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In discussing the Kaluza-Klein formalism for Type IIB Supergravity on $S^5$, or the AdS5xS5 compactification, one requires Killing spinors on $S^5$.

I read that the Dirac matrices on $S^5$ satisfy

$$\{\tau_a, \tau_b\} = 2 \delta_{ab}$$

Since the Euclidean metric appears here instead of the usual metric for $S^5$, is this because $a$ and $b$ are flat indices?

When is it okay to simply convert from curved to flat indices using the vielbein? It isn't clear why this should work for the Dirac matrices.

I see that $\tau_\alpha = \tau_a e^{a}_{\alpha}$ but why is this a correct equation at all? Furthermore, why should the Dirac matrices commute with the covariant derivative?


EDIT: Based on discussions in the comments,

(a) it is indeed possible to define $\tau_a = e^\mu_a(x) \tau_\mu(x)$ to go from curved to flat indices. (b) I should have been more explicit about the commutator part of my query.

Specifically,

$$[D_\alpha, \tau_\beta]\eta = e^{a}_{\alpha}[D_a, e^{b}_{\beta}]\tau_b \eta$$

Now the first vielbein postulate (cf. the book by Ortin) states that

$$D_\mu e^{\nu}_a = 0$$

so as to be able to go back and forth between curved and flat indices inside a covariant derivative. So, the question is: does this imply that $D_a e^b_\beta = 0$ as well?

$$D_\mu e^\nu_a = e^b_\mu D_b e^\nu_a = 0$$

Assuming $e^b_\mu$ is nonsingular, one can multiply by its inverse, yielding

$$D_b e^\nu_a = 0$$

whereas what I want is $D_b e^a_\nu = 0$. So, I use

$$e^\nu_a = g^{\nu\mu}\eta_{ac}e^{c}_{\mu}$$

to get

$$ D_b g^{\nu\mu}\eta_{ac}e^{c}_{\mu} = 0$$

Now, as pointed out by Ali Moh, I do seem to need metric compatibility to pull $g$ out of the covariant derivative and assert that

$$ D_b e^c_\mu = 0$$

which is what I need to show that the commutator $[D_a, e^b_\beta]$ is zero.

Does this make sense?

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The gamma matrices in a curved space-time satisfy $$ \{ \tau_\alpha(x),\tau_\beta(x)\} = 2 g_{\alpha\beta}(x) $$ Now if you "define" $\tau_a$ by $\tau_\alpha \equiv \tau_a e^a_\alpha$ you find \begin{align*} \{\tau_a,\tau_b\} &= \{\tau_\alpha e^\alpha_a, \tau_\beta e^\beta_b\} \\ &= \{\tau_\alpha , \tau_\beta \}e^\alpha_a e^\beta_b\\ &= 2 g_{\alpha\beta}e^\alpha_a e^\beta_b\\ &= 2 \eta_{ab} \end{align*} Therefore these are the numerical gamma matrices (independent of $x$, just like in flat space)

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  • $\begingroup$ Thanks for your answer. I see your point but can one always go from flat to curved indices or vice versa? Isn't it only possible or valid for quantities that transform as tenors? Even though Dirac matrices are defined with covariant or contravariant curved or flat indices, they don't transform under coordinate transformations. Hence my confusion. $\endgroup$ – leastaction Mar 22 '15 at 3:34
  • $\begingroup$ So the fact that the gamma matrices in flat space are independent of x is then adequate to show that the commutator of the covariant derivative with the gamma matrix vanishes identically? I suppose that was your point? $\endgroup$ – leastaction Mar 22 '15 at 3:42
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    $\begingroup$ I think that from the definition of the clifford algebra in curved spacetime you can see that the gamma matrices (with curved indices) must transform under coordinate transformations (because the metric does) $\endgroup$ – Ali Moh Mar 22 '15 at 3:45
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    $\begingroup$ I found Friedman's SUGRA has an informative concise chapter on this topic. But there are more comprehensive references out there. $\endgroup$ – Ali Moh Mar 22 '15 at 3:49
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    $\begingroup$ You are correct, but metric compatibility is usually something physicists learn before learning about the first postulate for vielbiens $\endgroup$ – Ali Moh Mar 22 '15 at 4:28

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