What Does One Non-Trivial Irreducible Representation of Clifford Algebra Mean?

[countable]

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:

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

where [..] is the anticommutator rather than the commutator

 

[Bill_K]

The γ_μ's are initially understood to be abstract objects satisfying the equation you've written. Representation means we assign a matrix to each γ_μ 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 γ_μ is a solution then the conjugate set γ_μ' = M γ_μ M^-1 is also a solution. Irreducible means that the γ_μ'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]

The γ_μ's are initially understood to be abstract objects satisfying the equation you've written. Representation means we assign a matrix to each γ_μ 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 γ_μ is a solution then the conjugate set γ_μ' = M γ_μ M^-1 is also a solution. Irreducible means that the γ_μ'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:)