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I'm trying to decompose tensor products of semisimple Lie group/algebra representations into direct sums of irreducible representations. I know this question has been asked many times before, with lots of different answers. See here (Steinberg formula, section 24 in Humphreys, + more options) and here (crystals, Weyl character formula, + more options).

Despite all these resources, I'm not sure where to start and I hope that since my representations aren't too complicated I can use the simplest of these tools.

Say, for example, I wanted to decompose $\wedge^2 V \otimes V^*$ (for $V$ the standard $\mathfrak{sl}_n(\mathbb{C})$ representation). What steps can I take to do this in the easiest way possible?

Chase
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    For small $n$ (1, 2, 3) the best way I think is the approach taken by Fulton and Harris: compute the weights of the representations you start with, draw them in some physical copy (i.e. picture on paper, or some nice 3D-model) of the weight lattice, then, using a different color (or a different copy of the weight lattice) color in the weights of the tensor product. These are just all the sums of one weight of the left rep and one weight of the right rep. Now stare at the resulting picture and see how it can be written as the union of the weights of irreps. These are your decompostion – Vincent Nov 09 '23 at 22:08
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    Calculating that particular example using LiE, it seems to decompose as $V \oplus V_{\omega_2 + \omega_{n-1}}$ where $V_\lambda$ means the irrep with highest weight $\lambda$ so $V = V_{\omega_1}$, $V^* = V_{\omega_n}$ and so on. You could likely prove this by observing that $V_{\lambda} \otimes V_{\omega}$ always contains $V_{\lambda+\omega}$ and then comparing dimensions (note $\mathfrak{sl}_n$ only has three reps of dimension $n$: $V,V^*$ and the trivial one). – Callum Nov 09 '23 at 23:08
  • @Callum What commands did you put into LiE to calculate this? (I haven't succeeded in downloading LiE but I think it's time I figure that out.) – Chase Nov 09 '23 at 23:46
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    simply "tensor([0,1,0,...,0],[0,...,0,1],An)" where n is replaced by 2,3,4, etc. and the remaining spaces in those vectors are filled with 0's so that they have length n. You can also use "setdefault(An)" beforehand and then omit the An from the command. – Callum Nov 10 '23 at 08:27
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    We have discussed this in your previous question. Instead of LiE you can use LieArt, and it has been calculated already. – Dietrich Burde Nov 12 '23 at 19:20

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