# Basic question about curved and flat indices, and the Dirac matrices on $S^5$

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?

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)