- #1
Michi123
- 2
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Hello,
i have here some identities for gamma matrices in n dimensions to prove and don't know how to do this. My problem is that I am 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
i have here some identities for gamma matrices in n dimensions to prove and don't know how to do this. My problem is that I am 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