of anomalous commutators, it's elementary to compute the anomalous conservation law: by using the Gauss law, we have
dQLdt∼[H,QL]=|QL=∫d3r∇⋅E|=[H,∫d3r∇⋅E(r)]=
=[12∫d3yE2(y)2,∫d3r∇⋅E(r)]=∫d3rEi(r)∂jΔij(A,r)