Products of gamma matrices in n dimensions

1. Aug 6, 2016

Michi123

Hello,

i have here some identities for gamma matrices in n dimensions to prove and dont know how to do this. My problem is that im not very familiar with the ⊗ in the equations. I think it should be the Kronecker-product. If someone could give me a explanation of how to work with this stuff it would be great.
here the exercise:
Let γ(n)μ1... μn be the totally antisymmetric products of n γ-matrices and γ(n)⊗γ(n) = γ(n)μ1... μn ⊗ γ(n)μ1... μn.

it should hold that:
1.)γμ γν γ(2)⊗γμ γν γ(2) = γ(4)⊗γ(4) +2(5μ -4) γ(2)⊗γ(2) +4μ(2μ-1) id⊗id
2.)γργμγσγν⊗γργνγσγμ= -γ(4)⊗γ(4) +4γ(2)⊗γ(4) + 4μ(3μ-1) id⊗id

id is the n dimensional identity matrix and μ =d/2 where d is the dimension
for the gamma matrices in n dimensions also holds the basic anticommutation relation and ημνημν = d

greetz mk

2. Aug 11, 2016

Greg Bernhardt

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Sep 14, 2016

Michi123

Ok, I think the antisymmetric product of the gamma-matrices is defined by:
$\gamma^{(n)} =\gamma^{[\mu_1\mu_2\ldots\mu_n]}= \frac{1}{n!}\epsilon_{\mu_1 \mu_2 \ldots \mu_n}\gamma^{\mu_1}\gamma^{\mu_2}\ldots\gamma^{\mu_n}$
It would be good to show that $\gamma^\nu\cdot\gamma^{[\mu_1\mu_2\ldots\mu_n]} = \gamma^{[\nu\mu_1\mu_2\ldots\mu_n]} + \sum\limits_{i=1} ^n (-1)^{i+1} g^{\nu\mu_i}\gamma^{[\mu_1\mu_2\ldots\hat\mu_i\ldots\mu_n]}.$The $\hat\mu_i$ means that this indice is deleted from the product because $\nu$ and $\mu_i$ were equal. So if $\nu$ is different than all other indices, I'm left with a n+1 matrices product. If $\nu$ matches with one indice, I'm left with a n-1 matrices product. Im not pretty sure how to do this. For the cases n=2 or n=3 one can do this simply and just form an antisymmetric product to see how this works. But I can't do this for the general case, i.e. for arbitrary n. Perhaps one can do this by induction or just by using some combinatorial stuff?! If one has this identity, the tensor-product identities in the first post should follow by using this relation.

Last edited: Sep 14, 2016