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Dimension of weight spaces of Lie algebra representation

asked 2022-01-03 14:07:51 +0100

slartibartfast gravatar image

Consider a Lie algebra $\frak{g}$. Let $\lambda$ be a dominant integral weight and $L(\lambda)$ be the unique irreducible representation of highest weight $\lambda$. (Since $\lambda$ is dominant and integral, $L(\lambda)$ is finite dimensional).

We know that $L(\lambda)$ decomposes into a direct sum $$L(\lambda)=\bigoplus_{\mu} L(\lambda)_\mu$$ where $L(\lambda)_\mu$ is a weight space of weight $\mu$.

Is there a way to compute $\dim L(\lambda)_\mu$ 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 2022-01-03 14:22:48 +0100

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 $\frak{sl}_2$, $\alpha_1=2\varpi_1$, where $\varpi_1$ is the unique fundamental weight and $\alpha_1$ 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 $(1,-1)\neq 2 \cdot (1,0)$. Am I missing something?

slartibartfast gravatar imageslartibartfast ( 2022-01-03 15:23:18 +0100 )edit
FrédéricC gravatar imageFrédéricC ( 2022-01-03 17:32:10 +0100 )edit

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 ( 2022-01-03 17:34:56 +0100 )edit

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Asked: 2022-01-03 14:07:51 +0100

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Last updated: Jan 03 '22