consider the case of spontaneously broken global symmetries. Suppose we have a multiplet of quantum fields $\phi_{i}(x)$ that transforms under the action of (some representation $D(g)$ of) a symmetry group $G$ according to $$\phi_i(x) \rightarrow \phi^g_i(x) = D(g)_{ij} \phi_j(x)$$ If this is a symmetry of the theory, then the quantum charges $Q^a_0$ commute with the full Hamiltonian: $$[Q^a_0, H] =0, \quad a= 1,2, \cdots {\rm dim}\,\, G.$$ These conventionally arise from quantum currents $J^{a,\mu}(x)$ which are conserved: $$\partial_\mu J^{a,\mu}_0 =0$$ where the currents $J^{a,\mu}_0$ are constructed from the quantum fields $\phi_{i,0}(x)$ and its derivatives.

Given a commutative quantum theory with conserved currents $J^{a,
  \mu}_0$, it is easy to see that in the corresponding noncommutative theory (obtained by replacing $\phi_{i,0}(x) \rightarrow
  \phi_{i,\theta}(x) = \phi_{i,0}(x) e^{\frac{1}{2}
  \overleftarrow{\partial} \wedge P}$), the noncommutative currents $J^{a,\mu}_\theta (x) = J^{a,\mu}_0(x)e^{\frac{1}{2}
  \overleftarrow{\partial} \wedge P}$ are also conserved: $$\partial_\mu J^{a,\mu}_\theta (x) = \partial_\mu J^{a,\mu}_0(x)e^{\frac{1}{2}
  \overleftarrow{\partial} \wedge P} = (\partial_\mu
  J^{a,\mu}_0(x))e^{\frac{1}{2} \overleftarrow{\partial} \wedge P} =
  0.$$ Interestingly, the charges $Q^a_\theta$ in the noncommutative theory $$Q^a_\theta = \int d^3 x J^{a,0}_\theta (x) = \int d^3 x J^{a,0}_0 (x)
  e^{\frac{1}{2} \overleftarrow{\partial} \wedge P} = \int d^3 x
  J^{a,0}_0 (x) = Q^a_0$$ are the same as in the commutative case. The last 