The Dirac Lagrangian in curved spacetime is usually given by \begin{equation} \mathcal{L} = i\bar{\Psi}\gamma^a e^{\mu}_a(\partial_\mu + \frac{1}{4}\omega_{\mu bc}\gamma^b\gamma^c)\Psi \end{equation} According to Nakahara's "Geometry, Topology and Physics" section 7.10.3. this Lagrangian is not Hermitian but we can rewrite this in Hermitian form as \begin{equation} \mathcal{L}' = \frac{1}{2}(\mathcal{L} + \mathcal{L}^{\dagger}) = i\bar{\Psi}\gamma^a e^{\mu}_a \overleftrightarrow{\partial}\Psi + \frac{1}{4}e^{\mu}_a\omega_{\mu bc}\bar{\Psi}\{\gamma^a, \gamma^{bc}\}\Psi \end{equation} up to total derivative. I can see why this is true for the derivative term but I really can't see how this can be done for the second term. I'm trying to commute $\gamma^a$ with $\gamma^{bc} = \frac{1}{2}[\gamma^b, \gamma^c]$ somehow, if $b,c \neq a$ then this is easy $\{\gamma^a, \gamma^{bc}\} = \gamma^a\gamma^{bc}$ and the second term is recovered. But the problem is if $a = b \neq c$ then $\{\gamma^a, \gamma^{bc}\} = 0$. I don't see how we can recover a term like $e^{\mu}_a\omega_{\mu ac}\gamma^a\gamma^a\gamma^c$ in the original Lagrangian (no summation in $a$). So my question is:

  • How do I show that the two Lagrangian are the same? (or equivalently, how do I show that $\mathcal{L} = \mathcal{L}^{\dagger}$ up to total derivative).
  • If $\mathcal{L}$ and $\mathcal{L}'$ are not actually the same up to total derivative (i.e. another typo in Nakahara). Then which one do I use as a Lagrangian for spin-1/2 particle in curved spacetime? I've seen people use the first one which also reproduce Dirac equation on a source like Wikipedia http://en.wikipedia.org/wiki/Dirac_equation_in_curved_spacetime. The Lagrangian was used without rewritten in Hermitian form in places like Freedman's "Supergravity" equation (9.1). Or in http://arxiv.org/abs/hep-th/0604198 for example.

I was hoping that $e^{\mu}_b \omega_{\mu bc} = \omega_{bbc}$ would be 0 by some antisymmetrical properties of $\omega$. But this apparently not true, only the last two indices of $\omega$ are antisymmetric. My guess is there's a reason why these two Lagrangian will ultimately give the same field theory but I just don't see why. Could someone please explain to me?

By the way, sorry if this question seems to be very similar to Dirac operator in curved spacetime in 2 dimensions – hermitian?. I read that entry but it seems to concern as of why we should rewrite it in Hermitian form, not how. I can't comment on that question either because my reputation is $< 50$ apparently.


1 Answer 1


Comments to the question (v3):

  1. Note that the gamma matrices are covariantly conserved$^1$
    $$\nabla_{\mu}\gamma^c~=~\omega_{\mu}{}^c{}_b\gamma^b+\frac{1}{4}\omega_{\mu}{}^{ab}[\gamma_{ab},\gamma^c] ~=~0,\qquad \nabla_{\mu}\gamma^{\nu}~=~0,\tag{1}$$ cf. e.g. Ref. 1.

  2. Consider the vector current $$J^{\mu}~:=~\bar{\psi}\gamma^{\mu}\psi,\tag{2}$$ where $\psi$ is a Dirac spinor.

  3. The divergence of the vector field (2) is $${\rm div}_e J~=~e^{-1}d_{\mu}(e J^{\mu})~=~ \nabla_{\mu}J^{\mu} ~\stackrel{(1)}{=}~\bar{\psi}\left(\stackrel{\leftarrow}{\nabla_{\mu}}\gamma^{\mu} + \gamma^{\mu}\nabla_{\mu}\right)\psi. \tag{3}$$

  4. Now define the Lagrangian density $${\cal L}~:=~ie \bar{\psi}\gamma^{\mu}\nabla_{\mu}\psi, \tag{4}$$ and the manifestly real Lagrangian density $${\cal L}^{\prime}~:=~\frac{1}{2}{\cal L} + {\rm c.c.} ~=~\frac{ie}{2}\bar{\psi}\left(\gamma^{\mu}\nabla_{\mu}-\stackrel{\leftarrow}{\nabla_{\mu}}\gamma^{\mu}\right)\psi\tag{5} .$$

  5. The difference is a total space-time derivative $${\cal L}-{\cal L}^{\prime} ~=~\frac{ie}{2}\bar{\psi}\left(\stackrel{\leftarrow}{\nabla_{\mu}}\gamma^{\mu} + \gamma^{\mu}\nabla_{\mu}\right)\psi~\stackrel{(3)}{=}~d_{\mu}\left(\frac{ie}{2} J^{\mu}\right),\tag{6}$$ so the Euler-Lagrange eqs. for ${\cal L}$ and ${\cal L}^{\prime}$ are the same.


  1. D.Z. Freedman & A. Van Proeyen, SUGRA, 2012; eq. (8.37) p. 180 Exercise 8.12.


$^1$ I use the same notation and conventions as in my Phys.SE answer here.

  • $\begingroup$ Thank you very much. I totally forgot the fact that $\nabla_{\mu} \gamma_{\nu} = 0$ $\endgroup$
    – user113988
    Commented Aug 8, 2015 at 3:18

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