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Eric Rowell
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I think you can probably prove this by induction using Littelmann paths. Since the (16 dimensional) spin rep is minuscule, you can at least figure out the tensor powers by adding up weights of the form $(\pm 1/2,\ldots,\pm 1/2)$ where you take an even number of $-$ signs and throwing away any sums that are not in the dominant Weyl chamber.

Edit: So you can calculate the weights of the irreps. in $V^{\otimes 2}$ this way. Then use they Weyl dimension formula to pick the reps. whose dimension adds up to $136$. This will give you the $k=2$ case anyway.

I think you can probably prove this by induction using Littelmann paths. Since the (16 dimensional) spin rep is minuscule, you can at least figure out the tensor powers by adding up weights of the form $(\pm 1/2,\ldots,\pm 1/2)$ where you take an even number of $-$ signs and throwing away any sums that are not in the dominant Weyl chamber.

I think you can probably prove this by induction using Littelmann paths. Since the (16 dimensional) spin rep is minuscule, you can at least figure out the tensor powers by adding up weights of the form $(\pm 1/2,\ldots,\pm 1/2)$ where you take an even number of $-$ signs and throwing away any sums that are not in the dominant Weyl chamber.

Edit: So you can calculate the weights of the irreps. in $V^{\otimes 2}$ this way. Then use they Weyl dimension formula to pick the reps. whose dimension adds up to $136$. This will give you the $k=2$ case anyway.

Source Link
Eric Rowell
  • 1.6k
  • 1
  • 14
  • 16

I think you can probably prove this by induction using Littelmann paths. Since the (16 dimensional) spin rep is minuscule, you can at least figure out the tensor powers by adding up weights of the form $(\pm 1/2,\ldots,\pm 1/2)$ where you take an even number of $-$ signs and throwing away any sums that are not in the dominant Weyl chamber.