|Mar19-11, 07:15 AM||#1|
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:
where [..] is the anticommutator rather than the commutator
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|Mar19-11, 09:49 AM||#2|
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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.
|Mar19-11, 09:32 PM||#3|
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