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How to compute the determinant of a representation of an element of the special linear group? How do I argue that it doesn't change?

(@Marek: @rschwieb: Yes well, given one represenation (with det=1), I should be able to find others. And if I take the abstract definition of the representation being a group homomorphism, then I don't immediatenly see why the value of det stays equal.)

Is the trace of the representation of a generator maybe automatically zero too?

(I added this suggestion because I thought maybe for the linear trace-operation, it's easier to push the value forward. And if the new generators have trace=0 too, then the representation of the group would have det=1 too.)

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Special linear group has determinant equal to one by definition, so I have no idea what you are asking. For the second part, if by generator you mean an element of the representation of the corresponding Lie algebra (the derived representation) then this vanishes since $\exp {\rm tr} A = \det \exp A$. – Marek Jan 18 '13 at 23:04
If we weren't talking about the special linear group here, you would really have to elaborate on what you mean when you say "doesn't change". I guess you are interested in knowing why equivalent reps have the same determinant. – rschwieb Jan 18 '13 at 23:44

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