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If $\gamma^m$ denotes a tangent space gamma matrix, and $\gamma^\mu$ denotes a curved space gamma matrix, then they are related by

$$\gamma^\mu(x) = \gamma^m e_{m}^{\mu}(x)$$

where $e_{m}^{\mu}(x)$ is the vielbein. Under a local Lorentz transformation,

$$\delta_{lL}e_m^\mu(x) = {\lambda_m}^n(x) e_n^\mu(x)$$

Also, the gamma matrices with flat space indices satisfy $\{\gamma^m, \gamma^n\} = 2\eta^{\mu\nu}$ whereas the gamma matrices with curved space indices satisfy $\{\gamma^\mu(x), \gamma^\nu(x)\} = 2g^{\mu\nu}(x)$.

To my mind, $\gamma^m$ and $e_{m}^{\mu}(x)$ should transform oppositely under local Lorentz transformations, and hence $\gamma^\mu(x)$ should stay inert.

But then so should arbitrary products of curved space gamma matrices, in particular something like $\gamma^{\mu\nu\rho}$. However, when proving that the spin-3/2 Rarita-Schwinger Lagrangian density is invariant under local Lorentz transformations, one does encounter a term $\bar{\psi}_\mu \gamma^{\mu\nu\rho}D_{\nu}\psi_\rho$, which would involve $\delta_{lL}(\gamma^{\mu\nu\rho})$ (among other things). Is such a term zero?

What is the flaw in this hypothesis, if any?

EDIT: It is not true that $\gamma^m$ and $e_{m}^{\mu}$ transform oppositely under local Lorentz transformations. In fact, as the answers below show, $\gamma^m$ does not transform at all.

Qmechanic
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2 Answers2

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Comments to the question (v1):

  1. There are three types of indices at play: (i) spinor indices, (ii) flat (vector) indices, and (iii) curved (vector) indices.

  2. The gamma matrices with flat indices are constants. They don't transform under local Lorentz transformations (LLTs). They can be viewed as intertwiners between spinor indices and flat indices. (LLTs are also discussed in e.g. this Phys.SE post.)

  3. Spinor indices and flat indices of (1) vielbeins and (2) spinor/vector/tensor fields (such as, e.g., the Rarita-Schwinger field) transform under LLTs, but not curved indices.

  4. One may show that the Rarita-Schwinger Lagrangian density $$\tag{A} {\cal L}_{RS}(e,\psi)~=~\bar{\psi}_\mu \gamma^{\mu\nu\rho}D_{\nu}\psi_\rho$$ is invariant under LLTs. Be aware that the spinor indices are implicitly understood in eq. (A).

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Qmechanic
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I posted the question a few hours ago, and realized the answer lies in the fact that ${(\gamma^m)^{\alpha}}_{\beta}$ has two kinds of indices. It is indeed true that so does ${(\gamma^\mu)^\alpha}_{\beta}$. But the fact is that the flat space gamma matrices are invariant tensors of the Lorentz group $SO(d-1,1)$.

The fact that $\delta_{lL}{(\gamma^m)^{\alpha}}_{\beta} = 0$ can is due to the fact that there are three terms in the transformation law: one orbital term (for the flat index $m$) and two spin terms (for the spinor indices $\alpha$ and $\beta$). The sum of these three terms is zero.

To perform similar manipulations on $\delta_{lL}\gamma^{\mu\nu\rho}$, one must write the curved space gamma matrix in terms of vielbeins (which do transform), and flat space gamma matrices (which do not transform).

The purpose for writing this answer is to emphasize that it is the flat space gamma matrices that are invariant tensors of the Lorentz group, and not the ones with curved space indices. I did not find this clearly spelled out in a book, so I thought it might be useful to someone who is new to supergravity and such manipulations.

Note: I just saw QMechanic's answer, which reinforces this.

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