Dirac operator in curved spacetime in 2 dimensions – hermitian? I'm currently trying to learn about the Dirac equation in curved spacetime and have come across an odd remark in Nakahara's well-known textbook "Geometry, Topology and Physics" that I would like to understand.
In section 7.10.2, Nakahara writes the Dirac in two-dimensional spacetime as
$$S[\bar{ψ,ψ}] = ∫_M d^2x\sqrt{-g}\, \bar{ψ}\,iγ^a e_a^µ\left[∂_µ+\frac12iΓ^{b\ c}_{\ µ}Σ_{bc}\right]ψ$$
where


*

*$Γ^{b\ c}_{\ µ}$ are the coefficients for the spin connection, $\nabla_µ e_c = Γ^{b\ }_{\ µc} e_b$

*$e_a = e^µ_a ∂_µ$ are the vielbein vectors

*and $Σ_{bc}=\frac14 i[γ_b,γ_c]$ are the spinor representations of the generators for the Lorentz Lie algebra $so(1,1)$.


(I have dropped the mass and $U(1)$ gauge field terms that are also present in the book.)
Then, he writes

It is interesting to note that the spin connection term vanishes if $\dim M=2$.

and argues that the Lagrangian has to be made hermitian first, so he ends up with
$$S[\bar{ψ,ψ}] = ∫_M d^2x\sqrt{-g}\, \bar{ψ}\,ie_a^µ\left[γ^a \overleftrightarrow{∂}_µ
+
\frac12iΓ^{b\ c}_{\ µ}\{γ^a,Σ_{bc}\}\right]ψ$$
In two dimensions, there is only one generator $Σ_{01}=γ_0γ_1$, and it anticommutes with both gamma matrices, $\{γ^0,Σ_{01}\}=\{γ^1,Σ_{01}\}=0$, so the spin connection term vanishes.
However, I do not understand why the Lagrangian has to be made hermitian first, so my questions are:

  
*
  
*Is the Dirac operator in curved spacetime hermitian?
  
*If not, why is making it hermitian like Nakahara does a good idea?

For the first question, I am currently trying to calculate the adjoint, but did not get the terms to cancel yet.
 A: 1) Dirac operator Pseudo-Hermitian in flat and not pseudo-hermitian in curved space if you define it as in your first equation (without symmetrizing).
because $(i\gamma^a e^\mu_a\partial_\mu)^\dagger = \gamma^0 (i\gamma^a e^\mu_a\partial_\mu)\gamma^0$, whereas $(\gamma^a \left[ \gamma^b,\gamma^c\right])^\dagger =\gamma^0 ( \left[ \gamma^b,\gamma^c\right]\gamma^a)\gamma^0 $
2) Lagrangian must always be hermitian so that the S-matrix is unitary (QM).
In flat space the kinetic term in the Lagrangian is not hermitian, so in principle we should add the hermitian conjugate
$\mathcal{L} = \bar{\psi}i\gamma\partial \psi + h.c.$, however if we do not its ok because the lagrangian will still be hermitian up to a total derivative.. This statement was actually redundant because hermiticity means $(\psi,A\psi) = (A^\dagger \psi, \psi) = (A\psi,\psi)$ so total derivatives don't matter for the definition.
This does not work anymore for a nonzero spin connection as demonstrated above. So you should explicitly add the hermitian conjugate. 
