Consider a spinor field $\psi$ on a general smooth Lorentzian manifold. Let $\Sigma_{ab}$ be a matrix representation of the Lorentz group, and let Greek/Latin letters represent world/Lorentz indices. Using the conventions of Parker and Toms we have ($\nabla$ means covariant derivative)

$$\nabla_\mu \psi = \partial_\mu \psi + i \omega_\mu^{ab} \Sigma_{ab} \psi$$ $$[\nabla_\mu, \nabla_\nu]\psi = -i R_{\mu \nu}^{ab} \Sigma_{ab} \psi $$

where $\omega_\mu^{ab}$ is the antisymmetric connection and $R^{ab}_{\mu \nu}$ is the Riemann tensor contracted with two n-biens.

I need to commute higher derivatives of the spinor, for example

$$[\nabla_a, \nabla_b] \nabla_c \psi $$ $$[\nabla_a, \nabla_b] \nabla_c \nabla_d \psi$$

and so on. For tensors (i.e. when the connection is symmetric) there is a very straightforward rule for doing this: you just write one Riemann tensor term for each slot the derivatives are going to be commuted through, and add signs depending on whether that slot was upstairs or downstairs. For example

$$[\nabla_a, \nabla_b] T_{cd} = -R^e_{cab} T_{ed} - R^e_{dab} T_{ce}. $$

I seek a similar prescription for spinors. Is there one?


1 Answer 1


I will be using a different notation to you, and all of what I write down may be found in Ref. [1] chapters 1.1 - 1.6, I highly recommend it (section 1.6 is particularly relevant, but you will need the prev sections too).

I will do everything in the tangent frame (so all indices are local Lorentz), you can just convert back using the vielbein if you want. I will also be working in four dimensions. The covariant derivative is written as $$\nabla_{a}=e_a+\frac{1}{2}\omega_a{}^{bc}M_{bc}~,$$ where $e_a=e_a{}^{m}\partial_m$ is the inverse vielbein, $\omega_a{}^{bc}$ is the spin connection and $M_{ab}$ are the Lorentz generators. The important difference between my expression and yours is that my Lorentz generators are in an arbitrary representation, where as yours are in what I think might be the Dirac representation (i.e. 4 component reducible spinors).

One can then show that, in the torsion free case, the commutator of covariant derivatives is $$[\nabla_a,\nabla_b]=\frac{1}{2}R_{ab}{}^{cd}M_{cd}~.$$ The beauty of this formula is that it can act on any type of spin-tensor, you only need to remember how the Lorentz generators act on different objects, the Leibniz rule for the generators will take you from there.

I will give some examples on how $M_{ab}$ act on certain objects, but you are probably familiar with these. Let $V_a$ be a vector, $\psi_{\alpha}$ and $\bar{\chi}^{\dot{\alpha}}$ a left and right handed 2 component spinor respectively and let $\Psi=\big(\psi_{\alpha},~\bar{\chi}^{\dot{\alpha}}\big)^T$ be a 4 component spinor. The Lorentz generators act on the as follows

\begin{align} M_{ab}V_c&=\eta_{ca}V_b-\eta_{cb}V_a~, \tag{1.a}\\ M_{ab}\psi_{\alpha}&=(\sigma_{ab})_{\alpha}{}^{\beta}\psi_{\beta}~, \tag{1.b}\\ M_{ab}\bar{\chi}^{\dot{\alpha}}&=(\tilde{\sigma}_{ab})^{\dot{\alpha}}{}_{\dot{\beta}}\bar{\chi}^{\dot{\beta}}~,\tag{1.c}\\ M_{ab}\Psi&=\Sigma_{ab}\Psi~. \tag{1.d} \end{align}

Here $\sigma_{ab}=-\frac{1}{4}\big(\sigma^a\tilde{\sigma}^b-\sigma^b\tilde{\sigma}^a\big)$, $\Sigma_{ab}=-\frac{1}{4}[\gamma_a,\gamma_b]$ and $\gamma_a=\begin{pmatrix} 0 & \sigma_a \\ \tilde{\sigma}_a &0\end{pmatrix}$ etc etc.

Of course, if you are working with spinors, it is usually much easier to convert everything into 2 component spinor notation, then there is only two rules to remember.

To conclude I will do one of your examples explicitly. I will assume that your $\psi$ is a 4 component spinor. \begin{align} [\nabla_a,\nabla_b]\nabla_c\nabla_d\psi&=\frac{1}{2}R_{ab}{}^{fg}M_{fg}\bigg(\nabla_c\nabla_d\psi\bigg) \\ &=\frac{1}{2}R_{ab}{}^{fg}\bigg(M_{fg}\nabla_c\nabla_d\psi+\nabla_cM_{fg}\nabla_d\psi+\nabla_c\nabla_dM_{fg}\psi\bigg)\\ &=\frac{1}{2}R_{ab}{}^{fg}\bigg(2\eta_{cf}\nabla_g\nabla_d\psi+2\nabla_c\eta_{df}\nabla_g\psi+\nabla_c\nabla_d\Sigma_{fg}\psi\bigg)\\ &=R_{abc}{}^{f}\nabla_f\nabla_d\psi+R_{abd}{}^{f}\nabla_c\nabla_f\psi+\frac{1}{2}R_{ab}{}^{fg}\Sigma_{fg}\nabla_c\nabla_d\psi~. \end{align}

[1] I.L. Buchbinder and S.M. Kuzenko, Ideas and Methods of Supersymmetry and Supergravity, Or a Walk Through Superspace, IOP, Bristol (1995) (Revised Edition 1998).

  • $\begingroup$ Thank you for this! I will be working in 2D, so my "Dirac" spinors are actually Pauli spinors with two indices. I assume the above remains essentially correct? $\endgroup$
    – AGML
    Nov 29, 2018 at 19:46
  • 1
    $\begingroup$ No Problem! Actually I have never worked with two dimensional spinors, so i'm not in a position to say what exactly will change. My guess is that everything will remain the same except for the identities (1.b,c,d). In addition, since the Riemann tensor is $R_{abcd}=\frac{1}{2}R(\eta_{ac}\eta_{bd}-\eta_{ad}\eta_{bc})$, the commutator would simplify to $$[\nabla_a,\nabla_b]=\frac{1}{2}RM_{ab}~.$$ $\endgroup$ Nov 30, 2018 at 1:36

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