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Let's consider $\mathfrak h$ the Cartan subalgebra of $\mathfrak{sl}_n(\mathbb C)$ consisting of diagonal matrices. This is just to fix ideas, I'm more interested in a method than in this case.

Now the goal is to find the weights of $\mathfrak{sl}_n(\mathbb C) $ as a $\mathfrak h$-module. I found an answer here.

But how to proceed apart from trying to guess maps that make sense ? For example I personally tried summing the diagonal entries at first. And in cases where we have a different subalgebra it may be less clear what to do.

Since it's a generalization of eigenvalues and eigenspaces is there a systematic way to do it with matrices, instead of guessing the weights ?

raisinsec
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    If you are just interested in the weights of the adjoint representation, which are by definition ($0$ and) the roots, you can of course just use matrices and get the straightforward generalization of this: https://math.stackexchange.com/a/2585175/96384 to general $n$. That's kind of what the root system theory is about. For the more general question, I would not know what kind of answer you expect that is not a course in the theory of representation theory of semisimple Lie algberas with weights. – Torsten Schoeneberg Nov 01 '22 at 16:33
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    In the case of roots, say, the important fact is that once you know the roots for one CSA (and that might be the one of diagonal matrices, which makes the matrix calculatins easy) then for any other choice of CSA you can find elements $H_\alpha, E_\alpha, F_\alpha$ (see Cartan basis or Chevalley basis) that behave just like certain nice matrices for the standard CSA. – Torsten Schoeneberg Nov 01 '22 at 16:35
  • So here we should get 6 weights, which will be given as in your example by subtractions of elements of the diagonal is this right ? – raisinsec Nov 01 '22 at 17:21
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    For $\mathfrak{sl}_4$ as in the linked example, actually 13 weights: One is zero (that's how the CSA acts on itself), and then there's six pairs, where in each pair the two are the negatives of each other. These are indeed subtractions of diagonal entries (and each's negative is just the one with switching the order of subtraction). – Torsten Schoeneberg Nov 01 '22 at 17:27
  • Yes sorry I had $\mathfrak{sl}_3(\mathbb C)$ in mind. – raisinsec Nov 01 '22 at 18:11
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    For $\mathfrak{sl}_3$ one has six roots (in three $\pm$ pairs), plus the zero weight. – Torsten Schoeneberg Nov 01 '22 at 20:04

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