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Gamma matrices

by countable
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countable
#1
Mar19-11, 07:15 AM
P: 13
I'm doing a course which assumes knowledge of Group Theory - unfortunately I don't have very much.

Can someone please explain this statement to me (particularly the bits in bold):

"there is only one non-trivial irreducible representation of the Cliford algebra, up to conjugacy"

FYI The Clifford algebra is just the the relationship between gamma matrices:

[tex][\gamma_\mu,\gamma_\nu]=2\eta_{\nu\mu}[/tex]

where [..] is the anticommutator rather than the commutator
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Bill_K
#2
Mar19-11, 09:49 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
The γμ's are initially understood to be abstract objects satisfying the equation you've written. Representation means we assign a matrix to each γμ and interpret the equation as a matrix equation, where the RHS contains the identity matrix I. Conjugacy refers to the fact that for any matrix M, if γμ is a solution then the conjugate set γμ' = M γμ M-1 is also a solution. Irreducible means that the γμ's are not simultaneously block diagonal, nor conjugate to a set that is block diagonal.

Putting that all together, it means that you can write down any set of γ matrices you can think of that solve the equation, and be assured that any other set γ' you might have chosen instead is related to your set by a conjugation.
countable
#3
Mar19-11, 09:32 PM
P: 13
Quote Quote by Bill_K View Post
The γμ's are initially understood to be abstract objects satisfying the equation you've written. Representation means we assign a matrix to each γμ and interpret the equation as a matrix equation, where the RHS contains the identity matrix I. Conjugacy refers to the fact that for any matrix M, if γμ is a solution then the conjugate set γμ' = M γμ M-1 is also a solution. Irreducible means that the γμ's are not simultaneously block diagonal, nor conjugate to a set that is block diagonal.

Putting that all together, it means that you can write down any set of γ matrices you can think of that solve the equation, and be assured that any other set γ' you might have chosen instead is related to your set by a conjugation.
thanks for the info Bill:)


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