From Spin Connection in 5 dimensions I can define a massless fermion's covariant derivative on a curved manifold as $$ \nabla_\mu \psi = (\partial_\mu - {i \over 4} \omega_\mu^{ab} \sigma_{ab}) \psi \tag{1} $$ where $\sigma_{ab}$ are the dirac bilinears and $\omega_\mu^{ab}$ is the spin connection with three indices.

In 5 dimensions I have a $4\times 4$ spinor space, giving me three sets of irreducible matrices: $I$ as identity, $\gamma^a$ as monolinears, and $\sigma_{ab}=[\gamma_a,\gamma_b]$ as bilinears. This give me a total of $1+5+10=16$ matrices forming a complete set.

In 9 dimensions I can have $9=2(4)+1$, giving me a spinor space of $2^{(4)}\times 2^{(4)}=16\times 16$ creating additional irreducibles: $\sigma^{abc}=[\gamma^a,\gamma^b,\gamma^c]$ as trilinears and $\sigma^{abcd}=[\gamma^a,\gamma^b,\gamma^c,\gamma^d]$ as quadrilinears. This gives me a total of $1+9+36+84+126=256$. These numbers were calculated from the binomial coefficients ( binomial[d,k] ) for the total number of kth-linears in $d$ spatial dimensions.

Since there are additional irreducibles in $9$ dimensions, not found in 5 dimensions, does my covariant derivative in Eq. 1 have additional terms? For example $$ \nabla_\mu \psi = \left(\partial_\mu - {i \over 4} \omega_\mu^{ab} \sigma_{ab} - {i \over 48} \omega_\mu^{abcd} \sigma_{abcd} \right) \psi \tag{2} $$ where $\omega_\mu^{abcd}$ is a new spin connection of 5 indices or is Eq. 1 still valid in $9$ dimensions?