Timeline for How to calculate symmetric tensor products of SO(10) representations?
Current License: CC BY-SA 2.5
12 events
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| Jun 6, 2010 at 20:28 | vote | accept | Osiris | ||
| Jun 6, 2010 at 20:25 | history | edited | Osiris | CC BY-SA 2.5 |
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| Jun 6, 2010 at 20:19 | history | edited | Osiris | CC BY-SA 2.5 |
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| Jun 6, 2010 at 1:17 | answer | added | Victor Protsak | timeline score: 9 | |
| Jun 6, 2010 at 0:12 | comment | added | Bruce Westbury | Thanks. The case k=2 (and all spin representations) can be found in Adams posthumous book. I still find the formula in the question surprising. | |
| Jun 6, 2010 at 0:00 | comment | added | José Figueroa-O'Farrill | (I wish I could edit comments!) Eric's guess is spot on: in Lie, with default Lie algebra of type D5, the representation with Dynkin labels [0,0,0,0,1] is one of the 16-dimensional complex half-spinor representations. The other, which is the complex conjugate representation, is [0,0,0,1,0]. | |
| Jun 5, 2010 at 23:54 | answer | added | Eric Rowell | timeline score: 2 | |
| Jun 5, 2010 at 23:35 | comment | added | Bruce Westbury | Could you clarify the notation? I take it that V=[0,0,0,0,1] is the 10 dimensional vector representation. Then I would expect [1,0,0,0,0] to be one of the two spin representations. Except you will not get a spinor in $Sym^k V$. The case k=2 is straightforward. You get [0,0,0,0,2] and the trivial representation which is [0,0,0,0,0]. | |
| Jun 5, 2010 at 23:32 | comment | added | Eric Rowell | I guess this is one of the two spin reps since $Sym^2(W)$ would contain the trivial rep. if $W$ is the $10$-dimensional rep. | |
| Jun 5, 2010 at 22:37 | history | edited | Victor Protsak |
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| Jun 5, 2010 at 22:25 | answer | added | Tom Church | timeline score: 2 | |
| Jun 5, 2010 at 21:38 | history | asked | Osiris | CC BY-SA 2.5 |