What is the spin connection in 9 dimensions as opposed to 5 dimensions? 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? 
 A: To answer this question, we have to make a few rather natural assumptions, which I give in a slightly different notation.
Let the spinor bundle $S$ be associated to a Euclidean vector bundle $E$.
Assume that both bundles are locally trivialised over some open coordinate patch $U$ by orthonormal / unitary frames $e_1,\dots, e_n$, $\psi_1,\dots,\psi_N$, such that Clifford multiplication has constant coefficients ("Pauli matrices"/"Dirac matrices") with respect to these frames, so $e_i\cdot\psi_\alpha=\sum_\beta c_{i\alpha}^\beta\psi_\beta$.
Let $\nabla^E$ be a metric connection on $E$, so the metric satisfies a Leibniz rule. Then $\nabla^E$ differs from partial derivation with respect to the given trivialisation by a one-form with values in skew-symmetric matrices,
so $\nabla^E_{\partial_i}e_j|_x=\sum_k\Gamma_{ij}^k(x)e_k$. The Christoffel symbols $\Gamma_{ij}^k\colon U\to\mathbb R$ are smooth functions with $\Gamma_{ij}^k=-\Gamma_{ik}^j$.
Assume that the spin connection $\nabla^S$ is compatible with $\nabla^E$ in the sense that Clifford multiplication satisfies a Leibniz rule.
This determines $\nabla^S$ uniquely.
Then $\nabla^S$ differs from partial derivation in the given trivialisation by
a one-form with values in the degree-2-part of the clifford algebra, more precisely $$\nabla^S_{\partial_i}\psi_\alpha=\frac14\sum_{j,k}\Gamma_{ij}^k(x)e_j\cdot e_k\cdot\psi_\alpha\;.$$
No terms of higher degree in the Clifford algebra are needed.
We check the Leibniz rule\begin{align}
\nabla^S_{\partial_i}(e_\ell\cdot \psi_\alpha)&=\frac14\sum_{j,k}\Gamma_{ij}^ke_j\cdot e_k\cdot e_\ell\cdot \psi_\alpha\\
&=-\frac12\sum_j\Gamma_{ij}^\ell e_j\cdot\psi_\alpha
+\frac12\sum_k\Gamma_{i\ell}^ke_k\cdot\psi_\alpha
+\frac14\sum_{j,k}\Gamma_{ij}^ke_\ell\cdot e_j\cdot e_k\cdot \psi_\alpha\\
&=(\nabla^E_{\partial_i}e_\ell)\cdot\psi_\alpha
+e_\ell\cdot(\nabla^S_{\partial_i}\psi_\alpha)\;.
\end{align}
You find the whole story in Lawson-Michelsohn, section II.4 or in Berline-Getzler-Vergne, Section 3.3.
