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Mar 13, 2019 at 13:08 comment added Sergei Akbarov @ZoranSkoda, excuse me, I don't understand something. The pairing between the universal enveloping algebra $U(G)$ and the algebra of functions, say ${\mathcal E}(G)$, is inherited from the pairing between the group algebra of distributions, ${\mathcal E}'(G)$, and ${\mathcal E}(G)$, if we represent $U(G)$ as the subalgebra in ${\mathcal E}'(G)$ consisting of distributions with the support located in $1_G$. What is the problem?
Apr 13, 2017 at 12:58 history edited CommunityBot
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Oct 17, 2011 at 17:42 comment added Alexander Chervov Hi, Zoran, nice question, it puzzled me all my life
Mar 15, 2011 at 23:53 comment added Bruce Westbury I would like to see this as a homomorphism from the enveloping algebra onto the Schur algebras. Can this be done explicitly?
Mar 15, 2011 at 21:09 comment added Zoran Skoda I mean one can try to guess the formulas by applying derivatives many times and trying to write general formulas and use induction, but I have not succeeded into controlling this. Of course, eventually I could have a geometric proof in my generalization, but it is good to know if one is on the right track at least.
Mar 15, 2011 at 21:04 comment added Zoran Skoda So what is the pairing in terms of monomials on both sides (it is not enough the pairing for generators, it is not multiplicative, instead the Hopf structure controls the value on the products, which should b controlled) ? I can think of kernel later. For quantum SL(n) depending on version there may be a small finite-dimensional kernel, by a result of Takeuchi.
Mar 15, 2011 at 19:49 comment added Theo Johnson-Freyd Rather, I think that to give the pairing $GL(n) \otimes U(gl(n)) \to k$ in generators and relations is not difficult (where by "$GL(n)$" I mean the Hopf algebra of $k$-valued functions on the algebraic group), but it is hard to show that the pairing has no kernels.
Mar 15, 2011 at 19:46 comment added Theo Johnson-Freyd If you have access to the groups as geometric objects, then the usual argument of course realizes $U(gl_n)$ as distributions supported near the identity in $GL(n)$; but you do not have access to these, and anyway I think the proof I know somewhere uses characteristic=0. In the quantum groups case, we were not able to find a proof even when n=2 for a Hopf pairing between $U_q(sl_2)$ and $SL_q(2)$ (or rather, we were not able to show the corresponding pairing was nondegenerate) in our quantum groups class (unedited notes at math.berkeley.edu/~theojf/QuantumGroups10.pdf )
Mar 15, 2011 at 19:00 history asked Zoran Skoda CC BY-SA 2.5