Timeline for Splitting of homomorphism from cactus group to permutation group
Current License: CC BY-SA 3.0
9 events
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| Jan 28, 2012 at 1:17 | history | edited | François G. Dorais |
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| May 18, 2011 at 20:21 | comment | added | David E Speyer | I don't have an answer, but you might want to look into Garside structures math.ucsb.edu/~mccammon/papers/intro-garside.pdf as a way of axiomatizing what properties you might want a splitting to have. | |
| May 18, 2011 at 16:17 | history | edited | MTS |
Retagged with operad
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| May 18, 2011 at 11:50 | comment | added | James Griffin | A curious question, my intuition says that the answer is no, but I think that the question is too open for a negative answer. If you were to look at the group which drops the squaring to the identity, call it BJ_n, then I suspect that the map BJ_n → J_n would 'split', but that's not the answer to your question. Also does this question deserve an operad tag? | |
| May 18, 2011 at 5:53 | comment | added | MTS | @Noah: I was led to this stuff by looking at Zwicknagl's papers! | |
| May 18, 2011 at 4:53 | comment | added | Noah Snyder | Also, if you're interested in quantum symmetrizers and anti-symmetrizers and their relation to the cactus group, then you might want to look at Zwicknagl's papers. | |
| May 18, 2011 at 4:51 | comment | added | Noah Snyder | Interesting question! My best guess was to try writing elements of S_n as a product of the generators s_{p,q} from Kamnitzer-Henriques where they generators are "nested" with the largest generators written first. However, a quick count shows that although this works for n=3, for n=4 there are only 22 products of that form so this doesn't work. | |
| May 18, 2011 at 4:51 | comment | added | Andy Putman | I think you went to a very different kindergarten than I did... | |
| May 17, 2011 at 23:58 | history | asked | MTS | CC BY-SA 3.0 |