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Dimension of weight spaces of Lie algebra representation

asked 3 years ago

slartibartfast gravatar image

Consider a Lie algebra g. Let λ be a dominant integral weight and L(λ) be the unique irreducible representation of highest weight λ. (Since λ is dominant and integral, L(λ) is finite dimensional).

We know that L(λ) decomposes into a direct sum L(λ)=μL(λ)μ where L(λ)μ is a weight space of weight μ.

Is there a way to compute dimL(λ)μ in Sage?

I know that Freudenthal formula can be used to find these dimensions by hand. But I want to verify if my calculations are correct. Thanks in advance!

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answered 3 years ago

FrédéricC gravatar image

Like this

sage: W = WeylCharacterRing(['A', 1])
sage: L4 = W([4])
sage: L4.weight_multiplicities()
{(4, 0): 1, (3, 1): 1, (2, 2): 1, (1, 3): 1, (0, 4): 1}
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Thanks a lot! I am afraid I have some further questions. My understanding is that for sl2, α1=2ϖ1, where ϖ1 is the unique fundamental weight and α1 is the unique positive root. But according to sage, the output of W.simple_roots() is Finite family {1: (1, -1)} and the output of W.fundamental_weights() is Finite family {1: (1, 0)}. But (1,1)2(1,0). Am I missing something?

slartibartfast gravatar imageslartibartfast ( 3 years ago )
FrédéricC gravatar imageFrédéricC ( 3 years ago )
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excerpt: "For type A (also G2, F4, E6 and E7) we will take as the weight lattice not the weight lattice of the semisimple group, but for a larger one. For type A, this means we are concerned with the representation theory of K=U(n) or G=GL(n,C) rather than SU(n) or SU(n,C)."

FrédéricC gravatar imageFrédéricC ( 3 years ago )

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Asked: 3 years ago

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Last updated: Jan 03 '22