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Mar 17, 2013 at 17:44 comment added Abtan Massini Sorry problem with the editing - should be now ok.
Mar 17, 2013 at 17:43 history edited Abtan Massini CC BY-SA 3.0
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Mar 17, 2013 at 16:34 comment added Abtan Massini @Adrien: Sorry, the question was very badly written, it should be better now.
Mar 17, 2013 at 16:33 history edited Abtan Massini CC BY-SA 3.0
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Mar 17, 2013 at 16:20 comment added Adrien Alexander> But $U_q$ is not a deformation of $U$, even if $q$ is a variable. Abtan> I'm confused about the objects you're considering: your $U_{\hbar}(\mathfrak g)$ is just the trivial deformation of $U(\mathfrak g)$, and not what people usually mean when they write $U_{\hbar}(\mathfrak g)$ (though they are indeed non canonically isomorphic as algebras). And if I understand correctly the second algebra you look at is $U_q(\mathfrak g)[[\hbar]]$ for some complex number $q$, and in particular without any relation between $q$ and $\hbar$, is it really what you meant ?
Mar 17, 2013 at 12:37 history edited Abtan Massini CC BY-SA 3.0
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Mar 17, 2013 at 11:49 comment added Alexander Chervov Second Cohomology of semisinple lie alg vanishes. So any deformation is trivial. So the two algs are isomorphic.
Mar 17, 2013 at 11:22 history asked Abtan Massini CC BY-SA 3.0