- #141

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I actually calculated from behind: took some irreps of ##\mathfrak{sl}_2##, wrote them as polynomials, and changed the basis to ##\mathfrak{su}_2## and finally choose the Pauli-matrices (##\sigma_j##) instead of proper Lie algebra vectors ##(i\sigma_j##), mixed the basis vectors, in order - yes - not to give away the solution for free, et voilà: the standard example in every book about Lie algebras became a standard example of what physicists use instead. I thought this is an interesting (and simple) example to demonstrate what physicists call "ladder-operators".julian said:I guess when you originally set the question you didn't want to give any clues away that the irreducible components are organised as ##\{1 \}##, ##\{x , y \}##, and ##\{ x^2 , xy , y^2 \}## as that was question 16 b).

I'll be adjusting the calculations I did before (by adjusting ##\varphi##) to get the appropriate homomorphism later when I'm less preoccupied.