I'm thinking about the Clifford algebra in arbitrary dimensions, and following "Supergravity" from Freedman and Van Proeyen. Specifically I am working on problem 22.15 therein.

The charge conjugation matrix is introduced as being a unitary matrix such that $$(C \Gamma^{(r)})^T = -t_r C \Gamma^{(r)}$$ with $t_r = \pm 1$ and $\Gamma^{(r)}$ being a rank $r$ gamma matrix (e.g. rank $1$: $\gamma^\mu$; rank $2$: $\gamma^{\mu\nu}$).

I am trying to prove that for arbitrary AdS$_D$ Killing spinors $\epsilon,\epsilon'$ (with $D_\mu \epsilon = \frac{1}{2L} \gamma_\mu \epsilon$) the following identify holds: $$ \nabla_\mu \left( \overline{\epsilon}' \gamma_\nu \epsilon \right) = -\frac{1}{L}\overline{\epsilon}' \gamma_{\mu\nu} \epsilon, $$ where $\overline{\epsilon} \equiv \epsilon^T C$.

Using that $\nabla_\mu = e_\mu^a D_a$, where $D_a$ is the covariant derivative connected to the spin connection, and the given that $t_0 t_1 = -1$, I found the asked right hand side, but with an extra term involving $e^a_\mu D_a (C\gamma_\nu)$. By the comments on this question, I know that $D_\mu \gamma_\nu = 0$. I could not, however come up with a reason why $D_\mu C = e^a_\mu D_a C$ would be zero. I have tried to work with the fact that $C C^{-1} = 1$ and with the fact that $(\gamma^\mu)^T = - C \gamma^\mu C^{-1}$. However, I could not find anything useful.

Can anybody help me understand why $D_\mu C = 0$, or where my reasoning is at fault?


1 Answer 1


All the objects $\gamma^a $ and $C$ and so on are just Clebsh-Gordan coefficients intertwining different spin representations, so they are constants and their covariant derivative is zero. For $\gamma^\mu= e_a^\mu\gamma^a$ the $e_a^\mu$ are the components of the identity operator ${\bf E}:TM\to TM$ where ${\bf E}={\bf e}_a \otimes {\bf e}^{*a}= {\bf e}_a e_\mu^{*a}\otimes dx^\mu= e_a^\mu {\boldsymbol \partial}_\mu \otimes {\bf e}^{*a}$ maps ${\bf x}=x^a {\bf e}_a=x^\mu {\boldsymbol \partial}_\mu$ to itelf ${\bf E}({\bf x})={\bf x}$. The identity must also have vanishing covariant derivative (this last is the badly named "tetrad postulate").

  • $\begingroup$ Thank you so much for your response! So, the $e$ is just a way to rewrite $\gamma$ (etc) in a different basis. So they come from your identity operator $E$ but take a non-trivial form as the $e$ because they correspond to a 'change of basis matrix' (but then on every tangent space of course). Is this the point you are making? Could you perhaps elaborate a bit on why this implies that the $C$ is covariantly constant, as I do not fully grasp your claim concerning the Clebsh-Gordan coefficients. $\endgroup$ Commented Jan 4, 2022 at 22:54
  • $\begingroup$ For example $\bar\psi \gamma^a \psi$ must transform as a four vector, but the $\bar\psi_\alpha $ and $\psi^\beta$ are in spinor reps $\bar S$ and $S$. This means that the array of numbers ${(\gamma^a)^\alpha}_\beta$ are Clebsh's that couple $\bar S\otimes S$ to the four vector rep. Similary $\psi^T C\psi$ is scalar so $C_{\alpha\beta} $ couples $S\otimes S$ to the singlet rep. $\endgroup$
    – mike stone
    Commented Jan 4, 2022 at 23:15
  • $\begingroup$ The various identities that look like the Covariant derivative of the $\gamma^\mu$ etc are zero are just statements that the Covariant derivative is a derivation. i.e it obeys Leibnitz rule for differentiating products. The effect is the all Clebsh's have to be treated as constants in covariant differentiation. $\endgroup$
    – mike stone
    Commented Jan 4, 2022 at 23:19
  • $\begingroup$ Thank you for your explanation of the Clebsh-Gordan idea here, I feel like I better understand the ideas behind it all now. $\endgroup$ Commented Jan 4, 2022 at 23:33
  • $\begingroup$ For the comment on derivations: do you mean that $D_a$ is only a derivation when we impose that $\gamma^b$ is covariantly constant? $\endgroup$ Commented Jan 4, 2022 at 23:34

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