For any metric $$g_{μν}$$ is there always a linearly independant spacetime algebra satisfying $$\{\bar{γ}_μ,\bar{γ}_ν\} = 2 g_{μν} I?$$

For a diagonal metric I was able to work out that $$\bar{γ}_μ=\sqrt{n_{μμ}*g_{μμ}}γ_μ$$ satisfied these conditions (the minkowski simply adds negatives to cancel the spacelike gammas). However for metrics which are not diagonalizable at every point in spacetime I've been having trouble.

$$+---$$ is being used here.

Playing around with Tetrads seems like the way to go but I havent had as much luck this far. Thanks in advance to any help!

  • 4
    $\begingroup$ The spacetime manifold should be a spin manifold. $\endgroup$
    – Qmechanic
    Oct 30, 2018 at 3:49
  • $\begingroup$ @Qmechanic Are you sure about that? Where there exist obstructions to the construction of a spin structure, the construction of the Clifford algebra bundle is functorial, so that all pseudo-Riemannian manifolds have a Clifford bundle. In any case, I think the OP is thinking of Minkowski space but representations by possibly non-orthonormal matrices $\gamma_\mu$. $\endgroup$
    – doetoe
    Oct 31, 2018 at 11:26
  • $\begingroup$ @doetoe I'm thinking more spacetimes like the Kerr metric. Metrics for which there is no coordinate system which can diagonalize the metric everywhere. $\endgroup$
    – Craig
    Oct 31, 2018 at 14:37
  • $\begingroup$ @Craig OK. Could we read your question as "Let a manifold be given with a pseudo-Riemannian metric $g_{\mu\nu}$. Is it possible to find a smoothly varying set of linearly independent matrices $\gamma_\mu$ for which $$\{\gamma_\mu,\gamma_\nu\} = 2 g_{\mu\nu} I$$ holds at every point?" $\endgroup$
    – doetoe
    Oct 31, 2018 at 14:51
  • $\begingroup$ @doetoe yes this is precisely what I meant :) $\endgroup$
    – Craig
    Oct 31, 2018 at 22:14

1 Answer 1


Yes, tetrads are the way to go. Suppose we have Dirac matrices for Minkowski space, $$ \{ \gamma^a, \gamma^b \} = 2 \eta^{ab} I, $$ and we also have a tetrad/vierbein, $e_a^\mu$, [1] $$ g_{\mu\nu}\ e_a^\mu\ e_b^\nu = \eta_{ab}, \qquad \eta^{ab}\ e_a^\mu\ e_b^\nu = g^{\mu\nu}. $$ Then we can define $\bar{\gamma}^\mu$: $$ \bar{\gamma}^\mu = e_a^\mu\ \gamma^a, \qquad \{ \bar{\gamma}^\mu, \bar{\gamma}^\nu \} = e_a^\mu\ e_b^\nu\ \{ \gamma^a, \gamma^b \} = e_a^\mu\ e_b^\nu\ 2 \eta^{ab} I = 2 g^{\mu\nu} I. $$ When dealing with spinors in curved space, people usually stick to the tetrad formalism.

Carroll, Spacetime and geometry (the last appendix).
Weinberg, The Quantum Theory of Fields: Volume 3 (Section 31.1)

[1] As alluded to by Qmechanic and doetoe in the comments above, there can be topological problems with finding a tetrad or defining spinors. I'd ask them for references if you want to know more about this.

  • $\begingroup$ Wow thank you for answering this months after it was posted! I really do appreciate it! :) $\endgroup$
    – Craig
    Mar 28, 2019 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.