Timeline for When are Jones-Wenzl projectors defined?
Current License: CC BY-SA 3.0
13 events
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| May 6, 2020 at 1:12 | comment | added | Ben Webster♦ | @rspencer I don't think it's written anywhere. Probably you can make an essentially equivalent argument using the cellular algebra structure (the corresponding fact you want to use is that that cell module is irreducible), with the binomial coefficients appearing as Gram determinants. Now that you put it that way, that's probably a "better" argument. | |
| May 1, 2020 at 16:00 | comment | added | rspencer | Is there a known argument that doesn't rely on $TL_n$'s life as an endomorphism ring: one that follows from a diagrammatic definition? | |
| Sep 16, 2014 at 12:49 | comment | added | Ben Webster♦ | $U_q(\mathfrak{sl}_2)$-modules. | |
| Sep 16, 2014 at 12:49 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Sep 16, 2014 at 1:28 | comment | added | darij grinberg | Sorry for being lame, but in what category is your $\operatorname{End}$ defined? (Endomorphisms of what-modules?) | |
| Sep 15, 2014 at 20:42 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Sep 13, 2014 at 22:12 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Aug 1, 2013 at 21:09 | comment | added | Ben Webster♦ | I certainly don't know of a reference; knowing the answer, it might not be too hard to check that in the JW is defined in this case. There are formulas for the coefficients in a paper of Morrison though I don't see instantly why they don't blow up if the qbc is non-zero. This would have the additional advantage of working over the integers. | |
| Aug 1, 2013 at 20:03 | comment | added | Ben | Finally, do you know of any reasonable paper where I could quote this result, or something similar? Is this fact well-known enough to be "folklorish?" | |
| Aug 1, 2013 at 19:57 | comment | added | Ben | For what its worth, when $[m]=0$, one can observe that $\binom{m-1}{k}_q$ is invertible for all $k$. So, as desired, $JW_{m-1}$ appears to be defined. | |
| Aug 1, 2013 at 19:21 | comment | added | Ben | Thanks Ben, that's a great answer. It's certainly believable that this could be the answer over $\mathbb{Z}$ too. I don't think this representation theory argument immediately implies this though - usually arguments involving cellular intersection forms require the base ring to be local in order to guarantee that, for instance, idempotents in the associated graded will lift to true idempotents. | |
| Aug 1, 2013 at 19:18 | vote | accept | Ben | ||
| Jul 31, 2013 at 20:14 | history | answered | Ben Webster♦ | CC BY-SA 3.0 |