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

+ 1 like - 0 dislike
1510 views

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
retagged Dec 18, 2015
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
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
@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
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
@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
@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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysi$\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification