What is the spin connection in 9 dimensions as opposed to 5 dimensions?

Question

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 dimenions 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 quadlinears. 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 $d$ spacial dimensions.

Since there are additional irreduciables 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 $9$ dimesions?

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user linuxfreebird

Comments

That wikipedia article does use 3+1 dimensional specific vocabulary where it does not need to.

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user AHusain

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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.

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user Sebastian Goette

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@SebastianGoette Thanks for the comment. I am unsure why this question was down voted. May you provide a reference to a paper or book I can read that outlines a proof of why bilinears form a complete set for all coordinate transformations on the spinor?

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user linuxfreebird

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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.

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user Sebastian Goette

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@SebastianGoette I read pages 101 to 135 of Spin Geometry as you suggested, and I was unable to find anything specific to the original topic in question. Sorry. May you suggest another book?

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user linuxfreebird

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@SebastianGoette I have a question. How does $\psi$ transform under coordinate transformation. Can we simplify its transformation using an infinitesimal transformation?

This post imported from StackExchange MathOverflow at 2015-12-18 20:42 (UTC), posted by SE-user linuxfreebird