Timeline for Does there exist any "quantum Lie algebra" embeded into the quantum enveloping algebra U_q(g)?
Current License: CC BY-SA 4.0
15 events
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| Nov 20, 2018 at 15:13 | history | edited | YCor |
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| S Nov 20, 2018 at 15:00 | history | suggested | Christoph Mark | CC BY-SA 4.0 |
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| Nov 20, 2018 at 13:50 | review | Suggested edits | |||
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| Apr 17, 2012 at 8:44 | answer | added | Nicola Ciccoli | timeline score: 4 | |
| Apr 16, 2012 at 14:59 | answer | added | Réamonn Ó Buachalla | timeline score: 8 | |
| Apr 12, 2012 at 6:33 | comment | added | Alexander Chervov | equivalent to category of U(g) modules with twisted tensor product, the twisting can be constructed with the Drinfeld's associator. | |
| Apr 12, 2012 at 6:32 | comment | added | Alexander Chervov | I also heard that U_q and U are isomorphic q=e^h and working over C[h], as I remember it mentioned in the Cartier's Bourbaki talks. As far as I understand reason is simple H^2(g) are trivial - so we cannot deform Lie algebra structure and this also implies we cannot deform U(g) is a reasonably non-trivial way. Another to look on it - let us look on the category of finite-dim representations of U(g) - it is semi-simple - so there is no deformation of the category structure, the only thing we can deform tensor product, this indeed can be done, and tensor category of U_q(g) modules is ... | |
| Apr 12, 2012 at 5:33 | comment | added | tzhang | U(g) and Uq(g) are different as algebras and coalgebras, the relationship between “different” quantum deformations are given in mathoverflow.net/questions/55647/… | |
| Apr 12, 2012 at 5:01 | history | edited | tzhang | CC BY-SA 3.0 |
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| Apr 11, 2012 at 20:33 | comment | added | Christopher Drupieski | @Allen, I think the isomorphism occurs when your coefficient ring is a power series ring in the indeterminate $q$. If your coefficient ring is only $\mathbb{C}[q]$, or if $q$ is specialized to a complex value, then there is no reason to expect an algebra isomorphism. | |
| Apr 11, 2012 at 20:10 | comment | added | Jim Humphreys | @Allen: I'm confused. How would you construct an algebra isomorphism here? @tzhang: What meaning do you attach to the symbol q here? It's used in more than one sense in the literature, sometimes as an indeterminate or arbitrary complex number or root of unity. Does it matter for your question? | |
| Apr 11, 2012 at 17:15 | answer | added | Jake | timeline score: 5 | |
| Apr 11, 2012 at 16:55 | comment | added | Allen Knutson | Aren't $U(\mathfrak g)$ and $U_q(\mathfrak g)$ isomorphic as algebras, only different as coalgebras? | |
| Apr 11, 2012 at 15:57 | history | edited | tzhang | CC BY-SA 3.0 |
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| Apr 11, 2012 at 15:51 | history | asked | tzhang | CC BY-SA 3.0 |