# 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?

• That wikipedia article does use 3+1 dimensional specific vocabulary where it does not need to. – AHusain Dec 15 '15 at 7:33
• 1. In odd dimension $n$, the Clifford volume element acts as $\pm 1$, so the products of up to $\frac{n-1}2$ elements span $\mathrm{End}\Sigma$, if $\Sigma$ is the spinor module. 2. If your connection comes from a metric connection of the underlying space, you only need to correct by elements of length $2$ in the Clifford algebra. – Sebastian Goette Dec 15 '15 at 10:20
• I would suggest the book by Lawson and Michelssohn. It is quite long though, so maybe you find a shorter one that does the job. – Sebastian Goette Dec 15 '15 at 13:51
• @SebastianGoette I have a question. How does $\psi$ transform under coordinate transformation. Can we simplify its transformation using an infinitesimal transformation? – linuxfreebird Dec 15 '15 at 17:58
• Meta discussion: meta.mathoverflow.net/questions/2656/learn-from-my-mistake – Todd Trimble Dec 18 '15 at 20:09

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.