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 $\mathbf E(\mathbf x)$:

$$

[E_{i}(\mathbf x), E_{j}(\mathbf y)] \sim \Delta (A_{i}(\mathbf x), A_{j}(\mathbf y)), \qquad (1)

$$

where $\mathbf A$ is canonical coordinate.

Is this information the only one which we need to conclude that corresponding fermion current $J_{\mu}^{L}$ isn't conserved, i.e.,

$$

\partial_{\mu}J^{\mu}_{L} \neq 0?

$$

In particular, if the answer is "yes", for the left charge $Q_{L} = \int d^{3}\mathbf r J_{L}^{0}$ there must be

$$

\frac{dQ_{L}}{dt} \sim [H, Q_{L}] \neq 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" $G(x) = \nabla \cdot \mathbf E - J^{0}$ don't commute with each other. This means, that physical states $|\psi\rangle$ of theory no longer satisfy the relation

$$

G(x)|\psi\rangle = 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

$$

\frac{dQ_{L}}{dt} \sim [H,Q_L] =|Q_L = \int d^{3}\mathbf r \nabla \cdot \mathbf E| = [H, \int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)] =

$$

$$

= \left[\frac{1}{2}\int d^{3}\mathbf y \frac{\mathbf E^{2}(\mathbf y )}{2},\int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)\right]= \int d^{3}\mathbf r E_{i}(\mathbf r)\partial_{j}\Delta^{ij}(\mathbf A,\mathbf r)

$$