chiral anomaly and translation symmetry in 1+1D A Luttinger liquid at low energies can be captured by Dirac fermions in 1+1D, where the two component fermion field is given by
$$\Psi(x)=\left(\begin{array}{c}\psi_R(x)\\\psi_L(x)\end{array}\right),$$ where $\psi_{R,L}$ are (slow-moving parts of) the right and left movers near the two Fermi points, i.e.,
$$\phi(x)\sim \psi_R(x)e^{ik_F x}+\psi_L(x)e^{-ik_F x}.$$ 
In this terminology a translation operation $x\to x+a$ is equivalent to a chiral transformation on the spinor field $\Psi(x)$
$$\Psi(x)\to e^{i\sigma^z\theta}\Psi(x)=\left(\begin{array}{c}e^{i\theta}\psi_R(x)\\e^{-i\theta}\psi_L(x)\end{array}\right),$$
where $\theta=k_F a$.
In 1+1D we know there is chiral anomaly, which means the above chiral transformation is not a symmetry of the system at quantum level. Does it mean then, translational symmetry in the Luttinger liquid is not a real symmetry at quantum level? 
That certainly sounds rediculous to me. Am I missing something obvious?
 A: More precisely, the chiral anomaly tells us that
$$ \nabla.j^5 \propto \epsilon^{\mu \nu}F_{\mu \nu} = E$$
At this point we have to distinguish between two cases. If $E$ is considered to be a dynamical field, we are studying QED$_2$ at finite density. In this case, the answer is yes: translation symmetry is actually broken! See for instance M. Metlitski's "Is Schwinger Model at Finite Density a Crystal?," arXiv:hep-th/0609046.
On the other hand, we can also consider $E$ to be an external probe field. In this case, if we just set $E=0$, there is translation symmetry. 
Indeed, the microscopic model of fermions at half-filling hopping on a lattice certainly has a fully regularized implementation of translation symmetry! If we externally change $E$, however, the non-conservation of momentum is quite real. For example, put the system on a ring, and then slowly thread an external flux $\Phi(t)$ through the ring, establishing a field $E = \partial_t \Phi / L$.  According to the argument of Lieb, Schultz and Mattis, at fractional-filling $\nu$ (which you implicitly assume by taking $2 \Theta = 2 k_F a \notin \mathbb{Z}$), the global momentum quantum number for translation must change by $\Delta P = \nu \, 2 \pi / a $. Hence the momentum, as usually defined, is not conserved in the presence of E-fields.
